Design of Machinery Solutions Manual Norton 5th Edition (1).pdf
November 23, 2017 | Author: Ricardo Alves Felippe | Category: N/A
Short Description
Download Design of Machinery Solutions Manual Norton 5th Edition (1).pdf...
Description
DESIGN OF MACHINERY - 5th Ed
SOLUTION MANUAL 2-1-1
PROBLEM 2-1 Statement:
Find three (or other number as assigned) of the following common devices. Sketch careful kinematic diagrams and find their total degrees of freedom. a. An automobile hood hinge mechanism b. An automobile hatchback lift mechanism c. An electric can opener d. A folding ironing board e. A folding card table f. A folding beach chair g. A baby swing h. A folding baby walker i. A fancy corkscrew as shown in Figure P2-9 j. A windshield wiper mechanism k. A dump-truck dump mechanism l. A trash truck dumpster mechanism m. A pickup tailgate mechanism n. An automobile jack o. A collapsible auto radio antenna
Solution:
See Mathcad file P0201.
Equation 2.1c is used to calculate the mobility (DOF) of each of the models below. a.
An automobile hood hinge mechanism. The hood (3) is linked to the body (1) through two rocker links (2 and 4). Number of links
L 4
Number of full joints
J1 4
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2 M1 b.
HOOD 3 2
4
1 BODY
An automobile hatchback lift mechanism. The hatch (2) is pivoted on the body (1) and is linked to the body by the lift arm, which can be modeled as two links (3 and 4) connected through a translating slider joint. HATCH Number of links L 4 Number of full joints
J1 4
Number of half joints
J2 0
2 3 1
M 3 ( L 1 ) 2 J1 J2
4
M1
1 BODY
c.
An electric can opener has 2 DOF.
d.
A folding ironing board. The board (1) itself has one pivot (full) joint and one pin-in-slot sliding (half) joint. The two legs (2 and 3) hav a common pivot. One leg connects to the pivot joint on the board and the other to the slider joint.
DESIGN OF MACHINERY - 5th Ed
SOLUTION MANUAL 2-1-2
Number of links
L 3
Number of full joints
J1 2
Number of half joints
J2 1
1 3
2
M 3 ( L 1 ) 2 J1 J2 M1 e.
A folding card table has 7 DOF: One for each leg, 2 for location in xy space, and one for angular orientation.
f.
A folding beach chair. The seat (3) and the arms (6) are ternary links. The seat is linked to the front leg(2), the back (5) and a coupling link (4). The arms are linked to the front leg (2), the rear leg (1), and the back (5). Links 1, 2, 4, and 5 are binar links. The analysis below is appropriate when the chair is not fully opened. When fully opened, one or more links are prevented from moving by a stop. Subtract 1 DOF when forced against the stop. Number of links
L 6
Number of full joints
J1 7
Number of half joints
J2 0
5 6 4
1
M 3 ( L 1 ) 2 J1 J2
2 3
M1 g.
A baby swing has 4 DOF: One for the angular orientation of the swing with respect to the frame, and 3 for the location and orientation of the frame with respect to a 2-D frame.
h.
A folding baby walker has 4 DOF: One for the degree to which it is unfolded, and 3 for the location and orientation of the walker with respect to a 2-D frame.
i.
A fancy corkscrew has 2 DOF: The screw can be rotated and the arms rotate to translate the screw.
j.
A windshield wiper mechanism has 1 DOF: The position of the wiper blades is defined by a single input.
k.
A dump-truck dump mechanism has 1 DOF: The angle of the dump body is determined by the length of the hydraulic cylinder that links it to the body of the truck.
l.
A trash truck dumpster mechanism has 2 DOF: These are generally a rotation and a translation.
m. A pickup tailgate mechanism has 1 DOF: n.
An automobile jack has 4 DOF: One is the height of the jack and the other 3 are the position and orientation o the jack with respect to a 2-D frame.
o.
A collapsible auto radio antenna has as many DOF as there are sections, less one.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-2-1
PROBLEM 2-2 Statement:
How many DOF do you have in your wrist and hand combined?
Solution:
See Mathcad file P0202.
1.
Holding the palm of the hand level and facing toward the floor, the hand can be rotated about an axis through the wrist that is parallel to the floor (and perpendicular to the forearm axis) and one perpendicular to the floor (2 DOF). The wrist can rotate about the forearm axis (1 DOF).
2.
Each finger (and thumb) can rotate up and down and side-to-side about the first joint. Additionally, each finger can rotate about each of the two remaining joints for a total of 4 DOF for each finger (and thumb).
3.
Adding all DOF, the total is Wrist Hand Thumb Fingers 4x4
1 2 4 16
TOTAL
23
DESIGN OF MACHINERY - 5th Ed.
PROBLEM 2-3 Statement:
How many DOF do the following joints have? a. Your knee b. Your ankle c. Your shoulder d. Your hip e. Your knuckle
Solution:
See Mathcad file P0203.
a.
Your knee. 1 DOF: A rotation about an axis parallel to the ground.
b.
Your ankle. 3 DOF: Three rotations about mutually perpendicular axes.
c.
Your shoulder. 3 DOF: Three rotations about mutually perpendicular axes.
d.
Your hip. 3 DOF: Three rotations about mutually perpendicular axes.
e
Your knuckle. 2 DOF: Two rotations about mutually perpendicular axes.
SOLUTION MANUAL 2-3-1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-4-1
PROBLEM 2-4 Statement:
How many DOF do the following have in their normal environment? a. A submerged submarine b. An earth-orbit satellite c. A surface ship d. A motorcycle (road bike) e. A two-button mouse f. A computer joy stick.
Solution:
See Mathcad file P0204.
a.
A submerged submarine. Using a coordinate frame attached to earth, or an inertial coordinate frame, a submarine has 6 DOF: 3 linear coordinates and 3 angles.
b.
An earth-orbit satellite. If the satellite was just a particle it would have 3 DOF. But, since it probably needs to be oriented with respect to the earth, sun, etc., it has 6 DOF.
c.
A surface ship. There is no difference between a submerged submarine and a surface ship, both have 6 DOF. One might argue that, for an earth-centered frame, the depth of the ship with respect to mean sea level is constant, however that is not strictly true. A ship's position is generally given by two coordinates (longitude and latitude). For a given position, a ship can also have pitch, yaw, and roll angles. Thus, for all practical purposes, a surface ship has 5 DOF.
d.
A motorcycle. At an intersection, the motorcycle's position is given by two coordinates. In addition, it will have some heading angle (turning a corner) and roll angle (if turning). Thus, there are 4 DOF.
e.
A two-button mouse. A two-button mouse has 4 DOF. It can move in the x and y directions and each button has 1 DOF.
f.
A computer joy stick. The joy stick has 2 DOF (x and y) and orientation, for a total of 3 DOF.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-5-1
PROBLEM 2-5 Statement:
Are the joints in Problem 2-3 force closed or form closed?
Solution:
See Mathcad file P0205.
They are force closed by ligaments that hold them together. None are geometrically closed.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-6-1
PROBLEM 2-6 Statement:
Describe the motion of the following items as pure rotation, pure translation, or complex planar motion. a. A windmill b. A bicycle (in the vertical plane, not turning) c. A conventional "double-hung" window d. The keys on a computer keyboard e. The hand of a clock f. A hockey puck on the ice g. A "casement" window
Solution:
See Mathcad file P0206.
a.
A windmill. Pure rotation.
b.
A bicycle (in the vertical plane, not turning). Pure translation for the frame, complex planar motion for the wheels.
c.
A conventional "double-hung" window. Pure translation.
d.
The keys on a computer keyboard. Pure translation.
e.
The hand of a clock. Pure rotation.
f.
A hockey puck on the ice. Complex planar motion.
g.
A "casement" window. Pure rotation.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-7-1
PROBLEM 2-7 Statement:
Calculate the mobility of the linkages assigned from Figure P2-1 part 1 and part 2.
Solution:
See Figure P2-1 and Mathcad file P0207.
1.
Use equation 2.1c (Kutzbach's modification) to calculate the mobility.
a.
Number of links
L 6
Number of full joints
J1 7
Number of half joints
J2 1
6 3 5 2
M 3 ( L 1 ) 2 J1 J2
4 1
M0
(a)
1 3
b.
Number of links
L 3
Number of full joints
J1 2
Number of half joints
J2 1
1
M 3 ( L 1 ) 2 J1 J2 M1
2 1
(b) 4
c.
Number of links
L 4
Number of full joints
J1 4
Number of half joints
J2 0
1
3
M 3 ( L 1 ) 2 J1 J2 2
M1 (c)
1
7
d.
Number of links
L 7
Number of full joints
J1 7
Number of half joints
J2 1
1
6 5
M 3 ( L 1 ) 2 J1 J2 M3
1
2 3
4
(d)
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-7-2
8
5
8 1
5 1
9 10
6
1
1 7
4
4
1
2
2 3
3
1
5 6
2 1
(e)
1
e.
g.
Number of links
L 10
Number of full joints Number of half joints
(f)
Number of links
L 6
J1 13
Number of full joints
J1 6
J2 0
Number of half joints
J2 2
f.
M 3 ( L 1 ) 2 J1 J2
M 3 ( L 1 ) 2 J1 J2
M1
M1
Number of links
L 8
Number of full joints
J1 9
Number of half joints
J2 2
M 3 ( L 1 ) 2 J1 J2
4 1
4
7
6
3
7 1
5
8 1
2
2 1
1
1
M1 (g)
2 h.
Number of links
L 4
Number of full joints
J1 4
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2 M1
1 3 1 4 (h)
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-8-1
PROBLEM 2-8 Statement:
Identify the items in Figure P2-1 as mechanisms, structures, or preloaded structures.
Solution:
See Figure P2-1 and Mathcad file P0208.
1.
Use equation 2.1c (Kutzbach's modification) to calculate the mobility and the definitions in Section 2.5 of the text to classify the linkages.
a.
Number of links
L 6
Number of full joints
J1 7
Number of half joints
J2 1
6 3 5 2
M 3 ( L 1 ) 2 J1 J2 M0
4 1
Structure
(a)
1 3
b.
Number of links
L 3
Number of full joints
J1 2
Number of half joints
J2 1
1
M 3 ( L 1 ) 2 J1 J2 M1
Mechanism
2 1
(b) 4
c.
Number of links
L 4
Number of full joints
J1 4
Number of half joints
J2 0
1
3
M 3 ( L 1 ) 2 J1 J2 M1
2
Mechanism (c)
1
7
d.
Number of links
L 7
Number of full joints
J1 7
Number of half joints
J2 1
1
6 5
M 3 ( L 1 ) 2 J1 J2 M3
Mechanism
1
2 3
4
(d)
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-9-1
PROBLEM 2-9 Statement:
Use linkage transformation on the linkage of Figure P2-1a to make it a 1-DOF mechanism.
Solution:
See Figure P2-1a and Mathcad file P0209.
1.
The mechanism in Figure P2-1a has mobility: Number of links
L 6
Number of full joints
J1 7
Number of half joints
J2 1
6 3 5 2
M 3 ( L 1 ) 2 J1 J2 M0
4 1 1
2.
Use rule 2, which states: "Any full joint can be replaced by a half joint, but this will increase the DOF by one." One way to do this is to replace one of the pin joints with a pin-in-slot joint such as that shown in Figure 2-3c. Choosing the joint between links 2 and 4, we now have mobility: Number of links
L 6
Number of full joints
J1 6
Number of half joints
J2 2
6 3 5
M 3 ( L 1 ) 2 J1 J2
2 4
M1
1 1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-10-1
PROBLEM 2-10 Statement:
Use linkage transformation on the linkage of Figure P2-1d to make it a 2-DOF mechanism.
Solution:
See Figure P2-1d and Mathcad file P0210.
1.
7
The mechanism in Figure P2-1d has mobility: Number of links
L 7
Number of full joints
J1 7
Number of half joints
J2 1
M 3 ( L 1 ) 2 J1 J2
1
6 5 1
2 3
4
M3 2.
Use rule 3, which states: "Removal of a link will reduce the DOF by one." One way to do this is to remove link 7 such that link 6 pivots on the fixed pin attached to the ground link (1). We now have mobility: Number of links
L 6
Number of full joints
J1 6
Number of half joints
J2 1
M 3 ( L 1 ) 2 J1 J2
1 6 5 1
2 3
M2
4
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-11-1
PROBLEM 2-11 Statement:
Use number synthesis to find all the possible link combinations for 2-DOF, up to 9 links, to hexagonal order, using only revolute joints.
Solution:
See Mathcad file P0211.
1.
Use equations 2.4a and 2.6 with DOF = 2 and iterate the solution for valid combinations. Note that the number of links must be odd to have an even DOF (see Eq. 2.4). The smallest possible 2-DOF mechanism is then 5 links since three will give a structure (the delta triplet, see Figure 2-7). L B T Q P H
L 3 M T 2 Q 3 P 4 H L 5 T 2 Q 3 P 4 H
2.
3.
For L 5 0 T 2 Q 3 P 4 H
0=T =Q=P=H
2 T 2 Q 3 P 4 H
H 0
For L 7
Case 1:
Case 2:
4.
B 5
Q 0
Q 1
P 0
T 2 2 Q 3 P 4 H
T2
B L T Q P H
B5
T 2 2 Q 3 P 4 H
T0
B L T Q P H
B6
T 0
P 0
For L 9 4 T 2 Q 3 P 4 H Case 1:
H 1
Q 0
B L T Q P H Case 2a:
H 0
B8
4 T 2 Q 3 P 9 B T Q P
Case 2b:
P 1
1 T 2 Q
Q 0
B L T Q P H Case 2c:
P 0
T 1 B7
4 T 2 Q 9 B T Q
Case 2c1:
Case 2c2:
Case 2c3:
Q 2 Q 1 Q 0
T 4 2 Q
T0
B 9 T Q
B7
T 4 2 Q
T2
B 9 T Q
B6
T 4 2 Q
T4
B 9 T Q
B5
M 2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-12-1
PROBLEM 2-12 Statement:
Find all of the valid isomers of the eightbar 1-DOF link combinations in Table 2-2 (p. 38) having a. Four binary and four ternary links. b. Five binaries, two ternaries, and one quaternary link. c. Six binaries and two quaternary links. d. Six binaries, one ternary, and one pentagonal link.
Solution:
See Mathcad file P0212.
1.
2.
a.
Table 2-3 lists 16 possible isomers for an eightbar chain. However, Table 2-2 shows that there are five possible link sets, four of which are listed above. Therefore, we expect that the 16 valid isomers are distributed among the five link sets and that there will be fewer than 16 isomers among the four link sets listed above. One method that is helpful in finding isomers is to represent the linkage in terms of molecules as defined in Franke's Condensed Notations for Structural Synthesis. A summary of the rules for obtaining Franke's molecules follows: (1) The links of order greater than 2 are represented by circles. (2) A number is placed within each circle (the "valence" number) to describe the type (ternary, quaternary, etc.) of link. (3) The circles are connected using straight lines. The number of straight lines emanating from a circle must be equal to its valence number. (4) Numbers (0, 1, 2, etc.) are placed on the straight lines to correspond to the number of binary links used in connecting the higher order links. (5) There is one-to-one correspondence between the molecule and the kinematic chain that it represents. Four binary and four ternary links. Draw 4 circles with valence numbers of 3 in each. Then find all unique combinations of straight lines that can be drawn that connect the circles such that there are exactly three lines emanating from each circle and the total of the numbers written on the lines is exactly equal to 4. In this case, there are three valid isomers as depicted by Franke's molecules and kinematic chains below.
8
1 3
3
5
1 0
0
1 3
6
3
3
1
4
2 1
7
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-12-2
8 0 3
3 2
0
5
0
1 3
7
6
3 1
3
4
1 2
8 5
0 3
3
4
2 0
6
0
2 3
3
3 0
7
1 2
The mechanism shown in Figure P2-5b is the same eightbar isomer as that depicted schematically above. b.
Five binaries, two ternaries, and one quaternary link. Draw 2 circles with valence numbers of 3 in each and one with a valence number of 4. Then find all unique combinations of straight lines that can be drawn that connect the circles such that there are exactly three lines emanating from each circle with valence of three and four lines from the circle with valence of four; and the total of the numbers written on the lines is exactly equal to 5. In this case, there are five valid isomers as depicted by Franke's molecules and kinematic chains below.
0
3 2
4
7
0
1
5
3
3
2
6 4
8
1
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-12-3
5 1
3 0
3 0
2
6 4
7
3
2 4
8
1
2
5 0
3 1
3
3
7
1
2
4
6
1
2
8
4
1
5 1
3 1
6
3
4
0
1
3 2
7
8 4
1
2
5 1
3 1
6
3
3
1
1
1
8
4
7
2
4
1
c.
Six binaries and two quaternary links. Draw 2 circles with valence numbers of 4 in each. Then find all unique combinations of straight lines that can be drawn that connect the circles such that there are exactly four lines emanating from each circle and the total of the numbers written on the lines is exactly equal to 6. In this case, there are two valid isomers as depicted by Franke's molecules and kinematic chains below.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-12-4
0 2
7 4
4
4
5
3
6
2
8
2
1
2
1
4
1
7 4
4
d.
3
6
2 2
5
8
2 1
Six binaries, one ternary, and one pentagonal link. There are no valid implementations of 6 binary links with 1 pentagonal link.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-13-1
PROBLEM 2-13 Statement:
Use linkage transformation to create a 1-DOF mechanism with two sliding full joints from a Stephenson's sixbar linkage as shown in Figure 2-14a (p. 47).
Solution:
See Figure 2-14a and Mathcad file P0213.
1.
The mechanism in Figure 2-14a has mobility: Number of links
L 6
Number of full joints
J1 7
Number of half joints
J2 0
A 4
3
5
B
2
6
M 3 ( L 1 ) 2 J1 J2 M1
2.
1
Use rule 1, which states: "Revolute joints in any loop can be replaced by prismatic joints with no change in DOF of the mechanism, provided that at least two revolute joints remain in the loop." One way to do this is to replace pin joints at A and B with translating full slider joints such as that shown in Figure 2-3b. Note that the sliders are attached to links 3 and 5 in such a way that they can not rotate relative to the links. The number of links and 1-DOF joints remains the same. There are no 2-DOF joints in either mechanism.
A 4
3
5 2
1
6 B
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-14-1
PROBLEM 2-14 Statement:
Use linkage transformation to create a 1-DOF mechanism with one sliding full joint a a half joint from a Stephenson's sixbar linkage as shown in Figure 2-14b (p. 48).
Solution:
See Figure 2-14a and Mathcad file P0213.
1.
The mechanism in Figure 2-14b has mobility: Number of links
L 6
Number of full joints
J1 7
Number of half joints
J2 0
3 5 4 2
6
M 3 ( L 1 ) 2 J1 J2 1
M1
2.
To get the sliding full joint, use rule 1, which states: "Revolute joints in any loop can be replaced by prismati joints with no change in DOF of the mechanism, provided that at least two revolute joints remain in the loop." One way to do this is to replace pin joint links 3 and 5 with a translating full slider joint such as that shown in Figure 2-3b. Note that the slider is attached to link 3 in such a way that it can not rotate relative to the link. The number of links and 1-DOF joints remains the same.
3
5 4
2
6
1
3.
To get the half joint, use rule 4 on page 42, which states: "The combination of rules 2 and 3 above will keep the original DOF unchanged." One way to do this is to remove link 6 (and its two nodes) and insert a half joint between links 5 and 1. Number of links
L 5
Number of full joints
J1 5
Number of half joints
3 5 4
J2 1 2
M 3 ( L 1 ) 2 J1 J2
1
M1 1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-15-1
PROBLEM 2-15 Statement:
Calculate the Grashof condition of the fourbar mechanisms defined below. Build cardboard models of the linkages and describe the motions of each inversion. Link lengths are in inches (or double given numbers for centimeters). Part 1. a. b. c.
2 2 2
4.5 3.5 4.0
7 7 6
9 9 8
Part 2. d. e. f.
2 2 2
4.5 4.0 3.5
7 7 7
9 9 9
Solution: 1.
See Mathcad file P0215
Use inequality 2.8 to determine the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
a.
Condition( 2 4.5 7 9 ) "Grashof"
b.
Condition( 2 3.5 7 9 ) "non-Grashof"
c.
Condition( 2 4.0 6 8 ) "Special Grashof" This is a special case Grashof since the sum of the shortest and longest is equal to the sum of the other two link lengths.
d.
Condition( 2 4.5 7 9 ) "Grashof"
e.
Condition( 2 4.9 7 9 ) "Grashof"
f.
Condition( 2 3.5 7 9 ) "non-Grashof"
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-16-1
PROBLEM 2-16 Statement:
Which type(s) of electric motor would you specify a. b. c.
Solution:
To drive a load with large inertia. To minimize variation of speed with load variation. To maintain accurate constant speed regardless of load variations.
See Mathcad file P0216.
a.
Motors with high starting torque are suited to drive large inertia loads. Those with this characteristic include series-wound, compound-wound, and shunt-wound DC motors, and capacitor-start AC motors.
b.
Motors with flat torque-speed curves (in the operating range) will minimize variation of speed with load variation. Those with this characteristic include shunt-wound DC motors, and synchronous and capacitor-start AC motors.
b.
Speed-controlled DC motors will maintain accurate constant speed regardless of load variations.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-17-1
PROBLEM 2-17 Statement:
Describe the difference between a cam-follower (half) joint and a pin joint.
Solution:
See Mathcad file P0217.
1.
A pin joint has one rotational DOF. A cam-follower joint has 2 DOF, rotation and translation. The pin joint also captures its lubricant in the annulus between pin and bushing while the cam-follower joint squeezes its lubricant out of the joint.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-18-1
PROBLEM 2-18 Statement:
Examine an automobile hood hinge mechanism of the type described in Section 2.14. Sketch it carefully. Calculate its DOF and Grashof condition. Make a cardboard model. Analyze it with a free-body diagram. Describe how it keeps the hood up.
Solution:
Solution of this problem will depend upon the specific mechanism modeled by the student.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-19-1
PROBLEM 2-19 Statement:
Find an adjustable arm desk lamp of the type shown in Figure P2-2. Sketch it carefully. Measure it and sketch it to scale. Calculate its DOF and Grashof condition. Make a cardboard model. Analyze it with a free-body diagram. Describe how it keeps itself stable. Are there any positions in which it loses stability? Why?
Solution:
Solution of this problem will depend upon the specific mechanism modeled by the student.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-20-1
PROBLEM 2-20 Statement:
The torque-speed curve for a 1/8 hp permanent magnet (PM) DC motor is shown in Figure P2-3. The rated speed for this fractional horsepower motor is 2500 rpm at a rated voltage of 130V. Determine: a) The rated torque in oz-in (ounce-inches, the industry standard for fractional hp motors) b) The no-load speed c) Plot the power-torque curve and determine the maximum power that the motor can deliver.
Given:
Rated speed, N
NR 2500 rpm
R
HR
Rated power, H
R
1 8
hp
4000 3500
Speed, rpm
3000 2500 2000 1500 1000 500 0
0
50
100
150
200
250
300
Torque, oz-in
Figure P2-3 Torque-speed Characteristic of a 1/8 HP, 2500 rpm PM DC Motor
Solution: a.
See Figure P2-3 and Mathcad file P0220.
The rated torque is found by dividing the rated power by the rated speed: TR
Rated torque, TR
HR NR
TR 50 ozf in
b.
The no-load speed occurs at T = 0. From the graph this is 3000 rpm.
c.
The power is the product of the speed and the torque. From the graph the equation for the torque-speed curve is: 3000 rpm N ( T ) T 3000 rpm 300 ozf in and the power, therefore, is: H ( T ) 10
rpm ozf in
2
T 3000 rpm T
Plotting the power as a function of torque over the range T 0 ozf in 10 ozf in 300 ozf in
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-20-2
0.25 0.225 0.2
Power, hp
0.175 0.15 0.125 0.1 0.075 0.05 0.025 0
0
50
100
150
200
250
300
Torque, oz-in
Maximum power occurs when dH/dT = 0. The value of T at maximum power is: ozf in
Value of T at Hmax
THmax 3000 rpm
Maximum power
Hmax H THmax
Hmax 0.223 hp
Speed at max power
NHmax N THmax
NHmax 1500 rpm
2 10 rpm
THmax 150 ozf in
Note that the curve goes through the rated power point of 0.125 hp at the rated torque of 50 oz-in.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-21-1
PROBLEM 2-21 Statement:
Find the mobility of the mechanisms in Figure P2-4.
Solution:
See Figure P2-4 and Mathcad file P0221.
1.
Use equation 2.1c (Kutzbach's modification) to calculate the mobility.
a.
This is a basic fourbar linkage. The input is link 2 and the output is link 4. The cross-hatched pivot pins at O2 and O4 are attached to the ground link (1).
Number of links
L 4
Number of full joints
J1 4
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2
A 2 3 O2
M1
b.
4
C
O4
This is a fourbar linkage. The input is link 2, which in this case is the wheel 2 with a pin at A, and the output is link 4. The cross-hatched pivot pins at O2 and O4 are attached to the ground link (1).
Number of links
L 4
Number of full joints
J1 4
Number of half joints
J2 0
A
2
O2
3
M 3 ( L 1 ) 2 J1 J2
4 B
M1
c.
O4
This is a 3-cylinder, rotary, internal combustion engine. The pistons (sliders) 6, 7, and 8 drive the output crank (2) through piston rods (couplers 3, 4, and 5). There are 3 full joints at the crank where rods 3, 4and 5 are pinned to crank 2. The cross-hatched crank-shaft at O2 is supported by the ground link (1) through bearings. Number of links
L 8
Number of full joints
J1 10
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2
6
3
2
4
M1 7
5
8
DESIGN OF MACHINERY - 5th Ed.
d.
SOLUTION MANUAL 2-21-2
This is a fourbar linkage. The input is link 2, which in this case is a wheel with a pin at A, and the output is the vertical member on the coupler, link 3. Since the lengths of links 2 and 4 (O2A and O4B) are the same, the coupler link (3) has curvilinear motion and AB remains parallel to O2O4 throughout the cycle. The cross-hatched pivot pins at O2 and O4 are attached to the ground link (1).
Number of links
L 4
Number of full joints
J1 4
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2
B O4
O2
M1
e.
3
A 2
4
This is a fourbar linkage with an output dyad. The input (rocker) is link 2 and the output (rocker) is link 8. Links 5 and 6 are redundant, i.e. the mechanism will have the same motion if they are removed. The input fourbar consists of links 1, 2, 3, and 4. The output dyad consists of links 7 and 8. The cross-hatched pivot pins at O2, O4 and O8 are attached to the ground link (1). In the calculation below, the redundant links and their joints are not counted (subtract 2 links and 4 joints from the totals). A
Number of links
L 6
Number of full joints
J1 7
Number of half joints
J2 0
O2
4 O4 G
E
3
2 D
5
C
6
7
M 3 ( L 1 ) 2 J1 J2 O8
M1 F
H 8
f.
This is a fourbar offset slider-crank linkage. The input is link 2 (crank) and the output is link 4 (slider block). The cross-hatched pivot pin at O2 is attached to the ground link (1).
Number of links
L 4
Number of full joints
J1 4
Number of half joints
J2 0
4
B
3
M 3 ( L 1 ) 2 J1 J2 A
M1 2 O2
DESIGN OF MACHINERY - 5th Ed.
g.
SOLUTION MANUAL 2-21-3
This is a fourbar linkage with an alternate output dyad. The input (rocker) is link 2 and the outputs (rockers) are links 4 and 6. The input fourbar consists of links 1, 2, 3, and 4. The alternate output dyad consists of links 5 and 6. The cross-hatched pivot pins at O2, O4 and O6 are attached to the ground link (1). Number of links
L 6
Number of full joints
J1 7
Number of half joints
J2 0
O6 3
B
A
2
M 3 ( L 1 ) 2 J1 J2
4
C
6
O2
M1
5 D O4
h.
This is a ninebar mechanism with three redundant links, which reduces it to a sixbar. Since this mechanism is symmetrical about a vertical centerline, we can split it into two mirrored mechanisms to analyze it. Either links 2, 3 and 5 or links 7, 8 and 9 are redundant. To analyze it, consider 7, 8 and 9 as the redundant links. Analyzing the ninebar, there are two full joints at the pins A, B and C for a total of 12 joints. Number of links
L 9
Number of full joints
J1 12
Number of half joints
J2 0
6
O2 2
8
7
5
C
B
M 3 ( L 1 ) 2 J1 J2
O8
A
9
3
M0 4
D
E
The result is that this mechanism seems to be a structure. By splitting it into mirror halves about the vertical centerline the mobility is found to be 1. Subtract the 3 redundant links and their 5 (6 minus the joint at A) associated joints to determine the mobility of the mechanism. Number of links
L 9 3
Number of full joints
J1 12 5
Number of half joints
J2 0
6 O2 2
5 B
M 3 ( L 1 ) 2 J1 J2
3
M1 D
4
A
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-22-1
PROBLEM 2-22 Statement: Solution: 1.
Find the Grashof condition and Barker classifications of the mechanisms in Figure P2-4a, b, and d. See Figure P2-4 and Mathcad file P0222.
Use inequality 2.8 to determine the Grashof condition and Table 2-4 to determine the Barker classification. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
a.
This is a basic fourbar linkage. The input is link 2 and the output is link 4. The cross-hatched pivot pins at O2 and O4 are attached to the ground link (1). L1 174
L2 116
L3 108
L4 110
A 2 3
Condition L1 L2 L3 L4 "non-Grashof"
O2
4
C
This is a Barker Type 5 RRR1 (non-Grashof, longest link grounded). b.
O4
This is a fourbar linkage. The input is link 2, which in this case is the wheel with a pin at A, and the output is link 4. The cross-hatched pivot pins at O2 and O4 are attached to the ground link (1). L1 162
L2 40
L3 96
L4 122
B
A
2
3
O2 4
Condition L1 L2 L3 L4 "Grashof" This is a Barker Type 2 GCRR (Grashof, shortest link is input). d.
This is a fourbar linkage. The input is link 2, which in this case is a wheel with a pin at A, and the output is the vertical member on the coupler, link 3. Since the lengths of links 2 and 4 (O2A and O4B) are the same, the coupler link (3) has curvilinear motion and AB remains parallel to O2O4 throughout the cycle. The cross-hatched pivot pins at O2 and O4 are attached to the ground link (1). L1 150 L2 30 L3 150
L4 30
Condition L1 L2 L3 L4 "Special Grashof" This is a Barker Type 13 S2X (special case Grashof, two equal pairs, parallelogram).
O4
A
3
2 O2
B O4
4
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-23-1
PROBLEM 2-23 Statement:
Find the rotability of each loop of the mechanisms in Figure P2-4e, f, and g.
Solution:
See Figure P2-4 and Mathcad file P0223.
1.
Use inequality 2.15 to determine the rotability of each loop in the given mechanisms.
e.
This is a fourbar linkage with an output dyad. The input (rocker) is link 2 and the output (rocker) is link 8. Links 5 and 6 are redundant, i.e. the mechanism will have the same motion if they are removed. The input fourbar consists of links 1, 2, 3, and 4. The output dyad consists of links 7 and 8. The cross-hatched pivot pins at O2, O4 and O8 are attached to the ground link (1). In the calculation below, the redundant links and their joints are not counted (subtract 2 links and 4 joints from the totals).
B
A O2
4 O4 G
E
3
2 D
5
C
6
7
O8
There are two loops in this mechanism. The first loop consists of links 1, 2, 3 (or 5), and 4. The second consists of links 1, 4, 7 (or 6), and 8. By inspection, we see that the sum of the shortest and longest in each loop is equal to the sum of the other two. Thus, both loops are Class III. f.
8
This is a fourbar offset slider-crank linkage. The input is link 2 (crank) and the output is link 4 (slider block). The cross-hatched pivot pin at O2 is attached to the ground link (1).
4
A 2 O2
O6
This is a fourbar linkage with an alternate output dyad. The input (rocker) is link 2 and the outputs (rockers) are links 4 and 6. The input fourbar consists of links 1, 2, 3, and 4. The alternate output dyad consists of links 5 and 6. The cross-hatched pivot pins at O2, O4 and O6 are attached to the ground link (1). r1 87
r2 49
r3 100
r4 153
B
3
We can analyze this linkage if we replace the slider ( 4) with an infinitely long binary link that is pinned at B to link 3 and pinned to ground (1). Then links 1 and 4 for are both infinitely long. Since these two links are equal in length and, if we say they are finite in length but very long, the rotability of the mechanism will be determined by the relative lengths of 2 and 3. Thus, this is a Class I linkage since link 2 is shorter than link 3.
g.
H
F
3
B
2 4
C
A 6
O2 5 D
Using the notation of inequality 2.15, N 4 LN r4 L1 r2 L2 r1 LN L1 202
O4
L3 r3 L2 L3 187
Since LN L1 L2 L3, this is a a class II mechanism.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-24-1
PROBLEM 2-24 Statement:
Find the mobility of the mechanisms in Figure P2-5.
Solution:
See Figure P2-5 and Mathcad file P0224.
1.
Use equation 2.1c (Kutzbach's modification) to calculate the mobility. In the kinematic representations of the linkages below, binary links are depicted as single lines with nodes at their end points whereas higher order links are depicted as 2-D bars.
a.
This is a sixbar linkage with 4 binary (1, 2, 5, and 6) and 2 ternary (3 and 4) links. The inverted U-shaped link at the top of Figure P2-5a is represented here as the binary link 6. Number of links
L 6
Number of full joints
J1 7
Number of half joints
J2 0
3
5
4
2
M 3 ( L 1 ) 2 J1 J2 O2
M1
b.
6
O4
This is an eightbar linkage with 4 binary (1, 4, 7, and 8) and 4 ternary (2, 3, 5, and 6) links. The inverted U-shaped link at the top of Figure P2-5b is represented here as the binary link 8. Number of links
L 8
Number of full joints
J1 10
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2 M1
5
6
2
3
7
8 2 O2
4 O4
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-25-1
PROBLEM 2-25 Statement:
Find the mobility of the ice tongs in Figure P2-6. a. When operating them to grab the ice block. b. When clamped to the ice block but before it is picked up (ice grounded). c. When the person is carrying the ice block with the tongs.
Solution:
See Figure P2-6 and Mathcad file P0225.
1.
Use equation 2.1c (Kutzbach's modification) to calculate the mobility.
a.
In this case there are two links and one full joint and 1 DOF. Number of links
L 2
Number of full joints
J1 1
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2 b.
When the block is clamped in the tongs another link and two more full joints are added reducing the DOF to zero (the tongs and ice block form a structure). Number of links
L 2 1
Number of full joints
J1 1 2
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2
c.
M1
M0
When the block is being carried the system has at least 4 DOF: x, y, and z position and orientation about a vertical axis.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-26-1
PROBLEM 2-26 Statement:
Find the mobility of the automotive throttle mechanism shown in Figure P2-7.
Solution:
See Figure P2-7 and Mathcad file P0226.
1.
This is an eightbar linkage with 8 binary links. It is assumed that the joint between the gas pedal (2) and the roller (3) that pivots on link 4 is a full joint, i.e. the roller rolls without slipping. The pivot pins at O2, O4, O6, and O8 are attached to the ground link (1). Use equation 2.1c (Kutzbach's modification) to calculate the mobility. Number of links
L 8
Number of full joints
J1 10
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2
M1
7 6 O6
8
FULL JOINT 5 4 O4 3 2
O2
O8
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-27-1
PROBLEM 2-27 Statement:
Sketch a kinematic diagram of the scissors jack shown in Figure P2-8 and determine its mobility. Describe how it works.
Solution:
See Figure P2-8 and Mathcad file P0227.
1.
The scissors jack depicted is a seven link mechanism with eight full and two half joints (see kinematic diagram below). Link 7 is a variable length link. Its length is changed by rotating the screw with the jack handle (not shown). The two blocks at either end of link 7 are an integral part of the link. The block on the left is threaded and acts like a nut. The block on the right is not threaded and acts as a bearing. Both blocks have pins that engage the holes in links 2, 3, 5, and 6. Joints A and B have 2 full joints apiece. For any given length of link 7 the jack is a structure (DOF = 0). When the screw is turned to give the jack a different height the jack has 1 DOF.
4
3
5
7
A
B 2
6 1
Number of links
L 7
Number of full joints
J1 8
Number of half joints
J2 2
M 3 ( L 1 ) 2 J1 J2
M0
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-28-1
PROBLEM 2-28 Statement:
Find the mobility of the corkscrew in Figure P2-9.
Solution:
See Figure P2-9 and Mathcad file P0228.
1.
The corkscrew is made from 4 pieces: the body (1), the screw (2), and two arms with teeth (3), one of which is redundant. The second arm is present to balance the forces on the assembly but is not necessary from a kinematic standpoint. So, kinematically, there are 3 links (body, screw, and arm), 2 full joints (sliding joint between the screw and the body, and pin joint where the arm rotates on the body), and 1 half joint where the arm teeth engage the screw "teeth". Using equation 2.1c, the DOF (mobility) is Number of links
L 3
Number of full joints
J1 2
Number of half joints
J2 1
M 3 ( L 1 ) 2 J1 J2
M1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-29-1
PROBLEM 2-29 Statement:
Figure P2-10 shows Watt's sun and planet drive that he used in his steam engine. The beam 2 is driven in oscillation by the piston of the engine. The planet gear is fixed rigidly to link 3 and its center is guided in the fixed track 1. The output rotation is taken from the sun gear 4. Sketch a kinematic diagram of this mechanism and determine its DOF. Can it be classified by the Barker scheme? If so, what Barker class and subclass is it?
Solution:
See Figure P2-10 and Mathcad file P0229.
1.
Sketch a kinematic diagram of the mechanism. The mechanism is shown on the left and a kinematic model of it is sketched on the right. It is a fourbar linkage with 1 DOF (see below).
A
2
2
1
3
3 1
4
B
4
1
2.
C
Use equation 2.1c to determine the DOF (mobility). There are 4 links, 3 full pin joints, 1 half pin-in-slot joint (at B), and 1 half joint (at the interface C between the two gears, shown above by their pitch circles). Links 1 and 3 are ternary. Kutzbach's mobility equation (2.1c) Number of links
L 4
Number of full joints
J1 3
Number of half joints
J2 2
M 3 ( L 1 ) 2 J1 J2 3.
M1
The Barker classification scheme requires that we have 4 link lengths. The motion of link 3 can be modeled by a basic fourbar if the half joint at B is replaced with a full pin joint and a link is added to connect B and the fixed pivot that is coincident with the center of curvature of the slot that guides pin B. L1 2.15
L2 1.25
L3 1.80
L4 0.54
This is a Grashof linkage and the Barker classification is I-4 (type 4) because the shortest link is the output.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-30-1
PROBLEM 2-30 Statement:
Figure P2-11 shows a bicycle hand brake lever assembly. Sketch a kinematic diagram of this device and draw its equivalent linkage. Determine its mobility. Hint: Consider the flexible cable to be a link.
Solution:
See Figure P2-11 and Mathcad file P0230.
1.
The motion of the flexible cable is along a straight line as it leaves the guide provided by the handle bar so it can be modeled as a translating full slider that is supported by the handlebar (link 1). The brake lever is a binary link that pivots on the ground link. Its other node is attached through a full pin joint to a third link, which drives the slider (link 4). Number of links
L 4
Number of full joints
J1 4
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2 M1
CABLE BRAKE LEVER 3 2
4 1 1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-31-1
PROBLEM 2-31 Statement:
Figure P2-12 shows a bicycle brake caliper assembly. Sketch a kinematic diagram of this device and draw its equivalent linkage. Determine its mobility under two conditions. a. b.
Brake pads not contacting the wheel rim. Brake pads contacting the wheel rim.
Hint: Consider the flexible cable to be replaced by forces in this case. Solution: 1.
See Figure P2-12 and Mathcad file P0231.
The rigging of the cable requires that there be two brake arms. However, kinematically they operate independently and can be analyzed that way. Therefore, we only need to look at one brake arm. When the brake pads are not contacting the wheel rim there is a single lever (link 2) that is pivoted on a full pin joint that is attached to the ground link (1). Thus, there are two links (frame and brake arm) and one full pin joint. Number of links
L 2
Number of full joints
J1 1
Number of half joints
J2 0
BRAKE ARM
FRAME
2
M 3 ( L 1 ) 2 J1 J2 M1
2.
1
When the brake pad contacts the wheel rim we could consider the joint between the pad, which is rigidly attached to the brake arm and is, therefore, a part of link 2, to be a half joint. The brake arm (with pad), wheel (which is constrained from moving laterally by the frame), and the frame constitute a structure. Number of links
L 2
Number of full joints
J1 1
Number of half joints
J2 1
BRAKE ARM
FRAME
2
M 3 ( L 1 ) 2 J1 J2 M0
1 1 HALF JOINT
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-32-1
PROBLEM 2-32 Statement:
Find the mobility, the Grashof condition, and the Barker classifications of the mechanism in Figure P2-13.
Solution:
See Figure P2-13 and Mathcad file P0232.
1.
Use equation 2.1c (Kutzbach's modification) to calculate the mobility. When there is no cable in the jaw or before the cable is crimped this is a basic fourbar mechanism with with 4 full pin joints: Number of links
L 4
Number of full joints
J1 4
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2
M1
When there is a cable in the jaw this is a threebar mechanism with with 3 full pin joints. While the cable is clamped the jaws are stationary with respect to each other so that link 4 is grounded along with link 1, leaving only three operational links.
2.
Number of links
L 3
Number of full joints
J1 3
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2
M0
Use inequality 2.8 to determine the Grashof condition and Table 2-4 to determine the Barker classification. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
L1 0.92
L2 0.27
L3 0.50
L4 0.60
Condition L1 L2 L3 L4 "non-Grashof" The Barker classification is II-1 (Type 5) RRR1 (non-Grashof, longest link grounded).
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-33-1
PROBLEM 2-33 Statement:
The approximate torque-speed curve and its equation for a 1/4 hp shunt-wound DC motor are shown in Figure P2-14. The rated speed for this fractional horsepower motor is 10000 rpm at a rated voltage of 130V. Determine: a) The rated torque in oz-in (ounce-inches, the industry standard for fractional hp motors) b) The no-load speed c) The operating speed range d) Plot the power-torque curve in the operating range and determine the maximum power that the motor can deliver in the that range.
Given:
Rated speed, N N ( T )
NR 10000 rpm
R
0.1 1.7
NR TR NR TR
HR
Rated power, H
R
1 4
hp
T 1.1 NR if T 62.5 ozf in T 5.1 NR otherwise
T 0 ozf in 2.5 ozf in 75 ozf in
12000
10000
Speed, rpm
8000
6000
4000
2000
0
0
25
50
75
100
Torque, oz-in
Figure P2-14 Torque-speed Characteristic of a 1/4 HP, 10000 rpm DC Motor Solution: a.
See Figure P2-3 and Mathcad file P0220.
The rated torque is found by dividing the rated power by the rated speed: Rated torque, TR
TR
HR NR
TR 25 ozf in
b.
The no-load speed occurs at T = 0. From the graph this is 11000 rpm.
c.
The operating speed range for a shunt-wound DC motor is the speed at which the motor begins to stall up to the no-load speed. For the approximate torque-speed curve given in this problem the minimum speed is defined as the speed at the knee of the curve. Nopmin N ( 62.5 ozf in)
Nopmin 8500 rpm
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-33-2
Nopmax N ( 0 ozf in)
The power is the product of the speed and the torque. From the graph the equation for the torque-speed curve over the operating range is: N ( T )
40 rpm ozf in
T 11000 rpm
and the power, therefore, is: H ( T ) N ( T ) T Plotting the power as a function of torque over the range T 0 ozf in 2.5 ozf in 62.5 ozf in 0.750 0.700 0.650 0.600 0.550 0.500 Power, hp
d.
Nopmax 11000 rpm
0.450 0.400 0.350 0.300 0.250 0.200 0.150 0.100 0.050 0.000 0.0
12.5
25.0
37.5
50.0
62.5
75.0
Torque, oz-in
Maximum power occurs at the maximum torque in the operating range. The value of T at maximum power is: Value of T at Hmax
THmax 62.5 ozf in
THmax 62.5 ozf in
Maximum power
Hmax H THmax
Hmax 0.527 hp
Speed at max power
NHmax N THmax
NHmax 8500 rpm
Note that the curve goes through the rated power point of 0.25 hp at the rated torque of 25 oz-in.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-34-1
PROBLEM 2-34 Statement:
Figure P2-15 shows a power hacksaw, used to cut metal. Link 5 pivots at O5 and its weight forces the sawblade against the workpiece while the linkage moves the blade (link 4) back and forth within link 5 to cut the part. Sketch its kinematic diagram, determine its mobility and its type (i.e., is it a fourbar, a Watt's sixbar, a Stephenson's sixbar, an eightbar, or what?) Use reverse linkage transformation to determine its pure revolute-jointed equivalent linkage.
Solution:
See Figure P2-15 and Mathcad file P0234.
1.
Sketch a kinematic diagram of the mechanism. The mechanism is shown on the left and a kinematic model of it is sketched on the right. It is a fivebar linkage with 1 DOF (see below). 5
3
5 3
4
4
2
2
2.
1
1
1
Use equation 2.1c to determine the DOF (mobility). There are 5 links, 4 full pin joints, 1 full sliding joint, and 1 half joint (at the interface between the hacksaw blade and the pipe being cut). Kutzbach's mobility equation (2.1c) Number of links
L 5
Number of full joints
J1 5
Number of half joints
J2 1
M 3 ( L 1 ) 2 J1 J2 3.
M1
Use rule 1 to transform the full sliding joint to a full pin joint for no change in DOF. Then use rules 2 and 3 by changing the half joint to a full pin joint and adding a link for no change in DOF. The resulting kinematically equivalent linkage has 6 links, 7 full pin joints, no half joints, and is shown below. Kutzbach's mobility equation (2.1c) Number of links
L 6
Number of full joints
J1 7
Number of half joints
J2 0
5 4
3 2
M 3 ( L 1 ) 2 J1 J2
1 6
M1
1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-35-1
PROBLEM 2-35 Statement:
Figure P2-16 shows a manual press used to compact powdered materials. Sketch its kinematic diagram, determine its mobility and its type (i.e., is it a fourbar, a Watt's sixbar, a Stephenson's sixbar, an eightbar, or what?) Use reverse linkage transformation to determine its pure revolute-jointed equivalent linkage.
Solution:
See Figure P2-16 and Mathcad file P0235.
1.
Sketch a kinematic diagram of the mechanism. The mechanism is shown on the left and a kinematic model of it is sketched on the right. It is a fourbar linkage with 1 DOF (see below).
4 3
4
3 2
2
O2
O2
2.
Use equation 2.1c to determine the DOF (mobility). There are 4 links, 3 full pin joints, 1 full sliding joint, and 0 half joints. This is a fourbar slider-crank. Kutzbach's mobility equation (2.1c) Number of links
L 4
Number of full joints
J1 4
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2 3.
M1
Use rule 1 to transform the full sliding joint to a full pin joint for no change in DOF. The resulting kinematically equivalent linkage has 4 links, 4 full pin joints, no half joints, and is shown below. 4 O4 3
2 O2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-36-1
PROBLEM 2-36 Statement:
Sketch the equivalent linkage for the cam and follower mechanism in Figure P2-17 in the position shown. Show that it has the same DOF as the original mechanism.
Solution:
See Figure P2-17 and Mathcad file P0236.
1.
The cam follower mechanism is shown on the left and a kinematically equivalent model of it is sketched on the right.
4 1
1 4
3
3
2
2
INSTANTANEOUS CENTER OF CURVATURE OF CAM SURFACE
1
1
2.
Use equation 2.1c to determine the DOF (mobility) of the original mechanism. There are 4 links, 2 full pin joints, 1 full sliding joint, 1 pure rolling joint and 0 half joints. Kutzbach's mobility equation (2.1c) Number of links
L 4
Number of full joints
J1 4
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2 3.
M1
Use equation 2.1c to determine the DOF (mobility) of the equivalent mechanism. There are 4 links, 3 full pin joints, 1 full sliding joint, and 0 half joints. This is a fourbar slider-crank. Kutzbach's mobility equation (2.1c) Number of links
L 4
Number of full joints
J1 4
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2
M1
DESIGN OF MACHINERY - 5th Ed.
SOLUTIONS MANUAL 2-37-1
PROBLEM 2-37 Statement:
Describe the motion of the following rides, commonly found at an amusement park, as pure rotation, pure translation, or complex planar motion. a. A Ferris wheel b. A "bumper" car c. A drag racer ride d. A roller coaster whose foundation is laid out in a straight line e. A boat ride through a maze f. A pendulum ride g. A train ride
Solution:
See Mathcad file P0211.
a.
A Ferris wheel Pure rotation.
b.
A "bumper car" Complex planar motion.
c.
A drag racer ride Pure translation.
d.
A roller coaster whose foundation is laid out in a straight line Complex planar motion.
e.
A boat ride through a maze Complex planar motion.
f.
A pendulum ride Pure rotation.
g.
A train ride Complex planar motion.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-38-1
PROBLEM 2-38 Statement:
Figure P2-1a is an example of a mechanism. Number the links, starting with 1. (Hint: Don't forget the "ground" link.) Letter the joints alphabetically, starting with A. a. Using the link numbers, describe each link as binary, ternary, etc. b. Using the joint letters, determine each joint's order. c. Using the joint letters, determine whether each is a half or full joint.
Solution:
See Figure P2-1a and Mathcad file P0238.
1.
Label the link numbers and joint letters for Figure P2-1a.
G B
3 C
5
2 A
4
1
1
6
F
E
D 1
a.
Using the link numbers, describe each link as binary, ternary, etc. Link No. 1 2 3 4 5 6
Link Order Ternary Ternary Binary Ternary Binary Binary
b,c. Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C D E F G
Joint Order 1 1 1 1 1 2 1
Half/Full Full Full Full Half Full Full Full
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-39-1
PROBLEM 2-39 Statement:
Figure P2-1b is an example of a mechanism. Number the links, starting with 1. (Hint: Don't forget the "ground" link.) Letter the joints alphabetically, starting with A. a. Using the link numbers, describe each link as binary, ternary, etc. b. Using the joint letters, determine each joint's order. c. Using the joint letters, determine whether each is a half or full joint.
Solution:
See Figure P2-1b and Mathcad file P0239.
1.
Label the link numbers and joint letters for Figure P2-1b.
3 C 1 B
2 A 1
a.
Using the link numbers, describe each link as binary, ternary, etc. Link No. 1 2 3
Link Order Binary Binary Binary
b,c. Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C
Joint Order 1 1 1
Half/Full Full Half Full
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-40-1
PROBLEM 2-40 Statement:
Figure P2-1c is an example of a mechanism. Number the links, starting with 1. (Hint: Don't forget the "ground" link.) Letter the joints alphabetically, starting with A. a. Using the link numbers, describe each link as binary, ternary, etc. b. Using the joint letters, determine each joint's order. c. Using the joint letters, determine whether each is a half or full joint.
Solution:
See Figure P2-1c and Mathcad file P0240.
1.
Label the link numbers and joint letters for Figure P2-1c. 4 1 C
D
3
2 B
A 1
a.
Using the link numbers, describe each link as binary, ternary, etc. Link No. 1 2 3 4
Link Order Binary Binary Binary Binary
b,c. Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C D
Joint Order 1 1 1 1
Half/Full Full Full Full Full
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-41-1
PROBLEM 2-41 Statement:
Figure P2-1d is an example of a mechanism. Number the links, starting with 1. (Hint: Don't forget the "ground" link.) Letter the joints alphabetically, starting with A. a. Using the link numbers, describe each link as binary, ternary, etc. b. Using the joint letters, determine each joint's order. c. Using the joint letters, determine whether each is a half or full joint.
Solution:
See Figure P2-1d and Mathcad file P0241.
1.
Label the link numbers and joint letters for Figure P2-1d. H 1
7 G
6
E 5
F
A 1
D
2 3 B
a.
4 C
Using the link numbers, describe each link as binary, ternary, etc. Link No. 1 2 3 4 5 6 7
Link Order Binary Binary Ternary Binary Binary Ternary Binary
b,c. Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C D E F G H
Joint Order 1 1 1 1 1 1 1 1
Half/Full Full Full Half Full Full Full Full Full
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-42-1
PROBLEM 2-42 Statement:
Find the mobility, Grashof condition and Barker classification of the oil field pump shown in Figure P2-18.
Solution:
See Figure P2-18 and Mathcad file P0242.
1.
Use inequality 2.8 to determine the Grashof condition and Table 2-4 to determine the Barker classification. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
4 O4 3
O2
2
This is a basic fourbar linkage. The input is the 14-in-long crank (link 2) and the output is the top beam (link 4). The mobility (DOF) is found using equation 2.1c (Kutzbach's modification): Number of links
L 4
Number of full joints
J1 4
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2
M1
The link lengths and Grashof condition are L1
2
( 76 12) 47.5
2
L1 79.701
Condition L1 L2 L3 L4 "Grashof" This is a Barker Type 2 GCRR (Grashof, shortest link is input).
L2 14
L3 80
L4 51.26
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-43-1
PROBLEM 2-43 Statement:
Find the mobility, Grashof condition and Barker classification of the aircraft overhead bin shown in Figure P2-19.
Solution:
See Figure P2-19 and Mathcad file P0243.
1.
Use inequality 2.8 to determine the Grashof condition and Table 2-4 to determine the Barker classification. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
2.79 O2
6.95
B
2
9.17
9.17 4 3 O4
9.57
A
9.17
This is a basic fourbar linkage. The input is the link 2 and the output is link 4. The mobility (DOF) is found using equation 2.1c (Kutzbach's modification): Number of links
L 4
Number of full joints
J1 4
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2
The link lengths and Grashof condition are 2
2
L1 7.489
L2 9.17
2
2
L3 12.968
L4 9.57
L1
2.79 6.95
L3
9.17 9.17
Condition L1 L2 L3 L4 "non-Grashof" This is a Barker Type 7 RRR3 (non-Grashof, longest link is coupler).
M1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-44-1
PROBLEM 2-44 Statement:
Figure P2-20 shows a "Rube Goldberg" mechanism that turns a light switch on when a room door is opened and off when the door is closed. The pivot at O2 goes through the wall. There are two spring-loaded piston-in cylinder devices in the assembly. An arrangement of ropes and pulleys inside the room transfers the door swing into a rotation of link 2. Door opening rotates link 2 CW, pushing the switch up as shown in the figure, and door closing rotates link 2 CCW, pulling the switch down. Find the mobility of the linkage.
Solution:
See Figure P2-20 and Mathcad file P0244.
1.
2.
Examination of the figure shows 20 links (including the the switch) and 28 full joints. The second piston-in cylinder that actuates the switch is counted as a single binary link of variable length with joints at its ends. The other cylinder consists of two binary links, each link having one pin joint and one slider joint. There are no half joints. Use equation 2.1c to determine the DOF (mobility). Kutzbach's mobility equation (2.1c) Number of links
L 20
Number of full joints
J1 28
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2 3.
M1
An alternative is to ignore the the first piston-in cylinder that acts on the third bellcrank from O2 since it does not affect the the motion of the linkage (it acts only as a damper.) In that case, subtract two links and three full joints, giving L = 18, J1 = 25 and M = 1.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-45-1
PROBLEM 2-45 Statement:
All of the eightbar linkages in Figure 2-11 part 2 have eight possible inversions. Some of these will give motions similar to others. Those that have distinct motions are called distinct inversions. How many distinct inversions does the linkage in row 4, column 1 have?
Solution:
See Figure 2-11, part 2 and Mathcad file P0245.
1.
This isomer has one quaternary, two ternary, and five binary links arranged in a symetrical fashion. Due to this symmetry, grounding link 2 or 7 gives the same inversion, as do grounding 3 or 6 and 4 or 5. This makes 3 of the possible 8 inversions the same leaving 5 distinct inversions. Distinct inversions are obtained by grounding link 1, 2, 3, 4, or 8 (or 1, 5, 6, 7, or 8) for a total of 5 distinct inversions.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-46-1
PROBLEM 2-46 Statement:
All of the eightbar linkages in Figure 2-11 part 2 have eight possible inversions. Some of these will give motions similar to others. Those that have distinct motions are called distinct inversions. How many distinct inversions does the linkage in row 4, column 2 have?
Solution:
See Figure 2-11, part 2 and Mathcad file P0246.
1.
This isomer has four ternary, and four binary links arranged in a symetrical fashion. Due to this symmetry, grounding link 1 or 5 gives the same inversion, as do grounding 2 or 8, 4 or 6, and 3 or 7. This makes 4 of the possible 8 inversions the same leaving 4 distinct inversions. Distinct inversions are obtained by grounding link 1, 2, 3, or 4 (or 5, 6, 7, or 8) for a total of 4 distinct inversions.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-47-1
PROBLEM 2-47 Statement:
All of the eightbar linkages in Figure 2-11 part 2 have eight possible inversions. Some of these will give motions similar to others. Those that have distinct motions are called distinct inversions. How many distinct inversions does the linkage in row 4, column 3 have?
Solution:
See Figure 2-11, part 2 and Mathcad file P0247.
1.
This isomer has four ternary, and four binary links arranged in a symetrical fashion. Due to this symmetry, grounding link 2 or 4 gives the same inversion, as does grounding 5 or 7. This makes 2 of the possible 8 inversions the same leaving 6 distinct inversions. Distinct inversions are obtained by grounding link 1, 2, 3, 5, 6 or 8 (or 1, 3, 4, 6, 7, or 8) for a total of 6 distinct inversions.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-48-1
PROBLEM 2-48 Statement:
Find the mobility of the mechanism shown in Figure 3-33.
Solution:
See Figure 3-33 and Mathcad file P0248.
1.
Use equation 2.1c to determine the DOF (mobility). There are 6 links, 7 full pin joints (two at B), and no half-joints.
Kutzbach's mobility equation (2.1c) Number of links
L 6
Number of full joints
J1 7
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2
M1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-49-1
PROBLEM 2-49 Statement:
Find the mobility of the mechanism shown in Figure 3-34.
Solution:
See Figure 3-34 and Mathcad file P0249.
1.
Use equation 2.1c to determine the DOF (mobility). There are 6 links, 7 full pin joints, and no half-joints. Kutzbach's mobility equation (2.1c) Number of links
L 6
Number of full joints
J1 7
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2
M1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-50-1
PROBLEM 2-50 Statement:
Find the mobility of the mechanism shown in Figure 3-35.
Solution:
See Figure 3-35 and Mathcad file P0250.
1.
Use equation 2.1c to determine the DOF (mobility). There are 6 links, 7 full pin joints, and no half-joints. Kutzbach's mobility equation (2.1c) Number of links
L 6
Number of full joints
J1 7
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2
M1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-51-1
PROBLEM 2-51 Statement:
Find the mobility of the mechanism shown in Figure 3-36.
Solution:
See Figure 3-36 and Mathcad file P0251.
1.
Use equation 2.1c to determine the DOF (mobility). There are 8 links, 10 full pin joints (two at O4), and no half-joints. Kutzbach's mobility equation (2.1c) Number of links
L 8
Number of full joints
J1 10
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2
M1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-52-1
PROBLEM 2-52 Statement:
Find the mobility of the mechanism shown in Figure 3-37.
Solution:
See Figure 3-37 and Mathcad file P0252.
1.
Use equation 2.1c to determine the DOF (mobility). There are 6 links, 7 full pin joints (two at O4), and no half-joints. Kutzbach's mobility equation (2.1c) Number of links
L 6
Number of full joints
J1 7
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2
M1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-53-1
PROBLEM 2-53 Statement:
Figure P2-1e is an example of a mechanism. Number the links, starting with 1. (Hint: Don't forget the "ground" link.) Letter the joints alphabetically, starting with A. a. Using the link numbers, describe each link as binary, ternary, etc. b. Using the joint letters, determine each joint's order. c. Using the joint letters, determine whether each is a half or full joint.
Solution:
See Figure P2-1e and Mathcad file P0253.
1.
Label the link numbers and joint letters for Figure P2-1e. J
K
8
I
8 1
9
1
L 10
M
7
a.
4 D C
A 1
2
2 3
E
H
B 5
F
Using the link numbers, describe each link as binary, ternary, etc. Link No. 1 2 3 4 5 6
Link Order 5 nodes Quaternary Binary Binary Binary Binary
Link No. 7 8 9 10
6
G 1
b,c. Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C D E F G H I J K L M
Joint Order 1 1 1 1 1 1 1 1 1 1 1 1 1
Half/Full Full Full Full Full Full Full Full Full Full Full Full Full Full
Joint Classification Grounded rotating joint Moving rotating joint Pure rolling joint Grounded rotating joint Moving rotating joint Moving translating joint Grounded rotating joint Moving rotating joint Moving rotating joint Grounded rotating joint Moving translating joint Moving rotating joint Grounded translating joint
Link Order Binary Ternary Binary Binary
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-54-1
PROBLEM 2-54 Statement:
Figure P2-1f is an example of a mechanism. Number the links, starting with 1. (Hint: Don't forget the "ground" link.) Letter the joints alphabetically, starting with A. a. Using the link numbers, describe each link as binary, ternary, etc. b. Using the joint letters, determine each joint's order. c. Using the joint letters, determine whether each is a half or full joint.
Solution:
See Figure P2-1f and Mathcad file P0254.
1.
Label the link numbers and joint letters for Figure P2-1f.
F 5
E
5 1 6
1
G
H
4
a.
C 3
1
B A 1
2
Using the link numbers, describe each link as binary, ternary, etc. Link No. 1 2 3 4 5 6
Link Order Quaternary Binary Ternary Binary Ternary Binary
b,c. Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C D E F G H
Joint Order 1 1 1 1 1 1 1 1
Half/Full Full Full Full Full Full Full Full Full
Joint Classification Grounded rotating joint Moving half joint Grounded translating joint Moving rotating joint Moving rotating joint Grounded rotating joint Moving half joint Grounded translating joint
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-55-1
PROBLEM 2-55 Statement:
Figure P2-1g is an example of a mechanism. Number the links, starting with 1. (Hint: Don't forget the "ground" link.) Letter the joints alphabetically, starting with A. a. Using the link numbers, describe each link as binary, ternary, etc. b. Using the joint letters, determine each joint's order. c. Using the joint letters, determine whether each is a half or full joint.
Solution:
See Figure P2-1g and Mathcad file P0255.
1.
Label the link numbers and joint letters for Figure P2-1g.
I
D 4 E
5
1
4 C 2
1
a.
B
G
A
A
1
F
7
6
3
H
7 1 J
2 1
8 1
K
Using the link numbers, describe each link as binary, ternary, etc. Link No. 1 2 3 4
Link Order 5 nodes Binary Binary Ternary
Link No. 5 6 7 8
Link Order Binary Binary Ternary Binary
b,c. Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C D E F G H I J K
Joint Order 1 1 1 1 1 1 1 1 1 1 1
Half/Full Full Full Full Full Half Full Full Full Full Half Full
Joint Classification Grounded rotating joint Moving rolling joint Moving rotating joint Grounded rotating joint Moving sliding joint Grounded translating joint Moving rolling joint Moving rotating joint Grounded rotating joint Moving sliding joint Grounded translating joint
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-56-1
PROBLEM 2-56 Statement:
For the example linkage shown in Figure 2-4 find the number of links and their respective link orders, the number of joints and their respective orders, and the mobility of the linkage.
Solution:
See Figure 2-4 and Mathcad file P0256.
1.
Label the link numbers and joint letters for Figure 2-4 example.
K
1
9 8
I G
J
6
7 H
D 1
4
E
3 B A 1
C 2
1
5 F 1
2.
Using the link numbers, describe each link as binary, ternary, etc. Link No. 1 2 3 4 5
3.
Link No. 6 7 8 9
Link Order Ternary Binary Binary Binary
Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C D E F G H I J K
4.
Link Order 5 nodes Binary Ternary Binary Binary
Joint Order 1 1 1 2 1 1 1 1 1 1 1
Half/Full Full Half Full Full Full Full Full Full Full Full Full
Joint Classification Grounded rotating joint Moving sliding joint Grounded rotating joint Moving rotating joint Moving translating joint Grounded rotating joint Moving rotating joint Grounded rotating joint Moving rotating joint Moving rotating joint Grounded translating joint
Use equation 2.1c to calculate the DOF (mobility). Kutzbach's mobility equation (2.1c) Number of links
L 9
M 3 ( L 1 ) 2 J1 J2
Number of full joints M1
J1 11
Number of half joints
J2 1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-57-1
PROBLEM 2-57 Statement:
For the linkage shown in Figure 2-5b find the number of joints and their respective orders, and mobility for: a) The condition of a finite load W in the direction shown and a zero F b) The condition of a finite load W and a finite load F both in the directions shown after link 6 is off the stop.
Solution:
See Figure 2-5b and Mathcad file P0257.
1.
Label the link numbers and joint letters for Figure 2-5b. 1
6
O6
W
D 3
A
B
1
F 5
4 O4
2 O2
1
C
1
a)
The condition of a finite load W in the direction shown and a zero F: Using the joint letters, determine each joint's order and whether each is a half or full joint. Link 6 is grounded so joint D is a grounded rotating joint and O6 is not a joint. For this condition there is a total of 6 full joints and no half joints. Joint Letter O2 B C D O4 A
Joint Order 1 1 1 1 1 1
Half/Full Full Full Full Full Full Full
Joint Classification Grounded rotating joint Moving rotating joint Moving rotating joint Grounded rotating joint Grounded rotating joint Moving rotating joint
Use equation 2.1c to calculate the DOF (mobility). Kutzbach's mobility equation (2.1c) Number of links L 5
Number of full joints
M 3 ( L 1 ) 2 J1 J2
J1 6
Number of half joints
J2 0
M0
b) The condition of a finite load W and a finite load F both in the directions shown after link 6 is off the stop. Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C D O2 O4 O6
Joint Order 1 1 1 1 1 1 1
Half/Full Full Full Full Full Full Full Full
Joint Classification Moving rotating joint Moving rotating joint Moving rotating joint Moving rotating joint Grounded rotating joint Grounded rotating joint Grounded rotating joint
Use equation 2.1c to calculate the DOF (mobility). Kutzbach's mobility equation (2.1c) Number of links L 6 M 3 ( L 1 ) 2 J1 J2
Number of full joints M1
J1 7
Number of half joints
J2 0
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-58-1
PROBLEM 2-58 Statement:
Figure P2-21a shows a "Nuremberg scissors" mechanism. Find its mobility.
Solution:
See Figure P2-21a and Mathcad file P0258.
1.
Use equation 2.1c to calculate the DOF (mobility). Kutzbach's mobility equation (2.1c) Number of links
L 10
Number of full joints
J1 13
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2
M1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-59-1
PROBLEM 2-59 Statement:
Figure P2-21b shows a mechanism. Find its mobility and classify its isomer type.
Solution:
See Figure P2-21b and Mathcad file P0259.
1.
Use equation 2.1c to calculate the DOF (mobility). Kutzbach's mobility equation (2.1c) Number of links
L 6
Number of full joints
J1 7
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2
2.
M1
Using Figure 2-9, we see that the mechanism is a Stephenson's sixbar isomer ( the two ternary links are connected with two binary links and one dyad).
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-60-1
PROBLEM 2-60 Statement:
Figure P2-21c shows a circular saw mounted on the coupler of a fourbar linkage. The centerline of the saw blade is at a coupler point that moves in an approximate straight line. Draw its kinematic diagram and determine its mobility.
Solution:
See Figure P2-21c and Mathcad file P0260.
1.
Draw a kinematic diagram of the mechanism. The saw's rotation axis is at point P and the saw is attached to link 3.
A
B 3 4
2 O2
O4
P
1
1 2.
Use equation 2.1c to calculate the DOF (mobility). Kutzbach's mobility equation (2.1c) Number of links
L 4
Number of full joints
J1 4
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2
M1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-61-1
PROBLEM 2-61 Statement:
Figure P2-21d shows a log transporter. Draw a kinematic diagram of the mechanism, specify the number of links and joints, and then determine its mobility: a) For the transporter wheels locked and no log in the "claw" of the mechanism b) For the transporter wheels locked with it lifting a log c) For the transporter moving a log to a destination in a straight line.
Solution:
See Figure P2-21d and Mathcad file P0261.
1.
Draw a kinematic diagram of the mechanism. Link 1 is the frame of the transporter. Joint B is of order 3. Actuators E and F provide two inputs (to get x-y motion) and actuator H provides an additional input for clamping logs. G
D
9 12 H
9
C
I 9
11
8
J 10
4 A
3
2 O2
B
5
7
E O4
1
F
1
6 O6 1
a)
Wheels locked, no log in "claw." Kutzbach's mobility equation (2.1c) Number of links
L 12
Number of full joints
J1 15
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2
M3
b) Wheels locked, log held tighly in the "claw." With a log held tightly between links 9 and 10 a structure will be formed by links 9 through 12 and the log so that there will only be 9 links and 11 joints active. Number of links
L 9
Number of full joints
J1 11
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2 c)
M2
Transporter moving in a straight line with the log holding mechanism inactive. There are two tires, the transporter frame, and the ground, making 4 links and two points of contact with the ground and two axels, making 4 joints.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-61-2
Number of links
L 4
Number of full joints
J1 4
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2
M1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-62-1
PROBLEM 2-62 Statement:
Figure P2-21d shows a plow mechanism attached to a tractor. Draw its kinematic diagram and find its mobility including the earth as a "link." a) When the tractor is stopped and the turnbuckle is fixed. (Hint: Consider the tractor and the wheel to be one with the earth.) b) When the tractor is stopped and the turnbuckle is being adjusted. (Same hint.) c) When the tractor is moving and the turnbuckle is fixed. (Hint: Add the moving tractor's DOF to those found in part a.)
Solution:
See Figure P2-21e and Mathcad file P0262.
1.
Draw a kinematic diagram of the mechanism with the ground, tractor wheels, and tractor frame as link 1. Joint O4 is of order 2 and joint F is a half joint. The plow and its truss structure attach at joints D and E. Since the turnbuckle is fixed it can be modeled as a single binary link (6).
C
6
5 O4 1
D
B
4
7 7
3 A 2
7
2 O2
E 1
F
7 1
a)
Tractor stopped and turnbuckle fixed. Kutzbach's mobility equation (2.1c) Number of links
L 7
Number of full joints
J1 8
Number of half joints
J2 1
M 3 ( L 1 ) 2 J1 J2
M1
b) When the tractor is stopped and the turnbuckle is being adjusted. Between joints C and D we now have 2 net links (2 links threaded LH and RH on one end and the turnbuckle body) and 1 additional helical full joint. Number of links L 8 Number of full joints
J1 9
Number of half joints
J2 1
M 3 ( L 1 ) 2 J1 J2 c)
M2
When the tractor is moving and the turnbuckle is fixed. If the tractor moved only in a straight line we would add 1 DOF to the 1 DOF that we got in part a for a total of M = 2. More realistically, the tractor can turn and move up and down hills so that we would add 3 DOF to the 1 DOF of part a to get a total of 4 DOF.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-63-1
PROBLEM 2-63 Statement:
Figure P2-22 shows a Hart's inversor sixbar linkage. a) Is it a Watt or Stephenson linkage? b) Determine its inversion, i.e. is it a type I, II, or III?
Solution:
See Figure P2-22, Figure 2-14, and Mathcad file P0263.
1.
From Figure 2-14 we see that the Watt's sixbar has the two ternary links connected with a common joint while the Stephenson's sixbar has the two ternary links connected by binary links. Thus, Hart's inversor is a Watt's sixbar (links 1 and 2, the ternary links, are connected at a common joint). Further, the Hart's linkage is a Watt's sixbar inversion I since neither of the ternary links is grounded.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-64-1
PROBLEM 2-64 Statement:
Figure P2-23 shows the top view of the partially open doors on one side of an entertainment center cabinet. The wooden doors are hinged to each other and one door is hinged to the cabinet. There is also a ternary, metal link attached to the cabinet and door through pin joints. A spring-loaded piston-in cylinder device attaches to the ternary link and the cabinet through pin joints. Draw a kinematic diagram of the door system and find the mobility of this mechanism.
Solution:
See Figure P2-23 and Mathcad file P0264.
1.
Draw the kinematic diagram of this sixbar mechanism. The spring-loaded piston is just an in-line sliding joint (links 5 and 6, and joint F). The doors are binary links (3 and 4), and the metal ternary link (2) has nodes at A, B, and C. Link 1 is the cabinet.
A Cabinet
1
2
4
E Cabinet
B
D
Door
1
5 Cylinder F 6
Door
2
3
Link
G Cabinet 1 C
2.
Use equation 2.1c (Kutzbach's modification) to calculate the mobility. Number of links
L 6
Number of full joints
J1 7
Number of half joints
J2 0 M 3 ( L 1 ) 2 J1 J2
M1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-65-1
PROBLEM 2-65 Statement:
Figure P2-24a shows the seat and seat-back of a reclining chair with the linkage that connects them to the chair frame. Draw its kinematic diagram and determine its mobility with respect to the frame of the chair.
Solution:
See Figure P2-24a and Mathcad file P0265.
1.
Draw a kinematic diagram of the mechanism. The chair-back attaches to link 2 and the seat with the attached slider slot is link 3. The node at at C is a half-joint as it allows two degrees of freedom.
A 1 2
3
B
C 2.
Determine the mobility of the mechanism. Kutzbach's mobility equation (2.1c) Number of links
L 3
Number of full joints
J1 2
Number of half joints
J2 1
M 3 ( L 1 ) 2 J1 J2
M1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-66-1
PROBLEM 2-66 Statement:
Figure P2-24b shows the mechanism used to extend the foot support on a reclining chair. Draw its kinematic diagram and determine its mobility with respect to the frame of the chair.
Solution:
See Figure P2-24b and Mathcad file P0266.
1.
Draw a kinematic diagram of the mechanism. Link 1 is the frame.
D E 1
J
4 6
3
C 4
5
G
5 6
A 3 1
7
F
H
2
B 2.
Determine the mobility of the mechanism. Kutzbach's mobility equation (2.1c) Number of links
L 8
Number of full joints
J1 10
Number of half joints
J2 0
M 3 ( L 1 ) 2 J1 J2
8
M1
K
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-67-1
PROBLEM 2-67 Statement:
Figure P2-24b shows the mechanism used to extend the foot support on a reclining chair. Number the links, starting with 1. (Hint: Don't forget the "ground" link.) Letter the joints alphabetically, starting with A. a. Using the link numbers, describe each link as binary, ternary, etc. b. Using the joint letters, determine each joint's order. c. Using the joint letters, determine whether each is a half or full joint.
Solution:
See Figure P2-24b and Mathcad file P0267.
1.
Label the link numbers and joint letters for Figure P2-24b.
D E 1
J
4 6
3
C 4
5
G
5 6
A 3 1
F
2
8
7
H
B a.
Using the link numbers, describe each link as binary, ternary, etc. Link No. 1 2 3 4 5 6 7 8
Link Order Binary Binary Ternary Ternary Ternary Ternary Binary Binary
b,c. Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C D E F G H J K
Joint Order 1 1 1 1 1 1 1 1 1 1
Half/Full Full Full Full Half Full Full Full Full Full Full
K
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 2-68-1
PROBLEM 2-68 Statement:
Figure P2-24 shows a sixbar linkage. a) Is it a Watt or Stephenson linkage? b) Determine its inversion, i.e. is it a type I, II, or III?
Solution:
See Figure P2-24, Figure 2-14, and Mathcad file P0268.
1.
From Figure 2-14 we see that the Watt's sixbar has the two ternary links connected with a common joint while the Stephenson's sixbar has the two ternary links connected by binary links. Thus, the sixbar linkage shown is a Watt's sixbar (links 3 and 4, the ternary links, are connected at a common joint). Further, the linkage shown is a Watt's sixbar inversion I since neither of the ternary links is grounded.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-1-1
PROBLEM 3-1 Statement:
Define the following examples as path, motion, or function generation cases. a. b. c. d. e.
Solution:
A telescope aiming (star tracking) mechanism A backhoe bucket control mechanism A thermostat adjusting mechanism A computer printing head moving mechanism An XY plotter pen control mechanism
See Mathcad file P0301.
a.
Path generation. A star follows a 2D path in the sky.
b.
Motion generation. To dig a trench, say, the position and orientation of the bucket must be controlled.
c.
Function generation. The output is some desired function of the input over some range of the input.
d.
Path generation. The head must be at some point on a path.
e.
Path generation. The pen follows a straight line from point to point.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-2-1
PROBLEM 3-2 Statement:
Design a fourbar Grashof crank-rocker for 90 deg of output rocker motion with no quick return. (See Example 3-1.) Build a cardboard model and determine the toggle positions and the minimum transmission angle.
Given:
Output angle
Solution:
See Example 3-1 and Mathcad file P0302.
Design choices: 1.
θ 90 deg
Link lengths:
L3 6.000
Link 3
L4 2.500
Link 4
2.
Draw the output link O4B in both extreme positions, B1 and B2, in any convenient location such that the desired angle of motion 4 is subtended. In this solution, link 4 is drawn such that the two extreme positions each make an angle of 45 deg to the vertical. Draw the chord B1B2 and extend it in any convenient direction. In this solution it was extended to the left.
3.
Layout the distance A1B1 along extended line B1B2 equal to the length of link 3. Mark the point A1.
4.
Bisect the line segment B1B2 and layout the length of that radius from point A1 along extended line B1B2. Mark the resulting point O2 and draw a circle of radius O2A1 with center at O2.
5.
Label the other intersection of the circle and extended line B1B2, A2.
6.
Measure the length of the crank (link 2) as O2A1 or O2A2. From the graphical solution, L2 1.76775
7.
Measure the length of the ground link (link 1) as O2O4. From the graphical solution, L1 6.2550 1.7677
6.0000
3.5355
2
A2
A1
3
B2
B1
O2 90.00° 1 4 6.2550
8.
O4
Find the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition L1 L2 L3 L4 "Grashof"
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-3-1
PROBLEM 3-3 Statement:
Design a fourbar mechanism to give the two positions shown in Figure P3-1 of output rocker motion with no quick-return. (See Example 3-2.) Build a cardboard model and determine the toggle positions and the minimum transmission angle.
Given:
Coordinates of A1, B1, A2, and B2 (with respect to A1):
Solution:
xA1 0.00
xB1 1.721
xA2 2.656
xB2 5.065
yA1 0.00
yB1 1.750
yA2 0.751
yB2 0.281
See Figure P3-1 and Mathcad file P0303.
Design choices:
Link length:
Link 3
L3 5.000
Link 4
L4 2.000
1.
Following the notation used in Example 3-2 and Figure 3-5, change the labels on points A and B in Figure P3-1 to C and D, respectively. Draw the link CD in its two desired positions, C1D1 and C2D2, using the given coordinates.
2.
Draw construction lines from C1 to C2 and D1 to D2.
3.
Bisect line C1C2 and line D1D2 and extend their perpendicular bisectors to intersect at O4.
4.
Using the length of link 4 (design choice) as a radius, draw an arc about O4 to intersect both lines O4C1 and O4C2. Label the intersections B1 and B2.
5.
Draw the chord B1B2 and extend it in any convenient direction. In this solution it was extended to the left.
6.
Layout the distance A1B1 along extended line B1B2 equal to the length of link 3. Mark the point A1.
7.
Bisect the line segment B1B2 and layout the length of that radius from point A1 along extended line B1B2. Mark the resulting point O2 and draw a circle of radius O2A1 with center at O2.
8.
Label the other intersection of the circle and extended line B1B2, A2.
9.
Measure the length of the crank (link 2) as O2A1 or O2A2. From the graphical solution, L2 0.9469
10. Measure the length of the ground link (link 1) as O2O4. From the graphical solution, L1 5.3013
5.3013 5.0000 0.9469 A1 2
O2
O4
1
A2
3
B1
C1
4
R2.000
B2 C2 D1
D2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-3-2
11. Find the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition L1 L2 L3 L4 "Grashof"
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-4-1
PROBLEM 3-4 Statement:
Design a fourbar mechanism to give the two positions shown in Figure P3-1 of coupler motion. (See Example 3-3.) Build a cardboard model and determine the toggle positions and the minimum transmission angle. Add a driver dyad. (See Example 3-4.)
Given:
Position 1 offsets:
Solution:
See figure below for one possible solution. Input file P0304.mcd from the solutions manual disk to the Mathcad program for this solution, file P03-04.4br to the program FOURBAR to see the fourbar solution linkage, and file P03-04.6br into program SIXBAR to see the complete sixbar with the driver dyad included.
xA1B1 1.721 in
yA1B1 1.750 in
1.
Connect the end points of the two given positions of the line AB with construction lines, i.e., lines from A1 to A2 and B1 to B2.
2.
Bisect these lines and extend their perpendicular bisectors in any convenient direction. In the solution below the bisector of A1A2 was extended downward and the bisector of B1B2 was extended upward.
3.
Select one point on each bisector and label them O4 and O6, respectively. In the solution below the distances O4A and O6B were each selected to be 4.000 in. This resulted in a ground-link-length O4O6 for the fourbar of 6.457 in.
4.
The fourbar stage is now defined as O4ABO6 with link lengths Link 5 (coupler) L5
2
xA1B1 yA1B1
Link 4 (input)
L4 4.000 in
Ground link 1b
L1b 6.457 in
2
L5 2.454 in Link 6 (output)
L6 4.000 in
5.
Select a point on link 4 (O4A) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it D. (Note that link 4 is now a ternary link with nodes at O4, D, and A.) In the solution below the distance O4D was selected to be 2.000 in.
6.
Draw a construction line through D1D2 and extend it to the left.
7.
Select a point on this line and call it O2. In the solution below the distance CD was selected to be 4.000 in.
8.
Draw a circle about O2 with a radius of one-half the length D1D2 and label the intersections of the circle with the extended line as C1 and C2. In the solution below the radius was measured as 0.6895 in.
9.
The driver fourbar is now defined as O2CDO4 with link lengths Link 2 (crank)
L2 0.6895 in
Link 4a (rocker) L4a 2.000 in
Link 3 (coupler) L3 4.000 in Link 1a (ground) L1a 4.418 in
10. Use the link lengths in step 9 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 2). Shortest link
S L2
S 0.6895 in
Longest link
L L1a
L 4.4180 in
Other links
P L3
P 4.0000 in
Q L4a
Q 2.0000 in
DESIGN OF MACHINERY - 5th Ed.
Condition( a b c d )
SOLUTION MANUAL 3-4-2
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( S L P Q) "Grashof" O6
6
Ground Link 1b
A1
6 50.231°
5 C1
A2
5
B2
47.893°
2 O2 3
4
C2
B1 D1
4
3 D2
Ground Link 1a
O4
11. Using the program FOURBAR and the link lengths given above, it was found that the fourbar O4ABO6 is non-Grashoff with toggle positions at 2 = -71.9 deg and +71.9 deg. The minimum transmission angle is 35.5 deg. The fourbar operates between 2 = +21.106 deg and -19.297 deg.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-5-1
PROBLEM 3-5 Statement:
Design a fourbar mechanism to give the three positions of coupler motion with no quick return shown in Figure P3-2. (See also Example 3-5.) Ignore the points O2 and O4 shown. Build a cardboard model and determine the toggle positions and the minimum transmission angle. Add a driver dyad.
Solution:
See Figure P3-2 and Mathcad file P0305.
Design choices: L5 4.250
Length of link 5:
L4b 1.375
Length of link 4b:
1.
Draw link CD in its three design positions C1D1, C2D2, C3D3 in the plane as shown.
2.
Draw construction lines from point C1 to C2 and from point C2 to C3.
3.
Bisect line C1C2 and line C2C3 and extend their perpendicular bisectors until they intersect. Label their intersection O2.
4.
Repeat steps 2 and 3 for lines D1D2 and D2D3. Label the intersection O4.
5.
Connect O2 with C1 and call it link 2. Connect O4 with D1 and call it link 4.
6.
Line C1D1 is link 3. Line O2O4 is link 1 (ground link for the fourbar). The fourbar is now defined as O2CDO4 and has link lengths of Ground link 1a
L1a 0.718
Link 2
L2 2.197
Link 3
L3 2.496
Link 4
L4 3.704 1.230
O6 0.718 1b
2.197 O4 O2
2
C1
6 A
5
1a
4.328 B
2.496
D3
C3
3 4 C2 D1
D2 3.704
7.
Check the Grashof condition. Note that any Grashof condition is potentially acceptable in this case.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-5-2
Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition L1a L2 L3 L4 "Grashof" 8.
9.
Select a point on link 4 (O4D) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, C, and B.) In the solution above the distance O4B was selected to be L4b 1.375 . Draw a construction line through B1B3 and extend it up to the right.
10. Layout the length of link 5 (design choice) along the extended line. Label the other end A. 11. Draw a circle about O6 with a radius of one-half the length B1B3 and label the intersections of the circle with the extended line as A1 and A3. In the solution below the radius was measured as L6 1.230. 12. The driver fourbar is now defined as O4BAO6 with link lengths Link 6 (crank)
L6 1.230
Link 5 (coupler) L5 4.250 Link 1b (ground) L1b 4.328 Link 4b (rocker) L4b 1.375 13. Use the link lengths in step 12 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition L6 L1b L4b L5 "Grashof"
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-6-1
PROBLEM 3-6 Statement:
Design a fourbar mechanism to give the three positions shown in Figure P3-2 using the fixed pivots O2 and O4 shown. Build a cardboard model and determine the toggle positions and the minimum transmission angle. Add a driver dyad.
Solution:
See Figure P3-2 and Mathcad file P0306.
Design choices: Length of link 5:
L5 5.000
L2b 2.000
Length of link 2b:
1.
Draw link CD in its three design positions C1D1, C2D2, C3D3 in the plane as shown.
2.
Draw the ground link O2O4 in its desired position in the plane with respect to the first coupler position C1D1.
3.
Draw construction arcs from point C2 to O2 and from point D2 to O2 whose radii define the sides of triangle C2O2D2. This defines the relationship of the fixed pivot O2 to the coupler line CD in the second coupler position.
4.
Draw construction arcs from point C2 to O4 and from point D2 to O4 whose radii define the sides of triangle C2O4D2. This defines the relationship of the fixed pivot O4 to the coupler line CD in the second coupler position.
5.
Transfer this relationship back to the first coupler position C1D1 so that the ground plane position O2'O4' bears the same relationship to C1D1 as O2O4 bore to the second coupler position C2D2.
6.
Repeat the process for the third coupler position and transfer the third relative ground link position to the first, or reference, position.
7.
The three inverted positions of the ground link that correspond to the three desired coupler positions are labeled O2O4, O2'O4', and O2"O4" in the first layout below and are renamed E1F1, E2F2, and E3F3, respectively, in the second layout, which is used to find the points G and H.
O2''
C1
D3
O2'
C3 C2
O4'' D1
O4'
8.
O2
Draw construction lines from point E1 to E2 and from point E2 to E3.
D2
O4
DESIGN OF MACHINERY - 5th Ed.
9.
SOLUTION MANUAL 3-6-2
Bisect line E1E2 and line E2E3 and extend their perpendicular bisectors until they intersect. Label their intersection G.
10. Repeat steps 2 and 3 for lines F1F2 and F2F3. Label the intersection H. 11. Connect E1 with G and label it link 2. Connect F1 with H and label it link 4. Reinverting, E1 and F1 are the original fixed pivots O2 and O4, respectively. 12. Line GH is link 3. Line O2O4 is link 1a (ground link for the fourbar). The fourbar is now defined as O2GHO4 and has link lengths of Ground link 1a
L1a 4.303
Link 2
L2 8.597
Link 3
L3 1.711
Link 4
L4 7.921
E3
G
3
H
E2 F3
4
2
F2 E1
1a O2
F1 O4
13. Check the Grashof condition. Note that any Grashof condition is potentially acceptable in this case. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition L1a L2 L3 L4 "Grashof" The fourbar that will provide the desired motion is now defined as a Grashof double crank in the crossed configuration. It now remains to add the original points C1 and D1 to the coupler GH and to define the driving dyad.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-6-3
14. Select a point on link 2 (O2G) at a suitable distance from O2 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 2 is now a ternary link with nodes at O2, B, and G.) In the solution below, the distance O2B was selected to be L2b 2.000 . 15. Draw a construction line through B1B3 and extend it up to the right. 16. Layout the length of link 5 (design choice) along the extended line. Label the other end A. 17. Draw a circle about O6 with a radius of one-half the length B1B3 and label the intersections of the circle with the extended line as A1 and A3. In the solution below the radius was measured as L6 0.412. 18. The driver fourbar is now defined as O2BAO6 with link lengths Link 6 (crank)
L6 0.412
Link 5 (coupler) L5 5.000 Link 1b (ground) L1b 5.369 Link 2b (rocker) L2b 2.000 19. Use the link lengths in step 18 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition L6 L1b L2b L5 "Grashof"
G2
G3
H1
G1 3
H2 H3
2
C1
A3 O6
6
D3
C3
A1 5
C2
D1 B3
O2
D2 B1
4
1a
O4
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-7-1
PROBLEM 3-7 Statement:
Given: Solution: 1.
Repeat Problem 3-2 with a quick-return time ratio of 1:1.4. (See Example 3.9). Design a fourbar Grashof crank-rocker for 90 degrees of output rocker motion with a quick-return time ratio of 1:1.4. 1 Time ratio Tr 1.4 See figure below for one possible solution. Also see Mathcad file P0307.
Determine the crank rotation angles and , and the construction angle from equations 3.1 and 3.2. Tr = Solving for , and
β
α
α β = 360 deg
β 360 deg
β 210 deg
1 Tr
α 360 deg β
α 150 deg
δ β 180 deg
δ 30 deg
2.
Start the layout by arbitrarily establishing the point O4 and from it layoff two lines of equal length, 90 deg apart. Label one B1 and the other B2. In the solution below, each line makes an angle of 45 deg with the horizontal and has a length of 2.000 in.
3.
Layoff a line through B1 at an arbitrary angle (but not zero deg). In the solution below, the line is 30 deg to the horizontal.
4.
Layoff a line through B2 that makes an angle with the line in step 3 (60 deg to the horizontal in this case). The intersection of these two lines establishes the point O2.
5.
From O2 draw an arc that goes through B1. Extend O2B2 to meet this arc. Erect a perpendicular bisector to the extended portion of the line and transfer one half of the line to O2 as the length of the input crank.
3.8637 = b
90.0000°
B2 B1
B2
2.0000 = c
1.0353 = a
LAYOUT
B1
4 3 O4
O4 A1 2 O2
O2
3.0119 = d
A2 LINKAGE DEFINITION
DESIGN OF MACHINERY - 5th Ed.
6.
SOLUTION MANUAL 3-7-2
For this solution, the link lengths are: Ground link (1)
d 3.0119 in
Crank (2)
a 1.0353 in
Coupler (3)
b 3.8637 in
Rocker (4)
c 2.000 in
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-8-1
PROBLEM 3-8 Statement:
Design a sixbar drag link quick-return linkage for a time ratio of 1:2, and output rocker motion of 60 degrees. (See Example 3-10.)
Given:
Time ratio
Solution: 1.
Tr
1 2
See figure below for one possible solution. Also see Mathcad file P0308.
Determine the crank rotation angles and from equation 3.1. Tr = Solving for and
β
α
α β = 360 deg
β 360 deg 1 Tr
α 360 deg β
β 240 deg α 120 deg
2.
Draw a line of centers XX at any convenient location.
3.
Choose a crank pivot location O2 on line XX and draw an axis YY perpendicular to XX through O2.
4.
Draw a circle of convenient radius O2A about center O2. In the solution below, the length of O2A is a 1.000 in.
5.
Lay out angle with vertex at O2, symmetrical about quadrant one.
6.
Label points A1 and A2 at the intersections of the lines subtending angle and the circle of radius O2A.
7.
8.
Set the compass to a convenient radius AC long enough to cut XX in two places on either side of O2 when swung from both A1 and A2. Label the intersections C1 and C2. In the solution below, the length of AC is b 1.800 in. The line O2A is the driver crank, link 2, and the line AC is the coupler, link 3.
9.
The distance C1C2 is twice the driven (dragged) crank length. Bisect it to locate the fixed pivot O4.
10. The line O2O4 now defines the ground link. Line O4C is the driven crank, link 4. In the solution below, O4C measures c 2.262 in and O2O4 measures d 0.484 in. 11. Calculate the Grashoff condition. If non-Grashoff, repeat steps 7 through 11 with a shorter radius in step 7. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( a b c d ) "Grashof" 12. Invert the method of Example 3-1 to create the output dyad using XX as the chord and O4C1 as the driving crank. The points B1 and B2 will lie on line XX and be spaced apart a distance that is twice the length of O4C (link 4). The pivot point O6 will lie on the perpendicular bisector of B1B2 at a distance from XX which subtends the specified output rocker angle, which is 60 degrees in this problem. In the solution below, the length BC was chosen to be e 5.250 in.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-8-2
LAYOUT OF SIXBAR DRAG LINK QUICK RETURN WITH TIME RATIO OF 1:2 a = 1.000 b = 1.800 c = 2.262 d = 0.484 e = 5.250 f = 4.524
13. For the design choices made (lengths of links 2, 3 and 5), the length of the output rocker (link 6) was measured as f 4.524 in.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-9-1
PROBLEM 3-9 Statement:
Design a crank-shaper quick-return mechanism for a time ratio of 1:3 (Figure 3-14, p. 112).
Given:
Time ratio
Solution:
See Figure 3-14 and Mathcad file P0309.
TR
1 3
Design choices:
1.
Length of link 2 (crank)
L2 1.000
Length of link 5 (coupler)
L5 5.000
S 4.000
Length of stroke
Calculate from equations 3.1. TR
α β
α β 360 deg
α
360 deg 1
α 90.000 deg
1 TR
2.
Draw a vertical line and mark the center of rotation of the crank, O2, on it.
3.
Layout two construction lines from O2, each making an angle /2 to the vertical line through O2.
4.
Using the chosen crank length (see Design Choices), draw a circle with center at O2 and radius equal to the crank length. Label the intersections of the circle and the two lines drawn in step 3 as A1 and A2.
5.
Draw lines through points A1 and A2 that are also tangent to the crank circle (step 2). These two lines will simultaneously intersect the vertical line drawn in step 2. Label the point of intersection as the fixed pivot center O4.
6.
Draw a vertical construction line, parallel and to the right of O2O4, a distance S/2 (one-half of the output stroke length) from the line O2O4.
7.
Extend line O4A1 until it intersects the construction line drawn in step 6. Label the intersection B1.
8.
Draw a horizontal construction line from point B1, either to the left or right. Using point B1 as center, draw an arc of radius equal to the length of link 5 (see Design Choices) to intersect the horizontal construction line. Label the intersection as C1.
9.
Draw the slider blocks at points A1 and C1 and finish by drawing the mechanism in its other extreme position.
STROKE 4.000 C2
2.000
6
C1
B2
B1
5 O2 4
2 A2
3 O4
A1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-10-1
PROBLEM 3-10 Statement:
Find the two cognates of the linkage in Figure 3-17 (p. 116). Draw the Cayley and Roberts diagrams. Check your results with program FOURBAR.
Given:
Link lengths:
Solution: 1.
Coupler point data:
Ground link
L1 2
Crank
L2 1
A1P 1.800
δ 34.000 deg
Coupler
L3 3
Rocker
L4 3.5
B1P 1.813
γ 33.727 deg
See Figure 3-17 and Mathcad file P0310.
Draw the original fourbar linkage, which will be cognate #1, and align links 2 and 4 with the coupler. A1
B1
3
2
A1
3
OA
P
B1
4
2 OA P 4 1
OB
OB
2.
Construct lines parallel to all sides of the aligned fourbar linkage to create the Cayley diagram (see Figure 3-24) OA
2
A1
B1
3
OB
4
10
5 A2
B3
P
9 B2
6 8 7
A3
OC A1
3.
4.
Return links 2 and 4 to their fixed pivots OA and OB and establish OC as a fixed pivot by making triangle OAOBOC similar to A1B1P. Separate the three cognates. Point P has the same path motion in each cognate.
2 4
OA
10
P
Calculate the cognate link lengths based on the geometry of the Cayley diagram (Figure 3-24c, p. 114). L5 B1P L6
L4 L3
B1P
9
8
6
A2 OC OB
L5 1.813 L6 2.115
B2
A3 7
5.
B1
3
5 B3
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-10-2
P
P
OA
B2
10 A3
9 OC
7
8 A2 OC
6 OB 5
Cognate #2
B3
Cognate #3
L10 A1P
L10 1.800
L7 L9
B1P
L8 L6
A1P
L9
L2 L3
A1P
L9 0.600
L7 0.604
A1P
L8 2.100
B1P
From the Roberts diagram, calculate the ground link lengths for cognates #2 and #3 L1BC L1AC
L1 L3 L1 L3
B1P
L1BC 1.209
A1P
L1AC 1.200
Calculate the coupler point data for cognates #2 and #3 A3P L8
A3P 2.100
δ 180 deg δ γ
δ 247.727 deg
A2P L2
A2P 1.000
δ δ
δ 34.000 deg
SUMMARY OF COGNATE SPECIFICATIONS:
6.
Cognate #1
Cognate #2
Cognate #3
Ground link length
L1 2.000
L1AC 1.200
L1BC 1.209
Crank length
L2 1.000
L10 1.800
L7 0.604
Coupler length
L3 3.000
L9 0.600
L6 2.115
Rocker length
L4 3.500
L8 2.100
L5 1.813
Coupler point
A1P 1.800
A2P 1.000
A3P 2.100
Coupler angle
δ 34.000 deg
δ 34.000 deg
δ 247.727 deg
Verify that the three cognates yield the same coupler curve by entering the original link lengths in program FOURBAR and letting it calculate the cognates.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-10-3
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-10-4
Note that cognate #2 is a Grashof double rocker and, therefore, cannot trace out the entire coupler curve.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-11-1
PROBLEM 3-11 Statement:
Find the three equivalent geared fivebar linkages for the three fourbar cognates in Figure 3-25a (p. 125). Check your results by comparing the coupler curves with programs FOURBAR and FIVEBAR.
Given:
Link lengths:
Solution: 1.
Coupler point data:
Ground link
L1 39.5
Crank
L2 15.5
Coupler
L3 14.0
Rocker
L4 20.0
A1P 26.0
See Figure 3-25a and Mathcad file P0311.
Calculate the length BP and the angle using the law of cosines on the triangle APB. B1P L3 A1P 2 L3 A1P cos δ 2
2
0.5
B1P 23.270
L32 B1P 2 A1P 2 γ acos 2 L3 B1P 2.
δ 63.000 deg
γ 84.5843 deg
Use the Cayley diagram (see Figure 3-24) to calculate the link lengths of the two cognates. Note that the diagram is made up of three parallelograms and three similar triangles L4
L5 B1P
L5 23.270
L6
L10 A1P
L10 26.000
L9
L7 25.763
L8 L6
L7 L9
B1P A1P
L3 L2 L3
B1P
L6 33.243
A1P
L9 28.786
A1P
L8 37.143
B1P
Calculate the coupler point data for cognates #2 and #3 A3P L4
A3P 20.000
A2P L2
A2P 15.500
δ γ
δ 84.584 deg
δ δ
δ 63.000 deg
From the Roberts diagram, calculate the ground link lengths for cognates #2 and #3 L1BC
3.
L1 L3
B1P
L1BC 65.6548
L1AC
L1 L3
A1P
L1AC 73.3571
Using the calculated link lengths, draw the Roberts diagram (see next page). SUMMARY OF COGNATE SPECIFICATIONS: Cognate #1
Cognate #2
Cognate #3
Ground link length
L1 39.500
L1AC 73.357
L1BC 65.655
Crank length
L2 15.500
L10 26.000
L7 25.763
Coupler length
L3 14.000
L9 28.786
L6 33.243
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-11-2
Rocker length
L4 20.000
L8 37.143
L5 23.270
Coupler point
A1P 26.000
A2P 15.500
A3P 20.000
Coupler angle
δ 63.000 deg
δ 63.000 deg
δ 84.584 deg
OC 8 B2
7
B3 9
P
6 A2 A3 10 3
A1
5
B1 4
2 1
OA
OB
4.
The three geared fivebar cognates can be seen in the Roberts diagram. They are: OAA2PA3OB, OAA1PB3OC, and OBB1PB2OC. They are shown individually below with their associated gears. P
A2 A3 10 5
OA OB
OC
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-11-3
OC
7
B3
OD P
A1 2 OA
OC 8 B2
P
OE
B1 4
OB
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-11-4
SUMMARY OF GEARED FIVEBAR COGNATE SPECIFICATIONS: Cognate #1
Cognate #2
Cognate #3
Ground link length
L1 39.500
L1AC 73.357
L1BC 65.655
Crank length
L10 26.000
L2 15.500
L4 20.000
Coupler length
A2P 15.500
A1P 26.000
L5 23.270
Rocker length
A3P 20.000
L8 37.143
L7 25.763
Crank length
L5 23.270
L7 25.763
L8 37.143
Coupler point
A2P 15.500
A1P 26.000
B1P 23.270
Coupler angle
δ 0.00 deg
δ 0.00 deg
δ 0.00 deg
5.
Enter the cognate #1 specifications into program FOURBAR to get a trace of the coupler path.
6.
Enter the geared fivebar cognate #1 specifications into program FIVEBAR to get a trace of the coupler path for the geared fivebar (see next page).
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-11-5
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-12-1
PROBLEM 3-12 Statement:
Design a sixbar, single-dwell linkage for a dwell of 90 deg of crank motion, with an output rocker motion of 45 deg.
Given:
Crank dwell period: 90 deg. Output rocker motion: 45 deg.
Solution:
See Figures 3-20, 3-21, and Mathcad file P0312.
Design choices: Ground link ratio, L1/L2 = 2.0: GLR 2.0 Common link ratio, L3/L2 = L4/L2 = BP/L2 = 2.5: CLR 2.5 Coupler angle, γ 72 deg Crank length, L2 2.000 1.
For the given design choices, determine the remaining link lengths and coupler point specification. Coupler link (3) length
L3 CLR L2
L3 5.000
Rocker link (4) length
L4 CLR L2
L4 5.000
Ground link (1) length
L1 GLR L2
L1 4.000
Angle PAB
δ
Length AP on coupler 2.
180 deg γ 2
AP 2 L3 cos δ
δ 54.000 deg AP 5.878
Enter the above data into program FOURBAR, plot the coupler curve, and determine the coordinates of the coupler curve in the selected range of crank motion, which in this case will be from 135 to 225 deg..
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-12-2
FOURBAR for Windows Angle Step Deg 135 140 145 150 155 160 165 170 175 180 185 190 195 200 205 210 215 220 225 3.
File P03-12.DAT
Coupler Pt X
Coupler Pt Y
Coupler Pt Mag
-1.961 -2.178 -2.393 -2.603 -2.809 -3.008 -3.201 -3.386 -3.563 -3.731 -3.890 -4.038 -4.176 -4.302 -4.417 -4.520 -4.610 -4.688 -4.753
7.267 7.128 6.977 6.813 6.638 6.453 6.257 6.052 5.839 5.617 5.389 5.155 4.915 4.671 4.424 4.175 3.924 3.673 3.424
7.527 7.453 7.375 7.293 7.208 7.119 7.028 6.935 6.840 6.744 6.646 6.548 6.450 6.351 6.252 6.153 6.054 5.956 5.858
Coupler Pt Ang
105.099 106.992 108.930 110.911 112.933 114.994 117.093 119.228 121.396 123.595 125.822 128.075 130.351 132.646 134.955 137.274 139.598 141.921 144.235
Layout this linkage to scale, including the coupler curve whose coordinates are in the table above. Use the points at crank angles of 135, 180, and 225 deg to define the pseudo-arc. Find the center of the pseudo-arc erecting perpendicular bisectors to the chords defined by the selected coupler curve points. The center will lie at the intersection of the perpendicular bisectors, label this point D. The radius of this circle is the length of link 5.
y 135 P
PSEUDO-ARC
180 B
225 3 D A
4
2 x O2
O4
DESIGN OF MACHINERY - 5th Ed.
4.
SOLUTION MANUAL 3-12-3
The position of the end of link 5 at point D will remain nearly stationary while the crank moves from 135 to 225 deg. As the crank motion causes the coupler point to move around the coupler curve there will be another extreme position of the end of link 5 that was originally at D. Since a symmetrical linkage was chosen, the other extreme position will be located along a line through the axis of symmetry (see Figure 3-20) a distance equal to the length of link 5 measured from the point where the axis of symmetry intersects the coupler curve near the 0 deg coupler point. Establish this point and label it E. FOURBAR for Windows Angle Step Deg 300 310 320 330 340 350 0 10 20 30 40 50 60
File P03-12.DAT
Coupler Pt X
Coupler Pt Y
Coupler Pt Mag
-4.271 -4.054 -3.811 -3.526 -3.159 -2.651 -1.968 -1.181 -0.441 0.126 0.478 0.631 0.617
0.869 0.926 1.165 1.628 2.343 3.286 4.336 5.310 6.085 6.654 7.068 7.373 7.598
4.359 4.158 3.985 3.883 3.933 4.222 4.762 5.440 6.101 6.656 7.085 7.400 7.623
Coupler Pt Ang
168.495 167.133 162.998 155.215 143.437 128.892 114.414 102.534 94.142 88.914 86.129 85.111 85.354
y 135 P
PSEUDO-ARC
180 B
5
4
225 3
AXIS OF SYMMETRY
D A
E
2
x O2 5.
O4
The line segment DE represents the maximum displacement that a link of the length equal to link 5, attached at P, will reach along the axis of symmetry. Construct a perpendicular bisector of the line segment DE and extend it to the right (or left, which ever is convenient). Locate fixed pivot O6 on the bisector of DE such that the lines O6D and O6E subtend the desired output angle, in this case 45 deg. Draw link 6 from D through O6 and extend it to any convenient length. This is the output link that will dwell during the specified motion of the crank. See next page for the completed layout and further linkage specifications.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-12-4
y 135 P
PSEUDO-ARC
180
45.000° B
5
4
225
O6 BISECTOR
3 D A
E
2
x O4
O2
SUMMARY OF LINKAGE SPECIFICATIONS Original fourbar: Ground link
L1 4.000
Crank
L2 2.000
Coupler
L3 5.000
Rocker
L4 5.000
Coupler point
AP 5.878
δ 54.000 deg
Added dyad: Coupler
L5 6.363
Output
L6 2.855
Pivot O6
x 3.833
y 3.375
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-13-1
PROBLEM 3-13 Statement:
Design a sixbar double-dwell linkage for a dwell of 90 deg of crank motion, with an output of rocker motion of 60 deg, followed by a second dwell of about 60 deg of crank motion.
Given:
Initial crank dwell period: 90 deg Final crank dwell period: 60 deg (approx.) Output rocker motion between dwells: 60 deg
Solution:
See Mathcad file P0313.
Design choices:
1.
Ground link length
L1 5.000
Crank length
L2 2.000
Coupler link length
L3 5.000
Rocker length
L2 5.500
Coupler point data:
AP 8.750
δ 50 deg
In the absence of a linkage atlas it is difficult to find a coupler curve that meets the specifications. One approach is to start with a symmetrical linkage, using the data in Figure 3-21. Then, using program FOURBAR and by trial-and-error, adjust the link lengths and coupler point data until a satisfactory coupler curve is found. The link lengths and coupler point data given above were found this way. The resulting coupler curve is shown below and a printout of the coupler curve coordinates taken from FOURBAR is also printed below.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-13-2
FOURBAR for Windows Angle Step Deg 0.000 10.000 20.000 30.000 40.000 50.000 60.000 70.000 80.000 90.000 100.000 110.000 120.000 130.000 140.000 150.000 160.000 170.000 180.000 190.000 200.000 210.000 220.000 230.000 240.000 250.000 260.000 270.000 280.000 290.000 300.000 310.000 320.000 330.000 340.000 350.000 360.000
File P03-13.DAT
Cpler Pt Cpler Pt Cpler Pt Cpler Pt X Y Mag Ang
9.353 9.846 10.167 10.286 10.226 10.031 9.746 9.406 9.039 8.665 8.301 7.958 7.647 7.376 7.151 6.977 6.853 6.778 6.748 6.755 6.792 6.847 6.912 6.976 7.031 7.073 7.099 7.112 7.120 7.137 7.184 7.288 7.481 7.792 8.233 8.779 9.353
4.742 4.159 3.491 2.840 2.274 1.815 1.457 1.180 0.963 0.787 0.637 0.507 0.391 0.291 0.209 0.151 0.126 0.140 0.201 0.316 0.488 0.719 1.008 1.351 1.741 2.170 2.626 3.098 3.570 4.030 4.458 4.834 5.131 5.312 5.332 5.147 4.742
10.487 10.688 10.750 10.671 10.476 10.194 9.854 9.480 9.090 8.701 8.325 7.974 7.657 7.382 7.154 6.978 6.854 6.779 6.751 6.763 6.809 6.885 6.985 7.105 7.243 7.398 7.569 7.757 7.965 8.196 8.455 8.746 9.072 9.430 9.809 10.177 10.487
26.886 22.900 18.951 15.437 12.537 10.257 8.503 7.152 6.081 5.187 4.391 3.644 2.928 2.256 1.671 1.242 1.051 1.182 1.708 2.678 4.110 5.996 8.300 10.963 13.911 17.057 20.302 23.536 26.632 29.448 31.819 33.555 34.446 34.286 32.931 30.384 26.886
2.
Layout this linkage to scale, including the coupler curve whose coordinates are in the table above. Fit tangent lines to the nearly straight portions of the curve. Label their intersection O6. The coordinates of O6 are (6.729, 0.046).
3.
Design link 6 to lie along these straight tangents, pivoted at O6. Provide a slot in link 6 to accommodate slider block 5, which pivots on the coupler point P. (See next page).
4.
The beginning and ending crank angles for the dwell portions of the motion are indicated on the layout and in the table above by boldface entries.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-13-3
y 6 B 60.000°
260 5 4
P
3
90 O2 2
A
170 x O4
150 O6
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-14-1
PROBLEM 3-14 Statement:
Figure P3-3 shows a treadle-operated grinding wheel driven by a fourbar linkage. Make a cardboard model of the linkage to any convenient scale. Determine its minimum transmission angles. Comment on its operation. Will it work? If so, explain how it does.
Given:
Link lengths:
Link 2
L2 0.60 m
Link 3
L3 0.75 m
Link 4
L4 0.13 m
Link 1
L1 0.90 m
Grashof condition function: Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Solution: 1.
See Mathcad file P0314.
Determine the Grashof condition of the mechanism from inequality 2.8 and its Barker classification from Table 2-4. Grashof condition: Barker classification:
Condition L1 L2 L3 L4 "Grashof" Class I-4, Grashof rocker-rocker-crank, GRRC, since the shortest link is the output link.
2.
As a Grashof rocker-crank, the minimum transmission angle will be 0 deg, twice per revolution of the output (link 4) crank.
3.
Despite having transmission angles of 0 deg twice per revolution, the mechanism will work. That is, one will be able to drive the grinding wheel from the treadle (link 2). The reason is that the grinding wheel will act as a flywheel and will carry the linkage through the periods when the transmission angle is low. Typically, the operator will start the motion by rotating the wheel by hand.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-15-1
PROBLEM 3-15 Statement:
Figure P3-4 shows a non-Grashof fourbar linkage that is driven from link O2A. All dimensions are in centimeters (cm). (a) (b) (c) (d)
Given:
Solution: 1.
Find the transmission angle at the position shown. Find the toggle positions in terms of angle AO2O4. Find the maximum and minimum transmission angles over its range of motion. Draw the coupler curve of point P over its range of motion.
Link lengths: Link 1 (ground)
L1 95 mm
Link 2 (driver)
L2 50 mm
Link 3 (coupler)
L3 44 mm
Link 4 (driven)
L4 50 mm
See Figure P3-4 and Mathcad file P0315.
To find the transmission angle at the position shown, draw the linkage to scale in the position shown and measure the transmission angle ABO4. P
y 77.097°
B
3 A 2
4
O4
50.000° 1
x
O2
The measured transmission angle at the position shown is 77.097 deg. 2.
The toggle positions will be symmetric with respect to the O2O4 axis and will occur when links 3 and 4 are colinear. Use the law of cosines to calculate the angle of link 2 when links 3 and 4 are in toggle.
L3 L42 L12 L22 2 L1 L2 cosθ where 2 is the angle AO2O4. Solving for 2,
L12 L22 L3 L4 2 θ acos 2 L1 L2 The other toggle position occurs at θ 73.558 deg
θ 73.558 deg
DESIGN OF MACHINERY - 5th Ed.
3.
SOLUTION MANUAL 3-15-2
Use the program FOURBAR to find the maximum and minimum transmission angles.
FOURBAR for Windows
File P03-15
Design #
1
Angle Step Deg
Theta2 Mag degrees
Theta3 Mag degrees
Theta4 Mag degrees
Trans Ang Mag degrees
-73.557 -58.846 -44.134 -29.423 -14.711 0.000 14.711 29.423 44.134 58.846 73.557
-73.557 -58.846 -44.134 -29.423 -14.711 0.000 14.711 29.423 44.134 58.846 73.557
30.861 64.075 77.168 83.147 80.604 68.350 50.145 32.106 16.173 0.566 -30.486
-149.490 -176.312 170.696 157.514 142.103 125.123 111.644 106.473 109.701 120.179 149.159
0.352 60.387 86.472 74.367 61.499 56.773 61.499 74.367 86.472 60.387 0.355
A partial output from FOURBAR is shown above. From it, we see that the maximum transmission angle is approximately 86.5 deg and the minimum is zero deg. 4.
Use program FOURBAR to draw the coupler curve with respect to a coordinate frame through O2O4.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-16-1
PROBLEM 3-16 Statement:
Draw the Roberts diagram for the linkage in Figure P3-4 and find its two cognates. Are they Grashof or non-Grashof?
Given:
Link lengths:
Solution: 1.
Coupler point data:
Ground link
L1 9.5
Crank
L2 5
Coupler
L3 4.4
Rocker
L4 5
A1P 8.90
See Figure P3-4 and Mathcad file P0316.
Calculate the length BP and the angle using the law of cosines on the triangle APB. B1P L3 A1P 2 L3 A1P cos δ 2
2
0.5
B1P 7.401
L32 B1P 2 A1P 2 γ acos 2 L3 B1P 2.
δ 56.000 deg
γ 94.4701 deg
Use the Cayley diagram (see Figure 3-24) to calculate the link lengths of the two cognates. Note that the diagram is made up of three parallelograms and three similar triangles L4
L5 B1P
L5 7.401
L6
L10 A1P
L10 8.900
L9
L7 8.410
L8 L6
L7 L9
B1P A1P
L3 L2 L3
B1P
L6 8.410
A1P
L9 10.114
A1P
L8 10.114
B1P
Calculate the coupler point data for cognates #2 and #3 A3P L4
A3P 5.000
A2P L2
A2P 5.000
δ γ
δ 94.470 deg
δ δ
δ 56.000 deg
From the Roberts diagram, calculate the ground link lengths for cognates #2 and #3 L1BC
3.
L1 L3
B1P
L1BC 15.9793
L1AC
L1 L3
A1P
L1AC 19.2159
Using the calculated link lengths, draw the Roberts diagram (see next page). SUMMARY OF COGNATE SPECIFICATIONS: Cognate #1
Cognate #2
Cognate #3
Ground link length
L1 9.500
L1AC 19.216
L1BC 15.979
Crank length
L2 5.000
L10 8.900
L7 8.410
Coupler length
L3 4.400
L9 10.114
L6 8.410
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-16-2
Rocker length
L4 5.000
L8 10.114
L5 7.401
Coupler point
A1P 8.900
A2P 5.000
A3P 5.000
Coupler angle
δ 56.000 deg
δ 56.000 deg
δ 94.470 deg
B2
OC
8
7
B3
P 9 6 A3
A2
5
3
B1
10
4
A1 2
OB 1
OA
6.
Determine the Grashof condition of each of the two additional cognates. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Cognate #2:
Condition L10 L1AC L8 L9 "non-Grashof"
Cognate #3:
Condition L5 L1BC L6 L7 "non-Grashof"
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-17-1
PROBLEM 3-17 Statement:
Design a Watt-I sixbar to give parallel motion that follows the coupler path of point P of the linkage in Figure P3-4.
Given:
Link lengths:
Solution: 1.
Coupler point data:
Ground link
L1 9.5
Crank
L2 5
Coupler
L3 4.4
Rocker
L4 5
A1P 8.90
See Figure P3-4 and Mathcad file P0317.
Calculate the length BP and the angle using the law of cosines on the triangle APB. B1P L3 A1P 2 L3 A1P cos δ 2
2
0.5
B1P 7.401
L32 B1P 2 A1P 2 γ acos 2 L3 B1P 2.
δ 56.000 deg
γ 94.4701 deg
Use the Cayley diagram (see Figure 3-24) to calculate the link lengths of the two cognates. Note that the diagram is made up of three parallelograms and three similar triangles L4
L5 B1P
L5 7.401
L6
L10 A1P
L10 8.900
L9
L7 8.410
L8 L6
L7 L9
B1P A1P
L3 L2 L3
B1P
L6 8.410
A1P
L9 10.114
A1P
L8 10.114
B1P
Calculate the coupler point data for cognates #2 and #3 A3P L4
A3P 5.000
A2P L2
A2P 5.000
δ γ
δ 94.470 deg
δ δ
δ 56.000 deg
From the Roberts diagram, calculate the ground link lengths for cognates #2 and #3 L1BC
3.
L1 L3
B1P
L1BC 15.9793
L1AC
L1 L3
A1P
L1AC 19.2159
Using the calculated link lengths, draw the Roberts diagram (see next page). SUMMARY OF COGNATE SPECIFICATIONS: Cognate #1
Cognate #2
Cognate #3
Ground link length
L1 9.500
L1AC 19.216
L1BC 15.979
Crank length
L2 5.000
L10 8.900
L7 8.410
Coupler length
L3 4.400
L9 10.114
L6 8.410
DESIGN OF MACHINERY - 5th Ed.
B2
SOLUTION MANUAL 3-17-2
Rocker length
L4 5.000
L8 10.114
L5 7.401
Coupler point
A1P 8.900
A2P 5.000
A3P 5.000
Coupler angle
δ 56.000 deg
δ 56.000 deg
δ 94.470 deg
OC
8
7
B3
P 9 6 A3
A2
P
5
3 10
B1 4
A1 2
OB 1 3
OA
4
A1
4.
5.
All three of these cognates are non-Grashof and will, therefore, have limited motion. However, following Example 3-11, discard cognate #2 and retain cognates #1 and #3. Draw line qq parallel to line OAOC and through point OB. Without allowing links 5, 6, and 7 to rotate, slide them as an assembly along lines OAOC and qq until the free end of link 7 is at OA. The free end of link 5 will then be at point O'B and point P on link 6 will be at P'. Add a new link of length OAOC between P and P'. This is the new output link 8 and all points on it describe the original coupler curve. Join links 2 and 7, making one ternary link. Remove link 5 and reduce link 6 to a binary link. The result is a Watt-I sixbar with links numbered 1, 2, 3, 4, 6, and 8 (see next page). Link 8 is in curvilinear translation and follows the coupler path of the original point P.
B1 q
2
OB 1
OA 7
B3 P'
6 A3 5
q
O'B
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-17-3
P
B1 3
4
A1 8 2
OB
1
OA
B3 P'
6
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-18-1
PROBLEM 3-18 Statement:
Design a Watt-I sixbar to give parallel motion that follows the coupler path of point P of the linkage in Figure P3-4 and add a driver dyad to drive it over its possible range of motion with no quick return. (The result will be an 8-bar linkage).
Given:
Link lengths:
Solution: 1.
Coupler point data:
Ground link
L1 9.5
Crank
L2 5
Coupler
L3 4.4
Rocker
L4 5
A1P 8.90
See Figure P3-4 and Mathcad file P0318.
Calculate the length BP and the angle using the law of cosines on the triangle APB. B1P L3 A1P 2 L3 A1P cos δ 2
2
0.5
B1P 7.401
L32 B1P 2 A1P 2 γ acos 2 L3 B1P 2.
δ 56.000 deg
γ 94.4701 deg
Use the Cayley diagram (see Figure 3-24) to calculate the link lengths of the two cognates. Note that the diagram is made up of three parallelograms and three similar triangles L4
L5 B1P
L5 7.401
L6
L10 A1P
L10 8.900
L9
L7 8.410
L8 L6
L7 L9
B1P A1P
L3 L2 L3
B1P
L6 8.410
A1P
L9 10.114
A1P
L8 10.114
B1P
Calculate the coupler point data for cognates #2 and #3 A3P L4
A3P 5.000
A2P L2
A2P 5.000
δ γ
δ 94.470 deg
δ δ
δ 56.000 deg
From the Roberts diagram, calculate the ground link lengths for cognates #2 and #3 L1BC 3.
L1 L3
B1P
L1BC 15.9793
L1AC
L1 L3
A1P
L1AC 19.2159
Using the calculated link lengths, draw the Roberts diagram (see next page). SUMMARY OF COGNATE SPECIFICATIONS: Cognate #1
Cognate #2
Cognate #3
Ground link length
L1 9.500
L1AC 19.216
L1BC 15.979
Crank length
L2 5.000
L10 8.900
L7 8.410
Coupler length
L3 4.400
L9 10.114
L6 8.410
DESIGN OF MACHINERY - 5th Ed.
B2
SOLUTION MANUAL 3-18-2
Rocker length
L4 5.000
L8 10.114
L5 7.401
Coupler point
A1P 8.900
A2P 5.000
A3P 5.000
Coupler angle
δ 56.000 deg
δ 56.000 deg
δ 94.470 deg
OC
8
7
B3
P 9 6 A3
A2
5
3 10
P
B1 4
A1 2
OB 1
OA
3
4
A1
4.
5.
6.
All three of these cognates are non-Grashof and will, therefore, have limited motion. However, following Example 3-11, discard cognate #2 and retain cognates #1 and #3. Draw line qq parallel to line OAOC and through point OB. Without allowing links 5, 6, and 7 to rotate, slide them as an assembly along lines OAOC and qq until the free end of link 7 is at OA. The free end of link 5 will then be at point O'B and point P on link 6 will be at P'. Add a new link of length OAOC between P P' and P'. This is the new output link 8 and all points on it describe the original coupler curve. Join links 2 and 7, making one ternary link. Remove link 5 and reduce link 6 to a binary link. The result is a Watt-I sixbar with links numbered 1, 2, 3, 4, 6, and 8 (see next page). Link 8 is in curvilinear translation and follows the coupler path of the original point P.
B1 q
2
OB 1
OA 7
B3
6 A3 5
Add a driver dyad following Example 3-4.
q
O'B
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-18-3
P
B1 3
4
A1 8 2
OB
1
OA
B3 P'
6
P
B1 3
4
A1 8 2
OB
1
OA
P'
6
B3
OC
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-19-1
PROBLEM 3-19 Statement:
Design a pin-jointed linkage that will guide the forks of the fork lift truck in Figure P3-5 up and down in an approximate straight line over the range of motion shown. Arrange the fixed pivots so they are close to some part of the existing frame or body of the truck.
Given:
Length of straight line motion of the forks: Δx 1800 mm
Solution:
See Figure P3-5 and Mathcad file P0319.
Design choices: Use a Hoeken-type straight line mechanism optimized for straightness. Maximum allowable error in straightness of line: ΔCy 0.096 % 1.
Using Table 3-1 and the required length of straight-line motion, determine the link lengths. Link ratios from Table 3-1 for ΔCy 0.096 %: L1overL2 2.200 L3overL2 2.800
ΔxoverL2 4.181
Link lengths:
2.
L2
Coupler
L3 L3overL2 L2
L3 1205.5 mm
Ground link
L1 L1overL2 L2
L1 947.1 mm
Rocker
L4 L3
L4 1205.5 mm
Coupler point
BP L3
BP 1205.5 mm
L2 430.5 mm
ΔxoverL2
Calculate the distance from point P to pivot O4 (Cy). Cy
3.
Δx
Crank
2 L32 L1 L22
Cy 1978.5 mm
Draw the fork lift truck to scale with the mechanism defined in step 1 superimposed on it..
1978.5mm
O4
P
620.0mm
4 947.1mm B
900.0mm
487.1mm 3
O2 2 A
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-20-1
PROBLEM 3-20 Statement:
Figure P3-6 shows a "V-link" off-loading mechanism for a paper roll conveyor. Design a pinjointed linkage to replace the air cylinder driver that will rotate the rocker arm and V-link through the 90 deg motion shown. Keep the fixed pivots as close to the existing frame as possible. Your fourbar linkage should be Grashof and be in toggle at each extreme position of the rocker arm.
Given:
Dimensions scaled from Figure P3-6: Rocker arm (link 4) distance between pin centers:
Solution:
L4 320 mm
See Figure P3-6 and Mathcad file P0320.
Design choices: 1. Use the same rocker arm that was used with the air cylinder driver. 2. Place the pivot O2 80 mm to the right of the right leg and on a horizontal line with the center of the pin on the rocker arm. 3. Design for two-position, 90 deg of output rocker motion with no quick return, similar to Example 3-2. 1.
Draw the rocker arm (link 4) O4B in both extreme positions, B1 and B2, in any convenient location such that the desired angle of motion 4 is subtended. In this solution, link 4 is drawn such that the two extreme positions each make an angle of 45 deg to the vertical.
2.
Draw the chord B1B2 and extend it in any convenient direction. In this solution it was extended horizontally to the left.
3.
Mark the center O2 on the extended line such that it is 80 mm to the right of the right leg. This will allow sufficient space for a supporting pillow block bearing.
4.
Bisect the line segment B1B2 and draw a circle of that radius about O2.
5.
Label the two intersections of the circle and extended line B1B2, A1 and A2.
6.
Measure the length of the coupler (link 3) as A1B1 or A2B2. From the graphical solution, L3 1045 mm
7.
Measure the length of the crank (link 2) as O2A1 or O2A2. From the graphical solution, L2 226.274 mm
8.
Measure the length of the ground link (link 1) as O2O4. From the graphical solution, L1 1069.217 mm 1045.000 80.000 1069.217 320.000
1 4 3
1045.000
9.
Find the Grashof condition.
2
226.274
DESIGN OF MACHINERY - 5th Ed.
Condition( a b c d )
SOLUTION MANUAL 3-20-2
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition L1 L2 L3 L4 "Grashof"
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-21-1
PROBLEM 3-21 Statement:
Figure P3-7 shows a walking-beam transport mechanism that uses a fourbar coupler curve, replicated with a parallelogram linkage for parallel motion. Note the duplicate crank and coupler shown ghosted in the right half of the mechanism - they are redundant and have been removed from the duplicate fourbar linkage. Using the same fourbar driving stage (links 1, 2, 3, 4 with coupler point P), design a Watt-I sixbar linkage that will drive link 8 in the same parallel motion using two fewer links.
Given:
Link lengths:
Solution: 1.
Coupler point data:
Ground link
L1 2.22
Crank
L2 1
Coupler
L3 2.06
Rocker
L4 2.33
A1P 3.06
See Figure P3-7 and Mathcad file P0321.
Calculate the length BP and the angle using the law of cosines on the triangle APB. B1P L3 A1P 2 L3 A1P cos δ 2
2
0.5
B1P 1.674
L32 B1P 2 A1P 2 γ acos 2 L3 B1P 2.
δ 31.000 deg
γ 109.6560 deg
Use the Cayley diagram (see Figure 3-24) to calculate the link lengths of the two cognates. Note that the diagram is made up of three parallelograms and three similar triangles L4
L5 B1P
L5 1.674
L6
L10 A1P
L10 3.060
L9
L7 0.812
L8 L6
L7 L9
B1P A1P
L3 L2 L3
B1P
L6 1.893
A1P
L9 1.485
A1P
L8 3.461
B1P
Calculate the coupler point data for cognates #2 and #3 A3P L8
A3P 3.461
δ 180 deg δ γ
A2P L2
A2P 1.000
δ δ
δ 31.000 deg
δ 39.344 deg From the Roberts diagram, calculate the ground link lengths for cognates #2 and #3 L1BC
3.
L1 L3
B1P
L1BC 1.8035
L1AC
L1 L3
A1P
L1AC 3.2977
Using the calculated link lengths, draw the Roberts diagram (see next page). SUMMARY OF COGNATE SPECIFICATIONS: Cognate #1 Ground link length
L1 2.220
Cognate #2
Cognate #3
L1AC 3.298
L1BC 1.804
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-21-2
Crank length
L2 1.000
L10 3.060
L7 0.812
Coupler length
L3 2.060
L9 1.485
L6 1.893
Rocker length
L4 2.330
L8 3.461
L5 1.674
Coupler point
A1P 3.060
A2P 1.000
A3P 3.461
Coupler angle
δ 31.000 deg
δ 31.000 deg
δ 39.344 deg
OC 7 6
B3
A3
8
5
OB
9
4
1
B2
A2
10
P
OA 2 A1 3 B1
4.
Determine the Grashof condition of each of the two additional cognates. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
5.
Cognate #2:
Condition L8 L9 L10 L1AC "Grashof"
Cognate #3:
Condition L5 L6 L7 L1BC "Grashof"
Both of these cognates are Grashof but cognate #3 is a crank rocker. Following Example 3-11, discard cognate #2 and retain cognates #1 and #3. Draw line qq parallel to line OAOC and through point OB. Without allowing links 5, 6, and 7 to rotate, slide them as an assembly along lines OAOC and qq until the free end of link 7 is at OA. The free end of link 5 will then be at point O'B and point P on link 6 will be at P'. Add a new link of length OAOC between P and P'. This is the new output link 8 and all points on it describe the original coupler curve.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-21-3
q OB 1 4 P
OA
2
7
B3
6 A3 A1
3 B1 8
5 O'B q P'
6.
Join links 2 and 7, making one ternary link. Remove link 5 and reduce link 6 to a binary link. The result is a Watt-I sixbar with links numbered 1, 2, 3, 4, 6, and 8 (see next page). Link 8 is in curvilinear translation and follows the coupler path of the original point P. The walking-beam (link 8 in Figure P3-7) is rigidly attached to link 8 below.
OB 1 4 P
OA
2
A3 A1 3 B1
6
8
P'
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-22-1
PROBLEM 3-22 Statement:
Find the maximum and minimum transmission angles of the fourbar driving stage (links L1, L2, L3, L4) in Figure P3-7 (to graphical accuracy).
Given:
Link lengths:
Link 2
L2 1.00
Link 3
L3 2.06
Link 4
L4 2.33
Link 1
L1 2.22
Grashof condition function: Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Solution: 1.
See Figure P3-7 and Mathcad file P0322.
Determine the Grashof condition of the mechanism from inequality 2.8 and its Barker classification from Table 2-4. Grashof condition: Barker classification:
2.
Condition L1 L2 L3 L4 "Grashof" Class I-2, Grashof crank-rocker-rocker, GCRR, since the shortest link is the input link.
It can be shown (see Section 4.10) that the minimum transmission angle for a fourbar GCRR linkage occurs when links 2 and 1 (ground link) are colinear. Draw the linkage in these two positions and measure the transmission angles. O4
O4
A O2
31.510°
O2 A 85.843°
B
3.
B
As measured from the layout, the minimum transmission angle is 31.5 deg. The maximum is 90 deg.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-23-1
PROBLEM 3-23 Statement:
Figure P3-8 shows a fourbar linkage used in a power loom to drive a comb-like reed against the thread, "beating it up" into the cloth. Determine its Grashof condition and its minimum and maximum transmission angles to graphical accuracy.
Given:
Link lengths:
Link 2
L2 2.00 in
Link 3
L3 8.375 in
Link 4
L4 7.187 in
Link 1
L1 9.625 in
Grashof condition function: Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Solution: 1.
See Figure P3-8 and Mathcad file P0323.
Determine the Grashof condition of the mechanism from inequality 2.8 and its Barker classification from Table 2-4. Grashof condition: Barker classification:
2.
Condition L1 L2 L3 L4 "Grashof" Class I-2, Grashof crank-rocker-rocker, GCRR, since the shortest link is the input link.
It can be shown (see Section 4.10) that the minimum transmission angle for a fourbar GCRR linkage occurs when links 2 and 1 (ground link) are colinear. Draw the linkage in these two positions and measure the transmission angles.
83.634°
58.078°
3.
As measured from the layout, the minimum transmission angle is 58.1 deg. The maximum is 90.0 deg.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-24-1
PROBLEM 3-24 Statement:
Draw the Roberts diagram and find the cognates for the linkage in Figure P3-9.
Given:
Link lengths:
Solution: 1.
Coupler point data:
Ground link
L1 2.22
Crank
L2 1.0
Coupler
L3 2.06
Rocker
L4 2.33
A1P 3.06
See Figure P3-9 and Mathcad file P0324.
Calculate the length BP and the angle using the law of cosines on the triangle APB. B1P L3 A1P 2 L3 A1P cos δ 2
2
0.5
B1P 1.674
L32 B1P 2 A1P 2 γ acos 2 L3 B1P 2.
δ 31.00 deg
γ 109.6560 deg
Use the Cayley diagram (see Figure 3-24) to calculate the link lengths of the two cognates. Note that the diagram is made up of three parallelograms and three similar triangles L4
L5 B1P
L5 1.674
L6
L10 A1P
L10 3.060
L9
L7 0.812
L8 L6
L7 L9
B1P A1P
L3 L2 L3
B1P
L6 1.893
A1P
L9 1.485
A1P
L8 3.461
B1P
Calculate the coupler point data for cognates #2 and #3 A3P L8
A3P 3.461
δ 180 deg δ γ
δ 39.344 deg
A2P L2
A2P 1.000
δ δ
δ 31.000 deg
From the Roberts diagram, calculate the ground link lengths for cognates #2 and #3 L1BC 3.
L1 L3
B1P
L1BC 1.8035
L1AC
L1 L3
A1P
L1AC 3.2977
Using the calculated link lengths, draw the Roberts diagram (see next page). SUMMARY OF COGNATE SPECIFICATIONS: Cognate #1
Cognate #2
Cognate #3
Ground link length
L1 2.220
L1AC 3.298
L1BC 1.804
Crank length
L2 1.000
L10 3.060
L7 0.812
Coupler length
L3 2.060
L9 1.485
L6 1.893
Rocker length
L4 2.330
L8 3.461
L5 1.674
Coupler point
A1P 3.060
A2P 1.000
A3P 3.461
Coupler angle
δ 31.000 deg
δ 31.000 deg
δ 39.344 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-24-2
B1 P B2 3
2
9
4
A1
A2
10
8 OB
5
1
OA
1BC 1AC
B3 6
A3
7
OC
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-25-1
PROBLEM 3-25 Statement:
Find the equivalent geared fivebar mechanism cognate of the linkage in Figure P3-9.
Given:
Link lengths:
Solution: 1.
Coupler point data:
Ground link
L1 2.22
Crank
L2 1.0
Coupler
L3 2.06
Rocker
L4 2.33
A1P 3.06
See Figure P3-9 and Mathcad file P0325.
Calculate the length BP and the angle using the law of cosines on the triangle APB. B1P L3 A1P 2 L3 A1P cos δ 2
2
0.5
B1P 1.674
L32 B1P 2 A1P 2 γ acos 2 L3 B1P 2.
δ 31.00 deg
γ 109.6560 deg
Use the Cayley diagram (see Figure 3-24) to calculate the link lengths of the two cognates. Note that the diagram is made up of three parallelograms and three similar triangles L4
L5 B1P
L5 1.674
L6
L10 A1P
L10 3.060
L9
L7 0.812
L8 L6
L7 L9
B1P A1P
L3 L2 L3
B1P
L6 1.893
A1P
L9 1.485
A1P
L8 3.461
B1P
Calculate the coupler point data for cognates #2 and #3 A3P L8
A3P 3.461
δ 180 deg δ γ
δ 39.344 deg
A2P L2
A2P 1.000
δ δ
δ 31.000 deg
From the Roberts diagram, calculate the ground link lengths for cognates #2 and #3 L1BC
3.
L1 L3
B1P
L1BC 1.8035
L1AC
L1 L3
A1P
L1AC 3.2977
Using the calculated link lengths, draw the Roberts diagram (see next page). SUMMARY OF COGNATE SPECIFICATIONS: Cognate #1
Cognate #2
Cognate #3
Ground link length
L1 2.220
L1AC 3.298
L1BC 1.804
Crank length
L2 1.000
L10 3.060
L7 0.812
Coupler length
L3 2.060
L9 1.485
L6 1.893
Rocker length
L4 2.330
L8 3.461
L5 1.674
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-25-2
Coupler point
A1P 3.060
A2P 1.000
A3P 3.461
Coupler angle
δ 31.000 deg
δ 31.000 deg
δ 39.344 deg
B1 P B2 3
2
9
4
A1
A2
10
8 OB
5
1
OA
1BC 1AC
B3 6
A3
4.
7
OC
The three geared fivebar cognates can be seen in the Roberts diagram. They are: OAA2PB3OB, OAA1PA3OC, and OBB1PB2OC. The three geared fivebar cognates are summarized in the table below. SUMMARY OF GEARED FIVEBAR COGNATE SPECIFICATIONS: Cognate #1
Cognate #2
Cognate #3
Ground link length
L1 2.220
L1AC 3.298
L1BC 1.804
Crank length
L10 3.060
L2 1.000
L4 2.330
Coupler length
A2P 1.000
A1P 3.060
L5 1.674
Rocker length
L4 2.330
L8 3.461
L7 0.812
Crank length
L5 1.674
L7 0.812
L8 3.461
Coupler point
A2P 1.000
A1P 3.060
B1P 1.674
Coupler angle
δ 0.00 deg
δ 0.00 deg
δ 0.00 deg
5.
Enter the cognate #1 specifications into program FOURBAR to get a trace of the coupler path (see next page)
6.
Enter the geared fivebar cognate #1 specifications into program FIVEBAR to get a trace of the coupler path for the geared fivebar (see next page).
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-25-3
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-26-1
PROBLEM 3-26 Statement:
Use the linkage in Figure P3-9 to design an eightbar double-dwell mechanism that has a rocker output through 45 deg.
Given:
Link lengths:
Solution: 1.
Coupler point data:
Ground link
L1 2.22
Crank
L2 1.0
Coupler
L3 2.06
Rocker
L4 2.33
A1P 3.06
δ 31.00 deg
See Figure P3-9 and Mathcad file P0326.
Enter the given data into program FOURBAR and print out the resulting coupler point coordinates (see table below). FOURBAR for Windows
File P03-26.DAT
Angle Step Deg
Cpler Pt Cpler Pt Cpler Pt Cpler Pt X Y Mag Ang
0.000 10.000 20.000 30.000 40.000 50.000 60.000 70.000 80.000 90.000 100.000 110.000 120.000 130.000 140.000 150.000 160.000 170.000 180.000 190.000 200.000 210.000 220.000 230.000 240.000 250.000 260.000 270.000 280.000 290.000 300.000 310.000 320.000 330.000 340.000 350.000 360.000
2.731 3.077 3.350 3.515 3.576 3.554 3.473 3.350 3.203 3.040 2.872 2.706 2.548 2.403 2.274 2.164 2.075 2.005 1.953 1.917 1.892 1.875 1.862 1.848 1.832 1.810 1.784 1.754 1.723 1.698 1.687 1.702 1.761 1.883 2.088 2.380 2.731
2.523 2.407 2.228 2.032 1.855 1.708 1.592 1.499 1.420 1.348 1.278 1.207 1.135 1.062 0.990 0.925 0.869 0.826 0.802 0.798 0.817 0.860 0.925 1.011 1.115 1.235 1.367 1.508 1.654 1.804 1.955 2.105 2.251 2.386 2.494 2.550 2.523
3.718 3.906 4.023 4.060 4.028 3.943 3.820 3.671 3.503 3.326 3.144 2.963 2.789 2.627 2.480 2.354 2.249 2.168 2.111 2.076 2.061 2.063 2.079 2.107 2.145 2.192 2.248 2.313 2.388 2.477 2.582 2.707 2.858 3.040 3.253 3.488 3.718
42.731 38.029 33.626 30.035 27.412 25.672 24.635 24.107 23.915 23.915 23.988 24.039 24.001 23.834 23.533 23.134 22.719 22.404 22.326 22.614 23.365 24.632 26.417 28.678 31.340 34.306 37.463 40.683 43.826 46.730 49.207 51.038 51.965 51.715 50.064 46.967 42.731
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-26-2
2.
Layout this linkage to scale, including the coupler curve whose coordinates are in the table above. Fit tangent lines to the nearly straight portions of the curve. Label their intersection O6.
3.
Design link 6 to lie along these straight tangents, pivoted at O6. Provide a guide on link 6 to accommodate slider block 5, which pivots on the coupler point P. O8
8 45.000° F
E D
7
B 70.140°
C 6
3 P 5 A O6 2
4
O4
O2
4.
Extend link 6 a convenient distance to point C. Draw an arc through point C with center at O6. Label the intersection of the arc with the other tangent line as point D. Attach link 7 to the pivot at C. The length of link 7 is CE, a design choice. Extend line CDE from point E a distance equal to CD. Label the end point F. Layout two intersecting lines through E and F such that they subtend an angle of 45 deg. Label their intersection O8. The link joining O8 and point E is link 8. The link lengths and locations of O6 and O8 are: Link 6
L6 2.330
Fixed pivot O6:
Link 7
x 1.892 y 0.762
L7 3.000 Fixed pivot O8:
Link 8 x 1.379 y 6.690
L8 3.498
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-27-1
PROBLEM 3-27 Statement:
Use the linkage in Figure P3-9 to design an eightbar double-dwell mechanism that has a slider output stroke of 5 crank units.
Given:
Link lengths:
Solution: 1.
Coupler point data:
Ground link
L1 2.22
Crank
L2 1.0
Coupler
L3 2.06
Rocker
L4 2.33
A1P 3.06
δ 31.00 deg
See Figure P3-9 and Mathcad file P0327.
Enter the given data into program FOURBAR and print out the resulting coupler point coordinates (see table below). FOURBAR for Windows
File P03-26.DAT
Angle Step Deg
Cpler Pt Cpler Pt Cpler Pt Cpler Pt X Y Mag Ang
0.000 10.000 20.000 30.000 40.000 50.000 60.000 70.000 80.000 90.000 100.000 110.000 120.000 130.000 140.000 150.000 160.000 170.000 180.000 190.000 200.000 210.000 220.000 230.000 240.000 250.000 260.000 270.000 280.000 290.000 300.000 310.000 320.000 330.000 340.000 350.000 360.000
2.731 3.077 3.350 3.515 3.576 3.554 3.473 3.350 3.203 3.040 2.872 2.706 2.548 2.403 2.274 2.164 2.075 2.005 1.953 1.917 1.892 1.875 1.862 1.848 1.832 1.810 1.784 1.754 1.723 1.698 1.687 1.702 1.761 1.883 2.088 2.380 2.731
2.523 2.407 2.228 2.032 1.855 1.708 1.592 1.499 1.420 1.348 1.278 1.207 1.135 1.062 0.990 0.925 0.869 0.826 0.802 0.798 0.817 0.860 0.925 1.011 1.115 1.235 1.367 1.508 1.654 1.804 1.955 2.105 2.251 2.386 2.494 2.550 2.523
3.718 3.906 4.023 4.060 4.028 3.943 3.820 3.671 3.503 3.326 3.144 2.963 2.789 2.627 2.480 2.354 2.249 2.168 2.111 2.076 2.061 2.063 2.079 2.107 2.145 2.192 2.248 2.313 2.388 2.477 2.582 2.707 2.858 3.040 3.253 3.488 3.718
42.731 38.029 33.626 30.035 27.412 25.672 24.635 24.107 23.915 23.915 23.988 24.039 24.001 23.834 23.533 23.134 22.719 22.404 22.326 22.614 23.365 24.632 26.417 28.678 31.340 34.306 37.463 40.683 43.826 46.730 49.207 51.038 51.965 51.715 50.064 46.967 42.731
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-27-2
2.
Layout this linkage to scale, including the coupler curve whose coordinates are in the table above. Fit tangent lines to the nearly straight portions of the curve. Label their intersection O6.
3.
Design link 6 to lie along these straight tangents, pivoted at O6. Provide a guide on link 6 to accommodate slider block 5, which pivots on the coupler point P. F
E
8
D 7
C
B 6 70.140° 3 P 5 A O6 2
4
O4
O2
4.
Extend link 6 and the other tangent line until points C and E are 5 units apart. Attach link 7 to the pivot at C. The length of link 7 is CD, a design choice. Extend line CDE from point D a distance equal to CE. Label the end point F. As link 6 travels from C to E, slider block 8 will travel from D to F, a distance of 5 units. The link lengths and location of O6: Link 6
L6 4.351
Fixed pivot O6:
Link 7
x 1.892 y 0.762
L7 2.000
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-28-1
PROBLEM 3-28 Statement:
Use two of the cognates in Figure 3-26 (p. 126) to design a Watt-I sixbar parallel motion mechanism that carries a link through the same coupler curve at all points. Comment on its similarities to the original Roberts diagram.
Given:
Link lengths:
Solution: 1.
Coupler point data:
Ground link
L1 45
Crank
L2 56
Coupler
L3 22.5
Rocker
L4 56
A1P 11.25 δ 0.000 deg
See Figure 3-26 and Mathcad file P0328.
Calculate the length BP and the angle using the law of cosines on the triangle APB. B1P L3 A1P 2 L3 A1P cos δ 2
2
0.5
B1P 11.250
L32 B1P 2 A1P 2 γ acos 2 L3 B1P 2.
γ 0.0000 deg
Use the Cayley diagram (see Figure 3-26) to calculate the link lengths of the two cognates. Note that the diagram is made up of three parallelograms and three similar triangles L4
L5 B1P
L5 11.250
L6
L10 A1P
L10 11.250
L9
L7 28.000
L8 L6
L7 L9
B1P A1P
L3 L2 L3
B1P
L6 28.000
A1P
L9 28.000
A1P
L8 28.000
B1P
Calculate the coupler point data for cognates #2 and #3 A3P L4
A3P 56.000
A2P L2
A2P 56.000
δ δ
δ 0.000 deg
δ δ
δ 0.000 deg
From the Roberts diagram, calculate the ground link lengths for cognates #2 and #3 L1BC
3.
L1 L3
B1P
L1BC 22.5000
L1AC
L1 L3
A1P
L1AC 22.5000
Using the calculated link lengths, draw the Roberts diagram (see next page). SUMMARY OF COGNATE SPECIFICATIONS: Cognate #1
Cognate #2
Cognate #3
Ground link length
L1 45.000
L1AC 22.500
L1BC 22.500
Crank length
L2 56.000
L10 11.250
L7 28.000
Coupler length
L3 22.500
L9 28.000
L6 28.000
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-28-2
Rocker length
L4 56.000
L8 28.000
L5 11.250
Coupler point
A1P 11.250
A2P 56.000
A3P 56.000
Coupler angle
δ 0.000 deg
δ 0.000 deg
δ 0.000 deg
P
3
B1
A1
B2
B3
6 9
8
2
A2
7
5
10 OA
4.
4
A3
OB
OC
Both of these cognates are identical. Following Example 3-11, discard cognate #2 and retain cognates #1 and #3. Without allowing links 5, 6, and 7 to rotate, slide them as an assembly along line OAOC until the free end of link 7 is at OA. The free end of link 5 will then be at point O'B and point P on link 6 will be at P'. Add a new link of length OAOC between P and P'. This is the new output link 8 and all points on it describe the original coupler curve. P'
B1
8
3
P
A1
B3
6
7 2
4
5 OA
O'B
A3 OB
DESIGN OF MACHINERY - 5th Ed.
5.
SOLUTION MANUAL 3-28-3
Join links 2 and 7, making one ternary link. Remove link 5 and reduce link 6 to a binary link. The result is a Watt-I sixbar with links numbered 1, 2, 3, 4, 6, and 8. Link 8 is in curvilinear translation and follows the coupler path of the original point P. Link 8 is a binary link with nodes at P and P'. It does not attach to link 4 at B1.
P'
8
B1
3
P
A1
6
B3
2
OA
4
OB
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-29-1
PROBLEM 3-29 Statement:
Find the cognates of the Watt straight-line mechanism in Figure 3-29a (p. 131).
Given:
Link lengths:
Solution:
Coupler point data:
Ground link
L1 4
Crank
L2 2
A1P 0.500
δ 0.00 deg
Coupler
L3 1
Rocker
L4 2
B1P 0.500
γ 0.00 deg
See Figure 3-29a and Mathcad file P0329.
1.
Input the link dimensions and coupler point data into program FOURBAR.
2.
Use the Cognate pull-down menu to get the link lengths for cognates #2 and #3 (see next page). Note that, for this mechanism, cognates #2 and #3 are identical. All three mechanisms are non-Grashof with limited crank angles.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-29-2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-30-1
PROBLEM 3-30 Statement:
Find the cognates of the Roberts straight-line mechanism in Figure 3-29b.
Given:
Link lengths:
Solution:
Coupler point data:
Ground link
L1 2
Crank
L2 1
A1P 1.000
δ 60.0 deg
Coupler
L3 1
Rocker
L4 1
B1P 1.000
γ 60.0 deg
See Figure 3-29b and Mathcad file P0330.
1.
Input the link dimensions and coupler point data into program FOURBAR.
2.
Note that, for this mechanism, cognates #2 and #3 are identical with cognate #1 because of the symmetry of the linkage (draw the Cayley diagram to see this). All three mechanisms are non-Grashof with limited crank angles.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-31-1
PROBLEM 3-31 Statement:
Design a Hoeken straight-line linkage to give minimum error in velocity over 22% of the cycle for a 15-cm-long straight line motion. Specify all linkage parameters.
Given:
Length of straight line motion: Δx 150 mm Percentage of cycle over which straight line motion takes place: 22%
Solution:
See Figure 3-30 and Mathcad file P0331.
1.
Using Table 3-1 and the required length of straight-line motion, determine the link lengths. Link ratios from Table 3-1 for 22% cycle: L1overL2 1.975
L3overL2 2.463
ΔxoverL2 1.845
Link lengths:
2.
L2
Coupler
L3 L3overL2 L2
L3 200.24 mm
Ground link
L1 L1overL2 L2
L1 160.57 mm
Rocker
L4 L3
L4 200.24 mm
Coupler point
AP 2 L3
AP 400.49 mm
L2 81.30 mm
ΔxoverL2
Calculate the distance from point P to pivot O4 (Cy) when crank angle is 180 deg. Cy
3.
Δx
Crank
2 L32 L1 L22
Cy 319.20 mm
Enter the link lengths into program FOURBAR to verify the design (see next page for coupler point curve). Using the PRINT facility, determine the x,y coordinates of the coupler curve and the x,y components of the coupler point velocity in the straight line region. A table of these values is printed below. Notice the small deviations over the range of crank angles from the y-coordinate and the x-velocity at a crank angle of 180 deg.
FOURBAR for Windows
File P03-31.DOC
Angle Step Deg
Cpler Pt X mm
Cpler Pt Y mm
Veloc CP X mm/sec
Veloc CP Y mm/sec
140 150 160 170 180 190 200 210 220
235.60 216.84 198.06 179.31 160.58 141.85 123.09 104.31 85.55
319.95 319.72 319.46 319.27 319.20 319.27 319.47 319.72 319.95
-1,072.61 -1,076.20 -1,075.51 -1,073.75 -1,072.93 -1,073.75 -1,075.52 -1,076.22 -1,072.63
-14.74 -13.54 -7.99 0.02 8.03 13.58 14.78 10.76
-10.73
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-31-2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-32-1
PROBLEM 3-32 Statement:
Design a Hoeken straight-line linkage to give minimum error in straightness over 39% of the cycle for a 20-cm-long straight line motion. Specify all linkage parameters.
Given:
Length of straight line motion: Δx 200 mm Percentage of cycle over which straight line motion takes place: 39%
Solution:
See Figure 3-30 and Mathcad file P0332.
1.
Using Table 3-1 and the required length of straight-line motion, determine the link lengths. Link ratios from Table 3-1 for 39% cycle: L1overL2 2.500
L3overL2 3.250
ΔxoverL2 3.623
Δx ΔxoverL2
L2 55.20 mm
Link lengths:
2.
Crank
L2
Coupler
L3 L3overL2 L2
L3 179.41 mm
Ground link
L1 L1overL2 L2
L1 138.01 mm
Rocker
L4 L3
L4 179.41 mm
Coupler point
AP 2 L3
AP 358.82 mm
Calculate the distance from point P to pivot O4 (Cy) when crank angle is 180 deg. Cy
3.
2 L32 L1 L22
Cy 302.36 mm
Enter the link lengths into program FOURBAR to verify the design (see next page for coupler point curve). Using the PRINT facility, determine the x,y coordinates of the coupler curve and the x,y components of the coupler point velocity in the straight line region. A table of these values is printed below. Notice the small deviations over the range of crank angles from the y-coordinate and the x-velocity from a crank angle of 180 deg. FOURBAR for Windows
File P03-32.DAT
Angle Step Deg
Coupler Pt X mm
Coupler Pt Y mm
Veloc CP X mm/sec
110 120 130 140 150 160 170 180 190 200 210 220 230 240 250
237.992 225.289 211.710 197.521 182.927 168.076 153.076 138.010 122.944 107.944 93.093 78.499 64.311 50.731 38.028
302.408 302.361 302.378 302.398 302.399 302.385 302.368 302.360 302.368 302.385 302.399 302.398 302.378 302.361 302.408
-696.591 -755.847 -797.695 -826.217 -844.774 -856.043 -861.994 -863.841 -861.994 -856.043 -844.774 -826.217 -797.695 -755.847 -696.591
Veloc CP Y mm/sec -6.416 -0.019 1.426 0.664 -0.483 -1.052 -0.800 0.000 0.800 1.052 0.483 -0.664 -1.426 0.019 6.416
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-32-2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-33-1
PROBLEM 3-33 Statement:
Design a linkage that will give a symmetrical "kidney bean" shaped coupler curve as shown in Figure 3-16 (p. 114 and 115). Use the data in Figure 3-21 (p. 120) to determine the required link ratios and generate the coupler curve with program FOURBAR.
Solution:
See Figures 3-16, 3-21, and Mathcad file P0333.
Design choices: Ground link ratio, L1/L2 = 2.0: GLR 2.0 Common link ratio, L3/L2 = L4/L2 = BP/L2 = 2.5: CLR 2.5 Coupler angle, γ 72 deg Crank length, L2 2.000 1.
For the given design choices, determine the remaining link lengths and coupler point specification. Coupler link (3) length
L3 CLR L2
L3 5.000
Rocker link (4) length
L4 CLR L2
L4 5.000
Ground link (1) length
L1 GLR L2
L1 4.000
Angle PAB
δ
Length AP on coupler 2.
180 deg γ 2
AP 2 L3 cos δ
δ 54.000 deg AP 5.878
Enter the above data into program FOURBAR and plot the coupler curve.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-34-1
PROBLEM 3-34 Statement:
Design a linkage that will give a symmetrical "double straight" shaped coupler curve as shown in Figure 3-16. Use the data in Figure 3-21 to determine the required link ratios and generate the coupler curve with program FOURBAR.
Solution:
See Figures 3-16, 3-21, and Mathcad file P0334.
Design choices: Ground link ratio, L1/L2 = 2.5: GLR 2.5 Common link ratio, L3/L2 = L4/L2 = BP/L2 = 2.5: CLR 2.5 Coupler angle, γ 252 deg Crank length, L2 2.000 1.
For the given design choices, determine the remaining link lengths and coupler point specification. Coupler link (3) length
L3 CLR L2
L3 5.000
Rocker link (4) length
L4 CLR L2
L4 5.000
Ground link (1) length
L1 GLR L2
L1 5.000
Angle PAB
δ
Length AP on coupler 2.
180 deg γ 2
AP 2 L3 cos δ
δ 36.000 deg AP 8.090
Enter the above data into program FOURBAR and plot the coupler curve.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-35-1
PROBLEM 3-35 Statement:
Design a linkage that will give a symmetrical "scimitar" shaped coupler curve as shown in Figure 3-16. Use the data in Figure 3-21 to determine the required link ratios and generate the coupler curve with program FOURBAR. Show that there are (or are not) true cusps on the curve.
Solution:
See Figures 3-16, 3-21, and Mathcad file P0334.
Design choices: Ground link ratio, L1/L2 = 2.0: GLR 2.0 Common link ratio, L3/L2 = L4/L2 = BP/L2 = 2.5: CLR 2.5 Coupler angle, γ 144 deg Crank length, L2 2.000 1.
For the given design choices, determine the remaining link lengths and coupler point specification. Coupler link (3) length
L3 CLR L2
L3 5.000
Rocker link (4) length
L4 CLR L2
L4 5.000
Ground link (1) length
L1 GLR L2
L1 4.000
Angle PAB
δ
Length AP on coupler 2.
180 deg γ 2
AP 2 L3 cos δ
δ 18.000 deg AP 9.511
Enter the above data into program FOURBAR and plot the coupler curve.
DESIGN OF MACHINERY - 5th Ed.
3.
SOLUTION MANUAL 3-35-2
The points at the ends of the "scimitar" will be true cusps if the velocity of the coupler point is zero at these points. Using FOURBAR's plotting utility, plot the magnitude and angle of the coupler point velocity vector. As seen below for the range of crank angle from 50 to 70 degrees, the magnitude of the velocity does not quite reach zero. Therefore, these are not true cusps.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-36-1
PROBLEM 3-36 Statement:
Find the Grashof condition, inversion, any limit positions, and the extreme values of the transmission angle (to graphical accuracy) of the linkage in Figure P3-10.
Given:
Link lengths:
Link 2
L2 0.785
Link 3
L3 0.356
Link 4
L4 0.950
Link 1
L1 0.544
Grashof condition function: Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ
Solution: 1.
return "non-Grashof" otherwise See Figure P3-10 and Mathcad file P0336.
Determine the Grashof condition of the mechanism from inequality 2.8 and its Barker classification from Table 2-4. Condition L1 L2 L3 L4 "Grashof"
Grashof condition: Barker classification:
2.
Class I-3, Grashof rocker-crank-rocker, GRCR, since the shortest link is the coupler link.
A GRCR linkage will have two toggle positions. Draw the linkage in these two positions and measure the input link angles. A
B
158.286° O4
O2
O4
O2 B A
158.286°
3.
As measured from the layout, the input link angles at the toggle positions are: +158.3 and -158.3 deg.
4.
Since the coupler link in a GRCR linkage can make a full rotation with respect to the input and output rockers, the minimum transmission angle is 0 deg and the maximum is 90 deg.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-37-1
PROBLEM 3-37 Statement:
Draw the Roberts diagram and find the cognates for the linkage in Figure P3-10.
Given:
Link lengths:
Solution: 1.
Coupler point data:
Ground link
L1 0.544 Crank
L2 0.785
Coupler
L3 0.356 Rocker
L4 0.950
δ 0.00 deg
See Figure P3-10 and Mathcad file P0337.
Calculate the length BP and the angle using the law of cosines on the triangle APB. B1P L3 A1P 2 L3 A1P cos δ 2
2
0.5
B1P 0.734
L32 B1P 2 A1P 2 γ acos 2 L3 B1P 2.
A1P 1.09
γ 180.0000 deg
Use the Cayley diagram (see Figure 3-24) to calculate the link lengths of the two cognates. Note that the diagram is made up of three parallelograms and three similar triangles L4
L5 B1P
L5 0.734
L6
L10 A1P
L10 1.090
L9
L7 1.619
L8 L6
L7 L9
B1P A1P
L3 L2 L3
B1P
L6 1.959
A1P
L9 2.404
A1P
L8 2.909
B1P
Calculate the coupler point data for cognates #2 and #3 A3P L4
A3P 0.950
A2P L2
A2P 0.785
δ 180 deg δ
δ 180.000 deg
δ δ
δ 0.000 deg
From the Roberts diagram, calculate the ground link lengths for cognates #2 and #3 L1BC
3.
L1 L3
B1P
L1BC 1.1216
L1AC
L1 L3
A1P
L1AC 1.6656
Using the calculated link lengths, draw the Roberts diagram (see next page). SUMMARY OF COGNATE SPECIFICATIONS: Cognate #1
Cognate #2
Cognate #3
Ground link length
L1 0.544
L1AC 1.666
L1BC 1.122
Crank length
L2 0.785
L10 1.090
L7 1.619
Coupler length
L3 0.356
L9 2.404
L6 1.959
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-37-2
Rocker length
L4 0.950
L8 2.909
L5 0.734
Coupler point
A1P 1.090
A2P 0.785
A3P 0.950
Coupler angle
δ 0.000 deg
δ 0.000 deg
δ 180.000 deg
B2
9
8
P B1 A1
3 4
A2
2
5
10 OA
1
A3 OC
OB 6
7
B3
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-38-1
PROBLEM 3-38 Statement:
Find the three geared fivebar cognates of the linkage in Figure P3-10.
Given:
Link lengths:
Solution: 1.
Coupler point data:
Ground link
L1 0.544 Crank
L2 0.785
Coupler
L3 0.356 Rocker
L4 0.950
δ 0.00 deg
See Figure P3-10 and Mathcad file P0338.
Calculate the length BP and the angle using the law of cosines on the triangle APB. B1P L3 A1P 2 L3 A1P cos δ 2
2
0.5
B1P 0.734
L32 B1P 2 A1P 2 γ acos 2 L3 B1P 2.
A1P 1.09
γ 180.0000 deg
Use the Cayley diagram (see Figure 3-24) to calculate the link lengths of the two cognates. Note that the diagram is made up of three parallelograms and three similar triangles L4
L5 B1P
L5 0.734
L6
L10 A1P
L10 1.090
L9
L7 1.619
L8 L6
L7 L9
B1P A1P
L3 L2 L3
B1P
L6 1.959
A1P
L9 2.404
A1P
L8 2.909
B1P
Calculate the coupler point data for cognates #2 and #3 A3P L4
A3P 0.950
A2P L2
A2P 0.785
δ 180 deg δ
δ 180.000 deg
δ δ
δ 0.000 deg
From the Roberts diagram, calculate the ground link lengths for cognates #2 and #3 L1BC 3.
L1 L3
B1P
L1BC 1.1216
L1AC
L1 L3
A1P
L1AC 1.6656
Using the calculated link lengths, draw the Roberts diagram (see next page). SUMMARY OF COGNATE SPECIFICATIONS: Cognate #1
Cognate #2
Cognate #3
Ground link length
L1 0.544
L1AC 1.666
L1BC 1.122
Crank length
L2 0.785
L10 1.090
L7 1.619
Coupler length
L3 0.356
L9 2.404
L6 1.959
Rocker length
L4 0.950
L8 2.909
L5 0.734
Coupler point
A1P 1.090
A2P 0.785
A3P 0.950
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-38-2
δ 0.000 deg
Coupler angle
δ 0.000 deg
δ 180.000 deg
B2
9
8
P B1 A1
3 4
A2
2
5
10 OA
1
A3 OC
OB 6
7
B3 4.
The three geared fivebar cognates can be seen in the Roberts diagram. They are: OAA2PA3OB, OAA1PB3OC, and OBB1PB2OC. They are specified in the summary table below. SUMMARY OF GEARED FIVEBAR COGNATE SPECIFICATIONS: Cognate #1
Cognate #2
Cognate #3
Ground link length
L1 0.544
L1AC 1.666
L1BC 1.122
Crank length
L10 1.090
L2 0.785
L4 0.950
Coupler length
A2P 0.785
A1P 1.090
L5 0.734
Rocker length
A3P 0.950
L8 2.909
L7 1.619
Crank length
L5 0.734
L7 1.619
L8 2.909
Coupler point
A2P 0.785
A1P 1.090
B1P 0.734
Coupler angle
δ 0.00 deg
δ 0.00 deg
δ 0.00 deg
5.
Enter the cognate #1 specifications into program FOURBAR to get a trace of the coupler path (see next page).
6.
Enter the geared fivebar cognate #1 specifications into program FIVEBAR to get a trace of the coupler path for the geared fivebar (see next page).
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-38-3
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-39-1
PROBLEM 3-39 Statement:
Find the Grashof condition, any limit positions, and the extreme values of the transmission angle (to graphical accuracy) of the linkage in Figure P3-11.
Given:
Link lengths:
Link 2
L2 0.86
Link 3
L3 1.85
Link 4
L4 0.86
Link 1
L1 2.22
Grashof condition function: Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Solution: 1.
See Figure P3-11 and Mathcad file P0339.
Determine the Grashof condition of the mechanism from inequality 2.8 and its Barker classification from Table 2-4. Condition L1 L2 L3 L4 "non-Grashof"
Grashof condition: Barker classification:
2.
Class II-1, non-Grashof triple rocker, RRR1, since the longest link is the ground link.
An RRR1 linkage will have two toggle positions. Draw the linkage in these two positions and measure the input link angles. 116.037° A B
O4
O2
O4
O2 B A 116.037°
88.2° B
A 67.3° O2
O4
3.
As measured from the layout, the input link angles at the toggle positions are: +116 and -116 deg.
4.
Since the coupler link in an RRR1 linkage cannot make a full rotation with respect to the input and output rockers, the minimum transmission angle is 0 deg and the maximum is 88 deg.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-40-1
PROBLEM 3-40 Statement:
Draw the Roberts diagram and find the cognates for the linkage in Figure P3-11.
Given:
Link lengths:
Solution: 1.
Coupler point data:
Ground link
L1 2.22
Crank
L2 0.86
Coupler
L3 1.85
Rocker
L4 0.86
δ 0.00 deg
See Figure P3-11 and Mathcad file P0340.
Calculate the length BP and the angle using the law of cosines on the triangle APB. B1P L3 A1P 2 L3 A1P cos δ 2
2
0.5
B1P 0.520
L32 B1P 2 A1P 2 γ acos 2 L3 B1P 2.
A1P 1.33
γ 0.0000 deg
Use the Cayley diagram (see Figure 3-24) to calculate the link lengths of the two cognates. Note that the diagram is made up of three parallelograms and three similar triangles L4
L5 B1P
L5 0.520
L6
L10 A1P
L10 1.330
L9
L7 0.242
L8 L6
L7 L9
B1P A1P
L3 L2 L3
B1P
L6 0.242
A1P
L9 0.618
A1P
L8 0.618
B1P
Calculate the coupler point data for cognates #2 and #3 A3P L8
A3P 0.618
A2P L7
A2P 0.242
δ 180 deg
δ 180.000 deg
δ 180 deg
δ 180.000 deg
From the Roberts diagram, calculate the ground link lengths for cognates #2 and #3 L1BC 3.
L1 L3
B1P
L1BC 0.6240
L1AC
L1 L3
A1P
L1AC 1.5960
Using the calculated link lengths, draw the Roberts diagram (see next page). SUMMARY OF COGNATE SPECIFICATIONS: Cognate #1
Cognate #2
Cognate #3
Ground link length
L1 2.220
L1AC 1.596
L1BC 0.624
Crank length
L2 0.860
L10 1.330
L7 0.242
Coupler length
L3 1.850
L9 0.618
L6 0.242
Rocker length
L4 0.860
L8 0.618
L5 0.520
Coupler point
A1P 1.330
A2P 0.242
A3P 0.618
Coupler angle
δ 0.000 deg
δ 180.000 deg
δ 180.000 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-40-2
B2
B1
9 A2
3
10
P
4
8
OC
OA
A3
7 6
2 A1
OB 5
B3
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-41-1
PROBLEM 3-41 Statement:
Find the three geared fivebar cognates of the linkage in Figure P3-11.
Given:
Link lengths:
Solution: 1.
Coupler point data:
Ground link
L1 2.22
Crank
L2 0.86
Coupler
L3 1.85
Rocker
L4 0.86
δ 0.00 deg
See Figure P3-11 and Mathcad file P0341.
Calculate the length BP and the angle using the law of cosines on the triangle APB. B1P L3 A1P 2 L3 A1P cos δ 2
2
0.5
B1P 0.520
L32 B1P 2 A1P 2 γ acos 2 L3 B1P 2.
A1P 1.33
γ 0.0000 deg
Use the Cayley diagram (see Figure 3-24) to calculate the link lengths of the two cognates. Note that the diagram is made up of three parallelograms and three similar triangles L4
L5 B1P
L5 0.520
L6
L10 A1P
L10 1.330
L9
L7 0.242
L8 L6
L7 L9
B1P A1P
L3 L2 L3
B1P
L6 0.242
A1P
L9 0.618
A1P
L8 0.618
B1P
Calculate the coupler point data for cognates #2 and #3 A3P L8
A3P 0.618
A2P L7
A2P 0.242
δ 180 deg
δ 180.000 deg
δ 180 deg
δ 180.000 deg
From the Roberts diagram, calculate the ground link lengths for cognates #2 and #3 L1BC
3.
L1 L3
B1P
L1BC 0.6240
L1AC
L1 L3
A1P
L1AC 1.5960
Using the calculated link lengths, draw the Roberts diagram (see next page). SUMMARY OF COGNATE SPECIFICATIONS: Cognate #1
Cognate #2
Cognate #3
Ground link length
L1 2.220
L1AC 1.596
L1BC 0.624
Crank length
L2 0.860
L10 1.330
L7 0.242
Coupler length
L3 1.850
L9 0.618
L6 0.242
Rocker length
L4 0.860
L8 0.618
L5 0.520
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-41-2
Coupler point
A1P 1.330
A2P 0.242
A3P 0.618
Coupler angle
δ 0.000 deg
δ 180.000 deg
δ 180.000 deg
B2
B1
9 A2
3
10
P
4
8
OC
OA
A3
7 6
2
5 B3
A1 4.
OB
The three geared fivebar cognates can be seen in the Roberts diagram. They are: OAB2PB3OB, OAA1PA3OC, and OBB1PA2OC. The three geared fivebar cognates are summarized in the table below. SUMMARY OF GEARED FIVEBAR COGNATE SPECIFICATIONS: Cognate #1
Cognate #2
Cognate #3
Ground link length
L1 2.220
L1AC 1.596
L1BC 0.624
Crank length
L10 1.330
L2 0.860
L4 0.860
Coupler length
L2 0.860
A1P 1.330
L5 0.520
Rocker length
L4 0.860
L8 0.618
L7 0.242
Crank length
L5 0.520
L7 0.242
L8 0.618
Coupler point
L2 0.860
A1P 1.330
B1P 0.520
Coupler angle
δ 0.00 deg
δ 0.00 deg
δ 0.00 deg
5.
Enter the cognate #1 specifications into program FOURBAR to get a trace of the coupler path (see next page)
6.
Enter the geared fivebar cognate #1 specifications into program FIVEBAR to get a trace of the coupler path for the geared fivebar (see next page).
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-41-3
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-42-1
PROBLEM 3-42 Statement:
Find the Grashof condition, any limit positions, and the extreme values of the transmission angle (to graphical accuracy) of the linkage in Figure P3-12.
Given:
Link lengths:
Link 2
L2 0.72
Link 3
L3 0.68
Link 4
L4 0.85
Link 1
L1 1.82
Grashof condition function: Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Solution: 1.
See Figure P3-12 and Mathcad file P0342.
Determine the Grashof condition of the mechanism from inequality 2.8 and its Barker classification from Table 2-4. Condition L1 L2 L3 L4 "non-Grashof"
Grashof condition: Barker classification:
2.
Class II-1, non-Grashof triple rocker, RRR1, since the longest link is the ground link.
An RRR1 linkage will have two toggle positions. Draw the linkage in these two positions and measure the input link angles.
A
55.4° B
O4
O2
O4
O2
B A
55.4°
B 88.8° O4 O2
A
3.
As measured from the layout, the input link angles at the toggle positions are: +55.4 and -55.4 deg.
4.
Since the coupler link in an RRR1 linkage it cannot make a full rotation with respect to the input and output rockers, the minimum transmission angle is 0 deg and the maximum is 88.8 deg.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-43-1
PROBLEM 3-43 Statement:
Draw the Roberts diagram and find the cognates for the linkage in Figure P3-12.
Given:
Link lengths:
Solution: 1.
Coupler point data:
Ground link
L1 1.82
Crank
L2 0.72
Coupler
L3 0.68
Rocker
L4 0.85
δ 54.0 deg
See Figure P3-12 and Mathcad file P0343.
Calculate the length BP and the angle using the law of cosines on the triangle APB. B1P L3 A1P 2 L3 A1P cos δ 2
2
0.5
B1P 0.792
L32 B1P 2 A1P 2 γ acos 2 L3 B1P 2.
A1P 0.97
γ 82.0315 deg
Use the Cayley diagram (see Figure 3-24) to calculate the link lengths of the two cognates. Note that the diagram is made up of three parallelograms and three similar triangles L4
L5 B1P
L5 0.792
L6
L10 A1P
L10 0.970
L9
L7 0.839
L8 L6
L7 L9
B1P A1P
L3 L2 L3
B1P
L6 0.990
A1P
L9 1.027
A1P
L8 1.212
B1P
Calculate the coupler point data for cognates #2 and #3 A3P L4
A3P 0.850
A2P L2
A2P 0.720
δ γ
δ 82.032 deg
δ δ
δ 54.000 deg
From the Roberts diagram, calculate the ground link lengths for cognates #2 and #3 L1BC 3.
L1 L3
B1P
L1BC 2.1208
L1AC
L1 L3
A1P
L1AC 2.5962
Using the calculated link lengths, draw the Roberts diagram (see next page). SUMMARY OF COGNATE SPECIFICATIONS: Cognate #1
Cognate #2
Cognate #3
Ground link length
L1 1.820
L1AC 2.596
L1BC 2.121
Crank length
L2 0.720
L10 0.970
L7 0.839
Coupler length
L3 0.680
L9 1.027
L6 0.990
Rocker length
L4 0.850
L8 1.212
L5 0.792
Coupler point
A1P 0.970
A2P 0.720
A3P 0.850
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-43-2
δ 54.000 deg
Coupler angle
δ 54.000 deg
OC
8 B2
7
9
B3
P 6 A2 1AC 10 3
OA
B1
5 4
A1 2
1BC
A3
1 OB
δ 82.032 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-44-1
PROBLEM 3-44 Statement:
Find the three geared fivebar cognates of the linkage in Figure P3-12.
Given:
Link lengths:
Solution: 1.
Coupler point data:
Ground link
L1 1.82
Crank
L2 0.72
Coupler
L3 0.68
Rocker
L4 0.85
δ 54.0 deg
See Figure P3-12 and Mathcad file P0344.
Calculate the length BP and the angle using the law of cosines on the triangle APB. B1P L3 A1P 2 L3 A1P cos δ 2
2
0.5
B1P 0.792
L32 B1P 2 A1P 2 γ acos 2 L3 B1P 2.
A1P 0.97
γ 82.0315 deg
Use the Cayley diagram (see Figure 3-24) to calculate the link lengths of the two cognates. Note that the diagram is made up of three parallelograms and three similar triangles L4
L5 B1P
L5 0.792
L6
L10 A1P
L10 0.970
L9
L7 0.839
L8 L6
L7 L9
B1P A1P
L3 L2 L3
B1P
L6 0.990
A1P
L9 1.027
A1P
L8 1.212
B1P
Calculate the coupler point data for cognates #2 and #3 A3P L4
A3P 0.850
A2P L2
A2P 0.720
δ γ
δ 82.032 deg
δ δ
δ 54.000 deg
From the Roberts diagram, calculate the ground link lengths for cognates #2 and #3 L1BC
3.
L1 L3
B1P
L1BC 2.1208
L1AC
L1 L3
A1P
L1AC 2.5962
Using the calculated link lengths, draw the Roberts diagram (see next page). SUMMARY OF COGNATE SPECIFICATIONS: Cognate #1
Cognate #2
Cognate #3
Ground link length
L1 1.820
L1AC 2.596
L1BC 2.121
Crank length
L2 0.720
L10 0.970
L7 0.839
Coupler length
L3 0.680
L9 1.027
L6 0.990
Rocker length
L4 0.850
L8 1.212
L5 0.792
Coupler point
A1P 0.970
A2P 0.720
A3P 0.850
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-44-2
δ 54.000 deg
Coupler angle
δ 54.000 deg
δ 82.032 deg
OC
8 B2
7
9
B3
P 6 A2 1AC 10 3
B1
5 4
A1 2
1BC
A3
1 OB
OA
4.
The three geared fivebar cognates can be seen in the Roberts diagram. They are: OAA2PA3OB, OAA1PB3OC, and OBB1PB2OC. SUMMARY OF GEARED FIVEBAR COGNATE SPECIFICATIONS: Cognate #1
Cognate #2
Cognate #3
Ground link length
L1 1.820
L1AC 2.596
L1BC 2.121
Crank length
L10 0.970
L2 0.720
L4 0.850
Coupler length
A2P 0.720
A1P 0.970
L5 0.792
Rocker length
A3P 0.850
L8 1.212
L7 0.839
Crank length
L5 0.792
L7 0.839
L8 1.212
Coupler point
A2P 0.720
A1P 0.970
B1P 0.792
Coupler angle
δ 0.00 deg
δ 0.00 deg
δ 0.00 deg
5.
Enter the cognate #1 specifications into program FOURBAR to get a trace of the coupler path (see next page).
6.
Enter the geared fivebar cognate #1 specifications into program FIVEBAR to get a trace of the coupler path for the geared fivebar (see next page).
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-44-3
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-45-1
PROBLEM 3-45 Statement:
Prove that the relationships between the angular velocities of various links in the Roberts diagram as shown in Figure 3-25 (p. 125) are true.
Given:
OAA1PA2, OCB2PB3, and OBB1PA3 are parallelograms for any position of link 2..
Proof: 1.
OAA1 and A2P are opposite sides of a parallelogram and are, therefore, always parallel.
2.
Any change in the angle of OAA1 (link 2) will result in an identical change in the angle of A2P.
3.
Angular velocity is the change in angle per unit time.
4.
Since OAA1 and A2P have identical changes in angle, their angular velocities are identical.
5.
A2P is a line on link 9 and all lines on a rigid body have the same angular velocity. Therefore, link 9 has the same angular velocity as link 2.
6.
OCB3 (link 7) and B2P are opposite sides of a parallelogram and are, therefore, always parallel.
7.
B2P is a line on link 9 and all lines on a rigid body have the same angular velocity. Therefore, link 7 has the same angular velocity as links 9 and 2.
8.
The same argument holds for links 3, 5, and 10; and links 4, 6, and 8.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-46-1
PROBLEM 3-46 Statement:
Design a fourbar linkage to move the object in Figure P3-13 from position 1 to 2 using points A and B for attachment. Add a driver dyad to limit its motion to the range of positions shown, making it a sixbar. All fixed pivots should be on the base.
Given:
Length of coupler link: L3 52.000
Solution:
See Figure P3-13 and Mathcad file P0346.
Design choices: Length of link 2
L2 130
Length of link 2b
L2b 40
L4 110
Length of link 4
1.
Connect the end points of the two given positions of the line AB with construction limes, i.e., lines from A1 to A2 and B1 to B2.
2.
Bisect these lines and extend their perpendicular bisectors into the base.
3.
Select one point on each bisector and label them O2 and O4, respectively. In the solution below the distances O2A was selected to be L2 130.000 and O4B to be L4 110.000 . This resulted in a ground-link-length O2O4 for the fourbar of 27.080.
4.
The fourbar stage is now defined as O2ABO4 with link lengths Ground link 1a
L1a 27.080
Link 3 (coupler)
L3 52.000
Link 2 (input)
L2 130.000
Link 4 (output)
L4 110.000
5.
Select a point on link 2 (O2A) at a suitable distance from O2 as the pivot point to which the driver dyad will be connected and label it C. (Note that link 2 is now a ternary link with nodes at O2, C, and A.) In the solution below the distance O2C was selected to be L2b 40.000 .
6.
Draw a construction line through C1C2 and extend it to the left.
7.
Select a point on this line and call it O6. In the solution below O6 was placed 20 units from the left edge of the base.
8.
Draw a circle about O6 with a radius of one-half the length C1C2 and label the intersections of the circle with the extended line as D1 and D2. In the solution below the radius was measured as 23.003 units.
A1
B1
3
A2
6
D1
O6
2 D2
C1 5
9.
The driver fourbar is now defined as O2CDO6 with link lengths Link 6 (crank)
L6 23.003
Link 5 (coupler) L5 106.866 Link 1b (ground) L1b 111.764 Link 2b (rocker) L2b 40.000
23.003 106.866 111.764
B2
4 C2
40.000 O2
O4 27.080
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-46-2
10. Use the link lengths in step 9 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition L1b L2b L5 L6 "Grashof" min L1b L2b L5 L6 23.003
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-47-1
PROBLEM 3-47 Statement:
Design a fourbar linkage to move the object in Figure P3-13 from position 2 to 3 using points A and B for attachment. Add a driver dyad to limit its motion to the range of positions shown, making it a sixbar. All fixed pivots should be on the base.
Given:
Length of coupler link: L3 52.000
Solution:
See Figure P3-13 and Mathcad file P0347.
Design choices: Length of link 2
L2 130
Length of link 4b
L4b 40
L4 225
Length of link 4
1.
Connect the end points of the two given positions of the line AB with construction limes, i.e., lines from A2 to A3 and B2 to B3.
2.
Bisect these lines and extend their perpendicular bisectors into the base.
3.
Select one point on each bisector and label them O2 and O4, respectively. In the solution below the distances O2A was selected to be L2 130.000 and O4B to be L4 225.000 . This resulted in a ground-link-length O2O4 for the fourbar of 111.758.
4.
