18424291 Kinematics of Machines

Kinematics of Machines {ME44} CHAPTER - I Mechanics: It is that branch of scientific analysis which deals with motion, time and force. Kinematics is the study of motion, without considering the forces which produce that motion. Kinematics of machines deals with the study of the relative motion of machine parts. It involves the study of position, displacement, velocity and acceleration of machine parts. Dynamics of machines involves the study of forces acting on the machine parts and the motions resulting from these forces. Plane motion: A body has plane motion, if all its points move in planes which are parallel to some reference plane. A body with plane motion will have only three degrees of freedom. I.e., linear along two axes parallel to the reference plane and rotational/angular about the axis perpendicular to the reference plane. (eg. linear along X and Z and rotational about Y.)The reference plane is called plane of motion. Plane motion can be of three types. 1) Translation 2) rotation and 3) combination of translation and rotation. Translation: A body has translation if it moves so that all straight lines in the body move to parallel positions. Rectilinear translation is a motion wherein all points of the body move in straight lie paths. Eg. The slider in slider crank mechanism has rectilinear translation. (link 4 in fig.1.1)

Fig.1.1 Translation, in which points in a body move along curved paths, is called curvilinear translation. The tie rod connecting the wheels of a steam locomotive has curvilinear translation. (link 3 in fig.1.2)

1

Fig.1.2 Rotation: In rotation, all points in a body remain at fixed distances from a line which is perpendicular to the plane of rotation. This line is the axis of rotation and points in the body describe circular paths about it. (Eg. link 2 in Fig.1.1 and links 2 & 4 in Fig.1.2) Translation and rotation: It is the combination of both translation and rotation which is exhibited by many machine parts. (Eg. link 3 in Fig.1.1) Link or element: It is the name given to any body which has motion relative to another. All materials have some elasticity. A rigid link is one, whose deformations are so small that they can be neglected in determining the motion parameters of the link.

Fig.1.3 Binary link: Link which is connected to other links at two points. (Fig.1.3 a) Ternary link: Link which is connected to other links at three points. (Fig.1.3 b) Quaternary link: Link which is connected to other links at four points. (Fig1.3 c) Pairing elements: the geometrical forms by which two members of a mechanism are joined together, so that the relative motion between these two is consistent are known as pairing elements and the pair so formed is called kinematic pair. Each individual link of a mechanism forms a pairing element.

Fig.1.4 Kinematic pair

Fig.1.5

2

Degrees of freedom (DOF): It is the number of independent coordinates required to describe the position of a body in space. A free body in space (fig 1.5) can have six degrees of freedom. I.e., linear positions along x, y and z axes and rotational/angular positions with respect to x, y and z axes. In a kinematic pair, depending on the constraints imposed on the motion, the links may loose some of the six degrees of freedom. Types of kinematic pairs: (i)

Based on nature of contact between elements:

(a) Lower pair. If the joint by which two members are connected has surface contact, the pair is known as lower pair. Eg. pin joints, shaft rotating in bush, slider in slider crank mechanism.

Fig.1.6 Lower pairs (b) Higher pair. If the contact between the pairing elements takes place at a point or along a line, such as in a ball bearing or between two gear teeth in contact, it is known as a higher pair.

Fig.1.7 Higher pairs (ii) Based on relative motion between pairing elements: (a) Siding pair. Sliding pair is constituted by two elements so connected that one is constrained to have a sliding motion relative to the other. DOF = 1

3

(b) Turning pair (revolute pair). When connections of the two elements are such that only a constrained motion of rotation of one element with respect to the other is possible, the pair constitutes a turning pair. DOF = 1 (c) Cylindrical pair. If the relative motion between the pairing elements is the combination of turning and sliding, then it is called as cylindrical pair. DOF = 2

Fig.1.8 Sliding pair

Fig.1.9 Turning pair

Fig.1.10 Cylindrical pair

(d) Rolling pair. When the pairing elements have rolling contact, the pair formed is called rolling pair. Eg. Bearings, Belt and pulley. DOF = 1

Fig.1.11 (a) Ball bearing

Fig.1.11(b) Belt and pulley

(e) Spherical pair. A spherical pair will have surface contact and three degrees of freedom. Eg. Ball and socket joint. DOF = 3 (f) Helical pair or screw pair. When the nature of contact between the elements of a pair is such that one element can turn about the other by screw threads, it is known as screw pair. Eg. Nut and bolt. DOF = 1

4

Fig.1.13 Screw pair

Fig.1.12 Ball and socket joint (iii) Based on the nature of mechanical constraint.

(a) Closed pair. Elements of pairs held together mechanically due to their geometry constitute a closed pair. They are also called form-closed or self-closed pair. (b) Unclosed or force closed pair. Elements of pairs held together by the action of external forces constitute unclosed or force closed pair .Eg. Cam and follower.

Fig.1.14 Closed pair

Fig. 1.15 Force closed pair (cam & follower)

Constrained motion: In a kinematic pair, if one element has got only one definite motion relative to the other, then the motion is called constrained motion. (a) Completely constrained motion. If the constrained motion is achieved by the pairing elements themselves, then it is called completely constrained motion.

5

Fig.1.16 completely constrained motion (b) Successfully constrained motion. If constrained motion is not achieved by the pairing elements themselves, but by some other means, then, it is called successfully constrained motion. Eg. Foot step bearing, where shaft is constrained from moving upwards, by its self weight. (c) Incompletely constrained motion. When relative motion between pairing elements takes place in more than one direction, it is called incompletely constrained motion. Eg. Shaft in a circular hole.

Fig.1.17 Foot strep bearing

Fig.1.18 Incompletely constrained motion

Kinematic chain: A kinematic chain is a group of links either joined together or arranged in a manner that permits them to move relative to one another. If the links are connected in such a way that no motion is possible, it results in a locked chain or structure.

Fig.1.19 Locked chain or structure

6

Mechanism: A mechanism is a constrained kinematic chain. This means that the motion of any one link in the kinematic chain will give a definite and predictable motion relative to each of the others. Usually one of the links of the kinematic chain is fixed in a mechanism.

Fig.1.20 Slider crank and four bar mechanisms. If, for a particular position of a link of the chain, the positions of each of the other links of the chain can not be predicted, then it is called as unconstrained kinematic chain and it is not mechanism.

Fig.1.21 Unconstrained kinematic chain Machine: A machine is a mechanism or collection of mechanisms, which transmit force from the source of power to the resistance to be overcome. Though all machines are mechanisms, all mechanisms are not machines. Many instruments are mechanisms but are not machines, because they do no useful work nor do they transform energy. Eg. Mechanical clock, drafter.

Fig.1.21 Drafter Planar mechanisms: When all the links of a mechanism have plane motion, it is called as a planar mechanism. All the links in a planar mechanism move in planes parallel to the reference plane.

7

Degrees of freedom/mobility of a mechanism: It is the number of inputs (number of independent coordinates) required to describe the configuration or position of all the links of the mechanism, with respect to the fixed link at any given instant. Grubler’s equation: Number of degrees of freedom of a mechanism is given by F = 3(n-1)-2l-h. Where, F = Degrees of freedom n = Number of links = n2 + n3 +……+nj, where, n2 = number of binary links, n3 = number of ternary links…etc. l = Number of lower pairs, which is obtained by counting the number of joints. If more than two links are joined together at any point, then, one additional lower pair is to be considered for every additional link. h = Number of higher pairs Examples of determination of degrees of freedom of planar mechanisms: (i) F = 3(n-1)-2l-h Here, n2 = 4, n = 4, l = 4 and h = 0. F = 3(4-1)-2(4) = 1 I.e., one input to any one link will result in definite motion of all the links.

