DESIGN OF MACHINERY SOLUTION MANUAL Chapter 6
March 4, 2017 | Author: Kagrath | Category: N/A
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DESIGN OF MACHINERY SOLUTION MANUAL Chapter 6...
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DESIGN OF MACHINERY
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
naut_mile hr
Given:
Vs 20 knots
Speed of ship
t 15 deg
Vt 30 knots
Speed of torpedo
s 90 deg di 0.5 naut_mile
Initial distance between ship and torpedo
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 t
tt_east 62.117 s
Distance traveled by the torpedo due north in that time
dt_north Vt sin t tt_east
dt_north 0.134naut_mile
Distance traveled by the ship due north in that time ds_north Vs tt_east
ds_north 0.345naut_mile
Distance by which the torpedo will miss the ship dmiss ds_north dt_north
dmiss 0.211naut_mile
-Vs 22.89° V ts
DESIGN OF MACHINERY
SOLUTION MANUAL 6-1b-1
PROBLEM 6-1b Statement:
Given:
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. Speed of first plane
V1 500 mph
Speed of second plane
V2 550 mph
Initial distance between planes
di 40 mi
1 270 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
V2 = 550 mph
3.
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.
4.
The angle can also be determined analytically from the velocity triangle as follows.
V1 acos V2 5.
24.620° V2
24.620 deg
V1
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
t 628.468 s
t 10.474 min
DESIGN OF MACHINERY
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:
Solution: 1.
Rotation speed
rad 100 sec
Vector angles
A 45 deg
Vector magnitude
R 6.5 in
B 75 deg
See Mathcad file P0602.
Calculate the magnitude of the velocity at points A and B using equation 6.3. V R
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
Polar form:
SOLUTION MANUAL 6-2-2 j 4
j A
RA R e
RA 6.5 e
j 4
j A
VA R j e Cartesian form:
VA 650e j
VA R j cos A j sin A VA ( 459.619 459.619j)
b.
in sec
Write an expression for the particle's velocity vector in position B using complex number notation, in both polar and Cartesian forms. Polar form:
j 4
j B
RB R e
RA 6.5 e
75 j 180
j B
VB R j e
Cartesian form:
VB 650e j
VB R j cos B j sin B VB ( 627.852 168.232j)
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. VBA VB VA
d.
in sec
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 in kv 250 sec
0
1.166
Horizontal component
Velocity scale:
VA VBA
VBAx 0.673 kv VBAx 168.3
500 in/sec
VB
in sec
Ov
X
0.673
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
SOLUTION MANUAL 6-3-1
PROBLEM 6-3 Statement:
Given:
Solution: 1.
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. Rotation speeds
rad 50 sec
rad 75 sec
Vector angles
A 45 deg
B 75 deg
Vector magnitude
R 6.5 in
See Mathcad file P0603.
Calculate the magnitude of the velocity at points A and B using equation 6.3. VA R
VA 325.000
VB R
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 0
8 VB
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
Polar form:
SOLUTION MANUAL 6-3-2 j 4
j A
RA R e
RA 6.5 e
j 4
j A
VA R j e Cartesian form:
VA 325e j
VA R j cos A j sin A VA ( 229.810 229.810j)
b.
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:
j 4
j B
RB R e
RA 6.5 e
75 j 180
j B
VB R j e Cartesian form:
VB 487.5e j
VB R j cos B j sin B VB ( 470.889 126.174j)
c.
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.
in
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 in kv 250 sec
0
Velocity scale:
Horizontal component VBAx 2.803 kv VBAx 700.7
500 in/sec
VB 1.424
Ov
VBA
VA
in sec
X
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
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
30 deg
Crank angle:
10 rad sec
Crank velocity:
1
Coupler point data: Rpa 6 in Solution: 1.
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 I 1,3
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
117.2861 deg
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
3.
SOLUTION MANUAL 6-4a-2
in
VA a
VA 20.0
VA 90 deg
VA 120.0 deg
sec
Determine the angular velocity of link 3 using equation 6.9a.
VA AI13
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 in VB 35.925 sec VB 90 deg VB 27.286 deg
5.
Use equation 6.9c to determine the angular velocity of link 4.
6.
VB c
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
VP 40.778
VP 148.2007 deg 90 deg
VP 58.201 deg
sec
DESIGN OF MACHINERY
SOLUTION MANUAL 6-5a-1
PROBLEM 6-5a Statement:
Given:
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. Link lengths: d 6
Link 1
Link 2
a 2
Rpa 6
Coupler point:
b 7
Link 3
c 9
Link 4
30 deg 30 deg
Link 2 position and velocity:
10
atan2 (x y) return 0.5 if x = 0 y 0
Two argument inverse tangent
return 1.5 if x = 0 y 0
y if x 0 return atan x Solution: 1.
y atan x otherwise
See Mathcad file P0605a.
Draw the linkage to scale and label it. y
2.
B
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K 1 3.0000 2
K3
OPEN
d 3
c
4
K 2 0.6667 2
2
a b c d
2
2 a c
88 .837°
K 3 2.0000
117.28 6°
A 2
O2
A cos K 1 K 2 cos K3
115.211°
B 2 sin
143.660 °
C K 1 K2 1 cos K 3 A 0.7113
B 1.0000
CROSSED
C 3.5566 B'
3.
Use equation 4.10b to find values of 4 for the open and crossed circuits. Open:
Crossed: 4.
2
2atan2 2 A B B 4 A C
477.286 deg
360 deg
117.286 deg
O4
2
2atan2 2 A B B 4 A C
216.340 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12.
x
DESIGN OF MACHINERY
K4
SOLUTION MANUAL 6-5a-2 2
d
K5
b
2
2
c d a b
2
K 4 0.8571
2 a b
D cos K 1 K 4 cos K 5
D 1.6774
E 2 sin
E 1.0000
F K 1 K4 1 cos K 5 5.
Crossed:
7.
F 2.5906
Use equation 4.13 to find values of 3 for the open and crossed circuits. Open:
6.
K 5 0.2857
2
2atan2 2 D E E 4 D F
448.837 deg
360 deg
88.837 deg
2
2atan2 2 D E E 4 D F
244.789 deg
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18.
a sin b sin
5.991
a sin c sin
3.992
Determine the velocity of points A and B for the open circuit using equations 6.19.
VA a sin j cos VA 10.000 17.321j
arg VA120 deg
VA 20
VB c sin j cos VB 31.928 16.470j 8.
arg VB27.286 deg
VB 35.926
Determine the velocity of the coupler point P for the open circuit using equations 6.36.
VPA Rpa sin j cos VPA 31.488 17.337j VP VA VPA VP 21.488 34.658j 9.
VP 40.779
arg VP 58.201 deg
Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18.
a sin b sin
0.662
a sin c sin
2.662
10. Determine the velocity of point B for the crossed circuit using equations 6.19.
DESIGN OF MACHINERY
SOLUTION MANUAL 6-5a-3
VB c sin j cos VB 14.195 19.295j
arg VB126.340 deg
VB 23.954
11. Determine the velocity of the coupler point P for the crossed circuit using equations 6.36.
VPA Rpa sin j cos VPA 3.960 0.332j VP VA VPA VP 13.960 16.989j
VP 21.989
arg VP 129.411 deg
DESIGN OF MACHINERY
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
45 deg
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
3
45.000°
Axis of slip and Direction of VB
B 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 VA 14.000 VA 90 deg sec
VA 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
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
VB 180 deg
Since the slip axis and the direction of the velocity of point B are parallel, Vslip = VB.
DESIGN OF MACHINERY
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
Offset (yB )
45 deg
Crank angular velocity
10
c 1
See Figure P6-2 and Mathcad file P0607a.
Draw the linkage to scale and label it. Y d2 = 3.010
d1 = 4.990
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:
sin c a asin b
0.144 deg
d2 a cos b cos
Open:
d2 3.010
sin c a asin b
d1 a cos b cos 3.
4.
180.144 deg d1 4.990
Determine the angular velocity of link 3 using equation 6.22a:
2.475
2.475
Open
a cos b cos
Crossed
a cos b cos
Determine the velocity of pin A using equation 6.23a:
VA a sin j cos VA 9.899 9.899i
VA 14.000
arg VA135.000 deg
DESIGN OF MACHINERY
5.
SOLUTION MANUAL 6-7a-2
Determine the velocity of pin B using equation 6.22b: Open
VB1 a sin b sin
VB1 9.875
Crossed
VB2 a sin b sin
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
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
90 deg
30 deg
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 30.000°
127.333° c 142. 666°
A
d
x 04
02
2.
3.
Use equation 6.7 to calculate the magnitude of the velocity at point A. in VA a VA 20.000 VA 90 deg sec
VA 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
52.667° VB 0.771
1
kv
10 in sec in
Y V BA
From the velocity triangle we have: Velocity scale factor:
10 in/sec
1.846
X
DESIGN OF MACHINERY
5.
SOLUTION MANUAL 6-8a-2
in
VB3 0.771 in kv
VB3 7.710
VBA3 1.846 in kv
VBA3 18.460
sec in sec
Determine the angular velocity of link 3 using equation 6.7. From the linkage layout above: b 1.793 in and
VBA3 b
10.296
rad
7.
142.666 deg CW
sec
The way in which link 3 slides in link 4 requires that 6.
VB3 52.667 deg
Determine the magnitude and sense of the vector VB4 using equation 6.7. in
VB4 c
VB4 41.182
VB4 90 deg
VB4 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
slip VB4 180 deg
slip 232.666 deg
sec
DESIGN OF MACHINERY
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
Link 2
c 4
Link 4
a 2
90 deg
30 deg
10
atan2 (x y) return 0.5 if x = 0 y 0
Two argument inverse tangent
return 1.5 if x = 0 y 0
y if x 0 return atan x y atan x otherwise Solution: 1.
See Mathcad file P0609a.
Draw the linkage to scale and label it. B y 90.0°
b
127.333° 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 4.8a and 4.10a.
Q a sin cos a cos d sin P a sin sin a cos d cos
3.
Q 4.268
R c sin
R 4.000
T 2 P
T 2.000
S R Q
S 0.268
U Q R
U 8.268
Use equation 4.22c to find values of 4 for the open and crossed circuits. OPEN CROSSED
4.
P 1.000
2 2 2 atan2 2 S T T 4 S U
2 atan2 2 S T T 4 S U
142.667 deg 169.041 deg
Use equation 4.18 to find values of 3 for the open and crossed circuits.
DESIGN OF MACHINERY
5.
6.
7.
SOLUTION MANUAL 6-9a-2
OPEN
232.667 deg
CROSSED
259.041 deg
Determine the magnitude of the instantaneous "length" of link 3 from equation 4.20a. OPEN
b1
CROSSED
b2
sin
a sin c sin
b1 1.793
sin
a sin c sin
b2 1.793
Determine the angular velocity of link 4 using equation 6.30c: OPEN
CROSSED
a cos b 1 c cos
10.292
a cos b 2 c cos
3.639
Determine the velocity of pin A using equation 6.23a:
VA a sin j cos VA 10.000 17.321i 8.
VA 20.000
Determine the velocity of point B on link 4 using equation 6.31: OPEN
VB4x1 c sin
VB4x1 24.966
VB4y1 32.734
VB4y1 c cos 2
VB41 VB4x1 VB4y1
2
VB41 41.168
VB1 atan2 VB4x1 VB4y1 CROSSED
VB4x2 c sin
2
VB42 VB4x2 VB4y2
VB1 52.667 deg VB4x2 2.767
VB4y2 c cos
VB4y2 14.289
2
VB42 14.555
VB2 atan2 VB4x2 VB4y2 9.
arg VA 120.000 deg
VB2 79.041 deg
Determine the slip velocity using equation 6.30a: OPEN
Vslip1
CROSSED
Vslip2
cos
cos
a sin b 1 sin c sin
a sin b 2 sin c sin
Vslip1 33.461
Vslip2 33.461
DESIGN OF MACHINERY
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
Given:
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. 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: 2 Solution: 1.
30 deg
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
f
r2 r5
r5 1.200in
3
r2 r 5
r2 2.400in
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
4
C
B 3
A
2
4.
1
Determine the angle of link 5 using the equation in Figure P6-4.
f r2 3 r5
3.
10 rad sec
See Figure P6-4 and Mathcad file P0610a.
2.
60 deg
5 O5
O2
Use equation 6.7 to calculate the magnitude of the velocity at points A and C. in
VA a
VA 10.00
20.000
rad
VC d
VC 80.00
in
sec
60 deg 90 deg
sec
sec
150 deg 90 deg
DESIGN OF MACHINERY
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:
0
50 in/sec
1
Velocity scale factor:
kv
Y
50 in sec
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.
VBA b VBC c
Both links are rotating CCW.
32.586 16.950
4.562
rad
3.051
sec rad sec
V BA V BC
DESIGN OF MACHINERY
SOLUTION MANUAL 6-11a-1
PROBLEM 6-11a Statement:
Given:
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
60 deg
Gear ratio
2.0
Phase angle
30 deg
10
atan2 (x y) return 0.5 if x = 0 y 0
Two argument inverse tangent
return 1.5 if x = 0 y 0
y if x 0 return atan x y atan x otherwise Solution: 1.
See Figure P6-4 and Mathcad file P0611a.
Draw the linkage to scale and label it. y C
4
B
1 77.715 2°
173.6 421° 3
5 12 4.050 1°
2
x
O2
3
150. 0000°
11 5.4074 °
O5
4
B`
2.
Determine the values of the constants needed for finding 3 and 4 from equations 4.24h and 4.24i.
B 2 c a sin dsin A 2 c d cos a cos f
2
2
2
2
2
B 20.412
C a b c d f 2 af cos 2 d a cos f cos 2 a d sin sin
A 36.646
C 37.431
D C A
D 0.785
E 2 B
E 40.823
F A C
F 74.077
DESIGN OF MACHINERY
SOLUTION MANUAL 6-11a-2
H 2 b d sin asin
G 2 b d cos a cos f
2
2
2
2
2
H 15.876
K a b c d f 2 af cos 2 da cos f cos
2 a d sin sin
3.
K 26.569
L K G
L 1.933
M 2 H
M 31.751
N G K
N 55.072
Use equations 4.24h and 4.24i to find values of 3 and 4 for the open and crossed circuits. OPEN
CROSSED
4.
G 28.503
2 atan22D E E2 4DF 2 atan22LM M2 4LN 2 atan22D E E2 4 DF 2
2atan2 2 L M M 4 L N
173.642 deg
177.715 deg
115.407 deg
124.050 deg
Determine the position and angular velocity of gear 5 from equations 4.23c and 6.32c
150.000 deg
20.000
Angular velocity of links 3 and 4 from equations 6.33 OPEN
32.585
CCW
c sin
a sin b sin d sin
16.948
CROSSED
2 sin a sin d sin bcos 2 cos
CCW
2 sin a sin d sin bcos 2 cos 75.191
CW
c sin
a sin b sin d sin
59.554
CW
CCW
DESIGN OF MACHINERY
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)
C 6
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
2.
n ( n 1) 2
C 6
2,3 at contact point (behind link 4) 1,3
Draw the linkage to scale and identify those ICs that can be found by inspection, which in this case, is all of them.
1 3 A 4
c.
This is a fourbar with n 4.
2 O
1.
Determine the number of instant centers for this mechanism using equation 6.8a.
2
1,2 and 2,4 and 1,4
DESIGN OF MACHINERY
C 2.
SOLUTION MANUAL 6-12-2
n ( n 1)
C 6
2
Draw the linkage to scale and identify those ICs that can be found by inspection. 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
4
O2
2,4 at infinity
B
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)
C 6
2
Draw the linkage to scale and identify those ICs that can be found by inspection. 2,3
3,4
1,2 3 O2 A
4
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
DESIGN OF MACHINERY
SOLUTION MANUAL 6-12-3 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. C
2.
n ( n 1)
C 3
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 O2
1,3 3
3
2
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) 2
C 6
Draw the linkage to scale and identify those ICs that can be found by inspection.
DESIGN OF MACHINERY
SOLUTION MANUAL 6-12-4 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
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)
C 6
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
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 1
2,3
2
I2,6: I1,2 -I1,6 and I2,5 -I5,6 1
I1,5: I1,6 -I5,6 and I1,2 -I2,5
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
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
n ( n 1)
C 15
2 2.
3 4
Draw the linkage to scale and identify those ICs that can be found by inspection.
1,3
1
1,4 at infinity
DESIGN OF MACHINERY
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 infinity
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
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)
C 6
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 3 2
2,3
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)
C 6
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. C
2.
n ( n 1) 2
4 3,4 3
C 6
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
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 3,4
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)
C 6
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)
C 6
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
I2,4: I1,2 -I1,4 and I2,3 -I3,4
3,4 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
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
2. 3.
n ( n 1)
C 6 2 Draw the linkage to scale and identify those ICs that can be found by inspection. 3,4
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
2,3
I2,4: I1,2 -I1,4 and I2,3 -I3,4
3
1,4 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)
C 6
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 2,4
1
2
3,4 3
4
2
4 3
1,2
2
2,3 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)
1,2
1,4 at infinity
C 6 2 Draw the linkage to scale and identify those ICs that can be found by inspection.
1
3 3,4 at infinity 4 1,4
1
DESIGN OF MACHINERY
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)
C 6 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
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)
C 6
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)
2,3
C 6
2 Draw the linkage to scale and identify those ICs that can be found by inspection, which in this case, is all of them.
2 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
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
1,6 at infinity
I2,7: I1,2 -I1,7 and I2,4 -I4,7
3,6
5
8
1,8 at infinity 1
3
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.
8
5
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
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)
E
C 6
2 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 6
2
2,3
3,4
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
3,4
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
2.
n ( n 1) 2
C 6
Draw the linkage to scale and identify those ICs that can be found by inspection.
DESIGN OF MACHINERY
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 4
3,4 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
2.