The fourbar stage is now defined as O2ABO4 with link lengths Ground link 1a
L1a 111.758
Link 2 (input)
L2 130.000
Link 3 (coupler)
L3 52.000
Link 4 (output)
L4 225.000
5.
Select a point on link 4 (O4B) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it C. (Note that link 4 is now a ternary link with nodes at O4, C, and B.) In the solution below the distance O4C was selected to be L4b 40.000 .
6.
Draw a construction line through C2C3 and extend it downward.
7.
Select a point on this line and call it O6. In the solution below O6 was placed 20 units from the bottom of the base.
8.
9.
Draw a circle about O6 with a radius of one-half the length C1C2 and label the intersections of the circle with the extended line as D2 and D3. In the solution below the radius was measured as 10.480 units.
A
2
B 111.758
C3
O4 83.977
Link 5 (coupler) L5 83.977
1b
6 O6
92.425
A
5 D2
10.480
Link 4b (rocker) L4b 40.000
O2
1a
L6 10.480
Link 1b (ground) L1b 92.425
4
C2
The driver fourbar is now defined as O4CDO6 with link lengths Link 6 (crank)
3
D3
B
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-47-2
10. Use the link lengths in step 9 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition L1b L4b L5 L6 "Grashof" min L1b L4b L5 L6 10.480
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-48-1
PROBLEM 3-48 Statement:
Design a fourbar linkage to move the object in Figure P3-13 through the three positions shown using points A and B for attachment. Add a driver dyad to limit its motion to the range of positions shown, making it a sixbar. All fixed pivots should be on the base.
Given:
Length of coupler link: L3 52.000
Solution:
See Figure P3-13 and Mathcad file P0348.
Design choices: Length of link 4b
L4b 50
1.
Draw link AB in its three design positions A1B1, A2B2, A3B3 in the plane as shown.
2.
Draw construction lines from point A1 to A2 and from point A2 to A3.
3.
Bisect line A1A2 and line A2A3 and extend their perpendicular bisectors until they intersect. Label their intersection O2.
4.
Repeat steps 2 and 3 for lines B1B2 and B2B3. Label the intersection O4.
5.
Connect O2 with A1 and call it link 2. Connect O4 with B1 and call it link 4.
6.
Line A1B1 is link 3. Line O2O4 is link 1 (ground link for the fourbar). The fourbar is now defined as O2ABO4 and has link lengths of Ground link 1a
L1a 20.736
Link 2
L2 127.287
Link 3
L3 52.000
Link 4
L4 120.254
B1
A1 3
A2 4
D3
B2
2
O6
D1
6
5 A3 C3 O2 C1
C2 O4
7.
B3
Check the Grashof condition. Note that any Grashof condition is potentially acceptable in this case.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-48-2
Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition L1a L2 L3 L4 "Grashof" 8.
Select a point on link 4 (O4B) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it C. (Note that link 4 is now a ternary link with nodes at O4, C, and B.) In the solution above the distance O4C was selected to be L4b 50.000 .
9.
Draw a construction line through C1C3 and extend it to the left.
10. Select a point on this line and call it O6. In the solution above O6 was placed 20 units from the left edge of the base. 11. Draw a circle about O6 with a radius of one-half the length C1C3 and label the intersections of the circle with the extended line as D1 and D3. In the solution below the radius was measured as L6 45.719. 12. The driver fourbar is now defined as O4CDO6 with link lengths Link 6 (crank)
L6 45.719
Link 5 (coupler) L5 126.875 Link 1b (ground) L1b 128.545 Link 4b (rocker) L4b 50.000 13. Use the link lengths in step 12 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition L6 L1b L4b L5 "Grashof" min L6 L1b L4b L5 45.719
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-49-1
PROBLEM 3-49 Statement:
Design a fourbar linkage to move the object in Figure P3-14 from position 1 to 2 using points A and B for attachment. Add a driver dyad to limit its motion to the range of positions shown, making it a sixbar. All fixed pivots should be on the base.
Given:
Length of coupler link: L3 86.000
Solution:
See Figure P3-14 and Mathcad file P0349.
Design choices: Length of link 2
L2 125
Length of link 2b
L4b 50
Length of link 4
L4 140
1.
Connect the end points of the two given positions of the line AB with construction limes, i.e., lines from A1 to A2 and B1 to B2.
2.
Bisect these lines and extend their perpendicular bisectors into the base.
3.
Select one point on each bisector and label them O2 and O4, respectively. In the solution below the distances O2A was selected to be L2 125.000 and O4B to be L4 140.000 . This resulted in a ground-link-length O2O4 for the fourbar of 97.195.
4.
The fourbar stage is now defined as O2ABO4 with link lengths Ground link 1a
L1a 97.195
Link 3 (coupler)
L3 86.000
Link 2 (input)
L2 125.000
Link 4 (output)
L4 140.000
5.
Select a point on link 4 (O4B) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it C. (Note that link 4 is now a ternary link with nodes at O4, C, and B.) In the solution below the distance O4C was selected to be L4b 50.000 .
6.
Draw a construction line through C1C2 and extend it to the left.
7.
Select a point on this line and call it O6. In the solution below O6 was placed 20 units from the left edge of the base.
8.
9.
Draw a circle about O6 with a radius of one-half the length C1C2 and label the intersections of the circle with the extended line as D1 and D2. In the solution below the radius was measured as 25.808 units.
A1
3 2 B1 B2
6 D1
O6
L6 25.808
Link 5 (coupler) L5 130.479 Link 1b (ground) L1b 137.327 Link 4b (rocker) L4b 50.000
4
D2
O2 1a
97.195
5
The driver fourbar is now defined as O4CDO6 with link lengths Link 6 (crank)
A2
1b
C1
C2
25.808 130.479
137.327
O4
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-49-2
10. Use the link lengths in step 9 to find the Grashof condition of the driving fourbar (it must be Grashof and the shortest link must be link 6). Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition L1b L4b L5 L6 "Grashof" min L1b L4b L5 L6 25.808
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-50-1
PROBLEM 3-50 Statement:
Design a fourbar linkage to move the object in Figure P3-14 from position 2 to 3 using points A and B for attachment. Add a driver dyad to limit its motion to the range of positions shown, making it a sixbar. All fixed pivots should be on the base.
Given:
Length of coupler link: L3 86.000
Solution:
See Figure P3-14 and Mathcad file P0350.
Design choices: Length of link 2
L2 130
Length of link 2b
L2b 50
L4 130
Length of link 4
1.
Connect the end points of the two given positions of the line AB with construction limes, i.e., lines from A2 to A3 and B2 to B3.
2.
Bisect these lines and extend their perpendicular bisectors into the base.
3.
Select one point on each bisector and label them O2 and O4, respectively. In the solution below the distances O2A was selected to be L2 130.000 and O4B to be L4 130.000 . This resulted in a ground-link-length O2O4 for the fourbar of 67.395.
4.
The fourbar stage is now defined as O2ABO4 with link lengths Ground link 1a Link 3 (coupler)
5.
6. 7.
8.
9.
L1a 67.395 L3 86.000
Link 2 (input)
L2 130.000
Link 4 (output)
L4 130.000
Select a point on link 2 (O2A) at a suitable distance from O2 as the pivot point to which the driver dyad will be connected and label it C. (Note that link 4 is now a ternary link with nodes at O2, C, and A.) In the solution below the distance O2C was selected to be L2b 50.000 and the link was extended away from A to give a better position for the driving dyad. Draw a construction line through C2C3 and extend it downward. Select a point on this line and call it O6. In the solution below O6 was placed 35 units from the bottom of the base. A2
Draw a circle about O6 with a radius of one-half the length C1C2 and label the intersections of the circle with the extended line as D2 and D3. In the solution below the radius was measured as 24.647 units. The driver fourbar is now defined as O2CDO6 with link lengths Link 6 (crank)
3 C3
155°
107.974
O2
C2
1a
4
67.395
1b 5
Link 5 (coupler) L5 98.822
Link 2b (rocker) L2b 50.000
B2 A3
L6 24.647
Link 1b (ground) L1b 107.974
2
D3
O4 6
98.822
B3
O6
24.647 D2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-50-2
10. Use the link lengths in step 9 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition L1b L2b L5 L6 "Grashof" min L1b L2b L5 L6 24.647
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-51-1
PROBLEM 3-51 Statement:
Design a fourbar linkage to move the object in Figure P3-14 through the three positions shown using points A and B for attachment. Add a driver dyad to limit its motion to the range of positions shown, making it a sixbar. All fixed pivots should be on the base.
Given:
Length of coupler link: L3 86.000
Solution:
See Figure P3-14 and Mathcad file P0351.
Design choices: Length of link 4b
L4b 50
1.
Draw link AB in its three design positions A1B1, A2B2, A3B3 in the plane as shown.
2.
Draw construction lines from point A1 to A2 and from point A2 to A3.
3.
Bisect line A1A2 and line A2A3 and extend their perpendicular bisectors until they intersect. Label their intersection O2.
4.
Repeat steps 2 and 3 for lines B1B2 and B2B3. Label the intersection O4.
5.
Connect O2 with A1 and call it link 2. Connect O4 with B1 and call it link 4.
6.
Line A1B1 is link 3. Line O2O4 is link 1 (ground link for the fourbar). The fourbar is now defined as O2ABO4 and has link lengths of Ground link 1a
L1a 61.667
Link 2
L2 142.357
Link 3
L3 86.000
Link 4
L4 124.668
A1 A2
3
2 B1
D3
B2 O6
O2 6
D1
4
1b
A3
1a
5 C3
O4
7.
B3 C2
C1
Check the Grashof condition. Note that any Grashof condition is potentially acceptable in this case.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-51-2
Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition L1a L2 L3 L4 "Grashof" 8.
Select a point on link 4 (O4B) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it C. (Note that link 4 is now a ternary link with nodes at O4, C, and B.) In the solution above the distance O4C was selected to be L4b 50.000 .
9.
Draw a construction line through C1C3 and extend it to the left.
10. Select a point on this line and call it O6. In the solution above O6 was placed 20 units from the left edge of the base. 11. Draw a circle about O6 with a radius of one-half the length C1C3 and label the intersections of the circle with the extended line as D1 and D3. In the solution below the radius was measured as L6 45.178. 12. The driver fourbar is now defined as O4CDO6 with link lengths Link 6 (crank)
L6 45.178
Link 5 (coupler) L5 140.583 Link 1b (ground) L1b 142.205 Link 4b (rocker) L4b 50.000 13. Use the link lengths in step 12 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition L6 L1b L4b L5 "Grashof" 14. Unfortunately, although the solution presented appears to meet the design specification, a simple cardboard model will quickly demonstrate that it has a branch defect. That is, in the first position shown, the linkage is in the "open" configuration, but in the 2nd and 3rd positions it is in the "crossed" configuration. The linkage cannot get from one circuit to the other without removing a pin and reassembling after moving the linkage. The remedy is to attach the points A and B to the coupler, but not at the joints between links 2 and 3 and links 3 and 4.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-52-1
PROBLEM 3-52 Statement:
Design a fourbar linkage to move the object in Figure P3-15 from position 1 to 2 using points A and B for attachment. Add a driver dyad to limit its motion to the range of positions shown, making it a sixbar. All fixed pivots should be on the base.
Given:
Length of coupler link: L3 52.000
Solution:
See Figure P3-15 and Mathcad file P0352.
Design choices: Length of link 2
L2 100
Length of link 4b
L4b 40
L4 160
Length of link 4
1.
Connect the end points of the two given positions of the line AB with construction limes, i.e., lines from A1 to A2 and B1 to B2.
2.
Bisect these lines and extend their perpendicular bisectors into the base.
3.
Select one point on each bisector and label them O2 and O4, respectively. In the solution below the distances O2A was selected to be L2 100.000 and O4B to be L4 160.000 . This resulted in a ground-link-length O2O4 for the fourbar of 81.463.
4.
The fourbar stage is now defined as O2ABO4 with link lengths Ground link 1a
L1a 81.463
Link 3 (coupler)
L3 52.000
Link 2 (input)
L2 100.000
Link 4 (output)
L4 160.000
5.
Select a point on link 4 (O4B) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it C. (Note that link 4 is now a ternary link with nodes at O4, C, and B.) In the solution below the distance O4C was selected to be L4b 40.000 .
6.
Draw a construction line through C1C2 and extend it to the left.
7.
Select a point on this line and call it O6. In the solution below O6 was placed 20 units from the left edge of the base.
8.
Draw a circle about O6 with a radius of one-half the length C1C2 and label the intersections of the circle with the extended line as D1 and D2. In the solution below the radius was measured as 14.351 units.
A2
B1
A1 3
B2
2 14.351
9.
The driver fourbar is now defined as O4CDO6 with link lengths Link 6 (crank)
L6 14.351
Link 5 (coupler) L5 132.962
1a
O2 C2
5
O6
1b
C1
6 132.962 O4
Link 1b (ground) L1b 138.105 Link 4b (rocker) L4b 40.000
138.105
81.463
4
D2
D1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-52-2
10. Use the link lengths in step 9 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition L1b L4b L5 L6 "Grashof" min L1b L4b L5 L6 14.351
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-53-1
PROBLEM 3-53 Statement:
Design a fourbar linkage to move the object in Figure P3-15 from position 2 to 3 using points A and B for attachment. Add a driver dyad to limit its motion to the range of positions shown, making it a sixbar. All fixed pivots should be on the base.
Given:
Length of coupler link: L3 52.000
Solution:
See Figure P3-15 and Mathcad file P0353.
Design choices: Length of link 2
L2 150
Length of link 4b
L4b 50
L4 200
Length of link 4
1.
Connect the end points of the two given positions of the line AB with construction limes, i.e., lines from A2 to A3 and B2 to B3.
2.
Bisect these lines and extend their perpendicular bisectors into the base.
3.
Select one point on each bisector and label them O2 and O4, respectively. In the solution below the distances O2A was selected to be L2 150.000 and O4B to be L4 200.000 . This resulted in a ground-link-length O2O4 for the fourbar of L1a 80.864.
4.
The fourbar stage is now defined as O2ABO4 with link lengths Ground link 1a Link 3 (coupler)
L1a 80.864 L3 52.000
Link 2 (input)
L2 150.000
Link 4 (output)
L4 200.000
5.
Select a point on link 4 (O4B) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it C. (Note that link 4 is now a ternary link with nodes at O4, C, and B.) In the solution below the distance O4C was selected to be L4b 50.000 .
6.
Draw a construction line through C2C3 and extend it downward.
7.
Select a point on this line and call it O6. In the solution below O6 was placed 25 units from the bottom of the base.
8.
9.
Draw a circle about O6 with a radius of one-half the length C1C2 and label the intersections of the circle with the extended line as D2 and D3. In the solution below the radius was measured as L6 12.763. The driver fourbar is now defined as O4CDO6 with link lengths
A2 3
C2
B2
A3
4 2
O4 C3 1a
Link 6 (crank)
L6 12.763
B3
Link 5 (coupler) L5 112.498 Link 1b (ground) L1b 122.445 Link 4b (rocker) L4b 50.000
5
1b
O2 112.498
80.864
D2
122.445 O6
D3
12.763
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-53-2
10. Use the link lengths in step 9 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition L1b L4b L5 L6 "Grashof" min L1b L4b L5 L6 12.763
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-54-1
PROBLEM 3-54 Statement:
Design a fourbar linkage to move the object in Figure P3-15 through the three positions shown using points A and B for attachment. Add a driver dyad to limit its motion to the range of positions shown, making it a sixbar. All fixed pivots should be on the base.
Given:
Length of coupler link: L3 52.000
Solution:
See Figure P3-15 and Mathcad file P0354.
Design choices: L2b 40
Length of link 2b 1.
Draw link AB in its three design positions A1B1, A2B2, A3B3 in the plane as shown.
2.
Draw construction lines from point A1 to A2 and from point A2 to A3.
3.
Bisect line A1A2 and line A2A3 and extend their perpendicular bisectors until they intersect. Label their intersection O2.
4.
Repeat steps 2 and 3 for lines B1B2 and B2B3. Label the intersection O4.
5.
Connect O2 with A1 and call it link 2. Connect O4 with B1 and call it link 4.
6.
Line A1B1 is link 3. Line O2O4 is link 1 (ground link for the fourbar). The fourbar is now defined as O2ABO4 and has link lengths of Ground link 1a
L1a 53.439
Link 2
L2 134.341
Link 3
L3 52.000
Link 4
L4 90.203
A1
A2
B1 3
B2
6 D1
4
O6 2
D3 5
C1
C2
A3
O4
1b C3 O2
7.
1a B3
Check the Grashof condition. Note that any Grashof condition is potentially acceptable in this case.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-54-2
Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition L1a L2 L3 L4 "non-Grashof" Although this fourbar is non-Grashof, there are no toggle points within the required range of motion. 8.
9.
Select a point on link 2 (O2A) at a suitable distance from O2 as the pivot point to which the driver dyad will be connected and label it C. (Note that link 2 is now a ternary link with nodes at O2, C, and A.) In the solution above the distance O2C was selected to be L2b 40.000 . Draw a construction line through C1C3 and extend it to the left.
10. Select a point on this line and call it O6. In the solution above O6 was placed 20 units from the left edge of the base. 11. Draw a circle about O6 with a radius of one-half the length C1C3 and label the intersections of the circle with the extended line as D1 and D3. In the solution below the radius was measured as L6 29.760. 12. The driver fourbar is now defined as O2CDO6 with link lengths Link 6 (crank)
L6 29.760
Link 5 (coupler) L5 119.665 Link 1b (ground) L1b 122.613 Link 2b (rocker) L2b 40.000 13. Use the link lengths in step 12 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition L6 L1b L2b L5 "Grashof"
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-55-1
PROBLEM 3-55 Statement:
Design a fourbar mechanism to move the link shown in Figure P3-16 from position 1 to position 2. Ignore the third position and the fixed pivots O2 and O4 shown. Build a cardboard model and add a driver dyad to limit its motion to the range of positions designed, making it a sixbar.
Given:
Position 1 offsets:
Solution:
See figure below for one possible solution. Input file P0355.mcd from the solutions manual disk to the Mathcad program for this solution, file P03-55.4br to the program FOURBAR to see the fourbar solution linkage, and file P03-55.6br into program SIXBAR to see the complete sixbar with the driver dyad included.
xC1D1 3.744 in
yC1D1 2.497 in
1.
Connect the end points of the two given positions of the line CD with construction lines, i.e., lines from C1 to C2 and D1 to D2.
2.
Bisect these lines and extend their perpendicular bisectors in any convenient direction. In the solution below the bisector of C1C2 was extended downward and the bisector of D1D2 was extended upward.
3.
Select one point on each bisector and label them O4 and O6, respectively. In the solution below the distances O4D and O6C were each selected to be 7.500 in. This resulted in a ground-link-length O4O6 for the fourbar of 15.366 in.
4.
The fourbar stage is now defined as O4CDO6 with link lengths Link 5 (coupler) L5
2
xC1D1 yC1D1
Link 4 (input)
L4 7.500 in
Ground link 1b
L1b 15.366 in
2
L5 4.500 in Link 6 (output)
L6 7.500 in
5.
Select a point on link 4 (O4D) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, B, and D.) In the solution below the distance O4B was selected to be 4.000 in.
6.
Draw a construction line through B1B2 and extend it to the right.
7.
Select a point on this line and call it O2. In the solution below the distance AB was selected to be 6.000 in.
8.
Draw a circle about O2 with a radius of one-half the length B1B2 and label the intersections of the circle with the extended line as A1 and A2. In the solution below the radius was measured as 1.370 in.
9.
The driver fourbar is now defined as O2ABO4 with link lengths Link 2 (crank)
L2 1.370 in
Link 4a (rocker) L4a 4.000 in
Link 3 (coupler) L3 6.000 in Link 1a (ground) L1a 7.080 in
10. Use the link lengths in step 9 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 2). Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-55-2
Condition L1a L2 L3 L4a "Grashof" min L1a L2 L3 L4a 1.370 in
O4 4.000 7.500
B2
B1
A1 2 O2
4
4 D1 5 5
A2
3
C2
D2 15.366
C1 6 6 7.500 O6
11. Using the program FOURBAR and the link lengths given above, it was found that the fourbar O4DCO6 is non-Grashoff with toggle positions at 4 = -49.9 deg and +49.9 deg. The fourbar operates between 4 = +28.104 deg and -11.968 deg.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-56-1
PROBLEM 3-56 Statement:
Design a fourbar mechanism to move the link shown in Figure P3-16 from position 2 to position 3. Ignore the third position and the fixed pivots O2 and O4 shown. Build a cardboard model and add a driver dyad to limit its motion to the range of positions designed, making it a sixbar.
Given:
Position 2 offsets:
Solution:
See figure below for one possible solution. Input file P0356.mcd from the solutions manual disk to the Mathcad program for this solution, file P03-56.4br to the program FOURBAR to see the fourbar solution linkage, and file P03-56.6br into program SIXBAR to see the complete sixbar with the driver dyad included.
xC2D2 4.355 in
yC2D2 1.134 in
1.
Connect the end points of the two given positions of the line CD with construction lines, i.e., lines from C2 to C3 and D2 to D3.
2.
Bisect these lines and extend their perpendicular bisectors in any convenient direction. In the solution below the bisector of C2C3 was extended downward and the bisector of D2D3 was extended upward.
3.
Select one point on each bisector and label them O4 and O6, respectively. In the solution below the distances O4D and O6C were each selected to be 6.000 in. This resulted in a ground-link-length O4O6 for the fourbar of 14.200 in.
4.
The fourbar stage is now defined as O4DCO6 with link lengths Link 5 (coupler) L5
2
xC2D2 yC2D2
Link 4 (input)
L4 6.000 in
Ground link 1b
L1b 14.200 in
2
L5 4.500 in Link 6 (output)
L6 6.000 in
5.
Select a point on link 4 (O4D) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, B, and D.) In the solution below the distance O4B was selected to be 4.000 in.
6.
Draw a construction line through B1B2 and extend it to the right.
7.
Select a point on this line and call it O2. In the solution below the distance AB was selected to be 6.000 in.
8.
Draw a circle about O2 with a radius of one-half the length B1B2 and label the intersections of the circle with the extended line as A1 and A2. In the solution below the radius was measured as 1.271 in.
9.
The driver fourbar is now defined as O2ABO4 with link lengths Link 2 (crank)
L2 1.271 in
Link 4a (rocker) L4a 4.000 in
Link 3 (coupler) L3 6.000 in Link 1a (ground) L1a 7.099 in
10. Use the link lengths in step 9 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 2). Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-56-2
Condition L1a L2 L3 L4a "Grashof" min L1a L2 L3 L4a 1.271 in
6.000 O4
4.000
7.099
4 4
B1 D2 C2
B2 3
5
A1
5 C3 6.000
6
D3
2
O2
A2
6
O6
14.200
11. Using the program FOURBAR and the link lengths given above, it was found that the fourbar O4DCO6 is non-Grashoff with toggle positions at 4 = -41.6 deg and +41.6 deg. The fourbar operates between 4 = +26.171 deg and -11.052 deg.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-57-1
PROBLEM 3-57 Statement:
Design a fourbar mechanism to give the three positions shown in Figure P3-16. Ignore the points O2 and O4 shown. Build a cardboard model and add a driver dyad to limit its motion to the range of positions designed, making it a sixbar.
Solution:
See Figure P3-16 and Mathcad file P0357.
Design choices: L3 10.000
Length of link 3:
Length of link 4b:
L4b 4.500
1.
Draw link CD in its three design positions C1D1, C2D2, C3D3 in the plane as shown.
2.
Draw construction lines from point C1 to C2 and from point C2 to C3.
3.
Bisect line C1C2 and line C2C3 and extend their perpendicular bisectors until they intersect. Label their intersection O6.
4.
Repeat steps 2 and 3 for lines D1D2 and D2D3. Label the intersection O4.
5.
Connect O6 with C1 and call it link 6. Connect O4 with D1 and call it link 4.
6.
Line C1D1 is link 5. Line O6O4 is link 1a (ground link for the fourbar). The fourbar is now defined as O6CDO4 and has link lengths of Ground link 1a
L1a 2.616
Link 6
L6 6.080
Link 5
L5 4.500
Link 4
L4 6.901
D2
D1 5 5 C1
C3
C2
D3
B2
B1 6
2.765
5
4
3
4 4
6
B3
6 O4
6.080
A1 O2
2
6.901
O6 10.611
2.616
7.
Check the Grashof condition. Note that any Grashof condition is potentially acceptable in this case. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
A3
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-57-2
Condition L1a L4 L5 L6 "Grashof" 8.
9.
Select a point on link 4 (O4D) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, D, and B.) In the solution above the distance O4B was selected to be L4b 4.500 . Draw a construction line through B1B3 and extend it up to the right.
10. Layout the length of link 3 (design choice) along the extended line. Label the other end A. 11. Draw a circle about O2 with a radius of one-half the length B1B3 and label the intersections of the circle with the extended line as A1 and A3. In the solution below the radius was measured as L2 2.765. 12. The driver fourbar is now defined as O4BAO2 with link lengths Link 2 (crank)
L2 2.765
Link 3 (coupler) L3 10.000 Link 1b (ground) L1b 10.611 Link 4b (rocker) L4b 4.500 13. Use the link lengths in step 12 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 2). Condition L2 L3 L1b L4b "Grashof" min L2 L3 L1b L4b 2.765
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-58-1
PROBLEM 3-58 Statement:
Design a fourbar mechanism to give the three positions shown in Figure P3-16 using the fixed pivots O2 and O4 shown. (See Example 3-7.) Build a cardboard model and add a driver dyad to limit its motion to the range of positions designed, making it a sixbar.
Solution:
See Figure P3-16 and Mathcad file P0358.
Design choices:
L5 5.000
Length of link 5:
L2b 2.500
Length of link 2b:
1.
Draw link CD in its three design positions C1D1, C2D2, C3D3 in the plane as shown.
2.
Draw the ground link O2O4 in its desired position in the plane with respect to the first coupler position C1D1.
3.
Draw construction arcs from point C2 to O2 and from point D2 to O2 whose radii define the sides of triangle C2O2D2. This defines the relationship of the fixed pivot O2 to the coupler line CD in the second coupler position.
4.
Draw construction arcs from point C2 to O4 and from point D2 to O4 whose radii define the sides of triangle C2O4D2. This defines the relationship of the fixed pivot O4 to the coupler line CD in the second coupler position.
5.
Transfer this relationship back to the first coupler position C1D1 so that the ground plane position O2'O4' bears the same relationship to C1D1 as O2O4 bore to the second coupler position C2D2.
6.
Repeat the process for the third coupler position and transfer the third relative ground link position to the first, or reference, position.
7.
The three inverted positions of the ground link that correspond to the three desired coupler positions are labeled O2O4, O2'O4', and O2"O4" in the first layout below and are renamed E1F1, E2F2, and E3F3, respectively, in the second layout, which is used to find the points G and H. D2 D1
C2 C3
C1
O2
O4''
O2''
O2'
O4
O4'
D3
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-58-2
8.
Draw construction lines from point E1 to E2 and from point E2 to E3.
9.
Bisect line E1E2 and line E2E3 and extend their perpendicular bisectors until they intersect. Label their intersection G.
10. Repeat steps 2 and 3 for lines F1F2 and F2F3. Label the intersection H. 11. Connect E1 with G and label it link 2. Connect F1 with H and label it link 4. Reinverting, E1 and F1 are the original fixed pivots O2 and O4, respectively. 12. Line GH is link 3. Line O2O4 is link 1a (ground link for the fourbar). The fourbar is now defined as O2GHO4 and has link lengths of Ground link 1a
L1a 3.000
Link 2
L2 8.597
Link 3
L3 1.711
Link 4
L4 7.921
G 3
H
2
4
F1
E 1 O2 1a
F3
E3
O4
F2
E2
13. Check the Grashof condition. Note that any Grashof condition is potentially acceptable in this case. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition L1a L2 L3 L4 "Grashof" The fourbar that will provide the desired motion is now defined as a Grashof double crank in the crossed configuration. It now remains to add the original points C1 and D1 to the coupler GH and to define the driving dyad.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-58-3
14. Select a point on link 2 (O2G) at a suitable distance from O2 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 2 is now a ternary link with nodes at O2, B, and G.) In the solution below, the distance O2B was selected to be L2b 2.500 . 15. Draw a construction line through B1B3 and extend it up to the left. 16. Layout the length of link 5 (design choice) along the extended line. Label the other end A. 17. Draw a circle about O6 with a radius of one-half the length B1B3 and label the intersections of the circle with the extended line as A1 and A3. In the solution below the radius was measured as L6 1.541. 18. The driver fourbar is now defined as O2BAO6 with link lengths Link 6 (crank)
L6 1.541
Link 5 (coupler) L5 5.000 Link 1b (ground) L1b 5.374 Link 2b (rocker) L2b 2.500 19. Use the link lengths in step 18 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition L6 L5 L1b L2b "Grashof"
D2
D1
C2
3
D3
C3 G3
G2 C1
3
H1
3
H2
2
G1 2
B3
2
4
H3
4 4
5 O6
B1
A3 O2
A1
6
1a
O4
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-59-1
PROBLEM 3-59 Statement:
Design a fourbar mechanism to move the link shown in Figure P3-17 from position 1 to position 2. Ignore the third position and the fixed pivots O2 and O4 shown. Build a cardboard model and add a driver dyad to limit its motion to the range of positions designed, making it a sixbar.
Given:
Position 1 offsets:
Solution:
See figure below for one possible solution. Input file P0359.mcd from the solutions manual disk to the Mathcad program for this solution, file P03-59.4br to the program FOURBAR to see the fourbar solution linkage, and file P03-59.6br into program SIXBAR to see the complete sixbar with the driver dyad included.
xC1D1 1.896 in
yC1D1 1.212 in
1.
Connect the end points of the two given positions of the line CD with construction lines, i.e., lines from C1 to C2 and D1 to D2.
2.
Bisect these lines and extend their perpendicular bisectors in any convenient direction. In the solution below the bisector of C1C2 was extended downward and the bisector of D1D2 was extended upward.
3.
Select one point on each bisector and label them O4 and O6, respectively. In the solution below the distances O6C and O4D were each selected to be 6.500 in. This resulted in a ground-link-length O4O6 for the fourbar of 14.722 in.
4.
The fourbar stage is now defined as O4DCO6 with link lengths Link 5 (coupler) L5
2
xC1D1 yC1D1
Link 4 (input)
L4 6.500 in
Ground link 1b
L1b 14.722 in
2
L5 2.250 in Link 6 (output)
L6 6.500 in
5.
Select a point on link 4 (O4D) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, B, and D.) In the solution below the distance O4B was selected to be 4.500 in.
6.
Draw a construction line through B1B2 and extend it to the right.
7.
Select a point on this line and call it O2. In the solution below the distance AB was selected to be 6.000 in.
8.
Draw a circle about O2 with a radius of one-half the length B1B2 and label the intersections of the circle with the extended line as A1 and A2. In the solution below the radius was measured as 1.037 in.
9.
The driver fourbar is now defined as O2ABO4 with link lengths Link 2 (crank)
L2 0.645 in
Link 4a (rocker) L4a 4.500 in
Link 3 (coupler) L3 6.000 in Link 1a (ground) L1a 7.472 in
10. Use the link lengths in step 9 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 2). Condition( a b c d ) S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-59-2
Condition L1a L2 L3 L4a "Grashof" min L1a L2 L3 L4a 0.645 in
6.500 4.500 B2
4 4
D2
B1
5
6.500
C2 5 C1
O6
7.472
3
6 6
O4
D1 14.722
A2 O2
2 A1
0.645
11. Using the program FOURBAR and the link lengths given above, it was found that the fourbar O4CDO6 is non-Grashoff with toggle positions at 4 = -17.1 deg and +17.1 deg. The fourbar operates between 4 = +5.216 deg and -11.273 deg.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-60-1
PROBLEM 3-60 Statement:
Design a fourbar mechanism to move the link shown in Figure P3-17 from position 2 to position 3. Ignore the third position and the fixed pivots O2 and O4 shown. Build a cardboard model and add a driver dyad to limit its motion to the range of positions designed, making it a sixbar.
Given:
Position 2 offsets:
Solution:
See figure below for one possible solution. Input file P0360.mcd from the solutions manual disk to the Mathcad program for this solution, file P03-60.4br to the program FOURBAR to see the fourbar solution linkage, and file P03-60.6br into program SIXBAR to see the complete sixbar with the driver dyad included.
xC2D2 0.834 in
yC2D2 2.090 in
1.
Connect the end points of the two given positions of the line CD with construction lines, i.e., lines from C2 to C3 and D2 to D3.
2.
Bisect these lines and extend their perpendicular bisectors in any convenient direction. In the solution below the bisector of C2C3 was extended downward and the bisector of D2D3 was extended upward.
3.
Select one point on each bisector and label them O4 and O6, respectively. In the solution below the distances O4D and O6C were each selected to be 6.000 in. This resulted in a ground-link-length O4O6 for the fourbar of 12.933 in.
4.
The fourbar stage is now defined as O4DCO6 with link lengths Link 5 (coupler) L5
2
xC2D2 yC2D2
Link 4 (input)
L4 5.000 in
Ground link 1b
L1b 12.933 in
2
L5 2.250 in Link 6 (output)
L6 5.000 in
5.
Select a point on link 4 (O4D) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, B, and D.) In the solution below the distance O4B was selected to be 4.000 in.
6.
Draw a construction line through B1B2 and extend it to the right.
7.
Select a point on this line and call it O2. In the solution below the distance AB was selected to be 6.000 in.
8.
Draw a circle about O2 with a radius of one-half the length B1B2 and label the intersections of the circle with the extended line as A1 and A2. In the solution below the radius was measured as 0.741 in.
9.
The driver fourbar is now defined as O2ABO4 with link lengths Link 2 (crank)
L2 0.741 in
Link 4a (rocker) L4a 4.000 in
Link 3 (coupler) L3 6.000 in Link 1a (ground) L1a 7.173 in
10. Use the link lengths in step 9 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 2). Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-60-2
Condition L1a L2 L3 L4 "Grashof"
O4 5.500
4.000 4 B3
7.173 4 B2
D3 D2
5 C3
5
3
A3
O2 2
A2
C2 6 6
12.933
O6
11. Using the program FOURBAR and the link lengths given above, it was found that the fourbar O4DCO6 is non-Grashoff with toggle positions at 4 = -14.9 deg and +14.9 deg. The fourbar operates between 4 = +12.403 deg and -8.950 deg.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-61-1
PROBLEM 3-61 Statement:
Design a fourbar mechanism to give the three positions shown in Figure P3-17. Ignore the points O2 and O4 shown. Build a cardboard model and add a driver dyad to limit its motion to the range of positions designed, making it a sixbar.
Solution:
See Figure P3-17 and Mathcad file P0361.
Design choices: L3 6.000
Length of link 3:
L4b 2.500
Length of link 4b:
1.
Draw link CD in its three design positions C1D1, C2D2, C3D3 in the plane as shown.
2.
Draw construction lines from point C1 to C2 and from point C2 to C3.
3.
Bisect line C1C2 and line C2C3 and extend their perpendicular bisectors until they intersect. Label their intersection O6.
4.
Repeat steps 2 and 3 for lines D1D2 and D2D3. Label the intersection O4.
5.
Connect O2 with C1 and call it link 2. Connect O4 with D1 and call it link 4.
6.
Line C1D1 is link 5. Line O2O4 is link 1a (ground link for the fourbar). The fourbar is now defined as O6CDO4 and has link lengths of Ground link 1a
L1a 1.835
Link 6
L6 2.967
Link 5
L5 2.250
Link 4
L4 3.323
D3 B3 3.323
D2
B2
4 5
4
5
C3 1.835
4 C2
O4 6 O6
6
D1
B1 3 5 C1
6
1.403 A3 O2 2
2.967
A1
6.347
7.
Check the Grashof condition. Note that any Grashof condition is potentially acceptable in this case.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-61-2
Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition L1a L4 L5 L6 "Grashof" 8.
Select a point on link 4 (O4D) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, D, and B.) In the solution above the distance O4B was selected to be L4b 2.500 .
9.
Draw a construction line through B1B3 and extend it up to the right.
10. Layout the length of link 3 (design choice) along the extended line. Label the other end A. 11. Draw a circle about O2 with a radius of one-half the length B1B3 and label the intersections of the circle with the extended line as A1 and A3. In the solution below the radius was measured as L2 1.403. 12. The driver fourbar is now defined as O2ABO4 with link lengths Link 2 (crank)
L2 1.403
Link 3 (coupler) L3 6.000 Link 1b (ground) L1b 6.347 Link 4b (rocker) L4b 2.500 13. Use the link lengths in step 12 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 2). Condition L1b L2 L3 L4b "Grashof" min L1b L2 L3 L4b 1.403
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-62-1
PROBLEM 3-62 Statement:
Design a fourbar mechanism to give the three positions shown in Figure P3-17 using the fixed pivots O2 and O4 shown. (See Example 3-7.) Build a cardboard model and add a driver dyad to limit its motion to the range of positions designed, making it a sixbar.
Solution:
See Figure P3-17 and Mathcad file P0362.
Design choices:
L5 4.000
Length of link 5:
Length of link 2b:
L2b 0.791
1.
Draw link CD in its three design positions C1D1, C2D2, C3D3 in the plane as shown.
2.
Draw the ground link O2O4 in its desired position in the plane with respect to the first coupler position C1D1.
3.
Draw construction arcs from point C2 to O2 and from point D2 to O2 whose radii define the sides of triangle C2O2D2. This defines the relationship of the fixed pivot O2 to the coupler line CD in the second coupler position.
4.
Draw construction arcs from point C2 to O4 and from point D2 to O4 whose radii define the sides of triangle C2O4D2. This defines the relationship of the fixed pivot O4 to the coupler line CD in the second coupler position.
5.
Transfer this relationship back to the first coupler position C1D1 so that the ground plane position O2'O4' bears the same relationship to C1D1 as O2O4 bore to the second coupler position C2D2.
6.
Repeat the process for the third coupler position and transfer the third relative ground link position to the first, or reference, position.
7.
The three inverted positions of the ground link that correspond to the three desired coupler positions are labeled O2O4, O2'O4', and O2"O4" in the first layout below and are renamed E1F1, E2F2, and E3F3, respectively, in the second layout, which is used to find the points G and H.
D3 D2
C3
D1 C2
C1 O2
O4 O2'' O2'
O4' O4''
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-62-2
8.
Draw construction lines from point E1 to E2 and from point E2 to E3.
9.
Bisect line E1E2 and line E2E3 and extend their perpendicular bisectors until they intersect. Label their intersection G.
10. Repeat steps 2 and 3 for lines F1F2 and F2F3. Label the intersection H. 11. Connect E1 with G and label it link 2. Connect F1 with H and label it link 4. Reinverting, E1 and F1 are the original fixed pivots O2 and O4, respectively. 12. Line GH is link 3. Line O2O4 is link 1a (ground link for the fourbar). The fourbar is now defined as O2GHO4 and has link lengths of Ground link 1a
L1a 3.000
Link 2
L2 0.791
Link 3
L3 1.222
Link 4
L4 1.950
E1 O2
F1 O4
1a 2
G
4
3 E3 E2
H
F2 F3
13. Check the Grashof condition. Note that any Grashof condition is potentially acceptable in this case. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition L1a L2 L3 L4 "non-Grashof" The fourbar that will provide the desired motion is now defined as a non-Grashof double rocker in the crossed configuration. It now remains to add the original points C1 and D1 to the coupler GH and to define the driving dyad, which in this case will drive link 4 rather than link 2. 14. Select a point on link 2 (O2G) at a suitable distance from O2 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 2 is now a ternary link with nodes at O2, B, and G.) In the solution below, the distance O2B was selected to be L2b 0.791 . Thus, in this case B and G coincide. 15. Draw a construction line through B1B3 and extend it up to the left. 16. Layout the length of link 5 (design choice) along the extended line. Label the other end A. 17. Draw a circle about O6 with a radius of one-half the length B1B3 and label the intersections of the circle
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-62-3
with the extended line as A1 and A3. In the solution below the radius was measured as L6 0.727.
18. The driver fourbar is now defined as O2BAO6 with link lengths Link 6 (crank)
L6 0.727
Link 5 (coupler) L5 4.000 Link 1b (ground) L1b 4.012 Link 2b (rocker) L2b 0.791 19. Use the link lengths in step 18 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition L6 L5 L1b L2b "Grashof"
D3 D2 A3 O6
D1
6 C3
A1
C2 3
5
3
1b
3
C1
4 O2
O4
4
2 G
H
4
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-63-1
PROBLEM 3-63 Statement:
Design a fourbar mechanism to move the link shown in Figure P3-18 from position 1 to position 2. Ignore the third position and the fixed pivots O2 and O4 shown. Build a cardboard model and add a driver dyad to limit its motion to the range of positions designed, making it a sixbar.
Given:
Position 1 offsets:
Solution:
See figure below for one possible solution. Input file P0363.mcd from the solutions manual disk to the Mathcad program for this solution, file P03-63.4br to the program FOURBAR to see the fourbar solution linkage, and file P03-63.6br into program SIXBAR to see the complete sixbar with the driver dyad included.
xC1D1 1.591 in
yC1D1 1.591 in
1.
Connect the end points of the two given positions of the line CD with construction lines, i.e., lines from C1 to C2 and D1 to D2.
2.
Bisect these lines and extend their perpendicular bisectors in any convenient direction. In the solution below the bisector of C1C2 was extended downward and the bisector of D1D2 was extended upward.
3.
Select one point on each bisector and label them O4 and O6, respectively. In the solution below the distances O4C and O6D were each selected to be 5.000 in. This resulted in a ground-link-length O4O6 for the fourbar of 10.457 in.
4.
The fourbar stage is now defined as O4CDO6 with link lengths Link 5 (coupler) L5
2
xC1D1 yC1D1
Link 4 (input)
L4 5.000 in
Ground link 1b
L1b 10.457 in
2
L5 2.250 in Link 6 (output)
L6 5.000 in
5.
Select a point on link 4 (O4C) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, B, and C.) In the solution below the distance O4B was selected to be 3.750 in.
6.
Draw a construction line through B1B2 and extend it to the right.
7.
Select a point on this line and call it O2. In the solution below the distance AB was selected to be 6.000 in.
8.
Draw a circle about O2 with a radius of one-half the length B1B2 and label the intersections of the circle with the extended line as A1 and A2. In the solution below the radius was measured as 0.882 in.
9.
The driver fourbar is now defined as O2ABO4 with link lengths Link 2 (crank)
L2 0.882 in
Link 4a (rocker) L4a 3.750 in
Link 3 (coupler) L3 6.000 in Link 1a (ground) L1a 7.020 in
10. Use the link lengths in step 9 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 2). Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition L1a L2 L3 L4a "Grashof"
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-63-2
O6
6 6
10.457
C2
5
5
A2 O2 2
C1
B2
B1
A1
3
4
4 5.000
D1
D2
3.750
7.020 O4
11. Using the program FOURBAR and the link lengths given above, it was found that the fourbar O4CDO6 is non-Grashoff with toggle positions at 4 = -38.5 deg and +38.5 deg. The fourbar operates between 4 = +15.206 deg and -12.009 deg.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-64-1
PROBLEM 3-64 Statement:
Design a fourbar mechanism to move the link shown in Figure P3-18 from position 2 to position 3. Ignore the third position and the fixed pivots O2 and O4 shown. Build a cardboard model and add a driver dyad to limit its motion to the range of positions designed, making it a sixbar.
Given:
Position 2 offsets:
Solution:
See figure below for one possible solution. Input file P0360.mcd from the solutions manual disk to the Mathcad program for this solution, file P03-60.4br to the program FOURBAR to see the fourbar solution linkage, and file P03-60.6br into program SIXBAR to see the complete sixbar with the driver dyad included.
xC2D2 2.053 in
yC2D2 0.920 in
1.
Connect the end points of the two given positions of the line CD with construction lines, i.e., lines from C2 to C3 and D2 to D3.
2.
Bisect these lines and extend their perpendicular bisectors in any convenient direction. In the solution below the bisector of C2C3 was extended downward and the bisector of D2D3 was extended upward.
3.
Select one point on each bisector and label them O4 and O6, respectively. In the solution below the distances O4D and O6C were each selected to be 5.000 in. This resulted in a ground-link-length O4O6 for the fourbar of 8.773 in.
4.
The fourbar stage is now defined as O4DCO6 with link lengths Link 5 (coupler) L5
2
xC2D2 yC2D2
Link 4 (input)
L4 5.000 in
Ground link 1b
L1b 8.773 in
2
L5 2.250 in Link 6 (output)
L6 5.000 in
5.
Select a point on link 4 (O4D) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, B, and D.) In the solution below the distance O4B was selected to be 3.750 in.
6.
Draw a construction line through B1B2 and extend it to the right.
7.
Select a point on this line and call it O2. In the solution below the distance AB was selected to be 6.000 in.
8.
Draw a circle about O2 with a radius of one-half the length B1B2 and label the intersections of the circle with the extended line as A1 and A2. In the solution below the radius was measured as 0.892 in.
9.
The driver fourbar is now defined as O2ABO4 with link lengths Link 2 (crank)
L2 0.892 in
Link 4a (rocker) L4a 3.750 in
Link 3 (coupler) L3 6.000 in Link 1a (ground) L1a 7.019 in
10. Use the link lengths in step 9 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 2). Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-64-2
Condition L1a L2 L3 L4a "Grashof"
7.019
O4
5.000
4
3.750 4 B3
B2 D2
D3 5 8.773
C3
A3 O2 2
A2
3
5 C2
6 6
O6
11. Using the program FOURBAR and the link lengths given above, it was found that the fourbar O4DCO6 is non-Grashoff with toggle positions at 4 = -55.7 deg and +55.7 deg. The fourbar operates between 4 = -7.688 deg and -35.202 deg.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-65-1
PROBLEM 3-65 Statement:
Design a fourbar mechanism to give the three positions shown in Figure P3-18. Ignore the points O2 and O4 shown. Build a cardboard model and add a driver dyad to limit its motion to the range of positions designed, making it a sixbar.
Solution:
See Figure P3-18 and Mathcad file P0365.
Design choices: L3 6.000
Length of link 3:
Length of link 4b:
L4b 5.000
1.
Draw link CD in its three design positions C1D1, C2D2, C3D3 in the plane as shown.
2.
Draw construction lines from point C1 to C2 and from point C2 to C3.
3.
Bisect line C1C2 and line C2C3 and extend their perpendicular bisectors until they intersect. Label their intersection O6.
4.
Repeat steps 2 and 3 for lines D1D2 and D2D3. Label the intersection O4.
5.
Connect O6 with C1 and call it link 6. Connect O4 with D1 and call it link 4.
6.
Line C1D1 is link 5. Line O6O4 is link 1a (ground link for the fourbar). The fourbar is now defined as O6CDO4 and has link lengths of Ground link 1a
L1a 8.869
Link 6
L6 1.831
Link 5
L5 2.250
Link 4
L4 6.953
7.646
O4 4
O2 A1
6.953 8.869
B3
4 B1
2 1.593
D3 D2
D 1 C2 5 C1
3
A3
5 C3
6 O6
6 1.831
7.
Check the Grashof condition. Note that any Grashof condition is potentially acceptable in this case.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-65-2
Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition L6 L1a L4 L5 "non-Grashof" 8.
9.
Select a point on link 4 (O4D) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, D, and B.) In the solution above the distance O4B was selected to be L4b 5.000 . Draw a construction line through B1B3 and extend it up to the right.
10. Layout the length of link 3 (design choice) along the extended line. Label the other end A. 11. Draw a circle about O2 with a radius of one-half the length B1B3 and label the intersections of the circle with the extended line as A1 and A3. In the solution below the radius was measured as L2 1.593. 12. The driver fourbar is now defined as O2ABO4 with link lengths Link 2 (crank)
L2 1.593
Link 3 (coupler) L3 6.000 Link 1b (ground) L1b 7.646 Link 4b (rocker) L4b 5.000 13. Use the link lengths in step 12 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 2). Condition L1b L2 L3 L4b "Grashof" min L1b L2 L3 L4b 1.593
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-66-1
PROBLEM 3-66 Statement:
Design a fourbar mechanism to give the three positions shown in Figure P3-18 using the fixed pivots O2 and O4 shown. (See Example 3-7.) Build a cardboard model and add a driver dyad to limit its motion to the range of positions designed, making it a sixbar.
Solution:
See Figure P3-18 and Mathcad file P0366.
Design choices:
Length of link 5:
L5 4.000
Length of link 2b:
L2b 2.000
1.
Draw link CD in its three design positions C1D1, C2D2, C3D3 in the plane as shown.
2.
Draw the ground link O2O4 in its desired position in the plane with respect to the first coupler position C1D1.
3.
Draw construction arcs from point C2 to O2 and from point D2 to O2 whose radii define the sides of triangle C2O2D2. This defines the relationship of the fixed pivot O2 to the coupler line CD in the second coupler position.
4.
Draw construction arcs from point C2 to O4 and from point D2 to O4 whose radii define the sides of triangle C2O4D2. This defines the relationship of the fixed pivot O4 to the coupler line CD in the second coupler position.
5.
Transfer this relationship back to the first coupler position C1D1 so that the ground plane position O2'O4' bears the same relationship to C1D1 as O2O4 bore to the second coupler position C2D2.
6.
Repeat the process for the third coupler position and transfer the third relative ground link position to the first, or reference, position.
7.
The three inverted positions of the ground link that correspond to the three desired coupler positions are labeled O2O4, O2'O4', and O2"O4" in the first layout below and are renamed E1F1, E2F2, and E3F3, respectively, in the second layout, which is used to find the points G and H.
D1 D2
D3
C1 C2
O2''
O4'
C3 O2
O4 O2'
O4''
8.
Draw construction lines from point E1 to E2 and from point E2 to E3.
9.
Bisect line E1E2 and line E2E3 and extend their perpendicular bisectors until they intersect. Label their intersection G.
10. Repeat steps 2 and 3 for lines F1F2 and F2F3. Label the intersection H.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-66-2
11. Connect E1 with G and label it link 2. Connect F1 with H and label it link 4. Reinverting, E1 and F1 are the original fixed pivots O2 and O4, respectively. 12. Line GH is link 3. Line O2O4 is link 1a (ground link for the fourbar). The fourbar is now defined as O2GHO4 and has link lengths of Ground link 1a
L1a 4.000
Link 2
L2 2.000
Link 3
L3 6.002
Link 4
L4 7.002
H
3
E3
4
F2
G 2 O2
O4 E 2 F1
E1
F3
13. Check the Grashof condition. Note that any Grashof condition is potentially acceptable in this case. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition L1a L2 L3 L4 "Grashof" The fourbar that will provide the desired motion is now defined as a non-Grashof crank rocker in the open configuration. It now remains to add the original points C1 and D1 to the coupler GH and to define the driving dyad, which in this case will drive link 4 rather than link 2.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-66-3
14. Select a point on link 2 (O2G) at a suitable distance from O2 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 2 is now a ternary link with nodes at O2, B, and G.) In the solution below, the distance O2B was selected to be L2b 2.000 . Thus, in this case B and G coincide. 15. Draw a construction line through B1B3 and extend it up to the left. 16. Layout the length of link 5 (design choice) along the extended line. Label the other end A. 17. Draw a circle about O6 with a radius of one-half the length B1B3 and label the intersections of the circle with the extended line as A1 and A3. In the solution below the radius was measured as L6 1.399. 18. The driver fourbar is now defined as O2BAO6 with link lengths L6 1.399
Link 6 (crank)
Link 5 (coupler) L5 4.000 Link 1b (ground) L1b 4.257 Link 2b (rocker) L2b 2.000 19. Use the link lengths in step 18 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition L6 L1b L2b L5 "Grashof" H1
H2
3 D1
3 D2
H3
D3 C1 C2 4 4 G2
2
3
C3 2
G1
O2
1a 2
1b
G3 5
A1 6 O6
A3
4 O4
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-67-1
PROBLEM 3-67 Statement:
Design a fourbar Grashof crank-rocker for 120 degrees of output rocker motion with a quick-return time ratio of 1:1.2. (See Example 3-9.)
Given:
Time ratio
Solution: 1.
Tr
1 1.2
See figure below for one possible solution. Also see Mathcad file P0367.
Determine the crank rotation angles and , and the construction angle from equations 3.1 and 3.2. Tr = Solving for , and
β
α
α β = 360 deg
β 360 deg
β 196 deg
1 Tr
α 360 deg β
α 164 deg
δ β 180 deg
δ 16 deg
2.
Start the layout by arbitrarily establishing the point O4 and from it layoff two lines of equal length, 90 deg apart. Label one B1 and the other B2. In the solution below, each line makes an angle of 45 deg with the horizontal and has a length of 1.000 in.
3.
Layoff a line through B1 at an arbitrary angle (but not zero deg). In the solution below the line is 60 deg to the horizontal.
0. 95 3
=
c
90.00°
B2 B2
B1
B1 4 O4
d
O4 3
4.4 91 =
b
3.8 33 =
0° .0 16
LAYOUT
A2
A1 2
O2
a
LINKAGE DEFINITION 0.2 55 =
O2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-67-2
4.
Layoff a line through B2 that makes an angle with the line in step 3 (76 deg to the horizontal in this case). The intersection of these two lines establishes the point O2.
5.
From O2 draw an arc that goes through B1. Extend O2B2 to meet this arc. Erect a perpendicular bisector to the extended portion of the line and transfer one half of the line to O2 as the length of the input crank.
6.
For this solution, the link lengths are: Ground link (1)
d 3.833 in
Coupler (3)
b 4.491 in
Crank (2)
a 0.255 in
Rocker (4)
c 0.953 in
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-68-1
PROBLEM 3-68 Statement:
Design a fourbar Grashof crank-rocker for 100 degrees of output rocker motion with a quick-return time ratio of 1:1.5. (See Example 3-9.)
Given:
Time ratio
Solution: 1.
Tr
1 1.5
See figure below for one possible solution. Also see Mathcad file P0368.
Determine the crank rotation angles and , and the construction angle from equations 3.1 and 3.2. Tr = Solving for , and
β
α
α β = 360 deg
β 360 deg
β 216 deg
1 Tr
α 360 deg β
α 144 deg
δ β 180 deg
δ 36 deg
2.
Start the layout by arbitrarily establishing the point O4 and from it layoff two lines of equal length, 100 deg apart. Label one B1 and the other B2. In the solution below, each line makes an angle of 40 deg with the horizontal and has a length of 2.000 in.
3.
Layoff a line through B1 at an arbitrary angle (but not zero deg). In the solution below the line is 20 deg to the horizontal.
4.
Layoff a line through B2 that makes an angle with the line in step 3 (56 deg to the horizontal in this case). The intersection of these two lines establishes the point O2.
5.
From O2 draw an arc that goes through B1. Extend O2B2 to meet this arc. Erect a perpendicular bisector to the extended portion of the line and transfer one half of the line to O2 as the length of the input crank.
B1
B2
B2
B1
3.0524 = b O4
3
O2
4 1.2694 = a
2
A1
LAYOUT O4
O2
2.0000 = c
A2 LINKAGE DEFINITION 2.5364 = d
DESIGN OF MACHINERY - 5th Ed.
6.
SOLUTION MANUAL 3-68-2
For this solution, the link lengths are: Ground link (1)
d 2.5364 in
Coupler (3)
b 3.0524 in
Crank (2)
a 1.2694 in
Rocker (4)
c 2.000 in
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-69-1
PROBLEM 3-69 Statement:
Design a fourbar Grashof crank-rocker for 80 degrees of output rocker motion with a quick-return time ratio of 1:1.33. (See Example 3-9.)
Given:
Time ratio
Solution: 1.
Tr
1 1.33
See figure below for one possible solution. Also see Mathcad file P0369.
Determine the crank rotation angles and , and the construction angle from equations 3.1 and 3.2. Tr = Solving for , and
β
α
α β = 360 deg
β 360 deg
β 205 deg
1 Tr
α 360 deg β
α 155 deg
δ β 180 deg
δ 25 deg
2.
Start the layout by arbitrarily establishing the point O4 and from it layoff two lines of equal length, 100 deg apart. Label one B1 and the other B2. In the solution below, each line makes an angle of 40 deg with the horizontal and has a length of 2.000 in.
3.
Layoff a line through B1 at an arbitrary angle (but not zero deg). In the solution below the line is 150 deg to the horizontal. 2. 00 0= c
90.00°
B2
B2
B1
B1 4
O4 25.0 0°
O4 3
6.2 32 =
b
4.7 63 =
d
LAYOUT
A2
A1 2
LINKAGE DEFINITION
a
O2
0.4 35 =
O2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-69-2
4.
Layoff a line through B2 that makes an angle with the line in step 3 (73 deg to the horizontal in this case). The intersection of these two lines establishes the point O2.
5.
From O2 draw an arc that goes through B1. Extend O2B2 to meet this arc. Erect a perpendicular bisector to the extended portion of the line and transfer one half of the line to O2 as the length of the input crank.
6.
For this solution, the link lengths are: Ground link (1)
d 4.763 in
Coupler (3)
b 6.232 in
Crank (2)
a 0.435 in
Rocker (4)
c 2.000 in
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-70-1
PROBLEM 3-70 Statement:
Design a sixbar drag link quick-return linkage for a time ratio of 1:4 and output rocker motion of 50 degrees. (See Example 3-10.)
Given:
Time ratio
Solution: 1.
Tr
1 4
See figure below for one possible solution. Also see Mathcad file P0370.
Determine the crank rotation angles and from equation 3.1. Tr = Solving for and
β
α
α β = 360 deg
β 360 deg 1 Tr
α 360 deg β
β 288 deg α 72 deg
2.
Draw a line of centers XX at any convenient location.
3.
Choose a crank pivot location O2 on line XX and draw an axis YY perpendicular to XX through O2.
4.
Draw a circle of convenient radius O2A about center O2. In the solution below, the length of O2A is a 1.000 in.
5.
Lay out angle with vertex at O2, symmetrical about quadrant one.
6.
Label points A1 and A2 at the intersections of the lines subtending angle and the circle of radius O2A.
7.
8.
Set the compass to a convenient radius AC long enough to cut XX in two places on either side of O2 when swung from both A1 and A2. Label the intersections C1 and C2. In the solution below, the length of AC is b 2.000 in. The line O2A is the driver crank, link 2, and the line AC is the coupler, link 3.
9.
The distance C1C2 is twice the driven (dragged) crank length. Bisect it to locate the fixed pivot O4.
10. The line O2O4 now defines the ground link. Line O4C is the driven crank, link 4. In the solution below, O4C measures c 2.282 in and O2O4 measures d 0.699 in. 11. Calculate the Grashoff condition. If non-Grashoff, repeat steps 7 through 11 with a shorter radius in step 7. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( a b c d ) "Grashof" 12. Invert the method of Example 3-1 to create the output dyad using XX as the chord and O4C1 as the driving crank. The points B1 and B2 will lie on line XX and be spaced apart a distance that is twice the length of O4C (link 4). The pivot point O6 will lie on the perpendicular bisector of B1B2 at a distance from XX which subtends the specified output rocker angle, which is 50 degrees in this problem. In the solution below, the length BC was chosen to be e 5.250 in.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-70-2
9.000°
72.000°
LAYOUT OF SIXBAR DRAG LINK QUICK RETURN WITH TIME RATIO OF 1:4 a = 1.000 b = 2.000 c = 2.282 d = 0.699 e = 5.250 f = 5.400
13. For the design choices made (lengths of links 2, 3 and 5), the length of the output rocker (link 6) was measured as f 5.400 in.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-71-1
PROBLEM 3-71 Statement:
Design a crank-shaper quick-return mechanism for a time ratio of 1:2.5 (Figure 3-14, p. 112).
Given:
Time ratio
Solution:
See Figure 3-14 and Mathcad file P0371.
TR
1 2.5
Design choices:
1.
Length of link 2 (crank)
L2 1.000
Length of link 5 (coupler)
L5 5.000
S 4.000
Length of stroke
Calculate from equations 3.1. TR
α β
α β 360 deg
α
360 deg 1
α 102.86 deg
1 TR
2.
Draw a vertical line and mark the center of rotation of the crank, O2, on it.
3.
Layout two construction lines from O2, each making an angle /2 to the vertical line through O2.
4.
Using the chosen crank length (see Design Choices), draw a circle with center at O2 and radius equal to the crank length. Label the intersections of the circle and the two lines drawn in step 3 as A1 and A2.
5.
Draw lines through points A1 and A2 that are also tangent to the crank circle (step 2). These two lines will simultaneously intersect the vertical line drawn in step 2. Label the point of intersection as the fixed pivot center O4.
6.
Draw a vertical construction line, parallel and to the right of O2O4, a distance S/2 (one-half of the output stroke length) from the line O2O4.
7.
Extend line O4A1 until it intersects the construction line drawn in step 6. Label the intersection B1.
8.
Draw a horizontal construction line from point B1, either to the left or right. Using point B1 as center, draw an arc of radius equal to the length of link 5 (see Design Choices) to intersect the horizontal construction line. Label the intersection as C1.
9.
Draw the slider blocks at points A1 and C1 and finish by drawing the mechanism in its other extreme position.
STROKE 4.000 2.000 C2
6
C1
B2
B1
5
O2 4
2 A2
3 O4
A1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-72-1
PROBLEM 3-72 Statement:
Design a sixbar, single-dwell linkage for a dwell of 70 deg of crank motion, with an output rocker motion of 30 deg using a symmetrical fourbar linkage with the following parameter values: ground link ratio = 2.0, common link ratio = 2.0, and coupler angle = 40 deg. (See Example 3-13.)
Given:
Crank dwell period: 70 deg. Output rocker motion: 30 deg. Ground link ratio, L1/L2 = 2.0: GLR 2.0 Common link ratio, L3/L2 = L4/L2 = BP/L2 = 2.0: CLR 2.0 Coupler angle, γ 40 deg
Design choice: Crank length, L2 2.000 Solution: 1.
See Figures 3-20 and 3-21 and Mathcad file P0372.
For the given design choice, determine the remaining link lengths and coupler point specification. Coupler link (3) length
L3 CLR L2
L3 4.000
Rocker link (4) length
L4 CLR L2
L4 4.000
Ground link (1) length
L1 GLR L2
L1 4.000
Angle PAB
δ
Length AP on coupler 2.
180 deg γ 2
AP 2 L3 cos δ
δ 70.000 deg AP 2.736
Enter the above data into program FOURBAR, plot the coupler curve, and determine the coordinates of the coupler curve in the selected range of crank motion, which in this case will be from 145 to 215 deg.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-72-2
FOURBAR for Windows
3.
File
P03-72
Angle Coupler Pt Step X Deg in
Coupler Pt Y in
Coupler Pt Mag in
Coupler Pt Ang in
145 150 155 160 165 170 175 180 185 190 195 200 205 210 215
3.818 3.661 3.494 3.319 3.135 2.945 2.749 2.547 2.342 2.133 1.923 1.711 1.499 1.289 1.080
4.422 4.360 4.295 4.226 4.156 4.083 4.009 3.935 3.859 3.783 3.707 3.631 3.555 3.479 3.403
120.297 122.895 125.549 128.259 131.025 133.846 136.723 139.655 142.639 145.674 148.757 151.886 155.055 158.261 161.498
-2.231 -2.368 -2.497 -2.617 -2.728 -2.829 -2.919 -2.999 -3.067 -3.124 -3.169 -3.202 -3.223 -3.232 -3.227
Layout this linkage to scale, including the coupler curve whose coordinates are in the table above. Use the points at crank angles of 145, 180, and 215 deg to define the pseudo-arc. Find the center of the pseudo-arc erecting perpendicular bisectors to the chords defined by the selected coupler curve points. The center will lie at the intersection of the perpendicular bisectors, label this point D. The radius of this circle is the length of link 5.
y
145 P
B
180
3 4 D
215 A 2
x
PSEUDO-ARC O2
4.
O4
The position of the end of link 5 at point D will remain nearly stationary while the crank moves from 145 to 215 deg. As the crank motion causes the coupler point to move around the coupler curve there will be another extreme position of the end of link 5 that was originally at D. Since a symmetrical linkage was chosen, the other extreme position will be located along a line through the axis of symmetry (see Figure 3-20) a distance equal to the length of link 5 measured from the point where the axis of symmetry intersects the coupler curve near the 0 deg coupler point. Establish this point and label it E.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-72-3
FOURBAR for Windows
File
P03-72
Angle Step Deg
Coupler Pt X in
Coupler Pt Y in
Coupler Pt Mag n
Coupler Pt Ang in
340.000 345.000 350.000 355.000 0.000 5.000 10.000 15.000
-0.718 -0.615 -0.506 -0.386 -0.255 -0.117 0.022 0.155
0.175 0.481 0.818 1.178 1.549 1.917 2.269 2.598
0.739 0.781 0.962 1.240 1.570 1.920 2.269 2.603
166.325 142.001 121.717 108.135 99.365 93.499 89.434 86.581
y 145 P
B 5
180
3 5
AXIS OF SYMMETRY
4 D
215
355 A
E
2
x
PSEUDO-ARC
O4
O2
5.
The line segment DE represents the maximum displacement that a link of the length equal to link 5, attached at P, will reach along the axis of symmetry. Construct a perpendicular bisector of the line segment DE and extend it to the right (or left, which ever is convenient). Locate fixed pivot O6 on the bisector of DE such that the lines O6D and O6E subtend the desired output angle, in this case 30 deg. Draw link 6 from D through O6 and extend it to any convenient length. This is the output link that will dwell during the specified motion of the crank. SUMMARY OF LINKAGE SPECIFICATIONS Original fourbar: O6
y 6
145 P
B 5
180
30.000° 3 5
L2 2.000
Coupler
L3 4.000
Rocker
L4 4.000
Coupler point
AP 2.736 δ 70.000 deg
E 2
x O2
Crank
Added dyad:
355 A
L1 4.000
4 D
215
Ground link
O4
Coupler
L5 3.840
Output
L6 5.595
Pivot O6
x 3.841 y 5.809
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-73-1
PROBLEM 3-73 Statement:
Design a sixbar, single-dwell linkage for a dwell of 100 deg of crank motion, with an output rocker motion of 50 deg using a symmetrical fourbar linkage with the following parameter values: ground link ratio = 2.0, common link ratio = 2.5, and coupler angle = 60 deg. (See Example 3-13.)
Given:
Crank dwell period: 100 deg. Output rocker motion: 50 deg. Ground link ratio, L1/L2 = 2.0: GLR 2.0 Common link ratio, L3/L2 = L4/L2 = BP/L2 = 2.0: CLR 2.5 Coupler angle, γ 60 deg
Design choice: Crank length, L2 2.000 Solution: 1.
See Figures 3-20 and 3-21 and Mathcad file P0373.
For the given design choice, determine the remaining link lengths and coupler point specification. Coupler link (3) length
L3 CLR L2
L3 5.000
Rocker link (4) length
L4 CLR L2
L4 5.000
Ground link (1) length
L1 GLR L2
L1 4.000
Angle PAB
δ
Length AP on coupler 2.
180 deg γ 2
AP 2 L3 cos δ
δ 60.000 deg AP 5.000
Enter the above data into program FOURBAR, plot the coupler curve, and determine the coordinates of the coupler curve in the selected range of crank motion, which in this case will be from 130 to 230 deg.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-73-2
FOURBAR for Windows
3.
File
P03-73
Angle Coupler Pt Step X Deg in
Coupler Pt Y in
Coupler Pt Mag in
Coupler Pt Ang in
130 140 150 160 170 180 190 200 210 220 230
6.449 6.171 5.840 5.464 5.047 4.598 4.123 3.631 3.130 2.629 2.138
6.812 6.695 6.559 6.408 6.244 6.071 5.892 5.709 5.523 5.336 5.146
108.774 112.833 117.078 121.493 126.060 130.765 135.588 140.504 145.482 150.482 155.454
-2.192 -2.598 -2.986 -3.347 -3.675 -3.964 -4.209 -4.405 -4.551 -4.643 -4.681
Layout this linkage to scale, including the coupler curve whose coordinates are in the table above. Use the points at crank angles of 130, 180, and 230 deg to define the pseudo-arc. Find the center of the pseudo-arc erecting perpendicular bisectors to the chords defined by the selected coupler curve points. The center will lie at the intersection of the perpendicular bisectors, label this point D. The radius of this circle is the length of link 5.
y 130 P
B
180
D
230
4
3
PSEUDO-ARC
A 2
x O2
4.
O4
The position of the end of link 5 at point D will remain nearly stationary while the crank moves from 130 to 230 deg. As the crank motion causes the coupler point to move around the coupler curve there will be another extreme position of the end of link 5 that was originally at D. Since a symmetrical linkage was chosen, the other extreme position will be located along a line through the axis of symmetry (see Figure 3-20) a distance equal to the length of link 5 measured from the point where the axis of symmetry intersects the coupler curve near the 0 deg coupler point. Establish this point and label it E.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-73-3
FOURBAR for Windows
File
P03-73
Angle Coupler Pt Step X Deg in
Coupler Pt Y in
Coupler Pt Mag in
Coupler Pt Ang in
340 350 0 10 20
1.429 2.316 3.316 4.265 5.047
3.013 3.237 3.746 4.414 5.078
151.688 134.332 117.727 104.920 96.371
-2.652 -2.262 -1.743 -1.137 -0.564
y 130 P 20 180
B
10 5
AXIS OF SYMMETRY
0
D
350
230
4
3 340 A
PSEUDO-ARC
E
2
x O4
O2
5.
The line segment DE represents the maximum displacement that a link of the length equal to link 5, attached at P, will reach along the axis of symmetry. Construct a perpendicular bisector of the line segment DE and extend it to the right (or left, which ever is convenient). Locate fixed pivot O6 on the bisector of DE such that the lines O6D and O6E subtend the desired output angle, in this case 30 deg. Draw link 6 from D through O6 and extend it to any convenient length. This is the output link that will dwell during the specified motion of the crank. SUMMARY OF LINKAGE y SPECIFICATIONS 130
Original fourbar: P
Ground link
L1 4.000
Crank
L2 2.000
Coupler
L3 5.000
Rocker
L4 5.000
Coupler point
AP 5.000
20 180
50.000°
B
10 5 0
O6 6
230
δ 60.000 deg
D
350
Added dyad:
4
3 340 PSEUDO-ARC
A
E
2
x O2
O4
Coupler
L5 5.395
Output
L6 2.998
Pivot O6
x 3.166 y 3.656
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-74-1
PROBLEM 3-74 Statement:
Design a sixbar, single-dwell linkage for a dwell of 80 deg of crank motion, with an output rocker motion of 45 deg using a symmetrical fourbar linkage with the following parameter values: ground link ratio = 2.0, common link ratio = 1.75, and coupler angle = 70 deg. (See Example 3-13.)
Given:
Crank dwell period: 80 deg. Output rocker motion: 45 deg. Ground link ratio, L1/L2 = 2.0: GLR 2.0 Common link ratio, L3/L2 = L4/L2 = BP/L2 = 2.0: CLR 1.75 Coupler angle, γ 70 deg
Design choice: Crank length, L2 2.000 Solution: 1.
See Figures 3-20 and 3-21 and Mathcad file P0374.
For the given design choice, determine the remaining link lengths and coupler point specification. Coupler link (3) length
L3 CLR L2
L3 3.500
Rocker link (4) length
L4 CLR L2
L4 3.500
Ground link (1) length
L1 GLR L2
L1 4.000
Angle PAB
δ
Length AP on coupler 2.
180 deg γ 2
AP 2 L3 cos δ
δ 55.000 deg AP 4.015
Enter the above data into program FOURBAR, plot the coupler curve, and determine the coordinates of the coupler curve in the selected range of crank motion, which in this case will be from 140 to 220 deg.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-74-2
FOURBAR for Windows
3.
File
P03-74
Angle Coupler Pt Step X Deg in
Coupler Pt Y in
Coupler Pt Mag in
Coupler Pt Ang in
140 150 160 170 180 190 200 210 220
5.208 4.940 4.645 4.332 4.005 3.668 3.322 2.969 2.613
5.252 5.032 4.804 4.578 4.359 4.152 3.958 3.779 3.612
97.395 100.971 104.781 108.860 113.242 117.942 122.946 128.210 133.663
-0.676 -0.958 -1.226 -1.480 -1.720 -1.945 -2.153 -2.337 -2.493
Layout this linkage to scale, including the coupler curve whose coordinates are in the table above. Use the points at crank angles of 140, 180, and 220 deg to define the pseudo-arc. Find the center of the pseudo-arc erecting perpendicular bisectors to the chords defined by the selected coupler curve points. The center will lie at the intersection of the perpendicular bisectors, label this point D. The radius of this circle is the length of link 5.
y
140 P 180
220
B
PSEUDO-ARC
4
3 A
O4
2 O2
4.
x D
The position of the end of link 5 at point D will remain nearly stationary while the crank moves from 140 to 220 deg. As the crank motion causes the coupler point to move around the coupler curve there will be another extreme position of the end of link 5 that was originally at D. Since a symmetrical linkage was chosen, the other extreme position will be located along a line through the axis of symmetry (see Figure 3-20) a distance equal to the length of link 5 measured from the point where the axis of symmetry intersects the coupler curve near the 0 deg coupler point. Establish this point and label it E.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-74-3
FOURBAR for Windows
File
P03-74
Angle Coupler Pt Step X Deg in
Coupler Pt Y in
Coupler Pt Mag in
Coupler Pt Ang in
340 350 0 10 20
1.658 2.360 3.147 3.886 4.490
2.158 2.562 3.185 3.887 4.530
129.810 112.856 98.919 88.916 82.372
-1.382 -0.995 -0.494 0.074 0.601
y
140 P 20 180
10
0
220
AXIS OF SYMMETRY
B
350 PSEUDO-ARC
A
4
340 3
O4
2 O2
x
D E
5.
The line segment DE represents the maximum displacement that a link of the length equal to link 5, attached at P, will reach along the axis of symmetry. Construct a perpendicular bisector of the line segment DE and extend it to the right (or left, which ever is convenient). Locate fixed pivot O6 on the bisector of DE such that the lines O6D and O6E subtend the desired output angle, in this case 30 deg. Draw link 6 from D through O6 and extend it to any convenient length. This is the output link that will dwell during the specified motion of the crank.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-74-4
y
SUMMARY OF LINKAGE SPECIFICATIONS Original fourbar:
140 P 20 180
10
0
220
Ground link
L1 4.000
Crank
L2 2.000
Coupler
L3 3.500
Rocker
L4 3.500
Coupler point
AP 4.015
B
δ 55.000 deg
350 PSEUDO-ARC
A
45.000°
340 3
4
O4
2 O2
O6
x
Added dyad:
D E
Coupler
L5 7.676
Output
L6 1.979
Pivot O6
x 6.217 y 0.653
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-75-1
PROBLEM 3-75 Statement:
Using the method of Example 3-11, show that the sixbar Chebychev straight-line linkage of Figure P2-5 is a combination of the fourbar Chebychev straight-line linkage of Figure 3-29d and its Hoeken's cognate of Figure 3-29e. See also Figure 3-26 for additional information useful to this solution. Graphically construct the Chebychev sixbar parallel motion linkage of Figure P2-5a from its two fourbar linkage constituents and build a physical or computer model of the result.
Solution:
See Figures P2-5, 3-29d, 3-29e, and 3-26 and Mathcad file P0375.
1.
Following Example 3-11and Figure 3-26 for the Chebyschev linkage of Figure 3-29d, the fixed pivot OC is found by laying out the triangle OAOBOC, which is similar to A1B1P. In this case, A1B1P is a striaght line with P halfway between A1 and B1 and therefore OAOBOC is also a straightline with OC halfway between OA and OB. As shown below and in Figure 3-26, cognate #1 is made up of links numbered 1, 2, 3, and 4. Cognate #2 is links numbered 1, 5, 6, and 7. Cognate #3 is links numbered 1, 8, 9, and 10.
3
P
3 3
B1
2
4
9
OC
B2
4
6
5 1
1
6
Links Removed
10 OA
2.
2
6
7
8
A3
4
B2
P2
A1
P1
6
9 B3
B1
A1
OB
A2
5 1
A2 OA
OC
OB
Discard cognate #3 and shift link 5 from the fixed pivot OB to OC and shift link 7 from OC to OB. Note that due to the symmetry of the figure above, L5 = 0.5 L3, L6 = L2, L7 = 0.5 L2 and OCOB = 0.5 OAOB. Thus, cognate #2 is, in fact, the Hoeken straight-line linkage. The original Chebyschev linkage with the Hoeken linkage superimposed is shown above right with the link 5 rotated to 180 deg. Links 2 and 6 will now have the same velocity as will 7 and 4. Thus, link 5 can be removed and link 6 can be reduced to a binary link supported and constrained by link 4. The resulting sixbar is the linkage shown in Figure P2-5.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-76-1
PROBLEM 3-76 Statement:
Design a driver dyad to drive link 2 of the Evans straigh-line linkage in Figure 3-29f from 150 deg to 210 deg. Make a model of the resulting sixbar linkage and trace the couple curve.
Given:
Output angle
Solution:
See Figjre 3-29f, Example 3-1, and Mathcad file P0376.
Design choices:
θ 60 deg
Link lengths:
L2 2.000
Link 2
Link 5
L5 3.000
1.
Draw the input link O2A in both extreme positions, A1 and A2, at the specified angles such that the desired angle of motion 2 is subtended.
2.
Draw the chord A1A2 and extend it in any convenient direction. In this solution it was extended downward.
3.
Layout the distance A1C1 along extended line A1A2 equal to the length of link 5. Mark the point C1.
4.
Bisect the line segment A1A2 and layout the length of that radius from point C1 along extended line A1A2. Mark the resulting point O6 and draw a circle of radius O6C1 with center at O6.
5.
Label the other intersection of the circle and extended line A1A2, C2.
6.
7.
A1
Measure the length of the crank (link 6) as O6C1 or O6C2. From the graphical solution, L6 1.000 Measure the length of the ground link (link 1) as O2O6. From the graphical solution, L1 3.073
P2 3 B1 , B 2
2 O2
4 A2
P1
1
5 C1
O4
6 O6
3.073" C2
8.
Find the Grashof condition. Condition( a b c d )
2.932"
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition L1 L2 L5 L6 "Grashof"
0.922" L1 = 2.4 L2 = 2 L3 = 3.2 L4 = 2.078 L5 = 3.00 L6 = 1.00 AP = 5.38
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-77-1
PROBLEM 3-77 Statement:
Design a driver dyad to drive link 2 of the Evans straigh-line linkage in Figure 3-29g from -40 deg to 40 deg. Make a model of the resulting sixbar linkage and trace the couple curve.
Given:
Output angle
Solution:
See Figjre 3-29G, Example 3-1, and Mathcad file P0377.
Design choices:
θ 80 deg
Link lengths:
L2 2.000
Link 2
Link 5
L5 3.000
1.
Draw the input link O2A in both extreme positions, A1 and A2, at the specified angles such that the desired angle of motion 2 is subtended.
2.
Draw the line A1C1 and extend it in any convenient direction. In this solution it was extended at a 30-deg angle from A1O2 (see note below) .
3.
Layout the distance A1C1 along extended line A1C1 equal to the length of link 5. Mark the point C1.
4.
Bisect the line segment A1A2 and layout the length of that radius from point C1 along extended line A1C1. Mark the resulting point O6 and draw a circle of radius O6C1 with center at O6.
5.
6.
7.
C2
O6 P2
Extend a line from A2 through O6. Label the other intersection of the circle and extended line A2O6, C2. Measure the length of the crank (link 6) as O6C1 or O6C2. From the graphical solution, L6 1.735
6
3.165"
C1
Measure the length of the ground link (link 1) as O2O6. From the graphical solution, L1 3.165
Find the Grashof condition. Condition( a b c d )
O2 B2
B1
3 2
4 A1
1
O4
P1
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition L1 L2 L5 L6 "Grashof"
A2 5
Note: If the angle between link 2 and link 5 is zero the resulting driving fourbar will be a special Grashof. For angles greater than zero but less than 33.68 degrees it is a Grashof crank-rocker. For angles greater than 33.68 it is a non-Grashof double rocker. 8.
L1 = 4.61 L2 = 2 L3 = 2.4 L4 = 2.334 L5 = 3.00 L6 = 1.735 AP = 3.00
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-78-1
PROBLEM 3-78 Statement:
Figure 6 on page ix of the Hrones and Nelson atlas of fourbar coupler curves (on the book DVD) shows a 50-point coupler that was used to generate the curves in the atlas. Using the definition of the vector R given in Figure 3-17b of the text, determine the 10 possible pairs of values of and R for the first row of points above the horizontal axis if the gridpoint spacing is one half the length of the unit crank.
Given:
Grid module g 0.5
Solution:
See Figure 6 H&N Atlas, Figure 3-17b, and Mathcad file P0378.
1.
The moving pivot point is located on the 3rd grid line from the bottom and the third grid line from the left when the crank angle is radians. Let the number of horizontal grid spaces from the left end of the coupler to the coupler point be n 2 1 7 and the number of vertical grid spaces from the coupler to the coupler point be m 2 1 2
2.
For the first row of points above the horizontal axis shown in Figure 6, n 2 1 7 and m 1.
3.
The angle, , between the coupler and the line from the coupler/crank pivot to the coupler point is
π π ϕ( m n ) if n 0 atan2( n m) if m = 0 0 if m 0
4.
2
2
The distance, R, from the pivot to the coupler point along the same line is 2
R( m n ) g m n
2
ϕ( m n ) n
5.
deg
R( m n )
-2.000
153.435
1.118
-1.000
135.000
0.707
0.000
90.000
0.500
1.000
45.000
0.707
2.000
26.565
1.118
3.000
18.435
1.581
4.000
14.036
2.062
5.000
11.310
2.550
6.000
9.462
3.041
7.000
8.130
3.536
The coupler point distance, R, like the link lengths A, B, and C is a ratio of the given length to the the length of the driving crank.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-79-1
PROBLEM 3-79 Statement:
The set of coupler curves in the Hrones and Nelson atlas of fourbar coupler curves (on the book DVD, page 16 of the PDF file) has A = B = C = 1.5. Model this linkage with program FOURBAR using the coupler point fartherest to the left in the row shown on page 1 and plot the resulting coupler curve.
Given:
A 1.5
Solution:
See Figure on page 1 H&N Atlas, Figure 3-17b, and Mathcad file P0379.
B 1.5
C 1.5
1.
The moving pivot point is located on the 3rd grid line from the bottom and the third grid line from the left when the crank angle is radians. Let the number of horizontal grid spaces from the left end of the coupler to the coupler point be n 2 1 7 and the number of vertical grid spaces from the coupler to the coupler point be m 2 1 2
2.
For the second column of points to the left of the coupler pivot and the second row of points above the horizontal axis n 2 and m 2. The grid spacing is g 0.5
3.
The angle, , between the coupler and the line from the coupler/crank pivot to the coupler point is
π π ϕ( m n ) if n 0 atan2( n m) if m = 0 0 if m 0
4.
2
6.
2
2
R( m n ) 1.414
Determine the values needed for input to FOURBAR. Link 2 (Crank)
a 1
Link 3 (Coupler)
b A a
b 1.500
Link 4 (Rocker)
c B a
c 1.500
Link 1 (Ground)
d C a
d 1.500
Distance to coupler point
R( m n ) 1.414
Angle from link 3 to coupler point
ϕ( m n ) 135.000 deg
Calculate the coordinates of O4. Let the angle between links 2 and 3 be , then
A 2 ( 1 C) 2 B2 2 A ( 1 C)
7.
2
The distance from the pivot to the coupler point, R, along the same line is R( m n ) g m n
5.
α acos
α 33.557 deg
xO4 C cos α
xO4 1.250
yO4 C sin α
yO4 0.829
Enter this data into FOURBAR and then plot the coupler curve. (See next page)
ϕ( m n ) 135.000 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-79-2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-80-1
PROBLEM 3-80 Statement:
The set of coupler curves on page 17 in the Hrones and Nelson atlas of fourbar coupler curves (on the book DVD, page 32 of the PDF file) has A = 1.5, B = C = 3.0. Model this linkage with program FOURBAR using the coupler point fartherest to the right in the row shown and plot the resulting coupler curve.
Given:
A 1.5
Solution:
See Figure on page 17 H&N Atlas, Figure 3-17b, and Mathcad file P0380.
B 3.0
C 3.0
1.
The moving pivot point is located on the 3rd grid line from the bottom and the third grid line from the left when the crank angle is radians. Let the number of horizontal grid spaces from the left end of the coupler to the coupler point be n 2 1 7 and the number of vertical grid spaces from the coupler to the coupler point be m 2 1 2
2.
For the fifth column of points to the right of the coupler pivot and the first row of points above the horizontal axis n 5 and m 1. The grid spacing is g 0.5
3.
The angle, , between the coupler and the line from the coupler/crank pivot to the coupler point is
π π ϕ( m n ) if n 0 atan2( n m) if m = 0 0 if m 0
4.
2
6.
2
2
R( m n ) 2.550
Determine the values needed for input to FOURBAR. Link 2 (Crank)
a 1
Link 3 (Coupler)
b A a
b 1.500
Link 4 (Rocker)
c B a
c 3.000
Link 1 (Ground)
d C a
d 3.000
Distance to coupler point
R( m n ) 2.550
Angle from link 3 to coupler point
ϕ( m n ) 11.310 deg
Calculate the coordinates of O4. Let the angle between links 2 and 3 be , then
A 2 ( 1 C) 2 B2 2 A ( 1 C)
7.
2
The distance from the pivot to the coupler point, R, along the same line is R( m n ) g m n
5.
α acos
α 39.571 deg
xO4 C cos α
xO4 2.313
yO4 C sin α
yO4 1.911
Enter this data into FOURBAR and then plot the coupler curve. (See next page)
ϕ( m n ) 11.310 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-80-2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-81-1
PROBLEM 3-81 Statement:
The set of coupler curves on page 21 in the Hrones and Nelson atlas of fourbar coupler curves (on the book DVD, page 36 of the PDF file) has A = 1.5, B = C = 3.5. Model this linkage with program FOURBAR using the coupler point fartherest to the right in the row shown and plot the resulting coupler curve.
Given:
A 1.5
Solution:
See Figure on page 21 H&N Atlas, Figure 3-17b, and Mathcad file P0381.
B 3.5
C 3.5
1.
The moving pivot point is located on the 3rd grid line from the bottom and the third grid line from the left when the crank angle is radians. Let the number of horizontal grid spaces from the left end of the coupler to the coupler point be n 2 1 7 and the number of vertical grid spaces from the coupler to the coupler point be m 2 1 2
2.
For the fourth column of points to the right of the coupler pivot and the second row of points above the horizontal axis n 4 and m 2. The grid spacing is g 0.5
3.
The angle, , between the coupler and the line from the coupler/crank pivot to the coupler point is
π π ϕ( m n ) if n 0 atan2( n m) if m = 0 0 if m 0
4.
2
6.
2
2
R( m n ) 2.236
Determine the values needed for input to FOURBAR. Link 2 (Crank)
a 1
Link 3 (Coupler)
b A a
b 1.500
Link 4 (Rocker)
c B a
c 3.500
Link 1 (Ground)
d C a
d 3.500
Distance to coupler point
R( m n ) 2.236
Angle from link 3 to coupler point
ϕ( m n ) 26.565 deg
Calculate the coordinates of O4. Let the angle between links 2 and 3 be , then
A 2 ( 1 C) 2 B2 2 A ( 1 C)
7.
2
The distance from the pivot to the coupler point, R, along the same line is R( m n ) g m n
5.
α acos
α 40.601 deg
xO4 C cos α
xO4 2.657
yO4 C sin α
yO4 2.278
Enter this data into FOURBAR and then plot the coupler curve. (See next page)
ϕ( m n ) 26.565 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-81-2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-82-1
PROBLEM 3-82 Statement:
The set of coupler curves on page 34 in the Hrones and Nelson atlas of fourbar coupler curves (on the book DVD, page 49 of the PDF file) has A = 2.0, B = 1.5, C = 2.0. Model this linkage with program FOURBAR using the coupler point fartherest to the right in the row shown and plot the resulting coupler curve.
Given:
A 2.0
Solution:
See Figure on page 34 H&N Atlas, Figure 3-17b, and Mathcad file P0382.
B 1.5
C 2.0
1.
The moving pivot point is located on the 3rd grid line from the bottom and the third grid line from the left when the crank angle is radians. Let the number of horizontal grid spaces from the left end of the coupler to the coupler point be n 2 1 7 and the number of vertical grid spaces from the coupler to the coupler point be m 2 1 2
2.
For the sixth column of points to the right of the coupler pivot and the first row of points below the horizontal axis n 6 and m 1. The grid spacing is g 0.5
3.
The angle, , between the coupler and the line from the coupler/crank pivot to the coupler point is
π π ϕ( m n ) if n 0 atan2( n m) if m = 0 0 if m 0
4.
2
6.
2
2
R( m n ) 3.041
Determine the values needed for input to FOURBAR. Link 2 (Crank)
a 1
Link 3 (Coupler)
b A a
b 2.000
Link 4 (Rocker)
c B a
c 1.500
Link 1 (Ground)
d C a
d 2.000
Distance to coupler point
R( m n ) 3.041
Angle from link 3 to coupler point
ϕ( m n ) 9.462 deg
Calculate the coordinates of O4. Let the angle between links 2 and 3 be , then
A 2 ( 1 C) 2 B2 2 A ( 1 C)
7.
2
The distance from the pivot to the coupler point, R, along the same line is R( m n ) g m n
5.
α acos
α 26.384 deg
xO4 C cos α
xO4 1.792
yO4 C sin α
yO4 0.889
Enter this data into FOURBAR and then plot the coupler curve. (See next page)
ϕ( m n ) 9.462 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-82-2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-83-1
PROBLEM 3-83 Statement:
The set of coupler curves on page 115 in the Hrones and Nelson atlas of fourbar coupler curves (on the book DVD, page 130 of the PDF file) has A = 2.5, B = 1.5, C = 2.5. Model this linkage with program FOURBAR using the coupler point fartherest to the right in the row shown and plot the resulting coupler curve.
Given:
A 2.5
Solution:
See Figure on page 115 H&N Atlas, Figure 3-17b, and Mathcad file P0383.
B 1.5
C 2.5
1.
The moving pivot point is located on the 3rd grid line from the bottom and the third grid line from the left when the crank angle is radians. Let the number of horizontal grid spaces from the left end of the coupler to the coupler point be n 2 1 7 and the number of vertical grid spaces from the coupler to the coupler point be m 2 1 2
2.
For the second column of points to the right of the coupler pivot and the second row of points below the horizontal axis n 2 and m 2. The grid spacing is g 0.5
3.
The angle, , between the coupler and the line from the coupler/crank pivot to the coupler point is
π π ϕ( m n ) if n 0 atan2( n m) if m = 0 0 if m 0
4.
2
6.
2
2
R( m n ) 1.414
Determine the values needed for input to FOURBAR. Link 2 (Crank)
a 1
Link 3 (Coupler)
b A a
b 2.500
Link 4 (Rocker)
c B a
c 1.500
Link 1 (Ground)
d C a
d 2.500
Distance to coupler point
R( m n ) 1.414
Angle from link 3 to coupler point
ϕ( m n ) 45.000 deg
Calculate the coordinates of O4. Let the angle between links 2 and 3 be , then
A 2 ( 1 C) 2 B2 2 A ( 1 C)
7.
2
The distance from the pivot to the coupler point, R, along the same line is R( m n ) g m n
5.
α acos
α 21.787 deg
xO4 C cos α
xO4 2.321
yO4 C sin α
yO4 0.928
Enter this data into FOURBAR and then plot the coupler curve. (See next page)
ϕ( m n ) 45.000 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-83-2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-84-1
PROBLEM 3-84 Statement:
Design a fourbar mechanism to move the link shown in Figure P3-19 from position 1 to position 2. Ignore the third position and the fixed pivots O2 and O4 shown. Build a cardboard model that demonstrates the required movement.
Given:
Position 1 offsets:
Solution:
See figure below and Mathcad file P0384 for one possible solution.
xC1D1 17.186 in
yC1D1 0.604 in
1.
Connect the end points of the two given positions of the line CD with construction lines, i.e., lines from C1 to C2 and D1 to D2.
2.
Bisect these lines and extend their perpendicular bisectors in any convenient direction. In the solution below the bisector of C1C2 was extended upward and the bisector of D1D2 was also extended upward.
3.
Select one point on each bisector and label them O2 and O4, respectively. In the solution below the distances O2C and O4D were selected to be 15.000 in. and 8.625 in, respectively. This resulted in a ground-link-length O2O4 for the fourbar of 9.351 in.
4.
The fourbar is now defined as O2CDO4 with link lengths Link 3 (coupler) L3
2
xC1D1 yC1D1
Link 2 (input)
L2 14.000 in
Ground link 1
L1 9.351 in
2
L3 17.197 in Link 4 (output)
L4 7.000 in
9.35 1
15 .00 0
O2
O4
17.197
8.6 25
D2
C1
D1
C2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-85-1
PROBLEM 3-85 Statement:
Design a fourbar mechanism to move the link shown in Figure P3-19 from position 2 to position 3. Ignore the first position and the fixed pivots O2 and O4 shown. Build a cardboard model that demonstrates the required movement.
Given:
Position 2 offsets:
Solution:
See figure below and Mathcad file P0385 for one possible solution.
xC2D2 15.524 in
yC2D2 7.397 in
1.
Connect the end points of the two given positions of the line CD with construction lines, i.e., lines from C2 to C3 and D2 to D3.
2.
Bisect these lines and extend their perpendicular bisectors in any convenient direction. In the solution below the bisector of C2C3 was extended upward and the bisector of D2D3 was also extended upward.
3.
Select one point on each bisector and label them O2 and O4, respectively. In the solution below the distances O2C and O4D were selected to be 15.000 in and 8.625 in, respectively. This resulted in a ground-link-length O2O4 for the fourbar of 9.470 in.
4.
The fourbar stage is now defined as O2CDO4 with link lengths Link 3 (coupler) L3
2
xC2D2 yC2D2
Link 2 (input)
L2 15.000 in
Ground link 1b
L1b 9.470 in
2
L3 17.196 in Link 4 (output)
L6 8.625 in
8.625
D3
9.47 0
O2 O4
15.000
D2 96 17.1
C3
C2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-85-2
11. Using the program FOURBAR and the link lengths given above, it was found that the fourbar O4DCO6 is non-Grashoff with toggle positions at 4 = -14.9 deg and +14.9 deg. The fourbar operates between 4 = +12.403 deg and -8.950 deg.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-86-1
PROBLEM 3-86 Statement:
Design a fourbar mechanism to give the three positions shown in Figure P3-19. Ignore the points O2 and O4 shown. Build a cardboard model that has stops to limit its motion to the range of positions designed.
Solution:
See Figure P3-19 and Mathcad file P0386.
1.
Draw link CD in its three design positions C1D1, C2D2, C3D3 in the plane as shown.
2.
Draw construction lines from point C1 to C2 and from point C2 to C3.
3.
Bisect line C1C2 and line C2C3 and extend their perpendicular bisectors until they intersect. Label their intersection O2.
4.
Repeat steps 2 and 3 for lines D1D2 and D2D3. Label the intersection O4.
5.
Connect O2 with C1 and call it link 2. Connect O4 with D1 and call it link 4.
6.
Line C1D1 is link 3. Line O2O4 is link 1 (ground link for the fourbar). The fourbar is now defined as O2CDO4 an has link lengths of Ground link 1
L1 9.187
Link 2
L2 14.973
Link 3
L3 17.197
Link 4
L4 8.815 D3
8.815
9.18 7 O2
14 .97 3
O4
2 4
D2
17.197 C1
D1
3
C3 C2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-87-1
PROBLEM 3-87 Statement:
Design a fourbar mechanism to give the three positions shown in Figure P3-17 using the fixed pivots O2 and O4 shown. (See Example 3-7.) Build a cardboard model that has stops to limit its motion to the range of positions designed.
Solution:
See Figure P3-19 and Mathcad file P0387.
1.
Draw link CD in its three design positions C1D1, C2D2, C3D3 in the plane as shown.
2.
Draw the ground link O2O4 in its desired position in the plane with respect to the first coupler position C1D1.
3.
Draw construction arcs from point C2 to O2 and from point D2 to O2 whose radii define the sides of triangle C2O2D2. This defines the relationship of the fixed pivot O2 to the coupler line CD in the second coupler position.
4.
Draw construction arcs from point C2 to O4 and from point D2 to O4 whose radii define the sides of triangle C2O4D2. This defines the relationship of the fixed pivot O4 to the coupler line CD in the second coupler position.
5.
Transfer this relationship back to the first coupler position C1D1 so that the ground plane position O2'O4' bears the same relationship to C1D1 as O2O4 bore to the second coupler position C2D2.
6.
Repeat the process for the third coupler position and transfer the third relative ground link position to the first, or reference, position.
7.
The three inverted positions of the ground link that correspond to the three desired coupler positions are labeled O2O4, O2'O4', and O2"O4" in the first layout below and are renamed E1F1, E2F2, and E3F3, respectively, in the second layout, which is used to find the points G and H. D3
O'2 O2
O"2
O4
O'4
D2
C1
O" 4
D1 C3
C2
First layout for steps 1 through 7
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-87-2
E2 O'2 E1 O2 E3 O" 2
F1 O4
O'4 F2 2 4 F3 O"4 3
G
H
Second layout for steps 8 through 12 8.
Draw construction lines from point E1 to E2 and from point E2 to E3.
9.
Bisect line E1E2 and line E2E3 and extend their perpendicular bisectors until they intersect. Label their intersection G.
10. Repeat steps 2 and 3 for lines F1F2 and F2F3. Label the intersection H. 11. Connect E1 with G and label it link 2. Connect F1 with H and label it link 4. Reinverting, E1 and F1 are the original fixed pivots O2 and O4, respectively. 12. Line GH is link 3. Line O2O4 is link 1a (ground link for the fourbar). The fourbar is now defined as O2GHO4 and has link lengths of Ground link 1a
L1a 9.216
Link 2
L2 16.385
Link 3
L3 18.017
Link 4
L4 8.786
13. Check the Grashof condition. Note that any Grashof condition is potentially acceptable in this case. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 3-87-3
Condition L1a L2 L3 L4 "non-Grashof" The fourbar that will provide the desired motion is now defined as a non-Grashof double rocker in the open configuration. It now remains to add the original points C1 and D1 to the coupler GH.
9.21 6
O2
16 .38 5
O4
4
C1
3
D1 H
G
18.017
8.786
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-7a-1
PROBLEM 4-7a Statement:
Given:
The link lengths and value of 2 for some fourbar linkages are defined in Table P4-1. The linkage configuration and terminology are shown in Figure P4-1. For row a, find all possible solutions (both open and crossed) for angles 3 and 4 using the vector loop method. Determine the Grashof condition. Link 2 Link 1 d 6 in a 2 in b 7 in
Link 3 Solution: 1.
c 9 in
Link 4
See Mathcad file P0407a.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 3.0000
2
d
K3
c
K2 0.6667
2 a c
B 1.0000
C K1 K2 1 cos θ K3
C 3.5566
Use equation 4.10b to find values of 4 for the open and crossed circuits. Open:
2
θ 2 atan2 2 A B
B 4 A C
θ 242.714 deg
θ θ 360 deg
θ 602.714 deg
2
Crossed: θ 2 atan2 2 A B
B 4 A C
θ 216.340 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2 a b
2
K4 0.8571 K5 0.2857
D cos θ K1 K4 cos θ K5
D 1.6774
E 2 sin θ
E 1.0000
F K1 K4 1 cos θ K5 4.
2
A 0.7113
B 2 sin θ
3.
2
a b c d
K3 2.0000
A cos θ K1 K2 cos θ K3
2.
θ 30 deg
F 2.5906
Use equation 4.13 to find values of 3 for the open and crossed circuits. Open:
θ 2 atan2 2 D E
2
E 4 D F
θ θ 360 deg
Crossed: θ 2 atan2 2 D E
θ 271.163 deg θ 631.163 deg
2
E 4 D F
θ 244.789 deg
2
DESIGN OF MACHINERY - 5th Ed.
5.
Check the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( a b c d ) "Grashof"
SOLUTION MANUAL 4-7a-2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-1-1
PROBLEM 4-1 Statement:
A position vector is defined as having length equal to your height in inches (or centimeters). The tangent of its angle is defined as your weight in lbs (or kg) divided by your age in years. Calculate the data for this vector and: a. Draw the position vector to scale on Cartesian axes. b. Write an expression for the position vector using unit vector notation. c. Write an expression for the position vector using complex number notation, in both polar and Cartesian forms.
Assumptions: Height 70, weight 160, age 20 Solution:
The magnitude of the vector is R Height. The angle that the vector makes with the x-axis is θ atan
weight
age
a.
θ 82.875 deg
θ 1.446 rad
Draw the position vector to scale on Cartesian axes. y 100 80
R
60 70.000
1.
See Mathcad file P0401.
40
82.875°
20 0
b.
x 20
40
60
80
100
Write an expression for the position vector using unit vector notation.
cos θ sin θ
R R
R
8.682 69.459
R = 8.682 i + 69.459 j
c. Write an expression for the position vector using complex number notation, in both polar and Cartesian forms. j 1.446
Polar form:
R 68 e
Cartesian form:
R 8.682 j 69.459
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-2-1
PROBLEM 4-2 Statement:
A particle is traveling along an arc of 6.5 inch radius. The arc center is at the origin of a coordinate system. When the particle is at position A, its position vector makes a 45-deg angle with the X axis. At position B, its vector makes a 75-deg angle with the X axis. Draw this system to some convenient scale and: a. b. c.
d.
Write an expression for the particle's position vector in position A using complex number notation, in both polar and Cartesian forms. Write an expression for the particle's position vector in position B using complex number notation, in both polar and Cartesian forms. Write a vector equation for the position difference between points B and A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. Check the result of part c with a graphical method.
Given:
Circle radius and vector magnitude, R 6.5 in; vector angles: θA 45 deg
Solution:
See Mathcad file P0402.
θB 75 deg
1.
Establish an X-Y coordinate frame and draw a circle with center at the origin and radius R.
2.
Draw lines from the origin that make angles of 45 and 75 deg with respect to the X axis. Label the intersections of the lines with the circles as A and B, respectively. Make the line segment OA a vector by putting an arrowhead at A, pointing away from the origin. Label the vector RA. Repeat for the line segment OB, labeling it RB.
Y 8 B 6 A 4
RB RA
2
0 a.
b.
X 2
4
6
8
Write an expression for the particle's position vector in position A using complex number notation, in both polar and Cartesian forms. j θA
Polar form:
RA R e
Cartesian form:
RA R cos θA j sin θA
j
RA 6.5 e
π 4
RA ( 4.596 4.596j) in
Write an expression for the particle's position vector in position B using complex number notation, in both polar and Cartesian forms.
DESIGN OF MACHINERY - 5th Ed.
c.
SOLUTION MANUAL 4-2-2
j
j θB
Polar form:
RB R e
RB 6.5 e
Cartesian form:
RB R cos θB j sin θB
180
RB ( 1.682 6.279j) in
Write a vector equation for the position difference between points B and A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. RBA RB RA
d.
75 π
RBA ( 2.914 1.682j) in
Check the result of part c with a graphical method.
Y 8
3.365 B
6
RBA
1.682
A 4
RB
2.914
2 RA 0
X 2
4
6
8
On the layout above the X and Y components of RBA are equal to the real and imaginary components calculated, confirming that the calculation is correct.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-3-1
PROBLEM 4-3 Statement:
Two particles are traveling along an arc of 6.5 inch radius. The arc center is at the origin of a coordinate system. When one particle is at position A, its position vector makes a 45-deg angle with the X axis. Simultaneously, the other particle is at position B, where its vector makes a 75deg angle with the X axis. Draw this system to some convenient scale and: a. b. c.
d.
Write an expression for the particle's position vector in position A using complex number notation, in both polar and Cartesian forms. Write an exp ession for the particle's position vector in position B using complex number notation, in both polar and Cartesian forms. Write a vector equation for the relative position of the particle at B with respect to the particle at A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. Check the result of part c with a graphical method.
Given:
Circle radius and vector magnitude, R 6.5 in; vector angles: θA 45 deg
Solution:
See Mathcad file P0403.
θB 75 deg
1.
Establish an X-Y coordinate frame and draw a circle with center at the origin and radius R.
2.
Draw lines from the origin that make angles of 45 and 75 deg with respect to the X axis. Label the intersections of the lines with the circles as A and B, respectively. Make the line segment OA a vector by putting an arrowhead at A, pointing away from the origin. Label the vector RA. Repeat for the line segment OB, labeling it RB.
Y 8 B 6 A 4
RB RA
2
0 a.
b.
X 2
4
6
8
Write an expression for the particle's position vector in position A using complex number notation, in both polar and Cartesian forms. j θA
Polar form:
RA R e
Cartesian form:
RA R cos θA j sin θA
j
RA 6.5 e
π 4
RA ( 4.596 4.596j) in
Write an expression for the particle's position vector in position B using complex number notation, in both polar and Cartesian forms.
DESIGN OF MACHINERY - 5th Ed.
c.
SOLUTION MANUAL 4-3-2
j
j θB
Polar form:
RB R e
RB 6.5 e
Cartesian form:
RB R cos θB j sin θB
180
RB ( 1.682 6.279j) in
Write a vector equation for the relative position of the particle at B with respect to the particle at A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. RBA RB RA
d.
75 π
RBA ( 2.914 1.682j) in
Check the result of part c with a graphical method.
Y 8
3.365 B
6
RBA
1.682
A 4
RB
2.914
2 RA 0
X 2
4
6
8
On the layout above the X and Y components of RBA are equal to the real and imaginary components calculated, confirming that the calculation is correct.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-4-1
PROBLEM 4-4 Statement:
A particle is traveling along the line y = -2x + 10. When the particle is at position A, its position vector makes a 45-deg angle with the X axis. At position B, its vector makes a 75-deg angle with the X axis. Draw this system to some convenient scale and: a. b. c.
d.
Write an expression for the particle's position vector in position A using complex number notation, in both polar and Cartesian forms. Write an expression for the particle's position vector in position B using complex number notation, in both polar and Cartesian forms. Write a vector equation for the position difference between points B and A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. Check the result of part c with a graphical method.
Given:
Vector angles: θA 45 deg
Solution:
See Mathcad file P0402.
θB 75 deg
1.
Establish an X-Y coordinate frame and draw the line y = -2x + 10.
2.
Draw lines from the origin that make angles of 45 and 75 deg with respect to the X axis. Label the intersections of the lines with the line in step 1 as A and B, respectively. Make the line segment OA a vector by putting an arrowhead at A, pointing away from the origin. Label the vector RA. Repeat for the line segment OB, labeling it RB.
Y 10 y = -2x + 10 8 B 6
4
RB
2
0
3.
A
RA X 2
4
6
8
Calculate the coordinates of points A and B. xA tan θA = 2 xA 10
xA
10
2 tan θA
yA xA tan θA xB tan θB = 2 xB 10
xB
10
2 tan θB
xA 3.333 yA 3.333 xB 1.745
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-4-2
yB xB tan θB 4.
a.
b.
yB 6.511
Calculate the distances of points A and B from the origin. 2
2
RA 4.714
2
2
RB 6.741
RA
xA yA
RB
xB yB
Write an expression for the particle's position vector in position A using complex number notation, in both polar and Cartesian forms. j θA
j
Polar form:
RA RA e
Cartesian form:
RA RA cos θA j sin θA
RA 4.714 e
π 4
RA 3.333 3.333j
Write an expression for the particle's position vector in position B using complex number notation, in both polar and Cartesian forms. j
j θB
Polar form:
RB RB e
RB 6.741 e
Cartesian form:
RB RB cos θB j sin θB
75 π 180
RB 1.745 6.511j
Y c.
10
Write a vector equation for the position difference between points B and A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically.
y = -2x + 10 8 B
RBA RB RA RBA 1.589 3.178j d.
Check the result of part c with a graphical method. On the layout above the X and Y components of RBA are equal to the real and imaginary components calculated, confirming that the calculation is correct.
6 3.178 4
3.553
RBA RB A
2
0
RA X 2
4
6 1.589
8
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-5-1
PROBLEM 4-5 Statement:
Two particles are traveling along the line y = -2x2 - 2x +10. When one particle is at position A, its position vector makes a 45-deg angle with the X axis. Simultaneously, the other particle is at position B, where its vector makes a 75-deg angle with the X axis. Draw this system to some convenient scale and: a. b. c.
d.
Write an expression for the particle's position vector in position A using complex number notation, in both polar and Cartesian forms. Write an expression for the particle's position vector in position B using complex number notation, in both polar and Cartesian forms. Write a vector equation for the relative position of the particle at B with respect to the particle at A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. Check the result of part c with a graphical method.
Given:
Vector angles: θA 45 deg
Solution:
See Mathcad file P0405.
θB 75 deg
1.
Establish an X-Y coordinate frame and draw the line y = -2x2 - 2x +10.
2.
Draw lines from the origin that make angles of 45 and 75 deg with respect to the X axis. Label the intersections of the lines with the line drawn in step 1 as A and B, respectively. Make the line segment OA a vector by putting an arrowhead at A, pointing away from the origin. Label the vector RA. Repeat for the line segment OB, labeling it RB.
Y 10 y = -2x^2 - 2x + 10 8
6 B 4 RB 2
A RA
0
3.
X 2
4
6
8
Calculate the coordinates of points A and B. xA tan θA = 2 xA 2 xA 10 2
2 tan θA tan θA xA 1 1 2 2 2
1
20
xA 1.608
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-5-2
yA xA tan θA
yA 1.608
xB tan θB = 2 xB 2 xB 10 2
2 tan θB tan θB xB 1 1 2 2 2
1
yB xB tan θB 4.
a.
b.
c.
xB 1.223
yB 4.564
Calculate the distances of points A and B from the origin. 2
2
RA 2.275
2
2
RB 4.725
RA
xA yA
RB
xB yB
Write an expression for the particle's position vector in position A using complex number notation, in both polar and Cartesian forms. j θA
Polar form:
RA R e
Cartesian form:
RA RA cos θA j sin θA
j
RA 2.275 e
π 4
RA 1.608 1.608j
Write an expression for the particle's position vector in position B using complex number notation, in both polar and Cartesian forms. j θB
Polar form:
RB R e
Cartesian form:
RB RB cos θB j sin θB
j
RB 4.725 e
75 π 180
RB 1.223 4.564j
Write a vector equation for the relative position of the particle at B with respect to the particle at A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. RBA RB RA
d.
20
RBA 0.386 2.955j
Check the result of part c with a graphical method.
On the layout on the next page the X and Y components of RBA are equal to the real and imaginary components calculated, confirming that the calculation is correct.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-5-3
Y 10 y = -2x^2 - 2x + 10 8
6 B 4 RB
2.955
RBA
2.980
2 A RA 0
X 4
2 0.386
6
8
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-6a-1
PROBLEM 4-6a Statement:
The link lengths and value of 2 for some fourbar linkages are defined in Table P4-1. The linkage configuration and terminology are shown in Figure P4-1. For row a, draw the linkage to scale and graphically find all possible solutions (both open and crossed) for angles 3 and 4. Determine the Grashoff condition.
Given:
Link 1
d 6 in
Link 2
a 2 in
Link 3
b 7 in
Link 4
c 9 in
Solution:
θ2 30 deg
See figure below for one possible solution. Also see Mathcad file P0406a.
1.
Lay out an xy-axis system. Its origin will be the link 2 pivot, O2.
2.
Draw link 2 to some convenient scale at its given angle.
3.
Draw a circle with center at the free end of link 2 and a radius equal to the given length of link 3.
4.
Locate pivot O4 on the x-axis at a distance from the origin equal to the given length of link 1.
5.
Draw a circle with center at O4 and a radius equal to the given length of link 4.
6.
The two intersections of the circles (if any) are the two solutions to the position analysis problem, crossed and open. If the circles don't intersect, there is no solution.
7.
Draw links 3 and 4 in their two possible positions (shown as solid for open and dashed for crossed in the figure) and measure their angles 3 and 4 with respect to the x-axis. From the solution below, OPEN
θ θ
CROSSED
θ θ
8.
31 41 32 42
88.84 deg 117.29 deg 360 deg 115.21 deg
θ
360 deg 143.66 deg
θ
42
244.790 deg 216.340 deg
y
Check the Grashof condition. Condition( a b c d )
32
B
S min ( a b c d )
OPEN
L max( a b c d ) SL S L
3
PQ a b c d SL
4
return "Grashof" if SL PQ 88.837°
return "Special Grashof" if SL = PQ
117.286°
A
return "non-Grashof" otherwise
2 O2
Condition( a b c d ) "Grashof"
115.211°
O4 143.660°
CROSSED B'
x
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-7a-1
PROBLEM 4-7a Statement:
Given:
The link lengths and value of 2 for some fourbar linkages are defined in Table P4-1. The linkage configuration and terminology are shown in Figure P4-1. For row a, find all possible solutions (both open and crossed) for angles 3 and 4 using the vector loop method. Determine the Grashof condition. Link 2 Link 1 d 6 in a 2 in b 7 in
Link 3
c 9 in
Link 4
θ 30 deg
Two argument inverse tangent atan2( x y )
return 0.5 π if x = 0 y 0 return 1.5 π if x = 0 y 0 return atan
y
if x 0 x
atan
y
π otherwise x
Solution: 1.
See Mathcad file P0407a.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 3.0000
2
d
K3
c
K2 0.6667
2 a c
A 0.7113
B 2 sin θ
B 1.0000
C K1 K2 1 cos θ K3
C 3.5566
Use equation 4.10b to find values of 4 for the open and crossed circuits.
Open:
2
θ 2 atan2 2 A B
B 4 A C
θ 477.286 deg
θ θ 360 deg
θ 117.286 deg
2
Crossed: θ 2 atan2 2 A B 3.
2
K3 2.0000
A cos θ K1 K2 cos θ K3
2.
2
a b c d
B 4 A C
θ 216.340 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2 a b
D cos θ K1 K4 cos θ K5
E 2 sin θ
2
K4 0.8571 K5 0.2857 D 1.6774 E 1.0000
F K1 K4 1 cos θ K5
F 2.5906
2
DESIGN OF MACHINERY - 5th Ed.
4.
SOLUTION MANUAL 4-7a-2
Use equation 4.13 to find values of 3 for the open and crossed circuits. Open:
θ 2 atan2 2 D E
2
E 4 D F
θ θ 360 deg
Crossed: θ 2 atan2 2 D E 5.
θ 88.837 deg 2
E 4 D F
Check the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( a b c d ) "Grashof"
θ 448.837 deg
θ 244.789 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-8-1
PROBLEM 4-8 Statement:
Expand equation 4.7b and prove that it reduces to equation 4.7c (p. 157).
Solution:
See Mathcad file P0408.
1.
Write equation 4.7b and expand the two terms that are squared. 2
2 a cosθ c cosθ d2
b a sin θ c sin θ
2.
(4.7b)
a sinθ c sinθ2 a2 sinθ2 2 a c sinθ sinθ c2 sinθ2
(a)
a cosθ c cosθ d2 a2 cosθ2 2 a c cosθ cosθ 2 a d cosθ 2 2 2 2 c d cos θ c cos θ d
(b)
Add the two expanded terms, equations a and b, noting the identity sin 2x + cos2x = 1. 2
2
2
2
b a c d 2 a d cos θ 2 c d cos θ 2 a c sin θ sin θ cos θ cos θ This is equation 4.7c.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-9a-1
PROBLEM 4-9a Statement:
The link lengths, value of 2, and offset for some fourbar slider-crank linkages are defined in Table P4-2. The linkage configuration and terminology are shown in Figure P4-2. For row a, draw the linkage to scale and graphically find all possible solutions (both open and crossed) for angles 3 and slider position d.
Given:
Link 2
a 1.4 in
Link 3
Offset
c 1 in
θ 45 deg
Solution:
b 4 in
See figure below for one possible solution. Also see Mathcad file P0409a.
1.
Lay out an xy-axis system. Its origin will be the link 2 pivot, O2.
2.
Draw link 2 to some convenient scale at its given angle.
3.
Draw a circle with center at the free end of link 2 and a radius equal to the given length of link 3.
4.
Draw a horizontal line through y = c (the offset).
5.
The two intersections of the circle with the horizontal line (if any) are the two solutions to the position analysis problem, crossed and open. If the circle and line don't intersect, there is no solution.
6.
Draw link 3 and the slider block in their two possible positions (shown as solid for open and dashed for crossed in the figure) and measure the angle 3 and length d for each circuit. From the solution below, θ31 360 deg 179.856 deg
θ31 180.144 deg
θ32 0.144 deg
d 3.010 in
d 4.990 in 1
2
Y d2 = 3.010
d1 = 4.990
3(CROSSED)
B'
A 2
0.144° O2
45.000°
3 (OPEN)
B 179.856° 1.000 X
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-10a-1
PROBLEM 4-10a Statement:
Given:
Solution:
The link lengths, value of 2, and offset for some fourbar slider-crank linkages are defined in Table P4-2. The linkage configuration and terminology are shown in Figure P4-2. For row a, using the vector loop method, find all possible solutions (both open and crossed) for angles 3 and slider position d. Link 2 Offset
a 1.4 in c 1 in
Link 3 b 4 in θ 45 deg
See Figure P4-2 and Mathcad file P0410a. Y d2 = 3.010
d1 = 4.990
3(CROSSED)
B'
A 2
0.144°
45.000°
3 (OPEN)
B 179.856° 1.000 X
O2
1.
Determine 3 and d using equations 4.16 and 4.17. Crossed:
a sin θ c b
θ 0.144 deg
d 2 3.010 in
θ asin
d 2 a cos θ b cos θ Open:
a sin θ c π b
θ 180.144 deg
d 1 4.990 in
θ asin
d 1 a cos θ b cos θ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-11a-1
PROBLEM 4-11a Statement:
The link lengths and the value of 2 and for some inverted fourbar slider-crank linkages are defined in Table P4-3. The linkage configuration and terminology are shown in Figure P4-3. For row a, draw the linkage to scale and graphically find both open and closed solutions for 3 and 4 and vector RB.
Given:
Link 1
d 6 in
Link 2
Link 4
c 4 in
γ 90 deg
Solution:
a 2 in θ 30 deg
See figure below for one possible solution. Also see Mathcad file P04011a.
1.
Lay out an xy-axis system. Its origin will be the link 2 pivot, O2.
2.
Draw link 2 to some convenient scale at its given angle.
3a. If = 90 deg, locate O4 on the x-axis at a distance equal the length of link 1 (d) from the origin. Draw a circle with center at O4 and radius equal to the length of link 4 (c). From point A, draw two lines that are tangent to the circle. The points of tangency define the location of the points B for the open and crossed circuits. 3b. When is not 90 deg there are two approaches to a graphical solution for link 3 and the location of point B: 1) establish the position of link 4 and the angle by trial and error, or 2) calculate the distance from point A to point B (the instantaneous length of link 3). Using the second approach, from triangle O2AO4
y
B
b
c
A a 2 d
x 04
02
2
2
2
AO4 = a d 2 a d cos θ
and, from triangle AO4B (for the open circuit)
AO4 = b c 2 b c cos π γ 2
2
2
where a, b, c, and d are the lengths of links 2, 3, 4, and 1, respectively. Eliminating AO4 and solving for the unknown distance b for the open branch, b 1
1 2
2 c cos π γ
2 c cos π γ 2 4 c2 a2 d2 2 a d cos θ
b 1 1.7932 in 2
2
2
For the closed branch: AO4 = b c 2 b c cos( γ) b 2
1 2
2 c cos γ
and
2 c cosγ 2 4 c2 a2 d2 2 a d cos θ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-11a-2
b 2 1.7932 in Draw a circle with center at point A and radius b 1. Draw a circle with center at O4 and radius equal to the length of link 4 (c). The intersections of these two circles is the solution for the open and crossed locations of the point B. 4.
Draw the complete linkage for the open and crossed circuits, including the slider. The results from the graphical solution below are: θ 127.333 deg
OPEN
CROSSED
θ 100.959 deg
θ 142.666 deg
θ 169.040 deg
RB1 3.719 at 40.708 deg
RB2 2.208 at -20.146 deg
B y 90.0°
b
127.333° c
A
142.666°
a 30.000°
d
x 04
02 B'
169.040° 79.041°
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-12a-1
PROBLEM 4-12a Statement:
Given:
The link lengths and the value of 2 and for some inverted fourbar slider-crank linkages are defined in Table P4-3. The linkage configuration and terminology are shown in Figure P4-3. For row a, using the vector loop method, find both open and closed solutions for 3 and 4 and vector RB. Link 1 Link 2 d 6 in a 2 in Link 4
Solution: 1.
c 4 in
γ 90 deg
θ 30 deg
See Mathcad file P0412a.
Determine the values of the constants needed for finding 4 from equations 4.25 and 4.26.
P a sin θ sin γ a cos θ d cos γ
2.
3.
4.
5.
P 1.000 in
Q a sin θ cos γ a cos θ d sin γ
Q 4.268 in
R c sin γ
R 4.000 in
T 2 P
T 2.000 in
S R Q
S 0.268 in
U Q R
U 8.268 in
Use equation 4.26c to find values of 4 for the open and crossed circuits.
T 4 S U
T 4 S U
OPEN
θ 2 atan2 2 S T
CROSSED
θ 2 atan2 2 S T
2
θ 142.667 deg
2
θ 169.041 deg
Use equation 4.22 to find values of 3 for the open and crossed circuits. OPEN
θ θ γ
θ 232.667 deg
CROSSED
θ θ γ
θ 79.041 deg
Determine the magnitude of the instantaneous "length" of link 3 from equation 4.24a. OPEN
b 1
CROSSED
b 2
sin θ γ
a sin θ c sin θ
b 1 1.793 in
sin θ γ
a sin θ c sin θ
b 2 1.793 in
Find the position vector RB from the definition given in the text. OPEN
b1 cosθ j sinθ
RB1 a cos θ j sin θ RB1 RB1
RB1 3.719 in
θ arg RB1
θ 40.707 deg
DESIGN OF MACHINERY - 5th Ed.
CROSSED
SOLUTION MANUAL 4-12a-2
b2 cosθ j sinθ
RB2 a cos θ j sin θ RB2 RB2
RB2 3.091 in
θ arg RB2
θ 63.254 deg
DESIGN OF MACHINERY
SOLUTION MANUAL 4-12c-1
PROBLEM 4-12c Statement:
Given:
The link lengths and the value of 2 and for some inverted fourbar slider-crank linkages are defined in Table P4-3. The linkage configuration and terminology are shown in Figure P4-3. For row c, using the vector loop method, find both open and closed solutions for 3 and 4 and vector RB . Link 1
d 3 in
Link 2
Link 4
c 6 in
45 deg
a 10 in 45 deg
Two argument inverse tangent atan2 (x y) return 0.5 if x = 0 y 0 return 1.5 if x = 0 y 0
y if x 0 return atan x y atan x otherwise Solution: 1.
See Mathcad file P0412c.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a.
P a sin sin a cos d cos
2.
P 7.879 in
Q a sin cos a cos d sin
Q 2.121 in
R c sin
R 4.243 in
T 2 P
T 15.757 in
S R Q
S 2.121 in
U Q R
U 6.364 in
Use equation 4.22c to find values of 4 for the open and crossed circuits. OPEN
2
46.400 deg
2
163.739 deg
2 atan2 2 S T T 4 S U 360 deg
CROSSED 2 atan2 2 S T T 4 S U 360 deg 3.
4.
Use equation 4.18 to find values of 3 for the open and crossed circuits. OPEN
91.400 deg
CROSSED
118.739 deg
Determine the magnitude of the instantaneous "length" of link 3 from equation 4.20a. OPEN
b1
CROSSED
b2
sin
a sin c sin
sin
a sin c sin
b1 2.727 in
b2 11.212 in
DESIGN OF MACHINERY
5.
SOLUTION MANUAL 4-12c-2
Find the position vector RB from the definition given on page 162 of the text. OPEN
CROSSED
RB1 acos j sin b 1cos j sin RB1 RB1
RB1 8.356in
arg RB1
31.331 deg
RB2 acos j sin b 2cos j sin RB2 RB2
RB2 12.764 in
arg RB2
12.488 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-13a-1
PROBLEM 4-13a Statement:
Find the transmission angles of the linkage in row a of Table P4-1.
Given:
Link 1
d 6 in
Link 2
a 2 in
Link 3
b 7 in
Link 4
c 9 in
Solution: 1.
See Mathcad file P0413a.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d a
2
d
K2
K1 3.0000
K3
c
K2 0.6667
2 a c
C 3.5566
Use equation 4.10b to find 4 for the open circuit.
2
θ 2 atan2 2 A B
B 4 A C
θ 242.714 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2 a b
2
K4 0.8571 K5 0.2857
D cos θ K1 K4 cos θ K5
D 1.6774
E 2 sin θ
E 1.0000
F K1 K4 1 cos θ K5
F 2.5906
Use equation 4.13 to find 3 for the open circuit.
θ 2 atan2 2 D E
2
E 4 D F
θ θ 360 deg 5.
2
B 1.0000
C K1 K2 1 cos θ K3
4.
2
A 0.7113
B 2 sin θ
3.
2
a b c d
K3 2.0000
A cos θ K1 K2 cos θ K3
2.
θ 30 deg
θ 271.163 deg θ 631.163 deg
Use equations 4.32 to find the transmission angle.
θtrans θ θ
t θ θ
θtrans θ θ 208.449 deg
return t if t 0.5 π π t otherwise 6.
It can be shown that the triangle ABO4 in Figure 4-17 is symmetric with respect to the line AO4 for the crossed branch and, therefore, the transmission angle for the crossed branch is identical to that for the open branch.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-14-1
PROBLEM 4-14 Statement:
Find the minimum and maximum values of the transmission angle for all the Grashof crankrocker linkages in Table P4-1.
Given:
Table P4-1 data:
i 1 2 14 Row i
"a" "b" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" Solution: 1.
d
a
b
c
6 7 3 8 8 5 6 20 4 20 4 9 9 9
2 9 10 5 5 8 8 10 5 10 6 7 7 7
7 3 6 7 8 8 8 10 2 5 10 10 11 11
9 8 8 6 6 9 9 10 5 10 7 7 8 6
i
i
i
i
See Table P4-1 and Mathcad file P0414.
Determine which of the linkages in Table P4-1 are Grashof. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Row a
Condition( 6 2 7 9 ) "Grashof"
Row b
Condition( 7 9 3 8 ) "Grashof"
Row c
Condition( 3 10 6 8 ) "Grashof"
Row d
Condition( 8 5 7 6 ) "Special Grashof"
Row e
Condition( 8 5 8 6 ) "Grashof"
Row f
Condition( 5 8 8 9 ) "Grashof"
Row g
Condition( 6 8 8 9 ) "Grashof"
Row h
Condition( 20 10 10 10) "non-Grashof"
Row i
Condition( 4 5 2 5 ) "Grashof"
Row j
Condition( 20 10 5 10) "non-Grashof"
Row k
Condition( 4 6 10 7 ) "non-Grashof"
Row l
Condition( 9 7 10 7 ) "non-Grashof"
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-14-2
Row m
Condition( 9 7 11 8 ) "non-Grashof"
Row n
Condition( 9 7 11 6 ) "non-Grashof"
2.
Determine which of the Grashof linkages are crank-rockers. To be a Grashof crank-rocker, the linkage must be Grashof and the shortest link is either 2 or 4. This is true of rows a, d, and e.
3.
Use equations 4.32 and 4.33 to calculate the maximum and minimum transmission angles.
Row a
i 1
b 2 c 2 d a 2 i i i i μ acos 2 b c i i
μ if μ
π 2
π μ μ
b 2 c 2 d a 2 i i i i μ acos 2 b c i i
Row d
i 4
π 2
π μ μ
b 2 c 2 d a 2 i i i i μ acos 2 b c i i
i 5
μ 25.209 deg
b 2 c 2 d a 2 i i i i μ acos 2 b c i i μ if μ
Row e
μ 58.412 deg
μ 0.000 deg
μ 25.209 deg
b 2 c 2 d a 2 i i i i μ acos 2 b c i i
μ if μ
π 2
π μ μ
b 2 c 2 d a 2 i i i i μ acos 2 b c i i
μ 44.049 deg
μ 18.573 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-15-1
PROBLEM 4-15 Statement:
Find the input angles corresponding to the toggle positions of the non-Grashof linkages in Table P4-1.
Given:
Table P4-1 data:
i 1 2 14 Row i
"a" "b" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" Solution: 1.
d
a
b
c
6 7 3 8 8 5 6 20 4 20 4 9 9 9
2 9 10 5 5 8 8 10 5 10 6 7 7 7
7 3 6 7 8 8 8 10 2 5 10 10 11 11
9 8 8 6 6 9 9 10 5 10 7 7 8 6
i
i
i
i
See Table P4-1 and Mathcad file P0415.
Determine which of the linkages in Table P4-1 are Grashof. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Row a
Condition( 6 2 7 9 ) "Grashof"
Row b
Condition( 7 9 3 8 ) "Grashof"
Row c
Condition( 3 10 6 8 ) "Grashof"
Row d
Condition( 8 5 7 6 ) "Special Grashof"
Row e
Condition( 8 5 8 6 ) "Grashof"
Row f
Condition( 5 8 8 9 ) "Grashof"
Row g
Condition( 6 8 8 9 ) "Grashof"
Row h
Condition( 20 10 10 10) "non-Grashof"
Row i
Condition( 4 5 2 5 ) "Grashof"
Row j
Condition( 20 10 5 10) "non-Grashof"
Row k
Condition( 4 6 10 7 ) "non-Grashof"
Row l
Condition( 9 7 10 7 ) "non-Grashof"
DESIGN OF MACHINERY - 5th Ed.
2.
SOLUTION MANUAL 4-15-2
Row m
Condition( 9 7 11 8 ) "non-Grashof"
Row n
Condition( 9 7 11 6 ) "non-Grashof"
There are six non-Grashof rows in the Table: Rows h, and j through n. For each row there are two possible arguments to the arccos function given in equation (4.37). They are: i 8
Row "h"
j 1
i
ai di bi ci 2
arg
j 1
2
2
2
b c
2 a d
i i
arg
j 2
i i
ai di bi ci 2
2
2
2
2 a d
Row "j"
i i
ai di bi ci 2
j 1
2
2
2
b c
2 a d
i i
j 2
2
2
2
2 a d
Row "k"
2
2
2
b c
2 a d
i i
j 2
2
2
2
b c
2 a d
i i
i 12
Row "l"
i i
ai di bi ci 2
j 1
2
2
2
b c
2 a d
i i
j 2
2
2
2
2 a d
Row "m"
2
2
2
b c
2 a d
i i
j 2
2
2
2 a d
i i
i i
a d
i i
ai di bi ci 2
arg
a d
i i
ai di bi ci 2
j 1
i i
j 5
i
arg
a d b c
i i
i 13
i i i i
ai di bi ci 2
arg
i i
a d
j 4
i
arg
i i
a d
i i
ai di bi ci 2
arg
a d
i i
ai di bi ci 2
j 1
i i
j 3
i
arg
a d b c
i i
i 11
i i i i
ai di bi ci 2
arg
i i
a d
j 2
i
arg
b c
i i
i 10
i i
a d
2
b c
i i
a d
i i
DESIGN OF MACHINERY - 5th Ed.
i 14
SOLUTION MANUAL 4-15-3
Row "n"
j 6
i
ai di bi ci 2
arg
j 1
2
2
2
b c
2 a d
i i
arg
j 2
2
2
2 a d
2
b c
i i
1.250 1.188 0.896 arg 0.960 0.960 0.833 3.
a d
i i
ai di bi ci 2
i i
i i
a d
i i
0.688 4.938 1.262 1.833 1.262 0.250
Choose the argument values that lie between plus and minus 1,
1 2
θ2h 75.5 deg
2 2
θ2j 46.6 deg
3 1
θ2k 26.4 deg
4 1
θ2l 16.2 deg
θ2h acos arg θ2j acos arg
θ2k acos arg θ2l acos arg
5 1
θ2m 16.2 deg
6 1
θ2n 33.6 deg
θ2m acos arg θ2n acos arg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-16a-1
PROBLEM 4-16a Statement:
The link lengths, gear ratio, phase angle, and the value of 2 for some geared fivebar linkages are defined in Table P4-4. The linkage configuration and terminology are shown in Figure P4-4. For row a, draw the linkage to scale and graphically find all possible solutions for angles 3 and 4.
Given:
Link 1
d 4 in
Link 2
a 1 in
Link 3
b 7 in
Link 4
c 9 in
Link 5
f 6 in
Gear ratio
λ 2.0
Phase angle
ϕ 30 deg
Input angle
θ 60 deg
Solution: 1.
See Mathcad file P0201.
Determine whether or not an idler is required. idler
"required" if λ 0 "not-required" otherwise
idler "required" 2.
Choose radii for gears 2 and 5 by making a design choice for their center distance (which must be increased if an idler is required). Let the standard center distance when no idler is required be C 0.5 c then C = r2 r5
and
λ =
r2 r5
Solving for r2 and r5, r5
C λ 1
r2 r5 λ
r5 1.500 in r2 3.000 in
If an idler is required, increase the center distance. C if idler = "required" C r5 C
C 6.000 in
Note that the amount by which C is increased if an idler is required is a design choice that is made based on the size of the gears and the space available. 3.
Using equation 4.27c, determine the angular position of link 5 corresponding to the position of link 2. θ λ θ ϕ
θ 150 deg
4.
Lay out an xy-axis system. Its origin will be the link 2 pivot, O2.
5.
Draw link 2 to some convenient scale at its given angle.
6.
Draw a circle with center at the free end of link 2 and a radius equal to the given length of link 3.
7.
Locate pivot O4 on the x-axis at a distance from the origin equal to the given length of link 1.
8.
Draw link 5 to some convenient scale at its calculated angle.
9.
Draw a circle with center at the free end of link 5 and a radius equal to the given length of link 4.
10. The two intersections of the circles (if any) are the two solutions to the position analysis problem, crossed and open. If the circles don't intersect, there is no solution. 11. Draw links 3 and 4 in their two possible positions (shown as solid for open and dashed for crossed in the figure) and measure their angles 3 and 4 with respect to the x-axis. From the solution below,
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-16a-2
θ 173.64 deg
OPEN
θ 360 deg 177.715 deg θ 182.285 deg
CROSSED
θ 360 deg 115.407 deg
θ 360 deg 124.050 deg
θ 244.593 deg
θ 235.950 deg
12. Draw gears 2 and 5 schematically at their calculated radii. If an idler is required, draw it tangent to gears 2 and 5. Its diameter is a design choice that will be made on strength and space requirements. It does not affect the gear ratio.
y C
4
B
177.7152°
173.6421° 3
5 2
B`
124.0501° x
O2
3
150.0000°
115.4074°
4
O5
DESIGN OF MACHINERY - 5th ed.
SOLUTION MANUAL 4-17a-1
PROBLEM 4-17a Statement:
Given:
Solution: 1.
The link lengths, gear ratio, phase angle, and the value of 2 for some geared fivebar linkages are defined in Table P4-4. The linkage configuration and terminology are shown in Figure P4-4. For row a, using the vector loop method, find all possible solutions for angles 3 and 4. Link 1
d 4 in
Link 2
a 1 in
Link 3
b 7 in
Link 4
c 9 in
Link 5
f 6 in
Gear ratio
λ 2.0
Phase angle
ϕ 30 deg
Input angle
θ 60 deg
See Mathcad file P0417a.
Determine the values of the constants needed for finding 3 and 4 from equations 4.27h and 4.27i.
A 2 c d cos λ θ ϕ a cos θ f
2
A 36.6462 in 2
B 2 c d sin λ θ ϕ a sin θ
2
2
2
2
2
B 20.412 in
C a b c d f 2 a f cos θ 2 d a cos θ f cos λ θ ϕ 2 a d sin θ sin λ θ ϕ
2
C 37.4308 in
2
D C A
D 0.78461 in
E 2 B
E 40.823 in
F A C
F 74.077 in
2 2
a cosθ f
G 28.503 in
a sinθ
H 15.876 in
2
G 2 b d cos λ θ ϕ
2
H 2 b d sin λ θ ϕ
2
2.
2
2
2
2
K a b c d f 2 a f cos θ 2 d a cos θ f cos λ θ ϕ 2 a d sin θ sin λ θ ϕ
K 26.569 in
L K G
L 1.933 in
M 2 H
M 31.751 in
N G K
N 55.072 in
2
2 2
2
Use equations 4.28h and 4.28i to find values of 3 and 4 for the open and crossed circuits. OPEN
M 4 L N
E 4 D F
M 4 L N
E 4 D F
θ 2 atan2 2 L M θ 2 atan2 2 D E
CROSSED
θ 2 atan2 2 L M θ 2 atan2 2 D E
2
2
2
2
θ 173.642 deg θ 177.715 deg θ 115.407 deg θ 124.050 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-18a-1
PROBLEM 4-18a Statement:
The angle between the X and x axes is 25 deg. Find the angular displacement of link 4 when link 2 rotates clockwise from the position shown (+37 deg) to horizontal (0 deg). How does the transmission angle vary and what is its minimum between those two positions? Find the toggle positions of this linkage in terms of the angle of link 2.
Given:
Link lengths: Crank
L2 116
Coupler
L3 108
Rocker
L4 110
Ground link
L1 174
Crank angle for position shown (relative to O2O4):
θ 62 deg
Y y
A
2
Crank rotation angle from position shown to horizontal:
37°
Δθ 37 deg 3 X O2
25°
B
O4
4
x
Solution: 1.
See Figure P4-5a and Mathcad file P0418a.
Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the crossed circuit). K1
L1
K2
L2
K1 1.5000
2
L1
K3
L3
K2 1.6111
2
2
L2 L3 L4 L1
2
2 L2 L4
K3 1.7307
A θ cos θ K1 K2 cos θ K3
C θ K1 K2 1 cos θ K3
B θ 2 sin θ
θ θ 2 atan2 2 A θ B θ 2.
Determine 4 for the position shown and after the crank has moved to the horizontal position.
θ θ θ
θ θ θ Δθ 3.
θ 183.5 deg
θ 212.8 deg
Subtract the two values of 4 to find the angular displacement of link 3 when link 2 rotates clockwise from the position shown to the horizontal. θ θ
4.
2 4 A θ Cθ
B θ
29.2 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
L1 L3
2
K5
2
2
L4 L1 L2 L3 2 L2 L3
2
K4 1.6111
K5 1.7280
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-18a-2
D θ cos θ K1 K4 cos θ K5
5.
F θ K1 K4 1 cos θ K5
E θ 2 sin θ
Use equation 4.13 to find values of 3 for the crossed circuit.
θ θ 2 atan2 2 D θ E θ 6.
Determine 3 for the position shown and after the crank has moved to the horizontal position.
θ θ θ
θ 275.1 deg
θ θ θ Δθ 7.
2 4 Dθ F θ
E θ
θ 256.1 deg
Use equations 4.28 to find the transmission angles. μ π θ θ
μ 88.4 deg
μ θ θ
μ 43.4 deg
The transmission angle is smaller when the crank is in the horizontal position. 8.
Check the Grashof condition of the linkage. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition L1 L2 L3 L4 "non-Grashof" 9.
Using equations 4.37, determine the crank angles (relative to the XY axes) at which links 3 and 4 are in toggle. 2
arg1
2
2
2 L2 L1 2
arg2
2
L2 L1 L3 L4
2
2
2
L2 L1 L3 L4 2 L2 L1
L3 L4 L2 L1 L3 L4 L2 L1
θ2toggle acos arg2 The other toggle angle is the negative of this.
arg1 1.083
arg2 0.094
θ2toggle 95.4 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-18b-1
PROBLEM 4-18b Statement:
Find and plot the angular position of links 3 and 4 and the transmission angle as a function of the angle of link 2 as it rotates through one revolution.
Given:
Link lengths: Wheel (crank)
L2 40
a L2
Coupler
L3 96
b L3
Rocker
L4 122
c L4
Ground link
L1 162
d L1
Two argument inverse tangent atan2( x y )
Y
return 0.5 π if x = 0 y 0 return atan
y
B
A
2
return 1.5 π if x = 0 y 0
y 3 X
if x 0 x
O2 4
y atan π otherwise x Solution: 1.
See Figure P4-5b and Mathcad file P0418b. O4
Check the Grashof condition of the linkage. Condition( a b c d )
x
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( a b c d ) "Grashof" 2.
Define one cycle of the input crank: θ 0 deg 1 deg 360 deg
3.
Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the open circuit). K1
d
K2
a
K1 4.0500
2
d
K3
c
K2 1.3279
2
2
a b c d
2
2 a c
K3 3.4336
A θ cos θ K1 K2 cos θ K3
C θ K1 K2 1 cos θ K3
B θ 2 sin θ
θ θ 2 atan2 2 A θ B θ 4.
2 4 A θ Cθ
B θ
If the calculated value of 4 is greater than 2, subtract 2 from it. If it is negative, make it positive.
θ θ if θ θ 2 π θ θ 2 π θ θ
θ θ if θ θ 0 θ θ 2 π θ θ
DESIGN OF MACHINERY - 5th Ed.
5.
SOLUTION MANUAL 4-18b-2
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
K4 1.6875
2 a b
K5 2.8875
D θ cos θ K1 K4 cos θ K5
6.
F θ K1 K4 1 cos θ K5
E θ 2 sin θ
Use equation 4.13 to find values of 3 for the crossed circuit.
θ θ 2 atan2 2 D θ E θ 7.
2 4 Dθ F θ
E θ
If the calculated value of 3 is greater than 2, subtract 2 from it. If it is negative, make it positive.
θ θ if θ θ 2 π θ θ 2 π θ θ
θ θ if θ θ 0 θ θ 2 π θ θ
8.
Plot 3 and 4 as functions of the crank angle 2 (measured from the ground link). Angular Displacement of Coupler & Rocker
Coupler or Rocker angle, deg
200
150
θ θ deg
100
θ θ deg
50
0
0
45
90
135
180
225
270
θ deg Crank angle, deg
9.
Use equations 4.32 to find the transmission angle.
Tran θ θ θ θ θ
Trans θ if Tran θ
π 2
π Tran θ Tran θ
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-18b-3
10. Plot the transmission angle.
Transmission Angle 90
Transmission Angle, deg
80
Trans θ
70
deg 60
50
40
0
45
90
135
180 θ deg
Wheel angle, deg
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-18c-1
PROBLEM 4-18c Statement:
Find and plot the position of any one piston as a function of the angle of crank 2 as it rotates through one revolution. Once one piston's motion is defined, find the motions of the other two pistons and their phase relationship to the first piston. Y
Given: L2 19
a L2
Piston-rod length L3 70
b L3
Crank length
6
3
2
5
8 X
c 0
Offset
4
Solution:
See Figure P4-5c and Mathcad file P0418c. 7
1.
Let pistons 1, 2, and 3 be links 7, 6, and 8, respectively.
2.
Solve first for piston 6. Establish 2 as a range variable: θ 0 deg 2 deg 360 deg
3.
Determine 3 and d using equations 4.16 and 4.17.
a sin θ b
θ θ asin
4.
c
d 1 θ a cos θ b cos θ θ
For each piston (slider) the crank angle is measured counter-clock-wise from the centerline of the piston, which goes through the O2 in all cases. Thus, when the crank angle for piston 1 is 0 deg, it is 120 deg for piston 2 and 240 deg for piston 3. Thus, the crank angles for pistons 2 and 3 are
θ θ θ 120 deg 5.
π
θ θ θ 240 deg
Determine 3 and d for pistons 2 and 3.
a sin θ θ b
θ θ asin
c
π
b cosθθ
d 2 θ a cos θ θ
a sin θ θ c π b
θ θ asin
b cosθθ
d 3 θ a cos θ θ
6.
Plot the piston displacements as a function of crank angle (referenced to line AC (see next page).
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-18c-2
Piston Displacement (1, 2, and 3) 90
Piston displacement, mm
80
d2 θ 70 d3 θ d1 θ
60
50
0
60
120
180
240
300
θ deg Piston 1 crank angle, deg.
The solid line is piston 1, the dotted line is piston 2, and the dashed line is piston 3.
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-18d-1
PROBLEM 4-18d Statement:
Find the total angular displacement of link 3 and the total stroke of the box as link 2 makes a complete revolution.
Given:
Ground link
L1 150
Input crank
L2 30
Coupler link
L3 150
Output crank
L4 30
Solution:
See Figure P4-5d and Mathcad file P0418d.
Y 3 2
B
A
O2
O4
X A 4
1.
This is a special-case Grashof mechanism in the parallelogram form (see Figure 2-17 in the text). As such, the coupler link 3 executes curvilinear motion and is always parallel to the ground link 1. Thus, the total angular motion of link 3 as crank 2 makes one complete revolution is zero degrees.
2.
The stroke of the box will be equal to twice the length of the crank link in one complete revolution of the crank stroke 2 L2
stroke 60
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-18e-1
PROBLEM 4-18e Statement:
Determine the ratio of angular displacement between links 8 and 2 as a function of angular displacement of input crank 2. Plot the transmission angle at point B for one revolution of crank 2. Comment on the behavior of this linkage. Can it make a full revolution as shown?
Given:
Link lengths:
Crank (O2A)
a 1 20
Coupler (L3)
b 1 160
Crank (O4B)
c1 20
B
A
Ground link (O2O4) d 1 160
O2
3
4 O4 G
E
2 D
5
C
6
7
Ground link (O4O8) d 2 120
Solution:
Crank (O4G)
a 2 30
Coupler (L6)
b 2 120
Crank (O8F)
c2 30
O8 H
F 8
See Figure P4-5e and Mathcad file P0418e.
1.
This is an eightbar, 1-DOF linkage with two redundant links (3 and 6 or 5 and 7) making it, effectively, a sixbar. It is composed of a fourbar (1, 2, 3, and 4) with an output dyad (7 and 8). The input fourbar is a special-case Grashof in the parallelogram configuration. Thus, the output angle is equal to the input angle and the couplers execute curvilinear motion with links 3 and 5 always parallel to the horizontal. The output dyad also behaves like a special-case Grashof with parallelogram configuration so that the angular motion of link 8 is equal to that of link 4. Therefore, the ratio of angular displacement between links 8 and 2 is unity. The mechanism is not capable of making a full revolution. The couplers 3 and 5 (also 6 and 7) cannot pass by each other near 2 = 0 and 180 deg because of interference with the pins that connect them to their cranks.
2.
Define the approximate range of motion of the input crank: θ 0 deg 2 deg 180 deg
3.
Define 3 and 4.
θ 0.deg
Use equations 4.32 to find and plot the transmission angle.
tran θ θ θ θ
Tran θ if tran θ π tran θ π tran θ
Trans θ if Tran θ
π 2
π Tran θ Tran θ
Transmission Angle at B Transmission Angle, deg
4.
θ θ θ
90 80 70 60 Trans θ 50 40 deg 30 20 10 0
0
45
90 θ deg Crank angle, deg
135
180
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-18f-1
PROBLEM 4-18f Statement:
Find and plot the displacement of piston 4 and the angular displacement of link 3 as a function of the angular displacement of crank 2.
Given:
Link lengths: Crank length, L2
Solution:
a 63
Piston-rod length, L3
b 130
Offset
c 52
4
B Y, x
3
See Figure P4-5f and Mathcad file P0418f.
1.
Establish 2 as a range variable: θ 0 deg 1 deg 360 deg
2.
Determine 3 and d in global XY coord using equations 4.16 and 4.17.
A
a sin θ 90 deg c θ θ asin π b
2 y
X
O2
d θ a cos θ 90 deg b cos θ θ
Plot the piston displacement (directly below) and rod angle (next page) as functions of crank angle in the global XY coordinate frame. Piston Displacement 200
150 Piston displacement, mm
3.
d θ 100
50
0
0
60
120
180 θ deg Crank angle, deg.
240
300
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-18f-2
Piston-Rod Angular Displacement 260
Angular displacement, deg
240
220
θ θ deg
200
180
160
0
60
120
180 θ deg Crank angle, deg.
240
300
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-18g-1
PROBLEM 4-18g Statement:
Find and plot the angular displacement of link 6 versus the angle of input link 2 as it is rotated from the position shown (+30 deg) to a vertical position (+90 deg). Find the toggle positions of this linkage in terms of the angle of link 2.
Given:
Link lengths:
Y
Input (L2)
a 49
Rocker (L4)
c 153
B
Coupler (L3)
b 100
Ground link (L1)
d 87
3 30° A
2 4
Angle from x axis to X axis:
α 121 deg
Starting angle:
θ 30 deg
Crank rotation angle from position shown to vertical:
Solution:
O6
X
6
C O2 5
D
y
O4
Δθ 60 deg
x
121°
See Figure P4-5g and Mathcad file P0418g.
1.
Define one cycle of the input crank in global coord: θ θ θ 1 deg θ Δθ
2.
Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the crossed circuit). K1
d
K2
a
K1 1.7755
2
d
K3
c
K2 0.5686
2
2
a b c d
2
2 a c
K3 1.5592
A θ cos θ α K1 K2 cos θ α K3
θ θ 2 atan2 2 A θ B θ 3.
C θ K1 K2 1 cos θ α K3
B θ 2 sin θ α
2 4 A θ Cθ α
B θ
If the calculated value of 4 is greater than 2, subtract 2 from it. If it is negative, make it positive.
θ θ if θ θ 2 π θ θ 2 π θ θ
θ θ if θ θ 0 θ θ 2 π θ θ
4.
Plot 4 as a function of the crank angle 2 (measured from the X-axis) as it rotates from the position shown to the vertical position.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-18g-2
Angular Displacement of Rocker Link 4 120
Rocker angle, deg
110
θ θ
100
deg
90
80 30
40
50
60
70
80
90
θ deg Crank angle, deg
4.
Check the Grashof condition of the linkage. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( a b c d ) "non-Grashof" 5.
Using equations 4.37, determine the crank angles (relative to the x-axis) at which links 3 and 4 are in toggle. 2
arg1
2
2
2 a d 2
arg2
2
a d b c
2
2
a d b c 2 a d
θ2toggle acos arg1
2
b c a d b c a d
arg1 0.840
arg2 6.338
θ2toggle 32.9 deg
The other toggle angle is the negative of this. Thus, in the global XY frame the toggle positions are:
θ2XYtoggle θ2toggle α
θ2XYtoggle 88.130 deg
θ2XYtoggle θ2toggle α
θ2XYtoggle 153.870 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-18h-1
PROBLEM 4-18h Statement:
Find link 4's maximum displacement vertically downward from the position shown. What will the angle of input link 2 be at that position?
Given:
Link lengths:
c2
a 19.8 mm
Crank length, L2 or L8
Solution:
Coupler length, L3 or L5
b1 19.4 mm
Offset of 1, 2, 3, 4
c1 4.5 mm
Distance from O2 to O8
L1 45.8 mm
Coupler length, L5 or L7
b2 13.3 mm
Offset of 1, 2, 5, 6
c2 22.9 mm
Angle of link 2 as shown
θ 47 deg
6
O2
O8
A
2
8
7
5
C
B
9
3
4
D
E
c1
See Figure P4-5h and Mathcad file P0418h.
1.
Links 1, 2, 3, 4, 5, and 6 make up two offset slider-cranks with a common crank, link 2. Links 7, 8, and 9 are kinematically redundant and contribute only to equalizing the forces in the mirror image links. Slider-crank 1, 2, 3, 4 is in the open circuit, and slider-crank 1, 2, 5, 6 is in the crossed circuit.
2.
Calculate the displacement of link 4 with respect to link 2 angle for the position shown in Figure P4-5h using equations 4.17 and 4.16b.
a sin θ c1 π b1
θ 149.038 deg
d 10 30.14 mm
θ asin
d 10 a cos θ b1 cos θ 3.
Link 4 will reach its maximum downward displacement when links 8 and 9 and links 2 and 3 are in the toggle position. However, it is possible that they may not be able to reach this position because links 5 and 7 may be too short to allow links 2 and 8 to rotate far enough to reach toggle with 3 and 9, respectively.
4.
Using equation 4.16a, determine the angle that the crank will make with the x axis (see layout below) when links 5 and 7 are horizontal (5 = -90 deg). This will be the least value of the angle 2.
a sin θ c2 θ asin b2
22.9 = c2 O2
y
θ 90 deg
A
a b2
29.0°
sin θ 1.000
a sin θ c2 b2
B
47°
A'
D' b1
1.000
c2 b2 θ asin a
D 5.90 D'
θ 29.00 deg 4.5= c1
DESIGN OF MACHINERY - 5th Ed.
5.
SOLUTION MANUAL 4-18h-2
Use equations 4.17 and 4.16b to determine the displacement of D' with respect to O2.
a sin θ c1 π b1
θ 164.759 deg
d 1 36.03 mm
θ asin
d 1 a cos θ b1 cos θ 6.
The maximum displacement of link 4 from the position shown in Figure P4-5h is the difference between the displacement found in step 5 and that found in step 2.
Δdmax d 1 d 10
Δdmax 5.90 mm
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-19-1
PROBLEM 4-19 Statement:
For one revolution of the driving link 2 of the walking-beam indexing and pick-and-place mechanism in Figure P4-6, find the horizontal stroke of link 3 for the portion of their motion where their tips are above the platen. Express the stroke as a percentage of the crank length O2B. What portion of a revolution of link 2 does this stroke correspond to? Also find the total angular displacement of link 6 over one revolution of link 2.
Given:
Measured lengths: Input crank length (O2A)
a 40
Coupler length (L3)
b 108
Output crank length (L4)
c 40
95
Q
3
A
64
2
4
O4
p 119.81
Coupler data (finger at Q) Distance from O2 to the platen surface
1.
D
C
Ground link length (O2O4) d 108
Solution:
73
E
6
δ 37.54 deg e 64
B
O2
7
O6 O5 185
See Figure P4-6 and Mathcad file P0419.
Links 1, 2, 3 and 4 are a special-case Grashof linkage in the parallelogram form. The tip of the finger at point Q (left end of the coupler) is used as the coupler point. The distance from the tip to the platen is . Top platen surface Q p
b D
A a
c d
x
O2
O4
y
2.
Define the crank angle as a range variable and define 3 ,which is constant because the coupler has curvilinear motion.. θ 0 deg 1 deg 360 deg θ 0 deg
3.
82
5
Use equations 4.27 to define the y-component of the vector RP. RP RA RPA
RA a cos θ j sin θ
RPA p cos θ δ j sin θ δ
RPy θ a sin θ p sin θ δ
DESIGN OF MACHINERY - 5th Ed.
4.
SOLUTION MANUAL 4-19-2
Define the distance of point Q above the platen (note the direction of the positive y axis in the figure above).
ε θ e RPy θ 5.
Plot as a function of crank angle 2. Height of Q Above Platen 60
14
168
Height Above Platen
40
20
ε θ
0
20
40
0
60
120
180
240
300
360
θ deg
6.
From the graph we see that the coupler point Q is above the platen when the crank angle is greater than 168 deg and less than 14 deg. To find the horizontal stroke during that range of 2, calculate the x-components of any point on the coupler, say point A, for those two crank angles and subtract them. Ax1 a cos( 14 deg)
Ax1 38.812
Ax2 a cos( 168 deg)
Ax2 39.126
Horizontal stroke when above the platen normalized by dividing by the crank length Stroke 7.
Ax1 Ax2
Stroke 1.95
a
times the crank length
Links 1, 4, 5, and 6 constitute a Grashoff crank-rocker-rocker. The extreme positions of the output rocker (link 6) occur when links 4 and 5 are in extended and overlapping toggle positions (see Figure 3-1b in the text for example, but in this case the mechanism is in the crossed circuit). C2
29.609°
C1
6
O6
B1
O5 5
B2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-19-3
Given link lengths: LO5B 13
L7 193
LO6C 92
LO5O6 128
In the first position (links 5 and 7 extended), the angle between link 6 and the ground link is:
L 2 L 2 L L 2 O6C O5O6 O5B 7 α acos L 2 L O6C O5O6
α 138.312 deg
In the second position (links 5 and 7 overlapping), the angle between link 6 and the ground link is:
LO6C2 LO5O62 L7 LO5B 2 α acos 2 LO6C LO5O6
α 108.702 deg
The total angular displacement of link 6 is the difference between these two angles.
Δ12 α α
Δ12 29.609 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-20-1
PROBLEM 4-20 Statement:
Figure P4-7 shows a power hacksaw, used to cut metal. Link 5 pivots at O5 and its weight forces the saw blade against the workpiece while the linkage moves the blade (link 4) back and forth on link 5 to cut the part. It is an offset slider-crank mechanism. The dimensions are shown in the figure. For one revolution of the driving link 2 of the hacksaw mechanism on the cutting stroke, find and plot the horizontal stroke of the saw blade as a function of the angle of link 2.
Given:
Link Lengths:
3
Crank length, L2
a 75 mm
Coupler length, L3
b 170 mm
Offset
c 45 mm
B
A 4
5
2 O2 O5
1
Assumptions: The arm that guides the slider (hacksaw blade carrier) remains horizontal throughout the stroke. Solution:
See Figure P4-7 and Mathcad file P0420.
1.
This is a slider-crank mechanism in the crossed circuit. The offset is the vertical distance from the horizontal centerline through O2 to point B.
2.
Establish 2 as a range variable: θ 0 deg 2 deg 360 deg
3.
Determine 3 and d using equations 4.16a and 4.17.
a sin θ c b
θ θ asin
d θ a cos θ b cos θ θ
Plot the blade (point B) displacement as a function of crank angle. Hacksaw Blade Stroke 50
100 Blade displacement, mm
4.
d θ
150
mm
200
250
0
60
120
180 θ deg Crank angle, deg.
240
300
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-21-1
PROBLEM 4-21 Statement:
Given:
For the linkage in Figure P4-8, find its limit (toggle) positions in terms of the angle of link O2A referenced to the line of centers O2O4 when driven from link O2A. Then calculate and plot the xy coordinates of coupler point P between those limits, referenced to the line of centers O2O4. P Link lengths: Input (O2A)
a 5.00 in
Coupler (AB)
b 4.40 in
Rocker (O4B)
c 5.00 in
Ground link
d 9.50 in
y
Y
Coupler point data:
B
p 8.90 in δ 56 deg
3 A
4
Coordinate transformation angle: x
2
α 14 deg
O4 1
14.000° X
O2
See Figure P4-8 and Mathcad file P0421. Solution: 1. Define the coordinate systems. The local frame has origin at O2 with the positive x axis going through O4. Let the global frame also have its origin at O2 with the positive X axis to the right. 2.
Check the Grashof condition of the linkage. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( a b c d ) "non-Grashof" 3.
Using equations 4.37, determine the crank angles (relative to the line AD) at which links 3 and 4 are in toggle. 2
arg1
2
2
2 a d 2
arg2
2
a d b c 2
2
a d b c 2 a d
2
b c a d b c a d
θ2toggle acos arg2
arg1 1.209 arg2 0.283
θ2toggle 73.6 deg
The other toggle angle is the negative of this. 4.
Define one cycle of the input crank between limit positions: θ θ2toggle θ2toggle 1 deg θ2toggle
5.
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12.
DESIGN OF MACHINERY - 5th Ed.
K1
SOLUTION MANUAL 4-21-2
d a
K1 1.9000
2
d
K4
K5
b
K4 2.1591
2
2
c d a b
2
2 a b
K5 2.4911
D θ cos θ K1 K4 cos θ K5
6.
F θ K1 K4 1 cos θ K5
E θ 2 sin θ
Use equation 4.13 to find values of 3 for the open circuit.
2 4 Dθ F θ
θ θ 2 atan2 2 D θ E θ 7.
E θ
Use equations 4.31 to define the x- and y-components of the vector RP. RP RA RPA
RA a cos θ j sin θ
RPA p cos θ δ j sin θ δ
RPx θ a cos θ p cos θ θ δ 8.
Transform the coupler point coordinates in the local frame to the global frame using coordinate transformation equations.
XP θ RPx θ cos α RPy θ sin α YP θ RPx θ sin α RPy θ cos α Plot the coordinates of the coupler point in the global system.
COUPLER CURVE 1.2
1
0.8
Y
9.
RPy θ a sin θ p sin θ θ δ
0.6
0.4
0.2
0 0.4
0.2
0
0.2 X
0.4
0.6
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-22-1
PROBLEM 4-22 Statement:
For the walking beam mechanism of Figure P4-9, calculate and plot the x and y components of the position of the coupler point P for one complete revolution of the crank O2A. Hint: Calculate them first with respect to the ground link O2O4 and then transform them into the global XY coordinate system (i.e., horizontal and vertical in the figure).
Given:
Link lengths:
Coupler point data:
Ground link
d 2.22
Crank
a 1
Coupler
b 2.06
Rocker
c 2.33
1.
δ 31.000 deg
α 26.5 deg
Coordinate transformation angle: Solution:
p 3.06
See Figure P4-9 and Mathcad file P0422.
Define the coordinate systems. The local frame has origin at O2 with the positive x axis going through O4. Let the global frame also have its origin at O2 with the positive X axis to the right. Y
x
y O4 4
1
26.500° X
O2
P
2 A 3 B
2.
Define one revolution of the input crank: θ 0 deg 2 deg 360 deg
3.
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K1
d
K4
a
K1 2.2200
2
d
K5
b
K4 1.0777
2
2 a b
K5 1.1512
D θ cos θ K1 K4 cos θ K5
4.
Use equation 4.13 to find values of 3 for the crossed circuit.
θ θ 2 atan2 2 D θ E θ 5.
F θ K1 K4 1 cos θ K5
E θ 2 sin θ
2 4 Dθ F θ
E θ
Use equations 4.31 to define the x- and y-components of the vector RP. RP RA RPA
2
c d a b
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-22-2
RA a cos θ j sin θ
RPA p cos θ δ j sin θ δ
RPx θ a cos θ p cos θ θ δ 6.
Transform the coupler point coordinates in the local frame to the global frame using coordinate transformation equations.
XP θ RPx θ cos α RPy θ sin α YP θ RPx θ sin α RPy θ cos α Plot the coordinates of the coupler point in the global system.
COUPLER CURVE 0.5
0
Y
7.
RPy θ a sin θ p sin θ θ δ
0.5
1
1.5
2
2.5
3
3.5 X
4
4.5
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-23-1
PROBLEM 4-23 Statement:
For the linkage in Figure P4-10, calculate and plot the angular displacement of links 3 and 4 and the path coordinates of point P with respect to the angle of the input crank O2A for one revolution.
Given:
Link lengths:
B
y 3
Ground link
d 2.22
Crank
a 1.0
Coupler
b 2.06
Rocker
c 2.33
b
Coupler point data: p 3.06
P p
A
δ 31.00 deg
2
a
4
2
4
c d
Solution:
x
1
O2
O4
See Figure P4-10 and Mathcad file P0423.
1.
Define one revolution of the input crank: θ 0 deg 1 deg 360 deg
2.
Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the open circuit). K1
d a
K1 2.2200
d
K2
2
K3
c
K2 0.9528
2
2
a b c d
2
2 a c
K3 1.5265
A θ cos θ K1 K2 cos θ K3
C θ K1 K2 1 cos θ K3
B θ 2 sin θ
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 3.
B θ
If the calculated value of 4 is greater than 2, subtract 2 from it.
θ θ if θ θ 2 π θ θ 2 π θ θ 4.
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
K4 1.0777
2 a b
D θ cos θ K1 K4 cos θ K5
5.
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ 6.
F θ K1 K4 1 cos θ K5
E θ 2 sin θ
2 4 Dθ F θ
E θ
If the calculated value of 3 is greater than 2, subtract 2 from it.
θ θ if θ θ 2 π θ θ 2 π θ θ
K5 1.1512
DESIGN OF MACHINERY - 5th Ed.
7.
SOLUTION MANUAL 4-23-2
Plot 3 and 4 as functions of the crank angle 2 (measured from the ground link). Angular Displacement of Coupler 240
Coupler angle, deg
260
280
θ θ deg
300 320 340
0
60
120
180
240
300
360
300
360
θ deg Crank angle, deg
Angular Displacement of Rocker 200
Rocker angle, deg
220
θ θ
240
deg 260
280
0
60
120
180
240
θ deg Crank angle, deg
8.
Use equations 4.31 to define the x- and y-components of the vector RP. RP RA RPA
RA a cos θ j sin θ
RPA p cos θ δ j sin θ δ
RPx θ a cos θ p cos θ θ δ 9.
RPy θ a sin θ p sin θ θ δ
Plot the coordinates of the coupler point in the local xy coordinate system.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-23-3
COUPLER POINT CURVE 3
2.5
Y
2
1.5
1
0.5 1.5
2
2.5
3 X
3.5
4
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-24-1
PROBLEM 4-24 Statement:
For the linkage in Figure P4-11, calculate and plot the angular displacement of links 3 and 4 with respect to the angle of the input crank O2A for one revolution.
Given:
Link lengths: Link 2
a 2.00 in
Link 3
b 8.375 in
Link 4
c 7.187 in
Link 1
d 9.625 in
A 3 2
B
2
O2 4
1
O4
Solution:
See Figure P4-11 and Mathcad file P0424.
1.
Define one revolution of the input crank: θ 0 deg 2 deg 360 deg
2.
Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the open circuit). K1
d a
K1 4.8125
2
d
K2
K3
c
K2 1.3392
2
2
a b c d
2
2 a c
K3 2.7186
A θ cos θ K1 K2 cos θ K3
C θ K1 K2 1 cos θ K3
B θ 2 sin θ
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 3.
B θ
If the calculated value of 4 is greater than 2, subtract 2 from it.
θ θ if θ θ 2 π θ θ 2 π θ θ 4.
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
K4 1.1493
2 a b
D θ cos θ K1 K4 cos θ K5
5.
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ 6.
F θ K1 K4 1 cos θ K5
E θ 2 sin θ
2 4 Dθ F θ
E θ
If the calculated value of 3 is greater than 2, subtract 2 from it.
θ θ if θ θ 2 π θ θ 2 π θ θ
K5 3.4367
DESIGN OF MACHINERY - 5th Ed.
Plot 3 and 4 as functions of the crank angle 2 (measured from the ground link).
Angular Displacement of Coupler
Coupler angle, deg
290
300
310
320
330
0
60
120
180
240
300
360
300
360
Crank angle, deg
Angular Displacement of Rocker 220
230 Rocker angle, deg
7.
SOLUTION MANUAL 4-24-2
240
250
260
0
60
120
180 Crank angle, deg
240
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-25-1
PROBLEM 4-25 Statement:
For the linkage in Figure P4-12, find its limit (toggle) positions in terms of the angle of link O2A referenced to the line of centers O2O4 when driven from link O2A. Then calculate and plot the angular displacement of links 3 and 4 and the path coordinates of point P with respect to the angle of the input crank O2A over its possible range of motion referenced to the line of centers O2O4.
Given:
Link lengths: Input (O2A)
a 0.785
Coupler (AB)
b 0.356
Rocker (O4B)
c 0.950
Ground link
d 0.544
A
1.
B
158.286° c
a O2
Coupler point data: p 1.09 Solution:
b
O4
d
δ 0 deg
See Figure P4-12 and Mathcad file P0425.
Check the Grashof condition of the linkage. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( a b c d ) "Grashof" 2.
double rocker
Using the geometry defined in Figure 3-1a in the text, determine the input crank angles (relative to the line O2O4) at which links 2 and 3, and 3 and 4 are in toggle.
d2 ( a b ) 2 c2 θ acos 2 d ( a b)
θ 55.937 deg
a2 d 2 ( b c) 2 2 a d
θ acos 3.
θ 158.286 deg
Define one cycle of the input crank between limit positions: θ θ θ 1 deg θ
4.
Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the open circuit). K1
d
K2
a
K1 0.6930
2
d
K3
c
K2 0.5726
K3 1.1317
A θ cos θ K1 K2 cos θ K3
C θ K1 K2 1 cos θ K3
B θ 2 sin θ
θ θ 2 atan2 2 A θ B θ
2 4 A θ Cθ
B θ
2
2
a b c d 2 a c
2
DESIGN OF MACHINERY - 5th Ed.
5.
SOLUTION MANUAL 4-25-2
If the calculated value of 4 is greater than 2, subtract 2 from it. If it is negative, make it positive.
θ θ if θ θ 2 π θ θ 2 π θ θ
θ θ if θ θ 0 θ θ 2 π θ θ 6.
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. 2
d
K4
K5
b
2
2
c d a b
2
K4 1.5281
2 a b
K5 0.2440
D θ cos θ K1 K4 cos θ K5
7.
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ 8.
F θ K1 K4 1 cos θ K5
E θ 2 sin θ
2 4 Dθ F θ
E θ
If the calculated value of 3 is greater than 2, subtract 2 from it. If it is negative, make it positive.
θ θ if θ θ 2 π θ θ 2 π θ θ
θ θ if θ θ 0 θ θ 2 π θ θ 9.
Plot 3 and 4 as functions of the crank angle 2 (measured from the ground link). Angular Displacement of Coupler & Rocker
Coupler or Rocker angle, deg
360 300 240 180 120 60 0 50
60
70
80
90
100
110
120
130
Crank angle, deg Coupler Rocker
10. Use equations 4.31 to define the x- and y-components of the vector RP. RP RA RPA
RA a cos θ j sin θ
140
150
160
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-25-3
RPA p cos θ δ j sin θ δ
RPx θ a cos θ p cos θ θ δ
11. Plot the coordinates of the coupler point in the local xy coordinate system.
COUPLER POINT PATH
Coupler Point Coordinate - y
2
1.5
1
0.5
0
0
0.5
RPy θ a sin θ p sin θ θ δ
1
Coupler Point Coordinate - x
1.5
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-26-1
PROBLEM 4-26 Statement:
For the linkage in Figure P4-13, find its limit (toggle) positions in terms of the angle of link O2A referenced to the line of centers O2O4 when driven from link O2A. Then calculate and plot the angular displacement of links 3 and 4 and the path coordinates of point P with respect to the angle of the input crank O2A over its possible range of motion referenced to the line of centers O2O4.
Given:
Link lengths: a 0.86
Input (O2A) Coupler (AB)
b 1.85
Rocker (O4B)
c 0.86
Ground link
d 2.22
1.
B c
a O2
Coupler point data: p 1.33 Solution:
116.037° b
A
O4
d
δ 0 deg
See Figure P4-13 and Mathcad file P0426.
Check the Grashof condition of the linkage. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( a b c d ) "non-Grashof" 2.
Using equations 4.33, determine the crank angles (relative to the line AD) at which links 3 and 4 are in toggle. 2
arg1
2
2
2 a d 2
arg2
2
a d b c
2
2
a d b c 2 a d
2
b c
arg1 1.228
a d b c
arg2 0.439
a d
θ2toggle acos arg2
θ2toggle 116.037 deg
The other toggle angle is the negative of this. 3.
Define one cycle of the input crank between limit positions: θ θ2toggle θ2toggle 1 deg θ2toggle
4.
Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the open circuit). K1
d
K2
a
K1 2.5814
d c
K2 2.5814
2
K3
K3 2.0181
A θ cos θ K1 K2 cos θ K3
B θ 2 sin θ
2
2
a b c d
C θ K1 K2 1 cos θ K3
2 a c
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-26-2
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 5.
B θ
If the calculated value of 4 is greater than 2, subtract 2 from it. If it is negative, make it positive.
θ θ if θ θ 2 π θ θ 2 π θ θ
θ θ if θ θ 0 θ θ 2 π θ θ 6.
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. 2
d
K4
K5
b
2
2
c d a b
2
K4 1.2000
2 a b
K5 2.6244
D θ cos θ K1 K4 cos θ K5
7.
F θ K1 K4 1 cos θ K5
E θ 2 sin θ
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ 8.
2 4 Dθ F θ
E θ
If the calculated value of 3 is greater than 2, subtract 2 from it. If it is negative, make it positive.
θ θ if θ θ 2 π θ θ 2 π θ θ
θ θ if θ θ 0 θ θ 2 π θ θ 9.
Plot 3 and 4 as functions of the crank angle 2 (measured from the ground link). Angular Displacement of Coupler & Rocker
Coupler or Rocker angle, deg
360 300 240 180 120 60 0 120
80
40
0
40
Crank angle, deg Coupler Rocker
10. Use equations 4.31 to define the x- and y-components of the vector RP. RP RA RPA
RA a cos θ j sin θ
80
120
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-26-3
RPA p cos θ δ j sin θ δ
RPx θ a cos θ p cos θ θ δ
11. Plot the coordinates of the coupler point in the local xy coordinate system.
COUPLER POINT PATH 1
Y
0.5
0
0.5 0.5
1
RPy θ a sin θ p sin θ θ δ
1.5 X
2
2.5
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-27-1
PROBLEM 4-27 Statement:
For the linkage in Figure P4-13, find its limit (toggle) positions in terms of the angle of link O4B referenced to the line of centers O4O2 when driven from link O4B. Then calculate and plot the angular displacement of links 2 and 3 and the path coordinates of point P with respect to the angle of the input crank O4B over its possible range of motion referenced to the line of centers O4O2.
Given:
Link lengths:
116.037° b 3 4
a 0.86
Input (O4B)
b 1.85
Coupler (AB) Rocker (O2A)
c 0.86
Ground link
d 2.22
x
Coupler point data: p 0.52
c A
O2
B a O4
d
δ 0 deg y
Solution: 1.
See Figure P4-13 and Mathcad file P0427.
Check the Grashof condition of the linkage. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( a b c d ) "non-Grashof" 2.
Using equations 4.33, determine the crank angles (relative to the line O4O2) at which links 2 and 3 are in toggle. 2
arg1
2
2
2 a d 2
arg2
2
a d b c
2
2
a d b c 2 a d
2
b c a d b c a d
θ4toggle acos arg2
arg1 1.228
arg2 0.439
θ4toggle 116.037 deg
The other toggle angle is the negative of this. 3.
Define one cycle of the input crank between limit positions: θ θ4toggle θ4toggle 1 deg θ4toggle
4.
Use equations 4.8a and 4.10 to calculate 2 as a function of 4 (for the open circuit). K1
d a
K2
d c
2
K3
K1 2.5814
K2 2.5814
B θ 2 sin θ
C θ K1 K2 1 cos θ K3
A θ cos θ K1 K2 cos θ K3
2
2
a b c d
K3 2.0181
2 a c
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-27-2
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 5.
B θ
If the calculated value of 2 is greater than 2, subtract 2 from it. If it is negative, make it positive.
θ θ if θ θ 2 π θ θ 2 π θ θ
θ θ if θ θ 0 θ θ 2 π θ θ 6.
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. 2
d
K4
b
K5
2
2
c d a b
2
K4 1.2000
2 a b
K5 2.6244
D θ cos θ K1 K4 cos θ K5
7.
F θ K1 K4 1 cos θ K5
E θ 2 sin θ
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ 8.
2 4 Dθ F θ
E θ
If the calculated value of 3 is greater than 2, subtract 2 from it. If it is negative, make it positive.
θ θ if θ θ 2 π θ θ 2 π θ θ
θ θ if θ θ 0 θ θ 2 π θ θ 9.
Plot 3 and 2 as functions of the crank angle 4 (measured from the ground link). Angular Displacement of Coupler & Rocker
Coupler or Rocker angle, deg
360 300 240 180 120 60 0 120
80
40
0
40
Crank angle, deg Coupler Rocker
10. Use equations 4.31 to define the x- and y-components of the vector RP. RP RB RPB
RB a cos θ j sin θ
80
120
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-27-3
RPB p cos θ δ j sin θ δ
RPx θ a cos θ p cos θ θ δ
11. Plot the coordinates of the coupler point in the local xy coordinate system. COUPLER POINT PATH 1
Y
0.5
0
0.5
1
0
0.25
0.5
RPy θ a sin θ p sin θ θ δ
0.75 X
1
1.25
1.5
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-28-1
PROBLEM 4-28 Statement:
For the rocker-crank linkage in Figure P4-14, find the maximum angular displacement possible for the treadle link (to which force F is applied). Determine the toggle positions. How does this work? Explain why the grinding wheel is able to fully rotate despite the presence of toggle positions when driven from the treadle. How would you get it started if it was in a toggle position?
Given:
Link lengths:
x
Input (O2A)
a 600 mm
Coupler (AB)
b 750 mm
Rocker (O4B)
c 130 mm
Ground link
d 900 mm
B
B''
c
O4
B'
b
d
43.331° A'' 25.182° a
A
O2
A'
y
Solution: 1.
See Figure P4-14 and Mathcad file P0428.
Use Figure 3-1(b) in the text to calculate the angles that link O2A makes with the ground link in the toggle positions.
a2 d 2 ( b c) 2 θ acos 2 a d
θ 43.331 deg
a2 d 2 ( b c) 2 2 a d
θ 68.513 deg
θ acos 2.
Subtract these two angles to get the maximum angular displacement of the treadle. θ θ
3.
25.182 deg
Despite having transmission angles of 0 deg twice per revolution, the mechanism will work. That is, one will be able to drive the grinding wheel from the treadle (link 2). The reason is that the grinding wheel will act as a flywheel and will carry the linkage through the periods when the transmission angle is low. Typically, the operator will start the motion by rotating the wheel by hand if it is in or near a toggle position.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-29-1
PROBLEM 4-29 Statement:
For the linkage in Figure P4-15, find its limit (toggle) positions in terms of the angle of link O2A referenced to the line of centers O2O4 when driven from link O2A. Then calculate and plot the angular displacement of links 3 and 4 and the path coordinates of point P with respect to the angle of the input crank O2A over its possible range of motion referenced to the line of centers O2O4.
Given:
Link lengths: Input (O2A)
a 0.72
Coupler (AB)
b 0.68
Rocker (O4B)
c 0.85
Ground link
d 1.82
P A B
Coupler point data: p 0.97
1.
O4
O2
δ 54 deg Solution:
55.355°
See Figure P4-15 and Mathcad file P0429.
Check the Grashof condition of the linkage. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( a b c d ) "non-Grashof" 2.
Using equations 4.37, determine the crank angles (relative to the line AD) at which links 3 and 4 are in toggle. 2
arg1
2
2
2 a d 2
arg2
2
a d b c 2
2
a d b c 2 a d
2
b c
arg1 1.451
a d b c
arg2 0.568
a d
θ2toggle acos arg2
θ2toggle 55.355 deg
The other toggle angle is the negative of this. 3.
Define one cycle of the input crank between limit positions: θ θ2toggle θ2toggle 0.5 deg θ2toggle
4.
Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the open circuit). K1
d a
K1 2.5278
K2
d c
K2 2.1412
2
K3
K3 3.3422
A θ cos θ K1 K2 cos θ K3
B θ 2 sin θ
2
2
a b c d
C θ K1 K2 1 cos θ K3
2 a c
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-29-2
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 5.
B θ
If the calculated value of 4 is greater than 2, subtract 2 from it. If it is negative, make it positive.
θ θ if θ θ 2 π θ θ 2 π θ θ
θ θ if θ θ 0 θ θ 2 π θ θ 6.
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
K4 2.6765
2 a b
K5 3.6465
D θ cos θ K1 K4 cos θ K5
7.
F θ K1 K4 1 cos θ K5
E θ 2 sin θ
Use equation 4.13 to find values of θ3 for the open circuit.
θ θ 2 atan2 2 D θ E θ 8.
2 4 Dθ F θ
E θ
If the calculated value of 3 is greater than 2, subtract 2 from it. If it is negative, make it positive.
θ θ if θ θ 2 π θ θ 2 π θ θ
θ θ if θ θ 0 θ θ 2 π θ θ 9.
Plot 3 and 4 as functions of the crank angle 2 (measured from the ground link).
Angular Displacement of Coupler & Rocker
Coupler or Rocker angle, deg
360 300 240 180 120 60 0 60
45
30
15
0
15
30
Crank angle, deg Coupler Rocker
10. Use equations 4.27 to define the x- and y-components of the vector RP. RP RA RPA
45
60
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-29-3
RA a cos θ j sin θ
RPA p cos θ δ j sin θ δ
RPx θ a cos θ p cos θ θ δ
11. Plot the coordinates of the coupler point in the local xy coordinate system.
COUPLER POINT PATH 1.4
1.2
Y
1
0.8
0.6
0.4
0.2 0.2
0.4
0.6
RPy θ a sin θ p sin θ θ δ
0.8 X
1
1.2
1.4
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-30-1
PROBLEM 4-30 Statement:
For the linkage in Figure P4-15, find its limit (toggle) positions in terms of the angle of link O4B referenced to the line of centers O4O2 when driven from link O4B. Then calculate and plot the angular displacement of links 2 and 3 and the path coordinates of point P with respect to the angle of the input crank O4B over its possible range of motion referenced to the line of centers O4O2.
Given:
Link lengths: Input (O4B)
a 0.85
Coupler (AB)
b 0.68
Rocker (O2A)
c 0.72
Ground link
d 1.82
P 47.885° c
Coupler point data: p 0.792 δ 82.032 deg Solution: 1.
B A
a
b
O4
O2
See Figure P4-15 and Mathcad file P0430.
Check the Grashof condition of the linkage. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( a b c d ) "non-Grashof" 2.
Using equations 4.37, determine the crank angles (relative to the line O4O2) at which links 2 and 3 are in toggle. 2
arg1
2
2
2 a d 2
arg2
2
a d b c 2
2
a d b c 2 a d
θ4toggle acos arg2
2
b c
arg1 1.304
a d b c
arg2 0.671
a d
θ4toggle 47.885 deg
The other toggle angle is the negative of this. 3.