(ii) F = 3(n-1)-2l-h Here, n2 = 5, n = 5, l = 5 and h = 0. F = 3(5-1)-2(5) = 2 I.e., two inputs to any two links are required to yield definite motions in all the links. (iii) F = 3(n-1)-2l-h Here, n2 = 4, n3 =2, n = 6, l = 7 and h = 0. F = 3(6-1)-2(7) = 1 I.e., one input to any one link will result in definite motion of all the links.

8

(iv) F = 3(n-1)-2l-h Here, n2 = 5, n3 =1, n = 6, l = 7 (at the intersection of 2, 3 and 4, two lower pairs are to be considered) and h = 0. F = 3(6-1)-2(7) = 1

(v) F = 3(n-1)-2l-h Here, n = 11, l = 15 (two lower pairs at the intersection of 3, 4, 6; 2, 4, 5; 5, 7, 8; 8, 10, 11) and h = 0. F = 3(11-1)-2(15) = 0 (vi) Determine the mobility of the following mechanisms.

(a)

(b)

(c)

F = 3(n-1)-2l-h Here, n = 4, l = 5 and h = 0. F = 3(4-1)-2(5) = -1 I.e., it is a structure

F = 3(n-1)-2l-h Here, n = 3, l = 2 and h = 1. F = 3(3-1)-2(2)-1 = 1

F = 3(n-1)-2l-h Here, n = 3, l = 2 and h = 1. F = 3(3-1)-2(2)-1 = 1

9

Inversions of mechanism: A mechanism is one in which one of the links of a kinematic chain is fixed. Different mechanisms can be obtained by fixing different links of the same kinematic chain. These are called as inversions of the mechanism. By changing the fixed link, the number of mechanisms which can be obtained is equal to the number of links. Excepting the original mechanism, all other mechanisms will be known as inversions of original mechanism. The inversion of a mechanism does not change the motion of its links relative to each other. Four bar chain:

Fig 1.22 Four bar chain One of the most useful and most common mechanisms is the four-bar linkage. In this mechanism, the link which can make complete rotation is known as crank (link 2). The link which oscillates is known as rocker or lever (link 4). And the link connecting these two is known as coupler (link 3). Link 1 is the frame. Inversions of four bar chain:

Fig.1.23 Inversions of four bar chain.

10

Crank-rocker mechanism: In this mechanism, either link 1 or link 3 is fixed. Link 2 (crank) rotates completely and link 4 (rocker) oscillates. It is similar to (a) or (b) of fig.1.23.

Fig.1.24 Drag link mechanism. Here link 2 is fixed and both links 1 and 4 make complete rotation but with different velocities. This is similar to 1.23(c).

Fig.1.25 Double crank mechanism. This is one type of drag link mechanism, where, links 1& 3 are equal and parallel and links 2 & 4 are equal and parallel.

Fig.1.26 11

Double rocker mechanism. In this mechanism, link 4 is fixed. Link 2 makes complete rotation, whereas links 3 & 4 oscillate (Fig.1.23d) Slider crank chain: This is a kinematic chain having four links. It has one sliding pair and three turning pairs. Link 2 has rotary motion and is called crank. Link 3 has got combined rotary and reciprocating motion and is called connecting rod. Link 4 has reciprocating motion and is called slider. Link 1 is frame (fixed). This mechanism is used to convert rotary motion to reciprocating and vice versa.

Fig1.27 Inversions of slider crank chain: Inversions of slider crank mechanism is obtained by fixing links 2, 3 and 4.

(a) crank fixed

(b) connecting rod fixed

(c) slider fixed

Fig.1.28 Rotary engine – I inversion of slider crank mechanism. (crank fixed)

Fig.1.29

12

Whitworth quick return motion mechanism–I inversion of slider crank mechanism.

Fig.1.30 Crank and slotted lever quick return motion mechanism – II inversion of slider crank mechanism (connecting rod fixed).

Fig.1.31

13

Oscillating cylinder engine–II inversion of slider crank mechanism (connecting rod fixed).

Fig.1.32 Pendulum pump or bull engine–III inversion of slider crank mechanism (slider fixed).

Fig.1.33

14

Double slider crank chain: It is a kinematic chain consisting of two turning pairs and two sliding pairs. Scotch –Yoke mechanism. Turning pairs – 1&2, 2&3; Sliding pairs – 3&4, 4&1.

Fig.1.34 Inversions of double slider crank mechanism: Elliptical trammel. This is a device which is used for generating an elliptical profile.

Fig.1.35 In fig. 1.35, if AC = p and BC = q, then, x = q.cosθ and y = p.sinθ. 2

2

x  y Rearranging,   +   = cos 2 θ + sin 2 θ = 1 . This is the equation of an ellipse. The q  p path traced by point C is an ellipse, with major axis and minor axis equal to 2p and 2q respectively.

15

Oldham coupling. This is an inversion of double slider crank mechanism, which is used to connect two parallel shafts, whose axes are offset by a small amount.

Fig.1.36 References: 1. Theory of Machines and Mechanisms by Joseph Edward Shigley and John Joseph Uicker,Jr. McGraw-Hill International Editions. 2. Kinematics and Dynamics of Machines by George H.Martin. McGraw-Hill Publications. 3. Mechanisms and Dynamics of Machinery by Hamilton H. Mabie and Fred W. Ocvirk. John Wiley and Sons. 4. Theory of Machines by V.P.Singh. Dhanpat Rai and Co. 5. The Theory of Machines through solved problems by J.S.Rao. New age international publishers. 6. A text book of Theory of Machines by Dr.R.K.Bansal. Laxmi Publications (P) Ltd.

16

Kinematics of Machines {ME44} CHAPTER – I (contd.) Quick return motion mechanisms. Quick return mechanisms are used in machine tools such as shapers and power driven saws for the purpose of giving the reciprocating cutting tool a slow cutting stroke and a quick return stroke with a constant angular velocity of the driving crank. Some of the common types of quick return motion mechanisms are discussed below. The ratio of time required for the cutting stroke to the time required for the return stroke is called the time ratio and is greater than unity. Drag link mechanism This is one of the inversions of four bar mechanism, with four turning pairs. Here, link 2 is the input link, moving with constant angular velocity in anti-clockwise direction. Point C of the mechanism is connected to the tool post E of the machine. During cutting stroke, tool post moves from E1 to E2. The corresponding positions of C are C1 and C2 as shown in the fig. 1.37. For the point C to move from C1 to C2, point B moves from B1 to B2, in anti-clockwise direction. IE, cutting stroke takes place when input link moves through angle B1AB2 in anti-clockwise direction and return stroke takes place when input link moves through angle B2AB1 in anti-clockwise direction.

Fig.1.37 The time ratio is given by the following equation. Timeforforwardstroke B1 Aˆ B2 ( anti − clockwise ) = Timeforreturnstroke B2 Aˆ B1 ( anti − clockwise ) Whitworth quick return motion mechanism: This is first inversion of slider mechanism, where, crank 1 is fixed. Input is given to link 2, which moves at constant speed. Point C of the mechanism is connected to the tool post

17

D of the machine. During cutting stroke, tool post moves from D1 to D11. The corresponding positions of C are C1 and C11 as shown in the fig. 1.38. For the point C to move from C1 to C11, point B moves from B1 to B11, in anti-clockwise direction. I.E., cutting stroke takes place when input link moves through angle B1O2B11 in anti-clockwise direction and return stroke takes place when input link moves through angle B 11O2B1 in anti-clockwise direction.