3.
3,4
1,6
2,3
3
n ( n 1)
C 15 2 Draw the linkage to scale and identify those ICs that can be found by inspection.
1,4 2 4
6
5
2,5
Use Kennedy's Rule and a linear graph to find the remaining 8 ICs. I1,3: I1,4 -I4,3 and I1,2 -I3,2
5,6
1,4
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
I4,5: I4,6 -I5,6 and I3,4 -I3,5
1,6 2,4 3
I3,6: I3,4 -I4,6 and I3,5 -I5,6
3,5 at infinity
1,2 2 4
2,3 and 1,5
6 5
4,5 and 1,3
1,4 4,6 at infinity
3,6
5,6 2,6
DESIGN OF MACHINERY
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
38.6 deg
Offset
c 0.38 in
Crank angle:
34.3 deg
Input crank angular velocity Solution:
Coupler point data: Distance from A to C
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° X
154.502°
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
3.
VA 12.00
in
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
4.
SOLUTION MANUAL 6-16a-2
From the velocity triangle we have: Velocity scale factor:
5.
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
5.692
rad sec
Determine the magnitude and sense of the vector VCA using equation 6.7. VCA p
VCA 7.570
in sec
CA ( 154.502 180 38.6 90) deg 7.
B 180 deg
sec
Determine the angular velocity of link 3 using equation 6.7.
6.
kv
CA 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.
From the velocity triangle we have: Velocity scale factor:
VC 1.130 in kv
1.130
kv
5 in sec
1
in
VC 5.650
in sec
C 153.28 deg
DESIGN OF MACHINERY
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, and
Given:
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
38.6 deg
Offset
c 0.38 in
Input crank angular velocity Solution: 1.
34.3 deg
Crank angle:
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
C (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 VA 12.000
B
in
1,4 at infinity
VA 124.3 deg
Determine the angular velocity of link 3 using equation 6.9a.
4.
1,4 at infinity
sec
VA 90 deg 3.
3,4
2,3
VA AI13
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
VB 11.488
in sec
VC 180 deg 5.
Determine the magnitude of the velocity at point C using equation 6.9b. Determine its direction by inspection. in
VC CI13
VC 5.650
VC C 90 deg
VC 153.268 deg
sec
DESIGN OF MACHINERY
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, and VC for the position shown for 2 = 15 rad/sec in the direction shown. Use an analytical method.
Given:
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
38.6 deg
Offset
c 0.38 in 15 rad sec
Input crank angular velocity Solution: 1.
1
CCW
See Figure P6-5a and Mathcad file P0616c.
C
Draw the linkage to scale and label it. Y
2.
34.3 deg
Crank angle:
Determine 3 and d using equation 4.17.
A
38.600°
34.300° X O2
154.502°
0.380"
B
sin c a asin b 154.502 deg
d1 a cos b cos 3.
d1 2.403 in
Determine the angular velocity of link 3 using equation 6.22a:
a cos b cos 4.
5.691
rad sec
Determine the velocity of pin A using equation 6.23a:
VA a sin j cos VA ( 6.762 9.913j) 5.
in sec
VA 12.000
arg VA124.300 deg
in sec
Determine the velocity of pin B using equation 6.22b:
VB a sin b sin VB 11.490 6.
in
VB 11.490
sec
arg VB180.000 deg
in sec
Determine the velocity of the coupler point C for the open circuit using equations 6.36.
VCA Rca sin j cos VCA (1.716 7.371j)
in sec
VC VA VCA VC ( 5.047 2.542j)
in sec
VC 5.651
in sec
arg VC153.268 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
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
30 deg
Link 4 (point of contact to B)
c 0.75 in
Link 1 (between contact points)
d 1.5 in
15 rad sec
Input crank angular velocity Solution: 1.
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
3.
77.000°
VA 11.250
in sec
77 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 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
SOLUTION MANUAL 6-17a-2
Direction of VB
Y
Direction of V BA
167.000° VA
VB X
0
4.
5 in/sec
From the velocity triangle we have: VB VA
in
VB 11.250
sec
167 deg
in VBA 0 sec 5.
Determine the angular velocity of links 3 and 4 using equation 6.7.
6.
VBA b VB c
rad sec rad
15.000
sec
Determine the magnitude of the vector VCA using equation 6.7. VCA p
7.
0.000
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
C
C 167.000 deg
sec
DESIGN OF MACHINERY
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, 4,
Given:
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
30 deg
Crank angle:
15 rad sec
Input crank angular velocity
Distance A to C
1
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 infinity
B
4
1,4
1,3 at infinity
is at infinity, link 3 is not rotating ( 0 rad sec
1
). Thus, the velocity of every point on link 3 is
the same. From the layout above: 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 VA 11.250 sec
VA 90 deg 3.
0.000 deg
VA 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
VB 90 deg
VB 167.000 deg
sec
DESIGN OF MACHINERY
4.
Use equation 6.9c to determine the angular velocity of link 4.
5.
SOLUTION MANUAL 6-17b-2
VB c
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
VC VA
VC 167.000 deg
sec
DESIGN OF MACHINERY
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, 4,
Given:
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
30 deg
Link 4 (point of contact to B)
c 0.75 in
Link 1 (between contact points)
d 1.5 in
Crank angle:
77 deg
15 rad sec
Input crank angular velocity
1
atan2 (x y) return 0.5 if x = 0 y 0
Two argument inverse tangent
return 1.5 if x = 0 y 0
y if x 0 return atan x Solution: 1.
y atan x otherwise
See Figure P6-5c and Mathcad file P0617c.
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
77.000°
d
X
O2
O4 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
K 1 2.0000
2
d
K3
c
K 2 2.0000
2
2 a c
K 3 1.0000
B 2 sin C K 1 K2 1 cos K 3
A cos K 1 K 2 cos K3
A 1.2250 3.
B 1.9487
C 2.3251
Use equation 4.10b to find values of 4 for the open circuit.
2
2atan2 2 A B B 4 A C
2
a b c d
437.000 deg
2
DESIGN OF MACHINERY
SOLUTION MANUAL 6-17c-2
360 deg 4.
77.000 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
K 4 1.0000
2 a b
K 5 2.0000
E 2 sin F K 1 K4 1 cos K 5
D cos K 1 K 4 cos K 5
5.
6.
D 3.5501 E 1.9487 F 0.0000
Use equation 4.13 to find values of 3 for the open circuit.
2
2atan2 2 D E E 4 D F
360.000 deg
360 deg
0.000deg
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18.
a sin c sin
7.
a sin b sin
0.000
rad sec
15.000
rad sec
Determine the velocity of points A and B for the open circuit using equations 6.19.
VA a sin j cos VA ( 10.962 2.531j)
in sec
VA 11.250
in sec
arg VA167.000 deg
VB c sin j cos VB ( 10.962 2.531j) 8.
in
VB 11.250
sec
in sec
arg VB167.000 deg
Determine the velocity of the coupler point C for the open circuit using equations 6.36.
VCA Rca sin j cos VCA 0.000
in sec
VC VA VCA VC ( 10.962 2.531j)
in sec
VC 11.250
in sec
arg VC167.000 deg
DESIGN OF MACHINERY
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
128 deg
Angle BAC
49 deg
Slider 4 angle
59 deg
Input slider velocity
VA 10 in sec
1
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
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
VBA 4.784 VB
3.436
38.000°
59.000° X
VA
DESIGN OF MACHINERY
4.
SOLUTION MANUAL 6-18a-2
From the velocity triangle we have: Velocity scale factor:
5.
5 in sec
1
in in
VB 3.438 in kv
VB 17.190
VBA 4.784 in kv
VBA 23.920
VBA
in
38 deg
sec
13.289
rad
VCA 19.136
in
b sec Determine the magnitude and sense of the vector VCA using equation 6.7. VCA p
sec
CA ( 128 49 90) deg 7.
59 deg
sec
Determine the angular velocity of link 3 using equation 6.7.
6.
kv
CA 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 VCA
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
C 22.572 deg
DESIGN OF MACHINERY
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
128 deg
Angle BAC
49 deg
Slider 4 angle
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
C 67.428 deg
4
0.716 67.428° 3
2.
4.
VA AI13
13.280
1,3
1.293
Determine the angular velocity of link 3 using equation 6.9a.
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
VB 17.17
VB
VB 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
VC 9.509
VC C 90 deg
VC 22.572 deg
sec
DESIGN OF MACHINERY
SOLUTION MANUAL 6-18c-1
PROBLEM 6-18c Statement:
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.
Given:
Link lengths and angles:
Solution: 1.
Coupler point:
Link 3 (A to B)
b 1.8 in
Distance A to C
Rca 1.44 in
Coupler angle
128 deg
Angle BAC
49 deg
Slider 4 angle
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 .0 00°
12 8.00 0°
59.0 00° 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 j
a e
j
b e
j
c e
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 d d j a j b e c e dt dt
Substituting the Euler equivalents, d d a b sin j cos c cos j sin dt dt Separating into real and imaginary components and solving for 3 and VB . Note that dc/dt = VB and da/dt = VA
VB
b sin tancos VA tan
cos
VA b sin
13.288
VB 17.180
rad sec in
sec
DESIGN OF MACHINERY
SOLUTION MANUAL 6-18c-2
arg VB59.000 deg
VB VB cos j sin 3.
Determine the velocity of the coupler point C using equations 6.36.
VCA Rca sin j cos 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 VC22.572 deg
DESIGN OF MACHINERY
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:
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
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
Vslip
Vtrans
V4
X
0.870
4.
From the velocity triangle we have: 1
Velocity scale factor:
Vslip 1.523 in kv
kv
10 in sec 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
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:
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
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:
5 in sec
1
1.599
in
VA2slip 5.670
Vtrans 1.599 in kv
Vtrans 7.995
VA3 2.598 in kv
sec
2.598
in sec
VA3slip 10.240 VA3 12.990
V trans
in
V A2
in
V A2slip
VA3
VA3 b
The relative slip velocity is
9.665
V A3slip
in sec
Vslip VA3slip VA2slip
rad
1.134
sec
Determine the angular velocity of link 3 using equation 6.7.
6.
kv
VA2slip 1.134 in kv
VA3slip 2.048 in kv
5.
5 in/sec
CCW
sec Vslip 4.570
in sec
2.048
X
DESIGN OF MACHINERY
SOLUTION MANUAL 6-21a-1
PROBLEM 6-21a Statement:
The linkage in Figure P6-6b has L1 = 61.9, L2 = 15, L 3 = 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:
45 deg
Coordinate rotation angle
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.
y 5,6 6
mm VA 10 sec
2,3 A
Direction of VCB C
4 5.0° 3
2
1
5
X
O2
VC 45 deg 90 deg
23 .3 °
B
1
VC 135.000 deg 3.
Direction of VC
Direction of VA Y
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
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
Y
Velocity scale factor:
kv
5 mm sec
1
5 mm/sec
2.053
in
VB 2.248 in kv VB 11.2
0
mm sec
VB ( 360 167.558 ) deg
V BA
X 167.553°
VB 2.248
DESIGN OF MACHINERY
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
VC
5 mm sec
1
in
VC 5.6
mm sec
VC 270 deg
0.564
VA
I2,3
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. Tin
Fin =
rin
Fout =
mA =
Pout Vout
Fout Fin
Pin
=
rin in =
=
=
Pin VA
Pout VC
Pout VA VC Pin
mA
VA VC
mA 1.773
DESIGN OF MACHINERY
SOLUTION MANUAL 6-21b-1
PROBLEM 6-21b Statement:
The linkage in Figure P6-6b has L1 = 61.9, L2 = 15, L 3 = 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:
45 deg
Coordinate rotation angle
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.
6 1,5
2,3 A
mm VA 10 sec
2
50.12 1
C
3
1
5
22.68 3
X
O2
VC 90 deg
B 1
VC 135.000 deg 3.
11 .3 77
44.59 4 Y
4
Determine the angular velocity of link 3 using equation 6.9a.
O4 1
VA
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 VB 11.239 sec
5.
Determine the angular velocity of link 5 using equation 6.9a.
6.
VB
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
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
Tin
Fin =
rin
Fout =
mA =
SOLUTION MANUAL 6-21b-2
Pout Vout
Fout Fin
Pin
=
rin in =
=
=
Pin VA
Pout VC
Pout VA VC Pin
mA
VA VC
mA 1.77
DESIGN OF MACHINERY
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:
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
2,3
1
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. mm VA 10 sec
3.
VC 90 deg
VC 114.200 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. VA Y 0
5 mm/sec
VBA X
VB 1.073
DESIGN OF MACHINERY
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
VB
5 mm sec
1
in
VB 5.4
mm sec
VB 180 deg
0.54
VA
I2,3
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. Tin
Fin =
rin
Fout =
mA =
Pout Vout
Fout Fin
Pin
=
rin in
=
=
=
Pin VA
Pout VB
Pout VA VB Pin
mA
VA VB
mA 1.86
DESIGN OF MACHINERY
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:
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 1
C
3
1,5
48.561
26.075
5,6 A
5
2
O2
6 B
2,3
1
1
From the layout: AI15 48.561 mm 2.
BI15 26.075 mm
Choose an arbitrary value for the magnitude of the velocity at I2,3 (point A). Let link 2 rotate CCW. mm VA 10 sec
3.
VA
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
5.
VC 114.200 deg
Determine the angular velocity of link 3 using equation 6.9a.
4.
VC 90 deg
is I2,3
VB
VB 5.370
mm
to the left
sec
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 =
mA =
Tin rin Fout Fin
=
=
Pin rin in
=
Pout VA VC Pin
Pin
Fout =
VA mA
VA VB
Pout Vout
mA 1.86
=
Pout VC
DESIGN OF MACHINERY
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
SOLUTION MANUAL 6-23-2
MOVING CENTRODE
DESIGN OF MACHINERY
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.
37 deg
Global XY system
Input crank angular velocity 15 rad sec
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
O2
X
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
3.
O4
VA 1740.0
mm sec
A 37 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 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
SOLUTION MANUAL 6-24-2
0
1000 mm/sec
Y
X 0.952 53.000° VB VBA
VA
1.498 112.319°
4.
From the velocity triangle we have: Velocity scale factor:
5.
1000 mm sec
mm
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. VBA b
13.87
rad sec
Determine the angular velocity of link 4 using equation 6.7.
VB c
8.655
1
in
VB 0.952 in kv
6.
kv
rad sec
B 112.319 deg
DESIGN OF MACHINERY
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: 37 deg
Global XY system
Input crank angular velocity 15 rad sec
25 deg
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 2
157.681 deg
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
VA 1740.0
VA 90 deg
VA 53.0 deg
sec
Determine the angular velocity of link 3 using equation 6.9a.
4.
68.638
VA
13.869
rad
CW AI13 sec Determine the magnitude of the velocity at point B using equation 6.9b. Determine its direction by inspection. mm
VB BI13
VB 951.915
VB 90 deg
VB 247.681 deg
sec
Use equation 6.9c to determine the angular velocity of link 4.
VB c
8.654
rad sec
CCW
DESIGN OF MACHINERY
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:
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
62 deg
Global XY system
Input crank angular velocity 15 rad sec
1
CW
atan2 (x y) return 0.5 if x = 0 y 0
Two argument inverse tangent
return 1.5 if x = 0 y 0
y if x 0 return atan x y atan x otherwise Solution:
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
K 1 1.5000
a d
3 62.000°
K 2 1.5818
c 2
K3
2 O2
2
2
a b c d
2
2
K 3 1.7307
2 a c
B
B 2 sin C K 1 K2 1 cos K 3
4
A cos K 1 K 2 cos K3
A 0.0424 3.
B 1.7659
x
C 2.0186
Use equation 4.10b to find values of 4 for the crossed circuit.
2
2atan2 2 A B B 4 A C 4.
182.681 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 F K 1 K4 1 cos K 5
D cos K 1 K 4 cos K 5
5.
O4
Use equation 4.13 to find values of 3 for the crossed circuit.
K 4 1.6111 D 2.0021 E 1.7659 F 0.0589
K 5 1.7280
DESIGN OF MACHINERY
SOLUTION MANUAL 6-26-2
2
2atan2 2 D E E 4 D F 6.
7.
275.133 deg
Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18.
13.869
8.654
a sin b sin
a sin c sin
rad sec
rad sec
Determine the velocity of points A and B for the crossed circuit using equations 6.19.
VA a sin j cos VA ( 1536.329 816.881j)
mm sec
VA 1740.000
mm sec
arg VA 28.000 deg
VB c sin j cos VB ( 44.524 950.875j)
mm sec
VB 951.917
mm sec
arg VB 87.319 deg
DESIGN OF MACHINERY
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 the
Given:
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 4 (B to O4)
c 110 mm 15 rad sec
Input crank angular velocity Two argument inverse tangent
Link 3 (A to B)
b 108 mm
Link 1 (O2 to O4)
d 174 mm
1
CW
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 Solution: 1.
y atan x otherwise
See Figure P6-8a and Mathcad file P0627.
Draw the linkage to scale and label it. y
2.
A
Determine the range of motion for this non-Grashof triple rocker using equations 4.33. 2
arg 1
2
2
2 a d 2
arg 2
2
a d b c
2
2
2
a d b c 2 a d
b c
2
a d
3
O2
b c
62.000°
2
a d
B
arg 1 1.083
arg 2 0.094 4
2toggle acos arg 2
2toggle 95.4deg x
The other toggle angle is the negative of this. Thus, 2toggle 2toggle 1 deg 2toggle 3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K 1 1.5000 2
K3
d c
K 2 1.5818 2
2
a b c d 2 a c
2
K 3 1.7307
B 2 sin C K 1 K 2 1 cos K 3
A cos K1 K 2 cos K 3
4.
Use equation 4.10b to find values of 4 for the crossed circuit.
O4
2
2 2 A B B 4 A C atan2
DESIGN OF MACHINERY
5.