Define one cycle of the input crank between limit positions: θ θ4toggle θ4toggle 0.5 deg θ4toggle
4.
Use equations 4.8a and 4.10 to calculate 2 as a function of 4 (for the open circuit). K1
d
K2
a
K1 2.1412
d c
K2 2.5278
2
K3
K3 3.3422
A θ cos θ K1 K2 cos θ K3
B θ 2 sin θ
2
2
a b c d
C θ K1 K2 1 cos θ K3
2 a c
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-30-2
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 5.
B θ
If the calculated value of 2 is greater than 2, subtract 2 from it. If it is negative, make it positive.
θ θ if θ θ 2 π θ θ 2 π θ θ
θ θ if θ θ 0 θ θ 2 π θ θ 6.
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
K4 2.6765
2 a b
K5 3.4420
D θ cos θ K1 K4 cos θ K5
7.
F θ K1 K4 1 cos θ K5
E θ 2 sin θ
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ 8.
2 4 Dθ F θ
E θ
If the calculated value of 3 is greater than 2, subtract 2 from it. If it is negative, make it positive.
θ θ if θ θ 2 π θ θ 2 π θ θ
θ θ if θ θ 0 θ θ 2 π θ θ 9.
Plot 3 and 2 as functions of the crank angle 4 (measured from the ground link). Angular Displacement of Coupler & Rocker
Coupler or Rocker angle, deg
360 300 240 180 120 60 0 60
45
30
15
0
15
30
Crank angle, deg Coupler Rocker
10. Use equations 4.27 to define the x- and y-components of the vector RP. RP RB RPB
45
60
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-30-3
RB a cos θ j sin θ
RPB p cos θ δ j sin θ δ
RPx θ a cos θ p cos θ θ δ
11. Plot the coordinates of the coupler point in the local xy coordinate system.
COUPLER POINT PATH 1.5
Y
1
0.5
0
0
0.2
0.4
0.6 X
RPy θ a sin θ p sin θ θ δ
0.8
1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-31-1
PROBLEM 4-31 Statement:
Write a computer program (or use an equation solver such as Mathcad, Matlab, or TKSolver) to find the roots of y = 9x2 + 50x - 40. Hint: Plot the function to determine good guess values.
Solution:
See Mathcad file P0431.
1.
Plot the function. 2
x 10 9.5 10
f ( x) 9 x 50 x 40
200
100
f ( x)
0
100
200 10
8
6
4
2
0
2
x
2.
From the graph, make guesses of x1 6 , x2 1
3.
Define the program using the pseudo code in the text. nroot( f df x)
y f ( x) y TOL
return x if
while y TOL xx
y df ( x)
y f ( x) x
where,
3
TOL 1.000 10
4.
Define the derivative of the given function. df ( x) 18 x 50
5.
Use the program to find the roots. r1 nroot f df x1
r1 6.265
r2 nroot f df x2
r2 0.709
4
6
8
10
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-32-1
PROBLEM 4-32 Statement:
Write a computer program (or use an equation solver such as Mathcad, Matlab, or TKSolver) to find the roots of y = -x3 - 4x2 + 80x - 40. Hint: Plot the function to determine good guess values.
Solution:
See Mathcad file P0432.
1.
Plot the function. 3
x 15 14.5 10
2
f ( x) x 4 x 80 x 40
200
100
f ( x)
0
100
200 20
15
10
5
0
x
2.
From the graph, make guesses of x1 11 , x2 0 , x3 7
3.
Define the program using the pseudo code in the text. nroot( f df x)
y f ( x) y TOL
return x if
while y TOL xx
y df ( x)
y f ( x) x
where,
3
TOL 1.000 10
2
4.
Define the derivative of the given function. df ( x) 3 x 8 x 80
5.
Use the program to find the roots. r1 nroot f df x1
r1 11.355
r2 nroot f df x2
r2 0.515
r3 nroot f df x3
r3 6.840
5
10
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-33-1
PROBLEM 4-33 Statement:
Figure 4-18 (p. 193) plots the cubic function from equation 4.34. Write a computer program (or use an equation solver such as Mathcad, Matlab, or TKSolver) to investigate the behavior of the Newton-Raphson algorithm as the initial guess value is varied from x = 1.8 to 2.5 in steps of 0.1. Determine the guess value at which the convergence switches roots. Explain this root-switching phenomenon based on your observations from this exercise.
Solution:
See Figure 4-18 and Mathcad file P0433.
1.
Define the range of the guess value, the function, and the derivative of the function. xguess 1.8 1.9 2.5 3
2
2
f ( x) x 2 x 50 x 60 2.
df ( x) 3 x 4 x 50
Define the root-finding program using the pseudo code in the text. nroot( f df x)
y f ( x) y TOL
return x if
while y TOL xx
y df ( x)
y f ( x) x 3.
Find the roots that correspond to the guess values. r( xguess) nroot( f df xguess) 1
1
f ( xguess) df ( xguess)
1
1
1
1.800
1
33.080
1
-2.362
1
-1.177
2
1.900
2
31.570
2
-2.564
2
-1.177
3
2.000
3
30.000
3
-2.800
3
-1.177
2.100 df ( xguess) 4
28.370
nextx( xguess) 4
-3.079
r( xguess) 4
-1.177
xguess 4
4.
nextx( xguess) xguess
5
2.200
5
26.680
5
-3.410
5
-1.177
6
2.300
6
24.930
6
-3.807
6
-1.177
7
2.400
7
23.120
7
-4.289
7
6.740
8
2.500
8
21.250
8
-4.882
8
-7.562
Find the roots of the derivative (values of x where the slope is zero). ddf ( x) 6 x 4
5.
xz1 nroot( df ddf 5 )
xz1 4.803
xz2 nroot( df ddf 4 )
xz2 3.470
For guess values up to 2.3, the root found is that whose slope is nearly the same as the slope of the function at the guess value. At 2.4, the value of x that is calculated next results in a slope that throws the next x-value to the right of the extreme function value at x = 3.470. Subsequent estimates of x then follow down the slope to x = 6.740. At a guess value of 2.5, the value of x that is calculated next is to the left of the extreme function value at x = -4.803. Subsequent estimates of x follow up the slope to x = -7.562.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-34-1
PROBLEM 4-34 Statement:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the angular position of link 4 and the position of slider 6 in Figure 3-33 as a function of the angle of input link 2.
Given:
Link lengths: Input crank (L2)
a 2.170
Fourbar coupler (L3)
b 2.067
Output crank (L4)
c 2.310
Sllider coupler (L5)
e 5.40
Fourbar ground link (L1) Solution:
d 1.000
See Figure 3-33 and Mathcad file P0434.
1.
This sixbar drag-link mechanism can be analyzed as a fourbar Grashof double crank in series with a slidercrank mechanism using the output of the fourbar, link 4, as the input to the slider-crank.
2.
Define one revolution of the input crank: θ 0 deg 1 deg 360 deg
3.
Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the open circuit) in the global XY system. K1
d
K2
a
K1 0.4608
2
d
K3
c
K2 0.4329
2
2
a b c d
2
2 a c
K3 0.6755
A θ cos θ K1 K2 cos θ K3
θ θ 2 atan2 2 A θ B θ 4.
C θ K1 K2 1 cos θ K3
B θ 2 sin θ
2 4 A θ Cθ 102 deg
B θ
If the calculated value of 4 is greater than 2, subtract 2 from it and if it is negative, make it positive.
θ θ if θ θ 2 π θ θ 2 π θ θ
θ θ if θ θ 0 θ θ 2 π θ θ 5.
Determine the slider-crank motion using equations 4.16 and 4.17 with 4 as the input angle.
c sin θ θ π e
e cosθθ
θ θ asin
f θ c cos θ θ 6.
Plot the angular position of link 4 and the position of link 6 as functions of the angle of input link 2. See next page.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-34-2
Angular Position of Link 4 360 315 270
θ θ
225 180
deg 135 90 45 0
0
45
90
135
180
225
270
315
360
315
360
θ deg
Position of Slider 6 With Respect to O4 8 7.167 6.333
f θ
5.5 4.667 3.833 3
0
45
90
135
180 θ deg
225
270
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-35-1
PROBLEM 4-35 Statement:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the transmission angles at points B and C of the linkage in Figure 3-33 as a function of the angle of input link 2.
Given:
Link lengths: Input crank (L2)
a 2.170
Fourbar coupler (L3)
b 2.067
Output crank (L4)
c 2.310
Sllider coupler (L5)
e 5.40
d 1.000
Fourbar ground link (L1) Solution:
See Figure 3-33 and Mathcad file P0435.
1.
This sixbar drag-link mechanism can be analyzed as a fourbar Grashof double crank in series with a slidercrank mechanism using the output of the fourbar, link 4, as the input to the slider-crank.
2.
Define one revolution of the input crank: θ 0 deg 1 deg 360 deg
3.
Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the open circuit). K1
d a
K1 0.4608
2
d
K2
K3
c
K2 0.4329
2
2
a b c d
2
2 a c
K3 0.6755
A θ cos θ K1 K2 cos θ K3
C θ K1 K2 1 cos θ K3
B θ 2 sin θ
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 4.
B θ
If the calculated value of 4 is greater than 2, subtract 2 from it and if it is negative, make it positive.
θ θ if θ θ 2 π θ θ 2 π θ θ
θ θ if θ θ 0 θ θ 2 π θ θ 5.
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d b
K5
2
2
c d a b
2
K4 0.4838
2 a b
K5 0.5178
D θ cos θ K1 K4 cos θ K5
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 6.
Use equation 4.13 to find 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ 7.
2 4 Dθ F θ
E θ
If the calculated value of 3 is greater than 2, subtract 2 from it and if it is negative, make it positive.
θ θ if θ θ 2 π θ θ 2 π θ θ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-35-2
θ θ if θ θ 0 θ θ 2 π θ θ 8.
Calculate (using equations 4.32) and plot the transmission angle at B.
θtransB1 θ θ θ θ θ
θtransB θ if θtransB1 θ
π 2
π θtransB1 θ θtransB1 θ
Transmission Angle at B 40 35 30
θtransB θ
25 20
deg 15 10 5 0
0
45
90
135
180
225
270
315
360
θ deg
9.
Determine the slider-crank motion using equations 4.16 and 4.17 with 4 as the input angle.
c sin θ θ π e
θ θ asin
10. Calculate (using equations 4.32) and plot the transmission angle at C.
θtransC1 θ θ θ
θtransC θ if θtransC1 θ
π 2
π θtransC1 θ θtransC1 θ
Transmission Angle at C 30 25
θtransC θ
20 15
deg 10 5 0
0
45
90
135
180 θ deg
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-36-1
PROBLEM 4-36 Statement:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the path of the coupler point of the approximate straight-line linkage shown in Figure 3-29f (p. 142). Use program Fourbar to check your result.
Given:
Link lengths: Input (O2A)
a 1.000
Coupler (AB)
b 1.600
Rocker (O4B)
c 1.039
Ground link
d 1.200
p 2.690
Coupler point data:
α 60 deg
Coordinate rotation angle: Solution: 1.
δ 0 deg
See Figure 3-29f and Mathcad file P0436.
Check the Grashof condition of the linkage and determine its Baker classification. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( a b c d ) "non-Grashof" Since b (link 3) is the longest link and the linkage is non-Grashof, this is a Class 3 triple rocker. Using Figure 3-1a as a guide, determine the limiting values of 2 at the toggle positions. For links 2 and 3 colinear:
( b a) 2 d2 c2 π 2 d ( b a)
θ acos
θ 240 deg
For links 3 and 4 colinear:
a2 d 2 ( b c) 2 θ acos 2 d a 2.
θ 27.683 deg
θ θ
Define one cycle of the input crank (driving through the links 2-3 toggle position): θ θ θ 1 deg 360 deg θ
3.
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K1
d
K4
a
K1 1.2000
2
d
K5
b
K4 0.7500
2
2 a b
K5 1.2251
D θ cos θ K1 K4 cos θ K5
E θ 2 sin θ 4.
F θ K1 K4 1 cos θ K5
Use equation 4.13 to find values of 3 for the open circuit.
2
c d a b
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-36-2
θ θ 2 atan2 2 D θ E θ 5.
2 4 Dθ F θ
E θ
Use equations 4.31 to define the x- and y-components of the vector RP. RP RA RPA
RPA p cos θ δ j sin θ δ RA a cos θ j sin θ
RPx θ a cos θ p cos θ θ δ
Plot the coordinates of the coupler point in the global X,Y coordinate system using equations 4.0b to rotate the local coordinates to a global frame.
PX θ RPx θ cos( α) RPy θ sin( α) PY θ RPx θ sin( α) RPy θ cos( α) COUPLER POINT PATH 2
1
Coupler Point Coordinate - Y
6.
RPy θ a sin θ p sin θ θ δ
0
1
2
3
4
0
1
2 Coupler Point Coordinate - X
3
4
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-37-1
PROBLEM 4-37 Statement:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the angular position of link 6 in Figure 3-34 as a function of the angle of input link 2.
Given:
Link lengths:
Solution: 1.
Input crank (L2)
g 1.556
First coupler (L3)
f 4.248
First rocker (L4)
c 2.125
Third coupler (CD)
b 2.158
Output rocker (L6)
a 1.542
Second ground link (O4O6) d 1.000
Angle CDB
δ 36 deg
Distance (BD)
O2O4 ground link offsets:
h X 3.259
h Y 2.905
See Figure 3-34 and Mathcad file P0437.
Calculate the length of the O2O4 ground link and the angle that it makes with the global XY system. h
2.
p 3.274
2
hX hY
2
hY hX
γ atan
h 4.366
γ 41.713 deg
Calculate the distance BC on link 5. This is the length of vector R51. Also, calculate the angle between vectors R51 and R52 e
b p 2 b p cos δ 2
2
e 1.986
Second coupler (BC)
b 2 e2 p 2 2 b e
α acos
α 104.305 deg
β π α
β 75.695 deg
3.
This is a Stephenson's sixbar linkage similar to the one shown in Figure 4-13. Since the output link 6 is known to rotate 180 deg and return for a full revolution of link 2 we can use links 6, 5, and 4 as a first-stage fourbar with known input (link 6) and then solve for vector loop equations to get the corresponding motion of link 2.
4.
Define the rotation of the output crank: θ 90 deg 91 deg 270 deg
5.
Use equations 4.8a and 4.10 to calculate 4 in the local xy coordinate system as a function of 6 (for the crossed circuit). K1
d
K2
a
K1 0.6485
2
d
K3
c
K2 0.4706
2
2
a b c d
2
2 a c
K3 0.4938
A θ cos θ K1 K2 cos θ K3
C θ K1 K2 1 cos θ K3
B θ 2 sin θ
θ θ 2 atan2 2 A θ B θ 6.
2 4 A θ Cθ
B θ
Use equations 4.12 and 4.13 to calculate 5 in the local xy coordinate system as a function of 6 (for the crossed circuit). K4
d b
K4 0.463
2
K5
2
2
c d a b
K5 0.529
2 a b
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-37-2
D θ cos θ K1 K4 cos θ K5
E θ 2 sin θ
F θ K1 K4 1 cos θ K5
θ θ 2 atan2 2 D θ E θ 7.
2 4 Dθ F θ
E θ
Transform the angles for 4 and 52 into the global XY system and define 51 in the global system.
θ θ θ θ 90 deg
θ θ θ θ 90 deg θ θ θ θ β 8.
Define a vector loop for the remaining links and solve the resulting vector equation by separating it into real and imaginary parts using the method of section 4.5 and the identities of equations 4.9.
Y X
O2 2 R2
R1
6
A
O6
y
D R 5 R52 3 C R4
3
O4 4 x
R51 B R1 + R4 + R51 + R3 - R2 = 0. In this equation the unknowns are 3 and 2. Following the method of Section 4.5, substitute the complex number notation for each position vector and separate the resulting equations into real and imaginary parts:
f cos θ = g cos θ G1 f sin θ = g sin θ G2 where
e cosθθ
G1 θ h cos γ c cos θ θ
e sinθθ
G2 θ h sin γ c sin θ θ 9.
Solve these equations in the manner of equations 4.11 and 4.12 using the identities of equations 4.9 gives:
2 G2θ2 f 2
2
g G1 θ
G3 θ
2 g
A' θ G1 θ G3 θ
B' θ 2 G2 θ
C' θ G1 θ G3 θ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-37-3
θ B' B'2 4 A' C' = 2 A' 2
tan
θ θ 2 atan2 2 A' θ B' θ
2 4 A' θ C'θ
B' θ
10. Plot 6 vs 2 in global XY coordinates: Rotation of Link 6 vs Link 2 320 300 280 260
θ θ
240 220
deg 200 180 160 140 120
0
20
40
60
80 θ deg
100 90
120
140
160
180
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-38-1
PROBLEM 4-38 Statement:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the transmission angles at points B, C, and D of the linkage in Figure 3-34 as a function of the angle of input link 2.
Given:
Link lengths:
Solution: 1.
Input crank (L2)
g 1.556
First coupler (L3)
f 4.248
First rocker (L4)
c 2.125
Third coupler (CD)
b 2.158
Output rocker (L6)
a 1.542
Second ground link (O4O6) d 1.000
Angle CDB
δ 36 deg
Distance (BD)
O2O4 ground link offsets:
h X 3.259
h Y 2.905
See Figure 3-34 and Mathcad file P0438.
Calculate the length of the O2O4 ground link and the angle that it makes with the global XY system. h
2.
p 3.274
2
hX hY
2
hY hX
γ atan
h 4.366
γ 41.713 deg
Calculate the distance BC on link 5. This is the length of vector R51. Also, calculate the angle between vectors R51 and R52 e
b p 2 b p cos δ 2
2
e 1.986
Second coupler (BC)
b 2 e2 p 2 2 b e
α acos
α 104.305 deg
β π α
β 75.695 deg
3.
This is a Stephenson's sixbar linkage similar to the one shown in Figure 4-13. Since the output link 6 is known to rotate 180 deg and return for a full revolution of link 2 we can use links 6, 5, and 4 as a first-stage fourbar with known input (link 6) and then solve for vector loop equations to get the corresponding motion of link 2.
4.
Define the rotation of the output crank: θ 90 deg 91 deg 270 deg
5.
Use equations 4.8a and 4.10 to calculate 4 in the local xy coordinate system as a function of 6 (for the crossed circuit). K1
d
K2
a
K1 0.6485
2
d
K3
c
K2 0.4706
2
2
a b c d
2
2 a c
K3 0.4938
A θ cos θ K1 K2 cos θ K3
C θ K1 K2 1 cos θ K3
B θ 2 sin θ
θ θ 2 atan2 2 A θ B θ 6.
2 4 A θ Cθ
B θ
Use equations 4.12 and 4.13 to calculate 5 in the local xy coordinate system as a function of 6 (for the crossed circuit). K4
d b
K4 0.463
2
K5
2
2
c d a b
K5 0.529
2 a b
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-38-2
D θ cos θ K1 K4 cos θ K5
E θ 2 sin θ
F θ K1 K4 1 cos θ K5
2 4 Dθ F θ
θ θ 2 atan2 2 D θ E θ 7.
E θ
Transform the angles for 4 and 52 into the global XY system and define 51 in the global system.
θ θ θ θ 90 deg
θ θ θ θ 90 deg θ θ θ θ β 8.
Define a vector loop for the remaining links and solve the resulting vector equation by separating it into real and imaginary parts using the method of section 4.5 and the identities of equations 4.9.
Y X
O2 2 R2
R1
6
A
O6
y
D R 5 R52 3 C R4
3
O4 4 x
R51 B R1 + R4 + R51 + R3 - R2 = 0. In this equation the unknowns are 3 and 2. Following the method of Section 4.5, substitute the complex number notation for each position vector and separate the resulting equations into real and imaginary parts:
g cos θ = f cos θ G1 g sin θ = f sin θ G2 where
G1 θ h cos γ c cos θ θ
e cosθθ
G2 θ h sin γ c sin θ θ
9.
e sinθθ
Solve these equations for 2 in the manner of equations 4.11 and 4.12 using the identities of equations 4.9 gives:
2 G2θ2 f 2
2
g G1 θ
G3 θ
2 g
A' θ G1 θ G3 θ
B' θ 2 G2 θ
C' θ G1 θ G3 θ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-38-3
θ B' B'2 4 A' C' = 2 A' 2
tan
θ θ 2 atan2 2 A' θ B' θ
2 4 A' θ C'θ
B' θ
10. Solve these equations for 3 in the manner of equations 4.11 and 4.12 using the identities of equations 4.9 gives:
2 G2θ2 f 2
2
g G1 θ
G4 θ
2 f
D' θ G1 θ G4 θ
E' θ 2 G2 θ
F' θ G1 θ G4 θ
θ E' E'2 4 D' F' tan = 2 D' 2
θ θ 2 atan2 2 D' θ E' θ
2 4 D'θ F'θ
E' θ
11. Calculate (using equations 4.32) and plot the transmission angle at B.
θtransB1 θ θ θ θ θ
θtransB θ if θtransB1 θ
π 2
π θtransB1 θ θtransB1 θ
Transmission Angle at B 90 80 70 60
θtransB θ 50 deg
40 30 20 10 0 125
150
175
200
225
θ θ deg
12. Calculate (using equations 4.32) and plot the transmission angle at C.
θtransC1 θ θ θ θ θ
250
275
300
325
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-38-4
θtransC θ if θtransC1 θ
π 2
π θtransC1 θ θtransC1 θ
Transmission Angle at C 60 55 50
θtransC θ
45
deg 40 35 30 125
150
175
200
225
250
275
300
325
θ θ deg
13. Calculate (using equations 4.32) and plot the transmission angle at D.
θtransD1 θ θ θ θ
θtransD θ if θtransD1 θ
π 2
π θtransD1 θ θtransD1 θ
Transmission Angle at D 60 50 40
θtransD θ
30
deg 20 10 0 125
150
175
200
225
θ θ deg
250
275
300
325
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-39-1
PROBLEM 4-39 Statement:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the path of the coupler point of the approximate straight-line linkage shown in Figure 3-29g (p. 142). Use program Fourbar to check your result.
Given:
Link lengths: Input (O2A)
a 1.000
Coupler (AB)
b 1.200
Rocker (O4B)
c 1.167
Ground link
d 2.305
p 1.5
Coupler point data:
α 30 deg
Coordinate rotation angle: Solution: 1.
δ 180 deg
See Figure 3-29g and Mathcad file P0439.
Check the Grashof condition of the linkage and determine its Baker classification. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( a b c d ) "non-Grashof" Since d (link 1) is the longest link and the linkage is non-Grashof, this is a Class 1 triple rocker. Using Figure 3-1a as a guide, determine the limiting values of 2 at the toggle positions. For links 3 and 4 colinear:
a2 d 2 ( b c) 2 2 d a
θ acos 1.
θ 81.136 deg
θ θ
Define one cycle of the input crank: θ θ θ 0.5 deg θ
2.
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K1
d
K4
a
K1 2.3050
2
d
K5
b
K4 1.9208
2
2 a b
K5 2.6630
D θ cos θ K1 K4 cos θ K5
3.
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ 4.
F θ K1 K4 1 cos θ K5
E θ 2 sin θ
2 4 Dθ F θ
E θ
Use equations 4.31 to define the x- and y-components of the vector RP.
2
c d a b
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-39-2
RP RA RPA
RPA p cos θ δ j sin θ δ RA a cos θ j sin θ
RPx θ a cos θ p cos θ θ δ
Plot the coordinates of the coupler point in the global X,Y coordinate system using equations 4.0b to rotate the local coordinates to a global frame.
PX θ RPx θ cos( α) RPy θ sin( α) PY θ RPx θ sin( α) RPy θ cos( α) COUPLER POINT PATH 2
1
Coupler Point Coordinate - Y
5.
RPy θ a sin θ p sin θ θ δ
0
1
2 2
1.5
1 Coupler Point Coordinate - X
0.5
0
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-40-1
PROBLEM 4-40 Statement:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the angular position of link 6 in Figure 3-35 as a function of the angle of input link 2.
Given:
Link lengths:
Solution:
Input crank (L2)
a 1.00
First coupler (L3)
b 3.80
Common rocker (O4B)
c 1.29
Second coupler (L5)
b' 1.29
First ground link (O2O4)
d 3.86
Common rocker (O4C)
a' 1.43
Output rocker (L6)
c' 0.77
Second ground link (O4O6) d' 0.78
Angle BO4C
α 157 deg
See Figure P3-35 and Mathcad file P0440.
1.
This sixbar drag-link mechanism can be analyzed as two fourbar linkages in series that use the output of the first fourbar, link 4, as the input to the second fourbar.
2.
Define one revolution of the input crank: θ 0 deg 1 deg 360 deg
3.
Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the crossed circuit). K1
d
K2
a
K1 3.8600
2
d
K3
c
K2 2.9922
2
2
a b c d
2
2 a c
K3 1.2107
A θ cos θ K1 K2 cos θ K3
C θ K1 K2 1 cos θ K3
B θ 2 sin θ
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 4.
B θ
Use equations 4.8a and 4.10 to calculate 6 as a function of 2 (for the open circuit).
K' 1
d'
K' 2
a'
K' 1 0.5455
θ θ θ θ α
Input angle to second fourbar:
2
d'
K' 3
c'
K' 2 1.0130
2
2
2 a' c'
K' 3 0.7184
K'1 K'2 cosθθ K'3
A' θ cos θ θ
θ θ 2 atan2 2 A' θ B' θ
5.
K'3
C' θ K' 1 K' 2 1 cos θ θ
B' θ 2 sin θ θ
2 4 A' θ C'θ
B' θ
Plot the angular position of link 6 as a function of the angle of input link 2.
2
a' b' c' d'
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-40-2
Angular Position of Link 6 100 75 50 25
θ θ
0 25
deg 50 75 100 125 150
0
45
90
135
180 θ deg
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-41-1
PROBLEM 4-41 Statement:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the transmission angles at points B, C, and D of the linkage in Figure 3-35 as a function of the angle of input link 2.
Given:
Link lengths:
Solution:
Input crank (L2)
a 1.00
First coupler (L3)
b 3.80
Common rocker (O4B)
c 1.29
Second coupler (L5)
b' 1.29
First ground link (O2O4)
d 3.86
Common rocker (O4C)
a' 1.43
Output rocker (L6)
c' 0.77
Second ground link (O4O6) d' 0.78
Angle BO4C
α 157 deg
See Figure P3-35 and Mathcad file P0441.
1.
This sixbar drag-link mechanism can be analyzed as two fourbar linkages in series that use the output of the first fourbar, link 4, as the input to the second fourbar.
2.
Define one revolution of the input crank: θ 0 deg 1 deg 360 deg
3.
Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the crossed circuit). K1
d
K2
a
K1 3.8600
2
d
K3
c
K2 2.9922
2
2 a c
K3 1.2107
A θ cos θ K1 K2 cos θ K3
C θ K1 K2 1 cos θ K3
B θ 2 sin θ
θ θ 2 atan2 2 A θ B θ 4.
2 4 A θ Cθ
B θ
Use equations 4.12 and 4.13 to calculate 3 as a function of 2 (for the crossed circuit). K4
2
d
K5
b
K4 1.016
2
2
c d a b
2
2 a b
K5 3.773
D θ cos θ K1 K4 cos θ K5
E θ 2 sin θ
F θ K1 K4 1 cos θ K5
θ θ 2 atan2 2 D θ E θ 5.
2 4 Dθ F θ
E θ
Calculate (using equations 4.32) and plot the transmission angle at B.
θtransB1 θ θ θ θ θ
θtransB θ if θtransB1 θ
π 2
2
a b c d
π θtransB1 θ θtransB1 θ
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-41-2
Transmission Angle at B 90 82.5 75 67.5
θtransB θ
60
deg 52.5 45 37.5 30
0
45
90
135
180
225
270
315
360
θ deg
6.
Use equations 4.8a and 4.10 to calculate 6 as a function of 2 (for the open circuit).
K' 1
d'
K' 2
a'
K' 1 0.5455
θ θ θ θ α
Input angle to second fourbar:
2
d'
K' 3
c'
K' 2 1.0130
2
2
2
a' b' c' d' 2 a' c'
K' 3 0.7184
K'1 K'2 cosθθ K'3
A' θ cos θ θ
2 4 A' θ C'θ
θ θ 2 atan2 2 A' θ B' θ 7.
K'3
C' θ K' 1 K' 2 1 cos θ θ
B' θ 2 sin θ θ
B' θ
Use equations 4.12 and 4.13 to calculate 5 as a function of 2 (for the crossed circuit). K' 4
2
d'
K' 5
b'
K' 4 0.605
2
2
2
c' d' a' b' 2 a' b'
K' 5 1.010
K'1 K'4 cosθθ K'5
D' θ cos θ θ
K'5
F' θ K' 1 K' 4 1 cos θ θ
θ θ 2 atan2 2 D' θ E' θ 8.
2 4 D'θ F'θ
E' θ
Calculate (using equations 4.32) and plot the transmission angle at C.
θtransC1 θ θ θ θ θ
θtransC θ if θtransC1 θ
π 2
E' θ 2 sin θ θ
π θtransC1 θ θtransC1 θ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-41-3
Transmission Angle at C 30
24
θtransC θ
18
deg 12
6
0
0
45
90
135
180
225
270
315
360
θ deg
9.
Calculate (using equations 4.32) and plot the transmission angle at D.
θtransD1 θ θ θ θ θ
θtransD θ if θtransD1 θ
π 2
π θtransD1 θ θtransD1 θ
Transmission Angle at D 120 100 80
θtransD θ
60
deg 40 20 0
0
45
90
135
180 θ deg
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-42-1
PROBLEM 4-42 Statement:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the path of the coupler point of the approximate straight-line linkage shown in Figure 3-29h (p. 142). Use program Fourbar to check your result.
Given:
Link lengths: Input (O2A)
a 1.000
Coupler (AB)
b 1.000
Rocker (O4B)
c 1.000
Ground link
d 2.000
p 2.0
Coupler point data: Solution: 1.
δ 0 deg
See Figure 3-29h and Mathcad file P0442.
Check the Grashof condition of the linkage and determine its Baker classification. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( a b c d ) "non-Grashof" Since d (link 1) is the longest link and the linkage is non-Grashof, this is a Class 1 triple rocker. Using Figure 3-1a as a guide, determine the limiting values of 2 at the toggle positions. For links 3 and 4 colinear:
a2 d 2 ( b c) 2 2 d a
θ acos 1.
θ 75.522 deg
θ θ
Define one cycle of the input crank: θ θ θ 0.25 deg θ
2.
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. d
K1
K4
a
K1 2.0000
2
d
K5
b
K4 2.0000
2
2 a b
K5 2.5000
D θ cos θ K1 K4 cos θ K5
3.
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ 4.
F θ K1 K4 1 cos θ K5
E θ 2 sin θ
2 4 Dθ F θ
E θ
Use equations 4.31 to define the x- and y-components of the vector RP. RP RA RPA
RA a cos θ j sin θ
2
c d a b
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-42-2
RPA p cos θ δ j sin θ δ
RPx θ a cos θ p cos θ θ δ
COUPLER POINT PATH 2
Coupler Point Coordinate - Y
1.5
1
0.5
0
1
1.5
RPy θ a sin θ p sin θ θ δ
2 Coupler Point Coordinate - X
2.5
3
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-43-1
PROBLEM 4-43 Statement:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the angular position of link 8 in Figure 3-36 as a function of the angle of input link 2.
Given:
Link lengths:
Solution:
Input crank (L2)
a 0.450
First coupler (L3)
b 0.990
Common rocker (O4B)
c 0.590
First ground link (O2O4)
d 1.000
Common rocker (O4C)
a' 0.590
Second coupler (CD)
b' 0.325
Output rocker (L6)
c' 0.325
Second ground link (O4O6) d' 0.419
Link 7 (L7)
e 0.938
Link 8 (L8)
f 0.572
Link 5 extension (DE)
p 0.823
Angle DCE
δ 7.0 deg
Angle BO4C
α 128.6 deg
See Figure P3-36 and Mathcad file P0443.
1.
This eightbar can be analyzed as a fourbar (links 1, 2, 3, and 4) with its output (link 4) as the input to another fourbar (links 1, 4, 5, and 6). Since links 1 and 4 are common to both, we have an eightbar linkage with links 7 & 8 included. Start by analyzing the input fourbar.
2.
Define one revolution of the input crank: θ 0 deg 1 deg 360 deg
3.
Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the open circuit). d
K1
K2
a
K1 2.2222
2
d
K3
c
K2 1.6949
2
2
a b c d
2
2 a c
K3 1.0744
A θ cos θ K1 K2 cos θ K3
C θ K1 K2 1 cos θ K3
B θ 2 sin θ
θ θ 2 atan2 2 A θ B θ 4.
2 4 A θ Cθ
B θ
Use equations 4.8a and 4.10 to calculate 6 as a function of 2 (for the open circuit).
K' 1
d'
K' 2
a'
K' 1 0.7102
θ θ θ θ α
Input angle to second fourbar:
2
d'
K' 3
c'
K' 2 1.2892
2
2
2 a' c'
K' 3 1.3655
K'1 K'2 cosθθ K'3
A' θ cos θ θ
θ θ 2 atan2 2 A' θ B' θ 5.
K'3
C' θ K' 1 K' 2 1 cos θ θ
B' θ 2 sin θ θ
2 4 A' θ C'θ
B' θ
Use equations 4.11b, 4.12 and 4.13 to calculate 5 as a function of 2.
2
a' b' c' d'
DESIGN OF MACHINERY - 5th Ed.
K' 4
SOLUTION MANUAL 4-43-2 2
d'
K' 5
b'
K' 4 1.289
2
2
2
c' d' a' b' 2 a' b'
K' 5 1.365
K'1 K'4 cosθθ K'5
D' θ cos θ θ
E' θ 2 sin θ θ
K'5
F' θ K' 1 K' 4 1 cos θ θ
θ θ 2 atan2 2 D' θ E' θ 6.
2 4 D'θ F'θ
E' θ
Define a vector loop for links 1, 5, 6, 7, and 8 as shown below and write the vector loop equation.
Y C
4 R12
O4
5 O6
X D
R6 6
8
R8 RDE
F
R7 7
E
R12 + R6 +RDE + R7 - R8 = 0. Solving for R7 gives R7 = R8 - R12 - R6 - RDE. In this equation the only unknowns are 7 and 8. Following the method of Section 4.5, substitute the complex number notation for each position vector and separate the resulting equation into real and imaginary parts:
e cos θ = f cos θ D1 e sin θ = f sin θ D2 where
p cosθθ δ
D1 θ d' c' cos θ θ
p sinθθ δ
D2 θ c' sin θ θ 7.
Solve these equations in the manner of equations 4.11 and 4.12 using the identities of equations 4.9 gives:
DESIGN OF MACHINERY - 5th Ed.
D3 θ
f
2
SOLUTION MANUAL 4-43-3
2 D2θ2 e2
D1 θ
2 f
A' θ D1 θ D3 θ
B' θ 2 D2 θ
C' θ D1 θ D3 θ
θ B' B'2 4 A' C' = 2 A' 2
tan
θ θ 2 atan2 2 A' θ B' θ 8.
2 4 A' θ C'θ
B' θ
Plot the angular position of link 8 as a function of the angle of input link 2. If 81 is greater than 360 deg, subtra 2 from it.
θ θ if θ
π
θ θ 0 θ θ 2 π θ θ
4
Angular Position of Link 8 360 315 270 225
180
θ θ
135
deg 90 45 0 45 90
0
45
90
135
180
225
270
θ deg
The graph shows that link 8 rotates 360 deg between 2 = 19 deg and 2 = 209 deg. θ( 19 deg) 42 deg θ( 209 deg) θ( 19 deg) 360.0 deg
θ( 209 deg) 2 π 42 deg
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-44-1
PROBLEM 4-44 Statement:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the transmission angles at points B, C, D, E, and F of the linkage in Figure 3-36 as a function of the angle of input link 2.
Given:
Link lengths:
Solution:
Input crank (L2)
a 0.450
First coupler (L3)
b 0.990
Common rocker (O4B)
c 0.590
First ground link (O2O4)
d 1.000
Common rocker (O4C)
a' 0.590
Second coupler (CD)
b' 0.325
Output rocker (L6)
c' 0.325
Second ground link (O4O6) d' 0.419
Link 7 (L7)
e 0.938
Link 8 (L8)
f 0.572
Link 5 extension (DE)
p 0.823
Angle DCE
δ 7.0 deg
Angle BO4C
α 128.6 deg
See Figure P3-36 and Mathcad file P0444.
1.
This eightbar can be analyzed as a fourbar (links 1, 2, 3, and 4) with its output (link 4) as the input to another fourbar (links 1, 4, 5, and 6). Since links 1 and 4 are common to both, we have an eightbar linkage with links 7 & 8 included. Start by analyzing the input fourbar.
2.
Define one revolution of the input crank: θ 0 deg 1 deg 360 deg
3.
Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the open circuit). d
K1
K2
a
K1 2.2222
2
d
K3
c
K2 1.6949
2
2
a b c d 2 a c
K3 1.0744
A θ cos θ K1 K2 cos θ K3
C θ K1 K2 1 cos θ K3
B θ 2 sin θ
θ θ 2 atan2 2 A θ B θ 4.
2 4 A θ Cθ
B θ
Use equations 4.12 and 4.13 to calculate 3 as a function of 2 (for the open circuit). K4
2
d
K5
b
K4 1.010
2
2
c d a b
2
2 a b
K5 2.059
D θ cos θ K1 K4 cos θ K5
F θ K1 K4 1 cos θ K5
θ θ 2 atan2 2 D θ E θ 5.
2 4 Dθ F θ
E θ
Use equations 4.8a and 4.10 to calculate 6 as a function of 2 (for the open circuit). Input angle to second fourbar:
E θ 2 sin θ
θ θ θ θ α
2
DESIGN OF MACHINERY - 5th Ed.
K' 1
SOLUTION MANUAL 4-44-2
d'
K' 2
a'
K' 1 0.7102
2
d'
K' 3
c'
K' 2 1.2892
2
2
2
a' b' c' d' 2 a' c'
K' 3 1.3655
K'1 K'2 cosθθ K'3
A' θ cos θ θ
θ θ 2 atan2 2 A' θ B' θ 6.
K'3
C' θ K' 1 K' 2 1 cos θ θ
B' θ 2 sin θ θ
2 4 A' θ C'θ
B' θ
Use equations 4.11b, 4.12 and 4.13 to calculate 5 as a function of 2. K' 4
2
d' b'
K' 4 1.289
2
2
2
c' d' a' b'
K' 5
2 a' b'
K' 5 1.365
K'1 K'4 cosθθ K'5
D' θ cos θ θ
E' θ 2 sin θ θ
K'5
F' θ K' 1 K' 4 1 cos θ θ
θ θ 2 atan2 2 D' θ E' θ
2 4 D'θ F'θ
E' θ
θ θ if θ θ 2 π θ θ 2 π θ θ
θ θ if 105 deg θ 322 deg θ θ 0 θ θ 2 π θ θ 7.
Define a vector loop for links 1, 5, 6, 7, and 8 as shown on the next page and write the vector loop equation. R12 + R6 +RDE + R7 - R8 = 0. Solving for R7 gives R7 = R8 - R12 - R6 - RDE. In this equation the only unknowns are 7 and 8. Following the method of Section 4.5, substitute the complex number notation for each position vector and separate the resulting equation into real and imaginary parts:
e cos θ = f cos θ D1 e sin θ = f sin θ D2 where
p cosθθ δ
D1 θ d' c' cos θ θ
p sinθθ δ
D2 θ c' sin θ θ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-44-3
Y C
4 R12
O4
5 X
O6
D
R6 6
R8
8
RDE
R7
F
E
7 8.
Solve these equations in the manner of equations 4.10 using the identities of equations 4.9 gives:
D3 θ
f
2
2 D2θ2 e2
D1 θ
2 f
A' θ D1 θ D3 θ
B' θ 2 D2 θ
C' θ D1 θ D3 θ
θ B' B'2 4 A' C' = 2 A' 2
tan
θ θ 2 atan2 2 A' θ B' θ
θ θ if θ 9.
2 4 A' θ C'θ
B' θ
π
θ θ 0 θ θ 2 π θ θ 4
Similarly, solve these equations in the manner of equations 4.11, 4.12 and 4.13 using the identities of equations 4.9 gives:
D4 θ
f
2
2 D2θ2 e2
D1 θ
2 e
D' θ D1 θ D4 θ
θ E' E'2 4 D' F' = 2 D' 2
tan
E' θ 2 D2 θ
F' θ D1 θ D4 θ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-44-4
2 4 D'θ F'θ
θ θ 2 atan2 2 D' θ E' θ
E' θ
10. Calculate (using equations 4.32) and plot the transmission angle at B.
θtransB1 θ θ θ θ θ
θtransB θ if θtransB1 θ
π 2
π θtransB1 θ θtransB1 θ
Transmission Angle at B 90 80 70
θtransB θ 60 deg
50 40 30 20
0
45
90
135
180
225
270
315
360
θ deg
11. Calculate (using equations 4.32) and plot the transmission angle at C.
θtransC1 θ θ θ θ θ θtransC2 θ if
π
2
θtransC1 θ 2 π π θtransC1 θ θtransC1 θ
Transmission Angle at C 150
100
θtransC θ deg 50
0
0
45
90
135
180 θ deg
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-44-5
12. Calculate (using equations 4.32) and plot the transmission angle at D.
θtransD1 θ θ θ θ θ π θtransD2 θ if
π
2
θtransD1 θ π π θtransD1 θ θtransD1 θ
Transmission Angle at D 90 80 70 60
θtransD θ 50 deg
40 30 20 10 0
0
45
90
135
180
225
270
315
360
315
360
θ deg
13. Calculate (using equations 4.32) and plot the transmission angle at E.
θtransE1 θ θ θ θ θ
θtransE θ if θtransE1 θ
π 2
π θtransE1 θ θtransE1 θ
Transmission Angle at E 90 80 70
θtransE θ
60
deg 50 40 30
0
45
90
135
180 θ deg
225
270
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-45-1
PROBLEM 4-45 Statement:
Model the linkage shown in Figure 3-37a in Fourbar. Export the coupler curve coordinates to Excel and calculate the error function versus a true circle.
Given:
Link lengths: Input (O2A)
a 0.136
Coupler (AB)
b 1.000
Rocker (O4B)
c 1.000
Ground link
d 1.414
Coupler point data: Solution: 1.
p 2.000
δ 0 deg
See Figure 3-37a and Mathcad file P0445.
Model the linkage in Fourbar.
2.
Write coupler point coordinates to a data file.
3.
Import the data file into Excell and add columns for the true circle coordinates and radius. Analyze the coupler point radius to determine its mean, maximum deviation from mean, and average absolute deviation from its mean (See next two pages, note that the last four columns were added to the imported data).
DESIGN OF MACHINERY - 5th Ed.
FOURBAR P0445 Tom Cook
SOLUTION MANUAL 4-45-2
Design #
Selected Linkage Parameters a = 0.136 b = 1.000 Angle Step Deg
2
8/9/2006
c = 1.000
d = 1.414
Coupler Pt Coupler Pt Coupler Pt Coupler Pt True Circ True Circ True Circ Coupler Pt X Y Mag Ang X Y R R in in in in in in in in 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105 110 115 120 125 130 135 140 145 150 155 160 165 170 175 180 185 190 195 200 205 210
1.414 1.4283 1.4423 1.4561 1.4693 1.4819 1.4937 1.5046 1.5145 1.5233 1.5309 1.5373 1.5425 1.5465 1.5493 1.5508 1.5512 1.5505 1.5488 1.546 1.5424 1.5379 1.5326 1.5266 1.52 1.5128 1.5052 1.4971 1.4887 1.4799 1.471 1.4618 1.4524 1.4429 1.4333 1.4237 1.414 1.4043 1.3947 1.3851 1.3756 1.3662 1.357
1.5384 1.5379 1.5363 1.5336 1.5299 1.5251 1.5195 1.5129 1.5055 1.4974 1.4885 1.479 1.469 1.4585 1.4475 1.4363 1.4248 1.4131 1.4014 1.3897 1.378 1.3665 1.3552 1.3442 1.3336 1.3235 1.314 1.3051 1.2969 1.2895 1.2829 1.2772 1.2725 1.2688 1.2661 1.2645 1.2639 1.2645 1.2661 1.2688 1.2725 1.2772 1.2829
2.0895 2.0988 2.1072 2.1147 2.1212 2.1265 2.1307 2.1337 2.1355 2.136 2.1353 2.1333 2.1301 2.1258 2.1203 2.1138 2.1063 2.0979 2.0887 2.0788 2.0683 2.0572 2.0458 2.0341 2.0221 2.0101 1.998 1.9861 1.9744 1.9629 1.9518 1.9411 1.931 1.9214 1.9124 1.9041 1.8965 1.8897 1.8836 1.8784 1.8739 1.8703 1.8674
47.413 47.1164 46.8058 46.4846 46.1562 45.8237 45.4901 45.1581 44.8303 44.5088 44.1957 43.8925 43.6007 43.3213 43.0555 42.8038 42.567 42.3456 42.1401 41.9507 41.7781 41.6225 41.4846 41.3647 41.2636 41.1819 41.1204 41.0799 41.0612 41.0654 41.0931 41.1453 41.2227 41.3257 41.455 41.6105 41.7924 42.0001 42.2329 42.49 42.7699 43.0708 43.3907
1.4140 1.4021 1.3904 1.3788 1.3675 1.3565 1.3460 1.3360 1.3266 1.3178 1.3098 1.3026 1.2962 1.2907 1.2862 1.2826 1.2801 1.2785 1.2780 1.2785 1.2801 1.2826 1.2862 1.2907 1.2962 1.3026 1.3098 1.3178 1.3266 1.3360 1.3460 1.3565 1.3675 1.3788 1.3904 1.4021 1.4140 1.4259 1.4376 1.4492 1.4605 1.4715 1.4820
1.5500 1.5495 1.5479 1.5454 1.5418 1.5373 1.5318 1.5254 1.5182 1.5102 1.5014 1.4920 1.4820 1.4715 1.4605 1.4492 1.4376 1.4259 1.4140 1.4021 1.3904 1.3788 1.3675 1.3565 1.3460 1.3360 1.3266 1.3178 1.3098 1.3026 1.2962 1.2907 1.2862 1.2826 1.2801 1.2785 1.2780 1.2785 1.2801 1.2826 1.2862 1.2907 1.2962
0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360
0.1244 0.1247 0.1255 0.1268 0.1284 0.1302 0.1322 0.1341 0.1359 0.1375 0.1386 0.1394 0.1398 0.1398 0.1394 0.1386 0.1376 0.1365 0.1354 0.1342 0.1334 0.1327 0.1324 0.1325 0.1330 0.1340 0.1353 0.1370 0.1389 0.1409 0.1430 0.1449 0.1466 0.1480 0.1492 0.1498 0.1501 0.1498 0.1492 0.1480 0.1466 0.1449 0.1430
DESIGN OF MACHINERY - 5th Ed.
215 220 225 230 235 240 245 250 255 260 265 270 275 280 285 290 295 300 305 310 315 320 325 330 335 340 345 350 355 360
1.3481 1.3393 1.3309 1.3228 1.3152 1.308 1.3014 1.2954 1.2901 1.2856 1.282 1.2792 1.2775 1.2768 1.2772 1.2787 1.2815 1.2855 1.2907 1.2971 1.3047 1.3135 1.3234 1.3343 1.3461 1.3587 1.3719 1.3857 1.3997 1.414
1.2895 1.2969 1.3051 1.314 1.3235 1.3336 1.3442 1.3552 1.3665 1.378 1.3897 1.4014 1.4131 1.4248 1.4363 1.4475 1.4585 1.469 1.479 1.4885 1.4974 1.5055 1.5129 1.5195 1.5251 1.5299 1.5336 1.5363 1.5379 1.5384
SOLUTION MANUAL 4-45-3
1.8655 1.8643 1.864 1.8645 1.8659 1.868 1.871 1.8747 1.8793 1.8846 1.8907 1.8975 1.905 1.9132 1.922 1.9314 1.9415 1.952 1.963 1.9744 1.9861 1.998 2.0101 2.0222 2.0342 2.0461 2.0577 2.0688 2.0795 2.0895
43.7272 44.0777 44.439 44.8081 45.1815 45.5556 45.9269 46.2915 46.6458 46.986 47.3085 47.61 47.8871 48.1368 48.3563 48.5433 48.6956 48.8117 48.8904 48.9308 48.9328 48.8966 48.823 48.713 48.5685 48.3915 48.1846 47.9504 47.6921 47.413
1.4920 1.5014 1.5102 1.5182 1.5254 1.5318 1.5373 1.5418 1.5454 1.5479 1.5495 1.5500 1.5495 1.5479 1.5454 1.5418 1.5373 1.5318 1.5254 1.5182 1.5102 1.5014 1.4920 1.4820 1.4715 1.4605 1.4492 1.4376 1.4259 1.4140
1.3026 1.3098 1.3178 1.3266 1.3360 1.3460 1.3565 1.3675 1.3788 1.3904 1.4021 1.4140 1.4259 1.4376 1.4492 1.4605 1.4715 1.4820 1.4920 1.5014 1.5102 1.5182 1.5254 1.5318 1.5373 1.5418 1.5454 1.5479 1.5495 1.5500
The mean value of the coupler point radius is
0.1366
The maximum deviation from the mean is
0.0135
The average absolute deviation from the mean is
0.005127
0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360
0.1409 0.1389 0.1370 0.1353 0.1340 0.1330 0.1325 0.1324 0.1327 0.1334 0.1342 0.1354 0.1365 0.1376 0.1386 0.1394 0.1398 0.1398 0.1394 0.1386 0.1375 0.1359 0.1341 0.1322 0.1302 0.1284 0.1268 0.1255 0.1247 0.1244
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-46-1
PROBLEM 4-46 Statement:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the path of point P in Figure 3-37a as a function of the angle of input link 2. Also plot the variation (error) in the path of point P versus that of point A.
Given:
Link lengths: Input (O2A)
a 0.136
Coupler (AB)
b 1.000
Rocker (O4B)
c 1.000
Ground link
d 1.414
p 2.000
Coupler point data: Solution: 1.
δ 0 deg
See Figure 3-37a and Mathcad file P0446.
Check the Grashof condition of the linkage. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ
return "non-Grashof" otherwise crank rocker Condition( a b c d ) "Grashof" 1.
Define one cycle of the input crank: θ 0 deg 1 deg 360 deg
2.
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. d
K1
K4
a
K1 10.3971
2
d
K5
b
K4 1.4140
2
2
c d a b
2
2 a b
K5 7.4187
D θ cos θ K1 K4 cos θ K5
3.
F θ K1 K4 1 cos θ K5
E θ 2 sin θ
Use equation 4.13 to find values of 3 for the crossed circuit.
θ θ 2 atan2 2 D θ E θ 4.
2 4 Dθ F θ
E θ
Use equations 4.31 to define the x- and y-components of the vector RP. RP RA RPA
RA a cos θ j sin θ
RPA p cos θ δ j sin θ δ
RPx θ a cos θ p cos θ θ δ 5.
RPy θ a sin θ p sin θ θ δ
Plot the coordinates of the coupler point in the local xy coordinate system.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-46-2
COUPLER POINT PATH
Coupler Point Coordinate - y
1.2
1.414
1.3
1.4
1.414
1.5
1.6 1.2
1.3
1.4
1.5
1.6
Coupler Point Coordinate - x
6.
Replot, transforming the coupler path to 0,0 and plot the path of point A.
XA θ a cos θ
YA θ a sin θ
PATHS OF POINTS A &P 0.2
0.1
0
0.1
0.2 0.2
0.1 Point P Point A
0
0.1
0.2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-47-1
PROBLEM 4-47 Statement:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the transmission angle at point B of the linkage in Figure 3-37a as a function of the angle of input link 2.
Given:
Link lengths: Input (O2A)
a 0.136
Coupler (AB)
b 1.000
Rocker (O4B)
c 1.000
Ground link
d 1.414
p 2.000
Coupler point data: Solution: 1.
δ 0 deg
See Figure 3-37a and Mathcad file P0447.
Check the Grashof condition of the linkage. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( a b c d ) "Grashof" 1.
crank rocker
Define one cycle of the input crank: θ 0 deg 1 deg 360 deg
2.
Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the crossed circuit). K1
d
K2
a
K1 10.3971
2
d
K3
c
K2 1.4140
2
2 a c
K3 7.4187
A θ cos θ K1 K2 cos θ K3
C θ K1 K2 1 cos θ K3
B θ 2 sin θ
θ θ 2 atan2 2 A θ B θ 3.
2 4 A θ Cθ
B θ
Use equations 4.12 and 4.13 to calculate 3 as a function of 2 (for the crossed circuit). K4
2
d
K5
b
K4 1.414
2
2
c d a b
2
2 a b
K5 7.419
D θ cos θ K1 K4 cos θ K5
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 θ θ 2 atan2 2 D θ E θ
2 4 Dθ F θ
E θ
2
a b c d
2
DESIGN OF MACHINERY - 5th Ed.
4.
SOLUTION MANUAL 4-47-2
Calculate (using equations 4.32) and plot the transmission angle at B.
θtransB1 θ θ θ θ θ
θtransB θ if θtransB1 θ
π 2
π θtransB1 θ θtransB1 θ
Transmission Angle at B 90
85
θtransB θ deg 80
75
0
45
90
135
180 θ deg
225
270
315
360
DESIGN OF MACHINERY - 5thEd.
SOLUTION MANUAL 4-48-1
PROBLEM 4-48 Statement:
Figure 3-29f shows Evan's approximate straight-line linkage #1. Determine the range of motion of link 2 for which the point P varies no more than 0.0025 from the straight line X = 1.690 (assuming that O2 is the origin of a global coordinate frame whose positive X axis is rotated 60 deg from O2O4).
Given:
Link lengths: Input (O2A)
a 1.000
Coupler (AB)
b 1.600
Rocker (O4B)
c 1.039
Ground link
d 1.200
p 2.690
Coupler point data:
α 60 deg
Coordinate rotation angle: Solution: 1.
δ 0 deg
See Figure 3-29f and Mathcad file P0448.
Check the Grashof condition of the linkage and determine its Baker classification. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( a b c d ) "non-Grashof" Since b (link 3) is the longest link and the linkage is non-Grashof, this is a Class 3 triple rocker. Using Figure 3-1a as a guide, determine the limiting values of 2 at the toggle positions. For links 2 and 3 colinear:
( b a) 2 d2 c2 π 2 d ( b a)
θ acos
θ 240 deg
For links 3 and 4 colinear:
a2 d 2 ( b c) 2 2 d a
θ acos 1.
θ 27.683 deg
θ θ
Define one cycle of the input crank (driving through the links 2-3 toggle position): θ θ θ 1 deg 360 deg θ
2.
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K1
d
K4
a
K1 1.2000
2
d
K5
b
K4 0.7500
2
2 a b
K5 1.2251
D θ cos θ K1 K4 cos θ K5
E θ 2 sin θ 3.
F θ K1 K4 1 cos θ K5
Use equation 4.13 to find values of 3 for the open circuit.
2
c d a b
2
DESIGN OF MACHINERY - 5thEd.
SOLUTION MANUAL 4-48-2
θ θ 2 atan2 2 D θ E θ 4.
2 4 Dθ F θ
E θ
Use equations 4.31 to define the x-components of the vector RP. RP RA RPA
RPA p cos θ δ j sin θ δ RA a cos θ j sin θ
RPx θ a cos θ p cos θ θ δ 5.
Plot the X coordinate of the coupler point in the global X,Y coordinate system using equations 4.0b to rotate the local coordinates to a global frame.
PX θ RPx θ cos( α) RPy θ sin( α) X COORDINATE 1.7 1.698 Coupler Point Coordinate - X
1.696 1.694 1.692 1.69 1.688 1.686 1.684 1.682 1.68 180
210
240
270
300
Input Angle - Theta2
6.
RPy θ a sin θ p sin θ θ δ
Using the graph for guess values, solve by trial and error to find 2 for X = 1.6900 +/-0.0025. PX ( 201.525 deg) 1.69250
θ2min 201.525 deg
PX ( 273.450 deg) 1.69250
θ2max 273.450 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-49-1
PROBLEM 4-49 Statement:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the path of point P in Figure 3-37b as a function of the angle of input link 2.
Given:
Link lengths:
Solution: 1.
Input crank (L2)
a 0.50
First coupler (AB)
b 1.00
Rocker 4 (O4B)
c 1.00
Rocker 5 (L5)
c' 1.00
Ground link (O2O4)
d 0.75
Second coupler 6 (CD)
b' 1.00
Coupler point (DP)
p 1.00
Distance to OP (O2OP)
d' 1.50
See Figure 3-37b and Mathcad file P0449.
Links 4, 5, BC, and CD form a parallelogram whose opposite sides remain parallel throughout the motion of the fourbar 1, 2, AB, 4. Define a position vector whose tail is at point D and whose tip is at point P and another whose tail is at O4 and whose tip is at point D. Then, since R5 = RAB and RDP = -R4, the position vector from O2 to P is P = R1 + RAB - R4. Separating this vector equation into real and imaginary parts gives the equations for the X and Y coordinates of the coupler point P.
XP = d b cos θ c cos θ
YP = b sin θ c sin θ
2.
Define one revolution of the input crank: θ 0 deg 1 deg 360 deg
3.
Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the open circuit). K1
d a
K1 1.5000
2
d
K2
K3
c
K2 0.7500
2
2
a b c d
2
2 a c
K3 0.8125
A θ cos θ K1 K2 cos θ K3
C θ K1 K2 1 cos θ K3
B θ 2 sin θ
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 4.
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
2 a b
D θ cos θ K1 K4 cos θ K5
K4 0.7500
K5 0.8125
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 5.
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ 6.
2 4 Dθ F θ
E θ
Define a local xy coordinate system with origin at OP and with the positive x axis to the right. The coordinates of P are transformed to xP = XP - d', yP = YP.
c cosθθ d'
xP θ d b cos θ θ
c sinθθ
yP θ b sin θ θ
DESIGN OF MACHINERY - 5th Ed.
7.
SOLUTION MANUAL 4-49-2
Plot the path of P as a function of the angle of link 2.
Path of Coupler Point P About OP 0.6 0.5 0.4 0.3 0.2 0.1
yP θ
0
0.1 0.2 0.3 0.4 0.5 0.6 0.6 0.5
0.4 0.3
0.2 0.1
0
xP θ
0.1
0.2
0.3
0.4
0.5
0.6
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-50-1
PROBLEM 4-50 Statement:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the transmission angles at points B, C, and D of the linkage in Figure 3-37b as a function of the angle of input link 2.
Given:
Link lengths:
Solution:
Input crank (L2)
a 0.50
First coupler (AB)
b 1.00
Rocker 4 (O4B)
c 1.00
Rocker 5 (L5)
c' 1.00
Ground link (O2O4)
d 0.75
Second coupler 6 (CD)
b' 1.00
Coupler point (DP)
p 1.00
Distance to OP (O2OP)
d' 1.50
See Figure 3-37b and Mathcad file P0450.
1.
Links 4, 5, BC, and CD form a parallelogram whose opposite sides remain parallel throughout the motion of the fourbar 1, 2, AB, 4. Therefore, the transmission angles at points B and D will be the same and the transmission angle at point C will be the complement of the angle at B.
2.
Define one revolution of the input crank: θ 0 deg 1 deg 360 deg
3.
Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the crossed circuit). K1
d a
K1 1.5000
2
d
K2
K3
c
K2 0.7500
2
2
a b c d
2
2 a c
K3 0.8125
A θ cos θ K1 K2 cos θ K3
C θ K1 K2 1 cos θ K3
B θ 2 sin θ
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 4.
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
K4 0.7500
2 a b
D θ cos θ K1 K4 cos θ K5
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 5.
Use equation 4.13 to find values of 3 for the crossed circuit.
θ θ 2 atan2 2 D θ E θ 6.
2 4 Dθ F θ
E θ
Calculate (using equations 4.32) and plot the transmission angles at B and D.
θtransB1 θ θ θ θ θ
θtransB θ if θtransB1 θ
π 2
K5 0.8125
π θtransB1 θ θtransB1 θ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-50-2
Transmission Angles at B and D 80
60
θtransB θ
40
deg
20
0
0
45
90
135
180
225
270
315
360
θ deg
6.
Calculate and plot the transmission angle at C.
θtransC1 θ 180 deg θtransB θ
θtransC θ if θtransC1 θ
π 2
π θtransC1 θ θtransC1 θ
Transmission Angle at C 80
60
θtransC θ
40
deg
20
0
0
45
90
135
180 θ deg
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-51-1
PROBLEM 4-51 Statement:
Figure 3-29g shows Evan's approximate straight-line linkage #2. Determine the range of motion of link 2 for which the point P varies no more than 0.005 from the straight line X = -0.500 (assuming that O2 is the origin of a global coordinate frame whose positive X axis is rotated 30 deg from O2O4).
Given:
Link lengths: Input (O2A)
a 1.000
Coupler (AB)
b 1.200
Rocker (O4B)
c 1.167
Ground link
d 2.305
p 1.50
Coupler point data:
α 30 deg
Coordinate rotation angle: Solution: 1.
δ 180 deg
See Figure 3-29g and Mathcad file P0451.
Check the Grashof condition of the linkage and determine its Baker classification. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( a b c d ) "non-Grashof" Since d (link 1) is the longest link and the linkage is non-Grashof, this is a Class 1 triple rocker. Using Figure 3-1a as a guide, determine the limiting values of 2 at the toggle positions. For links 3 and 4 colinear:
a2 d 2 ( b c) 2 2 d a
θ acos 1.
θ 81.136 deg
θ θ
Define one cycle of the input crank (driving through the links 2-3 toggle position): θ θ θ 0.5 deg θ
2.
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K1
d
K4
a
K1 2.3050
2
d
K5
b
K4 1.9208
2
2 a b
K5 2.6630
D θ cos θ K1 K4 cos θ K5
3.
F θ K1 K4 1 cos θ K5
E θ 2 sin θ
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ
2 4 Dθ F θ
E θ
2
c d a b
2
DESIGN OF MACHINERY - 5th Ed.
4.
SOLUTION MANUAL 4-51-2
Use equations 4.31 to define the x and y-components of the vector RP. RP RA RPA
RPA p cos θ δ j sin θ δ RA a cos θ j sin θ
RPx θ a cos θ p cos θ θ δ 5.
Plot the X coordinate of the coupler point in the global X,Y coordinate system using equations 4.0b to rotate the local coordinates to a global frame.
PX θ RPx θ cos( α) RPy θ sin( α) X COORDINATE 0.48
Coupler Point Coordinate - X
0.485 0.49 0.495
PX θ
0.5 0.505 0.51 0.515 0.52
0
15
30
45
60
θ deg Input Angle - Theta2
6.
RPy θ a sin θ p sin θ θ δ
Using the graph for guess values, solve by trial and error to find 2 for X = -0.500 +/-0.005 PX ( 11.59 deg) 0.49500
θ2min 11.59 deg
PX ( 57.80 deg) 0.50500
θ2max 57.80 deg
75
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-52-1
PROBLEM 4-52 Statement:
For the linkage in Figure P4-16, what are the angles that link 2 makes with the positive X-axis when links 2 and 3 are in toggle positions?
Given:
Link lengths:
Solution: 1.
Input (O2A)
a 14
Rocker (O4B)
c 51.26
b 80
Coupler (AB)
O4 offset in XY coordinates:
O4X 47.5
Ground link:
d
2
O4X O4Y
O4Y 76 12 2
O4Y 64.000
d 79.701
See Figure P4-16 and Mathcad file P0452.
Check the Grashof condition of the linkage. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( a b c d ) "Grashof" 2.
Define the coordinate frame transformation angle:
O4Y O4X
δ π atan 3.
crank rocker
δ 126.582 deg
Calculate the angle of link 2 in the XY system when links 2 and 3 are in the overlapped toggle position.
( b a) 2 d2 c2 δ 2 ( b a) d
θ21XY acos 4.
θ21XY 86.765 deg
Calculate the angle of link 2 in the XY system when links 2 and 3 are in the extended toggle position.
( b a) 2 d2 c2 δ 2 ( b a) d
θ22XY acos
θ22XY 93.542 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-53-1
PROBLEM 4-53 Statement:
The coordinates of the point P1 on link 4 in Figure P4-16 are (114.68, 33.19) with respect to the xy coordinate system when link 2 is in the position shown. When link 2 is in another position the coordinates of P2 with respect to the xy system are (100.41, 43.78). Calculate the coordinates of P1 and P2 in the XY system for the two positions of link 2. What is the salient feature of the coordinates of P1 and P2 in the XY system?
Given:
Vertical and horizontal offsets from O2 to O4. O2O4X 47.5 in
O2O4Y 64 in
Coordinates of P1 and P2 in the local system
Solution: 1.
P1x 114.68 in
P1y 33.19 in
P2x 100.41 in
P2y 43.78 in
See Figure P4-16 and Mathcad file P0453.
Calculate the angle from the global X axis to the local x axis.
O2O4Y O2O4X
δ 180 deg atan 2.
3.
δ 126.582 deg
Use equations 4.0b to transform the given coordinates from the local to the global system. P1X P1x cos( δ) P1y sin( δ)
P1X 95.00 in
P1Y P1x sin( δ) P1y cos( δ)
P1Y 72.31 in
P2X P2x cos( δ) P2y sin( δ)
P2X 95.00 in
P2Y P2x sin( δ) P2y cos( δ)
P2Y 54.54 in
In the global XY system the X-coordinates are the same for each point, which indicates that the head on the end of the rocker beam 4 is designed such that its tangent is always parallel to the Y-axis.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-54-1
PROBLEM 4-54 Statement:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the angular position of link 4 with respect to the XY coordinate frame and the transmission angle at point B of the linkage in Figure P4-16 as a function of the angle of input link 2 with respect to the XY frame.
Given:
Link lengths:
Solution: 1.
Input (O2A)
a 14
Coupler (AB)
b 80
Rocker (O4B)
c 51.26
Ground link
d 79.70
See Figure P4-16 and Mathcad file P0454.
Check the Grashof condition of the linkage. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( a b c d ) "Grashof" 2.
Define the coordinate frame transformation angle: δ 90 deg atan
47.5
64
3.
crank rocker
δ 126.582 deg
Define one cycle of the input crank with respect to the XY frame:
θ2 θ2XY θ2XY δ
θ2XY 0 deg 1 deg 360 deg 4.
Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the crossed circuit). K1
d
K2
a
K1 5.6929
2
d
K3
c
K2 1.5548
2
2
a b c d
2
2 a c
K3 1.9339
A θ2XY cos θ2 θ2XY K1 K2 cos θ2 θ2XY K3 B θ2XY 2 sin θ2 θ2XY
C θ2XY K1 K2 1 cos θ2 θ2XY K3
θ θ2XY 2 atan2 2 A θ2XY B θ2XY
2
θ θ2XY θ θ2XY δ 2 π 5.
Use equations 4.12 and 4.13 to calculate 3 as a function of 2 (for the crossed circuit). K4
d b
K4 0.996
2
K5
2
2
c d a b
K5 4.607
2 a b
B θ2XY 4 A θ2XY C θ2XY
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-54-2
D θ2XY cos θ2 θ2XY K1 K4 cos θ2 θ2XY K5 E θ2XY 2 sin θ2 θ2XY F θ2XY K1 K4 1 cos θ2 θ2XY K5
θ θ2XY 2 atan2 2 D θ2XY E θ2XY 6.
E θ2XY 4 D θ2XY F θ2XY 2
Plot the angular position of link 4 as a function of the input angle of link 2 with respect to the XY frame.
Angular Position of Link 4 40 35 30 θ θ2XY 25 20 deg 15 10 5
0
0
45
90
135
180
225
270
315
360
θ2XY deg
7.
Calculate (using equations 4.32) and plot the transmission angle at B.
θtransB1 θ2XY θ θ2XY θ θ2XY
θtransB θ2XY if θtransB1 θ2XY
π 2
π θtransB1 θ2XY θtransB1 θ2XY
Transmission Angle at B 90 85 80
θtransB θ2XY
75 70
deg 65 60 55 50
0
45
90
135
180
θ2XY deg
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-55-1
PROBLEM 4-55 Statement:
For the linkage in Figure P4-17, calculate the maximum CW rotation of link 2 from the position shown, which is -20.60 deg with respect to the local xy system. What angles do link 3 and link 4 rotate through for that excursion of link 2?
Given:
Link lengths: Input (O2A)
a 9.17
Coupler (AB)
b 12.97
Rocker (O4B)
c 9.57
Ground link
d 7.49
Initial position of link 2: θ 26.00 deg 2 π (with respect to xy system) Solution: 1.
See Figure P4-17 and Mathcad file P0455.
Check the Grashof condition of the linkage. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( d b a c) "non-Grashof" 2.
Using equations 4.37, determine the crank angles (relative to the line O2O4) at which links 3 and 4 are in toggle 2
arg1
2
2
2 a d 2
arg2
2
a d b c
2
2
a d b c
2
2 a d
b c
arg1 0.936
a d b c
arg2 2.678
a d
θ acos arg1
θ 20.55 deg
The other toggle angle is the negative of this. θ θ 2 π 3.
θ 339.45 deg
Calculate the CW rotation of link 2 from the initial position to the toggle position. Δ θ θ
4.
Δ 313.45 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K1
d
K4
a
K1 0.8168
2
d
K5
b
K4 0.5775
2
2 a b
K5 0.9115
D θ cos θ K1 K4 cos θ K5
6.
F θ K1 K4 1 cos θ K5
E θ 2 sin θ
Use equation 4.13 to find values of 3 for the crossed circuit.
θ θ 2 atan2 2 D θ E θ
2 4 Dθ F θ
E θ
2
c d a b
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-55-2
Initial angular position of link 3:
θ θ 2 π 647.755 deg
Final angular position of link 3:
θ θ 0.001 deg 250.764 deg
θ θ 0.001 deg θ θ 2 π 7.
898.518 deg
Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the crossed circuit). K1
d
K2
a
K1 0.8168
2
d
K3
c
K2 0.7827
2
2 a c
K3 0.3621
A θ cos θ K1 K2 cos θ K3
C θ K1 K2 1 cos θ K3
B θ 2 sin θ
θ θ 2 atan2 2 A θ B θ
2 4 A θ Cθ
B θ
Initial angular position of link 4:
θ θ 2 π 659.462 deg
Final angular position of link 4:
θ θ 0.001 deg 250.615 deg
θ θ 0.001 deg θ θ 2 π
2
a b c d
910.077 deg
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-56-1
PROBLEM 4-56 Statement:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the path of point P in Figure P4-17 with respect to the XY coordinate system as a function of the angle of input link 2 with respect to the XY coordinate system.
Given:
Link lengths: Input (O2A)
a 9.174
Coupler (AB)
b 12.971
Rocker (O4B)
c 9.573
Ground link
d 7.487
p 15.00
Coupler point data: Solution: 1.
δ 0 deg
See Figure P4-17 and Mathcad file P0456.
Check the Grashof condition of the linkage. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( a b c d ) "non-Grashof" 2.
Using equations 4.37, determine the crank angles (relative to the line O2O4) at which links 3 and 4 are in toggle 2
arg1
2
2
2 a d 2
arg2
2
a d b c
2
2
a d b c
2
2 a d
b c
arg1 0.937
a d b c
arg2 2.679
a d
θ2toggle acos arg1
θ2toggle 20.501 deg
The other toggle angle is the negative of this. 3.
Define the coordinate transformation angle. Transformation angle:
4.
α atan
6.95
2.79
α 68.128 deg
Define one cycle of the input crank between limit positions: θ θ2toggle θ2toggle 1 deg 2 π θ2toggle
5.
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K1
d
K4
a
K1 0.8161
2
d
K5
b
K4 0.5772
D θ cos θ K1 K4 cos θ K5
2
2
c d a b 2 a b
K5 0.9110
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-56-2
6.
F θ K1 K4 1 cos θ K5
E θ 2 sin θ
Use equation 4.13 to find values of 3 for the crossed circuit.
θ θ 2 atan2 2 D θ E θ 7.
2 4 Dθ F θ
E θ
Use equations 4.31 to define the x- and y-components of the vector RP. RP RA RPA
RPA p cos θ δ j sin θ δ RA a cos θ j sin θ
RPx θ a cos θ p cos θ θ δ 8.
Transform these local xy coordinates to the global XY coordinate system using equations 4.0b.
RPX θ RPx θ cos α RPy θ sin α RPY θ RPx θ sin α RPy θ cos α Plot the coordinates of the coupler point in the global XY coordinate system. COUPLER POINT PATH 5
0
Coupler Point Coordinate - Y
9.
5
10
15
20 10
5
0
RPy θ a sin θ p sin θ θ δ
5
10
Coupler Point Coordinate - X
15
20
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-57-1
PROBLEM 4-57 Statement:
For the linkage in Figure P4-17, calculate the coordinates of the point P in the XY coordinate syste if its coordinates in the xy system are (2.71, 10.54).
Given:
Vertical and horizontal offsets from O2 to O4. O2O4X 2.790 in
O2O4Y 6.948 in
Coordinates of P in the local system Px 12.816 in Solution: 1.
See Figure P4-17 and Mathcad file P0457.
Calculate the angle from the global X axis to the local x axis.
O2O4Y O2O4X
δ atan 2.
Py 10.234 in
δ 68.122 deg
Use equations 4.0b to transform the given coordinates from the local to the global system. PX Px cos( δ) Py sin( δ)
PX 14.273 in
PY Px sin( δ) Py cos( δ)
PY 8.079 in
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-58-1
PROBLEM 4-58 Statement:
The elliptical trammel in Figure P4-18 must be driven by rotating link 3 in a full circle. Derive analytical expressions for the positions of points A, B, and a point C on link 3 midway between A and B as a function of 3 and the length AB of link 3. Use a vector loop equation. (Hint: Place the global origin off the mechanism, preferably below and to the left and use a total of 5 vectors.) Code your solution in an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the path of point C for one revolution of link 3.
Solution:
See Figure P4-18 and Mathcad file P0458.
1.
Establish the global XY system such that the coordinates of the intersection of the slot centerlines is at (d X,d Y ). Then, define position vectors R1X, R1Y , R2, R3, and R4 as shown below. Y
4 B
3 R3
3
C
R2 2
R4
A 1
R1Y
X R1X
2.
Write the vector loop equation: R1Y + R2 + R3 - R1X - R4 = 0 then substitute the complex number notation for each position vector. The equation then becomes:
π π j j θ 3 2 j ( 0) j ( 0) 2 dY e a e c e dX e b e = 0 j
3.
Substituting the Euler identity into this equation gives: d Y j a c cos θ3 j sin θ3 d X b j = 0
4.
Separate this equation into its real (x component) and imaginary (y component) parts, setting each equal to zero. a c cos θ3 d X = 0
5.
Solve for the two unknowns a and b in terms of the constants d X and d Y and the independent variable 3. Where (a,d Y ) and (d X,b) are the coordinates of points A and B, respectively, and c is the length of link 3. With no loss of generality, let d X = d Y = d. Then, a = d c cos θ3
6.
b = d c sin θ3
The coordinates of the point C are: CX = d 0.5 cos θ3
7.
d Y c sin θ3 b = 0
CY = d 0.5 c sin θ3
Using a local coordinate system whose origin is located at the intersection of the centerlines of the two slots and transforming the above functions to the local xy system:
DESIGN OF MACHINERY - 5th Ed.
8.
SOLUTION MANUAL 4-58-2
a x = c cos θ3
ay = 0
bx = 0
b y = c sin θ3
To plot the path of point C as a function of 3, let c 1 and define a range function for 3
θ3 0 deg 1 deg 360 deg Cx θ3 0.5 c cos θ3
Cy θ3 0.5 sin θ3 Path of Point C
1
0.5
Cy θ3
0
0.5
1 1
0.5
0
Cx θ3
0.5
1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-59-1
PROBLEM 4-59 Statement:
Calculate and plot the angular position of link 6 in Figure P4-19 as a function of the angle of input link 2.
Given:
Link lengths:
Solution: 1.
Input crank (L2)
a 1.75
First coupler (AB)
b 1.00
First rocker (O4B)
c 1.75
Ground link (O2O4)
d 1.00
Second input (BC)
e 1.00
Second coupler (L5)
f 1.75
Output rocker (L6)
g 1.00
Third coupler (BE)
h 1.75
Ternary link (O4C)
i 2.60
See Figure P4-19 and Mathcad file P0459.
Because the linkage is symmetrical and composed of two parallelograms the analysis can be done with simple trigonometry.
39.582°
C B
5
D
3
3
A
2
4
6 E
O4
2.
1
O2
Calculate the fixed angle that line BC makes with the extension of line O4B using the law of cosines.
c2 e2 i 2 2 c e
δ π acos
δ 39.582 deg
3.
Define one revolution of the input crank: θ 0 deg 1 deg 360 deg
4.
Because links 1, 2, 3, and 4 are a parallelogram, link 4 will have the same angle as link 2 and AB will always be parallel to O2O4. And because links BC, 5, 6, and BE are also a parallelogram, link 6 will have the same angle as link BC. Thus,
θ θ θ δ 5.
Plot the angular position of link 6 as a function of the angle of input link 2 (see next page).
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-59-2
Angular Position of Link 6 400 360 320 280 240
θ θ
200
deg 160 120 80 40 0
0
45
90
135
180 θ deg
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-60a-1
PROBLEM 4-60a The link lengths, offset, and value of d for some fourbar slider-crank linkages are defined in Table P4-5. The linkage configuration and terminology are shown in Figure P4-2. For row a, draw the linkage to scale and graphically find all possible solutions (both open and crossed) for angles 2 and θ3.
Statement:
Given:
Link 2
a 1.4 in
Link 3
b 4 in
Offset
c 1 in
Slider position
d 2.5 in
See Figure P4-2, Table P4-5, and Mathcad file P0460a.
Solution: 1.
Lay out an xy-axis system. Its origin will be the link 2 pivot, O2.
2.
Draw a circle centered at the origin with radius equal to a at some convenient scale.
3.
Draw construction lines to define the point (d,c).
4.
From the point (d,c) draw an arc with radius equal to b.
5.
The two intersections of the circle and arc are the two solutions to the position analysis problem, crossed and open. If the circle and line don't intersect, there is no solution.
6.
Draw links 2 and 3 in their two possible positions (shown as solid for branch 1 and dashed for branch 2 in the figure) and measure the angles θ2 and 3 for each branch. From the solution below, Branch 1:
θ21 176.041 deg
θ31 ( 180 13.052) deg
θ31 193.052 deg
Branch 2:
θ22 132.439 deg
θ32 ( 180 30.551) deg
θ32 210.551 deg
13.052° 30.551°
176.041° 000 b = 4.
c = 1.000"
a = 1.400
132.439°
d = 2.500"
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-61a-1
PROBLEM 4-61a Statement:
Given:
Solution: 1.
The link lengths, offset, and value of d for some fourbar slider-crank linkages are defined in Table P4-5. The linkage configuration and terminology are shown in Figure P4-2. For row a, using the vector loop method, find all possible solutions (both open and crossed) for angles 2 and θ3. Link 2
a 1.4 in
Link 3
Offset
c 1 in
Slider position
d 2.5 in
See Figure P4-2, Table P4-5, and Mathcad file P0461a.
Determine both values of 2 using equations 4.20 and 4.21. 2
2
2
K1 a b c d
2
2
K1 6.790 in
2
K2 2 a c
K2 2.800 in
K3 2 a d
K3 7.000 in
A K1 K3
A 0.210 in
B 2 K2
B 5.600 in
C K1 K3
C 13.790 in
2
2 2 2
2 atan2 2 A B
θ21 2 atan2 2 A B θ22 2.
b 4 in
2 B 4 A C 2
B 4 A C
θ21 176.041 deg θ22 132.439 deg
Determine both values of 3 using equation 4.16a or 4.17.
β ( a b c d α)
θ asin
a sin( α) c
b
d 1 a cos( α) b cos( θ ) return θ if d 1 = d asin
a sin( α) c b
π otherwise
θ31 β a b c d θ21
θ31 193.052 deg
θ32 β a b c d θ22
θ32 210.551 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-61b-1
PROBLEM 4-61b Statement:
Given:
Solution: 1.
The link lengths, offset, and value of d for some fourbar slider-crank linkages are defined in Table P4-5. The linkage configuration and terminology are shown in Figure P4-2. For row b, using the vector loop method, find all possible solutions (both open and crossed) for angles 2 and θ3. Link 2
a 2 in
Link 3
Offset
c 3 in
Slider position
d 5 in
See Figure P4-2, Table P4-5, and Mathcad file P0461b.
Determine both values of 2 using equations 4.20 and 4.21. 2
2
2
K1 a b c d
2
2
K1 2.000 in
2
K2 2 a c
K2 12.000 in
K3 2 a d
K3 20.000 in
A K1 K3
A 22.000 in
B 2 K2
B 24.000 in
C K1 K3
C 18.000 in
2
2
2 2
2 atan2 2 A B
θ21 2 atan2 2 A B θ22 2.
b 6 in
2 B 4 A C 2
B 4 A C
θ21 54.117 deg θ22 116.045 deg
Determine both values of 3 using equation 4.16a or 4.17.
β ( a b c d α)
θ asin
a sin( α) c
b
d 1 a cos( α) b cos( θ ) return θ if d 1 = d asin
a sin( α) c b
π otherwise
θ31 β a b c d θ21
θ31 129.640 deg
θ32 β a b c d θ22
θ32 168.433 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-61c-1
PROBLEM 4-61c Statement:
Given:
Solution: 1.
The link lengths, offset, and value of d for some fourbar slider-crank linkages are defined in Table P4-5. The linkage configuration and terminology are shown in Figure P4-2. For row c, using the vector loop method, find all possible solutions (both open and crossed) for angles 2 and θ3. Link 2
a 3 in
Link 3
Offset
c 2 in
Slider position
d 8 in
See Figure P4-2, Table P4-5, and Mathcad file P0461c.
Determine both values of 2 using equations 4.20 and 4.21. 2
2
2
K1 a b c d
2
2
K1 13.000 in
2
K2 2 a c
K2 12.000 in
K3 2 a d
K3 48.000 in
A K1 K3
A 61.000 in
B 2 K2
B 24.000 in
C K1 K3
C 35.000 in
2
2 2 2
2 atan2 2 A B
θ21 2 atan2 2 A B θ22 2.
b 8 in
2 B 4 A C 2
B 4 A C
θ21 88.803 deg θ22 60.731 deg
Determine both values of 3 using equation 4.16a or 4.17.
β ( a b c d α)
θ asin
a sin( α) c
b
d 1 a cos( α) b cos( θ ) return θ if d 1 = d asin
a sin( α) c b
π otherwise
θ31 β a b c d θ21
θ31 172.824 deg
θ32 β a b c d θ22
θ32 215.249 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-61d-1
PROBLEM 4-61d Statement:
Given:
Solution: 1.
The link lengths, offset, and value of d for some fourbar slider-crank linkages are defined in Table P4-5. The linkage configuration and terminology are shown in Figure P4-2. For row d, using the vector loop method, find all possible solutions (both open and crossed) for angles 2 and θ3. Link 2
a 3.5 in
Link 3
Offset
c 1 in
Slider position
d 8 in
See Figure P4-2, Table P4-5, and Mathcad file P0461d.
Determine both values of 2 using equations 4.20 and 4.21. 2
2
2
K1 a b c d
2
2
K1 22.750 in 2
K2 2 a c
K2 7.000 in
K3 2 a d
K3 56.000 in
A K1 K3
A 78.750 in
B 2 K2
B 14.000 in
C K1 K3
C 33.250 in
2 2
2
2
2 atan2 2 A B
θ21 2 atan2 2 A B θ22 2.
b 10 in
2 B 4 A C 2
B 4 A C
θ21 286.648 deg θ22 300.898 deg
Determine both values of 3 using equation 4.16a or 4.17.
β ( a b c d α)
θ asin
a sin( α) c
b
d 1 a cos( α) b cos( θ ) return θ if d 1 = d asin
a sin( α) c b
π otherwise
θ31 β a b c d θ21
θ31 25.806 deg
θ32 β a b c d θ22
θ32 11.556 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-61e-1
PROBLEM 4-61e Statement:
Given:
Solution: 1.
The link lengths, offset, and value of d for some fourbar slider-crank linkages are defined in Table P4-5. The linkage configuration and terminology are shown in Figure P4-2. For row e, using the vector loop method, find all possible solutions (both open and crossed) for angles 2 and θ3. Link 2
a 5 in
Link 3
Offset
c 5 in
Slider position
d 15 in
See Figure P4-2, Table P4-5, and Mathcad file P0461e.
Determine both values of 2 using equations 4.20 and 4.21. 2
2
2
K1 a b c d
2
2
K1 125.000 in 2
K2 2 a c
K2 50.000 in
K3 2 a d
K3 150.000 in
A K1 K3
A 25.000 in
B 2 K2
B 100.000 in
C K1 K3
C 275.000 in
2
2 2 2
2 atan2 2 A B
θ21 2 atan2 2 A B θ22 2.
b 20 in
2 B 4 A C 2
B 4 A C
θ21 123.804 deg θ22 160.674 deg
Determine both values of 3 using equation 4.16a or 4.17.
β ( a b c d α)
θ asin
a sin( α) c
b
d 1 a cos( α) b cos( θ ) return θ if d 1 = d asin
a sin( α) c b
π otherwise
θ31 β a b c d θ21
θ31 152.759 deg
θ32 β a b c d θ22
θ32 170.371 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-61f-1
PROBLEM 4-61f Statement:
Given:
Solution: 1.
The link lengths, offset, and value of d for some fourbar slider-crank linkages are defined in Table P4-5. The linkage configuration and terminology are shown in Figure P4-2. For row f, using the vector loop method, find all possible solutions (both open and crossed) for angles 2 and θ3. Link 2
a 3 in
Link 3
Offset
c 0 in
Slider position
d 12 in
See Figure P4-2, Table P4-5, and Mathcad file P0461f.
Determine both values of 2 using equations 4.20 and 4.21. 2
2
2
K1 a b c d
2
2
K1 16.000 in 2
K2 2 a c
K2 0.000 in
K3 2 a d
K3 72.000 in
A K1 K3
A 88.000 in
B 2 K2
B 0.000 in
C K1 K3
C 56.000 in
2 2
2 2
2 atan2 2 A B
θ21 2 atan2 2 A B θ22 2.
b 13 in
2 B 4 A C 2
B 4 A C
θ21 282.840 deg θ22 282.840 deg
Determine both values of 3 using equation 4.16a or 4.17.
β ( a b c d α)
θ asin
a sin( α) c
b
d 1 a cos( α) b cos( θ ) return θ if d 1 = d asin
a sin( α) c b
π otherwise
θ31 β a b c d θ21
θ31 13.003 deg
θ32 β a b c d θ22
θ32 13.003 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 4-61g-1
PROBLEM 4-61g Statement:
Given:
Solution: 1.
The link lengths, offset, and value of d for some fourbar slider-crank linkages are defined in Table P4-5. The linkage configuration and terminology are shown in Figure P4-2. For row g, using the vector loop method, find all possible solutions (both open and crossed) for angles 2 and θ3. Link 2
a 7 in
Link 3
Offset
c 10 in
Slider position
d 25 in
See Figure P4-2, Table P4-5, and Mathcad file P0461g.
Determine both values of 2 using equations 4.20 and 4.21. 2
2
2
K1 a b c d
2
2
K1 149.000 in
2
K2 2 a c
K2 140.000 in
K3 2 a d
K3 350.000 in
A K1 K3
A 499.000 in
B 2 K2
B 280.000 in
C K1 K3
C 201.000 in
2
2 2 2
2 atan2 2 A B
θ21 2 atan2 2 A B θ22 2.
b 25 in
2 B 4 A C 2
B 4 A C
θ21 88.519 deg θ22 44.916 deg
Determine both values of 3 using equation 4.16a or 4.17.
β ( a b c d α)
θ asin
a sin( α) c
b
d 1 a cos( α) b cos( θ ) return θ if d 1 = d asin
a sin( α) c b
π otherwise
θ31 β a b c d θ21
θ31 186.898 deg
θ32 β a b c d θ22
θ32 216.705 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-1-1
PROBLEM 5-1 Statement:
Design a fourbar mechanism to give the two positions shown in Figure P3-1 of output rocker motion with no quick-return. (See Problem 3-3).
Given:
Coordinates of the points C1, D1, C2, and D2 with respect to C1: C1x 0.0
D1x 1.721
C2x 2.656
D2x 5.065
C1y 0.0
D1y 1.750
C2y 0.751
D2y 0.281
Assumptions: Use the pivot point between links 3 and 4 (C1 and C2 )as the precision points P1 and P2. Define position vectors in the global frame whose origin is at C1. Solution:
See solution to Problem 3-3 and Mathcad file P0501.
1.
Note that this is a two-position function generation (FG) problem because the output is specified as an angular displacement of the rocker, link 4. See Section 5.13 which details the 3-position FG solution. See also Section 5.3 in which the equations for the two-position motion generation problem are derived. These are really the same problem and have the same solution. The method of Section 5.3 will be used here.
2.
Two solution methods are derived in Section 5.3 and are presented in equations 5.7 and 5.8 for the left dyad and equations 5.11 and 5.12 for the right dyad. The first method (equations 5.7 and 5.11) looks like the better one to use in this case since it allows us to choose the link's angular positions and excursions and solve for the lengths of links 2 and 3 (w and z). Unfortunately, this method fails in this problem because of the requirement for a non-quick-return Grashof linkage, which requires the angular displacement of link 3 in going from position 1 to position 2 to be zero (2 = 0) causing a divide-by-zero error in equations 5.7d.
3.
Method 2 (equations 5.8 and 5.12) requires the choosing of two angles and a length for each dyad.
4.
To obtain the same solution as was done graphically in Problem 3-4, we need to know the location of the fixed pivot O4 with respect to the given CD. While we could take the results from Problem 3-3 and use them here to establish the location of O4, that won't be done. Instead, we will use the point C as the joint between links 3 and 4 as well as the precision point P.
5.
Define the position vectors R1 and R2 and the vector P21 using Figure 5-1 and equation 5.1. R1
C1x C1y
R2
C2x C2y
P21x R2 R1 P21y
P21x 2.656 P21y 0.751
p 21 6.
2
2
P21x P21y
p 21 2.760
From the trigonometric relationships given in Figure 5-1, determine 2. From the requirement for a non-quick-return, α 0. δ atan2 P21x P21y
7.
δ 15.789 deg
From a graphical solution (see figure below), determine the values necessary for input to equations 5.8. z 5.000
β 180 deg
ϕ δ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-1-2
3.483 5.0000
1.380
2.027 5.621
O4
1
A1 O2
2
4 2.095
0.989
3
56.519°
A2 C1
D2 C2 2.922
D1
8.
Solve for the WZ dyad using equations 5.8. Z1x z cos ϕ
Z1x 4.811
A 2.000
D sin α
B 0.000
E p 21 cos δ
C 0.000
F p 21 sin δ
A cos β 1 B sin β
C cos α 1
W1x
W1y w
Z1y z sin ϕ
D 0.000
E 2.656
F 0.751
A C Z1x D Z1y E B C Z1y D Z1x F
W1x 1.328
2 A A C Z1y D Z1x F B C Z1x D Z1y E
W1y 0.375
2 A 2
Z1y 1.360
2
W1x W1y
w 1.380
θ atan2 W1x W1y
θ 164.211 deg
This is the expected value of w (one half of p 21) based on the design choices made in the graphical solution and the assumptions made in this problem. 9.
From the graphical solution (see figure above), determine the values necessary for input to equations 5.12. s 0
γ 56.519 deg
ψ 0 deg
10. Solve for the US dyad using equations 5.12. S 1x s cos ψ
S 1x 0.000
S 1y s sin ψ
A 0.448
D sin α
B 0.834
E p 21 cos δ
A cos γ 1 B sin γ
S 1y 0.000 D 0.000
E 2.656
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-1-3
C cos α 1
C 0.000
F p 21 sin δ
A C S 1x D S 1y E B C S 1y D S 1x F
U1x
2 A A C S 1y D S 1x F B C S 1x D S 1y E
U1y
2
U1x 2.027
U1y 2.095
2 A
u
F 0.751
2
U1x U1y
u 2.915
σ atan2 U1x U1y
σ 134.048 deg
This is the expected value of u based on the design choices made in the graphical solution and the assumptions made in this problem. 11. Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
V1x z cos ϕ s cos ψ
V1x 4.811
V1y z sin ϕ s sin ψ
V1y 1.360
θ atan2 V1x V1y
θ 15.789 deg
v Link 1:
2
2
V1x V1y
v 5.000
G1x w cos θ v cos θ u cos σ
G1x 5.510
G1y w sin θ v sin θ u sin σ
G1y 1.110
θ atan2 G1x G1y
θ 11.391 deg
g
2
2
G1x G1y
g 5.621
12. Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 152.820 deg
θ2f θ2i β
θ2f 332.820 deg
13. Define the coupler point with respect to point A and the vector V. rp z
δp ϕ θ
rp 5.000
δp 0.000 deg
which is correct for the assumption that the precision point is at C. 14. Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ 0 deg R1
2
ρ 0.000 deg 2
C1x C1y
R1 0.000
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-1-4
O2x 3.483
O2y 0.985
O4x 2.027
O4y 2.095
O2x R1 cos ρ z cos ϕ w cos θ O2y R1 sin ρ z sin ϕ w sin θ O4x R1 cos ρ s cos ψ u cos σ O4y R1 sin ρ s sin ψ u sin σ
15. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis the line O2O4.
θrot atan2 O4x O2x O4y O2y
θrot 11.391 deg
16. Determine the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "Grashof" 17. DESIGN SUMMARY Link 2:
w 1.380
θ 164.211 deg
Link 3:
v 5.000
θ 15.789 deg
Link 4:
u 2.915
σ 134.048 deg
Link 1:
g 5.621
θ 11.391 deg
Coupler:
rp 5.000
δp 0.000 deg
Crank angles:
θ2i 152.820 deg θ2f 332.820 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-2-1
PROBLEM 5-2 Statement:
Design a fourbar mechanism to give the two positions shown in Figure P3-1 of coupler motion.
Given:
Coordinates of the points A1, B1, A2, and B2 with respect to A1: A1x 0.0
B1x 1.721
A2x 2.656
B2x 5.065
A1y 0.0
B1y 1.750
A2y 0.751
B2y 0.281
Assumptions: Use the points A1 and A2 as the precision points P1 and P2. Define position vectors in the global frame whose origin is at A1. Solution:
See solution to Problem 3-4 and Mathcad file P0502.
1.
Note that this is a two-position motion generation (MG) problem because the output is specified as a complex motion of the coupler, link 3. See Section 5.3 in which the equations for the two-position motion generation problem are derived.
2.
Two solution methods are derived in Section 5.3 and are presented in equations 5.7 and 5.8 for the left dyad and equations 5.11 and 5.12 for the right dyad.
3.
Method 1 (equations 5.7 and 5.11) requires the choosing of three angles for each dyad. Method 2 (equations 5.8 and 5.12) requires the choosing of two angles and a length for each dyad. Method 1 is used in this solution.
4.
In order to obtain the same solution as was done graphically in Problem 3-4, the necessary assumed values were taken from that solution as shown below.
5.
Define the position vectors R1 and R2 and the vector P21 using Figure 5-1 and equation 5.1. R1
A1x A1y
R2
A2x A2y
P21x R2 R1 P21y
P21x 2.656 P21y 0.751
p 21 6.
7.
2
2
P21x P21y
p 21 2.760
From the trigonometric relationships given in Figure 5-1, determine 2 and 2. α atan2 A2x B2x A2y B2y atan2 A1x B1x A1y B1y
α 303.481 deg
δ atan2 P21x P21y
δ 15.789 deg
From the graphical solution (see figure below), determine the values necessary for input to equations 5.7. θ 94.394 deg
β 40.366 deg
ϕ 45.479 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-2-2
O4 93.449° jY 54.330° 2.760 u
A1 X 45.479°
0.281
P 21
15.789°
B2
0.751
v 1.750
A2 134.521° 2.656
2.409
B1 w
40.366°
94.394° 75.124° O2
8.
Solve for the WZ dyad using equations 5.7.
A cos θ cos β 1 sin θ sin β
B cos ϕ cos α 1 sin ϕ sin α
C p 21 cos δ
D sin θ cos β 1 cos θ sin β
E sin ϕ cos α 1 cos ϕ sin α w
C E B F
F p 21 sin δ z
A E B D
w 4.000
A F C D A E B D
z 0.000
These are the expected values of w and z based on the design choices made in the graphical solution and the assumptions made in this problem. 9.
From the graphical solution (see figure above), determine the values necessary for input to equations 5.11. σ 93.449 deg
γ 54.330 deg
ψ 134.521 deg
10. Solve for the US dyad using equations 5.11.
A' cos σ cos γ 1 sin σ sin γ
B' cos ψ cos α 1 sin ψ sin α
C p 21 cos δ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-2-3
D' sin σ cos γ 1 cos σ sin γ
E' sin ψ cos α 1 cos ψ sin α u
C E' B' F
F p 21 sin δ s
A' E' B' D'
u 4.000
A' F C D' A' E' B' D'
s 2.455
These are the expected values of u and s based on the design choices made in the graphical solution and the assumptions made in this problem. 11. Solve for the links 3 and 1 using the vector definitions of V and G. Link 3:
V1x z cos ϕ s cos ψ
V1x 1.721
V1y z sin ϕ s sin ψ
V1y 1.750
θ atan2 V1x V1y
θ 45.479 deg
v
Link 1:
2
2
V1x V1y
v 2.455
G1x w cos θ v cos θ u cos σ
G1x 1.655
G1y w sin θ v sin θ u sin σ
G1y 6.231
θ atan2 G1x G1y
θ 75.123 deg
g
2
2
G1x G1y
g 6.447
12. Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 19.271 deg
θ2f θ2i β
θ2f 21.095 deg
13. Define the coupler point with respect to point A and the vector V. rp z
δp ϕ θ
rp 0.000
δp 0.000 deg
which is correct for the assumption that the precision point is at A. 14. Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ 0 deg R1
ρ 0.000 deg
2
A1x A1y
2
R1 0.000
O2x 0.306
O2y 3.988
O2x R1 cos ρ z cos ϕ w cos θ O2y R1 sin ρ z sin ϕ w sin θ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-2-4
O4x 1.962
O4y 2.243
O4x R1 cos ρ s cos ψ u cos σ O4y R1 sin ρ s sin ψ u sin σ
15. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.
θrot atan2 O4x O2x O4y O2y
θrot 75.123 deg
16. Determine the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "non-Grashof" 17. DESIGN SUMMARY Link 2:
w 4.000
θ 94.394 deg
Link 3:
v 2.455
θ 45.479 deg
Link 4:
u 4.000
σ 93.449 deg
Link 1:
g 6.447
θ 75.123 deg
Coupler:
rp 0.000
δp 0.000 deg
Crank angles:
θ2i 19.271 deg θ2f 21.095 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-3-1
PROBLEM 5-3 Statement:
Design a fourbar mechanism to give the three positions of coupler motion with no quick return shown in Figure P3-2. (See Problem 3-5). Ignore the fixed pivot points in the Figure.
Given:
Coordinates of points A and B with respect to point A1: A1x 0.0
A1y 0.0
B1x 0.741
B1y 2.383
A2x 2.019
A2y 1.905
B2x 4.428
B2y 2.557
A3x 3.933
A3y 1.035
B3x 6.304
B3y 0.256
Assumptions: Let points A1, A2, and A3 be the precision points P1, P2, and P3, respectively. Solution: 1.
2.
3.
See Figure P3-2 Mathcad file P0503.
Determine the magnitudes and orientation of the position difference vectors. 2
2
p 21 2.776
δ atan2 A2x A2y
δ 43.336 deg
2
2
p 31 4.067
δ atan2 A3x A3y
δ 14.744 deg
p 21
A2x A2y
p 31
A3x A3y
Determine the angle changes of the coupler between precision points.
θP1 atan2 A1x B1x A1y B1y
θP1 107.273 deg
θP2 atan2 A2x B2x A2y B2y
θP2 164.856 deg
θP3 atan2 A3x B3x A3y B3y
θP3 161.812 deg
α θP2 θP1
α 57.582 deg
α θP3 θP1
α 269.085 deg
The free choices for this linkage are (from the graphical solution to Problem 3-5): β 78.375 deg
4.
β 135.560 deg
γ 59.771 deg
γ 107.023 deg
Evaluate terms in the WZ coefficient matrix and constant vector from equations 5.26 and form the matrix and vector:
D sin α G sin β L p 31 cos δ
A cos β 1
A F AA B G
B C D
G H K A D C F K H
B sin β
H cos α 1 M p 21 sin δ E p 21 cos δ
E L CC M N
F cos β 1 K sin α N p 31 sin δ C cos α 1
W1x W1y AA 1 CC Z1x Z1y
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-3-2
The components of the W and Z vectors are: W1x 2.178
W1y 0.286
Z1x 0.000
Z1y 0.000
θ atan2 W1x W1y
θ 172.523 deg
ϕ atan2 Z1x Z1y
ϕ 174.344 deg
W1x2 W1y2 , w 2.197
The length of link 2 is: w
Z1x2 Z1y2 , z 0.000
The length of vector Z is: z 5.
Evaluate terms in the US coefficient matrix and constant vector from equations 5.31 and form the matrix and vector:
A' cos γ 1
B' sin γ
C cos α 1
E p 21 cos δ
H cos α 1
D sin α
G' sin γ
K sin α
L p 31 cos δ
A' F' AA B' G'
F' cos γ 1
M p 21 sin δ
B' C D
E L CC M N
G' H K
F' K H
A'
D C
N p 31 sin δ
U1x U 1y AA 1 CC S1x S1y
The components of the U and S vectors are: U1x 1.995
U1y 3.121
S 1x 0.741
S 1y 2.383
σ atan2 U1x U1y
σ 122.591 deg
ψ atan2 S 1x S 1y
ψ 107.267 deg
The length of link 4 is: u
U 2 U 2 , u 3.704 1y 1x
The length of vector S is: s 6.
S 1x2 S 1y2 , s 2.495
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
V1x Z1x S 1x
V1x 0.741
V1y Z1y S 1y
V1y 2.383
θ atan2 V1x V1y
θ 72.734 deg
v Link 1:
2
2
V1x V1y
G1x W1x V1x U1x
v 2.495 G1x 0.558
DESIGN OF MACHINERY - 5th Ed.
G1y W1y V1y U1y
G1y 0.452
θ atan2 G1x G1y
θ 39.009 deg
g 7.
8.
9.
SOLUTION MANUAL 5-3-3
2
2
G1x G1y
g 0.718
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 211.532 deg
θ2f θ2i β
θ2f 75.972 deg
Define the coupler point with respect to point A and the vector V. rp z
δp ϕ θ
rp 0.000
δp 247.078 deg
Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. O2x z cos ϕ w cos θ
O2x 2.178
O2y z sin ϕ w sin θ
O2y 0.286
O4x s cos ψ u cos σ
O4x 2.736
O4y s sin ψ u sin σ
O4y 0.738
10. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.
θrot atan2 O4x O2x O4y O2y
θrot 39.009 deg
11. Determine the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "Grashof" 12. DESIGN SUMMARY Link 2:
w 2.197
θ 172.523 deg
Link 3:
v 2.495
θ 72.734 deg
Link 4:
u 3.704
σ 122.591 deg
Link 1:
g 0.718
θ 39.009 deg
Coupler:
rp 0.000
δp 247.078 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-3-4
Crank angles:
θ2i 211.532 deg θ2f 75.972 deg 13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.
Y
0.718 2.197 O4 O2
2
A1
1a
B3
2.496
A3
3 4 A2 B1
B2 3.704
This is the same result as that found in Problem 3-5.
X
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-4-1
PROBLEM 5-4 Statement:
Design a fourbar mechanism to give the three positions shown in Figure P3-2 (see Problem 3-6). Use analytical synthesis and design it for the fixed pivots shown.
Given:
Link end points (with respect to A1): A1x 0.0
B1x 0.741
A2x 2.019
B2x 4.428
A3x 3.933
B3x 6.304
A1y 0.0
B1y 2.383
A2y 1.905
B2y 2.557
A3y 1.035
B3y 0.256
Fixed pivot points (with respect to A1): O2x 0.995 Solution: 1.
2.
3.
O2y 5.086
O4x 5.298
O4y 5.086
See Figure P3-2 and Mathcad file P0504.
Determine the angle changes between precision points from the body angles given.
θP1 atan2 A1x B1x A1y B1y
θP1 107.273 deg
θP2 atan2 A2x B2x A2y B2y
θP2 164.856 deg
θP3 atan2 A3x B3x A3y B3y
θP3 161.812 deg
α θP2 θP1
α 57.582 deg
α θP3 θP1
α 269.085 deg
Using Figure 5-6, determine the magnitudes of R1, R2, and R3 and their x and y components. P21x A2x
P21x 2.019
P31x A3x
P31x 3.933
P21y A2y
P21y 1.905
P31y A3y
P31y 1.035
R1x O2x
R1x 0.995
R1y O2y
R1y 5.086
R2x R1x P21x
R2x 1.024
R2y R1y P21y
R2y 3.181
R3x R1x P31x
R3x 2.938
R3y R1y P31y
R3y 4.051
2
2
R1 5.182
2
2
R2 3.342
2
2
R3 5.004
R1
R1x R1y
R2
R2x R2y
R3
R3x R3y
Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ atan2 R1x R1y
ζ 101.069 deg
ζ atan2 R2x R2y
ζ 72.156 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-4-2
ζ atan2 R3x R3y 4.
ζ 54.048 deg
Solve for 2 and 3 using equations 5.34
C3 8.007
C4 5.127
C5 5.851
C6 1.294
C1 R3 cos α ζ R2 cos α ζ C2 R3 sin α ζ R2 sin α ζ C3 R1 cos α ζ R3 cos ζ
C4 R1 sin α ζ R3 sin ζ
C5 R1 cos α ζ R2 cos ζ
C6 R1 sin α ζ R2 sin ζ 2
2
C2 3.679
A1 C3 C4
A1 90.406
A2 C3 C6 C4 C5
A2 19.633
A3 C4 C6 C3 C5
A3 53.487
A4 C2 C3 C1 C4
A4 22.524
A5 C4 C5 C3 C6
A5 19.633
A6 C1 C3 C2 C4
A6 29.689
K1 A2 A4 A3 A6
K1 1.146 10
K2 A3 A4 A5 A6
K2 1.788 10
2
K3
3 3
2
2
2
A1 A2 A3 A4 A6
2
2
3
K3 1.769 10
K K 2 K 2 K 2 2 1 2 3 β 2 atan K1 K3
β 90.915 deg
K K 2 K 2 K 2 2 1 2 3 β 2 atan K1 K3
β 23.770 deg
The first value is the same as 3, so use the second value
β β
A5 sin β A3 cos β A6 A1
β 16.790 deg
A3 sin β A2 cos β A4 A1
β 16.790 deg
β acos
β asin
Since both values are the same, 5.
C1 1.352
Repeat steps 2, 3, and 4 for the right-hand dyad to find 1 and 2.
β β
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-4-3
R1x O4x
6.
7.
R1x 5.298
R1y O4y
R2x R1x P21x
R2x 3.279
R2y R1y P21y
R2y 3.181
R3x R1x P31x
R3x 1.365
R3y R1y P31y
R3y 4.051
2
2
R1 7.344
2
2
R2 4.568
2
2
R3 4.275
R1
R1x R1y
R2
R2x R2y
R3
R3x R3y
R1y 5.086
Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ atan2 R1x R1y
ζ 136.170 deg
ζ atan2 R2x R2y
ζ 135.869 deg
ζ atan2 R3x R3y
ζ 108.621 deg
Solve for 2 and 3 using equations 5.34
C3 3.636
C4 9.430
C5 3.855
C6 4.927
C1 R3 cos α ζ R2 cos α ζ C2 R3 sin α ζ R2 sin α ζ C3 R1 cos α ζ R3 cos ζ
C4 R1 sin α ζ R3 sin ζ
C5 R1 cos α ζ R2 cos ζ
C6 R1 sin α ζ R2 sin ζ 2
2
C1 1.023 C2 4.349
A1 C3 C4
A1 102.135
A2 C3 C6 C4 C5
A2 18.434
A3 C4 C6 C3 C5
A3 60.472
A4 C2 C3 C1 C4
A4 25.460
A5 C4 C5 C3 C6
A5 18.434
A6 C1 C3 C2 C4
A6 37.287
K1 A2 A4 A3 A6
K1 1.785 10
K2 A3 A4 A5 A6
K2 2.227 10
3 3
DESIGN OF MACHINERY - 5th Ed.
2
K3
SOLUTION MANUAL 5-4-4
2
2
2
A1 A2 A3 A4 A6
2 3
K3 2.198 10
2
K K 2 K 2 K 2 2 1 2 3 γ 2 atan K1 K3
γ 90.915 deg
K K 2 K 2 K 2 2 1 2 3 2 atan K1 K3
11.643 deg
The first value is the same as 3, so use the second value
γ
A5 sin γ A3 cos γ A6 A1
11.069 deg
A3 sin γ A2 cos γ A4 A1
11.069 deg
acos
asin
γ
Since both angles are the same, 8.
Use the method of Section 5.7 to synthesize the linkage. Start by determining the magnitudes of the vectors P21 and P31 and their angles with respect to the X axis. p 21
2
2
P21x P21y
p 21 2.776
δ atan2 P21x P21y p 31
2
δ 43.336 deg
2
P31x P31y
p 31 4.067
δ atan2 P31x P31y 9.
δ 14.744 deg
Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector: A cos β 1 B sin β C cos α 1
E p 21 cos δ
H cos α 1
D sin α G sin β
L p 31 cos δ
A F AA B G
B C D
G H K A D C F K H
10. The components of the W and Z vectors are:
M p 21 sin δ
E L CC M N
F cos β 1
K sin α
N p 31 sin δ
W1x W1y AA 1 CC Z1x Z1y
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-4-5
W1x 3.594
W1y 7.810 2
w
11. The length of link 2 is:
Z1x 4.589
2
W1x W1y
Z1y 2.724
w 8.597
12. Evaluate terms in the US coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:
A' cos γ 1
B' sin γ
C cos α 1
E p 21 cos δ
H cos α 1
D sin α
G' sin γ
K sin α
L p 31 cos δ
A' F' AA B' G'
F' cos γ 1
M p 21 sin δ
B' C D
N p 31 sin δ
E L CC M N
G' H K A' D C F' K H
U1x U1y AA 1 CC S1x S1y
13. The components of the W and Z vectors are: U1x 2.400
14. The length of link 4 is:
U1y 7.549 2
u
U1x U1y
S1x 2.898
2
S1y 2.463
u 7.921
15. Solving for links 3 and 1 from equations 5.2a and 5.2b. V1x Z1x S1x
V1x 1.691
V1y Z1y S1y
V1y 0.261
The length of link 3 is:
v
2
2
V1x V1y
v 1.711
G1x W1x V1x U1x
G1x 4.303
G1y W1y V1y U1y
G1y 5.329 10
The length of link 1 is:
g
2
G1x G1y
2
14
g 4.303
16. Check the location of the fixed pivots with respect to the global frame using the calculated vectors W1, Z1, U1, and S1. O2x Z1x W1x
O2x 0.995
O2y Z1y W1y
O2y 5.086
O4x S1x U1x
O4x 5.298
O4y S1y U1y
O4y 5.086
These check with Figure P3-2.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-4-6
17. Determine the location of the coupler point with respect to point A and line AB. 2
2
z 5.336
2
2
s 3.803
Distance from A to P
z
Z1x Z1y
Angle BAP (p)
s
S1x S1y
rP z
ψ atan2( S1x S1y)
ψ 139.639 deg
ϕ atan2( Z1x Z1y )
ϕ 149.309 deg
θ atan2 z cos ϕ s cos ψ z sin ϕ s sin ψ θ 171.233 deg
δp ϕ θ
δp 21.924 deg
18. DESIGN SUMMARY Link 1:
g 4.303
Link 2:
w 8.597
Link 3:
v 1.711
Link 4:
u 7.921
Coupler point:
rP 5.336
δp 21.924 deg
19. VERIFICATION: The calculated values of g (length of the ground link) and of the coordinates of O2 and O4 give the same values as those on the problem statement, verifying that the calculated values for the other links and the coupler point are correct.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-5-1
PROBLEM 5-5 Statement:
See Project P3-8. Define three positions of the boat and analytically synthesize a linkage to move through them.
Assumptions: Launch ramp angle is 15 deg to the horizontal. Solution: 1.
See Project P3-8 and Mathcad file P0505.
This is an open-ended design problem that has many valid solutions. First define the problem more completely than is stated by deciding on three positions for the boat to move through. The figure below shows one such set of positions (dimensions are in mm). Y 3539 1453 1 deg. P2
15 deg.
725 X
P1 1179
1261
1431
0 deg. P3
O4
O2
WATER LEVEL 331 2694 RAMP
2.
From the figure, the design choices are: P21x 1453
P21y 725
P31x 3539
P31y 1261
O2x 331
O2y 1179
O4x 2694
O4y 1431
Body angles:
θP1 15 deg
θP2 1 deg
θP3 0 deg
3.
The methods of Section 5.8 are used to get a solution for this problem. The solution is sensitive to small changes in the design choices so a trial-and-error approach is warranted.
4.
Determine the angle changes between precision points from the body angles given.
5.
α θP2 θP1
α 14.000 deg
α θP3 θP1
α 15.000 deg
Using Figure 5-6, determine the magnitudes of R1, R2, and R3 and their x and y components. R1x O2x
R1x 331.000
R1y O2y 3
R2x R1x P21x
R2x 1.122 10
R2y R1y P21y
R2y 1.904 10
R3x R1x P31x
R3x 3.208 10
R3y R1y P31y
R3y 82.000
3 3
R1y 1.179 10
3
DESIGN OF MACHINERY - 5th Ed.
6.
7.
SOLUTION MANUAL 5-5-2
2
2
R1 1.225 10
3
2
2
R2 2.210 10
3
2
2
R3 3.209 10
3
R1
R1x R1y
R2
R2x R2y
R3
R3x R3y
Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ atan2 R1x R1y
ζ 74.318 deg
ζ atan2 R2x R2y
ζ 120.510 deg
ζ atan2 R3x R3y
ζ 178.536 deg
Solve for 2 and 3 using equations 5.34
C3 3.833 10
C4 1.135 10
C5 1.728 10
C6 840.097
C1 R3 cos α ζ R2 cos α ζ C2 R3 sin α ζ R2 sin α ζ
C4 R1 sin α ζ R3 sin ζ
C6 R1 sin α ζ R2 sin ζ 2
C2 1.433 10
3
3
3
C5 R1 cos α ζ R2 cos ζ
3
3
C3 R1 cos α ζ R3 cos ζ
C1 2.542 10
2
7
A1 C3 C4
A1 1.598 10
A2 C3 C6 C4 C5
A2 5.182 10
A3 C4 C6 C3 C5
A3 5.671 10
6
A4 C2 C3 C1 C4
A4 2.607 10
6
A5 C4 C5 C3 C6
A5 5.182 10
6
A6 C1 C3 C2 C4
A6 1.137 10
7
K1 A2 A4 A3 A6
K1 5.096 10
K2 A3 A4 A5 A6
K2 7.370 10
2
K3
6
13 13
2
2
2
A1 A2 A3 A4 A6
2
2
K K 2 K 2 K 2 2 1 2 3 β 2 atan K1 K3
13
K3 3.015 10
β 125.675 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-5-3
K K 2 K 2 K 2 2 1 2 3 β 2 atan K1 K3
β 15.000 deg
The second value is the same as 3, so use the first value
β β
A5 sin β A3 cos β A6 A1
β 39.836 deg
A3 sin β A2 cos β A4 A1
β 39.836 deg
β acos
β asin
β β
Since both values are the same, 8.
Repeat steps 2, 3, and 4 for the right-hand dyad to find 1 and 2. R1x O4x
9.
R1x 2.694 10
3
R1y O4y
R2x R1x P21x
R2x 1.241 10
3
R2y R1y P21y
R2y 2.156 10
3
R3x R1x P31x
R3x 845.000
R3y R1y P31y
R3y 170.000
2
2
R1 3.050 10
3
2
2
R2 2.488 10
3
2
2
R3 861.931
R1
R1x R1y
R2
R2x R2y
R3
R3x R3y
R1y 1.431 10
Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ atan2 R1x R1y
ζ 27.976 deg
ζ atan2 R2x R2y
ζ 60.075 deg
ζ atan2 R3x R3y
ζ 168.625 deg
10. Solve for 2 and 3 using equations 5.34
C3 3.818 10
C4 514.981
C5 1.719 10
C6 1.419 10
C1 R3 cos α ζ R2 cos α ζ C2 R3 sin α ζ R2 sin α ζ C3 R1 cos α ζ R3 cos ζ
C4 R1 sin α ζ R3 sin ζ
C5 R1 cos α ζ R2 cos ζ
C6 R1 sin α ζ R2 sin ζ
C1 2.536 10
3
C2 1.392 10
3
3
3 3
3
DESIGN OF MACHINERY - 5th Ed. 2
SOLUTION MANUAL 5-5-4
2
7
A1 C3 C4
A1 1.484 10
A2 C3 C6 C4 C5
A2 6.303 10
A3 C4 C6 C3 C5
A3 5.832 10
6
A4 C2 C3 C1 C4
A4 4.008 10
6
A5 C4 C5 C3 C6
A5 6.303 10
6
A6 C1 C3 C2 C4
A6 1.040 10
7
K1 A2 A4 A3 A6
K1 3.537 10
K2 A3 A4 A5 A6
K2 8.891 10
2
K3
6
13 13
2
2
2
A1 A2 A3 A4 A6
2
2
13
K3 1.115 10
K K 2 K 2 K 2 2 1 2 3 γ 2 atan K1 K3
γ 151.615 deg
K K 2 K 2 K 2 2 1 2 3 2 atan K1 K3
15.000 deg
The second value is the same as 3, so use the first value
γ γ
A5 sin γ A3 cos γ A6 A1
56.167 deg
A3 sin γ A2 cos γ A4 A1
56.167 deg
acos
asin
γ
Since both angles are the same,
11. Use the method of Section 5.7 to synthesize the linkage. Start by determining the magnitudes of the vectors P21 and P31 and their angles with respect to the X axis. p 21
2
2
P21x P21y
δ atan2 P21x P21y
p 31
2
2
P31x P31y
δ atan2 P31x P31y
3
p 21 1.624 10
δ 153.482 deg
3
p 31 3.757 10
δ 160.388 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-5-5
12. Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:
B sin β
E p 21 cos δ
H cos α 1
A cos β 1
D sin α
F cos β 1
G sin β
K sin α
L p 31 cos δ
M p 21 sin δ
B C D
A F AA B G
C cos α 1
N p 31 sin δ
E L CC M N
G H K A D C F K H
W1x W1y AA 1 CC Z1x Z1y
13. The components of the W and Z vectors are: W1x 1.331 10
3
w
14. The length of link 2 is:
W1y 1.653 10 2
3
Z1x 1.000 10
2
W1x W1y
3
Z1y 474.187
w 2122.473
15. Evaluate terms in the US coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:
A' cos γ 1
B' sin γ
C cos α 1
E p 21 cos δ
H cos α 1
D sin α
G' sin γ
K sin α
L p 31 cos δ
A' F' AA B' G'
F' cos γ 1
M p 21 sin δ
B' C D
N p 31 sin δ
E L CC M N
G' H K A' D C F' K H
U1x U1y AA 1 CC S1x S1y
16. The components of the W and Z vectors are: 3
U1x 1.690 10
17. The length of link 4 is:
u
U1y 951.273 2
U1x U1y
2
S1x 1.004 10
3
S1y 479.727
u 1939.291
18. Solving for links 3 and 1 from equations 5.2a and 5.2b. V1x Z1x S1x
V1x 2.004 10
V1y Z1y S1y
V1y 953.914
The length of link 3 is:
v
2
2
V1x V1y
v 2219.601
3
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-5-6
G1x W1x V1x U1x
G1x 2.363 10
G1y W1y V1y U1y
G1y 252.000
The length of link 1 is:
2
g
G1x G1y
2
3
g 2376.399
19. Check the location of the fixed pivots with respect to the global frame using the calculated vectors W1, Z1, U1, and S1. O2x Z1x W1x
O2x 331.000
O2y Z1y W1y
O2y 1179.000
O4x S1x U1x
O4x 2694.000
O4y S1y U1y
O4y 1431.000
These check with the design choices shown in the figure above. 20. Determine the location of the coupler point with respect to point A and line AB. 2
2
z 1106.834
2
2
s 1112.770
Distance from A to P
z
Z1x Z1y
Angle BAP (p)
s
S1x S1y
rP z
ψ atan2( S1x S1y)
ψ 25.538 deg
ϕ atan2( Z1x Z1y )
ϕ 154.633 deg
θ atan2 z cos ϕ s cos ψ z sin ϕ s sin ψ θ 154.547 deg
δp ϕ θ
δp 0.086 deg
21. DESIGN SUMMARY Link 1:
g 2376.4
Link 2:
w 2122.5
Link 3:
v 2219.6
Link 4:
u 1939.3
Coupler point:
rP 1106.8
δp 0.086 deg
22. VERIFICATION: The calculated values of g (length of the ground link) and of the coordinates of O2 and O4 give the same values as those on the problem statement, verifying that the calculated values for the other links and the coupler point are correct. The solution is drawn below to show the locations of the moving pivots for the three positions chosen (see next page). 23. The design needs to be checked for the presence of toggle positions within its desired range of motion. This design has none, but is close to toggle at position 1. This could be used as a locking feature. 24. The transmission angles need to be checked also. This design has poor transmission angles, especially in and near positions 1 and 3. Unfortunately, this is where a large overturning moment is created by the mass of the boat. A large mechanical advantage input device will need to be used here, such as a hydraulic cylinder or geared drive. Note that the drive mechanism must also resist being overdriven (back driven) by the load as the boat descends from its high point onto the trailer.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-5-7
A2 A1
B2
B1 A3 B3 WATER LEVEL
RAMP
O4
O2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-6-1
PROBLEM 5-6 Statement:
See Project P3-20. Define three positions of the dumpster and analytically synthesize a linkage to move through them. The fixed pivots must be located on the existing truck.
Solution:
See Project P3-20 and Mathcad file P0506.
1.
This is an open-ended design problem that has many valid solutions. First define the problem more completely than it is stated by deciding on three positions for the dumpster box to move through. The figure below shows one such set of positions (dimensions are in mm). 1999 59.1 deg. 590
30.3 deg.
P3 P2 1817 0 deg.
1202
226 311
P1 O2 O4
2036 2094
2.
From the figure, the design choices are: P21x 590
P21y 1202
P31x 1999
P31y 1817
O2x 2094
O2y 226
O4x 2036
O4y 311
Body angles:
θP1 0 deg
θP2 30.3 deg
θP3 59.1 deg
3.
The methods of Section 5.8 are used to get a solution for this problem. The solution is sensitive to small changes in the design choices so a trial-and-error approach is warranted.
4.
Determine the angle changes between precision points from the body angles given.
5.
α θP2 θP1
α 30.300 deg
α θP3 θP1
α 59.100 deg
Using Figure 5-6, determine the magnitudes of R1, R2, and R3 and their x and y components. R1x O2x
R1x 2.094 10
3
R1y O2y
R2x R1x P21x
R2x 1.504 10
3
R2y R1y P21y
R2y 1.428 10
3
R3x R1x P31x
R3x 95.000
R1y 226.000
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-6-2
R3y R1y P31y
6.
7.
R3y 2.043 10
3
2
2
R1 2.106 10
3
2
2
R2 2.074 10
3
2
2
R3 2.045 10
3
R1
R1x R1y
R2
R2x R2y
R3
R3x R3y
Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ atan2 R1x R1y
ζ 6.160 deg
ζ atan2 R2x R2y
ζ 43.515 deg
ζ atan2 R3x R3y
ζ 87.338 deg
Solve for 2 and 3 using equations 5.34
C3 786.433
C4 130.152
C5 189.927
C6 176.392
C1 R3 cos α ζ R2 cos α ζ C2 R3 sin α ζ R2 sin α ζ C3 R1 cos α ζ R3 cos ζ
C4 R1 sin α ζ R3 sin ζ
C5 R1 cos α ζ R2 cos ζ
C6 R1 sin α ζ R2 sin ζ 2
2
C1 495.777 C2 212.019
5
A1 C3 C4
A1 6.354 10
A2 C3 C6 C4 C5
A2 1.140 10
A3 C4 C6 C3 C5
A3 1.723 10
5
A4 C2 C3 C1 C4
A4 2.313 10
5
A5 C4 C5 C3 C6
A5 1.140 10
5
A6 C1 C3 C2 C4
A6 3.623 10
5
K1 A2 A4 A3 A6
K1 3.607 10
K2 A3 A4 A5 A6
K2 8.115 10
2
K3
5
10 10
2
2
2
A1 A2 A3 A4 A6
2
2
K K 2 K 2 K 2 2 1 2 3 β 2 atan K1 K3
10
K3 8.816 10
β 72.976 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-6-3
K K 2 K 2 K 2 2 1 2 3 β 2 atan K1 K3
β 59.100 deg
The second value is the same as 3, so use the first value
β β
A5 sin β A3 cos β A6 A1
β 34.802 deg
A3 sin β A2 cos β A4 A1
β 34.802 deg
β acos
β asin
β β
Since both values are the same, 8.
Repeat steps 2, 3, and 4 for the right-hand dyad to find 1 and 2. R1x O4x
9.
R1x 2.036 10
3
R1y O4y
R2x R1x P21x
R2x 1.446 10
3
R2y R1y P21y
R2y 1.513 10
3
R3x R1x P31x
R3x 37.000
R3y R1y P31y
R3y 2.128 10
3
2
2
R1 2.060 10
3
2
2
R2 2.093 10
3
2
2
R3 2.128 10
3
R1
R1x R1y
R2
R2x R2y
R3
R3x R3y
R1y 311.000
Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ atan2 R1x R1y
ζ 8.685 deg
ζ atan2 R2x R2y
ζ 46.297 deg
ζ atan2 R3x R3y
ζ 89.004 deg
10. Solve for 2 and 3 using equations 5.34
C3 741.712
C4 221.269
C5 154.965
C6 217.266
C1 R3 cos α ζ R2 cos α ζ C2 R3 sin α ζ R2 sin α ζ C3 R1 cos α ζ R3 cos ζ
C4 R1 sin α ζ R3 sin ζ
C5 R1 cos α ζ R2 cos ζ
C6 R1 sin α ζ R2 sin ζ
C1 486.018 C2 161.777
DESIGN OF MACHINERY - 5th Ed. 2
SOLUTION MANUAL 5-6-4
2
5
A1 C3 C4
A1 5.991 10
A2 C3 C6 C4 C5
A2 1.269 10
A3 C4 C6 C3 C5
A3 1.630 10
5
A4 C2 C3 C1 C4
A4 2.275 10
5
A5 C4 C5 C3 C6
A5 1.269 10
5
A6 C1 C3 C2 C4
A6 3.247 10
5
K1 A2 A4 A3 A6
K1 2.406 10
K2 A3 A4 A5 A6
K2 7.828 10
2
K3
5
10 10
2
2
2
A1 A2 A3 A4 A6
2
2
10
K3 7.953 10
K K 2 K 2 K 2 2 1 2 3 γ 2 atan K1 K3
γ 86.724 deg
K K 2 K 2 K 2 2 1 2 3 2 atan K1 K3
59.100 deg
The second value is the same as 3, so use the first value
γ γ
A5 sin γ A3 cos γ A6 A1
39.743 deg
A3 sin γ A2 cos γ A4 A1
39.743 deg
acos
asin
γ
Since both angles are the same,
11. Use the method of Section 5.7 to synthesize the linkage. Start by determining the magnitudes of the vectors P21 and P31 and their angles with respect to the X axis. p 21
2
2
P21x P21y
δ atan2 P21x P21y p 31
2
2
P31x P31y
δ atan2 P31x P31y
p 21 1338.994 δ 116.144 deg p 31 2701.387 δ 137.731 deg
12. Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-6-5
B sin β
E p 21 cos δ
H cos α 1
A cos β 1
D sin α
F cos β 1
G sin β
K sin α
L p 31 cos δ
M p 21 sin δ
B C D
A F AA B G
C cos α 1
E L CC M N
G H K A D C F K H
N p 31 sin δ
W1x W1y AA 1 CC Z1x Z1y
13. The components of the W and Z vectors are: W1x 3.194 10
3
W1y 829.763
Z1x 1.100 10
3
Z1y 603.763
14. The length of link 2 is: w
2
2
W1x W1y
w 3299.543
15. Evaluate terms in the US coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:
A' cos γ 1
B' sin γ
C cos α 1
E p 21 cos δ
H cos α 1
D sin α
G' sin γ
K sin α
L p 31 cos δ
A' F' AA B' G'
F' cos γ 1
M p 21 sin δ
B' C D
N p 31 sin δ
E L CC M N
G' H K A' D C F' K H
U1x U1y AA 1 CC S1x S1y
16. The components of the W and Z vectors are: 3
U1x 1.621 10
U1y 13.492
S1x 415.016
S1y 297.508
17. The length of link 4 is: u
2
U1x U1y
2
u 1621.040
18. Solving for links 3 and 1 from equations 5.2a and 5.2b. V1x Z1x S1x
V1x 1.515 10
V1y Z1y S1y
V1y 901.272
3
DESIGN OF MACHINERY - 5th Ed.
The length of link 3 is:
SOLUTION MANUAL 5-6-6 2
v
2
V1x V1y
v 1762.404
G1x W1x V1x U1x
G1x 58.000
G1y W1y V1y U1y
G1y 85.000
The length of link 1 is:
2
g
G1x G1y
2
g 102.903
19. Check the location of the fixed pivots with respect to the global frame using the calculated vectors W1, Z1, U1, and S1. O2x Z1x W1x
O2x 2094.000
O2y Z1y W1y
O2y 226.000
O4x S1x U1x
O4x 2036.000
O4y S1y U1y
O4y 311.000
These check with the design choices shown in the figure above. 20. Determine the location of the coupler point with respect to point A and line AB. 2
2
z 1254.370
2
2
s 510.637
Distance from A to P
z
Z1x Z1y
Angle BAP (p)
s
S1x S1y
rP z
ψ atan2( S1x S1y)
ψ 35.635 deg
ϕ atan2( Z1x Z1y )
ϕ 151.228 deg
θ atan2 z cos ϕ s cos ψ z sin ϕ s sin ψ θ 149.244 deg
δp ϕ θ
δp 1.984 deg
21. DESIGN SUMMARY Link 1:
g 102.9
Link 2:
w 3299.5
Link 3:
v 1762.4
Link 4:
u 1621.0
Coupler point:
rP 1254.4
δp 1.984 deg
22. VERIFICATION: The calculated values of g (length of the ground link) and of the coordinates of O2 and O4 give the same values as those on the problem statement, verifying that the calculated values for the other links and the coupler point are correct. The solution is drawn below to show the locations of the moving pivots for the three positions chosen (see next page). 23. The design needs to be checked for the presence of toggle positions within its desired range of motion. This design has none, but is close to toggle at position 3.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-6-7
24. The transmission angles need to be checked also. A large mechanical advantage input device will need to be used here, such as a hydraulic cylinder. Note that the drive mechanism must also resist being overdriven (back driven) by the load as the dumpster descends from its high point onto the truck.
A3
A2
B3
B2
O2
B1 O4
A1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-7-1
PROBLEM 5-7 Statement:
See Project P3-7. Define three positions of the computer monitor and analytically synthesize a linkage to move through them. The fixed pivots must be located on the floor or wall.
Solution:
See Project P3-7 and Mathcad file P0507.
1.
This is an open-ended design problem that has many valid solutions. First define the problem more completely than it is stated by deciding on three positions for the computer monitor to move through. The figure below shows one such set of positions (dimensions are in inches). Y 33.816
O2 97 deg. 11.580 P1
O4
X
90 deg. 7.812
1.272
14.472
P2
WALL 85 deg.
P3
2.148
5.736
2.
From the figure, the design choices are: P21x 2.148
P21y 7.812
P31x 5.736
P31y 14.472
O2x 33.816
O2y 11.580
O4x 33.816
O4y 1.272
Body angles:
θP1 97 deg
θP2 90 deg
θP3 85 deg
3.
The methods of Section 5.8 are used to get a solution for this problem. The solution is sensitive to small changes in the design choices so a trial-and-error approach is warranted.
4.
Determine the angle changes between precision points from the body angles given.
5.
α θP2 θP1
α 7.000 deg
α θP3 θP1
α 12.000 deg
Using Figure 5-6, determine the magnitudes of R1, R2, and R3 and their x and y components. R1x O2x
R1x 33.816
R1y O2y
R2x R1x P21x
R2x 31.668
R2y R1y P21y
R2y 19.392
R3x R1x P31x
R3x 28.080
R3y R1y P31y
R3y 26.052
R1y 11.580
DESIGN OF MACHINERY - 5th Ed.
6.
7.
SOLUTION MANUAL 5-7-2
2
2
R1 35.744
2
2
R2 37.134
2
2
R3 38.304
R1
R1x R1y
R2
R2x R2y
R3
R3x R3y
Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ atan2 R1x R1y
ζ 161.097 deg
ζ atan2 R2x R2y
ζ 148.519 deg
ζ atan2 R3x R3y
ζ 137.146 deg
Solve for 2 and 3 using equations 5.34
C3 7.405
C4 21.756
C5 3.307
C6 12.019
C1 R3 cos α ζ R2 cos α ζ C2 R3 sin α ζ R2 sin α ζ C3 R1 cos α ζ R3 cos ζ
C4 R1 sin α ζ R3 sin ζ
C5 R1 cos α ζ R2 cos ζ
C6 R1 sin α ζ R2 sin ζ 2
2
C1 3.962 C2 10.052
A1 C3 C4
A1 528.143
A2 C3 C6 C4 C5
A2 17.049
A3 C4 C6 C3 C5
A3 285.981
A4 C2 C3 C1 C4
A4 11.771
A5 C4 C5 C3 C6
A5 17.049
A6 C1 C3 C2 C4
A6 248.020
K1 A2 A4 A3 A6
K1 7.073 10
K2 A3 A4 A5 A6
K2 7.595 10
2
K3
4 3
2
2
2
A1 A2 A3 A4 A6
2
2
K K 2 K 2 K 2 2 1 2 3 β 2 atan K1 K3
4
K3 6.760 10
β 24.258 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-7-3
K K 2 K 2 K 2 2 1 2 3 β 2 atan K1 K3
β 12.000 deg
The second value is the same as 3, so use the first value
β β
A5 sin β A3 cos β A6 A1
β 12.435 deg
A3 sin β A2 cos β A4 A1
β 12.435 deg
β acos
β asin
β β
Since both values are the same, 8.
Repeat steps 2, 3, and 4 for the right-hand dyad to find 1 and 2. R1x O4x
9.
R1x 33.816
R1y O4y
R2x R1x P21x
R2x 31.668
R2y R1y P21y
R2y 6.540
R3x R1x P31x
R3x 28.080
R3y R1y P31y
R3y 13.200
2
2
R1 33.840
2
2
R2 32.336
2
2
R3 31.028
R1
R1x R1y
R2
R2x R2y
R3
R3x R3y
R1y 1.272
Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ atan2 R1x R1y
ζ 177.846 deg
ζ atan2 R2x R2y
ζ 168.331 deg
ζ atan2 R3x R3y
ζ 154.822 deg
10. Solve for 2 and 3 using equations 5.34
C3 4.733
C4 21.475
C5 1.741
C6 11.924
C1 R3 cos α ζ R2 cos α ζ C2 R3 sin α ζ R2 sin α ζ C3 R1 cos α ζ R3 cos ζ
C4 R1 sin α ζ R3 sin ζ
C5 R1 cos α ζ R2 cos ζ
C6 R1 sin α ζ R2 sin ζ
C1 2.856 C2 9.867
DESIGN OF MACHINERY - 5th Ed. 2
SOLUTION MANUAL 5-7-4
2
A1 C3 C4
A1 483.571
A2 C3 C6 C4 C5
A2 19.043
A3 C4 C6 C3 C5
A3 264.299
A4 C2 C3 C1 C4
A4 14.646
A5 C4 C5 C3 C6
A5 19.043
A6 C1 C3 C2 C4
A6 225.402
K1 A2 A4 A3 A6
K1 5.929 10
K2 A3 A4 A5 A6
K2 8.163 10
2
K3
4 3
2
2
2
A1 A2 A3 A4 A6
2
2
4
K3 5.630 10
K K 2 K 2 K 2 2 1 2 3 γ 2 atan K1 K3
γ 27.678 deg
K K 2 K 2 K 2 2 1 2 3 2 atan K1 K3
12.000 deg
The second value is the same as 3, so use the first value
γ γ
A5 sin γ A3 cos γ A6 A1
14.435 deg
A3 sin γ A2 cos γ A4 A1
14.435 deg
acos
asin
γ
Since both angles are the same,
11. Use the method of Section 5.7 to synthesize the linkage. Start by determining the magnitudes of the vectors P21 and P31 and their angles with respect to the X axis. p 21
2
2
P21x P21y
δ atan2 P21x P21y
p 31
2
2
P31x P31y
δ atan2 P31x P31y
p 21 8.102 δ 74.626 deg
p 31 15.567 δ 68.379 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-7-5
12. Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:
B sin β
E p 21 cos δ
H cos α 1
A cos β 1
D sin α
F cos β 1
G sin β
K sin α
L p 31 cos δ
M p 21 sin δ
B C D
A F AA B G
C cos α 1
N p 31 sin δ
E L CC M N
G H K A D C F K H
W1x W1y AA 1 CC Z1x Z1y
13. The components of the W and Z vectors are: W1x 36.030
W1y 8.098
Z1x 2.214
Z1y 3.482
14. The length of link 2 is: w
2
2
W1x W1y
w 36.929
15. Evaluate terms in the US coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:
A' cos γ 1
B' sin γ
C cos α 1
E p 21 cos δ
H cos α 1
D sin α
G' sin γ
K sin α
L p 31 cos δ
A' F' AA B' G'
F' cos γ 1
M p 21 sin δ
B' C D
N p 31 sin δ
E L CC M N
G' H K A' D C F' K H
U1x U1y AA 1 CC S1x S1y
16. The components of the W and Z vectors are: U1x 32.294
U1y 2.592
S1x 1.522
17. The length of link 4 is: u
2
U1x U1y
2
u 32.398
18. Solving for links 3 and 1 from equations 5.2a and 5.2b. V1x Z1x S1x
V1x 3.736
V1y Z1y S1y
V1y 7.346
S1y 3.864
DESIGN OF MACHINERY - 5th Ed.
The length of link 3 is:
SOLUTION MANUAL 5-7-6 2
v
2
V1x V1y
v 8.241
G1x W1x V1x U1x
G1x 0.000
G1y W1y V1y U1y
G1y 12.852
The length of link 1 is:
2
g
G1x G1y
2
g 12.852
19. Check the location of the fixed pivots with respect to the global frame using the calculated vectors W1, Z1, U1, and S1. O2x Z1x W1x
O2x 33.816
O2y Z1y W1y
O2y 11.580
O4x S1x U1x
O4x 33.816
O4y S1y U1y
O4y 1.272
These check with the design choices shown in the figure above. 20. Determine the location of the coupler point with respect to point A and line AB. 2
2
z 4.126
2
2
s 4.153
Distance from A to P
z
Z1x Z1y
Angle BAP (p)
s
S1x S1y
ψ atan2( S1x S1y)
ψ 111.494 deg
ϕ atan2( Z1x Z1y )
ϕ 57.551 deg
rP z
θ atan2 z cos ϕ s cos ψ z sin ϕ s sin ψ θ 63.046 deg
δp ϕ θ
δp 5.495 deg
21. DESIGN SUMMARY Link 1:
g 12.852
Link 2:
w 36.929
Link 3:
v 8.241
Link 4:
u 32.398
Coupler point:
rP 4.126
δp 5.495 deg
22. VERIFICATION: The calculated values of g (length of the ground link) and of the coordinates of O2 and O4 give the same values as those on the problem statement, verifying that the calculated values for the other links and the coupler point are correct. The solution is drawn below to show the locations of the moving pivots for the three positions chosen (see next page). 23. The design needs to be checked for the presence of toggle positions within its desired range of motion. This design has none.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-7-7
24. The transmission angles need to be checked also. A means to support the weight of the monitor must be provided. The figure below shows a spring placed between links to provide the balancing moment. Further design and analysis needs to be done to optimize the spring placement in order to compensate for its change in force with deflection and the change in moment arm as the linkage moves.
SPRING O2
A1
A2 O4
B1
WALL A3 B2
B3
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-8-1
PROBLEM 5-8 Statement:
Design a linkage to carry the body in Figure P5-1 through the two positions P1 and P2 at the angles shown in the figure. Use analytical synthesis without regard for the fixed pivots shown. Use the free choices given below.
Given:
Coordinates of the points P1 and P2 with respect to P1: P1x 0.0
P1y 0.0
P2x 1.236
P2y 2.138
Angles made by the body in positions 1 and 2:
θP1 210 deg
θP2 147.5 deg
Free choices for the WZ dyad : z 1.075
β 27.0 deg
ϕ 204.4 deg
γ 40.0 deg
ψ 74.0 deg
Free choices for the US dyad : s 1.240 Solution:
See Figure P5-1 and Mathcad file P0508.
1.
Note that this is a two-position motion generation (MG) problem because the output is specified as a complex motion of the coupler, link 3. Because of the data given in the hint, the second method of Section 5.3 will be used here.
2.
Define the position vectors R1 and R2 and the vector P21 using Figure 5-1 and equation 5.1. R1
P1x P1y
R2
P2x P2y
P21x R2 R1 P21y
P21x 1.236 P21y 2.138
p 21 3.
4.
2
2
P21x P21y
p 21 2.470
From the trigonometric relationships given in Figure 5-1, determine 2 and 2. α θP2 θP1
α 62.500 deg
δ atan2 P21x P21y
δ 120.033 deg
Solve for the WZ dyad using equations 5.8. Z1x z cos ϕ
Z1y z sin ϕ
A 0.109
D sin α
B 0.454
E p 21 cos δ
C 0.538
F p 21 sin δ
A cos β 1 B sin β
C cos α 1
W1x
Z1x 0.979
A C Z1x D Z1y E B C Z1y D Z1x F 2 A
Z1y 0.444 D 0.887
E 1.236
F 2.138
W1x 1.462
DESIGN OF MACHINERY - 5th Ed.
W1y w
SOLUTION MANUAL 5-8-2
A C Z1y D Z1x F B C Z1x D Z1y E
W1y 3.367
2 A 2
2
W1x W1y
w 3.670
θ atan2 W1x W1y 5.
θ 113.472 deg
Solve for the US dyad using equations 5.12. S 1x s cos ψ
S 1x 0.342
A 0.234
D sin α
B 0.643
E p 21 cos δ
C 0.538
F p 21 sin δ
A cos γ 1 B sin γ
C cos α 1
U1x
U1y u
S 1y s sin ψ
D 0.887
E 1.236
F 2.138
A C S 1x D S 1y E B C S 1y D S 1x F 2 A A C S 1y D S 1x F B C S 1x D S 1y E
2
U1x U1y
u 5.461
σ atan2 U1x U1y 6.
σ 125.619 deg
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
V1x z cos ϕ s cos ψ
V1x 1.321
V1y z sin ϕ s sin ψ
V1y 1.636
θ atan2 V1x V1y
θ 128.914 deg
v Link 1:
2
2
V1x V1y
v 2.103
G1x w cos θ v cos θ u cos σ
G1x 0.398
G1y w sin θ v sin θ u sin σ
G1y 0.564
θ atan2 G1x G1y
θ 54.796 deg
g 7.
U1x 3.180
U1y 4.439
2 A 2
S 1y 1.192
2
2
G1x G1y
g 0.690
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 58.677 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-8-3
θ2f θ2i β 8.
9.
θ2f 85.677 deg
Define the coupler point with respect to point A and the vector V. rp z
δp ϕ θ
rp 1.075
δp 333.314 deg
Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ 0 deg R1
ρ 0.000 deg
2
2
P1x P1y
R1 0.000
O2x 2.441
O2y 3.811
O4x 2.838
O4y 3.247
O2x R1 cos ρ z cos ϕ w cos θ O2y R1 sin ρ z sin ϕ w sin θ O4x R1 cos ρ s cos ψ u cos σ O4y R1 sin ρ s sin ψ u sin σ
10. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis the line O2O4.
θrot atan2 O4x O2x O4y O2y
θrot 54.796 deg
11. Determine the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "non-Grashof" 12. DESIGN SUMMARY Link 2:
w 3.670
θ 113.472 deg
Link 3:
v 2.103
θ 128.914 deg
Link 4:
u 5.461
σ 125.619 deg
Link 1:
g 0.690
θ 54.796 deg
Coupler:
rp 1.075
δp 333.314 deg
Crank angles:
θ2i 58.677 deg θ2f 85.677 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-8-4
13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.
O2
1.236
G1 Y
O4 U2
P2 S2
B2
V2 2.138
W2
U1
Z2 A2
Z1 62.5°
W1
A1 X
P1 V1
S1 B1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-9-1
PROBLEM 5-9 Statement:
Design a linkage to carry the body in Figure P5-1 through the two positions P2 and P3 at the angles shown in the figure. Use analytical synthesis without regard for the fixed pivots shown. Hint: First try a rough graphical solution to create realistic values for free choices.
Given:
Coordinates of the points P2 and P3 with respect to P1: P2x 1.236
P2y 2.138
P3x 2.500
P3y 2.931
Angles made by the body in positions 1 and 2:
θP2 147.5 deg Solution:
θP3 110.2 deg
See Figure P5-1 and Mathcad file P0509.
1.
Note that this is a two-position motion generation (MG) problem because the output is specified as a complex motion of the coupler, link 3. The second method of Section 5.3 will be used here.
2.
Define the position vectors R1 and R2 and the vector P21 using Figure 5-1 and equation 5.1.
P2x P2y
R1
R2
P3x P3y
P21x R2 R1 P21y
P21x 1.264 P21y 0.793
p 21 3.
2
2
P21x P21y
p 21 1.492
From the trigonometric relationships given in Figure 5-1, determine 2 and 2. α θP3 θP2
α 37.300 deg
δ atan2 P21x P21y
δ 147.897 deg O4
4.
From a graphical solution (see figure at right), determine the values necessary for input to equations 5.8.
Y
1.250
P3 43.806°
z 0.0 B3
β 43.806 deg
P2 32.500°
57.012° B2
O2
ϕ 32.500 deg
P1
5.
Solve for the WZ dyad using equations 5.8. Z1x z cos ϕ
Z1x 0.000
Z1y z sin ϕ
A 0.278
D sin α
B 0.692
E p 21 cos δ
C 0.205
F p 21 sin δ
A cos β 1 B sin β
C cos α 1
Z1y 0.000 D 0.606
E 1.264
F 0.793
X
DESIGN OF MACHINERY - 5th Ed.
W1x
W1y w
SOLUTION MANUAL 5-9-2
A C Z1x D Z1y E B C Z1y D Z1x F
W1x 1.618
2 A A C Z1y D Z1x F B C Z1x D Z1y E
W1y 1.175
2 A 2
2
W1x W1y
w 2.000
θ atan2 W1x W1y 5.
θ 35.994 deg
From the graphical solution (see figure above), determine the values necessary for input to equations 5.12. s 1.250
6.
γ 57.012 deg
ψ 147.5 deg
Solve for the US dyad using equations 5.12. S 1x s cos ψ
S 1x 1.054
A 0.456
D sin α
B 0.839
E p 21 cos δ
C 0.205
F p 21 sin δ
A cos γ 1 B sin γ
C cos α 1
U1x
U1y u
S 1y s sin ψ
A C S 1x D S 1y E B C S 1y D S 1x F 2 A A C S 1y D S 1x F B C S 1x D S 1y E 2 A 2
2
U1x U1y
E 1.264
F 0.793
U1x 0.675
U1y 1.883
σ 70.278 deg
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
V1x z cos ϕ s cos ψ
V1x 1.054
V1y z sin ϕ s sin ψ
V1y 0.672
θ atan2 V1x V1y
θ 32.500 deg
v Link 1:
2
2
V1x V1y
v 1.250
G1x w cos θ v cos θ u cos σ
G1x 1.997
G1y w sin θ v sin θ u sin σ
G1y 2.386
θ atan2 G1x G1y
θ 50.070 deg
g 8.
D 0.606
u 2.000
σ atan2 U1x U1y 7.
S 1y 0.672
2
2
G1x G1y
g 3.112
Determine the initial and final values of the input crank with respect to the vector G.
DESIGN OF MACHINERY - 5th Ed.
9.
SOLUTION MANUAL 5-9-3
θ2i θ θ
θ2i 14.077 deg
θ2f θ2i β
θ2f 29.729 deg
Define the coupler point with respect to point A and the vector V. rp z
δp ϕ θ
rp 0.000
δp 0.000 deg
which is correct for the assumption that the precision point is at C. 10. Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ atan2 P2x P2y R1
2
ρ 120.033 deg
2
P2x P2y
R1 2.470
O2x 2.854
O2y 0.963
O4x 0.857
O4y 3.349
O2x R1 cos ρ z cos ϕ w cos θ O2y R1 sin ρ z sin ϕ w sin θ O4x R1 cos ρ s cos ψ u cos σ O4y R1 sin ρ s sin ψ u sin σ
11. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.
θrot atan2 O4x O2x O4y O2y 12. Determine the Grashof condition. Condition( a b c d )
θrot 50.070 deg
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "non-Grashof" 13. DESIGN SUMMARY Link 2:
w 2.000
θ 35.994 deg
Link 3:
v 1.250
θ 32.500 deg
Link 4:
u 2.000
σ 70.278 deg
Link 1:
g 3.112
θ 50.070 deg
Coupler:
rp 0.000
δp 0.000 deg
Crank angles:
θ2i 14.077 deg θ2f 29.729 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-10-1
PROBLEM 5-10 Statement:
Design a linkage to carry the body in Figure P5-1 through the three positions P1, P2 and P3 at the angles shown in the figure. Use analytical synthesis without regard for the fixed pivots shown. Use the free choices given below.
Given:
Coordinates of the points P1 and P2 with respect to P1: P1x 0.0
P1y 0.0
P2x 1.236
P3x 2.500
P3y 2.931
P2y 2.138
Angles made by the body in positions 1, 2 and 3:
θP1 210 deg
θP2 147.5 deg
θP3 110.2 deg
Free choices for the WZ dyad : β 30.0 deg
β 60.0 deg
Free choices for the US dyad : γ 10.0 deg Solution: 1.
2.
3.
γ 25.0 deg
See Figure P5-1 and Mathcad file P0510.
Determine the magnitudes and orientation of the position difference vectors. 2
2
p 21 2.470
δ atan2 P2x P2y
δ 120.033 deg
2
2
p 31 3.852
δ atan2 P3x P3y
δ 130.463 deg
p 21
P2x P2y
p 31
P3x P3y
Determine the angle changes of the coupler between precision points. α θP2 θP1
α 62.500 deg
α θP3 θP1
α 99.800 deg
Evaluate terms in the WZ coefficient matrix and constant vector from equations 5.26 and form the matrix and vector:
B sin β
E p 21 cos δ
H cos α 1
A cos β 1
D sin α
K sin α
L p 31 cos δ
M p 21 sin δ
B C D
E L CC M N
G H K
F K H A
F cos β 1
G sin β
A F AA B G
C cos α 1
D C
N p 31 sin δ
W1x W 1y AA 1 CC Z1x Z1y
The components of the W and Z vectors are: W1x 2.920
W1y 1.720
Z1x 0.756
Z1y 0.442
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-10-2
θ atan2 W1x W1y
ϕ 149.697 deg
W1x2 W1y2 , w 3.389
The length of link 2 is: w
Z1x2 Z1y2 , z 0.876
The length of vector Z is: z 4.
ϕ atan2 Z1x Z1y
θ 30.493 deg
Evaluate terms in the US coefficient matrix and constant vector from equations 5.31 and form the matrix and vector:
A' cos γ 1
B' sin γ
C cos α 1
E p 21 cos δ
H cos α 1
D sin α
G' sin γ
K sin α
L p 31 cos δ
A' F' AA B' G'
F' cos γ 1
M p 21 sin δ
B' C D
E L CC M N
G' H K A' D C F' K H
N p 31 sin δ
U1x U 1y AA 1 CC S1x S1y
The components of the U and S vectors are: U1x 1.009
U1y 2.693
S 1x 0.792
S 1y 2.418
σ atan2 U1x U1y
σ 110.545 deg
ψ atan2 S 1x S 1y
ψ 108.125 deg
The length of link 4 is: u
U1x2 U1y2 , u 2.875
The length of vector S is: s 5.
S 1x2 S 1y2 , s 2.544
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
V1x Z1x S 1x
V1x 0.036
V1y Z1y S 1y
V1y 1.976
θ atan2 V1x V1y
θ 88.968 deg
v Link 1:
2
2
V1x V1y
v 1.977
G1x W1x V1x U1x
G1x 3.965
G1y W1y V1y U1y
G1y 1.003
θ atan2 G1x G1y
θ 14.202 deg
DESIGN OF MACHINERY - 5th Ed.
g 6.
7.
8.
9.
2
SOLUTION MANUAL 5-10-3
2
G1x G1y
g 4.090
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 16.291 deg
θ2f θ2i β
θ2f 76.291 deg
Define the coupler point with respect to point A and the vector V. rp z
δp ϕ θ
rp 0.876
δp 238.665 deg
Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. O2x z cos ϕ w cos θ
O2x 2.164
O2y z sin ϕ w sin θ
O2y 1.278
O4x s cos ψ u cos σ
O4x 1.801
O4y s sin ψ u sin σ
O4y 0.274
Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.
θrot atan2 O4x O2x O4y O2y
θrot 14.202 deg
10. Determine the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "Grashof" 11. DESIGN SUMMARY Link 2:
w 3.389
θ 30.493 deg
Link 3:
v 1.977
θ 88.968 deg
Link 4:
u 2.875
σ 110.545 deg
Link 1:
g 4.090
θ 14.202 deg
Coupler:
rp 0.876
δp 238.665 deg
Crank angles:
θ2i 16.291 deg θ2f 76.291 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-10-4
12. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.
Y
P3 S3
Z3 A3
B1
S2 P2
B3
V3
V2 S1
A2
W3
60.0°
B2
10.0° 25.0° V1
Z1
30.0°
A1
U2 U1 X
W2
P1 W1 G1
O2
O4
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-11-1
PROBLEM 5-11 Statement:
Given:
Solution: 1.
2.
Design a linkage to carry the body in Figure P5-1 through the three positions P1, P2 and P3 at the angles shown in the figure. Use analytical synthesis and design it for the fixed pivots shown. P21x 1.236 O2x 2.164
P21y 2.138 O2y 1.260
P31x 2.500 O4x 2.190
P31y 2.931 O4y 1.260
Body angles:
θP1 210 deg
θP2 147.5 deg
θP3 110.2 deg
See Figure P5-1 and Mathcad file P0511.
Determine the angle changes between precision points from the body angles given. α θP2 θP1
α 62.500 deg
α θP3 θP1
α 99.800 deg
Using Figure 5-6, determine the magnitudes of R1, R2, and R3 and their x and y components. R1x O2x
3.
4.
R1x 2.164
R1y O2y
R2x R1x P21x
R2x 0.928
R2y R1y P21y
R2y 3.398
R3x R1x P31x
R3x 0.336
R3y R1y P31y
R3y 4.191
2
2
R1 2.504
2
2
R2 3.522
2
2
R3 4.204
R1
R1x R1y
R2
R2x R2y
R3
R3x R3y
R1y 1.260
Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ atan2 R1x R1y
ζ 30.210 deg
ζ atan2 R2x R2y
ζ 74.725 deg
ζ atan2 R3x R3y
ζ 94.584 deg
Solve for 2 and 3 using equations 5.34
C3 1.209
C4 6.538
C5 1.189
C1 R3 cos α ζ R2 cos α ζ C2 R3 sin α ζ R2 sin α ζ C3 R1 cos α ζ R3 cos ζ
C4 R1 sin α ζ R3 sin ζ
C5 R1 cos α ζ R2 cos ζ
C1 0.372 C2 3.726
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-11-2
C6 R1 sin α ζ R2 sin ζ 2
C6 4.736
2
A1 C3 C4
A1 44.206
A2 C3 C6 C4 C5
A2 2.046
A3 C4 C6 C3 C5
A3 32.399
A4 C2 C3 C1 C4
A4 6.937
A5 C4 C5 C3 C6
A5 2.046
A6 C1 C3 C2 C4
A6 23.911
K1 A2 A4 A3 A6
K1 760.497
K2 A3 A4 A5 A6
K2 273.669
2
K3
2
2
2
A1 A2 A3 A4 A6
2
K3 140.232
2
K K 2 K 2 K 2 2 1 2 3 β 2 atan K1 K3
β 60.217 deg
K K 2 K 2 K 2 2 1 2 3 β 2 atan K1 K3
β 99.800 deg
The second value is the same as 3, so use the first value
β β
A5 sin β A3 cos β A6 A1
β 30.143 deg
A3 sin β A2 cos β A4 A1
β 30.143 deg
β acos
β asin
β β
Since both values are the same, 5.
Repeat steps 2, 3, and 4 for the right-hand dyad to find 1 and 2. R1x O4x
R1x 2.190
R1y O4y
R2x R1x P21x
R2x 3.426
R2y R1y P21y
R2y 3.398
R3x R1x P31x
R3x 4.690
R3y R1y P31y
R3y 4.191
R1
2
2
R1x R1y
R1 2.527
R1y 1.260
DESIGN OF MACHINERY - 5th Ed.
6.
7.
SOLUTION MANUAL 5-11-3
2
2
R2 4.825
2
2
R3 6.290
R2
R2x R2y
R3
R3x R3y
Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ atan2 R1x R1y
ζ 150.086 deg
ζ atan2 R2x R2y
ζ 135.235 deg
ζ atan2 R3x R3y
ζ 138.216 deg
Solve for 2 and 3 using equations 5.34
C3 6.304
C4 2.247
C5 3.532
C6 0.874
C1 R3 cos α ζ R2 cos α ζ C2 R3 sin α ζ R2 sin α ζ C3 R1 cos α ζ R3 cos ζ
C4 R1 sin α ζ R3 sin ζ
C5 R1 cos α ζ R2 cos ζ
C6 R1 sin α ζ R2 sin ζ 2
2
C1 2.380 C2 3.298
A1 C3 C4
A1 44.796
A2 C3 C6 C4 C5
A2 2.431
A3 C4 C6 C3 C5
A3 24.233
A4 C2 C3 C1 C4
A4 15.441
A5 C4 C5 C3 C6
A5 2.431
A6 C1 C3 C2 C4
A6 22.414
K1 A2 A4 A3 A6
K1 505.612
K2 A3 A4 A5 A6
K2 428.679
2
K3
2
2
2
A1 A2 A3 A4 A6
2
2
K3 336.363
K K 2 K 2 K 2 2 1 2 3 γ 2 atan K1 K3
γ 19.215 deg
K K 2 K 2 K 2 2 1 2 3 2 atan K1 K3
99.800 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-11-4
The second value is the same as 3, so use the first value
γ γ
A5 sin γ A3 cos γ A6 A1
6.628 deg
A3 sin γ A2 cos γ A4 A1
6.628 deg
acos
asin
Since 2 is not in the first quadrant , 8.
γ
Use the method of Section 5.7 to synthesize the linkage. Start by determining the magnitudes of the vectors P21 and P31 and their angles with respect to the X axis. 2
p 21
2
P21x P21y
p 21 2.470
δ atan2 P21x P21y 2
p 31
δ 120.033 deg
2
P31x P31y
p 31 3.852
δ atan2 P31x P31y 9.
δ 130.463 deg
Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:
B sin β
E p 21 cos δ
H cos α 1
A cos β 1
D sin α
F cos β 1
G sin β
K sin α
L p 31 cos δ
A F AA B G
C cos α 1
M p 21 sin δ
B C D
E L CC M N
G H K A D C F K H
N p 31 sin δ
W1x W1y AA 1 CC Z1x Z1y
10. The components of the W and Z vectors are: W1x 2.915 11. The length of link 2 is:
w
W1y 1.702 2
Z1x 0.751
2
W1x W1y
Z1y 0.442
w 3.376
12. Evaluate terms in the US coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:
A' cos γ 1
D sin α
B' sin γ
E p 21 cos δ
C cos α 1
F' cos γ 1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-11-5
G' sin γ
K sin α
L p 31 cos δ
A' F' AA B' G'
H cos α 1
M p 21 sin δ
B' C D
N p 31 sin δ
E L CC M N
G' H K A' D C F' K H
U1x U1y AA 1 CC S1x S1y
13. The components of the W and Z vectors are: U1x 1.371
14. The length of link 4 is:
U1y 3.634 2
u
U1x U1y
S1x 0.819
2
S1y 2.374
u 3.884
15. Solving for links 3 and 1 from equations 5.2a and 5.2b. V1x Z1x S1x
V1x 0.068
V1y Z1y S1y
V1y 1.932 v
The length of link 3 is:
2
2
V1x V1y
v 1.933
G1x W1x V1x U1x
G1x 4.354
G1y W1y V1y U1y
G1y 2.220 10
g
The length of link 1 is:
2
G1x G1y
2
15
g 4.354
16. Check the location of the fixed pivots with respect to the global frame using the calculated vectors W1, Z1, U1, and S1. O2x Z1x W1x
O2x 2.164
O2y Z1y W1y
O2y 1.260
O4x S1x U1x
O4x 2.190
O4y S1y U1y
O4y 1.260
These check with Figure P5-1. 17. Determine the location of the coupler point with respect to point A and line AB. 2
2
z 0.871
2
2
s 2.511
Distance from A to P
z
Z1x Z1y
Angle BAP (p)
s
S1x S1y
rP z
ψ atan2( S1x S1y)
ψ 109.037 deg
ϕ atan2( Z1x Z1y )
ϕ 149.555 deg
θ atan2 z cos ϕ s cos ψ z sin ϕ s sin ψ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-11-6
θ 87.994 deg
δp ϕ θ
δp 237.549 deg
18. DESIGN SUMMARY Link 1:
g 4.354
Link 2:
w 3.376
Link 3:
v 1.933
Link 4:
u 3.884
Coupler point:
rP 0.871
δp 237.549 deg
19. VERIFICATION: The calculated values of g (length of the ground link) and of the coordinates of O2 and O4 give the same values as those on the problem statement, verifying that the calculated values for the other links and the coupler point are correct.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-12-1
PROBLEM 5-12 Statement:
Design a linkage to carry the body in Figure P5-2 through the two positions P1 and P2 at the angles shown in the figure. Use analytical synthesis without regard for the fixed pivots shown. Use the free choices given below.
Given:
Coordinates of the points P1 and P2 with respect to P1: P1x 0.0
P1y 0.0
P2x 1.903
P2y 1.347
Angles made by the body in positions 1 and 2:
θP1 101.0 deg
θP2 62.0 deg
Free choices for the WZ dyad : z 2.000
β 30.0 deg
ϕ 150.0 deg
γ 40.0 deg
ψ 50.0 deg
Free choices for the US dyad : s 3.000 Solution:
See Figure P5-2 and Mathcad file P0512.
1.
Note that this is a two-position motion generation (MG) problem because the output is specified as a complex motion of the coupler, link 3. Because of the data given in the hint, the second method of Section 5.3 will be used here.
2.
Define the position vectors R1 and R2 and the vector P21 using Figure 5-1 and equation 5.1. R1
P1x P1y
R2
P2x P2y
P21x R2 R1 P21y
P21x 1.903 P21y 1.347
p 21 3.
4.
2
2
P21x P21y
p 21 2.331
From the trigonometric relationships given in Figure 5-1, determine 2 and 2. α θP2 θP1
α 39.000 deg
δ atan2 P21x P21y
δ 35.292 deg
Solve for the WZ dyad using equations 5.8. Z1x z cos ϕ
Z1y z sin ϕ
A 0.134
D sin α
B 0.500
E p 21 cos δ
E 1.903
C 0.223
F p 21 sin δ
F 1.347
B sin β
C cos α 1
Z1y 1.000
A cos β 1
W1x
Z1x 1.732
A C Z1x D Z1y E B C Z1y D Z1x F 2 A
D 0.629
W1x 0.452
DESIGN OF MACHINERY - 5th Ed.
W1y w
SOLUTION MANUAL 5-12-2
A C Z1y D Z1x F B C Z1x D Z1y E
W1y 1.896
2 A 2
2
W1x W1y
w 1.949
θ atan2 W1x W1y 5.
θ 76.607 deg
Solve for the US dyad using equations 5.12. S 1x s cos ψ
S 1x 1.928 A 0.234
D sin α
B 0.643
E p 21 cos δ
E 1.903
C 0.223
F p 21 sin δ
F 1.347
B sin γ
C cos α 1
U1y u
D 0.629
A C S 1x D S 1y E B C S 1y D S 1x F 2 A A C S 1y D S 1x F B C S 1x D S 1y E
2
2
U1x U1y
u 6.284 σ 81.540 deg
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
V1x z cos ϕ s cos ψ
V1x 3.660
V1y z sin ϕ s sin ψ
V1y 3.298
θ atan2 V1x V1y
θ 137.980 deg
v Link 1:
2
2
V1x V1y
v 4.927
G1x w cos θ v cos θ u cos σ
G1x 4.133
G1y w sin θ v sin θ u sin σ
G1y 7.617
θ atan2 G1x G1y
θ 118.485 deg
g 7.
U1x 0.924
U1y 6.216
2 A
σ atan2 U1x U1y 6.
S 1y 2.298
A cos γ 1
U1x
S 1y s sin ψ
2
2
G1x G1y
g 8.667
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 195.092 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-12-3
θ2f θ2i β 8.
9.
θ2f 165.092 deg
Define the coupler point with respect to point A and the vector V. rp z
δp ϕ θ
rp 2.000
δp 12.020 deg
Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ 0 deg R1
ρ 0.000 deg
2
2
P1x P1y
R1 0.000
O2x 1.281
O2y 0.896
O4x 2.853
O4y 8.514
O2x R1 cos ρ z cos ϕ w cos θ O2y R1 sin ρ z sin ϕ w sin θ O4x R1 cos ρ s cos ψ u cos σ O4y R1 sin ρ s sin ψ u sin σ
10. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.
θrot atan2 O4x O2x O4y O2y
θrot 118.485 deg
11. Determine the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "Grashof" 12. DESIGN SUMMARY Link 2:
w 1.949
θ 76.607 deg
Link 3:
v 4.927
θ 137.980 deg
Link 4:
u 6.284
σ 81.540 deg
Link 1:
g 8.667
θ 118.485 deg
Coupler:
rp 2.000
δp 12.020 deg
Crank angles:
θ2i 195.092 deg θ2f 165.092 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-12-4
13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.
O4
Y
1.903
U2 G1
U1
B2
S2
39.0° B1
V2 P2
V1 O2 S1
Z2 1.347
W1
W2
P1
X Z1 A1
A2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-13-1
PROBLEM 5-13 Statement:
Design a linkage to carry the body in Figure P5-2 through the two positions P2 and P3 at the angles shown in the figure. Use analytical synthesis without regard for the fixed pivots shown. Hint: First try a rough graphical solution to create realistic values for free choices.
Given:
Coordinates of the points P2 and P3 with respect to P1: P2x 1.903
P2y 1.347
P3x 1.389
P3y 1.830
Angles made by the body in positions 1 and 2:
θP2 62.0 deg Solution:
θP3 39.0 deg
See Figure P5-2 and Mathcad file P0513.
1.
Note that this is a two-position motion generation (MG) problem because the output is specified as a complex motion of the coupler, link 3. The second method of Section 5.3 will be used here.
2.
Define the position vectors R1 and R2 and the vector P21 using Figure 5-1 and equation 5.1.
P2x P2y
R1
R2
P3x P3y
P21x R2 R1 P21y
P21x 0.514 P21y 0.483
p 21 3.
4.
2
2
P21x P21y
From the trigonometric relationships given in Figure 5-1, determine 2 and 2. α θP3 θP2
α 23.000 deg
δ atan2 P21x P21y
δ 136.781 deg
From a graphical solution (see figure next page), determine the values necessary for input to equations 5.8. z 3
5.
p 21 0.705
β 45.0 deg
ϕ 100 deg
Solve for the WZ dyad using equations 5.8. Z1x z cos ϕ
Z1x 0.521
Z1y z sin ϕ
A 0.293
D sin α
B 0.707
E p 21 cos δ
A cos β 1 B sin β
W1x
W1y w
D 0.391
C 0.079 F p 21 sin δ A C Z1x D Z1y E B C Z1y D Z1x F
C cos α 1
2 A A C Z1y D Z1x F B C Z1x D Z1y E 2 A 2
2
W1x W1y
θ atan2 W1x W1y
Z1y 2.954
E 0.514 F 0.483 W1x 1.476
W1y 1.807 w 2.333 θ 50.759 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-13-2
Y
B3 2.000
O4
20.000°
P3
B2
P2
20.0°
3.000 X
P1
100.0°
23.0° A3 45.000° A2
O2
5.
From the graphical solution (see figure above), determine the values necessary for input to equations 5.12. s 2.000
6.
γ 20.0 deg
ψ 20.0 deg
Solve for the US dyad using equations 5.12. S 1x s cos ψ
S 1x 1.879
A 0.060
D sin α
B 0.342
E p 21 cos δ
C 0.079
F p 21 sin δ
A cos γ 1 B sin γ
C cos α 1
U1x
U1y
u
S 1y s sin ψ
A C S 1x D S 1y E B C S 1y D S 1x F 2 A A C S 1y D S 1x F B C S 1x D S 1y E 2 A 2
2
U1x U1y
σ atan2 U1x U1y 7.
S 1y 0.684 D 0.391
E 0.514
F 0.483
U1x 3.346
U1y 0.306
u 3.360 σ 5.216 deg
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
V1x z cos ϕ s cos ψ
V1x 2.400
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-13-3
V1y z sin ϕ s sin ψ θ atan2 V1x V1y v Link 1:
2
θ 123.413 deg
2
V1x V1y
v 4.359
G1x w cos θ v cos θ u cos σ
9.
G1x 4.271
G1y w sin θ v sin θ u sin σ
G1y 5.751
θ atan2 G1x G1y
θ 126.601 deg
g 8.
V1y 3.638
2
2
G1x G1y
g 7.163
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 75.842 deg
θ2f θ2i β
θ2f 30.842 deg
Define the coupler point with respect to point A and the vector V. rp z
δp ϕ θ
rp 3.000
δp 23.413 deg
10. Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ atan2 P2x P2y R1
2
ρ 35.292 deg
2
P2x P2y
R1 2.331
O2x 0.948
O2y 3.414
O4x 3.323
O4y 2.337
O2x R1 cos ρ z cos ϕ w cos θ O2y R1 sin ρ z sin ϕ w sin θ O4x R1 cos ρ s cos ψ u cos σ O4y R1 sin ρ s sin ψ u sin σ
11. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis the line O2O4.
θrot atan2 O4x O2x O4y O2y
θrot 126.601 deg
12. Determine the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "non-Grashof"
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-13-4
13. DESIGN SUMMARY Link 2:
w 2.333
θ 50.759 deg
Link 3:
v 4.359
θ 123.413 deg
Link 4:
u 3.360
σ 5.216 deg
Link 1:
g 7.163
θ 126.601 deg
Coupler:
rp 3.000
δp 23.413 deg
Crank angles:
θ2i 75.842 deg θ2f 30.842 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-14-1
PROBLEM 5-14 Statement:
Design a linkage to carry the body in Figure P5-2 through the three positions P1, P2 and P3 at the angles shown in the figure. Use analytical synthesis without regard for the fixed pivots shown.
Given:
Coordinates of the points P1 and P2 with respect to P1: P1x 0.0
P1y 0.0
P2x 1.903
P3x 1.389
P3y 1.830
P2y 1.347
Angles made by the body in positions 1, 2 and 3:
θP1 101 deg
θP2 62.0 deg
θP3 39.0 deg
Free choices for the WZ dyad : β 40.0 deg
β 75.0 deg
Free choices for the US dyad : γ 0.0 deg Solution: 1.
2.
3.
γ 30.0 deg
See Figure P5-2 and Mathcad file P0514.
Determine the magnitudes and orientation of the position difference vectors. 2
2
p 21 2.331
δ atan2 P2x P2y
δ 35.292 deg
2
2
p 31 2.297
δ atan2 P3x P3y
δ 52.801 deg
p 21
P2x P2y
p 31
P3x P3y
Determine the angle changes of the coupler between precision points. α θP2 θP1
α 39.000 deg
α θP3 θP1
α 62.000 deg
Evaluate terms in the WZ coefficient matrix and constant vector from equations 5.26 and form the matrix and vector:
B sin β
E p 21 cos δ
H cos α 1
A cos β 1 D sin α G sin β
L p 31 cos δ
A F AA B G
B C D
G H K A D C F K H
The components of the W and Z vectors are:
C cos α 1
M p 21 sin δ
E L CC M N
F cos β 1
K sin α
N p 31 sin δ
W1x W 1y AA 1 CC Z1x Z1y
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-14-2
W1x 3.110
W1y 1.061
Z1x 0.297
Z1y 3.201
θ atan2 W1x W1y
θ 18.843 deg
ϕ atan2 Z1x Z1y
ϕ 84.698 deg
W1x2 W1y2 , w 3.286
The length of link 2 is: w
Z1x2 Z1y2 , z 3.215
The length of vector Z is: z 4.
Evaluate terms in the US coefficient matrix and constant vector from equations 5.31 and form the matrix and vector:
A' cos γ 1
B' sin γ
C cos α 1
E p 21 cos δ
H cos α 1
D sin α
G' sin γ
K sin α
L p 31 cos δ
A' F' AA B' G'
F' cos γ 1
M p 21 sin δ
B' C D
E L CC M N
G' H K A' D C F' K H
N p 31 sin δ
U1x U 1y AA 1 CC S1x S1y
The components of the U and S vectors are: U1x 1.658
U1y 3.361
S 1x 2.853
S 1y 2.013
σ atan2 U1x U1y
σ 63.740 deg
ψ atan2 S 1x S 1y
ψ 144.792 deg
The length of link 4 is: u
U1x2 U1y2 , u 3.748
The length of vector S is: s 5.
S 1x2 S 1y2 , s 3.492
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
V1x Z1x S 1x
V1x 3.150
V1y Z1y S 1y
V1y 1.188
θ atan2 V1x V1y
θ 20.657 deg
v Link 1:
2
2
V1x V1y
v 3.367
G1x W1x V1x U1x
G1x 4.602
G1y W1y V1y U1y
G1y 3.234
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-14-3
θ atan2 G1x G1y
g 6.
7.
8.
9.
2
θ 35.099 deg
2
G1x G1y
g 5.625
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 16.256 deg
θ2f θ2i β
θ2f 91.256 deg
Define the coupler point with respect to point A and the vector V. rp z
δp ϕ θ
rp 3.215
δp 64.041 deg
Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. O2x z cos ϕ w cos θ
O2x 3.407
O2y z sin ϕ w sin θ
O2y 2.140
O4x s cos ψ u cos σ
O4x 1.195
O4y s sin ψ u sin σ
O4y 5.374
Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.
θrot atan2 O4x O2x O4y O2y
θrot 35.099 deg
10. Determine the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "non-Grashof" 11. DESIGN SUMMARY Link 2:
w 3.286
θ 18.843 deg
Link 3:
v 3.367
θ 20.657 deg
Link 4:
u 3.748
σ 63.740 deg
Link 1:
g 5.625
θ 35.099 deg
Coupler:
rp 3.215
δp 64.041 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-14-4
Crank angles:
θ2i 16.256 deg
θ2f 91.256 deg
12. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.
Y
P3 P2
Z3 A3 Z2 P1
V3 W3
X
S3
A2
S2
S1
75.0°
W2
B3
Z1
V2
B1 , B2
40.0° O2
W1 V1 30.0°
A1 G1
U1 U2
U3
O4
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-15-1
PROBLEM 5-15 Statement:
Given:
Solution: 1.
2.
Design a linkage to carry the body in Figure P5-2 through the three positions P1, P2 and P3 at the angles shown in the figure. Use analytical synthesis and design it for the fixed pivots shown. P21x 1.903
P21y 1.347
P31x 1.389
P31y 1.830
O2x 0.884
O2y 1.251
O4x 3.062
O4y 1.251
Body angles:
θP1 101 deg
θP2 62 deg
θP3 39 deg
See Figure P5-2 and Mathcad file P0515.
Determine the angle changes between precision points from the body angles given. α θP2 θP1
α 39.000 deg
α θP3 θP1
α 62.000 deg
Using Figure 5-6, determine the magnitudes of R1, R2, and R3 and their x and y components. R1x O2x
3.
4.
R1x 0.884
R1y O2y
R2x R1x P21x
R2x 2.787
R2y R1y P21y
R2y 2.598
R3x R1x P31x
R3x 2.273
R3y R1y P31y
R3y 3.081
2
2
R1 1.532
2
2
R2 3.810
2
2
R3 3.829
R1
R1x R1y
R2
R2x R2y
R3
R3x R3y
R1y 1.251
Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ atan2 R1x R1y
ζ 54.754 deg
ζ atan2 R2x R2y
ζ 42.990 deg
ζ atan2 R3x R3y
ζ 53.582 deg
Solve for 2 and 3 using equations 5.34
C3 0.753
C4 3.274
C5 1.313
C1 R3 cos α ζ R2 cos α ζ C2 R3 sin α ζ R2 sin α ζ C3 R1 cos α ζ R3 cos ζ
C4 R1 sin α ζ R3 sin ζ
C5 R1 cos α ζ R2 cos ζ
C1 0.103 C2 2.205
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-15-2
C6 R1 sin α ζ R2 sin ζ 2
C6 2.182
2
A1 C3 C4
A1 11.288
A2 C3 C6 C4 C5
A2 2.654
A3 C4 C6 C3 C5
A3 8.134
A4 C2 C3 C1 C4
A4 1.324
A5 C4 C5 C3 C6
A5 2.654
A6 C1 C3 C2 C4
A6 7.297
K1 A2 A4 A3 A6
K1 55.842
K2 A3 A4 A5 A6
K2 30.136
2
K3
2
2
2
A1 A2 A3 A4 A6
2
K3 0.392
2
K K 2 K 2 K 2 2 1 2 3 β 2 atan K1 K3
β 118.708 deg
K K 2 K 2 K 2 2 1 2 3 β 2 atan K1 K3
β 62.000 deg
The second value is the same as 3, so use the first value
β β
A5 sin β A3 cos β A6 A1
β 59.564 deg
A3 sin β A2 cos β A4 A1
β 59.564 deg
β acos
β asin
β β
Since both values are the same, 5.
Repeat steps 2, 3, and 4 for the right-hand dyad to find 1 and 2. R1x O4x
R1x 3.062
R1y O4y
R2x R1x P21x
R2x 1.159
R2y R1y P21y
R2y 2.598
R3x R1x P31x
R3x 1.673
R3y R1y P31y
R3y 3.081
2
2
R1 3.308
2
2
R2 2.845
R1
R1x R1y
R2
R2x R2y
R1y 1.251
DESIGN OF MACHINERY - 5th Ed.
R3 6.
7.
SOLUTION MANUAL 5-15-3
2
2
R3x R3y
R3 3.506
Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ atan2 R1x R1y
ζ 157.777 deg
ζ atan2 R2x R2y
ζ 114.042 deg
ζ atan2 R3x R3y
ζ 118.502 deg
Solve for 2 and 3 using equations 5.34
C3 1.340
C4 0.210
C5 0.433
C6 0.301
C1 R3 cos α ζ R2 cos α ζ C2 R3 sin α ζ R2 sin α ζ C3 R1 cos α ζ R3 cos ζ
C4 R1 sin α ζ R3 sin ζ
C5 R1 cos α ζ R2 cos ζ
C6 R1 sin α ζ R2 sin ζ 2
2
C1 1.111 C2 1.204
A1 C3 C4
A1 1.840
A2 C3 C6 C4 C5
A2 0.495
A3 C4 C6 C3 C5
A3 0.517
A4 C2 C3 C1 C4
A4 1.847
A5 C4 C5 C3 C6
A5 0.495
A6 C1 C3 C2 C4
A6 1.236
K1 A2 A4 A3 A6
K1 1.553
K2 A3 A4 A5 A6
K2 0.344
2
K3
2
2
2
A1 A2 A3 A4 A6
2
2
K3 1.033
K K 2 K 2 K 2 2 1 2 3 γ 2 atan K1 K3
γ 62.000 deg
K K 2 K 2 K 2 2 1 2 3 2 atan K1 K3
36.991 deg
The first value is the same as 3, so use the second value
γ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-15-4
A5 sin γ A3 cos γ A6 A1
73.415 deg
A3 sin γ A2 cos γ A4 A1
73.415 deg
acos
asin
Since 2 is not in the first quadrant , 8.
γ
Use the method of Section 5.7 to synthesize the linkage. Start by determining the magnitudes of the vectors P21 and P31 and their angles with respect to the X axis. 2
p 21
2
P21x P21y
p 21 2.331
δ atan2 P21x P21y 2
p 31
δ 35.292 deg
2
P31x P31y
p 31 2.297
δ atan2 P31x P31y 9.