Fig.1.38 The time ratio is given by the following equation. Timeforforwardstroke B′oˆ2 B′′ θ1 = = Timeforreturnstroke B′′oˆ2 B′ θ 2 Crank and slotted lever quick return motion mechanism This is second inversion of slider mechanism, where, connecting rod is fixed. Input is given to link 2, which moves at constant speed. Point C of the mechanism is connected to the tool post D of the machine. During cutting stroke, tool post moves from D1 to D11. The corresponding positions of C are C1 and C11 as shown in the fig. 1.39. For the point C to move from C1 to C11, point B moves from B1 to B11, in anti-clockwise direction. I.E., cutting stroke takes place when input link moves through angle B1O2B11 in anti-clockwise direction and return stroke takes place when input link moves through angle B 11O2B1 in anti-clockwise direction.

18

Fig.1.39 The time ratio is given by the following equation. Timeforforwardstroke B′oˆ2 B′′ θ1 = = Timeforreturnstroke B′′oˆ2 B′ θ 2 Straight line motion mechanisms Straight line motion mechanisms are mechanisms, having a point that moves along a straight line, or nearly along a straight line, without being guided by a plane surface. Condition for exact straight line motion: If point B (fig.1.40) moves on the circumference of a circle with center O and radius OA, then, point C, which is an extension of AB traces a straight line perpendicular to AO, provided product of AB and AC is constant.

19

Fig.1.40 Locus of pt.C will be a straight line, ┴ to AE if, AB × AC is constant Proof: ∆AEC ≡ ∆ABD AD AB ∴ = AC AE AB × AC ∴ AE = AD butAD = const. ∴ AE = const., ifAB × AC = const.

Peaucellier exact straight line motion mechanism:

Fig.1.41 Here, AE is the input link and point E moves along a circular path of radius AE = AB. Also, EC = ED = PC = PD and BC = BD. Point P of the mechanism moves along exact straight line, perpendicular to BA extended.

20

To prove B, E and P lie on same straight line: Triangles BCD, ECD and PCD are all isosceles triangles having common base CD and apex points being B, E and P. Therefore points B, E and P always lie on the perpendicular bisector of CD. Hence these three points always lie on the same straight line. To prove product of BE and BP is constant. In triangles BFC and PFC, BC 2 = FB 2 + FC 2 and PC 2 = PF 2 + FC 2 ∴ BC 2 − PC 2 = FB 2 − PF 2 = ( FB + PF )( FB − PF ) = BP × BE But since BC and PC are constants, product of BP and BE is constant, which is the condition for exact straight line motion. Thus point P always moves along a straight line perpendicular to BA as shown in the fig.1.41. Approximate straight line motion mechanism: A few four bar mechanisms with certain modifications provide approximate straight line motions. Robert’s mechanism

Fig.1.42 This is a four bar mechanism, where, PCD is a single integral link. Also, dimensions AC, BD, CP and PD are all equal. Point P of the mechanism moves very nearly along line AB.

21

Intermittent motion mechanisms An intermittent-motion mechanism is a linkage which converts continuous motion into intermittent motion. These mechanisms are commonly used for indexing in machine tools. Geneva wheel mechanism

Fig.1.43 In the mechanism shown (Fig.1.43), link A is driver and it contains a pin which engages with the slots in the driven link B. The slots are positioned in such a manner, that the pin enters and leaves them tangentially avoiding impact loading during transmission of motion. In the mechanism shown, the driven member makes one-fourth of a revolution for each revolution of the driver. The locking plate, which is mounted on the driver, prevents the driven member from rotating except during the indexing period. Ratchet and pawl mechanism

Fig.1.44 Ratchets are used to transform motion of rotation or translation into intermittent rotation or translation. In the fig.1.44, A is the ratchet wheel and C is the pawl. As lever B is made

22

to oscillate, the ratchet wheel will rotate anticlockwise with an intermittent motion. A holding pawl D is provided to prevent the reverse motion of ratchet wheel. Other mechanisms Toggle mechanism

Fig.1.45 Toggle mechanisms are used, where large resistances are to be overcome through short distances. Here, effort applied will be small but acts over large distance. In the mechanism shown in fig.1.45, 2 is the input link, to which, power is supplied and 6 is the output link, which has to overcome external resistance. Links 4 and 5 are of equal length. Considering the equilibrium condition of slider 6, F

2 P ∴ F = 2 P tan α tan α =

For small angles of α, F (effort) is much smaller than P(resistance). This mechanism is used in rock crushers, presses, riveting machines etc.

23

Pantograph Pantographs are used for reducing or enlarging drawings and maps. They are also used for guiding cutting tools or torches to fabricate complicated shapes.

Fig.1.46 In the mechanism shown in fig.1.46 path traced by point A will be magnified by point E to scale, as discussed below. In the mechanism shown, AB = CD; AD =BC and OAE lie on a straight line. When point A moves to A′ , E moves to E ′ and OA′E ′ also lies on a straight line. From the fig.1.46, ∆ODA ≡ ∆OCE and ∆OD ′A′ ≡ ∆OC ′E ′ . ∴

OD OA DA OD ′ OA′ D ′A′ = = = = and OC OE CE OC ′ OE ′ C ′E ′

But,

OD OD ′ OA OA′ = ;∴ = ;∴ ∆OAA′ ≡ ∆OEE ′. OC OC ′ OE OE ′

∴ EE ′ // AA′ And

EE ′ OE OC = = AA′ OA OD

 OC  ∴ EE ′ = AA′   OD   OC   is the magnification factor. Where   OD 

24

Hooke’s joint (Universal joints) Hooke’s joins is used to connect two nonparallel but intersecting shafts. In its basic shape, it has two U –shaped yokes ‘a’ and ‘b’ and a center block or cross-shaped piece, C. (fig.1.47(a)) The universal joint can transmit power between two shafts intersecting at around 300 angles (α). However, the angular velocity ratio is not uniform during the cycle of operation. The amount of fluctuation depends on the angle (α) between the two shafts. For uniform transmission of motion, a pair of universal joints should be used (fig.1.47(b)). Intermediate shaft 3 connects input shaft 1 and output shaft 2 with two universal joints. The angle α between 1 and 2 is equal to angle α between 2 and 3. When shaft 1 has uniform rotation, shaft 3 varies in speed; however, this variation is compensated by the universal joint between shafts 2 and 3. One of the important applications of universal joint is in automobiles, where it is used to transmit power from engine to the wheel axle.

Fig.1.47(a)

Fig.1.47(b) Steering gear mechanism The steering mechanism is used in automobiles for changing the directions of the wheel axles with reference to the chassis, so as to move the automobile in the desired path.

25

Usually, the two back wheels will have a common axis, which is fixed in direction with reference to the chassis and the steering is done by means of front wheels. In automobiles, the front wheels are placed over the front axles (stub axles), which are pivoted at the points A & B as shown in the fig.1.48. When the vehicle takes a turn, the front wheels, along with the stub axles turn about the pivoted points. The back axle and the back wheels remain straight. Always there should be absolute rolling contact between the wheels and the road surface. Any sliding motion will cause wear of tyres. When a vehicle is taking turn, absolute rolling motion of the wheels on the road surface is possible, only if all the wheels describe concentric circles. Therefore, the two front wheels must turn about the same instantaneous centre I which lies on the axis of the back wheel. Condition for perfect steering The condition for perfect steering is that all the four wheels must turn about the same instantaneous centre. While negotiating a curve, the inner wheel makes a larger turning angle θ than the angle φ subtended by the axis of the outer wheel. In the fig.1.48, a = wheel track, L = wheel base, w = distance between the pivots of front axles.