SOLUTION MANUAL 6-27-2
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
K 4 1.6111
2 a b
K 5 1.7280
E 2 sin F K1 K 4 1 cos K5
D cos K1 K 4 cos K5
6.
Use equation 4.13 to find values of 3 for the crossed circuit.
2
2 2 D E E 4 D F atan2 7.
Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18. a sin b sin
a sin c sin
8.
Determine the velocity of points A and B for the crossed circuit using equations 6.19.
VA a sin j cos
VAx Re VA
VAy Im VA
VB c sin j cos
VBx Re VB
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 Im VB
20 0
sec rad
20 40 60 80 100 100
75
50
25
0 deg
25
50
75
100
DESIGN OF MACHINERY
SOLUTION MANUAL 6-27-3
VELOCITY COMPONENTS, POINT A 2000
Joint Velocity, mm/sec
1000
sec VAx mm
sec VAy mm
0
1000
2000 100
75
50
25
0
25
50
75
100
deg
VELOCITY COMPONENTS, POINT B 4000
Joint Velocity, mm/sec
3000 2000 sec VBx mm 1000
sec VBy mm
0 1000 2000 3000 100
75
50
25
0 deg
25
50
75
100
DESIGN OF MACHINERY
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 VB
Given:
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
57 deg
Input crank angular velocity 20 rad sec
36 deg
Coordinate rotation angle
Global XY system
1
Global XY system to local xy system
Solution: See Figure P6-8b and Mathcad file P0628. 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. Y
VA a VA 800.000
mm sec
2
Direction of VB
y A
2
B 3 X
O2
A 90 deg 3.
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: 1
kv
VA
400 mm sec
Y
in
VB 1.790 in kv
VB 716.0
mm
1.293" V BA
sec X
B 186.406 deg VBA 1.293 in kv 5.
mm
6.406°
1.790"
sec
85.486°
Determine the angular velocity of link 3 using equation 6.7.
6.
VBA 517.2
VB
VBA b
5.388
rad sec
Determine the angular velocity of link 4 using equation 6.7.
VB c
5.869
rad sec
DESIGN OF MACHINERY
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 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 20 rad sec Link 4 (B to O4) d 162 mm Link 1 (O2 to O4) Coordinate rotation angle
Solution: 1.
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
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 VA 800.0 sec
VA 90 deg 3.
148.700
Y 2
A
2
VA 147.0 deg
y
B 3 X
O2 4
Determine the angular velocity of link 3 using equation 6.9a.
VA AI13
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
VB 716.6
VB 90 deg
VB 186.406 deg
sec
Use equation 6.9c to determine the angular velocity of link 4.
VB c
5.874
rad sec
CCW
DESIGN OF MACHINERY
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 VB for the position shown for 2 = 20 rad/sec counterclockwise (CCW). Use an analytical method.
Given:
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
57 deg
Global XY system
Input crank angular velocity 20 rad sec
1
Coordinate rotation angle
36 deg
Two argument inverse tangent
atan2 (x y) return 0.5 if x = 0 y 0
Global XY system to local xy system return 1.5 if x = 0 y 0
y if x 0 return atan x y atan x otherwise Solution:
See Figure P6-8b and Mathcad file P0630.
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
K 1 4.0500 2
K3
c
2
2
4
K 3 3.4336
2 a c
B 1.9973
O4 x
B 2 sin
C 7.6054
Use equation 4.10b to find values of 4 for the open circuit.
2
2
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
E 2 sin F K 1 K4 1 cos K 5
2
2 a b
D cos K 1 K 4 cos K 5
K 4 1.6875
K 5 2.8875
D 7.0782 E 1.9973 F 1.1265
Use equation 4.13 to find values of 3 for the open circuit.
2
2atan2 2 D E E 4 D F 2 6.
X 93.000
2
2atan2 2 A B B 4 A C
5.
B
3
O2
C K 1 K2 1 cos K 3
4.
A
2
d
A cos K 1 K 2 cos K3
3.
2
y
K 2 1.3279
a b c d
A 0.5992
Y
31.504 deg
Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18.
DESIGN OF MACHINERY
7.
SOLUTION MANUAL 6-30-2
5.385
5.868
a sin b sin
a sin c sin
rad sec
rad sec
Determine the velocity of points A and B for the open circuit using equations 6.19.
VA a sin j cos VA ( 798.904 41.869j)
mm sec
VA 800.000
mm
VB 715.900
mm
sec
arg VA177.000 deg
VB c sin j cos VB ( 528.774 482.608j)
mm sec
sec
arg VB137.614 deg
DESIGN OF MACHINERY
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
1
Input crank angular velocity
20 rad sec
Two argument inverse tangent
atan2 (x y) return 0.5 if x = 0 y 0
CCW
return 1.5 if x = 0 y 0
y if x 0 return atan x y atan x otherwise Solution:
See Figure P6-8b and Mathcad file P0631.
1.
Draw the linkage to scale and label it.
2.
Determine Grashof condition.
Y 2
Condition( S L P Q) 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
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
K 1 4.0500
2
d
K3
c
K 2 1.3279
2
2
2 a c
K 3 3.4336
A cos K1 K 2 cos K 3
2
a b c d
B 2 sin
C K 1 K 2 1 cos K 3 4.
Use equation 4.10b to find values of 4 for the open circuit.
2
2 atan2 2 A B B 4 A C 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
2 a b
D cos K1 K 4 cos K5
K 4 1.6875
K 5 2.8875
DESIGN OF MACHINERY
SOLUTION MANUAL 6-31-2
F K1 K 4 1 cos K5 E 2 sin
6.
Use equation 4.13 to find values of 3 for the open circuit.
2
2 atan2 2 D E E 4 D F 7.
Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18.
a sin b sin
a sin c sin
8.
Determine the velocity of points B and C for the open circuit using equations 6.19.
VA a sin j cos
VAx Re VA
VAy Im VA
VB c sin j cos
VBx Re VB
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 10 Angular Velocity, rad/sec
9.
VBy Im VB
5
sec rad
0
5
10
0
45
90
135
180 deg
225
270
315
360
DESIGN OF MACHINERY
SOLUTION MANUAL 6-31-3
VELOCITY COMPONENTS, POINT A
Joint Velocity, mm/sec
1000
sec VAx mm
sec VAy mm
500
0
500
1000
0
45
90
135
180
225
270
315
360
deg
VELOCITY COMPONENTS, POINT B
Joint Velocity, mm/sec
1000 600
sec VBx mm
200
200
sec VBy mm
600 1000
0
45
90
135
180 deg
225
270
315
360
DESIGN OF MACHINERY
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:
51 deg
Link 3
b 130 mm
Input crank angular velocity
Offset
c 52 mm
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
VA 1575.0
Direction of VB 4 B
mm
Direction of VBA
1
sec
3
A 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
VB 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
SOLUTION MANUAL 6-33-1
PROBLEM 6-33 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 the instant center graphical method.
Given:
Link lengths: Link 2
a 63 mm
Crank angle:
51 deg
Link 3
b 130 mm
Offset
c 52 mm
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 VA 1575.0 sec
VA 90 deg 3.
VA 39.0 deg
Determine the angular velocity of link 3 using equation 6.9a.
4.
BI13 166.309 mm
VA AI13
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
VB 270 deg
VB 2207.8
mm sec
DESIGN OF MACHINERY
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
141 deg
Crank angle:
Local xy coordinate system Input crank angular velocity 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
sin c a asin b
1 3
d 141.160 mm
A
52.000 2
Determine the angular velocity of link 3 using equation 6.22a:
a cos b cos 4.
B
44.828 deg
d a cos b cos 3.
1
13.276
rad
141.000°
sec
x
Determine the velocity of pin A using equation 6.23a:
X, y
O2
VA a sin j cos VA ( 991.180 1224.005i )
mm
mm sec
VA arg VA 90 deg
In the global coordinate system, 5.
VA 1575.000
sec
arg VA51.000 deg
VA 39.000 deg
Determine the velocity of pin B using equation 6.22b:
VB a sin b sin
VB 2207.849
mm sec
VB VB In the global coordinate system,
VB arg VB 90 deg
VB 90.000 deg
DESIGN OF MACHINERY
SOLUTION MANUAL 6-35-1
PROBLEM 6-35 Statement:
The offset slider-crank 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
Input crank angular velocity Solution:
c 52 mm
Offset
25 rad sec
1
See Figure P6-8f and Mathcad file P0635.
1.
Draw the linkage to a convenient scale. The coordinate rotation angle is 90 deg
2.
Determine the range of motion for this slider-crank linkage.
Y 4
0 deg 2 deg 360 deg 3.
B
1
Determine 3 using equations 4.16 for the crossed circuit.
3
sin c a asin b
4.
Determine the angular velocity of link 3 using equation 6.22a:
2
a cos b cos
5.
VA a sin j cos
VAx Re VA
VAy Im VA
In the global coordinate system,
VAX VAy 6.
VAY VAx
Determine the velocity of pin B using equation 6.22b:
VBx a sin b sin In the global coordinate system,
VBY VBx 7.
X, y
O2
Determine the x and y components of the velocity of pin A using equation 6.23a:
A
52.000
Plot the x and y components of the velocity of A. (See next page.)
2 x
DESIGN OF MACHINERY
SOLUTION MANUAL 6-35-2
VELOCITY OF POINT A 2000
Velocity, mm/sec
1000
sec VAX mm
sec VAY mm
0
1000
2000
0
60
120
180
240
300
360
deg
Plot the velocity of point B. VELOCITY OF POINT B 2000
1000 Velocity, mm/sec
7.
0
sec VBY mm 1000
2000
3000
0
60
120
180 deg
240
300
360
DESIGN OF MACHINERY
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:
58 deg
Global XY system 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. VA a
VA 900.000
mm
Vbox
sec
1
Direction of VBA
A 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
A
B
3
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
5.
in VBA 0 sec
VA
Determine the angular velocity of links 3 and 4 using equation 6.7.
6.
VB
sec
B 32 deg
500 mm/sec
3 2.000°
mm
VBA
0.000
b
rad sec
VB c
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 A Vbox 763.243 sec
DESIGN OF MACHINERY
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: Link 2
a 30 mm
Link 3
b 150 mm
Link 4
c 30 mm
Link 1
d 150 mm
Crank angle:
58 deg
Input crank angular velocity
30 rad sec
Two argument inverse tangent
atan2 (x y) return 0.5 if x = 0 y 0
1
return 1.5 if x = 0 y 0
y if x 0 return atan x y atan x otherwise Solution: 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
K 1 5.0000 2
K3
2
2
c O2
K 3 1.0000
C K 1 K2 1 cos K 3
A cos K 1 K 2 cos K3
3.
4.
B 1.6961
B 2 sin
C 2.8205
Use equation 4.10b to find values of 4 for the open circuit.
2
2atan2 2 A B B 4 A C
418.000 deg
360 deg
58.000 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
E 2 sin F K 1 K4 1 cos K 5
D cos K 1 K 4 cos K 5
5.
O4
2
2 a c
A 6.1197
B
2
K 2 5.0000
a b c d
3
A
d
Use equation 4.13 to find values of 3 for the open circuit.
2
K 4 1.0000 D 8.9402 E 1.6961 F 0.0000
K 5 5.0000
4
DESIGN OF MACHINERY
SOLUTION MANUAL 6-37-2
6.
7.
2
2atan2 2 D E E 4 D F
360.000 deg
360 deg
0.000deg
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18.
0.000
30.000
a sin b sin
a sin c sin
rad sec rad sec
Determine the velocity of points A and B for the open circuit using equations 6.19.
VA a sin j cos VA ( 763.243 476.927j)
mm sec
VA 900.000
mm sec
arg VA32.000 deg
VB c sin j cos
mm VB 900.000 arg VB32.000 deg sec sec 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, VB ( 763.243 476.927j)
8.
Vbox VA
mm
Vbox 900.000
mm sec
DESIGN OF MACHINERY
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
1
Input crank angular velocity
30 rad sec
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 Solution:
y atan x otherwise
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
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
K 1 5.0000 2
K3
d
B
O2
O4
c
K 2 5.0000 2
2
a b c d
2
K 3 1.0000
2 a c
C K 1 K 2 1 cos K 3
A cos K1 K 2 cos K 3
4.
3
A 2
B 2 sin
Use equation 4.10b to find values of 4 for the open circuit.
2
2 2 A B B 4 A C atan2 5.
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
K 4 1.0000
E 2 sin F K1 K 4 1 cos K5
D cos K1 K 4 cos K5
6.
Use equation 4.13 to find values of 3 for the open circuit.
2
2 2 D E E 4 D F atan2
K 5 5.0000
4
DESIGN OF MACHINERY
7.
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18.
a sin b sin
a sin c sin
8.
SOLUTION MANUAL 6-38-2
Determine the velocity of points A and B for the open circuit using equations 6.19.
VA a sin j cos
VA VA
VB c sin j cos
VB VB 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. ( 5 deg ) 30.000
rad sec
VA (5 deg) 900.000
mm
VB (5 deg) 900.000
mm
sec
sec
( 135 deg) 30.000
rad sec
VA (135 deg ) 900.000
mm
VB (135 deg ) 900.000
mm
sec
sec
DESIGN OF MACHINERY
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 VB
Given:
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
29 deg
Crank angle:
Solution: 1.
Global XY system
Input crank angular velocity
15 rad sec
Coordinate rotation angle
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
3.
VB 735.0
mm sec
B 29 deg 90 deg
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point A, the magnitude of the relative 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 0
SOLUTION MANUAL 6-39-2 500 mm/sec
Y 0.124°
1.517 VB
X VBA
VA
4.
From the velocity triangle we have: 1
Velocity scale factor:
VA 1.517 in kv
5.
kv
500 mm sec
VA 758.5
in mm sec
Determine the angular velocity of link 4 using equation 6.7.
VA c
4.958
rad sec
A 0.124 deg
DESIGN OF MACHINERY
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 VB
Given:
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
29 deg
Crank angle:
Solution: 1.
Global XY system
Input crank angular velocity
15 rad sec
Coordinate rotation angle
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
89.876 deg 2.
3.
5.
A 6
O2 89.876°5
D
O4
VA 61.0 deg
Determine the angular velocity of link 3 using equation 6.9a.
4.
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 VA 735.0 sec
VA 90 deg
2
4
VA AI13
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
VB 758.694
VB 90 deg
VB 0.124 deg
sec
Use equation 6.9c to determine the angular velocity of link 4.
VB c
4.959
rad sec
CW
X
DESIGN OF MACHINERY
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
148 deg Local xy system
Crank angle:
Input crank angular velocity
15 rad sec
1
atan2 (x y) return 0.5 if x = 0 y 0
Two argument inverse tangent
return 1.5 if x = 0 y 0
y if x 0 return atan x Solution:
y atan x otherwise
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
K 1 1.7755 2
K3
O6
d
B
3
c
K 2 0.5686 2
2
a b c d
2
K 3 1.5592
2 a c
C
B 2 sin C K 1 K2 1 cos K 3
O2 148.000°
A cos K 1 K 2 cos K3
A 0.5821 3.
B 1.0598
D O4 x
C 4.6650
Use equation 4.10b to find values of 4 for the crossed circuit.
2
d b
2
K5
2
2
c d a b
2
2 a b
E 2 sin F K 1 K4 1 cos K 5
D cos K 1 K 4 cos K 5
5.
208.876 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
Use equation 4.13 to find values of 3 for the crossed circuit.
K 4 0.8700 D 3.0104 E 1.0598 F 2.2367
X
6
5
2atan2 2 A B B 4 A C 4.
A
2
4
K 5 0.3509
y
DESIGN OF MACHINERY
SOLUTION MANUAL 6-41-2
2
2atan2 2 D E E 4 D F 6.
7.
266.892 deg
Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18.
a sin b sin
7.570
rad
a sin c sin
4.959
rad
sec
sec
Determine the velocity of points B and A for the crossed circuit using equations 6.19.
VA a sin j cos VA ( 389.491 623.315j )
mm sec
VA 735.000
mm sec
arg VA58.000 deg
VB c sin j cos VB ( 366.389 664.362j)
mm sec
VB 758.694
mm sec
arg VB118.876 deg
DESIGN OF MACHINERY
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
15 rad sec
Input crank angular velocity
1
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 Solution:
y atan x otherwise
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 4.33. 2
arg 1
2
2 a d 2
2
2toggle acos arg 1
2
a d b c 2 a d
O6
2
a d b c 2
arg 2
2
Y
b c a d b c a d
B
arg 1 0.840 arg 2 6.338
3
A
2
4 C
O2
2toggle 32.9deg
2
5
The other toggle angle is the negative of this. Thus,
D
2toggle 2toggle 1 deg 360 deg 2toggle
O4 x
3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K 1 1.7755
2
d
K3
c
K 2 0.5686
2
a b c d 2 a c
K 3 1.5592
C K 1 K 2 1 cos K 3
A cos K1 K 2 cos K 3
4.
2
B 2 sin
Use equation 4.10b to find values of 4 for the crossed circuit.
2
2 2 A B B 4 A C atan2 5.
X
6
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12.
2
y
DESIGN OF MACHINERY
K4
SOLUTION MANUAL 6-42-2 2
d
K5
b
2
2
c d a b
2
K 4 0.8700
2 a b
F K1 K 4 1 cos K5
D cos K1 K 4 cos K5
6.
K 5 0.3509
E 2 sin
Use equation 4.13 to find values of 3 for the crossed circuit.
2
2 2 D E E 4 D F atan2 7.
Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18. a sin b sin
a sin c sin
8.
Determine the velocity of points A and B for the crossed circuit using equations 6.19.