δ 52.801 deg
Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:
B sin β
E p 21 cos δ
H cos α 1
A cos β 1
D sin α
F cos β 1
G sin β
K sin α
L p 31 cos δ
M p 21 sin δ
B C D
A F AA B G
C cos α 1
E L CC M N
G H K A D C F K H
N p 31 sin δ
W1x W1y AA 1 CC Z1x Z1y
10. The components of the W and Z vectors are: W1x 1.262 w
11. The length of link 2 is:
W1y 1.109 2
Z1x 0.378
2
W1x W1y
Z1y 2.360
w 1.680
12. Evaluate terms in the US coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:
A' cos γ 1
B' sin γ
C cos α 1
E p 21 cos δ
H cos α 1
D sin α
G' sin γ
L p 31 cos δ
M p 21 sin δ
F' cos γ 1
K sin α
N p 31 sin δ
DESIGN OF MACHINERY - 5th Ed.
A' F' AA B' G'
SOLUTION MANUAL 5-15-5
B' C D
E L CC M N
G' H K A' D C F' K H
U1x U1y AA 1 CC S1x S1y
13. The components of the W and Z vectors are: U1x 0.326
14. The length of link 4 is:
U1y 0.830 2
u
U1x U1y
S1x 2.736
2
S1y 0.421
u 0.892
15. Solving for links 3 and 1 from equations 5.2a and 5.2b. V1x Z1x S1x
V1x 2.359
V1y Z1y S1y
V1y 1.939 v
The length of link 3 is:
2
2
V1x V1y
v 3.054
G1x W1x V1x U1x
G1x 3.946
G1y W1y V1y U1y
G1y 1.110 10
g
The length of link 1 is:
2
G1x G1y
2
15
g 3.946
16. Check the location of the fixed pivots with respect to the global frame using the calculated vectors W1, Z1, U1, and S1. O2x Z1x W1x
O2x 0.884
O2y Z1y W1y
O2y 1.251
O4x S1x U1x
O4x 3.062
O4y S1y U1y
O4y 1.251
These check with Figure P5-2. 17. Determine the location of the coupler point with respect to point A and line AB. 2
2
z 2.390
2
2
s 2.769
Distance from A to P
z
Z1x Z1y
Angle BAP (p)
s
S1x S1y
ψ atan2( S1x S1y)
ψ 171.262 deg
ϕ atan2( Z1x Z1y )
ϕ 99.095 deg
rP z
θ atan2 z cos ϕ s cos ψ z sin ϕ s sin ψ θ 39.430 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-15-6
δp ϕ θ
δp 59.666 deg
18. DESIGN SUMMARY Link 1:
g 3.946
Link 2:
w 1.680
Link 3:
v 3.054
Link 4:
u 0.892
Coupler point:
rP 2.390
δp 59.666 deg
19. VERIFICATION: The calculated values of g (length of the ground link) and of the coordinates of O2 and O4 give the same values as those on the problem statement, verifying that the calculated values for the other links and the coupler point are correct.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-16-1
PROBLEM 5-16 Statement:
Design a linkage to carry the body in Figure P5-3 through the two positions P1 and P2 at the angles shown in the figure. Use analytical synthesis without regard for the fixed pivots shown.
Given:
Coordinates of the points P1 and P2 with respect to P1: P1x 0.0
P1y 0.0
P2x 0.907
P2y 0.0
Angles made by the body in positions 1 and 2:
θP1 111.8 deg
θP2 191.1 deg
Free choices for the WZ dyad : z 1.500
β 44.0 deg
ϕ 50.0 deg
γ 55.0 deg
ψ 20.0 deg
Free choices for the US dyad : s 2.500 Solution:
See Figure P5-3 and Mathcad file P0516.
1.
Note that this is a two-position motion generation (MG) problem because the output is specified as a complex motion of the coupler, link 3. The second method of Section 5.3 will be used here.
2.
Define the position vectors R1 and R2 and the vector P21 using Figure 5-1 and equation 5.1. R1
P1x P1y
R2
P2x P2y
P21x R2 R1 P21y
P21x 0.907 P21y 0.000
p 21 3.
4.
2
2
P21x P21y
p 21 0.907
From the trigonometric relationships given in Figure 5-1, determine 2 and 2. α θP2 θP1
α 79.300 deg
δ atan2 P21x P21y
δ 180.000 deg
Solve for the WZ dyad using equations 5.8. Z1x z cos ϕ
B sin β
A cos β 1
C cos α 1 W1x
W1y
Z1x 0.964
Z1y z sin ϕ
A 0.281
D sin α
B 0.695
E p 21 cos δ
C 0.814
F p 21 sin δ
A C Z1x D Z1y E B C Z1y D Z1x F 2 A A C Z1y D Z1x F B C Z1x D Z1y E 2 A
Z1y 1.149 D 0.983
E 0.907
F 0.000 W1x 1.705
W1y 2.490
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-16-2
2
w
2
W1x W1y
w 3.018
θ atan2 W1x W1y 5.
θ 124.405 deg
Solve for the US dyad using equations 5.12. S 1x s cos ψ
S 1x 2.349
S 1y s sin ψ
A 0.426
D sin α
B 0.819
E p 21 cos δ
C 0.814
F p 21 sin δ
A cos γ 1 B sin γ
C cos α 1
D 0.983
E 0.907
F 0.000
A C S 1x D S 1y E B C S 1y D S 1x F
U1x
2 A A C S 1y D S 1x F B C S 1x D S 1y E
U1y
2
2
U1x U1y
u 2.654
σ atan2 U1x U1y 6.
σ 76.373 deg
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
V1x z cos ϕ s cos ψ
V1x 1.385
V1y z sin ϕ s sin ψ
V1y 2.004
θ atan2 V1x V1y
θ 124.648 deg
v Link 1:
2
2
V1x V1y
v 2.436
G1x w cos θ v cos θ u cos σ
8.
G1x 3.715
G1y w sin θ v sin θ u sin σ
G1y 2.094
θ atan2 G1x G1y
θ 150.596 deg
g 7.
U1x 0.625
U1y 2.579
2 A
u
S 1y 0.855
2
2
G1x G1y
g 4.265
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 275.001 deg
θ2f θ2i β
θ2f 319.001 deg
Define the coupler point with respect to point A and the vector V. rp z
δp ϕ θ
rp 1.500
δp 74.648 deg
DESIGN OF MACHINERY - 5th Ed.
9.
SOLUTION MANUAL 5-16-3
Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ 0 deg R1
ρ 0.000 deg
2
2
P1x P1y
R1 0.000
O2x 0.741
O2y 1.341
O4x 2.975
O4y 3.434
O2x R1 cos ρ z cos ϕ w cos θ O2y R1 sin ρ z sin ϕ w sin θ O4x R1 cos ρ s cos ψ u cos σ O4y R1 sin ρ s sin ψ u sin σ
10. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.
θrot atan2 O4x O2x O4y O2y
θrot 150.596 deg
11. Determine the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "non-Grashof" 12. DESIGN SUMMARY Link 2:
w 3.018
θ 124.405 deg
Link 3:
v 2.436
θ 124.648 deg
Link 4:
u 2.654
σ 76.373 deg
Link 1:
g 4.265
θ 150.596 deg
Coupler:
rp 1.500
δp 74.648 deg
Crank angles:
θ2i 275.001 deg
θ2f 319.001 deg
13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-16-4
Y
A1 Z1 P2
V1
P1
Z2 A2
X S1
44.0°
B1
W2 V2
S2
55.0° U1
G1
B2 U2
O4
W1 O2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-17-1
PROBLEM 5-17 Statement:
Design a linkage to carry the body in Figure P5-3 through the two positions P2 and P3 at the angles shown in the figure. Use analytical synthesis without regard for the fixed pivots shown.
Given:
Coordinates of the points P2 and P3 with respect to P1: P2x 0.907
P2y 0.0
P3x 1.447
P3y 0.0
Angles made by the body in positions 1 and 2:
θP2 191.1 deg
θP3 237.4 deg
Free choices for the WZ dyad : z 2.000
β 66.0 deg
ϕ 60.0 deg
γ 44.0 deg
ψ 30.0 deg
Free choices for the US dyad : s 3.000 Solution:
See Figure P5-3 and Mathcad file P0517.
1.
Note that this is a two-position motion generation (MG) problem because the output is specified as a complex motion of the coupler, link 3. The second method of Section 5.3 will be used here.
2.
Define the position vectors R1 and R2 and the vector P21 using Figure 5-1 and equation 5.1. R1
P2x P2y
R2
P3x P3y
P21x R2 R1 P21y
P21x 0.540 P21y 0.000
p 21 3.
4.
2
2
P21x P21y
p 21 0.540
From the trigonometric relationships given in Figure 5-1, determine 2 and 2. α θP3 θP2
α 46.300 deg
δ atan2 P21x P21y
δ 180.000 deg
Solve for the WZ dyad using equations 5.8. Z1x z cos ϕ
A 0.593
D sin α
B 0.914
E p 21 cos δ
C 0.309
F p 21 sin δ
B sin β
C cos α 1
W1y
Z1y z sin ϕ
A cos β 1
W1x
Z1x 1.000
A C Z1x D Z1y E B C Z1y D Z1x F 2 A A C Z1y D Z1x F B C Z1x D Z1y E 2 A
Z1y 1.732 D 0.723
E 0.540
F 0.000 W1x 0.227
W1y 1.771
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-17-2
2
w
2
W1x W1y
w 1.786
θ atan2 W1x W1y 5.
θ 97.314 deg
Solve for the US dyad using equations 5.12. S 1x s cos ψ
S 1x 2.598
S 1y s sin ψ
A 0.281
D sin α
B 0.695
E p 21 cos δ
C 0.309
F p 21 sin δ
A cos γ 1 B sin γ
C cos α 1
D 0.723
E 0.540
F 0.000
A C S 1x D S 1y E B C S 1y D S 1x F
U1x
2 A A C S 1y D S 1x F B C S 1x D S 1y E
U1y
2
2
U1x U1y
u 2.608
σ atan2 U1x U1y 6.
σ 65.609 deg
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
V1x z cos ϕ s cos ψ
V1x 1.598
V1y z sin ϕ s sin ψ
V1y 3.232
θ atan2 V1x V1y
θ 116.310 deg
v Link 1:
2
2
V1x V1y
v 3.606
G1x w cos θ v cos θ u cos σ
8.
G1x 2.902
G1y w sin θ v sin θ u sin σ
G1y 3.836
θ atan2 G1x G1y
θ 127.111 deg
g 7.
U1x 1.077
U1y 2.375
2 A
u
S 1y 1.500
2
2
G1x G1y
g 4.810
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 224.425 deg
θ2f θ2i β
θ2f 290.425 deg
Define the coupler point with respect to point A and the vector V. rp z
δp ϕ θ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-17-3
rp 2.000 9.
δp 56.310 deg
Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ atan2 P2x P2y R1
2
ρ 180.000 deg
2
P2x P2y
R1 0.907
O2x 1.680
O2y 0.039
O4x 4.582
O4y 3.875
O2x R1 cos ρ z cos ϕ w cos θ O2y R1 sin ρ z sin ϕ w sin θ O4x R1 cos ρ s cos ψ u cos σ O4y R1 sin ρ s sin ψ u sin σ
10. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.
θrot atan2 O4x O2x O4y O2y
θrot 127.111 deg
11. Determine the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "non-Grashof" 12. DESIGN SUMMARY Link 2:
w 1.786
θ 97.314 deg
Link 3:
v 3.606
θ 116.310 deg
Link 4:
u 2.608
σ 65.609 deg
Link 1:
g 4.810
θ 127.111 deg
Coupler:
rp 2.000
δp 56.310 deg
Crank angles:
θ2i 224.425 deg
θ2f 290.425 deg
13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-17-4
Y
A2 66.0° V1
Z1
A3
W1 Z2 W2
P3
O2
P2
S1
B2
U1
S2
V2
G1 44.0° B1 U2
O4
P1 X
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-18-1
PROBLEM 5-18 Statement:
Design a linkage to carry the body in Figure P5-3 through the three positions P1, P2 and P3 at the angles shown in the figure. Use analytical synthesis without regard for the fixed pivots shown.
Given:
Coordinates of the points P1 and P2 with respect to P1: P1x 0.0
P1y 0.0
P3x 1.447
P3y 0.0
P2x 0.907
P2y 0.0
Angles made by the body in positions 1, 2 and 3:
θP1 111.8 deg
θP2 191.1 deg
θP3 237.4 deg
Free choices for the WZ dyad : β 40.0 deg
β 80.0 deg
Free choices for the US dyad : γ 20.0 deg Solution: 1.
2.
3.
γ 50.0 deg
See Figure P5-3 and Mathcad file P0518.
Determine the magnitudes and orientation of the position difference vectors. 2
2
p 21 0.907
δ atan2 P2x P2y
δ 180.000 deg
2
2
p 31 1.447
δ atan2 P3x P3y
δ 180.000 deg
p 21
P2x P2y
p 31
P3x P3y
Determine the angle changes of the coupler between precision points. α θP2 θP1
α 79.300 deg
α θP3 θP1
α 125.600 deg
Evaluate terms in the WZ coefficient matrix and constant vector from equations 5.26 and form the matrix and vector:
B sin β
E p 21 cos δ
H cos α 1
A cos β 1
D sin α
K sin α
L p 31 cos δ
M p 21 sin δ
B C D
E L CC M N
G H K
F K H A
F cos β 1
G sin β
A F AA B G
C cos α 1
D C
N p 31 sin δ
W1x W 1y AA 1 CC Z1x Z1y
The components of the W and Z vectors are: W1x 1.696
W1y 0.038
Z1x 0.396
Z1y 0.872
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-18-2
θ atan2 W1x W1y
ϕ 114.406 deg
W1x2 W1y2 , w 1.696
The length of link 2 is: w
Z1x2 Z1y2 , z 0.958
The length of vector Z is: z 4.
ϕ atan2 Z1x Z1y
θ 1.280 deg
Evaluate terms in the US coefficient matrix and constant vector from equations 5.31 and form the matrix and vector:
A' cos γ 1
B' sin γ
C cos α 1
E p 21 cos δ
H cos α 1
D sin α
G' sin γ
M p 21 sin δ
B' C D
E L CC M N
G' H K
F' K H
A'
K sin α
L p 31 cos δ
A' F' AA B' G'
F' cos γ 1
D C
N p 31 sin δ
U1x U 1y AA 1 CC S1x S1y
The components of the U and S vectors are: U1x 1.483
U1y 0.409
S 1x 0.048
S 1y 0.650
σ atan2 U1x U1y
σ 15.412 deg
ψ atan2 S 1x S 1y
ψ 85.790 deg
The length of link 4 is: u
U1x2 U1y2 , u 1.538
The length of vector S is: s 5.
S 1x2 S 1y2 , s 0.652
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
V1x Z1x S 1x
V1x 0.444
V1y Z1y S 1y
V1y 0.222
θ atan2 V1x V1y
θ 153.426 deg
v Link 1:
2
2
V1x V1y
v 0.496
G1x W1x V1x U1x
G1x 0.230
G1y W1y V1y U1y
G1y 0.225
θ atan2 G1x G1y
θ 135.700 deg
DESIGN OF MACHINERY - 5th Ed.
g 6.
7.
8.
9.
2
SOLUTION MANUAL 5-18-3
2
G1x G1y
g 0.322
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 134.419 deg
θ2f θ2i β
θ2f 214.419 deg
Define the coupler point with respect to point A and the vector V. rp z
δp ϕ θ
rp 0.958
δp 39.020 deg
Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. O2x z cos ϕ w cos θ
O2x 1.300
O2y z sin ϕ w sin θ
O2y 0.834
O4x s cos ψ u cos σ
O4x 1.530
O4y s sin ψ u sin σ
O4y 1.059
Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.
θrot atan2 O4x O2x O4y O2y
θrot 135.700 deg
10. Determine the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "Grashof" 11. DESIGN SUMMARY Link 2:
w 1.696
θ 1.280 deg
Link 3:
v 0.496
θ 153.426 deg
Link 4:
u 1.538
σ 15.412 deg
Link 1:
g 0.322
θ 135.700 deg
Coupler:
rp 0.958
δp 39.020 deg
Crank angles:
θ2i 134.419 deg θ2f 214.419 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-18-4
12. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.
Y A3 B3
A2
P3 P2 W2 O2 O4
U1
B2 U2
B1 W1
X
P1 S1 A1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-19-1
PROBLEM 5-19 Statement: Given:
Solution: 1.
2.
Design a linkage to carry the body in Figure P5-3 through the three positions P1, P2 and P3 at the angles shown in the figure. Use analytical synthesis and design it for the fixed pivots shown. P21x 0.907
P21y 0.0
P31x 1.447
P31y 0.0
O2x 1.788
O2y 1.994
O4x 0.212
O4y 1.994
Body angles:
θP1 111.8 deg
θP2 191.1 deg
θP3 237.4 deg
See Figure P5-3 and Mathcad file P0519.
Determine the angle changes between precision points from the body angles given. α θP2 θP1
α 79.300 deg
α θP3 θP1
α 125.600 deg
Using Figure 5-6, determine the magnitudes of R1, R2, and R3 and their x and y components. R1x O2x
3.
4.
R1x 1.788
R1y O2y
R2x R1x P21x
R2x 0.881
R2y R1y P21y
R2y 1.994
R3x R1x P31x
R3x 0.341
R3y R1y P31y
R3y 1.994
2
2
R1 2.678
2
2
R2 2.180
2
2
R3 2.023
R1
R1x R1y
R2
R2x R2y
R3
R3x R3y
R1y 1.994
Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ atan2 R1x R1y
ζ 48.118 deg
ζ atan2 R2x R2y
ζ 66.163 deg
ζ atan2 R3x R3y
ζ 80.296 deg
Solve for 2 and 3 using equations 5.34
C3 3.003
C4 1.701
C5 2.508
C1 R3 cos α ζ R2 cos α ζ C2 R3 sin α ζ R2 sin α ζ C3 R1 cos α ζ R3 cos ζ
C4 R1 sin α ζ R3 sin ζ
C5 R1 cos α ζ R2 cos ζ
C1 0.238 C2 1.150
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-19-2
C6 R1 sin α ζ R2 sin ζ 2
C6 0.133
2
A1 C3 C4
A1 11.912
A2 C3 C6 C4 C5
A2 4.666
A3 C4 C6 C3 C5
A3 7.307
A4 C2 C3 C1 C4
A4 3.048
A5 C4 C5 C3 C6
A5 4.666
A6 C1 C3 C2 C4
A6 2.671
K1 A2 A4 A3 A6
K1 5.293
K2 A3 A4 A5 A6
K2 34.731
2
K3
2
2
2
A1 A2 A3 A4 A6
2
K3 25.158
2
K K 2 K 2 K 2 2 1 2 3 β 2 atan K1 K3
β 125.600 deg
K K 2 K 2 K 2 2 1 2 3 β 2 atan K1 K3
β 37.070 deg
The first value is the same as 3, so use the second value
β β
A5 sin β A3 cos β A6 A1
β 18.241 deg
A3 sin β A2 cos β A4 A1
β 18.241 deg
β acos
β asin
β β
Since both values are the same, 5.
Repeat steps 2, 3, and 4 for the right-hand dyad to find 1 and 2. R1x O4x
R1x 0.212
R1y O4y
R2x R1x P21x
R2x 1.119
R2y R1y P21y
R2y 1.994
R3x R1x P31x
R3x 1.659
R3y R1y P31y
R3y 1.994
R1
2
2
R1x R1y
R1 2.005
R1y 1.994
DESIGN OF MACHINERY - 5th Ed.
6.
7.
SOLUTION MANUAL 5-19-3
2
2
R2 2.287
2
2
R3 2.594
R2
R2x R2y
R3
R3x R3y
Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ atan2 R1x R1y
ζ 96.069 deg
ζ atan2 R2x R2y
ζ 119.300 deg
ζ atan2 R3x R3y
ζ 129.760 deg
Solve for 2 and 3 using equations 5.34
C3 0.161
C4 3.327
C5 0.880
C6 1.832
C1 R3 cos α ζ R2 cos α ζ C2 R3 sin α ζ R2 sin α ζ C3 R1 cos α ζ R3 cos ζ
C4 R1 sin α ζ R3 sin ζ
C5 R1 cos α ζ R2 cos ζ
C6 R1 sin α ζ R2 sin ζ 2
2
C1 1.297 C2 0.811
A1 C3 C4
A1 11.096
A2 C3 C6 C4 C5
A2 3.222
A3 C4 C6 C3 C5
A3 5.954
A4 C2 C3 C1 C4
A4 4.186
A5 C4 C5 C3 C6
A5 3.222
A6 C1 C3 C2 C4
A6 2.906
K1 A2 A4 A3 A6
K1 3.816
K2 A3 A4 A5 A6
K2 34.288
2
K3
2
2
2
A1 A2 A3 A4 A6
2
2
K3 25.658
K K 2 K 2 K 2 2 1 2 3 γ 2 atan K1 K3
γ 125.600 deg
K K 2 K 2 K 2 2 1 2 3 2 atan K1 K3
41.699 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-19-4
The first value is the same as 3, so use the second value
γ
A5 sin γ A3 cos γ A6 A1
31.159 deg
A3 sin γ A2 cos γ A4 A1
31.159 deg
acos
asin
γ
Since both values are the same , 8.
Use the method of Section 5.7 to synthesize the linkage. Start by determining the magnitudes of the vectors P21 and P31 and their angles with respect to the X axis. 2
p 21
2
P21x P21y
p 21 0.907
δ atan2 P21x P21y 2
p 31
δ 180.000 deg
2
P31x P31y
p 31 1.447
δ atan2 P31x P31y 9.
δ 180.000 deg
Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:
B sin β
E p 21 cos δ
H cos α 1
A cos β 1
D sin α
F cos β 1
G sin β
K sin α
L p 31 cos δ
A F AA B G
C cos α 1
M p 21 sin δ
E L CC M N
B C D
G H K A D C F K H
N p 31 sin δ
W1x W1y AA 1 CC Z1x Z1y
10. The components of the W and Z vectors are: W1x 1.943
11. The length of link 2 is:
w
W1y 1.529 2
Z1x 0.155
2
W1x W1y
Z1y 0.465
w 2.472
12. Evaluate terms in the US coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:
A' cos γ 1
D sin α
B' sin γ
E p 21 cos δ
C cos α 1
F' cos γ 1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-19-5
G' sin γ
K sin α
L p 31 cos δ
A' F' AA B' G'
H cos α 1
M p 21 sin δ
B' C D
N p 31 sin δ
E L CC M N
G' H K A' D C F' K H
U1x U1y AA 1 CC S1x S1y
13. The components of the W and Z vectors are: U1x 0.395 14. The length of link 4 is:
U1y 2.460 2
u
U1x U1y
S1x 0.183
2
S1y 0.466
u 2.491
15. Solving for links 3 and 1 from equations 5.2a and 5.2b. V1x Z1x S1x
V1x 0.338
V1y Z1y S1y
V1y 0.931 v
The length of link 3 is:
2
2
V1x V1y
v 0.991
G1x W1x V1x U1x
G1x 2.000
G1y W1y V1y U1y
G1y 0.000
g
The length of link 1 is:
2
G1x G1y
2
g 2.000
16. Check the location of the fixed pivots with respect to the global frame using the calculated vectors W1, Z1, U1, and S1. O2x Z1x W1x
O2x 1.788
O2y Z1y W1y
O2y 1.994
O4x S1x U1x
O4x 0.212
O4y S1y U1y
O4y 1.994
These check with Figure P5-3. 17. Determine the location of the coupler point with respect to point A and line AB. 2
2
z 0.491
2
2
s 0.501
Distance from A to P
z
Z1x Z1y
Angle BAP (p)
s
S1x S1y
ψ atan2( S1x S1y)
ψ 68.558 deg
ϕ atan2( Z1x Z1y )
ϕ 108.446 deg
rP z
θ atan2 z cos ϕ s cos ψ z sin ϕ s sin ψ θ 109.959 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-19-6
δp ϕ θ
δp 1.513 deg
18. DESIGN SUMMARY Link 1:
g 2.000
Link 2:
w 2.472
Link 3:
v 0.991
Link 4:
u 2.491
Coupler point:
rP 0.491
δp 1.513 deg
19. VERIFICATION: The calculated values of g (length of the ground link) and of the coordinates of O2 and O4 give the same values as those on the problem statement, verifying that the calculated values for the other links and the coupler point are correct.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-20-1
PROBLEM 5-20 Statement:
Write a program to generate and plot the circle-point and center-point circles for Problem 5-19 using an equation solver or any program language.
Given: P21x 0.907
P21y 0.0
P31x 1.447
P31y 0.0
O2x 1.788
O2y 1.994
O4x 0.212
O4y 1.994
Body angles:
θP1 111.8 deg
θP2 191.1 deg
θP3 237.4 deg
Assumptions: Let the position 1 to position 2 rotation angles be: β 18.241 deg and γ 31.159 deg Let the position 1 to position 2 coupler rotation angle be: α 79.3 deg Solution: 1.
See Figure P5-3 and Mathcad file P0520.
Use the method of Section 5.6 to synthesize the linkage. Start by determining the magnitudes of the vectors P21 and P31 and their angles with respect to the X axis. p 21
2
2
P21x P21y
p 21 0.907
δ atan2 P21x P21y p 31
2
δ 180.000 deg
2
P31x P31y
p 31 1.447
δ atan2 P31x P31y 2.
δ 180.000 deg
Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector to get the center-point and circle-point circles for the left dyad. α 125.6 deg
β 0 deg 1 deg 360 deg
B sin β
E p 21 cos δ
A cos β 1 D sin α
C cos α 1
F β cos β 1
K α sin α
N p 31 sin δ
H α cos α 1
M p 21 sin δ
G β sin β L p 31 cos δ
B C D A F β G β H α K α AA α β B A D C G β F β K α H α
E L CC M N
1 CC
W1y α β DD α β 2
Z1y α β DD α β 4
DD α β AA α β
W1x α β DD α β 1 Z1x α β DD α β 3
DESIGN OF MACHINERY - 5th Ed.
3.
SOLUTION MANUAL 5-20-2
Check this against the solutions in Problem 5-19:
W1y α 37.070 deg 1.529
Z1y α 37.070 deg 0.465
W1x α 37.070 deg 1.943 Z1x α 37.070 deg 0.155
These are the same as the values calculated in Problem 5-19. 4.
Form the vector N, whose tip describes the center-point circle for the WZ dyad.
Nx α β W1x α β Z1x α β Ny α β W1y α β Z1y α β 5.
Plot the center-point circle for the WZ dyad. Center-Point Circle for WZ Dyad 0
1
Ny α β 2
3
4 2
1
0
1
2
Nx α β
4.
Form the vector Z, whose tip describes the center-point circle for the WZ dyad. β 37.070 deg
α 0 deg 1 deg 360 deg
Zx α β Z1x α β
5.
Zy α β Z1y α β
Plot the circle-point circle for the WZ dyad (see next page).
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-20-3
Circle-Point Circle for WZ Dyad 0.1
0.2
Zy α β 0.3
0.4
0.5 0.2
0.1
0
0.1
0.2
0.3
Zx α β
6.
Evaluate terms in the US coefficient matrix and constant vector from equations (5.31) and form the matrix and vector to get the center-point and circle-point circles for the left dyad. α 125.6 deg
γ 0 deg 1 deg 360 deg
B sin γ
E p 21 cos δ
A cos γ 1 D sin α
C cos α 1
F γ cos γ 1
K α sin α
N p 31 sin δ
H α cos α 1
M p 21 sin δ
G γ sin γ L p 31 cos δ
B C D A F γ G γ H α K α AA α γ B A D C G γ F γ K α H α
E L CC M N
1 CC
U1y α γ DD α γ 2
S1y α γ DD α γ 4
DD α γ AA α γ
U1x α γ DD α γ 1 S1x α γ DD α γ 3 7.
Check this against the solutions in Problem 5-19:
U1x α 41.699 deg 0.395
U1y α 41.699 deg 2.460
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-20-4
S1x α 41.699 deg 0.183
S1y α 41.699 deg 0.466
These are the same as the values calculated in Problem 5-19. 8.
Form the vector M, whose tip describes the center-point circle for the US dyad.
Mx α γ U1x α γ S1x α γ My α γ U1y α γ S1y α γ 9.
Plot the center-point circle for the US dyad. Center-Point Circle for US Dyad 0
0.5
1
My α γ
1.5
2
2.5 1.5
1
0.5
Mx α γ
0
0.5
10. Form the vector S, whose tip describes the center-point circle for the US dyad. γ 41.699 deg
α 0 deg 1 deg 360 deg
Sx α γ S1x α γ
11. Plot the circle-point circle for the WZ dyad (see next page).
Sy α γ S1y α γ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-20-5
Circle-Point Circle for the US Dyad 1.5
1
Sy α γ
0.5
0 0.5
0
Sx α γ
0.5
1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-21-1
PROBLEM 5-21 Statement:
Design a fourbar linkage to carry the box in Figure P5-4 from position 1 to 2 without regard for the fixed pivots shown. Use points A and B for your attachment points. Determine the range of the transmission angle. The fixed pivots should be on the base.
Given:
Coordinates of the points P1 and P2 with respect to P1: P1x 0.0
P1y 0.0
P2x 184.0
P2y 17.0
Angles made by the body in positions 1 and 2:
θP1 90.0 deg
θP2 45.0 deg
Coordinates of the points A1 and B1 with respect to P1: A1x 17.0
A1y 43.0
B1x 69.0
B1y 43.0
Free choice for the WZ dyad : β 44.0 deg Free choice for the US dyad : γ 55.0 deg Solution:
See Figure P5-4 and Mathcad file P0521.
1.
Note that this is a two-position motion generation (MG) problem because the output is specified as a complex motion of the coupler, link 3. The second method of Section 5.3 will be used here.
2.
Define the position vectors R1 and R2 and the vector P21 using Figure 5-1 and equation 5.1. R1
P1x P1y
R2
P2x P2y
P21x R2 R1 P21y
P21x 184.000 P21y 17.000
p 21 3.
4.
2
2
P21x P21y
From the trigonometric relationships given in Figure 5-1, determine 2 and 2. α θP2 θP1
α 45.000 deg
δ atan2 P21x P21y
δ 5.279 deg
Using Figure P5-4, the given data, and the law of cosines, determine z, s, , and . z
P1x A1x2 P1y A1y 2
z 46.239
s
P1x B1x 2 P1y B1y 2
s 81.302
v
A1x B1x 2 A1y B1y2
v 52.000
v2 z2 s2 2 v z
ϕ acos
v2 s2 z2 2 v s
ψ π acos 5.
p 21 184.784
Solve for the WZ dyad using equations 5.8.
ϕ 111.571 deg
ψ 148.069 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-21-2
Z1x z cos ϕ
Z1x 17.000 A 0.281
D sin α
B 0.695
E p 21 cos δ
E 184.000
C 0.293
F p 21 sin δ
F 17.000
B sin β
C cos α 1
W1y w
D 0.707
A C Z1x D Z1y E B C Z1y D Z1x F
W1x 53.979
2 A A C Z1y D Z1x F B C Z1x D Z1y E
W1y 192.131
2 A 2
2
W1x W1y
w 199.570
θ atan2 W1x W1y 6.
θ 105.693 deg
Solve for the US dyad using equations 5.12. S 1x s cos ψ
S 1x 69.000
D sin α
B 0.819
E p 21 cos δ
E 184.000
C 0.293
F p 21 sin δ
F 17.000
C cos α 1
U1y u
S 1y 43.000
A 0.426
B sin γ
U1x
S 1y s sin ψ
A cos γ 1
D 0.707
A C S 1x D S 1y E B C S 1y D S 1x F 2 A A C S 1y D S 1x F B C S 1x D S 1y E
2
U1x 15.598
U1y 154.713
2 A 2
U1x U1y
u 155.497
σ atan2 U1x U1y 7.
Z1y 43.000
A cos β 1
W1x
Z1y z sin ϕ
σ 95.757 deg
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
V1x z cos ϕ s cos ψ
V1x 52.000
V1y z sin ϕ s sin ψ
V1y 0.000
θ atan2 V1x V1y
θ 0.000 deg
v Link 1:
2
2
V1x V1y
v 52.000
G1x w cos θ v cos θ u cos σ
G1x 13.619
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-21-3
G1y w sin θ v sin θ u sin σ
G1y 37.418
θ atan2 G1x G1y
θ 70.000 deg
g 8.
9.
2
2
G1x G1y
g 39.819
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 35.692 deg
θ2f θ2i β
θ2f 8.308 deg
Define the coupler point with respect to point A and the vector V. rp z
δp ϕ θ
rp 46.239
δp 111.571 deg
10. Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ 0 deg R1
ρ 0.000 deg
2
2
P1x P1y
R1 0.000
O2x 70.979
O2y 235.131
O4x 84.598
O4y 197.713
O2x R1 cos ρ z cos ϕ w cos θ O2y R1 sin ρ z sin ϕ w sin θ O4x R1 cos ρ s cos ψ u cos σ O4y R1 sin ρ s sin ψ u sin σ
These fixed pivot points fall on the base and are, therefore, acceptable. 11. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.
θrot atan2 O4x O2x O4y O2y
θrot 70.000 deg
12. Determine the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "non-Grashof"
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-21-4
13. DESIGN SUMMARY Link 2:
w 199.570
θ 105.693 deg
Link 3:
v 52.000
θ 0.000 deg
Link 4:
u 155.497
σ 95.757 deg
Link 1:
g 39.819
θ 70.000 deg
Coupler:
rp 46.239
δp 111.571 deg
Crank angles:
θ2i 35.692 deg
θ2f 8.308 deg
14. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.
Y
P1
X Z1 A1
P2
S1 B1
Z2
V1 A2
V2 55.0°
U1 44.0° W1
W2
O4
O2
U2
S2
B2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-22-1
PROBLEM 5-22 Statement:
Design a fourbar linkage to carry the box in Figure P5-4 from position 1 to 3 without regard for the fixed pivots shown. Use points A and B for your attachment points. Determine the range of the transmission angle. The fixed pivots should be on the base.
Given:
Coordinates of the points P1 and P3 with respect to P1: P1x 0.0
P1y 0.0
P3x 211.0
P3y 180.0
Angles made by the body in positions 1 and 3:
θP1 90.0 deg
θP3 0.0 deg
Coordinates of the points A1 and B1 with respect to P1: A1x 17.0
A1y 43.0
B1x 69.0
B1y 43.0
Free choice for the WZ dyad : β 70.0 deg Free choice for the US dyad : γ 95.0 deg Solution:
See Figure P5-4 and Mathcad file P0522.
1.
Note that this is a two-position motion generation (MG) problem because the output is specified as a complex motion of the coupler, link 3. The second method of Section 5.3 will be used here.
2.
Define the position vectors R1 and R2 and the vector P21 using Figure 5-1 and equation 5.1. R1
p 21 3.
4.
P1x P1y
R2
2
P3x P3y
P21x R2 R1 P21y
2
P21x P21y
P21y 180.000 p 21 277.346
From the trigonometric relationships given in Figure 5-1, determine 2 and 2. α θP3 θP1
α 90.000 deg
δ atan2 P21x P21y
δ 40.467 deg
Using Figure P5-4, the given data, and the law of cosines, determine z, s, , and . z
P1x A1x2 P1y A1y 2
z 46.239
s
P1x B1x 2 P1y B1y 2
s 81.302
v
A1x B1x 2 A1y B1y2
v 52.000
v2 z2 s2 2 v z
ϕ acos
ϕ 111.571 deg
v2 s2 z2 2 v s
ψ π acos 5.
P21x 211.000
ψ 148.069 deg
Solve for the WZ dyad using equations 5.8. Z1x z cos ϕ
Z1x 17.000
Z1y z sin ϕ
Z1y 43.000
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-22-2
A 0.658
D sin α
B 0.940
E p 21 cos δ
C 1.000
F p 21 sin δ
A cos β 1 B sin β
C cos α 1 W1x
W1y w
D 1.000
E 211.000
F 180.000
A C Z1x D Z1y E B C Z1y D Z1x F
W1x 34.467
2 A A C Z1y D Z1x F B C Z1x D Z1y E
W1y 184.825
2 A 2
2
W1x W1y
w 188.012
θ atan2 W1x W1y 6.
θ 79.436 deg
Solve for the US dyad using equations 5.12. S 1x s cos ψ
S 1x 69.000
A 1.087
D sin α
B 0.996
E p 21 cos δ
C 1.000
F p 21 sin δ
A cos γ 1 B sin γ
C cos α 1 U1x
U1y u
S 1y s sin ψ
D 1.000
E 211.000
F 180.000
A C S 1x D S 1y E B C S 1y D S 1x F 2 A A C S 1y D S 1x F B C S 1x D S 1y E
2
U1x U1y
u 154.999
σ atan2 U1x U1y 7.
U1x 44.882
U1y 148.358
2 A 2
S 1y 43.000
σ 73.168 deg
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
V1x z cos ϕ s cos ψ
V1x 52.000
V1y z sin ϕ s sin ψ
V1y 0.000
θ atan2 V1x V1y
θ 0.000 deg
v Link 1:
2
2
V1x V1y
v 52.000
G1x w cos θ v cos θ u cos σ
G1x 41.585
G1y w sin θ v sin θ u sin σ
G1y 36.467
θ atan2 G1x G1y
θ 41.248 deg
DESIGN OF MACHINERY - 5th Ed.
g
8.
9.
2
SOLUTION MANUAL 5-22-3 2
G1x G1y
g 55.310
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 38.188 deg
θ2f θ2i β
θ2f 31.812 deg
Define the coupler point with respect to point A and the vector V. rp z
δp ϕ θ
rp 46.239
δp 111.571 deg
10. Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ 0 deg R1
ρ 0.000 deg
2
2
P1x P1y
R1 0.000
O2x 17.467
O2y 227.825
O4x 24.118
O4y 191.358
O2x R1 cos ρ z cos ϕ w cos θ O2y R1 sin ρ z sin ϕ w sin θ O4x R1 cos ρ s cos ψ u cos σ O4y R1 sin ρ s sin ψ u sin σ
These fixed pivot points fall on the base and are, therefore, acceptable. 11. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.
θrot atan2 O4x O2x O4y O2y
θrot 41.248 deg
12. Determine the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "non-Grashof" 13. DESIGN SUMMARY Link 2:
w 188.012
θ 79.436 deg
Link 3:
v 52.000
θ 0.000 deg
Link 4:
u 154.999
σ 73.168 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-22-4
Link 1:
g 55.310
θ 41.248 deg
Coupler:
rp 46.239
δp 111.571 deg
Crank angles:
θ2i 38.188 deg θ2f 31.812 deg 14. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design. Y
P1
X Z1
S1 B1
A1
V1
U1 W1 70.0°
95.0° P3
O4
O2
W2
A3
U2
Z2
V2
S2 B3
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-23-1
PROBLEM 5-23 Statement:
Design a fourbar linkage to carry the box in Figure P5-4 from position 2 to 3 without regard for the fixed pivots shown. Use points A and B for your attachment points. Determine the range of the transmission angle. The fixed pivots should be on the base.
Given:
Coordinates of the points P2 and P3 with respect to P1: P2x 184.0
P2y 17.0
P3x 211.0
P3y 180.0
Angles made by the body in positions 1 and 3:
θP2 45.0 deg
θP3 0.0 deg
Coordinates of the points A1 and B1 with respect to P1: A1x 17.0
A1y 43.0
B1x 69.0
B1y 43.0
Free choice for the WZ dyad : β 60.0 deg Free choice for the US dyad : γ 45.0 deg Solution:
See Figure P5-4 and Mathcad file P0523.
1.
Note that this is a two-position motion generation (MG) problem because the output is specified as a complex motion of the coupler, link 3. The second method of Section 5.3 will be used here.
2.
Define the position vectors R1 and R2 and the vector P21 using Figure 5-1 and equation 5.1. R1
p 21 3.
4.
P2x P2y
R2
2
P3x P3y
P21x R2 R1 P21y
2
P21x P21y
P21y 163.000 p 21 165.221
From the trigonometric relationships given in Figure 5-1, determine 2 and 2. α θP3 θP2
α 45.000 deg
δ atan2 P21x P21y
δ 80.595 deg
Using Figure P5-4, the given data, and the law of cosines, determine z, s, , and . z
0.0 A1x 2 0.0 A1y 2
z 46.239
s
0.0 B1x2 0.0 B1y 2
s 81.302
v
A1x B1x 2 A1y B1y2
v 52.000
v2 z2 s2 45 deg 2 v z
ϕ acos
ϕ 66.571 deg
v2 s2 z2 45 deg 2 v s
ψ π acos 5.
P21x 27.000
ψ 103.069 deg
Solve for the WZ dyad using equations 5.8. Z1x z cos ϕ
Z1x 18.385
Z1y z sin ϕ
Z1y 42.426
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-23-2
A 0.500
D sin α
B 0.866
E p 21 cos δ
C 0.293
F p 21 sin δ
A cos β 1 B sin β
C cos α 1
W1x
W1y w
D 0.707
E 27.000
F 163.000
A C Z1x D Z1y E B C Z1y D Z1x F
W1x 117.950
2 A A C Z1y D Z1x F B C Z1x D Z1y E
W1y 70.852
2 A 2
2
W1x W1y
w 137.594
θ atan2 W1x W1y 6.
θ 30.993 deg
Solve for the US dyad using equations 5.12. S 1x s cos ψ
S 1x 18.385
A 0.293
D sin α
B 0.707
E p 21 cos δ
C 0.293
F p 21 sin δ
A cos γ 1 B sin γ
C cos α 1 U1x
U1y u
S 1y s sin ψ
D 0.707
E 27.000
F 163.000
A C S 1x D S 1y E B C S 1y D S 1x F 2 A A C S 1y D S 1x F B C S 1x D S 1y E
2
U1x U1y
u 204.640
σ atan2 U1x U1y 7.
U1x 201.643
U1y 34.896
2 A 2
S 1y 79.196
σ 9.818 deg
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
V1x z cos ϕ s cos ψ
V1x 36.770
V1y z sin ϕ s sin ψ
V1y 36.770
θ atan2 V1x V1y v Link 1:
2
θ 45.000 deg
2
V1x V1y
v 52.000
G1x w cos θ v cos θ u cos σ
G1y w sin θ v sin θ u sin σ
G1x 46.924 G1y 0.813
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-23-3
θ atan2 G1x G1y g 8.
9.
2
θ 179.007 deg
2
G1x G1y
g 46.931
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 210.000 deg
θ2f θ2i β
θ2f 150.000 deg
Define the coupler point with respect to point A and the vector V. rp z
δp ϕ θ
rp 46.239
δp 111.571 deg
10. Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ atan2 P2x P2y R1
2
ρ 5.279 deg
2
P2x P2y
R1 184.784
O2x 47.665
O2y 130.278
O4x 0.742
O4y 131.092
O2x R1 cos ρ z cos ϕ w cos θ O2y R1 sin ρ z sin ϕ w sin θ O4x R1 cos ρ s cos ψ u cos σ O4y R1 sin ρ s sin ψ u sin σ
These fixed pivot points fall on the base and are, therefore, acceptable. 11. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.
θrot atan2 O4x O2x O4y O2y
θrot 179.007 deg
12. Determine the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "non-Grashof" 13. DESIGN SUMMARY Link 2:
w 137.594
θ 30.993 deg
Link 3:
v 52.000
θ 45.000 deg
Link 4:
u 204.640
σ 9.818 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-23-4
Link 1:
g 46.931
θ 179.007 deg
Coupler:
rp 46.239
δp 111.571 deg
Crank angles:
θ2i 210.000 deg θ2f 150.000 deg 14. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design. Y
P1
X P2
A2
Z2
S2
V2
B2
W1 O4
U1 O2
W2 P3 U2 A3 V2 B3
Z2 S2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-24-1
PROBLEM 5-24 Statement:
Given:
Design a fourbar linkage to carry the box in Figure P5-4 through the three positions shown in their numbered order without regard for the fixed pivots shown. Use any points on the object as attachment points. Determine the range of the transmission angle. The fixed pivots should be on the base. Coordinates of the points P1 , P2 and P3 with respect to P1: P1x 0.0
P1y 0.0
P2x 184.0
P3x 211.0
P3y 180.0
P2y 17.0
Angles made by the body in positions 1, 2 and 3:
θP1 90.0 deg
θP2 45.0 deg
θP3 0.0 deg
Free choices for the WZ dyad : β 80.0 deg
β 160.0 deg
Free choices for the US dyad : γ 80.0 deg Solution: 1.
2.
3.
γ 170.0 deg
See Figure P5-4 and Mathcad file P0524.
Determine the magnitudes and orientation of the position difference vectors. 2
2
p 21 184.784
δ atan2 P2x P2y
δ 5.279 deg
2
2
p 31 277.346
δ atan2 P3x P3y
δ 40.467 deg
p 21
P2x P2y
p 31
P3x P3y
Determine the angle changes of the coupler between precision points. α θP2 θP1
α 45.000 deg
α θP3 θP1
α 90.000 deg
Evaluate terms in the WZ coefficient matrix and constant vector from equations 5.26 and form the matrix and vector: A cos β 1 B sin β C cos α 1
D sin α G sin β L p 31 cos δ A F AA B G
H cos α 1 M p 21 sin δ E p 21 cos δ
B C D
E L CC M N
G H K
F K H A
F cos β 1 K sin α N p 31 sin δ
D C
W1x W 1y AA 1 CC Z1x Z1y
The components of the W and Z vectors are: W1x 51.854
W1y 109.176
Z1x 43.555
Z1y 29.523
θ atan2 W1x W1y
θ 115.406 deg
ϕ atan2 Z1x Z1y
ϕ 145.869 deg
The length of link 2 is: w
W1x2 W1y2 , w 120.864
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-24-2
Z1x2 Z1y2 , z 52.618
The length of vector Z is: z 4.
Evaluate terms in the US coefficient matrix and constant vector from equations 5.31 and form the matrix and vector:
D sin α
E p 21 cos δ
A' cos γ 1
C cos α 1
G' sin γ
H cos α 1
M p 21 sin δ
B' C D
E L CC M N
G' H K
F' K H
A'
K sin α
L p 31 cos δ
A' F' AA B' G'
F' cos γ 1
B' sin γ
D C
N p 31 sin δ
U1x U 1y AA 1 CC S1x S1y
The components of the U and S vectors are: U1x 35.056
U1y 94.023
S 1x 62.812
S 1y 62.282
σ atan2 U1x U1y
σ 110.448 deg
ψ atan2 S 1x S 1y
ψ 135.242 deg
The length of link 4 is: u
U1x2 U1y2 , u 100.345
The length of vector S is: s 6.
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
V1x Z1x S 1x
V1x 19.256
V1y Z1y S 1y
V1y 32.759
θ atan2 V1x V1y
θ 59.552 deg
v Link 1:
2
2
V1x V1y
v 38.000
G1x W1x V1x U1x
G1x 2.458
G1y W1y V1y U1y
G1y 17.606
θ atan2 G1x G1y
θ 82.052 deg
g 7.
S 1x2 S 1y2 , s 88.456
2
2
G1x G1y
g 17.777
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 197.458 deg
θ2f θ2i β
θ2f 37.458 deg
DESIGN OF MACHINERY - 5th Ed.
8.
9.
SOLUTION MANUAL 5-24-3
Define the coupler point with respect to point A and the vector V. rp z
δp ϕ θ
rp 52.618
δp 205.422 deg
Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. O2x z cos ϕ w cos θ
O2x 95.410
O2y z sin ϕ w sin θ
O2y 138.699
O4x s cos ψ u cos σ
O4x 97.868
O4y s sin ψ u sin σ
O4y 156.305
10. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.
θrot atan2 O4x O2x O4y O2y
θrot 82.052 deg
11. Determine the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "non-Grashof" 12. DESIGN SUMMARY Link 2:
w 120.864
θ 115.406 deg
Link 3:
v 38.000
θ 59.552 deg
Link 4:
u 100.345
σ 110.448 deg
Link 1:
g 17.777
θ 82.052 deg
Coupler:
rp 52.618
δp 205.422 deg
Crank angles:
θ2i 197.458 deg
θ2f 37.458 deg
13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-24-4
Y
P1
X Z1 S1
P2
A1 V1 B1
Z2 S2
W1
A2 V2 B2
W2
U1
U2
O2 W3
O4
P3
U3
S3
V3 B3
Z3 A3
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-25-1
PROBLEM 5-25 Statement:
Design a fourbar linkage to carry the box in Figure P5-4 through the three positions shown in their numbered order without regard for the fixed pivots shown. Use points A and B for your attachment points. Determine the range of the transmission angle. The fixed pivots should be on the base.
Given:
Coordinates of the points P1 , P2 and P3 with respect to P1: P1x 0.0
P1y 0.0
P2x 184.0
P3x 211.0
P3y 180.0
P2y 17.0
Angles made by the body in positions 1, 2 and 3:
θP1 90.0 deg
θP2 45.0 deg
θP3 0.0 deg
Coordinates of the points A1 and B1 with respect to P1: A1x 17.0 Solution: 1.
2.
3.
A1y 43.0
B1y 43.0
See Figure P5-4 and Mathcad file P0525.
Determine the magnitudes and orientation of the position difference vectors. 2
2
p 21 184.784
δ atan2 P2x P2y
δ 5.279 deg
2
2
p 31 277.346
δ atan2 P3x P3y
δ 40.467 deg
p 21
P2x P2y
p 31
P3x P3y
Determine the angle changes of the coupler between precision points. α θP2 θP1
α 45.000 deg
α θP3 θP1
α 90.000 deg
Using Figure P5-4, the given data, and the law of cosines, determine z, s, , and . z
P1x A1x2 P1y A1y 2
z 46.239
s
P1x B1x 2 P1y B1y 2
s 81.302
v
A1x B1x 2 A1y B1y2
v 52.000
v2 z2 s2 2 v z
ϕ acos
v2 s2 z2 2 v s
4.
B1x 69.0
ϕ 111.571 deg
ψ π acos
ψ 148.069 deg
Z1x z cos ϕ
Z1x 17.000
Z1y z sin ϕ
Z1y 43.000
S 1x s cos ψ
S 1x 69.000
S 1y s sin ψ
S 1y 43.000
Use equations 5.24 to solve for w, , 2, and 3. Since the points A and B are to be used as pivots, z and are known from the calculations above.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-25-2
Guess:
W1x 50
W1y 200
β 80 deg
Given
W1x cos β 1 W1y sin β = p 21 cos δ Z1x cos α 1 Z1y sin α
β 160 deg
W1x cos β 1 W1y sin β = p 31 cos δ Z1x cos α 1 Z1y sin α W1y cos β 1 W1x sin β = p 21 sin δ Z1y cos α 1 Z1x sin α W1y cos β 1 W1x sin β = p 31 sin δ Z1y cos α 1 Z1x sin α
W1x W1y Find W1x W1y β β β β β 86.887 deg
β 165.399 deg
The components of the W vector are: W1x 65.636 The length of link 2 is: w 5.
θ atan2 W1x W1y
W1y 86.672
θ 127.136 deg
W1x2 W1y2 , w 108.720
Use equations 5.28 to solve for u, , 2, and 3. Since the points A and B are to be used as pivots, s and are known from the calculations above. Guess:
U1x 30
U1y 100
γ 80 deg
Given
U1x cos γ 1 U1y sin γ = p 21 cos δ S 1x cos α 1 S 1y sin α
U1x cos γ 1 U1y sin γ = p 31 cos δ S 1x cos α 1 S 1y sin α U1y cos γ 1 U1x sin γ = p 21 sin δ S 1y cos α 1 S 1x sin α U1y cos γ 1 U1x sin γ = p 31 sin δ S 1y cos α 1 S 1x sin α
γ 160 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-25-3
U1x U1y Find U1x U1y γ γ γ γ γ 76.700 deg
γ 161.878 deg
The components of the U vector are: U1x 33.074
U1y 110.894
The length of link 4 is: u 6.
Link 1:
V1x 52.000
V1y Z1y S 1y
V1y 0.000
θ atan2 V1x V1y
θ 0.000 deg
2
2
V1x V1y
v 52.000
G1x W1x V1x U1x
G1x 19.438
G1y W1y V1y U1y
G1y 24.222
θ atan2 G1x G1y
θ 51.253 deg
g
9.
U1x2 U1y2 , u 115.721
V1x Z1x S 1x
v
8.
σ 106.607 deg
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
7.
σ atan2 U1x U1y
2
2
G1x G1y
g 31.057
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 178.389 deg
θ2f θ2i β
θ2f 12.990 deg
Define the coupler point with respect to point A and the vector V. rp z
δp ϕ θ
rp 46.239
δp 111.571 deg
Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. O2x z cos ϕ w cos θ
O2x 82.636
O2y z sin ϕ w sin θ
O2y 129.672
O4x s cos ψ u cos σ
O4x 102.074
O4y s sin ψ u sin σ
O4y 153.894
10. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-25-4
to the line O2O4.
θrot atan2 O4x O2x O4y O2y
θrot 51.253 deg
11. Determine the Grashof condition.
Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "Grashof" 12. DESIGN SUMMARY Link 2:
w 108.720
θ 127.136 deg
Link 3:
v 52.000
θ 0.000 deg
Link 4:
u 115.721
σ 106.607 deg
Link 1:
g 31.057
θ 51.253 deg
Coupler:
rp 46.239
δp 111.571 deg
Crank angles:
θ2i 178.389 deg
Y
θ2f 12.990 deg P1
13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.
X Z1 A1
P2
S1 B1 V1
A2
U1
14. A driver dyad with a crank should be added to link 2 to control the motion of the fourbar so that it cannot move beyond positions 1 and 3.
Z2
S2
V2
B2
W2
W1
U2
O2 O4
W3
P3 A3
U3
Z3
V3
S3 B3
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-26-1
PROBLEM 5-26 Statement:
Given:
Solution: 1.
2.
Design a fourbar linkage to carry the box in Figure P5-4 through the three positions shown in their numbered order using the fixed pivots shown. Determine the range of the transmission angle. P21x 184.0
P21y 17.0
P31x 211.0
P31y 180.0
O2x 86.0
O2y 132.0
O4x 104.0
O4y 155.0
Body angles:
θP1 90 deg
θP2 45 deg
θP3 0 deg
See Figure P5-4 and Mathcad file P0526.
Determine the angle changes between precision points from the body angles given. α θP2 θP1
α 45.000 deg
α θP3 θP1
α 90.000 deg
Using Figure 5-6, determine the magnitudes of R1, R2, and R3 and their x and y components. R1x O2x
3.
4.
R1x 86.000
R1y O2y
R2x R1x P21x
R2x 98.000
R2y R1y P21y
R2y 115.000
R3x R1x P31x
R3x 125.000
R3y R1y P31y
R3y 48.000
2
2
R1 157.544
2
2
R2 151.093
2
2
R3 133.899
R1
R1x R1y
R2
R2x R2y
R3
R3x R3y
R1y 132.000
Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ atan2 R1x R1y
ζ 123.085 deg
ζ atan2 R2x R2y
ζ 49.563 deg
ζ atan2 R3x R3y
ζ 21.007 deg
Solve for 2 and 3 using equations 5.34
C2 R3 sin α ζ R2 sin α ζ C1 R3 cos α ζ R2 cos α ζ
C3 7.000
C4 134.000
C5 65.473
C6 39.149
C4 R1 sin α ζ R3 sin ζ
C5 R1 cos α ζ R2 cos ζ
C2 24.329
C3 R1 cos α ζ R3 cos ζ
C1 60.553
C6 R1 sin α ζ R2 sin ζ
DESIGN OF MACHINERY - 5th Ed. 2
SOLUTION MANUAL 5-26-2
A1 C3 C4
2
A1 1.800 10
4
A2 C3 C6 C4 C5
A2 9.047 10
3
A3 C4 C6 C3 C5
A3 4.788 10
3
A4 C2 C3 C1 C4
A4 7.944 10
A5 C4 C5 C3 C6
A5 9.047 10
A6 C1 C3 C2 C4
A6 3.684 10
3
K1 A2 A4 A3 A6
K1 5.423 10
7
K2 A3 A4 A5 A6
K2 7.136 10
7
2
K3
3 3
2
2
2
A1 A2 A3 A4 A6
2 7
K3 7.136 10
2
K K 2 K 2 K 2 2 1 2 3 β 2 atan K1 K3
β 90.000 deg
K K 2 K 2 K 2 2 1 2 3 β 2 atan K1 K3
β 164.466 deg
The first value is the same as 3, so use the second value
β β
A5 sin β A3 cos β A6 A1
β 85.240 deg
A3 sin β A2 cos β A4 A1
β 85.240 deg
β acos
β asin
Since 2 is not in the first quadrant, 5.
β β
Repeat steps 2, 3, and 4 for the right-hand dyad to find 1 and 2. R1x O4x
R1x 104.000
R1y O4y
R2x R1x P21x
R2x 80.000
R2y R1y P21y
R2y 138.000
R3x R1x P31x
R3x 107.000
R3y R1y P31y
R3y 25.000
2
2
R1 186.657
2
2
R2 159.512
R1
R1x R1y
R2
R2x R2y
R1y 155.000
DESIGN OF MACHINERY - 5th Ed.
R3 6.
7.
SOLUTION MANUAL 5-26-3
2
2
R3x R3y
R3 109.882
Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ atan2 R1x R1y
ζ 123.860 deg
ζ atan2 R2x R2y
ζ 59.899 deg
ζ atan2 R3x R3y
ζ 13.151 deg
Solve for 2 and 3 using equations 5.34
C3 48.000
C4 129.000
C5 43.938
C6 45.141
C1 R3 cos α ζ R2 cos α ζ C2 R3 sin α ζ R2 sin α ζ C3 R1 cos α ζ R3 cos ζ
C4 R1 sin α ζ R3 sin ζ
C5 R1 cos α ζ R2 cos ζ
C6 R1 sin α ζ R2 sin ζ 2
C2 13.338
A1 C3 C4
A1 1.894 10
4
A2 C3 C6 C4 C5
A2 7.835 10
3
A3 C4 C6 C3 C5
A3 3.714 10
3
A4 C2 C3 C1 C4
A4 9.682 10
A5 C4 C5 C3 C6
A5 7.835 10
A6 C1 C3 C2 C4
A6 5.561 10
3
K1 A2 A4 A3 A6
K1 5.520 10
7
K2 A3 A4 A5 A6
K2 7.953 10
7
2
K3
2
C1 80.017
3 3
2
2
2
A1 A2 A3 A4 A6
2
2
K K 2 K 2 K 2 2 1 2 3 γ 2 atan K1 K3 K K 2 K 2 K 2 2 1 2 3 2 atan K1 K3 The first value is the same as 3, so use the second value
7
K3 7.953 10
γ 90.000 deg
159.525 deg γ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-26-4
A5 sin γ A3 cos γ A6 A1
75.253 deg
A3 sin γ A2 cos γ A4 A1
75.253 deg
acos
asin
Since 2 is not in the first quadrant , 8.
γ
Use the method of Section 5.7 to synthesize the linkage. Start by determining the magnitudes of the vectors P21 and P31 and their angles with respect to the X axis. 2
p 21
2
P21x P21y
p 21 184.784
δ atan2 P21x P21y 2
p 31
δ 5.279 deg
2
P31x P31y
p 31 277.346
δ atan2 P31x P31y 9.
δ 40.467 deg
Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:
B sin β
E p 21 cos δ
H cos α 1
A cos β 1
D sin α
F cos β 1
G sin β
K sin α
L p 31 cos δ
A F AA B G
C cos α 1
M p 21 sin δ
B C D
E L CC M N
G H K A D C F K H
N p 31 sin δ
W1x W1y AA 1 CC Z1x Z1y
10. The components of the W and Z vectors are: W1x 62.394
11. The length of link 2 is:
w
W1y 91.663 2
Z1x 23.606
2
W1x W1y
Z1y 40.337
w 110.884
12. Evaluate terms in the US coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:
A' cos γ 1
B' sin γ
E p 21 cos δ
H cos α 1
D sin α
G' sin γ
C cos α 1
F' cos γ 1
K sin α
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-26-5
L p 31 cos δ
A' F' AA B' G'
M p 21 sin δ
B' C D
N p 31 sin δ
E L CC M N
G' H K A' D C F' K H
U1x U1y AA 1 CC S1x S1y
13. The components of the U and S vectors are: U1x 29.920
14. The length of link 4 is:
U1y 116.933 2
u
U1x U1y
2
S1x 74.080
S1y 38.067
u 120.700
15. Solving for links 3 and 1 from equations 5.2a and 5.2b. V1x Z1x S1x
V1x 50.474
V1y Z1y S1y
V1y 2.270 v
The length of link 3 is:
2
2
V1x V1y
v 50.525
G1x W1x V1x U1x
G1x 18.000
G1y W1y V1y U1y
G1y 23.000
g
The length of link 1 is:
2
G1x G1y
2
g 29.206
16. Check the location of the fixed pivots with respect to the global frame using the calculated vectors W1, Z1, U1, and S1. O2x Z1x W1x
O2x 86.000
O2y Z1y W1y
O2y 132.000
O4x S1x U1x
O4x 104.000
O4y S1y U1y
O4y 155.000
These check with Figure P5-4. 17. Determine the location of the coupler point with respect to point A and line AB. 2
2
z 46.736
2
2
s 83.288
Distance from A to P
z
Z1x Z1y
Angle BAP (p)
s
S1x S1y
ψ atan2( S1x S1y)
ψ 152.803 deg
ϕ atan2( Z1x Z1y )
ϕ 120.337 deg
rP z
θ atan2 z cos ϕ s cos ψ z sin ϕ s sin ψ θ 2.575 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-26-6
δp ϕ θ
δp 117.762 deg
18. DESIGN SUMMARY Link 1:
g 29.206
Link 2:
w 110.884
Link 3:
v 50.525
Link 4:
u 120.700
Coupler point:
rP 46.736
δp 117.762 deg
19. VERIFICATION: The calculated values of g (length of the ground link) and of the coordinates of O2 and O4 give the same values as those on the problem statement, verifying that the calculated values for the other links and the coupler point are correct.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-27-1
PROBLEM 5-27 Statement:
Design a fourbar linkage to carry the box in Figure P5-5 through the three positions shown in their numbered order without regard for the fixed pivots shown. Use any points on the object as attachment points. Determine the range of the transmission angle. The fixed pivots should be on the base.
Given:
Coordinates of the points P1 , P2 and P3 with respect to P1: P1x 0.0
P1y 0.0
P2x 421.0
P3x 184.0
P3y 1400.0
P2y 963.0
Angles made by the body in positions 1, 2 and 3:
θP1 0.0 deg
θP2 27.0 deg
θP3 88.0 deg
Free choices for the WZ dyad : β 50.0 deg
β 100.0 deg
Free choices for the US dyad : γ 50.0 deg Solution: 1.
2.
3.
γ 80.0 deg
See Figure P5-5 and Mathcad file P0527.
Determine the magnitudes and orientation of the position difference vectors. 3
δ atan2 P2x P2y
δ 66.386 deg
3
δ atan2 P3x P3y
δ 82.513 deg
2
2
p 21 1.051 10
2
2
p 31 1.412 10
p 21
P2x P2y
p 31
P3x P3y
Determine the angle changes of the coupler between precision points. α θP2 θP1
α 27.000 deg
α θP3 θP1
α 88.000 deg
Evaluate terms in the WZ coefficient matrix and constant vector from equations 5.26 and form the matrix and vector:
B sin β
E p 21 cos δ
H cos α 1
A cos β 1
D sin α
K sin α
L p 31 cos δ
M p 21 sin δ
B C D
E L CC M N
G H K
F K H A
F cos β 1
G sin β
A F AA B G
C cos α 1
D C
N p 31 sin δ
W1x W 1y AA 1 CC Z1x Z1y
The components of the W and Z vectors are: 3
W1x 784.602
W1y 362.803
Z1x 1.092 10
Z1y 39.947
θ atan2 W1x W1y
θ 155.184 deg
ϕ atan2 Z1x Z1y
ϕ 2.094 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-27-2
W1x2 W1y2 , w 864.423
The length of link 2 is: w
Z1x2 Z1y2 , z 1093.069
The length of vector Z is: z 4.
Evaluate terms in the US coefficient matrix and constant vector from equations 5.31 and form the matrix and vector:
A' cos γ 1
B' sin γ
C cos α 1
E p 21 cos δ
H cos α 1
D sin α
G' sin γ
M p 21 sin δ
B' C D
E L CC M N
G' H K
F' K H
A'
K sin α
L p 31 cos δ
A' F' AA B' G'
F' cos γ 1
D C
N p 31 sin δ
U1x U 1y AA 1 CC S1x S1y
The components of the U and S vectors are: U1x 924.539
U1y 281.738
σ atan2 U1x U1y The length of link 4 is: u
σ 163.052 deg
S 1x2 S 1y2 , s 806.978
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
V1x Z1x S 1x
V1x 289.619
V1y Z1y S 1y
V1y 42.852
θ atan2 V1x V1y
θ 8.416 deg
v Link 1:
2
2
V1x V1y
v 292.772
G1x W1x V1x U1x
G1x 429.556
G1y W1y V1y U1y
G1y 38.214
θ atan2 G1x G1y
θ 5.084 deg
g 7.
ψ atan2 S 1x S 1y
U1x2 U1y2 , u 966.514
The length of vector S is: s 6.
S 1x 802.719
2
2
G1x G1y
g 431.252
Determine the initial and final values of the input crank with respect to the vector G.
S 1y 82.798 ψ 5.889 deg
DESIGN OF MACHINERY - 5th Ed.
8.
9.
SOLUTION MANUAL 5-27-3
θ2i θ θ
θ2i 150.100 deg
θ2f θ2i β
θ2f 50.100 deg
Define the coupler point with respect to point A and the vector V. rp z
δp ϕ θ
rp 1093.069
δp 10.511 deg
Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. O2x z cos ϕ w cos θ
O2x 307.736
O2y z sin ϕ w sin θ
O2y 402.750
O4x s cos ψ u cos σ
O4x 121.820
O4y s sin ψ u sin σ
O4y 364.536
10. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.
θrot atan2 O4x O2x O4y O2y
θrot 5.084 deg
11. Determine the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "Grashof" 12. DESIGN SUMMARY Link 2:
w 864.423
θ 155.184 deg
Link 3:
v 292.772
θ 8.416 deg
Link 4:
u 966.514
σ 163.052 deg
Link 1:
g 431.252
θ 5.084 deg
Coupler:
rp 1093.069
δp 10.511 deg
Crank angles:
θ2i 150.100 deg θ2f 50.100 deg 13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-27-4
Y
P3
P2
B2
B3
A2 U2 A1
B1
A3
W2
U3 P1
W3 W1 O2
U1
O4
X
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-28-1
PROBLEM 5-28 Statement:
Design a fourbar linkage to carry the box in Figure P5-5 through the three positions shown in their numbered order without regard for the fixed pivots shown. Use points A and B for your attachment points. Determine the range of the transmission angle. The fixed pivots should be on the base.
Given:
Coordinates of the points P1 , P2 and P3 with respect to P1: P1x 0.0
P1y 0.0
P3x 184.0
P3y 1400.0
P2x 421.0
P2y 963.0
Angles made by the body in positions 1, 2 and 3:
θP1 0.0 deg
θP2 27.0 deg
θP3 88.0 deg
Coordinates of the points A1 and B1 with respect to P1: A1x 1080 Solution: 1.
2.
3.
A1y 0.0
B1y 60
See Figure P5-5 and Mathcad file P0528.
Determine the magnitudes and orientation of the position difference vectors. 3
δ atan2 P2x P2y
δ 66.386 deg
3
δ atan2 P3x P3y
δ 82.513 deg
2
2
p 21 1.051 10
2
2
p 31 1.412 10
p 21
P2x P2y
p 31
P3x P3y
Determine the angle changes of the coupler between precision points. α θP2 θP1
α 27.000 deg
α θP3 θP1
α 88.000 deg
Using Figure P5-4, the given data, and the law of cosines, determine z, s, , and . z
P1x A1x2 P1y A1y 2
z 1080
s
P1x B1x 2 P1y B1y 2
s 742.428
v
A1x B1x 2 A1y B1y2
v 345.254
ϕ 0
ϕ 0.000 deg
z2 s2 v2 2 z s
4.
B1x 740
ψ acos
ψ 4.635 deg
Z1x z cos ϕ
Z1x 1080
Z1y z sin ϕ
Z1y 0.000
S 1x s cos ψ
S 1x 740.000
S 1y s sin ψ
S 1y 60.000
Use equations 5.24 to solve for w, , 2, and 3. Since the points A and B are to be used as pivots, z and are known from the calculations above.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-28-2
Guess:
W1x 50
W1y 200
β 80 deg
Given
W1x cos β 1 W1y sin β = p 21 cos δ Z1x cos α 1 Z1y sin α
β 160 deg
W1x cos β 1 W1y sin β = p 31 cos δ Z1x cos α 1 Z1y sin α W1y cos β 1 W1x sin β = p 21 sin δ Z1y cos α 1 Z1x sin α W1y cos β 1 W1x sin β = p 31 sin δ Z1y cos α 1 Z1x sin α
W1x W1y Find W1x W1y β β β β
β 54.243 deg
β 107.466 deg
The components of the W vector are: W1x 730.785 The length of link 2 is: w 5.
θ atan2 W1x W1y
W1y 289.533
θ 158.387 deg
W1x2 W1y2 , w 786.051
Use equations 5.28 to solve for u, , 2, and 3. Since the points A and B are to be used as pivots, s and are known from the calculations above. Guess:
U1x 30
U1y 100
γ 80 deg
Given
U1x cos γ 1 U1y sin γ = p 21 cos δ S 1x cos α 1 S 1y sin α
γ 160 deg
U1x cos γ 1 U1y sin γ = p 31 cos δ S 1x cos α 1 S 1y sin α U1y cos γ 1 U1x sin γ = p 21 sin δ S 1y cos α 1 S 1x sin α U1y cos γ 1 U1x sin γ = p 31 sin δ S 1y cos α 1 S 1x sin α
U1x U1y Find U1x U1y γ γ γ γ
γ 411.378 deg
γ 437.949 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-28-3
The components of the U vector are: U1x 923.018
U1y 232.957
The length of link 4 is: u 6.
Link 1:
V1x 340.000
V1y Z1y S 1y
V1y 60.000
θ atan2 V1x V1y
θ 10.008 deg
2
2
V1x V1y
v 345.254
G1x W1x V1x U1x
G1x 532.233
G1y W1y V1y U1y
G1y 3.424
θ atan2 G1x G1y
θ 0.369 deg
2
g
9.
U1x2 U1y2 , u 951.962
V1x Z1x S 1x
v
8.
σ 165.835 deg
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
7.
σ atan2 U1x U1y
2
G1x G1y
g 532.244
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 158.755 deg
θ2f θ2i β
θ2f 51.289 deg
Define the coupler point with respect to point A and the vector V. rp z
δp ϕ θ
rp 1080.000
δp 10.008 deg
Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. O2x z cos ϕ w cos θ
O2x 349.215
O2y z sin ϕ w sin θ
O2y 289.533
O4x s cos ψ u cos σ
O4x 183.018
O4y s sin ψ u sin σ
O4y 292.957
10. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.
θrot atan2 O4x O2x O4y O2y 11. Determine the Grashof condition.
θrot 0.369 deg
DESIGN OF MACHINERY - 5th Ed.
Condition( a b c d )
SOLUTION MANUAL 5-28-4
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "Grashof" 12. DESIGN SUMMARY Link 2:
w 786.051
θ 158.387 deg
Link 3:
v 345.254
θ 10.008 deg
Link 4:
u 951.962
σ 165.835 deg
Link 1:
g 532.244
θ 0.369 deg
Coupler:
rp 1080.000
δp 10.008 deg
Crank angles:
θ2i 158.755 deg θ2f 51.289 deg 13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.
Y
P3
P2
B3
A2
B2 U2
A1
B1
A3
W2 W 3
U3 P1
W1 O2
U1
O4
X
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-29-1
PROBLEM 5-29 Statement: Given:
Solution: 1.
2.
Design a linkage to carry the object in Figure P5-5 through the three positions shown in their numbered order using the fixed pivots shown. Determine the range of the transmission angle. P21x 421.0
P21y 963.0
P31x 184.0
P31y 1400.0
O2x 362.0
O2y 291.0
O4x 182.0
O4y 291.0
Body angles:
θP1 0 deg
θP2 27 deg
θP3 88 deg
See Figure P5-5 and Mathcad file P0529.
Determine the angle changes between precision points from the body angles given. α θP2 θP1
α 27.000 deg
α θP3 θP1
α 88.000 deg
Using Figure 5-6, determine the magnitudes of R1, R2, and R3 and their x and y components. R1x O2x
3.
4.
R1x 362.000
R1y O2y
R2x R1x P21x
R2x 783.000
R2y R1y P21y
R2y 1.254 10
R3x R1x P31x
R3x 546.000
R3y R1y P31y
R3y 1.691 10
3
3
2
2
R1 464.462
2
2
R2 1.478 10
3
2
2
R3 1.777 10
3
R1
R1x R1y
R2
R2x R2y
R3
R3x R3y
R1y 291.000
Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ atan2 R1x R1y
ζ 38.795 deg
ζ atan2 R2x R2y
ζ 58.019 deg
ζ atan2 R3x R3y
ζ 72.105 deg
Solve for 2 and 3 using equations 5.34
C3 824.189
C4 1.319 10
C5 592.567
C6 830.373
C1 R3 cos α ζ R2 cos α ζ C2 R3 sin α ζ R2 sin α ζ C3 R1 cos α ζ R3 cos ζ
C4 R1 sin α ζ R3 sin ζ
C5 R1 cos α ζ R2 cos ζ
C6 R1 sin α ζ R2 sin ζ
C1 944.701 C2 928.284
3
DESIGN OF MACHINERY - 5th Ed. 2
SOLUTION MANUAL 5-29-2 2
A1 2.419 10
A2 C3 C6 C4 C5
A2 9.725 10
A3 C4 C6 C3 C5
A3 1.584 10
A4 C2 C3 C1 C4
A4 4.810 10
A5 C4 C5 C3 C6
A5 9.725 10
4
A6 C1 C3 C2 C4
A6 2.003 10
6
K1 A2 A4 A3 A6
K1 3.219 10
K2 A3 A4 A5 A6
K2 5.670 10
11
K3 4.543 10
11
2
K3
4 6
5
12
2
2
2
A1 A2 A3 A4 A6
2
2
K K 2 K 2 K 2 2 1 2 3 β 2 atan K1 K3
β 88.000 deg
K K 2 K 2 K 2 2 1 2 3 β 2 atan K1 K3
β 107.980 deg
The first value is the same as 3, so use the second value
β β
A5 sin β A3 cos β A6 A1
β 54.008 deg
A3 sin β A2 cos β A4 A1
β 54.008 deg
β acos
β asin
Since 2 is not in the first quadrant, 5.
6
A1 C3 C4
β β
Repeat steps 2, 3, and 4 for the right-hand dyad to find 1 and 2. R1x O4x
R1x 182.000
R1y O4y
R2x R1x P21x
R2x 239.000
R2y R1y P21y
R2y 1.254 10
R3x R1x P31x
R3x 2.000
R3y R1y P31y
R3y 1.691 10
2
2
R1 343.227
2
2
R2 1.277 10
R1
R1x R1y
R2
R2x R2y
3
3
3
R1y 291.000
DESIGN OF MACHINERY - 5th Ed.
R3 6.
7.
SOLUTION MANUAL 5-29-3
2
2
R3x R3y
R3 1.691 10
3
Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ atan2 R1x R1y
ζ 122.023 deg
ζ atan2 R2x R2y
ζ 79.209 deg
ζ atan2 R3x R3y
ζ 89.932 deg
Solve for 2 and 3 using equations 5.34
C3 299.174
C4 1.863 10
C5 533.274
C6 1.077 10
C1 R3 cos α ζ R2 cos α ζ C2 R3 sin α ζ R2 sin α ζ C3 R1 cos α ζ R3 cos ζ
C5 R1 cos α ζ R2 cos ζ
3
C6 R1 sin α ζ R2 sin ζ 2
3
C2 1.225 10
3
C4 R1 sin α ζ R3 sin ζ
C1 478.979
2
6
A1 C3 C4
A1 3.559 10
A2 C3 C6 C4 C5
A2 6.710 10
A3 C4 C6 C3 C5
A3 2.166 10
A4 C2 C3 C1 C4
A4 5.257 10
A5 C4 C5 C3 C6
A5 6.710 10
5
A6 C1 C3 C2 C4
A6 2.425 10
6
K1 A2 A4 A3 A6
K1 5.606 10
K2 A3 A4 A5 A6
K2 4.884 10
2
K3
5 6
5
12 11
2
2
2
A1 A2 A3 A4 A6
2
2
11
K3 6.838 10
K K 2 K 2 K 2 2 1 2 3 γ 2 atan K1 K3
γ 88.000 deg
K K 2 K 2 K 2 2 1 2 3 2 atan K1 K3
78.042 deg
The first value is the same as 3, so use the second value
γ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-29-4
A5 sin γ A3 cos γ A6 A1
51.463 deg
A3 sin γ A2 cos γ A4 A1
51.463 deg
acos
asin
Since 2 is not in the first quadrant , 8.
γ
Use the method of Section 5.7 to synthesize the linkage. Start by determining the magnitudes of the vectors P21 and P31 and their angles with respect to the X axis. 2
p 21
2
P21x P21y
p 21 1051.004
δ atan2 P21x P21y 2
p 31
δ 66.386 deg
2
P31x P31y
p 31 1412.040
δ atan2 P31x P31y 9.
δ 82.513 deg
Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:
B sin β
E p 21 cos δ
H cos α 1
A cos β 1
D sin α
F cos β 1
G sin β
K sin α
L p 31 cos δ
A F AA B G
C cos α 1
M p 21 sin δ
B C D
E L CC M N
G H K A D C F K H
N p 31 sin δ
W1x W1y AA 1 CC Z1x Z1y
10. The components of the W and Z vectors are: W1x 728.089 11. The length of link 2 is:
w
3
W1y 294.291 2
Z1x 1.090 10
2
W1x W1y
Z1y 3.291
w 785.316
12. Evaluate terms in the US coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:
A' cos γ 1
B' sin γ
E p 21 cos δ
H cos α 1
D sin α
G' sin γ
C cos α 1
F' cos γ 1
K sin α
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-29-5
L p 31 cos δ
A' F' AA B' G'
M p 21 sin δ
B' C D
N p 31 sin δ
E L CC M N
G' H K A' D C F' K H
U1x U1y AA 1 CC S1x S1y
13. The components of the W and Z vectors are: U1x 921.699 14. The length of link 4 is:
U1y 231.572 2
u
U1x U1y
2
S1x 739.699
S1y 59.428
u 950.344
15. Solving for links 3 and 1 from equations 5.2a and 5.2b. V1x Z1x S1x
V1x 350.390
V1y Z1y S1y
V1y 62.718 v
The length of link 3 is:
2
2
V1x V1y
v 355.959
G1x W1x V1x U1x
G1x 544.000
G1y W1y V1y U1y
G1y 5.116 10
g
The length of link 1 is:
2
G1x G1y
2
13
g 544.000
16. Check the location of the fixed pivots with respect to the global frame using the calculated vectors W1, Z1, U1, and S1. O2x Z1x W1x
O2x 362.000
O2y Z1y W1y
O2y 291.000
O4x S1x U1x
O4x 182.000
O4y S1y U1y
O4y 291.000
These check with Figure P5-5. 17. Determine the location of the coupler point with respect to point A and line AB. 2
2
z 1.090 10
2
2
s 742.082
Distance from A to P
z
Z1x Z1y
Angle BAP (p)
s
S1x S1y
3
ψ atan2( S1x S1y)
ψ 4.593 deg
ϕ atan2( Z1x Z1y )
ϕ 0.173 deg
rP z
θ atan2 z cos ϕ s cos ψ z sin ϕ s sin ψ θ 10.148 deg
δp ϕ θ
δp 9.975 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-29-6
18. DESIGN SUMMARY Link 1:
g 544.000
Link 2:
w 785.316
Link 3:
v 355.959
Link 4:
u 950.344
Coupler point:
rP 1090.094
δp 9.975 deg
19. VERIFICATION: The calculated values of g (length of the ground link) and of the coordinates of O2 and O4 give the same values as those on the problem statement, verifying that the calculated values for the other links and the coupler point are correct.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-30-1
PROBLEM 5-30 Statement:
To the linkage solution from Problem 5-29, add a driver dyad with a crank to control the motion of your fourbar so that it cannot move beyond positions one and three.
Given:
Solution to Problem 5-29:
Solution: 1.
Length of link 2
w 785.316
Angle of link 2 in first position
θ 157.992 deg
Rotation angles for link 2
β 54.008 deg
β 107.980 deg
Coordinates of O2
O2x 362.0
O2y 291.0
See Figure P5-5 and Mathcad file P0530.
Link 2 of the solution to Problem 5-29 will become the driven link for the driver dyad. The driver dyad will be links 5 and 6 and the fixed pivot for the dyad will be at O6. Select a point on link 2 of Problem 5-29 and label it C. Let the distance O2C be R2 200. The solution that follows uses the algorithm presented in Section 5.2 with changes in nomenclature to account for the fact that the driven link is link 2 and the points A and B are already defined on the fourbar of Problem 5-29..
2.
Determine the coordinates of the points C1 and C3 using equations 5.0a. Determine the vector M using 5.0b. C1x O2x R2 cos θ
C1x 547.426
C1y O2y R2 sin θ
C1y 216.053
C3x 233.475
C3y 137.764
C3x O2x R2 cos θ β C3y O2y R2 sin θ β RC1
C1x C1y
RC3
C3x C3y
M RC3 RC1
M
313.952 78.289
3.
Select a suitable value for the multiplier, K, in equation 5.0d say K 3.0.
4.
Determine the coordinates of the crank pivot, O6 using equation 5.0d. Place the pivot to the left of O2 (by subtracting KM from RC3) so that it will be on the base below and to the left of O2. RO6 RC3 K M
O6x RO6
1
O6x 1175.330 5.
2
O6y 372.630
Determine the length of the driving crank using equation 5.0e.
R6 R2 sin 0.5 β 6.
O6y RO6
R6 161.783
Determine the length of the driver dyad coupler, link 5, and the ground link from eqauation 5.0f. R5 RC3 RO6 R6 RO2
R5 808.914
O2x O2y
R1 RO2 RO6
R1 817.416
DESIGN OF MACHINERY - 5th Ed.
7.
SOLUTION MANUAL 5-30-2
Determine the Grashof condition. R1 817.416
R2 200.000
R5 808.914
R6 161.783
Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition R1 R2 R5 R6 "Grashof" 8.
Draw the linkage using the link lengths and fixed pivot coordinates calculated above to verify that the driver dyad will perform as required.
Y
P3
P2
A2
B3
B2 A3
A1
B1
C2
D2 D1
O6
D3
C1
X
P1 C3
O2
O4
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-31-1
PROBLEM 5-31 Statement:
Design a fourbar linkage to carry the box in Figure P5-6 through the three positions shown in their numbered order without regard for the fixed pivots shown. Use points A and B for your attachment points. Determine the range of the transmission angle. The fixed pivots should be on the base.
Given:
Coordinates of the points P1 , P2 and P3 with respect to P1: P1x 0.0
P1y 0.0
P3x 148.0
P3y 187.0
P2x 130.0
P2y 29.0
Angles made by the body in positions 1, 2 and 3:
θP1 90.0 deg
θP2 65.0 deg
θP3 11.0 deg
Coordinates of the points A1 and B1 with respect to P1: A1x 69.0 Solution: 1.
2.
3.
A1y 43.0
B1x 17.0
B1y 43.0
See Figure P5-6 and Mathcad file P0531.
Determine the magnitudes and orientation of the position difference vectors. 2
2
p 21 133.195
δ atan2 P2x P2y
δ 12.575 deg
2
2
p 31 238.481
δ atan2 P3x P3y
δ 51.640 deg
p 21
P2x P2y
p 31
P3x P3y
Determine the angle changes of the coupler between precision points. α θP2 θP1
α 25.000 deg
α θP3 θP1
α 101.000 deg
Using Figure P5-4, the given data, and the law of cosines, determine z, s, , and . z
P1x A1x2 P1y A1y 2
z 81.302
s
P1x B1x 2 P1y B1y 2
s 46.239
v
A1x B1x 2 A1y B1y2
v 52.000
v2 z2 s2 2 v z
ϕ acos
v2 s2 z2 2 v s
ϕ 31.931 deg
ψ π acos
ψ 68.429 deg
Z1x z cos ϕ
Z1x 69.000
Z1y z sin ϕ
Z1y 43.000
S 1x s cos ψ
S 1x 17.000
S 1y s sin ψ
S 1y 43.000
DESIGN OF MACHINERY - 5th Ed.
4.
SOLUTION MANUAL 5-31-2
Use equations 5.24 to solve for w, , 2, and 3. Since the points A and B are to be used as pivots, z and are known from the calculations above. Guess:
W1x 50
W1y 200
β 80 deg
Given
W1x cos β 1 W1y sin β = p 21 cos δ Z1x cos α 1 Z1y sin α
β 160 deg
W1x cos β 1 W1y sin β = p 31 cos δ Z1x cos α 1 Z1y sin α W1y cos β 1 W1x sin β = p 21 sin δ Z1y cos α 1 Z1x sin α W1y cos β 1 W1x sin β = p 31 sin δ Z1y cos α 1 Z1x sin α
W1x W1y Find W1x W1y β β β β
β 52.277 deg
β 96.147 deg
The components of the W vector are: W1x 63.415 The length of link 2 is: w 5.
θ atan2 W1x W1y
W1y 118.432
θ 118.167 deg
W1x2 W1y2 , w 134.341
Use equations 5.28 to solve for u, , 2, and 3. Since the points A and B are to be used as pivots, s and are known from the calculations above. Guess:
U1x 30
U1y 100
γ 80 deg
Given
U1x cos γ 1 U1y sin γ = p 21 cos δ S 1x cos α 1 S 1y sin α
U1x cos γ 1 U1y sin γ = p 31 cos δ S 1x cos α 1 S 1y sin α
U1y cos γ 1 U1x sin γ = p 31 sin δ S 1y cos α 1 S 1x sin α U1y cos γ 1 U1x sin γ = p 21 sin δ S 1y cos α 1 S 1x sin α
γ 160 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-31-3
U1x U1y Find U1x U1y γ γ γ γ γ 79.044 deg
γ 147.982 deg
The components of the U vector are: U1x 45.930
U1y 77.634
The length of link 4 is: u 6.
Link 1:
V1x 52.000
V1y Z1y S 1y
V1y 0.000
θ atan2 V1x V1y
θ 0.000 deg
2
2
V1x V1y
v 52.000
G1x W1x V1x U1x
G1x 34.515
G1y W1y V1y U1y
G1y 40.798
θ atan2 G1x G1y
θ 49.768 deg
g
9.
U1x2 U1y2 , u 90.203
V1x Z1x S 1x
v
8.
σ 120.609 deg
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
7.
σ atan2 U1x U1y
2
2
G1x G1y
g 53.439
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 68.399 deg
θ2f θ2i β
θ2f 27.748 deg
Define the coupler point with respect to point A and the vector V. rp z
δp ϕ θ
rp 81.302
δp 31.931 deg
Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. O2x z cos ϕ w cos θ
O2x 5.585
O2y z sin ϕ w sin θ
O2y 161.432
O4x s cos ψ u cos σ
O4x 28.930
O4y s sin ψ u sin σ
O4y 120.634
10. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-31-4
θrot atan2 O4x O2x O4y O2y
θrot 49.768 deg
11. Determine the Grashof condition.
Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "non-Grashof" 12. DESIGN SUMMARY Link 2:
w 134.341
θ 118.167 deg
Link 3:
v 52.000
θ 0.000 deg
Link 4:
u 90.203
σ 120.609 deg
Link 1:
g 53.439
θ 49.768 deg
Coupler:
rp 81.302
δp 31.931 deg
Crank angles:
θ2i 68.399 deg θ2f 27.748 deg 13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design. 14. A driver dyad with a crank should be added to link 2 to control the motion of the fourbar so that it cannot move beyond positions 1 and 3.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-31-5 Y P1 Z1
A1
X
S1
S2
B1
V2 B2
U1 W1
U2
W2 O4 W3 O2
P2
Z2
A2
V1
A3
U3 V3 B3
Z3 S3 P3
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-32-1
PROBLEM 5-32 Statement:
Design a fourbar linkage to carry the box in Figure P5-6 through the three positions shown in their numbered order without regard for the fixed pivots shown. Use any points on the object as attachment points. Determine the range of the transmission angle. The fixed pivots should be on the base.
Given:
Coordinates of the points P1 , P2 and P3 with respect to P1: P1x 0.0
P1y 0.0
P2x 130.0
P3x 148.0
P3y 187.0
P2y 29.0
Angles made by the body in positions 1, 2 and 3:
θP1 90.0 deg
θP2 65.0 deg
θP3 11.0 deg
Free choices for the WZ dyad : β 52.0 deg
β 95.0 deg
Free choices for the US dyad : γ 76.0 deg Solution: 1.
2.
3.
γ 145.0 deg
See Figure P5-6 and Mathcad file P0532.
Determine the magnitudes and orientation of the position difference vectors. 2
2
p 21 133.195
δ atan2 P2x P2y
δ 12.575 deg
2
2
p 31 238.481
δ atan2 P3x P3y
δ 51.640 deg
p 21
P2x P2y
p 31
P3x P3y
Determine the angle changes of the coupler between precision points. α θP2 θP1
α 25.000 deg
α θP3 θP1
α 101.000 deg
Evaluate terms in the WZ coefficient matrix and constant vector from equations 5.26 and form the matrix and vector:
B sin β
E p 21 cos δ
H cos α 1
A cos β 1
D sin α
F cos β 1
G sin β
K sin α
L p 31 cos δ
A F AA B G
C cos α 1
M p 21 sin δ
B C D
E L CC M N
G H K A D C F K H
N p 31 sin δ
W1x W 1y AA 1 CC Z1x Z1y
The components of the W and Z vectors are: W1x 63.498
W1y 118.196
Z1x 69.575
Z1y 44.896
θ atan2 W1x W1y
θ 118.246 deg
ϕ atan2 Z1x Z1y
ϕ 32.834 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-32-2
W1x2 W1y2 , w 134.173
The length of link 2 is: w
Z1x2 Z1y2 , z 82.803
The length of vector Z is: z 4.
Evaluate terms in the US coefficient matrix and constant vector from equations 5.31 and form the matrix and vector:
A' cos γ 1
B' sin γ
C cos α 1
E p 21 cos δ
H cos α 1
D sin α
G' sin γ
K sin α
L p 31 cos δ
A' F' AA B' G'
F' cos γ 1
M p 21 sin δ
B' C D
E L CC M N
G' H K A' D C F' K H
N p 31 sin δ
U1x U1y AA 1 CC S1x S1y
The components of the U and S vectors are: U1x 46.325
U1y 82.956
S 1x 17.754
S 1y 37.985
σ atan2 U1x U1y
σ 119.180 deg
ψ atan2 S 1x S 1y
ψ 64.949 deg
The length of link 4 is: u
U1x2 U1y2 , u 95.014
The length of vector S is: s 6.
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
V1x Z1x S 1x
V1x 51.820
V1y Z1y S 1y
V1y 6.910
θ atan2 V1x V1y
θ 7.596 deg
v Link 1:
2
2
V1x V1y
v 52.279
G1x W1x V1x U1x
G1x 34.647
G1y W1y V1y U1y
G1y 42.151
θ atan2 G1x G1y
θ 50.580 deg
g 7.
S 1x2 S 1y2 , s 41.930
2
2
G1x G1y
g 54.563
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 67.666 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-32-3
θ2f θ2i β 8.
9.
θ2f 27.334 deg
Define the coupler point with respect to point A and the vector V. rp z
δp ϕ θ
rp 82.803
δp 25.238 deg
Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. O2x z cos ϕ w cos θ
O2x 6.076
O2y z sin ϕ w sin θ
O2y 163.092
O4x s cos ψ u cos σ
O4x 28.571
O4y s sin ψ u sin σ
O4y 120.941
10. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.
θrot atan2 O4x O2x O4y O2y
θrot 50.580 deg
11. Determine the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "non-Grashof" 12. DESIGN SUMMARY Link 2:
w 134.173
θ 118.246 deg
Link 3:
v 52.279
θ 7.596 deg
Link 4:
u 95.014
σ 119.180 deg
Link 1:
g 54.563
θ 50.580 deg
Coupler:
rp 82.803
δp 25.238 deg
Crank angles:
θ2i 67.666 deg θ2f 27.334 deg 13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design. 14. A driver dyad with a crank should be added to link 2 to control the motion of the fourbar so that it cannot move beyond positions 1 and 3.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-32-4
Y P1 Z1 A1 V1
X
S1 B1
S2 V2 U1
W1
B2
U2
W2 O4 W3 O2
P2
Z2
A2
A3
U3 V3
Z3
B 3 S3 P3
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-33-1
PROBLEM 5-33 Statement: Given:
Solution: 1.
2.
Design a linkage to carry the object in Figure P5-6 through the three positions shown in their numbered order using the fixed pivots shown. Determine the range of the transmission angle. P21x 130.0
P21y 29.0
P31x 148.0
P31y 187.0
O2x 6.2
O2y 164.0
O4x 28.0
O4y 121.0
Body angles:
θP1 90 deg
θP2 65 deg
θP3 11 deg
See Figure P5-6 and Mathcad file P0533.
Determine the angle changes between precision points from the body angles given. α θP2 θP1
α 25.000 deg
α θP3 θP1
α 101.000 deg
Using Figure 5-6, determine the magnitudes of R1, R2, and R3 and their x and y components. R1x O2x
3.
4.
R1x 6.200
R1y O2y
R2x R1x P21x
R2x 136.200
R2y R1y P21y
R2y 135.000
R3x R1x P31x
R3x 154.200
R3y R1y P31y
R3y 23.000
2
2
R1 164.117
2
2
R2 191.769
2
2
R3 155.906
R1
R1x R1y
R2
R2x R2y
R3
R3x R3y
R1y 164.000
Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ atan2 R1x R1y
ζ 87.835 deg
ζ atan2 R2x R2y
ζ 44.746 deg
ζ atan2 R3x R3y
ζ 8.484 deg
Solve for 2 and 3 using equations 5.34
C3 5.604
C4 14.379
C5 61.271
C6 11.014
C1 R3 cos α ζ R2 cos α ζ C2 R3 sin α ζ R2 sin α ζ C3 R1 cos α ζ R3 cos ζ
C4 R1 sin α ζ R3 sin ζ
C5 R1 cos α ζ R2 cos ζ
C6 R1 sin α ζ R2 sin ζ
C1 23.501 C2 73.444
DESIGN OF MACHINERY - 5th Ed. 2
SOLUTION MANUAL 5-33-2 2
A1 C3 C4
A1 238.152
A2 C3 C6 C4 C5
A2 819.286
A3 C4 C6 C3 C5
A3 501.727
A4 C2 C3 C1 C4
A4 749.483
A5 C4 C5 C3 C6
A5 819.286
A6 C1 C3 C2 C4
A6 924.339
K1 A2 A4 A3 A6
K1 1.503 10
K2 A3 A4 A5 A6
K2 1.133 10
2
K3
5 6
2
2
2
A1 A2 A3 A4 A6
2
K3 1.141 10
2
K K 2 K 2 K 2 2 1 2 3 β 2 atan K1 K3
β 101.000 deg
K K 2 K 2 K 2 2 1 2 3 β 2 atan K1 K3
β 94.106 deg
The first value is the same as 3, so use the second value
β β
A5 sin β A3 cos β A6 A1
β 53.072 deg
A3 sin β A2 cos β A4 A1
β 53.072 deg
β acos
β asin
Since 2 is not in the first quadrant, 5.
6
β β
Repeat steps 2, 3, and 4 for the right-hand dyad to find 1 and 2. R1x O4x
R1x 28.000
R1y O4y
R2x R1x P21x
R2x 102.000
R2y R1y P21y
R2y 92.000
R3x R1x P31x
R3x 120.000
R3y R1y P31y
R3y 66.000
2
2
R1 124.197
2
2
R2 137.361
R1
R1x R1y
R2
R2x R2y
R1y 121.000
DESIGN OF MACHINERY - 5th Ed.
R3 6.
7.
SOLUTION MANUAL 5-33-3
2
2
R3x R3y
R3 136.953
Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ atan2 R1x R1y
ζ 103.029 deg
ζ atan2 R2x R2y
ζ 42.049 deg
ζ atan2 R3x R3y
ζ 28.811 deg
Solve for 2 and 3 using equations 5.34
C3 4.120
C4 70.398
C5 76.240
C6 29.497
C1 R3 cos α ζ R2 cos α ζ C2 R3 sin α ζ R2 sin α ζ C3 R1 cos α ζ R3 cos ζ
C4 R1 sin α ζ R3 sin ζ
C5 R1 cos α ζ R2 cos ζ
C6 R1 sin α ζ R2 sin ζ 2
C2 7.150
A1 C3 C4
A1 4.973 10
3
A2 C3 C6 C4 C5
A2 5.489 10
3
A3 C4 C6 C3 C5
A3 1.762 10
3
A4 C2 C3 C1 C4
A4 675.715
A5 C4 C5 C3 C6
A5 5.489 10
A6 C1 C3 C2 C4
A6 544.601
K1 A2 A4 A3 A6
K1 2.749 10
K2 A3 A4 A5 A6
K2 4.180 10
2
K3
2
C1 10.017
3
6 6
2
2
2
A1 A2 A3 A4 A6
2
2
K3 4.628 10
6
K K 2 K 2 K 2 2 1 2 3 γ 2 atan K1 K3
γ 145.661 deg
K K 2 K 2 K 2 2 1 2 3 2 atan K1 K3
101.000 deg
The second value is the same as 3, so use the first value
γ γ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-33-4
A5 sin γ A3 cos γ A6 A1
77.265 deg
A3 sin γ A2 cos γ A4 A1
77.265 deg
acos
asin
Since 2 is not in the first quadrant , 8.
γ
Use the method of Section 5.7 to synthesize the linkage. Start by determining the magnitudes of the vectors P21 and P31 and their angles with respect to the X axis. 2
p 21
2
P21x P21y
p 21 133.195
δ atan2 P21x P21y 2
p 31
δ 12.575 deg
2
P31x P31y
p 31 238.481
δ atan2 P31x P31y 9.
δ 51.640 deg
Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:
B sin β
E p 21 cos δ
H cos α 1
A cos β 1
D sin α
F cos β 1
G sin β
K sin α
L p 31 cos δ
M p 21 sin δ
B C D
A F AA B G
C cos α 1
E L CC M N
G H K A D C F K H
N p 31 sin δ
W1x W1y AA 1 CC Z1x Z1y
10. The components of the W and Z vectors are: W1x 61.917 w
11. The length of link 2 is:
W1y 112.415 2
Z1x 68.117
2
W1x W1y
Z1y 51.585
w 128.339
12. Evaluate terms in the US coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:
A' cos γ 1
B' sin γ
C cos α 1
E p 21 cos δ
H cos α 1
D sin α
G' sin γ
L p 31 cos δ
M p 21 sin δ
F' cos γ 1
K sin α
N p 31 sin δ
DESIGN OF MACHINERY - 5th Ed.
A' F' AA B' G'
SOLUTION MANUAL 5-33-5
B' C D
E L CC M N
G' H K A' D C F' K H
U1x U1y AA 1 CC S1x S1y
13. The components of the W and Z vectors are: U1x 46.374
14. The length of link 4 is:
U1y 80.382 2
u
U1x U1y
S1x 18.374
2
S1y 40.618
u 92.800
15. Solving for links 3 and 1 from equations 5.2a and 5.2b. V1x Z1x S1x
V1x 49.743
V1y Z1y S1y
V1y 10.967 v
The length of link 3 is:
2
2
V1x V1y
v 50.938
G1x W1x V1x U1x
G1x 34.200
G1y W1y V1y U1y
G1y 43.000
g
The length of link 1 is:
2
G1x G1y
2
g 54.942
16. Check the location of the fixed pivots with respect to the global frame using the calculated vectors W1, Z1, U1, and S1. O2x Z1x W1x
O2x 6.200
O2y Z1y W1y
O2y 164.000
O4x S1x U1x
O4x 28.000
O4y S1y U1y
O4y 121.000
These check with Figure P5-6. 17. Determine the location of the coupler point with respect to point A and line AB. 2
2
z 85.446
2
2
s 44.581
Distance from A to P
z
Z1x Z1y
Angle BAP (p)
s
S1x S1y
ψ atan2( S1x S1y)
ψ 65.660 deg
ϕ atan2( Z1x Z1y )
ϕ 37.136 deg
rP z
θ atan2 z cos ϕ s cos ψ z sin ϕ s sin ψ θ 12.433 deg
δp ϕ θ
δp 24.704 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-33-6
18. DESIGN SUMMARY Link 1:
g 54.942
Link 2:
w 128.339
Link 3:
v 50.938
Link 4:
u 92.800
Coupler point:
rP 85.446
δp 24.704 deg
19. VERIFICATION: The calculated values of g (length of the ground link) and of the coordinates of O2 and O4 give the same values as those on the problem statement, verifying that the calculated values for the other links and the coupler point are correct. 20. A driver dyad with a crank should be added to link 2 to control the motion of the fourbar so that it cannot move beyond positions 1 and 3.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-34-1
PROBLEM 5-34 Statement:
Design a fourbar linkage to carry the bolt in Figure P5-7 from positions 1 to 2 to 3 without regard for the fixed pivots shown. The bolt is fed into the gripper in the z direction (into the paper). The gripper grabs the bolt, and your linkage moves it to position 3 to be inserted into the hole. A second degree of freedom within the gripper assembly (not shown) pushes the bolt into the hole. The moving pivots should be on, or close to, the gripper assembly, and the fixed pivots should be on the base. Use the free choices given below.
Given:
Coordinates of the points P1 , P2 and P3 with respect to P1: P1x 0.0
P1y 0.0
P2x 99.0
P3x 111.3
P3y 151.8
P2y 13.0
Angles made by the body in positions 1, 2 and 3:
θP1 272.3 deg
θP2 301.7 deg
θP3 270.0 deg
Free choices for the WZ dyad : β 70 deg
β 140 deg
Free choices for the US dyad : γ 5 deg Solution: 1.
2.
3.
γ 49 deg
See Figure P5-7 and Mathcad file P0534.
Determine the magnitudes and orientation of the position difference vectors. 2
2
p 21 99.850
δ atan2 P2x P2y
δ 7.481 deg
2
2
p 31 188.231
δ atan2 P3x P3y
δ 53.751 deg
p 21
P2x P2y
p 31
P3x P3y
Determine the angle changes of the coupler between precision points. α θP2 θP1
α 29.400 deg
α θP3 θP1
α 2.300 deg
Evaluate terms in the WZ coefficient matrix and constant vector from equations 5.26 and form the matrix and vector:
B sin β
E p 21 cos δ
H cos α 1
A cos β 1 D sin α G sin β
L p 31 cos δ
A F AA B G
B C D
G H K A D C F K H
The components of the W and Z vectors are:
C cos α 1
M p 21 sin δ
E L CC M N
F cos β 1
K sin α
N p 31 sin δ
W1x W 1y AA 1 CC Z1x Z1y
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-34-2
W1x 86.624
W1y 50.030
Z1x 198.147
Z1y 233.314
θ atan2 W1x W1y
θ 149.991 deg
ϕ atan2 Z1x Z1y
ϕ 49.660 deg
W1x2 W1y2 , w 100.033
The length of link 2 is: w
Z1x2 Z1y2 , z 306.100
The length of vector Z is: z 4.
Evaluate terms in the US coefficient matrix and constant vector from equations 5.31 and form the matrix and vector:
D sin α
E p 21 cos δ
A' cos γ 1
C cos α 1
G' sin γ
H cos α 1
K sin α
L p 31 cos δ
A' F' AA B' G'
F' cos γ 1
B' sin γ
M p 21 sin δ
B' C D
E L CC M N
G' H K A' D C F' K H
N p 31 sin δ
U1x U 1y AA 1 CC S1x S1y
The components of the U and S vectors are: U1x 107.545
U1y 205.365
S 1x 3.375
S 1y 166.927
σ atan2 U1x U1y
σ 62.360 deg
ψ atan2 S 1x S 1y
ψ 88.842 deg
The length of link 4 is: u
U1x2 U1y2 , u 231.821
The length of vector S is: s 6.
S 1x2 S 1y2 , s 166.961
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
V1x Z1x S 1x
V1x 194.772
V1y Z1y S 1y
V1y 66.387
θ atan2 V1x V1y
θ 18.821 deg
v Link 1:
2
2
V1x V1y
v 205.775
G1x W1x V1x U1x
G1x 0.602
G1y W1y V1y U1y
G1y 221.722
θ atan2 G1x G1y
θ 89.844 deg
g
2
2
G1x G1y
g 221.723
DESIGN OF MACHINERY - 5th Ed.
7.
8.
9.
SOLUTION MANUAL 5-34-3
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 239.836 deg
θ2f θ2i β 2 π
θ2f 19.836 deg
Define the coupler point with respect to point A and the vector V. rp z
δp ϕ θ
rp 306.100
δp 30.838 deg
Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. O2x z cos ϕ w cos θ
O2x 111.523
O2y z sin ϕ w sin θ
O2y 183.284
O4x s cos ψ u cos σ
O4x 110.920
O4y s sin ψ u sin σ
O4y 38.438
10. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.
θrot atan2 O4x O2x O4y O2y
θrot 89.844 deg
11. Determine the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "Grashof" 12. DESIGN SUMMARY Link 2:
w 100.033
θ 149.991 deg
Link 3:
v 205.775
θ 18.821 deg
Link 4:
u 231.821
σ 62.360 deg
Link 1:
g 221.723
θ 89.844 deg
Coupler:
rp 306.100
δp 30.838 deg
Crank angles:
θ2i 239.836 deg θ2f 19.836 deg 13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-34-4
Y A1 O2
B1
B2
A2 A3 P2
B3 X
P1 O4
P3
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-35-1
PROBLEM 5-35 Statement: Given:
Solution: 1.
2.
Design a linkage to carry the bolt in Figure P5-7 from positions 1 to 2 to 3 using the fixed pivots shown. See Problem 5-34 for more details. P21x 99.0
P21y 13.0
P31x 111.3
P31y 151.8
O2x 111.5
O2y 183.2
O4x 111.5
O4y 38.8
Body angles:
θP1 272.3 deg
θP2 301.7 deg
θP3 270 deg
See Figure P5-7 and Mathcad file P0535.
Determine the angle changes between precision points from the body angles given. α θP2 θP1
α 29.400 deg
α θP3 θP1
α 2.300 deg
Using Figure 5-6, determine the magnitudes of R1, R2, and R3 and their x and y components. R1x O2x
3.
4.
R1x 111.500
R1y O2y
R2x R1x P21x
R2x 210.500
R2y R1y P21y
R2y 170.200
R3x R1x P31x
R3x 222.800
R3y R1y P31y
R3y 335.000
2
2
R1 214.463
2
2
R2 270.700
2
2
R3 402.324
R1
R1x R1y
R2
R2x R2y
R3
R3x R3y
R1y 183.200
Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ atan2 R1x R1y
ζ 58.674 deg
ζ atan2 R2x R2y
ζ 38.957 deg
ζ atan2 R3x R3y
ζ 56.373 deg
Solve for 2 and 3 using equations 5.34
C3 118.742
C4 147.473
C5 23.426
C6 65.329
C1 R3 cos α ζ R2 cos α ζ C2 R3 sin α ζ R2 sin α ζ C3 R1 cos α ζ R3 cos ζ
C4 R1 sin α ζ R3 sin ζ
C5 R1 cos α ζ R2 cos ζ
C6 R1 sin α ζ R2 sin ζ
C1 155.059 C2 3.973
DESIGN OF MACHINERY - 5th Ed.
2
SOLUTION MANUAL 5-35-2
2
A1 3.585 10
A2 C3 C6 C4 C5
A2 4.303 10
A3 C4 C6 C3 C5
A3 1.242 10
4
A4 C2 C3 C1 C4
A4 2.240 10
4
A5 C4 C5 C3 C6
A5 4.303 10
3
A6 C1 C3 C2 C4
A6 1.900 10
4
K1 A2 A4 A3 A6
K1 1.395 10
K2 A3 A4 A5 A6
K2 3.598 10
2
K3
3
8 8
2
2
2
A1 A2 A3 A4 A6
2 8
K3 1.250 10
2
K K 2 K 2 K 2 2 1 2 3 β 2 atan K1 K3
β 139.911 deg
K K 2 K 2 K 2 2 1 2 3 β 2 atan K1 K3
β 2.300 deg
The second value is the same as 3, so use the first value
β β
A5 sin β A3 cos β A6 A1
β 69.984 deg
A3 sin β A2 cos β A4 A1
β 69.984 deg
β acos
β asin
β β
Since both angles are the same, 5.
4
A1 C3 C4
Repeat steps 2, 3, and 4 for the right-hand dyad to find 1 and 2. R1x O4x
R1x 111.500
R1y O4y
R2x R1x P21x
R2x 210.500
R2y R1y P21y
R2y 51.800
R3x R1x P31x
R3x 222.800
R3y R1y P31y
R3y 113.000
2
2
R1 118.058
2
2
R2 216.780
R1
R1x R1y
R2
R2x R2y
R1y 38.800
DESIGN OF MACHINERY - 5th Ed.
R3 6.
7.
SOLUTION MANUAL 5-35-3
2
2
R3x R3y
R3 249.818
Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ atan2 R1x R1y
ζ 19.187 deg
ζ atan2 R2x R2y
ζ 13.825 deg
ζ atan2 R3x R3y
ζ 26.893 deg
Solve for 2 and 3 using equations 5.34
C3 109.833
C4 147.294
C5 132.407
C6 36.739
C1 R3 cos α ζ R2 cos α ζ C2 R3 sin α ζ R2 sin α ζ C3 R1 cos α ζ R3 cos ζ
C4 R1 sin α ζ R3 sin ζ
C5 R1 cos α ζ R2 cos ζ
C6 R1 sin α ζ R2 sin ζ 2
C2 32.384
A1 C3 C4
A1 3.376 10
4
A2 C3 C6 C4 C5
A2 1.547 10
4
A3 C4 C6 C3 C5
A3 1.995 10
4
A4 C2 C3 C1 C4
A4 1.918 10
3
A5 C4 C5 C3 C6
A5 1.547 10
A6 C1 C3 C2 C4
A6 8.852 10
K1 A2 A4 A3 A6
K1 2.063 10
K2 A3 A4 A5 A6
K2 9.865 10
2
K3
2
C1 37.169
4 3
8
2
2
2
A1 A2 A3 A4 A6
7
2
2
8
K3 2.101 10
K K 2 K 2 K 2 2 1 2 3 γ 2 atan K1 K3
γ 2.300 deg
K K 2 K 2 K 2 2 1 2 3 2 atan K1 K3
48.814 deg
The first value is the same as 3, so use the second value
γ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-35-4
A5 sin γ A3 cos γ A6 A1
4.951 deg
A3 sin γ A2 cos γ A4 A1
4.951 deg
acos
asin
Since 2 is not in the first quadrant , 8.
γ
Use the method of Section 5.7 to synthesize the linkage. Start by determining the magnitudes of the vectors P21 and P31 and their angles with respect to the X axis. 2
p 21
2
P21x P21y
p 21 99.850
δ atan2 P21x P21y 2
p 31
δ 7.481 deg
2
P31x P31y
p 31 188.231
δ atan2 P31x P31y 9.
δ 53.751 deg
Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:
B sin β
E p 21 cos δ
H cos α 1
A cos β 1
D sin α
F cos β 1
G sin β
K sin α
L p 31 cos δ
M p 21 sin δ
B C D
A F AA B G
C cos α 1
E L CC M N
G H K A D C F K H
N p 31 sin δ
W1x W1y AA 1 CC Z1x Z1y
10. The components of the W and Z vectors are: W1x 86.684 w
11. The length of link 2 is:
W1y 49.977 2
Z1x 198.184
2
W1x W1y
Z1y 233.177
w 100.059
12. Evaluate terms in the US coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:
A' cos γ 1
B' sin γ
C cos α 1
E p 21 cos δ
H cos α 1
D sin α
G' sin γ
L p 31 cos δ
M p 21 sin δ
F' cos γ 1
K sin α
N p 31 sin δ
DESIGN OF MACHINERY - 5th Ed.
A' F' AA B' G'
SOLUTION MANUAL 5-35-5
B' C D
E L CC M N
G' H K A' D C F' K H
U1x U1y AA 1 CC S1x S1y
13. The components of the W and Z vectors are: U1x 108.268
14. The length of link 4 is:
U1y 205.938
S1x 3.232
2
u 232.664
2
u
U1x U1y
S1y 167.138
15. Solving for links 3 and 1 from equations 5.2a and 5.2b. V1x Z1x S1x
V1x 194.953
V1y Z1y S1y
V1y 66.039 v
The length of link 3 is:
2
2
V1x V1y
v 205.834
G1x W1x V1x U1x
G1x 1.990 10
G1y W1y V1y U1y
G1y 222.000
g
The length of link 1 is:
2
G1x G1y
2
13
g 222.000
16. Check the location of the fixed pivots with respect to the global frame using the calculated vectors W1, Z1, U1, and S1. O2x Z1x W1x
O2x 111.500
O2y Z1y W1y
O2y 183.200
O4x S1x U1x
O4x 111.500
O4y S1y U1y
O4y 38.800
These check with Figure P5-7. 17. Determine the location of the coupler point with respect to point A and line AB. 2
2
z 306.020
2
2
s 167.169
Distance from A to P
z
Z1x Z1y
Angle BAP (p)
s
S1x S1y
ψ atan2( S1x S1y)
ψ 88.892 deg
ϕ atan2( Z1x Z1y )
ϕ 49.638 deg
rP z
θ atan2 z cos ϕ s cos ψ z sin ϕ s sin ψ θ 18.714 deg
δp ϕ θ
δp 30.924 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-35-6
18. DESIGN SUMMARY Link 1:
g 222.000
Link 2:
w 100.059
Link 3:
v 205.834
Link 4:
u 232.664
Coupler point:
rP 306.020
δp 30.924 deg
19. VERIFICATION: The calculated values of g (length of the ground link) and of the coordinates of O2 and O4 give the same values as those on the problem statement, verifying that the calculated values for the other links and the coupler point are correct.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-36-1
PROBLEM 5-36 Statement:
To the linkage solution from Problem 5-35, add a driver dyad with a crank to control the motion of your fourbar so that it cannot move beyond positions one and three.
Given:
Solution to Problem 5-35:
Solution: 1.
Length of link 4
u 232.664
Angle of link 4 in first position
θ 62.268 deg
Rotation angles for link 2
β 4.951 deg
β 48.814 deg
Coordinates of O4
O4x 111.5
O4y 38.8
See Figure P5-7 and Mathcad file P0536.
Link 4 of the solution to Problem 5-35 will become the driven link for the driver dyad. The driver dyad will be links 5 and 6 and the fixed pivot for the dyad will be at O6. Select a point on link 4 of Problem 5-35 and label it C. Let the distance O4C be R4 60. The solution that follows uses the algorithm presented in Section 5.2 with changes in nomenclature to account for the fact that the driven link is link 2 and the points A and B are already defined on the fourbar of Problem 5-35.
2.
Determine the coordinates of the points C1 and C3 using equations 5.0a. Determine the vector M using 5.0b. C1x O4x R4 cos θ
C1x 83.580
C1y O4y R4 sin θ
C1y 14.308
C3x 53.147
C3y 24.840
C3x O4x R4 cos θ β C3y O4y R4 sin θ β RC1
3. 4.
C1x C1y
RC3
C3x C3y
M RC3 RC1
Determine the coordinates of the crank pivot, O6 using equation 5.0d. Place the pivot to the left of O4 (by subtracting KM from RC3) so that it will be on the base above and to the left of O2. O6x RO6
1
O6x 144.446
O6y RO6
2
O6y 92.604
Determine the length of the driving crank using equation 5.0e.
R6 R4 sin 0.5 β 6.
30.433 39.148
Select a suitable value for the multiplier, K, in equation 5.0d say K 3.0.
RO6 RC3 K M
5.
M
R6 24.793
Determine the length of the driver dyad coupler, link 5, and the ground link from eqauation 5.0f. R5 RC3 RO6 R6 RO2
R5 123.965
O4x O4y
R1 RO2 RO6
R1 135.472
DESIGN OF MACHINERY - 5th Ed.
7.
SOLUTION MANUAL 5-36-2
Determine the Grashof condition. R1 135.472
R4 60.000
R5 123.965
R6 24.793
Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition R1 R4 R5 R6 "Grashof" 8.
Draw the linkage using the link lengths and fixed pivot coordinates calculated above to verify that the driver dyad will perform as required.
Y A1 D2
D1 O6
O2
B1
B2
D3
A2 A3 C1
P2
C2
B3 X
C3
P1
O4
P3
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-37-1
PROBLEM 5-37 Statement:
Figure P5-8 shows an off-loading mechanism for paper rolls. The V-link is rotated through 90 deg by an air-driven fourbar slider-crank linkage. Design a pin-jointed fourbar linkage to replace the existing off-loading station and perform essentially the same function. Choose three positions of the roll including its two end positions and synthesize a substitute mechanism. Use a link similar to the existing V-link as one of your links.
Given:
Coordinates of the points P1 , P2 and P3 with respect to P1: P1x 0.0
P1y 0.0
P3x 1000.0
P3y 1000.0
P2x 450
P2y 140
Angles made by the body in positions 1, 2 and 3:
θP1 90.0 deg
θP2 65.0 deg
θP3 0.0 deg
Coordinates of the points A1 and B1 with respect to P1: A1x 400.0 Solution: 1.
2.
3.
4.
A1y 1035.0
B1x 35.0
B1y 600.0
See Figure P5-8 and Mathcad file P0537.
Determine the magnitudes and orientation of the position difference vectors. 2
2
p 21 471.275
2
2
p 31 1.414 10
p 21
P2x P2y
p 31
P3x P3y
3
δ atan2 P2x P2y
δ 17.281 deg
δ atan2 P3x P3y
δ 45.000 deg
Determine the angle changes of the coupler between precision points. α θP2 θP1
α 25.000 deg
α θP3 θP1
α 90.000 deg
Using Figure P5-4, the given data, and the law of cosines, determine z, s, , and . z
P1x A1x2 P1y A1y 2
z 1109.606
s
P1x B1x 2 P1y B1y 2
s 601.020
v
A1x B1x 2 A1y B1y2
v 615.183
ϕ atan2 A1x A1y π
ϕ 291.130 deg
ψ π atan2 B1x B1y
ψ 93.338 deg
Z1x z cos ϕ
Z1x 400.000
Z1y z sin ϕ
Z1y 1035.000
S 1x s cos ψ
S 1x 35.000
S 1y s sin ψ
S 1y 600.000
Use equations 5.24 to solve for w, , 2, and 3. Since the points A and B are to be used as pivots, z and are known from the calculations above.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-37-2
Guess:
W1x 50
W1y 200
β 80 deg
Given
W1x cos β 1 W1y sin β = p 21 cos δ Z1x cos α 1 Z1y sin α
β 160 deg
W1x cos β 1 W1y sin β = p 31 cos δ Z1x cos α 1 Z1y sin α W1y cos β 1 W1x sin β = p 21 sin δ Z1y cos α 1 Z1x sin α W1y cos β 1 W1x sin β = p 31 sin δ Z1y cos α 1 Z1x sin α
W1x W1y Find W1x W1y β β β β β 11.094 deg
β 47.757 deg
The components of the W vector are: W1x 673.809 The length of link 2 is: w 5.
θ atan2 W1x W1y
W1y 194.748
θ 163.879 deg
W1x2 W1y2 , w 701.388
Use equations 5.28 to solve for u, , 2, and 3. Since the points A and B are to be used as pivots, s and are known from the calculations above. Guess:
U1x 30
U1y 100
γ 80 deg
Given
U1x cos γ 1 U1y sin γ = p 21 cos δ S 1x cos α 1 S 1y sin α
γ 160 deg
U1x cos γ 1 U1y sin γ = p 31 cos δ S 1x cos α 1 S 1y sin α U1y cos γ 1 U1x sin γ = p 21 sin δ S 1y cos α 1 S 1x sin α U1y cos γ 1 U1x sin γ = p 31 sin δ S 1y cos α 1 S 1x sin α
U1x U1y Find U1x U1y γ γ γ γ
γ 25.881 deg
The components of the U vector are:
γ 71.807 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-37-3
U1x 117.927
U1y 469.583
The length of link 4 is: u 6.
Link 1:
V1x 435.000
V1y Z1y S 1y
V1y 435.000
θ atan2 V1x V1y
θ 45.000 deg
2
2
V1x V1y
v 615.183
G1x W1x V1x U1x
G1x 356.736
G1y W1y V1y U1y
G1y 160.165
θ atan2 G1x G1y
θ 155.821 deg
2
g
9.
U1x2 U1y2 , u 484.164
V1x Z1x S 1x
v
8.
σ 75.903 deg
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
7.
σ atan2 U1x U1y
2
G1x G1y
g 391.042
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 8.058 deg
θ2f θ2i β
θ2f 39.699 deg
Define the coupler point with respect to point A and the vector V. rp z
δp ϕ θ
rp 1109.606
δp 336.130 deg
Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. O2x z cos ϕ w cos θ O2y z sin ϕ w sin θ
O4x s cos ψ u cos σ O4y s sin ψ u sin σ
O2x 273.809 O2y 1229.748 O4x 82.927 O4y 1069.583
10. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.
θrot atan2 O4x O2x O4y O2y 11. Determine the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
θrot 155.821 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-37-4
Condition( g w u v) "Grashof" 12. DESIGN SUMMARY Link 2:
w 701.388
θ 163.879 deg
Link 3:
v 615.183
θ 45.000 deg
Link 4:
u 484.164
σ 75.903 deg
Link 1:
g 391.042
θ 155.821 deg
Coupler:
rp 1109.606
δp 336.130 deg
Crank angles:
θ2i 8.058 deg θ2f 39.699 deg 13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design. 14. A driver dyad with a crank should be added to link 2 to control the motion of the fourbar so that it cannot move beyond positions 1 and 3.
1000.0 450.0 P1
140.0 P2
90 deg 65 deg
1000.0
A3 B1
B3
A2 A1
B2
O4 O2
0 deg
P3
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-38-1
PROBLEM 5-38 Statement:
Design a fourbar linkage to carry the object in Figure P5-9 through the three positions shown in their numbered order without regard for the fixed pivots shown. Use points C and D for your attachment points. Determine the range of the transmission angle.
Given:
Coordinates of the points P1 , P2 and P3 with respect to C1: P1x 0.0
P1y 0.0
P3x 7.600
P3y 1.000
P2x 4.500
P2y 1.900
Angles made by the body in positions 1, 2 and 3:
θP1 33.70 deg
θP2 14.60 deg
θP3 0.0 deg
Coordinates of the points C1 and D1 with respect to P1: C1x 0.0 Solution: 1.
2.
3.
4.
C1y 0.0
D1x 3.744
D1y 2.497
See Figure P5-9 and Mathcad file P0538.
Determine the magnitudes and orientation of the position difference vectors. 2
2
p 21 4.885
δ atan2 P2x P2y
δ 22.891 deg
2
2
p 31 7.666
δ atan2 P3x P3y
δ 7.496 deg
p 21
P2x P2y
p 31
P3x P3y
Determine the angle changes of the coupler between precision points. α θP2 θP1
α 19.100 deg
α θP3 θP1
α 33.700 deg
Using Figure P5-9, the given data, and the law of cosines, determine z, s, , and . z
P1x C1x2 P1y C1y 2
z 0.000
s
P1x D1x2 P1y D1y 2
s 4.500
v
C1x D1x2 C1y D1y2
v 4.500
ϕ θP1
ϕ 33.700 deg
ψ θP1 π
ψ 213.700 deg
Z1x z cos ϕ
Z1x 0.000
Z1y z sin ϕ
Z1y 0.000
S 1x s cos θP1
S 1x 3.744
S 1y s sin θP1
S 1y 2.497
Use equations 5.24 to solve for w, , 2, and 3. Since the points C and D are to be used as pivots, z and are known from the calculations above. Use guess values from a graphical solution such as that for Problem 3-57. Guess:
W1x 4
W1y 4
β 50 deg
β 80 deg
DESIGN OF MACHINERY - 5th Ed.
Given
SOLUTION MANUAL 5-38-2
W1x cos β 1 W1y sin β = p 21 cos δ Z1x cos α 1 Z1y sin α W1x cos β 1 W1y sin β = p 31 cos δ Z1x cos α 1 Z1y sin α W1y cos β 1 W1x sin β = p 21 sin δ Z1y cos α 1 Z1x sin α W1y cos β 1 W1x sin β = p 31 sin δ Z1y cos α 1 Z1x sin α
W1x W1y Find W1x W1y β β β β
β 47.370 deg
β 78.160 deg
The components of the W vector are: W1x 4.416 The length of link 2 is: w 5.
θ atan2 W1x W1y
W1y 4.179
θ 136.576 deg
W1x2 W1y2 , w 6.080
Use equations 5.28 to solve for u, , 2, and 3. Since the points C and D are to be used as pivots, s and are known from the calculations above. Guess:
U1x 4
U1y 4
γ 44 deg
Given
U1x cos γ 1 U1y sin γ = p 21 cos δ S 1x cos α 1 S 1y sin α
γ 76 deg
U1x cos γ 1 U1y sin γ = p 31 cos δ S 1x cos α 1 S 1y sin α U1y cos γ 1 U1x sin γ = p 21 sin δ S 1y cos α 1 S 1x sin α U1y cos γ 1 U1x sin γ = p 31 sin δ S 1y cos α 1 S 1x sin α
U1x U1y Find U1x U1y γ γ γ γ γ 43.852 deg
γ 76.170 deg
The components of the U vector are: U1x 3.223
U1y 6.080
σ atan2 U1x U1y
σ 117.929 deg
DESIGN OF MACHINERY - 5th Ed.
The length of link 4 is: u 6.
SOLUTION MANUAL 5-38-3
U1x2 U1y2 , u 6.881
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
V1x Z1x S 1x
V1x 3.744
V1y Z1y S 1y
V1y 2.497
θ atan2 V1x V1y v Link 1:
2
8.
9.
2
V1x V1y
v 4.500
G1x W1x V1x U1x
G1x 2.551
G1y W1y V1y U1y
G1y 0.596
θ atan2 G1x G1y
θ 13.155 deg
2
g 7.
θ 33.700 deg
2
G1x G1y
g 2.620
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 123.420 deg
θ2f θ2i β
θ2f 45.261 deg
Define the coupler point with respect to point C and the vector V. rp z
δp ϕ θ
rp 0.000
δp 0.000 deg
Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. O2x z cos ϕ w cos θ
O2x 4.416
O2y z sin ϕ w sin θ
O2y 4.179
O4x s cos ψ u cos σ
O4x 6.967
O4y s sin ψ u sin σ
O4y 3.583
10. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.
θrot atan2 O4x O2x O4y O2y 11. Determine the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "Grashof"
θrot 13.155 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-38-4
12. DESIGN SUMMARY Link 2:
w 6.080
θ 136.576 deg
Link 3:
v 4.500
θ 33.700 deg
Link 4:
u 6.881
σ 117.929 deg
Link 1:
g 2.620
θ 13.155 deg
Coupler:
rp 0.000
δp 0.000 deg
Crank angles:
θ2i 123.420 deg
7.600
θ2f 45.261 deg
4.500 Y
13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.
14. A driver dyad with a crank should be added to link 2 to control the motion of the fourbar so that it cannot move beyond positions 1 and 3.
4.500
D2
D1 C3
C2
D3
78.160° C1 47.370° y
X 43.852° 76.170° x 13.150°
6.080
2.620
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-39-1
PROBLEM 5-39 Statement:
Design a fourbar linkage to carry the object in Figure P5-9 through the three positions shown in their numbered order without regard for the fixed pivots shown. Use any points on the object as attachment points. Determine the range of the transmission angle.
Given:
Coordinates of the points P1 , P2 and P3 with respect to C1: P1x 0.0
P1y 0.0
P3x 7.600
P3y 1.000
P2x 4.500
P2y 1.900
Angles made by the body in positions 1, 2 and 3:
θP1 33.70 deg
θP2 14.60 deg
θP3 0.0 deg
Coordinates of the points C1 and E1 (used for attachment) with respect to P1: C1x 0.0 Solution: 1.
2.
3.
4.
C1y 0.0
E1x 3.744
E1y 0.000
See Figure P5-9 and Mathcad file P0539.
Determine the magnitudes and orientation of the position difference vectors. 2
2
p 21 4.885
δ atan2 P2x P2y
δ 22.891 deg
2
2
p 31 7.666
δ atan2 P3x P3y
δ 7.496 deg
p 21
P2x P2y
p 31
P3x P3y
Determine the angle changes of the coupler between precision points. α θP2 θP1
α 19.100 deg
α θP3 θP1
α 33.700 deg
Using Figure P5-9, the given data, and the law of cosines, determine z, s, , and . z
P1x C1x2 P1y C1y 2
z 0.000
s
P1x E1x 2 P1y E1y 2
s 3.744
v
C1x E1x 2 C1y E1y2
v 3.744
ϕ atan2 E1x C1x E1y C1y
ϕ 0.000 deg
ψ ϕ π
ψ 180.000 deg
Z1x z cos ϕ
Z1x 0.000
Z1y z sin ϕ
Z1y 0.000
S 1x s cos ψ
S 1x 3.744
S 1y s sin ψ
S 1y 0.000
Use equations 5.24 to solve for w, , 2, and 3. Since the points C and D are to be used as pivots, z and are known from the calculations above. Use guess values from a graphical solution such as that for Problem 3-57. Guess:
W1x 4
W1y 4
β 50 deg
β 80 deg
DESIGN OF MACHINERY - 5th Ed.
Given
SOLUTION MANUAL 5-39-2
W1x cos β 1 W1y sin β = p 21 cos δ Z1x cos α 1 Z1y sin α W1x cos β 1 W1y sin β = p 31 cos δ Z1x cos α 1 Z1y sin α W1y cos β 1 W1x sin β = p 21 sin δ Z1y cos α 1 Z1x sin α W1y cos β 1 W1x sin β = p 31 sin δ Z1y cos α 1 Z1x sin α
W1x W1y Find W1x W1y β β β β β 47.370 deg
β 78.160 deg
The components of the W vector are: W1x 4.416 The length of link 2 is: w 5.
θ atan2 W1x W1y
W1y 4.179
θ 136.576 deg
W1x2 W1y2 , w 6.080
Use equations 5.28 to solve for u, , 2, and 3. Since the points C and D are to be used as pivots, s and are known from the calculations above. Guess:
U1x 4
U1y 4
γ 44 deg
Given
U1x cos γ 1 U1y sin γ = p 21 cos δ S 1x cos α 1 S 1y sin α
γ 76 deg
U1x cos γ 1 U1y sin γ = p 31 cos δ S 1x cos α 1 S 1y sin α U1y cos γ 1 U1x sin γ = p 21 sin δ S 1y cos α 1 S 1x sin α U1y cos γ 1 U1x sin γ = p 31 sin δ S 1y cos α 1 S 1x sin α
U1x U1y Find U1x U1y γ γ γ γ γ 48.844 deg
γ 84.280 deg
The components of the U vector are: U1x 2.890
U1y 4.391
σ atan2 U1x U1y
σ 123.354 deg
DESIGN OF MACHINERY - 5th Ed.
The length of link 4 is: u 6.
V1x Z1x S 1x
V1x 3.744
V1y Z1y S 1y
V1y 0.000
θ atan2 V1x V1y
θ 0.000 deg
v Link 1:
2
9.
2
V1x V1y
v 3.744
G1x W1x V1x U1x
G1x 2.218
G1y W1y V1y U1y
G1y 0.211
θ atan2 G1x G1y
θ 5.444 deg
2
g
8.
U1x2 U1y2 , u 5.256
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
7.
SOLUTION MANUAL 5-39-3
2
G1x G1y
g 2.228
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 142.019 deg
θ2f θ2i β
θ2f 63.860 deg
Define the coupler point with respect to point C and the vector V. rp z
δp ϕ θ
rp 0.000
δp 0.000 deg
Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. O2x z cos ϕ w cos θ O2y z sin ϕ w sin θ
O4x s cos ψ u cos σ O4y s sin ψ u sin σ
O2x 4.416 O2y 4.179 O4x 6.634 O4y 4.391
10. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis the line O2O4.
θrot atan2 O4x O2x O4y O2y
θrot 5.444 deg
11. Determine the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-39-4
Condition( g u v w) "Grashof" 12. DESIGN SUMMARY Link 2:
w 6.080
θ 136.576 deg
Link 3:
v 3.744
θ 0.000 deg
Link 4:
u 5.256
σ 123.354 deg
Link 1:
g 2.228
θ 5.444 deg
Coupler:
rp 0.000
δp 0.000 deg
Crank angles:
θ2i 142.019 deg θ2f 63.860 deg
13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design. 14. A driver dyad with a crank should be added to link 2 to control the motion of the fourbar so that it cannot move beyond positions 1 and 3.
7.600 4.500 Y D2
D1
4.500
C2
D3
C3 E2
C1
X
E1
E3
6.080 O2 O4
5.257
2.228
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-40-1
PROBLEM 5-40 Statement:
Given:
Solution: 1.
2.
Design a fourbar linkage to carry the object in Figure P5-9 through the three positions shown in their numbered order using the fixed pivots shown. Determine the range of the transmission angle. P21x 4.500
P21y 1.900
P31x 7.600
P31y 1.000
O2x 2.900
O2y 5.100
O4x 5.900
O4y 5.100
Body angles:
θP1 33.70 deg
θP2 14.60 deg
θP3 0.0 deg
See Figure P5-9 and Mathcad file P0540.
Determine the angle changes between precision points from the body angles given. α θP2 θP1
α 19.100 deg
α θP3 θP1
α 33.700 deg
Using Figure 5-6, determine the magnitudes of R1, R2, and R3 and their x and y components. R1x O2x
3.
4.
R1x 2.900
R1y O2y
R2x R1x P21x
R2x 1.600
R2y R1y P21y
R2y 7.000
R3x R1x P31x
R3x 4.700
R3y R1y P31y
R3y 6.100
2
2
R1 5.867
2
2
R2 7.181
2
2
R3 7.701
R1
R1x R1y
R2
R2x R2y
R3
R3x R3y
R1y 5.100
Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ atan2 R1x R1y
ζ 119.624 deg
ζ atan2 R2x R2y
ζ 77.125 deg
ζ atan2 R3x R3y
ζ 52.386 deg
Solve for 2 and 3 using equations 5.34
C3 4.283
C4 0.248
C5 2.672
C1 R3 cos α ζ R2 cos α ζ C2 R3 sin α ζ R2 sin α ζ C3 R1 cos α ζ R3 cos ζ
C4 R1 sin α ζ R3 sin ζ
C5 R1 cos α ζ R2 cos ζ
C1 1.222 C2 0.710
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-40-2
C6 R1 sin α ζ R2 sin ζ 2
C6 1.232
2
A1 C3 C4
A1 18.405
A2 C3 C6 C4 C5
A2 4.613
A3 C4 C6 C3 C5
A3 11.748
A4 C2 C3 C1 C4
A4 3.343
A5 C4 C5 C3 C6
A5 4.613
A6 C1 C3 C2 C4
A6 5.059
K1 A2 A4 A3 A6
K1 44.009
K2 A3 A4 A5 A6
K2 62.605
2
K3
2
2
2
A1 A2 A3 A4 A6
2
K3 71.350
2
K K 2 K 2 K 2 2 1 2 3 β 2 atan K1 K3
β 33.700 deg
K K 2 K 2 K 2 2 1 2 3 β 2 atan K1 K3
β 76.089 deg
The first value is the same as 3, so use the second value
β β
A5 sin β A3 cos β A6 A1
β 47.808 deg
A3 sin β A2 cos β A4 A1
β 47.808 deg
β acos
β asin
β β
Use the negative value, 5.
Repeat steps 2, 3, and 4 for the right-hand dyad to find 1 and 2. R1x O4x
R1x 5.900
R1y O4y
R2x R1x P21x
R2x 1.400
R2y R1y P21y
R2y 7.000
R3x R1x P31x
R3x 1.700
R3y R1y P31y
R3y 6.100
R1
2
2
R1x R1y
R1 7.799
R1y 5.100
DESIGN OF MACHINERY - 5th Ed.
6.
7.
SOLUTION MANUAL 5-40-3
2
2
R2 7.139
2
2
R3 6.332
R2
R2x R2y
R3
R3x R3y
Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ atan2 R1x R1y
ζ 139.160 deg
ζ atan2 R2x R2y
ζ 101.310 deg
ζ atan2 R3x R3y
ζ 74.427 deg
Solve for 2 and 3 using equations 5.34
C3 3.779
C4 1.417
C5 2.506
C6 0.250
C1 R3 cos α ζ R2 cos α ζ C2 R3 sin α ζ R2 sin α ζ C3 R1 cos α ζ R3 cos ζ
C4 R1 sin α ζ R3 sin ζ
C5 R1 cos α ζ R2 cos ζ
C6 R1 sin α ζ R2 sin ζ 2
2
C1 0.883 C2 1.393
A1 C3 C4
A1 16.286
A2 C3 C6 C4 C5
A2 4.496
A3 C4 C6 C3 C5
A3 9.117
A4 C2 C3 C1 C4
A4 4.011
A5 C4 C5 C3 C6
A5 4.496
A6 C1 C3 C2 C4
A6 5.310
K1 A2 A4 A3 A6
K1 30.381
K2 A3 A4 A5 A6
K2 60.441
2
K3
2
2
2
A1 A2 A3 A4 A6
2
2
K3 58.811
K K 2 K 2 K 2 2 1 2 3 γ 2 atan K1 K3
γ 33.700 deg
K K 2 K 2 K 2 2 1 2 3 2 atan K1 K3
92.928 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-40-4
The first value is the same as 3, so use the second value
γ
A5 sin γ A3 cos γ A6 A1
55.029 deg
A3 sin γ A2 cos γ A4 A1
55.029 deg
acos
asin
γ
Use the negative value , 8.
Use the method of Section 5.7 to synthesize the linkage. Start by determining the magnitudes of the vectors P21 and P31 and their angles with respect to the X axis. 2
p 21
2
P21x P21y
p 21 4.885
δ atan2 P21x P21y 2
p 31
δ 22.891 deg
2
P31x P31y
p 31 7.666
δ atan2 P31x P31y 9.
δ 7.496 deg
Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:
B sin β
E p 21 cos δ
H cos α 1
A cos β 1
D sin α
F cos β 1
G sin β
K sin α
L p 31 cos δ
A F AA B G
C cos α 1
M p 21 sin δ
B C D
E L CC M N
G H K A D C F K H
N p 31 sin δ
W1x W1y AA 1 CC Z1x Z1y
10. The components of the W and Z vectors are: W1x 5.043 11. The length of link 2 is:
w
W1y 3.126 2
Z1x 2.143
2
W1x W1y
Z1y 1.974
w 5.933
12. Evaluate terms in the US coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:
A' cos γ 1
D sin α
B' sin γ
E p 21 cos δ
C cos α 1
F' cos γ 1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-40-5
G' sin γ
K sin α
L p 31 cos δ
A' F' AA B' G'
H cos α 1
M p 21 sin δ
B' C D
N p 31 sin δ
E L CC M N
G' H K A' D C F' K H
U1x U1y AA 1 CC S1x S1y
13. The components of the W and Z vectors are: U1x 2.773 14. The length of link 4 is:
U1y 2.998 2
u
U1x U1y
S1x 3.127
2
S1y 2.102
u 4.083
15. Solving for links 3 and 1 from equations 5.2a and 5.2b. V1x Z1x S1x
V1x 5.271
V1y Z1y S1y
V1y 0.128 v
The length of link 3 is:
2
2
V1x V1y
v 5.272
G1x W1x V1x U1x
G1x 3.000
G1y W1y V1y U1y
G1y 2.176 10
g
The length of link 1 is:
2
G1x G1y
2
14
g 3.000
16. Check the location of the fixed pivots with respect to the global frame using the calculated vectors W1, Z1, U1, and S1. O2x Z1x W1x
O2x 2.900
O2y Z1y W1y
O2y 5.100
O4x S1x U1x
O4x 5.900
O4y S1y U1y
O4y 5.100
These check with Figure P5-7. 17. Determine the location of the coupler point with respect to point A and line AB. 2
2
z 2.914
2
2
s 3.768
Distance from A to P
z
Z1x Z1y
Angle BAP (p)
s
S1x S1y
ψ atan2( S1x S1y)
ψ 146.092 deg
ϕ atan2( Z1x Z1y )
ϕ 42.651 deg
rP z
θ atan2 z cos ϕ s cos ψ z sin ϕ s sin ψ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-40-6
θ 1.392 deg
δp ϕ θ
δp 44.042 deg
18. DESIGN SUMMARY Link 1:
g 3.000
Link 2:
w 5.933
Link 3:
v 5.272
Link 4:
u 4.083
Coupler point:
rP 2.914
δp 44.042 deg
19. VERIFICATION: The calculated values of g (length of the ground link) and of the coordinates of O2 and O4 give the same values as those on the problem statement, verifying that the calculated values for the other links and the coupler point are correct.
20. A driver dyad with a crank should be added to link 2 to control the motion of the fourbar so that it cannot move beyond positions 1 and 3.
DESIGN OF MACHINERY - 5th Ed,
SOLUTION MANUAL 5-41-1
PROBLEM 5-41 Statement:
Design a fourbar linkage to carry the object in Figure P5-10 through the three positions shown in their numbered order without regard for the fixed pivots shown. Use points C and D for your attachment points. Determine the range of the transmission angle.
Given:
Coordinates of the points P1 , P2 and P3 with respect to C1: P1x 0.0
P1y 0.0
P3x 1.750
P3y 2.228
P2x 0.743
P2y 1.514
Angles made by the body in positions 1, 2 and 3:
θP1 62.59 deg
θP2 68.25 deg
θP3 90.0 deg
Coordinates of the points C1 and D1 with respect to P1: C1x 0.0 Solution: 1.
2.
3.
4.
C1y 0.0
D1x 1.036
D1y 1.998
See Figure P5-10 and Mathcad file P0541.
Determine the magnitudes and orientation of the position difference vectors. 2
2
p 21 1.686
δ atan2 P2x P2y
δ 116.140 deg
2
2
p 31 2.833
δ atan2 P3x P3y
δ 128.148 deg
p 21
P2x P2y
p 31
P3x P3y
Determine the angle changes of the coupler between precision points. α θP2 θP1
α 5.660 deg
α θP3 θP1
α 27.410 deg
Using Figure P5-9, the given data, and the law of cosines, determine z, s, , and . z
P1x C1x2 P1y C1y 2
z 0.000
s
P1x D1x2 P1y D1y 2
s 2.251
v
C1x D1x2 C1y D1y2
v 2.251
ϕ θP1
ϕ 62.590 deg
ψ θP1 π
ψ 242.590 deg
Z1x z cos ϕ
Z1x 0.000
Z1y z sin ϕ
Z1y 0.000
S 1x s cos θP1
S 1x 1.036
S 1y s sin θP1
S 1y 1.998
Use equations 5.24 to solve for w, , 2, and 3. Since the points C and D are to be used as pivots, z and are known from the calculations above. Use guess values from a graphical solution such as that for Problem 3-61.
DESIGN OF MACHINERY - 5th Ed,
SOLUTION MANUAL 5-41-2
Guess:
W1x 3
W1y 0.5
β 33.3 deg
Given
W1x cos β 1 W1y sin β = p 21 cos δ Z1x cos α 1 Z1y sin α
β 57.0 deg
W1x cos β 1 W1y sin β = p 31 cos δ Z1x cos α 1 Z1y sin α W1y cos β 1 W1x sin β = p 21 sin δ Z1y cos α 1 Z1x sin α W1y cos β 1 W1x sin β = p 31 sin δ Z1y cos α 1 Z1x sin α
W1x W1y Find W1x W1y β β β β β 33.028 deg
β 57.045 deg
The components of the W vector are: W1x 2.925 The length of link 2 is: w 5.
θ atan2 W1x W1y
W1y 0.496
θ 9.626 deg
W1x2 W1y2 , w 2.967
Use equations 5.28 to solve for u, , 2, and 3. Since the points C and D are to be used as pivots, s and are known from the calculations above. Guess:
U1x 3.2
U1y 0.8
γ 32.3 deg
Given
U1x cos γ 1 U1y sin γ = p 21 cos δ S 1x cos α 1 S 1y sin α
γ 68.4 deg
U1x cos γ 1 U1y sin γ = p 31 cos δ S 1x cos α 1 S 1y sin α U1y cos γ 1 U1x sin γ = p 21 sin δ S 1y cos α 1 S 1x sin α U1y cos γ 1 U1x sin γ = p 31 sin δ S 1y cos α 1 S 1x sin α
U1x U1y Find U1x U1y γ γ γ γ γ 32.559 deg
γ 68.259 deg
DESIGN OF MACHINERY - 5th Ed,
SOLUTION MANUAL 5-41-3
The components of the U vector are: U1x 3.223
U1y 0.815
The length of link 4 is: u 6.
Link 1:
V1x 1.036
V1y Z1y S 1y
V1y 1.998
θ atan2 V1x V1y
θ 62.590 deg
2
2
V1x V1y
v 2.251
G1x W1x V1x U1x
G1x 0.738
G1y W1y V1y U1y
G1y 1.679
θ atan2 G1x G1y
θ 66.275 deg
2
g
9.
U1x2 U1y2 , u 3.324
V1x Z1x S 1x
v
8.
σ 14.189 deg
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
7.
σ atan2 U1x U1y
2
G1x G1y
g 1.834
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 56.650 deg
θ2f θ2i β
θ2f 0.395 deg
Define the coupler point with respect to point C and the vector V. rp z
δp ϕ θ
rp 0.000
δp 0.000 deg
Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. O2x z cos ϕ w cos θ
O2x 2.925
O2y z sin ϕ w sin θ
O2y 0.496
O4x s cos ψ u cos σ
O4x 2.187
O4y s sin ψ u sin σ
O4y 1.183
10. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.
θrot atan2 O4x O2x O4y O2y 11. Determine the Grashof condition.
θrot 66.275 deg
DESIGN OF MACHINERY - 5th Ed,
Condition( a b c d )
SOLUTION MANUAL 5-41-4
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "Grashof" 12. DESIGN SUMMARY Link 2:
w 2.967
θ 9.626 deg
Link 3:
v 2.251
θ 62.590 deg
Link 4:
u 3.324
σ 14.189 deg
Link 1:
g 1.834
θ 66.275 deg
Coupler:
rp 0.000
δp 0.000 deg
Crank angles:
θ2i 56.650 deg
D3
θ2f 0.395 deg
B3 3.323
D2
B2
4 5
4
5
C3 1.835
O6
14. A driver dyad with a crank should be added to link 2 to control the motion of the fourbar so that it cannot move beyond positions 1 and 3.
C2
O4 6
13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.
4
6
D1
B1 3 5 C1
6
1.403 A3 O2 2
2.967
6.347
A1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-42-1
PROBLEM 5-42 Statement:
Design a fourbar linkage to carry the object in Figure P5-10 through the three positions shown in their numbered order without regard for the fixed pivots shown. Use any points on the object as attachment points. Determine the range of the transmission angle.
Given:
Coordinates of the points P1 , P2 and P3 with respect to C1: P1x 0.0
P1y 0.0
P3x 1.750
P3y 2.228
P2x 0.743
P2y 1.514
Angles made by the body in positions 1, 2 and 3:
θP1 62.59 deg
θP2 68.25 deg
θP3 90.0 deg
Coordinates of the points C1 and E1 (used for attachment) with respect to P1: C1x 0.0 Solution: 1.
2.
3.
4.
C1y 0.0
E1x 1.036
E1y 0.000
See Figure P5-10 and Mathcad file P0542.
Determine the magnitudes and orientation of the position difference vectors. 2
2
p 21 1.686
δ atan2 P2x P2y
δ 116.140 deg
2
2
p 31 2.833
δ atan2 P3x P3y
δ 128.148 deg
p 21
P2x P2y
p 31
P3x P3y
Determine the angle changes of the coupler between precision points. α θP2 θP1
α 5.660 deg
α θP3 θP1
α 27.410 deg
Using Figure P5-9, the given data, and the law of cosines, determine z, s, , and . z
P1x C1x2 P1y C1y 2
z 0.000
s
P1x E1x 2 P1y E1y 2
s 1.036
v
C1x E1x 2 C1y E1y2
v 1.036
ϕ atan2 E1x C1x E1y C1y
ϕ 0.000 deg
ψ ϕ π
ψ 180.000 deg
Z1x z cos ϕ
Z1x 0.000
Z1y z sin ϕ
Z1y 0.000
S 1x s cos ψ
S 1x 1.036
S 1y s sin ψ
S 1y 0.000
Use equations 5.24 to solve for w, , 2, and 3. Since the points C and D are to be used as pivots, z and are known from the calculations above. Use guess values from a graphical solution such as that for Problem 3-61. Guess:
W1x 3
W1y 0.5
β 33.3 deg
β 57.0 deg
DESIGN OF MACHINERY - 5th Ed.
Given
SOLUTION MANUAL 5-42-2
W1x cos β 1 W1y sin β = p 21 cos δ Z1x cos α 1 Z1y sin α W1x cos β 1 W1y sin β = p 31 cos δ Z1x cos α 1 Z1y sin α W1y cos β 1 W1x sin β = p 21 sin δ Z1y cos α 1 Z1x sin α W1y cos β 1 W1x sin β = p 31 sin δ Z1y cos α 1 Z1x sin α
W1x W1y Find W1x W1y β β β β β 33.028 deg
β 57.045 deg
The components of the W vector are: W1x 2.925 The length of link 2 is: w 5.
θ atan2 W1x W1y
W1y 0.496
θ 9.626 deg
W1x2 W1y2 , w 2.967
Use equations 5.28 to solve for u, , 2, and 3. Since the points C and D are to be used as pivots, s and are known from the calculations above. Guess:
U1x 3.2
U1y 0.8
γ 32.3 deg
Given
U1x cos γ 1 U1y sin γ = p 21 cos δ S 1x cos α 1 S 1y sin α
γ 68.4 deg
U1x cos γ 1 U1y sin γ = p 31 cos δ S 1x cos α 1 S 1y sin α U1y cos γ 1 U1x sin γ = p 21 sin δ S 1y cos α 1 S 1x sin α U1y cos γ 1 U1x sin γ = p 31 sin δ S 1y cos α 1 S 1x sin α
U1x U1y Find U1x U1y γ γ γ γ γ 22.323 deg
γ 41.857 deg
The components of the U vector are: U1x 4.470
U1y 1.088
σ atan2 U1x U1y
σ 13.676 deg
DESIGN OF MACHINERY - 5th Ed.
The length of link 4 is: u 6.
V1x Z1x S 1x
V1x 1.036
V1y Z1y S 1y
V1y 0.000
θ atan2 V1x V1y
θ 0.000 deg
v Link 1:
2
9.
2
V1x V1y
v 1.036
G1x W1x V1x U1x
G1x 0.509
G1y W1y V1y U1y
G1y 0.591
θ atan2 G1x G1y
θ 130.699 deg
2
g
8.
U1x2 U1y2 , u 4.600
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
7.
SOLUTION MANUAL 5-42-3
2
G1x G1y
g 0.780
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 140.325 deg
θ2f θ2i β
θ2f 197.370 deg
Define the coupler point with respect to point C and the vector V. rp z
δp ϕ θ
rp 0.000
δp 0.000 deg
Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. O2x z cos ϕ w cos θ
O2x 2.925
O2y z sin ϕ w sin θ
O2y 0.496
O4x s cos ψ u cos σ
O4x 3.434
O4y s sin ψ u sin σ
O4y 1.088
10. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.
θrot atan2 O4x O2x O4y O2y 11. Determine the Grashof condition.
θrot 130.699 deg
DESIGN OF MACHINERY - 5th Ed.
Condition( a b c d )
SOLUTION MANUAL 5-42-4
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "non-Grashof" 12. DESIGN SUMMARY Link 2:
w 2.967
θ 9.626 deg
Link 3:
v 1.036
θ 0.000 deg
Link 4:
u 4.600
σ 13.676 deg
Link 1:
g 0.780
θ 130.699 deg
Coupler:
rp 0.000
δp 0.000 deg
Crank angles:
D3
θ2i 140.325 deg θ2f 197.370 deg
D2 E3
13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.
C3
E2
D1
C2 0.780
14. A driver dyad with a crank should be added to link 2 to control the motion of the fourbar so that it cannot move beyond positions 1 and 3.
C1 O2
E1
O4 2.967 4.600
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-43-1
PROBLEM 5-43 Statement:
Given:
Solution: 1.
2.
Design a fourbar linkage to carry the object in Figure P5-10 through the three positions shown in their numbered order using the fixed pivots shown. Determine the range of the transmission angle. Add a driver dyad with a crank to control the motion of your fourbar so that it cannot move beyond positions 1 and 3. P21x 0.743
P21y 1.514
P31x 1.750
P31y 2.228
O2x 3.100
O2y 1.200
O4x 0.100
O4y 1.200
Body angles:
θP1 62.59 deg
θP2 68.25 deg
θP3 90.0 deg
See Figure P5-10 and Mathcad file P0543.
Determine the angle changes between precision points from the body angles given. α θP2 θP1
α 5.660 deg
α θP3 θP1
α 27.410 deg
Using Figure 5-6, determine the magnitudes of R1, R2, and R3 and their x and y components. R1x O2x
3.
4.
R1x 3.100
R1y O2y
R2x R1x P21x
R2x 2.357
R2y R1y P21y
R2y 2.714
R3x R1x P31x
R3x 1.350
R3y R1y P31y
R3y 3.428
2
2
R1 3.324
2
2
R2 3.595
2
2
R3 3.684
R1
R1x R1y
R2
R2x R2y
R3
R3x R3y
R1y 1.200
Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ atan2 R1x R1y
ζ 21.161 deg
ζ atan2 R2x R2y
ζ 49.027 deg
ζ atan2 R3x R3y
ζ 68.505 deg
Solve for 2 and 3 using equations 5.34
C3 0.850
C4 0.936
C5 0.610
C1 R3 cos α ζ R2 cos α ζ C2 R3 sin α ζ R2 sin α ζ C3 R1 cos α ζ R3 cos ζ
C4 R1 sin α ζ R3 sin ζ
C5 R1 cos α ζ R2 cos ζ
C1 0.162 C2 0.050
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-43-2
C6 R1 sin α ζ R2 sin ζ 2
C6 1.214
2
A1 C3 C4
A1 1.597
A2 C3 C6 C4 C5
A2 0.461
A3 C4 C6 C3 C5
A3 1.654
A4 C2 C3 C1 C4
A4 0.194
A5 C4 C5 C3 C6
A5 0.461
A6 C1 C3 C2 C4
A6 0.091
K1 A2 A4 A3 A6
K1 0.061
K2 A3 A4 A5 A6
K2 0.364
2
K3
2
2
2
A1 A2 A3 A4 A6
2
K3 0.221
2
K K 2 K 2 K 2 2 1 2 3 β 2 atan K1 K3
β 27.410 deg
K K 2 K 2 K 2 2 1 2 3 β 2 atan K1 K3
β 133.549 deg
The first value is the same as 3 so use the second value
β β
A5 sin β A3 cos β A6 A1
β 124.137 deg
A3 sin β A2 cos β A4 A1
β 55.863 deg
β acos
β asin
β β
Use the first value, 5.
Repeat steps 2, 3, and 4 for the right-hand dyad to find 1 and 2. R1x O4x
R1x 0.100
R1y O4y
R2x R1x P21x
R2x 0.643
R2y R1y P21y
R2y 2.714
R3x R1x P31x
R3x 1.650
R3y R1y P31y
R3y 3.428
R1
2
2
R1x R1y
R1 1.204
R1y 1.200
DESIGN OF MACHINERY - 5th Ed.
6.
7.
SOLUTION MANUAL 5-43-3
2
2
R2 2.789
2
2
R3 3.804
R2
R2x R2y
R3
R3x R3y
Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ atan2 R1x R1y
ζ 85.236 deg
ζ atan2 R2x R2y
ζ 103.329 deg
ζ atan2 R3x R3y
ζ 115.703 deg
Solve for 2 and 3 using equations 5.34
C3 1.186
C4 2.317
C5 0.624
C6 1.510
C1 R3 cos α ζ R2 cos α ζ C2 R3 sin α ζ R2 sin α ζ C3 R1 cos α ζ R3 cos ζ
C4 R1 sin α ζ R3 sin ζ
C5 R1 cos α ζ R2 cos ζ
C6 R1 sin α ζ R2 sin ζ 2
2
C1 0.160 C2 1.135
A1 C3 C4
A1 6.774
A2 C3 C6 C4 C5
A2 0.345
A3 C4 C6 C3 C5
A3 4.239
A4 C2 C3 C1 C4
A4 0.977
A5 C4 C5 C3 C6
A5 0.345
A6 C1 C3 C2 C4
A6 2.820
K1 A2 A4 A3 A6
K1 12.289
K2 A3 A4 A5 A6
K2 3.165
2
K3
2
2
2
A1 A2 A3 A4 A6
2
2
K3 9.452
K K 2 K 2 K 2 2 1 2 3 γ 2 atan K1 K3
γ 27.410 deg
K K 2 K 2 K 2 2 1 2 3 2 atan K1 K3
56.299 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-43-4
The first value is the same as 3, so use the second value
γ
A5 sin γ A3 cos γ A6 A1
43.866 deg
A3 sin γ A2 cos γ A4 A1
43.866 deg
acos
asin
γ
Use the negative value , 8.
Use the method of Section 5.7 to synthesize the linkage. Start by determining the magnitudes of the vectors P21 and P31 and their angles with respect to the X axis. 2
p 21
2
P21x P21y
p 21 1.686
δ atan2 P21x P21y 2
p 31
δ 116.140 deg
2
P31x P31y
p 31 2.833
δ atan2 P31x P31y 9.
δ 128.148 deg
Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:
B sin β
E p 21 cos δ
H cos α 1
A cos β 1
D sin α
F cos β 1
G sin β
K sin α
L p 31 cos δ
A F AA B G
C cos α 1
M p 21 sin δ
E L CC M N
B C D
G H K A D C F K H
N p 31 sin δ
W1x W1y AA 1 CC Z1x Z1y
10. The components of the W and Z vectors are: W1x 0.621
11. The length of link 2 is:
w
W1y 0.489 2
Z1x 2.479
2
W1x W1y
Z1y 1.689
w 0.790
12. Evaluate terms in the US coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:
A' cos γ 1
D sin α
B' sin γ
E p 21 cos δ
C cos α 1
F' cos γ 1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-43-5
G' sin γ
K sin α
L p 31 cos δ
A' F' AA B' G'
H cos α 1
M p 21 sin δ
B' C D
N p 31 sin δ
E L CC M N
G' H K A' D C F' K H
U1x U1y AA 1 CC S1x S1y
13. The components of the W and Z vectors are: U1x 1.459
14. The length of link 4 is:
U1y 1.294 2
u
U1x U1y
S1x 1.559
2
S1y 2.494
u 1.950
15. Solving for links 3 and 1 from equations 5.2a and 5.2b. V1x Z1x S1x
V1x 0.920
V1y Z1y S1y
V1y 0.805 v
The length of link 3 is:
2
2
V1x V1y
v 1.222
G1x W1x V1x U1x
G1x 3.000
G1y W1y V1y U1y
G1y 0.000
g
The length of link 1 is:
2
G1x G1y
2
g 3.000
16. Check the location of the fixed pivots with respect to the global frame using the calculated vectors W1, Z1, U1, and S1. O2x Z1x W1x
O2x 3.100
O2y Z1y W1y
O2y 1.200
O4x S1x U1x
O4x 0.100
O4y S1y U1y
O4y 1.200
These check with Figure P5-7. 17. Determine the location of the coupler point with respect to point A and line AB. 2
2
z 3.000
2
2
s 2.941
Distance from A to P
z
Z1x Z1y
Angle BAP (p)
s
S1x S1y
ψ atan2( S1x S1y)
ψ 57.983 deg
ϕ atan2( Z1x Z1y )
ϕ 34.270 deg
rP z
θ atan2 z cos ϕ s cos ψ z sin ϕ s sin ψ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-43-6
θ 41.182 deg
δp ϕ θ
δp 75.452 deg
18. DESIGN SUMMARY Link 1:
g 3.000
Link 2:
w 0.790
Link 3:
v 1.222
Link 4:
u 1.950
Coupler point:
rP 3.000
δp 75.452 deg
19. VERIFICATION: The calculated values of g (length of the ground link) and of the coordinates of O2 and O4 give the same values as those on the problem statement, verifying that the calculated values for the other links and the coupler point are correct. However, this mechanism as designed has a branch defect. The three positions can only be reached by changing the links from a crossed circuit to an open circuit.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-44-1
PROBLEM 5-44 Statement:
Design a fourbar linkage to carry the object in Figure P5-11 through the three positions shown in their numbered order without regard for the fixed pivots shown. Use points C and D for your attachment points. Determine the range of the transmission angle.
Given:
Coordinates of the points P1 , P2 and P3 with respect to C1: P1x 0.0
P1y 0.0
P3x 2.751
P3y 2.015
P2x 2.332
P2y 0.311
Angles made by the body in positions 1, 2 and 3:
θP1 45.0 deg
θP2 24.14 deg
θP3 86.84 deg
Coordinates of the points C1 and D1 with respect to P1: C1x 0.0 Solution: 1.
2.
3.
4.
C1y 0.0
D1x 1.591
D1y 1.591
See Figure P5-11 and Mathcad file P0544.
Determine the magnitudes and orientation of the position difference vectors. 2
2
p 21 2.353
δ atan2 P2x P2y
δ 172.404 deg
2
2
p 31 3.410
δ atan2 P3x P3y
δ 143.779 deg
p 21
P2x P2y
p 31
P3x P3y
Determine the angle changes of the coupler between precision points. α θP2 θP1
α 20.860 deg
α θP3 θP1
α 41.840 deg
Using Figure P5-9, the given data, and the law of cosines, determine z, s, , and . z
P1x C1x2 P1y C1y 2
z 0.000
s
P1x D1x2 P1y D1y 2
s 2.250
v
C1x D1x2 C1y D1y2
v 2.250
ϕ θP1
ϕ 45.000 deg
ψ θP1 π
ψ 225.000 deg
Z1x z cos ϕ
Z1x 0.000
Z1y z sin ϕ
Z1y 0.000
S 1x s cos θP1
S 1x 1.591
S 1y s sin θP1
S 1y 1.591
Use equations 5.24 to solve for w, , 2, and 3. Since the points C and D are to be used as pivots, z and are known from the calculations above. Use guess values from a graphical solution such as that for Problem 3-65.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-44-2
Guess:
W1x 1.75 W1y 0.47
β 81 deg
β 138 deg
Given
W1x cos β 1 W1y sin β = p 21 cos δ Z1x cos α 1 Z1y sin α
W1x cos β 1 W1y sin β = p 31 cos δ Z1x cos α 1 Z1y sin α W1y cos β 1 W1x sin β = p 21 sin δ Z1y cos α 1 Z1x sin α W1y cos β 1 W1x sin β = p 31 sin δ Z1y cos α 1 Z1x sin α
W1x W1y Find W1x W1y β β β β β 79.928 deg
β 137.178 deg
The components of the W vector are: W1x 0.980 The length of link 2 is: w 5.
θ atan2 W1x W1y
W1y 1.547
θ 57.632 deg
W1x2 W1y2 , w 1.831
Use equations 5.28 to solve for u, , 2, and 3. Since the points C and D are to be used as pivots, s and are known from the calculations above. Guess:
U1x 0.1
U1y 7.0
γ 17.5 deg
Given
U1x cos γ 1 U1y sin γ = p 21 cos δ S 1x cos α 1 S 1y sin α
γ 37 deg
U1x cos γ 1 U1y sin γ = p 31 cos δ S 1x cos α 1 S 1y sin α U1y cos γ 1 U1x sin γ = p 21 sin δ S 1y cos α 1 S 1x sin α U1y cos γ 1 U1x sin γ = p 31 sin δ S 1y cos α 1 S 1x sin α
U1x U1y Find U1x U1y γ γ γ γ
γ 17.457 deg
γ 37.139 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-44-3
The components of the U vector are: U1x 4.132
U1y 5.598
The length of link 4 is: u 6.
Link 1:
V1x 1.591
V1y Z1y S 1y
V1y 1.591
θ atan2 V1x V1y
θ 45.000 deg
2
2
V1x V1y
v 2.250
G1x W1x V1x U1x
G1x 1.561
G1y W1y V1y U1y
G1y 8.736
θ atan2 G1x G1y
θ 100.130 deg
2
g
9.
U1x2 U1y2 , u 6.958
V1x Z1x S 1x
v
8.
σ 53.567 deg
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
7.
σ atan2 U1x U1y
2
G1x G1y
g 8.874
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 42.498 deg
θ2f θ2i β
θ2f 94.681 deg
Define the coupler point with respect to point C and the vector V. rp z
δp ϕ θ
rp 0.000
δp 0.000 deg
Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. O2x z cos ϕ w cos θ
O2x 0.980
O2y z sin ϕ w sin θ
O2y 1.547
O4x s cos ψ u cos σ
O4x 2.541
O4y s sin ψ u sin σ
O4y 7.189
10. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.
θrot atan2 O4x O2x O4y O2y 11. Determine the Grashof condition.
θrot 100.130 deg
DESIGN OF MACHINERY - 5th Ed.
Condition( a b c d )
SOLUTION MANUAL 5-44-4
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "non-Grashof" 12. DESIGN SUMMARY Link 2:
w 1.831
θ 57.632 deg
Link 3:
v 2.250
θ 45.000 deg
Link 4:
u 6.958
σ 53.567 deg
Link 1:
g 8.874
θ 100.130 deg
Coupler:
rp 0.000
δp 0.000 deg
Crank angles:
θ2i 42.498 deg
7.646
θ2f 94.681 deg
O4 4
O2 A3
6.958 8.874
B1
4 B3
2 1.593
D1 D2
13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.
3
A1
D3 5
C2
C3
6
5 C1
6 O6 1.831
14. A driver dyad with a crank should be added to link 2 to control the motion of the fourbar so that it cannot move beyond positions 1 and 3.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-45-1
PROBLEM 5-45 Statement:
Design a fourbar linkage to carry the object in Figure P5-11 through the three positions shown in their numbered order without regard for the fixed pivots shown. Use any points on the object as attachment points. Determine the range of the transmission angle.
Given:
Coordinates of the points P1 , P2 and P3 with respect to C1: P1x 0.0 P1y 0.0 P2x 2.332 P3x 2.751
P2y 0.311
P3y 2.015
Angles made by the body in positions 1, 2 and 3:
θP1 45.0 deg
θP2 24.14 deg
θP3 86.84 deg
Coordinates of the points C1 and E1 (used for attachment) with respect to P1: C1x 0.0 Solution: 1.
2.
3.
4.
C1y 0.0
E1x 1.591
E1y 0.000
See Figure P5-11 and Mathcad file P0545.
Determine the magnitudes and orientation of the position difference vectors. 2
2
p 21 2.353
δ atan2 P2x P2y
δ 172.404 deg
2
2
p 31 3.410
δ atan2 P3x P3y
δ 143.779 deg
p 21
P2x P2y
p 31
P3x P3y
Determine the angle changes of the coupler between precision points. α θP2 θP1
α 20.860 deg
α θP3 θP1
α 41.840 deg
Using Figure P5-9, the given data, and the law of cosines, determine z, s, , and . z
P1x C1x2 P1y C1y 2
z 0.000
s
P1x E1x 2 P1y E1y 2
s 1.591
v
C1x E1x 2 C1y E1y2
v 1.591
ϕ atan2 E1x C1x E1y C1y
ϕ 0.000 deg
ψ ϕ π
ψ 180.000 deg
Z1x z cos ϕ
Z1x 0.000
Z1y z sin ϕ
Z1y 0.000
S 1x s cos ψ
S 1x 1.591
S 1y s sin ψ
S 1y 0.000
Use equations 5.24 to solve for w, , 2, and 3. Since the points C and D are to be used as pivots, z and are known from the calculations above. Use guess values from a graphical solution such as that for Problem 3-65. Guess:
W1x 1.75 W1y 0.47
β 81 deg
β 138 deg
DESIGN OF MACHINERY - 5th Ed.
Given
SOLUTION MANUAL 5-45-2
W1x cos β 1 W1y sin β = p 21 cos δ Z1x cos α 1 Z1y sin α W1x cos β 1 W1y sin β = p 31 cos δ Z1x cos α 1 Z1y sin α W1y cos β 1 W1x sin β = p 21 sin δ Z1y cos α 1 Z1x sin α W1y cos β 1 W1x sin β = p 31 sin δ Z1y cos α 1 Z1x sin α
W1x W1y Find W1x W1y β β β β β 79.928 deg
β 137.178 deg
The components of the W vector are: W1x 0.980 The length of link 2 is: w 5.
θ atan2 W1x W1y
W1y 1.547
θ 57.632 deg
W1x2 W1y2 , w 1.831
Use equations 5.28 to solve for u, , 2, and 3. Since the points C and D are to be used as pivots, s and are known from the calculations above. Guess:
U1x 1
U1y 5.0
γ 17.5 deg
Given
U1x cos γ 1 U1y sin γ = p 21 cos δ S 1x cos α 1 S 1y sin α
γ 37 deg
U1x cos γ 1 U1y sin γ = p 31 cos δ S 1x cos α 1 S 1y sin α U1y cos γ 1 U1x sin γ = p 21 sin δ S 1y cos α 1 S 1x sin α U1y cos γ 1 U1x sin γ = p 31 sin δ S 1y cos α 1 S 1x sin α
U1x U1y Find U1x U1y γ γ γ γ γ 21.548 deg
γ 27.543 deg
The components of the U vector are: U1x 3.524
U1y 5.963
σ atan2 U1x U1y
σ 59.417 deg
DESIGN OF MACHINERY - 5th Ed.
The length of link 4 is: u 6.
V1x Z1x S 1x
V1x 1.591
V1y Z1y S 1y
V1y 0.000
θ atan2 V1x V1y
θ 0.000 deg
v Link 1:
2
9.
2
V1x V1y
v 1.591
G1x W1x V1x U1x
G1x 0.952
G1y W1y V1y U1y
G1y 7.510
θ atan2 G1x G1y
θ 97.228 deg
2
g
8.
U1x2 U1y2 , u 6.926
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
7.
SOLUTION MANUAL 5-45-3
2
G1x G1y
g 7.570
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 39.596 deg
θ2f θ2i β
θ2f 97.582 deg
Define the coupler point with respect to point C and the vector V. rp z
δp ϕ θ
rp 0.000
δp 0.000 deg
Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. O2x z cos ϕ w cos θ
O2x 0.980
O2y z sin ϕ w sin θ
O2y 1.547
O4x s cos ψ u cos σ
O4x 1.933
O4y s sin ψ u sin σ
O4y 5.963
10. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.
θrot atan2 O4x O2x O4y O2y 11. Determine the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
θrot 97.228 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-45-4
Condition( g u v w) "non-Grashof" 12. DESIGN SUMMARY Link 2:
w 1.831
θ 57.632 deg
Link 3:
v 1.591
θ 0.000 deg
Link 4:
u 6.926
σ 59.417 deg
Link 1:
g 7.570
θ 97.228 deg
Coupler:
rp 0.000
δp 0.000 deg
Crank angles:
θ2i 39.596 deg θ2f 97.582 deg
O4 6.926
13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.
14. A driver dyad with a crank should be added to link 2 to control the motion of the fourbar so that it cannot move beyond positions 1 and 3.
7.570 D1 D2
D3 C2 C3
C1
E1
O2 1.831
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-46-1
PROBLEM 5-46 Statement:
Given:
Solution: 1.
2.
Design a fourbar linkage to carry the object in Figure P5-11 through the three positions shown in their numbered order using the fixed pivots shown. Determine the range of the transmission angle. P21x 2.332
P21y 0.311
P31x 2.751
P31y 2.015
O2x 3.679
O2y 3.282
O4x 0.321
O4y 3.282
Body angles:
θP1 45.0 deg
θP2 24.14 deg
θP3 86.84 deg
See Figure P5-11 and Mathcad file P0546.
Determine the angle changes between precision points from the body angles given. α θP2 θP1
α 20.860 deg
α θP3 θP1
α 41.840 deg
Using Figure 5-6, determine the magnitudes of R1, R2, and R3 and their x and y components. R1x O2x
3.
4.
R1x 3.679
R1y O2y
R2x R1x P21x
R2x 1.347
R2y R1y P21y
R2y 2.971
R3x R1x P31x
R3x 0.928
R3y R1y P31y
R3y 1.267
2
2
R1 4.930
2
2
R2 3.262
2
2
R3 1.571
R1
R1x R1y
R2
R2x R2y
R3
R3x R3y
R1y 3.282
Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ atan2 R1x R1y
ζ 41.736 deg
ζ atan2 R2x R2y
ζ 65.611 deg
ζ atan2 R3x R3y
ζ 53.780 deg
Solve for 2 and 3 using equations 5.34
C3 0.376
C4 3.632
C5 3.260
C6 1.214
C1 R3 cos α ζ R2 cos α ζ C2 R3 sin α ζ R2 sin α ζ C3 R1 cos α ζ R3 cos ζ
C4 R1 sin α ζ R3 sin ζ
C5 R1 cos α ζ R2 cos ζ
C6 R1 sin α ζ R2 sin ζ
C1 2.297 C2 2.258
DESIGN OF MACHINERY - 5th Ed. 2
SOLUTION MANUAL 5-46-2 2
A1 C3 C4
A1 13.335
A2 C3 C6 C4 C5
A2 11.382
A3 C4 C6 C3 C5
A3 5.637
A4 C2 C3 C1 C4
A4 7.492
A5 C4 C5 C3 C6
A5 11.382
A6 C1 C3 C2 C4
A6 9.068
K1 A2 A4 A3 A6
K1 136.387
K2 A3 A4 A5 A6
K2 60.979
2
K3
2
2
2
A1 A2 A3 A4 A6
2
K3 60.933
2
K K 2 K 2 K 2 2 1 2 3 β 2 atan K1 K3
β 90.019 deg
K K 2 K 2 K 2 2 1 2 3 β 2 atan K1 K3
β 41.840 deg
The second value is the same as 3 so use the first value
β β
A5 sin β A3 cos β A6 A1
β 99.989 deg
A3 sin β A2 cos β A4 A1
β 80.011 deg
β acos
β asin
β β
Use the first value, 5.
Repeat steps 2, 3, and 4 for the right-hand dyad to find 1 and 2. R1x O4x
R1x 0.321
R1y O4y
R2x R1x P21x
R2x 2.653
R2y R1y P21y
R2y 2.971
R3x R1x P31x
R3x 3.072
R3y R1y P31y
R3y 1.267
2
2
R1 3.298
2
2
R2 3.983
R1
R1x R1y
R2
R2x R2y
R1y 3.282
DESIGN OF MACHINERY - 5th Ed.
R3 6.
7.
SOLUTION MANUAL 5-46-3
2
2
R3x R3y
R3 3.323
Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ atan2 R1x R1y
ζ 95.586 deg
ζ atan2 R2x R2y
ζ 131.764 deg
ζ atan2 R3x R3y
ζ 157.587 deg
Solve for 2 and 3 using equations 5.34
C3 0.644
C4 0.964
C5 3.522
C6 0.210
C1 R3 cos α ζ R2 cos α ζ C2 R3 sin α ζ R2 sin α ζ C3 R1 cos α ζ R3 cos ζ
C4 R1 sin α ζ R3 sin ζ
C5 R1 cos α ζ R2 cos ζ
C6 R1 sin α ζ R2 sin ζ 2
2
C1 1.539 C2 1.834
A1 C3 C4
A1 1.343
A2 C3 C6 C4 C5
A2 3.260
A3 C4 C6 C3 C5
A3 2.469
A4 C2 C3 C1 C4
A4 0.303
A5 C4 C5 C3 C6
A5 3.260
A6 C1 C3 C2 C4
A6 2.758
K1 A2 A4 A3 A6
K1 7.799
K2 A3 A4 A5 A6
K2 8.243
2
K3
2
2
2
A1 A2 A3 A4 A6
2
2
K3 11.309
K K 2 K 2 K 2 2 1 2 3 γ 2 atan K1 K3
γ 41.840 deg
K K 2 K 2 K 2 2 1 2 3 K1 K3
51.335 deg
2 atan
The first value is the same as 3, so use the second value
γ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-46-4
A5 sin γ A3 cos γ A6 A1
8.321 deg
A3 sin γ A2 cos γ A4 A1
8.321 deg
acos
asin
γ
Both are the same so use the first value , 8.
Use the method of Section 5.7 to synthesize the linkage. Start by determining the magnitudes of the vectors P21 and P31 and their angles with respect to the X axis. 2
p 21
2
P21x P21y
p 21 2.353
δ atan2 P21x P21y 2
p 31
δ 172.404 deg
2
P31x P31y
p 31 3.410
δ atan2 P31x P31y 9.
δ 143.779 deg
Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:
B sin β
E p 21 cos δ
H cos α 1
A cos β 1
D sin α
F cos β 1
G sin β
K sin α
L p 31 cos δ
A F AA B G
C cos α 1
M p 21 sin δ
B C D
E L CC M N
G H K A D C F K H
N p 31 sin δ
W1x W1y AA 1 CC Z1x Z1y
10. The components of the W and Z vectors are: W1x 1.732
11. The length of link 2 is:
w
W1y 1.000 2
Z1x 1.947
2
W1x W1y
Z1y 2.282
w 2.000
12. Evaluate terms in the US coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:
A' cos γ 1
B' sin γ
E p 21 cos δ
H cos α 1
D sin α
G' sin γ
C cos α 1
F' cos γ 1
K sin α
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-46-5
L p 31 cos δ
A' F' AA B' G'
M p 21 sin δ
B' C D
N p 31 sin δ
E L CC M N
G' H K A' D C F' K H
U1x U1y AA 1 CC S1x S1y
13. The components of the W and Z vectors are: U1x 1.178 14. The length of link 4 is:
U1y 6.903 2
u
U1x U1y
S1x 0.857
2
S1y 3.621
u 7.002
15. Solving for links 3 and 1 from equations 5.2a and 5.2b. V1x Z1x S1x
V1x 1.090
V1y Z1y S1y
V1y 5.903 v
The length of link 3 is:
2
2
V1x V1y
v 6.002
G1x W1x V1x U1x
G1x 4.000
G1y W1y V1y U1y
G1y 1.776 10
g
The length of link 1 is:
2
G1x G1y
2
15
g 4.000
16. Check the location of the fixed pivots with respect to the global frame using the calculated vectors W1, Z1, U1, and S1. O2x Z1x W1x
O2x 3.679
O2y Z1y W1y
O2y 3.282
O4x S1x U1x
O4x 0.321
O4y S1y U1y
O4y 3.282
These check with Figure P5-7. 17. Determine the location of the coupler point with respect to point A and line AB. 2
2
z 3.000
2
2
s 3.721
Distance from A to P
z
Z1x Z1y
Angle BAP (p)
s
S1x S1y
ψ atan2( S1x S1y)
ψ 76.683 deg
ϕ atan2( Z1x Z1y )
ϕ 49.525 deg
rP z
θ atan2 z cos ϕ s cos ψ z sin ϕ s sin ψ θ 79.535 deg
δp ϕ θ
δp 30.010 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-46-6
18. DESIGN SUMMARY Link 1:
g 4.000
Link 2:
w 2.000
Link 3:
v 6.002
Link 4:
u 7.002
Coupler point:
rP 3.000
δp 30.010 deg
19. VERIFICATION: The calculated values of g (length of the ground link) and of the coordinates of O2 and O4 give the same values as those on the problem statement, verifying that the calculated values for the other links and the coupler point are correct.
20. A driver dyad with a crank should be added to link 2 to control the motion of the fourbar so that it cannot move beyond positions 1 and 3.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-47-1
PROBLEM 5-47 Statement: Given:
Write a program to generate and plot the circle-point and center-point circles for Problem 5-40 using an equation solver or any program language. P21x 4.500
P21y 1.900
P31x 7.600
P31y 1.000
O2x 2.900
O2y 5.100
O4x 5.900
O4y 5.100
Body angles:
θP1 33.70 deg
θP2 14.60 deg
θP3 0.0 deg
Assumptions: Let the position 1 to position 2 rotation angles be: β 47.808 deg and γ 55.029 deg Let the position 1 to position 2 coupler rotation angle be: α 19.100 deg Solution: 1.
See Figure P5-9 and Mathcad file P0547.
Use the method of Section 5.6 to synthesize the linkage. Start by determining the magnitudes of the vectors P21 and P31 and their angles with respect to the X axis. p 21
2
2
P21x P21y
p 21 4.885
δ atan2 P21x P21y p 31
2
δ 22.891 deg
2
P31x P31y
p 31 7.666
δ atan2 P31x P31y 2.