Fig.1.48 From ∆IAE , cotθ =

AE AE = and EI L

from ∆BEI , cotφ =

EB ( EA + AB ) ( EA + w) EA w w = = = + = cot θ + EI EI L L L L

26

w . This is the fundamental equation for correct steering. If this L condition is satisfied, there will be no skidding of the wheels when the vehicle takes a turn.

∴ cot φ − cot θ =

Ackermann steering gear mechanism

Fig.1.49 c R

S

A' A

d

B

x

x

P

B'

d

Q

fig.1.50 Ackerman steering mechanism, RSAB is a four bar chain as shown in fig.1.50. Links RA and SB which are equal in length are integral with the stub axles. These links are connected with each other through track rod AB. When the vehicle is in straight ahead position, links RA and SB make equal angles α with the center line of the vehicle. The dotted lines in fig.1.50 indicate the position of the mechanism when the vehicle is turning left.

27

Let AB=l, RA=SB=r; PRˆ A = QSˆB = α and in the turned position, ARˆ A1 = θ & BSˆB 1 = φ . IE, the stub axles of inner and outer wheels turn by θ and φ angles respectively. Neglecting the obliquity of the track rod in the turned position, the movements of A and B in the horizontal direction may be taken to be same (x). Then, sin ( α + θ ) =

d+x d−x and sin ( α − φ ) = r r

Adding, sin ( α + θ ) + sin ( α − φ ) =

2d = 2 sin α r

[1]

Angle α can be determined using the above equation. The values of θ and φ to be taken in w this equation are those found for correct steering using the equation cot φ − cot θ = . [2] L This mechanism gives correct steering in only three positions. One, when θ = 0 and other two each corresponding to the turn to right or left (at a fixed turning angle, as determined by equation [1]). The correct values of φ, [φc] corresponding to different values of θ, for correct steering can be determined using equation [2]. For the given dimensions of the mechanism, actual values of φ, [φa] can be obtained for different values of θ. T he difference between φc and φa will be very small for small angles of θ, but the difference will be substantial, for larger values of θ. Such a difference will reduce the life of tyres because of greater wear on account of slipping. But for larger values of θ, the automobile must take a sharp turn; hence is will be moving at a slow speed. At low speeds, wear of the tyres is less. Therefore, the greater difference between φc and φa larger values of θ ill not matter. As this mechanism employs only turning pairs, friction and wear in the mechanism will be less. Hence its maintenance will be easier and is commonly employed in automobiles. References: 7. Theory of Machines and Mechanisms by Joseph Edward Shigley and John Joseph Uicker,Jr. McGraw-Hill International Editions. 8. Kinematics and Dynamics of Machines by George H.Martin. McGraw-Hill Publications. 9. Mechanisms and Dynamics of Machinery by Hamilton H. Mabie and Fred W. Ocvirk. John Wiley and Sons. 10. Theory of Machines by V.P.Singh. Dhanpat Rai and Co. 11. The Theory of Machines through solved problems by J.S.Rao. New age international publishers. 6. A text book of Theory of Machines by Dr.R.K.Bansal. Laxmi Publications (P) Ltd. Chapter VI

28

CAMS INTRODUCTION A cam is a mechanical device used to transmit motion to a follower by direct contact. The driver is called the cam and the driven member is called the follower. In a cam follower pair, the cam normally rotates while the follower may translate or oscillate. A familiar example is the camshaft of an automobile engine, where the cams drive the push rods (the followers) to open and close the valves in synchronization with the motion of the pistons. Types of cams Cams can be classified based on their physical shape. a) Disk or plate cam (Fig. 6.1a and b): The disk (or plate) cam has an irregular contour to impart a specific motion to the follower. The follower moves in a plane perpendicular to the axis of rotation of the camshaft and is held in contact with the cam by springs or gravity.

Fig. 6.1 Plate or disk cam. b) Cylindrical cam (Fig. 6.2): The cylindrical cam has a groove cut along its cylindrical surface. The roller follows the groove, and the follower moves in a plane parallel to the axis of rotation of the cylinder.

Fig. 6.2 Cylindrical cam. c) Translating cam (Fig. 6.3a and b). The translating cam is a contoured or grooved plate sliding on a guiding surface(s). The follower may oscillate (Fig. 6.3a) or reciprocate

29

(Fig. 6.3b). The contour or the shape of the groove is determined by the specified motion of the follower.

Fig. 6.3 Translating cam Types of followers: (i) Based on surface in contact. (Fig.6.4) (a) Knife edge follower (b) Roller follower (c) Flat faced follower (d) Spherical follower

Fig. 6.4 Types of followers (ii) Based on type of motion: (Fig.6.5) (a) Oscillating follower (b) Translating follower

30

Fig.6.5 (iii) Based on line of motion: (a) Radial follower: The lines of movement of in-line cam followers pass through the centers of the camshafts (Fig. 6.4a, b, c, and d). (b) Off-set follower: For this type, the lines of movement are offset from the centers of the camshafts (Fig. 6.6a, b, c, and d).

Fig.6.6 Off set followers Cam nomenclature (Fig. 6.7):

31

Fig.6.7 Cam Profile

The contour of the working surface of the cam.

Tracer Point The point at the knife edge of a follower, or the center of a roller, or the center of a spherical face. Pitch Curve

The path of the tracer point.

Base Circle

The smallest circle drawn, tangential to the cam profile, with its center on the axis of the camshaft. The size of the base circle determines the size of the cam.

Prime Circle

The smallest circle drawn, tangential to the pitch curve, with its center on the axis of the camshaft.

Pressure Angle The angle between the normal to the pitch curve and the direction of motion of the follower at the point of contact. Types of follower motion: Cam follower systems are designed to achieve a desired oscillatory motion. Appropriate displacement patterns are to be selected for this purpose, before designing the cam surface. The cam is assumed to rotate at a constant speed and the follower raises, dwells, returns to its original position and dwells again through specified angles of rotation of the cam, during each revolution of the cam. Some of the standard follower motions are as follows: They are, follower motion with,

32

(a) Uniform velocity (b) Modified uniform velocity (c) Uniform acceleration and deceleration (d) Simple harmonic motion (e) Cycloidal motion Displacement diagrams: In a cam follower system, the motion of the follower is very important. Its displacement can be plotted against the angular displacement θ of the cam and it is called as the displacement diagram. The displacement of the follower is plotted along the y-axis and angular displacement θ of the cam is plotted along x-axis. From the displacement diagram, velocity and acceleration of the follower can also be plotted for different angular displacements θ of the cam. The displacement, velocity and acceleration diagrams are plotted for one cycle of operation i.e., one rotation of the cam. Displacement diagrams are basic requirements for the construction of cam profiles. Construction of displacement diagrams and calculation of velocities and accelerations of followers with different types of motions are discussed in the following sections. (a) Follower motion with Uniform velocity: Fig.6.8 shows the displacement, velocity and acceleration patterns of a follower having uniform velocity type of motion. Since the follower moves with constant velocity, during rise and fall, the displacement varies linearly with θ. Also, since the velocity changes from zero to a finite value, within no time, theoretically, the acceleration becomes infinite at the beginning and end of rise and fall.

33

Fig.6.8 (b) Follower motion with modified uniform velocity: It is observed in the displacement diagrams of the follower with uniform velocity that the acceleration of the follower becomes infinite at the beginning and ending of rise and return strokes. In order to prevent this, the displacement diagrams are slightly modified. In the modified form, the velocity of the follower changes uniformly during the beginning and end of each stroke. Accordingly, the displacement of the follower varies parabolically during these periods. With this modification, the acceleration becomes

34

constant during these periods, instead of being infinite as in the uniform velocity type of motion. The displacement, velocity and acceleration patterns are shown in fig.6.9.

fig.6.9 (c) Follower motion with uniform acceleration and retardation (UARM): Here, the displacement of the follower varies parabolically with respect to angular displacement of cam. Accordingly, the velocity of the follower varies uniformly with respect to angular displacement of cam. The acceleration/retardation of the follower becomes constant accordingly. The displacement, velocity and acceleration patterns are shown in fig. 6.10.