VBx Re VB
VB a sin j cos
VBy Im VB
VA c sin j cos
VAx Re VA
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 Im VA
20 0
sec rad
20 40 60 80 100
0
45
90
135
180 deg
225
270
315
360
DESIGN OF MACHINERY
SOLUTION MANUAL 6-42-3
VELOCITY COMPONENTS, POINT B 1000
Joint Velocity, mm/sec
500
sec VBx mm
sec VBy mm
0
500
1000
0
45
90
135
180
225
270
315
360
315
360
deg
VELOCITY COMPONENTS, POINT A 4000
Joint Velocity, mm/sec
3000 2000
sec VAx mm 1000
sec VAy mm
0 1000 2000 3000
0
45
90
135
180 deg
225
270
DESIGN OF MACHINERY
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:
53 deg Global XY system
Input crank angular velocity
15 rad sec
Cylinder angular spacing
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
V2 285.0
3
mm
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 5
sec
2 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: 1
Velocity scale factor:
kv
V7 1.046 in kv
Y
200 mm sec in
V62
V7 209.2
1.046
V82
V6 83.4
V7
mm sec
V6 150 deg V8 292.6
X
V8
sec
V2
V8 1.463 in kv
V6
mm
V7 270 deg V6 0.417 in kv
0.417
1.463
mm sec
V8 210 deg
V72
DESIGN OF MACHINERY
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
15 rad sec
Cylinder angular spacing
120 deg
1
Draw the linkage to scale and label it.
Crank angle:
37 deg
CW
Local x'y' system
Y
x''
Determine 4 and d' using equation 4.17.
170.599 deg
x'''
y''' 8 6
3
5 X, y'
2
(x'y' system)
d' a cos b cos
1
d' 3.316in
4 y''
Determine the angular velocity of link 4 using equation 6.22a:
7
a cos b cos 4.
c 0 mm
See Figure P6-8c and Mathcad file P0644.
sin c a asin b
3.
Offset
3.296
rad sec 37.000°
Determine the velocity of the rod pin on link 2 using equation 6.23a:
x'
V2 a sin j cos V2 ( 171.517 227.611i )
mm
V2 285.000
sec
sec
arg V253.000 deg
V2 arg V290 deg
In the global coordinate system, 5.
mm
V2 143.000 deg
Determine the velocity of piston 7 using equation 6.22b:
V7 a sin b sin V7 209.204
mm
V7 209.204
sec
In the global coordinate system, 6.
sec
arg V70.000deg
V7 arg V790 deg
V7 90.000 deg
Determine 3 and d'' using equation 4.17. 120 deg
157.000 deg
sin c a asin b
d'' a cos b cos 7.
mm
173.912 deg d'' 2.052 in
Determine the angular velocity of link 5 using equation 6.22a:
a cos b cos
3.769
rad sec
(x''y'' system)
DESIGN OF MACHINERY
8.
SOLUTION MANUAL 6-44-2
Determine the velocity of the rod pin on link 2 using equation 6.23a:
V2 a sin j cos V2 ( 111.358 262.344i)
mm sec
arg V267.000 deg
mm sec
V2 arg V2150 deg
In the global coordinate system, 9.
V2 285.000
V2 217.000 deg
Determine the velocity of piston 6 using equation 6.22b:
V6 a sin b sin V6 83.378
mm
V6 83.378
sec
arg V60.000deg
mm sec
V6 arg V6150 deg
In the global coordinate system,
V6 150.000 deg
10. Determine 5 and d''' using equation 4.17. 120 deg
277.000 deg
sin c a asin b
(x'''y''' system)
195.629 deg
d''' a cos b cos
d''' 2.745in
11. Determine the angular velocity of link 5 using equation 6.22a:
a cos b cos
0.515
rad sec
12. Determine the velocity of the rod pin on link 2 using equation 6.23a:
V2 a sin j cos V2 ( 282.876 34.733i)
mm sec
V2 285.000
mm sec
V2 arg V230 deg
In the global coordinate system,
arg V2173.000 deg
V2 143.000 deg
13. Determine the velocity of piston 8 using equation 6.22b:
V8 a sin b sin V8 292.592
mm sec
In the global coordinate system,
V8 292.592
mm sec
V8 arg V830 deg
arg V8180.000 deg
V8 210.000 deg
DESIGN OF MACHINERY
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
15 rad sec
Cylinder angular spacing
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'
0 deg 2 deg 360 deg 3.
2
Determine 4 using equation 4.17.
1 4
sin c a asin (x'y' system) b 4.
y'' 7
Determine the angular velocity of link 3 using equation 6.22a:
a cos b cos
5.
x'
Determine the velocity of piston 7 using equation 6.22b:
V7 a sin b sin 6.
Determine 3 using equation 4.17.
sin c a asin b
7.
(x''y'' system)
Determine the angular velocity of link 3 using equation 6.22a:
a cos b cos
8.
Determine the velocity of piston 6 using equation 6.22b:
V6 a sin b sin 9.
Determine 5 and d''' using equation 4.17.
sin 2 c a asin b
10. Determine the angular velocity of link 5 using equation 6.22a:
a cos 2 b cos
11. Determine the velocity of piston 8 using equation 6.22b:
(x'''y''' system)
DESIGN OF MACHINERY
SOLUTION MANUAL 6-45-2
V8 a sin 2 b sin 12. Plot the velocities of pistons 6, 7, and 8.
VELOCITY OF PISTONS 6, 7, AND 8 400
Velocity, mm/sec
sec 200 V6 mm
sec V7 mm
0
sec V8 mm
200
400
0
60
120
180 deg Crank Angle, deg
240
300
360
DESIGN OF MACHINERY
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
56 deg
Coupler point:
50 deg
Crank angle and speed:
40 rad sec
1
atan2 (x y) return 0.5 if x = 0 y 0
Two argument inverse tangent
return 1.5 if x = 0 y 0
y if x 0 return atan x Solution:
y atan x otherwise
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
K 1 1.9000 2
K3
P
d c
y
K 2 1.9000 2
2
a b c d
Y B
2
2 a c
K 3 2.4178
3 A
4
A cos K 1 K 2 cos K3 2
B 2 sin
14.000°
3.
4.
B 1.5321
X
O2
C 2.4537
Use equation 4.10b to find values of 4 for the open circuit.
2
2atan2 2 A B B 4 A C
473.008 deg
360 deg
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
E 2 sin F K 1 K4 1 cos K 5
2 a b
D cos K 1 K 4 cos K 5
5.
O4
1
C K 1 K2 1 cos K 3 A 0.0607
x 50.000°
2
K 4 2.1591 D 2.3605 E 1.5321 F 0.1539
Use equation 4.13 to find values of 3 for the open circuit.
2
2atan2 2 D E E 4 D F
370.105 deg
K 5 2.4911
DESIGN OF MACHINERY
SOLUTION MANUAL 6-46-2
360 deg 6.
10.105 deg
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18.
a sin c sin
7.
a sin b sin
41.552
rad sec
26.320
rad sec
Determine the velocity of points A and B for the open circuit using equations 6.19.
VA a sin j cos VA ( 153.209 128.558j)
in
VA 200.000
sec
in sec
arg VA40.000 deg
VB c sin j cos
in in VB 131.598 arg VB23.008 deg sec sec Determine the velocity of the coupler point P for the open circuit using equations 6.36. VB ( 121.130 51.437j)
8.
VPA Rpa sin j cos VPA ( 338.121 149.797j )
in sec
VP VA VPA VP (184.912 21.239j)
in sec
VP 186.128
in sec
arg VP 173.448 deg
DESIGN OF MACHINERY
SOLUTION MANUAL 6-47-1
PROBLEM 6-47 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: Link 2 (O2 to A)
a 1.00
Link 3 (A to B)
b 2.06
Link 4 (B to O4)
c 2.33
Link 1 (O2 to O4)
d 2.22
Coupler point:
Rpa 3.06
31 deg
Crank speed:
1 rad sec
1
atan2 (x y) return 0.5 if x = 0 y 0
Two argument inverse tangent
return 1.5 if x = 0 y 0
y if x 0 return atan x Solution:
y atan x otherwise
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
0 deg 2 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
K 1 2.2200 2
K3
2
a b c d
A
d c
2
4
c
x
1
A cos K1 K 2 cos K 3
2
O2
K 3 1.5265
a
4
d
2
2 a c
p
K 2 0.9528 2
P
O4
B 2 sin
C K 1 K 2 1 cos K 3 4.
Use equation 4.10b to find values of 4 for the open circuit.
2
2 2 A B B 4 A C atan2 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
2 a b
D cos K1 K 4 cos K5
K 4 1.0777
K 5 1.1512
E 2 sin
F K1 K 4 1 cos K5 6.
Use equation 4.13 to find values of 3 for the open circuit.
2
2 2 D E E 4 D F atan2
DESIGN OF MACHINERY
7.
SOLUTION MANUAL 6-47-2
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18. a sin b sin
a sin c sin
8.
Determine the velocity of point A using equations 6.19.
VA a sin j cos 9.
VA VA
Determine the velocity of the coupler point P using equations 6.36.
VP VA VPA
VPA Rpa sin cos j 10. Plot the magnitude and direction of the velocity at coupler point P.
Direction: VP1 arg VP
VP VP
Magnitude:
VP if VP1 0 VP1 2 VP1 MAGNITUDE
Velocity, mm/sec
3
2
VP
1
0
0
45
90
135
180
225
270
315
360
deg
DIRECTION
Vector Angle, deg
300
VP 200 deg 100
0
0
45
90
135
180 deg
225
270
315
360
DESIGN OF MACHINERY
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 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)
500 rpm
Input crank angular velocity
b 8.375 in
52.360 rad sec
d 9.625 in 1
atan2 (x y) return 0.5 if x = 0 y 0
Two argument inverse tangent
return 1.5 if x = 0 y 0
y if x 0 return atan x Solution:
y atan x otherwise
See Figure P6-11 and Mathcad file P0648.
1.
Draw the linkage to scale and label it.
2.
Determine the range of motion for this Grashof crank rocker.
A 3 2
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
K 1 4.8125 2
K3
1
d
2
2
O4
2
K 3 2.7186
2 a c
C K 1 K 2 1 cos K 3
B 2 sin
Use equation 4.10b to find values of 4 for the open circuit.
2
2 2 A B B 4 A C atan2 5.
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
K 4 1.1493
D cos K1 K 4 cos K5
K 5 3.4367
E 2 sin
F K1 K 4 1 cos K5 6.
Use equation 4.13 to find values of 3 for the open circuit.
4
c
A cos K1 K 2 cos K 3
4.
O2
K 2 1.3392
a b c d
B
2
2
2 2 D E E 4 D F atan2
DESIGN OF MACHINERY
7.
SOLUTION MANUAL 6-48-2
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18. a sin b sin
a sin c sin
8.
Determine the velocity of point B using equations 6.19.
VB c sin j cos
VB1 arg VB
VB VB
Plot the magnitude and angle of the velocity at point B. MAGNITUDE OF VELOCITY AT B
Velocity, in/sec
150
100
sec VB in 50
0
0
60
120
180
240
300
360
deg
VB if VB1 0 VB1 VB1 DIRECTION OF VELOCITY AT B 50 45 40 Angle, deg
9.
35
VB
30
deg 25 20 15 10
0
60
120
180 deg
240
300
360
DESIGN OF MACHINERY
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.
Units:
rpm 2 rad min
Given:
Link lengths:
1
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
0 deg
Crank speed:
20 rpm atan2 (x y) return 0.5 if x = 0 y 0
Two argument inverse tangent
return 1.5 if x = 0 y 0
y if x 0 return atan x y atan x otherwise Solution:
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.
2
2
a d ( b c) acos 2 a d
P 3 B 3
A
2
4 2
158.286 deg
2
2 deg 3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K 1 0.6930
O4
d c
2
K 2 0.5726
K3
C K 1 K 2 1 cos K 3
2
2
a b c d
A cos K1 K 2 cos K 3
4.
x O2
2 a c
2
K 3 1.1317
B 2 sin
Use equation 4.10b to find values of 4 for the open circuit.
2
2 2 A B B 4 A C atan2 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
K 4 1.5281
K 5 0.2440
DESIGN OF MACHINERY
SOLUTION MANUAL 6-49-2
F K1 K 4 1 cos K5
D cos K1 K 4 cos K5 6.
E 2 sin
Use equation 4.13 to find values of 3 for the open circuit.
2
2 2 D E E 4 D F atan2 7.
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18.
a sin c sin
8.
a sin b sin
Determine the velocity of point A using equations 6.19.
VA a sin j cos 9.
Determine the velocity of the coupler point P using equations 6.36.
VP VA VPA
VPA Rpa sin j cos 10. Plot the magnitude and direction of the coupler point P. Magnitude:
Direction: VP arg VP
VP VP
Velocity, mm/sec
MAGNITUDE
sec VP in
50
0 200
150
100
50
0
50
100
150
200
deg
DIRECTION Vector Angle, deg
200
VP
0
deg
200 200
150
100
50
0 deg
50
100
150
200
DESIGN OF MACHINERY
SOLUTION MANUAL 6-50-1
PROBLEM 6-50 Statement:
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.
Units:
rpm 2 rad min
Given:
Link lengths:
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
0 deg
Crank speed:
80 rpm atan2 (x y) return 0.5 if x = 0 y 0
Two argument inverse tangent
return 1.5 if x = 0 y 0
y if x 0 return atan x y atan x otherwise Solution: 1.
See Figure P6-13 and Mathcad file P0650.
Draw the linkage to scale and label it. y
B
3 4 P
O2
O4 3 2
A
2.
Determine the range of motion for this non-Grashof triple rocker using equations 4.33. 2
arg 1
2
2
2 a d 2
arg 2
2
a d b c 2
2
2
a d b c 2 a d
2toggle acos arg 2
b c a d b c a d
The other toggle angle is the negative of this. Thus,
arg 1 1.228 arg 2 0.439
2toggle 116.0deg
x
DESIGN OF MACHINERY
SOLUTION MANUAL 6-50-2
2toggle 2toggle 1 deg 2toggle 3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K 1 2.5814
2
d
K3
c
2
2
2 a c
K 2 2.5814
K 3 2.0181
B 2 sin
A cos K1 K 2 cos K 3
2
a b c d
C K 1 K 2 1 cos K 3 4.
Use equation 4.10b to find values of 4 for the open circuit.
2
2 2 A B B 4 A C atan2 5.
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
K 4 1.2000
2 a b
D cos K1 K 4 cos K5
K 5 2.6244
E 2 sin
F K1 K 4 1 cos K5 6.
Use equation 4.13 to find values of 3 for the open circuit.
2
2 2 D E E 4 D F atan2 7.
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18.
8.
a sin b sin
a sin c sin
Determine the velocity of point A using equations 6.19.
VA a sin j cos 9.
Determine the velocity of the coupler point P using equations 6.36.
VPA Rpa sin j cos
VP VA VPA
10. Plot the magnitude and direction of the coupler point P. Magnitude:
(See next page)
VP VP
Direction:
VP arg VP
DESIGN OF MACHINERY
SOLUTION MANUAL 6-50-3
MAGNITUDE 20
Velocity, mm/sec
15
sec VP 10 in
5
0 150
100
50
0
50
100
150
50
100
150
deg
DIRECTION 200
Vector Angle, deg
100
VP
0
deg
100
200 150
100
50
0 deg
DESIGN OF MACHINERY
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.
Units:
rpm 2 rad min
Given:
Link lengths:
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
54 deg
Crank speed:
80 rpm atan2 (x y) return 0.5 if x = 0 y 0
Two argument inverse tangent
return 1.5 if x = 0 y 0
y if x 0 return atan x y atan x otherwise Solution: 1.
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 4.33. 2
arg 1
2
2
2 a d 2
arg 2
2
a d b c 2
2
2
a d b c 2 a d
b c a d b c a d
2toggle acos arg 2 The other toggle angle is the negative of this. Thus, 2toggle 2toggle 1 deg 2toggle
arg 1 1.451 arg 2 0.568
2toggle 55.4deg
DESIGN OF MACHINERY
3.
SOLUTION MANUAL 6-51-2
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K 1 2.5278 2
K3
c
K 2 2.1412 2
2
a b c d
2
K 3 3.3422
2 a c
d
A cos K1 K 2 cos K 3
B 2 sin
C K 1 K 2 1 cos K 3 4.
Use equation 4.10b to find values of 4 for the open circuit.
2
2 2 A B B 4 A C atan2 5.
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
K 4 2.6765
2 a b
K 5 3.6465
D cos K1 K 4 cos K5
E 2 sin
F K1 K 4 1 cos K5 6.
Use equation 4.13 to find values of 3 for the open circuit.
2
2 2 D E E 4 D F atan2 7.
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18. a sin b sin
a sin c sin
8.
Determine the velocity of point A using equations 6.19.
VA a sin j cos 9.
Determine the velocity of the coupler point P using equations 6.36.
VPA Rpa sin j cos
VP VA VPA
DESIGN OF MACHINERY
SOLUTION MANUAL 6-51-3
10. Plot the magnitude and direction of the coupler point P.
Magnitude:
VP VP
Direction:
VP arg VP
MAGNITUDE 20
Velocity, in/sec
15
sec VP 10 in
5
0
60
40
20
0
20
40
60
deg
DIRECTION 200
Vector Angle, deg
150 100
VP
50
deg 0 50 100
60
40
20
0 deg Crank Angle, deg
20
40
60
DESIGN OF MACHINERY
SOLUTION MANUAL 6-52-1
PROBLEM 6-52 Statement:
Figure P6-15 shows a power hacksaw that is an offset slider-crank 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.
Units:
rpm 2 rad min
Given:
Link lengths:
1
Link 2
a 75 mm
Link 3
b 170 mm 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
O2
2.
Determine the range of motion for this slider-crank linkage. 0 deg 2 deg 360 deg
3.
Determine 3 using equations 4.16 for the crossed circuit.
sin c a asin b
4.
Determine the angular velocity of link 3 using equation 6.22a:
a cos b cos
5.
Determine the velocity of pin B using equation 6.22b:
VB a sin b sin
7.