δ 7.496 deg
Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector to get the center-point and circle-point circles for the left dyad. α 33.700 deg
β 0 deg 0.5 deg 360 deg
B sin β
E p 21 cos δ
A cos β 1 D sin α
C cos α 1
F β cos β 1
K α sin α
N p 31 sin δ
H α cos α 1
M p 21 sin δ
G β sin β L p 31 cos δ
B C D A F β G β H α K α AA α β B D A C G β F β K α H α
E L CC M N
1 CC
W1y α β DD α β 2
Z1y α β DD α β 4
DD α β AA α β
W1x α β DD α β 1 Z1x α β DD α β 3 3.
Check this against the solutions in Problem 5-40:
W1x α 76.089 deg 5.043
W1y α 76.089 deg 3.126
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-47-2
Z1x α 76.089 deg 2.143
Z1y α 76.089 deg 1.974
These are the same as the values calculated in Problem 5-40. 4.
Form the vector N, whose tip describes the center-point circle for the WZ dyad.
Nx α β W1x α β Z1x α β Ny α β W1y α β Z1y α β 5.
Plot the center-point circle for the WZ dyad. Center-Point Circle for WZ Dyad 4
6
8
Ny α β
10
12
14
0
2
4
6
8
10
Nx α β
4.
Form the vector Z, whose tip describes the circle-point circle for the WZ dyad. β 76.089 deg
α 0 deg 1 deg 360 deg
Zx α β Z1x α β
5.
Zy α β Z1y α β
Plot the circle-point circle for the WZ dyad (see next page).
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-47-3
Circle-Point Circle for WZ Dyad 2
1
0
Zy α β
1
2
3 4
3
2
Zx α β
6.
1
0
1
Evaluate terms in the US coefficient matrix and constant vector from equations (5.31) and form the matrix and vector to get the center-point and circle-point circles for the left dyad. α 33.700 deg
γ 0 deg 1 deg 360 deg
B sin γ
E p 21 cos δ
A cos γ 1 D sin α
C cos α 1
F γ cos γ 1
K α sin α
N p 31 sin δ
H α cos α 1
M p 21 sin δ
G γ sin γ L p 31 cos δ
B C D A F γ G γ H α K α AA α γ B D A C G γ F γ K α H α
E L CC M N
1 CC
U1y α γ DD α γ 2
S1y α γ DD α γ 4
DD α γ AA α γ
U1x α γ DD α γ 1 S1x α γ DD α γ 3 7.
Check this against the solutions in Problem 5-40:
U1x α 92.928 deg 2.773
U1y α 92.928 deg 2.998
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-47-4
S1x α 92.928 deg 3.128
S1y α 92.928 deg 2.102
These are the same as the values calculated in Problem 5-40. 8.
Form the vector M, whose tip describes the center-point circle for the US dyad.
Mx α γ U1x α γ S1x α γ My α γ U1y α γ S1y α γ 9.
Plot the center-point circle for the US dyad. Center-Point Circle for US Dyad 4
6
8
My α γ
10
12
14
0
2
4
6
8
10
Mx α γ
10. Form the vector S, whose tip describes the circle-point circle for the US dyad. γ 92.928 deg
α 0 deg 1 deg 360 deg
Sx α γ S1x α γ
11. Plot the circle-point circle for the WZ dyad (see next page).
Sy α γ S1y α γ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-47-5
Circle-Point Circle for the US Dyad 1
0
1
Sy α γ
2
3
4 1
0
1
Sx α γ
2
3
4
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-48-1
PROBLEM 5-48 Statement:
Write a program to generate and plot the circle-point and center-point circles for Problem 5-43 using an equation solver or any program language. P21x 0.743
P21y 1.514
P31x 1.750
P31y 2.228
O2x 3.100
O2y 1.200
O4x 0.100
O4y 1.200
Body angles:
θP1 62.59 deg
θP2 68.25 deg
θP3 90.0 deg
Given:
Assumptions: Let the position 1 to position 2 rotation angles be: β 124.137 deg and γ 43.866 deg Let the position 1 to position 2 coupler rotation angle be: α 5.660 deg Solution: 1.
See Figure P5-10 and Mathcad file P0548.
Use the method of Section 5.6 to synthesize the linkage. Start by determining the magnitudes of the vectors P21 and P31 and their angles with respect to the X axis. p 21
2
2
P21x P21y
p 21 1.686
δ atan2 P21x P21y p 31
2
δ 116.140 deg
2
P31x P31y
p 31 2.833
δ atan2 P31x P31y 2.
δ 128.148 deg
Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector to get the center-point and circle-point circles for the left dyad. α 27.410 deg
β 0 deg 0.5 deg 360 deg
B sin β
E p 21 cos δ
A cos β 1 D sin α
C cos α 1
F β cos β 1
K α sin α
N p 31 sin δ
H α cos α 1
M p 21 sin δ
G β sin β L p 31 cos δ
B C D A F β G β H α K α AA α β B D A C G β F β K α H α
E L CC M N
1 CC
W1y α β DD α β 2
Z1y α β DD α β 4
DD α β AA α β
W1x α β DD α β 1 Z1x α β DD α β 3 3.
Check this against the solutions in Problem 5-43:
W1x α 133.549 deg 0.621
W1y α 133.549 deg 0.489
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-48-2
Z1x α 133.549 deg 2.479
Z1y α 133.549 deg 1.689
These are the same as the values calculated in Problem 5-43. 4.
Form the vector N, whose tip describes the center-point circle for the WZ dyad.
Nx α β W1x α β Z1x α β Ny α β W1y α β Z1y α β 5.
Plot the center-point circle for the WZ dyad. Center-Point Circle for WZ Dyad 1
0
Ny α β 1
2
3 7
6
5
4
3
Nx α β
4.
Form the vector Z, whose tip describes the circle-point circle for the WZ dyad. β 133.549 deg
α 8 deg 9 deg 364 deg
Zx α β3 Z1x α β
5.
Plot the circle-point arc for the WZ dyad (see next page).
Zy α β Z1y α β
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-48-3
Circle-Point Circle for WZ Dyad 20
10
0
Zy α β
10
20
30 20
10
0
Zx α β
6.
10
20
30
Evaluate terms in the US coefficient matrix and constant vector from equations (5.31) and form the matrix and vector to get the center-point and circle-point circles for the left dyad. α 27.410 deg
γ 0 deg 1 deg 360 deg
B sin γ
E p 21 cos δ
A cos γ 1 D sin α
C cos α 1
F γ cos γ 1
K α sin α
N p 31 sin δ
H α cos α 1
M p 21 sin δ
G γ sin γ L p 31 cos δ
B C D A F γ G γ H α K α AA α γ B D A C G γ F γ K α H α
E L CC M N
1 CC
U1y α γ DD α γ 2
S1y α γ DD α γ 4
DD α γ AA α γ
U1x α γ DD α γ 1 S1x α γ DD α γ 3 7.
Check this against the solutions in Problem 5-43:
U1x α 56.299 deg 1.459
U1y α 56.299 deg 1.294
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-48-4
S1x α 56.299 deg 1.559
S1y α 56.299 deg 2.494
These are the same as the values calculated in Problem 5-43. 8.
Form the vector M, whose tip describes the center-point circle for the US dyad.
Mx α γ U1x α γ S1x α γ My α γ U1y α γ S1y α γ 9.
Plot the center-point circle for the US dyad. Center-Point Circle for US Dyad 0
2
4
My α γ
6
8
10 8
6
4
2
0
2
Mx α γ
10. Form the vector S, whose tip describes the circle-point circle for the US dyad. γ 56.299 deg
α 10 deg 11 deg 363 deg
Sx α γ S1x α γ
11. Plot the circle-point arc for the WZ dyad (see next page).
Sy α γ S1y α γ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-48-5
Circle-Point Circle for the US Dyad 20
10
Sy α γ
0
10
20 15
10
5
Sx α γ
0
5
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-49-1
PROBLEM 5-49 Statement:
Write a program to generate and plot the circle-point and center-point circles for Problem 5-46 using an equation solver or any program language.
Given: P21x 2.332
P21y 0.311
P31x 2.751
P31y 2.015
O2x 3.679
O2y 3.282
O4x 0.321
O4y 3.282
Body angles:
θP1 45.0 deg
θP2 24.14 deg
θP3 86.84 deg
Assumptions: Let the position 1 to position 2 rotation angles be: β 99.989 deg and γ 8.321 deg Let the position 1 to position 2 coupler rotation angle be: α 20.860 deg Solution: 1.
See Figure P5-11 and Mathcad file P0549.
Use the method of Section 5.6 to synthesize the linkage. Start by determining the magnitudes of the vectors P21 and P31 and their angles with respect to the X axis. p 21
2
2
P21x P21y
p 21 2.353
δ atan2 P21x P21y p 31
2
δ 172.404 deg
2
P31x P31y
p 31 3.410
δ atan2 P31x P31y 2.
δ 143.779 deg
Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector to get the center-point and circle-point circles for the left dyad. α 41.840 deg
β 0 deg 1 deg 360 deg
B sin β
E p 21 cos δ
A cos β 1 D sin α
C cos α 1
F β cos β 1
K α sin α
N p 31 sin δ
H α cos α 1
M p 21 sin δ
G β sin β L p 31 cos δ
B C D A F β G β H α K α AA α β B D A C G β F β K α H α
E L CC M N
1 CC
W1y α β DD α β 2
Z1y α β DD α β 4
DD α β AA α β
W1x α β DD α β 1 Z1x α β DD α β 3
DESIGN OF MACHINERY - 5th Ed.
3.
SOLUTION MANUAL 5-49-2
Check this against the solutions in Problem 5-46:
W1y α 90.019 deg 1.000
Z1y α 90.019 deg 2.282
W1x α 90.019 deg 1.732 Z1x α 90.019 deg 1.947
These are the same as the values calculated in Problem 5-46. 4.
Form the vector N, whose tip describes the center-point circle for the WZ dyad.
Nx α β W1x α β Z1x α β Ny α β W1y α β Z1y α β 5.
Plot the center-point circle for the WZ dyad. Center-Point Circle for WZ Dyad 0
2
Ny α β 4
6
8 4
2
0
2
Nx α β
4.
Form the vector Z, whose tip describes the circle-point circle for the WZ dyad. β 90.019 deg
α 0 deg 1 deg 360 deg
Zx α β Z1x α β
5.
Zy α β Z1y α β
Plot the circle-point circle for the WZ dyad (see next page).
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-49-3
Circle-Point Circle for WZ Dyad 4
2
Zy α β
0
2
4 10
8
6
Zx α β
6.
4
2
0
Evaluate terms in the US coefficient matrix and constant vector from equations (5.31) and form the matrix and vector to get the center-point and circle-point circles for the left dyad. α 41.840 deg
γ 0 deg 1 deg 360 deg
B sin γ
E p 21 cos δ
A cos γ 1 D sin α
C cos α 1
F γ cos γ 1
K α sin α
N p 31 sin δ
H α cos α 1
M p 21 sin δ
G γ sin γ L p 31 cos δ
B C D A F γ G γ H α K α AA α γ B D A C G γ F γ K α H α
E L CC M N
1 CC
U1y α γ DD α γ 2
S1y α γ DD α γ 4
DD α γ AA α γ
U1x α γ DD α γ 1 S1x α γ DD α γ 3 7.
Check this against the solutions in Problem 5-46:
U1x α 51.335 deg 1.178
U1y α 51.335 deg 6.903
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-49-4
S1x α 51.335 deg 0.857
S1y α 51.335 deg 3.621
These are the same as the values calculated in Problem 5-46. 8.
Form the vector M, whose tip describes the center-point circle for the US dyad.
Mx α γ U1x α γ S1x α γ My α γ U1y α γ S1y α γ 9.
Plot the center-point circle for the US dyad. Center-Point Circle for US Dyad 20
0
My α γ 20
40
60 60
40
20
0
20
Mx α γ
10. Form the vector S, whose tip describes the circle-point circle for the US dyad. γ 51.335 deg
α 0 deg 1 deg 360 deg
Sx α γ S1x α γ
11. Plot the circle-point circle for the WZ dyad (see next page).
Sy α γ S1y α γ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-49-5
Circle-Point Circle for the US Dyad 7
6
5
Sy α γ
4
3
2 2
1
0
Sx α γ
1
2
3
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-50-1
PROBLEM 5-50 Statement:
In Example 5-2 the precision points and rotation angles are specified while the input and output rotation angles and are free choices. Using the choices given for 2 and 2, determine the radii and center coordinates of the center-point circles for O2 and O4. Plot those circles (or portions of them) and show that the choices of 3 and 3 give a solution that falls on the center-point circles.
Given: P21x 2.394
P21y 1.449
P31x 3.761
P31y 1.103
O2x 1.234
O2y 7.772
O4x 2.737
O4y 0.338
Body angles:
θP1 38.565 deg
θP2 6.435 deg
θP3 47.865 deg
Assumptions: Let the position 1 to position 2 rotation angles be: β 342.3 deg and γ 30.9 deg Let the position 1 to position 2 coupler rotation angle be: α 45.0 deg Solution: 1.
See Figure 5-5 and Mathcad file P0550.
Use the method of Section 5.6 to synthesize the linkage. Start by determining the magnitudes of the vectors P21 and P31 and their angles with respect to the X axis. p 21
2
2
P21x P21y
p 21 2.798
δ atan2 P21x P21y p 31
2
δ 31.185 deg
2
P31x P31y
p 31 3.919
δ atan2 P31x P31y 2.
δ 16.345 deg
Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector to get the center-point and circle-point circles for the left dyad. Since this is a non-Grashof linkage, the range of 3 is limited to approximately -56 deg < 3 < -3 deg. α 9.3 deg
β 56 deg 55 deg 3 deg
B sin β
E p 21 cos δ
A cos β 1 D sin α
C cos α 1
F β cos β 1
K α sin α
N p 31 sin δ
H α cos α 1
M p 21 sin δ
G β sin β L p 31 cos δ
B C D A F β G β H α K α AA α β B A D C G β F β K α H α
E L CC M N
1 CC
W1y α β DD α β 2
Z1y α β DD α β 4
DD α β AA α β
W1x α β DD α β 1 Z1x α β DD α β 3
DESIGN OF MACHINERY - 5th Ed.
3.
SOLUTION MANUAL 5-50-2
Check this against the solutions in Example 5-2:
W1y α 324.8 deg 6.832
Z1y α 324.8 deg 0.940
W1x α 324.8 deg 0.055 Z1x α 324.8 deg 1.179
These are the same as the values calculated in Example 5-2. 4.
Form the vector N, whose tip describes the center-point circle for the WZ dyad.
Nx α β W1x α β Z1x α β Ny α β W1y α β Z1y α β 5.
Plot the center-point circle for the WZ dyad. Portion Center-Point Circle for WZ Dyad 5
10
Ny α β 15
20
25 15
10
5
0
Nx α β
4.
Find the center and radius of this arc by taking three points on it and, from them, determine the radius and center point of the circle. Three points: x1 Nx α 56 deg x1 0.470
y1 Ny α 56 deg x2 Nx α 10 deg y2 Ny α 10 deg x3 Nx α 3 deg y3 Ny α 3 deg
y1 6.430 x2 5.506 y2 14.302 x3 13.631 y3 23.695
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-50-3
2
a
Form four ratios:
b
2
d
4.
2
a 12.278
2 y2 y1
x2 x1
b 0.640
y2 y1 2
c
2
y2 y1 x2 x1
2
2
y3 y2 x3 x2
2
c 32.547
2 y2 y1
x3 x2
d 0.865
y3 y2 ac
Circle center x coordinate:
h
h 89.999
Circle center y coordinate:
k a b h
Center-point circle radius:
R
bd
k 45.303
x1 h 2 y1 k 2
R 103.40
Show that the point determined by 3 = 324.8 deg falls on this circle.
yb3 Ny α 324.8 deg xb3 Nx α 324.8 deg
Coordinates of this point:
xb3 1.234 yb3 7.772
Substitute into radius equation: R
xb3 h 2 yb3 k2
R 103.42
Since the radius is the same, this point does fall on the center=point circle for the WZ dyad. 6.
Evaluate terms in the US coefficient matrix and constant vector from equations (5.31) and form the matrix and vector to get the center-point and circle-point circles for the left dyad. α 9.3 deg
γ 0 deg 1 deg 360 deg
B sin γ
E p 21 cos δ
A cos γ 1 D sin α
C cos α 1
F γ cos γ 1
K α sin α
N p 31 sin δ
H α cos α 1
M p 21 sin δ
G γ sin γ L p 31 cos δ
B C D A F γ G γ H α K α AA α γ B D A C G γ F γ K α H α
E L CC M N
1 CC
DD α γ AA α γ
U1x α γ DD α γ 1
U1y α γ DD α γ 2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-50-4
S1x α γ DD α γ 3 7.
S1y α γ DD α γ 4
Check this against the solutions in Example 5-2:
U1y α 80.6 deg 1.825
S1y α 80.6 deg 1.487
U1x α 80.6 deg 2.628 S1x α 80.6 deg 0.109
These are the same as the values calculated in Example 5-2. 8.
Form the vector M, whose tip describes the center-point circle for the US dyad.
Mx α γ U1x α γ S1x α γ My α γ U1y α γ S1y α γ 9.
Plot the center-point circle for the US dyad. Center-Point Circle for US Dyad 60
40
20
My α γ
0
20
40
60
0
20
40
60
80
100
Mx α γ
10. Find the center and radius of this circle by taking three points on it and, from them, determine the radius and center point of the circle. Three points:
y1 My α 0 deg x1 Mx α 0 deg
x1 42.228 y1 44.088
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-50-5
y2 My α 90 deg x3 Mx α 350 deg y3 My α 350 deg x2 Mx α 90 deg
2
Form four ratios:
a b
2
d
y2 0.214 x3 39.862 y3 42.592 2
y2 y1 x2 x1 2 y2 y1
x2 x1
a 41.977 b 0.891
y2 y1 2
c
2
x2 2.744
2
2
y3 y2 x3 x2
2
2 y2 y1
x3 x2
c 38.321 d 0.876
y3 y2 ac
Circle center x coordinate:
h
Circle center y coordinate:
k a b h
Center-point circle radius:
R
bd
h 45.441 k 1.479
x1 h 2 y1 k 2
R 42.73
11. Show that the point determined by 3 = 80.6 deg falls on this circle. Coordinates of this point:
yg3 My α 80.6 deg xg3 Mx α 80.6 deg
xg3 2.738 yg3 0.338
Substitute into radius equation: R
xg3 h 2 yg3 k2
R 42.72
Since the radius is the same, this point does fall on the centerpoint circle for the US dyad.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-51-1
PROBLEM 5-51 Statement:
Design a driver dyad to move link 2 of Example 5-1 from position 1 to position 2 and return.
Given:
Solution to Example 5-1: Length of link 2
w 2.467
Angle of link 2 in first position
θ 71.6 deg
Rotation angle for link 2
β 38.4 deg
Coordinates of O2
O2x 0.00
Design Choice:
Solution: 1.
O2y 0.00
See Example 5-1, Figure 5-3 and Mathcad file P0551.
Link 2 of the solution to Example 5-1 will become the driven link for the driver dyad. The driver dyad will be links 5 and 6 and the fixed pivot for the dyad will be at O6. Select a point on link 2 of Example 5-1 and label it C. Let the distance O2C be R2 1.200. The solution that follows uses the algorithm presented in Section 5.2 with changes in nomenclature to account for the fact that the driven link is link 2 and the points A and B are already defined on the fourbar of Example 5-1.
2.
Determine the coordinates of the points C1 and C2 using equations 5.0a. Determine the vector M using 5.0b. C1x O2x R2 cos θ
C1x 0.379
C1y O2y R2 sin θ
C1y 1.139
C2x 0.410
C2y 1.128
C2x O2x R2 cos θ β C2y O2y R2 sin θ β RC1
C1x C1y
RC2
C2x C2y
M RC2 RC1
3.
Select a suitable value for the multiplier, K, in equation 5.0d say K 3.0.
4.
Determine the coordinates of the crank pivot, O6 using equation 5.0d. RO6 RC1 K M
O6x RO6
1
O6x 1.989 5.
2
O6y 1.106
R6 0.395
Determine the length of the driver dyad coupler, link 5, and the ground link from eqauation 5.0f. R5 RC1 RO6 R6 RO2
R5 1.973
O2x O2y
R1 RO2 RO6 7.
0.789 0.011
Determine the length of the driving crank using equation 5.0e. R6 R2 sin 0.5 β
6.
O6y RO6
M
Determine the Grashof condition.
R1 2.275
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-51-2
R1 2.275
R2 1.200
R5 1.973
R6 0.395
Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition R1 R2 R5 R6 "Grashof" 8.
Draw the linkage using the link lengths and fixed pivot coordinates calculated above to verify that the driver dyad will perform as required. Solve for the coordinates of O4 using equations 5.2: Given:
z 1.298
ϕ 26.5 deg
u 1.486
σ 15.4 deg
Z1x z cos ϕ S 1x s cos ψ
W1x w cos θ U1x u cos σ Z1
Z1x Z1y
ψ 104.1 deg
Z1y z sin ϕ
Z1x 1.162
Z1y 0.579
S 1y s sin ψ
S 1x 0.252
S 1y 1.004
W1y w sin θ
W1x 0.779
W1y 2.341
U1y u sin σ
U1x 1.433 S1
s 1.035
S1x S1y
V1 Z1 S1
G1 W1 V1 U1
g G1
g 1.701
U1y 0.395
W1x W1y 0.760 G1 1.522
U1
v V1
v 1.476
P2
P1
B2 A1 A2
38.40°
B1
O6
C2
O4
° .60 71
C1
D1
D2
y
O2
U1x U1y
W1
x
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-52-1
PROBLEM 5-52 Statement:
Design a driver dyad to move link 2 of Example 5-2 from position 1 to position 3 and return.
Given:
Solution to Example 5-2:
Solution: 1.
Length of link 2
w 6.832
Angle of link 2 in first position
θ atan
Rotation angle for link 2
β 35.20 deg
Coordinates of O2
O2x 1.234
6.832
0.055
θ 89.539 deg
O2y 7.772
See Example 5-2, Figure 5-5, Table 5-1, and Mathcad file P0552.
Link 2 of the solution to Example 5-2 will become the driven link for the driver dyad. The driver dyad will be links 5 and 6 and the fixed pivot for the dyad will be at O6. Select a point on link 2 of Example 5-2 and label it C. Let the distance O2C be R2 4.000. The solution that follows uses the algorithm presented in Section 5.2 with changes in nomenclature to account for the fact that the driven link is link 2 and the points A and B are already defined on the fourbar of Example 5-2.
2.
Determine the coordinates of the points C1 and C3 using equations 5.0a. Determine the vector M using 5.0b. C1x O2x R2 cos θ
C1x 1.202
C1y O2y R2 sin θ
C1y 3.772
C3x 1.098
C3y 4.522
C3x O2x R2 cos θ β C3y O2y R2 sin θ β RC1
C1x C1y
RC3
C3x C3y
M RC3 RC1
3.
Select a suitable value for the multiplier, K, in equation 5.0d say K 3.0.
4.
Determine the coordinates of the crank pivot, O6 using equation 5.0d. RO6 RC1 K M
O6x RO6
1
O6x 5.698 5.
2
O6y 6.022
R6 1.209
Determine the length of the driver dyad coupler, link 5, and the ground link from eqauation 5.0f. R5 RC1 RO6 R6 RO2
R5 6.047
O2x O2y
R1 RO2 RO6 7.
2.300 0.750
Determine the length of the driving crank using equation 5.0e. R6 R2 sin 0.5 β
6.
O6y RO6
M
Determine the Grashof condition.
R1 7.149
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-52-2
R1 7.149
R2 4.000
R5 6.047
R6 1.209
Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition R1 R2 R5 R6 "Grashof" 8.
Draw the linkage using the link lengths and fixed pivot coordinates calculated above to verify that the driver dyad will perform as required. Solve for the vectors W, Z, U, and S. Given:
y
W1x 0.055
O4 P1
x
W1y 6.832 Z1x 1.179
A1 P3
Z1y 0.940
B1
U1x 2.628
A3
U1y 1.825
B3
S 1x 0.109 S 1y 1.487
C1
W1x W1 W1y
D1
S1x S1y
w W1
° 39 .5 89
Z1x Z1 Z1y U1x U1 U1y S1
C3
35.2 00°
O6 D3
O2
w 6.832
u U1
u 3.200
θ atan2 W1x W1y
θ 89.539 deg
σ atan2 U1x U1y
σ 145.222 deg
z Z1
z 1.508
s S1
s 1.491
ϕ atan2 Z1x Z1y
ϕ 38.565 deg
ψ atan2 S 1x S 1y
ψ 94.192 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-53-1
PROBLEM 5-53 Statement:
Design a driver dyad to move link 2 of Example 5-3 from position 1 to position 3 and return.
Given:
Solution to Example 5-3:
Solution: 1.
Length of link 2
w 1.000
Angle of link 2 in first position
θ atan
Rotation angle for link 2
β 23.96 deg
Coordinates of O2
O2x 1.712
0.500
0.866
θ 30.001 deg
O2y 0.033
See Example 5-3, Figure 5-7, Table 5-2, and Mathcad file P0553.
Link 2 of the solution to Example 5-3 will become the driven link for the driver dyad. The driver dyad will be links 5 and 6 and the fixed pivot for the dyad will be at O6. Select a point on link 2 of Example 5-3 and label it C. Let the distance O2C be R2 0.500. The solution that follows uses the algorithm presented in Section 5.2 with changes in nomenclature to account for the fact that the driven link is link 2 and the points A and B are already defined on the fourbar of Example 5-3.
2.
Determine the coordinates of the points C1 and C3 using equations 5.0a. Determine the vector M using 5.0b. C1x O2x R2 cos θ
C1x 1.279
C1y O2y R2 sin θ
C1y 0.283
C3x 1.418
C3y 0.437
C3x O2x R2 cos θ β C3y O2y R2 sin θ β RC1
C1x C1y
RC3
C3x C3y
M RC3 RC1
3.
Select a suitable value for the multiplier, K, in equation 5.0d say K 3.0.
4.
Determine the coordinates of the crank pivot, O6 using equation 5.0d. RO6 RC1 K M
O6x RO6
1
O6x 1.696 5.
2
O6y 0.746
R6 0.104
Determine the length of the driver dyad coupler, link 5, and the ground link from eqauation 5.0f. R5 RC1 RO6 R6 RO2
R5 0.519
O2x O2y
R1 RO2 RO6 7.
0.139 0.154
Determine the length of the driving crank using equation 5.0e. R6 R2 sin 0.5 β
6.
O6y RO6
M
Determine the Grashof condition.
R1 0.713
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-53-2
R1 0.713
R2 0.500
R5 0.519
R6 0.104
Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition R1 R2 R5 R6 "Grashof" 8.
Draw the linkage using the link lengths and fixed pivot coordinates calculated above to verify that the driver dyad will perform as required. Solve for the vectors W, Z, U, and S. Given: W1x 0.866
Z1x 0.846
U1x 0.253
S 1x 0.035
W1y 0.500
Z1y 0.533
U1y 0.973
S 1y 1.006
W1
W1x W1y
Z1
w W1
Z1x Z1y
U1
w 1.000
U1x U1y
S1
u U1
S1x S1y
u 1.005
θ atan2 W1x W1y
θ 30.001 deg
σ atan2 U1x U1y
σ 104.575 deg
z Z1
z 1.000
s S1
s 1.007
ϕ atan2 Z1x Z1y
ψ atan2 S 1x S 1y
ϕ 32.212 deg
ψ 91.993 deg
y
B1
B3 A3
D3 O6
D1
A1
C3 C1 O2
P3
P1
O4
x
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-54-1
PROBLEM 5-54 Statement:
Design a fourbar linkage to carry the object in Figure P5-12 from position 1 to 2 using points C and D for your attachment points. The fixed pivots should be within the indicated area.
Given:
Coordinates of the points P1 (C1) and P2 (C2) : P1x 0.0
P1y 0.0
P2x 13.871
P2y 3.299
v 12.387
Length of the coupler (link 3):
Angles made by the body in positions 1 and 2:
θP1 0.0 deg
θP2 24.0 deg
Coordinates of the points C1 and D1 with respect to P1: C1y 0.0 D1x C1x v cos θP1
C1x 0.0
Free choice for the WZ dyad : β 45 deg
D1x 12.387 D1y 0.000
Free choice for the US dyad : γ 70 deg Solution:
D1y C1y v sin θP1
See Figure P5-12 and Mathcad file P0554.
1.
Note that this is a two-position motion generation (MG) problem because the output is specified as a complex motion of the coupler, link 3. The second method of Section 5.3 will be used here.
2.
Define the position vectors R1 and R2 and the vector P21 using Figure 5-1 and equation 5.1. R1
P1x P1y
R2
P2x P2y
P21x R2 R1 P21y
P21x 13.871 P21y 3.299
p 21 3.
4.
2
2
P21x P21y
From the trigonometric relationships given in Figure 5-1, determine 2 and 2. α θP2 θP1
α 24.000 deg
δ atan2 P21x P21y
δ 13.378 deg
Using Figure P5-4, the given data, and the law of cosines, determine z, s, , and . z
P1x C1x2 P1y C1y 2
z 0.000
s
P1x D1x2 P1y D1y 2
s 12.387
v
C1x D1x2 C1y D1y2
v 12.387
v2 z2 s2 θP1 2 v z
5.
p 21 14.258
ϕ acos
ϕ 90.000 deg
v2 s2 z2 θP1 ψ π acos 2 v s
ψ 180.000 deg
Solve for the WZ dyad using equations 5.8.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-54-2
Z1x z cos ϕ
Z1x 0.000 A 0.293
D sin α
B 0.707
E p 21 cos δ
E 13.871
C 0.086
F p 21 sin δ
F 3.299
B sin β
C cos α 1
W1y w
D 0.407
A C Z1x D Z1y E B C Z1y D Z1x F
W1x 10.918
2 A A C Z1y D Z1x F B C Z1x D Z1y E
W1y 15.094
2 A 2
2
W1x W1y
w 18.629
θ atan2 W1x W1y 6.
θ 125.878 deg
Solve for the US dyad using equations 5.12. S 1x s cos ψ
S 1x 12.387
D sin α
B 0.940
E p 21 cos δ
E 13.871
C 0.086
F p 21 sin δ
F 3.299
C cos α 1
U1y u
S 1y 0.000
A 0.658
B sin γ
U1x
S 1y s sin ψ
A cos γ 1
D 0.407
A C S 1x D S 1y E B C S 1y D S 1x F 2 A A C S 1y D S 1x F B C S 1x D S 1y E
2
U1x 5.158
U1y 10.010
2 A 2
U1x U1y
u 11.261
σ atan2 U1x U1y 7.
Z1y 0.000
A cos β 1
W1x
Z1y z sin ϕ
σ 117.262 deg
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
V1x z cos ϕ s cos ψ
V1x 12.387
V1y z sin ϕ s sin ψ
V1y 0.000
θ atan2 V1x V1y
θ 0.000 deg
v Link 1:
2
2
V1x V1y
v 12.387
G1x w cos θ v cos θ u cos σ
G1x 6.627
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-54-3
G1y w sin θ v sin θ u sin σ
G1y 5.084
θ atan2 G1x G1y
θ 37.495 deg
g 8.
9.
2
2
G1x G1y
g 8.353
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 88.383 deg
θ2f θ2i β
θ2f 43.383 deg
Define the coupler point with respect to point A and the vector V. rp z
δp ϕ θ
rp 0.000
δp 90.000 deg
10. Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ 0.0 deg R1
ρ 0.000 deg
2
2
P1x P1y
R1 0.000
O2x 10.918
O2y 15.094
O4x 17.545
O4y 10.010
O2x R1 cos ρ z cos ϕ w cos θ O2y R1 sin ρ z sin ϕ w sin θ O4x R1 cos ρ s cos ψ u cos σ O4y R1 sin ρ s sin ψ u sin σ
These fixed pivot points fall on the base and are, therefore, acceptable. 11. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.
θrot atan2 O4x O2x O4y O2y
θrot 37.495 deg
12. Determine the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "non-Grashof"
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-54-4
13. DESIGN SUMMARY Link 2:
w 18.629
θ 125.878 deg
Link 3:
v 12.387
θ 0.000 deg
Link 4:
u 11.261
σ 117.262 deg
Link 1:
g 8.353
θ 37.495 deg
Coupler:
rp 0.000
δp 90.000 deg
Crank angles:
θ2i 88.383 deg
θ2f 43.383 deg
14. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.
8.353 O2
O4
88.383
18.629
43.383 11.261
Y
45.000°
70.000°
D2
D1 X
C1
24.000
12.387
C2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-55-1
PROBLEM 5-55 Statement:
Design a fourbar linkage to carry the object in Figure P5-12 from position 1 to 3 using points C and D for your attachment points. The fixed pivots should be within the indicated area.
Given:
Coordinates of the points P1 (C1) and P2 (C3) : P1x 0.0
P1y 0.0
P2x 19.544
P2y 0.373
v 12.387
Length of the coupler (link 3)
Angles made by the body in positions 1 and 3:
θP1 00.0 deg
θP2 90.0 deg
Coordinates of the points C1 and D1 : C1x 0.0
Solution:
C1y 0.0
D1x C1x v cos θP1
D1y C1y v sin θP1
Free choice for the WZ dyad : β 80 deg
D1x 12.387
Free choice for the US dyad : γ 180 deg
D1y 0.000
See Figure P5-12 and Mathcad file P0555.
1.
Note that this is a two-position motion generation (MG) problem because the output is specified as a complex motion of the coupler, link 3. The second method of Section 5.3 will be used here.
2.
Define the position vectors R1 and R2 and the vector P21 using Figure 5-1 and equation 5.1. R1
p 21 3.
4.
P1x P1y
R2
2
P2x P2y
P21x R2 R1 P21y
2
P21x P21y
P21y 0.373 p 21 19.548
From the trigonometric relationships given in Figure 5-1, determine 2 and 2. α θP2 θP1
α 90.000 deg
δ atan2 P21x P21y
δ 1.093 deg
Using Figure P5-4, the given data, and the law of cosines, determine z, s, , and . z
P1x C1x2 P1y C1y 2
z 0.000
s
P1x D1x2 P1y D1y 2
s 12.387
v
C1x D1x2 C1y D1y2
v 12.387
v2 z2 s2 θP1 2 v z
ϕ acos
ϕ 90.000 deg
v2 s2 z2 θP1 2 v s
ψ π acos 5.
P21x 19.544
ψ 180.000 deg
Solve for the WZ dyad using equations 5.8. Z1x z cos ϕ
Z1x 0.000
Z1y z sin ϕ
Z1y 0.000
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-55-2
A 0.826
D sin α
B 0.985
E p 21 cos δ
E 19.544
C 1.000
F p 21 sin δ
F 0.373
A cos β 1 B sin β
C cos α 1 W1x
W1y w
D 1.000
A C Z1x D Z1y E B C Z1y D Z1x F
W1x 9.994
2 A A C Z1y D Z1x F B C Z1x D Z1y E
W1y 11.459
2 A 2
2
W1x W1y
w 15.205
θ atan2 W1x W1y 6.
θ 131.093 deg
Solve for the US dyad using equations 5.12. S 1x s cos ψ
S 1x 12.387 A 2.000
D sin α
B 0.000
E p 21 cos δ
E 19.544
C 1.000
F p 21 sin δ
F 0.373
B sin γ
C cos α 1
U1y u
D 1.000
A C S 1x D S 1y E B C S 1y D S 1x F 2 A A C S 1y D S 1x F B C S 1x D S 1y E
2
U1x 3.579
U1y 6.007
2 A 2
U1x U1y
u 6.992
σ atan2 U1x U1y 7.
S 1y 0.000
A cos γ 1
U1x
S 1y s sin ψ
σ 120.783 deg
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
V1x z cos ϕ s cos ψ
V1x 12.387
V1y z sin ϕ s sin ψ
V1y 0.000
θ atan2 V1x V1y
θ 0.000 deg
v Link 1:
2
2
V1x V1y
v 12.387
G1x w cos θ v cos θ u cos σ
G1x 5.971
G1y w sin θ v sin θ u sin σ
G1y 5.452
θ atan2 G1x G1y
θ 42.399 deg
DESIGN OF MACHINERY - 5th Ed.
g 8.
9.
2
SOLUTION MANUAL 5-55-3 2
G1x G1y
g 8.086
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 88.694 deg
θ2f θ2i β
θ2f 8.694 deg
Define the coupler point with respect to point A and the vector V. rp z
δp ϕ θ
rp 0.000
δp 90.000 deg
10. Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ 0.0 deg R1
ρ 0.000 deg
2
2
P1x P1y
R1 0.000
O2x 9.994
O2y 11.459
O4x 15.966
O4y 6.007
O2x R1 cos ρ z cos ϕ w cos θ O2y R1 sin ρ z sin ϕ w sin θ O4x R1 cos ρ s cos ψ u cos σ O4y R1 sin ρ s sin ψ u sin σ
These fixed pivot points fall on the base and are, therefore, acceptable. 11. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.
θrot atan2 O4x O2x O4y O2y
θrot 42.399 deg
12. Determine the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "non-Grashof" 13. DESIGN SUMMARY Link 2:
w 15.205
θ 131.093 deg
Link 3:
v 12.387
θ 0.000 deg
Link 4:
u 6.992
σ 120.783 deg
Link 1:
g 8.086
θ 42.399 deg
Coupler:
rp 0.000
δp 90.000 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-55-4
Crank angles:
θ2i 88.694 deg θ2f 8.694 deg 14. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.
O2
Y
8.086
D3
15.205 O4
80.000 88.694
D1 X
C1
C3
12.387
6.992
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-56-1
PROBLEM 5-56 Statement:
Design a fourbar linkage to carry the object in Figure P5-12 from position 2 to 3 using points C and D for your attachment points. The fixed pivots should be within the indicated area.
Given:
Coordinates of the points P1 (C2) and P2 (C3) : P1x 13.871
P1y 3.299
P2x 19.544
P2y 0.373
v 12.387
Length of the coupler (link 3):
Angles made by the body in positions 2 and 3:
θP1 24.0 deg
θP2 90.0 deg
Coordinates of the points C2 and D2 : C2x P1x
Solution:
C2y P1y
D2x C2x v cos θP1
D2y C2y v sin θP1
Free choice for the WZ dyad : β 35 deg
D2x 25.187
Free choice for the US dyad : γ 90 deg
D2y 1.739
See Figure P5-12 and Mathcad file P0556.
1.
Note that this is a two-position motion generation (MG) problem because the output is specified as a complex motion of the coupler, link 3. The second method of Section 5.3 will be used here.
2.
Define the position vectors R1 and R2 and the vector P21 using Figure 5-1 and equation 5.1. R1
p 21 3.
4.
P1x P1y
R2
2
P2x P2y
P21x R2 R1 P21y
2
P21x P21y
P21y 2.926 p 21 6.383
From the trigonometric relationships given in Figure 5-1, determine 2 and 2. α θP2 θP1
α 66.000 deg
δ atan2 P21x P21y
δ 27.284 deg
Using Figure P5-4, the given data, and the law of cosines, determine z, s, , and . z
P1x C2x2 P1y C2y 2
z 0.000
s
P1x D2x2 P1y D2y 2
s 12.387
v
C2x D2x2 C2y D2y2
v 12.387
v2 z2 s2 θP1 2 v z
ϕ acos
ϕ 114.000 deg
v2 s2 z2 θP1 2 v s
ψ π acos 5.
P21x 5.673
ψ 204.000 deg
Solve for the WZ dyad using equations 5.8. Z1x z cos ϕ
Z1x 0.000
Z1y z sin ϕ
Z1y 0.000
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-56-2
A 0.181
D sin α
B 0.574
E p 21 cos δ
E 5.673
C 0.593
F p 21 sin δ
F 2.926
A cos β 1 B sin β
C cos α 1
W1x
W1y w
D 0.914
A C Z1x D Z1y E B C Z1y D Z1x F
W1x 1.804
2 A A C Z1y D Z1x F B C Z1x D Z1y E
W1y 10.459
2 A 2
2
W1x W1y
w 10.614
θ atan2 W1x W1y 6.
θ 80.216 deg
Solve for the US dyad using equations 5.12. S 1x s cos ψ
S 1x 11.316 A 1.000
D sin α
B 1.000
E p 21 cos δ
E 5.673
C 0.593
F p 21 sin δ
F 2.926
B sin γ
C cos α 1
U1y u
D 0.914
A C S 1x D S 1y E B C S 1y D S 1x F 2 A A C S 1y D S 1x F B C S 1x D S 1y E
2
U1x 7.959
U1y 2.316
2 A 2
U1x U1y
u 8.289
σ atan2 U1x U1y 7.
S 1y 5.038
A cos γ 1
U1x
S 1y s sin ψ
σ 16.224 deg
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
V1x z cos ϕ s cos ψ
V1x 11.316
V1y z sin ϕ s sin ψ
V1y 5.038
θ atan2 V1x V1y v Link 1:
2
θ 24.000 deg
2
V1x V1y
v 12.387
G1x w cos θ v cos θ u cos σ
G1x 5.161
G1y w sin θ v sin θ u sin σ
G1y 3.105
θ atan2 G1x G1y
θ 31.035 deg
DESIGN OF MACHINERY - 5th Ed.
g 8.
9.
2
SOLUTION MANUAL 5-56-3
2
G1x G1y
g 6.023
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 49.181 deg
θ2f θ2i β
θ2f 14.181 deg
Define the coupler point with respect to point C and the vector V. rp z
δp ϕ θ
rp 0.000
δp 90.000 deg
10. Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ atan2 P1x P1y R1
2
ρ 13.378 deg
2
P1x P1y
R1 14.258
O2x 12.067
O2y 7.160
O4x 17.228
O4y 4.055
O2x R1 cos ρ z cos ϕ w cos θ O2y R1 sin ρ z sin ϕ w sin θ O4x R1 cos ρ s cos ψ u cos σ O4y R1 sin ρ s sin ψ u sin σ
These fixed pivot points fall on the base and are, therefore, acceptable. 11. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.
θrot atan2 O4x O2x O4y O2y
θrot 31.035 deg
12. Determine the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "Grashof" 13. DESIGN SUMMARY Link 2:
w 10.614
θ 80.216 deg
Link 3:
v 12.387
θ 24.000 deg
Link 4:
u 8.289
σ 16.224 deg
Link 1:
g 6.023
θ 31.035 deg
Coupler:
rp 0.000
δp 90.000 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-56-4
Crank angles:
θ2i 49.181 deg θ2f 14.181 deg 14. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.
D3
8.289
Y O2
90.000 O4
D2 10.614 C3 35.000
12.387 C2
X
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-57-1
PROBLEM 5-57 Statement:
Design a fourbar linkage to carry the object in Figure P5-12 through the three positions shown in their numbered order using points C and D for your attachment points. The fixed pivots should be within the indicated area.
Given:
Coordinates of the points P1 , P2 and P3 : P1x 0.0
P1y 0.0
P2x 13.871
P3x 19.544
P3y 0.373
P2y 3.299
Angles made by the body in positions 1, 2 and 3:
θP1 0.0 deg
θP2 24.0 deg
θP3 90.0 deg
Coordinates of the points C1 and D1 with respect to P1: C1x 0.0 Solution: 1.
2.
3.
C1y 0.0
D1y 0.0
See Figure P5-12 and Mathcad file P0557.
Determine the magnitudes and orientation of the position difference vectors. 2
2
p 21 14.258
δ atan2 P2x P2y
δ 13.378 deg
2
2
p 31 19.548
δ atan2 P3x P3y
δ 1.093 deg
p 21
P2x P2y
p 31
P3x P3y
Determine the angle changes of the coupler between precision points. α θP2 θP1
α 24.000 deg
α θP3 θP1
α 90.000 deg
Using Figure P5-4, the given data, and the law of cosines, determine z, s, , and . z
P1x C1x2 P1y C1y 2
z 0.000
s
P1x D1x2 P1y D1y 2
s 12.387
v
C1x D1x2 C1y D1y2
v 12.387
v2 z2 s2 θP1 2 v z
ϕ acos
v2 s2 z2 θP1 2 v s
4.
D1x 12.387
ϕ 90.000 deg
ψ π acos
ψ 180.000 deg
Z1x z cos ϕ
Z1x 0.000
Z1y z sin ϕ
Z1y 0.000
S 1x s cos ψ
S 1x 12.387
S 1y s sin ψ
S 1y 0.000
Use equations 5.24 to solve for w, , 2, and 3. Since the points C and D are to be used as pivots, z and are known from the calculations above.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-57-2
Guess:
W1x 2
W1y 15
β 45 deg
β 80 deg
Given
W1x cos β 1 W1y sin β = p 21 cos δ Z1x cos α 1 Z1y sin α
W1x cos β 1 W1y sin β = p 31 cos δ Z1x cos α 1 Z1y sin α W1y cos β 1 W1x sin β = p 21 sin δ Z1y cos α 1 Z1x sin α W1y cos β 1 W1x sin β = p 31 sin δ Z1y cos α 1 Z1x sin α
W1x W1y Find W1x W1y β β β β β 56.754 deg
β 81.324 deg
The components of the W vector are: W1x 9.989 The length of link 2 is: w 5.
θ atan2 W1x W1y
W1y 11.190
θ 131.755 deg
W1x2 W1y2 , w 15.000
Use equations 5.28 to solve for u, , 2, and 3. Since the points C and D are to be used as pivots, s and are known from the calculations above. Guess:
U1x 3
U1y 4
γ 70 deg
Given
U1x cos γ 1 U1y sin γ = p 21 cos δ S 1x cos α 1 S 1y sin α
U1x cos γ 1 U1y sin γ = p 31 cos δ S 1x cos α 1 S 1y sin α U1y cos γ 1 U1x sin γ = p 21 sin δ S 1y cos α 1 S 1x sin α U1y cos γ 1 U1x sin γ = p 31 sin δ S 1y cos α 1 S 1x sin α
γ 90 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-57-3
U1x U1y Find U1x U1y γ γ γ γ γ 119.119 deg
γ 222.077 deg
The components of the U vector are: U1x 5.889
U1y 4.631
The length of link 4 is: u 6.
Link 1:
V1x 12.387
V1y Z1y S 1y
V1y 0.000
θ atan2 V1x V1y
θ 0.000 deg
2
2
V1x V1y
v 12.387
G1x W1x V1x U1x
G1x 8.287
G1y W1y V1y U1y
G1y 6.559
θ atan2 G1x G1y
θ 38.362 deg
g
9.
U1x2 U1y2 , u 7.492
V1x Z1x S 1x
v
8.
σ 141.822 deg
Solve for links 3 and 1 using the vector definitions of V and G. Link 3:
7.
σ atan2 U1x U1y
2
2
G1x G1y
g 10.569
Determine the initial and final values of the input crank with respect to the vector G.
θ2i θ θ
θ2i 93.393 deg
θ2f θ2i β
θ2f 12.069 deg
Define the coupler point with respect to point A and the vector V. rp z
δp ϕ θ
rp 0.000
δp 90.000 deg
Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. O2x z cos ϕ w cos θ
O2x 9.989
O2y z sin ϕ w sin θ
O2y 11.190
O4x s cos ψ u cos σ
O4x 18.276
O4y s sin ψ u sin σ
O4y 4.631
10. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 5-57-4
θrot atan2 O4x O2x O4y O2y
θrot 38.362 deg
11. Determine the Grashof condition. Condition( a b c d )
S min ( a b c d ) L max( a b c d ) SL S L PQ a b c d SL return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise
Condition( g u v w) "Grashof" 12. DESIGN SUMMARY Link 2:
w 15.000
θ 131.755 deg
Link 3:
v 12.387
θ 0.000 deg
Link 4:
u 7.492
σ 141.822 deg
Link 1:
g 10.569
θ 38.362 deg
Coupler:
rp 0.000
δp 90.000 deg
Crank angles:
θ2i 93.393 deg θ2f 12.069 deg 13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.
D3
O2 10.569 Y
15.000 7.492 O4
93.393 56.754
12.069 D2
81.324 D1
C3
C1
C2
12.387
X
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-1a-1
PROBLEM 6-1a Statement:
A ship is steaming due north at 20 knots (nautical miles per hour). A submarine is laying in wait 1/2 mile due west of the ship. The sub fires a torpedo on a course of 85 degrees. The torpedo travels at a constant speed of 30 knots. Will it strike the ship? If not, by how many nautical miles will it miss? Hint: Use the relative velocity equation and solve graphically or analytically.
Units:
naut_mile 1
knots
Given:
Speed of ship
Vs 20 knots
naut_mile
Vt 30 knots
Speed of torpedo
hr
θs 90 deg
θt 15 deg
Initial distance between ship and torpedo
d i 0.5 naut_mile
Note that, for compass headings, due north is 0 degrees, due east 90 degrees, and the angle increases clockwise. However, in a right-handed Cartesian system, due north is 90 degrees (up) and due east is 0 degrees (to the right). The Cartesian system has been used above to define the ship and torpedo headings. Solution:
See Mathcad file P0601a.
1.
The key to this solution is to recognize that the only information of interest is the relative velocity of one vessel to the other. The ship captain wants to know the relative velocity of the torpedo versus the ship, Vts = Vt - Vs. In effect, we want to resolve the situation with respect to a moving coordinate system attached to the ship.
2.
The figure below shows the initial positions of the torpedo and ship and their velocities. Vs = 20 knots Vt = 30 knots
torpedo
0.5 n. mi.
ship
3.
The figure below shows the vector diagram that solves the relative velocity equation Vts = Vt - Vs. For the torpedo to hit the moving ship, the relative velocity vector has to be perpendicular to the ship's velocity vector (if you were on the ship observing the torpedo, it would appear to be headed directly for you). As the velocity diagram shows, the relative velocity vector is not perpendicular to the ship's velocity vector so it will miss and pass behind the ship.
4.
Determine the distance by which the torpedo will miss the ship.
Vt
Time required for torpedo to travel 0.5 nautical miles due east tt_east
di
Vt cos θt
tt_east 62.117 s
Distance traveled by the torpedo due north in that time d t_north Vt sin θt tt_east
d t_north 0.134 naut_mile
Distance traveled by the ship due north in that time d s_north Vs tt_east
d s_north 0.345 naut_mile
Distance by which the torpedo will miss the ship d miss d s_north d t_north
d miss 0.211 naut_mile
-Vs V ts
22.89°
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-1b-1
PROBLEM 6-1b Statement:
A plane is flying due south at 500 mph at 35,000 feet altitude, straight and level. A second plane is initially 40 miles due east of the first plane, also at 35,000 feet altitude, flying straight and level at 550 mph. Determine the compass angle at which the second plane would be on a collision course with the first. How long will it take for the second plane to catch the first? Hint: Use the relative velocity equation and solve graphically or analytically.
Given: Speed of first plane
V1 500 mph
Speed of second plane
V2 550 mph
Initial distance between planes
d i 40 mi
θ1 270 deg
Note that, for compass headings, due north is 0 degrees, due east 90 degrees, and the angle increases clockwise. However, in a right-handed Cartesian system, due north is 90 degrees (up) and due east is 0 degrees (to the right). The Cartesian system has been used above to define the plane headings. Solution:
See Mathcad file P0601b.
1.
The key to this solution is to recognize that the only information of interest is the relative velocity of one plane to the other. For a collision to occur, the relative velocity of the second plane with respect to the first must be perpendicular to the velocity vector of the first.
2.
The figure below shows the initial positions of the two planes and their velocities.
plane 1
40 mi.
plane 2
V1 = 500 mph
3.
4.
5.
V2 = 550 mph
The figure below shows the vector diagram that solves the relative velocity equation V2 = V1 + V21. To construct this diagram, chose a convenient velocity scale and draw V1 to its correct length with the arrow head pointing straight down (indicating due south). From the tip of the vector, layoff a horizontal construction line to the left (due west) an undetermined length. From the tail of the V1 vector, construct a circle whose radius is equal to the scaled length of vector V2. The intersection of the circle and the horizontal construction line determines the length of V21. Draw the arrowheads for V2 and V21 pointing toward the intersection of the circle and construction line. Label the horizontal vector V21 and the vector that joins the tail of V1 with the head of V21 as V2. The angle between V1 and V2 is the required direction for V2 in order that plane 2 collides with plane 1. The angle can also be determined analytically from the velocity triangle as 24.620° follows. V1 V2 V1 θ acos θ 24.620 deg V2 V 21 The time it will take for the second plane to catch the first is the time that it will take plane 2 to travel the 40 miles to the west. t
di
V2 sin θ
t 628.468 s
t 10.474 min
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-2-1
PROBLEM 6-2 Statement:
A point is at a 6.5-in radius on a body in pure rotation with = 100 rad/sec. The rotation center is at the origin of a coordinate system. When the point is at position A, its position vector makes a 45 deg angle with the X axis. At position B, its position vector makes a 75 deg angle with the X axis. Draw this system to some convenient scale and: a. Write an expression for the particle's velocity vector in position A using complex number notation, in both polar and Cartesian forms. b. Write an expression for the particle's velocity vector in position B using complex number notation, in both polar and Cartesian forms. c. Write a vector equation for the velocity difference between points B and A. Substitute the complex number notation for the vectors in this equation and solve for the velocity difference numerically. d. Check the result of part c with a graphical method.
Given: ω 100
Rotation speed
Solution: 1.
rad sec
Vector angles
θA 45 deg
Vector magnitude
R 6.5 in
θB 75 deg
See Mathcad file P0602.
Calculate the magnitude of the velocity at points A and B using equation 6.3. V R ω
V 650.000
in sec
2.
Establish an X-Y coordinate frame and draw a circle with center at the origin and radius R.
3.
Draw lines from the origin that make angles of 45 and 75 deg with respect to the X axis. Label the intersections of the lines with the circles as A and B, respectively. Make the line segment OA a vector by putting an arrowhead at A, pointing away from the origin. Label the vector RA. Repeat for the line segment OB, labeling it RB.
4.
Choose a convenient velocity scale and draw the two velocity vectors VA and VB at the tips of RA and RB, respectively. The velocity vectors will be perpendicular to their respective position vectors.
Y 0
8
1
2 in
Distance scale:
VB B
0
VA
500 in/sec
Velocity scale:
6 A 4
RB RA
2
0 a.
X 2
4
6
8
Write an expression for the particle's velocity vector in position A using complex number notation, in both polar and Cartesian forms.
DESIGN OF MACHINERY - 5th Ed.
Polar form:
SOLUTION MANUAL 6-2-2
RA R e
VA R j ω e Cartesian form:
j
j θA
j
j θA
4
VA 650 j e
VA R j ω cos θA j sin θA
RB R e
in sec
Cartesian form:
j
j θB
VB R j ω e
RA 6.5 e
π 4 j
j θB
VB 650 j e
75 π 180
VB R j ω cos θB j sin θB VB ( 627.852 168.232j)
in sec
Write a vector equation for the velocity difference between points B and A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. VBA VB VA
d.
π
Write an expression for the particle's velocity vector in position B using complex number notation, in both polar and Cartesian forms. Polar form:
c.
4
RA 6.5 e
VA ( 459.619 459.619j) b.
π
VBA ( 168.232 291.387j)
in sec
Check the result of part c with a graphical method. Solve the equation VB = VA + VBA using a velocity scale of 250 in/sec per drawing unit.
Y
Velocity scale factor kv 250
in 0
sec
1.166
Horizontal component
VA VBA
VBAx 0.673 kv VBAx 168.3
500 in/sec
Velocity scale:
VB
in
Ov
X
0.673
sec
Vertical component VBAy 1.166 kv
VBAy 291.5
in sec
On the layout above the X and Y components of VBA are equal to the real and imaginary components calculated, confirming that the calculation is correct.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-3-1
PROBLEM 6-3 Statement:
Given:
A point A is at a 6.5-in radius on a body in pure rotation with = -50 rad/sec. The rotation center is at the origin of a coordinate system. At the instant considered its position vector makes a 45 deg angle with the X axis. A point B is at a 6.5-in radius on another body in pure rotation with = +75 rad/sec. Its position vector makes a 75 deg angle with the X axis. Draw this system to some convenient scale and: a. Write an expression for the particle's velocity vector in position A using complex number notation, in both polar and Cartesian forms. b. Write an expression for the particle's velocity vector in position B using complex number notation, in both polar and Cartesian forms. c. Write a vector equation for the velocity difference between points B and A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. d. Check the result of part c with a graphical method. ω 50
Rotation speeds
Solution: 1.
rad
ω 75
sec
Vector angles
θA 45 deg
Vector magnitude
R 6.5 in
rad sec
θB 75 deg
See Mathcad file P0603.
Calculate the magnitude of the velocity at points A and B using equation 6.3. VA R ω
VA 325.000
VB R ω
VB 487.500
in sec in
sec
1.
Establish an X-Y coordinate frame and draw a circle with center at the origin and radius R.
2.
Draw lines from the origin that make angles of 45 and 75 deg with respect to the X axis. Label the intersections of the lines with the circles as A and B, respectively. Make the line segment OA a vector by putting an arrowhead at A, pointing away from the origin. Label the vector RA. Repeat for the line segment OB, labeling it RB.
3.
Choose a convenient velocity scale and draw the two velocity vectors VA and VB at the tips of RA and RB, respectively. The velocity vectors will be perpendicular to their respective position vectors. Y 8 VB
0
1
2 in
Distance scale: B
0
500 in/sec
Velocity scale:
6 A 4
RB RA
2
0
a.
VA
X 2
4
6
8
Write an expression for the particle's velocity vector on body A using complex number notation, in both polar and Cartesian forms.
DESIGN OF MACHINERY - 5th Ed.
Polar form:
SOLUTION MANUAL 6-3-2
RA R e
j
j θA
Cartesian form:
j
j θA
VA R j ω cos θA j sin θA in sec
Write an expression for the particle's velocity vector on body B using complex number notation, in both polar and Cartesian forms. Polar form:
RB R e
j
j θB
RA 6.5 e
VB R j ω e Cartesian form:
j θB
4
VB 487.5 j e
75 π 180
VB R j ω cos θB j sin θB in sec
Write a vector equation for the velocity difference between the points on bodies B and A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. VBA VB VA
d.
π
j
VB ( 470.889 126.174j) c.
π 4
VA 325 j e
VA ( 229.810 229.810j) b.
4
RA 6.5 e
VA R j ω e
π
VBA ( 700.699 355.984j)
in sec
Check the result of part c with a graphical method. Solve the equation VB = VA + VBA using a velocity scale of 250 in/sec per drawing unit.
Y
Velocity scale factor kv 250
in
0
Horizontal component VBAx 2.803 kv VBAx 700.7
500 in/sec
Velocity scale:
sec
VB 1.424
Ov
VBA
VA
X
in sec
2.803
Vertical component VBAy 1.424 kv
VBAy 356.0
in sec
On the layout above the X and Y components of VBA are equal to the real and imaginary components calculated, confirming that the calculation is correct.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-4a-1
PROBLEM 6-4a Statement:
For the fourbar defined in Table P6-1, line a, find the velocities of the pin joints A and B, and of the instant centers I1,3 and I2,4. Then calculate 3 and 4 and find the velocity of point P. Use a graphical method.
Given:
Link lengths: Link 1
d 6 in
Link 2
a 2 in
Link 3
b 7 in
Link 4
c 9 in
θ 30 deg
Crank angle:
ω 10 rad sec
Crank velocity:
1
Coupler point data: Rpa 6 in Solution: 1.
δ 30 deg
See Figure P6-1 and Mathcad file P0604a.
Draw the linkage to scale in the position given, find the instant centers, distances from the pin joints to the instant centers and the angles that links 3 and 4 make with the x axis. B 0
1 IN
SCALE
P 5.9966 6.8067
148.2007°
3.3384 I1,3
A
117.2861°
88.8372° O2
O4
I 2,4 1.7118
4.2882
From the layout above:
2.
O2I24 1.7118 in
O4I24 4.2882 in
AI13 3.3384 in
BI13 5.9966 in
θ 117.2861 deg
θ 88.8372 deg
PI13 6.8067 in
Use equation 6.7 and inspection of the layout to determine the magnitude and direction of the velocity at point A.
DESIGN OF MACHINERY - 5th Ed.
3.
SOLUTION MANUAL 6-4a-2
in
VA a ω
VA 20.0
θVA θ 90 deg
θVA 120.0 deg
sec
Determine the angular velocity of link 3 using equation 6.9a. ω
VA AI13
ω 5.991
rad
CW
sec
4.
Determine the magnitude of the velocity at point B using equation 6.9b. Determine its direction by inspection. VB BI13 ω in VB 35.925 sec θVB θ 90 deg θVB 27.286 deg
5.
Use equation 6.9c to determine the angular velocity of link 4. ω
6.
VB c
ω 3.99
rad
CW
sec
Use equation 6.9d and inspection of the layout to determine the magnitude and direction of the velocity at point P. in
VP PI13 ω
VP 40.778
θVP 148.2007 deg 90 deg
θVP 58.201 deg
sec
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-5a-1
PROBLEM 6-5a Statement:
A general fourbar linkage configuration and its notation are shown in Figure P6-1. The link lengths, coupler point location, and the values of 2 and 2 for the same fourbar linkages as used for position analysis in Chapter 4 are redefined in Table P6-1, which is the same as Table P4-1. For row a, find the velocities of the pin joints A and B, and coupler point P. Calculate 3 and 4. Draw the linkage to scale and label it before setting up the equations.
Given:
Link lengths: d 6
Link 1
a 2
Link 2
Rpa 6
Coupler point:
c 9
Link 4
δ 30 deg
Link 2 position and velocity: θ 30 deg Solution:
b 7
Link 3
ω 10
See Mathcad file P0605a. y
1.
Draw the linkage to scale and label it.
2.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 3.0000 2
K3
B
OPEN
3
d
4
c
K2 0.6667 2
2
a b c d
88.837°
2
K3 2.0000
2 a c
2 O2
117.286°
A
O4
115.211°
A cos θ K1 K2 cos θ K3
143.660°
B 2 sin θ
C K1 K2 1 cos θ K3 A 0.7113 3.
B 1.0000
CROSSED
C 3.5566
B'
Use equation 4.10b to find values of 4 for the open and crossed circuits. Open:
2
θ 2 atan2 2 A B
B 4 A C
θ 242.714 deg
θ θ 360 deg Crossed: 4.
θ 602.714 deg
2
θ 2 atan2 2 A B
B 4 A C
θ 216.340 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2 a b
D cos θ K1 K4 cos θ K5
E 2 sin θ
2
K4 0.8571 D 1.6774 E 1.0000
F K1 K4 1 cos θ K5
F 2.5906
K5 0.2857
x
DESIGN OF MACHINERY - 5th Ed.
5.
SOLUTION MANUAL 6-5a-2
Use equation 4.13 to find values of 3 for the open and crossed circuits. Open:
2
θ 2 atan2 2 D E
E 4 D F
θ 271.163 deg
θ θ 360 deg Crossed: 6.
7.
θ 631.163 deg
2
θ 2 atan2 2 D E
E 4 D F
θ 244.789 deg
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18. ω
a ω sin θ θ b sin θ θ
ω 5.991
ω
a ω sin θ θ c sin θ θ
ω 3.992
Determine the velocity of points A and B for the open circuit using equations 6.19.
VA a ω sin θ j cos θ VA 10.000 17.321j
arg VA 120 deg
VA 20
VB c ω sin θ j cos θ VB 31.928 16.470j 8.
arg VB 27.286 deg
VB 35.926
Determine the velocity of the coupler point P for the open circuit using equations 6.36.
VPA Rpa ω sin θ δ j cos θ δ VPA 31.488 17.337j VP VA VPA VP 21.488 34.658j 9.
arg VP 58.201 deg
VP 40.779
Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18. ω
a ω sin θ θ b sin θ θ
ω 0.662
ω
a ω sin θ θ c sin θ θ
ω 2.662
10. Determine the velocity of point B for the crossed circuit using equations 6.19.
VB c ω sin θ j cos θ VB 14.195 19.295j
arg VB 126.340 deg
VB 23.954
11. Determine the velocity of the coupler point P for the crossed circuit using equations 6.36.
VPA Rpa ω sin θ δ j cos θ δ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-5a-3
VPA 3.960 0.332j VP VA VPA VP 13.960 16.989j
VP 21.989
arg VP 129.411 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-6a-1
PROBLEM 6-6a Statement:
The general linkage configuration and terminology for an offset fourbar slider-crank linkage are shown in Fig P6-2. The link lengths and the values of 2 and 2 are defined in Table P6-2. For row a, find the velocities of the pin joints A and B and the velocity of slip at the sliding joint using a graphical method.
Given:
Link lengths:
Solution: 1.
Link 2
a 1.4 in
Link 3
b 4 in
Offset
c 1 in
θ 45 deg
ω 10 rad sec
1
See Figure P6-2 and Mathcad file P0606a.
Draw the linkage to scale and indicate the axes of slip and transmission as well as the directions of velocities of interest. Direction of VBA Y
Axis of transmission
Direction of VA
A 2
45.000°
Axis of slip and Direction of VB
B
3
179.856°
1.000 X
O2
2.
3.
Use equation 6.7 to calculate the magnitude of the velocity at point A. in VA a ω VA 14.000 θVA θ 90 deg sec
θVA 135 deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B. The equation to be solved graphically is VB = VA + VBA a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line.
4.
0
5 units/sec
VA Y
135.000° V BA X
VB
From the velocity triangle we have: 1.975
DESIGN OF MACHINERY - 5th Ed.
Velocity scale factor:
VB 1.975 in kv
5.
SOLUTION MANUAL 6-6a-2
kv
5 in sec
1
in
VB 9.875
in sec
θVB 180 deg
Since the slip axis and the direction of the velocity of point B are parallel, Vslip = VB.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-7a-1
PROBLEM 6-7a Statement:
The general linkage configuration and terminology for an offset fourbar slider-crank linkage are shown in Fig P6-2. The link lengths and the values of 2 and 2 are defined in Table P6-2. For row a, find the velocities of the pin joints A and B and the velocity of slip at the sliding joint using the analytic method. Draw the linkage to scale and label it before setting up the equations.
Given:
Link lengths: Link 2 (O2 A) Crank angle
Solution: 1.
a 1.4
Link 3 (AB)
b 4
θ 45 deg
Crank angular velocity
Offset (yB) ω 10
See Figure P6-2 and Mathcad file P0607a.
Draw the linkage to scale and label it. Y d1 = 4.990
d2 = 3.010
3(CROSSED)
B'
A 2
0.144°
45.000°
B
3 (OPEN)
179.856° 1.000 X
O2
2.
Determine 3 and d using equations 4.16 and 4.17. Crossed:
a sin θ b
θ asin
c
θ 0.144 deg
d 2 a cos θ b cos θ
Open:
d 2 3.010
a sin θ c π b
θ 180.144 deg
d 1 4.990
θ asin
d 1 a cos θ b cos θ 3.
4.
Determine the angular velocity of link 3 using equation 6.22a:
ω 2.475
ω 2.475
Open
ω
a cos θ ω b cos θ
Crossed
ω
a cos θ ω b cos θ
Determine the velocity of pin A using equation 6.23a:
VA a ω sin θ j cos θ VA 9.899 9.899i
VA 14.000
arg VA 135.000 deg
c 1
DESIGN OF MACHINERY - 5th Ed.
5.
SOLUTION MANUAL 6-7a-2
Determine the velocity of pin B using equation 6.22b: Open
VB1 a ω sin θ b ω sin θ
VB1 9.875
Crossed
VB2 a ω sin θ b ω sin θ
VB2 9.924
The angle of VB is 0 deg if VB is positive and 180 deg if VB negative. 6.
The velocity of slip is the same as the velocity of pin B.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-8a-1
PROBLEM 6-8a Statement:
The general linkage configuration and terminology for an inverted fourbar slider-crank linkage are shown in Fig P6-3. The link lengths and the values of 2 and 2 and are defined in Table P6-3. For row a, using a graphical method, find the velocities of the pin joints A and B and the velocity of slip at the sliding joint. Draw the linkage to scale.
Given:
Link lengths: Link 1 d 6 in c 4 in
Link 4 Solution: 1.
Link 2
a 2 in
γ 90 deg
θ 30 deg
ω 10 rad sec
1
See Figure P6-3 and Mathcad file P0608a.
Draw the linkage to scale and indicate the axes of slip and transmission as well as the directions of velocities of interest. Direction of VA Axis of transmission and direction of VBA Axis of slip and Direction of VB
B y
1.793 90.0°
b a
127.333° c 142.666°
A
30.000°
d
x 04
02
2.
3.
Use equation 6.7 to calculate the magnitude of the velocity at point A. in VA a ω VA 20.000 θVA θ 90 deg sec
θVA 120 deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B on link 3. The equation to be solved graphically is VB3 = VA3 + VBA3 a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line.
4.
0
VA
V BA 52.667° VB 0.771
From the velocity triangle we have: Velocity scale factor:
kv
10 in sec in
10 in/sec
1.846 Y
1
X
DESIGN OF MACHINERY - 5th Ed.
5.
SOLUTION MANUAL 6-8a-2
in
VB3 0.771 in kv
VB3 7.710
VBA3 1.846 in kv
VBA3 18.460
sec
θVB3 52.667 deg
in sec
Determine the angular velocity of link 3 using equation 6.7. From the linkage layout above:b 1.793 in and ω
VBA3 b
ω 10.296
rad
θ 142.666 deg CW
sec
The way in which link 3 slides in link 4 requires that ω ω 6.
7.
Determine the magnitude and sense of the vector VB4 using equation 6.7. in
VB4 c ω
VB4 41.182
θVB4 θ 90 deg
θVB4 52.666 deg
sec
Note that VB3 and VB4 are in the same direction in this case. The velocity of slip is in
Vslip VB3 VB4
Vslip 33.472
θslip θVB4 180 deg
θslip 232.666 deg
sec
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-9a-1
PROBLEM 6-9a Statement:
The general linkage configuration and terminology for an inverted fourbar slider-crank linkage are shown in Fig P6-3. The link lengths and the values of 2 and 2 and are defined in Table P6-3. For row a, using an analytic method, find the velocities of the pin joints A and B and the velocity of slip at the sliding joint. Draw the linkage to scale and label it before setting up the equations.
Given:
Link lengths: Link 1 d 6 c 4
Link 4 Solution: 1.
Link 2
a 2
γ 90 deg
θ 30 deg
ω 10
See Mathcad file P0609a.
Draw the linkage to scale and label it. B y 90.0°
127.333°
b
c
A
142.666°
a 30.000°
d
x 04
02 B'
169.040° 79.041°
2.
Determine the values of the constants needed for finding 4 from equations 4.25 and 4.26. P a sin θ sin γ a cos θ d cos γ
Q a sin θ cos γ a cos θ d sin γ
3.
4.
5.
P 1.000 Q 4.268
R c sin γ
R 4.000
T 2 P
T 2.000
S R Q
S 0.268
U Q R
U 8.268
Use equation 4.26 to find values of 4 for the open and crossed circuits.
2 atan2 2 S T
OPEN
θ 2 atan2 2 S T
CROSSED
θ
2 T 4 S U 2
T 4 S U
θ 142.667 deg θ 169.041 deg
Use equation 4.22 to find values of 3 for the open and crossed circuits. OPEN
θ θ γ
θ 232.667 deg
CROSSED
θ θ γ
θ 79.041 deg
Determine the magnitude of the instantaneous "length" of link 3 from equation 4.20a.
DESIGN OF MACHINERY - 5th Ed.
6.
7.
OPEN
b 1
CROSSED
b 2
SOLUTION MANUAL 6-9a-2
sin θ γ
a sin θ c sin θ
b 1 1.793
sin θ γ
a sin θ c sin θ
b 2 1.793
Determine the angular velocity of link 4 using equation 6.30c: OPEN
ω
CROSSED
ω
a ω cos θ θ
b 1 c cos γ
a ω cos θ θ b 2 c cos γ
ω 10.292
ω 3.639
Determine the velocity of pin A using equation 6.23a:
VA a ω sin θ j cos θ VA 10.000 17.321i 8.
VA 20.000
arg VA 120.000 deg
Determine the velocity of point B on link 4 using equation 6.31: OPEN
VB4x1 c ω sin θ
VB4x1 24.966
VB4y1 32.734
VB4y1 c ω cos θ VB41
2
VB4x1 VB4y1
2
VB41 41.168
θVB1 atan2 VB4x1 VB4y1 CROSSED
θVB1 52.667 deg
VB4x2 c ω sin θ
VB4x2 2.767
VB4y2 14.289
VB4y2 c ω cos θ VB42
2
VB4x2 VB4y2
2
VB42 14.555
θVB2 atan2 VB4x2 VB4y2 9.
θVB2 100.959 deg
Determine the slip velocity using equation 6.30a: OPEN
CROSSED
Vslip1
Vslip2
a ω sin θ ω b 1 sin θ c sin θ
cos θ
a ω sin θ ω b 2 sin θ c sin θ
cos θ
Vslip1 33.461
Vslip2 4.351
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-10a-1
PROBLEM 6-10a Statement:
The link lengths, gear ratio (), phase angle (), and the values of 2 and 2 for a geared fivebar from row a of Table P6-4 are given below. Draw the linkage to scale and graphically find 3 and 4, using a graphical method.
Given:
Link lengths: Link 1
f 6 in
Link 3
b 7 in
Link 2
a 1 in
Link 4
c 9 in
Link 5
d 4 in
Gear ratio, phase angle, and crank angle: λ 2 Solution: 1.
ϕ 30 deg
θ 150 deg
Choose the pitch radii of the gears. Since the gear ratio is positive, an idler must be used between gear 2 and gear 5. Let the idler be the same diameter as gear 5, and let all three gears be in line.
r5
λ
f
r2 r5
r5 1.200 in
λ3
r2 λ r5
r2 2.400 in
Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest. Direction of VBC Direction of VBA
Direction of VC
Direction of VA
C
4 B 3
A
2
4.
1
Determine the angle of link 5 using the equation in Figure P6-4.
f r2 3 r5
3.
ω 10 rad sec
See Figure P6-4 and Mathcad file P0610a.
θ λ θ ϕ 2.
θ 60 deg
5 O5
O2
Use equation 6.7 to calculate the magnitude of the velocity at points A and C. in
VA a ω
VA 10.00
ω λ ω
ω 20.000
rad
VC 80.00
in
VC d ω
sec
θ 60 deg 90 deg
sec
sec
θ 150 deg 90 deg
DESIGN OF MACHINERY - 5th Ed.
5.
SOLUTION MANUAL 6-10a-2
Use equation 6.5 to (graphically) determine the magnitudes of the relative velocity vectors VBA and VBC. The equation to be solved graphically is the last of the following three. VB = VA + VBA
VB = VC + VBC
VA + VBA = VC + VBC
a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, layout the known vector VC. d. From the tip of VC, draw a construction line with the direction of VBC, magnitude unknown. e. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VBC construction line and drawing VBC from the tip of VC to the intersection of the VBA construction line. 6.
From the velocity triangle we have: Velocity scale factor:
0
kv
50 in sec
50 in/sec
1
Y VA
in
X
7.
VBA 4.562 in kv
VBA 228.100
VBC 3.051 in kv
VBC 152.550
in sec VC
in sec
Determine the angular velocity of links 3 and 4 using equation 6.7.
ω
VBA b
ω 32.586
4.562
rad
3.051
sec V BA
ω
VBC c
Both links are rotating CCW.
ω 16.950
rad sec
V BC
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-11a-1
PROBLEM 6-11a Statement:
Given:
Solution: 1.
The general linkage configuration and terminology for a geared fivebar linkage are shown in Figure P6-4. The link lengths, gear ratio (), phase angle (), and the values of 2 and 2 are defined in Table P6-4. For row a, find 3 and 4, using an analytic method. Draw the linkage to scale and label it before setting up the equations. Link lengths: Link 1
d 4
Link 2
a 1
Link 3
b 7
Link 4
c 9
Link 5
f 6
Input angle
θ 60 deg
Gear ratio
λ 2.0
Phase angle
ϕ 30 deg
ω 10
See Figure P6-4 and Mathcad file P0611a.
Draw the linkage to scale and label it. y C
4
B
177.7152°
173.6421° 3
5 124.0501°
2
x
O2
3
150.0000°
115.4074°
O5
4
B`
2.
Determine the values of the constants needed for finding 3 and 4 from equations 4.28h and 4.28i.
B 2 c d sin λ θ ϕ a sin θ
A 2 c d cos λ θ ϕ a cos θ f
2
2
2
2
2
A 36.646 B 20.412
C a b c d f 2 a f cos θ 2 d a cos θ f cos λ θ ϕ 2 a d sin θ sin λ θ ϕ
C 37.431
D C A
D 0.785
E 2 B
E 40.823
F A C
F 74.077
a cosθ f H 2 b d sin λ θ ϕ a sin θ G 2 b d cos λ θ ϕ
2
2
2
2
2
K a b c d f 2 a f cos θ 2 d a cos θ f cos λ θ ϕ 2 a d sin θ sin λ θ ϕ L K G
G 28.503 H 15.876
K 26.569 L 1.933
DESIGN OF MACHINERY - 5th Ed.
3.
M 2 H
M 31.751
N G K
N 55.072
Use equations 4.28h and 4.28i to find values of 3 and 4 for the open and crossed circuits. OPEN
CROSSED
2 atan2 2 D E 2 atan2 2 L M 2 atan2 2 D E
θ 2 atan2 2 L M θ θ θ
4.
SOLUTION MANUAL 6-11a-2
2 E 4 D F 2 M 4 L N 2 E 4 D F 2
M 4 L N
θ 173.642 deg θ 177.715 deg θ 115.407 deg θ 124.050 deg
Determine the position and angular velocity of gear 5 from equations 4.27c and 6.32c θ λ θ ϕ
θ 150.000 deg
ω λ ω
ω 20.000
Angular velocity of links 3 and 4 from equations 6.33 OPEN
ω
b cos θ 2 θ cos θ
ω 32.585 ω
c sin θ
a ω sin θ b ω sin θ d ω sin θ
ω
CCW
b cos θ 2 θ cos θ
2 sin θ a ω sin θ θ d ω sin θ θ
ω 75.191 ω
CCW
ω 16.948
CROSSED
2 sin θ a ω sin θ θ d ω sin θ θ
CW
c sin θ
a ω sin θ b ω sin θ d ω sin θ
ω 59.554
CW
CCW
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-12-1
PROBLEM 6-12 Statement:
Find all of the instant centers of the linkages shown in Figure P6-5.
Solution:
See Figure P6-5 and Mathcad file P0612.
a.
This is a fourbar slider-crank with n 4.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
2.
n ( n 1)
C6
2
Draw the linkage to scale and identify those ICs that can be found by inspection. C
2,3 Y
A
1,2
2
3
3,4
X O2
B 4
1,4 at infinity
3.
Use Kennedy's Rule and a linear graph to find the remaining 2 ICs, I1,3 and I2,4 I1,3: I1,2-I2,3 and I1,4-I3,4
I2,4: I1,2-I1,4 and I2,3-I3,4 1,3 1 4
2,4
C
2 3
A 1,2
2
3
3,4
2,3
O2
4 B 1,4 at infinity 1,4 at infinity
b.
This is a fourbar with planetary motion (roller 3), n 4.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
n ( n 1) 2
C6
2,3 at contact point (behind link 4) 1,3
2.
Draw the linkage to scale and identify those ICs that can be found by inspection, which in this case, is all of them. c. This is a fourbar with n 4. 1.
3 A 4
Determine the number of instant centers for this mechanism using equation 6.8a. C
2.
1
n ( n 1) 2
2 O2
C6
Draw the linkage to scale and identify those ICs that can be found by inspection.
1,2 and 2,4 and 1,4
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-12-2 C 2,3
3,4 3 A
B 4
2
O2
O4 1,2
3.
1,4
Use Kennedy's Rule and a linear graph to find the remaining 2 ICs, I1,3 and I2,4 I1,3: I1,2-I2,3 and I1,4-I3,4
I2,4: I1,2-I1,4 and I2,3-I3,4 1,3 at infinity
1,3 at infinity
1
C 2,3
4
3,4
2
3 3 2,4 at infinity A
2
2,4 at infinity
O2
2,4 at infinity
B
4
2,4 at infinity
O4 1,2
1,4
1,3 at infinity
1,3 at infinity
d.
This is a fourbar with n 4.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
2.
n ( n 1)
C6
2
Draw the linkage to scale and identify those ICs that can be found by inspection. 2,3
3,4
1,2 3 O2
4
A
2 1,4 at infinity
3.
Use Kennedy's Rule and a linear graph to find the remaining 2 ICs, I1,3 and I2,4 I1,3: I1,2-I2,3 and I1,4-I3,4
I2,4: I1,2-I1,4 and I2,3-I3,4 2,3
3,4
1,2 1 3 O2
4
A
2,4
4
2 3
2
1,3
1,4 at infinity 1,4 at infinity
e.
This is a threebar with n 3.
1.
Determine the number of instant centers for this mechanism using equation 6.8a.
DESIGN OF MACHINERY - 5th Ed.
C 2.
SOLUTION MANUAL 6-12-3
n ( n 1)
C3
2
Draw the linkage to scale and identify those ICs that can be found by inspection. 1,2 2
1,3 3
O2
3.
O3
Use Kennedy's Rule and a linear graph to find the remaining IC, I2,3 I2,3: I1,2-I1,3 Common normal 1 2,3 1,2 2
2
3
1,3 3
O2
O3
f.
This is a fourbar with n 4.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
2.
n ( n 1)
C6
2
Draw the linkage to scale and identify those ICs that can be found by inspection. 3,4
1,4 at infinity
3.
C
B
Use Kennedy's Rule and a linear graph to find the remaining 2 ICs, I1,3 and I2,4
4 3
I1,3: I1,2-I2,3 and I1,4-I3,4
I2,4: I1,2-I1,4 and I2,3-I3,4 2,3 VA
A
2
2,4 at infinity 3,4
1,4 at infinity
1,2 at infinity C
B
1
4 3 1,3
4
2 3
2,3 VA
A
2
1,2 at infinity
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-13-1
PROBLEM 6-13 Statement:
Find all of the instant centers of the linkages shown in Figure P6-6.
Solution:
See Figure P6-6 and Mathcad file P0613.
a.
This is a fourbar inverted slider-crank with n 4.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
2.
n ( n 1)
C6
2
Draw the linkage to scale and identify those ICs that can be found by inspection. 2,3
2
3 3,4 at infinity 1 1,2 4
1 1,4
3.
Use Kennedy's Rule and a linear graph to find the remaining 2 ICs, I1,3 and I2,4 I1,3: I1,2-I2,3 and I1,4-I3,4
I2,4: I1,2-I1,4 and I2,3-I3,4 3,4 at infinity 2,3
2,4 2
1 4
3
2
3,4 at infinity 1
3
1,2 4
1 1,4
1,3
b.
This is a sixbar with slider, n 6.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
n ( n 1) 2
C 15
2.
Draw the linkage to scale and identify those ICs that can be found by inspection.
3.
Use Kennedy's Rule and a linear graph to find the remaining 7 ICs. I1,3: I1,2-I2,3 and I1,4-I3,4
5,6 6 2,3 2
1,6 at infinity 3
1
5
I1,5: I1,6-I5,6 and I1,4-I4,5
3,4; 3,5; 4,5
I2,5: I1,2-I1,5 and I2,4-I4,5
1 4
1,2
I3,6: I1,6-I1,3 and I3,4-I4,6
I4,6: I1,6-I1,4 and I4,5-I5,6
I2,4: I1,2-I1,4 and I2,3-I3,4
I2,6: I1,2-I1,6 and I2,5-I5,6
1,4
1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-13-2
3,6
1,3
1,6 at infinity 2,5
to 2,4
5,6 6 to 2,4
1,5
2,3
1 1,6 at infinity
2 3
6
2
1
5
5
1,6 at infinity
3 4
3,4; 3,5; and 4,5 1 4
1,2
1,4
2,6 4,6
1,6 at infinity 1
c.
This is a sixbar with slider and roller with n 6.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
2.
n ( n 1)
C 15
2
Draw the linkage to scale and identify those ICs that can be found by inspection. 2,5 5,6
3.
Use Kennedy's Rule and a linear graph to find the remaining 8 ICs.
1,2
5
6
2,3
2
1
I2,6: I1,2-I1,6 and I2,5-I5,6 I1,5: I1,6-I5,6 and I1,2-I2,5
1
1,6
3
I4,5: I1,4-I1,5 and I2,4-I2,5 I3,6: I3,6-I5,6 and I2,3-I2,6 I1,3: I1,2-I2,3 and I1,4-I3,4
I3,5: I3,4-I4,5 and I2,5-I2,3
I2,4: I1,2-I1,4 and I2,5-I2,4
I4,6: I4,5-I5,6 and I3,4-I3,6
4 3,4
3,5 2,5 4,5
1,4 at infinity
1,5
5,6
1,2
5
1 6
2,6
1
2,3
2
6
2,4
2
1,4 at infinity 5
1
1,6
3
4,6 3,6 4 3,4
1,4 at infinity
1,3
1
d.
This is a sixbar with slider and roller with n 6.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
2.
n ( n 1) 2
3 4
C 15
Draw the linkage to scale and identify those ICs that can be found by inspection.
1
1,4 at infinity
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-13-3
3,4 4 1 1,4 3
5,6 5
1,2
2
6 2,3; 2,5; and 3,5
1
1
1,6 at infinit
3.
Use Kennedy's Rule and a linear graph to find the remaining 7 ICs. I1,3: I1,2-I2,3 and I1,4-I3,4
I3,6: I1,6-I1,3 and I3,5-I5,6
I2,6: I1,2-I1,6 and I2,5-I5,6
I1,5: I1,6-I5,6 and I1,2-I2,5
I4 ,5: I1,4-I1,5 and I3,5-I3,4
I2,4: I1,2-I1,4 and I2,3-I3,4
I4,6: I1,6-I1,4 and I4,5-I5,6 2,4 4,6 3,4 4,5
4
1,3
1 1
1,4 6 3
1,5
1,6 at infinity
2
5
3 4
2,6
5,6 5
1,2
2 1
6 2,3; 2,5; and 3,5
3,6
1
1,6 at infinity 1,6 at infinity
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-14-1
PROBLEM 6-14 Statement:
Find all of the instant centers of the linkages shown in Figure P6-7.
Solution:
See Figure P6-7 and Mathcad file P0614.
a.
This is a pin-jointed fourbar with n 4.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
n ( n 1)
C6
2
2.
Draw the linkage to scale and identify those ICs that can be found by inspection.
3.
Use Kennedy's Rule and a linear graph to find the remaining 2 ICs, I1,3 and I2,4
3,4
2,3
3
I1,3: I1,2-I2,3 and I1,4-I3,4
I2,4: I1,2-I1,4 and I2,3-I3,4
2
4
1,2
1,4
3,4 1 2,3
3 2
4
1 2
4 3
1,2 2,4 1
1,4
1,3
b.
This is a fourbar inverted slider-crank, n 4.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
n ( n 1)
C6
2
2.
Draw the linkage to scale and identify those ICs that can be found by inspection, which in this case, is all of them.
3.
Use Kennedy's Rule and a linear graph to find the remaining 2 ICs, I1,3 and I2,4 I1,3: I1,2-I2,3 and I1,4-I3,4
2,3 3 1,4
4 2
I2,4: I1,2-I1,4 and I2,3-I3,4 1
3,4 at infinity 1,2
2,3 1
3 1,4
4
4
2
2
2,4 3
1
1,2
3,4 at infinity
1,3 3,4 at infinity
1,4 at infinity
c.
This is a fourbar double slider with n 4.
1.
Determine the number of instant centers for this mechanism using equation 6.8a.
4 3,4 3
C 2.
n ( n 1) 2
C6
2,3 2 1
Draw the linkage to scale and identify those ICs that can be found by inspection. 1,2 at infinity
DESIGN OF MACHINERY - 5th Ed.
3.
SOLUTION MANUAL 6-14-2
Use Kennedy's Rule and a linear graph to find the remaining 2 ICs, I1,3 and I2,4 1,4 at infinity
I1,3: I1,2-I2,3 and I1,4-I3,4
2,4 at infinity
1,3
I2,4: I1,2-I1,4 and I2,3-I3,4 1
4 4
3,4
2
3 3
2,3 2 1 1,2 at infinity
d.
This is a pin-jointed fourbar with n 4.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
n ( n 1)
C6
2
2.
Draw the linkage to scale and identify those ICs that can be found by inspection.
3.
Use Kennedy's Rule and a linear graph to find the remaining 2 ICs, I1,3 and I2,4 I1,3: I1,2-I2,3 and I1,4-I3,4
2,3
3,4
1,2
I2,4: I1,2-I1,4 and I2,3-I3,4
3
2
4 1,4
1 2,3
1,3
1
3,4 4
3
2
2
4 1,4
3
2,4 1 1,2
e.
This is a fourbar effective slider-crank with n 4.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
n ( n 1)
C6
2
2.
Draw the linkage to scale and identify those ICs that can be found by inspection.
3.
Use Kennedy's Rule and a linear graph to find the remaining 2 ICs, I1,3 and I2,4 I1,3: I1,2-I2,3 and I1,4-I3,4
2,3
3,4
I2,4: I1,2-I1,4 and I2,3-I3,4
3
2
4
1,3
2,3 1 1,2
1
2,4 4
3,4 3
2
2 3
4 1 1,2 1,4 at infinity
1,4 at infinity
1,4 at infinity
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-14-3
f.
This is a fourbar cam-follower with n 4.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
n ( n 1)
C6
2
2.
Draw the linkage to scale and identify those ICs that can be found by inspection.
3.
Use Kennedy's Rule and a linear graph to find the remaining 2 ICs, I1,3 and I2,4 I1,3: I1,2-I2,3 and I1,4-I3,4
3,4 2,3 3
1,4
I2,4: I1,2-I1,4 and I2,3-I3,4
4 2
2,4
1
3,4 1,3
1
1,2 at infinity
2,3 4
3
1,4
2
4 3
2 1 1,2 at infinity 1,2 at infinity
g.
This is a fourbar slider-crank with n 4.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
n ( n 1)
C6
2
2.
Draw the linkage to scale and identify those ICs that can be found by inspection.
3.
Use Kennedy's Rule and a linear graph to find the remaining 2 ICs, I1,3 and I2,4
2,3
2 3,4
I1,3: I1,2-I2,3 and I1,4-I3,4
3
4
1,2
I2,4: I1,2-I1,4 and I2,3-I3,4
1,4 at infinity
1 1,3 2,3
1,4 at infinity
1
2
2,4
3,4 3
4
2
4 3
1,2
2
2,3 1
1
h.
This is a fourbar inverted slider-crank with n 4.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
2.
n ( n 1) 2
1,2
1,4 at infinity
3
C6
3,4 at infinity
Draw the linkage to scale and identify those ICs that can be found by inspection.
4 1,4
1
DESIGN OF MACHINERY - 5th Ed.
3.
SOLUTION MANUAL 6-14-4
Use Kennedy's Rule and a linear graph to find the remaining 2 ICs, I1,3 and I2,4 I1,3: I1,2-I2,3 and I1,4-I3,4
3,4 at infinity
2,4
I2,4: I1,2-I1,4 and I2,3-I3,4
2
2,3
1
1,2 4
2
1 3
3,4 at infinity
3 3,4 at infinity
1,3 4 1
1,4
i.
This is a fourbar slider-crank (hand pump) with n 4.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
2.
n ( n 1)
C6
2
Draw the linkage to scale and identify those ICs that can be found by inspection. 2,3
3
3,4 4 1,2 at infinity
2
1
1,4
3.
Use Kennedy's Rule and a linear graph to find the remaining 2 ICs, I1,3 and I2,4 I1,3: I1,2-I2,3 and I1,4-I3,4
I2,4: I1,2-I1,4 and I2,3-I3,4 2,3 3 1,3
1,2 at infinity
3,4
1 4
1,2 at infinity
2
4
1
2,4
1,2 at infinity 1,4
2 3
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-15-1
PROBLEM 6-15 Statement:
Find all of the instant centers of the linkages shown in Figure P6-8.
Solution:
See Figure P6-8 and Mathcad file P0615.
a.
This is a pin-jointed fourbar with n 4.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
n ( n 1)
C6
2
2.
Draw the linkage to scale and identify those ICs that can be found by inspection.
3.
Use Kennedy's Rule and a linear graph to find the remaining 2 ICs, I1,3 and I2,4
2,3 B
I1,3: I1,2-I2,3 and I1,4-I3,4 2 3
I2,4: I1,2-I1,4 and I2,3-I3,4 O2
1,2
B 4
O4
3,4 1,4
2,3
1
2
1,2
4
2
3 3 4 1,3 3,4 and 2,4 1,4
b.
This is a pin-jointed fourbar with n 4.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
2.
n ( n 1) 2
2,3
C6 2
Draw the linkage to scale and identify those ICs that can be found by inspection, which in this case, is all of them.
O2
B
A
3
3,4 4
3.
Use Kennedy's Rule and a linear graph to find the remaining 2 ICs, I1,3 and I2,4 I1,3: I1,2-I2,3 and I1,4-I3,4
1,4
1,2
O4
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-15-2
I2,4: I1,2-I1,4 and I2,3-I3,4 1,3
1
2,3 B
A
2,4
4
3
2 3
2
O2
3,4 4 1,4
1,2
O4
c.
This is an eightbar (three slider-cranks with a common crank) with n 8.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
2.
3.
n ( n 1)
C 28
2
Draw the linkage to scale and identify those ICs that can be found by inspection.
7
Use Kennedy's Rule and a linear graph to find the remaining 15 ICs.
1,7 at infinity
4
2
I2,7: I1,2-I1,7 and I2,4-I4,7
3,6
8
1 3
1,6 at infinity
5
1,8 at infinity 2,3; 2,4; 2,5; 3,4; 3,5; and 4,5
I1,5: I1,2-I2,5 and I1,8-I5,8 I2,6: I1,2-I1,6 and I2,3-I3,6
5,8
1,2
4,7
6
I3,7: I1,3-I1,7 and I3,4-I4,7
I6,8: I2,7-I2,8 and I3,7-I3,8
I3,8: I1,3-I3,8 and I3,5-I5,8
I4,6: I1,6-I1,4 and I3,4-I3,6
I2,8: I1,2-I1,8 and I2,5-I5,8
I4,8: I3,8-I3,4 and I4,5-I5,8
I5,6: I2,5-I2,6 and I3,5-I3,6
I1,4: I1,7-I4,7 and I1,2-I2,4
I5,7: I2,5-I2,7 and I3,5-I3,7
I6,7: I2,6-I2,7 and I3,6-I3,7
I1,3: I1,2-I2,3 and I1,6-I3,6
I6,8: I4,6-I4,8 and I3,6-I3,8
4,7
7
4
2
3 2,8
1,6 at infinity 3,6
Note that, for clarity, not all ICs are shown.
5
8 1,8 at infinity 2,6 3,8 2,3; 2,4; 2,5; 3,4; 3,5; and 4,5 3,7
1,7 at infinity 2,7
5,8
1,2 4,6
1,4
6 1
1,3
To 1,5 To 1,5
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-15-3
d.
This is a pin-jointed fourbar with n 4.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
2.
n ( n 1)
C6
2
E
Draw the linkage to scale and identify those ICs that can be found by inspection. 1
2,3
3.
3,4
Use Kennedy's Rule and a linear graph to find the remaining 2 ICs, I1,3 and I2,4
3 2
I1,3: I1,2-I2,3 and I1,4-I3,4
4 1,2
I2,4: I1,2-I1,4 and I2,3-I3,4
1,4
E
1,3 at infinity
1
1
2,3
4
3,4
2
3 3
2 2,4 at infinity 4 1,2
1,4
e.
This is an eightbar with n 8.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
2.
n ( n 1)
C 28
2
3,4
2,3 1,2
Draw the linkage to scale and identify those ICs that can be found by inspection.
4
4,6 2
4,7
5 4,5
2,5
3.
1,4
3
The remaining 17 ICs are at infinity.
6
7
7,8
6,8 8 4 1,4 at infinity 3
1
f.
This is an offset crank-slider with n 4.
1.
Determine the number of instant centers for this mechanism using equation 6.8a.
2,3
C 2 1,2
1,8
3,4
2.
n ( n 1) 2
C6
Draw the linkage to scale and identify those ICs that can be found by inspection.
DESIGN OF MACHINERY - 5th Ed.
3.
SOLUTION MANUAL 6-15-4
Use Kennedy's Rule and a linear graph to find the remaining 2 ICs, I1,3 and I2,4 I1,3: I1,2-I2,3 and I1,4-I3,4
I2,4: I1,2-I1,4 and I2,3-I3,4 3,4
4
1,3
1,4 at infinity
1 3
1
4
2
2,3
2
3
2,4
1,4 at infinity 1,2
g.
This is a sixbar with n 6.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
3,4
n ( n 1)
1,6
C 15
2
2,3
3 1,4 2
2.
Draw the linkage to scale and identify those ICs that can be found by inspection.
3.
Use Kennedy's Rule and a linear graph to find the remaining 8 ICs.
6
4
5
2,5
5,6
1,4
I1,3: I1,4-I4,3 and I1,2-I3,2 I1,5: I1,6-I5,6 and I1,2-I2,5
4,5
I3,5: I1,3-I1,5 and I3,2-I2,5 I2,4: I1,2-I1,4 and I3,4-I2,3 I2,6: I1,6-I1,2 and I2,5-I5,6 I4,6: I1,4-I1,6 and I2,4-I2,6
3,5 at infinity 3,4
1,6
I4,5: I4,6-I5,6 and I3,4-I3,5
2,4 3
I3,6: I3,4-I4,6 and I3,5-I5,6
3,5 at infinity
1,2 2
2,3 and 1,5
6
4 5 4,5 and 1,3
1,4 4,6 at infinity
3,6
2,6
5,6
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-16a-1
PROBLEM 6-16a Statement:
The linkage in Figure P6-5a has the dimensions and crank angle given below. Find 3, VA, VB, and VC for the position shown for 2 = 15 rad/sec in the direction shown. Use the velocity difference graphical method.
Given:
Link lengths: Link 2 (O2 to A)
a 0.80 in
1.
p 1.33 in
Link 3 (A to B)
b 1.93 in
Angle BAC
δ 38.6 deg
Offset
c 0.38 in
Crank angle:
θ 34.3 deg
Input crank angular velocity Solution:
Coupler point data: Distance from A to C
ω 15 rad sec
1
CCW
See Figure P6-5a and Mathcad file P0616a.
Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest. Direction of VCA Direction of VA
C
Y
A
38.600°
34.300° 154.502°
X O2
Direction of VB B
Direction of VBA
2.
Use equation 6.7 to calculate the magnitude of the velocity at point A. VA a ω
3.
VA 12.00
in
θ 34.3 deg 90 deg
sec
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B. The equation to be solved graphically is VB = VA + VBA a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line.
0
VA
5 in/sec
Y
2.197
124.300° V BA X
VB 2.298
DESIGN OF MACHINERY - 5th Ed.
4.
From the velocity triangle we have: Velocity scale factor:
5.
kv
5 in sec
1
in in
VB 2.298 in kv
VB 11.490
VBA 2.197 in kv
VBA 10.985
VBA
in sec
b
ω 5.692
rad sec
Determine the magnitude and sense of the vector VCA using equation 6.7. VCA p ω
VCA 7.570
in sec
θCA ( 154.502 180 38.6 90) deg 7.
θB 180 deg
sec
Determine the angular velocity of link 3 using equation 6.7. ω
6.
SOLUTION MANUAL 6-16a-2
θCA 76.898 deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point C. The equation to be solved graphically is VC = VA + VCA
VA
a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, layout the (now) known vector VCA. c. Complete the vector triangle by drawing VC from the tail of VA to the tip of the VCA vector.
Y 0
V CA
5 in/sec
153.280° VC X
8.
1.130
From the velocity triangle we have: Velocity scale factor:
VC 1.130 in kv
kv
5 in sec
1
in
VC 5.650
in sec
θC 153.28 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-16b-1
PROBLEM 6-16b Statement:
The linkage in Figure P6-5a has the dimensions and crank angle given below. Find 3, VA, VB,
Given:
and VC for the position shown for 2 = 15 rad/sec in the direction shown. Use the instant center graphical method. Link lengths: Coupler point data: Link 2 (O2 to A)
a 0.80 in
Distance from A to C
p 1.33 in
Link 3 (A to B)
b 1.93 in
Angle BAC
δ 38.6 deg
Offset
c 0.38 in
Input crank angular velocity Solution: 1.
θ 34.3 deg
Crank angle:
ω 15 rad sec
1
CCW
See Figure P6-5a and Mathcad file P0616b.
Draw the linkage to scale in the position given, find the instant centers, distances from the pin joints to the instant centers and the angles that links 3 and 4 make with the x axis. From the layout: AI13 2.109 in
1,3 2.109
BI13 2.019 in
0.993 116.732°
CI13 0.993 in 2,4
θC ( 360 116.732 ) deg 2.
Use equation 6.7 and inspection of the layout to determine the magnitude and direction of the velocity at point A.
C 2.019
A 1,2
2
3
O2
4
VA a ω VA 12.000
B
in
1,4 at infinity
θVA 124.3 deg
Determine the angular velocity of link 3 using equation 6.9a. ω
4.
1,4 at infinity
sec
θVA θ 90 deg 3.
3,4
2,3
VA AI13
ω 5.690
rad
CW
sec
Determine the magnitude of the velocity at point B using equation 6.9b. Determine its direction by inspection. VB BI13 ω
VB 11.488
in sec
θVC 180 deg 5.
Determine the magnitude of the velocity at point C using equation 6.9b. Determine its direction by inspection. in
VC CI13 ω
VC 5.650
θVC θC 90 deg
θVC 153.268 deg
sec
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-16c-1
PROBLEM 6-16c Statement:
The linkage in Figure P6-5a has the dimensions and crank angle given below. Find 3, VA, VB,
Given:
and VC for the position shown for 2 = 15 rad/sec in the direction shown. Use an analytical method. Link lengths: Coupler point data: Link 2 (O2 to A)
a 0.80 in
Distance from A to C
Rca 1.33 in
Link 3 (A to B)
b 1.93 in
Angle BAC
δ 38.6 deg
Offset
c 0.38 in
Input crank angular velocity Solution: 1.
ω 15 rad sec
1
CCW
See Figure P6-5a and Mathcad file P0616c.
C
Draw the linkage to scale and label it. Y
2.
θ 34.3 deg
Crank angle:
Determine 3 and d using equation 4.17.
A
38.600°
34.300° 154.502°
X O2
0.380"
B
a sin θ c π b
θ asin
θ 154.502 deg
d 1 a cos θ b cos θ 3.
Determine the angular velocity of link 3 using equation 6.22a: ω
4.
d 1 2.403 in
a cos θ ω b cos θ
ω 5.691
rad sec
Determine the velocity of pin A using equation 6.23a:
VA a ω sin θ j cos θ VA ( 6.762 9.913j) 5.
in sec
VA 12.000
arg VA 124.300 deg
in sec
Determine the velocity of pin B using equation 6.22b:
VB a ω sin θ b ω sin θ VB 11.490 6.
in
VB 11.490
sec
arg VB 180.000 deg
in sec
Determine the velocity of the coupler point C for the open circuit using equations 6.36.
VCA Rca ω sin π θ δ j cos π θ δ VCA ( 1.716 7.371j)
in sec
VC VA VCA VC ( 5.047 2.542j)
in sec
VC 5.651
in sec
arg VC 153.268 deg
Note that 3 is defined at point B for the slider-crank and at point A for the pin-jointed fourbar. Thus, to use equation 6.36a for the slider-crank, 180 deg must be added to the calculated value of 3.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-17a-1
PROBLEM 6-17a Statement:
The linkage in Figure P6-5c has the dimensions and effective crank angle given below. Find 3, 4, VA, VB, and VC for the position shown for 2 = 15 rad/sec in the direction shown. Use the velocity difference graphical method.
Given:
Link lengths:
Coupler point:
Link 2 (point of contact to A)
a 0.75 in
Distance A to C
p 1.2 in
Link 3 (A to B)
b 1.5 in
Angle BAC
δ 30 deg
Link 4 (point of contact to B)
c 0.75 in
Link 1 (between contact points)
d 1.5 in
ω 15 rad sec
Input crank angular velocity Solution: 1.
θ 77 deg
Crank angle: 1
See Figure P6-5c and Mathcad file P0617a.
Although the mechanism shown in Figure P6-5c is not entirely pin-jointed, it can be analyzed for the position shown by its effective pin-jointed fourbar, which is shown below. Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest. 0
0.5
1 in
Direction of VCA Y
C Direction of VA Direction of VBA
30.000°
Direction of VB
3
A
B b 4
2 a
77.000°
c d
X
O2
O4 Effective link 2
2.
Effective link 4
Use equation 6.7 to calculate the magnitude of the velocity at point A. VA a ω
3.
77.000°
VA 11.250
in sec
θ 77 deg 90 deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B, the magnitude of the relati velocity VBA, and the angular velocity of link 3. The equation to be solved graphically is VB = VA + VBA a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-17a-2
Direction of V B
Y
Direction of V BA
167.000° VA
VB X
0
4.
From the velocity triangle we have: VB VA VBA 0
5.
ω
sec
θ 167 deg
in sec
VBA b VB c
ω 0.000
rad sec rad
ω 15.000
sec
Determine the magnitude of the vector VCA using equation 6.7. VCA p ω
7.
in
VB 11.250
Determine the angular velocity of links 3 and 4 using equation 6.7. ω
6.
5 in/sec
VCA 0.000
in sec
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point C. The equation to be solved graphically is VC = VA + VCA Normally, we would draw the velocity triangle represented by this equation. However, since VCA is zero, in
VC VA
VC 11.250
θC θ
θC 167.000 deg
sec
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-17b-1
PROBLEM 6-17b Statement:
The linkage in Figure P6-5c has the dimensions and effective crank angle given below. Find 3,
Given:
4, VA, VB, and VC for the position shown for 2 = 15 rad/sec in the direction shown. Use the instant center graphical method. Link lengths: Coupler point: Link 2 (point of contact to A)
a 0.75 in
Link 3 (A to B)
b 1.5 in
Link 4 (point of contact to B)
c 0.75 in
Link 1 (between contact points)
d 1.5 in
Solution: 1.
p 1.2 in
Angle BAC
δ 30 deg
Crank angle:
ω 15 rad sec
Input crank angular velocity
Distance A to C
1
θ 77 deg
CCW
See Figure P6-5c and Mathcad file P0617b.
Although the mechanism shown in Figure P6-5c is not entirely pin-jointed, it can be analyzed for the position shown by its effective pin-jointed fourbar, which is shown below. Draw the linkage to scale in the position given, find the instant centers, distances from the pin joints to the instant centers and the angles that links 3 and 4 make with the x axis. 1,3 at infinity
1,3 at infinity C
2,3
3,4 30.000°
3
2,4 at infinity A
2
77.000° 2,4 at infinity
O2
77.000° 2,4 at infinity
O4 1,2
1,3 at infinity
Since I
1,3
2,4 at infi
B
4
1,4
1,3 at infinity
is at infinity, link 3 is not rotating ( ω 0 rad sec
1
). Thus, the velocity of every point on
link 3 is the same. From the layout above: θ 77.000 deg 2.
Use equation 6.7 and inspection of the layout to determine the magnitude and direction of the velocity at point A. in VA a ω VA 11.250 sec
θVA θ 90 deg 3.
θ 0.000 deg
θVA 167.0 deg
Determine the magnitude of the velocity at point B knowing that it is the same as that of point A. in
VB VA
VB 11.250
θVB θ 90 deg
θVB 167.000 deg
sec
DESIGN OF MACHINERY - 5th Ed.
4.
Use equation 6.9c to determine the angular velocity of link 4. ω
5.
SOLUTION MANUAL 6-17b-2
VB c
ω 15
rad
CCW
sec
Determine the magnitude of the velocity at point C knowing that it is the same as that of point A. in
VC VA
VC 11.250
θVC θVA
θVC 167.000 deg
sec
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-17c-1
PROBLEM 6-17c Statement:
The linkage in Figure P6-5c has the dimensions and effective crank angle given below. Find 3,
Given:
4, VA, VB, and VC for the position shown for 2 = 15 rad/sec in the direction shown. Use an analytical method. Link lengths: Coupler point: Link 2 (point of contact to A)
a 0.75 in
Distance A to C
Rca 1.2 in
Link 3 (A to B)
b 1.5 in
Angle BAC
δ 30 deg
Link 4 (point of contact to B)
c 0.75 in
Link 1 (between contact points)
d 1.5 in
Crank angle:
θ 77 deg
ω 15 rad sec
Input crank angular velocity Solution: 1.
1
See Figure P6-5c and Mathcad file P0617c.
Draw the linkage to scale and label it. Y
C
0
0.5
1 in
3
30.000° A
B b 4
2 a
77.000°
c
77.000°
d
X O4
O2 Effective link 2
2.
Effective link 4
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 2.0000
2
d
K3
c
K2 2.0000
2
2
a b c d
2
2 a c
K3 1.0000
B 2 sin θ C K1 K2 1 cos θ K3
A cos θ K1 K2 cos θ K3
A 1.2250 3.
B 1.9487
C 2.3251
Use equation 4.10b to find values of 4 for the open circuit.
θ 2 atan2 2 A B
2
B 4 A C
θ 283.000 deg
θ θ 360 deg 4.
θ 643.000 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
d b
2
K5
2
2
c d a b 2 a b
2
K4 1.0000
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-17c-2
K5 2.0000
E 2 sin θ F K1 K4 1 cos θ K5
D cos θ K1 K4 cos θ K5
5.
D 3.5501 E 1.9487 F 0.0000
Use equation 4.13 to find values of 3 for the open circuit.
2
θ 2 atan2 2 D E
E 4 D F
θ 360.000 deg
θ θ 360 deg 6.
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18. ω ω
7.
θ 0.000 deg
a ω sin θ θ c sin θ θ a ω sin θ θ b sin θ θ
ω 0.000
rad sec
ω 15.000
rad sec
Determine the velocity of points A and B for the open circuit using equations 6.19.
VA a ω sin θ j cos θ VA ( 10.962 2.531j)
in sec
in
VA 11.250
sec
arg VA 167.000 deg
VB c ω sin θ j cos θ VB ( 10.962 2.531j) 8.
in
in
VB 11.250
sec
sec
arg VB 167.000 deg
Determine the velocity of the coupler point C for the open circuit using equations 6.36.
VCA Rca ω sin θ δ j cos θ δ VCA 0.000
in sec
VC VA VCA VC ( 10.962 2.531j)
in sec
VC 11.250
in sec
arg VC 167.000 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-18a-1
PROBLEM 6-18a Statement:
Given:
Solution: 1.
The linkage in Figure P6-5f has the dimensions and coupler angle given below. Find 3, VA, VB, and VC for the position shown for VA = 10 in/sec in the direction shown. Use the velocity difference graphical method. Link lengths and angles: Coupler point: Link 3 (A to B)
b 1.8 in
Distance A to C
p 1.44 in
Coupler angle
θ 128 deg
Angle BAC
δ 49 deg
Slider 4 angle
θ 59 deg
Input slider velocity
VA 10 in sec
See Figure P6-5f and Mathcad file P0618a.
Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest.
Direction of VB 0
0.5
1 in
Y
Direction of VBA
C
B 4
Direction of VCA
3 b
49.000°
128.000°
59.000°
VA
X A
2.
The magnitude and sense of the velocity at point A. VA 10.000
3.
2
in sec
θ 180 deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B, the magnitude of the relative velocity VBA, and the angular velocity of link 3. The equation to be solved graphically is VB = VA + VBA a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line.
0
5 in/sec
Y
V BA 4.784 VB
3.436
38.000°
59.000° X
VA
1
DESIGN OF MACHINERY - 5th Ed.
4.
From the velocity triangle we have: Velocity scale factor:
5.
kv
5 in sec
1
in in
VB 3.438 in kv
VB 17.190
VBA 4.784 in kv
VBA 23.920
sec in sec
VBA
ω 13.289
b
θ 38 deg
rad sec
Determine the magnitude and sense of the vector VCA using equation 6.7. VCA p ω
VCA 19.136
in sec
θCA ( 128 49 90) deg 7.
θ 59 deg
Determine the angular velocity of link 3 using equation 6.7. ω
6.
SOLUTION MANUAL 6-18a-2
θCA 11.000 deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point C. The equation to be solved graphically is VC = VA + VCA a. b. c.
Choose a convenient velocity scale and layout the known vector VA. From the tip of VA, layout the (now) known vector VCA. Complete the vector triangle by drawing VC from the tail of VA to the tip of the VCA vector. 0
5 in/sec
Y 1.902 VA X 11.000°
22.572° VC V CA
3.827
8.
From the velocity triangle we have:
Velocity scale factor:
VC 1.902 in kv
kv
5 in sec
1
in
VC 9.510
in sec
θC 22.572 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-18b-1
PROBLEM 6-18b Statement:
Given:
Solution: 1.
The linkage in Figure P6-5f has the dimensions and coupler angle given below. Find 3, VA, VB, and VC for the position shown for VA = 10 in/sec in the direction shown. Use the instant center graphical method. Link lengths and angles: Coupler point: Link 3 (A to B)
b 1.8 in
Distance A to C
p 1.44 in
Coupler angle
θ 128 deg
Angle BAC
δ 49 deg
Slider 4 angle
θ 59 deg
VA 10 in sec
Input slider velocity
See Figure P6-5f and Mathcad file P0618b.
Draw the linkage to scale in the position given, find the instant centers, distances from the pin joints to the instant centers and the angles that links 3 and 4 make with the x axis. 2,4 at infinity
From the layout:
3,4
1,4 at infinity
C
B
AI13 0.753 in
BI13 1.293 in
CI13 0.716 in
θC 67.428 deg
4
0.716 67.428° 3
2.
4.
VA AI13
ω 13.280
1,3
1.293
Determine the angular velocity of link 3 using equation 6.9a. ω
3.
1
0.753
rad
VA
CW
sec
A
2 2,3
Determine the magnitude of the velocity at point B using equation 6.9b. Determine its direction by inspection.
1,2 at infinity
in
VB BI13 ω
VB 17.17
θVB θ
θVB 59.000 deg
sec
Determine the magnitude of the velocity at point C using equation 6.9b. Determine its direction by inspection. in
VC CI13 ω
VC 9.509
θVC θC 90 deg
θVC 22.572 deg
sec
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-18c-1
PROBLEM 6-18c Statement:
Given:
Solution: 1.
The linkage in Figure P6-5f has the dimensions and coupler angle given below. Find 3, VA, VB, and VC for the position shown for VA = 10 in/sec in the direction shown. Use an analytical method. Link lengths and angles: Coupler point: Link 3 (A to B)
b 1.8 in
Distance A to C
Rca 1.44 in
Coupler angle
θ 128 deg
Angle BAC
δ 49 deg
Slider 4 angle
θ 59 deg
VA 10 in sec
Input slider velocity
1
See Figure P6-5f and Mathcad file P0618c.
Draw the mechanism to scale and define a vector loop using the fourbar slider-crank derivation in Section 6.7 as a model. 0
0.5
1 in
Y
C
B 4 R3 R4 3 b
49.000°
128.000°
59.000° VA
X
R2
2.
A
2
Write the vector loop equation, differentiate it, expand the result and separate into real and imaginary parts to solve for 3 and VB. R2 R3 R4 a e
j θ
b e
j θ
c e
j θ
where a is the distance from the origin to point A, a variable; b is the distance from A to B, a constant; and c is the distance from the origin to point B, a variable. Angle 2 is zero, 3 is the angle that AB makes with the x axis, and 4 is the constant angle that slider 4 makes with the x axis. Differentiating, j θ d d j θ a j b ω e c e dt dt
Substituting the Euler equivalents, d d a b ω sin θ j cos θ c cos θ j sin θ dt dt Separating into real and imaginary components and solving for 3 and VB. Note that dc/dt = VB and da/dt = VA
ω
b sin θ tan θ cos θ VA tan θ
ω 13.288
rad sec
DESIGN OF MACHINERY - 5th Ed.
VB
SOLUTION MANUAL 6-18c-2
VA b ω sin θ
VB 17.180
cos θ
sec
arg VB 59.000 deg
VB VB cos θ j sin θ 3.
in
Determine the velocity of the coupler point C using equations 6.36.
VCA Rca ω sin θ δ j cos θ δ VCA ( 18.783 3.651j)
in sec
VA VA VC VA VCA VC ( 8.783 3.651j)
in sec
VC 9.512
in sec
arg VC 22.572 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-19-1
PROBLEM 6-19 Statement:
The cam-follower in Figure P6-5d has O2A = 0.853 in. Find V4, Vtrans, and Vslip for the position shown with 2 = 20 rad/sec in the direction (CCW) shown.
Given:
ω 20 rad sec
1
O A a 0.853 in 2
Assumptions: Rolling contact (no sliding) Solution: 1.
See Figure P6-5d and Mathcad file P0619.
Draw the linkage to scale and indicate the axes of slip and transmission as well as the directions of velocities of interest. Direction of VA Axis of transmission
Direction of V4 3 O2
4
B A 2
Axis of slip 0.853
2.
Use equation 6.7 to calculate the magnitude of the velocity at point A. VA a ω
3.
VA 17.060
in sec
Use equation 6.5 to (graphically) determine the magnitude of the velocity components at point A. The equation to be solved graphically is VA = Vtrans + Vslip a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of Vslip, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of Vtrans, magnitude unknown. d. Complete the vector triangle by drawing Vslip from the tip of Vtrans to the tip of VA and drawing Vtrans from the tail of VA to the intersection of the Vslip construction line.
Y 0.768 0
VA
10 in/sec
1.523
V slip
V trans
V4
X
0.870
4.
From the velocity triangle we have: Velocity scale factor:
Vslip 1.523 in kv
kv
10 in sec
1
in
Vslip 15.230
in sec
Vtrans 0.768 in kv
Vtrans 7.680
V4 0.870 in kv
V4 8.700
in sec
in sec
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-20-1
PROBLEM 6-20 Statement:
The cam-follower in Figure P6-5e has O2A = 0.980 in and O3A = 1.344 in. Find 3, Vtrans, and Vslip for the position shown with 2 = 10 rad/sec in the direction (CW) shown.
Given:
ω 10 rad sec
1
Distance from O to A: a 0.980 in 2
Distance from O3 to A: b 1.344 in Assumptions: Roll-slide contact Solution:
See Figure P6-5e and Mathcad file P0620.
1.
Draw the linkage to scale and indicate the axes of slip and transmission as well as the directions of velocities of interest.
2.
Use equation 6.7 to calculate the magnitude of the velocity at point A.
Direction of VA2 Axis of transmission Direction of VA3
VA a ω
VA 9.800
in
Axis of slip
sec 2
3.
O2
Use equation 6.5 to (graphically) determine the magnitude of the velocity components at point A. The equation to be solved graphically is
3
A
O3
1.344
VA2 = Vtrans + VA2slip
0.980
a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of Vslip, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of Vtrans, magnitude unknown. d. Complete the vector triangle by drawing Vslip from the tip of Vtrans to the tip of VA and drawing Vtrans from the tail of VA to the intersection of the Vslip construction line. Y
0
4.
From the velocity triangle we have: Velocity scale factor: VA2slip 1.134 in kv Vtrans 1.599 in kv VA3slip 2.048 in kv VA3 2.598 in kv
5.
5 in sec
1
1.599
in
VA2slip 5.670 Vtrans 7.995
sec
2.598
in sec
VA3slip 10.240 VA3 12.990
V trans
in
V A2
in
V A3
VA3 b
The relative slip velocity is
ω 9.665 Vslip VA3slip VA2slip
V A2slip
1.134
sec V A3slip
in sec
Determine the angular velocity of link 3 using equation 6.7. ω
6.
kv
rad
CCW
sec Vslip 4.570
in sec
2.048
X
5 in/sec
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-21a-1
PROBLEM 6-21a Statement:
The linkage in Figure P6-6b has L1 = 61.9, L2 = 15, L3 = 45.8, L4 = 18.1, L5 = 23.1 mm. 2 is 68.3 deg in the xy coordinate system, which is at -23.3 deg in the XY coordinate system. The X component of O2C is 59.2 mm. Find, for the position shown, the velocity ratio VI5,6/VI2,3 and the mechanical advantage from link 2 to link 6. Use the velocity difference graphical method.
Given:
Link lengths:
Solution:
Link 1
d 61.9 mm
Link 2
a 15.0 mm
Link 3
b 45.8 mm
Link 4
c 18.1 mm
Link 5
e 23.1 mm
Offset
f 59.2 mm
from O2
Crank angle:
θ 45 deg
Coordinate rotation angle
α 23.3 deg Global XY system to local xy system
Global XY system
See Figure P6-6b and Mathcad file P0621a.
1.
Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest.
2.
Choose an arbitrary value for the magnitude of the velocity at I2,3 (point A). Let link 2 rotate CCW. VA 10
y 5,6 6 2,3
mm
A
sec
Direction of VCB C
45.0° 3
2
θVC 45 deg 90 deg
1
5
X
O2 23.3°
B Direction of VBA
1
θVC 135.000 deg 3.
Direction of VC
Direction of VA Y
4
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B, the magnitude of the relative velocity VBA, and the angular velocity of link 3. The equation to be solved graphically is
O4 x
1 Direction of VB
VB = VA + VBA a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line. 4.
From the velocity triangle we have: VA
Velocity scale factor:
kv
5 mm sec
1
0
5 mm/sec
2.053
in
VB 2.248 in kv VB 11.2
Y
mm sec
θVB ( 360 167.558 ) deg
VBA
X 167.553°
VB 2.248
DESIGN OF MACHINERY - 5th Ed.
5.
SOLUTION MANUAL 6-21a-2
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point C. The equation to be solved graphically is VC = VB + VCB a. Choose a convenient velocity scale and layout the (now) known vector VB. b. From the tip of VB, draw a construction line with the direction of VCB, magnitude unknown. c. From the tail of VB, draw a construction line with the direction of VC, magnitude unknown. d. Complete the vector triangle by drawing VCB from the tip of VB to the intersection of the VC construction line and drawing VC from the tail of VB to the intersection of the VCB construction line. VA
Y 0
V BA
5 mm/sec
X VB 1.128 VC V CB
6.
From the velocity triangle we have: kv
Velocity scale factor:
VC 1.128 in kv
7.
The ratio V
/V I5,6
8.
is I2,3
VC
5 mm sec
1
in
VC 5.6
mm sec
θVC 270 deg
0.564
VA
Use equations 6.12 and 6.13 to derive an expression for the mechanical advantage for this linkage where the input is a rotating crank and the output is a slider. Fin =
Fout =
mA =
Tin rin
Pout Vout Fout Fin
Pin
=
rin ωin =
=
=
Pin VA
Pout VC Pout VA VC Pin
mA
VA VC
mA 1.773
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-21b-1
PROBLEM 6-21b Statement:
The linkage in Figure P6-6b has L1 = 61.9, L2 = 15, L3 = 45.8, L4 = 18.1, L5 = 23.1 mm. 2 is 68.3 deg in the xy coordinate system, which is at -23.3 deg in the XY coordinate system. The X component of O2C is 59.2 mm. Find, for the position shown, the velocity ratio VI5,6/VI2,3 and the mechanical advantage from link 2 to link 6. Use the instant center graphical method.
Given:
Link lengths:
Solution: 1.
Link 1
d 61.9 mm
Link 2
a 15.0 mm
Link 3
b 45.8 mm
Link 4
c 18.1 mm
Link 5
e 23.1 mm
Offset
f 59.2 mm
from O2
Crank angle:
θ 45 deg
Coordinate rotation angle
α 23.3 deg Global XY system to local xy system
Global XY system
See Figure P6-6b and Mathcad file P0621b.
Draw the linkage to scale in the position given, find the instant centers, distances from the pin joints to the instant centers and the angles that links 3 and 4 make with the x axis. From the layout:
2.
AI13 44.594 mm
BI13 50.121 mm
BI15 22.683 mm
CI15 11.377 mm
1,3
5,6
Choose an arbitrary value for the magnitude of the velocity at I2,3 (point A). Let link 2 rotate CCW. VA 10
6 1,5 2,3 A
mm 2
sec
50.121
C
3
1
5
22.683
X
O2
θVC θ 90 deg
B 1
θVC 135.000 deg 3.
11.377
44.594 Y
4
Determine the angular velocity of link 3 using equation 6.9a.
O4 1
ω
VA
ω 0.224
AI13
rad
CW
sec
4.
Determine the magnitude of the velocity at point B using equation 6.9b. Determine its direction by inspection. mm VB BI13 ω VB 11.239 sec
5.
Determine the angular velocity of link 5 using equation 6.9a. ω
6.
VB
ω 0.495
BI15
The ratio V
/V I5,6
8.
CW
sec
Determine the magnitude of the velocity at point C using equation 6.9b. Determine its direction by inspection. VC CI15 ω
7.
rad
is I2,3
VC VA
VC 5.637
mm
downward
sec
0.56
Use equations 6.12 and 6.13 to derive an expression for the mechanical advantage for this linkage where the input is a rotating crank and the output is a slider.
DESIGN OF MACHINERY - 5th Ed.
Fin =
Fout =
mA =
Tin rin
Pout Vout Fout Fin
Pin
=
rin ωin =
=
SOLUTION MANUAL 6-21b-2
=
Pin VA
Pout VC Pout VA VC Pin
mA
VA VC
mA 1.77
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-22a-1
PROBLEM 6-22a Statement:
Given:
Solution: 1.
The linkage in Figure P6-6d has L2 = 15, L3 = 40.9, L5 = 44.7 mm. 2 is 24.2 deg in the XY coordinate system. Find, for the position shown, the velocity ratio VI5,6/VI2,3 and the mechanical advantage from link 2 to link 6. Use the velocity difference graphical method. Link lengths: Link 2
a 15.0 mm
Link 3
b 40.9 mm
Link 5
c 44.7 mm
Offset
f 0 mm
Crank angle:
θ 24.2 deg
from O2
See Figure P6-6d and Mathcad file P0622a.
Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest.
4 1
C
Direction of VBA
3
Direction of VA
5,6 A
5
2
O2
6 B 1
2,3
1
Direction of VB
2.
Choose an arbitrary value for the magnitude of the velocity at I2,3 (point A). Let link 2 rotate CCW. VA 10
3.
mm
θVC θ 90 deg
sec
θVC 114.200 deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B, the magnitude of the relati velocity VBA, and the angular velocity of link 3. The equation to be solved graphically is VB = VA + VBA a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line. VA Y 0
5 mm/sec
V BA X
VB 1.073
DESIGN OF MACHINERY - 5th Ed.
4.
SOLUTION MANUAL 6-22a-2
From the velocity triangle we have: kv
Velocity scale factor:
VB 1.073 in kv
5.
The ratio V
/V I5,6
6.
is I2,3
VB
5 mm sec
1
in
VB 5.4
mm sec
θVB 180 deg
0.54
VA
Use equations 6.12 and 6.13 to derive an expression for the mechanical advantage for this linkage where the input is a rotating crank and the output is a slider. Fin =
Fout =
mA =
Tin rin
Pout Vout
Fout Fin
Pin
=
rin ωin
=
=
=
Pin VA
Pout VB
Pout VA VB Pin
mA
VA VB
mA 1.86
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-22b-1
PROBLEM 6-22b Statement:
Given:
Solution: 1.
The linkage in Figure P6-6d has L2 = 15, L3 = 40.9, L5 = 44.7 mm. 2 is 24.2 deg in the XY coordinate system. Find, for the position shown, the velocity ratio VI5,6/VI2,3 and the mechanical advantage from link 2 to link 6. Use the instant center graphical method. Link lengths: Link 2 Link 3 a 15.0 mm b 40.9 mm Link 5
c 44.7 mm
Crank angle:
θ 24.2 deg
f 0 mm
Offset
from O2
See Figure P6-6d and Mathcad file P0622b.
Draw the linkage to scale in the position given, find the instant centers, distances from the pin joints to the instant centers and the angles that links 3 and 4 make with the x axis. 4 C
1
3
1,5
48.561
26.075
5,6 A
5
2
O2
6 B 1
2,3
1
From the layout: AI15 48.561 mm 2.
Choose an arbitrary value for the magnitude of the velocity at I2,3 (point A). Let link 2 rotate CCW. VA 10
3.
mm
θVC θ 90 deg
sec
VA
ω 0.206
AI15
The ratio V
/V I5,6
6.
rad
CW
sec
Determine the magnitude of the velocity at point B using equation 6.9b. Determine its direction by inspection. VB BI15 ω
5.
θVC 114.200 deg
Determine the angular velocity of link 3 using equation 6.9a. ω
4.
BI15 26.075 mm
is I2,3
VB VA
VB 5.370
mm
to the left
sec
0.54
Use equations 6.12 and 6.13 to derive an expression for the mechanical advantage for this linkage where the input is a rotating crank and the output is a slider. Fin =
mA =
Tin rin Fout Fin
=
=
Pin rin ωin
=
Pout VA VC Pin
Pin
Fout =
VA mA
VA VB
Pout Vout
mA 1.86
=
Pout VC
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-23-1
PROBLEM 6-23 Statement:
Generate and draw the fixed and moving centrodes of links 1 and 3 for the linkage in Figure P6-7a
Solution:
See Figure P6-7a and Mathcad file P0623.
1.
Draw the linkage to scale, find the instant center I1,3, and repeat for several positions of the linkage. The locus of points I1,3 is the fixed centrode.
FIXED CENTRODE
2.
Invert the linkage, grounding link 3. Draw the linkage to scale, find the instant center I1,3, and repeat for several positions of the linkage. The locus of points I1,3 is the moving centrode. (See next page.)
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-23-2
MOVING CENTRODE
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-24-1
PROBLEM 6-24 Statement:
The linkage in Figure P6-8a has the dimensions and crank angle given below. Find f, VA, and VB for the position shown for = 15 rad/sec clockwise (CW). Use the velocity difference graphical method.
Given:
Link lengths:
Crank angle:
Link 2 (O2 to A)
a 116 mm
Link 3 (A to B)
b 108 mm
Link 4 (B to O4)
c 110 mm
Link 1 (O2 to O4)
d 174 mm
Coordinate rotation angle Solution: 1.
θ 37 deg
Global XY system
Input crank angular velocity ω 15 rad sec
α 25 deg
1
Global XY system to local xy system
See Figure P6-8a and Mathcad file P0624.
Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest. Y
Direction of VA A
37.000° 70.133°
2 3
X
O2 d
22.319° B
Direction of VBA 4
Direction of VB
2.
Use equation 6.7 to calculate the magnitude of the velocity at point A. VA a ω
3.
O4
VA 1740.0
mm sec
θA 37 deg 90 deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B, the magnitude of the relati velocity VBA, and the angular velocity of link 3. The equation to be solved graphically is VB = VA + VBA a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-24-2
1000 mm/sec
0
Y
X 0.952 53.000° VB VA
VBA
1.498 112.319°
4.
From the velocity triangle we have: Velocity scale factor:
5.
1000 mm sec
1
in mm
VB 0.952 in kv
VB 952.0
VBA 1.498 in kv
VBA 1498.0
sec mm sec
Determine the angular velocity of link 3 using equation 6.7. ω
6.
kv
VBA b
ω 13.87
rad sec
Determine the angular velocity of link 4 using equation 6.7. ω
VB c
ω 8.655
rad sec
θB 112.319 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-25-1
PROBLEM 6-25 Statement:
The linkage in Figure P6-8a has the dimensions and crank angle given below. Find f, VA, and VB for the position shown for = 15 rad/sec clockwise (CW). Use the instant center graphical method.
Given:
Link lengths: Link 2 (O2 to A)
a 116 mm
Link 3 (A to B)
b 108 mm
Link 4 (B to O4)
c 110 mm
Link 1 (O2 to O4)
d 174 mm
Coordinate rotation angle Solution: 1.
Crank angle:
α 25 deg
θ 37 deg
Global XY system
Input crank angular velocity ω 15 rad sec
1
Global XY system to local xy system
See Figure P6-8a and Mathcad file P0625.
Draw the linkage to scale in the position given, find the instant centers, distances from the pin joints to the instant centers and the angles that links 3 and 4 make with the x axis. Y
From the layout:
2,3
A
AI13 125.463 mm
125.463
BI13 68.638 mm θ 157.681 deg
2
3
O2 1,2
X 2,4 157.681°
1,3 B
2.
3.
Use equation 6.7 and inspection of the layout to determine the magnitude and direction of the velocity at point A.
5.
4
O4
3,4 1,4
mm
VA a ω
VA 1740.0
θVA θ 90 deg
θVA 53.0 deg
sec
Determine the angular velocity of link 3 using equation 6.9a. ω
4.
68.638
VA AI13
ω 13.869
rad
CW
sec
Determine the magnitude of the velocity at point B using equation 6.9b. Determine its direction by inspection. mm
VB BI13 ω
VB 951.915
θVB θ 90 deg
θVB 247.681 deg
sec
Use equation 6.9c to determine the angular velocity of link 4. ω
VB c
ω 8.654
rad sec
CCW
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-26-1
PROBLEM 6-26 Statement:
The linkage in Figure P6-8a has the dimensions and crank angle given below. Find 4, VA, and VB for the position shown for = 15 rad/sec clockwise (CW). Use an analytical method.
Given:
Solution:
Link lengths:
Crank angle:
Link 2 (O2 to A)
a 116 mm
Link 3 (A to B)
b 108 mm
Link 4 (B to O4)
c 110 mm
Link 1 (O2 to O4)
d 174 mm
θ 62 deg
Global XY system
Input crank angular velocity ω 15 rad sec
1
CW
See Figure P6-8a and Mathcad file P0626.
1.
Draw the linkage to scale and label it.
2.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a.
y A
K1 K2
d
K1 1.5000
a
3 O2
d
62.000°
K2 1.5818
c 2
K3
2
2
2
a b c d
2
2
K3 1.7307
2 a c
B
B 2 sin θ C K1 K2 1 cos θ K3
4
A cos θ K1 K2 cos θ K3
A 0.0424 3.
B 1.7659
4.
x
C 2.0186
Use equation 4.10b to find values of 4 for the crossed circuit.
2
θ 2 atan2 2 A B
B 4 A C
θ 182.681 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d b
2
2
c d a b
K5
2
2 a b
E 2 sin θ F K1 K4 1 cos θ K5
D cos θ K1 K4 cos θ K5
5.
6.
K4 1.6111
K5 1.7280
D 2.0021 E 1.7659 F 0.0589
Use equation 4.13 to find values of 3 for the crossed circuit.
2
θ 2 atan2 2 D E
E 4 D F
θ 275.133 deg
Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18. ω
a ω sin θ θ b sin θ θ
O4
ω 13.869
rad sec
DESIGN OF MACHINERY - 5th Ed.
ω
7.
SOLUTION MANUAL 6-26-2
a ω sin θ θ c sin θ θ
ω 8.654
rad sec
Determine the velocity of points A and B for the crossed circuit using equations 6.19.
VA a ω sin θ j cos θ VA ( 1536.329 816.881i)
mm sec
VA 1740.000
mm sec
arg VA 28.000 deg
VB c ω sin θ j cos θ VB ( 44.524 950.875j)
mm sec
VB 951.917
mm sec
arg VB 87.319 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-27-1
PROBLEM 6-27 Statement:
The linkage in Figure P6-8a has the dimensions given below. Find and plot 4, VA, and VB in
Given:
the local coordinate system for the maximum range of motion that this linkage allows if 2 = 15 rad/sec clockwise (CW). Link lengths: Link 2 (O2 to A)
a 116 mm
Link 3 (A to B)
b 108 mm
Link 4 (B to O4)
c 110 mm
Link 1 (O2 to O4)
d 174 mm
ω 15 rad sec
Input crank angular velocity Solution:
1
CW
See Figure P6-8a and Mathcad file P0627.
1.
Draw the linkage to scale and label it.
2.
Determine the range of motion for this non-Grashof triple rocker using equations 4.33.
y A
2
arg1
2
2
2 a d 2
arg2
2
a d b c
2
2
a d b c
2
2 a d
arg1 1.083
b c
2 3
a d
O2
b c
2
a d
62.000°
B
arg2 0.094
θ2toggle acos arg2
4
O4
θ2toggle 95.4 deg
x
The other toggle angle is the negative of this. Thus, θ θ2toggle 1 deg θ2toggle 2 deg θ2toggle 1 deg 3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 1.5000 2
K3
d c
K2 1.5818 2
2
a b c d
2
K3 1.7307
2 a c
B θ 2 sin θ C θ K1 K2 1 cos θ K3
A θ cos θ K1 K2 cos θ K3
4.
Use equation 4.10b to find values of 4 for the crossed circuit.
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 5.
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b 2 a b
D θ cos θ K1 K4 cos θ K5
2
K4 1.6111
K5 1.7280
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-27-2
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 6.
Use equation 4.13 to find values of 3 for the crossed circuit.
θ θ 2 atan2 2 D θ E θ 7.
Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18.
a ω
a ω
ω θ
ω θ 8.
2 4 Dθ F θ
E θ
b
c
sin θ θ θ θ
sin θ θ θ θ
sin θ θ θ
sin θ θ θ
Determine the velocity of points A and B for the crossed circuit using equations 6.19.
VA θ a ω sin θ j cos θ
VAx θ Re VA θ
VAy θ Im VA θ
j cosθθ
VB θ c ω θ sin θ θ
VBx θ Re VB θ
Plot the angular velocity of the output link, 4, and the x and y components of the velocities at points A and B. ANGULAR VELOCITY OF LINK 4 60 40 Angular Velocity, rad/sec
9.
VBy θ Im VB θ
20 0
ω θ
sec rad
20 40 60 80 100 100
75
50
25
0 θ deg
25
50
75
100
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-27-3
VELOCITY COMPONENTS, POINT A 0.5
Joint Velocity, m/sec
0
VAy θ
0.5 sec m 1
1.5
2 2
1.5
1
0.5
0
0.5
VAx θ
1
1.5
2
sec m
VELOCITY COMPONENTS, POINT B 4
Joint Velocity, m/sec
3 2
VBy θ
sec 1 m
0 1 2 3 4
3
2
1
0
VBx θ
1 sec m
2
3
4
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-28-1
PROBLEM 6-28 Statement:
The linkage in Figure P6-8b has the dimensions and crank angle given below. Find 4, VA, and
Given:
VB for the position shown for 2 = 20 rad/sec counterclockwise (CCW). Use the velocity difference graphical method. Link lengths: Crank angle: Link 2 (O2 to A)
a 40 mm
Link 3 (A to B)
b 96 mm
Link 4 (B to O4)
c 122 mm
Link 1 (O2 to O4)
d 162 mm
θ 57 deg
Input crank angular velocity ω 20 rad sec
α 36 deg
Coordinate rotation angle
Global XY system
1
Global XY system to local xy system
See Figure P6-8b and Mathcad file P0628. Solution: 1. Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest. 2.
Direction of VA
Use equation 6.7 to calculate the magnitude of the velocity at point A.
VA 800.000
mm sec
2
y
B 3
A
2
X O2
θA θ 90 deg 3.
Direction of VB
Y
VA a ω
Direction of VBA
4
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B, the magnitude of the relative velocity VBA, and the angular velocity of link 3. The equation to be solved graphically is VB = VA + VBA
O4 x
a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line. 0
4.
400 mm/sec
From the velocity triangle we have: 147.000°
Velocity scale factor: kv
400 mm sec
1
VA Y
in
VB 1.790 in kv
VB 716.0
mm
1.293" V BA
sec X
θB 186.406 deg VBA 1.293 in kv 5.
mm sec
6.406°
1.790" 85.486°
Determine the angular velocity of link 3 using equation 6.7. ω
6.
VBA 517.2
VB
VBA b
ω 5.388
rad sec
Determine the angular velocity of link 4 using equation 6.7. ω
VB c
ω 5.869
rad sec
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-29-1
PROBLEM 6-29 Statement:
Given:
The linkage in Figure P6-8b has the dimensions and crank angle given below. Find 4, VA, and VB for the position shown for 2 = 20 rad/sec counterclockwise (CCW). Use the instant center graphical method. Link lengths: Crank angle: a 40 mm θ 57 deg Global XY system Link 2 (O2 to A) Input crank angular velocity b 96 mm Link 3 (A to B) 1 c 122 mm ω 20 rad sec Link 4 (B to O4) d 162 mm Link 1 (O2 to O4) Coordinate rotation angle
Solution: 1.
α 36 deg
Global XY system to local xy system
See Figure P6-8b and Mathcad file P0629.
Draw the linkage to scale in the position given, find the instant centers, distances from the pin joints to the instant centers and the angles that links 3 and 4 make with the x axis. From the layout: AI13 148.700 mm
1,3
BI13 133.192 mm
θ 96.406 deg 2.
133.192
Use equation 6.7 and inspection of the layout to determine the magnitude and direction of the velocity at point A. mm VA a ω VA 800.0 sec
θVA θ 90 deg 3.
148.700
Y 2
A
2
θVA 147.0 deg
y
B 3 X
O2 4
Determine the angular velocity of link 3 using equation 6.9a. ω
VA AI13
ω 5.380
rad
96.406°
CW
sec
O4 x
4.
5.
Determine the magnitude of the velocity at point B using equation 6.9b. Determine its direction by inspection. mm
VB BI13 ω
VB 716.6
θVB θ 90 deg
θVB 186.406 deg
sec
Use equation 6.9c to determine the angular velocity of link 4. ω
VB c
ω 5.874
rad sec
CCW
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-30-1
PROBLEM 6-30 Statement:
The linkage in Figure P6-8b has the dimensions and crank angle given below. Find 4, VA, and
Given:
VB for the position shown for 2 = 20 rad/sec counterclockwise (CCW). Use an analytical method. Link lengths: Crank angle: Link 2 (O2 to A)
a 40 mm
Link 3 (A to B)
b 96 mm
Link 4 (B to O4)
c 122 mm
Link 1 (O2 to O4)
d 162 mm α 36 deg
Coordinate rotation angle Solution:
Draw the linkage to scale and label it.
2.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 4.0500 2
K3
1
Global XY system to local xy system
Y
y
2
c
B
3
A
X 93.000 4
K2 1.3279 2
2
a b c d
2
K3 3.4336
2 a c
A 0.5992
B 1.9973
θ θ α
O4 x
B 2 sin θ
C 7.6054
Use equation 4.10b to find values of 4 for the open circuit.
θ 2 atan2 2 A B
2
B 4 A C
2 π
θ 587.614 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
E 2 sin θ F K1 K4 1 cos θ K5
2
2 a b
D cos θ K1 K4 cos θ K5
K4 1.6875
K5 2.8875
D 7.0782 E 1.9973 F 1.1265
Use equation 4.13 to find values of 3 for the open circuit.
θ 2 atan2 2 D E 6.
ω 20 rad sec
O2
C K1 K2 1 cos θ K3
5.
Input crank angular velocity
2
d
A cos θ K1 K2 cos θ K3
4.
Global XY system
See Figure P6-8b and Mathcad file P0630.
1.
3.
θ 57 deg
2
E 4 D F
2 π
θ 688.496 deg
Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18.
ω
a ω sin θ θ b sin θ θ
ω 5.385
ω
a ω sin θ θ c sin θ θ
ω 5.868
rad sec
rad sec
DESIGN OF MACHINERY - 5th Ed.
7.
SOLUTION MANUAL 6-30-2
Determine the velocity of points A and B for the open circuit using equations 6.19.
VA a ω sin θ j cos θ VA ( 798.904 41.869i )
mm sec
VA 800.000
mm
VB 715.900
mm
sec
arg VA 177.000 deg
VB c ω sin θ j cos θ VB ( 528.774 482.608j)
mm sec
sec
arg VB 137.614 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-31-1
PROBLEM 6-31 Statement:
The linkage in Figure P6-8b has the dimensions and crank angle given below. Find and plot 4, VA, and VB in the local coordinate system for the maximum range of motion that this linkage allows if 2 = 20 rad/sec counterclockwise (CCW).
Given:
Link lengths: Link 2 (O2 to A)
a 40 mm
Link 3 (A to B)
b 96 mm
Link 4 (B to O4)
c 122 mm
Link 1 (O2 to O4)
d 162 mm
ω 20 rad sec
Input crank angular velocity Solution:
1
CCW
See Figure P6-8b and Mathcad file P0631.
1.
Draw the linkage to scale and label it.
2.
Determine Grashof condition. Condition( S L P Q)
Y 2
SL S L
y B
3
A
93.000 X
2
PQ P Q
O2 4
return "Grashof" if SL PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise O4
Condition( a d b c) "Grashof"
x
Crank-rocker
θ 0 deg 1 deg 360 deg 3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 4.0500
2
d
K3
c
K2 1.3279
2
2
2 a c
K3 3.4336
A θ cos θ K1 K2 cos θ K3
2
a b c d
B θ 2 sin θ
C θ K1 K2 1 cos θ K3 4.
Use equation 4.10b to find values of 4 for the open circuit.
θ θ 2 atan2 2 A θ B θ 5.
2 4 A θ Cθ
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
K4 1.6875
2 a b
K5 2.8875
E θ 2 sin θ F θ K1 K4 1 cos θ K5
D θ cos θ K1 K4 cos θ K5
6.
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ 7.
2 4 Dθ F θ
E θ
Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18.
DESIGN OF MACHINERY - 5th Ed.
a ω
a ω
ω θ
ω θ 8.
b
c
SOLUTION MANUAL 6-31-2
sin θ θ θ θ
sin θ θ θ θ
sin θ θ θ
sin θ θ θ
Determine the velocity of points B and C for the open circuit using equations 6.19.
VA θ a ω sin θ j cos θ
VAx θ Re VA θ
VAy θ Im VA θ
j cosθθ
VB θ c ω θ sin θ θ
VBx θ Re VB θ
Plot the angular velocity of the output link, 4, and the magnitudes of the velocities at points B and C. ANGULAR VELOCITY OF LINK 4
Angular Velocity, rad/sec
10
5
ω θ
sec 0
rad
5
10
0
45
90
135
180
225
270
315
360
θ deg
VELOCITY COMPONENTS, POINT A
3
1 10 Joint Velocity, mm/sec
9.
VBy θ Im VB θ
VAx θ
VAy θ
sec
500
mm sec mm
0 500 3
1 10
0
45
90
135
180 θ deg
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-31-3
VELOCITY COMPONENTS, POINT B
3
Joint Velocity, mm/sec
1 10
600
VBx θ
VBy θ
sec mm
200
sec
200
mm 600 1 10
3
0
45
90
135
180 θ deg
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-32-1
PROBLEM 6-32 Statement:
The offset slider-crank linkage in Figure P6-8f has the dimensions and crank angle given below. Find VA, and VB for the position shown if 2 = 25 rad/sec CW. Use the velocity difference graphical method.
Given:
Link lengths:
Solution:
Link 2
a 63 mm
Crank angle:
θ 51 deg
Link 3
b 130 mm
Input crank angular velocity
Offset
c 52 mm
ω 25 rad sec
1
CW
See Figure P6-8f and Mathcad file P0632.
1.
Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest.
2.
Use equation 6.7 to calculate the magnitude of the velocity at point A. VA a ω
VA 1575.0
Direction of VB 4 B
mm
Direction of VBA
1
sec
3
θA 51 deg 90 deg 3.
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B. The equation to be solved graphically is
Direction of VA A 2 51.000°
VB = VA + VBA O2
a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line. 4.
0
From the velocity triangle we have: Velocity scale factor:
kv
VB 2.208 in kv
θVB 270 deg
1000 mm sec
1000 mm/sec
Y
1 X
in VB 2208
mm VA
sec 2.208
VB V BA
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-33-1
PROBLEM 6-33 Statement:
The offset crank-slider linkage in Figure P6-8f has the dimensions and crank angle given below. Find VA, and VB for the position shown for 2 = 25 rad/sec CW. Use the instant center graphical method.
Given:
Link lengths: Link 2
a 63 mm
Crank angle:
θ 51 deg
Link 3
b 130 mm
Offset
c 52 mm
ω 25 rad sec
Input crank angular velocity Solution: 1.
1
CW
See Figure P6-8f and Mathcad file P0633.
Draw the linkage to scale in the position given, find the instant centers, distances from the pin joints to the instant centers and the angles that links 3 and 4 make with the x axis. 166.309 4 B
1,3
1 3
A
118.639
2
O2
From the layout above: AI13 118.639 mm 2.
Use equation 6.7 and inspection of the layout to determine the magnitude and direction of the velocity at point A. mm VA a ω VA 1575.0 sec
θVA θ 90 deg 3.
θVA 39.0 deg
Determine the angular velocity of link 3 using equation 6.9a. ω
4.
BI13 166.309 mm
VA AI13
ω 13.276
rad
CCW
sec
Determine the magnitude of the velocity at point B using equation 6.9b. Determine its direction by inspection. VB BI13 ω
θVB 270 deg
VB 2207.8
mm sec
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-34-1
PROBLEM 6-34 Statement:
The offset slider-crank linkage in Figure P6-8f has the dimensions and crank angle given below. Find VA, and VB for the position shown for 2 = 25 rad/sec CW. Use an analytical method.
Given:
Link lengths: Link 2
a 63 mm
Link 3
b 130 mm
Offset
c 52 mm
θ 141 deg
Crank angle:
Local xy coordinate system Input crank angular velocity ω 25 rad sec
Solution:
See Figure P6-8f and Mathcad file P0634.
Y
1.
Draw the linkage to a convenient scale.
2.
Determine 3 and d using equations 4.16 for the crossed circuit.
4
a sin θ b
θ asin
c
1 3
d 141.160 mm
A
52.000 2
Determine the angular velocity of link 3 using equation 6.22a: ω
4.
B
θ 44.828 deg
d a cos θ b cos θ 3.
1
a cos θ ω b cos θ
ω 13.276
rad
141.000°
sec
x
Determine the velocity of pin A using equation 6.23a:
X, y
O2
VA a ω sin θ j cos θ VA ( 991.180 1224.005i )
mm sec
In the global coordinate system, 5.
VA 1575.000
mm sec
θVA arg VA 90 deg
arg VA 51.000 deg
θVA 39.000 deg
Determine the velocity of pin B using equation 6.22b:
VB a ω sin θ b ω sin θ
VB 2207.849
mm sec
VB VB In the global coordinate system,
θVB arg VB 90 deg
θVB 90.000 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-35-1
PROBLEM 6-35 Statement:
The offset crank-slider linkage in Figure P6-8f has the dimensions and crank angle given below. Find and plot VA, and VB in the global coordinate system for the maximum range of motion that this linkage allows if 2 = 25 rad/sec CW.
Given:
Link lengths: Link 2
a 63 mm
Link 3
b 130 mm ω 25 rad sec
Input crank angular velocity Solution:
c 52 mm
Offset 1
See Figure P6-8f and Mathcad file P0635.
1.
Draw the linkage to a convenient scale. The coordinate rotation angle is α 90 deg
2.
Determine the range of motion for this slider-crank linkage. 4
θ 0 deg 2 deg 360 deg 3.
a sin θ b
3
c
a b
2
ω cos θ θ cos θ
O2
Determine the x and y components of the velocity of pin A using equation 6.23a:
VA θ a ω sin θ j cos θ
VAx θ Re VA θ
VAy θ Im VA θ
In the global coordinate system,
VAX θ VAy θ 6.
VAY θ VAx θ
Determine the velocity of pin B using equation 6.22b:
VBx θ a ω sin θ b ω θ sin θ θ In the global coordinate system,
VBY θ VBx θ 7.
A
52.000
Determine the angular velocity of link 3 using equation 6.22a: ω θ
5.
B
1
Determine 3 using equations 4.16 for the crossed circuit. θ θ asin
4.
Y
Plot the x and y components of the velocity of A. (See next page.)
2 x
X, y
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-35-2
VELOCITY OF POINT A 2000
Velocity, mm/sec
1000
VAX θ
VAY θ
sec mm 0
sec mm
1000
2000
0
60
120
180
240
300
360
θ deg
Plot the velocity of point B. VELOCITY OF POINT B 2000
1000 Velocity, mm/sec
7.
0
VBY θ
sec mm 1000
2000
3000
0
60
120
180 θ deg
240
300
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-36-1
PROBLEM 6-36 Statement:
Given:
The linkage in Figure P6-8d has the dimensions and crank angle given below. Find VA, VB, and Vbox for the position shown for 2 = 30 rad/sec clockwise (CW). Use the velocity difference graphical method. Link lengths: Link 2 Link 3 a 30 mm b 150 mm Link 4
c 30 mm
Crank angle:
θ 58 deg
Global XY system ω 30 rad sec
Input crank angular velocity Solution:
d 150 mm
Link 1 1
CW
See Figure P6-8d and Mathcad file P0636.
1.
Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest.
2.
Use equation 6.7 to calculate the magnitude of the velocity at point A. mm VA a ω VA 900.000 sec
Vbox 1
Direction of VBA
θA 58 deg 90 deg 3.
Direction of VA
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B, the magnitude of the relative velocity VBA, and the angular velocity of link 3. The equation to be solved graphically is
B
3
A
Direction of VB
2 O4
O2
4
VB = VA + VBA a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line. Y
4.
From the velocity triangle we have:
Direction of VBA
VB VA
X 0
VB 900.000
θB 32 deg 5.
VB
sec VBA 0
in
VA
sec
Determine the angular velocity of links 3 and 4 using equation 6.7. ω
6.
500 mm/sec
32.000°
mm
VBA b
ω 0.000
rad sec
ω
VB c
ω 30.000
rad sec
Determine the magnitude of the vector Vbox . This is a special case Grashof mechanism in the parallelogram configuration. Link 3 does not rotate, therefore all points on link 3 have the same velocity. The velocity Vbox is the horizontal component of VA. mm Vbox VA cos θA Vbox 763.243 sec
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-37-1
PROBLEM 6-37 Statement:
The linkage in Figure P6-8d has the dimensions and effective crank angle given below. Find VA,
Given:
VB, and Vbox in the global coordinate system for the position shown for 2 = 30 rad/sec CW. Use an analytical method. Link lengths:
Solution: 1.
Link 2
a 30 mm
Link 3
b 150 mm
Link 4
c 30 mm
Link 1
d 150 mm
Crank angle:
θ 58 deg
Input crank angular velocity
ω 30 rad sec
1
See Figure P6-8d and Mathcad file P0637. Vbox
Draw the linkage to scale and label it. 1
2.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 5.0000 2
K3
2
c
2
2
2
K3 1.0000
2 a c
C K1 K2 1 cos θ K3
A cos θ K1 K2 cos θ K3 A 6.1197 3.
B 1.6961
B 2 sin θ
C 2.8205
Use equation 4.10b to find values of 4 for the open circuit.
2
θ 2 atan2 2 A B
B 4 A C
θ 302.000 deg
θ θ 360 deg 4.
θ 662.000 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b 2 a b
E 2 sin θ F K1 K4 1 cos θ K5
D cos θ K1 K4 cos θ K5
5.
2
K4 1.0000
K5 5.0000
D 8.9402 E 1.6961 F 0.0000
Use equation 4.13 to find values of 3 for the open circuit.
2
θ 2 atan2 2 D E
E 4 D F
θ θ 360 deg 6.
O4
O2
K2 5.0000
a b c d
B
3
A
d
θ 360.000 deg θ 0.000 deg
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18. ω
a ω sin θ θ b sin θ θ
ω 0.000
rad sec
4
DESIGN OF MACHINERY - 5th Ed.
ω 7.
SOLUTION MANUAL 6-37-2
a ω sin θ θ c sin θ θ
ω 30.000
rad sec
Determine the velocity of points A and B for the open circuit using equations 6.19.
VA a ω sin θ j cos θ VA ( 763.243 476.927j)
mm sec
VA 900.000
mm
VB 900.000
mm
sec
arg VA 32.000 deg
VB c ω sin θ j cos θ VB ( 763.243 476.927j) 8.
mm sec
sec
arg VB 32.000 deg
Determine the velocity Vbox. Since link 3 does not rotate (this is a special case Grashof linkage in the parallelogram mode), all points on it have the same velocity. Therefore, Vbox VA
Vbox 900.000
mm sec
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-38-1
PROBLEM 6-38 Statement:
The linkage in Figure P6-8d has the dimensions and effective crank angle given below. Find and plot VA, VB, and Vbox in the global coordinate system for the maximum range of motion that this linkage allows if 2 = 30 rad/sec CW.
Given:
Link lengths: Link 2
a 30 mm
Link 3
b 150 mm
Link 4
c 30 mm
Link 1
d 150 mm
ω 30 rad sec
Input crank angular velocity Solution:
1
See Figure P6-8d and Mathcad file P0638.
1.
Draw the linkage to scale and label it.
2.
Determine the range of motion for this special-case Grashof double crank.
Vbox 1
θ 0 deg 2 deg 360 deg 3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 5.0000 2
K3
B
3
2 O4
O2
c
K2 5.0000 2
2
a b c d
2
K3 1.0000
2 a c
d
A
A θ cos θ K1 K2 cos θ K3
B θ 2 sin θ
C θ K1 K2 1 cos θ K3 4.
Use equation 4.10b to find values of 4 for the open circuit.
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 5.
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
2 a b
K4 1.0000
K5 5.0000
E θ 2 sin θ F θ K1 K4 1 cos θ K5
D θ cos θ K1 K4 cos θ K5
6.
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ 7.
2 4 Dθ F θ
E θ
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18.
a ω
a ω
ω θ
ω θ
b
c
sin θ θ θ θ
sin θ θ θ θ
sin θ θ θ
sin θ θ θ
4
DESIGN OF MACHINERY - 5th Ed.
8.
SOLUTION MANUAL 6-38-2
Determine the velocity of points A and B for the open circuit using equations 6.19.
VA θ a ω sin θ j cos θ VA θ VA θ
j cosθθ
VB θ c ω θ sin θ θ
VB θ VB θ 9.
Plot the angular velocity of the output link, 4, and the magnitudes of the velocities at points A and B. Since this is a special-case Grashof linkage in the parallelogram configuration, 3 = 0 and 4 = 2 for all values of 2. Similarly, VA, VB, and Vbox all have the same constant magnitude through all values of 2. ω( 5 deg) 30.000
rad sec
VA( 5 deg) 900.000
mm
VB( 5 deg) 900.000
mm
sec
sec
ω( 135 deg) 30.000
rad sec
VA( 135 deg) 900.000
mm
VB( 135 deg) 900.000
mm
sec
sec
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-39-1
PROBLEM 6-39 Statement:
The linkage in Figure P6-8g has the dimensions and crank angle given below. Find 4, VA, and
Given:
VB for the position shown for 2 = 15 rad/sec clockwise (CW). Use the velocity difference graphical method. Link lengths: Link 2 (O2 to A)
a 49 mm
Link 2 (O2 to C)
a' 49 mm
Link 3 (A to B)
b 100 mm
Link 5 (C to D)
b' 100 mm
Link 4 (B to O4)
c 153 mm
Link 6 (D to O6)
c' 153 mm
Link 1 (O2 to O4)
d 87 mm
Link 1 (O2 to O6)
d' 87 mm
θ 29 deg
Crank angle:
Solution: 1.
Global XY system
Input crank angular velocity
ω 15 rad sec
Coordinate rotation angle
α 119 deg
1
CW
Global XY system to local xy system
See Figure P6-8g and Mathcad file P0639.
Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest.
Y
Direction of VB
Direction of VBA O6
B 3 29.000°
A
2
4 C
6
O2
X
Direction of VA
5
D
O4 2.
Use equation 6.7 to calculate the magnitude of the velocity at point B. VB a ω
3.
VB 735.0
mm sec
θB 29 deg 90 deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point A, the magnitude of the relati velocity VAB, and the angular velocity of link 3. The equation to be solved graphically is VA = VB + VAB a. Choose a convenient velocity scale and layout the known vector VB. b. From the tip of VB, draw a construction line with the direction of VAB, magnitude unknown. c. From the tail of VB, draw a construction line with the direction of VA, magnitude unknown. d. Complete the vector triangle by drawing VAB from the tip of VB to the intersection of the VA construction line and drawing VA from the tail of VB to the intersection of the VAB construction line.
DESIGN OF MACHINERY - 5th Ed.
0
SOLUTION MANUAL 6-39-2
500 mm/sec
Y 0.124°
1.517 VB
X V BA
VA
4.
From the velocity triangle we have: Velocity scale factor:
VA 1.517 in kv
5.
kv
500 mm sec
VA 758.5
1
in mm sec
Determine the angular velocity of link 4 using equation 6.7. ω
VA c
ω 4.958
rad sec
θA 0.124 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-40-1
PROBLEM 6-40 Statement:
The linkage in Figure P6-8g has the dimensions and crank angle given below. Find 4, VA, and
Given:
VB for the position shown for 2 = 15 rad/sec clockwise (CW). Use the instant center graphical method. Link lengths: Link 2 (O2 to A)
a 49 mm
Link 2 (O2 to C)
a' 49 mm
Link 3 (A to B)
b 100 mm
Link 5 (C to D)
b' 100 mm
Link 4 (B to O4)
c 153 mm
Link 6 (D to O6)
c' 153 mm
Link 1 (O2 to O4)
d 87 mm
Link 1 (O2 to O6)
d' 87 mm
θ 29 deg
Crank angle:
Solution: 1.
Global XY system
Input crank angular velocity
ω 15 rad sec
Coordinate rotation angle
α 119 deg
1
CW
Global XY system to local xy system
See Figure P6-8g and Mathcad file P0640.
Y
Draw the linkage to scale in the position given, find the instant centers, distances from the pin joints to the instant centers and the angles that links 3 and 4 make with the x axis.
97.094 O6 B 3
From the layout: AI13 97.094 mm
BI13 100.224 mm
1,3
100.224
θ 89.876 deg 2.
3.
5.
6
89.876° 5
X
D
O4
θVA 61.0 deg
Determine the angular velocity of link 3 using equation 6.9a. ω
4.
A
O2
C
Use equation 6.7 and inspection of the layout to determine the magnitude and direction of the velocity at point A. mm VA a ω VA 735.0 sec
θVA θ 90 deg
2
4
VA AI13
ω 7.570
rad
CW
sec
Determine the magnitude of the velocity at point B using equation 6.9b. Determine its direction by inspection. mm
VB BI13 ω
VB 758.694
θVB θ 90 deg
θVB 0.124 deg
sec
Use equation 6.9c to determine the angular velocity of link 4. ω
VB c
ω 4.959
rad sec
CW
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-41-1
PROBLEM 6-41 Statement:
The linkage in Figure P6-8g has the dimensions and crank angle given below. Find 4, VA, and VB for the position shown for 2 = 15 rad/sec clockwise (CW). Use an analytical method.
Given:
Link lengths: Link 2 (O2 to A)
a 49 mm
Link 2 (O2 to C)
a' 49 mm
Link 3 (A to B)
b 100 mm
Link 5 (C to D)
b' 100 mm
Link 4 (B to O4)
c 153 mm
Link 6 (D to O6)
c' 153 mm
Link 1 (O2 to O4)
d 87 mm
Link 1 (O2 to O6)
d' 87 mm
θ 148 deg Local xy system
Crank angle:
ω 15 rad sec
Input crank angular velocity Solution:
1
See Figure P6-8g and Mathcad file P0641.
1.
Draw the linkage to scale and label it.
2.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
Y
d
K2
a
K1 1.7755 2
K3
O6
d
B 3
c
K2 0.5686 2
2
a b c d
2
K3 1.5592
2 a c
2
4 C
B 2 sin θ C K1 K2 1 cos θ K3 3.
B 1.0598
D O4 x
Use equation 4.10b to find values of 4 for the crossed circuit.
2
θ 2 atan2 2 A B 4.
5
C 4.6650
B 4 A C
θ 208.876 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
2 a b
E 2 sin θ F K1 K4 1 cos θ K5
D cos θ K1 K4 cos θ K5
5.
K5 0.3509
D 3.0104 E 1.0598 F 2.2367
Use equation 4.13 to find values of 3 for the crossed circuit.
2
θ 2 atan2 2 D E 6.
K4 0.8700
E 4 D F
θ 266.892 deg
Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18. ω
a ω sin θ θ b sin θ θ
ω 7.570
rad sec
X
6
O2 148.000°
A cos θ K1 K2 cos θ K3
A 0.5821
A
y
DESIGN OF MACHINERY - 5th Ed.
ω 7.
SOLUTION MANUAL 6-41-2
a ω sin θ θ c sin θ θ
ω 4.959
rad sec
Determine the velocity of points B and A for the crossed circuit using equations 6.19.
VA a ω sin θ j cos θ VA ( 389.491 623.315i)
mm sec
VA 735.000
mm
VB 758.694
mm
sec
arg VA 58.000 deg
VB c ω sin θ j cos θ VB ( 366.389 664.362j)
mm sec
sec
arg VB 118.876 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-42-1
PROBLEM 6-42 Statement:
The linkage in Figure P6-8g has the dimensions and crank angle given below. Find and plot 4, VA, and VB in the local coordinate system for the maximum range of motion that this linkage allows if 2 = 15 rad/sec clockwise (CW).
Given:
Link lengths: Link 2 (O2 to A)
a 49 mm
Link 2 (O2 to C)
a' 49 mm
Link 3 (A to B)
b 100 mm
Link 5 (C to D)
b' 100 mm
Link 4 (B to O4)
c 153 mm
Link 6 (D to O6)
c' 153 mm
Link 1 (O2 to O4)
d 87 mm
Link 1 (O2 to O6)
d' 87 mm
ω 15 rad sec
Input crank angular velocity Solution:
1
Y
See Figure P6-8g and Mathcad file P0642.
1.
Draw the linkage to scale and label it.
2.
Determine the range of motion for this non-Grashof triple rocker using equations 4.37. 2
arg1
2
2
a d b c
2
2 a d 2
arg2
O6
2
2
a d b c
2
2 a d
θ2toggle acos arg1
b c
B 3
arg1 0.840
a d
C
b c
O2
arg2 6.338
a d
A
2
4
2
5 D
θ2toggle 32.9 deg
O4 x
The other toggle angle is the negative of this. Thus,
θ θ2toggle 1 deg θ2toggle 2 deg 359 deg θ2toggle
3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 1.7755
2
d
K3
c
2
a b c d
K3 1.5592
B θ 2 sin θ
2
2 a c
K2 0.5686
A θ cos θ K1 K2 cos θ K3
2
C θ K1 K2 1 cos θ K3 4.
Use equation 4.10b to find values of 4 for the crossed circuit.
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 5.
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b 2 a b
D θ cos θ K1 K4 cos θ K5
F θ K1 K4 1 cos θ K5
2
K4 0.8700
K5 0.3509
E θ 2 sin θ
X
6
y
DESIGN OF MACHINERY - 5th Ed.
6.
SOLUTION MANUAL 6-42-2
Use equation 4.13 to find values of 3 for the crossed circuit.
θ θ 2 atan2 2 D θ E θ 7.
Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18.
a ω
a ω
ω θ
ω θ
8.
2 4 Dθ F θ
E θ
b
c
sin θ θ θ θ
sin θ θ θ θ
sin θ θ θ
sin θ θ θ
Determine the velocity of points A and B for the crossed circuit using equations 6.19.
VBx θ Re VB θ
VB θ a ω sin θ j cos θ
VBy θ Im VB θ
j cosθθ
VA θ c ω θ sin θ θ
VAx θ Re VA θ
Plot the angular velocity of the output link, 4, and the x and y components of the velocities at points B and A. ANGULAR VELOCITY OF LINK 4 60 40 Angular Velocity, rad/sec
9.
VAy θ Im VA θ
20 0
ω θ
sec rad
20 40 60 80 100
0
45
90
135
180 θ deg
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-42-3
VELOCITY COMPONENTS, POINT B 1000
Joint Velocity, mm/sec
500
VBx θ
VBy θ
sec mm 0
sec mm
500
1000
0
45
90
135
180
225
270
315
360
315
360
θ deg
VELOCITY COMPONENTS, POINT A 4000
Joint Velocity, mm/sec
3000 2000
VAx θ
VAy θ
sec mm 1000 sec mm
0
1000 2000 3000
0
45
90
135
180 θ deg
225
270
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-43-1
PROBLEM 6-43 Statement:
The 3-cylinder radial compressor in Figure P6-8c has the dimensions and crank angle given below Find piston velocities V6, V7, and V8 for the position shown for 2 = 15 rad/sec clockwise (CW). Use the velocity difference graphical method.
Given:
Link lengths:
Solution: 1. 2.
Link 2
a 19 mm
Links 3, 4, and 5
b 70 mm
c 0 mm
Offset
Crank angle:
θ 53 deg Global XY system
Input crank angular velocity
ω 15 rad sec
Cylinder angular spacing
α 120 deg
1
CW
See Figure P6-8c and Mathcad file P0643.
Direction of V62 Direction of V8
Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest. Use equation 6.7 to calculate the magnitude of the velocity at rod pin on link 2.
8 6
V2 a ω
V2 285.0
3
mm
5 X
Direction of V6
1 4
Use equation 6.5 to (graphically) determine the magnitude of the velocity at pistons 6, 7, and 8.. The equations to be solved graphically are V7 = V2 + V72
2
sec
θ2 53 deg 90 deg 3.
Direction of V82
Y
V6 = V2 + V62
Direction of V72 7
V8 = V2 + V82 Direction of V7
a. Choose a convenient velocity scale and layout the known vector V2. b. From the tip of V2, draw a construction line with the direction of V72, magnitude unknown. c. From the tail of V2, draw a construction line with the direction of V7, magnitude unknown. d. Complete the vector triangle by drawing V72 from the tip of V2 to the intersection of the V7 construction line and drawing V7 from the tail of V2 to the intersection of the V72 construction line. e. Repeat for V6 and V8. 0
4.
200 mm/sec
From the velocity triangle we have: Velocity scale factor:
kv
V7 1.046 in kv
200 mm sec
1
Y
in
V62
V7 209.2
1.046
V82
V6 83.4
V7
mm sec
θV6 150 deg V8 292.6
X
V8
sec
V2
V8 1.463 in kv
V6
mm
θV7 270 deg V6 0.417 in kv
0.417
1.463
mm sec
θV8 210 deg
V72
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-44-1
PROBLEM 6-44 Statement:
The 3-cylinder radial compressor in Figure P6-8c has the dimensions and crank angle given below. Find V6, V7, and V8 for the position shown for 2 = 15 rad/sec clockwise (CW). Use an analytical method.
Given:
Link lengths: Link 2
a 19 mm b 70 mm
Links 3, 4, and 5
Solution: 1. 2.
Input crank angular velocity
ω 15 rad sec
Cylinder angular spacing
α 120 deg
Draw the linkage to scale and label it.
Crank angle:
θ 37 deg
CW
Local x'y' system
Y
x''
θ 170.599 deg
x'''
y'''
Determine 4 and d' using equation 4.17.
a sin θ b
8 6
c
3
π
2 1
d' 3.316 in
4 y''
Determine the angular velocity of link 4 using equation 6.22a:
5 X, y'
(x'y' system)
d' a cos θ b cos θ
7
a cos θ ω ω b cos θ 4.
c 0 mm
See Figure P6-8c and Mathcad file P0644.
θ asin
3.
1
Offset
ω 3.296
rad sec 37.000°
Determine the velocity of the rod pin on link 2 using equation 6.23a:
x'
V2 a ω sin θ j cos θ V2 ( 171.517 227.611i)
mm sec
In the global coordinate system, 5.
V2 285.000
mm sec
arg V2 53.000 deg
θV2 arg V2 90 deg
θV2 143.000 deg
Determine the velocity of piston 7 using equation 6.22b:
V7 a ω sin θ b ω sin θ V7 209.204
mm
V7 209.204
sec
In the global coordinate system, 6.
sec
arg V7 0.000 deg
θV7 arg V7 90 deg
θV7 90.000 deg
Determine 3 and d'' using equation 4.17. θ θ 120 deg
θ 157.000 deg
a sin θ c π b
θ asin
d'' a cos θ b cos θ 7.
mm
θ 173.912 deg d'' 2.052 in
Determine the angular velocity of link 5 using equation 6.22a: ω
a cos θ ω b cos θ
ω 3.769
rad sec
(x''y'' system)
DESIGN OF MACHINERY - 5th Ed.
8.
SOLUTION MANUAL 6-44-2
Determine the velocity of the rod pin on link 2 using equation 6.23a:
V2 a ω sin θ j cos θ V2 ( 111.358 262.344i)
mm sec
In the global coordinate system, 9.
V2 285.000
arg V2 67.000 deg
mm sec
θV2 arg V2 150 deg
θV2 217.000 deg
Determine the velocity of piston 6 using equation 6.22b:
V6 a ω sin θ b ω sin θ mm
V6 83.378
V6 83.378
sec
In the global coordinate system,
arg V6 0.000 deg
mm sec
θV6 arg V6 150 deg
θV6 150.000 deg
10. Determine 5 and d''' using equation 4.17. θ θ 120 deg
θ 277.000 deg
a sin θ b
θ asin
c
π
(x'''y''' system)
θ 195.629 deg
d''' a cos θ b cos θ
d''' 2.745 in
11. Determine the angular velocity of link 5 using equation 6.22a: ω
a cos θ ω b cos θ
ω 0.515
rad sec
12. Determine the velocity of the rod pin on link 2 using equation 6.23a:
V2 a ω sin θ j cos θ V2 ( 282.876 34.733i )
mm sec
In the global coordinate system,
V2 285.000
mm sec
θV2 arg V2 30 deg
arg V2 173.000 deg
θV2 143.000 deg
13. Determine the velocity of piston 8 using equation 6.22b:
V8 a ω sin θ b ω sin θ V8 292.592
mm sec
In the global coordinate system,
V8 292.592
mm sec
θV8 arg V8 30 deg
arg V8 180.000 deg
θV8 210.000 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-45-1
PROBLEM 6-45 Statement:
The 3-cylinder radial compressor in Figure P6-8c has the dimensions and crank angle given below. Find and plot V6, V7, and V8 for one revolution of the crank if 2 = 15 rad/sec clockwise (CW).
Given:
Link lengths:
Solution: 1. 2.
Link 2
a 19 mm
Links 3, 4, and 5
b 70 mm
c 0 mm
Offset
Input crank angular velocity
ω 15 rad sec
Cylinder angular spacing
α 120 deg
1
CW
See Figure P6-8c and Mathcad file P0645.
Draw the linkage to scale and label it. Note that there are three local coordinate systems.
Y
x''
x'''
y''' 8
Determine the range of motion for this slider-crank linkage. This will be the same in each coordinate frame.
6
3
5 X, y'
θ 0 deg 2 deg 360 deg 3.
1
a sin θ θ θ asin b 4.
c
4
π
(x'y' system)
y'' 7
Determine the angular velocity of link 3 using equation 6.22a:
ω θ 5.
2
Determine 4 using equation 4.17.
a b
ω cos θ θ cos θ
x'
Determine the velocity of piston 7 using equation 6.22b:
V7 θ a ω sin θ b ω θ sin θ θ 6.
Determine 3 using equation 4.17.
a sin θ α c π b
θ θ asin 7.
Determine the angular velocity of link 3 using equation 6.22a:
ω θ 8.
(x''y'' system)
a cos θ α ω b cos θ θ
Determine the velocity of piston 6 using equation 6.22b:
V6 θ a ω sin θ α b ω θ sin θ θ 9.
Determine 5 and d''' using equation 4.17.
a sin θ 2 α c π b
θ θ asin
10. Determine the angular velocity of link 5 using equation 6.22a:
ω θ
a cos θ 2 α ω b cos θ θ
(x'''y''' system)
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-45-2
11. Determine the velocity of piston 8 using equation 6.22b:
V8 θ a ω sin θ 2 α b ω θ sin θ θ 12. Plot the velocities of pistons 6, 7, and 8.
VELOCITY OF PISTONS 6, 7, AND 8 400
Velocity, mm/sec
V6 θ
V7 θ
V8 θ
sec 200 mm sec mm
0
sec mm 200
400
0
60
120
180 θ deg Crank Angle, deg
240
300
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-46-1
PROBLEM 6-46 Statement:
Figure P6-9 shows a linkage in one position. Find the instantaneous velocities of points A, B, and P if link O2A is rotating CW at 40 rad/sec.
Given:
Link lengths: Link 2
a 5.00 in
Link 3
b 4.40 in
Link 4
c 5.00 in
Link 1
d 9.50 in
Rpa 8.90 in
δ 56 deg
Coupler point:
θ 50 deg
Crank angle and speed: Solution:
ω 40 rad sec
1
See Figure P6-9 and Mathcad file P0646.
1.
Draw the linkage to scale and label it. All calculated angles are in the local xy system.
2.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 1.9000 2
K3
d c
K2 1.9000 2
2
a b c d
P
y
Y B
2
K3 2.4178
2 a c
3
4
A
A cos θ K1 K2 cos θ K3 2
B 2 sin θ
14.000°
A 0.0607
B 1.5321
C 2.4537
2
B 4 A C
θ 246.992 deg
θ θ 360 deg
θ 606.992 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d b
2
2
c d a b
K5
E 2 sin θ F K1 K4 1 cos θ K5
2 a b
D cos θ K1 K4 cos θ K5
5.
2
K4 2.1591
K5 2.4911
D 2.3605 E 1.5321 F 0.1539
Use equation 4.13 to find values of 3 for the open circuit.
2
θ 2 atan2 2 D E
E 4 D F
θ θ 360 deg 6.
X
O2
Use equation 4.10b to find values of 4 for the open circuit. θ 2 atan2 2 A B
4.
O4
1
C K1 K2 1 cos θ K3 3.
x 50.000°
θ 349.895 deg θ 709.895 deg
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18. ω
a ω sin θ θ b sin θ θ
ω 41.552
rad sec
DESIGN OF MACHINERY - 5th Ed.
ω 7.
SOLUTION MANUAL 6-46-2
a ω sin θ θ c sin θ θ
ω 26.320
rad sec
Determine the velocity of points A and B for the open circuit using equations 6.19.
VA a ω sin θ j cos θ VA ( 153.209 128.558i)
in sec
VA 200.000
in sec
arg VA 40.000 deg
VB c ω sin θ j cos θ VB ( 121.130 51.437i ) 8.
in
VB 131.598
sec
in sec
arg VB 23.008 deg
Determine the velocity of the coupler point P for the open circuit using equations 6.36.
VPA Rpa ω sin θ δ j cos θ δ VPA ( 338.121 149.797i)
in sec
VP VA VPA VP ( 184.912 21.239i )
in sec
VP 186.128
in sec
arg VP 173.448 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-47-1
PROBLEM 6-47 Statement:
Figure P6-10 shows a linkage and its coupler curve. Write a computer program or use an equation solver to calculate and plot the magnitude and direction of the velocity of the coupler point P at 2-deg increments of crank angle for ω 2 = 100 rpm. Check your results with program FOURBAR.
Given:
Link lengths:
Solution:
Link 2 (O2 to A)
a 1.00 in
Link 3 (A to B)
b 2.06 in
Link 4 (B to O4)
c 2.33 in
Link 1 (O2 to O4)
d 2.22 in
Coupler point:
Rpa 3.06 in
31 deg
Crank speed:
100 rpm
See Figure P6-10 and Mathcad file P0647.
1.
Draw the linkage to scale and label it.
2.
Determine the range of motion for this Grashof crank rocker.
B
y 3
θ 0 deg 0.5 deg 360 deg 3.
b
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 2.2200 2
K3
2
a b c d
d c
2
4
c
x
1
A θ cos θ K1 K2 cos θ K3
2
O2
K3 1.5265
a
4
d
2
2 a c
p A
K2 0.9528 2
P
O4
B θ 2 sin θ
C θ K1 K2 1 cos θ K3 4.
Use equation 4.10b to find values of 4 for the open circuit.
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 5.
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d b
K5
2
2
c d a b
2
2 a b
D θ cos θ K1 K4 cos θ K5
K4 1.0777
K5 1.1512
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 6.
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ 7.
2 4 Dθ F θ
E θ
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18.
ω θ
a b
sin θ θ θ θ sin θ θ θ
DESIGN OF MACHINERY - 5th Ed.
ω θ 8.
a c
SOLUTION MANUAL 6-47-2
sin θ θ θ θ sin θ θ θ
Determine the velocity of point A using equations 6.19.
VA θ a sin θ j cos θ 9.
VA θ VA θ
Determine the velocity of the coupler point P using equations 6.36.
VP θ VA θ VPA θ
VPA θ Rpa ω θ sin θ θ j cos θ θ 10. Plot the magnitude and direction of the velocity at coupler point P.
VP θ VP θ
Magnitude:
Direction: θVP1 θ arg VP θ
θVP θ if θVP1 θ 0 θVP1 θ θVP1 θ 2 π MAGNITUDE
Velocity, in/sec
30
20
VP θ
s in 10
0
0
45
90
135
180
225
270
315
360
θ deg
DIRECTION
Vector Angle, deg
360
270
θVP θ
180
deg 90
0
0
45
90
135
180 θ deg
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-48-1
PROBLEM 6-48 Statement:
Figure P6-11 shows a linkage that operates at 500 crank rpm. Write a computer program or use an equation solver to calculate and plot the magnitude and direction of the velocity of point B at 2-deg increments of crank angle. Check your result with program FOURBAR.
Units:
rpm 2 π rad min
Given:
Link lengths:
1
Link 2 (O2 to A)
a 2.000 in
Link 3 (A to B)
Link 4 (B to O4)
c 7.187 in
Link 1 (O2 to O4)
ω 500 rpm
Input crank angular velocity Solution:
ω 52.360 rad sec
Draw the linkage to scale and label it.
2.
Determine the range of motion for this Grashof crank rocker.