35

Fig.6.10 s = Stroke of the follower θo and θr = Angular displacement of the cam during outstroke and return stroke. ω = Angular velocity of cam. Time required for follower outstroke = to =

θo ω

Time required for follower return stroke = tr =

θr ω

36

Average velocity of follower =

s t s

Average velocity of follower during outstroke = t o

2 = s = vomin + vomax to 2 2

vomin = 0 ∴ vomax =

2s 2ωs = = Max. velocity during outstroke. to θo s

Average velocity of follower during return stroke = t r

2 = s = vrmin + vrmax tr 2 2

vrmin = 0 ∴ vrmax =

2 s 2ωs = = Max. velocity during return stroke. tr θr

vomax 4ω 2 s a = = 2 Acceleration of the follower during outstroke = o to θo 2 Similarly acceleration of the follower during return stroke = a r =

4ω 2 s

θr

2

(d) Simple Harmonic Motion: In fig.6.11, the motion executed by point Pl, which is the projection of point P on the vertical diameter is called simple harmonic motion. Here, P moves with uniform angular velocity ωp, along a circle of radius r (r = s/2).

y

p' y

p r a

x

37

Fig.6.11 Displacement = y = r sin α = r sin ω p t ; y max = r

[d1]

Velocity = y = ω p r cos ω p t ; y max = rω p

[d2]

2 2 2 Acceleration = y = −ω p r sin ω p t = −ω p y ; y max = − rω p

[d3]

Fig.6.11 s= Stroke or displacement of the follower. θo = Angular displacement during outstroke. θr = Angular displacement during return stroke ω = Angular velocity of cam. to = Time taken for outstroke =

θo ω

tr = Time taken for return stroke =

θr ω

Max. velocity of follower during outstroke = vomax = rωp (from d2)

38

vomax =

s π πωs = 2 t o 2θ o

Similarly Max. velocity of follower during return stroke = , vrmax =

s π πωs = 2 t r 2θ r

sπ Max. acceleration during outstroke = aomax = rω p (from d3) =  2  to

 π 2ω 2 s  = 2 2θ o 

2

sπ Similarly, Max. acceleration during return stroke = armax =  2  tr

2

2

 π 2ω 2 s  = 2θ 2 r 

39

(e) Cycloidal motion: Cycloid is the path generated by a point on the circumference of a circle, as the circle rolls without slipping, on a straight/flat surface. The motion executed by the follower here, is similar to that of the projection of a point moving along a cyloidal curve on a vertical line as shown in figure 6.12. a7 F O L L O W E R

a6 CYCLOIDAL MOTION

a5 a4

66

M O T I O N

a3 a2 a1

a 21

Fig.6.12 The construction of displacement diagram and the standard patterns of velocity and acceleration diagrams are shown in fig.6.13. Compared to all other follower motions, cycloidal motion results in smooth operation of the follower. The expressions for maximum values of velocity and acceleration of the follower are shown below. s = Stroke or displacement of the follower. d = dia. of cycloid generating circle =

s π

θo = Angular displacement during outstroke. θr = Angular displacement during return stroke ω = Angular velocity of cam. to = Time taken for outstroke =

θo ω

tr = Time taken for return stroke =

θr ω

vomax = Max. velocity of follower during outstroke =

2ωs θo

40

vrmax = Max. velocity of follower during return stroke = aomax = Max. acceleration during outstroke =

2ωs θr

2πω 2 s

θo

2

2πω 2 s armax = Max. acceleration during return stroke = θ 2r

Fig. 6.13 41

42

Solved problems (1)

Draw the cam profile for following conditions:

Follower type = Knife edged, in-line; lift = 50mm; base circle radius = 50mm; out stroke with SHM, for 600 cam rotation; dwell for 450 cam rotation; return stroke with SHM, for 900 cam rotation; dwell for the remaining period. Determine max. velocity and acceleration during out stroke and return stroke if the cam rotates at 1000 rpm in clockwise direction. Displacement diagram: 6

5

g

f

e

h

d

4

i

b

2

LIFT =50mm

j

c

3

k

a 1

0

l 1

2

3

4

OUT STROKE

5

6

7 DWELL

8

9

10

11

RETURNSTROKE

12 DWELL

Cam profile: Construct base circle. Mark points 1,2,3…..in direction opposite to the direction of cam rotation. Transfer points a,b,c…..l from displacement diagram to the cam profile and join them by a smooth free hand curve. This forms the required cam profile.

43

d

e

c b

a

f 4 5

1

60°

45°

6

2

3

50

7 8

g

90°

9 10

11

h

12 l

i

k

j

Calculations: Angular velocity of cam = ω =

2πN 2 × π × 1000 = =104.76 rad/sec 60 60

Max. velocity of follower during outstroke = vomax = =

πωs = 2θ o

π × 104.76 × 50 =7857mm/sec =7.857m/sec 2×π 3

Similarly Max. velocity of follower during return stroke = , vrmax =

πωs = 2θ r

π × 104.76 × 50 = = 5238mm/sec = 5.238m/sec 2×π 2 π 2ω 2 s Max. acceleration during outstroke = aomax = rω p (from d3) = = 2 2θ o 2

44

π 2 × (104.76 ) 2 × 50 = 2469297.96mm/sec2 = 2469.3m/sec2 2 = 2× π 3

( )

π 2ω 2 s Similarly, Max. acceleration during return stroke = armax = = 2θ 2 r π 2 × (104.76 ) 2 × 50 = 1097465.76mm/sec2 = 1097.5m/sec2 2 = 2× π 2

( )

45

(2) Draw the cam profile for the same operating conditions of problem (1), with the follower off set by 10 mm to the left of cam center. Displacement diagram: Same as previous case. Cam profile: Construction is same as previous case, except that the lines drawn from 1,2,3…. are tangential to the offset circle of 10mm dia. as shown in the fig.

e

d

c

b

a

f

2

3

1

10

4 5

° 60

6

45°

50mm

7

90°

8

g

9

10

11

h

12 l

i

k j

46

(3)

Draw the cam profile for following conditions:

Follower type = roller follower, in-line; lift = 25mm; base circle radius = 20mm; roller radius = 5mm; out stroke with UARM, for 1200 cam rotation; dwell for 600 cam rotation; return stroke with UARM, for 900 cam rotation; dwell for the remaining period. Determine max. velocity and acceleration during out stroke and return stroke if the cam rotates at 1200 rpm in clockwise direction. Displacement diagram:

e

g

f

h

d

j

c b

LIFT k l

a 0

1

2

3

OUT STROKE

4

5

6

7

DWELL

8

9

10

11

RETURNSTROKE

12

DWELL

Cam profile: Construct base circle and prime circle (25mm radius). Mark points 1,2,3…..in direction opposite to the direction of cam rotation, on prime circle. Transfer points a,b,c…..l from displacement diagram. At each of these points a,b,c… draw circles of 5mm radius, representing rollers. Starting from the first point of contact between roller and base circle, draw a smooth free hand curve, tangential to all successive roller positions. This forms the required cam profile.