Plot the magnitude of the velocity of B. (See next page.)
2 x
DESIGN OF MACHINERY
SOLUTION MANUAL 6-52-2
VELOCITY OF POINT B 600
Velocity, mm/sec
400
200
sec VB mm 0
200
400
0
60
120
180 deg
240
300
360
DESIGN OF MACHINERY
SOLUTION MANUAL 6-53-1
PROBLEM 6-53 Statement:
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.
Units:
rpm 2 rad min
Given:
Link lengths ( walking-beam linkage):
1
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:
10 rpm
Rocker point E:
143 deg
Gears 4 & 5 phase angle
atan2 (x y) return 0.5 if x = 0 y 0
Two argument inverse tangent
return 1.5 if x = 0 y 0
y if x 0 return atan x y atan x otherwise 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. 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.
DESIGN OF MACHINERY
SOLUTION MANUAL 6-53-2
VA a' sin j cos
VAx Re VA In the global X-Y coordinate frame,
VP VAx 4.
Draw the pick and place linkage to scale and label it.
E
C
7 6
b
c
O6 5.
and
Determine the values of the constants needed for finding 6 from equations 4.8a and 4.10a. K1
d
K 1 9.8462
a 2
K3
2
2
a b c d
K2
d c
2
2 a c
K 3 5.1137
A cos K 1 K2 cos K 3
B 2 sin
C K 1 K 2 1 cos K3 7.
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,
6.
5
d 1
Use equation 4.10b to find values of 6 for the crossed circuit.
K 2 1.3913
DESIGN OF MACHINERY
SOLUTION MANUAL 6-53-3
2
2 atan2 2 A B B 4 A C 8.
Determine the values of the constants needed for finding 7 from equations 4.11b and 4.12. K4
2
d
K5
b
2
2
c d a b
2
K 4 0.6632
2 a b
D cos K 1 K 4 cos K 5
E 2 sin
F K1 K 4 1 cos K 5 9.
Use equation 4.13 to find values of 7 for the crossed circuit.
2
2 atan2 2 D E E 4 D F 10. Determine the angular velocity of links 6 and 7 using equations 6.18.
a sin b sin
a sin c sin
10. Determine the velocity of the rocker point E using equations 6.34.
VE u sin j cos 11. Transform this into the global XY system.
VEx Re VE
VEX VEx
12. Calculate and plot the velocity of E relative to P.
VEY VEx
VEXY VEX j VEY
VEy Im VE
VEP VEXY VP
K 5 9.0351
DESIGN OF MACHINERY
SOLUTION MANUAL 6-53-4
RELATIVE VELOCITY 100
Velocity, mm/sec
80
sec VEP mm
60 40
20
0
0
30
60
90
120
150
180
210
deg Crank Angle, deg
240
270
300
330
360
DESIGN OF MACHINERY
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
181 deg
Link 4 initial angle
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 4
c 46.000°
A
1
O4
d
4
3
2 b
O2
R4
R1 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 j
d e
j
c e
(a) j
a e
j
b e
(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 j sin fcos j sin
(c)
Separating into real and imaginary components and solving for 2 and f,
atan2 d c cos c sin
(d)
DESIGN OF MACHINERY
SOLUTION MANUAL 6-54-2
f
sin c sin
(e)
Differentiate equation b. j d j f e j f e dt
j
j c e
(f)
Substituting the Euler equivalents,
d c sin j cos f cos j sin f sin j cos dt
(g)
Separating into real and imaginary components and solving for 4 . Note that df/dt = adot.
3.
adot
(h)
c sin
Plot 4 over a range of 4 of 1 deg 90 deg
ANGULAR VELOCITY, LINK 4 1.2
Angular Velocity, rad/sec
1 0.8
sec 0.6 rad 0.4 0.2 0 60
80
100
120
140
160
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 arg VU
VU u sin j cos VU VU 5.
Plot the magnitude and direction of the velocity of the paper roll center. (See next page.)
DESIGN OF MACHINERY
SOLUTION MANUAL 6-54-3
MAGNITUDE OF PAPER CENTER VELOCITY 800
Velocity, mm/sec
600
sec VU 400 mm
200
0 60
80
100
120
140
160
deg Rocker Arm Position, deg
DIRECTION OF PAPER CENTER VELOCITY 80
Vector Angle, deg
60
40
VU
20
deg 0
20
40 60
80
100
120 deg
Rocker Arm Position, deg
140
160
DESIGN OF MACHINERY
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:
Solution: 1.
Link 2 (AC)
rin 301 mm
Position of link 2:
44 deg
Let
1
1 rad sec
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
sin c a asin b
C
164.509 deg 3.
3
Determine the angular velocity of link 3 using equation 6.22a: 44.000°
a cos b cos 4.
B 2
Determine the velocity of pin D using equation 6.22b:
VD a sin b sin 5.
Y
Positive upward
Calculate the velocity of point C using equation 6.23a:
VC rin sin j cos VC VC 6.
Calculate the mechanical advantage using equation 6.13. mA
VC VD
mA 3.206
A
DESIGN OF MACHINERY
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: Solution: 1.
15 deg
60 deg
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. 1 deg
3.
Determine 3 using equation 4.17.
sin c a asin b
4.
Determine the angular velocity of link 3 using equation 6.22a:
a cos b cos
5.
Determine the velocity of pin D using equation 6.22b:
VD a sin b sin 6.
Calculate the velocity of point C using equation 6.23a:
1
Let 1 rad sec
Positive upward
DESIGN OF MACHINERY
SOLUTION MANUAL 6-55b-2
VC rin sin j cos
VC VC
Calculate the mechanical advantage using equation 6.13.
mA
VD VC
MECHANICAL ADVANTAGE 12 11 10 9 8 Mechanical Advantage
7.
7
m A 6 5 4 3 2 1 0 15
20
25
30
35
40 deg
Crank Angle, deg
45
50
55
60
DESIGN OF MACHINERY
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.
Units:
rpm 2 rad min
Given:
Link lengths:
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
31 deg
Crank speed:
100 rpm atan2 (x y) return 0.5 if x = 0 y 0
Two argument inverse tangent
return 1.5 if x = 0 y 0
y if x 0 return atan x Solution: 1.
y atan x otherwise
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. 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
K 1 2.2200 2
K3
c
K 2 0.9528 2
2
a b c d
d
2 a c
2
K 3 1.5265
A cos K1 K 2 cos K 3
DESIGN OF MACHINERY
SOLUTION MANUAL 6-56-2
B 2 sin
C K 1 K 2 1 cos K 3 4.
Use equation 4.10b to find values of 4 for the open circuit.
2
2 2 A B B 4 A C atan2 5.
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
K 4 1.0777
2 a b
K 5 1.1512
D cos K1 K 4 cos K5
E 2 sin
F K1 K 4 1 cos K5 6.
Use equation 4.13 to find values of 3 for the open circuit.
2
2 2 D E E 4 D F atan2 7.
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18. a sin b sin
a sin c sin
8.
Determine the velocity of point A using equations 6.19.
VA a sin j cos 9.
Determine the velocity of the coupler point P using equations 6.36.
VPA Rpa sin j cos
VP VA VPA
10. Plot the X-component (global coordinate system) of the velocity of the coupler point P. Coordinate rotation angle:
26 deg
Vout Re VP cos Im VP sin
DESIGN OF MACHINERY
SOLUTION MANUAL 6-56-3
Vout 15
Velocity, in/sec
10
5
sec Vout in 0
5
10
0
45
90
135
180 deg
Crank Angle, deg
225
270
315
360
DESIGN OF MACHINERY
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)
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
F in
C F out
1 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
K 1 3.0000 2
K3
d c
K 2 1.5484 2
2
a b c d
2
K 3 2.9394
2 a c
A cos K 1 K 2 cos K3
B 2 sin
C K 1 K2 1 cos K 3 A 0.4204 3.
B 1.5094
C 4.2675
Use equation 4.10b to find value of 4 for the open circuit.
2
2atan2 2 A B B 4 A C 4.
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
K 4 1.9512
K 5 2.8000
DESIGN OF MACHINERY
SOLUTION MANUAL 6-57a-2
D cos K 1 K 4 cos K 5
D 3.8638
E 2 sin
E 1.5094
F K 1 K4 1 cos K 5 5.
F 0.8241
Use equation 4.13 to find values of 3 for the open circuit.
2
2atan2 2 D E E 4 D F 6.
325.961 deg
Referring to Figure 6-10, calculate the values of the angles and .
374.961 deg
If > 360 deg, subtract 360 deg from it. if 360 deg 360 deg
14.961 deg
89.479 deg
If > 90 deg, subtract it from 180 deg. if 90 deg 180 deg 7.
89.479 deg
Using equation 6.13e, calculate the mechanical advantage of the linkage in the position shown. mA
c sin rin a sin rout
mA 31.969
DESIGN OF MACHINERY
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)
60 deg
Range of positions of link 2: Solution: 1.
45 deg
See Figure P6-20 and Mathcad file P0657b.
Draw the mechanism to scale and label it.
A
2
B
2
3
F in
C F out
1 4
2
D
2.
Calculate the range of 2 in the local coordinate system (required to calculate 3 and 4). 1 deg
3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K 1 3.0000 2
K3
c
K 2 1.5484 2
2
a b c d
d
2 a c
2
K 3 2.9394
A cos K1 K 2 cos K 3
B 2 sin
C K 1 K 2 1 cos K 3 4.
Use equation 4.10b to find value of 4 for the open circuit.
2
2 atan2 2 A B B 4 A C 5.
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12.
DESIGN OF MACHINERY
K4
SOLUTION MANUAL 6-57b-2 2
d
K5
b
2
2
c d a b
2
K 4 1.9512
2 a b
K 5 2.8000
D cos K1 K 4 cos K5
E 2 sin
F K1 K 4 1 cos K5 6.
Use equation 4.13 to find values of 3 for the open circuit.
2
2 atan2 2 D E E 4 D F 7.
Referring to Figure 6-10, calculate the values of the angles and .
Using equation 6.13e, calculate and plot the mechanical advantage of the linkage over the given range.
mA
rin a sin rout c sin
MECHANICAL ADVANTAGE vs HANDLE ANGLE 60
50 Mechanical Advantage
8.
40
m A
30
20
10 45
48
51
54 deg
Handle Angle, deg
57
60
DESIGN OF MACHINERY
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
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)
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 from 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' 135.0 69°
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:
3.
135.069 deg
31.069 deg
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K 1 0.6857
a 2
K3
2
2
a b c d 2 a c
K2
d c
2
K 3 1.4989
K 2 1.4118
DESIGN OF MACHINERY
SOLUTION MANUAL 6-59a-2
A cos K 1 K 2 cos K3
B 2 sin
C K 1 K2 1 cos K 3 A 0.4605 4.
B 1.0321
C 0.1189
Use equation 4.10b to find value of 4 for the open circuit.
2
2atan2 2 A B B 4 A C 5.
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
K 5 1.4843
D cos K 1 K 4 cos K 5
D 0.1388
E 2 sin
E 1.0321
F K 1 K4 1 cos K 5 6.
F 0.4804
Use equation 4.13 to find values of 3 for the open circuit.
2
2atan2 2 D E E 4 D F 7.
K 4 1.3714
196.400 deg
Referring to Figure 6-10, calculate the values of the angles and .
165.331 deg
If > 90 deg, subtract it from 180 deg. if 90 deg 180 deg
14.669 deg
66.920 deg
If > 90 deg, subtract it from 180 deg. if 90 deg 180 deg 8.
66.920 deg
Using equation 6.13e, calculate the mechanical advantage of the linkage in the position shown. mA
c sin rin a sin rout
mA 2.970
DESIGN OF MACHINERY
SOLUTION MANUAL 6-59b-1
PROBLEM 6-59b Statement:
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).
Given:
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)
120 deg
Range of positions of link 2: Solution: 1.
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 from O2 and 34 mm from B' (see layout). Linkage in initial position Y C'
C Linkage in final position
A' A
3
4
B
2
x
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:
135.069 deg
1 deg 3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d a
K 1 0.6857
K2
d c
K 2 1.4118
DESIGN OF MACHINERY 2
K3
SOLUTION MANUAL 6-59b-2 2
2
a b c d
2
K 3 1.4989
2 a c
A cos K1 K 2 cos K 3
B 2 sin
C K 1 K 2 1 cos K 3 4.
Use equation 4.10b to find value of 4 for the open circuit.
2
2 atan2 2 A B B 4 A C 5.
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
K 4 1.3714
2 a b
K 5 1.4843
D cos K1 K 4 cos K5
E 2 sin
F K1 K 4 1 cos K5 6.
Use equation 4.13 to find values of 3 for the open circuit.
2
2 atan2 2 D E E 4 D F 7.
Referring to Figure 6-10, calculate the values of the angles and .
8.
Using equation 6.13e, calculate and plot the mechanical advantage of the linkage over the given range.
mA
rin a sin rout c sin
DESIGN OF MACHINERY
SOLUTION MANUAL 6-59b-3
50
40
30
m A
20
10
0 90
95
100
105 deg
110
115
120
DESIGN OF MACHINERY
SOLUTION MANUAL 6-60-1
PROBLEM 6-60 Statement:
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.
Units:
rpm 2 rad min
Given:
Link lengths:
Solution: 1.
1
Link 2 (O2 to A)
a 22 mm
Link 3 (A to B)
b 157 mm
Offset
Grinding wheel diameter
d 90 mm
Input crank angular velocity
120 rpm
CCW
Grinding wheel angular velocity
3450 rpm
CCW
c 40 mm
See Figure P6-23 and Mathcad file P0660.
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. 0 deg 2 deg 360 deg
3.
Determine 3 using equation 4.17.
sin c a asin b
4.
Determine the angular velocity of link 3 using equation 6.22a:
a cos b cos
5.
Determine the velocity of pin B using equation 6.22b:
VB a sin b sin 6.
Calculate the velocity of the grinding wheel contact point using equation 6.7:
Positive to the right
DESIGN OF MACHINERY
SOLUTION MANUAL 6-60-2
d VG 2
m
Directed to the right
sec
The velocity of the grinding wheel contact point relative to the workpiece, which has velocity VB , is
VGB VG VB
RELATIVE VELOCITY AT CONTACT POINT 16.6
16.5
16.4
Velocity, m/sec
7.
VG 16.258
16.3
sec VGB m 16.2
16.1
16
15.9
0
60
120
180 deg Crank Angle, deg
240
300
360
DESIGN OF MACHINERY
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:
trans 26.5 deg
75.5 deg
Velocity of point A on links 2 and 3: Solution: 1.
slip 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.
3.
VA2 a
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
SOLUTION MANUAL 6-61-2
Y
0
10 in/sec
1.686
X
Axis of slip
VA4
V trans
V A4slip 1.509
V A3
Axis of transmission VA3slip
2.064
4.
From the velocity triangle we have: 1
Velocity scale factor:
5.
kv
10 in sec 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.
VA4 c
Because link 3 slides within link 4, 3 = 4.
4.1
rad sec
CW
DESIGN OF MACHINERY
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
1
Input crank angular velocity
1 rad sec
Two argument inverse tangent
atan2 (x y) return 0.5 if x = 0 y 0
CW
return 1.5 if x = 0 y 0
y if x 0 return atan x Solution: 1.
y atan x otherwise
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. 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
K 1 0.4928 2
K3
c
K 2 0.4198 2
2
a b c d
d
2 a c
2
K 3 0.7834
A cos K1 K 2 cos K 3
B 2 sin
C K 1 K 2 1 cos K 3
DESIGN OF MACHINERY
4.
SOLUTION MANUAL 6-62-2
Use equation 4.10b to find values of 4 for the open circuit.
2
2 2 A B B 4 A C atan2 5.
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
K 4 0.5574
2 a b
K 5 0.3655
D cos K1 K 4 cos K5
E 2 sin
F K1 K 4 1 cos K5 6.
Use equation 4.13 to find values of 3 for the open circuit.
2
2 2 D E E 4 D F atan2 7.
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18. a sin b sin
a sin c sin
Plot the angular velocity of link 4.
ANGULAR VELOCITY, LINK 4 0.5
Angular Velocity, rad/sec
9.
1
sec 1.5 rad
2
2.5
0
60
120
180 deg Crank Angle, deg
240
300
360
DESIGN OF MACHINERY
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
1
Input crank angular velocity
1 rad sec
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 Solution: 1.
y atan x otherwise
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 4
x
2.
O4
Determine the range of motion for this Grashof double rocker. From Figure 3-1a on page 80, one toggle angle is
a d ( b c) acos 2 a d 2
2
2
124.294 deg
The other toggle angle is the negative of this. The range of motion is 1 deg 3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d a
K 1 1.1311
K2
d c
K 2 2.0294
DESIGN OF MACHINERY
SOLUTION MANUAL 6-63-2
2
K3
2
2
a b c d
2
K 3 0.7418
2 a c
A cos K1 K 2 cos K 3
B 2 sin
C K 1 K 2 1 cos K 3 4.
Use equation 4.10b to find values of 4 for the open circuit.
2
2 2 A B B 4 A C atan2 5.
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
Line AD:
y 0 ( x d) tan
Eliminating y and solving for the x- and y-coordinates of the intersection,
x 24
6.
tan tan d tan
y24 x 24 tan
Plot the fixed centrode. FIXED CENTRODE 10
y-Coordinate, in
5
y24
0
in 5
10
10
5
0
x24 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. 0 deg 0.5 deg 360 deg
d L4
5
10
DESIGN OF MACHINERY
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
K2
a
K 1 2.3824 2
K3
c
K 2 1.3279 2
2
a b c d
d
2
K 3 1.6097
2 a c
A cos K1 K 2 cos K 3
B 2 sin
C K 1 K 2 1 cos K 3 10. Use equation 4.10b to find values of 4 for the open circuit.