A 3 2
θ 0 deg 0.5 deg 360 deg
d
K2
a
K1 4.8125 2
K3
1
d
2
2
O4
2
K3 2.7186
2 a c
C θ K1 K2 1 cos θ K3
B θ 2 sin θ
Use equation 4.10b to find values of 4 for the open circuit.
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
2 a b
K4 1.1493
D θ cos θ K1 K4 cos θ K5
K5 3.4367
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 6.
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ 7.
2 4 Dθ F θ
E θ
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18.
ω θ
a ω b
sin θ θ θ θ sin θ θ θ
4
c
A θ cos θ K1 K2 cos θ K3
5.
2
K2 1.3392
a b c d
B
O2
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
4.
d 9.625 in 1
See Figure P6-11 and Mathcad file P0648.
1.
3.
b 8.375 in
DESIGN OF MACHINERY - 5th Ed.
ω θ 8.
a ω c
SOLUTION MANUAL 6-48-2
sin θ θ θ θ sin θ θ θ
Determine the velocity of point B using equations 6.19.
j cosθθ
VB θ c ω θ sin θ θ
θVB1 θ arg VB θ
VB θ VB θ
Plot the magnitude and angle of the velocity at point B. MAGNITUDE OF VELOCITY AT B
Velocity, in/sec
150
100
VB θ
sec in 50
0
0
60
120
180
240
300
360
θ deg
θVB θ if θVB1 θ 0 θVB1 θ π θVB1 θ DIRECTION OF VELOCITY AT B 60
50
40 Angle, deg
9.
θVB θ
30
deg 20
10
0
0
60
120
180 θ deg
240
300
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-49-1
PROBLEM 6-49 Statement:
Figure P6-12 shows a linkage and its coupler curve. Write a computer program or use an equation solver to calculate and plot the magnitude and direction of the velocity of the coupler point P at 2-deg increments of crank angle over the maximum range of motion possible. Check your results with program FOURBAR.
Given:
Link lengths:
Solution:
Link 2 (O2 to A)
a 0.785 in
Link 3 (A to B)
b 0.356 in
Link 4 (B to O4)
c 0.950 in
Link 1 (O2 to O4)
d 0.544 in
Coupler point:
Rpa 1.09 in
δ 0 deg
Crank speed:
ω 20 rpm
See Figure P6-12 and Mathcad file P0649. y
1.
Draw the linkage to scale and label it.
2.
Using the geometry defined in Figure 3-1a in the text, determine the input crank angles (relative to the line O2O4) at which links 2 and 3, and 3 and 4 are in toggle.
a2 d 2 ( b c) 2 2 a d
P 3 B 3
A
θ acos
4 2
θ 158.286 deg
2
θ θ 1 deg θ 2 deg θ 1 deg 3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 0.6930
O4
d c
2
2
2
a b c d
K2 0.5726
K3
B θ 2 sin θ
A θ cos θ K1 K2 cos θ K3
x O2
2
2 a c
K3 1.1317
C θ K1 K2 1 cos θ K3 4.
Use equation 4.10b to find values of 4 for the open circuit.
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 5.
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
2 a b
F θ K1 K4 1 cos θ K5
D θ cos θ K1 K4 cos θ K5
6.
K5 0.2440
E θ 2 sin θ
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ 7.
K4 1.5281
2 4 Dθ F θ
E θ
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18.
DESIGN OF MACHINERY - 5th Ed.
ω θ
ω θ 8.
SOLUTION MANUAL 6-49-2
b sin θ θ θ θ a ω sin θ θ θ c sin θ θ θ θ a ω
sin θ θ θ
Determine the velocity of point A using equations 6.19.
VA θ a ω sin θ j cos θ 9.
Determine the velocity of the coupler point P using equations 6.36.
VP θ VA θ VPA θ
VPA θ Rpa ω θ sin θ θ δ j cos θ θ δ
10. Plot the magnitude and direction of the coupler point P.
VP θ VP θ
Magnitude:
Direction: θVP θ arg VP θ
Velocity, mm/sec
MAGNITUDE
60
VP θ
sec in
40 20 0 200
150
100
50
0
50
100
50
100
150
200
θ deg
DIRECTION 200
Vector Angle, deg
100
θVP θ
0
deg 100
200 200
150
100
50
0 θ deg
150
200
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-50-1
PROBLEM 6-50 Statement:
Given:
Solution: 1.
Figure P6-13 shows a linkage and its coupler curve. Write a computer program or use an equation solver to calculate and plot the magnitude and direction of the velocity of the coupler point P at 2-deg increments of crank angle over the maximum range of motion possible. Check your results with program FOURBAR. Link lengths: Link 2 (O2 to A)
a 0.86 in
Link 3 (A to B)
b 1.85 in
Link 4 (B to O4)
c 0.86 in
Link 1 (O2 to O4)
d 2.22 in
Coupler point:
Rpa 1.33 in
δ 0 deg
Crank speed:
ω 80 rpm
See Figure P6-13 and Mathcad file P0650.
Draw the linkage to scale and label it. y
B
3 4 P
x
O4
O2 3 2
A
2.
Determine the range of motion for this non-Grashof triple rocker using equations 4.37. 2
arg1
2
2
2 a d 2
arg2
2
a d b c 2
2
a d b c 2 a d
θ2toggle acos arg2
2
b c
arg1 1.228
a d b c
arg2 0.439
a d
θ2toggle 116.0 deg
The other toggle angle is the negative of this. Thus, θ θ2toggle 1 deg θ2toggle 2 deg θ2toggle 1 deg 3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 2.5814
d c
2
K3
2
2
a b c d 2 a c
K2 2.5814
K3 2.0181
B θ 2 sin θ
A θ cos θ K1 K2 cos θ K3
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-50-2
C θ K1 K2 1 cos θ K3 4.
Use equation 4.10b to find values of 4 for the open circuit.
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 5.
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
K4 1.2000
2 a b
D θ cos θ K1 K4 cos θ K5
K5 2.6244
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 6.
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ 7.
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18.
ω θ
8.
2 4 Dθ F θ
E θ
a ω b
sin θ θ θ θ
sin θ θ θ
ω θ
a ω c
sin θ θ θ θ sin θ θ θ
Determine the velocity of point A using equations 6.19.
VA θ a ω sin θ j cos θ 9.
Determine the velocity of the coupler point P using equations 6.36.
VPA θ Rpa ω θ sin θ θ δ j cos θ θ δ
VP θ VA θ VPA θ
10. Plot the magnitude and direction of the coupler point P.
VP θ VP θ
Magnitude:
Direction:
θVP θ arg VP θ
MAGNITUDE 20
Velocity, mm/sec
15
VP θ
sec in
10
5
0 120
90
60
30
0 θ deg
30
60
90
120
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-50-3
DIRECTION 270 240
Vector Angle, deg
210 180
θ' VP θ 150 deg
120 90 60 30 0 120
90
60
30
0 θ deg
30
60
90
120
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-51-1
PROBLEM 6-51 Statement:
Figure P6-14 shows a linkage and its coupler curve. Write a computer program or use an equation solver to calculate and plot the magnitude and direction of the velocity of the coupler point P at 2-deg increments of crank angle over the maximum range of motion possible. Check your results with program FOURBAR.
Given:
Link lengths:
Solution: 1.
Link 2 (O2 to A)
a 0.72 in
Link 3 (A to B)
b 0.68 in
Link 4 (B to O4)
c 0.85 in
Link 1 (O2 to O4)
d 1.82 in
Coupler point:
Rpa 0.97 in
δ 54 deg
Crank speed:
ω 80 rpm
See Figure P6-14 and Mathcad file P0651.
Draw the linkage to scale and label it. P y
B
3 4
A 2
x O2
2.
O4
Determine the range of motion for this non-Grashof triple rocker using equations 4.37. 2
arg1
2
2
2 a d 2
arg2
2
a d b c 2
2
a d b c 2 a d
2
b c
arg1 1.451
a d b c
arg2 0.568
a d
θ2toggle acos arg2
θ2toggle 55.4 deg
The other toggle angle is the negative of this. Thus, θ θ2toggle 1 deg θ2toggle 2 deg θ2toggle 1 deg 3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 2.5278 2
K3
d c
K2 2.1412 2
2
a b c d 2 a c
2
K3 3.3422
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-51-2
A θ cos θ K1 K2 cos θ K3
B θ 2 sin θ
C θ K1 K2 1 cos θ K3 4.
Use equation 4.10b to find values of 4 for the open circuit.
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 5.
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
K4 2.6765
2 a b
K5 3.6465
D θ cos θ K1 K4 cos θ K5
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 6.
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ 7.
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18.
a ω
a ω
ω θ
ω θ 8.
2 4 Dθ F θ
E θ
b
c
sin θ θ θ θ
sin θ θ θ θ
sin θ θ θ
sin θ θ θ
Determine the velocity of point A using equations 6.19.
VA θ a ω sin θ j cos θ 9.
Determine the velocity of the coupler point P using equations 6.36.
VPA θ Rpa ω θ sin θ θ δ j cos θ θ δ
VP θ VA θ VPA θ
10. Plot the magnitude and direction of the coupler point P.
Magnitude:
VP θ VP θ
Direction:
θVP θ arg VP θ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-51-3
MAGNITUDE 20
Velocity, in/sec
15
VP θ
sec in
10
5
0 60
30
0
30
60
θ deg
DIRECTION 200
Vector Angle, deg
150 100
θVP θ
50
deg 0 50 100 60
30
0 θ deg Crank Angle, deg
30
60
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-52-1
PROBLEM 6-52 Statement:
Figure P6-15 shows a power hacksaw that is an offset crank-slider mechanism that has the dimensions given below. Draw an equivalent linkage diagram; then calculate and plot the velocity of the saw blade with respect to the piece being cut over one revolution of the crank, which rotates at 50 rpm.
Given:
Link lengths: Link 2
a 75 mm
Link 3
b 170 mm ω 50 rpm
Input crank angular velocity Solution: 1.
c 45 mm
Offset
See Figure P6-15 and Mathcad file P0652.
Draw the equivalent linkage to a convenient scale and label it. y
B
3
b
A
4 a 2
c
2
O2
2.
Determine the range of motion for this crank-slider linkage. θ 0 deg 2 deg 360 deg
3.
Determine 3 using equations 4.16 for the crossed circuit.
a sin θ b
θ θ asin 4.
Determine the angular velocity of link 3 using equation 6.22a:
ω θ
5.
c
a b
ω cos θ θ cos θ
Determine the velocity of pin B using equation 6.22b:
VB θ a ω sin θ b ω θ sin θ θ
7.
Plot the magnitude of the velocity of B. (See next page.)
x
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-52-2
VELOCITY OF POINT B 600
Velocity, mm/sec
400
200
VB θ
sec mm 0
200
400
0
60
120
180 θ deg
240
300
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-53-1
PROBLEM 6-53 Statement:
Given:
Figure P6-16 shows a walking-beam indexing and pick-and-place mechanism that can be analyzed as two fourbar linkages driven by a common crank. Calculate and plot the absolute velocities of points E and P and the relative velocity between points E and P for one revolution of gear 2. Link lengths ( walking-beam linkage): Link 2 (O2 to A)
a' 40 mm
Link 3 (A to D)
b' 108 mm
Link 4 (O4 to D)
c' 40 mm
Link 1 (O2 to O4)
d' 108 mm
Link lengths (pick and place linkage): Link 5 (O5 to B)
a 13 mm
Link 7 (B to C)
b 193 mm
Link 6 (C to O6)
c 92 mm
Link 1 (O5 to O6)
d 128 mm
u 164 mm
Crank speed:
ω 10 rpm
Rocker point E:
ϕ 143 deg
Gears 4 & 5 phase angle Solution: 1.
See Figure P6-16 and Mathcad file P0653.
Draw the walking-beam linkage to scale and label it. 30 mm Y
P
58 mm
80°
A
D c'
b'
a' d'
x'
O2
O4
2.
Determine the range of motion for this mechanism. θ 0 deg 2 deg 360 deg
3.
X
y'
(local x'y' coordinate system)
This part of the mechanism is a special-case Grashof in the parallelogram configuration. As such, the coupler does not rotate, but has curvilinear motion with ever point on it having the same velocity. Therefore, it is only necessary to calculate the X-component of the velocity at point A in order to determine the velocity of the cylinder center, P.
VA θ a' ω sin θ j cos θ
VAx θ Re VA θ
In the global X-Y coordinate frame,
VP θ VAx θ
DESIGN OF MACHINERY - 5th Ed.
4.
SOLUTION MANUAL 6-53-2
Draw the pick and place linkage to scale and label it.
E
C
7 6
b
c
O6 5.
B
a O5
Establish the relationship between 5 and 2. Note that gear 5 is driven by gear 4 and that their ratio is -1 (i.e., they rotate in opposite directions with the same speed). Also, because the walking beam fourbar is special Grashof, 4 = 2. Thus,
θ θ θ ϕ 6.
5
d 1
ω ω
and
Determine the values of the constants needed for finding 6 from equations 4.8a and 4.10a. K1
d
K1 9.8462
a 2
K3
2
2
a b c d
d c
K2 1.3913
2
K3 5.1137
2 a c
K2
K1 K2 cosθθ K3
A θ cos θ θ
B θ 2 sin θ θ
K3
C θ K1 K2 1 cos θ θ 7.
Use equation 4.10b to find values of 6 for the crossed circuit.
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 8.
B θ
Determine the values of the constants needed for finding 7 from equations 4.11b and 4.12. K4
d b
2
K5
2
2
c d a b 2 a b
2
K4 0.6632
K5 9.0351
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-53-3
K1 K4 cosθθ K5
D θ cos θ θ
E θ 2 sin θ θ
K5
F θ K1 K4 1 cos θ θ 9.
Use equation 4.13 to find values of 7 for the crossed circuit.
θ θ 2 atan2 2 D θ E θ
2 4 Dθ F θ
E θ
10. Determine the angular velocity of links 6 and 7 using equations 6.18.
ω θ
a ω sinθ θ θ θ b sin θ θ θ θ
a ω sin θ θ θ θ c sin θ θ θ θ
ω θ
10. Determine the velocity of the rocker point E using equations 6.34.
j cosθθ
VE θ u ω θ sin θ θ 11. Transform this into the global XY system.
VEx θ Re VE θ
VEX θ VEx θ
VEy θ Im VE θ
VEY θ VEx θ
VEXY θ VEX θ j VEY θ
12. Calculate and plot the velocity of E relative to P.
VEP θ VEXY θ VP θ
RELATIVE VELOCITY 100
Velocity, mm/sec
80
V EP θ
sec
60
mm 40
20
0
0
30
60
90
120
150
180
210
θ deg Crank Angle, deg
240
270
300
330
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-54-1
PROBLEM 6-54 Statement:
Figure P6-17 shows a paper roll off-loading mechanism driven by an air cylinder. In the position shown, it has the dimensions given below. The V-links are rigidly attached to O4A. The air cylinder is retracted at a constant velocity of 0.2 m/sec. Draw a kinematic diagram of the mechanism, write the necessary equations, and calculate and plot the angular velocity of the paper roll and the linear velocity of its center as it rotates through 90 deg CCW from the position shown.
Given:
Link lengths and angles:
Paper roll location from O4:
Link 4 (O4 to A)
c 300 mm
u 707.1 mm
Link 1 (O2 to O4)
d 930 mm
δ 181 deg
Link 4 initial angle
θ 62.8 deg
adot 200 mm sec
Input cylinder velocity Solution: 1.
with respect to local x axis 1
See Figure P6-17 and Mathcad file P0654.
Draw the mechanism to scale and define a vector loop using the fourbar derivation in Section 6.7 as a model. V-Link
Roll Center
707.107
45.000° x
O4 c
4
46.000° A
1
O4
d
4
3
O2
2 b
R1
R4 2 A
R3
R2
O2
a f y
2.
Write the vector loop equation, differentiate it, expand the result and separate into real and imaginary parts to solve for f, 2, and 4. R1 R4 R2 R3 d e
j θ
c e
j θ
(a)
a e
j θ
b e
j θ
(b)
where a is the distance from the origin to the cylinder piston, a variable; b is the distance from the cylinder piston to A, a constant; and c is the distance from 4 to point A, a constant. Angle 1 is zero, 3 = 2, and 4 is the variable angle that the rocker arm makes with the x axis. Solving the position equations: Let
f a b
then, making this substitution and substituting the Euler equivalents,
d c cos θ j sin θ f cos θ j sin θ
(c)
Separating into real and imaginary components and solving for 2 and f,
θ θ atan2 d c cos θ c sin θ
(d)
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-54-2
sin θ θ c sin θ
f θ
(e)
Differentiate equation b. j c ω e
j θ
d f ej θ j f ω ej θ dt
(f)
Substituting the Euler equivalents,
c ω sin θ j cos θ
d f cos θ j sin θ f ω sin θ j cos θ dt
(g)
Separating into real and imaginary components and solving for 4 . Note that df/dt = adot.
ω θ
3.
adot
c sin θ θ θ
(h)
Plot 4 over a range of 4 of θ θ θ 1 deg θ 90 deg
ANGULAR VELOCITY, LINK 4 1.2
Angular Velocity, rad/sec
1 0.8
ω θ
sec rad
0.6 0.4 0.2 0 60
80
100
120
140
160
θ deg Link 4 Angle, deg
4.
Determine the velocity of the center of the paper roll using equation 6.35. The direction is in the local xy coordinate system.
VU θ u ω θ sin θ δ j cos θ δ VU θ VU θ 5.
θVU θ arg VU θ
Plot the magnitude and direction of the velocity of the paper roll center. (See next page.)
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-54-3
MAGNITUDE OF PAPER CENTER VELOCITY
Velocity, mm/sec
600
VU θ
sec mm
400
200
0 60
80
100
120
140
160
θ deg Rocker Arm Position, deg
DIRECTION OF PAPER CENTER VELOCITY 80
Vector Angle, deg
60
40
θVU θ
20
deg 0
20
40 60
80
100
120 θ deg
Rocker Arm Position, deg
140
160
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-55a-1
PROBLEM 6-55a Statement:
Figure P6-18 shows a powder compaction mechanism. Calculate its mechanical advantage for the position shown.
Given:
Link lengths: Link 2 (A to B)
a 105 mm
Link 3 (B to D)
b 172 mm
c 27 mm
Offset
Distance to force application: rin 301 mm
Link 2 (AC)
θ 44 deg
Position of link 2: Solution: 1.
Let
ω 1 rad sec
1
See Figure P6-18 and Mathcad file P0655a.
Draw the linkage to scale and label it. X
2.
Determine 3 using equation 4.17.
4 D
a sin θ c π b
θ asin
C
θ 164.509 deg 3.
3
Determine the angular velocity of link 3 using equation 6.22a: 44.000°
a cos θ ω ω b cos θ 4.
B 2
Determine the velocity of pin D using equation 6.22b:
VD a ω sin θ b ω sin θ 5.
Y
Positive upward
Calculate the velocity of point C using equation 6.23a:
VC rin ω sin θ j cos θ VC VC 6.
Calculate the mechanical advantage using equation 6.13. mA
VC VD
mA 3.206
A
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-55b-1
PROBLEM 6-55b Statement:
Figure P6-18 shows a powder compaction mechanism. Calculate and plot its mechanical advantage as a function of the angle of link AC as it rotates from 15 to 60 deg.
Given:
Link lengths: Link 2 (A to B)
a 105 mm
Link 3 (B to D)
b 172 mm
Offset
c 27 mm
Distance to force application: rin 301 mm
Link 2 (AC)
Initial and final positions of link 2: θ 15 deg Solution: 1.
θ 60 deg
Let ω 1 rad sec
See Figure P6-18 and Mathcad file P0655b.
Draw the linkage to scale and label it. X
4 D C
3
2 B 2
Y
2.
A
Determine the range of motion for this slider-crank linkage. θ θ θ 1 deg θ
3.
Determine 3 using equation 4.17.
a sin θ b
θ θ asin 4.
π
Determine the angular velocity of link 3 using equation 6.22a:
ω θ
5.
c
a b
ω cos θ θ cos θ
Determine the velocity of pin D using equation 6.22b:
VD θ a ω sin θ b ω θ sin θ θ 6.
Calculate the velocity of point C using equation 6.23a:
Positive upward
1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-55b-2
VC θ rin ω sin θ j cos θ VC θ VC θ
Calculate the mechanical advantage using equation 6.13.
mA θ
VD θ VC θ
MECHANICAL ADVANTAGE 12 11 10 9 Mechanical Advantage
7.
8 7
m A θ 6 5 4 3 2 1 0 15
20
25
30
35
40 θ deg
Crank Angle, deg
45
50
55
60
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-56-1
PROBLEM 6-56 Statement:
Figure P6-19 shows a walking beam mechanism. Calculate and plot the velocity Vout for one revolution of the input crank 2 rotating at 100 rpm.
Given:
Link lengths:
Solution: 1.
Link 2 (O2 to A)
a 1.00 in
Link 3 (A to B)
b 2.06 in
Link 4 (B to O4)
c 2.33 in
Link 1 (O2 to O4)
d 2.22 in
Coupler point:
Rpa 3.06 in
δ 31 deg
Crank speed:
ω 100 rpm
See Figure P6-19 and Mathcad file P0656.
Draw the linkage to scale and label it. Y
x
y O4 4
1 26.00°
X
O2
P
2 A 3 B
2.
Determine the range of motion for this Grashof crank rocker. θ 0 deg 1 deg 360 deg
3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 2.2200 2
K3
c
K2 0.9528 2
2
a b c d
d
2
K3 1.5265
2 a c
A θ cos θ K1 K2 cos θ K3
B θ 2 sin θ
C θ K1 K2 1 cos θ K3 4.
Use equation 4.10b to find values of 4 for the open circuit.
θ θ 2 atan2 2 A θ B θ
2 4 A θ Cθ
B θ
DESIGN OF MACHINERY - 5th Ed.
5.
SOLUTION MANUAL 6-56-2
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
K4 1.0777
2 a b
K5 1.1512
D θ cos θ K1 K4 cos θ K5
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 6.
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ 7.
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18.
a ω
a ω
ω θ
ω θ 8.
2 4 Dθ F θ
E θ
b
c
sin θ θ θ θ
sin θ θ θ θ
sin θ θ θ
sin θ θ θ
Determine the velocity of point A using equations 6.19.
VA θ a ω sin θ j cos θ 9.
Determine the velocity of the coupler point P using equations 6.36.
VP θ VA θ VPA θ
VPA θ Rpa ω θ sin θ θ δ j cos θ θ δ
10. Plot the X-component (global coordinate system) of the velocity of the coupler point P. Coordinate rotation angle: α 26 deg
Vout θ Re VP θ cos α Im VP θ sin α Vout
Velocity, in/sec
20
10
0
10
0
45
90
135
180
Crank Angle, deg
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-57a-1
PROBLEM 6-57a Statement:
Figure P6-20 shows a crimping tool. For the dimensions given below, calculate its mechanical advantage for the position shown.
Given:
Link lengths: Link 2 (AB)
a 0.80 in
Link 3 (BC)
b 1.23 in
Link 4 (CD)
c 1.55 in
Link 1 (AD)
d 2.40 in
Link 4 (CD)
rout 1.00 in
Distance to force application: rin 4.26 in
Link 2 (AB)
θ 49 deg
Initial position of link 2: Solution: 1.
See Figure P6-20 and Mathcad file P0657a.
Draw the mechanism to scale and label it. A
2
B
2
3
C Fout
1
Fin
4
49.000°
D
2.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 3.0000 2
K3
d c
K2 1.5484 2
2
a b c d
2
K3 2.9394
2 a c
A cos θ K1 K2 cos θ K3
B 2 sin θ
C K1 K2 1 cos θ K3 A 0.4204 3.
B 1.5094
Use equation 4.10b to find value of 4 for the open circuit.
θ 2 atan2 2 A B 4.
C 4.2675
2
B 4 A C
θ 236.482 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
d b
2
K5
2
2
c d a b 2 a b
2
K4 1.9512
K5 2.8000
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-57a-2
D cos θ K1 K4 cos θ K5
D 3.8638
E 2 sin θ
E 1.5094
F K1 K4 1 cos θ K5 5.
Use equation 4.13 to find values of 3 for the open circuit.
θ 2 atan2 2 D E 6.
F 0.8241
2
E 4 D F
θ 325.961 deg
Referring to Figure 6-10, calculate the values of the angles and . ν θ θ
ν 374.961 deg
If > 360 deg, subtract 360 deg from it. ν if ν 360 deg ν 360 deg ν
ν 14.961 deg
μ θ θ
μ 89.479 deg
If > 90 deg, subtract it from 180 deg. μ if μ 90 deg 180 deg μ μ 7.
μ 89.479 deg
Using equation 6.13e, calculate the mechanical advantage of the linkage in the position shown. mA
c sin μ rin a sin ν rout
mA 31.969
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-57b-1
PROBLEM 6-57b Statement:
Figure P6-20 shows a crimping tool. For the dimensions given below, calculate and plot its mechanical advantage as a function of the angle of link AB as it rotates from 60 to 45 deg.
Given:
Link lengths: Link 2 (AB)
a 0.80 in
Link 3 (BC)
b 1.23 in
Link 4 (CD)
c 1.55 in
Link 1 (AD)
d 2.40 in
Link 4 (CD)
rout 1.00 in
Distance to force application: rin 4.26 in
Link 2 (AB)
θ 60 deg
Range of positions of link 2: Solution: 1.
θ 45 deg
See Figure P6-20 and Mathcad file P0657b.
Draw the mechanism to scale and label it.
A
2
B
2
3
C Fout
1
Fin
4
2
D
2.
Calculate the range of 2 in the local coordinate system (required to calculate 3 and 4). θ θ θ 1 deg θ
3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 3.0000
c
K2 1.5484
2
K3
d
2
2
a b c d
2
K3 2.9394
2 a c
A θ cos θ K1 K2 cos θ K3
B θ 2 sin θ
C θ K1 K2 1 cos θ K3 4.
Use equation 4.10b to find value of 4 for the open circuit.
θ θ 2 atan2 2 A θ B θ 5.
2 4 A θ Cθ
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12.
DESIGN OF MACHINERY - 5th Ed.
K4
SOLUTION MANUAL 6-57b-2
2
d
K5
b
2
2
c d a b
2
K4 1.9512
2 a b
K5 2.8000
D θ cos θ K1 K4 cos θ K5
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 6.
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ 7.
2 4 Dθ F θ
E θ
Referring to Figure 6-10, calculate the values of the angles and .
ν θ θ θ θ
μ θ θ θ θ θ
Using equation 6.13e, calculate and plot the mechanical advantage of the linkage over the given range.
mA θ
rin a sin ν θ rout
c sin μ θ
MECHANICAL ADVANTAGE vs HANDLE ANGLE 60
50 Mechanical Advantage
8.
40
m A θ
30
20
10 45
48
51
54 θ deg
Handle Angle, deg
57
60
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-58-1
PROBLEM 6-58 Statement:
Figure P6-21 shows a locking pliers. Calculate its mechanical advantage for the position shown. Scale any dimensions needed from the diagram.
Solution:
See Figure P6-21 and Mathcad file P0658.
1.
Draw the linkage to scale in the position given and find the instant centers.
F
1,4
1,2
P 1
O4
O2 2 3
B
A F
2.
2,3
4
P
3,4 and 1,3
Note that the linkage is in a toggle position (links 2 and 3 are in line) and the angle between links 2 and 3 is 0 deg. From the discussion below equation 6.13e in the text, we see that the mechanical advantage for this linkage in this position is theoretically infinite.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-59a-1
PROBLEM 6-59a Statement:
Given:
Figure P6-22 shows a fourbar toggle clamp used to hold a workpiece in place by clamping it at D. The linkage will toggle when link 2 reaches 90 deg. For the dimensions given below, calculate its mechanical advantage for the position shown. Link lengths: Link 2 (O2A) a 70 mm c 34 mm
Link 4 (O4B)
Link 3 (AB)
b 35 mm
Link 1 (O2O4)
d 48 mm
Link 4 (O4D)
rout 82 mm
Distance to force application: rin 138 mm
Link 2 (O2C)
θ 104 deg
Initial position of link 2: Solution: 1.
Global XY system
See Figure P6-22 and Mathcad file P0659a.
Draw the mechanism to scale and label it. To establish the position of O4 with respect to O2 (in the global coordinate frame), draw the linkage in the toggle position with 2 = 90 deg. The fixed pivot O4 is then 48 mm fr O2 and 34 mm from B' (see layout). Y C' C
Linkage in toggle position
A'
A 3 2
x
B
D
4
D'
B' 135.069°
O4
X O2 y
2.
Calculate the value of 2 in the local coordinate system (required to calculate 3 and 4). Rotation angle of local xy system to global XY system: θ θ α
3.
α 135.069 deg
θ 31.069 deg
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K1 0.6857
a 2
K3
2
2
a b c d 2 a c
K2
d c
2
K3 1.4989
K2 1.4118
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-59a-2
A cos θ K1 K2 cos θ K3
B 2 sin θ
C K1 K2 1 cos θ K3 A 0.4605 4.
B 1.0321
Use equation 4.10b to find value of 4 for the open circuit.
2
θ 2 atan2 2 A B 5.
C 0.1189
B 4 A C
θ 129.480 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
2 a b
K5 1.4843
D cos θ K1 K4 cos θ K5
D 0.1388
E 2 sin θ
E 1.0321
F K1 K4 1 cos θ K5 6.
F 0.4804
Use equation 4.13 to find values of 3 for the open circuit.
θ 2 atan2 2 D E 7.
K4 1.3714
2
E 4 D F
θ 196.400 deg
Referring to Figure 6-10, calculate the values of the angles and . ν θ θ
ν 165.331 deg
If > 90 deg, subtract it from 180 deg. ν if ν 90 deg 180 deg ν ν
ν 14.669 deg
μ θ θ
μ 66.920 deg
If > 90 deg, subtract it from 180 deg. μ if μ 90 deg 180 deg μ μ 8.
μ 66.920 deg
Using equation 6.13e, calculate the mechanical advantage of the linkage in the position shown. mA
c sin μ rin a sin ν rout
mA 2.970
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-59b-1
PROBLEM 6-59b Statement:
Given:
Figure P6-22 shows a fourbar toggle clamp used to hold a workpiece in place by clamping it at D. The linkage will toggle when link 2 reaches 90 deg. For the dimensions given below, calculate and plot its mechanical advantage as a function of the angle of link AB as link 2 rotates from 120 to 90 deg (in the global coordinate system). Link lengths: Link 2 (O2A)
a 70 mm
Link 3 (AB)
b 35 mm
Link 4 (O4B)
c 34 mm
Link 1 (O2O4)
d 48 mm
Link 4 (O4D)
rout 82 mm
Distance to force application: rin 138 mm
Link 2 (O2C)
θ 120 deg
Range of positions of link 2: Solution: 1.
θ 90.1 deg Global XY system
See Figure P6-22 and Mathcad file P0659b.
Draw the mechanism to scale and label it. To establish the position of O4 with respect to O2 (in the global coordinate frame), draw the linkage in the toggle position with 2 = 90 deg. The fixed pivot O4 is then 48 mm fr O2 and 34 mm from B' (see layout). Linkage in initial position Y C'
C Linkage in final position
A' A
3
x
4
B
2
D
B' 135.069°
O4
D'
X O2 y
2.
Calculate the range of 2 in the local coordinate system (required to calculate 3 and 4). Rotation angle of local xy system to global XY system:
α 135.069 deg
θ θ α θ α 1 deg θ α 3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K1 0.6857
a 2
K3
2
2
a b c d 2 a c
K2
d c
2
K3 1.4989
K2 1.4118
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-59b-2
A θ cos θ K1 K2 cos θ K3
B θ 2 sin θ
C θ K1 K2 1 cos θ K3 4.
Use equation 4.10b to find value of 4 for the open circuit.
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 5.
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
K4 1.3714
2 a b
K5 1.4843
D θ cos θ K1 K4 cos θ K5
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 6.
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ 7.
2 4 Dθ F θ
E θ
Referring to Figure 6-10, calculate the values of the angles and .
μ θ θ θ θ θ ν θ θ θ θ
8.
Using equation 6.13e, calculate and plot the mechanical advantage of the linkage over the given range.
mA θ
rin a sin ν θ rout
c sin μ θ
MECHANICAL ADVANTAGE 50
40
30
m A θ
20
10
0 90
95
100
105 θ α deg
110
115
120
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-60-1
PROBLEM 6-60 Statement:
Given:
Figure P6-23 shows a surface grinder. The workpiece is oscillated under the spinning grinding wheel by the slider-crank linkage that has the dimensions given below. Calculate and plot the velocity of the grinding wheel contact point relative to the workpiece over one revolution of the crank. Link lengths: Link 2 (O2 to A)
a 22 mm
Link 3 (A to B)
b 157 mm
Grinding wheel diameter
d 90 mm
Input crank angular velocity
ω 120 rpm
Grinding wheel angular velocity ω 3450 rpm Solution: 1.
Offset
c 40 mm
CCW CCW
See Figure P6-23 and Mathcad file P0660.
Draw the linkage to scale and label it.
5
4 2 3
A
B
c
2 O2
2.
Determine the range of motion for this slider-crank linkage. θ 0 deg 1 deg 360 deg
3.
Determine 3 using equation 4.17.
a sin θ c π b
θ θ asin 4.
Determine the angular velocity of link 3 using equation 6.22a:
ω θ
5.
a b
ω cos θ θ cos θ
Determine the velocity of pin B using equation 6.22b:
VB θ a ω sin θ b ω θ sin θ θ 6.
Positive to the right
Calculate the velocity of the grinding wheel contact point using equation 6.7: VG
d 2
ω
VG 16.258
m sec
Directed to the right
DESIGN OF MACHINERY - 5th Ed.
The velocity of the grinding wheel contact point relative to the workpiece, which has velocity VB, is
VGB θ VG VB θ
RELATIVE VELOCITY AT CONTACT POINT 16.6
16.4
Velocity, m/sec
7.
SOLUTION MANUAL 6-60-2
VGB θ
sec m
16.2
16
15.8
0
60
120
180 θ deg Crank Angle, deg
240
300
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-61-1
PROBLEM 6-61 Statement:
Figure P6-24 shows an inverted slider-crank mechanism. Given the dimensions below, find 2, 3, 4, VA4, Vtrans, and Vslip for the position shown with VA2 = 20 in/sec in the direction shown.
Given:
Link lengths: a 2.5 in
Link 2 (O2A)
Link 4 (O4A)
c 4.1 in
Link 1 (O2O4)
d 3.9 in
Measured angles: θ 75.5 deg
θtrans 26.5 deg
Velocity of point A on links 2 and 3: Solution: 1.
θslip 116.5 deg
VA2 20 in sec
1
VA3 VA2
See Figure P6-24 and Mathcad file P0661.
Draw the linkage to scale and indicate the axes of slip and transmission as well as the directions of velocities of interest. Y
Axis of slip A
O2
X
11.500° 2 3
Direction of VA4
1
Axis of transmission 4 Direction of VA2 O4
2.
Use equation 6.7 to calculate the angular velocity of link 2. ω
3.
VA2 a
ω 8.000
rad
CW
sec
Use equation 6.5 to (graphically) determine the magnitude of the velocity components at point A on links 2 and 3.. The equation to be solved graphically is VA3 = Vtrans + VA3slip a. Choose a convenient velocity scale and layout the known vector VA3. b. From the tip of VA3, draw a construction line with the direction of VA3slip, magnitude unknown. c. From the tail of VA3, draw a construction line with the direction of Vtrans, magnitude unknown. d. Complete the vector triangle by drawing VA3slip from the tip of Vtrans to the tip of VA3 and drawing Vtrans from the tail of VA3 to the intersection of the VA3slip construction line.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-61-2
Y
0
10 in/sec
1.686
X
Axis of slip
V A4
V trans
VA4slip 1.509
VA3
Axis of transmission V A3slip
2.064
4.
From the velocity triangle we have: Velocity scale factor:
5.
kv
10 in sec
1
in in
VA4 1.686 in kv
VA4 16.9
Vslip 2.064 in kv
Vslip 20.6
Vtrans 1.509 in kv
Vtrans 15.1
sec in sec in sec
Determine the angular velocity of link 4 using equation 6.7. ω
VA4 c
ω 4.1
Because link 3 slides within link 4, 3 = 4.
rad sec
CW
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-62-1
PROBLEM 6-62 Statement:
Figure P6-25 shows a drag-link mechanism with dimensions. Write the necessary equations and solve them to calculate and plot the angular velocity of link 4 for an input of 2 = 1 rad/sec. Comment on the uses for this mechanism.
Given:
Link lengths: Link 2 (L2)
a 1.38 in
Link 3 (L3)
b 1.22 in
Link 4 (L4)
c 1.62 in
Link 1 (L1)
d 0.68 in
ω 1 rad sec
Input crank angular velocity Solution: 1.
1
CW
See Figure P6-25 and Mathcad file P0662.
Draw the linkage to scale and label it. y
A
3 B
2 2 4 x O2
2.
O4
Determine the range of motion for this Grashof double crank. θ 0 deg 2 deg 360 deg
3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 0.4928 2
K3
c
K2 0.4198 2
2
a b c d
d
2
K3 0.7834
2 a c
A θ cos θ K1 K2 cos θ K3
B θ 2 sin θ
C θ K1 K2 1 cos θ K3 4.
Use equation 4.10b to find values of 4 for the open circuit.
θ θ 2 atan2 2 A θ B θ 5.
2 4 A θ Cθ
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12.
DESIGN OF MACHINERY - 5th Ed.
K4
SOLUTION MANUAL 6-62-2
2
d
K5
b
2
2
2
c d a b
K4 0.5574
2 a b
K5 0.3655
D θ cos θ K1 K4 cos θ K5
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 6.
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ 7.
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18.
a ω
a ω
ω θ
ω θ
b
c
sin θ θ θ θ
sin θ θ θ θ
sin θ θ θ
sin θ θ θ
Plot the angular velocity of link 4.
ANGULAR VELOCITY, LINK 4 0.5
Angular Velocity, rad/sec
9.
2 4 Dθ F θ
E θ
1
ω θ
sec rad
1.5
2
2.5
0
60
120
180 θ deg Crank Angle, deg
240
300
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-63-1
PROBLEM 6-63 Statement:
Figure P6-25 shows a drag-link mechanism with dimensions. Write the necessary equations and solve them to calculate and plot the centrodes of instant center I2,4.
Given:
Measured link lengths: Link 2 (L2)
L2 1.38 in
Link 3 (L3)
L3 1.22 in
Link 4 (L4)
L4 1.62 in
Link 1 (L1)
L1 0.68 in
ω 1 rad sec
Input crank angular velocity Solution: 1.
1
See Figure P6-25 and Mathcad file P0663.
Draw the linkage to scale and label it. Instant center I2,4 is at the intersection of line AB with line O2O4. To get the first centrode, ground link 2 and let link 3 be the input. Then we have a L3
b L4
c L1 A
d L2 3 B
2 y 2 4 O2
O4 4
x
2.
Determine the range of motion for this Grashof double rocker. From Figure 3-1a on page 80, one toggle angle is
a2 d 2 ( b c) 2 2 a d
θ acos
θ 124.294 deg
The other toggle angle is the negative of this. The range of motion is θ θ 1 deg θ 2 deg θ 1 deg 3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 1.1311 2
K3
d c
K2 2.0294 2
2
a b c d 2 a c
2
K3 0.7418
A θ cos θ K1 K2 cos θ K3
B θ 2 sin θ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-63-2
C θ K1 K2 1 cos θ K3 4.
Use equation 4.10b to find values of 4 for the open circuit.
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 5.
B θ
Calculate the coordinates of the intersection of line BC with line AD. This will be the instant center I2,4.
Line BC:
y x tan θ
Line AD:
y 0 ( x d ) tan θ
Eliminating y and solving for the x- and y-coordinates of the intersection,
x24 θ
6.
tan θ θ tan θ d tan θ θ
y24 θ x24 θ tan θ
Plot the fixed centrode.
FIXED CENTRODE 10
y-Coordinate, in
5
y24 θ
0
in 5
10 10
5
0
x24 θ in
x-Coordinate, in
7.
Invert the linkage, making C and D the fixed pivots. Then, a L1
8.
b L2
c L3
Determine the range of motion for this Grashof crank rocker. θ 0 deg 0.5 deg 360 deg
d L4
5
10
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-63-3 4 A
x
3 B
y
2
2
4 O2
9.
O4
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d a
K1 2.3824 2
K3
d
K2
K2 1.3279 2
2
a b c d
c
2
K3 1.6097
2 a c
A θ cos θ K1 K2 cos θ K3
B θ 2 sin θ
C θ K1 K2 1 cos θ K3 10. Use equation 4.10b to find values of 4 for the open circuit.
θ θ 2 atan2 2 A θ B θ
2 4 A θ Cθ
B θ
11. Calculate the coordinates of the intersection of line BC with line AD. This will be the instant center I2,4.
Line AD:
y x tan θ
Line BC:
y 0 ( x d ) tan θ
Eliminating y and solving for the x- and y-coordinates of the intersection,
x24 θ
6.
tan θ θ tan θ d tan θ θ
Plot the moving centrode. (See next page.)
y24 θ x24 θ tan θ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-63-4
MOVING CENTRODE 10
y-Coordinate, in
5
y24 θ
0
in 5
10 10
5
0
x24 θ in
x-Coordinate, in
5
10
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-64-1
PROBLEM 6-64 Statement:
Figure P6-26 shows a mechanism with dimensions. Use a graphical method to calculate the velocities of points A, B, and C and the velocity of slip for the position shown.
Given:
Link lengths and angles: Link 1 (O2O4)
d 1.22 in
Angle O2O4 makes with X axis
θ 56.5 deg
Link 2 (O2A)
a 1.35 in
Angle 2 makes with X axis
θ 14 deg
Link 4 (O4B)
e 1.36 in
Link 5 (BC)
f 2.69 in
Link 6 (O6C)
g 1.80 in
Angle O6C makes with X axis
θ 88 deg
Angular velocity of link 2 Solution: 1.
ω 20 rad sec
1
CW
See Figure P6-26 and Mathcad file P0664.
Draw the linkage to scale and indicate the axes of slip and transmission as well as the directions of velocities of interest.
B
Axis of slip Y
Axis of transmission O4
4
0.939
132.661° A 3 2 X O2
Direction of VA3
2.
Use equation 6.7 to calculate the magnitude of the velocity at point A on links 2 and 3. VA3 a ω
3.
VA3 27.000
in sec
θVA3 θ 90 deg
θVA3 76.0 deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity components at point A on link 3. The equation to be solved graphically is VA3 = Vtrans + Vslip a. Choose a convenient velocity scale and layout the known vector VA3. b. From the tip of VA3, draw a construction line with the direction of Vslip, magnitude unknown. c. From the tail of VA3, draw a construction line with the direction of Vtrans, magnitude unknown. d. Complete the vector triangle by drawing Vslip from the tip of Vtrans to the tip of VA3 and drawing Vtrans from the tail of VA3 to the intersection of the Vslip construction line.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-64-2
Y 0
12 in/sec
1.295" X Vtrans
VA3 2.369"
4.
From the velocity triangle we have: Velocity scale factor:
5.
kv
in in
Vtrans 1.295 in kv
Vtrans 15.540
sec in sec
The true velocity of point A on link 4 is Vtrans, VA4 15.54
in sec
Determine the angular velocity of link 4 using equation 6.7.
ω
VA4 c
c 0.939 in and
ω 16.550
rad
θ 132.661 deg
CW
sec
Determine the magnitude and sense of the vector VB using equation 6.7. VB e ω
θVA4 θ 90 deg 8.
1
Vslip 28.428
From the linkage layout above:
7.
12 in sec
Vslip 2.369 in kv
VA4 Vtrans 6.
Vslip
VB 22.507
in sec
θVA4 42.661 deg
Draw links 1, 4, 5, and 6 to a convenient scale. Indicate the directions of the velocity vectors of interest. (See next page.)
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-64-3
Direction of VCB
Direction of VB
B
Y O4
4
C A 3
Direction of VC 2
X O2
O6
9.
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point C. The equation to be solved graphically is VC = VB + VCB a. Choose a convenient velocity scale and layout the known vector VB. b. From the tip of VB, draw a construction line with the direction of VCB, magnitude unknown. c. From the tail of VB, draw a construction line with the direction of VC, magnitude unknown. d. Complete the vector triangle by drawing VCB from the tip of VB to the intersection of the VC construction line and drawing VC from the tail of VB to the intersection of the VCB construction line. VB
0
12 in/sec
Y
VCB X VC 2.088"
10. From the velocity triangle we have: Velocity scale factor: VC 2.088 in kv
kv
12 in sec
VC 25.1
in in sec
1
θVC θ 90 deg θVC 2.0 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-65-1
PROBLEM 6-65 Statement:
Figure P6-27 shows a cam and follower. Distances are given below. Find the velocities of points A and B, the velocity of transmission, velocity of slip, and 3 if 2 = 50 rad/sec (CW). Use a graphical method.
Given:
ω 50 rad sec
1
Distance from O2 to A:
a 1.890 in
Distance from O3 to B:
b 1.645 in
Assumptions: Roll-slide contact Solution: 1.
See Figure P6-27 and Mathcad file P0665.
Draw the linkage to scale and indicate the axes of slip and transmission as well as the directions of velocities of interest. Axis of slip Direction of VB
1.890
Axis of transmission A
B
3
2
O2
O3
1.645
Direction of VA
2.
Use equation 6.7 to calculate the magnitude of the velocity at point A. VA a ω
3.
VA 94.500
in sec
Use equation 6.5 to (graphically) determine the magnitude of the velocity components at point A. The equation to be solved graphically is VA = Vtrans + VAslip a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of Vslip, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of Vtrans, magnitude unknown. d. Complete the vector triangle by drawing Vslip from the tip of Vtrans to the tip of VA and drawing Vtrans from the tail of VA to the intersection of the Vslip construction line.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-65-2
50 in/sec
0
VB 2.304
V Bslip
Y 1.317 X
3.256
Vtrans
1.890 VA
4.
From the velocity triangle we have: Velocity scale factor:
5.
V A2slip
kv
50 in sec
1
in in
VA 1.890 in kv
VA 94.5
VB 2.304 in kv
VB 115.2
Vslip 3.256 in kv
Vslip 162.8
Vtrans 1.317 in kv
Vtrans 65.8
sec in sec in sec in sec
Determine the angular velocity of link 3 using equation 6.7. ω
VB b
ω 70.0
rad sec
CW
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-66-1
PROBLEM 6-66 Statement:
Figure P6-28 shows a quick-return mechanism with dimensions. Use a graphical method to calcula the velocities of points A, B, and C and the velocity of slip for the position shown.
Given:
Link lengths and angles: Link 1 (O2O4)
d 1.69 in
Angle O2O4 makes with X axis
Link 2 (L2)
a 1.00 in
Angle link 2 makes with X axis θ 99 deg
Link 4 (L4)
e 4.76 in
Link 5 (L5)
f 4.55 in
Offset (O2C)
g 2.86 in
Angular velocity of link 2 Solution: 1.
ω 10 rad sec
1
θ 15.5 deg
CCW
See Figure P6-28 and Mathcad file P0666.
Draw the linkage to scale and indicate the axes of slip and transmission as well as the directions of velocities of interest.
B
Axis of transmission
Direction of VA3
4
Y Axis of slip A 3 2.068
2 44.228° O2
X O4
2.
Use equation 6.7 to calculate the magnitude of the velocity at point A on links 2 and 3. VA2 a ω
3.
VA2 10.000
in sec
θVA2 θ 90 deg
θVA2 189.0 deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity components at point A on link 3. The equation to be solved graphically is VA3 = Vtrans + Vslip
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-66-2
a. Choose a convenient velocity scale and layout the known vector VA3. b. From the tip of VA3, draw a construction line with the direction of Vslip, magnitude unknown. c. From the tail of VA3, draw a construction line with the direction of Vtrans, magnitude unknown. d. Complete the vector triangle by drawing Vslip from the tip of Vtrans to the tip of VA3 and drawing Vtrans from the tail of VA3 to the intersection of the Vslip construction line. Y 0
5 in/sec
X
VA3 V trans
1.154
4.
kv
in in
Vtrans 1.154 in kv
Vtrans 5.770
sec in sec
The true velocity of point A on link 4 is Vtrans, VA4 5.77
in sec
Determine the angular velocity of link 4 using equation 6.7.
ω
VA4 c
c 2.068 in and
ω 2.790
rad
θ 44.228 deg
CCW
sec
Determine the magnitude and sense of the vector VB using equation 6.7. VB e ω
θVB θ 90 deg 8.
1
Vslip 8.170
From the linkage layout above:
7.
5 in sec
Vslip 1.634 in kv
VA4 Vtrans 6.
1.634
From the velocity triangle we have: Velocity scale factor:
5.
Vslip
VB 13.281
in sec
θVB 134.228 deg
Draw links 1, 4, 5, and 6 to a convenient scale. Indicate the directions of the velocity vectors of interest. (See next page.)
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-66-3
Direction of VCB Direction of VB B 5 5.805°
C 6
4 Direction of VC
Y
A 3
2 44.228° O2
X O4
9.
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point C. The equation to be solved graphically is VC = VB + VCB a. Choose a convenient velocity scale and layout the known vector VB. b. From the tip of VB, draw a construction line with the direction of VCB, magnitude unknown. c. From the tail of VB, draw a construction line with the direction of VC, magnitude unknown. d. Complete the vector triangle by drawing VCB from the tip of VB to the intersection of the VC construction line and drawing VC from the tail of VB to the intersection of the VCB construction line.
10. From the velocity triangle we have:
Velocity scale factor: VC 1.659 in kv
kv
5 in sec
VC 8.30
VB
1
0
5 in/sec
in in sec
Y
θVC 180 deg V CB X
VC 1.659
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-67-1
PROBLEM 6-67 Statement:
Given:
Figure P6-29 shows a drum pedal mechanism . For the dimensions given below, find and plot the mechanical advantage and the velocity ratio of the linkage over its range of motion. If the input velocity Vin is a constant and Fin is constant, find the output velocity, output force, and power in over the range of motion. Link lengths: Link 2 (O2A)
a 100 mm
Link 3 (AB)
b 28 mm
Link 4 (O4B)
c 64 mm
Link 1 (O2O4)
d 56 mm
Link 3 (AP)
rout 124 mm
Distance to force application: rin 48 mm
Link 2
Solution: 1.
1
Input force and velocity:
Fin 50 N
Vin 3 m sec
Range of positions of link 2:
θ 162 deg
θ 171 deg
See Figure P6-29 and Mathcad file P0667.
Draw the mechanism to scale and label it. P
3
B 3
Fin
4
Vin
A 2
2
x
1
O4
O2 y
2.
Calculate the range of 2 in the local coordinate system (required to calculate 3 and 4). Rotation angle of local xy system to global XY system:
α 180 deg
θ θ α θ α 1 deg θ α 3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 0.5600 2
K3
d c
K2 0.8750 2
2
a b c d 2 a c
2
K3 1.2850
A θ cos θ K1 K2 cos θ K3
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-67-2
B θ 2 sin θ
C θ K1 K2 1 cos θ K3 4.
Use equation 4.10b to find value of 4 for the open circuit.
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 5.
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. 2
d
K4
K5
b
2
2
c d a b
2
K4 2.0000
2 a b
K5 1.7543
D θ cos θ K1 K4 cos θ K5
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 6.
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ
Using equations 6.13d and 6.18a, where out is 3, calculate and plot the mechanical advantage of the linkage over the given range.
mA θ
rin rout a sin θ θ θ
b sin θ θ θ θ
MECHANICAL ADVANTAGE 0.14
Mechanical Advantage
7.
2 4 Dθ F θ
E θ
0.13
mA θ 0.12
0.11
0.1 162
164
166
168 θ α deg
Pedal Angle, deg
170
172
DESIGN OF MACHINERY - 5th Ed.
8.
SOLUTION MANUAL 6-67-3
Calculate and plot the velocity ratio using equation 6.13d,
1
mV θ
mA θ
VELOCITY RATIO 10
Velocity Ratio
8
6
mV θ
4 2 0 162
164
166
168
170
172
θ α deg Pedal Angle, deg
Calculate and plot the output velocity using equation 6.13a.
Vout θ Vin mV θ
OUTPUT VELOCITY
Velocity, m/sec
9.
20
Vout θ
sec m 10
0 162
164
166
168 θ α deg
Pedal Angle, deg
170
172
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-67-4
10. Calculate and plot the output force using equation 6.13a.
Fout θ Fin mA θ
OUTPUT FORCE 8
Force, N
6
Fout θ
4
N 2
0 162
164
166
168 θ α deg
Pedal Angle, deg
170
172
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-68-1
PROBLEM 6-68 Statement:
Figure 3-33 shows a sixbar slider crank linkage. Find all of its instant centers in the position shown:
Given:
Number of links n 6
Solution:
See Figure 3-33 and Mathcad file P0668.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
2.
n ( n 1)
C 15
2
Draw the linkage to scale and identify those ICs that can be found by inspection (8).
Y 2,3 3
3,4; 3,5; 4,5 1,6 at infinity
2 4 1,2 O2
5 X 6
1,4 O4 2.
1,5
5,6
Use Kennedy's Rule and a linear graph to find the remaining 7 ICs: I1,3; I1,5; I2,4; I2,5; I2,6; I3,6; and I4,6
1
I1,5: I1,6-I5,6 and I1,4-I4,5
6
2
5
3
I2,5: I1,2-I1,5 and I2,3-I3,5 I1,3: I1,2-I2,3 and I1,5-I3,5 I3,6: I1,6-I1,3 and I3,5-I5,6 I2,4: I2,3-I3,4 and I2,5-I4,5
4
2,5
I2,6: I1,2-I1,6 and I2,5-I5,6
2,6
I4,6: I1,4-I1,6 and I4,5-I5,6
Y 2,3
4,6 3,6
3 3,4; 3,5; 4,5; 2,4
2 4 1,2 O2
1,6 at infinity
5 X 6
1,4 O4 1,3
5,6
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-69-1
PROBLEM 6-69 Statement:
Calculate and plot the centrodes of instant center I24 of the linkage in Figure 3-33 so that a pair of noncircular gears can be made to replace the driver dyad 23.
Given:
Link lengths:
Solution: 1.
Input crank (L2)
L2 2.170
Fourbar coupler (L3)
L3 2.067
Output crank (L4)
L4 2.310
Fourbar ground link (L1)
L1 1.000
See Figure 3-33 and Mathcad file P0669.
Invert the linkage, grounding link 2 such that the input link is 3, the coupler is 4, and the output link is 1. a L3
b L4
c L1
d L2
2.
Define the input crank motion for this inversion: θ 47 deg 47.5 deg 102.5 deg
3.
Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the open circuit). K1
d
K2
a
K1 1.0498
2
d
K3
c
K2 2.1700
2
2
a b c d
2
2 a c
K3 1.1237
A θ cos θ K1 K2 cos θ K3
C θ K1 K2 1 cos θ K3
B θ 2 sin θ
θ θ 2 atan2 2 A θ B θ
2 4 A θ Cθ
B θ
θ θ if θ θ 2 π θ θ 2 π θ θ 5.
Calculate the coordinates of the intersection of links 1 and 3 in the xy coordinate system.
x242 θ
6.
tan θ tan θ θ
y242 θ x242 θ tan θ
Invert the linkage, grounding link 4 such that the input link is 1, the coupler is 2, and the output link is 3. a L1
8.
d tan θ θ
b L2
c L3
d L4
Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the open circuit). K1
d
K2
a
K1 2.3100
2
d
K3
c
K2 1.1176
2
2
a b c d 2 a c
K3 1.4271
A θ cos θ K1 K2 cos θ K3
θ θ 2 atan2 2 A θ B θ
C θ K1 K2 1 cos θ K3
B θ 2 sin θ
2 4 A θ Cθ
B θ
θ θ if θ θ 2 π θ θ 2 π θ θ 5.
Calculate the coordinates of the intersection of links 1 and 3 in the xy coordinate system.
2
DESIGN OF MACHINERY - 5th Ed.
x244 θ
SOLUTION MANUAL 6-69-2
tan θ tan θ θ d tan θ θ
y244 θ x244 θ tan θ
LINK 2 GROUNDED 5 4 3 2
1
y242 θ 0 1 2 3 4 5 4
3
2
1
0
1
x242 θ
7.
Define the input crank motion for this inversion: θ 41 deg 42 deg 241 deg LINK 4 GROUNDED 10 8 6 4 2
y244 θ
0 2 4 6 8
10 10 8 6 4 2
0
x244 θ
2
4
6
8
10
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-70a-1
PROBLEM 6-70a Statement:
Find the velocity of the slider in Figure 3-33 for 2 = 110 deg with respect to the global X axis assuming 2 = 1 rad/sec CW. Use a graphical method.
Given:
Link lengths: Link 2 (O2 to A)
a 2.170 in
Link 3 (A to B)
b 2.067 in
Link 4 (O4 to B)
c 2.310 in
Link 1 (O2 to O4)
d 1.000 in
Link 5 (B to C)
e 5.400 Crank angle:
θ2 110 deg
102 deg
Coordinate angle
ω 1 rad sec
Input crank angular velocity Solution: 1.
1
CW
See Figure P6-33 and Mathcad file P0670a.
Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest.
Direction of VA Y Direction of VBA A
147.635° 3 Direction of VB
B
2 4 O2
5 58.950°
158.818°
X
6
O4 C Direction of VC Direction of VCB 2.
3.
Use equation 6.7 to calculate the magnitude of the velocity at point A. in
VA a ω
VA 2.170
θVA θ2 90 deg
θVA 20.000 deg
sec
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B, the magnitude of the relati velocity VBA, and the angular velocity of link 3. The equation to be solved graphically is VB = VA + VBA a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line.
DESIGN OF MACHINERY - 5th Ed.
0
SOLUTION MANUAL 6-70a-2
VA
1 in/sec
Y
X V BA
VB 1.325
4.
From the velocity triangle we have: Velocity scale factor:
VB 1.325 in kv 5.
kv
1 in sec
1
in
VB 1.325
in
θVB 31.050 deg
sec
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point C, the magnitude of the relative velocity VCB, and the angular velocity of link 3. The equation to be solved graphically is VC = VB + VCB a. Choose a convenient velocity scale and layout the known vector VB. b. From the tip of VB, draw a construction line with the direction of VCB, magnitude unknown. c. From the tail of VB, draw a construction line with the direction of VC, magnitude unknown. d. Complete the vector triangle by drawing VCB from the tip of VB to the intersection of the VC construction line and drawing VC from the tail of VB to the intersection of the VCB construction line.
0
1.400
1 in/sec
Y VC X
VB 4.
V CB
From the velocity triangle we have: Velocity scale factor:
VC 1.400 in kv
kv
1 in sec
1
in
VC 1.400
in sec
θVC 0.0 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-70b-1
PROBLEM 6-70b Statement:
Find the velocity of the slider in Figure 3-33 for 2 = 110 deg with respect to the global X axis assuming 2 = 1 rad/sec CW. Use the method of instant centers.
Given:
Link lengths: Link 2 (O2 to A)
a 2.170 in
Link 3 (A to B)
b 2.067 in
Link 4 (O4 to B)
c 2.310 in
Link 1 (O2 to O4)
d 1.000 in
Link 5 (B to C)
e 5.400 Crank angle:
θ2 110 deg
102 deg
Coordinate angle
ω 1 rad sec
Input crank angular velocity Solution: 1.
1
CW
See Figure 3-33 and Mathcad file P0670b.
Draw the linkage to scale in the position given, find instant centers I1,3 and I1,5, and the distances from the pin joints to the instant centers. See Problem 6-68 for the determination of IC locations.
1,5
Y A 3 B
2 4
5 X
O2
C
6
O4 1,3 From the layout above: AI13 2.609 in 2.
BI13 1.641 in
BI15 9.406 in
CI15 9.896 in
Use equation 6.7 and inspection of the layout to determine the magnitude and direction of the velocity at point A.
DESIGN OF MACHINERY - 5th Ed.
3.
VA 2.170
θVA θ2 90 deg
θVA 20.0 deg
VA AI13
ω 0.832
rad
CW
sec
Determine the magnitude of the velocity at point B using equation 6.9b. The direction of VB is down and to the right VB 1.365
in sec
Use equation 6.9c to determine the angular velocity of link 5. ω
6.
sec
Determine the angular velocity of link 3 using equation 6.9a.
VB BI13 ω 5.
in
VA a ω
ω 4.
SOLUTION MANUAL 6-70b-2
VB BI15
ω 0.145
rad
CCW
sec
Determine the magnitude of the velocity at point C using equation 6.9b. VC CI15 ω
VC 1.436
in sec
to the right
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-70c-1
PROBLEM 6-70c Statement:
Find the velocity of the slider in Figure 3-33 for 2 = 110 deg with respect to the global X axis assuming 2 = 1 rad/sec CW. Use an analytical method.
Given:
Link lengths: Link 2 (O2 to A)
a 2.170 in
Link 3 (A to B)
b 2.067 in
Link 4 (O4 to B)
c 2.310 in
Link 1 (O2 to O4)
d 1.000 in
Link 5 (B to C)
e 5.400 in Crank angle:
θ2XY 110 deg
102 deg
Coordinate angle
ω2 1 rad sec
Input crank angular velocity Solution: 1.
1
CW
See Figure P6-33 and Mathcad file P0670c.
Draw the linkage to scale and label it.
Y A 3 B
2 4
5 X
O2
y 6
O4 x 2.
C
102°
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. Transform crank angle to the local xy coordinate system:
θ2 θ2XY K1
d
K1 0.4608
a 2
K3
θ2 212.000 deg
2
2
a b c d
K2
d c
2
2 a c
K3 0.6755
A cos θ2 K1 K2 cos θ2 K3 B 2 sin θ2 C K1 K2 1 cos θ2 K3 A 0.2662 3.
B 1.0598
C 2.3515
Use equation 4.10b to find values of 4 for the open circuit.
K2 0.4329
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-70c-2
2
θ4xy 2 atan2 2 A B 4.
2
d b
2
2
K4 0.4838
2 a b
E 2 sin θ2
E 1.0598
F K1 K4 1 cos θ2 K5
F 0.3808
Use equation 4.13 to find values of 3 for the open circuit.
2
E 4 D F
2 π
a ω2 sin θ2 θ3xy c sin θ4xy θ3xy
ω4 0.591
rad sec
Transform 4 back to the global XY system.
θ4 662.365 deg
Determine 5 and d, with respect to O4, using equation 4.17. cc 0 in
c sin θ4 cc π e
θ5 asin
θ5 158.818 deg
dd c cos θ4 e cos θ5
dd 6.272 in
Determine the angular velocity of link 5 using equation 6.22a:
ω5 7.
θ3xy 649.050 deg
Determine the angular velocity of link 4 for the open circuit using equations 6.18.
Offset:
6.
K5 0.5178
D 2.2370
θ4 θ4xy 5.
θ4xy 560.365 deg
D cos θ2 K1 K4 cos θ2 K5
ω4 7.
2
c d a b
K5
θ3xy 2 atan2 2 D E 6.
2 π
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
5.
B 4 A C
c cos θ4 ω4 e cos θ5
ω5 0.145
rad sec
Determine the velocity of pin C using equation 6.22b: VC c ω4 sin θ4 e ω5 sin θ5 VC 1.436
in sec
VC 1.436
in sec
arg VC 0.000 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-71-1
PROBLEM 6-71 Statement:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the angular velocity of link 4 and the linear velocity of slider 6 in the sixbar slider-crank linkage of Figure 3-33 as a function of the angle of input link 2 for a constant 2 = 1 rad/sec CW. Plot Vc both as a function of 2 and separately as a function of slider position as shown in the figure. What is the percent deviation from constant velocity over 240 deg < 2 < 270 deg and over 190 < 2 < 315 deg?
Given:
Link lengths: Input crank (L2)
a 2.170
Fourbar coupler (L3)
b 2.067
Output crank (L4)
c 2.310
Sllider coupler (L5)
e 5.40
d 1.000
Fourbar ground link (L1)
1
Crank velocity: Solution:
rad sec
See Figure 3-33 and Mathcad file P0671.
1.
This sixbar drag-link mechanism can be analyzed as a fourbar Grashof double crank in series with a crankslider mechanism using the output of the fourbar, link 4, as the input to the crank-slider.
2.
Define one revolution of the input crank: θ 0 deg 1 deg 360 deg
3.
Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the open circuit) in the global XY coordinate system. 2
K1
d
K2
a
K1 0.4608
K3
d c
K2 0.4329
2
2
a b c d
2
2 a c
K3 0.6755
A θ cos θ K1 K2 cos θ K3
C θ K1 K2 1 cos θ K3
B θ 2 sin θ
2 4 A θ Cθ 102 deg
θ θ 2 atan2 2 A θ B θ 4.
B θ
If the calculated value of 4 is greater than 2, subtract 2 from it and if it is negative, make it positive.
θ θ if θ θ 2 π θ θ 2 π θ θ
θ θ if θ θ 0 θ θ 2 π θ θ 5.
Determine the slider-crank motion using equations 4.16 and 4.17 with 4 as the input angle.
c sin θ θ π e
e cosθθ
θ θ asin
f θ c cos θ θ 5.
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
d b
2
K5
2
2
c d a b 2 a b
2
K4 0.4838
K5 0.5178
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-71-2
D θ cos θ K1 K4 cos θ K5
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 6.
Use equation 4.13 to find values of 3 for the open circuit.
2 4 Dθ F θ
θ θ 2 atan2 2 D θ E θ 7.
E θ
Determine the angular velocity of link 4 for the open circuit using equations 6.18.
θ
a c
sin θ θ θ
sin θ θ 102 deg θ θ
0.5 0.75 1
θ 1.25 1.5 1.75 2
0
45
90
135
180
225
270
θ deg
8.
Determine the angular velocity of link 5 using equation 6.22a:
ω θ
9.
c cos θ θ θ e cos θ θ
Determine the velocity of pin C using equation 6.22b:
e ωθ sinθθ
VC θ c θ sin θ θ
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-71-3
2 1 0
VC θ 1 2 3 4
0
45
90
135
180
225
270
315
360
θ deg
2
1
0
VC θ 1 2 3 4
3
4
5
f θ
6
7
8
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-72-1
PROBLEM 6-72 Statement:
Figure 3-34 shows a Stephenson's sixbar mechanism. Find all of its instant centers in the position shown:
Given:
Number of links n 6
Solution:
See Figure 3-34 and Mathcad file P0672.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
n ( n 1)
C 15
2
a.
In part (a) of the figure.
1.
Draw the linkage to scale and identify those ICs that can be found by inspection (7).
1,2 O2
2 4,6
1,6
5,6
2,3
6 O6 3
5
1,4 O4
4,5 4 5 2.
3,5
Use Kennedy's Rule and a linear graph to find the remaining 8 ICs: I1,3; I1,5; I2,4; I2,5; I2,6; I3,4; I3,6; and I4,6 I1,5: I1,6-I5,6 and I1,4-I4,5 I2,5: I1,2-I1,5 and I2,3-I3,5 I1,3: I1,2-I2,3 and I1,5-I3,5 I3,4: I1,4-I1,3 and I4,5-I3,5 I2,8: I2,3-I3,4 and I2,5-I4,5 I2,6: I1,2-I1,6 and I2,5-I5,6 I3,6: I1,3-I1,6 and I3,5-I5,6
1,2; 2,5; 2,4; 2,6
I4,6: I1,4-I1,6 and I4,5-I5,6 O2
1 6
2
2 6 3
5
3 4
1,6
5,6
2,3
5
O6
1,5
1,4 O4
4,5 4 5 3,5; 1,3; 3,4; 3,6
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-72-2
b.
In part (b) of the figure.
1.
Draw the linkage to scale and identify those ICs that can be found by inspection (7).
1,2 5,6 2
O2
2,3
5
6
4,5 3
4
5
1,6 O6 1,4 O4
3,5
2.
Use Kennedy's Rule and a linear graph to find the remaining 8 ICs: I1,3; I1,5; I2,4; I2,5; I2,6; I3,4; I3,6; and I4,6 I1,5: I1,6-I5,6 and I1,4-I4,5
3,6
1,2
I2,5: I1,2-I1,5 and I2,3-I3,5
5,6; 4,6 2 2,3
I1,3: I1,2-I2,3 and I1,5-I3,5
O2
5 4,5
3
5
I3,4: I1,4-I1,3 and I4,5-I3,5
6 1,6 O6 1,4; 1,5 O4
4
I2,4: I2,3-I3,4 and I2,5-I4,5 I2,6: I1,2-I1,6 and I2,5-I5,6
2,6
I3,6: I1,3-I1,6 and I3,5-I5,6 I4,6: I1,4-I1,6 and I4,5-I5,6
3,5; 3,4 To 1,3 1 6
2
5
3 4
2,5; 2,4; 2,8
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-72-3
c.
In part (c) of the figure.
1.
Draw the linkage to scale and identify those ICs that can be found by inspection (7).
2,3 1,6
2 1,2
4,5
O2 5 5
3
6 O6 1,4 O4
4 3,5
2.
5,6
Use Kennedy's Rule and a linear graph to find the remaining 8 ICs: I1,3; I1,5; I2,4; I2,5; I2,6; I3,4; I3,6; and I4,6 I1,5: I1,6-I5,6 and I1,4-I4,5
I2,4: I2,3-I3,4 and I2,5-I4,5
I2,5: I1,2-I1,5 and I2,3-I3,5
I2,6: I1,2-I1,6 and I2,5-I5,6
I1,3: I1,2-I2,3 and I1,5-I3,5
I3,6: I1,3-I1,6 and I3,5-I5,6
I3,4: I1,4-I1,3 and I4,5-I3,5
I4,6: I1,4-I1,6 and I4,5-I5,6
2,3 4,6
1
1,6
2 6
2
5
3
1,2; 2,5; 2,4: 2,6 4,5 O2
4
5 5
3
4 3,5; 1,3; 3,4; 3,6
1,5
6 O6 1,4 O4
5,6
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-73a-1
PROBLEM 6-73a Statement:
Find the angular velocity of link 6 in Figure 3-34b for 6 = 90 deg with respect to the global X axis assuming 2 = 10 rad/sec CW. Use a graphical method.
Given:
Link lengths: Link 2 (O2 to A)
g 1.556 in
Link 3 (A to B)
f 4.248 in
Link 4 (O4 to C)
c 2.125 in
Link 5 (C to D)
b 2.158 in
Link 6 (O6 to D)
a 1.542 in
Link 5 (B to D)
p 3.274 in
Link 1 X-offset
d X 3.259 in
Link 1 Y-offset
d Y 2.905 in
Angle CDB
δ5 36.0 deg
Output rocker angle:
θ 90 deg Global XY system ω 10 rad sec
Input crank angular velocity Solution: 1.
1
CW
See Figure 3-34b and Mathcad file P0673a.
Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest.
Direction of VCD Y
Direction of VD
Direction of VA X 2 A
D
O2
5
6
C Direction of VAB
O6 3
5
Direction of VC
4 O4 B
Direction of VBD 2.
Since this linkage is a Stephenson's II sixbar, we will have to start at link 6 and work back to link 2. We will assume a value for 6 and eventually find a value for 2. We will then multiply the magnitudes of all velocities by the ratio of the actual 2 to the found 2. Use equation 6.7 to calculate the magnitude of the velocity at point D. 1 Assume: CW ω 1 rad sec VD a ω
3.
VD 1.542
in sec
θVD 0 deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point C, the magnitude of the relative velocity VCD, and the angular velocity of link 5. The equation to be solved graphically is VC = VD + VCD a. b.
Choose a convenient velocity scale and layout the known vector VD. From the tip of VD, draw a construction line with the direction of VCD, magnitude unknown.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-73a-2
c. From the tail of VD, draw a construction line with the direction of VC, magnitude unknown. d. Complete the vector triangle by drawing VCD from the tip of VD to the intersection of the VC construction line and drawing VC from the tail of VD to the intersection of the VCD construction line. 0
1 in/sec
1.289
VC
VCD
1.309 127.003°
54.195° VD
4.
From the velocity triangle we have: kv
Velocity scale factor:
5.
1
in in
VC 1.289 in kv
VC 1.289
VCD 1.309 in kv
VCD 1.309
θVC 54.195 deg
sec in
θVCD 127.003 deg
sec
Determine the angular velocity of links 5 and 4 using equation 6.7. ω ω
6.
1 in sec
VCD b VC c
ω 0.607
rad
ω 0.607
rad
sec
sec
Determine the magnitude and sense of the vector VBD using equation 6.7. VBD p ω
VBD 1.986
in sec
θVBD θVCD δ5 7.
θVBD 163.003 deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B. The equation to be solved graphically is VB = VD + VBD a. b. c.
Choose a convenient velocity scale and layout the known vector VD. From the tip of VD, layout the (now) known vector VBD. Complete the vector triangle by drawing VB from the tail of VD to the tip of the VBD vector. 0
VBD VB 0.682 121.607°
VD
1 in/sec
DESIGN OF MACHINERY - 5th Ed.
8.
SOLUTION MANUAL 6-73a-3
From the velocity triangle we have:
kv
Velocity scale factor:
VB 0.682 in kv 9.
1 in sec
1
in
VB 0.682
in
θVB 121.607 deg
sec
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point A, the magnitude of the relati velocity VAB, and the angular velocity of link 2. The equation to be solved graphically is VA = VB + VAB a. Choose a convenient velocity scale and layout the known vector VB. b. From the tip of VB, draw a construction line with the direction of VAB, magnitude unknown. c. From the tail of VB, draw a construction line with the direction of VA, magnitude unknown. d. Complete the vector triangle by drawing VAB from the tip of VB to the intersection of the VA construction line and drawing VA from the tail of VB to the intersection of the VAB construction line. VAB
0
VB
1 in/sec
VA 0.645
131.690°
10. From the velocity triangle we have:
Velocity scale factor:
VA 0.645 in kv
kv
1 in sec
1
in
VA 0.645
in
θVA 131.690 deg
sec
11. Determine the angular velocity of link 2 with respect to the assumed value of 6 using equation 6.7.
ω
VA g
ω 0.415
rad sec
12. Calculate the actual value of the angular velocity of link 6.
ω rad ω sec
24.124
rad sec
CW
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-73b-1
PROBLEM 6-73b Statement:
Find the angular velocity of link 6 in Figure 3-34b for 6 = 90 deg with respect to the global X axis assuming 2 = 10 rad/sec CW. Use the method of instant centers.
Given:
Link lengths: Link 2 (O2 to A)
g 1.556 in
Link 3 (A to B)
f 4.248 in
Link 4 (O4 to C)
c 2.125 in
Link 5 (C to D)
b 2.158 in
Link 6 (O6 to D)
a 1.542 in
Link 5 (B to D)
p 3.274 in
Link 1 X-offset
d X 3.259 in
Link 1 Y-offset
d Y 2.905 in
Angle CDB
δ5 36.0 deg
Output rocker angle:
θ 90 deg Global XY system
Input crank angular velocity Solution: 1.
10 rad sec
1
CW
See Figure 3-34b and Mathcad file P0673b.
Draw the linkage to scale in the position given, find instant centers I1,3 and I1,5, and the distances from the pin joints to the instant centers. See Problem 6-72b for determination of IC locations. D A
2
O2
5
6
C 3
5
4 B
To 1,3
From the layout above: AI13 22.334 in
BI13 23.650 in
BI15 1.124 in
DI15 2.542 in
O6 1,5 O4
DESIGN OF MACHINERY - 5th Ed.
2.
SOLUTION MANUAL 6-73b-2
Start from point D with an assumed value for 6 and work to find 2. Then, use the ratio of the actual value of 2 to the found value to calculate the actual value of 6. Use equation 6.7 and inspection of the layout to determine the magnitude and direction of the velocity at point D. rad 1 sec in VD a VD 1.542 sec
θVD θ 90 deg 3.
Determine the angular velocity of link 3 using equation 6.9a. ω
4.
VD DI15
VB BI13
VB 0.682
in sec
ω 0.029
rad
CCW
sec
VA 0.644
in
to the left
sec
Use equation 6.9c to determine the angular velocity of link 2 based on the assumed value of 6.
8.
CW
sec
Determine the magnitude of the velocity at point A using equation 6.9b. VA AI13 ω
7.
rad
Use equation 6.9c to determine the angular velocity of link 3. ω
6.
ω 0.607
Determine the magnitude of the velocity at point B using equation 6.9b. VB BI15 ω
5.
θVD 0.0 deg
VA g
0.414
rad
CW
sec
Multiply the assumed vaue of 6 by the ratio of 21 over 22 to get the value of 6 for 2 = 10 rad/sec.
rad sec
24.166
rad sec
CW
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-73c-1
PROBLEM 6-73c Statement:
Find the angular velocity of link 6 in Figure 3-34b for 6 = 90 deg with respect to the global X axis assuming 2 = 10 rad/sec CW. Use an analytic method.
Given:
Solution: 1.
Link lengths: Link 2 (O2 to A)
g 1.556 in
Link 3 (A to B)
f 4.248 in
Link 4 (O4 to C)
c 2.125 in
Link 5 (C to D)
b 2.158 in
Link 6 (O6 to D)
a 1.542 in
Link 5 (B to D)
p 3.274 in
Link 1 X-offset
d X 3.259 in
Link 1 Y-offset
d Y 2.905 in
Angle CDB
δ5 36.0 deg
Link 1 (O4 to O6)
d 1.000 in
Output rocker angle:
θ6XY 90 deg
Global XY system
Input crank angular velocity
ω 10 rad sec
Coordinate rotationm angle
δ 90 deg
1
CW
See Figure 3-34b and Mathcad file P0673c.
Transform the crank angle to the local coordinate system. Draw the linkage to scale and label it.
θ6 θ6XY δ
θ6 180.000 deg
Y
2 A
D
O2
5
X
6
C
y O6
3
5
4 O4 B x
2.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d a
K1 0.6485
3.
K2
d c
K2 0.4706
2
K3
K3 0.4938
A cos θ6 K1 K2 cos θ6 K3
A 0.6841
B 2 sin θ6
B 0.0000
C K1 K2 1 cos θ6 K3
C 2.6129
Use equation 4.10b to find values of 4 for the crossed circuit.
2
2
a b c d 2 a c
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-73c-2
θ4 2 atan2 2 A B 4.
2
d
K5
b
θ4 234.195 deg
2
2
c d a b
2
2 a b
K4 0.4634
D cos θ6 K1 K4 cos θ6 K5
D 2.6407
E 2 sin θ6
E 0.0000
F K1 K4 1 cos θ6 K5
F 0.6563
K5 0.5288
Use equation 4.13 to find values of 5 for the crossed circuit.
θ51 2 atan2 2 D E 6.
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
5.
2
B 4 A C
2
E 4 D F
θ51 307.003 deg
Determine the angular velocity of links 4 and 5 for the open circuit using equations 6.18. Initially assume 6 = 1 rad/sec. then by trial and error, change it to make 2 = 10 rad/ sec CW.
ω6 1 rad sec
1
ω5
a ω6 sin θ4 θ6 b sin θ51 θ4
ω5 0.607
rad
ω4
a ω6 sin θ6 θ51 c sin θ4 θ51
ω4 0.607
rad
sec
sec
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-74-1
PROBLEM 6-74 Statement:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the angular velocity of link 6 in the sixbar linkage of Figure 3-34 as a function of 2 for a constant 2 = 1 rad/sec CW.
Given:
Link lengths:
Solution: 1.
Link 2 (O2 to A)
g 1.556 in
Link 3 (A to B)
f 4.248 in
Link 4 (O4 to C)
c 2.125 in
Link 5 (C to D)
b 2.158 in
Link 6 (O6 to D)
a 1.542 in
Link 5 (B to D)
p 3.274 in
Link 1 X-offset
d X 3.259 in
Link 1 Y-offset
d Y 2.905 in
Angle CDB
δ5 36.0 deg
Link 1 (O4 to O6)
d 1.000 in
Output rocker angle:
θ6XY 90 deg
Global XY system
Input crank angular velocity
ω 10 rad sec
Coordinate rotationm angle
δ 90 deg
1
CW
See Figure P6-34 and Mathcad file P0674.
This problem is long and may be more appropriate for a project assignment. The solution involves defining vector loops and solving the resulting equations using a method such as Newton-Raphson.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-75-1
PROBLEM 6-75 Statement:
Figure 3-35 shows a Stephenson's sixbar mechanism. Find all of its instant centers in the position shown:
Given:
Number of links n 6
Solution:
See Figure 3-35 and Mathcad file P0675.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
n ( n 1)
C 15
2
a.
In part (a) of the figure.
1.
Draw the linkage to scale and identify those ICs that can be found by inspection (7). 4,5
5 5,6
4
2 1,2 O2
O4
3
2,3
1,4
6
1,6 O6
3,4
2.
Use Kennedy's Rule and a linear graph to find the remaining 8 ICs: I1,3; I1,5; I2,4; I2,5; I2,6; I3,5; I3,6; and I4,6 4,5
5 4
2 2,3
1,2; 2,5; 2,4; 2,6 O2
3
O4
5,6 1,4
6
1,6 O6
4,6 1,3; 3,5; 3,6
1
3,4
I1,5: I1,6-I5,6 and I1,4-I4,5 I1,3: I1,2-I2,3 and I1,4-I3,4 I3,5: I1,5-I1,3 and I3,4-I4,5 I2,5: I1,2-I1,5 and I2,3-I3,5 I2,4: I2,3-I3,4 and I2,5-I4,5 I2,6: I1,2-I1,6 and I2,5-I5,6 I3,6: I1,3-I1,6 and I3,5-I5,6 I4,6: I1,4-I1,6 and I4,5-I5,6
1,5
6
2
5
3 4
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-75-2
b.
In part (b) of the figure.
1.
Draw the linkage to scale and identify those ICs that can be found by inspection (7). 1,6 2 2,3
1,4
1,2
3
O2
5
O4 5,6
3,4
2.
4,5 6
4
Use Kennedy's Rule and a linear graph to find the remaining 8 ICs: I1,3; I1,5; I2,4; I2,5; I2,6; I3,5; I3,6; and I4,6 I1,5: I1,6-I5,6 and I1,4-I4,5
I2,4: I2,3-I3,4 and I2,5-I4,5
I1,3: I1,2-I2,3 and I1,4-I3,4
I2,6: I1,2-I1,6 and I2,5-I5,6
I3,5: I1,5-I1,3 and I3,4-I4,5
I3,6: I1,3-I1,6 and I3,5-I5,6
I2,5: I1,2-I1,5 and I2,3-I3,5
I4,6: I1,4-I1,6 and I4,5-I5,6 1,5
1,2
1
2 6
2
2,3 3
5
3
4,5
1,4 O2
6
4 O4
5,6 4
1,2; 2,5; 2,4; 2,6
3,4; 1,3; 3,5; 3,6
4,6 5
1,6
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-76a-1
PROBLEM 6-76a Statement:
Use a compass and straightedge to draw the linkage in Figure 3-35 with link 2 at 90 deg and find the angular velocity of link 6 assuming 2 = 10 rad/sec CCW. Use a graphical method.
Given:
Link lengths: Link 2 (O2 to A)
a 1.000 in
Link 3 (A to B)
b 3.800 in
Link 4 (O4 to B)
c 1.286 in
Link 1 (O2 to O4)
d 3.857 in
Link 4 (O4 to D)
e 1.429 in
Link 5 (D to E)
f 1.286
Link 6 (O6 to E)
g 0.771 in
Crank angle:
θ2 90 deg
ω 10 rad sec
Input crank angular velocity Solution: 1.
1
CCW
See Figure 3-35 and Mathcad file P0676a.
Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest. Direction of VA Direction of VD
Y A
D
122.085°
100.938°
33.359°
2
5
3 4 O2
X
6 O4 O6 E
35.228°
B Direction of VE Direction of VBA
Direction of VED
Direction of VB
2.
3.
Use equation 6.7 to calculate the magnitude of the velocity at point A. in
VA a ω
VA 10.000
θVA θ2 90 deg
θVA 180.000 deg
sec
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B, the magnitude of the relati velocity VBA, and the angular velocity of link 3. The equation to be solved graphically is VB = VA + VBA a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-76a-2
Y
1.671 1.063 0
5 in/sec
VB
VBA
X VA
4.
From the velocity triangle we have: kv
Velocity scale factor:
VBA 1.063 in kv
VBA 5.315 6.497
c
θVB 147.915 deg
sec in
θVBA 56.641 deg
sec
rad
CW
sec
Calculate the magnitude and direction of VD. VD e
6.
in
VB 8.355
VB
1
in
VB 1.671 in kv
5.
5 in sec
VD 9.284
in
θVD 55.085 deg
sec
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point E, the magnitude of the relati velocity VED, and the angular velocity of link 3. The equation to be solved graphically is VE = VD + VED a. Choose a convenient velocity scale and layout the known vector VD. b. From the tip of VD, draw a construction line with the direction of VED, magnitude unknown. c. From the tail of VD, draw a construction line with the direction of VE, magnitude unknown. d. Complete the vector triangle by drawing VED from the tip of VD to the intersection of the VE construction line and drawing VE from the tail of VD to the intersection of the VED construction line.
4.
Y
From the velocity triangle we have: Velocity scale factor:
VE 2.450 in kv ω
kv
5 in sec
5 in/sec
1
in
VE 12.250
0
X
in sec
2.450
VE VD
g
ω 15.888
rad
CW
VE
sec V DE
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-76b-1
PROBLEM 6-76b Statement:
Use a compass and straightedge to draw the linkage in Figure 3-35 with link 2 at 90 deg and find the angular velocity of link 6 assuming 2 = 10 rad/sec CCW. Use the method of instant centers.
Given:
Link lengths: Link 2 (O2 to A)
a 1.000 in
Link 3 (A to B)
b 3.800 in
Link 4 (O4 to B)
c 1.286 in
Link 1 (O2 to O4)
d 3.857 in
Link 4 (O4 to D)
e 1.429 in
Link 5 (D to E)
f 1.286
Link 6 (O6 to E)
g 0.771 in
Crank angle:
θ2 90 deg
Input crank angular velocity Solution: 1.
ω 10 rad sec
1
CCW
See Figure 3-35 and Mathcad file P0676a.
Draw the linkage to scale in the position given, find instant centers I1,3 and I1,5, and the distances from the pin joints to the instant centers. See Problem 6-68 for the determination of IC locations.
1,5 A 2
D 5
3 4 O2
6 O4 O6 E
B
1,3 From the layout above: AI13 7.152 in 2.
BI13 5.975 in
DI15 0.947 in
EI15 1.249 in
Use equation 6.7 and inspection of the layout to determine the magnitude and direction of the velocity at point A. in VA a ω VA 10.000 sec
θVA θ2 90 deg 3.
BI15 1.740 in
θVA 180.0 deg
Determine the angular velocity of link 3 using equation 6.9a. ω
VA AI13
ω 1.398
rad sec
CCW
DESIGN OF MACHINERY - 5th Ed.
4.
Determine the magnitude of the velocity at point B using equation 6.9b. VB BI13 ω
5.
VB c
VD DI15
ω 6.496
rad
CW
sec
VD 9.283
in sec
ω 9.803
rad
CW
sec
Determine the magnitude of the velocity at point E using equation 6.9b. VE EI15 ω
9.
sec
Use equation 6.9c to determine the angular velocity of link 5. ω
8.
in
Determine the magnitude of the velocity at point D using equation 6.9b. VD e ω
7.
VB 8.354
Use equation 6.9c to determine the angular velocity of link 4. ω
6.
SOLUTION MANUAL 6-76b-2
VE 12.244
in sec
Use equation 6.9c to determine the angular velocity of link 6. ω
VE g
ω 15.88
rad sec
CW
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-76c-1
PROBLEM 6-76c Statement:
Use a compass and straightedge to draw the linkage in Figure 3-35 with link 2 at 90 deg and find the angular velocity of link 6 assuming 2 = 10 rad/sec CCW. Use an analytical method.
Given:
Link lengths: Link 2 (O2 to A)
a 1.000 in
Link 3 (A to B)
b 3.800 in
Link 4 (O4 to B)
c 1.286 in
Link 1 (O2 to O4)
d 3.857 in
Link 4 (O4 to D)
e 1.429 in
Link 5 (D to E)
f 1.286 in
Link 6 (O6 to E)
g 0.771 in
Link 1 (O4 to O6)
h 0.786 in
θ2 90 deg
Crank angle:
ω2 10 rad sec
Input crank angular velocity Solution: 1.
1
CCW
See Figure 3-35 and Mathcad file P0676c.
Draw the linkage to scale and label it.
Y A
D
122.085° 33.359°
2
100.938° 5
3 4 O2
X
6 O4 O6 E
35.228°
B 2.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a 2
K3
2
2
a b c d
d
K1 3.8570
c
K2 2.9992
2
K3 1.2015
2 a c
A cos θ2 K1 K2 cos θ2 K3 B 2 sin θ2 C K1 K2 1 cos θ2 K3 A 2.6555 3.
B 2.0000
Use equation 4.10b to find values of 4 for the crossed circuit.
θ41 2 atan2 2 A B 4.
C 5.0585
2
B 4 A C
θ41 237.915 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
d b
2
K5
2
2
c d a b 2 a b
D cos θ2 K1 K4 cos θ2 K5
2
K4 1.0150 D 7.6284
K5 3.7714
DESIGN OF MACHINERY - 5th Ed.
5.
SOLUTION MANUAL 6-76c-2
E 2 sin θ2
E 2.0000
F K1 K4 1 cos θ2 K5
F 0.0856
Use equation 4.13 to find values of 3 for the crossed circuit.
θ3 2 atan2 2 D E 6.
7.
2
E 4 D F
θ3 326.641 deg
Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18.
ω3
a ω2 sin θ41 θ2 b sin θ3 θ41
ω3 1.398
ω4
a ω2 sin θ2 θ3 c sin θ41 θ3
ω4 6.496
sec rad sec
Link 4 will be the input to the second fourbar (links 4, 5, 6, and 1). Let the angle that link 4 makes in the second fourbar be
θ42 θ41 157 deg 360 deg 8.
rad
θ42 34.915 deg
Determine the values of the constants needed for finding 6 from equations 4.8a and 4.10a. K1
h
K2
e 2
K3
e f
2
2
g h
h
K1 0.5500
g
K2 1.0195
2
K3 0.7263
2 e g
A cos θ42 K1 K2 cos θ42 K3 B 2 sin θ42
C K1 K2 1 cos θ42 K3 A 0.1603 9.
B 1.1447
C 0.3796
Use equation 4.10b to find values of 6 for the open circuit.
θ6 2 atan2 2 A B
2
B 4 A C
θ6 35.228 deg
10. Determine the values of the constants needed for finding 5 from equations 4.11b and 4.12. K4
2
h
K5
f
2
2
g h e f 2 e f
2
K4 0.6112
D cos θ42 K1 K4 cos θ42 K5
D 0.2408
E 2 sin θ42
E 1.1447
F K1 K4 1 cos θ42 K5
F 0.7807
K5 1.0119
11. Use equation 4.13 to find values of 5 for the open circuit.
θ5 2 atan2 2 D E
2
E 4 D F
θ5 280.938 deg
12. Determine the angular velocity of link6 for the open circuit using equations 6.18.
e ω4 sin θ42 θ5 g sin θ6 θ5
ω6
ω6 15.885
rad sec
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-77-1
PROBLEM 6-77 Statement:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the angular velocity of link 6 in the sixbar linkage of Figure 3-35 as a function of 2 for a constant 2 = 1 rad/sec CCW.
Given:
Link lengths: Link 2 (O2 to A)
a 1.000 in
Link 3 (A to B)
b 3.800 in
Link 4 (O4 to B)
c 1.286 in
Link 1 (O2 to O4)
d 3.857 in
Link 4 (O4 to D)
e 1.429 in
Link 5 (D to E)
f 1.286 in
Link 6 (O6 to E)
g 0.771 in
Link 1 (O4 to O6)
h 0.786 in
1 rad sec
Input crank angular velocity Solution:
1
CCW
See Figure 3-35 and Mathcad file P0677. 0 deg 1 deg 360 deg
1.
Define the range of the input angle:
2.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d a 2
K3
K2
d
2
2
2
a b c d
K1 3.8570
c
K3 1.2015
2 a c
K2 2.9992
A cos K1 K2 cos K3
B 2 sin
C K1 K2 1 cos K3 3.
Use equation 4.10b to find values of 4 for the crossed circuit.
2 atan2 2 A B 4.
2 4 A C
B
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
K4 1.0150
2 a b
D cos K1 K4 cos K5
K5 3.7714
E 2 sin
F K1 K4 1 cos K5 5.
Use equation 4.13 to find values of 3 for the crossed circuit.
2 atan2 2 D E 6.
Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18.
7.
2 4 D F
E
b sin a sin c sin a
sin
Link 4 will be the input to the second fourbar (links 4, 5, 6, and 1). Let the angle that link 4 makes in the second fourbar be
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-77-2
157 deg 360 deg 8.
Determine the values of the constants needed for finding 6 from equations 4.8a and 4.10a. h
K1
K2
e 2
2
e f
K3
2
g h
h
K1 0.5500
g 2
K3 0.7263
2 e g
K2 1.0195
K1 K2 cos K3 B' 2 sin
A' cos
K3
C' K1 K2 1 cos 9.
Use equation 4.10b to find values of 6 for the open circuit.
2 4 A' C'
2 atan2 2 A' B'
B'
10. Determine the values of the constants needed for finding 5 from equations 4.11b and 4.12. 2
h
K4
K5
f
2
2
g h e f
2
K4 0.6112
2 e f
K5 1.0119
K1 K4 cos K5 E' 2 sin
D' cos
K5
F' K1 K4 1 cos
11. Use equation 4.13 to find values of 5 for the open circuit.
2 atan2 2 D' E'
2 4 D' F'
E'
12. Determine the angular velocity of link6 for the open circuit using equations 6.18.
e sin g sin
ANGULAR VELOCITY - LINK 6
Angular velocity, rad/sec
3 2 1 0 1 2
0
45
90
135
180
225
Crank angle, deg
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-78-1
PROBLEM 6-78 Statement:
Figure 3-36 shows an eightbar mechanism. Find all of its instant centers in the position shown in part (a) of the figure:
Given:
Number of links n 8
Solution:
See Figure 3-36a and Mathcad file P0678.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
1.
n ( n 1)
C 28
2
Draw the linkage to scale and identify those ICs that can be found by inspection (11). 1,4; 1,8; 4,8 3,4
4,5
3 4 1,2 2
O4
O2
8
5,6
6
1,6
5
O6
2,3 7
7,8
5,7
2.
Use Kennedy's Rule and a linear graph to find the remaining 17 ICs. I1,3: I1,2-I2,3 and I1,4-I3,4
1,4; 1,8; 4,8; 5,8; 1,5
1,7
I4,6: I1,4-I1,6 and I4,5-I5,6 I2,4: I1,2-I1,4 and I2,3-I3,4
3,4; 1,3
4,5; 3,5
3
I2,8: I2,4-I4,8 and I1,2-I1,8
4 2
O2
1,2; 2,4
6
1,6
O4
8
O6 7,8
2,6; 2,8; 6,8; 2,7 at infinity
1 8
I2,6: I1,2-I1,6 and I2,4-I4,6
3,6; 3,7; 3,8 6,7
7
3
I2,5: I2,6-I5,6 and I2,3-I4,5
6
4 5
I3,5: I3,4-I4,5 and I2,3-I2,5 I3,6: I2,3-I2,6 and I3,5-I5,6 I3,7: I3,6-I6,7 and I2,3-I2,7
I5,8: I5,6-I6,8 and I4,5-I4,8
I3,8: I3,7-I7,8 and I2,3-I2,8
I1,5: I1,6-I5,6 and I1,4-I4,5
I4,7: I4,6-I6,7 and I3,4-I3,7
I1,7: I1,8-I7,8 and I1,6-I6,7
7 4,7
2
I6,8: I2,8-I2,6 and I1,8-I1,6
I2,7: I2,8-I7,8 and I2,6-I6,7
5
2,5
2,3
I6,7: I5,6-I5,7 and I6,8-I7,8
5,6; 4,6
5,7
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-79-1
PROBLEM 6-79 Statement:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the angular velocity of link 8 in the linkage of Figure 3-36 as a function of 2 for a constant 2 = 1 rad/sec CCW.
Given:
Link lengths: Input crank (L2)
a 0.450
First coupler (L3)
b 0.990
Common rocker (O4B)
c 0.590
First ground link (O2O4)
d 1.000
Common rocker (O4C)
a' 0.590
Second coupler (CD)
b' 0.325
Output rocker (L6)
c' 0.325
Second ground link (O4O6) d' 0.419
Link 7 (L7)
e 0.938
Link 8 (L8)
f 0.572
Link 5 extension (DE)
p 0.823
Angle DCE
δ 7.0 deg
Angle BO4C
α 128.6 deg
Input crank angular velocity Solution: 1.
1 rad sec
1
CCW
See Figure 3-36 and Mathcad file P0679.
See problem 4-43 for the position solution. The velocity solution will use the same vector loop equations for links 5, 6, 7, and 8, differentiated with respect to time. This problem is suitable for a project assignment and is probably too long for an overnight assignment.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-80-1
PROBLEM 6-80 Statement:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the magnitude and direction of the velocity of point P in Figure 3-37a as a function of 2 for a constant 2 = 1 rad/sec CCW. Also calculate and plot the velocity of point P versus point A.
Given:
Link lengths:
Solution: 1.
Link 2 (O2 to A)
a 0.136
Link 3 (A to B)
b 1.000
Link 4 (B to O4)
c 1.000
Link 1 (O2 to O4)
d 1.414
Coupler point:
Rpa 2.000
0 deg
Crank speed:
1 rad sec
1
See Figure 3-37a and Mathcad file P0680.
Determine the range of motion for this Grashof crank rocker. θ 0 deg 2 deg 360 deg
2.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 10.3971 2
K3
c
K2 1.4140
2
2
a b c d
2
K3 7.4187
2 a c
d
A θ cos θ K1 K2 cos θ K3
B θ 2 sin θ
C θ K1 K2 1 cos θ K3 3.
Use equation 4.10b to find values of 4 for the open circuit.
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 4.
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
2 a b
K4 1.4140
K5 7.4187
E θ 2 sin θ F θ K1 K4 1 cos θ K5
D θ cos θ K1 K4 cos θ K5
5.
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ 6.
2 4 Dθ F θ
E θ
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18.
DESIGN OF MACHINERY - 5th Ed.
a
a
ω θ
ω θ 7.
b
c
SOLUTION MANUAL 6-80-2
sin θ θ θ θ
sin θ θ θ θ
sin θ θ θ
sin θ θ θ
Determine the velocity of point A using equations 6.19.
VA θ a sin θ j cos θ 8.
VA θ VA θ
Determine the velocity of the coupler point P using equations 6.36.
VPA θ Rpa ω θ sin θ θ j cos θ θ
VP θ VA θ VPA θ
Plot the magnitude and direction of the velocity at coupler point P.
Magnitude:
VP θ VP θ
Direction:
θVP1 θ arg VP θ θVP θ if θVP1 θ 0 θVP1 θ 2 π θVP1 θ MAGNITUDE
0.18
0.16 Velocity, mm/sec
9.
0.14
0.12
0.1
0
45
90
135
180
Crank Angle, deg
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-80-3
DIRECTION
Vector Angle, deg
300
200
100
0
0
45
90
135
180
Crank Angle, deg
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-81-1
PROBLEM 6-81 Statement:
Calculate the percent error of the deviation from constant velocity magnitude of point P in Figure 3-37a.
Given:
Link lengths:
Solution: 1.
Link 2 (O2 to A)
a 0.136
Link 3 (A to B)
b 1.000
Link 4 (B to O4)
c 1.000
Link 1 (O2 to O4)
d 1.414
Coupler point:
Rpa 2.000
0 deg
Crank speed:
1 rad sec
1
See Figure 3-37a and Mathcad file P0681.
Determine the range of motion for this Grashof crank rocker. θ 0 deg 2 deg 360 deg
2.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 10.3971 2
K3
c
K2 1.4140
2
2
a b c d
2
K3 7.4187
2 a c
d
A θ cos θ K1 K2 cos θ K3
B θ 2 sin θ
C θ K1 K2 1 cos θ K3 3.
Use equation 4.10b to find values of 4 for the open circuit.
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 4.
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
2 a b
K4 1.4140
K5 7.4187
E θ 2 sin θ F θ K1 K4 1 cos θ K5
D θ cos θ K1 K4 cos θ K5
5.
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ 6.
2 4 Dθ F θ
E θ
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18.
ω θ
a b
sin θ θ θ θ sin θ θ θ
DESIGN OF MACHINERY - 5th Ed.
ω θ 7.
a c
SOLUTION MANUAL 6-81-2
sin θ θ θ θ sin θ θ θ
Determine the velocity of point A using equations 6.19.
VA θ a sin θ j cos θ 8.
VA θ VA θ
Determine the velocity of the coupler point P using equations 6.36.
VPA θ Rpa ω θ sin θ θ j cos θ θ
VP θ VA θ VPA θ
Calculate the magnitude of the velocity at coupler point P.
VP θ VP θ
Magnitude:
err θ
Deviation from constant speed:
VP θ VA θ
VA θ
DEVIATION FROM CONSTANT SPEED 30
20 Percent Error
9.
10
0
10
20
0
45
90
135
180
Crank Angle, deg
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-82-1
PROBLEM 6-82 Statement:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the magnitude and the direction of the velocity of point P in Figure 3-37b as a function of 2. Also calculate and plot the velocity of point P versus point A.
Given:
Link lengths: Input crank (L2)
a 0.50
First coupler (AB)
b 1.00
Rocker 4 (O4B)
c 1.00
Rocker 5 (L5)
c' 1.00
Ground link (O2O4)
d 0.75
Second coupler 6 (CD)
b' 1.00
Coupler point (DP)
p 1.00
Distance to OP (O2OP)
d' 1.50
1
Crank speed: Solution: 1.
rad sec
See Figure 3-37b and Mathcad file P0682.
Links 4, 5, BC, and CD form a parallelogram whose opposite sides remain parallel throughout the motion of the fourbar 1, 2, AB, 4. Define a position vector whose tail is at point D and whose tip is at point P and another whose tail is at O4 and whose tip is at point D. Then, since R5 = RAB and RDP = -R4, the position vector from O2 to P is P = R1 + RAB - R4. Separating this vector equation into real and imaginary parts gives the equations for the X and Y coordinates of the coupler point P.
XP = d b cos θ c cos θ
YP = b sin θ c sin θ
2.
Define one revolution of the input crank: θ 0 deg 0.5 deg 360 deg
3.
Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the crossed circuit). K1
d a
K1 1.5000
2
d
K2
K3
c
K2 0.7500
2
2
a b c d
2
2 a c
K3 0.8125
A θ cos θ K1 K2 cos θ K3
C θ K1 K2 1 cos θ K3
B θ 2 sin θ
2 4 A θ Cθ 2 π
θ θ 2 atan2 2 A θ B θ 4.
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
2 a b
D θ cos θ K1 K4 cos θ K5
K4 0.7500
K5 0.8125
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 5.
Use equation 4.13 to find values of 3 for the crossed circuit.
θ θ 2 atan2 2 D θ E θ 6.
2 4 Dθ F θ 2 π
E θ
Define a local xy coordinate system with origin at OP and with the positive x axis to the right. The coordinates of P are transformed to xP = XP - d', yP = YP.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-82-2
c cosθθ d'
xP θ d b cos θ θ
c sinθθ
yP θ b sin θ θ 7.
Use equations 6.18 to calculate 3 and 4.
a
a
θ
b
θ
sin θ θ θ θ
sin θ θ θ θ
sin θ θ θ
sin θ θ θ
Differentiate the position equations with respect to time to get the velocity components.
c θ sinθθ
VPx θ b θ sin θ θ
c θ cosθθ
VPy θ b θ cos θ θ
2 VPyθ2
VP θ
VPx θ
MAGNITUDE 0.6 0.4 Velocity, mm/sec
8.
c
0.2 0 0.2 0.4 0.6
0
45
90
135
180
Crank Angle, deg x Component y Component
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-83-1
PROBLEM 6-83 Statement:
Find all instant centers of the linkage in Figure P6-30 in the position shown.
Given:
Number of links n 4
Solution:
See Figure P6-30 and Mathcad file P0683.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
1.
n ( n 1)
C6
2
Draw the linkage to scale and identify those ICs that can be found by inspection (4). 3,4
4
P1
1,4
O4
3
P2
Y
2,3 2 1,2
X O2
1,3
2.
Use Kennedy's Rule and a linear graph to find the remaining 2 ICs. I1,3: I1,2-I2,3 and I1,4-I3,4
I2,4: I1,2-I1,4 and I2,3-I3,4
3,4
4
P1 O4 P2
1,4 3 Y
2,3 2 1,2
X O2 2,4
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-84a-1
PROBLEM 6-84a Statement:
Find the angular velocities of links 3 and 4 and the linear velocity of points A, B and P1 in the XY coordinate system for the linkage in Figure P6-30 in the position shown. Assume that 2 = 45 deg in the XY coordinate system and 2 = 10 rad/sec. The coordinates of the point P1 on link 4 are (114.68, 33.19) with respect to the xy coordinate system. Use a graphical method.
Given:
Link lengths: Link 2 (O2 to A)
a 14.00 in
Link 3 (A to B)
b 80.00 in
Link 4 (O4 to B)
c 51.26 in
Link 1 X-offset
d X 47.5 in
Link 1 Y-offset
d Y 76.00 in 12.00 in
Coupler point x-offset
p x 114.68 in
Coupler point y-offset p y 33.19 in
Crank angle:
θ 45 deg ω 10 rad sec
Input crank angular velocity
1
CCW
See Figure P6-30 and Mathcad file P0684a. Solution: 1. Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest. Direction of VB Direction of VBA B
x
4
P1 O4
29.063°
P2
3
Y
99.055° 45.000° A 2
X O2
Direction of VA
y
2.
Use equation 6.7 to calculate the magnitude of the velocity at point A. VA a ω
3.
VA 140.000
in sec
θVA 45 deg 90 deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B, the magnitude of the relati velocity VBA, and the angular velocity of link 3. The equation to be solved graphically is VB = VA + VBA a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line.
DESIGN OF MACHINERY - 5th Ed.
0
SOLUTION MANUAL 6-84a-2
100 in/sec
0.409" Y 1.206" 119.063° VBA VB
VA
135.000°
9.055°
4.
X
From the velocity triangle we have: kv
Velocity scale factor:
5.
VB 120.600
VBA 0.409 in kv
VBA 40.900
in sec
θVB 119.063 deg
in sec
Determine the angular velocity of links 3 and 4 using equation 6.7.
ω
VBA
ω 0.511
b VB
ω 2.353
c
rad sec
rad sec
Transform the xy coordinates of point P1 into the XY system using equations 4.0b. Coordinate transformation angle
7.
1
in
VB 1.206 in kv
ω
6.
100 in sec
δ atan2 d X d Y
δ 126.582 deg
PX p x cos δ p y sin δ
PX 94.998 in
PY p x sin δ p y cos δ
PY 72.308 in
Calculate the distance from O4 to P1 and use equation 6.7 to calculate the velocity at P1. Distance from O4 to P1: VP1 e ω
e
PX d X 2 PY d Y 2 VP1 113.446
in sec
e 48.219 in
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-84b-1
PROBLEM 6-84b Statement:
Find the angular velocities of links 3 and 4 and the linear velocity of points A, B and P1 in the XY coordinate system for the linkage in Figure P6-30 in the position shown. Assume that 2 = 45 deg in the XY coordinate system and 2 = 10 rad/sec. The coordinates of the point P1 on link 4 are (114.68, 33.19) with respect to the xy coordinate system. Use the method of instant centers.
Given:
Link lengths: Link 2 (O2 to A)
a 14.00 in
Link 3 (A to B)
b 80.00 in
Link 4 (O4 to B)
c 51.26 in
Link 1 X-offset
d X 47.5 in
Link 1 Y-offset
d Y 76.00 in 12.00 in
Coupler point x-offset
p x 114.68 in
Coupler point y-offset p y 33.19 in
Crank angle:
θ 45 deg
Input crank angular velocity Solution: 1.
ω 10 rad sec
1
CCW
See Figure P6-30 and Mathcad file P0684b.
Draw the linkage to scale in the position given, find instant center I1,3 and the distance from the pin joints to the instant center.
1,3
B
4
P1 O4 P2
3 Y
A 2
X O2
From the layout above: AI13 273.768 in 2.
BI13 235.874 in
Use equation 6.7 and inspection of the layout to determine the magnitude and direction of the velocity at point A.
DESIGN OF MACHINERY - 5th Ed.
3.
VA 140.000
θVA θ 90 deg
θVA 135.0 deg
VA
ω 0.511
AI13
CW
sec
VB 120.622
in sec
Use equation 6.9c to determine the angular velocity of link 4. VB
ω 2.353
c
rad
CCW
sec
Transform the xy coordinates of point P1 into the XY system using equations 4.0b. Coordinate transformation angle
7.
rad
Determine the magnitude of the velocity at point B using equation 6.9b.
ω 6.
sec
Determine the angular velocity of link 3 using equation 6.9a.
VB BI13 ω 5.
in
VA a ω
ω 4.
SOLUTION MANUAL 6-84b-2
δ atan2 d X d Y
PX p x cos δ p y sin δ
PX 94.998 in
PY p x sin δ p y cos δ
PY 72.308 in
δ 126.582 deg
Calculate the distance from O4 to P1 and use equation 6.7 to calculate the velocity at P1. Distance from O4 to P1: VP1 e ω
e
PX d X 2 PY d Y 2
VP1 113.467
in sec
e 48.219 in
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-84c-1
PROBLEM 6-84c Statement:
Find the angular velocities of links 3 and 4 and the linear velocity of points A, B and P1 in the XY coordinate system for the linkage in Figure P6-30 in the position shown. Assume that 2 = 45 deg in the XY coordinate system and 2 = 10 rad/sec. The coordinates of the point P1 on link 4 are (114.68, 33.19) with respect to the xy coordinate system. Use an analytical method.
Given:
Link lengths: Link 2 (O2 to A)
a 14.00 in
Link 3 (A to B)
b 80.00 in
Link 4 (O4 to B)
c 51.26 in
Link 1 X-offset
d X 47.5 in
Link 1 Y-offset
d Y 76.00 in 12.00 in
Coupler point x-offset
p x 114.68 in
Coupler point y-offset p y 33.19 in
Crank angle:
θ2XY 45 deg
δ 126.582 deg
Coordinate transformation angle: Input crank angular velocity Solution: 1.
XY coord system
ω2 10 rad sec
1
CCW
See Figure P6-30 and Mathcad file P0684c.
Draw the linkage to scale and label it. x
B
97.519°
4
P1 O4
27.528°
P2
3 81.582°
Y
126.582° 45.000° A 2
X O2
y
Transform the crank angle from global to local coordinate system:
θ2 θ2XY δ
2.
2
d
Distance O2O4:
θ2 81.582 deg dX dY
2
d 79.701 in
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d a 2
K3
K2
d
2
2
2
a b c d 2 a c
c
K1 5.6929
K2 1.5548
K3 1.9340
A cos θ2 K1 K2 cos θ2 K3
B 2 sin θ2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-84c-2
C K1 K2 1 cos θ2 K3 A 3.8401 3.
B 1.9785
Use equation 4.10b to find values of 4 for the crossed circuit.
2
θ4 2 atan2 2 A B 4.
C 7.2529
B 4 A C
θ4 262.482 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
K4 0.9963
2 a b
D cos θ2 K1 K4 cos θ2 K5
D 10.0081
E 2 sin θ2
E 1.9785
F K1 K4 1 cos θ2 K5 5.
7.
F 1.0849
Use equation 4.13 to find values of 3 for the crossed circuit.
2
θ3 2 atan2 2 D E 6.
K5 4.6074
E 4 D F
θ3 332.475 deg
Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18.
ω3
a ω2 sin θ4 θ2 b sin θ3 θ4
ω3 0.511
ω4
a ω2 sin θ2 θ3 c sin θ4 θ3
ω4 2.353
rad sec
rad sec
Determine the velocity of points A and B for the crossed circuit using equations 6.19. VA a ω2 sin θ2 j cos θ2 VA ( 138.492 20.495i )
θVAXY arg VA δ
in sec
VA 140.000
in sec
θVAXY 135.000 deg
VB c ω4 sin θ4 j cos θ4 VB ( 119.588 15.781j )
in sec
θVBXY arg VB δ 8.
in sec
θVBXY 119.064 deg
Calculate the distance from O4 to P1 and the angle BO4P1. e
Distance from O4 to P1:
py px d
δ4 180 deg atan 9.
VB 120.624
px d 2 py 2
δ4 136.503 deg
Determine the velocity of point P1 using equations 6.35. VP1 e ω4 sin θ4 δ j cos θ4 δ VP1 ( 55.122 99.181j )
θVPXY arg VP1 δ
in sec
VP1 113.469
in sec
θVPXY 245.646 deg
e 48.219 in
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-85-1
PROBLEM 6-85 Statement:
Given:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the absolute velocity of point P1 in Figure P6-30 as a function of 2 for 2 = 10 rad/sec. The coordinates of the point P1 on link 4 are (114.68, 33.19) with respect to the xy coordinate system. Link lengths: Link 2 (O2 to A)
a 14.00 in
Link 3 (A to B)
b 80.00 in
Link 4 (O4 to B)
c 51.26 in
Link 1 X-offset
d X 47.5 in
Link 1 Y-offset
d Y 76.00 in 12.00 in
Coupler point x-offset
p x 114.68 in
Coupler point y-offset p y 33.19 in
Crank angle:
θ2XY 0 deg 1 deg 360 deg 126.582 deg
Coordinate transformation angle: Input crank angular velocity Solution: 1.
XY coord system
10 rad sec
1
CCW
See Figure P6-30 and Mathcad file P0685.
Draw the linkage to scale and label it. x
B
97.519°
4
P1 O4
27.528°
P2
3 81.582°
Y
126.582° 45.000° A 2
X O2
y
Transform the crank angle from global to local coordinate system: θ2XY θ2XY d
Distance O2O4: 2.
2
dX dY
2
d 79.701 in
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d a 2
K3
K2
d
2
2
2
a b c d
c
K2 1.5548
K3 1.9340
2 a c
K1 5.6929
A θ2XY cos θ2XY K1 K2 cos θ2XY K3
C θ2XY K1 K2 1 cos θ2XY K3 B θ2XY 2 sin θ2XY
DESIGN OF MACHINERY - 5th Ed.
3.
SOLUTION MANUAL 6-85-2
Use equation 4.10b to find values of 4 for the crossed circuit.
θ2XY 2 atan2 2 A θ2XY B θ2XY 4.
B θ2XY 4 A θ2XY C θ2XY 2
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
K4 0.9963
2 a b
K5 4.6074
D θ2XY cos θ2XY K1 K4 cos θ2XY K5
F θ2XY K1 K4 1 cos θ2XY K5 E θ2XY 2 sin θ2XY 5.
Use equation 4.13 to find values of 3 for the crossed circuit.
θ2XY 2 atan2 2 D θ2XY E θ2XY 6.
a sin θ2XY θ2XY c sin θ2XY θ2XY
Calculate the distance from O4 to P1 and the angle BO4P1. e
Distance from O4 to P1:
px d 2 py 2
e 48.219 in
py px d
180 deg atan
136.503 deg
Determine and plot the velocity of point P1 using equations 6.35.
VP1 θ2XY e θ2XY sin θ2XY j cos θ2XY VP1 θ2XY VP1 θ2XY
θVPXY θ2XY arg VP1 θ2XY
P1 VELOCITY MAGNITUDE 200
150 Velocity - in/sec
8.
2
Determine the angular velocity of link 4 for the crossed circuit using equations 6.18. θ2XY
7.
E θ2XY 4 D θ2XY F θ2XY
100
50
0
0
45
90
135
180
225
270
Crank Angle - Global XY System
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-85-3
P1 VELOCITY DIRECTION
Velocity angle - Global XY System
300
200
100
0
0
45
90
135
180
225
270
Crank Angle - Global XY System
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-86-1
PROBLEM 6-86 Statement:
Find all instant centers of the linkage in Figure P6-31 in the position shown.
Given:
Number of links n 4
Solution:
See Figure P6-31 and Mathcad file P0686.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
1.
n ( n 1)
C6
2
Draw the linkage to scale and identify those ICs that can be found by inspection (4).
Y y
O2 1,2
X 2 1 O4 1,4 3,4
4
P
2,3 x 3 2.
Use Kennedy's Rule and a linear graph to find the remaining 2 ICs. I1,3: I1,2-I2,3 and I1,4-I3,4
I2,4: I1,2-I1,4 and I2,3-I3,4
Y y
O2 1,2
X 2 1 O4 1,4
1,3
3,4
4
P
2,3 x 3 2,4
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-87a-1
PROBLEM 6-87a Statement:
Find the angular velocities of links 3 and 4 and the linear velocity of point P in the XY coordinate system for the linkage in Figure P6-31 in the position shown. Assume that 2 = -94.121 deg in the XY coordinate system and 2 = 1 rad/sec. The position of the coupler point P on link 3 with respect to point A is: p = 15.00, 3 = 0 deg. Use a graphical method.
Given:
Link lengths: Link 2 (O2 to A)
a 9.17 in
Link 3 (A to B)
b 12.97 in
Link 4 (O4 to B)
c 9.57 in
Link 1 X-offset
d X 2.79 in
Link 1 Y-offset
d Y 6.95 in
Coupler point data:
p 15.00 in
δ 0 deg
Crank angle:
θ 94.121 deg ω 1 rad sec
Input crank angular velocity Solution: 1.
1
CW
See Figure P6-31 and Mathcad file P0687a.
Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest. Y y
O2
X Direction VBA 2 1
68.121°
Direction VPA
26.000° O4 B
4 Direction VA
P
A 72.224° 3 60.472°
Direction VB
x
2.
3.
Use equation 6.7 to calculate the magnitude of the velocity at point A. in
VA a ω
VA 9.170
θVA θ 90 deg
θVA 184.121 deg
sec
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B, the magnitude of the relati velocity VBA, and the angular velocity of link 3. The equation to be solved graphically is VB = VA + VBA a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-87a-2
Y 0
25 in/sec VA X
1.798"
1.783" VBA
4.
From the velocity triangle we have: Velocity scale factor:
5.
kv
25 in sec
1
in in
VB 1.783 in kv
VB 44.575
VBA 1.798 in kv
VBA 44.950
ω
VBA b VB
in
θVBA 274.103 deg
sec
c
ω 3.466
rad
ω 4.658
rad
sec
sec
Determine the magnitude and sense of the vector VPA using equation 6.7. VPA p ω
VPA 51.985
in sec
θVPA θVBA 7.
θVB 262.352 deg
sec
Determine the angular velocity of links 3 and 4 using equation 6.7. ω
6.
VB
θVPA 274.103 deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point P. The equation to be solved graphically is VP = VA + VPA a. b. c.
8.
Choose a convenient velocity scale and layout the known vector VA. From the tip of VA, layout the (now) known vector VPA. Complete the vector triangle by drawing VP from the tail of VA to the tip of the VPA vector. Y
From the velocity triangle we have:
0
25 in/sec VA
Velocity scale factor:
VP 2.059 in kv
θP 96.051 deg
kv
25 in sec
1
X
in
VP 51.475
96.051°
in 2.059"
sec VPA
VP
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-87b-1
PROBLEM 6-87b Statement:
Find the angular velocities of links 3 and 4 and the linear velocity of point P in the XY coordinate system for the linkage in Figure P6-31 in the position shown. Assume that 2 = -94.121 deg in the XY coordinate system and 2 = 1 rad/sec. The position of the coupler point P on link 3 with respect to point A is: p = 15.00, 3 = 0 deg. Use the method of instant centers.
Given:
Link lengths: Link 2 (O2 to A)
a 9.174 in
Link 3 (A to B)
b 12.971 in
Link 4 (O4 to B)
c 9.573 in
Link 1 X-offset
d X 2.790 in
Link 1 Y-offset
d Y 6.948 in
Coupler point data:
p 15.00 in
δ 0 deg
Crank angle:
θ 94.121 deg ω 1 rad sec
Input crank angular velocity Solution: 1.
1
CW
See Figure P6-31 and Mathcad file P0687b.
Draw the linkage to scale in the position given, find instant center I1,3 and the distance from the pin joints to the instant center.
Y y
O2
X 2 1 O4
1,3
B
4
P
A x 3 From the layout above: AI13 2.647 in 2.
θVA 184.121 deg
Determine the angular velocity of link 3 using equation 6.9a. ω
4.
PI13 14.825 in
Use equation 6.7 and inspection of the layout to determine the magnitude and direction of the velocity at point A. in VA a ω VA 9.174 sec
θVA θ 90 deg 3.
BI13 12.862 in
VA AI13
ω 3.466
rad
CW
sec
Determine the magnitude of the velocity at point B using equation 6.9b. in VB BI13 ω VB 44.577 sec
DESIGN OF MACHINERY - 5th Ed.
5.
Use equation 6.9c to determine the angular velocity of link 4. ω
6.
SOLUTION MANUAL 6-87b-2
VB c
ω 4.657
rad
CW
sec
Determine the magnitude of the velocity at point P using equation 6.9b. VP PI13 ω
VP 51.381
in sec
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-87c-1
PROBLEM 6-87c Statement:
Find the angular velocities of links 3 and 4 and the linear velocity of point P in the XY coordinate system for the linkage in Figure P6-31 in the position shown. Assume that 2 = -94.121 deg in the XY coordinate system and 2 = 1 rad/sec. The position of the coupler point P on link 3 with respect to point A is: p = 15.00, 3 = 0 deg. Use an analytical method.
Given:
Link lengths: Link 2 (O2 to A)
a 9.174 in
Link 3 (A to B)
b 12.971 in
Link 4 (O4 to B)
c 9.573 in
Link 1 X-offset
d X 2.790 in
Link 1 Y-offset
d Y 6.948 in
Coupler point data:
p 15.00 in
δ 0 deg
Crank angle:
θ2XY 94.121 deg Global XY coord system δ 68.121 deg
Coordinate transformation angle:
ω2 1 rad sec
Input crank angular velocity Solution: 1.
1
CW
See Figure P6-31 and Mathcad file P0687c.
Draw the linkage to scale and label it.
Y y
O2
X 2 1
68.121°
26.000° O4 B
4
P
A 72.224° 3 60.472° Transform crank angle into local xy coordinate system:
θ2 θ2XY δ
θ2 26.000 deg d
Calculate distance O2O4: 2.
x
2
dX dY
2
d 7.487 in
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a 2
K3
2
2
d c
a b c d
K1 0.8161
2
2 a c
K3 0.3622
A cos θ2 K1 K2 cos θ2 K3 C K1 K2 1 cos θ2 K3 A 0.2581
K2 0.7821
B 0.8767
C 0.4234
B 2 sin θ2
DESIGN OF MACHINERY - 5th Ed.
3.
SOLUTION MANUAL 6-87c-2
Use equation 4.10b to find values of 4 for the crossed circuit.
2
θ4 2 atan2 2 A B 4.
B 4 A C
2 π
θ4 60.488 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d b
2
2
c d a b
K5
2
K4 0.5772
2 a b
D cos θ2 K1 K4 cos θ2 K5
D 0.3096
E 2 sin θ2
E 0.8767
F K1 K4 1 cos θ2 K5 5.
7.
F 0.4749
Use equation 4.13 to find values of 3 for the crossed circuit.
2
θ3 2 atan2 2 D E 6.
K5 0.9111
E 4 D F
2 π
θ3 72.236 deg
Determine the angular velocity of links 3 and 4 using equations 6.18.
ω3
a ω2 sin θ4 θ2 b sin θ3 θ4
ω3 3.467
rad
ω4
a ω2 sin θ2 θ3 c sin θ4 θ3
ω4 4.658
rad
sec
sec
Determine the velocity of points A and B using equations 6.19. VA a ω2 sin θ2 j cos θ2 VA ( 4.022 8.246i)
in
VA 9.174
sec
θVAXY arg VA δ
in sec
θVAXY 184.121 deg
VB c ω4 sin θ4 j cos θ4 VB ( 38.807 21.967i )
in
VB 44.593
sec
θVBXY arg VB δ 8.
in sec
θVBXY 97.633 deg
Determine the velocity of the coupler point P for the crossed circuit using equations 6.36.
VPA p ω3 sin θ3 δ j cos θ3 δ VPA ( 49.529 15.867i )
in s
VP VA VPA
VP ( 45.507 24.113i )
θVPXY arg VP δ
θVPXY 96.039 deg
in sec
VP 51.501
in sec
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-88-1
PROBLEM 6-88 Statement:
Figure P6-32 shows a fourbar double slider known as an elliptical trammel. F ind all its instant centers in the position shown.
Given:
Number of links n 4
Solution:
See Figure P6-32 and Mathcad file P0688.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
1.
n ( n 1)
C6
2
Draw the linkage to scale and identify those ICs that can be found by inspection (4). Y
3,4 4
2,3
To 1,4 at infinity
3 2
1
X
To 1,2 at infinity
2.
Use Kennedy's Rule and a linear graph to find the remaining 2 ICs. I1,3: I1,2-I2,3 and I1,4-I3,4
I2,4: I1,2-I1,4 and I2,3-I3,4
Y
2,3
3,4 4 1,3
To 1,4 at infinity
3 2
1
To 2,4 at infinity
X
To 1,2 at infinity
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-89-1
PROBLEM 6-89 Statement:
The elliptical trammel in Figure P6-32 must be driven by rotating link 3 in a full circle. Points on line AB describe ellipses. Find and draw (mannually or with a computer) the fixed and moving centrodes of instant center I13. (Hint: These are called the Cardan circles.)
Solution:
See Figure P6-32 and Mathcad file P0689.
1.
The instant center I13 is shown as the point P in the diagram below. The length of link 3, c, is constant and it is a diagonal of the rectangle OBPA. Therefore, the other diagonal, OP, has a constant length, c, regardless of the current lengths a and b. Thus, the point P (I13) travels on a circle of radius c. This circle is the fixed centrode.
c 4 B
P 3
b O
2 A
1
a
2.
Now, invert the mechanism by holding link 3 fixed and allowing the slots to move, which can only be in a circle about O. The locus of points P will then be a circle of radius 0.5c with center at O'. This is the moving centrode.
c 4 B
P 3 O'
b O
2 A
1
a
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-90-1
PROBLEM 6-90 Statement:
Derive analytical expressions for the velocities of points A and B in Figure P6-32 as a function of 3, 3 and the length AB of link 3. Use a vector loop equation.
Solution:
See Figure P6-32 and Mathcad file P0690.
1.
Establish the global XY system such that the coordinates of the intersection of the slot centerlines is at (d X,d Y ). Then, define position vectors R1X, R1Y , R2, R3, and R4 as shown below. Y 4 3 R3 R2
B
3
C 2
R4
A 1
R 1Y
X R 1X
2.
Write the vector loop equation: R1Y + R2 + R3 - R1X - R4 = 0 then substitute the complex number notation for each position vector. The equation then becomes:
π π j j θ 3 2 j ( 0) j ( 0) 2 dY e a e c e dX e b e = 0 j
3.
Differentiate this equation with respect to time. j d a j c e dt
4.
j π 2 d b e = 0 dt
Substituting the Euler identity into this equation gives: Va jc cos θ3 j sin θ3 Vb j = 0
5.
Separate this equation into its real (x component) and imaginary (y component) parts, setting each equal to zero. Va c sin θ3 = 0
6.
c cos θ3 Vb = 0
Solve for the two unknowns Va and Vb in terms of the independent variables 3 and 3 Va = c sin θ3
Vb = c cos θ3
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-91-1
PROBLEM 6-91 Statement:
The linkage in Figure P6-33a has link 2 at 120 deg in the global XY coordinate system. Find 6
Given:
and VD in the global XY coordinate system for the position shown if 2 = 10 rad/sec CCW. Use the velocity difference graphical method. Link lengths: Link 2 (O2 to A)
a 6.20 in
Link 3 (A to B)
b 4.50 in
Link 4 (O4 to B)
c 3.00 in
Link 5 (C to D)
e 5.60 in
Link 3 (A to C)
p 2.25 in
Link 4 (O4 to D)
f 3.382 in
Link 1 X-offset
d X 7.80 in
Link 1 Y-offset
d Y 0.62 in
Angle ACB
δ3 0.0 deg
Angle BO4D
δ4 110.0 deg
Input rocker angle:
θ 120 deg
Global XY system
ω 10 rad sec
Input crank angular velocity Solution: 1.
1
CW
See Figure P6-33a and Mathcad file P0691.
Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest. Direction of VA Direction of VCA Axis of slip
A Y
87.972° 113.057° 5
D
C
110°
2
3
120°
Axis of transmission Direction of VDC
411.709° B O4
O2 Direction of VB
2.
Direction of VBA
Use equation 6.7 to calculate the magnitude of the velocity at point A. VA a ω
3.
X
VA 62.000
in sec
θVA 120 deg 90 deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B, the magnitude of the relative velocity VBA, and the angular velocity of link 5. The equation to be solved graphically is VB = VA + VBA a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-91-2
Y 0
50 in/sec
X VA
1.206 VB VBA 78.291°
1.433 4.
From the velocity triangle we have: Velocity scale factor:
5.
8.
1
in in
VB 60.300
VBA 1.433 in kv
VBA 71.650
θVB 78.291 deg
sec in
θVBA 23.057 deg
sec
Determine the angular velocity of links 3 and 4 using equation 6.7.
ω
7.
50 in sec
VB 1.206 in kv
ω
6.
kv
VBA b VB c
ω 15.922
rad
CCW
sec
ω 20.100
rad
CW
sec
Determine the magnitude and sense of the vector VCA using equation 6.7. in
VCA p ω
VCA 35.825
θVCA θVBA δ3
θVCA 23.057 deg
sec
Determine the magnitude and sense of the vector Vtrans using equation 6.7. in
Vtrans f ω
Vtrans 67.978
θtrans θVB δ4
θtrans 31.709 deg
sec
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point C. The equation to be solved graphically is VC = VA + VCA a. b. c.
Choose a convenient velocity scale and layout the known vector VA. From the tip of VA, layout the (now) known vector VCA. Complete the vector triangle by drawing VC from the tail of VA to the tip of the VCA vector.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-91-3
Y 0
50 in/sec
X VA VC
0.991
VCA 114.720° 9.
From the velocity triangle we have:
Velocity scale factor:
VC 0.991 in kv 9.
kv
50 in sec
1
in
VC 49.550
in
θVC 114.720 deg
sec
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point D, the magnitude of the relative velocity VDC. The equations to be solved (simultaneously) graphically are VD = VC + VDC
and
VD = Vtrans + Vslip
a. Choose a convenient velocity scale and layout the known vector VC. b. From the tip of VC, draw a construction line with the direction of VDC, magnitude unknown. c. Repeat steps a and b with Vtrans and Vslip. c. From the tail of VC, draw a construction line to the intersection of the two construction lines. d. Complete the vector polygon by drawing VDC from the tip of VC to the intersection of the VD construction line and drawing VD from the tail of VC to the intersection of the VDC construction line. 10. From the velocity polygon we have:
Y VD Vslip
VDC
Velocity scale factor: kv
50 in sec
3.045
95.189°
1
in
Vtrans VD 3.045 in kv
VD 152.250
in
X
sec
θVD 95.189 deg
0 VC
ω
20.100
rad sec
50 in/sec
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-92-1
PROBLEM 6-92 Statement:
The linkage in Figure P6-33a has link 2 at 120 deg in the global XY coordinate system. Find 6
Given:
and VD in the global XY coordinate system for the position shown if 2 = 10 rad/sec CCW. Use the instant center graphical method. Link lengths: Link 2 (O2 to A)
a 6.20 in
Link 3 (A to B)
b 4.50 in
Link 4 (O4 to B)
c 3.00 in
Link 5 (C to D)
e 5.60 in
Link 3 (A to C)
p 2.25 in
Link 4 (O4 to D)
f 3.382 in
Link 1 X-offset
d X 7.80 in
Link 1 Y-offset
d Y 0.62 in
Angle ACB
δ3 0.0 deg
Angle BO4D
δ4 110.0 deg
Input rocker angle:
θ 120 deg
Global XY system
ω 10 rad sec
Input crank angular velocity Solution: 1.
1
CW
See Figure P6-33a and Mathcad file P0692.
Draw the linkage to scale in the position given, find instant centers I1,3 and I1,5, and the distances from the pin joints to the instant centers. 2,3 A 6
5,6 5
D
Y
1,6
3
1,5
3,5 C
3,6 2 1,3
To 4,6 at infinity
B 3,4
4 O4
1,2 1,4
O2
X
From the layout above: AI13 3.893 in 2.
BI13 3.788 in
θVA 210.0 deg
Determine the angular velocity of link 3 using equation 6.9a. ω
4.
CI15 1.411 in DI15 4.333 in DI16 7.573 in
Use equation 6.7 and inspection of the layout to determine the magnitude and direction of the velocity at point A. in VA a ω VA 62.000 sec
θVA θ 90 deg 3.
CI13 3.113 in
VA AI13
ω 15.926
rad sec
Determine the magnitude of the velocity at point B using equation 6.9b. VB BI13 ω
VB 60.328
in sec
CCW
DESIGN OF MACHINERY - 5th Ed.
5.
Use equation 6.9c to determine the angular velocity of links 4 and 6.
VB c
w6 6.
rad
w6 20.109
rad
CW
sec CW
sec
VC 49.578
in sec
Determine the angular velocity of link 5 using equation 6.9a. ω
8.
20.109
Determine the magnitude of the velocity at point C using equation 6.9b. VC CI13 ω
7.
SOLUTION MANUAL 6-92-2
VC CI15
ω 35.137
rad
CW
sec
Determine the magnitude of the velocity at point D using equation 6.9b. VD DI15 ω
VD 152.247
in sec
@ 95.185 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-93-1
PROBLEM 6-93 Statement:
The linkage in Figure P6-33a has link 2 at 120 deg in the global XY coordinate system. Find 6
Given:
and VD in the global XY coordinate system for the position shown if 2 = 10 rad/sec CCW. Use an analytical method. Link lengths: Link 2 (O2 to A)
a 6.20 in
Link 3 (A to B)
b 4.50 in
Link 4 (O4 to B)
c 3.00 in
Link 5 (C to D)
e 5.60 in
Link 3 (A to C)
p 2.25 in
Link 4 (O4 to D)
f 3.382 in
Link 1 X-offset
d X 7.80 in
Link 1 Y-offset
d Y 0.62 in
Angle ACB
δ3 0.0 deg
Angle BO4D
δ4 110.0 deg
Input rocker angle:
θ2XY 120 deg
Global XY system
δ 175.455 deg
Coordinate transformation angle:
ω2 10 rad sec
Input crank angular velocity Solution: 1.
1
CCW
See Figure P6-33a and Mathcad file P0691.
Draw the linkage to scale and label it. A
6
Y
71.487° D
5
C
110°
120°
55.455° B 16.254°
4
x
2
3
O4
175.455°
O2 y
Calculate the distance O2O4:
2.
d
2
dX dY
2
d 7.825 in
Transform 2XY into the local coordinate θ2 θ2XY δ θ2 55.455 deg system: Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d a 2
K3
K2
d
2
2
2
a b c d
c
2 a c
K1 1.2620 K3 2.3767
A cos θ2 K1 K2 cos θ2 K3 C K1 K2 1 cos θ2 K3 A 0.2028 3.
B 1.6474
K2 2.6082
C 1.5927
Use equation 4.10b to find values of 4 for the open circuit.
B 2 sin θ2
X
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-93-2
θ41 2 atan2 2 A B 4.
2
B 4 A C
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
2 a b
D cos θ2 K1 K4 cos θ2 K5
K5 1.9877
E 1.6474
F K1 K4 1 cos θ2 K5
F 0.3067
Use equation 4.13 to find values of 3 for the open circuit.
θ3 2 atan2 2 D E 6.
K4 1.7388 D 1.6967
E 2 sin θ2 5.
θ41 163.746 deg
2
E 4 D F
2 π
θ3 648.512 deg
At this point write vector loop equations for links 3, 4, 5, and 6 to solve for the position and velocty of link 6.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-94-1
PROBLEM 6-94 Statement:
The linkage in Figure P6-33b has link 3 perpendicular to the X-axis and links 2 and 4 are parallel to each other. Find 3, VA, VB, and VP if 2 = 15 rad/sec CW. Use the velocity difference graphical method.
Given:
Link lengths: Link 2 (O2 to A)
a 2.75 in
Link 3 (A to B)
b 3.26 in
Link 4 (O4 to B)
c 2.75 in
Link 1 (O4 to B)
d 4.43 in
Coupler point data:
p 1.63 in
δ 0 deg ω 15 rad sec
Input crank angular velocity Solution: 1.
1
CW
See Figure P6-33b and Mathcad file P0694.
Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest.
Direction of VB Direction of VBA B
O4
O2 36.351° A
Direction of VA θ 36.352 deg 2.
3.
Use equation 6.7 to calculate the magnitude of the velocity at point A. in
VA a ω
VA 41.250
θVA θ 90 deg
θVA 126.352 deg
sec
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B, the magnitude of the relative velocity VBA, and the angular velocity of link 3. The equation to be solved graphically is VB = VA + VBA a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line.
4.
Without actually drawing the vector polygon, we see that VA and VB have the same angle with respect to the X axis. Therefore,
DESIGN OF MACHINERY - 5th Ed.
VB VA 5.
ω
7.
and
VBA 0
in sec
θVB θVA
θVBA 0 deg
Determine the angular velocity of links 3 and 4 using equation 6.7. ω
6.
SOLUTION MANUAL 6-94-2
VBA b VB c
ω 0.000
rad sec
ω 15.000
rad sec
Determine the magnitude and sense of the vector VPA using equation 6.7. in
VPA p ω
VPA 0.000
θVPA θVBA
θVPA 0.000 deg
sec
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point P. The equation to be solved graphically is VP = VA + VPA a. b. c.
8.
Choose a convenient velocity scale and layout the known vector VA. From the tip of VA, layout the (now) known vector VPA. Complete the vector triangle by drawing VP from the tail of VA to the tip of the VPA vector.
Again, without drawing the vector polygon we see that, VP VA At this instant, link 3 is in pure translation with every point on it having the same velocity.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-95-1
PROBLEM 6-95 Statement:
The linkage in Figure P6-33b has link 3 perpendicular to the X-axis and links 2 and 4 are parallel to each other. Find 3, VA, VB, and VP if 2 = 15 rad/sec CW. Use the instant center graphical method.
Given:
Link lengths: Link 2 (O2 to A)
a 2.75 in
Link 3 (A to B)
b 3.26 in
Link 4 (O4 to B)
c 2.75 in
Link 1 (O4 to B)
d 4.43 in
Coupler point data:
p 1.63 in
δ 0 deg ω 15 rad sec
Input crank angular velocity
1
CW
See Figure P6-33b and Mathcad file P0695. Solution: 1. Draw the linkage to scale in the position given, find instant center I1,3 and the distance from the pin joints to the instant center.
Y
B
4 P
O2 2
O4
X
3 36.351° A
Since O2A and O4B are parallel, I1,3 is at infinity and link 3 does not rotate for this instantaneous position of the linkage. Thus, link 3 is in pure translation and all points on it will have the same velocity. 2.
Use equation 6.7 and inspection of the layout to determine the magnitude and direction of the velocity at points A, B, and P, which will be the same. VA a ω
VA 41.250
in sec
36.351 deg
θVA 90 deg
θVA 126.351 deg
The velocity vectors for points B and P are identical to VA.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-96-1
PROBLEM 6-96 Statement:
The linkage in Figure P6-33b has link 3 perpendicular to the X-axis and links 2 and 4 are parallel to each other. Find 3, VA, VB, and VP if 2 = 15 rad/sec CW. Use an analytical method.
Given:
Link lengths: Link 2 (O2 to A)
a 2.75 in
Link 3 (A to B)
b 3.26 in
Link 4 (O4 to B)
c 2.75 in
Link 1 (O4 to B)
d 4.43 in
Coupler point data:
p 1.63 in
δ 0 deg w2 15 rad sec
Input crank angular velocity Solution: 1.
1
CW
See Figure P6-33b and Mathcad file P0696.
Draw the linkage to a convenient scale and label it.
B
P
O2
O4 36.351°
A θ 36.351 deg 2.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d a 2
K3
K2
d
2
2
2
a b c d
K1 1.6109
c
K3 1.5949
2 a c
B 2 sin θ C K1 K2 1 cos θ K3
A cos θ K1 K2 cos θ K3
3.
A 0.5081 B 1.1855 C 1.1029
Use equation 4.10b to find values of 4 for the open circuit.
2
2 atan2 2 A B 4.
B 4 A C
2 π
143.649 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
d b
2
K5
2
2
c d a b
2
2 a b
E 2 sin θ F K1 K4 1 cos θ K5
D cos θ K1 K4 cos θ K5
5.
K2 1.6109
Use equation 4.13 to find values of 3 for the open circuit.
K4 1.3589 D 1.3983 E 1.1855 F 0.2127
K5 1.6873
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-96-2
2
2 atan2 2 D E 6.
7.
E 4 D F
2 π
89.995 deg
Determine the angular velocity of links 3 and 4 using equations 6.18.
a w2 sin θ b sin
0.000
rad
a w2 sin θ c sin
15.000
rad
sec
sec
Determine the velocity of points A and B using equations 6.19.
VA a w2 sin θ j cos θ VA ( 24.450 33.223i )
in
VA 41.250
sec
θVAXY arg VA
in sec
θVAXY 126.351 deg
VB c sin j cos VB ( 24.450 33.223i )
in
VB 41.250
sec
θVBXY arg VB 8.
in sec
θVBXY 126.351 deg
Determine the velocity of the coupler point P using equations 6.36.
VPA p sin δ j cos δ
VPA 1.936 10 VP VA VPA
θVPXY arg VP
4
in sec
8
1.676i 10
VP 41.250
in sec
θVPXY 126.351 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-97-1
PROBLEM 6-97 Statement:
The crosshead linkage shown in Figure P6-33c has 2 DOF with inputs at crossheads 2 and 5. Find instant centers I1,3 and I1,4.
Given:
Number of links n 5
Solution:
See Figure P6-33c and Mathcad file P06976.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
1.
n ( n 1)
C 10
2
Draw the linkage to scale and identify those ICs that can be found by inspection (4). 2,5 at infinity 2 3,4
1,2 at infinity
3 2,3 1
4 5
4,5 1,5 at infinity
2.
Use Kennedy's Rule and a linear graph to find the remaining ICs required to find I1,3 and I1,4. I2,4: I1,2-I1,4 and I2,3-I3,4
2,5 at infinity 3,4
I3,5: I2,3-I2,5 and I4,5-I3,4
2
I1,4: I1,2-I2,4 and I1,5-I4,5 I1,3: I1,2-I2,3 and I1,4-I3,4
1,3; 1,4
1,2 at infinity
3 2,3; 2,4 1
4 5
4,5; 3,5 1,5 at infinity
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-98-1
PROBLEM 6-98 Statement:
The crosshead linkage shown in Figure P6-33c has 2 DOF with inputs at cross heads 2 and 5. Find VB, VP3, and VP4 if the crossheads are each moving toward the origin of the XY coordinate system with a speed of 20 in/sec. Use a graphical method.
Given:
Link lengths: Link 3 (A to B)
b 34.32 in
Link 4 (B to C)
c 50.4 in
Link 2 Y-offset
a 59.5 in
Link 5 X-offset
d 57 in
AP3 31.5 in
BP3 22.2 in
BP4 41.52 in
CP 4 27.0 in
Coupler point data:
V2Y 20 in sec
Input velocities: Solution: 1.
1
V5X 20 in sec
1
See Figure P6-33c and Mathcad file P0698.
Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest.
Y
Direction of VP3A Direction of VBA Direction of VBC
2 P3
A Direction of VA
3
Direction of VP4C B
P4
4
5 X C Direction of VC 2.
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B, the magnitude of the relative velocities VBA, VBC and the angular velocities of links 3 and 4. The equations to be solved (simultaneously) graphically are VB = VA + VBA
VC = VB + VCB
a. Choose a convenient velocity scale and layout the known vectors VA and VC b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tip of VC, draw a construction line with the direction of VCB, magnitude unknown. d. From the origin, draw a construction line to the intersection of the two construction lines in steps b and c. e. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line. And then do the same with the VCB vector (See next page)
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-98-2
5.932 Y 0
VBC
10 in/sec
VB VBA
60.123° 32.718°
46.990°
X
VC
6.004
4.385
VA
3.
From the velocity triangles we have: Velocity scale factor:
4.
10 in sec
1
in in
VB 4.385 in kv
VB 43.850
VBA 6.004 in kv
VBA 60.040
VBC 5.932 in kv
VBC 59.320
θVB 46.990 deg
sec in
θVBA 60.123 deg
sec in
θVBC 32.718 deg
sec
Determine the angular velocity of links 3 and 4 using equation 6.7. ω ω
6.
kv
VBA b VBC c
ω 1.749
rad
CCW
sec
ω 1.177
rad
CW
sec
Determine the magnitudes of the vectors VP3A and VP4C using equation 6.7. VP3A AP3 ω
VP3A 55.107
in sec
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-98-3
VP4C CP4 ω
7.
in
VP4C 47.234
sec
Use equation 6.5 to (graphically) determine the magnitude of the velocities at points P3 and P4. The equations be solved graphically are VP3 = VA + VP3A a. b. c. d.
VP4 = VC + VP4C
Choose a convenient velocity scale and layout the known vector VA. From the tip of VA, layout the (now) known vector VP3A. Complete the vector triangle by drawing VP3 from the tail of VA to the tip of the VP3A vector. Repeat for VP4.
V P3A
V P3 Y
0
3.537
10 in/sec
V P4 164.011°
V P4C
104.444°
X
VC 6.598
VA
8.
From the velocity triangles we have:
Velocity scale factor:
kv
10 in sec
1
in
VP3 3.537 in kv
VP3 35.370
VP4 6.598 in kv
VP4 65.980
in sec in sec
θP3 104.444 deg θP4 164.011 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-99-1
PROBLEM 6-99 Statement:
The linkage in Figure P6-33d has the path of slider 6 perpendicular to the global X-axis and link 2 aligned with the global X-axis. Find VA in the position shown if the velocity of the slider is 20 in/sec downward. Use the velocity difference graphical method.
Given:
Link lengths: Link 2 (O2 to A)
a 12 in
Link 3 (A to B)
b 24 in
Link 4 (O4 to B)
c 18 in
Link 5 (C to D)
f 24 in
Link 4 (O4 to C)
e 18 in
Link 1 X-offset
d X 19 in
Link 1 Y-offset
d Y 28 in
δ4 0.0 deg
Angle BO4C
Solution: 1.
Output crank angle:
θ 0 deg
Input slider velocity
VD 20 in sec
Global XY system 1
θVD 90.0 deg
See Figure P6-33d and Mathcad file P0699.
Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest. Direction of VC
Direction of VCD C
Direction of VB O4 19.963°
4
116.161°
B 5
Y
D 3
O2
2
114.410°
A
6 X
Direction of VD
Direction of VAB Direction of VA
2.
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point C. The equation to be solved graphically is VC = VD + VCD a. Choose a convenient velocity scale and layout the known vector VD. b. From the tip of VD, draw a construction line with the direction of VCD, magnitude unknown. c. From the tail of VD, draw a construction line with the direction of VC, magnitude unknown. d. Complete the vector triangle by drawing VCD from the tip of VD to the intersection of the VC construction line and drawing VC from the tail of VD to the intersection of the VCD construction line.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-99-2
Y
X 1.806 0
10 in/sec 70.037°
VC VD V CD
3.
From the velocity triangle we have: Velocity scale factor:
VC 1.806 in kv 4.
kv
10 in sec
1
in
VC 18.060
in
θVC 70.037 deg
sec
Determine the magnitude and sense of the vector VB . VB VC
VB 18.060
in sec
θVB θVC 180 deg 5.
θVB 109.963 deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point A. The equation to be solved graphically is VA = VB + VAB a. Choose a convenient velocity scale and layout the known vector VB. b. From the tip of VB, draw a construction line with the direction of VAB, magnitude unknown. c. From the tail of VB, draw a construction line with the direction of VA, magnitude unknown. d. Complete the vector triangle by drawing VAB from the tip of VB to the intersection of the VA construction line and drawing VA from the tail of VB to the intersection of the VAB construction line.
6.
From the velocity triangle we have:
Y
Velocity scale factor: kv
10 in sec
1
VA VB
in
VA 1.977 in kv VA 19.770
V AB
1.977
0
in sec
θVA 90.0 deg
X
10 in/sec
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-100-1
PROBLEM 6-100 Statement:
The linkage in Figure P6-33a has the path of slider 6 perpendicular to the global X-axis and link 2 aligned with the global X-axis. Find VA in the position shown if the velocity of the slider is 20 in/sec downward. Use the instant center graphical method.
Given:
Link lengths:
Solution: 1.
Link 2 (O2 to A)
a 12 in
Link 3 (A to B)
b 24 in
Link 4 (O4 to B)
c 18 in
Link 5 (C to D)
f 24 in
Link 3 (A to C)
p 2.25 in
Link 4 (O4 to C)
e 18 in
Link 1 X-offset
d X 19 in
Link 1 Y-offset
d Y 28 in
Angle BO4C
δ4 0.0 deg
Output crank angle:
θ 0 deg
Input slider velocity
VD 20 in sec
Global XY system 1
θVD 90.0 deg
See Figure P6-33d and Mathcad file P06100.
Draw the linkage to scale in the position given, find instant centers I1,3 and I1,5, and the distances from the pin joints to the instant centers.
C
O4
4
5
B 1,5
D 6
3
From the layout above:
1,3
AI13 70.085 in 2.
VD DI15
VC e
CW
sec
VC 18.057
ω 1.003
in sec
rad
CW
sec
VB 18.057
Determine the angular velocity of link 3 using equation 6.9a. ω
7.
rad
Determine the magnitude of the velocity at point B using equation 6.9b. VB c ω
6.
ω 0.286
Use equation 6.9c to determine the angular velocity of link 4. ω
5.
CI15 63.095 in
Determine the magnitude of the velocity at point C using equation 6.9b. VC CI15 ω
4.
BI13 64.013 in
2
Use equation 6.9c to determine the angular velocity of link 5. ω
3.
O2
VB BI13
ω 0.282
in sec
rad
CCW
sec
Determine the magnitude of the velocity at point A using equation 6.9b. in Upward VA AI13 ω VA 19.769 sec
A
DI15 69.886 in
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-101-1
PROBLEM 6-101 Statement:
For the linkage of Figure P6-33e, write a computer program or use an equation solver to find and plot VD in the global coordinate system for one revolution of link 2 if 2 = 10 rad/sec CW.
Given:
Link lengths: a 5.00
Input crank (L2)
Output crank (O4B) c 6.00 d 2.500
Fourbar ground link (L1)
10
Crank velocity: Solution:
Fourbar coupler (L3)
b 5.00
Slider coupler (L5)
e 15.00
Angle BO4C
83.621 deg
rad sec
See Figure P6-33e and Mathcad file P06101.
1.
This sixbar drag-link mechanism can be analyzed as a fourbar Grashof double crank in series with a crankslider mechanism using the output of the fourbar, link 4, as the input to the crank-slider.
2.
Define one revolution of the input crank: θ 0 deg 1 deg 360 deg
3.
Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the open circuit) in the global XY coordinate system. K1
d
K2
a
K1 0.5000
2
d
K3
c
K2 0.4167
2
2
a b c d
2
2 a c
K3 0.7042
A θ cos θ K1 K2 cos θ K3
C θ K1 K2 1 cos θ K3
B θ 2 sin θ
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 4.
B θ
If the calculated value of 4 is greater than 2, subtract 2 from it and if it is negative, make it positive.
θ θ if θ θ 2 π θ θ 2 π θ θ
θ θ if θ θ 0 θ θ 2 π θ θ 5.
Determine the slider-crank motion using equations 4.16 and 4.17 with 4 as the input angle.
c sin θ θ π e
θ θ asin
f θ c cos θ θ e cos θ θ 5.
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
K4 0.5000
2
2
c d a b
2
2 a b
K5 0.4050
D θ cos θ K1 K4 cos θ K5
E θ 2 sin θ
F θ K1 K4 1 cos θ K5
DESIGN OF MACHINERY - 5th Ed.
6.
SOLUTION MANUAL 6-101-2
Use equation 4.13 to find values of 3 for the open circuit.
2 4 Dθ F θ
θ θ 2 atan2 2 D θ E θ 7.
Determine the angular velocity of link 4 for the open circuit using equations 6.18.
θ 8.
a c
sin θ θ θ θ sin θ θ θ
Determine the angular velocity of link 5 using equation 6.22a:
ω θ
c cos θ θ θ e cos θ θ
Determine the velocity of pin D using equation 6.22b:
VD θ c θ sin θ θ e ω θ sin θ θ
VELOCITY OF PIN D 50
25
0 Velocity
9.
E θ
25 50 75 100
0
45
90
135
180
Crank angle, deg
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-102-1
PROBLEM 6-102 Statement:
For the linkage of Figure P6-33f, locate and identify all instant centers.
Given:
Number of links n 8
Solution:
See Figure P6-33f and Mathcad file P06102.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
1.
n ( n 1)
C 28
2
To 1,8 at infinity 7,8
8
Draw the linkage to scale and identify those ICs that can be found by inspection (10).
7 5,6 6
2.
Use Kennedy's Rule and a linear graph to find the remaining 18 ICs.
1,6 O6
5,7
I1,3: I1,2-I2,3 and I1,4-I3,4
5
I1,5: I1,6-I5,6 and I1,3-I3,5 2,3
I2,4: I1,2-I1,4 and I2,3-I3,4 I4,5: I1,4-I1,5 and I3,5-I3,4
3,4
I2,5: I1,2-I1,5 and I2,4-I4,5
4
I2,6: I1,2-I1,6 and I2,5-I5,6
3
2
3,5
1,4
O4
1,2
O2
I3,6: I2,3-I2,6 and I1,2-I1,6 5,8
I5,8: I5,7-I7,8 and I1,5-I1,8 I4,8: I4,5-I5,8 and I1,8-I1,4
To 1,8 at infinity
I2,8: I2,4-I4,8 and I1,2-I1,8 I3,8: I3,4-I4,8 and I1,3-I1,8
1,5
8
7,8
I6,8: I5,8-I5,6 and I2,8-I2,6
6,7 1,7
I2,7: I2,8-I7,8 and I2,5-I5,7
7 5,6
1,3; 3,8
I6,7: I5,6-I5,7 and I6,8-I7,8
6
I3,7: I3,6-I6,7 and I2,3-I2,7
1,6 O6
5,7
I4,6: I1,4-I1,6 and I4,5-I5,6 5
I4,7: I4,6-I6,7 and I3,4-I3,7
3,7
6,8 2,7
I1,7: I1,8-I7,8 and I1,6-I6,7
2,3 4,5 4,7
3 1,4 O4
2,4
3,5
2,5
2
4 3,4 O2
1,2
2,8 4,8
To 2,6 and 3,6 4,6
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-103-1
PROBLEM 6-103 Statement:
The linkage of Figure P6-33f has link 2 at 130 deg in the global XY coordinate system. Find VD in the global coordinate system for the position shown if 2 = 15 rad/sec CW. Use any graphical method.
Given:
Link lengths: Link 2 (O2 to A)
a 5.0 in
Link 3 (A to B)
b 8.4 in
Link 4 (O4 to B)
c 2.5 in
Link 1 (O2 to O4)
d 1 12.5 in
Link 5 (C to E)
e 8.9 in
Link 5 (C to D)
h 5.9 in
Link 6 (O6 to E)
f 3.2 in
Link 7 (D to F)
k 6.4 in
Link 3 (A to C)
g 2.4 in
Link 1 (O2 to O6)
d 2 10.5 in
θ2 130 deg
Crank angle:
Slider axis offset and angle:
s 11.7 in
ω 15 rad sec
Input crank angular velocity Solution: 1.
150 deg 1
CW
See Figure P6-33f and Mathcad file P06103.
Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest.
Direction of VF 8
Y
F
Direction of VFD
Direction of VE
7 E
Direction of VEC
6 Direction of VDC
5
Direction of VCA Direction VBA
C
Direction of VB
O6
D
Direction of VA
A
3
2
B 4 O4
X
O2
Angles measured from layout:
2.
θVB 24.351 deg
θVBA 79.348 deg
θVCA θVBA
θVE 64.594 deg
θVEC 162.461 deg
θVDC θVEC
θVF 150.00 deg
θVFD 198.690 deg
Use equation 6.7 to calculate the magnitude of the velocity at point A. in
VA a ω
VA 75.000
θVA θ2 90 deg
θVA 40.000 deg
sec
DESIGN OF MACHINERY - 5th Ed.
3.
SOLUTION MANUAL 6-103-2
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B, the magnitude of the relative velocity VBA, and the angular velocity of link 3. The equations to be solved graphically are VB = VA + VBA
and
VC = VA + VCA
a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line.
2.668
VA 0.943 V CA
0
VC
25 in/sec
3.302
Y 22.053° X
V BA VB 3.192
4.
From the velocity polygon we have: Velocity scale factor:
25 in sec
in
VB 79.800
VBA 3.302 in kv
VBA 82.550
VBA b
1
in
VB 3.192 in kv
5.
kv
9.827
θVB 24.351 deg
sec in
θVBA 79.348 deg
sec
rad
CCW
sec
Calculate the magnitude and direction of VCA and determine the magnitude and velocity of VC from the velocity polygon above. VCA g
VCA 23.586
Length of VCA on velocity polygon:vCA VC 2.668 in kv
in sec
VCA
vCA 0.943 in
kv
VC 66.700
θVCA 79.348 deg
in sec
θVC 22.053deg
DESIGN OF MACHINERY - 5th Ed.
6.
SOLUTION MANUAL 6-103-3
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point E, the magnitude of the relative velocity VED, and the angular velocity of link 5. The equations to be solved graphically are VE = VC + VEC
and
VD = VC + VDC
a. Choose a convenient velocity scale and layout the known vector VC. b. From the tip of VC, draw a construction line with the direction of VEC, magnitude unknown. c. From the tail of VC, draw a construction line with the direction of VE, magnitude unknown. d. Complete the vector triangle by drawing VEC from the tip of VC to the intersection of the VE construction line and drawing VE from the tail of VC to the intersection of the VEC construction line. 1.821
1.207 V EC VE
1.716 0
VD
V DC
25 in/sec VC Y
45.934°
X 1.900
7.
From the velocity polygon we have: Velocity scale factor:
25 in sec
VEC 1.821 in kv
VEC 45.525
e
5.115
θVE 64.594 deg
sec in
θVEC 162.461 deg
sec
rad
CCW
sec
Calculate the magnitude and direction of VDE and determine the magnitude and velocity of VD from the velocity polygon above. VDC h
VDC 30.179
Length of VDC on velocity polygon: vDC VD 1.900 in kv 9.
in
VE 42.900
VEC
1
in
VE 1.716 in kv
8.
kv
in sec
VDC
vDC 1.207 in
kv
VD 47.500
θVDC 162.461 deg
in sec
θVD 45.934deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point F, the magnitude of the relative velocity VFD, and the angular velocity of link 5. The equation to be solved graphically is VF = VD + VFD
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-103-4
a. Choose a convenient velocity scale and layout the known vector VD. b. From the tip of VD, draw a construction line with the direction of VFD, magnitude unknown. c. From the tail of VD, draw a construction line with the direction of VF, magnitude unknown. d. Complete the vector triangle by drawing VFD from the tip of VD to the intersection of the VF construction line and drawing VF from the tail of VD to the intersection of the VFD construction line. Y 0
VD
25 in/sec
V FD 150.000° VF
45.934°
X 1.158
10. From the velocity polygon we have: Velocity scale factor:
VF 1.158 in kv
kv
25 in sec
1
in
VF 28.950
in sec
θVF 150.000 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-75-1
PROBLEM 6-104 Statement:
For the linkage of Figure P6-34, locate and identify all instant centers.
Given:
Number of links n 6
Solution:
See Figure P6-34 and Mathcad file P06104.
1.
Determine the number of instant centers for this mechanism using equation 6.8a. C
2.
n ( n 1)
C 15
2
Draw the linkage to scale and identify those ICs that can be found by inspection (7). 4,5 C
3,4
2,3
B
5
A
3
5,6 3
D
2
4
6 E 3,6 O4
1,4
2.
1,2 O2
1
Use Kennedy's Rule and a linear graph to find the remaining 8 ICs: I1,3; I1,5; I1,6; I2,4; I2,5; I2,6; I3,5; and I4,6 I2,4: I1,2-I1,4 and I2,3-I3,4
I3,5: I2,3-I2,5 and I3,4-I4,5
I4,6: I3,4-I3,6 and I4,5-I5,6
I1,3: I1,2-I2,3 and I1,4-I3,4
I2,6: I1,4-I4,6 and I2,3-I3,6
I1,5: I1,2-I2,5 and I1,3-I3,5
I2,5: I2,6-I5,6 and I2,4-I4,5
I1,6: I1,3-I3,6 and I1,4-I4,6
1 6
2
5
3 4
1,3
1,5 3,5
4,6
4,5
2,5
C
2,3
B 3,4
5
2,6
A
3
5,6 D
3
2
4
2,4
6 E 3,6 1,4
1,6
O4
1
1,2 O2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-105-1
PROBLEM 6-105 Statement:
For the linkage of Figure P6-34, show that I1,6 is stationary for all positions of link 2.
Given:
Link lengths:
Solution: 1.
Input crank (L2)
a 1.75
First coupler (AB)
b 1.00
First rocker (O4B)
c 1.75
Ground link (O2O4)
d 1.00
Second input (BC)
e 1.00
Second coupler (L5)
f 1.75
Output rocker (L6)
g 1.00
Third coupler (BE)
h 1.75
Ternary link (AE)
k 2.60
See Figure P6-34 and Mathcad file P06105.
Because the linkage is symmetrical and composed of two parallelograms the analysis can be done with simple trigonometry. We will show that the path of point E on link 6 is a circle and that the center of the circle is the instant center I1,6.
C e
f 5
= 25.396°
3
h
B
RP k
R2
4
D g 6
A
b 3
2 a
c E
RE
x
O4
1 d
O2 y
2.
Calculate the fixed angle that line AB (b) makes with line AE (k) using the law of cosines. This is the angle that the coupler point, E, makes with link 3 in the fourbar made up of links 1, 2, 3, and 4. Because links 1, 2, 3, and 4 are a parallelogram, link 4 will have the same angle as link 2 and AB will always be parallel to O2O4.
b 2 k2 h 2 2 b k
δ acos 3.
δ 25.396 deg
Define the vectors R2, RP , and RE as shown above. Then, RE = R2 + RP or,
k cosδ jsin δ
RE θ = a cos θ jsin θ
Separating into real and imaginary parts, the coordinates of point E for all positions of link 2 are:
xE θ a cos θ k cos δ yE θ a sin θ k sin δ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-105-2
This pair of equations represent the parametric equation of a circle with radius a and center at xEC k cos δ
xEC 2.349
yEC k sin δ
yEC 1.115
4.
Define one revolution of the input crank: θ 0 deg 1 deg 360 deg
5.
Plot the position of point E as a function of the angle of input link 2.
PATH OF POINT E 4
2.349
3
2
yE θ
1.115
1
0
1
0
1
2
3
4
5
xE θ
6.
From Problem 6-104, the instant center I1,6 is located at the intersection of a line through O4 that is parallel to BE (angle ) with a line through E that is parallel to O2A (angle 2). (See figure on next page). The coordinates of the intersection of these two lines for the position shown ( θ 64.036 deg ) are:
b 2 h 2 k2 2 b h
Angle BAE:
α π acos
Coordinates of point E:
xE a cos θ k cos δ
α 39.582 deg
xE 3.115
yE 0.458
yE a sin θ k sin δ Line through E with angle 2:
y ( x) x xE tan θ yE
Coordinates of point O4:
xO4 d
yO4 0
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-105-3
Line through O4 with angle :
y ( x) x xO4 tan α
Solving for the intersection (point F, the instant center I1,6) of these two lines: xF
tan θ tan α
xE tan θ yE d tan α
xF 2.349
yF xF d tan α
yF 1.115
These are the coordinates of the instant center I1,6 and the center of the circle through which point E passes. Thus, the instant center I1,6 is stationary and is a hinge point that link 6 rotates about with respect to link 1.
1,3
1,5 3,5
4,6
4,5
2,5
C
2,3
B 3,4
5 5,6
2,6
A
3
39.582° D
3
2
4
6
2,4
64.036°
E 3,6
1,4
O4
1
1,2 O2 1.115"
1.750" 1,6
F
2.349"
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-106-1
PROBLEM 6-106 Statement:
Figure P6-26 shows a mechanism with dimensions. Use a graphical method to determine the velocities of points A and B and the velocity of slip for the position shown. Ignore links 5 and 6. ω 2 = 24 rad/s clock-wise.
Given:
Link lengths and angles: Link 1 (O2O4)
d 1.22 in
Angle O2O4 makes with X axis
θ 56.5 deg
Link 2 (O2A)
a 1.35 in
Angle 2 makes with X axis
θ 14 deg
Link 4 (O4B)
e 1.36 in
Angular velocity of link 2 Solution: 1.
ω 24 rad sec
1
CW
See Figure P6-26 and Mathcad file P06106.
Draw the linkage to scale and indicate the axes of slip and transmission as well as the directions of velocities of interest.
B
Axis of slip Y
Axis of transmission O4
4
0.939
132.661° A 3 2 X O2
Direction of VA3
2.
Use equation 6.7 to calculate the magnitude of the velocity at point A on links 2 and 3. VA3 a ω
3.
VA3 32.400
in sec
θVA3 θ 90 deg
θVA3 76.0 deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity components at point A on link 3. The equation to be solved graphically is VA3 = Vtrans + Vslip a. Choose a convenient velocity scale and layout the known vector VA3. b. From the tip of VA3, draw a construction line with the direction of Vslip, magnitude unknown. c. From the tail of VA3, draw a construction line with the direction of Vtrans, magnitude unknown. d. Complete the vector triangle by drawing Vslip from the tip of Vtrans to the tip of VA3 and drawing Vtrans from the tail of VA3 to the intersection of the Vslip construction line.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-106-2
Y 0
10 in/sec
1.295 X V trans
V A3 2.369
4.
From the velocity triangle we have: Velocity scale factor:
5.
kv
1
in in
Vslip 28.43
Vtrans 1.295 in kv
Vtrans 15.54
sec in sec
The true velocity of point A on link 4 is Vtrans, VA4 15.54
in sec
Determine the angular velocity of link 4 using equation 6.7. From the linkage layout above: ω
7.
12 in sec
Vslip 2.369 in kv
VA4 Vtrans 6.
V slip
VA4 c
c 0.939 in and
ω 16.55
rad
θ 132.661 deg
CW
sec
Determine the magnitude and sense of the vector VB using equation 6.7. VB e ω
θVA4 θ 90 deg
VB 22.51
in sec
θVA4 42.66 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-107-1
PROBLEM 6-107 Statement:
Figure P6-26 shows a mechanism with dimensions. Use an analytic method to calculate the velocities of points A and B and the velocity of slip for the position shown. Ignore links 5 and 6. ω 2 = 24 rad/s clock-wise.
Given:
Link lengths and angles: Link 1 (O2O4)
d 1.22 in
Angle O2O4 makes with X axis
θ 56.5 deg
Link 2 (O2A)
a 1.35 in
Angle 2 makes with X axis
θ 14 deg
Link 4 (O4B)
e 1.36 in ω 24 rad sec
Angular velocity of link 2 Solution: 1.
1
CW
See Figure P6-26 and Mathcad file P06107.
Draw the linkage to scale and indicate the axes of slip and transmission as well as the directions of velocities of interest.
B
Axis of slip Y
Axis of transmission O4
4
0.939
132.661° A 3 2 X O2
Direction of VA3
2.
Establish an x-y frame with origin at O2 and positive x-axis through O4. Using the given data, calculate b, θ2, θ3, and θ4 with respect to the x-y frame. From the vector loop equation for this geometry, θ θ θ
θ 42.500 deg
a sin θ d a cos θ
θ 180 deg atan
b a 3.
sin θ sin θ
θ 256.161 deg
b 0.939 in
Using Example 6-5, determine VA2, Vtrans, and Vslip. Velocity on link 2 at A: VA2 a ω
at an angle of
θVA2 θ sign ω 90 deg
θ θ
DESIGN OF MACHINERY - 5th Ed.
VA2 32.400
in s
SOLUTION MANUAL 6-107-2
θVA2 132.500 deg
Angle between VA2 vector and link 4 axis:
4.
α θVA2 θ 2 π
α 28.661 deg
Vslip VA2 cos( α)
Vslip 28.430
Vtrans VA2 sin( α)
Vtrans 15.540
in s in s
Calculate the angular velocity of link 4 and the velocity at point B. Angular velocity of link 4: Velocity at point B:
ω
Vtrans b
VB e ω
ω 16.544
rad
VB 22.500
in
s s
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-108-1
PROBLEM 6-108 Statement:
Figure P6-28 shows a quick-return mechanism with dimensions. Use a graphical method to determine the velocities of points A and B and the velocity of slip for the position shown. Ignore links 5 and 6. ω 2 = 16 rad/s.
Given:
Link lengths and angles: Link 1 (O2O4)
d 1.69 in
Angle O2O4 makes with X axis
Link 2 (L2)
a 1.00 in
Angle link 2 makes with X axis θ 99 deg
Link 4 (L4)
e 4.76 in
Angular velocity of link 2 Solution: 1.
ω 16 rad sec
1
θ 15.5 deg
CCW
See Figure P6-28 and Mathcad file P06108.
Draw the linkage to scale and indicate the axes of slip and transmission as well as the directions of velocities of interest.
B
Axis of transmission
Direction of VA3
4
Y Axis of slip A 3 2.068
2 44.228° O2
X O4
2.
Use equation 6.7 to calculate the magnitude of the velocity at point A on links 2 and 3. VA2 a ω
3.
VA2 16.000
in sec
θVA2 θ 90 deg
θVA2 189.0 deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity components at point A on link 3. The equation to be solved graphically is VA3 = Vtrans + Vslip a. b.
Choose a convenient velocity scale and layout the known vector VA3. From the tip of VA3, draw a construction line with the direction of Vslip, magnitude unknown.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-108-2
c. From the tail of VA3, draw a construction line with the direction of Vtrans, magnitude unknown. d. Complete the vector triangle by drawing Vslip from the tip of Vtrans to the tip of VA3 and drawing Vtrans from the tail of VA3 to the intersection of the Vslip construction line. Y 0
5 in/sec
X
VA3 V trans
1.154
4.
kv
1
in in
Vslip 8.17
Vtrans 1.154 in kv
Vtrans 5.77
sec in sec
The true velocity of point A on link 4 is Vtrans, VA4 5.77
in sec
Determine the angular velocity of link 4 using equation 6.7. From the linkage layout above: ω
7.
5 in sec
Vslip 1.634 in kv
VA4 Vtrans 6.
1.634
From the velocity triangle we have: Velocity scale factor:
5.
Vslip
VA4 c
c 2.068 in and
ω 2.790
rad
θ 44.228 deg
CCW
sec
Determine the magnitude and sense of the vector VB using equation 6.7. VB e ω
θVB θ 90 deg
VB 13.28
in sec
θVB 134.228 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-109-1
PROBLEM 6-109 Statement:
Figure P6-28 shows a quick-return mechanism with dimensions. Use an analytic method to calculate the velocities of points A and B and the velocity of slip for the position shown. Ignore links 5 and 6. ω 2 = 16 rad/s.
Given:
Link lengths and angles: Link 1 (O2O4)
d 1.69 in
Angle O2O4 makes with X axis
Link 2 (L2)
a 1.00 in
Angle link 2 makes with X axis θ 99 deg
Link 4 (L4)
e 4.76 in
Angular velocity of link 2 Solution: 1.
ω 16 rad sec
1
θ 15.5 deg
CCW
See Figure P6-28 and Mathcad file P06108.
Draw the linkage to scale and indicate the axes of slip and transmission as well as the directions of velocities of interest.
B
Axis of transmission Direction of VA3
4
Y
Axis of slip 99.00°
A 3 2.06"
2 X
O2 15.5 x
O4 y
2.
Establish an x-y frame with origin at O2 and positive x-axis through O4. Using the given data, calculate b, θ2, θ3, and θ4 with respect to the x-y frame. From the vector loop equation for this geometry,
θ θ θ 180 deg
a sin θ d a cos θ
θ 180 deg atan
b a
sin θ sin θ
b 2.059 in
θ 96.500 deg θ 208.855 deg
θ θ
DESIGN OF MACHINERY - 5th Ed.
3.
SOLUTION MANUAL 6-109-2
Using Example 6-5, determine VA2, Vtrans, and Vslip. Velocity on link 2 at A: VA2 a ω VA2 16.000
at an angle of in s
θVA2 θ sign ω 90 deg θVA2 6.500 deg
Angle between VA2 vector and link 4 axis:
α θVA2 θ 2 π
α 144.645 deg
Vslip VA2 cos( α)
Vslip 13.049
Vtrans VA2 sin( α) 4.
Vtrans 9.258
in s
in s
Calculate the angular velocity of link 4 and the velocity at point B. Angular velocity of link 4: Velocity at point B:
ω
Vtrans b
VB e ω
ω 4.497
rad
VB 21.405
in
s s
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-110a-1
PROBLEM 6-110a Statement:
Given:
Solution: 1.
The general linkage configuration and terminology for an offset fourbar slider-crank linkage are shown in Fig P6-2. The link lengths and the values of d and d-dot are defined in Table P6-5. For row a, find the velocity of the pin joint A and the angular velocity of the crank using a graphical method. Link lengths: Link 2
a 1.4 in
Link 3
b 4 in
Offset
c 1 in
θ 176.041 deg
ddot 10 in sec
1
See Figure P6-2, Table P6-5, and Mathcad file P06110a.
Draw the linkage to scale and indicate the axes of slip and transmission as well as the directions of velocities of interest.
Direction of VBA 13.052° B
. d = VB
176.041° A
000 b = 4. a = 1.400
c = 1.000"
O2
d = 2.500"
0
10 inches/sec
Direction of VA 2.
Use equation 6.6.24b to (graphically) determine the magnitude of the velocity at point A. The equation to be solved graphically is
VA
VA = VB + VBA a. Choose a convenient velocity scale and layout the known vector VB. b. From the tip of VB, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VB, draw a construction line with the direction of VA, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VB to the intersection of the VA construction line and drawing VA from the tail of VB to the intersection of the VBA construction line. 3.
VBA
3.414" 3.330"
Y
86.041°
103.052°
VB
From the velocity triangle we have: 1.000"
X
DESIGN OF MACHINERY - 5th Ed.
Velocity scale factor:
VA 3.330 in kv
4.
SOLUTION MANUAL 6-110a-2
kv
10 in sec
1
in
VA 33.300
in
θVB 86.041 deg
sec
Use equation 6.7 to find the angular velocity of the crank (link 2).
ω2
VA a
ω2 23.786
rad sec
CW
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 6-111a-1
PROBLEM 6-111a Statement:
Given:
Solution: 1.
The general linkage configuration and terminology for an offset fourbar slider-crank linkage are shown in Fig P6-2. The link lengths and the values of d and d-dot are defined in Table P6-5. For row a, find the velocity of pin joint A and the angular velocity of the crank using the analytic method. Draw the linkage to scale and label it before setting up the equations. Link lengths: Link 2 (O2 A)
a 1.4 in
Link 3 (AB)
b 4 in
Slider position
d 2.5 in
Slider velocity
ddot 10
Offset (yB)
c 1 in
in s
See Figure P6-2, Table P6-5, and Mathcad file P06111a.
Draw the linkage to scale and label it.
Y
13.052° . d
B
A
c = 1.000
000 b = 4.
176.041°
O2
X
a = 1.400 2.
Determine the open value of 2 using equations 4.20 and 4.21. 2
2
2
K1 a b c d
2
2
K2 2 a c
K2 2.8 in
K3 2 a d
K3 7 in
A K1 K3
A 0.21 in
B 2 K2
B 5.6 in
C K1 K3
C 13.79 in
2 2
2 2
2
θ2 2 atan2 2 A B 3.
2
K1 6.79 in
B 4 A C
θ2 176.041 deg
Determine the value of 3 using equation 4.16a or 4.17.
β ( a b c d α)
θ asin
a sin( α) c
b
d 1 a cos( α) b cos( θ ) return θ if d 1 = d asin
θ3 β a b c d θ2
a sin( α) c b
π otherwise
θ3 193.052 deg
DESIGN OF MACHINERY - 5th Ed.
4.
Determine the value of ω 2 using equation 6.24b.
ω2 5.
SOLUTION MANUAL 6-111a-2
ddot cos θ3
ω2 23.785
a cos θ2 sin θ3 sin θ2 cos θ3
Determine the value of VA using equation 6.7. in
VA a ω2
VA 33.299
θVA θ2 sign ω2 90 deg
θVA 86.041 deg
s
rad s
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-1-1
PROBLEM 7-1 Statement:
A point at a 6.5-in radius is on a body that is in pure rotation with = 100 rad/sec and a constant = -500 rad/sec2 at point A. The rotation center is at the origin of a coordinate system. When the point is at position A, its position vector makes a 45 deg angle with the X axis. It takes 0.01 sec to reach point B. Draw this system to some convenient scale, calculate the and of position B, and: a. Write an expression for the particle's acceleration vector in position A using complex number notation, in both polar and Cartesian forms. b. Write an expression for the particle's acceleration vector in position B using complex number notation, in both polar and Cartesian forms. c. Write a vector equation for the acceleration difference between points B and A. Substitute the complex number notation for the vectors in this equation and solve for the acceleration difference numerically. d. Check the result of part c with a graphical method.
Given: Initial rotation speed Rotation acceleration
ω 100
rad
α 500
rad
sec
sec
Solution: 1.
Time to reach point B
t 0.01 sec
Vector angle
θ 45 deg
Vector magnitude
R 6.5 in
See Mathcad file P0701.
Calculate the position and angular velocity at point B using equations 6.1 and 7.1. ω α t ω θ
2.
2
α t
ω 95.000
rad sec
2
2
ω t θ
θ 100.863 deg
Calculate the magnitudes and directions of the normal and tangential components of acceleration at points A and B using equations 7.2. 2
a An R ω
a An 65000
in sec
a At R α
2
sec
θAn 225.000 deg
θAt θ sign α 90 deg
in
a At 3250
θAn θ 180 deg
2
θAt 45.000 deg a Bn R ω
2
a Bn 58663
in sec
a Bt R α
a Bt 3250
in sec
2
2
θBn θ 180 deg θBt θ sign α 90 deg θBt 10.863 deg
3.
Establish an X-Y coordinate frame and draw a circle with center at the origin and radius R.
θBn 280.863 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-1-2
4.
Draw lines from the origin that make angles of 45 and 100.863 deg with respect to the X axis. Label the intersections of the lines with the circles as A and B, respectively.
5.
Choose a convenient acceleration scale and draw the two acceleration component vectors at points A and B.
Y 0
8 B
1
2 in
Distance scale:
ABt
0
10000 in/sec^2
Acceleration scale:
6 A 4
AAt
2 ABn
AAn
0
a.
X 2
4
8
6
Write an expression for the particle's acceleration vector in position A using complex number notation, in both polar and Cartesian forms.
Polar form:
RA R e
j
j θ
VA R j ω e AA R j α e
AA 6.50 j α e
j
j θ
j
Cartesian form:
π 4
VA 650 j e 2 j θ
R j ω e π 4
2
6.50 ω e
j
π 4
R ω2 cosθ j sinθ
AA R α sin θ j cos θ AA ( 43664 48260i)
in sec
b.
4
RA 6.5 e
j θ
π
2
Write an expression for the particle's velocity vector in position B using complex number notation, in both polar and Cartesian forms.
Polar form:
RB R e
j θ
VB R j ω e
π
j
RB 6.5 e j θ
100.863
180 j
VA 650 j e
π 180
100.863
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-1-3
AB R j α e
j θ j
AB 6.50 j α e
Cartesian form:
2 j θ
R j ω e π
100.863
180
6.50 ω e
π
R ω2 cosθ j sinθ
in sec
2
Write a vector equation for the velocity difference between points B and A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. ABA AB AA
ABA ( 57912 8739i)
in sec
d.
100.863
180
AB R α sin θ j cos θ AB ( 14248 56999i)
c.
2
j
2
Check the result of part c with a graphical method. Solve the equation ABt + ABn = AAt + AAn + ABA using an acceleration scale of 10000 in/sec2 per drawing unit. Acceleration scale factor
ka 10000
in sec
Horizontal component
2
ABAx 5.791 ka
ABAx 57910
in sec
ABAy 0.874 ka
Vertical component
ABAy 8740
in sec
AAt
ABt
0
2
2
10000 IN/S/S
Acceleration Scale
AAn ABA
0.874
ABn
5.791
On the layout above the X and Y components of ABA are equal (to three significant figures) to the real and imaginary components calculated, confirming that the calculation is correct.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-2-1
PROBLEM 7-2 Statement:
In Problem 7-1 let A and B represent points on separate, rotating bodies both having the given and at t = 0, A = 45 deg, and B = 120 deg. Find their relative acceleration.
Given: ω 100
Initial rotation speed
rad sec
α 500
Rotation acceleration
rad sec
Solution: 1.