47

25

i

a

b c

0

1 2

3 20mm

120 °

4 5

60°

e

90 °

d

12 l

6

11 7

f

8

9

k

10

j

i

g

h

Calculations: Angular velocity of the cam = ω =

2πN 2 × π × 1200 = = 125.71rad/sec 60 60

Max. velocity during outstroke = vomax = =

2 s 2ωs = = to θo

2 × 125.71 × 25 = 2999.9mm/sec =2.999m/sec 2×π 3

2 s 2ωs 2 × 125.71 × 25 = Max. velocity during return stroke = vrmax = t = θ = π r r 2 = 3999.86mm/sec = 3.999m/sec vomax 4ω 2 s a = = 2 Acceleration of the follower during outstroke = o to θo = 2 4 × (125.71) × 25 2

=

(2 × π 3 )

2

= 359975mm/sec2 = 359.975m/sec2

Similarly acceleration of the follower during return stroke = a r =

4ω 2 s

θr

2

=

48

4 × (125.71) × 25 2

=

(π 2 )

2

= 639956mm/sec2 = 639.956m/sec2

49

(4) Draw the cam profile for conditions same as in (3), with follower off set to right of cam center by 5mm and cam rotating counter clockwise. Displacement diagram: Same as previous case. Cam profile: Construction is same as previous case, except that the lines drawn from 1,2,3…. are tangential to the offset circle of 10mm dia. as shown in the fig.

a 1

5

3

d

4

l 12

90°

11

60°

k 10 9

j

c

2

120°

20mm

b

5

e

6 8

7

f

i h g

50

(5)

Draw the cam profile for following conditions:

Follower type = roller follower, off set to the right of cam axis by 18mm; lift = 35mm; base circle radius = 50mm; roller radius = 14mm; out stroke with SHM in 0.05sec; dwell for 0.0125sec; return stroke with UARM, during 0.125sec; dwell for the remaining period. During return stroke, acceleration is 3/5 times retardation. Determine max. velocity and acceleration during out stroke and return stroke if the cam rotates at 240 rpm. Calculations: Cam speed = 240rpm. Therefore, time for one rotation =

60 = 0.25 sec 240

0.05 × 360 = 72 0 0.25 0.0125 × 360 = 18 0 Angle of first dwell = θ w1 = 0.25 0.125 × 360 = 180 0 Angle of return stroke = θ r = 0.25 0 Angle of second dwell = θ w 2 = 90 Since acceleration is 3/5 times retardation during return stroke, 3 a 3 a = r (from acceleration diagram) ∴ = 5 r 5 v max v a t 3 ; r = max ∴ = r = But a = ta tr r ta 5 Displacement diagram is constructed by selecting ta and tr accordingly. Angle of out stroke = θ o =

51

6

5

g

f

e

h

i j

d

4

k

c

3

LIFT =35 mm l

b

2

m

a 1

0

n 2

1

3

4

5

6

OUT STROKE

7 DWELL

8

9

10

11

12

13

14

RETURNSTROKE

DWELL

v ta

tr

vr-max a

r

a

2πN 2 × π × 240 = =25.14 rad/sec 60 60 πωs Max. velocity of follower during outstroke = vomax = = 2θ o π × 25.14 × 35 = = 1099.87mm/sec =1.1m/sec 2× 2×π 5 2ωs 2 × 25.14 × 35 = = Similarly Max. velocity during return stroke = vrmax = θr π = 559.9 mm/sec = 0.56m/sec π 2ω 2 s Max. acceleration during outstroke = aomax = rω2p (from d3) = = 2 2θ o Angular velocity of cam = ω =

(

)

π 2 × ( 25.14 ) 2 × 35 = 69127.14mm/sec2 = 69.13m/sec2 2 = 2 × π 2× 5

(

)

52

acceleration of the follower during return stroke = 2ωs 2 vrmax θr 16 × ω 2 × s 16 × ( 25.14) × 35 = ar = = = = 5×π ta 5 × π ×θr 5×π ×π 8×ω 7.17m/sec2 similarly retardation of the follower during return stroke = 2ωs 2 vrmax θr 16 × ω 2 × s 16 × ( 25.14 ) × 35 = rr = = = = 3×π tr 3× π ×θr 3×π ×π 8×ω 11.94m/sec2

7166.37

mm/sec2

=

11943.9

mm/sec2

=

18

b

a

c

1

d 2

3

e 4

72°

f

5

n

6

14

18°

g

7 13 m

180°

8

12 11 l

9

h

10

i

k j

53

(6)

Draw the cam profile for following conditions:

Follower type = knife edged follower, in line; lift = 30mm; base circle radius = 20mm; out stroke with uniform velocity in 1200 of cam rotation; dwell for 600; return stroke with uniform velocity, during 900 of cam rotation; dwell for the remaining period. Displacement diagram:

a 1

b 2

c

d

4

3

e

5

OUT STROKE

g

f

h

6

7

i

j

30mm l

10 11 12

8 9

DWELL

k

DWELL

RETURNSTROKE

Cam profile:

e

f

3 4 5

1 ° 90

d

2 120 °

c

a

b

6

° 90

60°

7 8

9

12 l 11 10 k j i

g

h

54

55

(7)

Draw the cam profile for following conditions:

Follower type = oscillating follower with roller as shown in fig.; base circle radius = 20mm; roller radius = 7mm; follower to rise through 400 during 900 of cam rotation with cycloidal motion; dwell for 300; return stroke with cycloidal motion during 1200 of cam rotation; dwell for the remaining period. Also determine the max. velocity and acceleration during outstroke and return stroke, if the cam rotates at 600 rpm.

76

36

B

O

40°

76

A

Lift of the follower = S = length AB ≈ arc AB = OA × θ = 76 × 40 × Radius of cycloid generating circle =

π = 53 mm. 180

53 = 8.4 mm 2×π

Displacement diagram;

56

8

9 16.8 10

7 11

12 e

f

g

h

d

i j

c b

53

k

a 4

5

1

l 2

3

4

5

6

7

8

9

10

11

12

6

3 2

1 OUT STROKE

DWELL

RETURNSTROKE

DWELL

57

2πN 2 × π × 600 = = 62.86 rad/sec 60 60 2ωs 2 × 62.86 × 53 vomax = Max. velocity of follower during outstroke = θ = = 4240.2 π o 2 mm/sec 2ωs 2 × 62.86 × 53 vrmax = Max. velocity of follower during return stroke = θ = = 3180 2×π r 3 mm/sec 2 2πω 2 s 2 × π × ( 62.86 ) × 53 = 2 2 aomax = Max. acceleration during outstroke = = 533077 θo π 2 mm/sec2 = 533.1 m/sec2. 2 2πω 2 s 2 × π × ( 62.86 ) × 53 = 2 armax = Max. acceleration during return stroke = θ 2 r = 2×π 3 = 299855.8mm/sec2 = 299.8 m/sec2. Cam profile: Draw base circle and prime circle. Draw another circle of radius equal to the distance between cam center and follower pivot point. Take the line joining cam center and pivot point as reference and draw lines indicating successive angular displacements of cam. Divide these into same number of divisions as in the displacement diagram. Show points 1’, 2’, 3’… on the outer circle. With these points as centers and radius equal to length of follower arm, draw arcs, cutting the prime circle at 1,2,3…. Transfer points a,b,c.. on to these arcs from displacement diagram. At each of these points a,b,c… draw circles of 7mm radius, representing rollers. Starting from the first point of contact between roller and base circle, draw a smooth free hand curve, tangential to all successive roller positions. This forms the required cam profile. Angular velocity of cam = ω =

( )

(

)

58

76

c b

36

1'

4 5

3

6 f

2

12' 120°

90°

2'

e

a 1

7 8

120 °

d

30°

9 10

3'

11'

12 m

11

10'

l 4'

g

5'

k

h

9'

i 6'

8' 7'

59

(8)

Draw the cam profile for following conditions:

Follower type = knife edged follower, in line; follower rises by 24mm with SHM in 1/4 rotation, dwells for 1/8 rotation and then raises again by 24mm with UARM in 1/4 rotation and dwells for 1/16 rotation before returning with SHM. Base circle radius = 30mm. Angle of out stroke (1) = θ01 =

1 × 360 0 = 90 0 4

1 × 360 0 = 45 0 8 1 0 0 Angle of out stroke (2) = θ02 = × 360 = 90 4 1 × 360 0 = 22.5 0 Angle of dwell (2) = 16   1 1 1 1  5 0 0 Angle of return stroke = θr = 1 −  + + +  × 360 = × 360 = 112.5 4 8 4 16 16    Displacement diagram: Angle of dwell (1) =

m k

d

g

f

c b

l

o

j

24 e

n

h

p

i q

24mm

r s

a 1 2 3 4 5 6 OUT STROKE 1

7 8 9 10 11 12 13 DWELL 1

OUT STROKE 2

14 15 16 17 18 19

DWELL 2

RETURNSTROKE

Cam profile:

60

b

c d

3

e

4

1

2

19

90°

q

18

112.5°

5

f

r

s

a

p

17 16

6

45°

22.5°

7 8

60

15

90°

12

9 10

o

14

13

11

n

g h m

i j

k

l

61

(9)

Draw the cam profile for following conditions:

Follower type = flat faced follower, in line; follower rises by 20mm with SHM in 1200 of cam rotation, dwells for 300 of cam rotation; returns with SHM in 1200 of cam rotation and dwells during the remaining period. Base circle radius = 25mm.

b

a 1

2

c 3

d

4

OUT STROKE

e

5

f

g

6 DWELL

7

h

8

i

9

j 10

k

20

Displacement diagram:

l

11 12

RETURNSTROKE

DWELL

Cam profile: Construct base circle. Mark points 1,2,3…..in direction opposite to the direction of cam rotation, on prime circle. Transfer points a,b,c…..l from displacement diagram. At each of these points a,b,c… draw perpendicular lines to the radials, representing flat faced followers. Starting from the first point of contact between follower and base circle, draw a smooth free hand curve, tangential to all successive follower positions. This forms the required cam profile.

62

25 a

b c

2

1

4

120 °

d

° 90

3 5 6 f

° 30

e

7

8

° 120 10

9

12 l 11

k

j

g h

i

63

(10)

Draw the cam profile for following conditions:

Follower type = roller follower, in line; roller dia. = 5mm; follower rises by 25mm with SHM in 1800 of cam rotation, falls by half the distance instantaneously; returns with Uniform velocity in 1800 of cam rotation. Base circle radius = 20m. Displacement diagram: f d

e

c

25

g h

b

a 1

3

2

4

5

i

7

6

j

k

9

10

8

OUT STROKE

12.5

l 11

RETURNSTROKE

Cam profile:

a b

l 1

11

2 c

k 10 9

3

20 R

45

8

4

d

j

5

i

7 6 h

e

g f

64

(11)

Draw the cam profile for following conditions:

Follower type = roller follower, off-set to the right by 5mm; lift = 30mm; base circle radius = 25mm; roller radius = 5mm; out stroke with SHM, for 1200 cam rotation; dwell for 600 cam rotation; return stroke during 1200 cam rotation; first half of return stroke with Uniform velocity and second half with UARM; dwell for the remaining period. Displacement diagram: f d

h

e

c b a 1

2

3

4

5

g

6

i

j k

l

m

n

o

30 p

q r 7 8 9 10 11 12 13 14 15 16 17 18

Cam profile:

65

5 1

p

3

120°

5

13

60°

12 11

m

d

4

120°

14

n

c

60°

17 16 15

o

b 2

r 18 q

a

l

e

6 10

9

8

7 f

k j i h g

(12) A push rod of valve of an IC engine ascends with UARM, along a path inclined to the vertical at 600. The same descends with SHM. The base circle diameter of the cam is 50mm and the push rod has a roller of 60mm diameter, fitted to its end. The axis of the roller and the cam fall on the same vertical line. The stroke of the follower is 20mm. The angle of action for the outstroke and the return stroke is 600 each, interposed by a dwell period of 600. Draw the profile of the cam. Displacement diagram:

66

e

f

g h

d

i

c

a 1

20

j k

b 2

3

4

OUT STROKE

5

6

7 DWELL

8

9

10

11

l 12 DWELL

RETURNSTROKE

Cam profile:

60°

16

d e c f b a 1 2 3 4 5 6 60°

66

50

60°

g 60°

7

8 9 12 11 10 j k l

h i

67

4.0 Gears: Introduction: The slip and creep in the belt or rope drives is a common phenomenon, in the transmission of motion or power between two shafts. The effect of slip is to reduce the velocity ratio of the drive. In precision machine, in which a definite velocity ratio is importance (as in watch mechanism, special purpose machines..etc), the only positive drive is by means of gears or toothed wheels. Friction Wheels: Kinematiclly, the motion and power transmitted by gears is equivalent to that transmitted by friction wheels or discs in contact with sufficient friction between them. In order to understand motion transmitted by two toothed wheels, let us consider the two discs placed together as shown in the figure 4.1.

Figure 4.1

When one of the discs is rotated, the other disc will be rotate as long as the tangential force exerted by the driving disc does not exceed the maximum frictional resistance between the two discs. But when the tangential force exceeds the frictional resistance, slipping will take place between the two discs. Thus the friction drive is not positive a drive, beyond certain limit. Gears are machine elements that transmit motion by means of successively engaging teeth. The gear teeth act like small levers. Gears are highly efficient (nearly 95%) due to primarily rolling contact between the teeth, thus the motion transmitted is considered as positive. Gears essentially allow positive engagement between teeth so high forces can be transmitted while still undergoing essentially rolling contact. Gears do not depend on friction and do best when friction is minimized. Some common places that gears can normally be found are: Printing machinery parts Rotary die cutting machines Blow molding machinery Agricultural equipment High volume vacuum pumps Boat out drives Hoists and Cranes

Newspaper Industry Plastics machinery builders

Book binding machines Injection molding machinery

Motorcycle Transmissions (street Heavy earth moving to and race applications) personal vehicles Polymer pumps High volume water pumps for municipalities Turbo boosters for automotive Marine applications applications Special offshore racing drive Canning and bottling systems machinery builders Commercial and Military Military offroad vehicles 68

Automotive prototype and reproduction Diesel engine builders

operations Low volume automotive production Special gear box builders

Stamping presses Many different special machine tool builders

4.1 Gear Classification: Gears may be classified according to the relative position of the axes of revolution. The axes may be 1. Gears for connecting parallel shafts, 2. Gears for connecting intersecting shafts, 3. Gears for neither parallel nor intersecting shafts. Gears for connecting parallel shafts 1. Spur gears: Spur gears are the most common type of gears. They have straight teeth, and are mounted on parallel shafts. Sometimes, many spur gears are used at once to create very large gear reductions. Each time a gear tooth engages a tooth on the other gear, the teeth collide, and this impact makes a noise. It also increases the stress on the gear teeth. To reduce the noise and stress in the gears, most of the gears in your car are helical.

Spur gears are the most commonly used gear type. They are characterized by teeth, which Spur gears (Emerson Power Transmission Corp) External contact are perpendicular to the face of the gear. Spur gears are most commonly available, and are generally the least expensive. •

Limitations: Spur gears generally cannot be used when a direction change between the two shafts is required.

•

Advantages: Spur gears are easy to find, inexpensive, and efficient.