2
2 2 A B B 4 A C atan2 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
Line BC:
y 0 ( x d) tan
Eliminating y and solving for the x- and y-coordinates of the intersection,
x 24
6.
tan tan d tan
Plot the moving centrode. (See next page.)
y24 x 24 tan
DESIGN OF MACHINERY
SOLUTION MANUAL 6-63-4
MOVING CENTRODE 10
y-Coordinate, in
5
y24
0
in 5
10
10
5
0
x24 in
x-Coordinate, in
5
10
DESIGN OF MACHINERY
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
56.5 deg
Link 2 (O2A)
a 1.35 in
Angle 2 makes with X axis
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
88 deg
Angular velocity of link 2 Solution: 1.
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
3.
VA3 27.000
in sec
VA3 90 deg
VA3 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
SOLUTION MANUAL 6-64-2
Y
0
10 in/sec
1.295 X Vtrans
V A3 2.369
4.
Vslip
From the velocity triangle we have: 1
Velocity scale factor:
5.
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.
VA4 c
c 0.939 in and
13.791
rad
132.661 deg
CW
sec
Determine the magnitude and sense of the vector VB using equation 6.7. VB e
VA4 90 deg 8.
in
Vslip 23.690
From the linkage layout above:
7.
10 in sec
Vslip 2.369 in kv
VA4 Vtrans 6.
kv
VB 18.756
in sec
VA4 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
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
10 in/sec
Y
VCB X VC 2.088
10. From the velocity triangle we have: 1
Velocity scale factor: VC 2.088 in kv
kv
10 in sec
VC 20.9
in in sec
VC 90 deg VC 2.0 deg
DESIGN OF MACHINERY
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:
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
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
SOLUTION MANUAL 6-65-2
0
50 in/sec
VB 2.304
VBslip
Y 1.317 X
3.256
Vtrans
1.890 VA
4.
VA2slip
From the velocity triangle we have: 1
Velocity scale factor:
5.
kv
50 in sec 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.
VB b
70.0
rad sec
CW
DESIGN OF MACHINERY
SOLUTION MANUAL 6-66-1
PROBLEM 6-66 Statement:
Figure P6-28 shows a quick-return 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 1.69 in
Angle O2O4 makes with X axis
15.5 deg
Link 2 (L2)
a 1.00 in
Angle link 2 makes with X axis
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.
10 rad sec
1
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
3.
VA2 10.000
in sec
VA2 90 deg
VA2 189.0deg
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
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
V A3 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.
VA4 c
c 2.068 in and
2.790
rad
44.228 deg
CCW
sec
Determine the magnitude and sense of the vector VB using equation 6.7. VB e
VB 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.
V slip
VB 13.281
in sec
VB 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
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
VC 180 deg V CB X
VC 1.659
DESIGN OF MACHINERY
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:
162 deg
171 deg
See Figure P6-29 and Mathcad file P0667.
Draw the mechanism to scale and label it. P
3
B 3
Fin Vin
4
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:
180 deg
1 deg 3.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d
K2
a
K 1 0.5600 2
K3
c
K 2 0.8750 2
2
a b c d
d
2 a c
2
K 3 1.2850
A cos K1 K 2 cos K 3
DESIGN OF MACHINERY
SOLUTION MANUAL 6-67-2
B 2 sin
C K 1 K 2 1 cos K 3 4.
Use equation 4.10b to find value of 4 for the open circuit.
2
2 atan2 2 A B B 4 A C 5.
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
K 4 2.0000
2 a b
K 5 1.7543
D cos K1 K 4 cos K5
E 2 sin
F K1 K 4 1 cos K5 6.
Use equation 4.13 to find values of 3 for the open circuit.
2
2 atan2 2 D E E 4 D F 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
rin a sin rout
b sin
MECHANICAL ADVANTAGE
Mechanical Advantage
7.
0.13
m A 0.12
0.11
0.1 162
164
166
168 deg
Pedal Angle, deg
170
172
DESIGN OF MACHINERY
8.
SOLUTION MANUAL 6-67-3
Calculate and plot the velocity ratio using equation 6.13d,
mV
1
mA
VELOCITY RATIO 10
Velocity Ratio
8 6
m V
4 2 0 162
164
166
168
170
172
deg Pedal Angle, deg
Calculate and plot the output velocity using equation 6.13a.
Vout Vin mV
OUTPUT VELOCITY 30
Velocity, m/sec
9.
20
sec Vout m 10
0 162
164
166
168 deg
Pedal Angle, deg
170
172
DESIGN OF MACHINERY
SOLUTION MANUAL 6-67-4
10. Calculate and plot the output force using equation 6.13a.
Fout Fin mA
OUTPUT FORCE 8
Force, N
6
F out
4
N 2
0 162
164
166
168 deg
Pedal Angle, deg
170
172
DESIGN OF MACHINERY
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
5
1,2 O2
X
1,4
6 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 I2,6: I1,2 -I1,6 and I2,5 -I5,6 I4,6: I1,4 -I1,6 and I4,5 -I5,6
4
2,5 2,6
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
1,4
6 O4 1,3
5,6
DESIGN OF MACHINERY
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: 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
atan2 (x y) return 0.5 if x = 0 y 0
Two argument inverse tangent:
return 1.5 if x = 0 y 0
y if x 0 return atan x y atan x otherwise Solution: 1.
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: 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
K 1 1.0498
2
d
K3
c
K 2 2.1700
2
2
a b c d
2
2 a c
K 3 1.1237
A cos K1 K 2 cos K 3
B 2 sin
C K 1 K 2 1 cos K 3
2
2 2 A B B 4 A C atan2
if 2 2 5.
Calculate the coordinates of the intersection of links 1 and 3 in the xy coordinate system.
d tan x 242 tan tan
6.
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.
y242 x 242 tan
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
K 1 2.3100
d c
K 2 1.1176
A cos K1 K 2 cos K 3
2
K3
2
2
a b c d
K 3 1.4271
2 a c
2
DESIGN OF MACHINERY
SOLUTION MANUAL 6-69-2
C K 1 K 2 1 cos K 3
B 2 sin
2
2 2 A B B 4 A C atan2
if 2 2 5.
Calculate the coordinates of the intersection of links 1 and 3 in the xy coordinate system.
d tan x 244 tan tan
y244 x 244 tan
LINK 2 GROUNDED 5 4 3 2 1
y242 0 1 2 3 4 5
4
3
2
1
0
1
x242
7.
Define the input crank motion for this inversion: 41 deg 42 deg 241 deg LINK 4 GROUNDED 10 8 6 4 2
0
y244
2 4 6 8 10
10
8
6
4
2
0
2
x244
4
6
8
10
DESIGN OF MACHINERY
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:
2 110 deg
102 deg
Coordinate angle
1
1 rad sec
Input crank angular velocity Solution: 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
VA 2.170
VA 2 90 deg
VA 20.000 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.
DESIGN OF MACHINERY
0
SOLUTION MANUAL 6-70a-2
VA
1 in/sec
Y
X VBA
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
VB 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
VC 0.0 deg
DESIGN OF MACHINERY
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:
2 110 deg
102 deg
Coordinate angle
1
1 rad sec
Input crank angular velocity Solution: 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
3.
VA 2.170
VA 2 90 deg
VA 20.0 deg
VA AI13
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.
6.
sec
Determine the angular velocity of link 3 using equation 6.9a.
VB BI13 5.
in
VA a
4.
SOLUTION MANUAL 6-70b-2
VB BI15
0.145
rad
CCW
sec
Determine the magnitude of the velocity at point C using equation 6.9b. VC CI15
VC 1.436
in sec
to the right
DESIGN OF MACHINERY
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:
2XY 110 deg
102 deg
Coordinate angle
2 1 rad sec
Input crank angular velocity
1
CW
atan2 (x y) return 0.5 if x = 0 y 0
Two argument inverse tangent
return 1.5 if x = 0 y 0
y if x 0 return atan x y atan x otherwise Solution: 1.
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:
2 2XY K1
d
K 1 0.4608
a 2
K3
2 212.000 deg
2
2
a b c d 2 a c
K2
d c
2
K 3 0.6755
A cos 2 K 1 K 2 cos 2 K3
K 2 0.4329
DESIGN OF MACHINERY
SOLUTION MANUAL 6-70c-2
B 2 sin 2
C K 1 K2 1 cos 2 K 3 A 0.2662 3.
B 1.0598
C 2.3515
Use equation 4.10b to find values of 4 for the open circuit.
2
2
4xy 2atan2 2 A B B 4 A C 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
K 4 0.4838
2 a b
K 5 0.5178
D cos 2 K 1 K 4 cos 2 K 5
D 2.2370
E 2 sin 2
E 1.0598
F K 1 K4 1 cos 2 K 5 5.
F 0.3808
Use equation 4.13 to find values of 3 for the open circuit.
2
3xy 2atan2 2 D E E 4 D F 2 6.
a 2 sin 2 3xy c sin 4xy 3xy
4 0.591
4 57.635 deg
cc 0 in
sin 4cc c 5 asin e
5 158.818 deg
dd c cos 4 e cos 5
dd 6.272in
Determine the angular velocity of link 5 using equation 6.22a:
4 c cos 5 4 e cos 5 7.
sec
Determine 5 and d, with respect to O4, using equation 4.17. Offset:
6.
rad
Transform 4 back to the global XY system.
4 4xy 5.
3xy 70.950 deg
Determine the angular velocity of link 4 for the open circuit using equations 6.18.
4 7.
4xy 159.635 deg
5 0.145
rad sec
Determine the velocity of pin C using equation 6.22b:
VC c 4 sin 4 e 5 sin 5 VC 1.436
in sec
VC 1.436
in sec
arg VC 0.000deg
DESIGN OF MACHINERY
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 (L 5)
e 5.40
d 1.000
Fourbar ground link (L1)
atan2 (x y) return 0.5 if x = 0 y 0
Two argument inverse tangent: Crank velocity:
return 1.5 if x = 0 y 0
rad 1 sec
y if x 0 return atan x y atan x otherwise
Solution:
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 slider-crank mechanism using the output of the fourbar, link 4, as the input to the slider-crank.
2.
Define one revolution of the input crank: 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
K 1 0.4608
K3
d c
K 2 0.4329
2
2
a b c d
2
2 a c
K 3 0.6755
A cos K1 K 2 cos K 3
C K 1 K 2 1 cos K 3
B 2 sin
2
2 2 A B B 4 A C deg atan2 102 4.
If the calculated value of 4 is greater than 2, subtract 2from it and if it is negative, make it positive.
if 2 2
if 0 2 5.
Determine the slider-crank motion using equations 4.16 and 4.17 with 4 as the input angle.
sin c asin e
f c cos e cos
DESIGN OF MACHINERY
5.
SOLUTION MANUAL 6-71-2
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
K 4 0.4838
2 a b
K 5 0.5178
D cos K1 K 4 cos K5
E 2 sin
F K1 K 4 1 cos K5 6.
Use equation 4.13 to find values of 3 for the open circuit.
2
2 atan2 2 D E E 4 D F 7.
Determine the angular velocity of link 4 for the open circuit using equations 6.18.
a sin c sin 102 deg
0.5 0.75 1
1.25 1.5 1.75 2
0
45
90
135
180
225
270
deg
8.
Determine the angular velocity of link 5 using equation 6.22a:
c cos e cos
9.
Determine the velocity of pin C using equation 6.22b:
VC c sin e sin
315
360
DESIGN OF MACHINERY
SOLUTION MANUAL 6-71-3
2 1 0
VC 1 2 3 4
0
45
90
135
180
225
270
315
360
deg
2
1
0
VC 1
2
3
4
3
4
5
f
6
7
8
DESIGN OF MACHINERY
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 2,3
1,6
5,6
4,6
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
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
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 2
1,6 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
1,2; 2,5; 2,4: 2,6 4,5
2
O2 5
3 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
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
dX 3.259 in
Link 1 Y-offset
dY 2.905 in
Angle CDB
5 36.0 deg
Output rocker angle:
90 deg Global XY system 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 1 rad sec VD a
3.
VD 1.542
in sec
VD 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
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
VC 54.195 deg
sec in
VCD 127.003 deg
sec
Determine the angular velocity of links 5 and 4 using equation 6.7.
6.
1 in sec
VCD b VC c
0.607
rad
0.607
rad
sec
sec
Determine the magnitude and sense of the vector VBD using equation 6.7. VBD p
VBD 1.986
in sec
VBD VCD 5 7.
VBD 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
V BD VB 0.682 121.607°
VD
1 in/sec
DESIGN OF MACHINERY
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
VB 121.607 deg
sec
Use equation 6.5 to (graphically) determine the magnitude of the velocity at point A, the magnitude of the relative 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
VA 131.690 deg
sec
11. Determine the angular velocity of link 2 with respect to the assumed value of 6 using equation 6.7.
VA g
0.415
rad sec
12. Calculate the actual value of the angular velocity of link 6.
rad sec
24.124
rad sec
CW
DESIGN OF MACHINERY
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
dX 3.259 in
Link 1 Y-offset
dY 2.905 in
Angle CDB
5 36.0 deg
Output rocker angle:
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
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 vale 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
VD 90 deg 3.
Determine the angular velocity of link 3 using equation 6.9a.
4.
VD DI15
VB BI13
VB 0.682
in sec
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
7.
rad
Use equation 6.9c to determine the angular velocity of link 3.
6.
0.607
Determine the magnitude of the velocity at point B using equation 6.9b. VB BI15
5.
VD 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
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:
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
dX 3.259 in
Link 1 Y-offset
dY 2.905 in
Angle CDB
5 36.0 deg
Link 1 (O4 to O6)
d 1.000 in
Output rocker angle:
6XY 90 deg
Global XY system
Input crank angular velocity
10 rad sec
Coordinate rotationm angle
90 deg
Two argument inverse tangent
1
CW
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 Solution: 1.
y atan x otherwise
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.
6 6XY
6 180.000 deg
Y
2 A
D
O2
5
X
6
C O6 3
5
y
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
K 1 0.6485
K2
d c
K 2 0.4706
2
K3
2
2
a b c d
K 3 0.4938
2 a c
2
DESIGN OF MACHINERY
SOLUTION MANUAL 6-73c-2
A cos 6 K 1 K 2 cos 6 K3
A 0.6841
B 2 sin 6
B 0.0000
C K 1 K2 1 cos 6 K 3 3.
C 2.6129
Use equation 4.10b to find values of 4 for the crossed circuit.
2
4 2atan2 2 A B B 4 A C 4.
4 234.195 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
K 5 0.5288
D cos 6 K 1 K 4 cos 6 K 5
D 2.6407
E 2 sin 6
E 0.0000
F K 1 K4 1 cos 6 K 5 5.
K 4 0.4634
F 0.6563
Use equation 4.13 to find values of 5 for the crossed circuit.
2
51 2atan2 2 D E E 4 D F 6.
51 307.003 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. 1
6 1 rad sec
5 0.607
rad
4 0.607
rad
5
a 6 sin 4 6 b sin 51 4
4
a 6 sin 6 51 c sin 4 51
sec
sec
DESIGN OF MACHINERY
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
dX 3.259 in
Link 1 Y-offset
dY 2.905 in
Angle CDB
5 36.0 deg
Link 1 (O4 to O6)
d 1.000 in
Output rocker angle:
6XY 90 deg
Global XY system
Input crank angular velocity
10 rad sec
Coordinate rotationm angle
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
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
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
1
6
2
5
3 4
DESIGN OF MACHINERY
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
4
O2
5
O4 5,6
3,4
2.
4,5 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; I3,5; 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
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:
2 90 deg
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° 33.359°
2
100.938° 5
3 4 O2
X
6 O 4 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
VA 10.000
VA 2 90 deg
VA 180.000 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.
DESIGN OF MACHINERY
SOLUTION MANUAL 6-76a-2
Y 1.671 1.063 0
5 in/sec
VB
V BA
X
VA
4.
From the velocity triangle we have: kv
Velocity scale factor:
VBA 1.063 in kv
VBA 5.315 6.497
c
VB 147.915 deg
sec in
VBA 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
VD 55.085 deg
sec
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 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: kv
Velocity scale factor:
VE 2.450 in kv
5 in sec
5 in/sec
1
in
VE 12.250
0
X
in sec
2.450
VE VD
g
15.888
rad
CW
VE
sec VDE
DESIGN OF MACHINERY
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:
2 90 deg
Input crank angular velocity Solution: 1.
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 VA 10.000 sec
VA 2 90 deg 3.
BI15 1.740 in
VA 180.0 deg
Determine the angular velocity of link 3 using equation 6.9a.
VA AI13
1.398
rad sec
CCW
DESIGN OF MACHINERY
4.
Determine the magnitude of the velocity at point B using equation 6.9b. VB BI13
5.
VB c
VD DI15
6.496
rad
CW
sec
VD 9.283
in sec
9.803
rad
CW
sec
Determine the magnitude of the velocity at point E using equation 6.9b. VE EI15
9.
sec
Use equation 6.9c to determine the angular velocity of link 5.
8.
in
Determine the magnitude of the velocity at point D using equation 6.9b. VD e
7.
VB 8.354
Use equation 6.9c to determine the angular velocity of link 4.
6.
SOLUTION MANUAL 6-76b-2
VE 12.244
in sec
Use equation 6.9c to determine the angular velocity of link 6.
VE g
15.88
rad sec
CW
DESIGN OF MACHINERY
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
2 90 deg
Crank angle:
2 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° 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 2
K3
2
2
a b c d
d
K 1 3.8570
c
K 2 2.9992
2
K 3 1.2015
2 a c
A cos 2 K 1 K 2 cos 2 K3
B 2 sin 2
C K 1 K2 1 cos 2 K 3 A 2.6555 3.
B 2.0000
C 5.0585
Use equation 4.10b to find values of 4 for the crossed circuit.
2
41 2atan2 2 A B B 4 A C 4.