Vector angles
θ 45 deg
Vector magnitude
R 6.5 in
2
θ 120 deg
See Mathcad file P0702.
Calculate the acceleration of points A and B using equation 7.3.
R ω2 cosθ j sinθ
AA R α sin θ j cos θ AA ( 43664 48260i)
in sec
2
R ω2 cosθ j sinθ
AB R α sin θ j cos θ AB ( 35315 54667i)
in sec
2.
2
Calculate the relative acceleration of point B with respect to A using equation 7.4. ABA AB AA
ABA ( 78978 6407i)
in sec
Magnitude:
ABA ABA
ABA 79238
in sec
Direction:
θ arg ABA
2
θ 4.638 deg
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-3a-1
PROBLEM 7-3a Statement:
The link lengths, coupler point location, and the values of 2, 2, and 2 for the same fourbar linkages as used for position and velocity analysis in Chapters 4 and 6 are redefined in Table P7-1, which is the same as Table P6-1. The general linkage configuration and terminology are shown in Figure P7-1. For row a, draw the linkage to scale and graphically find the accelerations of points A and B. Then, calculate 3 and 4 and the acceleration of point P.
Given:
Link lengths:
Link 1
d 6 in
Link 2
a 2 in
Link 3
b 7 in
Link 4
c 9 in
RPA 6 in
Coupler point:
δ 30 deg θ 30 deg
Link 2 position, velocity, and acceleration: Solution: 1.
ω 10
rad
α 0
sec
rad sec
SeeFigure P7-1 and Mathcad file P0703a.
In order to solve for the accelerations at points A and B, we will need 3, 3, and 4. From the graphical position solution below, θ 88.837 deg y
θ 117.286 deg Using equation (6.18), ω
OPEN
P
a ω sin θ θ b sin θ θ
ω 5.991 rad sec ω
B
3 4
1 88.837°
a ω sin θ θ c sin θ θ
ω 3.992 rad sec
2 O4
O2
x
1
2.
The graphical solution for accelerations uses equation 7.4:
3.
For point B, this becomes: ABn c ω
117.286°
A
(APt + APn) = (AAt + AAn) + (APAt + APAn)
(ABt + ABn) = (AAt + AAn) + (ABAt + ABAn) , where
2
ABn 143.403 in sec
θABn θ 180 deg 2
θABn 297.286 deg
AAn a ω
AAn 200.000 in sec
θAAn θ 180 deg
θAAn 210.000 deg
AAt a α
AAt 0.000 in sec 2
2
2
2
ABAn b ω
ABAn 251.239 in sec
θABAn θ 180 deg
θABAn 268.837 deg
2
2
DESIGN OF MACHINERY - 5th Ed.
4.
SOLUTION MANUAL 7-3a-2
Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ABn at an angle of 4 + 180 deg and AAn at an angle of 2 + 180 deg. From the tip of AAn, draw ABAn at an angle of 3 + 180 deg. Now that the vectors with known magnitudes are drawn, from the tips of ABn and ABAn, draw construction lines in the directions of ABt and ABAt, respectively. The intersection of these two lines are the tips of ABt, ABAt, and AB. 0
100 IN/S/S
Acceleration Scale AA
136.080° n
AB
5.009
AB t
AB
n
ABA
t ABA
4.800 1.826
5.
From the graphical solution above, Acceleration scale factor
ka 100
in sec
6.
2
ABAt 1.826 ka
ABAt 182.6 in sec
ABt 4.800 ka
ABt 480.0 in sec
AB 5.009 ka
AB 500.9 in sec
2
2
2
at an angle of -136.08 deg
Calculate 3 and 4using equation 7.6. α α
ABAt
α 26.086 rad sec
b ABt
α 53.333 rad sec
c
2
2
CCW
CCW
From the graphical solution we see that both of the tangential acceleration vectors for links 3 and 4 point to the left, so both of the angular accelerations are positive. 7.
For point P, equation 7.4 becomes: APAn RPA ω
2
AP = (AAt + AAn) + (APAt + APAn) , where APAn 215.348 in sec
θAPAn θ δ 180 deg
θAPAn 298.837 deg
APAt RPA α
APAt 156.514 in sec
2
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-3a-3
θAPAt θ δ 90 deg 8.
θAPAt 208.837 deg
Repeat proceedure of step 4 for the equation in step 7. 0
100 IN/S/S
Acceleration Scale AA 119.550°
n
APA 4.186 AP t
APA
9.
From the graphical solution above, Acceleration scale factor
ka 100
in sec
AP 4.186 ka
2
AP 418.6 in sec
2
at an angle of -119.6 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-4a-1
PROBLEM 7-4a Statement:
The link lengths, coupler point location, and the values of 2, 2, and 2 for the same fourbar linkages as used for position and velocity analysis in Chapters 4 and 6 are redefined in Table P7-1, which is the same as Table P6-1. The general linkage configuration and terminology are shown in Figure P7-1. For row a, find the accelerations of points A and B using the analytic method. Then, calculate 3 and 4 and the acceleration of point P.
Given:
Link lengths: Link 1
d 6
Link 2
a 2
Link 3
b 7
Link 4
c 9
RPA 6
Coupler point:
δ 30 deg θ 30 deg
Link 2 position, velocity, and acceleration: Solution: 1.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. d
K2
a
K1 3.0000
2
d
K3
c
K2 0.6667
2
2
2 a c
A 0.7113
B 2 sin θ
B 1.0000
C K1 K2 1 cos θ K3
C 3.5566
Use equation 4.10b to find values of 4 for the open and crossed circuits. Open:
2
θ 2 atan2 2 A B
B 4 A C
θ 242.714 deg
θ θ 360 deg
θ 602.714 deg
2
Crossed: θ 2 atan2 2 A B
B 4 A C
θ 216.340 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
2 a b
D cos θ K1 K4 cos θ K5
K4 0.8571 D 1.6774
E 2 sin θ
E 1.0000
F K1 K4 1 cos θ K5 4.
2
a b c d
K3 2.0000
A cos θ K1 K2 cos θ K3
3.
α 0
See Figure P7-1 and Mathcad file P0704a.
K1
2.
ω 10
F 2.5906
Use equation 4.13 to find values of 3 for the open and crossed circuits. Open:
θ 2 atan2 2 D E θ θ 360 deg
2
E 4 D F
θ 271.163 deg θ 631.163 deg
K5 0.2857
DESIGN OF MACHINERY - 5th Ed.
Crossed: 5.
2
θ 2 atan2 2 D E
E 4 D F
θ 244.789 deg
Determine the angular velocity of links 3 and 4 for the open and crossed circuits using equations 6.18. ω
a ω sin θ θ b sin θ θ
ω 5.991
ω
a ω sin θ θ c sin θ θ
ω 3.992
CROSSED ω
a ω sin θ θ b sin θ θ
ω 0.662
ω
a ω sin θ θ c sin θ θ
ω 2.662
OPEN
6.
SOLUTION MANUAL 7-4a-2
Using the Euler identity to expand equation 7.13a for AA. Separate into real and imaginary parts to give the x and y components
a ω2 cosθ j sinθ
2
AA = a α sin θ j cos θ
AAx a α sin θ a ω cos θ
AAx 173.205
AAy 100.000
2
AAy a α cos θ a ω sin θ AA
2
AAx AAy
θAA atan2 AAx AAy
2
The acceleration of pin A is 7.
AA 200.000
θAA 150 deg
at
Use equations 7.12 to determine the angular accelerations of links 3 and 4 for the open and crossed circuits.
OPEN A c sin θ A 7.999
B b sin θ
D c cos θ
E b cos θ
B 6.999
D 4.126
E 0.142
2
2
2
2
2
2
C a α sin θ a ω cos θ b ω cos θ c ω cos θ C 244.045
F a α cos θ a ω sin θ b ω sin θ c ω sin θ F 223.741 α
C D A F A E B D
CROSSED A c sin θ A 5.333
α 26.080
α
C E B F A E B D
α 53.331
B b sin θ
D c cos θ
E b cos θ
B 6.333
D 7.250
E 2.982
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-4a-3
2
2
2
2
2
2
C a α sin θ a ω cos θ b ω cos θ c ω cos θ C 223.253
F a α cos θ a ω sin θ b ω sin θ c ω sin θ F 135.002 α 8.
C D A F
α 77.920
A E B D
α
C E B F
Use equation 7.13c to determine the acceleration of point B for the open and crossed circuits. OPEN ABx1 c α sin θ ω cos θ
2
ABx1 360.826
ABy1 c α cos θ ω sin θ
AB1
2
ABx1 ABy1
2
ABy1 347.485
2
AB1 500.941
θAB1 atan2 ABx1 ABy1
θAB1 136.1 deg
CROSSED ABx2 c α sin θ ω cos θ
2
ABx2 321.587
ABy2 c α cos θ ω sin θ
AB2
2
ABx2 ABy2
2
ABy2 329.551
2
AB2 460.459
θAB2 atan2 ABx2 ABy2 9.
α 50.669
A E B D
θAB2 45.7 deg
Use equations 7.32 to find the acceleration of the point P for the open and crossed circuits.
a ω2 cosθ j sinθ
OPEN AA a α sin θ j cos θ
RPA ω cos θ δ j sin θ δ
APA1 RPA α sin θ δ j cos θ δ 2
AP1 AA APA1 AP1 AP1
AP1 418.556
arg AP1 119.548 deg
a ω2 cosθ j sinθ
CROSSED AA a α sin θ j cos θ
RPA ω cos θ δ j sin θ δ
APA2 RPA α sin θ δ j cos θ δ 2
AP2 AA APA2 AP2 AP2
AP2 298.225
arg AP2 11.282 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-5a-1
PROBLEM 7-5a Statement:
The link lengths and the values of 2, 2, and 2 for the some non inverted offset fourbar slider-crank linkages are defined in Table P7-2. The general linkage configuration and termin- olo are shown in Figure P7-2. For row a, draw the linkage to scale and graphically find the accelerations of the pin joints A and B the acceleration of slip at the sliding joint.
Given: a 1.4 in
Link lengths:
Link 2
Offset:
c 1 in θ 45 deg
Link 2 position, velocity, and acceleration: Solution: 1.
b 4 in
Link 3
ω 10
rad sec
α 0
rad sec
2
See Figure P7-2 and Mathcad file P0705a.
In order to solve for the accelerations at point B, we will need 3, and 3. From the graphical position solution below, θ 360 deg 179.856 deg ω
Using equation 6.22a,
θ 180.144 deg
a ω cos θ b cos θ
ω 2.475 rad sec
1
Direction of ABAt Y
Axis of transmission
Direction of AAt
A 2
45.000°
Axis of slip and Direction of AB
B
3
179.856°
1.000 X
O2
2.
The graphical solution for accelerations uses equation 7.4:
3.
For point B, this becomes:
(APt + APn) = (AAt + AAn) + (APAt + A
AB = (AAt + AAn) + (ABAt + ABAn) , where
2
AAn a ω
AAn 140.000 in sec
θAAn θ 180 deg
θAAn 225.000 deg
AAt a α
AAt 0.000 in sec 2
2
2
ABAn b ω
ABAn 24.500 in sec
θABAn θ
θABAn 180.144 deg
2
DESIGN OF MACHINERY - 5th Ed.
4.
SOLUTION MANUAL 7-5a-2
Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw AAn at an angle of AAn. From the tip of AAn, draw ABAn at an angle of ABAn. Now that the vectors with known magnitudes are drawn, from the origin and the tip of ABAn, draw construction lines in the directions of AB (horizontal) and ABAt, respectively. The intersection of these two lines are the tips of ABAt, and AB. 4.950 Y AB X t BA
A
0
25 IN/S/S
Acceleration Scale AA
n
ABA
5.
From the graphical solution above, Acceleration scale factor
ka 25
in sec
AB 4.950 ka 6.
2
AB 123.8 in sec
2
The acceleration of slip is equal to AB since link 1 is stationary.
at an angle of 180 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-6a-1
PROBLEM 7-6a Statement:
The link lengths and the values of 2, 2, and 2 for the some non inverted offset fourbar slider-crank linkages are defined in Table P7-2. The general linkage configuration and terminology are shown in Figure P7-2. For row a, draw the linkage to scale and find the accelerations of the pin joints A and B the acceleration of slip at the sliding joint using an analytical method.
Given:
Link lengths: Link 2
a 1.4 in
b 4 in
Link 3
c 1 in
Offset:
Link 2 position, velocity, and acceleration: Solution: 1.
θ 45 deg
ω 10
rad sec
α 0
sec
See Figure P7-2 and Mathcad file P0706a.
Draw the linkage to scale and label it. Y d2 = 3.010
d1 = 4.990
3(CROSSED)
B'
A 2
0.144°
45.000°
3 (OPEN)
B 179.856° 1.000 X
O2
2.
Determine 3 and d using equations 4.16 and 4.17. Open:
a sin θ c π b
θ 180.144 deg
d 2 4.990 in
θ asin
d 2 a cos θ b cos θ Crossed:
a sin θ c b
θ 0.144 deg
d 1 3.010 in
θ asin
d 1 a cos θ b cos θ 3.
Determine the angular velocity of link 3 using equation 6.22a. Open
Crossed
4.
ω 2.475
ω 2.475
ω
a cos θ ω b cos θ
ω
a cos θ ω b cos θ
rad
rad sec
rad sec
Using the Euler identity to expand equation 7.15b for AA, determine its magnitude, and direction.
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-6a-2
a ω2 cosθ j sinθ
AA a α sin θ j cos θ in
AA ( 98.995 98.995i )
sec
AA 140.0
The acceleration of pin A is
in sec
5.
at 2
θAA 135.0 deg
Determine the angular acceleration of link 3 using equation 7.16d.
α
Open
α
2
b cos θ
2
rad sec
Crossed
b cos θ 2
a α cos θ a ω sin θ b ω sin θ
α 24.764
2
2
a α cos θ a ω sin θ b ω sin θ
α 24.764
rad sec
6.
θAA arg AA
AA AA
2
2
Use equation 7.16e for the acceleration of pin B. Open:
2
2
AB2 a α sin θ a ω cos θ b α sin θ b ω cos θ AB2 123.7
in sec
2
A negative sign means that AB is to the left
Crossed:
2
2
AB1 a α sin θ a ω cos θ b α sin θ b ω cos θ AB1 74.2
in sec
2
A negative sign means that AB is to the left
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-7a-1
PROBLEM 7-7a Statement:
The link lengths and the values of 2, 2, and for an inverted fourbar slider-crank linkage are defined in Table P7-3, row a, and are given below. Find the acceleration of the pin joints A and B and the acceleration of slip at the sliding joint. Solve by the analytic vector loop method of Section 7.3 for the open configuration of the linkage.
Given:
Link lengths: a 2 in
Link 2
Link 2 position, velocity, and accel.
1.
Link 1
d 6 in
γ 90 deg
Angle between links 3 and 4
Solution:
c 4 in
Link 4
θ 30 deg
ω 10
rad sec
α 25
sec
See Figure P7-3 and Mathcad file P0707a.
Draw the linkage to scale and label it. B y 90.0°
127.333°
b
c 142.666°
A
a 30.000°
d
x 04
02
2.
Use the equations in Section 4.7 to solve for the positions of links 3 and 4 and for the length b.
P a sin θ sin γ a cos θ d cos γ
P 1.000 in
Q a sin θ cos γ a cos θ d sin γ
Q 4.268 in
R c sin γ
S R Q
T 2 P
U Q R
R 4.000 in
S 0.268 in
T 2.000 in
U 8.268 in
θ 2 atan2( 2 S ) T θ θ γ b
2
T 4 S U
θ 142.667 deg θ 232.667 deg
a sin θ c sin θ
b 1.793 in
sin θ
3.
Calculate the angular velocity of links 3 and 4 and the slip velocity using equations 6.30.
rad 2
DESIGN OF MACHINERY - 5th Ed.
ω
SOLUTION MANUAL 7-7a-2
a ω cos θ θ b c cos γ
bdot
cos θ
a ω sin θ ω b sin θ c sin θ
ω 10.292
rad
bdot 33.461
in
sec
sec
ω ω 4.
Solve for the accelerations using equations (7.26) and (7.27).
P a α cos θ θ
P 46.138 in sec
2
Q 77.075 in sec
2
R 423.705 in sec
Q a ω sin θ θ R c ω sin θ θ S 2 bdot ω
α 130.56 rad sec
T
K 639.230 in sec
2
N 2 bdot c ω sin θ θ bddot
2
L 150.000 in sec
M 2.035 10 in sec
2
M ω b c 2 b c cos θ θ 2
2
L a α b sin θ θ c sin θ θ 2
2
2
K a ω b cos θ θ c cos θ θ
2
T 1.793 in
P Q R S 2
2
S 688.757 in sec
T b c cos θ θ α
2
2
3
2
3
2
bddot 128.48
in
N 2.755 10 in sec
K L M N T
sec
2
2
2
The acceleration of slip is bddot. It is directed along link 3, positive inward from B towards A, so that its angle is 3. For the acceleration of the pin joints A and B,
a ω2 cosθ j sinθ
AA a α sin θ j cos θ AA AA
AA 206.155
in sec
2
θAA arg AA
θAA 135.964 deg
c ω2 cosθ j sinθ
AB c α sin θ j cos θ AB AB
AB 672.505
in sec
2
θAB arg AB
θAB 88.280 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-8a-1
PROBLEM 7-8a Statement:
The link lengths and the values of 2, 2, and for an inverted fourbar slider-crank linkage are defined in Table P7-3, row a, and are given below. Find the acceleration of the pin joints A and B and the acceleration of slip at the sliding joint. Solve by the analytic vector loop method of Section 7.3 for the crossed configuration of the linkage.
Given:
Link lengths: a 2 in
Link 2
θ 30 deg
Link 2 position, velocity, and accel.
1.
Link 1
d 6 in
γ 90 deg
Angle between links 3 and 4
Solution:
c 4 in
Link 4
ω 10
rad
rad
α 25
sec
sec
See Figure P7-3 and Mathcad file P0708a.
Draw the linkage to scale and label it. y
A a b
30.000°
d
x 04
02 c B
169.040° 79.041°
2.
Use the equations in Section 4.7 to solve for the positions of links 3 and 4 and for the length b.
P a sin θ sin γ a cos θ d cos γ
P 1.000 in
Q a sin θ cos γ a cos θ d sin γ
Q 4.268 in
R c sin γ
S R Q
T 2 P
U Q R
R 4.000 in
S 0.268 in
T 2.000 in
U 8.268 in
θ 2 atan2( 2 S ) T θ θ γ b
2
T 4 S U
θ 169.041 deg θ 79.041 deg
a sin θ c sin θ
b 1.793 in
sin θ 3.
Calculate the angular velocity of links 3 and 4 and the slip velocity using equations 6.30.
2
DESIGN OF MACHINERY - 5th Ed.
ω
SOLUTION MANUAL 7-8a-2
a ω cos θ θ b c cos γ
bdot
ω 3.639
cos θ
a ω sin θ ω b sin θ c sin θ
rad sec
bdot 33.461
in sec
ω ω 4.
Solve for the accelerations using equations (7.26) and (7.27).
P a α cos θ θ
P 16.312 in sec
2
Q 189.057 in sec
2
R 52.961 in sec
Q a ω sin θ θ R c ω sin θ θ S 2 bdot ω
α 9.93 rad sec
T
K 639.230 in sec L 46.352 in sec
M 254.418 in sec
2
M ω b c 2 b c cos θ θ 2
2
N 2 bdot c ω sin θ θ bddot
2
L a α b sin θ θ c sin θ θ 2
2
2
K a ω b cos θ θ c cos θ θ
2
T 1.793 in
P Q R S 2
2
2
S 243.508 in sec
T b c cos θ θ α
2
2
2
2
N 974.033 in sec
K L M N
bddot 18.98
T
2
2
in sec
2
The acceleration of slip is bddot. It is directed along link 3, positive inward from B towards A, so that its angle is 3. For the acceleration of the pin joints A and B,
a ω2 cosθ j sinθ
AA a α sin θ j cos θ AA AA
AA 206.155
in sec
2
θAA arg AA
θAA 135.964 deg
c ω2 cosθ j sinθ
AB c α sin θ j cos θ AB AB
AB 66.195
in sec
2
θAB arg AB
θAB 47.822 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-9a-1
PROBLEM 7-9a Statement:
The link lengths, gear ratio (), phase angle (), and the values of 2, 2, and 2 for a geared fivebar from row a of Table P7-4 are given below. Find 3 and 4 and the linear acceleration of point P.
Given:
Link lengths: Link 1 f 6 in
Link 3
b 7 in
a 1 in
Link 4
c 9 in
Link 2
Link 5
d 4 in
Gear ratio, phase angle, and crank angle:
Solution: 1.
λ 2
ϕ 30 deg
θ 60 deg
Coupler data:
Rpa 6 in
δ 30 deg
α 0 rad sec
θ 150 deg
Choose the pitch radii of the gears. Since the gear ratio is positive, an idler must be used between gear 2 and gear 5. Let the idler be the same diameter as gear 5, and let all three gears be in line.
r5
λ
f
r2 r5
r5 1.200 in
λ3
r2 λ r5
r2 2.400 in
Draw the linkage to scale and label it. 4
C
B 3
A
5 O5
O2
2 P
4.
2
Determine the angle of link 5 using the equation in Figure P6-4.
f r2 3 r5
3.
1
See Figure P7-4 and Mathcad file P0709a.
θ λ θ ϕ 2.
ω 10 rad sec
Determine the values of the constants needed for finding 3 and 4 from equations 4.24h and 4.24i.
A 2 c d cos λ θ ϕ a cos θ f
2
B 2 c d sin λ θ ϕ a sin θ
2
2
2
2
2
B 20.412 in
C a b c d f 2 a f cos θ 2 d a cos θ f cos λ θ ϕ 2 a d sin θ sin λ θ ϕ
2
A 36.6462 in
2
C 37.4308 in
2
D C A
D 0.78461 in
E 2 B
E 40.823 in
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-9a-2
2
F A C
F 74.077 in
a cosθ f
G 28.503 in
a sinθ
H 15.876 in
2
G 2 b d cos λ θ ϕ
2
H 2 b d sin λ θ ϕ
2
5.
2
2
2
2
K a b c d f 2 a f cos θ 2 d a cos θ f cos λ θ ϕ 2 a d sin θ sin λ θ ϕ
K 26.569 in
L K G
L 1.933 in
M 2 H
M 31.751 in
N G K
N 55.072 in
2
2 2
2
Use equations 4.24h and 4.24i to find values of 3 and 4 for the open and crossed circuits. OPEN
M 4 L N
E 4 D F
M 4 L N
E 4 D F
2
θ 2 atan2 2 L M
2
θ 2 atan2 2 D E
CROSSED
2
θ 2 atan2 2 D E 6.
Use equation 6.32c to find 5. ω λ ω
7.
Calculate 3 and 4 using equations 6.33.
OPEN
ω
ω
θ 177.715 deg θ 115.407 deg
θ 124.050 deg
b cos θ 2 θ cos θ rad sec
c sin θ
a ω sin θ b ω sin θ d ω sin θ
ω 16.948
CROSSED
θ 173.642 deg
2 sin θ a ω sin θ θ d ω sin θ θ
ω 32.585
ω
2
θ 2 atan2 2 L M
rad sec
2 sin θ a ω sin θ θ d ω sin θ θ
ω 75.191
b cos θ 2 θ cos θ rad sec
DESIGN OF MACHINERY - 5th Ed.
ω
SOLUTION MANUAL 7-9a-3
c sin θ
a ω sin θ b ω sin θ d ω sin θ
ω 59.554 8.
Use equation 7.28c to find 5. α λ α
9.
Calculate 3 and 4 using equations 7.29.
rad sec
2
a α sin θ θ a ω cos θ θ
2
2
b ω cos θ θ d ω cos θ θ OPEN
α
2
d α sin θ θ c ω
b sin θ θ
α 3191
rad sec
2
2
a α sin θ θ a ω cos θ θ
2
2
c ω cos θ θ d ω cos θ θ α
2
d α sin θ θ b ω
c sin θ θ
α 2492
rad sec
2
2
a α sin θ θ a ω cos θ θ
2
2
b ω cos θ θ d ω cos θ θ CROSSED
α
2
d α sin θ θ c ω
b sin θ θ
α 6648
rad sec
2
2
a α sin θ θ a ω cos θ θ
2
2
c ω cos θ θ d ω cos θ θ α
2
d α sin θ θ b ω
α 5950
c sin θ θ rad sec
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-10-1
PROBLEM 7-10 Statement:
An automobile driver took a curve too fast. The car spun out of control about its CG and slid off the road in a northeasterly direction. The friction of the skidding tires provided a 0.25g linear deceleration. The car rotated at 100 rpm. When the car hit the tree head-on at 30 mph, it took 0.1 sec to come to rest.
Given:
V 30 mph
Solution:
See Mathcad file P0710.
a.
Ac 0.25 g
ω 100 rpm
What was the acceleration experienced by the child seated on the middle of the rear seat, 2 ft behind the car's CG, just prior to impact? r 2 ft
ω 10.472 rad sec
1
Ac 8.044 ft sec
2
We want to solve the vector equation Ap = Ac + Apcn + Apct where P is the position of the child and C is the CG. Since is zero, Apct is zero. Ac and Apcn both have the same direction, but are opposite in sense (see diagram). 2
Apcn r ω
Apcn 6.817 g Ac
Ap Ac Apcn
V
Ap 6.567 g
n
A pc
This acceleration is toward the tree. Ap 211.3 ft sec Ap 2535 in sec b.
t
A pc
2
2
What force did the 100 lb child exert on her seat belt harness as a result of the acceleration just prior to impact? W 100 lbf The seat back must provide a force of
c.
2'
W
F
g
Ap
F 657 lbf
Assuming a constant deceleration during the impact, what was the magnitude of the average deceleration felt by the passengers in that interval? t 0.1 sec Assume that the rotation has ceased. Then, the average velocity during impact is Vavg
V
Vavg 22.0
2
and the average deceleration is
Aavg
Adding the deceleration due to skidding, Adec 7.088 g
Vavg t
ft sec
Aavg 6.838 g
Aavg 220.0 ft sec
Adec Ac Aavg Adec 228 ft sec
2
Adec 2737 in sec
2
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-11a-1
PROBLEM 7-11a Statement:
For row a in Table P7-1, find the angular jerk of links 3 and 4 and the linear jerk of the pin between links 3 and 4 (point B). Assume an angular jerk of zero on link 2. The linkage configuration and terminology are shown in Figure P7-1.
Given:
Link lengths: Link 1
d 6
Link 2
a 2
Link 3
b 7
Link 4
c 9
RPA 6
Coupler point:
δ 30 deg
Link 2 position, velocity, and acceleration:θ 30 deg Solution: 1.
d
K2
a
2
d
K3
c
K2 0.6667
2
2
a b c d
2
2 a c
K3 2.0000
A cos θ K1 K2 cos θ K3
A 0.7113
B 2 sin θ
B 1.0000
C K1 K2 1 cos θ K3
C 3.5566
Use equation 4.10b to find values of 4 for the open and crossed circuits. Open:
2
θ 2 atan2 2 A B
B 4 A C
θ 242.714 deg
θ θ 360 deg
θ 602.714 deg
2
Crossed: θ 2 atan2 2 A B
B 4 A C
θ 216.340 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
2 a b
D cos θ K1 K4 cos θ K5
K4 0.8571 D 1.6774
E 2 sin θ
E 1.0000
F K1 K4 1 cos θ K5 4.
ϕ 0
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a.
K1 3.0000
3.
α 0
See Figure P7-1 and Mathcad file P0711a.
K1
2.
ω 10
F 2.5906
Use equation 4.13 to find values of 3 for the open and crossed circuits. Open:
θ 2 atan2 2 D E
2
E 4 D F
θ θ 360 deg Crossed:
θ 2 atan2 2 D E
θ 271.163 deg θ 631.163 deg
2
E 4 D F
θ 244.789 deg
K5 0.2857
DESIGN OF MACHINERY - 5th Ed.
5.
6.
SOLUTION MANUAL 7-11a-2
Determine the angular velocity of links 3 and 4 for the open and crossed circuits using equations 6.18. OPEN ω
a ω sin θ θ b sin θ θ
ω 5.991
ω
a ω sin θ θ c sin θ θ
ω 3.992
CROSSED ω
a ω sin θ θ b sin θ θ
ω 0.662
ω
a ω sin θ θ c sin θ θ
ω 2.662
Use equations 7.12 to determine the angular accelerations of links 3 and 4 for the open and crossed circuits.
OPEN A c sin θ A 7.999
B b sin θ
D c cos θ
E b cos θ
B 6.999
D 4.126
E 0.142
2
2
2
2
2
2
C a α sin θ a ω cos θ b ω cos θ c ω cos θ C 244.045
F a α cos θ a ω sin θ b ω sin θ c ω sin θ F 223.741 α
C D A F
α 26.080
A E B D
α
CROSSED A c sin θ A 5.333
C E B F A E B D
α 53.331
B b sin θ
D c cos θ
E b cos θ
B 6.333
D 7.250
E 2.982
2
2
2
2
2
2
C a α sin θ a ω cos θ b ω cos θ c ω cos θ C 223.253
F a α cos θ a ω sin θ b ω sin θ c ω sin θ F 135.002 α 7.
C D A F
α 77.920
A E B D
α
C E B F A E B D
α 50.669
Use equations 7.36 and 7.37 to determine the angular jerk of links 3 and 4 for the open and crossed circuits. OPEN
3
3
A a ω sin θ
B 3 a ω α cos θ
C a ϕ sin θ
D b ω sin θ
G 3 c ω α cos θ
E 3 b ω α cos θ 3
F c ω sin θ
H c sin θ
K b sin θ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-11a-3
3
3
L a ω cos θ
Q 3 b ω α sin θ
N a ϕ cos θ
ϕ
T 3 c ω α sin θ U c cos θ
ϕ 749.012
K U H R A B C D E F G H ϕ
ϕ 1242.6
K
R b cos θ
K ( N L M P Q S T ) R ( A B C D E F G)
3
3
S c ω cos θ
M 3 a ω α sin θ
ϕ
P b ω cos θ
3 a ω α cosθ j sinθ
J A a ω sin θ j cos θ a ϕ sin θ j cos θ 3
3 b ω α cosθ j sinθ
J BA1 b ω sin θ j cos θ b ϕ sin θ j cos θ J B1 J A J BA1 JB1 J B1
JB1x Re J B1
JB1y Im J B1
JB1 9301.9
JB1x 9134.7
JB1y 1755.5
3
3
CROSSED A a ω sin θ
E 3 b ω α cos θ
C a ϕ sin θ
K b sin θ
3
S c ω cos θ
3
R b cos θ
K ( N L M P Q S T ) R ( A B C D E F G) K U H R A B C D E F G H ϕ K 3
Q 3 b ω α sin θ
N a ϕ cos θ
ϕ
P b ω cos θ
M 3 a ω α sin θ
H c sin θ
3
F c ω sin θ
L a ω cos θ
ϕ
G 3 c ω α cos θ
B 3 a ω α cos θ
3
D b ω sin θ
T 3 c ω α sin θ
U c cos θ
ϕ 246.639 ϕ 740.2
3 b ω α cosθ j sinθ
J BA2 b ω sin θ j cos θ b ϕ sin θ j cos θ J B2 J A J BA2 JB2 J B2
JB2x Re J B2
JB2y Im J B2
JB2 4178.7
JB2x 4147.9
JB2y 506.4
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-12-1
PROBLEM 7-12 Statement:
You are riding on a carousel that is rotating at a constant 12 rpm. It has an inside radius of 4 ft and an outside radius of 12 ft. You begin to run from the inside to the outside along a radius. Your peak velocity with respect to the carousel is 4 mph and occurs at a radius of 8 ft. What is your maximum Coriolis acceleration magnitude and its direction with respect to the carousel?
Given:
Carousel angular velocity Peak velocity Radius at peak velocity
Solution: 1.
rad
ω 12 rpm
ω 1.257
Vslip 4 mph
Vslip 5.867
sec ft sec
r 8 ft
See Mathcad file P0712.
Draw a plan view of the carousel floor showing your position at peak velocity and the velocity and Coriolis acceleration vectors. Axis of Transmission
Axis of Slip
Ac V slip
r
2.
The direction of your path defines the axis of slip. The transmission axis is perpendicular to the axis of slip and positive in the direction of the tangential velocity of the carousel. Thus, the direction of the Coriolis acceleration vector is along the positive transmission axis.
3.
Use equation 7.19 to calculate the magnitude of the Coriolis component of your acceleration. Ac 2 Vslip ω
Ac 14.745
ft sec
2
Ac 176.93
in sec
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-13a-1
PROBLEM 7-13a Statement:
The linkage in Figure P7-5a has the dimensions and crank angle given below. Find 3, AA, AB, and AC for the position shown for 2 = 15 rad/sec and 2 = 10 rad/sec2 in the direction shown. Use the acceleration difference graphical method.
Given:
a 0.8 in
Link lengths:
Link 2
Offset:
c 0.38 in
Coupler point data:
p 1.33 in
b 1.93 in
Link 3
δ 38.6 deg
Link 2 position, velocity, and acceleration:θ 34.3 deg Solution: 1.
ω 15
rad sec
α 10
rad sec
2
See Figure P7-5a and Mathcad file P0713a.
In order to solve for the accelerations at points A, B and C, we will need 3, and 3. From the graphical position solution below, θ 154.502 deg ω
Using equation 6.22a,
a ω cos θ b cos θ
ω 5.691 rad sec
1
Direction of ACAt Direction of AAt
C
Y
A
38.600°
34.300° 154.502°
X O2
Direction of AB B
Direction of ABAt
2.
The graphical solution for accelerations uses equation 7.4:
3.
For point B, this becomes: AB = (AAt + AAn) + (ABAt + ABAn) , where 2
(APt + APn) = (AAt + AAn) + (APAt + APAn)
AAn a ω
AAn 180.000 in sec
θAAn θ 180 deg
θAAn 214.300 deg
AAt a α
AAt 8.000 in sec
θAAt θ 90 deg
θAAt 124.300 deg
2
2
2
ABAn b ω
ABAn 62.500 in sec
θABAn θ
θABAn 154.502 deg
2
DESIGN OF MACHINERY - 5th Ed.
4.
SOLUTION MANUAL 7-13a-2
Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw AAn at an angle of AAn. From the tip of AAn, draw AAt at an angle of AAt. From the tip of AAt, draw ABAn at an angle of ABAn. Now that the vectors with known magnitudes are drawn, from the origin and the tip of ABAn, draw construction lines in the directions of AB (horizontal) and ABAt, respectively. The intersection of these two lines are the tips of ABAt, and AB. 7.089 Y AB X t BA
A 3.010
0
25 IN/S/S
Acceleration Scale n BA
A
n
t
AA
5.
AA
From the graphical solution above, Acceleration scale factor
ka 25
in sec
6.
AB 177.2 in sec
ABAt 3.010 ka
ABAt 75.3 in sec
at an angle of 180 deg
2
Calculate 3 using equation 7.6. ABAt
α 38.990 rad sec
b
2
CCW
For point C, equation 7.4 becomes: AC = (AAt + AAn) + (ACAt + ACAn) , where 2
8.
2
AB 7.089 ka
α 7.
2
ACAn p ω
ACAn 43.070 in sec
θAPAn θ δ
θAPAn 193.102 deg
ACAt p α
ACAt 51.856 in sec
θAPAt θ δ 90 deg
θAPAt 103.102 deg
Repeat procedure of step 4 for the equation in step 7.
2
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-13a-3
Y
8.554
X
165.351°
AC
0 t ACA
25 IN/S/S
Acceleration Scale
n
t
n CA
A
9.
AA
AA
From the graphical solution above, Acceleration scale factor
ka 25
in sec
AC 8.554 ka
2
AC 213.9 in sec
2
at an angle of -165.35 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-13b-1
PROBLEM 7-13b Statement:
The linkage in Figure P7-5a has the dimensions and crank angle given below. Find 3, AA, AB, and AC for the position shown for 2 = 15 rad/sec and 2 = 10 rad/sec2 in the direction shown. Use an analytical method.
Given:
Link lengths: Link 2 Offset:
a 0.8 in
Link 3
b 1.93 in
c 0.38 in
Coupler point data:
p 1.33 in
δ 38.6 deg
Link 2 position, velocity, and acceleration:θ 34.3 deg Solution: 1.
ω 15
rad sec
α 10
See Figure P7-5a and Mathcad file P0713b.
Draw the linkage to scale and label it. C
Y
A
38.600°
34.300° 154.502°
X O2
B
2.
Determine 3 and d using equations 4.16 and 4.17.
a sin θ b
θ asin
c
π
θ 154.502 deg
d a cos θ b cos θ 3.
Determine the angular velocity of link 3 using equation 6.22a. ω
4.
d 2.403 in
a cos θ ω b cos θ
ω 5.691
rad sec
Using the Euler identity to expand equation 7.15b for AA, determine its magnitude, and direction.
a ω2 cosθ j sinθ
AA a α sin θ j cos θ AA ( 153.206 94.826i )
in sec
The acceleration of pin A is
2
AA 180
in sec
5.
θAA arg AA
AA AA
at 2
Determine the angular acceleration of link 3 using equation 7.16d.
θAA 148.2 deg
rad sec
2
DESIGN OF MACHINERY - 5th Ed.
α
6.
SOLUTION MANUAL 7-13b-2
b cos θ 2
2
a α cos θ a ω sin θ b ω sin θ
α 38.990
rad sec
2
Use equation 7.16e for the acceleration of pin B.
2
2
AB a α sin θ a ω cos θ b α sin θ b ω cos θ in
AB 177.2
sec 7.
2
A negative sign means that AB is to the left
Determine the acceleration of the coupler point C using equations 7.32. Note that 3 is defined from point B in Figure 7-6 and from point A in Figure 7-9. To use equation 7.32 for a slider-crank we must redefine 3. θ θ 180 deg
θ 25.498 deg
p ω cos θ δ j sin θ δ
ACA p α sin θ δ j cos θ δ 2
AC AA ACA AC ( 206.910 54.083i )
in sec
2
θAC arg AC
AC AC
The acceleration of point C is AC 213.861
in sec
at 2
θAC 165.352 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-14a-1
PROBLEM 7-14a Statement:
The linkage in Figure P7-5b has the dimensions and effective crank angle given below. Find 3, AA, AB, and AC for the position shown for 2 = 15 rad/sec and 2 = 10 rad/sec2 in the directions shown. Use the acceleration difference graphical method.
Given:
Solution: 1.
Link lengths: Link 2 (point of contact to A)
a 0.75 in
Link 3 (A to B)
b 1.5 in
Link 4 (point of contact to B)
c 0.75 in
Link 1 (between contact points)
d 1.5 in
Coupler point: Distance A to C
p 1.2 in
Angle BAC
δ 30 deg
Crank angle:
θ 77 deg
Input crank angular velocity
ω 15 rad sec
1
α 10 rad sec
2
See Figure P7-5b and Mathcad file P0714a.
Although the mechanism shown in Figure P6-5b is not entirely pin-jointed, it can be analyzed for the position shown by its effective pin-jointed fourbar, which is shown below. Draw the linkage to a convenient scale. Indicate the directions of the acceleration vectors of interest. 0
0.5
1 in
Direction of ACAt Y
C Direction of AAt Direction of ABAt
30.000°
Direction of ABt
3
A
B b 4
2 a
77.000°
c d
O2
X O4
Effective link 2
2.
77.000°
Effective link 4
In order to solve for the accelerations at points A and B, we will need 3, 3, and 4. From the graphical position solution above, θ 0.0 deg
θ 77.0 deg
This is a special-case Grashof in the parallelogram configuration. Therefore, ω 0.0 rad sec
1
ω ω
3.
The graphical solution for accelerations uses equation 7.4:
4.
For point B, this becomes:
(APt + APn) = (AAt + AAn) + (APAt + APAn)
(ABt + ABn) = (AAt + AAn) + (ABAt + ABAn) , where
DESIGN OF MACHINERY - 5th Ed.
ABn c ω
2
ABn 168.750 in sec
θABn θ 180 deg 2
AAn 168.750 in sec
θAAn θ 180 deg
θAAn 257.000 deg
AAt a α
AAt 7.500 in sec
θAt θ 90 deg
θAt 167.000 deg
2
2
θABn 257.000 deg
AAn a ω
ABAn b ω 5.
SOLUTION MANUAL 7-14a-2
2
2
ABAn 0.000 in sec
2
Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ABn at an angle of 4 + 180 deg and AAn at an angle of 2 + 180 deg. From the tip of AAn, draw AAt at an angle of 2 + 90 deg. Now that the vectors with known magnitudes are drawn, from the tips of ABn and AAt, draw construction lines in the directions of ABt and ABAt, respectively. The intersection of these two lines are the tips of ABt, ABAt, and AB. 0
25 IN/S/S
Acceleration Scale 105.545°
From the layout ABt = AAt and ABAt = 0. Also, AB = AA. Acceleration scale factor ka 25
in sec
6.757
2
AB 6.757 ka AB 168.9 in sec
2
at an angle of -105.55 deg
AB n
AA
AB n
AA
AAt ABt
6.
Since ABAt = 0, 3 = 0 and, since AB = AA, 4 = 2.
7.
For point P, equation 7.4 becomes: AP = (AAt + AAn) + (APAt + APAn) , where APAt and APAn are both zero since and 3 are zero. Therefore, AP = AA.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-14b-1
PROBLEM 7-14b Statement:
The linkage in Figure P7-5b has the dimensions and effective crank angle given below. Find
Given:
3, AA, AB, and AC for the position shown for 2 = 15 rad/sec and 2 = 10 rad/sec2 in the directions shown. Use an analytical method. Link lengths: Link 2 (point of contact to A)
a 0.75 in
Link 3 (A to B)
b 1.5 in
Link 4 (point of contact to B)
c 0.75 in
Link 1 (between contact points)
d 1.5 in
Coupler point:
Solution: 1.
Rca 1.2 in
Crank angle:
θ 77 deg
α 10 rad sec
Angle BAC
δ 30 deg
Input crank angular velocity
ω 15 rad sec
See Figure P7-5b and Mathcad file P0714b.
Draw the linkage to scale and label it. Y
C
30.000°
0
0.5
1 in
3
A
B b 4
2 77.000°
a
c d
77.000°
X O4
O2 Effective link 2
2.
Effective link 4
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 2.0000 2
K3
d c
K2 2.0000 2
2
a b c d
2
K3 1.0000
2 a c
A cos θ K1 K2 cos θ K3
B 2 sin θ
C K1 K2 1 cos θ K3 A 1.2250 3.
2
Distance A to C
B 1.9487
C 2.3251
Use equation 4.10b to find values of 4 for the open circuit.
θ 2 atan2 2 A B θ θ 360 deg
2
B 4 A C
θ 283.000 deg θ 77.000 deg
1
DESIGN OF MACHINERY - 5th Ed.
4.
SOLUTION MANUAL 7-14b-2
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
K4 1.0000
2 a b
D cos θ K1 K4 cos θ K5
D 3.5501
E 2 sin θ
E 1.9487
F K1 K4 1 cos θ K5 5.
F 0.0000
Use equation 4.13 to find values of 3 .
2
θ 2 atan2 2 D E
E 4 D F
θ 360.000 deg
θ θ 360 deg 6.
7.
K5 2.0000
θ 0.000 deg
Determine the angular velocity of links 3 and 4 using equations 6.18. ω
a ω sin θ θ b sin θ θ
ω 0.000
ω
a ω sin θ θ c sin θ θ
ω 15.000
rad sec rad sec
Using the Euler identity to expand equation 7.13a for AA. Determine the magnitude, and direction.
a ω2 cosθ j sinθ
AA a α sin θ j cos θ AA ( 45.268 162.738i)
in sec
The acceleration of pin A is
in
AA 168.917
sec 8.
θAA arg AA
AA AA
2
at 2
θAA 105.5 deg
Use equations 7.12 to determine the angular accelerations of links 3 and 4.
A c sin θ
B b sin θ
D c cos θ
E b cos θ
A 0.731 in
B 0.000 in
D 0.169 in
E 1.500 in
2
2
2
2
2
2
C a α sin θ a ω cos θ b ω cos θ c ω cos θ C 7.308 in sec
2
F a α cos θ a ω sin θ b ω sin θ c ω sin θ F 1.687 in sec α
9.
2
C D A F A E B D
α 0.000
rad sec
2
Use equation 7.13c to determine the acceleration of point B.
α
C E B F A E B D
α 10.000
rad sec
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-14b-3
c ω2 cosθ j sinθ
AB c α sin θ j cos θ AB ( 45.268 162.738i)
in sec
The acceleration of pin B is
2
θAB arg AB
AB AB
AB 168.917
in sec
at 2
θAB 105.5 deg
10. Use equations 7.32 to find the acceleration of the point C.
2 Rca ω cos θ δ j sin θ δ
ACA Rca α sin θ δ j cos θ δ
AC AA ACA AC AC
AC 168.917
in sec
2
arg AC 105.545 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-15a-1
PROBLEM 7-15a Statement:
The linkage in Figure P7-5c has the dimensions and coupler angle given below. Find 3, AB, and AC for the position shown for VA = 10 in/sec and AA = 15 in/sec2 in the directions shown. Use the acceleration difference graphical method.
Given:
Link lengths and angles: Link 3 (A to B)
b 1.8 in
Coupler angle
θ 128 deg
Slider 4 angle
θ 59 deg
p 1.44 in
Angle BAC
δ 49 deg
Coupler point: Distance A to C Input slider motion Solution: 1.
VA 10 in sec
1
AA 15 in sec
2
See Figure P7-5c and Mathcad file P0715a.
In order to solve for the accelerations at points B and C, we will need 3. From the layout below and Problem 6-18, rad θ 128 deg ω 13.288 sec Direction of AB 0
0.5
1 in
Y
Direction of ABAt
C
B 4
Direction of ACAt
3 b
49.000°
128.000°
59.000°
AA VA
X A
2
2.
The graphical solution for accelerations uses equation 7.4:
3.
For point B, this becomes: AB = AA + (ABAt + ABAn) , where AA 15.000 in sec
2
2
4.
(APt + APn) = (AAt + AAn) + (APAt + A
θAA 0 deg
ABAn b ω
ABAn 317.828 in sec
θABAn θ 180 deg
θABAn 308.000 deg
2
Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw AA at an angle of AA. From the tip of AA, draw ABAn at an angle of ABAn. Now that the vectors with known magnitudes are drawn, from the origin and the tip of ABAn, draw construction lines in the directions of AB and ABAt, respectively. The intersection of these two lines are the tips of ABAt, and AB.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-15a-2 Y AA X
0
100 IN/S/S
Acceleration Scale n
ABA
9.126
8.638 AB t ABA
5.
From the graphical solution above, Acceleration scale factor
ka 100
in sec
6.
2
AB 9.126 ka
AB 912.6 in sec
ABAt 8.638 ka
ABAt 863.8 in sec
at an angle of 239 deg
2
Calculate 3 using equation 7.6. α
7.
2
ABAt
α 479.9 rad sec
b
2
CW
For point C, equation 7.4 becomes: AC = AA + (ACAt + ACAn) , where 2
ACAn p ω
ACAn 254.262 in sec
θAPAn θ δ 180 deg
θAPAn 259.000 deg
ACAt p α
ACAt 691.040 in sec
θAPAt θ δ 90 deg
θAPAt 169.000 deg
2
2
DESIGN OF MACHINERY - 5th Ed.
8.
SOLUTION MANUAL 7-15a-3
Repeat procedure of step 4 for the equation in step 7.
Y 7.215 AA
0
100 IN/S/S
X
Acceleration Scale
AC 170.609° t ACA
n
ACA
9.
From the graphical solution above, Acceleration scale factor
ka 100
in sec
AC 7.215 ka
2
AC 721.5 in sec
2
at an angle of -170.61 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-15b-1
PROBLEM 7-15b Statement:
The linkage in Figure P7-5c has the dimensions and coupler angle given below. Find 3, AB, and AC for the position shown for VA = 10 in/sec and AA = 15 in/sec2 in the directions shown. Use an analytical method.
Given:
Link lengths and angles: Link 3 (A to B)
b 1.8 in
Coupler angle
θ 128 deg
Slider 4 angle
θ 59 deg
Coupler point: Distance A to C
Rca 1.44 in
Angle BAC
δ 49 deg VA 10 in sec
Input slider motion Solution: 1.
1
AA 15 in sec
2
See Figure P7-5c and Mathcad file P0715b.
Draw the mechanism to scale and define a vector loop using the fourbar slider-crank derivation in Section 7.3 as a model. 0
0.5
1 in
Y
C
B 4 R3 R4 3 b
49.000°
128.000°
59.000° VA
X
R2
2.
A
2
Write the vector loop equation, differentiate it, expand the result and separate into real and imaginary parts to solve for 3 and VB. R2 R3 R4
a e
j θ
b e
j θ
c e
j θ
where a is the distance from the origin to point A, a variable; b is the distance from A to B, a constant; and c is the distance from the origin to point B, a variable. Angle 2 is zero, 3 is the angle that AB makes with the x axis, and 4 is the constant angle that slider 4 makes with the x axis. Differentiating, j θ d d j θ a j b ω e c e dt dt
Substituting the Euler equivalents,
d c cosθ j sinθ
d a b ω sin θ j cos θ dt
dt
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-15b-2
Separating into real and imaginary components and solving for 3. Note that dc/dt = VB and da/dt = VA ω
3.
b sin θ tan θ cos θ VA tan θ
ω 13.288
rad sec
Differentiate the velocity equation, expand it and solve for 3 and AB. d
a b α j ej θ b ω 2 j ej θ d2 c ej θ 2 2
2
dt
dt
Substituting the Euler equivalents, d
2
dt
2
a b α sin θ j cos θ 2
d
b ω cos θ j sin θ
2
dt
2
0
c cos θ j sin θ
Separating into real and imaginary components and solving for 3 and AB. Note that d 2c/dt2 = AB and d 2a/dt2 = AA
α
AB
4.
b cos θ θ 2
AA sin θ b ω sin θ θ
α 479.924
sec
2
b α cos θ b ω sin θ
rad
AB 912.662
sin θ
2
in sec
2
Determine the acceleration of the coupler point C using equations 7.32.
Rca ω2 cosθ δ j sinθ δ
ACA Rca α sin θ δ j cos θ δ
ACA ( 726.910 117.729j)
in sec
AA AA AC ( 711.910 117.729j)
2
AC AA ACA in sec
2
AC 721.579
in sec
2
arg AC 170.610 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-16-1
PROBLEM 7-16 Statement:
For the linkage shown in Figure P7-6a, write the vector loop equations; differentiate them, and do a complete position, velocity, and acceleration analysis of the linkage. Measure the linkage geometry from the figure. Assume 2 = 10 rad/sec and 2 = 20 rad/sec2.
Given:
Link lengths and angles: Link 2 (O2 to A)
a 5.6 mm
Link 4 (O4 to C)
c 9.5 mm
Link 3 offset (A to B)
f 9.5 mm
Link 1 (O2 to O4)
d 38.8 mm
Link 2 position, velocity and acceleration Solution: 1.
θ 135 deg
ω 10
rad
α 20
sec
rad sec
2
See Figure P7-6a and Mathcad file P0716.
Draw the mechanism to scale and define a vector loop using the fourbar inverted slider-crank derivation in Section 7.3 as a model. 9.5 y
B
B 135.00°
A
2
2
5.6 O2
2
3 R3
O2
3
1
R5 A R2
R1 4
4 R4
C
38.8 O4
O4
1
C 4 x
9.5
2.
Write the vector loop equation, expand the result and separate into real and imaginary parts to solve for 3 and b. R2 R5 R1 R4 R3 a e
j θ
f e
j θ
d e
j θ
c e
j θ
b e
j θ
where a is the distance from pivot O2 to point A, a constant; f is the distance from A to B, a constant; b is the distance from B to C, a variable; and c is the distance from pivot O4 to point C, a constant. Angle 2 is the input crank angle, 3 is the angle that BC makes with the x axis (measured from C), and 4 is the variable angle that link 4 makes with the x axis. The angular relationships are θ θ
and
θ θ 90 deg
Substituting the Euler equivalents,
f cosθ j sinθ d c cosθ j sinθ b cos θ j sin θ
a cos θ j sin θ
If we stipulate that f = c, this reduces to
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-16-2
d b cosθ j sinθ
a cos θ j sin θ
Separating into real and imaginary components and solving for 3 and b.
θ 174.7 deg
θ θ 90 deg
θ 84.7 deg
b
3.
θ atan2 a cos θ d a sin θ
sin θ
a sin θ
b 42.9 mm
Differentiate the position equation to get the velocity equation. a e
Position (f and c terms cancel) a j ω e
Velocity
j θ
bdot e
j θ
j θ
d b e
b j ω e
j θ
j θ
Separating into real and imaginary components and solving for 3 and bdot. ω
a cos θ θ b
ω
ω 1.00
b ω sin θ a ω sin θ
bdot
rad sec mm
bdot 35.8
sec
cos θ 4.
Differentiate the velocity equation to get the acceleration equation. Velocity
a j ω e
Acceleration
a j α e
j θ
j θ
bdot e
j θ
b j ω e
2 j θ
2
a j ω e
j θ
bddot e
j θ
b j α e
bdot j ω e
j θ
1
2
b j ω e
b
a α cos θ θ a ω sin θ θ bdot ω
α 9.50
2
rad sec
bddot
2
2 bdot ω cos θ b α cos θ ω sin θ
1
sin θ
bddot 316.0
a α cos θ ω sin θ
mm sec
2
2
2 j θ
Separating into real and imaginary components and solving for 3 and bddot. α
j θ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-17-1
PROBLEM 7-17 Statement:
For the linkage shown in Figure P7-6b, write the vector loop equations; differentiate them, and do a complete position, velocity, and acceleration analysis of the linkage. Use the linkage geometry given below. Assume 2 = 10 rad/sec and 2 = 20 rad/sec2.
Given:
Measured lengths and angles: Link 1
d 61.9 mm
Link 2
a 15.0 mm
Link 3
b 45.8 mm
Link 4
c 18.1 mm
Link 5
e 23.1 mm
Offset
f 2.6 mm
from x' axis
Crank angle:
θ 45 deg
Coordinate rotation angles
α 23.3 deg Global XY system to local xy system
Global XY system
β 113.3 deg Local xy system to local x'y' system γ 90 deg Link 2 position, velocity and acceleration Solution: 1.
Global XY system to local x'y' system θ 68.3 deg
ω 10
rad sec
α 20
rad sec
2
See Figure P7-6b and Mathcad file P0717.
Draw the mechanism to scale and label it. x'
Y
y
6 A 2 O2
C 3
1
5
2
X 23.300°
B
1
113.300° 4
4 y' O4 1
x
2.
This mechanism can be analyzed as a pin-jointed fourbar (links 1, 2, 3, and 4) and a fourbar slider-crank (links 1, 4, 5, and 6). Link 4 is the common, non-stationary link in the two branches and is the input to the slider-crank. Since the vector loops are the same as those defined for the pin-jointed and slider-crank linkages, we can use the equations derived in the text with slight modification of variable names for the slider-crank. There are three coordinate systems: the global XY frame, the local xy frame for the pin-jointed fourbar, and the local x'y' frame for the slider crank. Inputs to the fourbars must be in their respective local coordinates. All output will be given in local coordinate system.
3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d a
K1 4.1267
K2
d c
K2 3.4199
DESIGN OF MACHINERY - 5th Ed.
2
K3
2
SOLUTION MANUAL 7-17-2
2
a b c d
2
K3 4.2110
2 a c
A cos θ K1 K2 cos θ K3
A 0.8104
B 2 sin θ
B 1.8583
C K1 K2 1 cos θ K3 4.
Use equation 4.10b to find values of 4 for the open circuit.
2
θ 2 atan2 2 A B 5.
C 6.7034
B 4 A C 360 deg
θ 125.692 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
K4 1.3515
2 a b
D cos θ K1 K4 cos θ K5
D 7.4978
E 2 sin θ
E 1.8583
F K1 K4 1 cos θ K5 6.
8.
F 0.0160
Use equation 4.13 to find values of 3 for the open circuit.
2
θ 2 atan2 2 D E 7.
K5 4.2406
E 4 D F
360deg
θ 0.955 deg
Determine the angular velocity of links 3 and 4 using equations 6.18. ω
a ω sin θ θ b sin θ θ
ω 3.357
ω
a ω sin θ θ c sin θ θ
ω 9.307
rad sec
rad sec
Using the Euler identity to expand equation 7.13a for AB. Determine the magnitude, and direction (in the local coordinate system).
a ω2 cosθ j sinθ
AB a α sin θ j cos θ AB ( 833 1283i)
mm sec
The acceleration of pin B is
AB 1530
mm sec
9.
θAB arg AB
AB AB
2
2
at
θAB 123.01 deg
Use equations 7.12 to determine the angular accelerations of links 3 and 4 for the crossed circuit.
A c sin θ
B b sin θ
D c cos θ
E b cos θ
A 14.700 mm
B 0.763 mm
D 10.560 mm
E 45.794 mm
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-17-3
2
2
2
2
2
C a α sin θ a ω cos θ b ω cos θ c ω cos θ 3
2
2
C 2.264 10 mm sec
F a α cos θ a ω sin θ b ω sin θ c ω sin θ F 18.160 mm sec α
9.
2
C D A F
α 34.706
A E B D
rad sec
α
2
C E B F
α 152.221
A E B D
rad sec
Use equation 7.13c to determine the acceleration of point C.
c ω2 cosθ j sinθ
AC c α sin θ j cos θ AC ( 1323 2881i)
mm sec
θAC arg AC
AC AC
2
AC 3170
The acceleration of pin C is
mm sec
at
2
θAC 114.7 deg
10. Determine 5 and the vertical distance from O4 to C (d') for the slider-crank using equations 4.16 and 4.17.
c sin θ β f π e
θ 163.7 deg
d' 39.8 mm
θ asin
d' c cos θ β e cos θ
11. Determine the angular velocity of link 5 using equation 6.22a.
ω
c cos θ β ω e cos θ
ω 7.421
rad sec
12. Determine the angular acceleration of link 5 using equation 7.16d.
α
2
2
c α cos θ β c ω sin θ β e ω sin θ
e cos θ
α 122.305
sec
13. Use equation 7.16e for the acceleration of pin C.
2
2
AC c α sin θ β c ω cos θ β e α sin θ e ω cos θ AC 162.9
in sec
2
rad
A negative sign means that AC is downward
2
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-18-1
PROBLEM 7-18 Statement:
For the linkage shown in Figure P7-6c, write the vector loop equations; differentiate them, and do a complete position, velocity, and acceleration analysis of the linkage. Use the linkage geometry given below. Assume 2 = 10 rad/sec and 2 = 20 rad/sec2.
Given:
Measured lengths and angles: Link 21
a1 11.7 mm
Link 22
a2 20.0 mm
Link 3
b1 25.0 mm
Link 5
b2 25.9 mm
Offset
c1 3.7 mm
Offset'
c2 24.7 mm from x' axis
θ 61.3 deg
Crank angle:
Link 2 velocity and acceleration Solution: 1.
θ 13.3 deg ω 10
rad
α 20
sec
rad sec
2
See Figure P7-6c and Mathcad file P0718.
Draw the mechanism to scale and label it. x'
24.7
Y 25.9 20.0 C
6 1
5 11.7
2
D 13.3°
X, y O2
A
1 3 61.3° 25.0 B
y'
1 4
x 3.7
2.
This mechanism can be analyzed as a fourbar slider-crank (links 1, 2, 3, and 4) and a fourbar slider-crank (links 1, 2, 5, and 6). Link 2 is the common, non-stationary link in the two branches and is the input to both slider-cranks. Since the vector loops are the same as those defined for the slider-crank linkage, we can use the equations derived in the text with slight modification of variable names. There are three coordinate systems: the global XY frame, the local xy frame for the first slider-crank, and the local x'y' frame for the second slider crank. Inputs to the fourbars must be in their respective local coordinates. All output will be given in local coordinate system.
3.
Determine and the vertical distance from O2 to B (d1) for the first slider-crank using equations 4.16 and 4.17.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-18-2
a1 sin θ c1 π b1
θ asin
θ 164.8 deg
d1 a1 cos θ b1 cos θ 4.
Determine the angular velocity of link 3 using equation 6.22a. ω
5.
d1 29.7 mm
a1 cos θ ω b1 cos θ
ω 2.329
rad sec
Using the Euler identity to expand equation 7.15b for AA, determine its magnitude, and direction.
a1 ω2 cosθ j sinθ
AA a1 α sin θ j cos θ AA ( 767.114 913.889i)
mm sec
2
AA 1193
The acceleration of pin A is
mm sec
6.
θAA 130.0 deg
at
2
Determine the angular acceleration of link 3 using equation 7.16d.
α
7.
θAA arg AA
AA AA
b1 cos θ 2
2
a1 α cos θ a1 ω sin θ b1 ω sin θ
α 36.4
rad sec
2
Use equation 7.16e for the acceleration of pin B.
2
2
AB a1 α sin θ a1 ω cos θ b1 α sin θ b1 ω cos θ AB 659
mm sec
8.
A negative sign means that AB is upward
2
Determine and the distance perpendicular to the x' axis from O2 to BD (d2) for the second slider-crank using equations 4.16 and 4.17.
a2 sin θ c2 π b2
θ asin
θ 230.9 deg
d2 a2 cos θ b2 cos θ 9.
d2 35.8 mm
Determine the angular velocity of link 5 using equation 6.22a. ω
a2 cos θ ω b2 cos θ
ω 11.915
rad sec
10. Using the Euler identity to expand equation 7.15b for AC, determine its magnitude, and direction.
a2 ω2 cosθ j sinθ
AC a2 α sin θ j cos θ AC ( 2038.378 70.828i )
mm sec
2
AC AC
θAC arg AC
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-18-3
AC 2040
The acceleration of pin C is
mm sec
at deg θAC 178.0
2
11. Determine the angular acceleration of link 5 using equation 7.16d.
α
b2 cos θ 2
2
a2 α cos θ a2 ω sin θ b2 ω sin θ
α 179.0
rad sec
12. Use equation 7.16e for the acceleration of pin D.
2
2
AD a2 α sin θ a2 ω cos θ b2 α sin θ b2 ω cos θ AD 7956
mm sec
2
A negative sign means that AD is downward and to the right
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-19-1
PROBLEM 7-19 Statement:
For the linkage shown in Figure P7-6d, write the vector loop equations; differentiate them, and do a complete position, velocity, and acceleration analysis of the linkage. Use the linkage geometry given below. Assume 2 = 10 rad/sec and 2 = 20 rad/sec2.
Given:
Measured lengths and angles: Link 21
a1 15.0 mm
Link 22
a2 15 mm
Link 3
b1 40.9 mm
Link 5
b2 44.7 mm
Offset
c1 1.0 mm
Offset'
c2 0.0 mm
θ 65.8 deg
Crank angle:
Solution: 1.
θ 24.2 deg ω 10
Link 2 velocity and acceleration
from x' axis
rad
α 20
sec
rad sec
2
See Figure P7-6d and Mathcad file P0719.
Draw the mechanism to scale and label it. Y, x 1.0
1
4
B 40.9 65.8°
3
44.7 A 2
y
24.2°
5
6 C
O2
X
1
1 15.0
2.
This mechanism can be analyzed as a fourbar slider-crank (links 1, 2, 3, and 4) and a fourbar slider-crank (links 1, 2, 5, and 6). Link 2 is the common, non-stationary link in the two branches and is the input to both slider-cranks. Since the vector loops are the same as those defined for the slider-crank linkage, we can use the equations derived in the text with slight modification of variable names. There are two coordinate systems: the global XY frame and the local xy frame for the first slider-crank. Inputs to the fourbars must be in their respective local coordinates. All output will be given in local coordinate system.
3.
Determine and the vertical distance from O2 to B (d1) for the first slider-crank using equations 4.16 and 4.17.
a1 sin θ c1 π b1
θ asin
d1 a1 cos θ b1 cos θ
θ 198.1 deg d1 45.0 mm
DESIGN OF MACHINERY - 5th Ed.
4.
Determine the angular velocity of link 3 using equation 6.22a. ω
5.
SOLUTION MANUAL 7-19-2
a1 cos θ ω b1 cos θ
ω 1.581
rad sec
Using the Euler identity to expand equation 7.15b for AA, determine its magnitude, and direction.
a1 ω2 cosθ j sinθ
AA a1 α sin θ j cos θ mm
AA ( 341.249 1491.157i )
sec
2
AA 1530
The acceleration of pin A is
mm sec
6.
θAA 102.9 deg
at
2
Determine the angular acceleration of link 3 using equation 7.16d.
α
7.
θAA arg AA
AA AA
b1 cos θ 2
2
a1 α cos θ a1 ω sin θ b1 ω sin θ
α 37.5
rad sec
2
Use equation 7.16e for the acceleration of pin B.
2
2
AB a1 α sin θ a1 ω cos θ b1 α sin θ b1 ω cos θ AB 37.5
mm sec
8.
A positive sign means that AB is upward
2
Determine and the distance perpendicular to the x' axis from O2 to BD (d2) for the second slider-crank using equations 4.16 and 4.17.
a2 sin θ c2 π b2
θ asin
θ 172.1 deg
d2 a2 cos θ b2 cos θ 9.
d2 58.0 mm
Determine the angular velocity of link 5 using equation 6.22a. ω
a2 cos θ ω b2 cos θ
ω 3.090
rad sec
10. Determine the angular acceleration of link 5 using equation 7.16d. α
b2 cos θ 2
2
a2 α cos θ a2 ω sin θ b2 ω sin θ
α 6.4
sec
11. Use equation 7.16e for the acceleration of pin C.
2
2
AC a2 α sin θ a2 ω cos θ b2 α sin θ b2 ω cos θ AC 1875
mm sec
2
rad
A negative sign means that AC is to the left
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-20-1
PROBLEM 7-20 Statement:
Figure P7-7 shows a sixbar linkage with the dimensions and crank angle given below. Find the angular acceleration of link 6 if 2 is a constant 1 rad/sec.
Given:
Link lengths: Link 2
a 1.00 in
Link 3
b 5.00 in
Link 6
d 3.00 in
Distance DB
LDB 1.50 in
Link 2 position, velocity, and acceleration:θ 45 deg Solution: 1.
c 0.0 in
Offset:
ω 1
rad
α 0
sec
rad sec
2
See Figure P7-7 and Mathcad file P0720.
Links 1, 2, 3, and 5 constitute a fourbar slider-crank. In order to solve for the accelerations at points B, C and D, we will need 3, and 3. From the graphical position solution below, θ 171.870 deg ω
Using equation 6.22a,
a ω cos θ b cos θ
ω 0.143 rad sec
1
CW
Axis of Slip Axis of Transmission 15.566°
Y
O6 6
B
171.870°
2
4
D
3
C 5
X
O2
2.
The graphical solution for accelerations of pin-jointed linkages uses equation 7.4: (APt + APn) = (AAt + AAn) + (APAt + APAn)
3.
For point C, this becomes: AC = (ABt + ABn) + (ACBt + ACBn) , where 2
2
ABn a ω
ABn 1.000 in sec
θABn θ 180 deg
θABn 225.000 deg
ABt a α
ABt 0.000 in sec
ACBn b ω
θACBn θ
2
2
ACBn 0.102 in sec
2
θACBn 171.870 deg
DESIGN OF MACHINERY - 5th Ed.
4.
SOLUTION MANUAL 7-20-2
Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ABn at an angle of ABn. From the tip of ABn, draw ACBn at an angle of ACBn. Now that the vectors with known magnitudes are drawn, from the origin and the tip of ACBn, draw construction lines in the directions of AC (horizontal) and ACBt, respectively. The intersection of these two lines are the tips of ACBt, and AC. 2.837 Y AC X t ACB
2.799
0
0.25 IN/S/S
Acceleration Scale n
AB n
ACB
5.
From the graphical solution above, Acceleration scale factor
in
ka 0.25
sec
6.
2
AC 2.837 ka
AC 0.709 in sec
ACBt 2.799 ka
ACBt 0.700 in sec
at an angle of 180 deg
2
Calculate 3 using equation 7.6. α
7.
2
ACBt
α 0.140 rad sec
b
2
CCW
In order to solve for the acceleration at points D, we will need 6, and 6. From the graphical position solution above, θ 15.566 deg 180 deg
θ 195.566 deg
From the vector diagram in Figure P7-7, VD 0.40
in sec
Vslip 0.65 8.
and
ω
VD
ω 0.133
d
in sec
For point D, use equation 7.19, referenced to point B instead of O2: (ADt + ADn)= AB + (ADBt + ADBn + ADBcor + ADBslip) , where ADn d ω
2
θADn θ 180 deg
ADn 0.053 in sec
θADn 15.566 deg
2
rad sec
DESIGN OF MACHINERY - 5th Ed.
AB a ω
SOLUTION MANUAL 7-20-3
2
AB 1.000 in sec
θAB θ 180 deg
θAB 225.000 deg
ADBt LDB α
ADBt 0.210 in sec
θADBt θ 90 deg
θADBt 81.870 deg
2
9.
2
2
2
ADBn LDB ω
ADBn 0.031 in sec
θADBn θ
θADBn 171.870 deg
ADBcor 2 Vslip ω
ADBcor 0.186 in sec
θADBcor θ 90 deg
θADBcor 261.870 deg
2
Repeat procedure of step 4 for the equation in step 8. Y ADn 0
X
0.25 IN/S/S
Acceleration Scale 3.483 n DB
A
t ADB cor ADB
AB
ADt
slip ADB
10. From the graphical solution above, Acceleration scale factor
ka 0.25
in sec
ADt 3.483 ka
2
ADt 0.871 in sec
2
at an angle of -74.434 deg
The angular acceleration of link 6 is α
ADt d
α 0.290
rad sec
2
CCW
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-21-1
PROBLEM 7-21 Statement:
The linkage in Figure P7-8a has the dimensions and crank angle given below. Find 4, AA, and AB in the global coordinate system for the position shown for 2 = 15 rad/sec clockwise (CW) and 2 = 25 rad/sec2 CCW. Use the acceleration difference graphical method.
Given:
Solution: 1.
Link lengths: Link 2 (O2 to A)
a 116 mm
Link 3 (A to B)
b 108 mm
Link 4 (B to O4)
c 110 mm
Link 1 (O2 to O4)
d 174 mm
Crank angle:
θ 62 deg
Input crank angular velocity
ω 15 rad sec
Coordinate rotation angle
α 25 deg
Local xy system 1
α 25 rad sec
2
Global XY system to local xy system
See Figure P7-8a and Mathcad file P0721.
Draw the linkage to a convenient scale. Indicate the directions of the acceleration vectors of interest. Y
Direction of AAt
y A
37.000° 70.133°
2
3
O2 X 2
22.319° B
Direction of ABAt 4
Direction of ABt
O4
x
2.
In order to solve for the accelerations at points A and B, we will need 3, 3, and 4. From the graphical position solution above (in the local xy coordinate system), θ 180 deg 70.133 deg α
θ 275.133 deg
θ 180 deg 22.319 deg α
θ 182.681 deg
Using equation (6.18),
3.
ω 13.869 rad sec
ω 8.654 rad sec
ω
a ω sin θ θ b sin θ θ
ω
a ω sin θ θ c sin θ θ
The graphical solution for accelerations uses equation 7.4:
1
1
(APt + APn) = (AAt + AAn) + (APAt + APAn)
DESIGN OF MACHINERY - 5th Ed.
4.
SOLUTION MANUAL 7-21-2
For point B, this becomes: (ABt + ABn) = (AAt + AAn) + (ABAt + ABAn) , where ABn c ω
2
ABn 8237.9 mm sec
θABn θ 180 deg
θABn 2.681 deg
2
AAn a ω
AAn 26100.0 mm sec
θAAn θ 180 deg
θAAn 242.000 deg
AAt a α
AAt 2900.0 mm sec
θAAt θ 90 deg
θAAt 152.000 deg
2
5.
2
2
2
ABAn b ω
ABAn 20773 mm sec
θABAn θ 180 deg
θABAn 95.133 deg
2
Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ABn at an angle of ABn + and AAn at an angle of AAn + . From the tip of AAn, draw AAt at an angle of AAt + . From the tip of AAt, draw ABAn at an angle of BAn + . Now that the vectors with known magnitudes are drawn, from the tips of ABn and ABAn, draw construction lines in the directions of ABt and ABAt, respectively. The intersection of these two lines are the tips of ABt, ABt, and ABAt.
n
ABA
41.409
Y
y
t
ABA
X 131.308
AB x 149.340°
16.401° n
AB t
AB
AA
t
AA
6.
0
200 mm/s/s
Acceleration Scale n
AA
From the graphical solution above, Acceleration scale factor
ka 200
mm sec
2
AA 131.308 ka
AA 26262 mm sec
AB 41.409 ka
AB 8282 mm sec
2
2
at an angle of -149.34 deg at an angle of -16.40 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-22-1
PROBLEM 7-22 The linkage in Figure P7-8a has the dimensions and crank angle given below. Find 4, AA, and
Statement:
AB in the global coordinate system for the position shown for 2 = 15 rad/sec clockwise (CW) and 2 = 25 rad/sec2 CCW. Use an analytical method. Given:
Link lengths:
Solution: 1. 2.
Link 2 (O2 to A)
a 116 mm
Link 3 (A to B)
b 108 mm
Link 4 (B to O4)
c 110 mm
Link 1 (O2 to O4)
d 174 mm
Crank angle:
θ 62 deg
Input crank angular velocity
ω 15 rad sec
Coordinate rotation angle
β 25 deg
Local xy system 1
α 25 rad sec
2
Global XY system to local xy system
See Figure P7-8a and Mathcad file P0722.
Draw the linkage to scale and label it. Y
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 1.5000 2
K3
y A
d c 2
K2 1.5818 2
2
a b c d
3 2
O2
2
X
K3 1.7307
2 a c
2
A cos θ K1 K2 cos θ K3
B
B 2 sin θ
O4
4
C K1 K2 1 cos θ K3 A 0.0424 3.
B 1.7659
Use equation 4.10b to find values of 4 for the crossed circuit.
2
θ 2 atan2 2 A B 4.
x
C 2.0186
B 4 A C
θ 182.681 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2 a b
D cos θ K1 K4 cos θ K5
E 2 sin θ
2
K4 1.6111 D 2.0021 E 1.7659
F K1 K4 1 cos θ K5
F 0.0589
K5 1.7280
DESIGN OF MACHINERY - 5th Ed.
5.
SOLUTION MANUAL 7-22-2
Use equation 4.13 to find values of 3 for the crossed circuit.
2
θ 2 atan2 2 D E 6.
7.
E 4 D F
θ 275.133 deg
Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18.
ω 13.869
ω 8.654
ω
a ω sin θ θ b sin θ θ
ω
a ω sin θ θ c sin θ θ
rad sec
rad sec
Using the Euler identity to expand equation 7.13a for AA. Determine the magnitude, and direction (in the global coordinate system).
a ω2 cosθ j sinθ
AA a α sin θ j cos θ mm
AA ( 14814 21683i)
sec
AA 26261
The acceleration of pin A is
mm sec
8.
θAA arg AA β
AA AA
2
at
2
θAA 149.3 deg
Use equations 7.12 to determine the angular accelerations of links 3 and 4 for the crossed circuit.
A c sin θ
B b sin θ
D c cos θ
E b cos θ
A 5.145 mm
B 107.567 mm
D 109.880 mm
E 9.662 mm
2
2
2
2
2
C a α sin θ a ω cos θ b ω cos θ c ω cos θ 4
2
2
C 2.490 10 mm sec
F a α cos θ a ω sin θ b ω sin θ c ω sin θ 3
F 1.380 10 mm sec α
9.
C D A F
2
α 231.119
A E B D
rad sec
α
2
C E B F A E B D
α 7.768
sec
Use equation 7.13c to determine the acceleration of point B for the crossed circuit.
c ω2 cosθ j sinθ
AB c α sin θ j cos θ AB ( 8189 1239i)
mm sec
2
The acceleration of pin B is
θAB arg AB β
AB AB AB 8282
mm sec
2
at
θAB 16.4 deg
rad
(Global)
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-23-1
PROBLEM 7-23 Statement:
The linkage in Figure P7-8a has the dimensions and crank angle given below. Find and plot 4, AA, and AB in the local coordinate system for the maximum range of motion that this linkage allows if 2 = 15 rad/sec CW and 2 = 25 rad/sec2 CCW.
Given:
Link lengths: Link 2 (O2 to A)
a 116 mm
Link 4 (B to O4)
c 110 mm
Solution: 1.
d 174 mm
Link 1 (O2 to O4)
ω 15 rad sec
Input crank angular velocity
b 108 mm
Link 3 (A to B) 1
α 25 rad sec
2
See Figure P7-8a and Mathcad file P0723.
Draw the linkage to scale and label it. Y y A
2 3 2
O2
X 2
B O4
4
x
2.
Determine the range of motion for this non-Grashof triple rocker using equations 4.37. 2
arg1
2
2
2 a d 2
arg2
2
a d b c
2
2
a d b c 2 a d
2
b c
arg1 1.083
a d b c
arg2 0.094
a d
θ2toggle acos arg2
θ2toggle 95.4 deg
The other toggle angle is the negative of this. Thus, θ θ2toggle 0.5 deg θ2toggle 1 deg θ2toggle 0.5 deg 3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d a
K1 1.5000
K2
d c
K2 1.5818
DESIGN OF MACHINERY - 5th Ed. 2
SOLUTION MANUAL 7-23-2
2
2
a b c d
K3
2
K3 1.7307
2 a c
A θ cos θ K1 K2 cos θ K3
B θ 2 sin θ
C θ K1 K2 1 cos θ K3 4.
Use equation 4.10b to find values of 4 for the crossed circuit.
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 5.
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. 2
d
K4
K5
b
2
2
c d a b
2
K4 1.6111
2 a b
K5 1.7280
D θ cos θ K1 K4 cos θ K5
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 6.
Use equation 4.13 to find values of 3 for the crossed circuit.
θ θ 2 atan2 2 D θ E θ 7.
Determine the angular velocity of links 3 and 4 using equations 6.18.
a ω
a ω
ω θ
ω θ
8.
2 4 Dθ F θ
E θ
b
c
sin θ θ θ θ
sin θ θ θ θ
sin θ θ θ
sin θ θ θ
Using the Euler identity to expand equation 7.13a for AA. Determine the x and y components in the local coordinate system.
a ω2 cosθ j sinθ
AA θ a α sin θ j cos θ
AAx θ Re AA θ 9.
AAy θ Im AA θ
Use equations 7.12 to determine the angular acceleration of link 4.
B θ b sin θ θ
E θ b cos θ θ
A θ c sin θ θ
D θ c cos θ θ
2
2
c ωθ2 cosθθ
2
2 c ωθ2 sinθθ
C θ a α sin θ a ω cos θ b ω θ cos θ θ F θ a α cos θ a ω sin θ b ω θ sin θ θ
DESIGN OF MACHINERY - 5th Ed.
α θ
SOLUTION MANUAL 7-23-3
A θ E θ B θ D θ C θ E θ B θ F θ
10. Use equation 7.13c to determine the acceleration of point B. Determine the x and y components in the local coordinate system.
j cosθθ c ωθ2 cosθθ j sinθθ
AB θ c α θ sin θ θ
ABx θ Re AB θ
ABy θ Im AB θ
11. Plot the angular acceleration for link 4 and the acceleration components for pins B and C.
Angular Accel., rad/sec^2
ANGULAR ACCELERATION OF LINK 4
0 1000 2000 3000 4000 100
50
0
50
100
Crank Angle, deg
ACCELERATION OF POINT A 30
Acceleration, m/sec^2
20 10 0 10 20 30 100
50
0 Crank Angle, deg
x component y component
50
100
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-23-4
ACCELERATION OF POINT B 300
Acceleration, m/sec^2
200 100 0 100 200 300 100
50
0 Crank Angle, deg
x component y component
50
100
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-24-1
PROBLEM 7-24 Statement:
The linkage in Figure P7-8b has the dimensions and crank angle given below. Find 4, AA, and
Given:
AB in the global coordinate system for the position shown for = 20 rad/sec CCW. Use the acceleration difference graphical method. Link lengths:
Solution: 1.
Link 2 (A to B)
a 40 mm
Link 3 (B to C)
b 96 mm
Link 4 (C to D)
c 122 mm
Link 1 (A to D)
d 162 mm
Crank angle:
θ 93 deg
Input crank angular velocity
ω 20 rad sec
Coordinate rotation angle
α 36 deg
Local xy system 1
α 0 rad sec
2
Global XY system to local xy system
See Figure P7-8b and Mathcad file P0724.
Draw the linkage to a convenient scale. Indicate the directions of the acceleration vectors of interest. Direction of AAt
Direction of ABt
Y 2
Direction of ABAt
y
B
57.0° 3
A
2
X O2
36.0° 4
O4 x
2.
In order to solve for the accelerations at points A and B, we will need 3, 3, and 4. From the graphical position solution above (in the local xy coordinate system), θ 31.486 deg
θ 132.406 deg
Using equation (6.18), ω
a ω sin θ θ b sin θ θ
ω 5.388 rad sec
ω
a ω sin θ θ c sin θ θ
ω 5.870 rad sec
1
1
(APt + APn) = (AAt + AAn) + (APAt + APAn)
3.
The graphical solution for accelerations uses equation 7.4:
4.
For point B, this becomes: (ABt + ABn) = (AAt + AAn) + (ABAt + ABAn) , where ABn c ω
2
ABn 4203 mm sec
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-24-2
θABn θ 180 deg
θABn 47.594 deg
2
AAn a ω
AAn 16000 mm sec
θAAn θ 180 deg
θAAn 273.000 deg
AAt a α
AAt 0.000 mm sec 2
5.
2
2 2
ABAn b ω
ABAn 2787 mm sec
θABAn θ 180 deg
θABAn 148.514 deg
Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ABn at an angle of ABn and AAn at an angle of AAn . From the tip of AAn, draw AAt at an angle of AAt . From the tip of AAt, draw ABAn at an angle of BAn . Now that the vectors with known magnitudes are drawn, from the tips of ABn and ABAn, draw construction lines in the directions of ABt and ABAt, respectively. The intersection of these two lines are the tips of ABt, ABt, and ABAt.
Y
0
500 mm/s/s y
Acceleration Scale X 24.340
n
AB
AB x
t
AB
AA
t ABA n
22.244
ABA 6.
From the graphical solution above, Acceleration scale factor
ka 500
mm sec
7.
2 2
AA AAn
AA 16000 mm sec
AB 24.340 ka
AB 12170 mm sec
ABt 22.244 ka
ABt 11122 mm sec
2
at an angle of 273.0 deg (Global) at an angle of 242.610 deg
2
Calculate 4 using equation 7.6. α
ABt c
α 91.2 rad sec
2
CCW
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-25-1
PROBLEM 7-25 Statement:
The linkage in Figure P7-8b has the dimensions and crank angle given below. Find , AA, and
Given:
AB in the global coordinate system for the position shown for = 20 rad/sec CCW constant. Use an analytical method. Link lengths:
Solution: 1.
Link 2 (A to B)
a 40 mm
Link 3 (B to C)
b 96 mm
Link 4 (C to D)
c 122 mm
Link 1 (A to D)
d 162 mm
Crank angle:
θ 93 deg
Input crank angular velocity
ω 20 rad sec
Coordinate rotation angle
β 36 deg
Local xy system 1
α 0 rad sec
Global XY system to local xy system
See Figure P7-8b and Mathcad file P0725.
Draw the linkage to scale and label it.
y
Y 2.
2
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 4.0500 2
K3
2
d
2
2
57.0°
2
c
B
3 X
O2
K2 1.3279
a b c d
A
36.0° 4
2
K3 3.4336
2 a c
A cos θ K1 K2 cos θ K3
O4
B 2 sin θ
x
C K1 K2 1 cos θ K3 A 0.5992 3.
B 1.9973
C 7.6054
Use equation 4.10b to find values of 4 for the open circuit in the local xy coordinate system.
2
θ 2 atan2 2 A B 4.
B 4 A C
θ 132.386 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2 a b
D cos θ K1 K4 cos θ K5
E 2 sin θ
2
K4 1.6875
K5 2.8875
D 7.0782 E 1.9973
F K1 K4 1 cos θ K5 5.
2 π
F 1.1265
Use equation 4.13 to find values of 3 for the open circuit in the local xy coordinate system.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-25-2
2
θ 2 atan2 2 D E 6.
7.
E 4 D F
2 π
θ 31.504 deg
Determine the angular velocity of links 3 and 4 using equations 6.18. ω
a ω sin θ θ b sin θ θ
ω 5.385
ω
a ω sin θ θ c sin θ θ
ω 5.868
rad sec
rad sec
Using the Euler identity to expand equation 7.13a for AA. Determine the magnitude, and direction (in the global coordinate system).
a ω2 cosθ j sinθ
AA a α sin θ j cos θ AA ( 837 15978i)
mm sec
AA 16000
The acceleration of pin A is
mm sec
8.
θAA arg AA β
AA AA
2
at
2
θAA 123.0 deg
Use equations 7.12 to determine the angular accelerations of links 3 and 4 for the crossed circuit.
A c sin θ
B b sin θ
D c cos θ
E b cos θ
A 90.111 mm
B 50.166 mm
D 82.244 mm
E 81.850 mm
2
2
2
2
2
C a α sin θ a ω cos θ b ω cos θ c ω cos θ 3
2
2
C 4.368 10 mm sec
F a α cos θ a ω sin θ b ω sin θ c ω sin θ 4
F 1.433 10 mm sec α
9.
C D A F A E B D
2
α 81.037
rad sec
α
2
C E B F A E B D
α 93.586
sec
Use equation 7.13c to determine the acceleration of point B for the open circuit.
c ω2 cosθ j sinθ
AB c α sin θ j cos θ
AB ( 5601 10800i)
mm sec
The acceleration of pin B is
2
θAB arg AB β
AB AB AB 12166
mm sec
2
at
θAB 153.4 deg
rad
(Global)
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-26-1
PROBLEM 7-26 Statement:
Given:
The linkage in Figure P7-8b has the dimensions and crank angle given below. Find and plot , AA, and AB in the local coordinate system for the maximum range of motion that this linkage allows if = 20 rad/sec CCW constant. Link lengths: Link 2 (A to B)
a 40 mm
Link 3 (B to C)
b 96 mm
Link 4 (C to D)
c 122 mm
Link 1 (A to D)
d 162 mm
ω 20 rad sec
Input crank angular velocity Solution: 1. 2.
1
α 0 rad sec
See Figure P7-8b and Mathcad file P0726.
Draw the linkage to scale and label it.
y
Y
The range of 2 for this Grashof crank-rocker is:
2
θ 0 deg 1 deg 360 deg
A
57.0°
2 3.
2
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 4.0500 2
2
a b c d
O2
36.0° 4
d c
2
K3 3.4336
2 a c
X
K2 1.3279
2
K3
B
3
O4 x
A θ cos θ K1 K2 cos θ K3
4.
C θ K1 K2 1 cos θ K3
B θ 2 sin θ
Use equation 4.10b to find values of 4 for the open circuit.
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 5.
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
2 a b
K4 1.6875
D θ cos θ K1 K4 cos θ K5
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 6.
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ
2 4 Dθ F θ
E θ
K5 2.8875
DESIGN OF MACHINERY - 5th Ed.
7.
Determine the angular velocity of links 3 and 4 using equations 6.18.
a ω
a ω
ω θ
ω θ
8.
SOLUTION MANUAL 7-26-2
b
c
sin θ θ θ θ
sin θ θ θ θ
sin θ θ θ
sin θ θ θ
Using the Euler identity to expand equation 7.13a for AA. Determine the x and y components in the local coordinate system.
a ω2 cosθ j sinθ
AA θ a α sin θ j cos θ
AAx θ Re AA θ 9.
AAy θ Im AA θ
Use equations 7.12 to determine the angular acceleration of link 4.
B θ b sin θ θ
E θ b cos θ θ
A θ c sin θ θ
D θ c cos θ θ
2
2
c ωθ2 cosθθ
2
2 c ωθ2 sinθθ
C θ a α sin θ a ω cos θ b ω θ cos θ θ F θ a α cos θ a ω sin θ b ω θ sin θ θ
α θ
A θ E θ B θ D θ C θ E θ B θ F θ
10. Use equation 7.13c to determine the acceleration of point B. Determine the x and y components in the local coordinate system.
j cosθθ c ωθ2 cosθθ j sinθθ
AB θ c α θ sin θ θ
ABx θ Re AB θ
ABy θ Im AB θ
11. Plot the angular acceleration for link 4 and the acceleration components for pins A and B. ANGULAR ACCELERATION OF LINK 4 Angular Accel., rad/sec^2
200 100 0 100 200
0
45
90
135
180
225
Crank Angle, deg
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-26-3
ACCELERATION OF POINT A
Acceleration, m/sec^2
20
10
0
10
20
0
45
90
135
180
225
270
315
360
Crank Angle, deg x component y component
ACCELERATION OF POINT B 20
Acceleration, m/sec^2
10
0
10
20
0
45
90
135
180
Crank Angle, deg x component y component
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-27-1
PROBLEM 7-27 Statement:
The offset crank-slider linkage in Figure P7-8f has the dimensions and crank angle given below. Find AA, and AB in the global coordinate system for the position shown for = 25 rad/sec CW, constant. Use the acceleration difference graphical method.
Given:
Link lengths: Link 2 (D to E)
a 63 mm
Offset
c 52 mm
Link 2 position, velocity, and acceleration: θ 141 deg
1.
ω 25
rad sec
α 0
α 90 deg
Coordinate rotation angle: Solution:
b 130 mm
Link 3 (E to F)
rad sec
See Figure P7-8f and Mathcad file P0727.
In order to solve for the accelerations at point F, we will need 3, and 3. From the graphical position solution below (in the local coordinate system), θ 44.828 deg ω
Using equation 6.22a,
a ω cos θ b cos θ
ω 13.276 rad sec
1
Direction of AB 4 B
Direction of ABAt
1 3 44.828°
Direction of AAt
Y
A 2 51.000°
O2
X, y
x
2.
The graphical solution for accelerations uses equation 7.4: (APt + APn) = (AAt + AAn) + (APAt + APAn)
3.
For point F, this becomes: AF = (AEt + AEn) + (AFEt + AFEn) , where (angles in the global coordinate system) AEn a ω
2
AEn 39375 mm sec
θAEn θ α 180 deg
θAEn 231.000 deg
AEt a α
AEt 0.000 mm sec 2
2
2
AFEn b ω
AFEn 22911 mm sec
θAFEn θ α
θAFEn 45.172 deg
2
2
DESIGN OF MACHINERY - 5th Ed.
4.
SOLUTION MANUAL 7-27-2
Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw AEn at an angle of AEn. From the tip of AEn, draw AFEn at an angle of AFEn. Now that the vectors with known magnitudes are drawn, from the origin and the tip of AFEn, draw construction lines in the directions of AF (vertical) and AFEt, respectively. The intersection of these two lines are the tips of AFEt, and AF. Y
X,y
0
500 mm/s/s
Acceleration Scale x 76.546 n
AE AF t AFE n
AFE
5.
From the graphical solution above, Acceleration scale factor
ka 500
mm sec
AF 76.546 ka
2
AF 38273 mm sec
2
at an angle of -90 deg (global)
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-28-1
PROBLEM 7-28 Statement:
The offset crank-slider linkage in Figure P7-8f has the dimensions and crank angle given below. Find AA, and AB in the global coordinate system for the position shown for = 25 rad/sec CW, constant. Use an analytical method.
Given:
Link lengths: Link 2 (O2 to A)
a 63 mm
Offset
c 52 mm
Link 2 position, velocity, and acceleration: θ 141 deg
1.
ω 25
rad
rad
α 0
sec
sec
α 90 deg
Coordinate rotation angle: Solution:
b 130 mm
Link 3 (A to B)
2
See Figure P7-8f and Mathcad file P0728.
Draw the linkage to a convenient scale. 4
2.
Determine 3 (in the local coordinate system) and d using equations 4.16 for the crossed circuit.
a sin θ b
θ asin
c
d 141.160 mm
Y 52.000
Determine the angular velocity of link 3 using equation 6.22a:
A 2 51.000°
a cos θ ω ω b cos θ 4.
3
θ 44.828 deg
d a cos θ b cos θ 3.
B 1
ω 13.276
rad O2
sec
Using the Euler identity to expand equation 7.15b for AB, determine its magnitude, and direction (global).
X, y
x
a ω2 cosθ j sinθ
AA a α sin θ j cos θ
AA ( 30600.122 24779.490i)
mm sec
AA 39375
The acceleration of pin A is
mm sec
5.
at
2
θAA 129.0 deg
Determine the angular acceleration of link 3 using equation 7.16d.
α
6.
θAA arg AA α
AA AA
2
b cos θ 2
2
a α cos θ a ω sin θ b ω sin θ
α 93.574
sec
Use equation 7.16e for the acceleration of pin B.
2
2
AB a α sin θ a ω cos θ b α sin θ b ω cos θ AB 38274
mm sec
2
rad
A positive sign means that AB is downward
2
DESIGN OF MACHINERY
SOLUTION MANUAL 7-29-1
PROBLEM 7-29 Statement:
The offset crank-slider linkage in Figure P7-8f has the dimensions and crank angle given below. Find and plot AA, and AB in the global coordinate system for the maximum range of motion that this linkage allows if = 25 rad/sec CW, constant..
Given:
Link lengths: Link 2 (O2 to A)
a 63 mm
Offset
c 52 mm ω 25
Link 2 velocity, and acceleration:
1. 2.
rad
rad
α 0
sec
sec
2
α 90 deg
Coordinate rotation angle: Solution:
See Figure P7-8f and Mathcad file P0729.
Y
Draw the linkage to a convenient scale.
4
Determine the range of motion for this slider-crank linkage.
B
1 3
θ 0 deg 2 deg 360 deg 3.
Determine 3 using equations 4.16 for the crossed circuit.
a sin θ c b
2
Determine the angular velocity of link 3 using equation 6.22a:
ω θ
5.
A
52.000
θ θ asin 4.
b 130 mm
Link 3 (A to B)
a ω b cos θ θ
O2
cos θ
2
X, y
x
Using the Euler identity to expand equation 7.15b for AA, determine its X and Y components in the global coordinate system.
a ω2 cosθ j sinθ
AA θ a α sin θ j cos θ
AAX θ Re AA θ cos α Im AA θ sin α AAY θ Re AA θ sin α Im AA θ cos α 6.
Determine the angular acceleration of link 3 using equation 7.16d.
α θ 7.
2 b cos θ θ 2
a α cos θ a ω sin θ b ω θ sin θ θ
Use equation 7.16e for the acceleration of pin B.
2 2 b α θ sin θ θ b ω θ cos θ θ
AB θ a α sin θ a ω cos θ
8.
Plot the components of AA and the magnitude of AB. A positive value for AB is downward, a negative value, upward.
DESIGN OF MACHINERY
SOLUTION MANUAL 7-29-2
PIN A ACCELERATION COMPONENTS
Acceleration, m/sec^2
40
20
0
20
40
0
45
90
135
180
225
270
315
360
Crank Angle, deg x component y component
PIN B ACCELERATION MAGNITUDE 100
Acceleration, m/sec^2
50
0
50
100
0
45
90
135
180
Crank Angle, deg
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-30-1
PROBLEM 7-30 Statement:
The linkage in Figure P7-8d has the dimensions and crank angle given below. Find AA, AB, and
Given:
Abox in the global coordinate system for the position shown for = 30 rad/sec CW, constant . Use the acceleration difference graphical method. Link lengths:
Solution: 1.
Link 2 (O2 to A)
a 30 mm
Link 3 (A to B)
b 150 mm
Link 4 (O4 to B)
c 30 mm
Link 1 (O2 to O4)
d 150 mm
Crank angle:
θ 58 deg
Input crank angular velocity
ω 30 rad sec
Global XY system 1
α 0 rad sec
2
See Figure P7-8d and Mathcad file P0730.
Draw the linkage to a convenient scale. Indicate the directions of the acceleration vectors of interest.
Vbox 1
Direction of ABAt
A 2
58°
B
3
58° Direction of ABt
O4
O2
4
Direction of AAt 2.
In order to solve for the accelerations at points A and B, we will need 3, 3, and 4. From the graphical position solution above (in the global XY coordinate system), θ 0 deg
θ 0.000 deg
θ θ
θ 58.000 deg
This is a special-case Grashof in the parallelogram configuration. Therefore, ω 0.0 rad sec
1
ω ω (APt + APn) = (AAt + AAn) + (APAt + APAn)
3.
The graphical solution for accelerations uses equation 7.4:
4.
For point B, this becomes: (ABt + ABn) = (AAt + AAn) + (ABAt + ABAn) , where ABn c ω
2
ABn 27000 mm sec
θABn θ 180 deg 2
θABn 238.000 deg
AAn a ω
AAn 27000 mm sec
θAAn θ 180 deg
θAAn 238.000 deg
AAt a α
AAt 0.0 mm sec 2
ABAn b ω
2
ABAn 0 mm sec
2
2
2
DESIGN OF MACHINERY - 5th Ed.
5.
SOLUTION MANUAL 7-30-2
In this case, there is no need to draw an acceleration diagram since AAt and ABAn are zero. This means that ABAt and ABt will also be zero and AA = AAn, AB = ABn.
6.
Since ABAt = 0, 3 = 0 and, since AB = AA, 4 = 2= 0.
7.
For the box, equation 7.4 becomes: Abox = (AAt + AAn) + (AboxAt + AboxAn) , where AboxAt and AboxAn are both zero since 3 and 3 are zero. Therefore, Abox = AAx.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-31-1
PROBLEM 7-31 Statement:
The linkage in Figure P7-8d has the dimensions and crank angle given below. Find AA, AB, and
Given:
Abox in the global coordinate system for the position shown for = 30 rad/sec CW, constant. Use an analytical method. Link lengths:
Solution: 1.
Link 2 (O2 to A)
a 30 mm
Link 3 (A to B)
b 150 mm
Link 4 (O4 to B)
c 30 mm
Link 1 (O2 to O4)
d 150 mm
Crank angle:
θ 58 deg
Input crank angular velocity
ω 30 rad sec
1
α 0 rad sec
2
See Figure P7-8d and Mathcad file P0731.
Draw the linkage to a convenient scale and label it.
E Vbox 1
A
58°
2
3
58° B
O2
2.
O4
4
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 5.0000 2
K3
d c
K2 5.0000 2
2
a b c d
2
K3 1.0000
2 a c
A cos θ K1 K2 cos θ K3
B 2 sin θ
C K1 K2 1 cos θ K3 A 6.1197 3.
B 1.6961
Use equation 4.10b to find values of 4 for the open circuit.
θ 2 atan2 2 A B θ θ 360 deg 4.
C 2.8205
2
B 4 A C
θ 302.000 deg θ 58.000 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12.
DESIGN OF MACHINERY - 5th Ed.
K4
SOLUTION MANUAL 7-31-2
2
d b
2
2
c d a b
K5
2
K4 1.0000
2 a b
D cos θ K1 K4 cos θ K5
D 8.9402
E 2 sin θ
E 1.6961
F K1 K4 1 cos θ K5 5.
F 0.0000
Use equation 4.13 to find values of 3 .
2
θ 2 atan2 2 D E
E 4 D F
θ 360.000 deg
θ θ 360 deg 6.
7.
K5 5.0000
θ 0.000 deg
Determine the angular velocity of links 3 and 4 using equations 6.18.
ω 0.000
ω 30.000
ω
a ω sin θ θ b sin θ θ
ω
a ω sin θ θ c sin θ θ
rad sec rad sec
Using the Euler identity to expand equation 7.13a for AA. Determine the magnitude, and direction.
a ω2 cosθ j sinθ
AA a α sin θ j cos θ AA ( 14308 22897i)
mm sec
The acceleration of pin A is
AA 27000
mm sec
8.
θAA arg AA
AA AA
2
θAA 122.0 deg
at
2
Use equations 7.12 to determine the angular accelerations of links 3 and 4.
A c sin θ
B b sin θ
D c cos θ
E b cos θ
A 1.002 in
B 0.000 in
D 0.626 in
E 5.906 in
2
2
2
2
2
2
C a α sin θ a ω cos θ b ω cos θ c ω cos θ C 0.000 mm sec
2
F a α cos θ a ω sin θ b ω sin θ c ω sin θ F 0.000 mm sec α 9.
C D A F A E B D
2
α 0.000
rad sec
2
Use equation 7.13c to determine the acceleration of point B.
α
C E B F A E B D
α 0.000
rad sec
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-31-3
c ω2 cosθ j sinθ
AB c α sin θ j cos θ AB ( 14308 22897i)
mm sec
The acceleration of pin B is
2
θAB arg AB
AB AB AB 27000
mm sec
at
2
θAB 122.0 deg
10. This is a special case Grashof in the parallelogram configuration. All points on link 3 have the same velocity and acceleration. The acceleration of the box will be equal to the X-component of the acceleration of any point on link 3. Abox Re AA The acceleration of the box is Abox 14308
mm sec
2
a negative sign means Abox is to the left
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-32-1
PROBLEM 7-32 Statement:
The linkage in Figure P7-8d has the dimensions and crank angle given below. Write a computer program or use an equation solver to find and plot AA, AB, and Abox in the global coordinate system for the maximum range of motion that this linkage allows if = 30 rad/sec CW, constant .
Given:
Link lengths: Link 2 (O2 to A)
a 30 mm
Link 3 (A to B)
Link 4 (O4 to B)
c 30 mm
Link 1 (O2 to O4)
ω 30 rad sec
Input crank angular velocity Solution: 1.
1
α 0 rad sec
b 150 mm d 150 mm 2
See Figure P7-8d and Mathcad file P0732.
Draw the linkage to a convenient scale and label it.
Vbox 1
A
3
58°
2
58° B
O2
2.
O4
4
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 5.0000 2
2
a b c d
2
K3 1.0000
2 a c
c
K2 5.0000
2
K3
d
A θ cos θ K1 K2 cos θ K3
B θ 2 sin θ
C θ K1 K2 1 cos θ K3 3.
Use equation 4.10b to find values of 4 for the open circuit.
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 4.
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b 2 a b
D θ cos θ K1 K4 cos θ K5
E θ 2 sin θ
2
K4 1.0000
K5 5.0000
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-32-2
F θ K1 K4 1 cos θ K5 5.
Use equation 4.13 to find values of 3 .
2 4 Dθ F θ
θ θ 2 atan2 2 D θ E θ 6.
Determine the angular velocity of links 3 and 4 using equations 6.18.
a ω
a ω
ω θ
ω θ
7.
E θ
b
c
sin θ θ θ θ
sin θ θ θ θ
sin θ θ θ
sin θ θ θ
Using the Euler identity to expand equation 7.13a for AA. Determine the XY components.
a ω2 cosθ j sinθ
AA θ a α sin θ j cos θ
AAX θ Re AA θ 8.
AAY θ Im AA θ
Use equations 7.12 to determine the angular accelerations of links 3 and 4 for the crossed circuit.
B' θ b sin θ θ
E' θ b cos θ θ
A' θ c sin θ θ
D' θ c cos θ θ
2
2 c ωθ2 cosθθ
2
2 c ωθ2 sinθθ
C' θ a α sin θ a ω cos θ b ω θ cos θ θ F' θ a α cos θ a ω sin θ b ω θ sin θ θ
α θ
9.
A' θ E' θ B' θ D' θ C' θ D' θ A' θ F' θ
α θ
A' θ E' θ B' θ D' θ C' θ E' θ B' θ F' θ
Use equation 7.13c to determine the acceleration of point B.
j cosθθ 2 c ω θ cos θ θ j sin θ θ
AB θ c α θ sin θ θ
ABX θ Re AB θ
ABY θ Im AB θ
10. This is a special case Grashof in the parallelogram configuration. All points on link 3 have the same velocity an acceleration. The acceleration of the box will be equal to the X-component of the acceleration of any point on link 3.
Abox θ Re AA θ
11. Plot the accelerations over the range: θ 0 deg 2 deg 360 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-32-3
ACCELERATION OF PINS A & B
Acceleration, m/sec^2
40
20
0
20
40
0
45
90
135
180
225
270
315
360
Input Angle, deg x component y component
ACCELERATION OF THE BOX 40
Acceleration, m/sec^2
20
0
20
40
0
45
90
135
180
Input Angle, deg
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-33-1
PROBLEM 7-33 Statement:
The linkage in Figure P7-8g has the dimensions and crank angle given below. Find 4, AA, and
Given:
AB in the global coordinate system for the position shown if = 15 rad/sec CW, constant. Use the acceleration difference graphical method. Link lengths:
Solution: 1.
Link 2 (O2 to A)
a 49 mm
Link 3 (A to B)
b 100 mm
Link 4 (O4 to B)
c 153 mm
Link 1 (O2 to O4)
d 87 mm
Crank angle:
θ 148 deg Local xy system
Input crank angular velocity
ω 15 rad sec
Coordinate rotation angle
β 119 deg Global XY system to local xy system
1
α 0 rad sec
2
See Figure P7-8e and Mathcad file P0733.
Draw the linkage to a convenient scale. Indicate the directions of the acceleration vectors of interest. Y
Direction of ABt
Direction of ABAt O6
B 3 29.000°
2
4 C
A
5
X
6
O2
Direction of A y
D
O4 119.000° x 2.
In order to solve for the accelerations at points A and B, we will need 3, 3, and 4. From the graphical position solution above (in the local xy coordinate system), θ 266.812 deg
θ 208.876 deg
Using equation (6.18), ω
a ω sin θ θ b sin θ θ
ω 7.576 rad sec
ω
a ω sin θ θ c sin θ θ
ω 4.967 rad sec
1
1
(APt + APn) = (AAt + AAn) + (APAt + APAn)
3.
The graphical solution for accelerations uses equation 7.4:
4.
For point B, this becomes: (ABt + ABn) = (AAt + AAn) + (ABAt + ABAn) , where (angles in the global coordinate system)
DESIGN OF MACHINERY - 5th Ed. 2
2
ABn 3775 mm sec
θABn θ 180 deg β
θABn 90.124 deg
AAn a ω
2
AAn 11025 mm sec
θAAn θ 180 deg β
θAAn 209.000 deg
AAt a α
AAt 0.000 mm sec
ABn c ω
5.
SOLUTION MANUAL 7-33-2
2
2
ABAn b ω
2
ABAn 5740 mm sec
θAABn θ 180 deg β
θAABn 32.188 deg
2
Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ABn at an angle of ABn and AAn at an angle of AAn . From the tip of AAn, draw AAt at an angle of AAt. From the tip of AAt, draw ABAn at an angle of ABAn . Now that the vectors with known magnitudes are drawn, from the tips of ABn and ABAn, draw construction lines in the directions of ABt and ABAt, respectively. The intersection of these two lines are the tips of ABt, ABt, and ABAt. Y
X 42.029 y x 116.604° AB A
n A
t BA
A
n
0
AB
100 mm/s/s
t B
A
Acceleration Scale 18.740mm n
ABA
6.
From the graphical solution above, Acceleration scale factor ka 100
mm sec
7.
2
AA AAn
AA 11025 mm sec
AB 42.029 ka
AB 4203 mm sec
ABt 18.740 ka
ABt 1874 mm sec
2
2
at an angle of 209 deg at an angle of -116.6 deg
2
Calculate 4 using equation 7.6. α
ABt c
α 12.2 rad sec
2
CCW
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-34-1
PROBLEM 7-34 The linkage in Figure P7-8g has the dimensions and crank angle given below. Find 4, AA, and
Statement:
AB in the global coordinate system for the position shown if = 15 rad/sec CW and 2 = 10 rad/sec CCW, constant. Use an analytical method. Link lengths: Link 2 (O2 to A) Link 3 (A to B) a 49 mm b 100 mm
Given:
c 153 mm
Link 4 (O4 to B)
Solution: 1. 2.
Crank angle:
θ 148 deg Local xy system (see layout below)
Input crank angular velocity Coordinate rotation angle
ω 15 rad sec α 10 rad sec 119 deg Global XY system to local xy system
1
2
See Figure P7-8g and Mathcad file P0734. Y
Draw the linkage to scale and label it. Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 1.7755 2
K3
O6 B
d
3
c 29.000°
K2 0.5686 2
2
a b c d
C
K3 1.5592
2 a c
5
y
B 1.0598
119.000° x
C 4.6650
Use equation 4.10b to find values of 4 for the crossed circuit.
2
θ 2 atan2 2 A B
B 4 A C
θ 208.876 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
2 a b
D cos θ K1 K4 cos θ K5
K4 0.8700 D 3.0104
E 2 sin θ
E 1.0598
F K1 K4 1 cos θ K5 5.
D
O4
C K1 K2 1 cos θ K3
4.
X
6
O2
B 2 sin θ
A 0.5821
A
2
4
2
A cos θ K1 K2 cos θ K3
3.
d 87 mm
Link 1 (O2 to O4)
F 2.2367
Use equation 4.13 to find values of 3 for the crossed circuit.
θ 2 atan2 2 D E
2
E 4 D F
θ 266.892 deg
K5 0.3509
DESIGN OF MACHINERY - 5th Ed.
6.
7.
SOLUTION MANUAL 7-34-2
Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18. ω
a ω sin θ θ b sin θ θ
ω 7.570
rad
ω
a ω sin θ θ c sin θ θ
ω 4.959
rad
sec
sec
Using the Euler identity to expand equation 7.13a for AA. Determine the magnitude, and direction (in the local coordinate system).
a ω2 cosθ j sinθ
AA a α sin θ j cos θ AA ( 9090 6258i)
mm sec
AA 11036
The acceleration of pin A is
mm sec
8.
θAA arg AA
AA AA
2
at
2
θAA 84.46 deg
Use equations 7.12 to determine the angular accelerations of links 3 and 4 for the crossed circuit.
A c sin θ
B b sin θ
D c cos θ
E b cos θ
A 73.887 mm
B 99.853 mm
D 133.977 mm
E 5.422 mm
2
2
2
2
2
C a α sin θ a ω cos θ b ω cos θ c ω cos θ 3
C 6.106 10 mm sec
2
2
F a α cos θ a ω sin θ b ω sin θ c ω sin θ 3
F 2.353 10 mm sec α
9.
C D A F
2
α 49.646
A E B D
rad sec
α
2
C E B F A E B D
α 15.552
Use equation 7.13c to determine the acceleration of point B for the crossed circuit.
c ω2 cosθ j sinθ
AB c α sin θ j cos θ AB ( 4444 267i)
mm sec
2
The acceleration of pin B is
θAB arg AB
AB AB AB 4452
mm sec
2
at
θAB 115.6 deg
rad sec
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-35-1
PROBLEM 7-35 Statement:
At t = 0, the non-Grashof linkage in Figure P7-8g has the local axis at -119 deg and O2A at 29 deg in the global XY coordinate system and 2 = 0. Write a computer program or use an equation solver to find and plot 4, VA, AA, VB, and AB in the local coordinate system for the maximum range of motion that this linkage allows if 2 = 15 rad/sec CCW, constant. Link lengths:
Given:
Solution: 1. 2.
Link 2 (O2 to A)
a 49 mm
Link 3 (A to B)
Link 4 (O4 to B)
c 153 mm
Link 1 (O2 to O4)
1
b 100 mm d 87 mm
α 15 rad sec
2
Input crank angular velocity
ω 0 rad sec
Initial crank angle:
148 deg Local xy system (see layout below)
See Figure P7-8g and Mathcad file P0735. Y
Draw the linkage to scale and label it. Determine the range of motion for this non-Grashof triple rocker using equations 4.37. 2
arg1
2
a d b c
2
2
a d b c
O6
2
2 a d 2
arg2
2
2
2 a d
b c a d b c a d
θ2toggle acos arg1
B
arg1 0.840
3
arg2 6.338
2
4
A
θ2toggle 32.9 deg
O2
2
5
The other toggle angle is the negative of this. Thus,
y
θ 0.5 deg 1 deg 360 deg θ2toggle 0.5 deg
O4
D
x
3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 1.7755
d c
K2 0.5686
2
K3
2
2
a b c d
2
K3 1.5592
2 a c
A θ cos θ K1 K2 cos θ K3
B θ 2 sin θ
C θ K1 K2 1 cos θ K3 4.
Use equation 4.10b to find values of 4 for the crossed circuit.
θ θ 2 atan2 2 A θ B θ 5.
X
6
C
2 4 A θ Cθ
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-35-2 2
d
K4
b
2
2
c d a b
K5
2
K4 0.8700
2 a b
K5 0.3509
D θ cos θ K1 K4 cos θ K5
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 6.
Use equation 4.13 to find values of 3 for the crossed circuit.
θ θ 2 atan2 2 D θ E θ 7.
Determine the angular velocity of links 2, 3, and 4 using equations 6.18.
ω θ
2 θ α
a ω θ
a ω θ
ω θ
ω θ
8.
2 4 Dθ F θ
E θ
b
c
sin θ θ θ θ
sin θ θ θ θ
sin θ θ θ
sin θ θ θ
Using the Euler identity to expand equation 7.13a for AA. and determine its magnitude.
a ωθ2 cosθ j sinθ
AA θ a α sin θ j cos θ AA θ AA θ 9.
Use equations 7.12 to determine the angular acceleration of link 4.
B θ b sin θ θ
E θ b cos θ θ
A θ c sin θ θ
D θ c cos θ θ
2
2
c ωθ2 cosθθ
C θ a α sin θ a ω θ cos θ b ω θ cos θ θ
2 2 c ωθ2 sinθθ C θ E θ B θ F θ A θ E θ B θ D θ
F θ a α cos θ a ω θ sin θ b ω θ sin θ θ
α θ
10. Use equation 7.13c to determine the acceleration of point B and determine its magnitude.
j cosθθ c ωθ2 cosθθ j sinθθ
AB θ c α θ sin θ θ
AB θ AB θ
11. Plot the angular velocity and acceleration for link 4 and the velocity and acceleration components for pins A and B.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-35-3
ANGULAR VELOCITY OF LINK 4
Angular Accel., rad/sec^2
20
0
20
40 100
150
200
250
300
350
300
350
300
350
Crank Angle, deg
ANGULAR ACCELERATION OF LINK 4 Angular Accel., rad/sec^2
100
50
0 50 100 100
150
200
250
Crank Angle, deg
VELOCITY OF POINT A 500
Velocity, mm/sec
400 300 200 100 0 100
150
200
250
Crank Angle, deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-35-4
ACCELERATION OF POINT A
Acceleration, mm/sec^2
5 4 3 2 1 0 100
150
200
250
300
350
300
350
300
350
Crank Angle, deg
VELOCITY OF POINT B
Velocity, m/sec
2
0
2
4 100
150
200
250
Crank Angle, deg
ACCELERATION OF POINT B
Acceleration, m/sec^2
20
15
10
5
0 100
150
200
250
Crank Angle, deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-36-1
PROBLEM 7-36 Statement:
The 3-cylinder radial compressor in Figure P7-8c has the dimensions and crank angle given below. Find the piston accelerations A6, A7, and A8 for the position shown for = 15 rad/sec CW, constant. Use the acceleration difference graphical method.
Given:
Link lengths: Link 2
a 19 mm
Offset
c 0 mm
Link 2 position, velocity, and acceleration: θ 53 deg
1.
ω 15
rad sec
α 0
β 120 deg
Cylinder angular spacing: Solution:
b 70 mm
Link 3
rad sec
See Figure P7-8c and Mathcad file P0736.
In order to solve for the accelerations at piston 7, we will need 4, and 4. From the graphical position solution below (in the local coordinate system), α 90 deg
Coordinate system rotation angle: θ θ α
θ 37.000 deg ω
Using equation 6.22a,
θ 180 deg 9.401 deg
a ω cos θ b cos θ
ω 3.296 rad sec
θ 170.599 deg
1
Direction of ACBt Direction of A8 Direction of AEB
Y
6.088°
6
E 8
C 3
2
O2 Direction of A6
5
15.629°
X
B
1 4
Direction of ADBt 7 D
9.401°
Direction of A7
(APt + APn) = (AAt + AAn) + (APAt + APAn)
2.
The graphical solution for accelerations uses equation 7.4:
3.
For point D, this becomes: AD = (ABt + ABn) + (ADBt + ADBn) , where (angles in the global coordinate system) 2
2
ABn a ω
ABn 4275 mm sec
θABn θ α 180 deg
θABn 127.000 deg
ABt a α
ABt 0.000 mm sec
ADBn b ω
2
θADBn θ α
ADBn 760 mm sec
θADBn 80.599 deg
2
2
2
DESIGN OF MACHINERY - 5th Ed.
4.
SOLUTION MANUAL 7-36-2
Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ABn at an angle of ABn. From the tip of ABn, draw ADBn at an angle of ADBn. Now that the vectors with known magnitudes are drawn, from the origin and the tip of ADBn, draw construction lines in the directions of AD (vertical) and ADBt, respectively. The intersection of these two lines are the tips of ADBt, and AD. t ADB
n
ADB
AD n
0
AB Y
75.171
50 mm/s/s
Acceleration Scale
X,y
x
5.
From the graphical solution above, ka 50
Acceleration scale factor
mm sec
AD 75.171 ka 6.
AD 3759 mm sec
2
at an angle of 90 deg (global)
In order to solve for the accelerations at piston 6, we will need 3, and 3. From the graphical position solution above (in the local coordinate system), α 90 deg β
Coordinate system rotation angle: θ θ α
Using equation 6.22a,
7.
2
θ 157.000 deg ω
θ 180 deg 6.088 deg
a ω cos θ b cos θ
ω 3.769 rad sec
1
For point C, equation 7.4 becomes: AC = (ABt + ABn) + (ACBt + ACBn) , where (angles in the global coordinate system) ACBn b ω
n
2
ACBn 994.4 mm sec
ACB A
n B
Y
2
θACBn θ α
x
t ACB
0
Acceleration Scale
X
62.266
Repeat the procedure of step 4 for the equation in step 7 using ABn and ABt from step 3.
50 mm/s/s
150.000°
AC
θACBn 36.088 deg 8.
θ 173.912 deg
y
DESIGN OF MACHINERY - 5th Ed.
9.
SOLUTION MANUAL 7-36-3
From the graphical solution above, mm
ka 50
Acceleration scale factor
sec AC 62.266 ka
2
AC 3113 mm sec
2
at an angle of 150 deg (global)
10. In order to solve for the accelerations at piston 8, we will need 5, and 5. From the graphical position solution above (in the local coordinate system), α 360 deg 90 deg 2 β
Coordinate system rotation angle: θ θ α
θ 83.000 deg ω
Using equation 6.22a,
θ 180 deg 15.629 deg
a ω cos θ b cos θ
ω 0.515 rad sec
θ 195.629 deg
1
11. For point E, equation 7.4 becomes: AE = (ABt + ABn) + (AEBt + AEBn) , where (angles in the global coordinate system) 2
AEBn b ω
AEBn 18.583 mm sec
θAEBn θ α
θAEBn 225.629 deg
2
12. Repeat the procedure of step 4 for the equation in step 11 using ABn and ABt from step 3. n
AEB
0 A
50 mm/s/s
n B
Acceleration Scale
Y
x
y t EB
A
30.000° AE
X 12.934
Note that, at the acceleration scale chosen, AEBn is so small that it can hardly be seen in the layout above. 13. From the graphical solution above, Acceleration scale factor
ka 50
mm sec
AE 12.934 ka
2
AE 647 mm sec
2
at an angle of 30 deg (global)
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-37-1
PROBLEM 7-37 Statement:
The 3-cylinder radial compressor in Figure P7-8c has the dimensions and crank angle given below. Find the piston accelerations A6, A7, and A8 for the position shown for = 15 rad/sec CW, constant. Use an analytical method.
Given:
Link lengths: Link 2
a 19 mm
Offset
c 0 mm
b 70 mm
Link 3
Link 2 position, velocity, and acceleration: θ 37 deg
1.
α 0
sec
rad sec
See Figure P7-8c and Mathcad file P0737. Y
Draw the linkage to scale and label it. x''
2.
rad
β 120 deg
Cylinder angular spacing: Solution:
ω 15
Determine 4 and d' using equation 4.17. Coordinate rotation angle α 90 deg (x'y' system)
a sin θ θ4 asin b
c
π
d' a cos θ b cos θ4
x'''
y''' 6
E 8
C 3
2
5 X, y'
θ4 170.599 deg
B 1
d' 3.316 in
4 y''
3.
Determine the angular velocity of link 4 using equation 6.22a:
ω4 4.
a cos θ ω b cos θ4
ω4 3.296
7 D
rad
37.000°
sec
x'
Using the Euler identity to expand equation 7.15b for AB, determine its magnitude, and direction (global).
a ω2 cosθ j sinθ
AB a α sin θ j cos θ AB ( 3414.167 2572.759i )
mm sec
2
AB 4275
The acceleration of pin B is
mm sec
5.
at
2
θAB 127 deg
Determine the angular acceleration of link 4 using equation 7.16d.
α4 6.
θAB arg AB α 360 deg
AB AB
b cos θ4
a α cos θ a ω sin θ b ω4 sin θ4 2
2
α4 35.456
sec
Use equation 7.16e for the acceleration of pin D.
AD a α sin θ a ω cos θ b α4 sin θ4 b ω4 cos θ4 2
rad
2
2
2
DESIGN OF MACHINERY - 5th Ed.
mm
AD 3759
sec 7.
SOLUTION MANUAL 7-37-2
A negative sign means that AD is inward
2
Determine 3 and d'' using equation 4.17. θ θ 120 deg
θ 157.000 deg
a sin θ c π b
asin
173.912 deg
d'' a cos θ b cos 8.
d'' 2.052 in
Determine the angular velocity of link 3 using equation 6.22a:
9.
(x''y'' system)
a cos θ ω b cos
3.769
rad sec
Determine the angular acceleration of link 3 using equation 7.16d.
b cos 2
2
a α cos θ a ω sin θ b sin
22.483
rad sec
2
10. Use equation 7.16e for the acceleration of pin C.
2
2
AC a α sin θ a ω cos θ b sin b cos AC 3113
mm sec
A positive sign means that AC is outward
2
11. Determine 5 and d''' using equation 4.17. θ θ 120 deg
θ 277.000 deg
a sin θ c π b
θ5 asin
(x'''y''' system)
θ5 195.629 deg
d''' a cos θ b cos θ5
d''' 2.745 in
12. Determine the angular velocity of link 5 using equation 6.22a:
ω5
a cos θ ω b cos θ5
ω5 0.515
rad sec
13. Determine the angular acceleration of link 5 using equation 7.16d.
α5
b cos θ5
a α cos θ a ω sin θ b ω5 sin θ5 2
2
α5 62.869
sec
14. Use equation 7.16e for the acceleration of pin E.
AE a α sin θ a ω cos θ b α5 sin θ5 b ω5 cos θ5 AE 646.72
mm sec
2
2
rad
2
A positive sign means that AE is outward
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-38-1
PROBLEM 7-38 Statement:
The 3-cylinder radial compressor in Figure P7-8c has the dimensions and crank angle given below. Find and plot the piston accelerations A6, A7, and A8 for one revolution of the crank if = 15 rad/sec CW, constant.
Given:
Link lengths: Link 2
a 19 mm
Offset
c 0 mm
Link 2 velocity, and acceleration: ω 15
1. 2.
rad sec
sec
2
Draw the linkage to scale and label it. Note that there are three local coordinate systems. Determine the range of motion for this slider-crank linkage. This will be the same in each coordinate frame.
Y
x'' 6
E 8
C 3
a b
2
5 X, y'
(x'y' system)
Determine the angular velocity of link 4 using equation 6.22a:
ω4 θ
x'''
y'''
Determine 4 using equation 4.17.
a sin θ c θ4 θ asin π b 4.
rad
See Figure P7-8c and Mathcad file P0738.
θ 0 deg 2 deg 360 deg 3.
α 0
α 120 deg
Cylinder angular spacing: Solution:
b 70 mm
Link 3
B 1 4 y''
ω cos θ4 θ
7 D
cos θ
37.000° 5.
Determine the angular acceleration of link 4 using equation 7.16d.
α4 θ 6.
x'
2 b cos θ4 θ 2
a α cos θ a ω sin θ b ω4 θ sin θ4 θ
Use equation 7.16e for the acceleration of pin D.
2 2 b α4 θ sin θ4 θ b ω4 θ cos θ4 θ
AD θ a α sin θ a ω cos θ
7.
Determine 3 using equation 4.17.
a sin θ α c π b
θ3 θ asin 8.
Determine the angular velocity of link 3 using equation 6.22a:
9.
a cos θ α ω b cos θ3 θ Determine the angular acceleration of link 3 using equation 7.16d.
ω3 θ
(x''y'' system)
DESIGN OF MACHINERY - 5th Ed.
α3 θ
SOLUTION MANUAL 7-38-2
b cos θ3 θ
2
2
a α cos θ α a ω sin θ α b ω3 θ sin θ3 θ
10. Use equation 7.16e for the acceleration of pin C.
2 2 b α3 θ sin θ3 θ b ω3 θ cos θ3 θ
AC θ a α sin θ α a ω cos θ α
11. Determine 5 and d''' using equation 4.17.
a sin θ 2 α b
θ5 θ asin
c
π
(x'''y''' system)
12. Determine the angular velocity of link 5 using equation 6.22a:
ω5 θ
a cos θ 2 α ω b cos θ5 θ
13. Determine the angular acceleration of link 5 using equation 7.16d.
α5 θ
b cos θ5 θ 2
2
a α cos θ 2 α a ω sin θ 2 α b ω5 θ sin θ5 θ
14. Use equation 7.16e for the acceleration of pin E.
2 2 b α5 θ sin θ5 θ b ω5 θ cos θ5 θ
AE θ a α sin θ 2 α a ω cos θ 2 α
15. Plot the piston accelerations. PISTON ACCELERATIONS 4
Acceleration, m/sec^2
2
0 2 4 6
0
30
60
90
120
150
180
210
Piston D Crank Angle, deg Piston C Piston D Piston E
240
270
300
330
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-39-1
PROBLEM 7-39 Statement:
Figure P7-9 shows a linkage in one position. Find the instantaneous accelerations of points A, B, and P if link O2A is rotating CW at 40 rad/sec.
Given:
Link lengths: Link 2 (A to B)
a 5.00 in
Link 3 (B to C)
b 4.40 in
Link 4 (C to D)
c 5.00 in
Link 1 (A to D)
d 9.50 in
Rpa 8.90 in
δ 56 deg
Coupler point:
Crank angle and motion: θ 50 deg Solution:
ω 40 rad sec
1
α 0 rad sec
2
See Figure P7-9 and Mathcad file P0739.
1.
Draw the linkage to scale and label it.
2.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a.
P
K1
d
K1 1.9000 2
K3
d
K2
a
c
K2 1.9000 2
2
a b c d
y
Y
2
K3 2.4178
2 a c
B
3
A cos θ K1 K2 cos θ K3
B 2 sin θ
2
x 50.000°
3.
B 1.5321
14.000°
C 2.4537
2
B 4 A C
θ 246.992 deg
θ θ 360 deg
θ 113.008 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2 a b
D cos θ K1 K4 cos θ K5
Use equation 4.13 to find values of 3 .
θ 2 atan2 2 D E θ θ 360 deg
K4 2.1591
E 1.5321
F K1 K4 1 cos θ K5
2
D 2.3605
E 2 sin θ
5.
X
O2
Use equation 4.10b to find values of 4 for the open circuit. θ 2 atan2 2 A B
4.
O4
1
C K1 K2 1 cos θ K3 A 0.0607
4
A
2
E 4 D F
F 0.1539
θ 349.895 deg θ 10.105 deg
K5 2.4911
DESIGN OF MACHINERY - 5th Ed.
6.
7.
SOLUTION MANUAL 7-39-2
Determine the angular velocity of links 3 and 4 using equations 6.18. ω
a ω sin θ θ b sin θ θ
ω 41.552
ω
a ω sin θ θ c sin θ θ
ω 26.320
rad sec rad sec
Using the Euler identity to expand equation 7.13a for AA. Determine the magnitude, and direction.
a ω2 cosθ j sinθ
AA a α sin θ j cos θ in
AA ( 5142.3 6128.4i )
sec The acceleration of pin A is
in
AA 8000
sec 8.
θAA arg AA
AA AA
2
at 2
θAA 130.0 deg
Use equations 7.12 to determine the angular accelerations of links 3 and 4.
A c sin θ
B b sin θ
D c cos θ
E b cos θ
A 4.602 in
B 0.772 in
D 1.954 in
E 4.332 in
2
2
2
4
2
2
2
α 356.538
rad
C a α sin θ a ω cos θ b ω cos θ c ω cos θ C 1.398 10 in sec
2
3
2
F a α cos θ a ω sin θ b ω sin θ c ω sin θ F 4.273 10 in sec α 9.
C D A F A E B D
sec
α
2
C E B F A E B D
α 2.977 10
Use equation 7.13c to determine the acceleration of point B.
c ω2 cosθ j sinθ
AB c α sin θ j cos θ AB ( 12346.3 9005.8i )
in sec
The acceleration of pin B is
2
θAB arg AB
AB AB in
AB 15282
sec
at 2
θAB 143.9 deg
10. Use equations 7.32 to find the acceleration of the point C.
Rpa ω cos θ δ j sin θ δ
ACA Rpa α sin θ δ j cos θ δ 2
AC AA ACA AC AC
AC 23073
in sec
2
arg AC 111.525 deg
3 rad
sec
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-40-1
PROBLEM 7-40 Statement:
Figure P7-10 shows a linkage and its coupler curve. Write a computer program or use an equation solver to calculate and plot the magnitude and direction of the acceleration of the coupler point P at 2-deg increments of crank angle. Check your results with program FOURBAR.
Units:
rpm 2 π rad min
Given:
Link lengths:
Solution: 1. 2.
1
Link 2 (O2 to A)
a 10 mm
Link 3 (A to B)
b 20.6 mm
Link 4 (B to O4)
c 23.3 mm
Link 1 (O2 to O4)
d 22.2 mm
Coupler point:
Rpa 30.6 mm
δ 31 deg
Crank motion:
ω 100 rpm
α 0 rad sec
See Figure P7-10 and Mathcad file P0740.
Draw the linkage to scale and label it. Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 2.2200 2
K3
B
y
d
3
c
K2 0.9528 2
2
a b c d
b p
K3 1.5265
2 a c
A
2
a
B θ 2 sin θ
4
2
1
O2
2 4 A θ Cθ
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
2 a b
K4 1.0777
D θ cos θ K1 K4 cos θ K5
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 Use equation 4.13 to find values of 3 .
θ θ 2 atan2 2 D θ E θ 6.
x O4
Use equation 4.10b to find values of 4 for the open circuit. θ θ 2 atan2 2 A θ B θ
5.
4
c d
C θ K1 K2 1 cos θ K3
4.
P
2
A θ cos θ K1 K2 cos θ K3
3.
2
2 4 Dθ F θ
E θ
Determine the angular velocity of links 3 and 4 using equations 6.18.
K5 1.1512
DESIGN OF MACHINERY - 5th Ed.
a ω
a ω
ω θ
b
ω θ 7.
c
SOLUTION MANUAL 7-40-2
sin θ θ θ θ
sin θ θ θ θ
sin θ θ θ
sin θ θ θ
Using the Euler identity to expand equation 7.13a for AA.
a ω2 cosθ j sinθ
AA θ a α sin θ j cos θ 8.
Use equations 7.12 to determine the angular accelerations of link 3.
B' θ b sin θ θ
E' θ b cos θ θ
A' θ c sin θ θ
D' θ c cos θ θ
2
2 c ωθ2 cosθθ
2
2 c ωθ2 sinθθ
C' θ a α sin θ a ω cos θ b ω θ cos θ θ F' θ a α cos θ a ω sin θ b ω θ sin θ θ
α θ 9.
A' θ E' θ B' θ D' θ C' θ D' θ A' θ F' θ
Use equations 7.32 to find the acceleration of the point P.
2 Rpa ω θ cos θ θ δ j sin θ θ δ AP θ AA θ APA θ APA θ Rpa α θ sin θ θ δ j cos θ θ δ
θAP θ arg AP θ
AP θ AP θ
10. Plot the magnitude and direction of the coupler point P. (See next page.) θ 0 deg 1 deg 360 deg MAGNITUDE OF COUPLER POINT ACCELERATION
Acceleration, in/sec^2
200
150
100
50
0
0
45
90
135
180
Crank Angle, deg
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-40-3
DIRECTION OF COUPLER POINT ACCELERATION 200
Direction Angle, deg
100
0
100
200
0
45
90
135
180
Crank Angle, deg
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-41-1
PROBLEM 7-41 Statement:
Figure P7-11 shows a linkage that operates at 500 crank rpm. Find and plot the magnitude and direction of the acceleration of point B at 2-deg increments of crank angle. Check your result with program FOURBAR.
Given:
Link lengths: Link 2 (O2 to A)
a 2.000 in
Link 3 (A to B)
b 8.375 in
Link 4 (B to O4)
c 7.187 in
Link 1 (O2 to O4)
d 9.625 in
ω 500 rpm
Input crank angular velocity
α 0 rad sec Solution:
2
1
See Figure P7-11 and Mathcad file P0741.
1.
Draw the linkage to scale and label it.
2.
Determine the range of motion for this Grashof crank rocker.
A
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 4.8125 2
2
a b c d
O2 d c
4
1
2
K3 2.7186
2 a c
O4
A θ cos θ K1 K2 cos θ K3
C θ K1 K2 1 cos θ K3
B θ 2 sin θ 4.
Use equation 4.10b to find values of 4 for the open circuit.
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 5.
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
2 a b
K4 1.1493
K5 3.4367
D θ cos θ K1 K4 cos θ K5
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 6.
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ 7.
B
2
K2 1.3392
2
K3
3 2
θ 0 deg 2 deg 360 deg 3.
ω 52.360 rad sec
2 4 Dθ F θ
E θ
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18.
DESIGN OF MACHINERY - 5th Ed.
ω θ
ω θ 8.
SOLUTION MANUAL 7-41-2
b sin θ θ θ θ a ω sin θ θ θ c sin θ θ θ θ a ω
sin θ θ θ
Use equations 7.12 to determine the angular acceleration of link 4.
D θ c cos θ θ
E θ b cos θ θ
A θ c sin θ θ
B θ b sin θ θ
2
2
c ωθ2 cosθθ
2
2 c ωθ2 sinθθ
C θ a α sin θ a ω cos θ b ω θ cos θ θ F θ a α cos θ a ω sin θ b ω θ sin θ θ
α θ 9.
A θ E θ B θ D θ C θ E θ B θ F θ
Determine the acceleration of point B using equations 7.13c.
j cosθθ c ωθ2 cosθθ j sinθθ
AB θ c α θ sin θ θ
θAB θ arg AB θ
AB θ AB θ
10. Plot the magnitude and angle of the acceleration at point B. MAGNITUDE OF ACCELERATION AT B Acceleration, in/sec^2
8000 6000 4000 2000 0
0
60
120
180
240
300
360
300
360
DIRECTION OF ACCELERATION AT B
Angle, deg
100
0
100
200
0
60
120
180
240
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-42-1
PROBLEM 7-42 Statement:
Figure P7-12 shows a linkage and its coupler curve. Write a computer program or use an equation solver to calculate and plot the magnitude and direction of the acceleration of the coupler point P at 2-deg increments of crank angle over the maximum range of motion possible. Check your results with program FOURBAR. Link lengths:
Given:
Solution:
Link 2 (O2 to A)
a 0.785 in
Link 3 (A to B)
b 0.356 in
Link 4 (B to O4)
c 0.950 in
Link 1 (O2 to O4)
d 0.544 in
Coupler point:
Rpa 1.09 in
δ 0 deg
Crank speed:
ω 20 rpm
α 0 rad sec
2
See Figure P7-12 and Mathcad file P0742.
1.
Draw the linkage to scale and label it.
2.
Using the geometry defined in Figure 3-1a in the text, determine the input crank angles (relative to the line O2O4) at which links 2 and 3, and 3 and 4 are in toggle.
y
a2 d 2 ( b c) 2 θ acos 2 a d
3 B 3 A
θ 158.286 deg 4
θ θ 0.5 deg θ 1 deg θ 0.5 deg 3.
d
K2
a
K1 0.6930 2
K3
2
d
x O2
c
2
2
a b c d
2
K3 1.1317
2 a c
A θ cos θ K1 K2 cos θ K3
C θ K1 K2 1 cos θ K3
Use equation 4.10b to find values of 4 for the open circuit.
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 5.
O4
K2 0.5726
B θ 2 sin θ 4.
2
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
P
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
2 a b
K4 1.5281
E θ 2 sin θ F θ K1 K4 1 cos θ K5
D θ cos θ K1 K4 cos θ K5
6.
Use equation 4.13 to find values of 3 .
θ θ 2 atan2 2 D θ E θ
2 4 Dθ F θ
E θ
K5 0.2440
DESIGN OF MACHINERY - 5th Ed.
7.
Determine the angular velocity of links 3 and 4 using equations 6.18.
a ω
a ω
ω θ
ω θ 8.
SOLUTION MANUAL 7-42-2
b
c
sin θ θ θ θ
sin θ θ θ θ
sin θ θ θ
sin θ θ θ
Using the Euler identity to expand equation 7.13a for AA.
a ω2 cosθ j sinθ
AA θ a α sin θ j cos θ 9.
Use equations 7.12 to determine the angular accelerations of link 3.
D' θ c cos θ θ
E' θ b cos θ θ
A' θ c sin θ θ
2
B' θ b sin θ θ
2 c ωθ2 cosθθ
C' θ a α sin θ a ω cos θ b ω θ cos θ θ
2 c ωθ2 sinθθ C' θ D' θ A' θ F' θ A' θ E' θ B' θ D' θ 2
F' θ a α cos θ a ω sin θ b ω θ sin θ θ
α θ
10. Use equations 7.32 to find the acceleration of the point P.
2 Rpa ω θ cos θ θ δ j sin θ θ δ AP θ AA θ APA θ APA θ Rpa α θ sin θ θ δ j cos θ θ δ
θAP θ arg AP θ
AP θ AP θ
11. Plot the magnitude and direction of the coupler point P. MAGNITUDE OF COUPLER POINT ACCELERATION
Acceleration, in/sec^2
40
30
20
10
0 200
150
100
50
0
50
100
150
200
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-42-3
DIRECTION OF COUPLER POINT ACCELERATION 200
Direction Angle, deg
100
0
100
200 200
150
100
50
0
Crank Angle, deg
50
100
150
200
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-43-1
PROBLEM 7-43 Statement:
Figure P7-13 shows a linkage and its coupler curve. Write a computer program or use an equation solver to calculate and plot the magnitude and direction of the acceleration of the coupler point P at 2-deg increments of crank angle over the maximum range of motion possible. Check your results with program FOURBAR.
Given:
Link lengths:
Solution: 1.
Link 2 (O2 to A)
a 0.86 in
Link 3 (A to B)
b 1.85 in
Link 4 (B to O4)
c 0.86 in
Link 1 (O2 to O4)
d 2.22 in
Coupler point:
Rpa 1.33 in
δ 0 deg
Crank motion:
ω 80 rpm
α 0 rad sec
2
See Figure P7-13 and Mathcad file P0743.
Draw the linkage to scale and label it.
y
B
3 4 P
O4
O2 3 2
A
2.
Determine the range of motion for this non-Grashof triple rocker using equations 4.37. 2
arg1
2
2
2 a d 2
arg2
2
a d b c
2
2
a d b c 2 a d
2
b c a d b c a d
θ2toggle acos arg2
arg1 1.228
arg2 0.439
θ2toggle 116.0 deg
The other toggle angle is the negative of this. Thus, θ θ2toggle 0.5 deg θ2toggle 1 deg θ2toggle 0.5 deg 3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a.
x
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-43-2
d
K1
K2
a
K1 2.5814 2
2
2
2
K3 2.0181
2 a c
c
K2 2.5814
a b c d
K3
d
A θ cos θ K1 K2 cos θ K3
B θ 2 sin θ
C θ K1 K2 1 cos θ K3 4.
Use equation 4.10b to find values of 4 for the open circuit.
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 5.
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. 2
d
K4
K5
b
2
2
c d a b
2
K4 1.2000
2 a b
K5 2.6244
D θ cos θ K1 K4 cos θ K5
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 6.
Use equation 4.13 to find values of 3 .
θ θ 2 atan2 2 D θ E θ 7.
Determine the angular velocity of links 3 and 4 using equations 6.18.
a ω
a ω
ω θ
ω θ 8.
2 4 Dθ F θ
E θ
b
c
sin θ θ θ θ
sin θ θ θ θ
sin θ θ θ
sin θ θ θ
Using the Euler identity to expand equation 7.13a for AA.
a ω2 cosθ j sinθ
AA θ a α sin θ j cos θ 9.
Use equations 7.12 to determine the angular accelerations of link 3.
B' θ b sin θ θ
E' θ b cos θ θ
A' θ c sin θ θ
D' θ c cos θ θ
2
2 c ωθ2 cosθθ
2
2 c ωθ2 sinθθ
C' θ a α sin θ a ω cos θ b ω θ cos θ θ F' θ a α cos θ a ω sin θ b ω θ sin θ θ
DESIGN OF MACHINERY - 5th Ed.
α θ
SOLUTION MANUAL 7-43-3
A' θ E' θ B' θ D' θ C' θ D' θ A' θ F' θ
10. Use equations 7.32 to find the acceleration of the point P.
2 Rpa ω θ cos θ θ δ j sin θ θ δ
APA θ Rpa α θ sin θ θ δ j cos θ θ δ
AP θ AA θ APA θ
θAP θ arg AP θ
AP θ AP θ
11. Plot the magnitude and direction of the coupler point P.
Acceleration, in/sec^2
MAGNITUDE OF COUPLER POINT ACCELERATION
150
100
50
0 200
100
0
100
200
DIRECTION OF COUPLER POINT ACCELERATION 200
Direction Angle, deg
100
0
100
200 200
100
0
100
200
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-44-1
PROBLEM 7-44 Statement:
Figure P7-14 shows a linkage and its coupler curve. Write a computer program or use an equation solver to calculate and plot the magnitude and direction of the acceleration of the coupler point P at 2-deg increments of crank angle over the maximum range of motion possible. Check your results with program FOURBAR.
Given:
Link lengths:
Solution: 1.
Link 2 (O2 to A)
a 0.72 in
Link 3 (A to B)
b 0.68 in
Link 4 (B to O4)
c 0.85 in
Link 1 (O2 to O4)
d 1.82 in
Coupler point:
Rpa 0.97 in
δ 54 deg
Crank motion:
ω 80 rpm
α 0 rad sec
2
See Figure P7-14 and Mathcad file P0744.
Draw the linkage to scale and label it. P y
B
3 4
A 2
x O2
2.
O4
Determine the range of motion for this non-Grashof triple rocker using equations 4.37. 2
arg1
2
2
2 a d 2
arg2
2
a d b c
2
2
a d b c 2 a d
2
b c
arg1 1.451
a d b c
arg2 0.568
a d
θ2toggle acos arg2
θ2toggle 55.4 deg
The other toggle angle is the negative of this. Thus, θ θ2toggle 0.5 deg θ2toggle 1 deg θ2toggle 0.5 deg 3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d a
K1 2.5278
K2
d c
K2 2.1412
DESIGN OF MACHINERY - 5th Ed. 2
SOLUTION MANUAL 7-44-2
2
2
a b c d
K3
2
K3 3.3422
2 a c
A θ cos θ K1 K2 cos θ K3
B θ 2 sin θ
C θ K1 K2 1 cos θ K3 4.
Use equation 4.10b to find values of 4 for the open circuit.
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 5.
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. 2
d
K4
K5
b
2
2
c d a b
2
K4 2.6765
2 a b
D θ cos θ K1 K4 cos θ K5
K5 3.6465
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 6.
Use equation 4.13 to find values of 3 .
θ θ 2 atan2 2 D θ E θ 7.
Determine the angular velocity of links 3 and 4 using equations 6.18.
a ω
a ω
ω θ
ω θ 8.
2 4 Dθ F θ
E θ
b
c
sin θ θ θ θ
sin θ θ θ θ
sin θ θ θ
sin θ θ θ
Using the Euler identity to expand equation 7.13a for AA.
a ω2 cosθ j sinθ
AA θ a α sin θ j cos θ 9.
Use equations 7.12 to determine the angular accelerations of link 3.
B' θ b sin θ θ
E' θ b cos θ θ
A' θ c sin θ θ
D' θ c cos θ θ
2
2 c ωθ2 cosθθ
2
2 c ωθ2 sinθθ
C' θ a α sin θ a ω cos θ b ω θ cos θ θ F' θ a α cos θ a ω sin θ b ω θ sin θ θ
α θ
A' θ E' θ B' θ D' θ C' θ D' θ A' θ F' θ
10. Use equations 7.32 to find the acceleration of the point P.
2 Rpa ω θ cos θ θ δ j sin θ θ δ AP θ AA θ APA θ APA θ Rpa α θ sin θ θ δ j cos θ θ δ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-44-3
θAP θ arg AP θ
AP θ AP θ
11. Plot the magnitude and direction of the coupler point P.
Acceleration, in/sec^2
MAGNITUDE OF COUPLER POINT ACCELERATION
200
100
0 60
40
20
0
20
40
60
Crank Angle, deg
DIRECTION OF COUPLER POINT ACCELERATION
Direction Angle, deg
50
0
50
100 60
40
20
0 Crank Angle, deg
20
40
60
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-45-1
PROBLEM 7-45 Statement:
Given:
Figure P7-15 shows a power hacksaw that is an offset crank-slider mechanism that has the dimensions given below. Draw an equivalent linkage diagram and then calculate and plot the acceleration of the saw blade with respect to the piece being cut over one revolution of the crank, which rotates at 50 rpm. Link lengths: Link 2 (O2 to A)
a 75 mm
Link 3 (A to B)
b 170 mm ω 50 rpm
Input crank angular velocity Solution: 1.
c 45 mm
Offset
α 0 rad sec
2
See Figure P7-15 and Mathcad file P0745.
Draw the equivalent linkage to a convenient scale and label it.
y
B
3
b
A
4 a 2
c
2
O2 2.
Determine the range of motion for this slider-crank linkage. θ 0 deg 2 deg 360 deg
3.
Determine 3 using equations 4.16 for the crossed circuit.
a sin θ c b
θ θ asin 4.
Determine the angular velocity of link 3 using equation 6.22a:
ω θ 5.
b
ω cos θ θ cos θ
Determine the angular acceleration of link 3 using equation 7.16d.
α θ 6.
a
2 b cos θ θ 2
a α cos θ a ω sin θ b ω θ sin θ θ
Use equation 7.16e for the acceleration of pin B.
2 2 b α θ sin θ θ b ω θ cos θ θ
AB θ a α sin θ a ω cos θ
x
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-45-2
ACCELERATION OF POINT B 4
Acceleration, m/sec^2
2
0
2
4
0
60
120
180 Crank Angle, deg
240
300
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-46-1
PROBLEM 7-46 Statement:
Figure P7-16 shows a walking beam indexing and pick-and-place mechanism that can be analyzed as two fourbar linkages driven by a common crank. Calculate and plot the relative acceleration between points E and P for one revolution of gear 2.
Given:
Link lengths (walking-beam linkage): Link 2 (O2 to A)
a' 40 mm
Link 3 (A to D)
b' 108 mm
Link 4 (O4 to D)
c' 40 mm
Link 1 (O2 to O4)
d' 108 mm
Link lengths (pick and place linkage): Link 2 (O5 to B)
a 13 mm
Link 7 (B to C)
b 193 mm
Link 6 (C to O6)
c 92 mm
Link 1 (O5 to O6)
d 128 mm
Rocker point E:
u 164 mm
Crank speed:
ω 10 rpm
1.
2
ϕ 143 deg
Gears 4 & 5 phase angle Solution:
α 0 rad sec
See Figure P7-16 and Mathcad file P0746.
Draw the walking-beam linkage to scale and label it. 30 mm Y
P
58 mm
80°
A
D b'
c'
a' d'
x'
O2
O4
X
y'
2.
Determine the range of motion for this mechanism. θ 0 deg 2 deg 360 deg
3.
(local x'y' coordinate system)
This part of the mechanism is a special-case Grashof in the parallelogram configuration. As such, the coupler does not rotate, but has curvilinear motion with every point on it having the same velocity and acceleration. Therefore, it is only necessary to calculate the X-component of the acceleration at point A in order to determine the acceleration of the cylinder center, P.
a ω2 cosθ j sinθ
AA θ a α sin θ j cos θ
AAx θ Re AA θ In the global coordinate frame,
AP θ AAx θ 4.
Draw the pick-and-place linkage to scale and label it.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-46-2
E
C
7 6
b
c
O6 5.
B
a O5
Establish the relationship between 5 and 2. Note that gear 5 is driven by gear 4 and that their ratio is -1 (i.e., they rotate in opposite directions with the same speed). Also, because the walking beam fourbar is a special Grashof, 4 = 2. Thus,
θ θ θ ϕ 6.
5
d 1
ω ω
and
α α
Determine the values of the constants needed for finding 6 from equations 4.8a and 4.10a. K1
d
K2
a
K1 9.8462 2
2
a b c d
2
K3 5.1137
2 a c
c
K2 1.3913
2
K3
d
K1 K2 cosθθ K3
A θ cos θ θ
B θ 2 sin θ θ
K3
C θ K1 K2 1 cos θ θ 7.
Use equation 4.10b to find values of 6 for the crossed circuit.
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 8.
B θ
Determine the values of the constants needed for finding 7 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
2 a b
K1 K4 cosθθ K5
D θ cos θ θ
2
c d a b
E θ 2 sin θ θ
K4 0.6632
K5 9.0351
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-46-3
K5
F θ K1 K4 1 cos θ θ 9.
Use equation 4.13 to find values of 7 for the crossed circuit.
θ θ 2 atan2 2 D θ E θ
2 4 Dθ F θ
E θ
10. Determine the angular velocity of links 7 and 6 using equations 6.18.
ω θ
a ω sinθ θ θ θ b sin θ θ θ θ
a ω sin θ θ θ θ c sin θ θ θ θ
ω θ
11. Use equations 7.12 to determine the angular acceleration of link 6.
B θ b sin θ θ
E θ b cos θ θ
A θ c sin θ θ
D θ c cos θ θ
a ω2 cosθθ 2 2 b ω θ θ cos θ θ c ω θ cos θ θ
a ω2 sinθθ 2 2 b ω θ θ sin θ θ c ω θ sin θ θ
C θ a α sin θ θ
F θ a α cos θ θ
α θ
A θ E θ B θ D θ C θ E θ B θ F θ
12. Determine the acceleration of the rocker point E using equations 7.13c (substituting the distance to point E from O6 for the distance c).
j cosθθ u ωθ2 cosθθ j sinθθ
AE θ u α θ sin θ θ
13. Transform this into the global XY system.
AEX θ AEx θ
AEY θ AEx θ
AEx θ Re AE θ
AEy θ Im AE θ
AEXY θ AEX θ j AEY θ
14. Calculate and plot the acceleration of E relative to P. (See next page).
AEP θ AEXY θ AP θ
AEPX θ Re AEP θ
AEPY θ Im AEP θ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-46-4
RELATIVE ACCELERATION of E with respect to P 80 70
Acceleration, mm/sec^2
60 50 40 30 20 10 0
0
30
60
90
120
150
180
210
Crank Angle, deg
240
270
300
330
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-47-1
PROBLEM 7-47 Statement:
Figure P6-17 shows a paper roll off-loading mechanism driven by an air cylinder. In the position shown, it has the dimensions given below. The V-links are rigidly attached to O4A. The air cylinder is retracted at a constant acceleration of 0.1 m/sec2. Draw a kinematic diagram of the mechanism, write the necessary equations, and calculate and plot the angular acceleration of the paper roll and the linear acceleration of its center as it rotates through 90 deg CCW from the position shown.
Given:
Link lengths and angles:
Paper roll location from O4:
Link 4 (O4 to A)
c 300 mm
u 707.1 mm
Link 1 (O2 to O4)
d 930 mm
δ 181 deg
Link 4 initial angle
θ 62.8 deg
addot 100 mm sec
Input cylinder acceleration Solution: 1.
with respect to local x axis 2
See Figure P7-17 and Mathcad file P0747.
Draw the mechanism to scale and define a vector loop using the fourbar derivation in Section 6.7 as a model. V-Link
Roll Center
707.107
45.000° x
O4 4
c 46.000° A
1
O4
d
4
3
O2
2 b
R1
R4 2 A
R3
R2
O2
a f y
2.
Write the vector loop equation, differentiate it, expand the result and separate into real and imaginary parts to solve for f, 2, and 4. R1 R4 R2 R3 d e
j θ
c e
j θ
(a)
a e
j θ
b e
j θ
(b)
where a is the distance from the origin to the cylinder piston, a variable; b is the distance from the cylinder piston to A, a constant; and c is the distance from 4 to point A, a constant. Angle 1 is zero, 3 = 2, and 4 is the variable angle that the rocker arm makes with the x axis. Solving the position equations: Let
f a b
then, making this substitution and substituting the Euler equivalents,
d c cos θ j sin θ f cos θ j sin θ
(c)
Separating into real and imaginary components and solving for 2 and f,
θ θ atan2 d c cos θ c sin θ
(d)
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-47-2
sin θ θ c sin θ
f θ
(e)
For constant deceleration of the hydraulic cylinder rod, the velocity of the rod is
adot θ addot
2 f θ f θ addot
Differentiate equation b. j c ω e
j θ
d f ej θ j f ω ej θ dt
(f)
Substituting the Euler equivalents,
c ω sin θ j cos θ
d f dt
cos θ j sin θ f ω sin θ j cos θ
(g)
Separating into real and imaginary components and solving for 4 and 2. Note that df/dt = adot.
ω θ
ω θ 3.
c sin θ θ θ adot θ
(h)
f θ sin θ θ
c ω θ adot θ cos θ θ
(i)
Differentiate equation f, expand the result and separate into real and imaginary parts to solve for 4. j c α e
j θ
2
2 j θ
j c ω e
addot e
j θ
j adot ω e
j adot ω e
j θ
j θ
j f α e
j θ
(f) 2
Separating into real and imaginary components and solving for 4.
addot 2 adot θ ω θ f θ ω θ 2 sin 2 θ θ
α θ
c sin θ θ θ
Plot 4 over a range of 4 of θ θ θ 1 deg θ 90 deg ANGULAR ACCELERATION, LINK 4 Angular Acceleration, rad/sec^2
4.
2
c ω θ sin θ θ θ
1 0 1 2 3 4 60
80
100
120
Link 4 Angle, deg
140
2 j θ
j f ω e
160
DESIGN OF MACHINERY - 5th Ed.
5.
SOLUTION MANUAL 7-47-3
Determine the acceleration of the center of the paper roll using equation 7.31. The direction is in the local xy coordinate system.
AU θ u α θ sin θ δ j cos θ δ
θAU θ arg AU θ
AU θ AU θ
Plot the magnitude and direction of the acceleration of the paper roll center.
Acceleration, m/sec^2
MAGNITUDE OF PAPER CENTER ACCELERATION
2
1
0 60
80
100
120
140
160
Rocker Arm Position, deg
DIRECTION OF PAPER CENTER ACCELERATION 200
100 Vector Angle, deg
6.
u ωθ2 cosθ δ j sinθ δ
0
100
200 60
80
100
120
Rocker Arm Position, deg
140
160
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-48-1
PROBLEM 7-48 Statement:
Figure P7-18 shows a mechanism and its dimensions. Find the accelerations of points A, B, and C for the position shown.
Given:
Lengths and angles:
Solution: 1.
Link 2 (O2 to A)
a 0.80 in
Link 1 (O2 to O4)
d 1.85 in
Link 4 (O4 to B)
a' 2.97 in
Link 5 (B to C)
b' 2.61 in
Crank angle
θ 37.5 deg
Coordinate rotation angle
β 81.5 deg
Input crank motion
ω 40
Local xy system c' 3.25 in
Slider-crank offset
rad
rad
α 1500
min
min
2
See Figure P7-18 and Mathcad file P0748.
Draw the mechanism to scale and define a vector loop for the input portion (links 1, 2, 3, and 4) using the fourbar crank-slider derivation in Section 7.3 as a model. 6 C
5 B
Y 4
O2
y R2
2 A
3
y
O2
A
R1 R3
37.5° X
O4 x
2.
O4 x
81.5°
Write the vector loop equation, expand the result and separate into real and imaginary parts to solve for 3 and b (distance from O4 to A). R2 R1 R3 a e
j θ
d b e
j θ
Substituting the Euler equivalents,
d bcosθ j sinθ
a cos θ j sin θ
Separating into real and imaginary components and solving for 3 and b.
θ atan2 a cos θ d a sin θ
θ 158.2 deg
DESIGN OF MACHINERY - 5th Ed.
b 3.
SOLUTION MANUAL 7-48-2
sin θ
a sin θ
b 1.31 in
Differentiate the position equation, expand it and solve for 3 and bdot. a j ω e ω
j θ
a ω b
bdot
b j ω e
j θ
bdot e
j θ
cos θ θ
ω 0.208
a ω cos θ b ω cos θ
rad sec
bdot 0.459
sin θ 4.
sec
Differentiate the velocity equation, expand it and solve for 3. a j e
j θ
2 j θ
2
a j ω e
bdot j ω e
j θ
b j e
2 j θ
2
b j ω e α
1 b
j θ
bddot e
j θ
bdot j ω e
j θ
a α cos θ θ a ω sin θ θ 2 bdot ω
2
rad
α 0.249
sec 5.
in
2
Determine the acceleration of point A on link 2 (in the local xy coordinate system) using equations 7.13a.
a ω2 cosθ j sinθ
AA a α sin θ j cos θ AA AA
in
AA 0.487
sec 6.
2
θAA arg AA
θAA 174.348 deg
Determine the acceleration of points A and B on link 3 (in the Global XY coordinate system) using equations 7.13a. Transform 3 to the global coordinate frame:
θ θ β 360 deg
θ 120.337 deg
b ω2 cosθ j sinθ
AA3 b α sin θ j cos θ AA3 AA3
AA3 0.331
in sec
2
θAA3 arg AA3
θAA3 20.518 deg
a' ω2 cosθ j sinθ
AB a' α sin θ j cos θ AB AB
AB 0.752
in sec
7.
2
θAB arg AB
Determine 5 using equation 4.16.
a' sin θ c' π b'
θ asin
θ 195.254 deg
θAB 20.518 deg
DESIGN OF MACHINERY - 5th Ed.
8.
Determine the angular velocity of link 5 using equation 6.22a. ω
9.
SOLUTION MANUAL 7-48-3
a' cos θ ω b' cos θ
ω 0.124
rad sec
Determine the angular acceleration of link 5 using equation 7.16d.
α
b' cos θ 2
2
a' α cos θ a' ω sin θ b' ω sin θ
α 0.100
sec
10. Use equation 7.16e for the acceleration of pin C.
2
2
AC a' α sin θ a' ω cos θ b' α sin θ b' ω cos θ AC 0.73
in sec
2
rad
A positive sign means that AC is to the right
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-49-1
PROBLEM 7-49 Statement:
Figure P7-19 shows a walking beam mechanism. Calculate and plot the acceleration Aout for one revolution of the input crank 2 rotating at 100 rpm.
Given:
Link lengths:
Solution: 1.
Link 2 (O2 to A)
a 1.00 in
Link 3 (A to B)
b 2.06 in
Link 4 (B to O4)
c 2.33 in
Link 1 (O2 to O4)
d 2.22 in
Coupler point:
p 3.06 in
δ 31 deg
Crank speed:
ω 100 rpm
α 0 rad sec
2
See Figure P7-19 and Mathcad file P0749.
Draw the linkage to scale and label it. Y
x
y O4 4
1 26.00°
X
O2
P
2 A 3 B
2.
Determine the range of motion for this Grashof crank rocker. θ 0 deg 2 deg 360 deg
3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 2.2200 2
K3
d c
K2 0.9528 2
2
a b c d 2 a c
2
K3 1.5265
A θ cos θ K1 K2 cos θ K3
B θ 2 sin θ
C θ K1 K2 1 cos θ K3 4.
Use equation 4.10b to find values of 4 for the crossed circuit.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-49-2
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 5.
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. 2
d
K4
K5
b
2
2
c d a b
2
K4 1.0777
2 a b
D θ cos θ K1 K4 cos θ K5
K5 1.1512
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 6.
Use equation 4.13 to find values of 3 for the crossed circuit.
θ θ 2 atan2 2 D θ E θ 7.
Determine the angular velocity of links 3 and 4 using equations 6.18.
a ω
a ω
ω θ
ω θ 8.
2 4 Dθ F θ
E θ
b
c
sin θ θ θ θ
sin θ θ θ θ
sin θ θ θ
sin θ θ θ
Determine the acceleration of point A using equation 7.13a.
a ω2 cosθ j sinθ
AA θ a α sin θ j cos θ 9.
Use equations 7.12 to determine the angular acceleration of link 3.
B θ b sin θ θ
E θ b cos θ θ
A θ c sin θ θ
D θ c cos θ θ
2
2
c ωθ2 cosθθ
2
2 c ωθ2 sinθθ
C θ a α sin θ a ω cos θ b ω θ cos θ θ F θ a α cos θ a ω sin θ b ω θ sin θ θ
α θ
A θ E θ B θ D θ
C θ D θ A θ F θ
10. Determine the acceleration of the coupler point P using equations 7.32.
2 p ω θ cos θ θ δ j sin θ θ δ
APA θ p α θ sin θ θ δ j cos θ θ δ
AP θ AA θ APA θ
11. Plot the X-component (global coordinate system) of the acceleration of the coupler point P. (See next page). Coordinate rotation angle:
α 26 deg
Aout θ Re AP θ cos α Im AP θ sin α
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-49-3
Aout
Acceleration, in/sec^2
400
200
0
200
400
0
45
90
135
180
Crank Angle, deg
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-50-1
PROBLEM 7-50 Statement:
Given:
Figure P7-20 shows a surface grinder. The workpiece is oscillated under the spinning grinding wheel by the slider-crank linkage that has the dimensions given below. Calculate and plot the acceleration of the grinding wheel contact point relative to the workpiece over one revolution of the crank. Link lengths: Link 2 (O2 to A)
a 22 mm
Link 3 (A to B)
b 157 mm
Grinding wheel diameter
d 90 mm
Input crank angular velocity
ω 30 rpm
Offset
CCW
Grinding wheel angular velocity ω 3450 rpm Solution: 1.
CCW
See Figure P7-20 and Mathcad file P0750.
Draw the linkage to scale and label it.
5
4 2 3
A
B
2
c
O2 2.
Determine the range of motion for this slider-crank linkage. θ 0 deg 2 deg 360 deg
3.
Determine 3 using equation 4.17.
a sin θ b
θ θ asin 4.
a b
ω cos θ θ cos θ
Determine the angular acceleration of link 3 using equation 7.16d.
α θ 6.
π
Determine the angular velocity of link 3 using equation 6.22a:
ω θ 5.
c
2 b cos θ θ 2
a α cos θ a ω sin θ b ω θ sin θ θ
Use equation 7.16e for the acceleration of pin B.
2 2 b α θ sin θ θ b ω θ cos θ θ
AB θ a α sin θ a ω cos θ
c 40 mm
α 0 rad sec
2
DESIGN OF MACHINERY - 5th Ed.
The acceleration of the grinding wheel contact point relative to the workpiece, which has zero tangential acceleration due to constant angular velocity, is AB (positive to the right). RELATIVE ACCELERATION AT CONTACT POINT 200
100
Acceleration, mm/sec^2
7.
SOLUTION MANUAL 7-50-2
0
100
200
300
0
60
120
180 Crank Angle, deg
240
300
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-51-1
PROBLEM 7-51 Statement:
Figure P7-21 shows a drag-link mechanism with dimensions. Write the necessary equations and solve them to calculate and plot the angular acceleration of link 4 for an input of 2 = 1 rad/sec. Comment on the uses for this mechanism.
Given:
Link lengths: Link 2 (L2)
a 1.38 in
Link 3 (L3)
b 1.22 in
Link 4 (L4)
c 1.62 in
Link 1 (L1)
d 0.68 in
ω 1 rad sec
Input crank angular velocity Solution:
1
α 0 rad sec
CW
2
See Figure P7-21 and Mathcad file P0751.
1.
Draw the linkage to scale and label it.
2.
Determine the range of motion for this Grashof double crank.
y
A
3 B
θ 0 deg 2 deg 360 deg 3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 0.4928 2
K3
2 2
d
4
c
K2 0.4198 2
2
a b c d
x O2
2
K3 0.7834
2 a c
A θ cos θ K1 K2 cos θ K3 4.
O4
2 4 A θ Cθ
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
2 a b
K4 0.5574
D θ cos θ K1 K4 cos θ K5
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 6.
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ 7.
C θ K1 K2 1 cos θ K3
Use equation 4.10b to find values of 4 for the open circuit. θ θ 2 atan2 2 A θ B θ
5.
B θ 2 sin θ
2 4 Dθ F θ
E θ
Determine the angular velocity of links 3 and 4 using equations 6.18.
ω θ
a ω b
sin θ θ θ θ sin θ θ θ
K5 0.3655
DESIGN OF MACHINERY - 5th Ed.
ω θ 8.
a ω c
SOLUTION MANUAL 7-51-2
sin θ θ θ θ sin θ θ θ
Use equations 7.12 to determine the angular acceleration of link 4.
B θ b sin θ θ
E θ b cos θ θ
A θ c sin θ θ
D θ c cos θ θ
2
2
c ωθ2 cosθθ
C θ a α sin θ a ω cos θ b ω θ cos θ θ
2 c ωθ2 sinθθ C θ E θ B θ F θ A θ E θ B θ D θ 2
F θ a α cos θ a ω sin θ b ω θ sin θ θ
α θ
Plot the angular acceleration of link 4.
ANGULAR ACCELERATION, LINK 4 2
Angular Acceleration, rad/sec^2
9.
1
0
1
2
0
60
120
180 Crank Angle, deg
240
300
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-52-1
PROBLEM 7-52 Statement:
Figure P7-22 shows a mechanism with dimensions. Use a graphical method to calculate the accelerations of points A, B, and C for the position shown.
Given:
Link lengths and angles: Link 1 (O2O4)
d 1.22 in
Angle O2O4 makes with X axis
θ 56.5 deg
Link 2 (O2A)
a 1.35 in
Angle O2A makes with X axis
θ 14 deg
Link 4 (O4B)
e 1.36 in
Link 5 (BC)
f 2.69 in
Link 6 (CO6)
g 1.80 in
Angle CO6 makes with X axis
θ 88 deg
ω 20 rad sec
Motion of link 2 Solution: 1.
1
α 0 rad sec
CW
2
See Figure P7-22 and Mathcad file P0752.
Draw the linkage to scale and indicate the axes of slip and transmission as well as the directions of velocities of interest.
B
Axis of slip Y
Axis of transmission O4
4
0.939
132.661° A 3 2 X O2
Direction of VA3
2.
Use equation 6.7 to calculate the magnitude of the velocity at point A on links 2 and 3. VA3 a ω
3.
VA3 27.000
in sec
θVA3 θ 90 deg
θVA3 76.0 deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity components at point A on link 3. The equation to be solved graphically is VA3 = Vtrans + Vslip a. Choose a convenient velocity scale and layout the known vector VA3. b. From the tip of VA3, draw a construction line with the direction of Vslip, magnitude unknown. c. From the tail of VA3, draw a construction line with the direction of Vtrans, magnitude unknown. d. Complete the vector triangle by drawing Vslip from the tip of Vtrans to the tip of VA3 and drawing Vtrans from the tail of VA3 to the intersection of the Vslip construction line.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-52-2
Y 0
10 in/sec
1.295 X V trans
V A3 2.369
4.
From the velocity triangle we have: Velocity scale factor:
5.
kv
10 in sec
in
Vtrans 1.295 in kv
Vtrans 12.950
sec in sec
The true velocity of point A on link 4 is Vtrans, VA4 12.95
in sec
Determine the angular velocity of link 4 using equation 6.7.
ω
VA4 c
c 0.939 in and
ω 13.791
rad
θ 132.661 deg
CW
sec
Determine the magnitude and sense of the vector VB using equation 6.7. VB e ω
θVA4 θ 90 deg 8.
in
Vslip 23.690
From the linkage layout above:
7.
1
Vslip 2.369 in kv
VA4 Vtrans 6.
V slip
VB 18.756
in sec
θVA4 42.661 deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point C. The equation to be solved graphically is VC = VB + VCB a. Choose a convenient velocity scale and layout the known vector VB. b. From the tip of VB, draw a construction line with the direction of VCB, magnitude unknown. c. From the tail of VB, draw a construction line with the direction of VC, magnitude unknown. d. Complete the vector triangle by drawing VCB from the tip of VB to the intersection of the VC construction line and drawing VC from the tail of VB to the intersection of the VCB construction line.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-52-3
Draw links 1, 4, 5, and 6 to a convenient scale. Indicate the directions of the velocity vectors of interest.
Direction of VCB
Direction of VB
B
Y O4
4
C A 3
Direction of VC 2
X O2
O6
VB
0
10 in/sec
Y
V CB X VC 2.088
9.
From the velocity triangle we have: Velocity scale factor:
kv
10 in sec
1
θVC θ 90 deg
in in
VC 2.088 in kv
VC 20.9
VCB 1.519 in kv
VCB 15.2
θVC 2.0 deg
sec in sec
10. Determine the angular velocity of links 5 and 6 using equation 6.7. ω
ω
VCB f VC g
ω 5.647
rad
CCW
sec
ω 11.600
rad sec
CCW
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-52-4
11. For the acceleration of point A, use equation 7.19: (AA2t + AA2n) = AA4t + AA4n + AAcor + AAslip , where AA2n a ω
2
AA2n 540.000 in sec
θAA θ 180 deg
θAA 194.000 deg
AA2t a α
AA2t 0.000 in sec 2
2
2 2
AA4n c ω
AA4n 178.6 in sec
θAA4n θ
θAA4n 132.661 deg
AAcor 2 Vslip ω
AAcor 653.430 in sec
θAAcor θ 90 deg
θAAcor 222.661 deg
2
12. Repeat procedure of steps 3 and 7 for the equation in step 11.
1.751 slip
AA t
AA4 AA4
0
2.501 n A4
A
Y
100 IN/S/S
Acceleration Scale
88.220°
cor
AA
X
AA2
13. From the acceleration polygon above, Acceleration scale factor
in
ka 100
sec
2 2
AA4 2.501 ka
AA4 250.1 in sec
AA4t 1.751 ka
AA4t 175.1 in sec
at an angle of 88.22 deg
2
The angular acceleration of link 4 is α
AA4t c
α 186.5
rad sec
CCW
2
14. The graphical solution for accelerations in pin-jointed fourbars uses equation 7.4: (APt + APn) = (AAt + AAn) + (APAt + APAn)
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-52-5
15. For point C, this becomes: (ACt + ACn) = (ABt + ABn) + (ACBt + ACBn) , where ACn g ω
2
ACn 242.2 in sec
θACn θ 180 deg ABn e ω
2
θACn 268.000 deg
2
ABn 258.7 in sec
2
θABn θ 180 deg
θABn 312.661 deg
ABt e α
ABt 253.6 in sec
θABt θ 90 deg
θABt 222.661 deg
2
ACBn f ω
2
ACBn 85.78 in sec
2
θ 27.5 deg
θACBn θ
θACBn 27.500 deg
16. Repeat procedure of step 12 for the equation in step 15. Y
Y X
X 84.453°
91.769°
2.443 A
n B
0
100 IN/S/S
n B
A
Acceleration Scale
3.623
ACn
ABt
AC ACt
t ACB
ABt
AB n ACB
17. From the acceleration polygon above, Acceleration scale factor
ka 100
in sec
2 2
AB 3.623 ka
AB 362.3 in sec
AC 2.443 ka
AC 244.3 in sec
2
at an angle of -91.8 deg at an angle of -84.5 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-53-1
PROBLEM 7-53 Statement:
Figure P7-23 shows a quick-return mechanism with dimensions. Use a graphical method to calculate the accelerations of points A, B, and C for the position shown.
Given:
Link lengths and angles: Link 1 (O2O4)
d 1.69 in
Angle O2O4 makes with X axis
Link 2 (L2)
a 1.00 in
Angle link 2 makes with X axis θ 99 deg
Link 4 (L4)
e 4.76 in
Link 5 (L5)
f 4.55 in
Offset (O2C)
g 2.86 in
Angular velocity of link 2 Solution: 1.
ω 10 rad sec
1
CCW
θ 195.5 deg
α 0 rad sec
2
See Figure P7-23 and Mathcad file P0753.
Draw the linkage to scale and indicate the axes of slip and transmission as well as the directions of velocities of interest.
B
Axis of transmission
Direction of VA3
4
Y Axis of slip A 3 2.068
2 44.228° O2
X O4
2.
Use equation 6.7 to calculate the magnitude of the velocity at point A on links 2 and 3. VA3 a ω
3.
VA3 10.000
in sec
θVA3 θ 90 deg
θVA3 189.0 deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity components at point A on link 3. The equation to be solved graphically is VA3 = Vtrans + Vslip
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-53-2
a. Choose a convenient velocity scale and layout the known vector VA3. b. From the tip of VA3, draw a construction line with the direction of Vslip, magnitude unknown. c. From the tail of VA3, draw a construction line with the direction of Vtrans, magnitude unknown. d. Complete the vector triangle by drawing Vslip from the tip of Vtrans to the tip of VA3 and drawing Vtrans from the tail of VA3 to the intersection of the Vslip construction line. Y 0
5 in/sec
X
VA3 V trans
1.154
4.
kv
5 in sec
1
in in
Vslip 1.634 in kv
Vslip 8.170
Vtrans 1.154 in kv
Vtrans 5.770
sec in sec
The true velocity of point A on link 4 is Vtrans, VA4 Vtrans
6.
1.634
From the velocity triangle we have: Velocity scale factor:
5.
Vslip
VA4 5.77
in sec
Determine the angular velocity of link 4 using equation 6.7. From the linkage layout above: ω
VA4 c
c 2.068 in and
ω 2.790
rad
θ 44.228 deg
CCW
sec
7.
Draw links 1, 4, 5, and 6 to a convenient scale. Indicate the directions of the velocity vectors of interest. (See next page.)
8.
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point C. The equation to be solved graphically is VC = VB + VCB a. Choose a convenient velocity scale and layout the known vector VB. b. From the tip of VB, draw a construction line with the direction of VCB, magnitude unknown. c. From the tail of VB, draw a construction line with the direction of VC, magnitude unknown. d. Complete the vector triangle by drawing VCB from the tip of VB to the intersection of the VC construction line and drawing VC from the tail of VB to the intersection of the VCB construction line.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-53-3
Direction of VCB Direction of VB B 5 5.805°
C 6
4 Direction of VC
Y
A 3
2 44.228° O2
X O4
9.
From the velocity triangle we have: VB
kv
Velocity scale factor: VC 1.659 in kv
5 in sec
VC 8.30
0
1
5 in/sec
in in
Y
sec V CB
θVC 180 deg VCB 1.913 in kv
VCB 9.57
X
VC
in
1.659
sec
10. Determine the angular velocity of link 5 using equation 6.7. From the linkage layout above: ω
VCB
θ 5.805 deg
ω 2.102
f
rad
CCW
sec
11. For the acceleration of point A, use equation 7.19: (AA2t + AA2n) = AA4t + AA4n + AAcor + AAslip , where AA2n a ω
2
AA2n 100.000 in sec
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-53-4
θAA θ 180 deg
θAA 279.000 deg
AA2t a α
AA2t 0.000 in sec 2
2 2
AA4n c ω
AA4n 16.099 in sec
θAA4n θ 180 deg
θAA4n 224.228 deg
AAcor 2 Vslip ω
AAcor 45.591 in sec
θAAcor θ 90 deg
θAAcor 45.772 deg
2
(Vslip is negative)
12. Repeat procedure of steps 3 and 7 for the equation in step 11. Y
X
n
AA4
1.581 69.810° AA4
t AA4
0
25 IN/S/S
slip
AA
Acceleration Scale
1.444
AA2 cor
AA
13. From the acceleration polygon above, Acceleration scale factor
in
ka 25
sec
2 2
AA4 1.581 ka
AA4 39.5 in sec
AA4t 1.444 ka
AA4t 36.1 in sec
at an angle of -69.81 deg
2
The angular acceleration of link 4 is α
AA4t
α 17.5
c
rad sec
CW
2
14. The graphical solution for accelerations in pin-jointed fourbars uses equation 7.4: (APt + APn) = (AAt + AAn) + (APAt + APAn) 15. For point C, this becomes: AC = (ABt + ABn) + (ACBt + ACBn) , where ABn e ω
2
θABn θ 180 deg
ABn 37.056 in sec
θABn 224.228 deg
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-53-5 2
ABt e α
ABt 83.09 in sec
θABt θ 90 deg
θABt 45.772 deg
2
ACBn f ω
ACBn 20.108 in sec
θACBn θ
θACBn 5.805 deg
2
16. Repeat procedure of step 12 for the equation in step 15. 1.717
Y
AC
n
AB
X
69.808°
t ACB
0
25 IN/S/S
Acceleration Scale 3.639
AB t B
A
n
ACB
17. From the acceleration polygon above, Acceleration scale factor
ka 25
in sec
2 2
AB 3.639 ka
AB 91.0 in sec
AC 1.717 ka
AC 42.9 in sec
2
at an angle of -69.81 deg at an angle of 0 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-54-1
PROBLEM 7-54 Statement:
Figure P7-23 shows a quick-return mechanism with dimensions. Use an analytical method to calculate the accelerations of points A, B, and C for one revolution of the input link.
Given:
Link lengths and angles:
Solution: 1.
Link 1 (O2O4)
d 1.69 in
Link 4 (L4)
a' 4.76 in
Link 2 (L2)
a 1.00 in
Link 5 (L5)
b' 4.55 in
Input crank motion
ω 10 rad sec
Coordinate rotation angle
β 164.5 deg
1
α 0 rad sec
CCW Offset (AE)
2
c' 2.86 in
See Figure P7-23 and Mathcad file P0754.
Draw the mechanism to scale and define a vector loop for the input portion (links 1, 2, 3, and 4) using the fourbar crank-slider derivation in Section 7.3 as a model. θ 0 deg 0.5 deg 360 deg B 5 C 6
4
Y
A 3
2 O2
15.5°
X x
O4 y
R2
R3
R1 x y
DESIGN OF MACHINERY - 5th Ed.
2.
SOLUTION MANUAL 7-54-2
Write the vector loop equation, expand the result and separate into real and imaginary parts to solve for 3 and b (distance from O4 to A). R2 R1 R3 a e
j θ
d b e
j θ
Substituting the Euler equivalents,
d bcosθ j sinθ
a cos θ j sin θ
Separating into real and imaginary components and solving for 3 and b.
sin θ θ
θ θ atan2 a cos θ d a sin θ a sin θ
b θ
3.
Differentiate the position equation, expand it and solve for 3 and bdot. a j ω e
j θ
ω θ
a ω
b θ
bdot θ
4.
b j ω e
j θ
bdot e
j θ
cos θ θ θ
sin θ θ
a ω cos θ b θ ω θ cos θ θ
Differentiate the velocity equation, expand it and solve for 3. a j e
j θ
2
2 j θ
a j ω e
bdot j ω e 2
j θ
2 j θ
b j ω e
α θ 5.
1
b θ
b j e
j θ
bddot e
j θ
bdot j ω e
j θ
a α cos θ θ θ
a ω2 sinθθ θ 2 bdotθ ωθ
Determine the acceleration of point A on link 2 (in the global XY coordinate system) using equations 7.13a.
a ω2 cosθ j sinθ
AA2 θ a α sin θ j cos θ AA2 θ AA2 θ 6.
θAA2 θ arg AA2 θ β
Determine the acceleration of points A and B on link 4 (in the Global XY coordinate system) using equations 7.13a. Transform and rename 3 to the global coordinate frame:
θ θ θ θ β
Rename 3 to 4:
α θ α θ
ω θ ω θ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-54-3
j cosθθ 2 b θ ω θ cos θ θ j sin θ θ
AA4 θ b θ α θ sin θ θ
θAA4 θ arg AA4 θ
AA4 θ AA4 θ
j cosθθ 2 a' ω θ cos θ θ j sin θ θ
AB θ a' α θ sin θ θ
θAB θ arg AB θ
AB θ AB θ 7.
Determine 5 using equation 4.16.
a' sin θ θ c' b'
θ θ asin 8.
Determine the angular velocity of link 5 using equation 6.22a:
ω θ
9.
a' cos θ θ ω θ b' cos θ θ
Determine the angular acceleration of link 5 using equation 7.16d.
α θ
a' ωθ2 sinθθ b' ωθ2 sinθθ b' cos θ θ
a' α θ cos θ θ
10. Use equation 7.16e for the acceleration of pin C.
a' ωθ2 cosθθ 2 b' α θ sin θ θ b' ω θ cos θ θ
AC θ a' α θ sin θ θ
ACCELERATION OF POINT A 250
Acceleration, in/sec^2
200
150
100
50
0
0
60
120
180 Crank Angle, deg
240
300
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-54-4
Acceleration, in/sec^2
ACCELERATION OF POINT B
1500
1000
500
0
0
60
120
180
240
300
360
300
360
Crank Angle, deg
ACCELERATION OF POINT C
Acceleration, in/sec^2
1000
500
0
500
1000
0
60
120
180 Crank Angle, deg
240
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-55-1
PROBLEM 7-55 Statement:
Figure P7-24 shows a drum pedal mechanism . For the dimensions given below, find and plot the output acceleration of the linkage over its range of motion if the input velocity Vin is a constant.