41 237.915 deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4
2
d b
K5
2
2 a b
D cos 2 K 1 K 4 cos 2 K 5
2
c d a b
2
K 4 1.0150 D 7.6284
K 5 3.7714
DESIGN OF MACHINERY
SOLUTION MANUAL 6-76c-2
F K 1 K4 1 cos 2K 5 E 2 sin 2
5.
E 2.0000 F 0.0856
Use equation 4.13 to find values of 3 for the crossed circuit.
2
3 2atan2 2 D E E 4 D F 6.
Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18.
3
a 2 sin 41 2 b sin 3 41
3 1.398
42 41 157 deg 360 deg 8.
rad sec
a 2 sin 2 3 rad 4 6.496 c sin 41 3 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
4
7.
3 326.641 deg
42 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
2
2
e f g h
h
K 1 0.5500
g
K 2 1.0195
2
K 3 0.7263
2g e
B 2 sin 42 C K 1 K2 1 cos 42K 3
A cos 42 K 1 K 2 cos 42 K 3
A 0.1603 9.
B 1.1447
C 0.3796
Use equation 4.10b to find values of 6 for the open circuit.
2
6 2atan2 2 A B B 4 A C
6 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 ef
D cos 42 K 1 K4 cos 42 K 5
E 2 sin 42
2
K 4 0.6112
K 5 1.0119
D 0.2408 E 1.1447
F K 1 K4 1 cos 42 K 5
F 0.7807
11. Use equation 4.13 to find values of 5 for the open circuit.
2
5 2atan2 2 D E E 4 D F
5 280.938 deg
12. Determine the angular velocity of link6 for the open circuit using equations 6.18.
4 sin 42 5 e 6 6 5 g sin
6 15.885
rad sec
DESIGN OF MACHINERY
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
K2
a 2
K3
2
2
a b c d
d
K 1 3.8570
c
2
K 3 1.2015
2 a c
K 2 2.9992
A cos K 1 K 2 cos K3
B 2 sin
C K1 K 2 1 cos K 3 3.
Use equation 4.10b to find values of 4 for the crossed circuit.
2
2 atan2 2 A B B 4 A C 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
K 4 1.0150
2 a b
D cos K 1 K4 cos K 5
K 5 3.7714
E 2 sin
F K 1 K 4 1 cos K 5 5.
Use equation 4.13 to find values of 3 for the crossed circuit.
2
2 atan2 2 D E E 4 D F 6.
Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18.
a sin c sin
7.
a sin b 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
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. K1
h
K2
e 2
K3
2
2
e f g h
h
K 1 0.5500
g
2
K 3 0.7263
2g e
K 2 1.0195
A' cos K1 K 2 cos K 3 B' 2 sin
C' K 1 K 2 1 cos K 3 9.
Use equation 4.10b to find values of 6 for the open circuit.
2
2 atan2 2 A' B' B' 4 A' C' 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
K 4 0.6112
2 ef
K 5 1.0119
D' cos K1 K 4 cos K 5 E' 2 sin
F' K 1 K 4 1 cos K5 11. Use equation 4.13 to find values of 5 for the open circuit.
2
2 atan2 2 D' E' E' 4 D' F' 12. Determine the angular velocity of link6 for the open circuit using equations 6.18.
sin e g sin
ANGULAR VELOCITY - LINK 6 3 2
1
0 1 2
0
45
90
135
180 deg
225
270
315
360
DESIGN OF MACHINERY
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
5,6
6
1,6
5
O6
8
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
2
4 O2
1,2; 2,4
O4
8
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
5
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
3,6; 3,7; 3,8 6,7
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
O6
5,6; 4,6
2,5
2,3
I6,7: I5,6 -I5,7 and I6,8 -I7,8
6
1,6
5,7
DESIGN OF MACHINERY
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
7.0 deg
Angle BO4C
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
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: 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
atan2 (x y) return 0.5 if x = 0 y 0
Two argument inverse tangent
return 1.5 if x = 0 y 0
y if x 0 return atan x Solution: 1.
y atan x otherwise
See Figure 3-37a and Mathcad file P0680.
Determine the range of motion for this Grashof crank rocker. 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
K 1 10.3971 2
K3
c
K 2 1.4140
2
2
a b c d
2
K 3 7.4187
2 a c
d
A cos K1 K 2 cos K 3
B 2 sin
C K 1 K 2 1 cos K 3 3.
Use equation 4.10b to find values of 4 for the open circuit.
2
2 2 A B B 4 A C atan2 4.
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
E 2 sin F K1 K 4 1 cos K5
D cos K1 K 4 cos K5
5.
Use equation 4.13 to find values of 3 for the open circuit.
2
K 4 1.4140
K 5 7.4187
DESIGN OF MACHINERY
SOLUTION MANUAL 6-80-2
2
2 2 D E E 4 D F atan2 6.
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18. a sin b sin
a sin c sin
7.
Determine the velocity of point A using equations 6.19.
VA a sin j cos 8.
VA VA
Determine the velocity of the coupler point P using equations 6.36.
VPA Rpa sin cos j
VP VA VPA
Plot the magnitude and direction of the velocity at coupler point P.
Magnitude:
VP VP
Direction:
VP1 arg VP VP if VP1 0 VP1 2 VP1 MAGNITUDE 0.18
0.16 Velocity, mm/sec
9.
0.14 VA VP
0.12
0.1
0
45
90
135
180 deg
Crank Angle, deg
225
270
315
360
DESIGN OF MACHINERY
SOLUTION MANUAL 6-80-3
DIRECTION
Vector Angle, deg
300
VP 200 deg
100
0
0
45
90
135
180 deg
Crank Angle, deg
225
270
315
360
DESIGN OF MACHINERY
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: 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
atan2 (x y) return 0.5 if x = 0 y 0
Two argument inverse tangent
return 1.5 if x = 0 y 0
y if x 0 return atan x Solution: 1.
y atan x otherwise
See Figure 3-37a and Mathcad file P0681.
Determine the range of motion for this Grashof crank rocker. 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
K 1 10.3971 2
K3
c
K 2 1.4140
2
2
a b c d
2
K 3 7.4187
2 a c
d
A cos K1 K 2 cos K 3
B 2 sin
C K 1 K 2 1 cos K 3 3.
Use equation 4.10b to find values of 4 for the open circuit.
2
2 2 A B B 4 A C atan2 4.
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
E 2 sin F K1 K 4 1 cos K5
D cos K1 K 4 cos K5
5.
Use equation 4.13 to find values of 3 for the open circuit.
2
K 4 1.4140
K 5 7.4187
DESIGN OF MACHINERY
SOLUTION MANUAL 6-81-2
2
2 2 D E E 4 D F atan2 6.
Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18. a sin b sin
a sin c sin
7.
Determine the velocity of point A using equations 6.19.
VA a sin j cos 8.
VA VA
Determine the velocity of the coupler point P using equations 6.36.
VPA Rpa sin cos j
VP VA VPA
Calculate the magnitude of the velocity at coupler point P.
VP VP
Magnitude:
err
Deviation from constant speed:
VA
VP VA
DEVIATION FROM CONSTANT SPEED 30
20
Percent Error
9.
err
10
% 0
10
20
0
45
90
135
180 deg
Crank Angle, deg
225
270
315
360
DESIGN OF MACHINERY
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
atan2 (x y) return 0.5 if x = 0 y 0
Two argument inverse tangent:
return 1.5 if x = 0 y 0
rad 1 sec
Crank speed:
y if x 0 return atan x y atan x otherwise
Solution: 1.
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 c cos
YP = b sin c sin
2.
Define one revolution of the input crank: 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
K2
a
K 1 1.5000
2
d
K3
c
K 2 0.7500
2
2
a b c d
2
2 a c
K 3 0.8125
A cos K1 K 2 cos K 3
C K 1 K 2 1 cos K 3
B 2 sin
2
2 atan2 2 A B B 4 A C 2 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
2 a b
D cos K1 K 4 cos K5
F K1 K 4 1 cos K5 5.
Use equation 4.13 to find values of 3 for the crossed circuit.
K 4 0.7500
K 5 0.8125
E 2 sin
DESIGN OF MACHINERY
SOLUTION MANUAL 6-82-2
2
2 atan2 2 D E E 4 D F 2 6.
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 x P = XP - d ', y P = YP .
x P d b cos c cos d'
yP b sin c sin 7.
Use equations 6.18 to calculate 3 and 4.
a sin b sin
a sin c sin
Differentiate the position equations with respect to time to get the velocity components.
VPx b sin c sin VPy b cos c cos
2
2
VP VPx VPy
MAGNITUDE 0.6 0.4
Velocity, mm/sec
8.
0.2
VPy VPx
0
0.2 0.4 0.6
0
45
90
135
180 deg
Crank Angle, deg
225
270
315
360
DESIGN OF MACHINERY
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)
C 6
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
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
Given:
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. 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
dX 47.5 in
Link 1 Y-offset
dY 76.00 in 12.00 in
Coupler point x-offset
px 114.68 in
Coupler point y-offset
py 33.19 in
Crank angle:
45 deg 10 rad sec
Input crank angular velocity
1
CCW
Solution: See Figure P6-30 and Mathcad file P0684a. 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°
P 2
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
3.
VA 140.000
in sec
VA 45 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 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
SOLUTION MANUAL 6-84a-2
0
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
VB 119.063 deg
in sec
Determine the angular velocity of links 3 and 4 using equation 6.7.
VBA
0.511
b VB
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
6.
100 in sec
atan2 dX d Y
126.582 deg
PX px cos p y sin
PX 94.998 in
PY px sin py cos
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
e
PX dX2 PY dY2 VP1 113.446
in sec
e 48.219 in
DESIGN OF MACHINERY
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
Given:
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. 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
dX 47.5 in
Link 1 Y-offset
dY 76.00 in 12.00 in
Coupler point x-offset
px 114.68 in
Coupler point y-offset
py 33.19 in
Crank angle:
45 deg 10 rad sec
Input crank angular velocity Solution: 1.
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. in VA a VA 140.000 sec
DESIGN OF MACHINERY
SOLUTION MANUAL 6-84b-2
VA 90 deg 3.
Determine the angular velocity of link 3 using equation 6.9a.
4.
VA 135.0 deg
VA
0.511
AI13
VB 120.622
VB
2.353
c
sec
rad
CCW
sec
Transform the xy coordinates of point P1 into the XY system using equations 4.0b. Coordinate transformation angle
7.
in
Use equation 6.9c to determine the angular velocity of link 4.
6.
CW
sec
Determine the magnitude of the velocity at point B using equation 6.9b. VB BI13
5.
rad
atan2 dX d Y
PX px cos p y sin
PX 94.998 in
PY px sin py cos
PY 72.308 in
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
e
PX dX2 PY dY2
VP1 113.467
in sec
e 48.219 in
DESIGN OF MACHINERY
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
Given:
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. 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
dX 47.5 in
Link 1 Y-offset
dY 76.00 in 12.00 in
Coupler point x-offset
px 114.68 in
Coupler point y-offset
py 33.19 in
Crank angle:
2XY 45 deg
126.582 deg
Coordinate transformation angle:
2 10 rad sec
Input crank angular velocity Solution: 1.
XY coord system 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:
2 2XY 2
2
d d X dY
Distance O2O4: 2.
2 81.582 deg d 79.701 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
a b c d 2 a c
d
K 1 5.6929
c
2
K 3 1.9340
A cos 2 K 1 K 2 cos 2 K3
C K 1 K2 1 cos 2 K 3 A 3.8401
K 2 1.5548
B 1.9785
C 7.2529
B 2 sin 2
DESIGN OF MACHINERY
3.
SOLUTION MANUAL 6-84c-2
Use equation 4.10b to find values of 4 for the crossed circuit.
2
4 2atan2 2 A B B 4 A C 4.
4 262.482 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
K 4 0.9963
2 a b
E 2 sin 2 F K 1 K4 1 cos 2K 5
D cos 2 K 1 K 4 cos 2 K 5
5.
D 10.0081 E 1.9785 F 1.0849
Use equation 4.13 to find values of 3 for the crossed circuit.
2
3 2atan2 2 D E E 4 D F 6.
7.
K 5 4.6074
3 332.475 deg
Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18.
3 0.511
4 2.353
3
a 2 sin 4 2 b sin 3 4
4
a 2 sin 2 3 c sin 4 3
rad sec
rad sec
Determine the velocity of points A and B for the crossed circuit using equations 6.19.
VA a 2sin 2 j cos 2 VA ( 138.492 20.495j)
in
VA 140.000
sec
VAXY arg VA
in sec
VAXY 135.000 deg
VB c 4sin 4 j cos 4 VB ( 119.588 15.781j)
in
VBXY arg VB 8.
9.
VB 120.624
sec
in sec
VBXY 119.064 deg
Calculate the distance from O4 to P1 and the angle BO4P1.
px d2 py2
Distance from O4 to P1:
e
py 4 180 deg atan px d
4 136.503 deg
Determine the velocity of point P1 using equations 6.35.
VP1 e 4sin 4 j cos 4 VP1 ( 55.122 99.181j)
VPXY arg VP1
in sec
VP1 113.469
in sec
VPXY 245.646 deg
e 48.219 in
DESIGN OF MACHINERY
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
dX 47.5 in
Link 1 Y-offset
dY 76.00 in 12.00 in
Coupler point x-offset
px 114.68 in
Coupler point y-offset
py 33.19 in
Crank angle:
2XY 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:
2XY 2XY 2
2.
2
d d X dY
Distance O2O4:
d 79.701 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
a b c d
d c
K 1 5.6929
K 2 1.5548
2
2 a c
K 3 1.9340
K 1 K 2cos2XY K 3
A 2XY cos 2XY
C 2XY K 1 K 2 1 cos 2XY K 3 B 2XY 2 sin 2XY
DESIGN OF MACHINERY
3.
SOLUTION MANUAL 6-85-2
Use equation 4.10b to find values of 4 for the crossed circuit.
4A2XY C2XY
2XY 2 2 A 2XY B 2XY B 2XY atan2 4.
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
K 4 0.9963
2 a b
K 5 4.6074
K 1 K4cos2XY K 5 E 2XY 2 sin 2XY F 2XY K 1 K 4 1 cos 2XY K 5 D 2XY cos 2XY
5.
Use equation 4.13 to find values of 3 for the crossed circuit.
4D2XY F2XY
2XY 2 2 D 2XY E 2XY E 2XY atan2 6.
Determine the angular velocity of link 4 for the crossed circuit using equations 6.18.
2XY 7.
a sin 2XY 2XY c sin 2XY 2XY
Calculate the distance from O4 to P1 and the angle BO4P1. e
Distance from O4 to P1:
px d2 py2
e 48.219 in
py 180 deg atan px d
136.503 deg
Determine and plot the velocity of point P1 using equations 6.35.
VP1 2XY e 2XY sin 2XY cos 2XY j
VPXY 2XY arg VP1 2XY
VP1 2XY VP1 2XY
P1 VELOCITY MAGNITUDE 200
150 Velocity - in/sec
8.
2
sec VP1 2XY 100 in
50
0
0
45
90
135
180
225
2XY deg Crank Angle - Global XY System
270
315
360
DESIGN OF MACHINERY
SOLUTION MANUAL 6-85-3
P1 VELOCITY DIRECTION
Velocity angle - Global XY System
300 250 200
VPXY 2XY
150
deg 100 50 0
0
45
90
135
180
225
2XY deg Crank Angle - Global XY System
270
315
360
DESIGN OF MACHINERY
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)
C 6
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
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
dX 2.79 in
Link 1 Y-offset
dY 6.95 in
Coupler point data:
p 15.00 in
0 deg
Crank angle:
94.121 deg
Solution: 1.
1
1 rad sec
Input crank angular velocity
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
VA 9.170
VA 90 deg
VA 184.121 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.
DESIGN OF MACHINERY
SOLUTION MANUAL 6-87a-2
Y 0
25 in/sec VA X
1.798"
1.783" VBA
4.
VB
From the velocity triangle we have: 1
Velocity scale factor:
5.
25 in sec in
in
VB 1.783 in kv
VB 44.575
VBA 1.798 in kv
VBA 44.950
VBA b VB c
in
VBA 274.103 deg
sec
3.466
rad
4.658
rad
sec
sec
Determine the magnitude and sense of the vector VPA using equation 6.7. VPA p
VPA 51.985
in sec
VPA VBA 7.
VB 262.352 deg
sec
Determine the angular velocity of links 3 and 4 using equation 6.7.
6.
kv
VPA 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
1
Velocity scale factor:
VP 2.059 in kv
P 96.051 deg
kv
X
25 in sec in
VP 51.475
96.051°
in 2.059"
sec VPA
VP
DESIGN OF MACHINERY
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
dX 2.790 in
Link 1 Y-offset
dY 6.948 in
Coupler point data:
p 15.00 in
0 deg
Crank angle:
94.121 deg
Solution: 1.
1
1 rad sec
Input crank angular velocity
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.
VA 184.121 deg
Determine the angular velocity of link 3 using equation 6.9a.
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 VA 9.174 sec
VA 90 deg 3.
BI13 12.862 in
VA AI13
3.466
rad sec
Determine the magnitude of the velocity at point B using equation 6.9b. in VB BI13 VB 44.577 sec
CW
DESIGN OF MACHINERY
5.
Use equation 6.9c to determine the angular velocity of link 4.
6.
SOLUTION MANUAL 6-87b-2
VB c
4.657
rad
CW
sec
Determine the magnitude of the velocity at point P using equation 6.9b. VP PI13
VP 51.381
in sec
DESIGN OF MACHINERY
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
dX 2.790 in
Link 1 Y-offset
dY 6.948 in
Coupler point data:
p 15.00 in
0 deg
Crank angle:
2XY 94.121 deg Global XY coord system 68.121 deg
Coordinate transformation angle:
2 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
26.000°
68.121°
O4 B
4
P
A 72.224° 3 60.472° Transform crank angle into local xy coordinate system:
2 2XY
2.
2 26.000 deg 2
2
d d X dY
Calculate distance O2O4:
x
d 7.487in
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
K 1 0.8161
2
2 a c
K 3 0.3622
C K 1 K2 1 cos 2K 3
A cos 2 K 1 K 2 cos 2 K3 A 0.2581
K 2 0.7821
B 0.8767
C 0.4234
B 2 sin 2
DESIGN OF MACHINERY
3.
SOLUTION MANUAL 6-87c-2
Use equation 4.10b to find values of 4 for the crossed circuit.
2
2
4 2atan2 2 A B B 4 A C 4.
4 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
E 2 sin 2 F K 1 K4 1 cos 2K 5
2
K 4 0.5772
2 a b
D cos 2 K 1 K 4 cos 2 K 5
5.
D 0.3096 E 0.8767 F 0.4749
Use equation 4.13 to find values of 3 for the crossed circuit.
2
3 2atan2 2 D E E 4 D F 2 6.
7.
K 5 0.9111
3 72.236 deg
Determine the angular velocity of links 3 and 4 using equations 6.18.
3 3.467
rad
4 4.658
rad
3
a 2 sin 4 2 b sin 3 4
4
a 2 sin 2 3 c sin 4 3
sec
sec
Determine the velocity of points A and B using equations 6.19.
VA a 2sin 2 j cos 2 VA ( 4.022 8.246i)
in
VA 9.174
sec
VAXY arg VA
in sec
VAXY 184.121 deg
VB c 4sin 4 j cos 4 VB ( 38.807 21.967i)
in sec
VBXY arg VB 8.
VB 44.593
in sec
VBXY 97.633 deg
Determine the velocity of the coupler point P for the crossed circuit using equations 6.36.
VPA p 3sin 3 j cos 3 VPA ( 4.127 1.322i)
ft s
VP VA VPA
VP ( 3.792 2.009i)
VPXY arg VP
VPXY 96.039 deg
ft s
VP 51.501
in sec
DESIGN OF MACHINERY
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)
C 6
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
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
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 (dX ,dY). Then, define position vectors R1X , R1Y, R2, R3, and R4 as shown below. Y 4 B
3 R3 R2
3
C R4
2 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:
j 2
dY e 3.
j( 0)
a e
j 2
d e j (0 ) be
j 3
c e
X
=0
Differentiate this equation with respect to time. j 2 d d a j c e b e = 0 j
dt 4.
dt
Substituting the Euler identity into this equation gives:
Vbj = 0
Va jc cos 3 j sin 3 5.
Separate this equation into its real (x component) and imaginary (y component) parts, setting each equal to zero.
Va c sin 3 = 0 6.
c cos 3 Vb = 0
Solve for the two unknowns Va and Vb in terms of the independent variables 3 and 3
Va = c sin 3
Vb = c cos 3
DESIGN OF MACHINERY
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 and
Given:
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
dX 7.80 in
Link 1 Y-offset
dY 0.62 in
Angle ACB
3 0.0 deg
Angle BO4D
4 110.0 deg
Input rocker angle:
120 deg
Global XY system
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
3.
X
VA 62.000
in sec
VA 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
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: 1
Velocity scale factor:
5.
8.
in in
VB 60.300
VBA 1.433 in kv
VBA 71.650
VB 78.291 deg
sec in
VBA 23.057 deg
sec
Determine the angular velocity of links 3 and 4 using equation 6.7.
7.
50 in sec
VB 1.206 in kv
6.
kv
VBA b VB c
15.922
rad
CCW
sec
20.100
rad
CW
sec
Determine the magnitude and sense of the vector VCA using equation 6.7. in
VCA p
VCA 35.825
VCA VBA 3
VCA 23.057 deg
sec
Determine the magnitude and sense of the vector Vtrans using equation 6.7. in
Vtrans f
Vtrans 67.978
trans VB 4
trans 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
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: 1
Velocity scale factor:
VC 0.991 in kv 9.
kv
50 in sec in
VC 49.550
in
VC 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 V slip
VDC
Velocity scale factor:
3.045
95.189°
1
kv
50 in sec in
V trans VD 3.045 in kv
VD 152.250
in
X
sec
VD 95.189 deg
0 VC
20.100
rad sec
50 in/sec
DESIGN OF MACHINERY
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 and
Given:
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
dX 7.80 in
Link 1 Y-offset
dY 0.62 in
Angle ACB
3 0.0 deg
Angle BO4D
4 110.0 deg
Input rocker angle:
120 deg
Global XY system
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 1,5
5,6 5
D
Y
1,6
3 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
VA 210.0 deg
Determine the angular velocity of link 3 using equation 6.9a.
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 VA 62.000 sec
VA 90 deg 3.
CI13 3.113 in
VA AI13
15.926
rad sec
Determine the magnitude of the velocity at point B using equation 6.9b. VB BI13
VB 60.328
in sec
CCW
DESIGN OF MACHINERY
5.
Use equation 6.9c to determine the angular velocity of links 4 and 6.
VB c
w6 6.
rad
w 6 20.109
rad
CW
sec CW
sec
VC 49.578
in sec
Determine the angular velocity of link 5 using equation 6.9a.
8.
20.109
Determine the magnitude of the velocity at point C using equation 6.9b. VC CI13
7.
SOLUTION MANUAL 6-92-2
VC CI15
35.137
rad
CW
sec
Determine the magnitude of the velocity at point D using equation 6.9b. VD DI15
VD 152.247
in sec
@ 95.185 deg
DESIGN OF MACHINERY
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 and
Given:
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
dX 7.80 in
Link 1 Y-offset
dY 0.62 in
Angle ACB
3 0.0 deg
Angle BO4D
4 110.0 deg
Input rocker angle:
2XY 120 deg
Global XY system
175.455 deg
Coordinate transformation angle:
2 10 rad sec
Input crank angular velocity
1
CCW
atan2 (x y) return 0.5 if x = 0 y 0
Two argument inverse tangent
return 1.5 if x = 0 y 0
y if x 0 return atan x y atan x otherwise Solution: 1.
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
2
d d X dY
d 7.825in
Transform 2XY into the local coordinate system: 2 2XY 2 55.455 deg 2.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1
d a
K2
d c
K 1 1.2620
K 2 2.6082
X
DESIGN OF MACHINERY
SOLUTION MANUAL 6-93-2
2
K3
2
2
a b c d
2
K 3 2.3767
2 a c
C K 1 K2 1 cos 2K 3
A cos 2 K 1 K 2 cos 2 K3 A 0.2028
3.
B 1.6474
B 2 sin 2
C 1.5927
Use equation 4.10b to find values of 4 for the open circuit.
2
41 2atan2 2 A B B 4 A C 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
2 a b
E 2 sin 2 F K 1 K4 1 cos 2K 5
D cos 2 K 1 K 4 cos 2 K 5
5.
K 4 1.7388
K 5 1.9877
D 1.6967 E 1.6474 F 0.3067
Use equation 4.13 to find values of 3 for the open circuit.
2
3 2atan2 2 D E E 4 D F 2 6.
41 163.746 deg
3 71.488 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
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
0 deg 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 36.352 deg 2.
3.
Use equation 6.7 to calculate the magnitude of the velocity at point A. in
VA a
VA 41.250
VA 90 deg
VA 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
SOLUTION MANUAL 6-94-2
VB VA 5.
7.
in VBA 0 sec
VB VA
VBA 0 deg
Determine the angular velocity of links 3 and 4 using equation 6.7.
6.
and
VBA b VB c
0.000
rad sec
15.000
rad sec
Determine the magnitude and sense of the vector VPA using equation 6.7. in
VPA p
VPA 0.000
VPA VBA
VPA 0.000deg
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
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
0 deg 15 rad sec
Input crank angular velocity
1
CW
Solution: See Figure P6-33b and Mathcad file P0695. 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, I 1,3 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
VA 41.250
in sec
36.351 deg
VA 90 deg
VA 126.351 deg
The velocity vectors for points B and P are identical to VA .
DESIGN OF MACHINERY
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
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 36.351 deg 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
K 1 1.6109
c
2
K 3 1.5949
2 a c
B 2 sin C K 1 K2 1 cos K 3
A cos K 1 K 2 cos 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 B 4 A C 4.
2
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 F K 1 K4 1 cos K 5
D cos K 1 K 4 cos K 5
5.
K 2 1.6109
Use equation 4.13 to find values of 3 for the open circuit.
K 4 1.3589 D 1.3983 E 1.1855 F 0.2127
K 5 1.6873
DESIGN OF MACHINERY
SOLUTION MANUAL 6-96-2
2
2
2 atan2 2 D E E 4 D F 6.
7.
89.995 deg
Determine the angular velocity of links 3 and 4 using equations 6.18.
0.000
rad
15.000
rad
a w 2 sin b sin
a w 2 sin c sin
sec
sec
Determine the velocity of points A and B using equations 6.19.
VA a w 2sin j cos VA ( 24.450 33.223i)
in
VA 41.250
sec
VAXY arg VA
in sec
VAXY 126.351 deg
VB c sin j cos VB ( 24.450 33.223i)
in
VB 41.250
sec
VBXY arg VB 8.
in sec
VBXY 126.351 deg
Determine the velocity of the coupler point P using equations 6.36.
VPA p sin j cos
VPA 1.936 10
4
VP VA VPA
VPXY arg VP
1.676i 10
8
in sec
VP 41.250
in sec
VPXY 126.351 deg
DESIGN OF MACHINERY
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
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
CP4 27.0 in
Coupler point data:
1
V2Y 20 in sec
Input velocities: Solution: 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. 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. And then do the same with the VCB vector (See next page)
DESIGN OF MACHINERY
SOLUTION MANUAL 6-98-2
5.932
V BC
Y 0
10 in/sec
VB V BA
60.123° 32.718°
46. 990°
X
VC
6.004
4. 385
VA
3.
From the velocity triangles we have: 1
Velocity scale factor:
4.
10 in sec 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
VB 46.990 deg
sec in
VBA 60.123 deg
sec in
VBC 32.718 deg
sec
Determine the angular velocity of links 3 and 4 using equation 6.7.
6.
kv
VBA b VBC c
1.749
rad
CCW
sec
1.177
rad
CW
sec
Determine the magnitudes of the vectors VP3A and VP4C using equation 6.7. VP3A AP3
VP3A 55.107
in sec
DESIGN OF MACHINERY
SOLUTION MANUAL 6-98-3
VP4C CP4 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 to 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
VP4 164.011°
V P4C
104.444°
X
VC 6.598
VA
8.
From the velocity triangles we have: 1
Velocity scale factor:
kv
10 in sec in
VP3 3.537 in kv
VP3 35.370
VP4 6.598 in kv
VP4 65.980
in sec in sec
P3 104.444 deg P4 164.011 deg
DESIGN OF MACHINERY
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
dX 19 in
Link 1 Y-offset
dY 28 in
4 0.0 deg
Angle BO4C
0 deg
Output crank angle:
VD 20 in sec
Input slider velocity Solution: 1.
Global XY system 1
VD 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
SOLUTION MANUAL 6-99-2
Y
X 1.806 0
10 in/sec 70.037°
VC VD VCD
3.
From the velocity triangle we have: 1
Velocity scale factor:
VC 1.806 in kv 4.
kv
10 in sec in
VC 18.060
in
VC 70.037 deg
sec
Determine the magnitude and sense of the vector VB . VB VC
VB 18.060
in sec
VB VC 180 deg 5.
VB 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:
V AB
1
kv
10 in sec
VB
in
VA 1.977 in kv VA 19.770
VA
1.977
0
in sec
VA 90.0 deg
X
10 in/sec
DESIGN OF MACHINERY
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
dX 19 in
Link 1 Y-offset
dY 28 in
Angle BO4C
4 0.0 deg
Output crank angle:
0 deg
Input slider velocity
VD 20 in sec
Global XY system 1
VD 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 3 1,3
2
O2
6
A
From the layout above: AI13 70.085 in 2.
VD DI15
VC e
CW
sec
VC 18.057
in sec
1.003
rad
CW
sec
Determine the magnitude of the velocity at point B using equation 6.9b. VB c
6.
rad
Use equation 6.9c to determine the angular velocity of link 4.
5.
0.286
Determine the magnitude of the velocity at point C using equation 6.9b. VC CI15
4.
CI15 63.095 in
Use equation 6.9c to determine the angular velocity of link 5.
3.
BI13 64.013 in
VB 18.057
Determine the angular velocity of link 3 using equation 6.9a.
in sec
DI15 69.886 in
DESIGN OF MACHINERY
7.
VB BI13
SOLUTION MANUAL 6-100-2
0.282
rad
CCW
sec
Determine the magnitude of the velocity at point A using equation 6.9b. in Upward VA AI13 VA 19.769 sec
DESIGN OF MACHINERY
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)
b 5.00
Slider coupler (L 5)
e 15.00
Angle BO4C
83.621 deg
atan2 (x y) return 0.5 if x = 0 y 0
Two argument inverse tangent: Crank velocity:
Fourbar coupler (L3)
return 1.5 if x = 0 y 0
rad 10 sec
y if x 0 return atan x y atan x otherwise
Solution:
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 slider-crank mechanism using the output of the fourbar, link 4, as the input to the slider-crank.
2.
Define one revolution of the input crank: 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
K 1 0.5000
2
d
K3
c
K 2 0.4167
2
2
a b c d
2
2 a c
K 3 0.7042
A cos K1 K 2 cos K 3
C K 1 K 2 1 cos K 3
B 2 sin
2
2 2 A B B 4 A C atan2 4.
If the calculated value of 4 is greater than 2, subtract 2from it and if it is negative, make it positive.
if 2 2
if 0 2 5.
Determine the slider-crank motion using equations 4.16 and 4.17 with 4 as the input angle.
sin c asin e
f c cos e cos 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
DESIGN OF MACHINERY
SOLUTION MANUAL 6-101-2
K 4 0.5000
K 5 0.4050
D cos K1 K 4 cos K5
E 2 sin 6.
F K1 K 4 1 cos K5
Use equation 4.13 to find values of 3 for the open circuit.
2
2 atan2 2 D E E 4 D F 7.
Determine the angular velocity of link 4 for the open circuit using equations 6.18.
8.
a sin c sin
Determine the angular velocity of link 5 using equation 6.22a:
c cos e cos
9.
Determine the velocity of pin D using equation 6.22b:
VD c sin e sin 50
25
0
VD 25
50
75
100
0
45
90
135
180 deg
225
270
315
360
DESIGN OF MACHINERY
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
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)
d1 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
d2 10.5 in
Link 1 (O2 to O6)
2 130 deg
Crank angle:
Slider axis offset and angle:
s 11.7 in
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 Direction of VFD
Y
F
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.
VB 24.351 deg
VBA 79.348 deg
VCA VBA
VE 64.594 deg
VEC 162.461 deg
VDC VEC
VF 150.00 deg
VFD 198.690 deg
Use equation 6.7 to calculate the magnitude of the velocity at point A. in
VA a
VA 75.000
VA 2 90 deg
VA 40.000 deg
sec
DESIGN OF MACHINERY
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: 1
Velocity scale factor:
25 in sec in
in
VB 3.192 in kv
VB 79.800
VBA 3.302 in kv
VBA 82.550
5.
kv
VBA b
9.827
VB 24.351 deg
sec in
VBA 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: v CA VC 2.668 in kv
in sec
VCA
v CA 0.943in
kv
VC 66.700
VCA 79.348 deg
in sec
VC 22.053deg
DESIGN OF MACHINERY
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: 1
Velocity scale factor:
in in
VE 42.900
VEC 1.821 in kv
VEC 45.525
VEC e
5.115
VE 64.594 deg
sec in
VEC 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
Length of VDC on velocity polygon: VD 1.900 in kv 9.
25 in sec
VE 1.716 in kv
8.
kv
VDC 30.179 v DC
in sec
VDC
v DC 1.207 in
kv
VD 47.500
VDC 162.461 deg
in sec
VD 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
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: 1
Velocity scale factor:
VF 1.158 in kv
kv
25 in sec in
VF 28.950
in sec
VF 150.000 deg
DESIGN OF MACHINERY
SOLUTION MANUAL 6-104-1
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
4
2
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 3
D
2
4
2,4
6 E 3,6 1,4
1,6
O4
1
1,2 O2
DESIGN OF MACHINERY
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
B
5
= 25.396°
3
h
A
b 3
RP k
R2
4
D g 6
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 k h acos b k 2 2
3.
2
2
25.396 deg
Define the vectors R2, RP, and RE as shown above. Then, RE = R2 + RP or,
cos = acos jsin k jsin Separating into real and imaginary parts, the coordinates of point E for all positions of link 2 are:
x E a cos k cos yE a sin k sin
DESIGN OF MACHINERY
SOLUTION MANUAL 6-105-2
This pair of equations represent the parametric equation of a circle with radius a and center at x EC k cos
x EC 2.349
yEC k sin
yEC 1.115
4.
Define one revolution of the input crank: 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
1.115
1
0
1
0
1
2
3
4
5
xE
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 ( 64.036 deg ) are:
Angle BAE:
b h k acos b h 2
Coordinates of point E:
xE a cos k cos
2
2
2
x E 3.115
yE 0.458
yE a sin k sin Line through E with angle 2:
Coordinates of point O4:
y( x ) x x E tan yE x O4 d
39.582 deg
yO4 0
DESIGN OF MACHINERY
SOLUTION MANUAL 6-105-3
Line through O4 with angle :
y( x ) x x O4 tan
Solving for the intersection (point F, the instant center I1,6 ) of these two lines: xF
tan tan
xE tan yE d tan
x F 2.349
yF xF d tan
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
39.582° D
3
2
4
6
2,6
A
3
2,4
64.036°
E 3,6
1,4
O4
1
1,2 O2 1.115"
1.750" 1,6
F 2.349"
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