Given:
Link lengths: Link 2 (O2A)
a 100 mm
Link 3 (AB)
b 28 mm
Link 4 (O4B)
c 64 mm
Link 1 (O2O4)
d 56 mm
Link 3 (AP)
rout 124 mm
Distance to force application: rin 48 mm
Link 2
Vin 3 m sec
Input velocity:
1
α 0 rad sec
θ 162 deg
Range of positions of link 2:
2
δ 0 deg
θ 171 deg
See Figure P7-24 and Mathcad file P0755. Solution: 1. Draw the mechanism to scale and label it. 2.
P
Calculate the range of 2 in the local coordinate system (required to calculate 3 and 4). Rotation angle of local xy system to global XY system:
3
α 180 deg
θ θ α θ α 1 deg θ α 3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K1 0.5600 2
K3
3
2
2
2
2
1
O4
A θ cos θ K1 K2 cos θ K3
2
x
K3 1.2850
2 a c
Vin
A
c
K2 0.8750
a b c d
Fin
4
d
K2
a
B
y
B θ 2 sin θ
C θ K1 K2 1 cos θ K3 4.
Use equation 4.10b to find value of 4 for the open circuit.
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 5.
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
2 a b
K4 2.0000
D θ cos θ K1 K4 cos θ K5
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ
K5 1.7543
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 6.
2 4 Dθ F θ
E θ
O2
DESIGN OF MACHINERY - 5th Ed.
7.
Determine the angular velocity of link 2 using equation 6.7 and links 3 and 4 using equations 6.18. Vin
ω
ω 62.500
rin
a ω
a ω
ω θ
ω θ
8.
SOLUTION MANUAL 7-55-2
b
c
rad sec
sin θ θ θ θ
sin θ θ θ θ
sin θ θ θ
sin θ θ θ
Determine the acceleration of point A using equation 7.13a.
a ω2 cosθ j sinθ
AA θ a α sin θ j cos θ 9.
Use equations 7.12 to determine the angular acceleration of link 3.
B θ b sin θ θ
E θ b cos θ θ
A θ c sin θ θ
D θ c cos θ θ
2
2
c ωθ2 cosθθ
2
2 c ωθ2 sinθθ
C θ a α sin θ a ω cos θ b ω θ cos θ θ F θ a α cos θ a ω sin θ b ω θ sin θ θ
α θ
A θ E θ B θ D θ
C θ D θ A θ F θ
10. Determine the acceleration of the coupler point P using equations 7.32.
2 rout ω θ cos θ θ δ j sin θ θ δ AP θ AA θ APA θ APA θ rout α θ sin θ θ δ j cos θ θ δ
11. Plot the magnitude of the acceleration of the coupler point P. (See next page).
Aout θ AP θ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-55-3
Aout 222000
Acceleration, in/sec^2
220000
218000
216000
214000
212000 18
16.3
14.7
13
11.3
9.7
8
Crank Angle, deg
Note the acceleration scale values. If the scale went to zero, Aout would be a horizontal (constant) line over the range of 2 plotted.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-56-1
PROBLEM 7-56 Statement:
A tractor-trailer tipped over while negotiating an on-ramp to the New York Thruway. The road has a 50-ft radius at that point and tilts 3 deg toward the outside of the curve. The 45-ft-long by 8-ft-wide by 8.5-ft-high trailer box (13 ft from ground to top) was loaded with 44 415 lb of paper rolls in two rows by two high as shown in Figure P7-25. The rolls are 40-in-dia by 38-in-long and weigh about 900 lb each. They are wedged against rolling backward but not against sliding sidewards. The empty trailer weighed 14 000 lb. The driver claims that he was traveling at less than 15 mph and that the load of paper shifted inside the trailer, struck the trailer sidewall, and tipped the truck. The paper company that loaded the truck claims the load was properly stowed and would not shift at that speed. Independent test of the coefficient of friction between similar paper rolls and a similar trailer floor give a value of 0.43 +/- 0.08. The composite center of gravity of the loaded trailer is estimated to be 7.5 ft above the road. Determine the truck speed that would cause the truck to just begin to tip and the speed at which the rolls will just begin to slide sidways. What do you think caused the accident?
Given:
Weight of paper
Wp 44415 lbf
Weight of trailer
Wt 14000 lbf
Radius of curve
r 50 ft
Nominal coefficient of friction
μnom 0.43
Coefficient of friction uncertainty
u μ 0.08
Trailer width
w 8 ft
Height of CG from pavement
h 7.5 ft
Assumptions: 1. The paper rolls act as a monolith since they are tightly strapped together with steel bands. 2. The tractor has about 15 deg of potential roll freedom versus the trailer due to their relative angle in plan during the turn combined with the substantial pitch freedom in the fifth wheel. So the trailer can tip independently of the tractor. 3. The outside track width of the trailer tires is equal to the width of the trailer. Solution: 1.
See Figure P7-25 and Mathcad file P0756.
First, calculate the location of the trailer's CG with respect to the outside wheel when it is on the reverse-banked curve. From the figure at right, Tilt angle
θ 3 deg
a h tan( θ )
a 0.393 ft
w
b 3.607 ft
b
2
a
xbar b cos( θ )
xbar 3.602 ft
ybar b sin( θ )
h
3° 7.500'
ybar
cos( θ )
ybar 7.699 ft The coordinates of the CG of the loaded trailer with respect to the lower outside corner of the tires are: xbar 3.602 ft
a
b
ybar 7.699 ft
xbar 4.000'
DESIGN OF MACHINERY - 5th Ed.
2.
SOLUTION MANUAL 7-56-2
The trailer is on the verge of tipping over when the couple due to centrifugal force is equal to the couple formed by the weight of the loaded trailer acting through its CG and the vertical reaction at the outside edge of the tires. At this instant, it is assumed that the entire weight of the trailer is reacted at the outside tires with the inside tires carrying none of the weight. Any increase in tangential velocity of the tractor and trailer will result in tipping. Summing moments about the tire edge (see figure below),
M
Fw xbar Fc ybar = 0
(1)
where Fc is the centrifugal force due to the normal acceleration as the tractor and trailer go through the curve. The normal acceleration is a tip =
vtip
Fc
2
(2)
r
and the force necessary to keep the tractor trailer following a circular path is Fc = mtot a tip
Fw ybar (3)
where mtot is the total mass of the trailer and its payload. Combining equations (2) and (3) and solving for vtip, we have vtip =
Fc r
Rx Ry xbar
(4)
mtot
or, vtip =
Fc r g
(5)
Fw
3. Calculate the minimum tipping velocity of the tractor/trailer. From equations (1) and (5), Total weight
Fw Wt Wp
Centrifugal force required to tip the trailer
Fc
Minimum tipping speed
vtip
xbar ybar
Fw
Fc r g Fw
Fw 58415 lbf Fc 27329 lbf
vtip 18.7 mph
Thus, with the assumptions that we have made, the trailer would not begin to tip over until it reached a speed of vtip 18.7 mph 4.
The load will slip when the friction force between the paper rolls and the trailer floor is no longer sufficient to react the centrifugal force on the paper rolls. Looking at the FBD of the paper rolls below, we see that Normal force between paper and floor
Fn = Wp cos( θ ) Fcp sin( θ )
(6)
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-56-3
Tangential force tending to slide the paper Ft = Wp sin( θ ) Fcp cos( θ )
Fcp
(7)
Wp
Centrifugal force on the paper 2
Wp vs Fcp = as = g g r Wp
But, the maximum friction force is
Ft
(8)
Fn
Ff = μ Fn = Ft
(9)
Substituting equation (9) into (7), then combining (6) and (7) to eliminate Fn, and solving for Fcp yields
Fcp =
Wp ( μ cos( θ ) sin( θ ) )
(10)
μ sin( θ ) cos( θ )
Substituting equation (10) into (8), to eliminate Fcp, and solving for vs yields
vs =
5.
μ cos( θ ) sin( θ ) r g μ sin( θ ) cos( θ )
(11)
Use the upper and lower limit on the coefficient of friction to determine an upper and lower limit on the speed necessary to cause sliding. Maximum coefficient
μmax μnom u μ
μmax 0.510
Minimum coefficient
μmin μnom u μ
μmin 0.350
Maximum velocity to cause sliding
vsmax
μmax cos( θ ) sin( θ ) μ sin( θ ) cos( θ ) r g max
vsmax 18.3 mph
Minimum velocity to cause sliding
vsmin
6.
μmin cos( θ ) sin( θ ) μ sin( θ ) cos( θ ) r g min
vsmin 14.8 mph
This very rough analysis shows that , if the coefficient of friction was at or near the low end of its measured value, the paper load could slide at a tractor/trailer speed of 15 mph, which would lead to the trailer tipping over. In any case, it appears that the paper load would slide before the truck would tip with the load in place.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-57-1
PROBLEM 7-57 Statement:
Figure P7-26 shows a V-belt drive. The sheaves (pulleys) have pitch diameters of 150 and 300 mm, respectively. The smaller sheave is driven at a constant 1750 rpm. For a cross-sectional, differential element of the belt, write the equations of its acceleration for one complete trip around both pulleys including its travel between the pulleys. Compute and plot the acceleration of the differential element versus time for one circuit around the belt path. What does your analysis tell you about the dynamic behavior of the belt?
Given:
Sheave radii and speed: r2 75 mm
r4 150 mm
ω 1750 rpm
Assumptions: Center distance: C 450 mm Solution: 1.
See Figure P7-26 and Mathcad file P0757.
Draw a schematic representation of the V-belt drive to scale and label it.
443.706 B 3 9.594°
A 4
2
R150.000 D
R75.000
C
From the layout,
2.
Angle to point of tangency
β 9.594 deg
Distance from A to B
LAB 443.706 mm
Distance from B to C
LBC π 2 β r4
Distance from C to D
LCD 443.706 mm
Distance from D to A
LDA π 2 β r2
LDA 210.502 mm
Total path length
L LAB LBC LCD LDA
L 1619.387 mm
Calculate the angular velocity of each sheave. Sheave 2 Sheave 4
3.
LBC 521.473 mm
ω 183.260 ω
r2 r4
ω
rad sec ω 91.630
rad sec
Calculate the acceleration of an element on the belt for each portion of the path, starting at A.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-57-2
mm
AAB 0
From A to B the acceleration is zero since the belt velocity is constant
sec
2
From B to C the tangential acceleration is zero since the belt velocity is constant The normal component is ABC r4 ω
6 mm
2
ABC 1.259 10
sec
2
ACD 0
From C to D the acceleration is zero since the belt velocity is constant
mm sec
2
From D to A the tangential acceleration is zero since the belt velocity is constant The normal component is ADA r2 ω
6 mm
2
ADA 2.519 10
sec 4.
2
Plot the acceleration over the entire path using a path variable of s. s 0 mm 5 mm L Define a range function that will plot a variable only between two limits using the Heaviside step function . R( s a b ) Φ ( s a ) Φ ( s b ) The acceleration function for the entire path is. Let
s1 LAB
s2 LAB LBC
s3 LAB LBC LCD
A ( s) AAB R s 0 mm s1 ABC R s s1 s2 ACD R s s2 s3 ADA R s s3 L
BELT ACCELERATION ALONG PATH
Acceleration, m/sec^2
3000
2000
1000
0
0
0.25
0.5
0.75
1
1.25
1.5
1.75
2
Distance Along Path, m
5.
The graph shows the sudden change in acceleration as the belt enters and leaves a sheave. This results in infinite jerk at these points. This results in belt "hop" at these points, a condition that will cause eventual fatigue failure and wear in the belt. Because of the belt hop caused by infinite jerk at the initial belt/sheave tangency points the span of the belt is continuously vibrating.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-58-1
PROBLEM 7-58 Statement:
Write a program using an equation solver or any computer language to solve for the displacements, velocities, and accelerations in an offset crank-slider linkage as shown in Figure P7-2. Plot the variation in all link's angular and pin's linear positions, velocities, and accelerations with a constant angular velocity input to the crank over one revolution for both open and crossed configurations of the linkage. To test the program, use data from row a of Table P7-2. Check your results with program SLIDER.
Enter:
Link lengths: a 1.4 in
Link 2
Link 3
c 1 in
Offset:
ω 10
Link 2 velocity and acceleration: Crank position range: Solution: 1.
b 4 in rad
α 0
sec
rad sec
2
θ 0 deg 2 deg 360 deg
See Figure P7-2 and Mathcad file P0758.
Draw the linkage and label it.
Y 4 B 3 b
A
c 2 a
2 X
O2 d 2.
Determine 3 and d using equations 4.16 and 4.17. Open:
a sin θ b
θ θ asin
c
π
d 2 θ a cos θ b cos θ θ
Crossed:
a sin θ b
θ θ asin
c
d 1 θ a cos θ b cos θ θ 3.
Determine the angular velocity of link 3 using equation 6.22a. Open
ω θ
a b
ω cos θ θ cos θ
DESIGN OF MACHINERY - 5th Ed.
Crossed
4.
SOLUTION MANUAL 7-58-2
ω θ
a b
ω cos θ θ cos θ
Determine the velocity of pin A using equation 6.23a:
VA θ a ω sin θ j cos θ
VAX θ Re VA θ 5.
6.
VAY θ Im VA θ
Determine the velocity of pin B using equation 6.22b:
Open
VB1 θ a ω sin θ b ω θ sin θ θ
Crossed
VB2 θ a ω sin θ b ω θ sin θ θ
Using the Euler identity to expand equation 7.15b for AA, determine its x and y components.
a ω2 cosθ j sinθ
AA θ a α sin θ j cos θ
AAX θ Re AA θ 7.
Determine the angular acceleration of link 3 using equation 7.16d.
Open
Crossed
8.
AAY θ Im AA θ
2
2 b cos θ θ
2
a α cos θ a ω sin θ b ω θ sin θ θ
a α cos θ a ω sin θ b ω θ sin θ θ
α θ
α θ
2 b cos θ θ
Use equation 7.16e for the acceleration of pin B.
2 2 b α θ sin θ θ b ω θ cos θ θ
2 2 b α θ sin θ θ b ω θ cos θ θ
Open:
AB2 θ a α sin θ a ω cos θ
Crossed:
AB1 θ a α sin θ a ω cos θ
(See plots on the following pages.)
DESIGN OF MACHINERY - 5th Ed.
Plot the angular position of link 3 and the distance of pin B from the y axis for open (solid) and crossed (dashed) configurations.
ANGULAR POSITION OF LINK 3 250 200
Angle, deg
150 100 50 0 50
0
45
90
135
180
225
270
315
360
270
315
360
Crank Angle, deg Open Crossed
POSITION OF PIN B
Distance d, in
9.
SOLUTION MANUAL 7-58-3
6 5 4 3 2 1 0 1 2 3 4 5 6
0
45
90
135
180
Crank Angle, deg Open Crossed
225
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-58-4
10. Plot the angular velocity of link 3 and the velocities of pins A and B for open (solid) and crossed (dashed) configurations. ANGULAR VELOCITY OF LINK 3 4
Anglular Velocity, rad/sec
3 2 1 0 1 2 3 4
0
45
90
135
180
225
270
315
360
270
315
360
270
315
360
Crank Angle, deg
X & Y VELOCITY COMPONENTS OF PIN A
Velocity, in/sec
20
10
0 10 20
0
45
90
135
180
225
Crank Angle, deg
VELOCITY OF PIN B
Velocity, in/sec
20
10
0 10 20
0
45
90
135
180
Crank Angle, deg
225
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-58-5
11. Plot the angular acceleration of link 3 and the accelerations of pins A and B for open (solid) and crossed (dashed configurations. ANGULAR ACCELERATION OF LINK 3
Anglular Acceleration, rad/sec^2
60
30
0
30
60
0
45
90
135
180
225
270
315
360
270
315
360
270
315
360
Crank Angle, deg
X & Y ACCELERATION COMPONENTS OF PIN A
Acceleration, in/sec^2
200
100
0 100 200
0
45
90
135
180
225
Crank Angle, deg
ACCELERATION OF PIN B
Acceleration, in/sec^2
200 100 0 100 200
0
45
90
135
180
Crank Angle, deg
225
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-59-1
PROBLEM 7-59 Statement:
Write a program using an equation solver or any computer language to solve for the displacements, velocities, and accelerations in an inverted slider-crank linkage as shown in Figure P7-3. Plot the variation in all link's angular and pin's linear positions, velocities, and accelerations with a constant angular velocity input to the crank over one revolution for both open and crossed configurations of the linkage. To test the program, use data from row e of Table P7-3 except for the value of 2, which will be set to zero for this exercise.
Enter:
Link lengths: Link 2
a 4 in
Angle between links 3 and 4
γ 30 deg
Link 2 velocity and accel.
ω 45
1.
rad
α 0
sec
d 8 in
Link 1
rad sec
2
θ 0 deg 2 deg 360 deg
Crank position range: Solution:
c 2 in
Link 4
See Figure P7-3 and Mathcad file P0759.
Draw the linkage and label it.
B y
4
3 b A
a 2
c
2 d
02 2.
x
1
04
Use the equations in Section 4.7 to solve for the positions of links 3 and 4 and for the length b.
P θ a sin θ sin γ a cos θ d cos γ
Q θ a sin θ cos γ a cos θ d sin γ
R c sin γ
Open:
S θ R Q θ
θ θ 2 atan2 2 S θ T θ
θ θ θ θ γ
b 1 θ
T θ 2 P θ
sin θ θ
a sin θ c sin θ θ
2 4 Sθ U θ
T θ
U θ Q θ R
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-59-2
θ θ 2 atan2 2 S θ T θ
Crossed:
2 4 Sθ U θ
T θ
θ θ θ θ γ
b 2 θ
3.
sin θ θ
a sin θ c sin θ θ
Calculate the angular velocity of links 3 and 4 and the slip velocity using equations 6.30. Open:
ω θ
bdot1 θ
b 1 θ c cos γ
a ω cos θ θ θ
c sinθθ cos θ θ
a ω sin θ ω θ b 1 θ sin θ θ
ω θ ω θ
Crossed: ω θ
bdot2 θ
b 2 θ c cos γ
a ω cos θ θ θ
c sinθθ cos θ θ
a ω sin θ ω θ b 2 θ sin θ θ
ω θ ω θ 4.
Solve for the accelerations using equations (7.26) and (7.27). Open:
P1 θ a α cos θ θ θ
2
R1 θ c ω θ sin θ θ θ θ
2
Q1 θ a ω sin θ θ θ
T1 θ b 1 θ c cos θ θ θ θ
α θ
P1 θ Q1 θ R1 θ S 1 θ
T1 θ
2
K1 θ a ω b 1 θ cos θ θ θ c cos θ θ θ
c sinθθ θ
L1 θ a α b 1 θ sin θ θ θ M1 θ ω θ b 1 θ
2
2 c2 2 b1θ c cosθθ θθ
S 1 θ 2 bdot1 θ ω θ
N1 θ 2 bdot1 θ c ω θ sin θ θ θ θ
DESIGN OF MACHINERY - 5th Ed.
bddot1 θ
SOLUTION MANUAL 7-59-3
K1 θ L1 θ M1 θ N1 θ
T1 θ
a ω2 cosθ j sinθ
AA θ a α sin θ j cos θ
AAX θ Re AA θ
AAY θ Im AA θ
j cosθθ 2 c ω θ cos θ θ j sin θ θ
AB1 θ c α θ sin θ θ
θAB1 θ arg AB1 θ
AB1 θ AB1 θ Crossed:
P2 θ a α cos θ θ θ
2
R2 θ c ω θ sin θ θ θ θ
2
Q2 θ a ω sin θ θ θ
T2 θ b 2 θ c cos θ θ θ θ
α θ
P2 θ Q2 θ R2 θ S 2 θ
T2 θ
2
K2 θ a ω b 2 θ cos θ θ θ c cos θ θ θ
c sinθθ θ
L2 θ a α b 2 θ sin θ θ θ M2 θ ω θ b 2 θ
2 c2 2 b2θ c cosθθ θθ
2
N2 θ 2 bdot2 θ c ω θ sin θ θ θ θ
bddot2 θ
K2 θ L2 θ M2 θ N2 θ
T2 θ
j cosθθ 2 c ω θ cos θ θ j sin θ θ
AB2 θ c α θ sin θ θ
AB2 θ AB2 θ
S 2 θ 2 bdot2 θ ω θ
θAB2 θ arg AB2 θ
DESIGN OF MACHINERY - 5th Ed.
Plot the angular position of link 4 and the distance b for open (solid) and crossed (dashed) configurations.
ANGULAR POSITION OF LINK 4 400
Angle, deg
300
200
100
0
100
0
45
90
135
180
225
270
315
360
225
270
315
360
Crank Angle, deg
LENGTH b 20
10 Distance b, in
5.
SOLUTION MANUAL 7-59-4
0
10
20
0
45
90
135
180
Crank Angle, deg
DESIGN OF MACHINERY - 5th Ed.
Plot the angular acceleration of link 4 and the accelerations of points A and B for open (solid) and crossed (dashed) configurations. ANGULAR ACCELERATION OF LINK 4
Anglular Acceleration, rad/sec^2
5000 3750 2500 1250 0 1250 2500 3750 5000
0
45
90
135
180
225
270
315
360
270
315
360
270
315
360
Crank Angle, deg
X & Y ACCELERATION COMPONENTS OF PIN A
Acceleration, in/sec^2
10000
5000
0 5000 10000
0
45
90
135
180
225
Crank angle, deg
ACCELERATION OF PIN B 10000 Acceleration, in/sec^2
6.
SOLUTION MANUAL 7-59-5
7500
5000
2500
0
0
45
90
135
180
Crank Angle, deg
225
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-60-1
PROBLEM 7-60 Statement:
Write a program using an equation solver or any computer language to solve for the displacements, velocities, and accelerations in a geared fivebar linkage as shown in Figure P7-4. Plot the variation in all link's angular and pin's linear positions, velocities, and accelerations with a constant angular velocity input to the crank over one revolution for both open and crossed configurations of the linkage. To test the program, use data from row a of Table P7-4. Check your results with program FIVEBAR.
Enter:
Link lengths: Link 1
f 6 in
Link 3
b 7 in
Link 2
a 1 in
Link 4
c 9 in
d 4 in
Link 5
Gear ratio, phase angle, and crank velocity and acceleration:
Solution: 1.
λ 2
ϕ 30 deg
ω 10 rad sec
Coupler data:
Rpa 6 in
δ 30 deg
1
α 0 rad sec
2
See Figure P7-4 and Mathcad file P0760.
Determine the angle of link 5 using the equation in Figure P6-4.
θ θ λ θ ϕ 2.
Determine the values of the constants needed for finding 3 and 4 from equations 4.28h and 4.28i.
A θ 2 c d cos λ θ ϕ a cos θ f
B θ 2 c d sin λ θ ϕ a sin θ
2
D θ C θ A θ E θ 2 B θ 2
2
2
2
C θ a b c d f 2 a f cos θ 2 d a cos θ f cos λ θ ϕ 2 a d sin θ sin λ θ ϕ
a cosθ f
a sinθ
F θ A θ C θ
G θ 2 b d cos λ θ ϕ H θ 2 b d sin λ θ ϕ
2
2
2
2
2
K θ a b c d f 2 a f cos θ 2 d a cos θ f cos λ θ ϕ 2 a d sin θ sin λ θ ϕ
L θ K θ G θ M θ 2 H θ 3.
N θ G θ K θ
Use equations 4.28h and 4.28i to find values of 3 and 4 for the open and crossed circuits. OPEN
M θ
E θ
M θ
E θ
θ θ 2 atan2 2 L θ M θ θ θ 2 atan2 2 D θ E θ
CROSSED
θ θ 2 atan2 2 L θ M θ θ θ 2 atan2 2 D θ E θ
2 4 Lθ N θ
2 4 Dθ F θ 2 4 Lθ N θ
2 4 Dθ F θ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-60-2
4.
Use equation 6.32c to find 5. ω λ ω
5.
Calculate 3 and 4 and the pin velocities using equations 6.33. OPEN
ω θ
d ω sinθθ θθ b cos θ θ 2 θ θ cos θ θ
2 sin θ θ a ω sin θ θ θ
d ω sinθθ c sin θ θ VA θ a ω sin θ j cos θ
ω θ
a ω sin θ b ω θ sin θ θ
VAx θ Re VA θ
VAy θ Im VA θ
j cosθθ
VBA1 θ b ω θ sin θ θ
j cosθθ
VC θ d ω sin θ θ
VB1 θ VA θ VBA1 θ
VB1x θ Re VB1 θ
VB1y θ Im VB1 θ
CROSSED
ω θ
d ω sinθθ θθ b cos θ θ 2 θ θ cos θ θ
2 sin θ θ a ω sin θ θ θ
d ω sinθθ c sin θ θ VBA2 θ b ω θ sin θ θ j cos θ θ
ω θ
a ω sin θ b ω θ sin θ θ
VB2 θ VA θ VBA2 θ
VB2x θ Re VB2 θ
VB2y θ Im VB2 θ
6.
Use equation 7.28c to find 5. α λ α
7.
Calculate 3 and 4 and the pin accelerations using equations 7.29. OPEN
a ω2 cosθ θθ 2 2 b ω θ cos θ θ θ θ d ω cos θ θ θ θ 2 d α sin θ θ θ θ c ω θ b sin θ θ θ θ
a α sin θ θ θ
α θ
a ω2 cosθ θθ 2 2 c ω θ cos θ θ θ θ d ω cos θ θ θ θ 2 d α sin θ θ θ θ b ω θ c sin θ θ θ θ a α sin θ θ θ
α θ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-60-3
a ω2 cosθ j sinθ
AA θ a α sin θ j cos θ
AAx θ Re AA θ
AAy θ Im AA θ
j cosθθ 2 b ω θ cos θ θ j sin θ θ AB1 θ AA θ ABA1 θ ABA1 θ b α θ sin θ θ
AB1x θ Re AB1 θ
AB1y θ Im AB1 θ
j cosθθ 2 c ω cos θ θ j sin θ θ ACx θ Re AC θ ACy θ Im AC θ AC θ c α sin θ θ
CROSSED
a ω2 cosθ θθ 2 2 b ω θ cos θ θ θ θ d ω cos θ θ θ θ 2 d α sin θ θ θ θ c ω θ b sin θ θ θ θ
a α sin θ θ θ
α θ
a ω2 cosθ θθ 2 2 c ω θ cos θ θ θ θ d ω cos θ θ θ θ 2 d α sin θ θ θ θ b ω θ c sin θ θ θ θ a α sin θ θ θ
α θ
j cosθθ 2 b ω θ cos θ θ j sin θ θ AB2 θ AA θ ABA2 θ ABA2 θ b α θ sin θ θ
AB2x θ Re AB2 θ
AB2y θ Im AB2 θ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-61a-1
PROBLEM 7-61a Statement:
Find the acceleration of the slider in Figure 3-33 for 2 = 110 deg with respect to the global X axis assuming a constant 2 = 1 rad/sec CW. Use a graphical method.
Given:
Link lengths: Link 2 (O2 to A)
a 2.170 in
Link 3 (A to B)
b 2.067 in
Link 4 (O4 to B)
c 2.310 in
Link 1 (O2 to O4)
d 1.000 in
Link 5 (B to C)
e 5.400 in Crank angle:
θ2XY 110 deg
102 deg
Coordinate angle
Solution: 1.
Input crank angular velocity
1 rad sec
Input crank angular acceleration
0 rad sec
1
CW
2
See Figure P6-33 and Mathcad file P0761a.
Draw the linkage to scale and label it.
Direction of AAt Y Direction of ABAt
148.950°
A 3
Direction of ABt
B
2 4 O2
5 X
57.635° y
158.818° 6
O4 C x
102.000° Direction of AC Direction of ACBt
θ2XY 2.
212.000 deg
In order to solve for the acceleration at point B, we will need 3, 3, and 4. From the graphical position solution above (in the local xy coordinate system), θ 148.950 deg 180 deg
θ 70.950 deg
θ 57.635 deg
θ 159.635 deg
Using equation (6.18), ω
a sin θ b sin θ θ
ω 0.832 rad sec
ω
a sin θ c sin θ θ
ω 0.591 rad sec
1
1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-61a-2
(APt + APn) = (AAt + AAn) + (APAt + APAn)
3.
The graphical solution for accelerations uses equation 7.4:
4.
For point B, this becomes: (ABt + ABn) = (AAt + AAn) + (ABAt + ABAn) , where ABn c ω
2
ABn 0.806 in sec
θABn θ 180 deg
5.
2
θABn 122.365 deg 2
AAn a
2
AAn 2.170 in sec
θAAn 180 deg
θAAn 290.000 deg
AAt a
AAt 0.0 in sec
θAAt 90 deg
θAAt 200.000 deg
2
2
ABAn b ω
2
ABAn 1.429 in sec
θABAn θ 180 deg
θABAn 148.950 deg
Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ABn at an angle of ABn + and AAn at an angle of AAn + . From the tip of AAn, draw AAt at an angle of AAt + . From the tip of AAt, draw ABAn at an angle of BAn + . Now that the vectors with known magnitudes are drawn, from the tips of ABn and ABAn, draw construction lines in the directions of ABt and ABAt, respectively. The intersection of these two lines are the tips of ABt, ABt, and ABAt. Y
0
1 IN/S/S
X
n
0.850 A B AB
Acceleration Scale
t B
103.340°
A
t BA
A
AA n
AA
n
A BA
6.
From the graphical solution above, Acceleration scale factor
ka 1.000
in sec
7.
2
AA AAn
AA 2.170 in sec
AB 0.850 ka
AB 0.850 in sec
2 2
at an angle of 290.0 deg at an angle of -103.340 deg
From Problem 6-70a, the angular velocity of link 5 is 0.145 rad sec
1
CCW. From the graphical
layout, 158.818 deg in the global XY coordinate system. 8.
For point C, equation 7.4 becomes: AC = AB + (ACBt + ACBn) , where AB is known and 2
ACBn e
ACBn 0.114 in sec
2
DESIGN OF MACHINERY - 5th Ed.
θACBn
SOLUTION MANUAL 7-61a-3
θACBn 158.818 deg
Y 0.005 AC X
0
0.5 IN/S/S t ACB
Acceleration Scale
AB n
A CB 9.
From the graphical solution above, Acceleration scale factor
ka 0.50
in sec
AC 0.005 ka
2
AC 0.0025 in sec
2
at an angle of 0.000 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-61b-1
PROBLEM 7-61b Find the acceleration of the slider in Figure 3-33 for 2 = 110 deg with respect to the global X
Statement:
axis assuming a constant 2 = 1 rad/sec CW. Use an analytical method. Given:
Link lengths: Link 2 (O2 to A)
a 2.170 in
Link 3 (A to B)
b 2.067 in
Link 4 (O4 to B)
c 2.310 in
Link 1 (O2 to O4)
d 1.000 in
Link 5 (B to C)
e 5.400 in Crank angle:
θ2XY 110 deg
102 deg
Coordinate angle
Solution: 1.
Input crank angular velocity
ω2 1 rad sec
Input crank angular acceleration
α2 0 rad sec
1
CW
2
See Figure P6-33 and Mathcad file P0761b.
Draw the linkage to scale and label it.
Y A 3 B
2 4
5 X
O2
y 6
O4 x
2.
C
102°
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. Transform crank angle to the local xy coordinate system:
θ2 θ2XY K1
θ2 212.000 deg
d
K2
a
K1 0.4608 2
K3
d c
K2 0.4329 2
2
a b c d
2
2 a c
K3 0.6755
A cos θ2 K1 K2 cos θ2 K3 C K1 K2 1 cos θ2 K3 A 0.2662
B 1.0598
C 2.3515
B 2 sin θ2
DESIGN OF MACHINERY - 5th Ed.
3.
SOLUTION MANUAL 7-61b-2
Use equation 4.10b to find values of 4 for the open circuit.
2
θ4xy 2 atan2 2 A B 4.
B 4 A C
θ4xy 200.365 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
2 a b
K5 0.5178
D cos θ2 K1 K4 cos θ2 K5
5.
7.
D 2.2370
E 2 sin θ2
E 1.0598
F K1 K4 1 cos θ2 K5
F 0.3808
Use equation 4.13 to find values of 3 for the open circuit.
2
θ3xy 2 atan2 2 D E 6.
K4 0.4838
E 4 D F
θ3xy 289.050 deg
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18.
ω3
a ω2 sin θ4xy θ2 b sin θ3xy θ4xy
ω3 0.832
rad
ω4
a ω2 sin θ2 θ3xy c sin θ4xy θ3xy
ω4 0.591
rad
sec
sec
Use equations 7.12 to determine the angular accelerations of links 3 and 4. A c sin θ4xy
B b sin θ3xy
D c cos θ4xy
E b cos θ3xy
A 0.804 in
B 1.954 in
D 2.166 in
E 0.675 in
C a α2 sin θ2 a ω2 cos θ2 b ω3 cos θ3xy c ω4 cos θ4xy 2
C 0.618
2
2
in sec
2
F a α2 cos θ2 a ω2 sin θ2 b ω3 sin θ3xy c ω4 sin θ4xy 2
F 0.079
8.
2
C D A F A E B D
α3 0.267
rad sec
2
Transform 4 back to the global XY system.
θ4 θ4xy 9.
Determine 5 using equation 4.17. Offset:
2
in sec
α3
2
cc 0 in
θ4 302.365 deg
α4
C E B F A E B D
α4 0.120
rad sec
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-61b-3
c sin θ4 cc π e
θ5 asin
θ5 158.818 deg
10. Determine the angular velocity of link 5 using equation 6.22a:
ω5
c cos θ4 ω4 e cos θ5
ω5 0.145
rad sec
11. Determine the angular acceleration of link 5 using equation 7.16d. c α4 cos θ4 c ω4 sin θ4 e ω5 sin θ5 2
α5
2
α5 0.156
e cos θ5
sec
12. Use equation 7.16e for the acceleration of pin C. AC c α4 sin θ4 c ω4 cos θ4 e α5 sin θ5 e ω5 cos θ5 2
AC 0.00161
in sec
2
rad
2
A positive sign means that AC is to the right.
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-62-1
PROBLEM 7-62 Statement:
Given:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the angular acceleration of link 4 and the linear acceleration of slider 6 in the sixbar crank-slider linkage of Figure 3-33 as a function of the angle of input link 2 for a constant 2 = 1 rad/sec CW. Plot Ac both as a function of 2 and separately as a function of slider position as shown in the figure. Link lengths: Link 2 (O2 to A)
a 2.170 in
Link 3 (A to B)
b 2.067 in
Link 4 (O4 to B)
c 2.310 in
Link 1 (O2 to O4)
d 1.000 in
Link 5 (B to C)
e 5.400 in 102 deg
Coordinate angle
Solution:
θ2XY 0 deg 1 deg 360 deg
Crank angle:
Input crank angular velocity
ω2 1 rad sec
Input crank angular acceleration
α2 0 rad sec
1
CW
2
See Figure 3-33 and Mathcad file P0762.
1.
Transform crank angle to the local xy coordinate system: θ2 θ2XY θ2XY
2.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K1 0.4608
a 2
K3
2
2
a b c d
K2
d
K2 0.4329
c
2
K3 0.6755
2 a c
A θ2XY cos θ2 θ2XY K1 K2 cos θ2 θ2XY K3 B θ2XY 2 sin θ2 θ2XY C θ2XY K1 K2 1 cos θ2 θ2XY K3 3.
Use equation 4.10b to find values of 4 for the open circuit.
θ4xy θ2XY 2 atan2 2 A θ2XY B θ2XY 4.
B θ2XY 4 A θ2XY C θ2XY 2
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
K4 0.4838
2 a b
K5 0.5178
D θ2XY cos θ2 θ2XY K1 K4 cos θ2 θ2XY K5 E θ2XY 2 sin θ2 θ2XY F θ2XY K1 K4 1 cos θ2 θ2XY K5 5.
Use equation 4.13 to find values of 3 for the open circuit.
θ3xy θ2XY 2 atan2 2 D θ2XY E θ2XY 6.
E θ2XY 4 D θ2XY F θ2XY 2
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18.
DESIGN OF MACHINERY - 5th Ed.
7.
SOLUTION MANUAL 7-62-2
ω3 θ2XY
a ω2 sin θ4xy θ2XY θ2 θ2XY b sin θ3xy θ2XY θ4xy θ2XY
ω4 θ2XY
a ω2 sin θ2 θ2XY θ3xy θ2XY c sin θ4xy θ2XY θ3xy θ2XY
Use equations 7.12 to determine the angular acceleration of link 4. A θ2XY c sin θ4xy θ2XY
B θ2XY b sin θ3xy θ2XY
D θ2XY c cos θ4xy θ2XY
E θ2XY b cos θ3xy θ2XY
C θ2XY a α2 sin θ2 θ2XY a ω2 cos θ2 θ2XY 2
b ω3 θ2XY cos θ3xy θ2XY c ω4 θ2XY cos θ4xy θ2XY 2
2
F θ2XY a α2 cos θ2 θ2XY a ω2 sin θ2 θ2XY 2
b ω3 θ2XY sin θ3xy θ2XY c ω4 θ2XY sin θ4xy θ2XY 2
α4 θ2XY
2
C θ2XY E θ2XY B θ2XY F θ2XY A θ2XY E θ2XY B θ2XY D θ2XY ANGULAR ACCELERATION - LINK 4
Angular Acceleration, rad/s
2
1
0
1
0
45
90
135
180
225
Crank Angle, deg
8.
Transform 4 back to the global XY system.
θ4 θ2XY θ4xy θ2XY
9.
Determine 5 using equation 4.17. Offset: cc 0 in
c sin θ4 θ2XY cc π e
θ5 θ2XY asin
10. Determine the angular velocity of link 5 using equation 6.22a:
ω5 θ2XY
c cos θ4 θ2XY ω4 θ2XY e cos θ5 θ2XY
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-62-3
11. Determine the angular acceleration of link 5 using equation 7.16d. c α4 θ2XY cos θ4 θ2XY c ω4 θ2XY sin θ4 θ2XY 2
α5 θ2XY
e ω5 θ2XY sin θ5 θ2XY 2
e cos θ5 θ2XY
12. Use equation 7.16e for the acceleration of pin C. AC θ2XY c α4 θ2XY sin θ4 θ2XY c ω4 θ2XY cos θ4 θ2XY 2
e α5 θ2XY sin θ5 θ2XY e ω5 θ2XY cos θ5 θ2XY 2
ACCELERATION - PIN C 6
Acceleration, in/sec^2
4
2
0
2
4
0
45
90
135
180
Crank Angle, deg
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-63a-1
PROBLEM 7-63a Statement:
Find the angular acceleration of link 6 in Figure 3-34b for 6 = 90 deg with respect to the global
Given:
Link lengths:
X axis assuming constant 2 = 10 rad/sec CW. Use a graphical method. Link 2 (O2 to A)
g 1.556 in
Link 3 (A to B)
f 4.248 in
Link 4 (O4 to C)
c 2.125 in
Link 5 (C to D)
b 2.158 in
Link 6 (O6 to D)
a 1.542 in
Link 5 (B to D)
p 3.274 in
Link 1 X-offset
d X 3.259 in
Link 1 Y-offset
d Y 2.905 in
Angle CDB
36.0 deg
Output rocker angle:
θ 90 deg Global XY system ω 10 rad sec
Input crank angular velocity Solution: 1.
1
CW
See Figure 3-34b and Mathcad file P0763a.
Draw the linkage to a convenient scale. Indicate the directions of the acceleration vectors of interest.
Direction of ACDt Y
Direction of ADt
Direction of AAt X 2 A
D
O2
5
6
C
y
Direction of AABt
O6 3
5
4 x
Direction of ACt
O4
B Direction of ABDt
2.
Since this linkage is a Stephenson's II sixbar, we will have to start at link 6 and work back to link 2. We will assume a value for 6 and eventually find a value for 2. We will then multiply the magnitudes of all velocities by the ratio of the actual 2 to the found 2. Use Problem 6-73 for the link angular velocities. Assume:
100 rad sec
2
CCW
From Problem 6-73 and the graphical layout (angles in the global XY coordinate system): 24.124 rad sec 14.64 rad sec 14.64 rad sec 3.016 rad sec
1
1 1 1
CW
90.000 deg
CW
217.003 deg
CW
144.195 deg
CW
144.628 deg
253.003 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-63a-2
10.000 rad sec
1
221.690 deg
CW
(APt + APn) = (AAt + AAn) + (APAt + APAn)
3.
The graphical solution for accelerations uses equation 7.4:
4.
For point C, this becomes: (ACt + ACn) = (ADt + ADn) + (ACDt + ACDn) , where 2
ACn c
ACn 455.450 in sec
θACn 180 deg
θACn 324.195 deg
ADn a
2
ADn 897.394 in sec
θADn 180 deg
θADn 270.000 deg
ADt a
ADt 154.200 in sec
θADt 90 deg
θADt 180.000 deg
2
5.
2
2
2
ACDn b
ACDn 462.523 in sec
θACDn 180 deg
θACDn 37.003 deg
2
Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ACn at an angle of ACn and ADn at an angle of ADn. From the tip of ADn, draw ADt at an angle of ADt + . From the tip of ADt, draw ACDn at an angle of CDn. Now that the vectors with known magnitudes are drawn, from the tips of ACn and ACDn, draw construction lines in the directions of ACt and ACDt, respectively. The intersection of these two lines are the tips of ACt, ACt, and ACDt.
n
AC
0
200 IN/S/S
Acceleration Scale
AC
t
AC 1.755
t
ACD AD n
ACD n
AD t
AD 0.425
6.
From the graphical solution above, Acceleration scale factor
ka 200
in sec
ACt 1.755 ka
2
ACt 351.00 in sec
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-63a-3
ACt
165.176 rad sec
c
ACDt 0.425 ka ACDt
7.
ACDt 85.00 in sec
2
CW
For point B, equation 7.4 becomes: AB = AD + (ABDt + ABDn) , where 2
8.
CW
2
39.388 rad sec
b
2
ABDn p
ABDn 701.715 in sec
θABDn 180 deg
θABDn 73.003 deg
ABDt p
ABDt 128.957 in sec
θABDt θABDn 90 deg
θABDt 163.003 deg
2
1.010
2
AB
0
200 IN/S/S
Acceleration Scale
For point A, equation 7.4 becomes: (AAt + AAn) = AB + (AABt + AABn) , where 2
110.996°
t ABD
AAn g
AAn 155.600 in sec
θAAn 180 deg
θAAn 41.690 deg
2
AD
AABn f
2
AABn 38.641 in sec
θAABn 180 deg
2 n
ABD
θAABn 35.372 deg
0.153
t
AA
n
AA AA
2.021
AB
0 t AAB
100 IN/S/S
Acceleration Scale
n
AAB
9.
The requirement for this problem is that 2 = 0 (constant 2). For this to be so AAt must be zero. To do this the magnitude of AB must be smaller as shown above. Correction ratio:
r
Corrected value of 6:
0.153
r 0.076
2.021 r
7.571
rad sec
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-63b-1
PROBLEM 7-63b Statement:
Find the angular acceleration of link 6 in Figure 3-34b for 6 = 90 deg with respect to the global X axis assuming constant 2 = 10 rad/sec CW. Use an analytic method.
Given:
Solution: 1.
Link lengths: Link 2 (O2 to A)
g 1.556 in
Link 3 (A to B)
f 4.248 in
Link 4 (O4 to C)
c 2.125 in
Link 5 (C to D)
b 2.158 in
Link 6 (O6 to D)
a 1.542 in
Link 5 (B to D)
p 3.274 in
Link 1 X-offset
d X 3.259 in
Link 1 Y-offset
d Y 2.905 in
Angle CDB
δ5 36.0 deg
Output rocker angle:
θ6 90 deg
Global XY system
Input crank angular velocity
ω2 10 rad sec
Input crank angular acceleration
α2 0.0 rad sec
1
CW
2
See Figure P6-34b and Mathcad file P0763b.
Since this linkage is a Stephenson's II sixbar an iterative solution must be used. This problem is suited to a longer-term project rather than a daily homework problem.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-64-1
PROBLEM 7-64 Statement:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the angular acceleration of link 6 in Figure 3-34b as a function of 2 for a constant 2 = 10 rad/sec CW.
Given:
Solution: 1.
Link lengths: Link 2 (O2 to A)
g 1.556 in
Link 3 (A to B)
f 4.248 in
Link 4 (O4 to C)
c 2.125 in
Link 5 (C to D)
b 2.158 in
Link 6 (O6 to D)
a 1.542 in
Link 5 (B to D)
p 3.274 in
Link 1 X-offset
d X 3.259 in
Link 1 Y-offset
d Y 2.905 in
Angle CDB
δ5 36.0 deg
Output rocker angle:
θ6 90 deg
Global XY system
Input crank angular velocity
ω2 10 rad sec
Input crank angular acceleration
α2 0.0 rad sec
1
CW
2
See Figure P6-34b and Mathcad file P0764.
Since this linkage is a Stephenson's II sixbar an iterative solution must be used. This problem is suited to a longer-term project rather than a daily homework problem.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-65a-1
PROBLEM 7-65a Statement:
Use a compass and straightedge to draw the linkage in Figure 3-35 with link 2 at 90 deg and find the angular acceleration of link 6 assuming a constant 2 = 10 rad/sec CCW. Use a graphical method.
Given:
Link lengths: Link 2 (O2 to A)
a 1.000 in
Link 3 (A to B)
b 3.800 in
Link 4 (O4 to B)
c 1.286 in
Link 1 (O2 to O4)
d 3.857 in
Link 4 (O4 to D)
e 1.429 in
Link 5 (D to E)
f 1.286 in
Link 6 (O6 to E)
g 0.771 in
Link 1 (O4 to O6)
h 0.786 in
90 deg
Angle BO4D
157 deg
Crank angle:
Solution: 1.
Input crank angular velocity
10 rad sec
Input crank angular acceleration
0 rad sec
1
CCW
2
See Figure 3-35 and Mathcad file P0765a.
Draw the linkage to a convenient scale. Indicate the directions of the acceleration vectors of interest. Direction of AAt Direction of ADt
Y A
D
122.085° 33.359°
2
100.938° 5
3 4 O2
X
6 O4 O6 E
35.228°
B Direction of AEt Direction of ABAt
Direction of AEDt
Direction of ABt
2.
In order to solve for the accelerations at points A and B, we will need 3, 3, and 4. From the graphical position solution above, θ 33.359 deg
θ 33.359 deg
θ 122.085 deg
θ 122.085 deg
θ 122.085 deg
θ 34.915 deg
Using equation (6.18),
ω 1.398 rad sec
ω 6.496 rad sec
ω
a sin θ b sin θ θ
ω
a sin θ c sin θ θ
1
1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-65a-2
(APt + APn) = (AAt + AAn) + (APAt + APAn)
3.
The graphical solution for accelerations uses equation 7.4:
4.
For point B, this becomes: (ABt + ABn) = (AAt + AAn) + (ABAt + ABAn) , where ABn c ω
2
ABn 54.275 in sec
θABn θ 180 deg
θABn 57.915 deg
2
AAn a
AAn 100.000 in sec
θAAn 180 deg
θAAn 270.000 deg
AAt a
AAt 0.000 in sec
θAAt 90 deg
θAAt 0.000 deg
2
5.
2
2
2
2
ABAn b ω
ABAn 7.429 in sec
θABAn θ 180 deg
θABAn 146.641 deg
Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ABn at an angle of ABn and AAn at an angle of AAn. From the tip of AAn, draw AAt at an angle of AAt + . From the tip of AAt, draw ABAn at an angle of BAn. Now that the vectors with known magnitudes are drawn, from the tips of ABn and ABAn, draw construction lines in the directions of ABt and ABAt,
6.
respectively. The intersection of these two lines are the tips of AB, ABt, and ABAt. From the graphical solution at right, in
ka 25
Acceleration scale factor
sec AB 2.919 ka
AB 73.0 in sec
1.951
2 n
AB
2
ABt AB
at an angle of 15.969 deg ABt 1.951 ka 6.
ABt 48.8 in sec
ABt
2
37.928 rad sec
c
15.969° t ABA
2
CCW 2.919
From the graphical layout, 100.938 deg 180 deg
79.062 deg
35.228 deg
35.228 deg
0
AA n
AA
Acceleration Scale
Using equation (6.18), ω
ω
e ω sin θ f sin
e ω sin θ g sin
25 IN/S/S
n
ABA
ω 9.804 rad sec
1
ω 15.885 rad sec
1
DESIGN OF MACHINERY - 5th Ed.
7.
SOLUTION MANUAL 7-65a-3
For point E, this becomes: (AEt + AEn) = (ADt + ADn) + (AEDt + AEDn) , where 2
AEn g ω
AEn 194.560 in sec
θAEn 180 deg
θAEn 144.772 deg
2
ADn e ω
ADn 60.310 in sec
θADn θ 180 deg
θADn 214.915 deg
ADt e
ADt 54.199 in sec
θADt θ 90 deg
θADt 124.915 deg
2
θAEDn 180 deg
θAEDn 100.938 deg
AE
AEt
n
AED
n
AE
0
50 IN/S/S
Acceleration Scale
AD ADt n
AD
From the graphical solution above, Acceleration scale factor
in
ka 50
sec AEt 0.239 ka AEt g
2
AEDn 123.597 in sec
t AED
2
AEDn f ω
0.239
8.
2
AEt 11.950 in sec
2
2
15.499 rad sec
2
CCW
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-65b-1
PROBLEM 7-65b Statement:
Use a compass and straightedge to draw the linkage in Figure 3-35 with link 2 at 90 deg and find the angular acceleration of link 6 assuming a constant 2 = 10 rad/sec CCW. Use an analytical method.
Given:
Link lengths: Link 2 (O2 to A)
a 1.000 in
Link 3 (A to B)
b 3.800 in
Link 4 (O4 to B)
c 1.286 in
Link 1 (O2 to O4)
d 3.857 in
Link 4 (O4 to D)
e 1.429 in
Link 5 (D to E)
f 1.286 in
Link 6 (O6 to E)
g 0.771 in
Link 1 (O4 to O6)
h 0.786 in
Angle BO4D
δ4 157 deg
θ2 90 deg
Crank angle:
Solution: 1.
Input crank angular velocity
ω2 10 rad sec
Input crank angular acceleration
α2 0 rad sec
1
CCW
2
See Figure P6-35 and Mathcad file P0765b.
Draw the linkage to scale and label it.
Y A
D
122.085° 2
3
33.359°
100.938° 5
4 O2
X
6 O4 O6 E
35.228°
B 2.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 3.8570 2
K3
d c
K2 2.9992 2
2
a b c d
2
K3 1.2015
2 a c
A cos θ2 K1 K2 cos θ2 K3 B 2 sin θ2 C K1 K2 1 cos θ2 K3 A 2.6555 3.
B 2.0000
Use equation 4.10b to find values of 4 for the crossed circuit.
θ41 2 atan2 2 A B 4.
C 5.0585
2
B 4 A C
θ41 237.915 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12.
DESIGN OF MACHINERY - 5th Ed.
K4
5.
SOLUTION MANUAL 7-65b-2
2
d
K5
b
7.
2
2
K4 1.0150
2 a b
D cos θ2 K1 K4 cos θ2 K5
D 7.6284
E 2 sin θ2
E 2.0000
F K1 K4 1 cos θ2 K5
F 0.0856
2
E 4 D F
θ3 326.641 deg
Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18.
ω3
a ω2 sin θ41 θ2 b sin θ3 θ41
ω3 1.398
ω4
a ω2 sin θ2 θ3 c sin θ41 θ3
ω4 6.496
rad sec rad sec
Link 4 will be the input to the second fourbar (links 4, 5, 6, and 1). Let the angle that link 4 makes in the second fourbar be
θ42 θ41 157 deg 360 deg 8.
K5 3.7714
Use equation 4.13 to find values of 3 for the crossed circuit.
θ3 2 atan2 2 D E 6.
2
c d a b
θ42 34.915 deg
Determine the values of the constants needed for finding 6 from equations 4.8a and 4.10a. K1
h
K2
e
K1 0.5500 2
K3
e f
h g
K2 1.0195 2
2
g h
2
K3 0.7263
2 e g
A cos θ42 K1 K2 cos θ42 K3 B 2 sin θ42 C K1 K2 1 cos θ42 K3 A 0.1603 9.
B 1.1447
C 0.3796
Use equation 4.10b to find values of 6 for the open circuit.
θ6 2 atan2 2 A B
2
B 4 A C
θ6 35.228 deg
10. Determine the values of the constants needed for finding 5 from equations 4.11b and 4.12. K4
h f
2
K5
2
2 e f
D cos θ42 K1 K4 cos θ42 K5 E 2 sin θ42
2
g h e f
2
K4 0.6112 K5 1.0119 D 0.2408 E 1.1447
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-65b-3
F K1 K4 1 cos θ42 K5
F 0.7807
11. Use equation 4.13 to find values of 5 for the open circuit.
2
θ5 2 atan2 2 D E
E 4 D F
θ5 280.938 deg
12. Determine the angular velocity of links 5 and 6 for the open circuit using equations 6.18.
e ω4 sin θ6 θ42 f sin θ5 θ6
ω5 9.804
e ω4 sin θ42 θ5 g sin θ6 θ5
ω6 15.885
ω5 ω6
rad sec rad sec
13. Use equations 7.12 to determine the angular accelerations of links 3 and 4 for the crossed circuit. A c sin θ41
B b sin θ3
D c cos θ41
E b cos θ3
A 1.090 in
B 2.090 in
D 0.683 in
E 3.174 in
C a α2 sin θ2 a ω2 cos θ2 b ω3 cos θ3 c ω4 cos θ41 2
2
2
in
C 35.034
sec
2
F a α2 cos θ2 a ω2 sin θ2 b ω3 sin θ3 c ω4 sin θ41 2
2
in
F 141.900
sec
α3
2
2
C D A F
α3 36.545
A E B D
rad sec
2
α4
C E B F A E B D
α4 37.931
rad sec
14. Use equations 7.12 to determine the angular accelerations of links 5 and 6 for the open circuit. Use link 4 as the input to the second fourbar stage. A g sin θ6
B f sin θ5
D g cos θ6
E f cos θ5
A 0.037 ft
B 0.105 ft
D 0.052 ft
E 0.020 ft
C e α4 sin θ42 e ω4 cos θ42 f ω5 cos θ5 g ω6 cos θ6 2
2
2
ft
C 4.583
2.000
s
F e α4 cos θ42 e ω4 sin θ42 f ω5 sin θ5 g ω6 sin θ6 2
F 1.588
2
2
ft 2.000
s
α5
C D A F A E B D
α5 38.104
rad sec
2
α6
C E B F A E B D
α6 15.486
rad sec
2
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-66-1
PROBLEM 7-66 Statement:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the angular acceleration of link 6 in the sixbar linkage of Figure 3-35 as a function of 2 for a constant 2 = 1 rad/sec CCW.
Given:
Link lengths: Link 2 (O2 to A)
a 1.000 in
Link 3 (A to B)
b 3.800 in
Link 4 (O4 to B)
c 1.286 in
Link 1 (O2 to O4)
d 3.857 in
Link 4 (O4 to D)
e 1.429 in
Link 5 (D to E)
f 1.286 in
Link 6 (O6 to E)
g 0.771 in
Link 1 (O4 to O6)
h 0.786 in
1 rad sec
Input crank angular velocity Solution:
1
CCW
0 rad sec
2
See Figure 3-35 and Mathcad file P0766. 0 deg 1 deg 360 deg
1.
Define the range of the input angle:
2.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 3.8570 2
2
a b c d
2
K3 1.2015
2 a c
c
K2 2.9992
2
K3
d
A cos K1 K2 cos K3
B 2 sin
C K1 K2 1 cos K3 3.
Use equation 4.10b to find values of 4 for the crossed circuit.
2 atan2 2 A B 4.
2 4 A C
B
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
2 a b
D cos K1 K4 cos K5
K4 1.0150
K5 3.7714
E 2 sin
F K1 K4 1 cos K5 5.
Use equation 4.13 to find values of 3 for the crossed circuit.
2 atan2 2 D E 6.
2 4 D F
E
Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18.
a
a
b
c
sin
sin
sin
sin
DESIGN OF MACHINERY - 5th Ed.
7.
SOLUTION MANUAL 7-66-2
Link 4 will be the input to the second fourbar (links 4, 5, 6, and 1). Let the angle that link 4 makes in the second fourbar be
157 deg 360 deg 8.
Determine the values of the constants needed for finding 6 from equations 4.8a and 4.10a. K1
h
K2
e
K1 0.5500 2
e f
2
g h
2
K3 0.7263
2 e g
g
K2 1.0195
2
K3
h
K1 K2 cos K3
A' cos
B' 2 sin
K3
C' K1 K2 1 cos 9.
Use equation 4.10b to find values of 6 for the open circuit.
2 4 A' C'
2 atan2 2 A' B'
B'
10. Determine the values of the constants needed for finding 5 from equations 4.11b and 4.12. K4
2
h
K5
f
2
2
g h e f
2
K4 0.6112
2 e f
K1 K4 cos K5
D' cos
E' 2 sin
K5
F' K1 K4 1 cos
11. Use equation 4.13 to find values of 5 for the open circuit.
2 atan2 2 D' E'
2 4 D' F'
E'
12. Determine the angular velocity of link 6 for the open circuit using equations 6.18.
e sin f sin
e sin g sin
13. Use equations 7.12 to determine the angular acceleration of link 4.
B b sin
E b cos
A c sin
D c cos
2 2 2 b cos c cos
C a sin a cos
K5 1.0119
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-66-3
2 2 2 b sin c sin
F a cos a sin
α4
A E B D C E B F
14. Use equations 7.12 to determine the angular accelerations of links 5 and 6 for the open circuit. Use link 4 as the input to the second fourbar stage.
B' f sin
E' f cos
A' g sin
D' g cos
e 2 cos 2 2 f cos g cos
e 2 sin 2 2 f sin g sin
C' e α4 sin
F' e α4 cos
α6
A' E' B' D' C' E' B' F'
ANGULAR ACCELERATION - LINK 6
Angular Acceleration, rad/sec^2
4
2
0
2
4
0
45
90
135
180
Crank Angle, deg
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-67-1
PROBLEM 7-67 Statement:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the angular acceleration of link 8 in the linkage of Figure 3-36 as a function of 2 for a constant 2 = 1 rad/sec CCW.
Given:
Link lengths: Input crank (L2)
a 0.450
First coupler (L3)
b 0.990
Common rocker (O4B)
c 0.590
First ground link (O2O4)
d 1.000
Common rocker (O4C)
a' 0.590
Second coupler (CD)
b' 0.325
Output rocker (L6)
c' 0.325
Second ground link (O4O6) d' 0.419
Link 7 (L7)
e 0.938
Link 8 (L8)
f 0.572
Link 5 extension (DE)
p 0.823
Angle DCE
δ 7.0 deg
Angle BO4C
α 128.6 deg
Input crank angular velocity Solution: 1.
1 rad sec
1
CCW
See Figure 3-36 and Mathcad file P0767.
See problem 4-43 for the position solution. The velocity and acceleration solutions will use the same vector loop equations for links 5, 6, 7, and 8, differentiated with respect to time. This problem is suitable for a project assignment and is probably too long for an overnight assignment.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-68-1
PROBLEM 7-68 Statement:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the magnitude and direction of the acceleration of point P in Figure 3-37a as a function of 2 for a constant 2 = 1 rad/sec CCW. Also calculate and plot the acceleration of point P versus point A.
Given:
Link lengths:
Solution: 1.
Link 2 (O2 to A)
a 0.136
Link 3 (A to B)
b 1.000
Link 4 (B to O4)
c 1.000
Link 1 (O2 to O4)
d 1.414
Coupler point:
Rpa 2.000
0 deg
Crank speed:
1 rad sec
1
0 rad sec
2
See Figure 3-37a and Mathcad file P0768.
Determine the range of motion for this Grashof crank rocker. θ 0 deg 1 deg 360 deg
2.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 10.3971 2
K3
c
K2 1.4140
2
2
a b c d
2
K3 7.4187
2 a c
d
A θ cos θ K1 K2 cos θ K3
B θ 2 sin θ
C θ K1 K2 1 cos θ K3 3.
Use equation 4.10b to find values of 4 for the open circuit.
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 4.
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d b
K5
2
2
c d a b
2
2 a b
D θ cos θ K1 K4 cos θ K5
K4 1.4140
K5 7.4187
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 5.
Use equation 4.13 to find values of 3 for the open circuit.
θ θ 2 atan2 2 D θ E θ 6.
2 4 Dθ F θ
E θ
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18.
a
a
ω θ
ω θ
b
c
sin θ θ θ θ
sin θ θ θ θ
sin θ θ θ
sin θ θ θ
DESIGN OF MACHINERY - 5th Ed.
7.
SOLUTION MANUAL 7-68-2
Using the Euler identity to expand equation 7.13a for AA. Determine the magnitude, and direction.
a 2 cosθ j sinθ
AA θ a sin θ j cos θ 8.
Use equations 7.12 to determine the angular accelerations of links 3 and 4.
B' θ b sin θ θ
E' θ b cos θ θ
A' θ c sin θ θ
D' θ c cos θ θ
2
2 c ωθ2 cosθθ
C' θ a sin θ a cos θ b ω θ cos θ θ
2 C' θ D' θ A' θ F' θ A' θ E' θ B' θ D' θ
2 c ωθ2 sinθθ C' θ E' θ B' θ F' θ α θ A' θ E' θ B' θ D' θ
F' θ a cos θ a sin θ b ω θ sin θ θ
α θ
Use equations 7.32 to find the acceleration of the point P.
2 Rpa ω θ cos θ θ j sin θ θ AP θ AA θ APA θ APA θ Rpa α θ sin θ θ j cos θ θ
θAp θ arg AP θ
AP θ AP θ
ACCELERATION MAGNITUDE - POINTS P and A 2.2 2.0 1.8 1.6 1.4 1.2 1.0 0.0
45.0
90.0
135.0
180.0
225.0
Crank Angle, deg Point P Point A
θAA θ arg AA θ
270.0
AA θ AA θ
θAp θ if θ 91 deg θAp θ 2 π θAp θ
Acceleration, in/sec^2
9.
315.0
360.0
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-68-3
ACCELERATION VECTOR DIRECTION - POINTS P and A 360
Acceleration Vector Angle, deg
315 270 225 180 135 90 45 0 45 90 135 180
0
45
90
135
180
Crank Angle, deg Point P Point A
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-69-1
PROBLEM 7-69 Statement:
Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the magnitude and the direction of the acceleration of point P in Figure 3-37b as a function of 2. Also calculate and plot the velocity of point P versus point A.
Given:
Link lengths: Input crank (L2)
a 0.50
First coupler (AB)
b 1.00
Rocker 4 (O4B)
c 1.00
Rocker 5 (L5)
c' 1.00
Ground link (O2O4)
d 0.75
Second coupler 6 (CD)
b' 1.00
Coupler point (DP)
p 1.00
Distance to OP (O2OP)
d' 1.50
Crank speed: Solution: 1.
1 rad sec
1
0 rad sec
2
See Figure 3-37b and Mathcad file P07-69.
Links 4, 5, BC, and CD form a parallelogram whose opposite sides remain parallel throughout the motion of the fourbar 1, 2, AB, 4. Define a position vector whose tail is at point D and whose tip is at point P and another whose tail is at O4 and whose tip is at point D. Then, since R5 = RAB and RDP = -R4, the position vector from O2 to P is P = R1 + RAB - R4. Separating this vector equation into real and imaginary parts gives the equations for the X and Y coordinates of the coupler point P.
XP = d b cos θ c cos θ
YP = b sin θ c sin θ
2.
Define one revolution of the input crank: θ 0 deg 0.5 deg 360 deg
3.
Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the crossed circuit). K1
d a
K1 1.5000
2
d
K2
K3
c
K2 0.7500
2
2
a b c d
2
2 a c
K3 0.8125
A θ cos θ K1 K2 cos θ K3
C θ K1 K2 1 cos θ K3
B θ 2 sin θ
2 4 A θ Cθ 2 π
θ θ 2 atan2 2 A θ B θ 4.
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
2 a b
D θ cos θ K1 K4 cos θ K5
K4 0.7500
K5 0.8125
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 5.
Use equation 4.13 to find values of 3 for the crossed circuit.
θ θ 2 atan2 2 D θ E θ 6.
2 4 Dθ F θ 2 π
E θ
Define a local xy coordinate system with origin at OP and with the positive x axis to the right. The coordinates of P are transformed to xP = XP - d', yP = YP.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-69-2
c cosθθ d'
xP θ d b cos θ θ
c sinθθ
yP θ b sin θ θ 7.
Use equations 6.18 to calculate 3 and 4.
a
a
θ
θ 8.
b
c
sin θ θ θ θ
sin θ θ θ θ
sin θ θ θ
sin θ θ θ
Use equations 7.12 to determine the angular accelerations of links 3 and 4.
B' θ b sin θ θ
E' θ b cos θ θ
2
2 c θ2 cosθθ
2
2 c θ2 sinθθ
A' θ c sin θ θ
D' θ c cos θ θ
C' θ a sin θ a cos θ b θ cos θ θ F' θ a cos θ a sin θ b θ sin θ θ
α θ
α θ
A' θ E' θ B' θ D' θ C' θ E' θ B' θ F' θ
Differentiate the position equations twice with respect to time to get the acceleration components. APx θ b α θ sin θ θ
θ2 cosθθ 2 c α θ sin θ θ θ cos θ θ 2 APy θ b α θ cos θ θ θ sin θ θ 2 c α θ cos θ θ θ sin θ θ
AP θ
2 APyθ2
APx θ
VELOCITY MAGNITUDE of POINT P 0.6 0.4 Velocity, 1/sec
9.
A' θ E' θ B' θ D' θ C' θ D' θ A' θ F' θ
0.2 0 0.2 0.4 0.6
0
45
90
135
180
Crank Angle, deg x component y component
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-70a-1
PROBLEM 7-70a Statement:
Find the angular accelerations of links 3 and 4 and the linear accelerations of points A, B and P1 in the XY coordinate system for the linkage in Figure P7-27 in the position shown. Assume that 2 = 45 deg in the XY coordinate system and 2 = 10 rad/sec, constant. The coordinates of the point P1 on link 4 are (114.68, 33.19) with respect to the xy coordinate system. Use a graphical method.
Given:
Link lengths:
Solution: 1.
Link 2 (O2 to A)
a 14.00 in
Link 3 (A to B)
b 80.00 in
Link 4 (O4 to B)
c 51.26 in
Link 1 X-offset
d X 47.5 in
Link 1 Y-offset
d Y 76.00 in 12.00 in
Coupler point x-offset
p x 114.68 in
Coupler point y-offset p y 33.19 in
Crank angle:
45 deg
Coordinate rotation angle
α 126.582 deg
Input crank angular velocity
ω 10 rad sec
Global XY system to local xy system
1
α 0 rad sec
CCW
2
See Figure P7-27 and Mathcad file P0770a.
Draw the linkage to a convenient scale. Indicate the directions of the velocity acceleration of interest. Direction of ABt Direction of ABAt B
x
4
P1 O4
29.063° Y
3
126.582°
99.055° Direction of AP1t A
45.000°
2
X O2
45 deg α 81.582 deg 2.
y
In order to solve for the accelerations at points A and B, we will need 3, 3, and 4. From the graphical position solution above (in the local xy coordinate system), θ 99.055 deg α
θ 27.527 deg
θ 29.063 deg α
θ 97.519 deg
Using equation (6.18),
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-70a-2
ω
a ω sin θ b sin θ θ
ω 0.511 rad sec
ω
a ω sin θ c sin θ θ
ω 2.353 rad sec
The graphical solution for accelerations uses equation 7.4:
4.
For point B, this becomes: (ABt + ABn) = (AAt + AAn) + (ABAt + ABAn) , where 2
ABn 283.838 in sec
θABn θ 180 deg
2
θABn 277.519 deg
2
AAn a ω
AAn 1400.0 in sec
θAAn 180 deg
θAAn 98.418 deg
AAt a α
AAt 0.0 in sec
θAAt 90 deg
θAAt 8.418 deg
2
5.
1
(APt + APn) = (AAt + AAn) + (APAt + APAn)
3.
ABn c ω
1
2
2
2
ABAn b ω
ABAn 20.921 in sec
θABAn θ 180 deg
θABAn 207.527 deg
Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ABn at an angle of ABn + and AAn at an angle of AAn + . From the tip of AAn, draw AAt at an angle of AAt + . From the tip of AAt, draw ABAn at an angle of BAn + . Now that the vectors with known magnitudes are drawn, from the tips of ABn and ABAn, draw construction lines in the directions of ABt and ABAt, respectively. The intersection of these two lines are the tips of ABt, ABt, and ABAt. x
0
Y
n
X
AB
20 in/s/s y
80.620°
Acceleration Scale 39.675 ABt
42.136
AB
AA t ABA n
ABA
6.
From the graphical solution above, Acceleration scale factor
ka 20.0
in sec
AA AAn
2
AA 1400 in sec
2
at an angle of -135.00 deg
DESIGN OF MACHINERY - 5th Ed.
4.
2
AB 42.136 ka
AB 842.7 in sec
ABt 39.675 ka
ABt 793.500 in sec
7.
SOLUTION MANUAL 7-70a-3
ABt
15.480 rad sec
c
at an angle of -80.62 deg 2
at an angle of -60.937 deg
2
Determine the distance from O4 to P1 and the angle O4P1makes with the x axis. 2
2
Distance O2O4:
d
dX dY
Distance P1O4:
u
px d2 py2
Angle O4P1:
atan
d 79.701 in u 48.219 in
py px d
43.497 deg
Determine the acceleration at point P1. AP1t u
AP1t 746.431 in sec
AP1n u ω
2
AP1
2
AP1t AP1n
AP1n 267.001 in sec 2
AP1 792.748 in sec
AP1t 9.921 deg 60.397 deg AP1n
atan
2 2
2
at an angle of 80.079 deg at an angle of -9.921 deg
at an angle of 60.397 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-70b-1
PROBLEM 7-70b Statement:
Find the angular accelerations of links 3 and 4 and the linear accelerations of points A, B and P1 in the XY coordinate system for the linkage in Figure P7-27 in the position shown. Assume that 2 = 45 deg in the XY coordinate system and 2 = 10 rad/sec, constant. The coordinates of the point P1 on link 4 are (114.68, 33.19) with respect to the xy coordinate system. Use an analytical method.
Given:
Link lengths:
Solution: 1.
Link 2 (O2 to A)
a 14.00 in
Link 3 (A to B)
b 80.00 in
Link 4 (O4 to B)
c 51.26 in
Link 1 X-offset
d X 47.5 in
Link 1 Y-offset
d Y 76.00 in 12.00 in
Coupler point x-offset
p x 114.68 in
Coupler point y-offset p y 33.19 in
Crank angle:
θ2XY 45 deg
Coordinate rotation angle
α 126.582 deg
Input crank angular velocity
ω 10 rad sec
Global XY system to local xy system
1
CCW
α 0 rad sec
2
See Figure P7-27 and Mathcad file P0770b.
Draw the linkage to scale and label it. B
x
4
P1 O4
29.063° Y
3
126.582°
99.055°
A
θ2XY α 81.582 deg d
2
dX dY
X O2
2 y
d 79.701 in 2.
45.000°
2
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 5.6929 2
K3
d c
K2 1.5548 2
2
a b c d
2
2 a c
K3 1.9340
A cos K1 K2 cos K3
C K1 K2 1 cos K3
B 2 sin
DESIGN OF MACHINERY - 5th Ed.
A 3.8401 3.
SOLUTION MANUAL 7-70b-2
B 1.9785
Use equation 4.10b to find values of 4 for the crossed circuit.
2
θ 2 atan2 2 A B 4.
C 7.2529
B 4 A C
θ 262.482 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d b
2
2
c d a b
K5
2
K4 0.9963
2 a b
D cos K1 K4 cos K5
D 10.0081
E 2 sin
E 1.9785
F K1 K4 1 cos K5 5.
7.
F 1.0849
Use equation 4.13 to find values of 3 for the crossed circuit.
2
θ 2 atan2 2 D E 6.
K5 4.6074
E 4 D F
θ 332.475 deg
Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18. ω
a ω sin θ b sin θ θ
ω 0.511
ω
a ω sin θ c sin θ θ
ω 2.353
rad sec
rad sec
Using the Euler identity to expand equation 7.13a for AA. Determine the magnitude, and direction (in the global coordinate system).
a ω2 cos j sin
AA a α sin j cos AA ( 205 1385i)
in
The acceleration of pin A is
AA 1400
in sec
8.
θAA arg AA α
AA AA
2
sec
at 2
θAA 225 deg
Use equations 7.12 to determine the angular accelerations of links 3 and 4 for the crossed circuit.
A c sin θ
B b sin θ 3
A 1.291 10 mm B 939.044 mm
D c cos θ
E b cos θ
D 170.343 mm
E 1.802 10 mm
2
2
2
2
2
2
C a α sin a ω cos b ω cos θ c ω cos θ C 260.638 in sec
2
F a α cos a ω sin b ω sin θ c ω sin θ 3
F 1.113 10 in sec
2
3
DESIGN OF MACHINERY - 5th Ed.
α
9.
SOLUTION MANUAL 7-70b-3
C D A F
α 14.227
A E B D
rad sec
α
2
C E B F A E B D
α 15.479
sec
Use equation 7.13c to determine the acceleration of point B for the crossed circuit.
c ω2 cosθ j sinθ
AB c α sin θ j cos θ AB ( 749 385i)
in sec
θAB arg AB α
AB AB
2
The acceleration of pin B is
in
AB 842.7
sec
at 2
θAB 279.4 deg
(Global)
10. Use equation 7.31 to determine the acceleration of the point P1 on link 4. u
px d2 py2
u 48.219 in
py px d
141.014 deg
u ω2 cosθ j sinθ
360 deg θ atan
AP1 u α sin θ j cos θ AP1 ( 320 725i)
in sec
2
θAP1 arg AP1 α
AP1 AP1
The acceleration of point P1 is AP1 792.7
in sec
at 2
θAP1 60.39 deg
rad
(Global)
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-71-1
PROBLEM 7-71 Statement:
Find the angular accelerations of links 3 and 4 and the linear accelerations of points A, B and P1 in the XY coordinate system for the linkage in Figure P7-27 in the position shown. Assume that 2 = 45 deg in the XY coordinate system and 2 = 10 rad/sec, constant. The coordinates of the point P1 on link 4 are (114.68, 33.19) with respect to the xy coordinate system. Use an analytical method.
Given:
Link lengths:
Solution: 1.
Link 2 (O2 to A)
a 14.00 in
Link 4 (O2 to B)
c 51.26 in
Link 1 X-offset
d X 47.5 in
Link 1 (O2 to O2)
d
2
dX dY
2
Link 3 (A to B)
b 80.00 in
Link 1 Y-offset
d Y 76.00 in 12.00 in
d 79.701 in
Coupler point x-offset
p x 114.68 in
Coupler point y-offset p y 33.19 in
Crank angle:
0 deg 1 deg 360 deg
Coordinate rotation angle
α 126.582 deg
Input crank angular velocity
ω 10 rad sec
1
Global XY system to local xy system α 0 rad sec
CCW
See Figure P7-27 and Mathcad file P0771.
Draw the linkage to scale and label it.
B
x
4
P1 O4
Y
3
A 2
X O2
y
2.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 5.6929 2
K3
d c
K2 1.5548 2
2
a b c d 2 a c
2
K3 1.9340
A cos K1 K2 cos K3
B 2 sin
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-71-2
C K1 K2 1 cos K3 3.
Use equation 4.10b to find values of 4 for the crossed circuit.
θ 2 atan2 2 A B 4.
2 4 A C
B
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
K4 0.9963
2 a b
K5 4.6074
D cos K1 K4 cos K5
E 2 sin
F K1 K4 1 cos K5 5.
Use equation 4.13 to find values of 3 for the crossed circuit.
θ 2 atan2 2 D E 6.
Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18.
a ω
a ω
ω
ω
7.
2 4 D F
E
b
c
sin θ θ
sin θ θ
sin θ
sin θ
Use equations 7.12 to determine the angular acceleration of link 4.
D' c cos θ
E' b cos θ
A' c sin θ
B' b sin θ
2
2 c ω2 cosθ
C' a α sin a ω cos b ω cos θ
2 C' E' B' F' A' E' B' D'
2 c ω2 sinθ
F' a α cos a ω sin b ω sin θ
α 8.
Use equation 7.31 to determine and plot the acceleration of the point P1 on link 4. u
px d2 py2
u 48.219 in
141.067 deg
2 u ω cos θ j sin θ
AP1 u α sin θ j cos θ
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-71-3
θAP1 arg AP1 α
AP1 AP1
θAP1 if θAP1 100 deg θAP1 2 π θAP1 MAGNITUDE OF Ap 2000
Acceleration, in/sec^2
1500
1000
500
0
0
45
90
135
180
225
270
315
360
270
315
360
Crank Angle, deg
DIRECTION OF Ap (Global)
Direction Angle, deg
100
0
100
200
0
45
90
135
180
225
Crank Angle, deg
AP1 ( 81.582 deg) 792.706
in sec
2
θAP1( 81.582 deg) 60.447 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-72a-1
PROBLEM 7-72a Statement:
Find the angular accelerations of links 3 and 4 and the linear acceleration of point P in the XY coordinate system for the linkage in Figure P7-28 in the position shown. Assume that 2 = -94.121 deg in the XY coordinate system, 2 = 1 rad/sec, and 2 = 10 rad/sec2. The position of the coupler point P on link 3 with respect to point A is: p = 15.00, 3 = 0 deg. Use a graphical method.
Given:
Link lengths: Link 2 (O2 to A)
a 9.17 in
Link 3 (A to B)
b 12.97 in
Link 4 (O4 to B)
c 9.57 in
Link 1 X-offset
d X 2.79 in
Link 1 Y-offset
d Y 6.95 in
Coupler point data:
p 15.00 in
δ 0 deg
Crank angle:
θ 26.000 deg α 68.121 deg
Coordinate rotation angle
ω 1 rad sec
Input crank angular velocity Solution: 1.
Global XY system to local xy system 1
10 rad sec
CW
2
See Figure P7-28 and Mathcad file P0772a.
Draw the linkage to a convenient scale. Indicate the directions of the acceleration vectors of interest. Y y
O2
X Direction ABAt 2 1
68.121°
Direction APAt
26.000° O4 B
4
P
A
Direction AAt
72.224° 3 60.472°
Direction ABt
x
2.
In order to solve for the accelerations at points A and B, we will need 3, 3, and 4. From the graphical position solution above (in the local xy coordinate system), θ 72.224 deg
θ 72.224 deg
θ 60.472 deg
θ 60.472 deg
Using equation (6.18), ω
a ω sin θ θ b sin θ θ
ω 3.465 rad sec
ω
a ω sin θ θ c sin θ θ
ω 4.656 rad sec
1
1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-72a-2
(APt + APn) = (AAt + AAn) + (APAt + APAn)
3.
The graphical solution for accelerations uses equation 7.4:
4.
For point B, this becomes: (ABt + ABn) = (AAt + AAn) + (ABAt + ABAn) , where ABn c ω
2
ABn 207.476 in sec
θABn θ 180 deg α
θABn 172.351 deg
AAn a ω
2
AAn 9.170 in sec
θAAn θ 180 deg α
θAAn 85.879 deg
AAt a
AAt 91.700 in sec
θAAt θ 90 deg α
θAAt 4.121 deg
2
5.
2
2
2
ABAn b ω
ABAn 155.694 in sec
θABAn θ 180 deg α
θABAn 184.103 deg
2
Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ABn at an angle of ABn + and AAn at an angle of AAn + . From the tip of AAn, draw AAt at an angle of AAt + . From the tip of AAt, draw ABAn at an angle of BAn + . Now that the vectors with known magnitudes are drawn, from the tips of ABn and ABAn, draw construction lines in the directions of ABt and ABAt, respectively. The intersection of these two lines are the tips of ABt, ABt, and ABAt.
6.
From the graphical solution at right, Acceleration scale factor
ka 10.0
in sec
AA 0.922 ka
2
AA 9.220 in sec
2
at an angle of 1.590 deg AB 7.156 ka
AB 71.560 in sec
2
at an angle of 99.204 deg t
ABAt 7.169 ka 7.
ABAt
ABAt 71.690 in sec 5.527 rad sec
b
2
AB
7.156
t
ABA
7.169
2 AB
For point P, equation 7.4 becomes: AP = AA + (APAt + APAn) , where APAn p ω
2
APAn 180.062 in sec
99.204°
Y
2
y
n
AA
1.590° t
AA
n
AB
X
n
A BA
θAPAn θ δ 180 deg α
AA 0.922
θAPAn 184.103 deg APAt p
APAt 82.911 in sec
θAPAt θAPAn 90 deg
2
θAPAt 94.103 deg
x
0
20 IN/S/S
Acceleration Scale
DESIGN OF MACHINERY - 5th Ed.
8.
SOLUTION MANUAL 7-72a-3
Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. n
A PA
t
APA AP
157.174°
Y y
9.065
0
20 IN/S/S
X AA
Acceleration Scale x
9.
From the graphical solution above, Acceleration scale factor
ka 10.0
in sec
AP 9.065 ka
2
AP 90.650 in sec
2
at an angle of 157.174 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-72b-1
PROBLEM 7-72b Statement:
Find the angular accelerations of links 3 and 4 and the linear acceleration of point P in the XY coordinate system for the linkage in Figure P7-28 in the position shown. Assume that 2 = -94.121 deg in the XY coordinate system, 2 = 1 rad/sec, and 2 = 10 rad/sec2. The position
Given:
of the coupler point P on link 3 with respect to point A is: p = 15.00, 3 = 0 deg. Use an analytical method. Link lengths: Link 2 (O2 to A)
a 9.174 in
Link 4 (O4 to B)
c 9.573 in
Link 1 X-offset
d X 2.790 in
Link 1(O2 to O4)
d
2
dX dY
Rpa 15.00 in
Crank angle:
θ 26.000 deg
Link 1 Y-offset
d Y 6.948 in
d 7.487 in δ 0 deg
α 68.121 deg
Global XY system to local xy system
ω 1 rad sec
Input crank angular velocity
1.
b 12.971 in
2
Coupler point data:
Coordinate rotation angle
Solution:
Link 3 (A to B)
1
10 rad sec
CW
See Figure P7-28 and Mathcad file P0772b.
Draw the linkage to scale and label it. Y y
O2
X 2 68.121°
1 26.000°
O4 B
4
P
A 72.224° 3 60.472° x
2.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 0.8161 2
K3
c
K2 0.7821 2
2
a b c d
d
2
2 a c
K3 0.3622
A cos θ K1 K2 cos θ K3
C K1 K2 1 cos θ K3
B 2 sin θ
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-72b-2
A 0.2581 3.
B 0.8767
C 0.4234
Use equation 4.10b to find values of 4 for the crossed circuit.
2
θ 2 atan2 2 A B
B 4 A C
θ 299.512 deg
θ θ 360 deg 4.
θ 60.488 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
K4 0.5772
2 a b
D cos θ K1 K4 cos θ K5
D 0.3096
E 2 sin θ
E 0.8767
F K1 K4 1 cos θ K5 5.
F 0.4749
Use equation 4.13 to find values of 3 .
2
θ 2 atan2 2 D E
E 4 D F
θ 287.764 deg
θ θ 360 deg 6.
7.
K5 0.9111
θ 72.236 deg
Determine the angular velocity of links 3 and 4 using equations 6.18. ω
a ω sin θ θ b sin θ θ
ω 3.467
rad
ω
a ω sin θ θ c sin θ θ
ω 4.658
rad
sec
sec
Using the Euler identity to expand equation 7.13a for AA. Determine the magnitude, and direction.
a ω2 cosθ j sinθ
AA a sin θ j cos θ AA ( 32.0 86.5i )
in sec
The acceleration of pin A is
AA 92.198
in sec
8.
θAA arg AA α
AA AA
2
at 2
θAA 1.590 deg
(Global)
Use equations 7.12 to determine the angular accelerations of links 3 and 4.
A c sin θ
B b sin θ
D c cos θ
E b cos θ
A 8.331 in
B 12.353 in
D 4.716 in
E 3.957 in
2
2
2
2
2
2
C a sin θ a ω cos θ b ω cos θ c ω cos θ C 86.722 in sec
2
F a cos θ a ω sin θ b ω sin θ c ω sin θ F 118.753 in sec
2
DESIGN OF MACHINERY - 5th Ed.
α 9.
SOLUTION MANUAL 7-72b-3
C D A F
α 55.307
A E B D
rad sec
α
2
C E B F A E B D
α 71.596
Use equation 7.13c to determine the acceleration of point B.
c ω2 cosθ j sinθ
AB c α sin θ j cos θ AB ( 698.8 156.9i)
in sec
The acceleration of pin B is
θAB arg AB α
AB AB
2
AB 716.18
in sec
at 2
θAB 99.228 deg
10. Use equations 7.32 to find the acceleration of the point P.
Rpa ω cos θ δ j sin θ δ
APA Rpa α sin θ δ j cos θ δ 2
AP AA APA AP AP
AP 830
in sec
2
arg AP α 100.214 deg
rad sec
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-73-1
PROBLEM 7-73 Statement:
For the linkage in Figure P7-28, write a computer program or use an equation solver to calculate and plot the angular velocity and acceleration of links 3 and 4, and the magnitude and direction of the velocity and acceleration of point P as a function of 2 through its possible range of motion starting at the position shown. The position of the coupler point P on link 3 with respect to point A is: p = 15.00, 3 = 0 deg. Assume that, at t = 0, 2 = -94.121 deg in the XY coordinate system, 2 = 0, and 2 = 10 rad/sec2, constant.
Given:
Link lengths: Link 2 (O2 to A)
a 9.174 in
Link 4 (O4 to B)
c 9.573 in
Link 1 X-offset
d X 2.790 in
Link 1(O2 to O4)
d
2
dX dY
Link 1 Y-offset
d Y 6.948 in
d 7.487 in
Rpa 15.00 in
Crank angle:
θ 26.000 deg 25.9 deg 15 deg
δ 0 deg
α 68.121 deg
θ 26 deg
Global XY system to local xy system
ω 0 rad sec
Input crank angular velocity
1.
b 12.971 in
Coupler point data:
Coordinate rotation angle
Solution:
2
Link 3 (A to B)
1
10 rad sec
See Figure P7-28 and Mathcad file P0773.
Draw the linkage to scale and label it.
Y y
O2
X 2 1 26.000° O4 B
4
P
A
3
x
2.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 0.8161 2
K3
d c
K2 0.7821 2
2
a b c d 2 a c
2
K3 0.3622
A θ cos θ K1 K2 cos θ K3
C θ K1 K2 1 cos θ K3
B θ 2 sin θ
2
DESIGN OF MACHINERY - 5th Ed.
3.
SOLUTION MANUAL 7-73-2
Use equation 4.10b to find values of 4 for the crossed circuit.
2 4 A θ Cθ
θ θ 2 atan2 2 A θ B θ 4.
B θ
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
K4 0.5772
2 a b
K5 0.9111
D θ cos θ K1 K4 cos θ K5
E θ 2 sin θ
F θ K1 K4 1 cos θ K5 5.
Use equation 4.13 to find values of 3 .
θ θ 2 atan2 2 D θ E θ 6.
2 4 Dθ F θ
E θ
Determine the angular velocity of links 3 and 4 using equations 6.18.
ω θ
2 θ θ
a ω θ
a ω θ
ω θ
ω θ
b
c
sin θ θ θ θ
sin θ θ θ θ
sin θ θ θ
sin θ θ θ
2
50 40
1.5
ω θ
sec rad
ω θ
1
0.5
rad
20 10
0 30
7.
sec 30
25
20
15
0 26
10
22
θ
θ
deg
deg
Using the Euler identity to expand equation 7.13a for AA. Determine the magnitude, and direction.
a ωθ2 cosθ j sinθ
AA θ a sin θ j cos θ 8.
24
Use equations 7.12 to determine the angular accelerations of links 3 and 4.
B' θ b sin θ θ
E' θ b cos θ θ
A' θ c sin θ θ
D' θ c cos θ θ
20
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-73-3
2
2 c ωθ2 cosθθ
2 C' θ D' θ A' θ F' θ A' θ E' θ B' θ D' θ
2 c ωθ2 sinθθ C' θ E' θ B' θ F' θ α θ A' θ E' θ B' θ D' θ
C' θ a sin θ a ω θ cos θ b ω θ cos θ θ
F' θ a cos θ a ω θ sin θ b ω θ sin θ θ
α θ
14 150
12 2
sec
α θ
10
rad
2
sec
100
rad
8
50
6 30
9.
25
20
15
0 26
10
24
22
θ
θ
deg
deg
20
Use equations 7.32 to find the acceleration of the point P.
2 Rpa ω θ cos θ θ δ j sin θ θ δ AP θ AA θ APA θ APA θ Rpa α θ sin θ θ δ j cos θ θ δ
θAp θ arg AP θ α
AP θ AP θ
θAp θ if θAp θ 0 θAp θ 2 π θAp θ
4
MAGNITUDE
DIRECTION
1 10
3
8 10
AP θ
2
sec in
3
θAp θ
3
deg
6 10 4 10
150
100 3
2 10
0 26
24
22
20
50 26
24
22
θ
θ
deg
deg
20
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-74-1
PROBLEM 7-74 Statement:
Derive analytical expressions for the accelerations of points A and B in Figure P7-29 as a function of 3, 3, 3 and the length AB of link 3. Use a vector loop equation.
Solution:
See Figure P7-29 and Mathcad file P0774.
1.
Establish the global XY system such that the coordinates of the intersection of the slot centerlines is at (d X,d Y ). Then, define position vectors R1X, R1Y , R2, R3, and R4 as shown below.
Y 4 3 R3 R2
B
3
C 2
R4
A 1
R 1Y
X R 1X 2.
Write the vector loop equation: R1Y + R2 + R3 - R1X - R4 = 0 then substitute the complex number notation for each position vector. The equation then becomes:
π π j j θ 2 j ( 0) 3 j ( 0) 2 dY e a e c e dX e b e = 0 j
3.
Differentiate this equation with respect to time.
d a j c e dt 3.
j
j π 2 d b e = 0 dt
Substituting the Euler identity into this equation gives: Va jc cos θ3 j sin θ3 Vb j = 0
4.
Separate this equation into its real (x component) and imaginary (y component) parts, setting each equal to zero. Va c sin θ3 = 0
5.
Solve for the two unknowns Va and Vb in terms of the independent variables 3 and 3 Va = c sin θ3
6.
c cos θ3 Vb = 0 Vb = c cos θ3
Differentiate again with respect to time to get the accelerations. Aa = c cos θ3 2
Ab = c sin θ3 2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-75-1
PROBLEM 7-75 Statement:
The linkage in Figure P7-30a has link 2 at 120 deg in the global XY coordinate system. Find 6 and AD in the global XY coordinate system for the position shown if 2 = 10 rad/sec CCW and 2 = 50 rad/sec2 CW. Use the acceleration difference graphical method.
Given:
Link lengths: Link 2 (O2 to A)
a 6.20 in
Link 3 (A to B)
b 4.50 in
Link 4 (O4 to B)
c 3.00 in
Link 5 (C to D)
e 5.60 in
Link 3 (A to C)
p 2.25 in
Link 4 (O4 to D)
f 3.382 in
Link 1 X-offset
d X 7.80 in
Link 1 Y-offset
d Y 0.62 in
Angle ACB
0.0 deg
Angle BO4D
110.0 deg
Input rocker angle:
θ 120 deg
Global XY system
ω 10 rad sec
Input crank angular velocity Solution: 1.
1
50 rad sec
CCW
2
See Figure P7-30a and Mathcad file P0775.
Draw the linkage to a convenient scale. Indicate the directions of the acceleration vectors of interest. Direction of AAt Direction of ACAt Axis of slip
A Y
87.972° 113.057° 5
D
C
110°
2
3
Axis of transmission
120° 175.455°
411.709° B
x Direction of ADCt
O4
113.057 deg
X
O2 Direction of ABt
Direction of ABAt
11.709 deg y
Coordinate rotation angle: 175.455 deg 2.
In order to solve for the accelerations at points A and B, we will need 3, 3, and 4. From the graphical position solution above (in the local xy coordinate system),
θ2xy θ
θ2xy 295.455 deg
θ3xy
θ3xy 62.398 deg
θ4xy
θ4xy 187.164 deg
Using equation (6.18), ω
a ω sin θ4xy θ2xy b sin θ3xy θ4xy
ω 15.924 rad sec
1
DESIGN OF MACHINERY - 5th Ed.
ω
SOLUTION MANUAL 7-75-2
a ω sin θ2xy θ3xy c sin θ4xy θ3xy
ω 20.107 rad sec
(APt + APn) = (AAt + AAn) + (APAt + APAn)
3.
The graphical solution for accelerations uses equation 7.4:
4.
For point B, this becomes: (ABt + ABn) = (AAt + AAn) + (ABAt + ABAn) , where ABn c ω
2
ABn 1212.852 in sec
θABn 180 deg
1
2
θABn 191.709 deg
2
AAn a ω
AAn 620.000 in sec
θAAn θ 180 deg
θAAn 300.000 deg
AAt a
AAt 310.000 in sec
θAAt θ 90 deg
θAAt 30.000 deg
2
2
2
ABAn b ω
ABAn 1141.131 in sec
θABAn 180 deg
θABAn 66.943 deg
t ABA
2
5.727
t B
A
AB
4.158
Y n
A BA
131.968°
4.814 X
n
AB 0
AA
500 IN/S/S
A Acceleration Scale
1.386
n A
t A
A
33.435°
DESIGN OF MACHINERY - 5th Ed.
6.
SOLUTION MANUAL 7-75-3
From the graphical solution above, Acceleration scale factor
in
ka 500
sec
2
AA 1.386 ka
AA 693.0 in sec
AB 4.814 ka
AB 2407.0 in sec
ABAt 5.727 ka
ABAt 2863.5 in sec
ABAt b
ABt
ABt 2079.0 in sec
at an angle of 131.968 deg
2 2
CW
2
693.000 rad sec
c
at an angle of -33.435 deg
2
636.333 rad sec
ABt 4.158 ka
7.
2
2
CCW
For point C, equation 7.4 becomes: AC = AA + (ACAt + ACAn) , where 2
ACAn p ω
ACAn 570.566 in sec
θACAn 180 deg
θACAn 66.943 deg
ACAt p
ACAt 1431.8 in sec
θACAt θACAn 90 deg
θACAt 156.943 deg t
ACA
2
2
Y
AC 126.220° 0
500 IN/S/S
n
A CA
1.745
X
Acceleration Scale
AA 8.
From the graphical solution above, Acceleration scale factor
ka 500
in sec
AC 1.745 ka 9.
2
AC 872.50 in sec
2
at an angle of 126.220 deg
Determining the acceleration of point D requires the (graphical) solution of two equations simultaneously. They are: AD = AC + (ADCt + ADCn) and AD = ADt + ADn + ADcor + ADslip , where AC is known from the above analysis and From the velocity analysis (see Problem 6-91) ADCn e
2
35.13 rad sec
1
ADCn 6911 in sec
Vslip 136.20 in sec 2
1
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-75-4
θADCn 177.972 deg 180 deg
θADCn 357.972 deg
ADt f
ADt 2343.7 in sec
θADt 90 deg
θADt 211.709 deg
ADn f ω
2
ADn 1367.3 in sec
θADn θADt 90 deg
θADn 301.709 deg
ADcor 2 Vslip ω
ADcor 5477.1 in sec
θADcor θADt 180 deg
θADcor 31.709 deg
2
2
2
Y
n
A DC AC
X
ADt 33.485° A A
0
cor D
ADslip
n D
2000 IN/S/S
AD
3.700
t ADC
Acceleration Scale
8.
From the graphical solution above, Acceleration scale factor
ka 2000
in sec
AD 3.700 ka
2
AD 7400 in sec
2
at an angle of -33.485 deg 693.0 rad sec
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-76-1
PROBLEM 7-76 Statement:
The linkage in Figure P7-30a has link 2 at 120 deg in the global XY coordinate system. Find 6 and AD in the global XY coordinate system for the position shown if 2 = 10 rad/sec CCW and 2 = 50 rad/sec2 CW. Use an analytical method.
Given:
Solution: 1.
Link lengths: Link 2 (O2 to A)
a 6.20 in
Link 3 (A to B)
b 4.50 in
Link 4 (O4 to B)
c 3.00 in
Link 5 (C to D)
h 5.60 in
Link 3 (A to C)
p 2.25 in
Link 4 (O4 to D)
f 3.382 in
Link 1 X-offset
d X 7.80 in
Link 1 Y-offset
d Y 0.62 in
Angle ACB
δ3 0.0 deg
Angle BO4D
δ4 110.0 deg
Input rocker angle:
θ2XY 120 deg
Global XY system
Coordinate transformation angle:
δ 175.455 deg
Input crank angular velocity
ω2 10 rad sec
Input crank angular acceleration
α2 50 rad sec
1
(CCW)
2
(CW)
See Figure P7-30a and Mathcad file P0776.
Draw the linkage to scale and label it.
A
6
Y
71.487° D
5
C
110°
3
120°
55.455° B 16.254°
4
x
2
O4
175.455° X
O2 y d
Calculate the distance O2O4:
Transform 2XY into the local coordinate system: 2.
2
dX dY
2
d 7.825 in
θ2 θ2XY δ
θ2 55.455 deg
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K1 1.2620 2
K3
d c
K2 2.6082 2
2
a b c d 2 a c
2
K3 2.3767
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-76-2
A cos θ2 K1 K2 cos θ2 K3 B 2 sin θ2 C K1 K2 1 cos θ2 K3 A 0.2028 3.
B 1.6474
Use equation 4.10b to find values of 4 for the open circuit.
2
θ4 2 atan2 2 A B 4.
B 4 A C
2
d
K5
b
7.
θ4 163.746 deg
2
2
c d a b
2
K4 1.7388
2 a b
D cos θ2 K1 K4 cos θ2 K5
D 1.6967
E 2 sin θ2
E 1.6474
F K1 K4 1 cos θ2 K5
F 0.3067
K5 1.9877
Use equation 4.13 to find values of 3 for the open circuit.
2
θ3 2 atan2 2 D E 6.
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
5.
C 1.5927
E 4 D F
2 π
θ3 648.512 deg
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18.
ω3
a ω2 sin θ4 θ2 b sin θ3 θ4
ω3 15.924
ω4
a ω2 sin θ2 θ3 c sin θ4 θ3
ω4 20.107
rad sec rad sec
Use equations 7.12 to determine the angular accelerations of links 3 and 4 for the open circuit. A c sin θ4
B b sin θ3
D c cos θ4
E b cos θ3
A 21.328 mm
B 108.386 mm
D 73.154 mm
E 36.291 mm
C a α2 sin θ2 a ω2 cos θ2 b ω3 cos θ3 c ω4 cos θ4 2
3
C 2.134 10 in sec
2
2
2
F a α2 cos θ2 a ω2 sin θ2 b ω3 sin θ3 c ω4 sin θ4 2
3
F 1.087 10 in sec
α3 8.
C D A F A E B D
2
2
2
α3 636.368
rad sec
2
α4
C E B F A E B D
α4 692.987
rad sec
Establish a vector loop that defines the position of point D with respect to O4 and write the loop equation.
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-76-3
A 2.517°
Y
71.487° R5
D
C 5
R BC
3 110° R4
RD
2 120° 175.455°
55.455° B
x O4
4
X
O2
16.254° y
RD = R4 RBC R5
f e
= c ejθ4 p ej θ3π h ejθ5
j θ4 δ4
where f and the angles are functions of time and c, p, h, and 4 are constants. 9.
Solve the vector loop equation for 5 by substituting the Euler identity for the unit vectors and then separating into real and imaginary parts. f cos θ4 δ4 j sin θ4 δ4 = c cos θ4 j sin θ4 p cos θ3 π j sin θ3 π h cos θ5 j sin θ5 f cos θ4 δ4 = c cos θ4 p cos θ3 π h cos θ5 f sin θ4 δ4 = c sin θ4 p sin θ3 π h sin θ5 tan θ5 =
f sin θ4 δ4 c sin θ4 p sin θ3 π
f cos θ4 δ4 c cos θ4 p cos θ3 π
θ5 atan2 f cos θ4 δ4 c cos θ4 p cos θ3 π f sin θ4 δ4 c sin θ4 p sin θ3 π f 3.382 in
θ5 2.518 deg
10. Differentiate the position equaton and solve for 5 and fdot by substituting the Euler identity for the unit vectors and then separating into real and imaginary parts. f ω4 sin θ4 δ4 j cos θ4 δ4 fdot cos θ4 δ4 j sin θ4 δ4
=
c ω4 sin θ4 j cos θ4 p ω3 sin θ3 π j cos θ3 π h ω5 sin θ5 j cos θ5 f ω4 sin θ4 δ4 fdot cos θ4 δ4 = c ω4 sin θ4 p ω3 sin θ3 π h ω5 sin θ5 f ω4 cos θ4 δ4 fdot sin θ4 δ4 = c ω4 cos θ4 p ω3 cos θ3 π hω5 cos θ5 Let
A f ω4 sin θ4 δ4 c ω4 sin θ4 p ω3 sin θ3 π
A 71.929
in
B f ω4 cos θ4 δ4 c ω4 cos θ4 p ω3 cos θ3 π
B 86.747
in
s s
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-76-4
then fdot cos θ4 δ4 = A h ω5 sin θ5
fdot sin θ4 δ4 = B h ω5 cos θ5
Eliminating fdot and solving for 5,
ω5
A tan θ4 δ4 B
h cos θ5 sin θ5 tan θ4 δ4
and
A h ω5 sin θ5
fdot
ω5 35.145
rad s in
fdot 136.252
cos θ4 δ4
s
11. Differentiate the velocity equaton and solve for 5 and fddot by substituting the Euler identity for the unit vectors and then separating into real and imaginary parts. f α4 j e
j θ 4 δ 4
f ω 2 ej θ4δ4 2 fdot ω j ej θ4δ4 fddot ej θ4δ4 4
c j α4 e
j θ4
4
=
j θ3π ω 2 ej θ3π h j α ej θ5 ω ej θ5 3 5 p j e 5
2 j θ4
ω4 e
f α4 sin θ4 δ4 j cos θ4 δ4 f ω4 cos θ4 δ4 j sin θ4 δ4 2 fdot ω4 sin θ4 δ4 j cos θ4 δ4 fddot cos θ4 δ4 j sin θ4 δ4 2
=
c α4 sin θ4 j cos θ4 c ω4 cos θ4 j sin θ4 2
p α3 sin θ3 π j cos θ3 π p ω3 cos θ3 π j sin θ3 π 2
h α5 sin θ5 j cos θ5 h ω5 cos θ5 j sin θ5 2
Let
C f α4 sin θ4 δ4 f ω4 cos θ4 δ4 2 fdot ω4 sin θ4 δ4 2
c α4 sin θ4 ω4 cos θ4 p α3 sin θ3 π ω3 cos θ3 π 2
2
h ω5 cos θ5 2
C 3003.51
in 2.00
s
D f α4 cos θ4 δ4 f ω4 sin θ4 δ4 2 fdot ω4 cos θ4 δ4 2
c α4 cos θ4 ω4 sin θ4 p α3 cos θ3 π ω3 sin θ3 π 2
2
h ω5 sin θ5 2
D 1295.03
in 2.00
s
then, similar to the velocity solution,
α5
C tan θ4 δ4 D
h cos θ5 sin θ5 tan θ4 δ4
fddot
C h α5 sin θ5 cos θ4 δ4
α5 1025.02
rad 2
s
fddot 5505.35
in 2.00
s
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-76-5
12. Using the terms in the acceleration equation in step 10, define the components of AD and then combine them to determine the total acceleration of point D.
ADtangential f α4 sin θ4 δ4 j cos θ4 δ4 ADnormal f ω42 cos θ4 δ4 j sin θ4 δ4 ADcoriolis 2 fdot ω4 sin θ4 δ4 j cos θ4 δ4 ADslip fddot cos θ4 δ4 j sin θ4 δ4 AD ADtangential ADnormal ADcoriolis ADslip AD ( 6592.71 3687.93i)
in
in the local coordinate frame
2.00
s Magnitude:
AD
AD 7554.11
AD
in 2.00
s Angle in local coordinate frame:
θAD arg AD
θAD 150.778 deg
Angle in global coordinate frame:
θADXY θAD δ
θADXY 326.233 deg
The angular accelaration of block 6 is the same as link 4, that is,
α6 α4
α6 692.987
rad 2
s
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-77-1
PROBLEM 7-77 Statement:
The linkage in Figure P7-30b has link 3 perpendicular to the X-axis and links 2 and 4 are parallel to each other. Find 4, AA, AB, and AP if 2 = 15 rad/sec CW and 2 = 100 rad/sec2 CW. Use the acceleration difference graphical method.
Given:
Link lengths: Link 2 (O2 to A)
a 2.75 in
Link 3 (A to B)
b 3.26 in
Link 4 (O4 to B)
c 2.75 in
Link 1 (O4 to B)
d 4.43 in
Coupler point data:
p 1.63 in
δ 0 deg
ω 15 rad sec
Input crank angular velocity Solution: 1.
1
100 rad sec
CW
2
See Figure P7-30b and Mathcad file P0777.
Draw the linkage to a convenient scale. Indicate the directions of the acceleration vectors of interest.
Direction of ABt Direction of ABA B
P
O2
O4 36.351°
A
θ 36.352 deg
Direction of AAt 2.
In order to solve for the accelerations at points A and B, we will need 3, 3, and 4. From the graphical position solution above (in the local xy coordinate system), θ 90.000 deg
θ 90.000 deg
θ 180 deg θ
θ 143.648 deg
Using equation (6.18),
ω 1.924 10
ω 15.000 rad sec
ω
a ω sin θ θ b sin θ θ
ω
a ω sin θ θ c sin θ θ
15
The graphical solution for accelerations uses equation 7.4:
4.
For point B, this becomes: (ABt + ABn) = (AAt + AAn) + (ABAt + ABAn) , where 2
1
1
(APt + APn) = (AAt + AAn) + (APAt + APAn)
3.
ABn c ω
rad sec
ABn 618.750 in sec
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-77-2
θABn θ 180 deg
θABn 323.648 deg
2
AAn a ω
AAn 618.750 in sec
θAAn θ 180 deg
θAAn 143.648 deg
AAt a
AAt 275.000 in sec
θAAt θ 90 deg
θAAt 126.352 deg
2
5.
2
2
2
ABAn b ω
ABAn 0.000 in sec
θABAn θ 180 deg
θABAn 270.000 deg
Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ABn at an angle of ABn + and AAn at an angle of AAn + . From the tip of AAn, draw AAt at an angle of AAt + . From the tip of AAt, draw ABAn at an angle of BAn + . Now that the vectors with known magnitudes are drawn, from the tips of ABn and ABAn, draw construction lines in the directions of ABt and ABAt, respectively. The intersection of these two lines are the tips of ABt, ABt, and ABAt. 3.841 167.610° t A
A
0
AB
Y
n
AA
t ABA
400 IN/S/S
9.425°
AA X
t
AB
celeration Scale 1.693
n
AB 2.218
6.
From the graphical solution above, Acceleration scale factor
ka 400
in sec
AA 677.2 in sec
AB 2.218 ka
AB 887.2 in sec
ABAt 3.841 ka
ABAt 1536.4 in sec
ABAt
at an angle of 167.610 deg
2
at an angle of 9.425 deg
2
471.288 rad sec
b
2
CW
For point P, equation 7.4 becomes: AP = AA + (APAt + APAn) , where 2
2
APAn 0.000 in sec
θAPAn θ δ 180 deg
θAPAn 270.000 deg
APAt p
APAt 768.2 in sec
θAPAt θAPAn 90 deg
θAPAt 360.000 deg
APAn p ω
8.
2
AA 1.693 ka
7.
2
2
Since 3 = 0 the acceleration at P is one-half that at B and is in the same direction as AB.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-78-1
PROBLEM 7-78 Statement:
The linkage in Figure P6-33b has link 3 perpendicular to the X-axis and links 2 and 4 are parallel to each other. Find 4, AA, AB, and AP if 2 = 15 rad/sec CW and 2 = 100 rad/sec2 CW. Use an analytical method.
Given:
Link lengths: Link 2 (O2 to A)
a 2.75 in
Link 3 (A to B)
b 3.26 in
Link 4 (O4 to B)
c 2.75 in
Link 1 (O4 to B)
d 4.43 in
RPA 1.63 in
δ3 0 deg
Coupler point data:
Solution: 1.
1
Input crank angular velocity
ω2 15 rad sec
Input crank angular acceleration
α2 100 rad sec
CW
2
CW
See Figure P6-33b and Mathcad file P0778.
Draw the linkage to scale and label it.
Y
B
O4
O2
X
36.351° A From the figure, 2.
θ2 36.351 deg
From the problem statement we have:
θ3 90 deg 3.
4.
θ4 θ2 180 deg
θ4 216.351 deg
Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18.
ω3
a ω2 sin θ4 θ2 b sin θ3 θ4
ω3 1.924 10
ω4
a ω2 sin θ2 θ3 c sin θ4 θ3
ω4 15.000
15 rad
sec
rad
Using the Euler identity to expand equation 7.13a for AA. Determine the magnitude, and direction. AA a α2 sin θ2 j cos θ2 a ω2 cos θ2 j sin θ2 2
AA ( 335 588i)
in sec
2
The acceleration of pin A is
θAA arg AA
AA AA AA 677.1
in sec
5.
CCW
sec
at 2
θAA 119.7 deg
Use equations 7.12 to determine the angular accelerations of links 3 and 4.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-78-2
A c sin θ4
B b sin θ3
D c cos θ4
E b cos θ3
A 1.630 in
B 3.260 in
D 2.215 in
E 0.0 in
C a α2 sin θ2 a ω2 cos θ2 b ω3 cos θ3 c ω4 cos θ4 2
C 833.683 in sec
2
2
2
F a α2 cos θ2 a ω2 sin θ2 b ω3 sin θ3 c ω4 sin θ4 F 954.989 in sec
α3 6.
2
2
2
α3 471.320
rad
2
C D A F A E B D
sec
α4
2
C E B F A E B D
α4 431.175
Use equation 7.13c to determine the acceleration of point B. AB c α4 sin θ4 j cos θ4 c ω4 cos θ4 j sin θ4 2
AB ( 30509 14941i)
mm sec
The acceleration of pin B is
AB 1337.5
in sec
7.
θAB arg AB
AB AB
2
at 2
θAB 26.092 deg
Use equations 7.32 to find the acceleration of the point P. APA RPA α3 sin θ3 δ3 j cos θ3 δ3 RPA ω3 cos θ3 δ3 j sin θ3 δ3 2
AP AA APA AP AP
AP 730.37
in sec
2
arg AP 53.649 deg
rad sec
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-79-1
PROBLEM 7-79 Statement:
The crosshead linkage shown in Figure P7-30c has 2 DOF with inputs at cross heads 2 and 5. Find AB, AP3, and AP4 if the crossheads are each moving toward the origin of the XY coordinate system with a speed of 20 in/sec and are decelerating at 75 in/sec2. Use a graphical method.
Given:
Link lengths: Link 3 (A to B)
b 34.32 in
Link 4 (B to C)
c 50.4 in
Link 2 Y-offset
a 59.5 in
Link 5 X-offset
d 57 in
AP3 31.5 in
BP3 22.2 in
BP4 41.52 in
CP 4 27.0 in
Coupler point data:
Solution:
Input velocities:
V2Y 20 in sec
Input accelerations
AA 75 in sec
1
2
V5X 20 in sec AC 75 in sec
1
2
See Figure P7-30c and Mathcad file P0779.
1.
Draw the linkage to a convenient scale. Indicate the directions of the acceleration vectors of interest.
2.
In order to solve for the accelerations at point B, we will need 3, 3, 4, and 3. From the velocity solution (Problem 6-98),
Y
θ 320.123 deg 1.749 rad sec
1
P3
A
CCW Direction of AA
3
θ 122.718 deg 1.177 rad sec 3.
Direction of AP3At Direction of ABAt Direction of ABCt
2
1
Direction of AP4Ct B
5
Use equation 7.4 to (graphically) determine the magnitude of the acceleration at point B, the magnitude of the relative accelerations ABA, ABC and the angular accelerations of links 3 and 4. The equations to be solved (simultaneously) graphically are
X C Direction of AC
AB = AA + ABA
AB = AC + ABC
Eliminating AB by combing equations and expanding, AA + (ABAt + ABAn) = AC + (ABCt + ABCn) AA 75.000 in sec
P4
4
CW
2
AC 75.000 in sec
2
2
θAA 270 deg θAC 180 deg
ABAn b
ABAn 104.985 in sec
θABAn θ 180 deg
θABAn 140.123 deg
2
ABCn c
ABCn 69.821 in sec
θABCn θ 180 deg
θABCn 57.282 deg
2
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-79-2
a.
Choose a convenient velocity scale and layout the known vectors AA and AC
b.
From the tip of AA, draw known vector ABAn.
c.
From the tip of AC, draw known vector ABCn.
d. From the tips of ABAn and ABCn draw construction lines in the known directions of these vectors. e. From the origin, draw a construction line to the intersection of the two construction lines in step d. f. Complete the vector triangle by drawing ABA from the tip of AA to the intersection of the AB construction line and drawing AB from the tail of AA to the intersection of the ABA construction line. And then do the same with the ACB vector
Y
AC X
5.701
A
n BC
141.390° A
n BA
AA
4.436
t ABC
AB 0
t ABA
50 IN/S/S
4.409 Acceleration Scale
6.
From the graphical solution above, Acceleration scale factor
ka 50
in sec
2
AB 5.701 ka
AB 285.1 in sec
ABAt 4.409 ka
ABAt 220.5 in sec
ABAt b
ABCt 4.436 ka 7.
2
ABCt c
6.423 rad sec
2
2
ABCt 221.8 in sec 4.401 rad sec
at an angle of -141.390 deg
CW
2
2
From the graphical solution, the angles from A to P3 and C to P4 are
CCW
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-79-3
θP3 9.250 deg 8.
The accelerations of points P3 and P4 with respect to A are: 2
2
AP3An AP3
AP3An 96.359 in sec
AP3At AP3
AP3At 202.336 in sec
AP4Cn CP4
2
AP4Cn 37.404 in sec
AP4Ct CP4 3.
θP4 67.313 deg
2
2
AP4Ct 118.821 in sec
2
θAP3An θP3 180 deg
θAP3An 170.750 deg
θAP3At θP3 90 deg
θAP3At 80.750 deg
θAP4Cn θP4 180 deg
θAP4Cn 112.687 deg
θAP4Ct θP4 90 deg
θAP4Ct 157.313 deg
Use equation 7.4 to (graphically) determine the magnitude of the acceleration at points P3 and P4.
2.832
AP3 = AA + (AP3At + AP3An)
Y
AP4 = AC + (AP4Ct + AP4Cn)
6.
A P4
164.000°
t AP4C
From the graphical solution at right,
n
A P4C
Acceleration scale factor ka 200
X
AC n
A P3A
AA 87.517°
in sec
AP3 4.577 ka
2
AP3 915.4 in sec
2
AP4 566.4 in sec
200 IN/S/S
Acceleration Scale
at an angle of -87.517 deg AP4 2.832 ka
0
2
4.577
at an angle of 164.000 deg
t AP3A
A P3
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-80-1
PROBLEM 7-80 Statement:
The crosshead linkage shown in Figure P7-30c has 2 DOF with inputs at cross heads 2 and 5. Find AB, AP3, and AP4 if the crossheads are each moving toward the origin of the XY coordinate system with a speed of 20 in/sec and are decelerating at 75 in/sec2. Use an analytical method.
Given:
Link lengths: Link 3 (A to B)
b 34.32 in
Link 4 (B to C)
c 50.4 in
Link 2 Y-offset
a 59.5 in
Link 5 X-offset
d 57 in
AP3 31.5 in
BP3 22.2 in
BP4 41.52 in
CP 4 27.0 in
Coupler point data:
Solution:
Input velocities:
V2Y 20 in sec
Input accelerations
AA 75 in sec
1
2
V5X 20 in sec AC 75 in sec
1
2
See Figure P7-30c and Mathcad file P0780.
1.
Draw the linkage to a convenient scale and lable it.
2.
In order to solve for the accelerations at point B, we will need 3, 3, 4, and 3. From the velocity solution (Problem 6-98),
Y
θ 320.123 deg 1.749 rad sec 3.
2
θ 122.718 deg 1
1.177 rad sec
CCW
1
CW
P3
A 3
Use equation 7.4 to (analytically) determine the magnitude of the acceleration at point B, the magnitude of the relative accelerations ABA, ABC and the angular accelerations of links 3 and 4. The equations to be solved simultaneously are AB = AA + ABA
B
C
AA + (ABAt + ABAn) = AC + (ABCt + ABCn) 2
AC 75.000 in sec
2
2
Let
θAA 270 deg θAC 180 deg
ABAn b
ABAn 104.985 in sec
θABAn θ 180 deg
θABAn 140.123 deg
2
5 X
AB = AC + ABC
Eliminating AB by combing equations and expanding,
AA 75.000 in sec
P4
4
2
2
ABCn c
ABCn 69.821 in sec
θABCn θ 180 deg
θABCn 57.282 deg
AAy AA
AAy 75.000 in sec
ABAnx ABAn cos θABAn
ABAnx 80.568 in sec
ABAny ABAn sin θABAn
ABAny 67.310 in sec
2 2
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-80-2 2
ACx AC
ACx 75.000 in sec
ABCnx ABCn cos θABCn
ABCnx 37.738 in sec
ABCny ABCn sin θABCn
ABCny 58.743 in sec
θABAt θABAn 90 deg
θABAt 50.123 deg
θABCt θABCn 90 deg
θABCt 32.718 deg
mBCt tan θABCt
mBCt 0.642
mBAt tan θABAt
2 2
mBAt 1.197
Write the equations for the two lines that represent the vectors ABAt and ABCt and solve them simultaneously to find the coordinates of AB: ABx
mBCt ACx ABCnx mBAt ABAnx ABCny AAy ABAny mBCt mBAt
ABy mBCt ABx ACx ABCnx ABCny ABx 222.804 in sec AB
2
ABx ABy
2
2
θAB atan2 ABx ABy
ABy 177.941 in sec AB 285.140 in sec
θAB 141.388 deg
A similar approach can be taken to find AP3 and AP4.
2
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-81-1
PROBLEM 7-81 Statement:
The crosshead linkage shown in Figure P7-30c has 2 DOF with inputs at cross heads 2 and 5. At t = 0, crosshead 2 is at rest at the origin of the global XY coordinate system and crosshead 5 is at rest at (70,0). Write a computer program to find and plot AP3 and AP4 for the first 5 sec of motion if A2 = 0.5 in/sec2 upward and A5 = 0.5 in/sec2 to the left.
Given:
Link lengths: Link 3 (A to B)
b 34.32 in
Link 4 (B to C)
c 50.4 in
Link 2 Y-offset
a 59.5 in
Link 5 X-offset
d 57 in
AP3 31.5 in
BP3 22.2 in
39.128 deg
BP4 41.52 in
CP 4 27.0 in
55.405 deg
Coupler point data:
A2 0.5 in sec
Input accelerations Solution: 1.
2
A5 0.5 in sec
2
See Figure P7-30c and Mathcad file P0781.
Draw the linkage to a convenient scale and lable it.
Y
2 P3
A 3
R3
b B a 1
c
R2
P4
4
3
5
R4 X
4
R5
C d t 0 sec 0.1 sec 5 sec a ( t) a 0 0.5 A2 t
a 0 0 in
2
d ( t) d 0 0.5A5 t
V2( t) A2 t 2.
d 0 70 in 2
V5( t) A5 t
Define a vector loop as shown above. Summing the vectors around the loop, R2 R3 R4 R5 = 0
a b sin c sin = 0
b cos c cos d = 0
3.
Substitute the polar notation and Euler equivalents and solve for 3 and 4 using the method of Section 4.5 and the identities in equations 4.9. 2
K1( t)
2
2
a( t) b c d( t) 2 c
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-81-2
A ( t) K1( t) d ( t)
B( t) 2 a ( t)
2
( t) 2 atan2 2 A ( t) B( t) 2
K2( t)
2
B( t ) 4 A ( t ) C ( t )
2
a ( t) b c d ( t)
2
2 b
D( t) K2( t) d ( t)
E( t) 2 a ( t)
F ( t) K2( t) d ( t)
2
( t) 2 atan2 2 D( t) E( t) 4.
C( t) K1( t) d ( t)
E( t) 4 D( t) F ( t)
Differentiate the position equations with respect to time and solve for 3 and 4.
b sin c sin V5 = 0
V2 b cos c cos = 0 ( t)
( t)
5.
V2( t) sin ( t) V5( t) cos ( t) b sin ( t) ( t)
V2( t) sin ( t) V5( t) cos ( t)
c sin ( t) ( t)
Differentiate the velocity equations with respect to time and solve for 3 and 4. b sin cos
c sin 2 cos A5 = 0
2
A2 b cos sin
c cos 2 sin = 0
2
A' ( t) b sin ( t)
B'( t) c sin ( t)
2
2
C' ( t) b ( t) cos ( t) c ( t) cos ( t) A5
D'( t) b cos ( t) 2
E'( t) c cos ( t)
2
F'( t) b ( t) sin ( t) c ( t) sin ( t) A2
( t)
6.
B'( t) F'( t) C' ( t) E'( t)
( t)
A' ( t) E'( t) B'( t) D'( t)
C' ( t) D'( t) A' ( t) F'( t) A' ( t) E'( t) B'( t) D'( t)
Using equations 6.32, solve for the accelerations at P3.
2 ( t) cos ( t) j sin ( t)
AP3A( t) AP3 ( t) sin ( t) j cos ( t)
AP3( t) j A2 AP3A( t)
AP3 ( t) AP3( t)
θAP3( t) if θAP3( t) 0 θAP3( t) 2 π θAP3( t)
θAP3( t) arg AP3( t)
DESIGN OF MACHINERY - 5th Ed.
7.
SOLUTION MANUAL 7-81-3
Plot the magnitude and direction of the acceleration at P3. MAGNITUDE - P3 80
60
2
AP3 ( t )
sec
40
in
20
0
0
1
2
3
4
5
t sec
DIRECTION - P3 200
190
θAP3( t )
180
deg
170
160
0
1
2
3
4
5
t
8.
sec
Using equations 6.32, solve for the accelerations at P4.
2 ( t) cos ( t) j sin ( t)
AP4A( t) CP4 ( t) sin ( t) j cos ( t)
AP4( t) j A5 AP4A( t)
AP4 ( t) AP4( t)
θAP4( t) if θAP4( t) 0 θAP4( t) 2 π θAP4( t)
θAP4( t) arg AP4( t)
DESIGN OF MACHINERY - 5th Ed.
9.
SOLUTION MANUAL 7-81-4
Plot the magnitude and direction of the acceleration at P4. MAGNITUDE - P4 40
30
2
AP4 ( t )
sec
20
in
10
0
0
1
2
3
4
5
t sec
DIRECTION - P3 350
348
346
θAP4( t ) deg 344
342
340
0
1
2
3 t sec
4
5
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-82-1
PROBLEM 7-82 Statement:
The linkage in Figure P7-30d has the path of slider 6 perpendicular to the global X-axis and link 2 aligned with the global X-axis. Find 2 and AA in the position shown if the velocity of the slider is constant at 20 in/sec downward. Use the acceleration difference graphical method.
Given:
Link lengths: Link 2 (O2 to A)
a 12 in
Link 3 (A to B)
b 24 in
Link 4 (O4 to B)
c 18 in
Link 5 (C to D)
f 24 in
Link 4 (O4 to C)
e 18 in
Link 1 X-offset
d X 19 in
Link 1 Y-offset
d Y 28 in
0.0 deg
Angle BO4C
VD 20 in sec
Input slider velocity Solution: 1.
1
θVD 90.0 deg
See Figure P7-30d and Mathcad file P0782.
Draw the linkage to a convenient scale. Indicate the directions of the acceleration vectors of interest. Direction of ACt
Direction of ACDt C
Direction of ABt O4 19.963°
4
116.161°
B 5
Y
D 3
O2
2
114.410°
6 X
A
Direction of AD
Direction of AABt Direction of AAt
2.
In order to solve for the accelerations at points A, B and C, we will need 2, 2, 3, 3, and 4 4. From the graphical position solution above and the velocity solution (see Problem 6-99), 0.000 deg
1.648 rad sec
θ 114.410 deg
0.282 rad sec
θ 199.963 deg
1.003 rad sec
1 1 1
CCW CCW CW
θ 19.963 deg 116.161 deg
0.286 rad sec
1
CW
(APt + APn) = (AAt + AAn) + (APAt + APAn)
3.
The graphical solution for accelerations uses equation 7.4:
4.
For point C, this becomes: (ACt + ACn) = AD + (ACDt + ACDn) , where
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-82-2 2
ACn e
ACn 18.108 in sec
θACn θ 180 deg
θACn 199.963 deg
AD 0 in sec ACDn f
2
2
ACDn 1.963 in sec
θACDn 180 deg 5.
2
2
θACDn 296.161 deg
Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ACn at an angle of ACn + and ADn at an angle of ADn + . From the tip of ADn, draw ADt at an angle of ADt + . From the tip of ADt, draw ACDn at an angle of CDn + . Now that the vectors with known magnitudes are drawn, from the tips of ACn and ACDn, draw construction lines in the directions of ACt and ACDt, respectively. The intersection of these two lines are the tips of ACt, AC, and ACDt. 0
5 IN/S/S Y
Acceleration Scale 3.706
X
n
A CD n
AC 147.759°
AC 0.788
6.
t CD
A
AtC
From the graphical solution above, ka 5.0
Acceleration scale factor
in sec
2
AC 3.706 ka
AC 18.530 in sec
ACt 0.788 ka
ACt 3.940 in sec
ACt
at an angle of -147.759 deg
2
0.219 rad sec
e
2
2
CW
Since link 4 is straight and the distance from O4 to B is the same as to C, AB AC 7.
AB 18.530 in sec
2
at an angle of 32.241 deg
For point A, equation 7.4 becomes: (AAt + AAn) = AB + (AABt + AABn) , where 2
AAn a
AAn 32.591 in sec
θAAn 180 deg
θAAn 180.000 deg
2
2
2
AABn b
AABn 1.909 in sec
θAABn θ
θAABn 114.410 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-82-3
0
10 IN/S/S
n
A AB Y
Acceleration Scale
AB
n
AA X 0.993
AAt
t AAB
AA
163.062°
3.407
8.
From the graphical solution above, Acceleration scale factor
ka 10
in sec
2
AA 3.407 ka
AA 34.070 in sec
AAt 0.993 ka
AAt 9.930 in sec
AAt a
2
at an angle of -163.062 deg
2
0.828 rad sec
2
CW
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-83-1
PROBLEM 7-83 Statement:
The linkage in Figure P7-30d has the path of slider 6 perpendicular to the global X-axis and link 2 aligned with the global X-axis. Find 2 and AA in the position shown if the velocity of the slider is constant at 20 in/sec downward. Use an analytical method.
Given:
Link lengths: Link 2 (O2 to A)
c 12 in
Link 3 (A to B)
b 24 in
Link 4 (O4 to B)
a 18 in
Link 5 (C to D)
f 24 in
Link 4 (O4 to C)
e 18 in
Link 1 X-offset
d X 19 in
Link 1 Y-offset
d Y 28 in
Link 1 (O2 to O4)
d
2
dX dY
Axis rotation angle:
124.160 deg
Input slider velocity
VD 20 in sec
Input slider acceleration AD 0 in sec Solution: 1.
2
1
d 33.838 in
Downward (positive x' direction)
2
See Figure P7-30d and Mathcad file P0783.
Draw the linkage to a convenient scale and label it. C O4 y'
19.963°
4
116.161°
B 5
Y x'
y" D
3
O2 x"
2.
2
114.410°
6 X
A
124.160°
From the graphical position solution above and the velocity solution (see Problem 6-99), Local x"y"coordinate system: Output
Input
1
124.160 deg
1.648 rad sec
θ 58.570 deg
0.282 rad sec
θ 324.123 deg
1.003 rad sec
1 1
CCW CCW CW
Local x'y' coordinate system:
3.
θ 109.963 deg
1.003 rad sec
206.161 deg
0.286 rad sec
Solve equations 7.16b and c for 2 (which is 4 in this case):
1 1
CW CW
DESIGN OF MACHINERY - 5th Ed.
4.
SOLUTION MANUAL 7-83-2
2
AD cos e cos θ f
e sin θ
2
0.219 rad sec
2
Use equations 7.12 to determine the angular accelerations of link 2. A c sin
B b sin θ
D c cos
E b cos θ
A 9.930 in
B 20.479 in
D 6.738 in
E 12.515 in
2
2
2
2
2
2
C a sin θ a cos θ b cos θ c cos C 36.278 in sec
2
F a cos θ a sin θ b sin θ c sin F 32.758 in sec α
9.
2
C E B F A E B D
α 0.827
rad sec
2
Use equation 7.13c to determine the acceleration of point A.
c 2 cos j sin
AA c α sin j cos AA ( 26.510 21.397i )
in sec
The acceleration of pin A is
2
θAA arg AA
AA AA AA 34.068
in sec
at 2
θAA 163.068 deg
(Global)
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-84-1
PROBLEM 7-84 Statement:
The linkage in Figure P7-30a has the path of slider 6 perpendicular to the global X-axis and link 2 aligned with the global X-axis at t = 0. Write a computer program or use an equation solver to find and plot AD as a function of 2, over the possible range of motion of link 2, in the global XY coordinate system.
Given:
Link lengths:
Solution: 1.
Link 2 (O2 to A)
a 12 in
Link 3 (A to B)
b 24 in
Link 4 (O4 to B)
c 18 in
Link 5 (C to D)
f 24 in
Link 4 (O4 to C)
e 18 in
Link 1 X-offset
d X 19 in
Link 1 Y-offset
d Y 28 in
Link 1 (O2 to O4)
d
2
dX dY
2
d 33.838 in
Axis rotation angle:
55.840 deg
90.000 deg
Input velocity & accel:
ω2 1 rad sec
Crank angle:
θ2XY 70 deg 69 deg 180 deg
1
α2 0 rad sec
2
θ2 θ2XY θ2XY
See Figure P7-30d and Mathcad file P0784.
Draw the linkage to a convenient scale and label it. C
x' O4 y" 4 B
5
Y x"
D y' 3 2
O2
2.
55.840°
6 X
A
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K1 2.8198
a 2
K3
2
2
a b c d
K2
d c
K2 1.8799
2
2 a c
K3 2.4005
A θ2XY cos θ2 θ2XY K1 K2 cos θ2 θ2XY K3 B θ2XY 2 sin θ2 θ2XY C θ2XY K1 K2 1 cos θ2 θ2XY K3 3.
Use equation 4.10b to find values of 4 for the open circuit.
θ4xy θ2XY 2 atan2 2 A θ2XY B θ2XY
B θ2XY 4 A θ2XY C θ2XY 2
DESIGN OF MACHINERY - 5th Ed.
4.
SOLUTION MANUAL 7-84-2
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d b
2
2
c d a b
K5
2
K4 1.4099
2 a b
K5 2.6753
D θ2XY cos θ2 θ2XY K1 K4 cos θ2 θ2XY K5 E θ2XY 2 sin θ2 θ2XY F θ2XY K1 K4 1 cos θ2 θ2XY K5 5.
Use equation 4.13 to find values of 3 for the open circuit.
θ3xy θ2XY 2 atan2 2 D θ2XY E θ2XY 6.
7.
E θ2XY 4 D θ2XY F θ2XY 2
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18.
ω3 θ2XY
a ω2 sin θ4xy θ2XY θ2 θ2XY b sin θ3xy θ2XY θ4xy θ2XY
ω4 θ2XY
a ω2 sin θ2 θ2XY θ3xy θ2XY c sin θ4xy θ2XY θ3xy θ2XY
Use equations 7.12 to determine the angular acceleration of link 4. A θ2XY c sin θ4xy θ2XY
B θ2XY b sin θ3xy θ2XY
D θ2XY c cos θ4xy θ2XY
E θ2XY b cos θ3xy θ2XY
C θ2XY a α2 sin θ2 θ2XY a ω2 cos θ2 θ2XY 2
b ω3 θ2XY cos θ3xy θ2XY c ω4 θ2XY cos θ4xy θ2XY 2
2
F θ2XY a α2 cos θ2 θ2XY a ω2 sin θ2 θ2XY 2
b ω3 θ2XY sin θ3xy θ2XY c ω4 θ2XY sin θ4xy θ2XY 2
α4 θ2XY 8.
2
C θ2XY E θ2XY B θ2XY F θ2XY A θ2XY E θ2XY B θ2XY D θ2XY
Transform 4 to the local x"y"coordinate system and add 180 deg to get the link from O4 to C, which is the input angle to the crank-slider portion of the linkage.
θ4 θ2XY θ4xy θ2XY 180 deg 9.
Determine 5 in the x"y" coordinate system using equation 4.17. Offset: cc 27.5 in
e sin θ4 θ2XY cc π f
θ5 θ2XY asin
10. Determine the angular velocity of link 5 using equation 6.22a:
ω5 θ2XY
e cos θ4 θ2XY ω4 θ2XY f cos θ5 θ2XY
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-84-3
11. Determine the angular acceleration of link 5 using equation 7.16d. e α4 θ2XY cos θ4 θ2XY e ω4 θ2XY sin θ4 θ2XY 2
α5 θ2XY
f ω5 θ2XY sin θ5 θ2XY 2
f cos θ5 θ2XY
12. Use equation 7.16e for the acceleration of pin D. AD θ2XY e α4 θ2XY sin θ4 θ2XY e ω4 θ2XY cos θ4 θ2XY 2
f α5 θ2XY sin θ5 θ2XY f ω5 θ2XY cos θ5 θ2XY 2
ACCELERATION - PIN D
20
AD θ2XY
2
sec
10
in
0
100
50
0
50
100
θ2XY deg
Positive is up (positive Y direction in the global coordinate system).
150
200
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-85-1
PROBLEM 7-85 Statement:
For the linkage of Figure P7-30e, write a computer program or use an equation solver to find and plot AD in the global coordinate system for one revolution of link 2 if 2 is constant at 10 rad/sec CW.
Given:
Link lengths: Link 2 (O2 to A)
a 5.0 in
Link 3 (A to B)
b 5.0 in
Link 4 (O4 to B)
c 6.0 in
Link 1 (O2 to O4)
d 2.5 in
Link 5 (B to C)
e 15 in
Angle BO4C
83.621 deg
θ2 0 deg 1 deg 360 deg
Crank angle:
ω2 10 rad sec
Input crank angular velocity Solution:
1
α2 0 rad sec
2
See Figure P7-30e and Mathcad file P0785.
1.
This sixbar drag-link mechanism can be analyzed as a fourbar Grashof double crank in series with a crankslider mechanism using the output of the fourbar, link 4, as the input to the slider-crank.
2.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K1 0.5000
a 2
K3
2
2
a b c d
K2
d
K2 0.4167
c
2
K3 0.7042
2 a c
A θ2 cos θ2 K1 K2 cos θ2 K3 B θ2 2 sin θ2 C θ2 K1 K2 1 cos θ2 K3 3.
Use equation 4.10b to find values of 4 for the open circuit.
θ41 θ2 2 atan2 2 A θ2 B θ2 4.
B θ2 4 A θ2 C θ2 2
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
K4 0.5000
2 a b
D θ2 cos θ2 K1 K4 cos θ2 K5 E θ2 2 sin θ2 F θ2 K1 K4 1 cos θ2 K5 5.
Use equation 4.13 to find values of 3 for the open circuit.
θ3 θ2 2 atan2 2 D θ2 E θ2 6.
E θ2 4 D θ2 F θ2 2
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18.
ω3 θ2
a ω2 b
sin θ41 θ2 θ2
sin θ3 θ2 θ41 θ2
K5 0.4050
DESIGN OF MACHINERY - 5th Ed.
a ω2
ω4 θ2 7.
c
SOLUTION MANUAL 7-85-2
sin θ2 θ3 θ2
sin θ41 θ2 θ3 θ2
Use equations 7.12 to determine the angular acceleration of link 4. A θ2 c sin θ41 θ2
B θ2 b sin θ3 θ2
D θ2 c cos θ41 θ2
E θ2 b cos θ3 θ2
C θ2 a α2 sin θ2 a ω2 cos θ2 2
b ω3 θ2 cos θ3 θ2 c ω4 θ2 cos θ41 θ2 2
2
F θ2 a α2 cos θ2 a ω2 sin θ2 2
b ω3 θ2 sin θ3 θ2 c ω4 θ2 sin θ41 θ2 2
α4 θ2
2
C θ2 E θ2 B θ2 F θ2
A θ2 E θ2 B θ2 D θ2
ANGULAR ACCEL - LINK 4 200
100
α4 θ2
2
sec
0
rad 100
200
0
45
90
135
180
225
270
θ2 deg
8.
Transform 4 to the slider-crank coordinate system (origin at O4) system.
θ42 θ2 θ41 θ2 9.
Determine 5 using equation 4.17. Offset: cc 0 in
c sin θ42 θ2 cc π e
θ5 θ2 asin
10. Determine the angular velocity of link 5 using equation 6.22a:
ω5 θ2
c cos θ42 θ2 ω4 θ2 e cos θ5 θ2
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-85-3
11. Determine the angular acceleration of link 5 using equation 7.16d. c α4 θ2 cos θ42 θ2 c ω4 θ2 sin θ42 θ2 2
α5 θ2
e ω5 θ2 sin θ5 θ2 2
e cos θ5 θ2
12. Use equation 7.16e for the acceleration of pin D. AD θ2 c α4 θ2 sin θ42 θ2 c ω4 θ2 cos θ42 θ2 2
e α5 θ2 sin θ5 θ2 e ω5 θ2 cos θ5 θ2 2
ACCELERATION - PIN D
3
2 10
3
1 10
AD θ2
2
sec in
0
3
1 10
0
45
90
135
180
θ2 deg
225
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-86-1
PROBLEM 7-86 Statement:
The linkage of Figure P6-33f has link 2 at 130 deg in the global XY coordinate system. Find AD in the global coordinate system for the position shown if 2 = 15 rad/sec CW and 2 = 50 rad/sec2 CW. Use the acceleration difference graphical method.
Given:
Link lengths: Link 2 (O2 to A)
a 5.0 in
Link 3 (A to B)
b 8.4 in
Link 4 (O4 to B)
c 2.5 in
Link 1 (O2 to O4)
d 1 12.5 in
Link 5 (C to E)
e 8.9 in
Link 5 (C to D)
h 5.9 in
Link 6 (O6 to E)
f 3.2 in
Link 7 (D to F)
k 6.4 in
Link 3 (A to C)
g 2.4 in
d 2 10.5 in
Link 1 (O2 to O6)
θ2 130 deg
Crank angle:
Slider axis offset and angle:
Solution: 1.
s 11.7 in
150 deg
Input crank angular velocity
ω2 15 rad sec
Input crank angular acceleration
α2 50 rad sec
1
2
CW CW
See Figure P6-33f and Mathcad file P0786.
Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest.
Direction of VF 8
Y
F
Direction of VFD
Direction of VE
7 E
Direction of VEC
6 Direction of VDC
5
Direction of VCA Direction VBA
C
Direction of VB
O6
D
Direction of VA
A
3
2
B 4 O4
O2
X
Angles measured from layout:
2.
θVB 24.351 deg
θVBA 79.348 deg
θVCA θVBA
θVE 64.594 deg
θVEC 162.461 deg
θVDC θVEC
θVF 150.00 deg
θVFD 198.690 deg
θ3 169.348 deg
θ4 65.649 deg
θ5 72.461 deg
θ6 154.594 deg
Use equation 6.7 to calculate the magnitude of the velocity at point A. VA a ω2
VA 75.000
in sec
θ7 108.690 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-86-2
θVA θ2 90 deg 3.
θVA 40.000 deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B, the magnitude of the relative velocity VBA, and the angular velocity of link 3. The equations to be solved graphically are VB = VA + VBA
and
VC = VA + VCA
a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line.
2.668
VA 0.943 V CA
0
VC
25 in/sec
3.302
Y
22.053° X
V BA VB 3.192
4.
From the velocity polygon we have: Velocity scale factor: VB 3.192 in kv
ω4
VB c
VBA 3.302 in kv
ω3 5.
VBA b
kv
25 in sec
1
in
VB 79.800
ω4 31.920
in rad
CW
sec
VBA 82.550
ω3 9.827
θVB 24.351 deg
sec
in
θVBA 79.348 deg
sec
rad
CCW
sec
Calculate the magnitude and direction of VCA and determine the magnitude and velocity of VC from the velocity polygon above. in VCA g ω3 VCA 23.586 θVCA 79.348 deg sec VCA Length of VCA on velocity polygon: vCA vCA 0.943 in kv VC 2.668 in kv
VC 66.700
in sec
θVC 22.053deg
DESIGN OF MACHINERY - 5th Ed.
6.
SOLUTION MANUAL 7-86-3
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point E, the magnitude of the relative velocity VED, and the angular velocity of link 5. The equations to be solved graphically are VE = VC + VEC
and
VD = VC + VDC
a. Choose a convenient velocity scale and layout the known vector VC. b. From the tip of VC, draw a construction line with the direction of VEC, magnitude unknown. c. From the tail of VC, draw a construction line with the direction of VE, magnitude unknown. d. Complete the vector triangle by drawing VEC from the tip of VC to the intersection of the VE construction line and drawing VE from the tail of VC to the intersection of the VEC construction line. 1.821
1.207 V EC VE
1.716 0
V DC
VD
25 in/sec VC Y
45.934°
X 1.900
7.
From the velocity polygon we have: Velocity scale factor: VE 1.716 in kv
ω6
VE f
VEC 1.821 in kv
ω5 8.
VEC e
kv
25 in sec
1
in
VE 42.900
ω6 13.406
in rad in
θVEC 162.461 deg
sec
rad
CCW
sec
Calculate the magnitude and direction of VDE and determine the magnitude and velocity of VD from the velocity polygon above. VDC h ω5
VDC 30.179
in
VD 1.900 in kv
VD 47.500
θVDC 162.461 deg
sec vDC
Length of VDC on velocity polygon:
9.
CW
sec
VEC 45.525
ω5 5.115
θVE 64.594 deg
sec
in sec
VDC kv
vDC 1.207 in
θVD 45.934deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point F, the magnitude of the relative velocity VFD, and the angular velocity of link 5. The equation to be solved graphically is VF = VD + VFD
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-86-4
a. Choose a convenient velocity scale and layout the known vector VD. b. From the tip of VD, draw a construction line with the direction of VFD, magnitude unknown. c. From the tail of VD, draw a construction line with the direction of VF, magnitude unknown. d. Complete the vector triangle by drawing VFD from the tip of VD to the intersection of the VF construction line and drawing VF from the tail of VD to the intersection of the VFD construction line.
2.454 Y 0
VD
25 in/sec
V FD 150.000° VF
45.934°
X 1.158
10. From the velocity polygon we have: Velocity scale factor:
kv
25 in sec in
in
VF 1.158 in kv
VF 28.950
VFD 2.454 in kv
VFD 61.350
ω7
VFD
ω7 9.586
k
1
sec
θVF 150.000 deg
in sec
rad
CCW
sec
11. The graphical solution for accelerations uses equation 7.4:
(APt + APn) = (AAt + AAn) + (APAt + APAn)
12. For point B, this becomes: (ABt + ABn) = (AAt + AAn) + (ABAt + ABAn) , where ABn c ω4
2
ABn 2547.2 in sec
θABn θ4 180 deg 2
θABn 114.351 deg
AAn a ω2
AAn 1125.0 in sec
θAAn θ2 180 deg
θAAn 50.000 deg
AAt a α2
AAt 250.0 in sec
θAAt θ2 90 deg
θAAt 40.000 deg
2
2
2
2
2
ABAn b ω3
ABAn 811.3 in sec
θABAn θ3 180 deg
θABAn 349.348 deg
13. Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ABn at an angle of ABn and AAn at an angle of AAn. From the tip of AAn, draw AAt at an angle of AAt. From the tip of AAt, draw ABAn at an angle of BAn. Now that the vectors with known magnitudes are drawn, from the tips of ABn and ABAn, draw construction lines in the directions of ABt and ABAt, respectively. The intersection of these two lines are the tips of AB, ABt, and ABAt.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-86-5
Y 0
1000 in/s/s
1.152 X
cceleration Scale
AA A
n
ABA 37.471°
AAt
n A
n
AB
3.358 t ABA
AB t B
A
14. From the graphical solution above, Acceleration scale factor
ka 1000
in sec
2 2
AA 1.152 ka
AA 1152 in sec
ABAt 3.358 ka
ABAt 3358 in sec
α3
ABAt
α3 399.762
b
θAA 37.471 deg
2
rad sec
CW
2
15. For point C, : AC = AA + (ACAt + ACAn) , where Y
X 3
AA 1.152 10 in sec
2
θAA 37.471 deg
AA 50.502°
θAAn θ2 180 deg
θAAn 50.000 deg
ACAt g α3
ACAt 959.4 in sec
θACAt θ3 90 deg
θACAt 79.348 deg
ACAn g ω3
2
θACAn θ3 180 deg
n
ACA
2
ACAn 231.8 in sec
4.150
2
θACAn 349.348 deg
0
500 in/s/s
Acceleration Scale
16. From the graphical solution above,
AC
t ACA
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-86-6
in
ka 500
Acceleration scale factor
sec AC 4.150 ka
2
AC 2075 in sec
2
θAC 50.502 deg
17. For point E: (AEt + AEn) = AC + (AECt + AECn) , where AEn f ω6
2
AEn 575.1 in sec
θAEn θ6 180 deg 2
Acceleration Scale
θAEn 25.406 deg
n
2
θAECn θ5 180 deg
θAECn 107.539 deg ka 500
in sec
α5
2
AECt 1349 in sec
AECt
sec
AEt
AE
2
AC
rad
α5 151.517
e
X
AE
AECn 232.9 in sec
AECt 2.697 ka
500 in/s/s
2
AECn e ω5
Acceleration scale factor
0
Y
CCW
t AEC
2
n
AEC 2.697
18. For point D, : AD = AC + (ADCt + ADCn) , where AC 2075.0 in sec
2
θAC 50.502 deg 2
ADCt h α5
ADCt 893.9 in sec
θADCt θ5 90 deg
θADCt 162.461 deg
2
θADCn θ5 180 deg
θADCn 107.539 deg
AD 3.194 ka
3.194
in sec
125.841°
AD 2
AC 2075 in sec
AE 2
θAD 125.841 deg 19. For point F, : AF = AD + (AFDt + AFDn) , where 2
Acceleration Scale
2
ADCn 154.4 in sec
ka 500
AFDn k ω7
AFDn 588.10 in sec
θAFDn θ7 180 deg
θAFDn 71.310 deg
500 in/s/s
X
ADCn h ω5
Acceleration scale factor
0
Y
2
t
ADC
n
ADC
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-86-7
Y 0
500 in/s/s
4.033 X
cceleration Scale
30.000°
AF AD
t AFD
n
AFD
Acceleration scale factor
ka 500
in sec
AF 4.033 ka
2
AF 2017 in sec
2
θAD 30.00 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-87-1
PROBLEM 7-87 Statement:
Figure 3-14 shows a crank-shaper quick-return mechanism with the following dimensions: L2 = 4.80 in, L4 = 24.00 in, L5 = 19.50 in, the distance from link 4's fixed pivot (O4) to link 2's fixed pivot (O2) is 16.50 in, and the vertical distance from O2 to the pivot point on link 6 is 6.465 in. Use a graphical method to calculate the acceleration of link 6 when 2 = 45 deg. Assume that link 2 has a constant angular velocity of 2 rad/sec CW.
Given:
Link lengths and angles: Link 2
L2 4.80 in
Link 4
L4 24.00 in
Link 5
L5 19.50 in ω 2 rad sec
Angular velocity of link 2 Solution: 1.
Angle link 2 makes with X axis θ 45 deg
1
CW
α 0 rad sec
2
See Figure P3-14 and Mathcad file P0787.
Draw the linkage to scale and indicate the axes of slip and transmission as well as the directions of velocities of interest.
C 6
5
B
6.465" 4
O2
14.519° y
2 A 3
Axis of slip
13.538"
Axis of transmission
4 Direction of VA3 45.000°
O4 x
2.
Use equation 6.7 to calculate the magnitude of the velocity at point A on links 2 and 3. VA3 L2 ω
3.
VA3 9.600
in sec
θVA3 θ 90 deg
θVA3 45.0 deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity components at point A on link 3. The equation to be solved graphically is VA3 = Vtrans + Vslip
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-87-2
a. Choose a convenient velocity scale and layout the known vector VA3. b. From the tip of VA3, draw a construction line with the direction of Vslip, magnitude unknown. c. From the tail of VA3, draw a construction line with the direction of Vtrans, magnitude unknown. d. Complete the vector triangle by drawing Vslip from the tip of Vtrans to the tip of VA3 and drawing Vtrans from the tail of VA3 to the intersection of the Vslip construction line. 4.
From the velocity triangle we have: kv
Velocity scale factor:
5 in sec
1
0.975"
Vtrans
in
y
in
Vslip 1.654 in kv
Vslip 8.270
Vtrans 0.975 in kv
Vtrans 4.875
1.654"
sec in
x
Vslip
sec
0
5 in/sec
VA3 5.
The true velocity of point A on link 4 is Vtrans, VA4 Vtrans
6.
VA4 4.875
sec
Determine the angular velocity of link 4 using equation 6.7. From the linkage layout above: ω
VA4 c
VB L4 ω 7.
in
c 13.538 in and
ω 0.360
rad
VB 8.642
in
θ 180 deg 14.519 deg θ 165.481 deg
CCW
sec sec
Draw links 1, 4, 5, and 6 to a convenient scale. Indicate the directions of the velocity vectors of interest. Direction of VCB C 6
5
B Direction of VB
Direction of VC
0.806° 4
O2
y
2 A 3 4 45.000°
O4 x
DESIGN OF MACHINERY - 5th Ed.
8.
SOLUTION MANUAL 7-87-3
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point C. The equation to be solved graphically is VC = VB + VCB a. Choose a convenient velocity scale and layout the known vector VB. b. From the tip of VB, draw a construction line with the direction of VCB, magnitude unknown. c. From the tail of VB, draw a construction line with the direction of VC, magnitude unknown. d. Complete the vector triangle by drawing VCB from the tip of VB to the intersection of the VC construction line and drawing VC from the tail of VB to the intersection of the VCB construction line.
9.
From the velocity triangle we have:
Velocity scale factor: VC 1.669 in kv
kv
5 in sec in
VC 8.345
1.730"
0.434"
1
VB
VCB
in
y
VC
sec
0
θVC 180 deg
5 in/sec
x
VCB 0.434 in kv
VCB 2.170
in
1.669"
sec
10. Determine the angular velocity of link 5 using equation 6.7. θ 270.806 deg
From the linkage layout above: ω
VCB
ω 0.111
L5
rad
CCW
sec
11. For the acceleration of point A, use equation 7.19: (AA2t + AA2n) = AA4t + AA4n + AAcor + AAslip , where 2
AA2n L2 ω
AA2n 19.200 in sec
θAA θ 180 deg
θAA 225.000 deg
2
2
AA4n c ω
AA4n 1.755 in sec
θAA4n θ 180 deg
θAA4n 345.481 deg
AAcor 2 Vslip ω
AAcor 5.956 in sec
θAAcor θ 90 deg
θAAcor 75.481 deg
(Vslip is negative)
2
2
AA2t L2 α
AA2t 0.000 in sec
AAcor AAslip
AA2
2.259" AA4 t AA4
n AA4
13. From the acceleration polygon above,
y
2.254" x
12. Repeat procedure of steps 3 and 7 for the equation in step 11. (See next page).
2
0
10 in/sec
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-87-4
ka 10
Acceleration scale factor
in sec
2 2
AA4 2.259 ka
AA4 22.6 in sec
AA4t 2.254 ka
AA4t 22.5 in sec
at an angle of -100.285 deg
2
The angular acceleration of link 4 is α
AA4t
α 1.665
c
rad sec
CCW
2
14. The graphical solution for accelerations in pin-jointed fourbars uses equation 7.4: (APt + APn) = (AAt + AAn) + (APAt + APAn) 15. For point C, this becomes: AC = (ABt + ABn) + (ACBt + ACBn) , where ABn L4 ω
2
ABn 3.112 in sec
2
θABn θ 180 deg
θABn 345.481 deg
ABt L4 α
ABt 39.96 in sec
θABt θ 90 deg
θABt 75.481 deg
2
2
2
ACBn L5 ω
ACBn 0.241 in sec
θACBn θ
θACBn 270.806 deg
16. Repeat procedure of step 12 for the equation in step 15.
3.748" AC
ABn
y t ACB
x 0
10 in/sec
ABt n ACB
17. From the acceleration polygon above, Acceleration scale factor
ka 10
in sec
AC 3.748 ka
2
AC 37.5 in sec
2
at an angle of 90 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-88-1
PROBLEM 7-88 Statement:
Repeat Problem 7-87 using an analytical method to calculate the acceleration of point C for one revolution of the input crank (link 2).
Given:
Link lengths and angles: Link 1 (O2O4)
d 16.50 in
Link 2 (L2)
a 4.80 in
Link 4 (L4)
a' 24.00 in
Link 5 (L5)
b' 19.50 in ω 2 rad sec
Angular velocity of link 2 Solution: 1.
d' 22.965 in
Vertical offset from O4 to C
1
α 0 rad sec
2
See Figure 3-14 and Mathcad file P0788.
Draw the mechanism to scale and define a vector loop for the input portion (links 1, 2, 3, and 4) using the fourbar crank-slider derivation in Section 7.3 as a model. θ 0 deg 0.5 deg 360 deg
c'
C 6
5
B
4
O2
y
2
A 3
d' 4
2 y R2
O4 R1
x R3
x
DESIGN OF MACHINERY - 5th Ed.
2.
SOLUTION MANUAL 7-88-2
Write the vector loop equation, expand the result and separate into real and imaginary parts to solve for 3 and b (distance from O4 to A). R2 R1 R3 a e
j θ
d b e
j θ
Substituting the Euler equivalents,
d bcosθ j sinθ
a cos θ j sin θ
Separating into real and imaginary components and solving for 3 and b.
a sin θ sin θ θ
θ θ atan2 a cos θ d a sin θ
b θ 3.
Differentiate the position equation, expand it and solve for 3 and bdot. a j ω e
j θ
ω θ
a ω
b θ
j θ
bdot e
j θ
cos θ θ θ
sin θ θ
a ω cos θ b θ ω θ cos θ θ
bdot θ 4.
b j ω e
Differentiate the velocity equation, expand it and solve for 3. a j e
j θ
2
2 j θ
a j ω e
bdot j ω e 2
j θ
2 j θ
b j ω e
α θ 5.
1
b θ
j θ
bddot e
j θ
bdot j ω e
j θ
a α cos θ θ θ
a ω2 sinθθ θ 2 bdotθ ωθ
The position, angular velocity, and angular acceleration are the same for links 3 and 4.
θ θ θ θ 6.
b j e
ω θ ω θ
α θ α θ
Write a vector loop equation for links 1, 4, 5, and 6, expand the result and separate into real and imaginary parts to solve for 5 and c' (horizontal distance from O4 to C).
3 π j θ j θ j π 2 c' e d' e = a' e b' e j
R6 R7 = R4 R5 Substituting the Euler equivalents,
b' cosθ j sinθ
j c' d' = a' cos θ j sin θ
Separating into real and imaginary components and solving for 5 and c'.
a' cos θ θ d' b'
θ θ acos
b' sinθθ
c' θ a' sin θ θ
DESIGN OF MACHINERY - 5th Ed.
7.
SOLUTION MANUAL 7-88-3
Differentiate the position equation, expand it and solve for 5 and VC.
3 π j θ j θ 2 VC e = a' ω j e b' ω j e j
Substituting the Euler equivalents,
b' ω sinθ j cosθ
VC = a' ω sin θ j cos θ
Separating into real and imaginary components and solving for 5 and VC.
VC θ a' ω θ cos θ θ b' ω θ cos θ θ a' sin θ θ ω θ ω θ b' sin θ θ
Differentiate the position equation, expand it and solve for 5 and AC.
3 π 2 j θ α ej θ b' j j ω 2 ej θ α ej θ 2 AC e = a' j j ω e j
Substituting the Euler equivalents,
a' sinθ j cosθ 2 b' ω cos θ j sin θ b' sin θ j cos θ 2
AC = a' ω cos θ j sin θ
Separating into real and imaginary components and solving for 5 and AC.
a' ωθ2 sinθθ b' ωθ2 cosθθ b' sin θ θ
a' α θ cos θ θ
a' ωθ2 sinθθ 2 b' α θ cos θ θ b' ω θ sin θ θ
α θ
AC θ a' α θ cos θ θ
ACCELERATION OF POINT C 100
Acceleration, in/sec^2
8.
50
AC θ
2
sec
0
in
50
100
0
45
90
135
180
225
θ deg Crank Angle, deg
270
315
360
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-89-1
PROBLEM 7-89 Statement:
Figure P7-22 shows a mechanism with dimensions. Use a graphical method to determine the accelerations of points A and B for the position shown. Ignore links 5 and 6. ω 2 = 24 rad/s clock-wise.
Given:
Link lengths and angles: Link 1 (O2O4)
d 1.22 in
Angle O2O4 makes with X axis
θ 56.5 deg
Link 2 (O2A)
a 1.35 in
Angle O2A makes with X axis
θ 14 deg
Link 4 (O4B)
e 1.36 in ω 24 rad sec
Motion of link 2 Solution: 1.
1
α 0 rad sec
CW
2
See Figure P7-22 and Mathcad file P0789.
Draw the linkage to scale and indicate the axes of slip and transmission as well as the directions of velocities of interest.
B
Axis of slip Y
Axis of transmission O4
4
0.939
132.661° A 3 2 X O2
Direction of VA3
2.
Use equation 6.7 to calculate the magnitude of the velocity at point A on links 2 and 3. VA3 a ω
3.
VA3 32.400
in sec
θVA3 θ 90 deg
θVA3 76.0 deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity components at point A on link 3. The equation to be solved graphically is VA3 = Vtrans + Vslip a. Choose a convenient velocity scale and layout the known vector VA3. b. From the tip of VA3, draw a construction line with the direction of Vslip, magnitude unknown. c. From the tail of VA3, draw a construction line with the direction of Vtrans, magnitude unknown. d. Complete the vector triangle by drawing Vslip from the tip of Vtrans to the tip of VA3 and drawing Vtrans from the tail of VA3 to the intersection of the Vslip construction line.
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-89-2
Y 0
12 in/sec
1.295" X Vtrans
VA3 2.369"
4.
From the velocity triangle we have: Velocity scale factor:
5.
kv
12 in sec in
in
Vslip 28.428
Vtrans 1.295 in kv
Vtrans 15.540
sec in sec
The true velocity of point A on link 4 is Vtrans, VA4 15.54
in sec
Determine the angular velocity of link 4 using equation 6.7. From the linkage layout above: ω
7.
1
Vslip 2.369 in kv
VA4 Vtrans 6.
Vslip
VA4
c 0.939 in and
ω 16.550
c
rad
CW
sec
Determine the magnitude and sense of the vector VB using equation 6.7. VB e ω
VB 22.507
in sec
θVA4 θ 90 deg 8.
θ 132.661 deg
θVA4 42.661 deg
For the acceleration of point A, use equation 7.19: (AA2t + AA2n) = AA4t + AA4n + AAcor + AAslip , where AA2n a ω
2
AA2n 777.600 in sec
θAA θ 180 deg
θAA 194.000 deg
AA2t a α
AA2t 0.000 in sec 2
AA4n c ω
2
2
AA4n 257.2 in sec
2
DESIGN OF MACHINERY - 5th Ed.
9.
SOLUTION MANUAL 7-89-3
θAA4n θ
θAA4n 132.661 deg
AAcor 2 Vslip ω
AAcor 940.940 in sec
θAAcor θ 90 deg
θAAcor 222.661 deg
2
Repeat procedure of steps 3 and 7 for the equation in step 8.
Y cor
AA
2.572" t
A A4 0
A A4
100 IN/S/S
3.647"
n A A4
87.504° X
cor AA
A A2 10. From the acceleration polygon above, Acceleration scale factor
ka 100
in sec
2 2
AA4 3.647 ka
AA4 364.7 in sec
AA4t 2.572 ka
AA4t 257.2 in sec
at an angle of 87.50 deg
2
The angular acceleration of link 4 is α
AA4t
α 273.9
c
rad sec
CCW
2
11. For point B: ABn e ω
2
ABn 372.5 in sec
2
θABn θ 180 deg
θABn 312.661 deg
ABt e α
ABt 372.5 in sec
θABt θ 90 deg
θABt 222.661 deg
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-90-1
PROBLEM 7-90 Statement:
Figure P7-22 shows a mechanism with dimensions. Use an analytical method to calculate the accelerations of points A and B for the position shown. Ignore links 5 and 6. ω 2 = 24 rad/s clock-wise.
Given:
Link lengths and angles: Link 1 (O2O4)
d 1.22 in
Angle link 2 makes with X axis θ 42.5 deg
Link 2 (O2A)
a 1.35 in
Angle link 3 makes with X axis θ 103.839 deg
Link 4 (O4B)
c 1.36 in ω 24 rad sec
Motion of link 2 Solution: 1.
1
α 0 rad sec
2
See Figure P7-22 and Mathcad file P0790.
Draw the mechanism to scale and define a vector loop for links 1, 2, 3, and 4 using the four- bar crank-slider derivation in Section 7.3 as a model.
B Axis of slip X b = 0.939"
O4
4
103.839°
R3
d = 1.220"
Y
Axis of transmission
R1 42.500°
A 3
2
R2 O2 a = 1.350"
2.
Write the vector loop equation, expand the result and separate into real and imaginary parts to solve for 3 and b (distance from O4 to A). R2 = R1 R3 a e
j θ
= d b e
j θ
Substituting the Euler equivalents,
= d bcosθ j sinθ
a cos θ j sin θ
Separating into real and imaginary components and solving for b.
DESIGN OF MACHINERY - 5th Ed.
b 3.
SOLUTION MANUAL 7-90-2
sin θ
a sin θ
b 0.939 in
Differentiate the position equation, expand it and solve for 3 and bdot. a j ω e ω
j θ
a ω b
bdot
= b j ω e
j θ
bdot e
j θ
cos θ θ
a ω cos θ b ω cos θ
ω 16.544
rad
bdot 28.430
in
sin θ 4.
s
Differentiate the velocity equation, expand it and solve for 3. a j e
j θ
2
2 j θ
a j ω e
= bdot j ω e
j θ
α
1 b
b j e
2 j θ
2
b j ω e
5.
s
j θ
bddot e
j θ
bdot j ω e
a α cos θ θ a ω sin θ θ 2 bdot ω
2
j θ
α 275.059
rad 2
s
Determine the acceleration of point A on link 2 (in the local XY coordinate system) using equations 7.13a.
a ω2 cosθ j sinθ
AA2 a α sin θ j cos θ AA2 AA2
AA2 777.600
θAA2 arg AA2
in 2
θAA2 137.500 deg
s 6.
Determine the acceleration of points A and B on link 4 (in the local XY coordinate system) using equations 7.13a. Rename 3 to θ4:
θ θ
Rename 3 to 4:
α α ω ω
b ω2 cosθ j sinθ
AA4 b α sin θ j cos θ AA4 AA4
AA4 364.484
in 2
θAA4 arg AA4
θAA4 31.019 deg
s
a ω2 cosθ j sinθ
AB a α sin θ j cos θ AB AB
AB 523.844
in 2
s
θAB arg AB
θAB 31.019 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-91-1
PROBLEM 7-91 Statement:
Figure P7-23 shows a quick-return mechanism with dimensions. Use a graphical method to calculate the accelerations of points A, B, and C for the position shown. 2 = 16 rad/s. Ignore links 5 and 6.
Given:
Link lengths and angles: Link 1 (O2O4)
d 1.69 in
Angle O2O4 makes with X axis
Link 2 (L2)
a 1.00 in
Angle link 2 makes with X axis θ 99 deg
Link 4 (L4)
e 4.76 in
Angular velocity of link 2 Solution: 1.
ω 16 rad sec
1
CCW
θ 195.5 deg
α 0 rad sec
2
See Figure P7-23 and Mathcad file P0791.
Draw the linkage to scale and indicate the axes of slip and transmission as well as the directions of velocities of interest.
B
Axis of transmission
Direction of VA3
4
Y Axis of slip A 3 2.068
2 44.228° O2
X O4
2.
Use equation 6.7 to calculate the magnitude of the velocity at point A on links 2 and 3. VA3 a ω
3.
VA3 16.000
in sec
θVA3 θ 90 deg
θVA3 189.0 deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity components at point A on link 3. The equation to be solved graphically is VA3 = Vtrans + Vslip
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-91-2
a. Choose a convenient velocity scale and layout the known vector VA3. b. From the tip of VA3, draw a construction line with the direction of Vslip, magnitude unknown. c. From the tail of VA3, draw a construction line with the direction of Vtrans, magnitude unknown. d. Complete the vector triangle by drawing Vslip from the tip of Vtrans to the tip of VA3 and drawing Vtrans from the tail of VA3 to the intersection of the Vslip construction line.
Y 0
8 in/sec
X
VA3
Vtrans
4.
From the velocity triangle we have: Velocity scale factor:
5.
kv
in
Vtrans 1.154 in kv
Vtrans 9.232
in sec in sec
The true velocity of point A on link 4 is Vtrans, VA4 9.23
in sec
Determine the angular velocity of link 4 using equation 6.7.
ω
VA4
c 2.068 in and
ω 4.464
c
rad
θ 44.228 deg
CCW
sec
Determine the magnitude and sense of the vector VB using equation 6.7. VB e ω
VB 21.250
θVA4 θ 90 deg 8.
1
Vslip 13.072
From the linkage layout above:
7.
8 in sec
Vslip 1.634 in kv
VA4 Vtrans 6.
1.634"
Vslip
1.154"
in sec
θVA4 45.772 deg
For the acceleration of point A, use equation 7.19: (AA2t + AA2n) = AA4t + AA4n + AAcor + AAslip , where AA2n a ω
2
AA2n 256.000 in sec
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-91-3
θAA θ 180 deg
θAA 279.000 deg
AA2t a α
AA2t 0.000 in sec 2
9.
2 2
AA4n c ω
AA4n 41.214 in sec
θAA4n θ 180 deg
θAA4n 224.228 deg
AAcor 2 Vslip ω
AAcor 116.712 in sec
θAAcor θ 90 deg
θAAcor 134.228 deg
2
Repeat procedure of steps 3 and 7 for the equation in step 8. 0
Y
25 IN/S/S
Acceleration Scale
n
X
AA4
4.042" 69.810°
AA4 t A4
A 3.711"
slip
AA
AA2 cor
AA
10. From the acceleration polygon above, Acceleration scale factor
ka 25
in sec
2
AA4 4.042 ka
AA4 101.0 in sec
AA4t 3.711 ka
AA4t 92.8 in sec
2
at an angle of -69.81 deg
2
The angular acceleration of link 4 is α
AA4t c
α 44.9
rad sec
2
CW
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-91-4
11. For point B: ABn e ω
2
ABn 94.9 in sec
2
θABn θ 180 deg
θABn 224.228 deg
ABt e α
ABt 213.5 in sec
θABt θ 90 deg
θABt 134.228 deg
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-92-1
PROBLEM 7-92 Statement:
Figure P7-23 shows a quick-return mechanism with dimensions. Use an analytical method to calculate the accelerations of points A and B for the position shown. Ignore links 5 and 6. ω 2 = 16 rad/s.
Given:
Link lengths and angles: Link 1 (O2O4)
d 1.69 in
Angle link 2 makes with X axis θ 96.517 deg
Link 2 (O2A)
a 1.00 in
Angle link 3 makes with X axis θ 151.289 deg
Link 4 (O4B)
e 4.76 in
Motion of link 2 Solution: 1.
ω 16 rad sec
1
α 0 rad sec
2
See Figure P7-23 and Mathcad file P0792.
Draw the mechanism to scale and define a vector loop for links 1, 2, 3, and 4 using the four- bar crank-slider derivation in Section 7.3 as a model.
B
Axis of transmission 4
Axis of slip A 3 2.068" 1.000"
R2 151.289°
96.517° R3
2
O2 R1
X O4
1.690
Y
2.
Write the vector loop equation, expand the result and separate into real and imaginary parts to solve for 3 and b (distance from O4 to A). R2 = R1 R3
DESIGN OF MACHINERY - 5th Ed.
a e
j θ
= d b e
SOLUTION MANUAL 7-92-2 j θ
Substituting the Euler equivalents,
= d bcosθ j sinθ
a cos θ j sin θ
Separating into real and imaginary components and solving for b. b 3.
sin θ
a sin θ
b 2.068 in
Differentiate the position equation, expand it and solve for 3 and bdot. a j ω e ω
j θ
a ω b
bdot
= b j ω e
j θ
bdot e
j θ
cos θ θ
ω 4.463
a ω cos θ b ω cos θ
rad s
bdot 13.070
in
sin θ 4.
Differentiate the velocity equation, expand it and solve for 3. a j e
j θ
2
2 j θ
a j ω e
= bdot j ω e
j θ
α
1 b
b j e
2 j θ
2
b j ω e
5.
s
j θ
bddot e
j θ
bdot j ω e
a α cos θ θ a ω sin θ θ 2 bdot ω
2
j θ
α 44.710
rad 2
s
Determine the acceleration of point A on link 2 (in the local XY coordinate system) using equations 7.13a.
a ω2 cosθ j sinθ
AA2 a α sin θ j cos θ AA2 AA2
AA2 256.000
θAA2 arg AA2
in 2
θAA2 83.483 deg
s 6.
Determine the acceleration of points A and B on link 4 (in the local XY coordinate system) using equations 7.13a. Rename 3 to θ4:
θ θ
Rename 3 to 4:
α α
ω ω
b ω2 cosθ j sinθ
AA4 b α sin θ j cos θ AA4 AA4
AA4 101.226
in 2
θAA4 arg AA4
θAA4 94.703 deg
s
a ω cosθ j sinθ
AB a α sin θ j cos θ AB AB
AB 48.944
2
in 2
s
θAB arg AB
θAB 94.703 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-93a-1
PROBLEM 7-93a Statement:
Given:
The general linkage configuration and terminology for an offset fourbar slider-crank linkage are shown in Fig P7-2. The link lengths and the values of d, d-dot, and d-double-dot are defined in Table P7-5. For row a, find the acceleration of the pin joint A and the angular acceleration of the crank using a graphical method. Link lengths: a 1.4 in
Link 2 Offset Solution: 1.
c 1 in
Link 3
b 4 in
θ2 176.041 deg
ddot 10 in sec
1
dddot 0 in sec
2
See Figure P7-2, Table P7-5, and Mathcad file P0793a.
Draw the linkage to scale and indicate the axes of slip and transmission as well as the directions of velocities of interest.
Direction of VBA 13.052° B
. d = VB
176.041° A
000 b = 4. a = 1.400
c = 1.000"
O2
d = 2.500" 0
10 inches/sec
Direction of VA 2.
Use equation 6.6.24b to (graphically) determine the magnitude of the velocity at point A. The equation to be solved graphically is
VA
VA = VB + VBA a. Choose a convenient velocity scale and layout the known vector VB. b. From the tip of VB, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VB, draw a construction line with the direction of VA, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VB to the intersection of the VA construction line and drawing VA from the tail of VB to the intersection of the VBA construction line. 3.
VBA
3.414" 3.330"
Y
86.041°
From the velocity triangle we have:
103.052°
VB
1.000"
X
DESIGN OF MACHINERY - 5th Ed.
Velocity scale factor:
SOLUTION MANUAL 7-93a-2
kv
10 in sec
1
in in
VA 3.330 in kv
VA 33.300
VBA 3.414 in kv
VBA 34.140
θVB 86.041 deg
sec in
θVBA 103.052 deg 180deg
sec
θVBA 76.948 deg 4.
Use equation 6.7 to find the angular velocity of the crank (link 2).
ω2 ω3
VA
ω2 23.786
a VBA
ω3 8.535
b
rad
CW
sec
rad
CW
sec
5.
The graphical solution for accelerations uses equation 7.4:
6.
For point A, this becomes:
AAt + AAn = ABt - (ABAt + ABAn) , where
2
AAn a ω2
AAn 792.064 in sec
θAAn θ2 180 deg
θAAn 356.041 deg
ABt dddot
ABt 0.000 in sec
θABt 0 deg
θABt 0.000 deg
2
4.
(APt + APn) = (AAt + AAn) + (APAt + APAn)
2
2
ABAn b ω3
ABAn 291.385 in sec
θABAn 13.052 deg 180 deg
θABAn 193.052 deg
2
Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw AAn at an angle of AAn. From the origin, draw -ABAt at an angle of ABt. Now that the vectors with known magnitudes are drawn, from the tip AAn and the tip of -ABAt, draw construction lines in the directions of AAt and ABAn, respectively. The intersection of these two lines are the tips of AAt and ABAn.
5.
From the graphical solution below, in
Acceleration scale factor ka 200
sec
2 2
AA 8.895 ka
AA 1779 in sec
AAt 7.965 ka
AAt 1593 in sec
α2
AAt a
α2 1138
rad sec
2
2
at an angle of -67.52 deg
DESIGN OF MACHINERY - 5th Ed.
Y
SOLUTION MANUAL 7-93a-3
0
200 IN/S/S
Acceleration Scale
n -A BA
X
n
AA
67.521°
t
-A A4
AA t
AA
7.965" 8.895"
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-94a-1
PROBLEM 7-94a The general linkage configuration and terminology for an offset fourbar slider-crank linkage are shown in Fig P7-2. The link lengths and the values of d, d-dot and d-double-dot are defined in Table P7-5. For row a, find the acceleration of pin joint A and the angular acceleration of the crank using the analytic method. Draw the linkage to scale and label it before setting up the equations. Link lengths:
Statement:
Given:
Link 2 (O2 A)
a 1.4 in
Slider position
d 2.5 in dddot 0
Slider acceleration:
b 4 in
Link 3 (AB) Slider velocity in
ddot 10
Offset (yB) in s
2
s
See Figure P7-2, Table P7-5, and Mathcad file P0794a.
Solution: 1.
Draw the linkage to scale and label it. 13.052° B
. .. d d
176.041° 000 b = 4.
A
a = 1.400
c = 1.000"
O2
d = 2.500"
2.
Determine the open value of 2 using equations 4.20 and 4.21. 2
2
2
K1 a b c d
2
2
K2 2 a c
K2 2.8 in
K3 2 a d
K3 7 in
A K1 K3
A 0.21 in
B 2 K2
B 5.6 in
C K1 K3
C 13.79 in
2 2
2 2
2
θ2 2 atan2 2 A B 3.
2
K1 6.79 in
B 4 A C
θ2 176.041 deg
Determine the value of 3 using equation 4.16a or 4.17.
β ( a b c d α)
θ asin
a sin( α) c
b
d 1 a cos( α) b cos( θ ) return θ if d 1 = d asin
a sin( α) c b
π otherwise
c 1 in
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 7-94a-2
θ3 β a b c d θ2 4.
Determine the value of ω 2 and ω 3 using equations 6.24.
ω2
ω3 5.
6.
θ3 193.052 deg
ddot cos θ3
ω2 23.785
a cos θ2 sin θ3 sin θ2 cos θ3 a ω2 cos θ2
ω3 8.525
b cos θ3
s
rad s
Determine the value of VA using equation 6.7. in
VA a ω2
VA 33.299
θVA θ2 sign ω2 90 deg
θVA 86.041 deg
s
Determine the value of α2 using equation 7.17b. a ω2 cos θ2 cos θ3 sin θ2 sin θ3 b ω3 dddot cos θ3 2
α2
α2 1139.37
2
a cos θ2 sin θ3 sin θ2 cos θ3
rad 2
s 7.
rad
Determine the value of AA using equation 7.3. AA a α2 sin θ2 j cos θ2 a ω2 cos θ2 j sin θ2 2
AA AA
AA 1780.93
in sec
θAA arg AA
2
θAA 67.553 deg
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 8-1-1
PROBLEM 8-1 Statement:
Figure P8-1 shows the cam and follower from Problem 6-65. Using graphical methods, find and sketch the equivalent fourbar linkage for this position of the cam and follower.
Solution:
See Figure P8-1 and Mathcad file P0801.
1.
Draw the cam and follower to scale. Axis of slip
Axis of transmission
4
2
O2
2.
O4
As described in Section 8.1 and Figure 8-1, find the centers of curvature of the cams, which are located on the common normal (axis of transmission). They will be the locations of the moving pivots of the effective pin-jointed fourbar linkage.
A 3 B
2
O2
4
O4
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 8-2-1
PROBLEM 8-2 Statement:
Figure P8-1 shows the cam and follower from Problem 6-65. Using graphical methods, find the pressure angle at the position shown.
Solution:
See Figure P8-1 and Mathcad file P0802.
1.
Draw the cam and follower to scale. Axis of slip
Axis of transmission
2
4
O2
2.
O4
As defined in Section 8.6 and Figure 8-41, the pressure angle is the angle between the direction of motion of the follower and the direction of the axis of transmission. Establish those two directions and measure the angle between them. Direction of motion, link 4
55.3° Axis of transmission
2
O2
4
O4
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 8-3-1
PROBLEM 8-3 Statement:
Figure P8-2 shows the cam and follower. Using graphical methods, find and sketch the equivalent fourbar linkage for this position of the cam and follower.
Solution:
See Figure P8-2 and Mathcad file P0803.
1.
Draw the cam and follower to scale.
4 Common Normal
3
2
2.
As described in Section 8.1 and Figure 8-1, find the centers of curvature of the cams, which are located on the common normal (axis of transmission). They will be the locations of the moving pivots of the effective pin-jointed fourbar linkage.
4
O4
B
3
A 2 O2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 8-4-1
PROBLEM 8-4 Statement:
Figure P8-2 shows the cam and follower. Using graphical methods, find the pressure angle at the position shown.
Solution:
See Figure P8-2 and Mathcad file P0804.
1.
Draw the cam and follower to scale.
4 Common Normal
3
2
2.
As defined in Section 8.6 and Figure 8-41, the pressure angle is the angle between the direction of motion of the follower and the direction of the axis of transmission. Establish those two directions and measure the angle between them. Direction of motion, link 4 Common Normal 4.9° 4
3
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 8-5-1
PROBLEM 8-5 Statement:
Figure P8-3 shows the cam and follower. Using graphical methods, find and sketch the equivalent fourbar linkage for this position of the cam and follower.
Solution:
See Figure P8-3 and Mathcad file P0805.
1.
Draw the cam and follower to scale. Common Normal
4
3
2
2.
As described in Section 8.1 and Figure 8-1, find the centers of curvature of the cams, which are located on the common normal (axis of transmission). They will be the locations of the moving pivots of the effective pin-jointed fourbar linkage.
Slider 1
4
3 2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 8-6-1
PROBLEM 8-6 Statement:
Figure P8-3 shows the cam and follower. Using graphical methods, find the pressure angle at the position shown.
Solution:
See Figure P8-3 and Mathcad file P0806.
1.
Draw the cam and follower to scale. Common Normal
4
3
2
2.
As defined in Section 8.6 and Figure 8-41, the pressure angle is the angle between the direction of motion of the follower and the direction of the axis of transmission. Establish those two directions and measure the angle between them.
Direction of motion, link 4 13.8° Common Normal 4
3
2
DESIGN OF MACHINERY - 5th Ed.
SOLUTION MANUAL 8-7-1
PROBLEM 8-7 Statement:
Given:
Design a double-dwell cam to move a follower from 0 to 2.5 in in 60 deg, dwell for 120 deg, fall 2.5 in in 30 deg and dwell for the remainder. The total cycle must take 4 sec. Choose suitable programs for rise and fall to minimize accelerations. Plot the s v a j diagrams. RISE
DWELL
β1 60 deg
β2 120 deg
h 1 2.5 in Cycle time: Solution: 1.
3.
DWELL
β3 30 deg
h 2 0 in
β4 150 deg
h 3 2.5 in
h 4 0 in
τ 4 sec
See Mathcad file P0807.
The camshaft turns 2 rad during the time for one cycle. Thus, its speed is ω
2.
FALL
2 π rad
ω 1.571
rad
sec τ From Table 8-3, the motion program with lowest acceleration that does not have infinite jerk is the modified trapezoidal. The modified trapezoidal motion is defined in local coordinates by equations 8.15 through 8.19. The numerical constants in these SCCA equations are given in Table 8-2. b 0.25
c 0.50
d 0.25
Cv 2.0000
Ca 4.8881
Cj 61.426
The SCCA equations for the rise or fall interval () are divided into 5 subintervals. These are: for 0
View more...
Comments