Iinternal contact 2. Parallel helical gears: The teeth on helical gears are cut at an angle to the face of the gear. When two teeth on a helical gear system engage, the contact starts at one end of the tooth and gradually spreads as the gears rotate, until the two teeth are in full engagement.

69

Helical gears (EmersonPower Transmission Corp)

Herringbone gears (or double-helical gears)

This gradual engagement makes helical gears operate much more smoothly and quietly than spur gears. For this reason, helical gears are used in almost all car transmission. Because of the angle of the teeth on helical gears, they create a thrust load on the gear when they mesh. Devices that use helical gears have bearings that can support this thrust load. One interesting thing about helical gears is that if the angles of the gear teeth are correct, they can be mounted on perpendicular shafts, adjusting the rotation angle by 90 degrees. Helical gears to have the following differences from spur gears of the same size: o o o

Tooth strength is greater because the teeth are longer, Greater surface contact on the teeth allows a helical gear to carry more load than a spur gear The longer surface of contact reduces the efficiency of a helical gear relative to a spur gear

Rack and pinion (The rack is like a gear whose axis is at infinity.): Racks are straight gears that are used to convert rotational motion to translational motion by means of a gear mesh. (They are in theory a gear with an infinite pitch diameter). In theory, the torque and angular velocity of the pinion gear are related to the Force and the velocity of the rack by the radius of the pinion gear, as is shown. Perhaps the most well-known application of a rack is the rack and pinion steering system used on many cars in the past Gears for connecting intersecting shafts: Bevel gears are useful when the direction of a shaft's rotation needs to be changed. They are usually mounted on shafts that are 90 degrees apart, but can be designed to work at other angles as well. The teeth on bevel gears can be straight, spiral or hypoid. Straight bevel gear teeth actually have the same problem as straight spur gear teeth, as each tooth engages; it impacts the corresponding tooth all at once.

70

Just like with spur gears, the solution to this problem is to curve the gear teeth. These spiral teeth engage just like helical teeth: the contact starts at one end of the gear and progressively spreads across the whole tooth.

Straight bevel gears

Spiral bevel

gears On straight and spiral bevel gears, the shafts must be perpendicular to each other, but they must also be in the same plane. The hypoid gear, can engage with the axes in different planes. This feature is used in many car differentials. The ring gear of the differential and the input pinion gear are both hypoid. This allows the input pinion to be mounted lower than the axis of the ring gear. Figure shows the input pinion engaging the ring gear of the differential. Since the driveshaft of the car is connected to the input pinion, this also Hypoid gears (Emerson Power Transmission Corp) lowers the driveshaft. This means that the driveshaft doesn't pass into the passenger compartment of the car as much, making more room for people and cargo. Neither parallel nor intersecting shafts: Helical gears may be used to mesh two shafts that are not parallel, although they are still primarily use in parallel shaft applications. A special application in which helical gears are used is a crossed gear mesh, in which the two shafts are perpendicular to each other.

Crossed-helical gears

to

Worm and worm gear: Worm gears are used when large gear reductions are needed. It is common for worm gears to have reductions of 20:1, and even up 300:1 or greater.

71

Many worm gears have an interesting property that no other gear set has: the worm can easily turn the gear, but the gear cannot turn the worm. This is because the angle on the worm is so shallow that when the gear tries to spin it, the friction between the gear and the worm holds the worm in place. This feature is useful for machines such as conveyor systems, in which the locking feature can act as a brake for the conveyor when the motor is not turning. One other very interesting usage of worm gears is in the Torsen differential, which is used on some high-performance cars and trucks. 4.3 Terminology for Spur Gears

72

Figure 4-4 Spur Gear

73

Terminology:

Addendum: The radial distance between the Pitch Circle and the top of the teeth. Arc of Action: Is the arc of the Pitch Circle between the beginning and the end of the engagement of a given pair of teeth. Arc of Approach: Is the arc of the Pitch Circle between the first point of contact of the gear teeth and the Pitch Point. Arc of Recession: That arc of the Pitch Circle between the Pitch Point and the last point of contact of the gear teeth. Backlash: Play between mating teeth. Base Circle: The circle from which is generated the involute curve upon which the tooth profile is based. Center Distance: The distance between centers of two gears. Chordal Addendum: The distance between a chord, passing through the points where the Pitch Circle crosses the tooth profile, and the tooth top. Chordal Thickness: The thickness of the tooth measured along a chord passing through the points where the Pitch Circle crosses the tooth profile. Circular Pitch: Millimeter of Pitch Circle circumference per tooth. Circular Thickness: The thickness of the tooth measured along an arc following the Pitch Circle Clearance: The distance between the top of a tooth and the bottom of the space into which it fits on the meshing gear.

74

Contact Ratio: The ratio of the length of the Arc of Action to the Circular Pitch. Dedendum: The radial distance between the bottom of the tooth to pitch circle. Diametral Pitch: Teeth per mm of diameter. Face: The working surface of a gear tooth, located between the pitch diameter and the top of the tooth. Face Width: The width of the tooth measured parallel to the gear axis. Flank: The working surface of a gear tooth, located between the pitch diameter and the bottom of the teeth Gear: The larger of two meshed gears. If both gears are the same size, they are both called "gears". Land: The top surface of the tooth. Line of Action: That line along which the point of contact between gear teeth travels, between the first point of contact and the last. Module: Millimeter of Pitch Diameter to Teeth. Pinion: The smaller of two meshed gears. Pitch Circle: The circle, the radius of which is equal to the distance from the center of the gear to the pitch point. Diametral pitch: Teeth per millimeter of pitch diameter. Pitch Point: The point of tangency of the pitch circles of two meshing gears, where the Line of Centers crosses the pitch circles. Pressure Angle: Angle between the Line of Action and a line perpendicular to the Line of Centers. Profile Shift: An increase in the Outer Diameter and Root Diameter of a gear, introduced to lower the practical tooth number or acheive a non-standard Center Distance. Ratio: Ratio of the numbers of teeth on mating gears. Root Circle: The circle that passes through the bottom of the tooth spaces. Root Diameter: The diameter of the Root Circle.

75

Working Depth: The depth to which a tooth extends into the space between teeth on the mating gear. 4.2 Gear-Tooth Action 4.2.1 Fundamental Law of Gear-Tooth Action Figure 5.2 shows two mating gear teeth, in which •

Tooth profile 1 drives tooth profile 2 by acting at the instantaneous contact point K.

•

N1N2 is the common normal of the two profiles.

•

N1 is the foot of the perpendicular from O1 to N1N2 N2 is the foot of the perpendicular from O2 to N1N2.

•

φ

Although the two profiles have different velocities V1 and V2 at point K, their velocities along N1N2 are equal in both magnitude and direction. Otherwise the two tooth profiles would separate from each other. Therefore, we have O1 N 1 ω1 = O2 N 2 ω 2

( 4.1)

or

ω1 O2 N 2 = ω 2 O1 N 1

( 4.2)

Figure 5-2 Two gearing tooth profiles

We notice that the intersection of the tangency N1N2 and the line of center O1O2 is point P, and from the similar triangles, ∆O1 N 1 P = ∆O2 N 2 P

( 4.3)

Thus, the relationship between the angular velocities of the driving gear to the driven gear, or velocity ratio, of a pair of mating teeth is

76

ω1 O2 P = ω 2 O1 P

( 4.4)

Point P is very important to the velocity ratio, and it is called the pitch point. Pitch point divides the line between the line of centers and its position decides the velocity ratio of the two teeth. The above expression is the fundamental law of gear-tooth action. From the equations 4.2 and 4.4, we can write,

ω1 O2 P O2 N 2 = = ω 2 O1 P O1 N 1

( 4.5)

which determines the ratio of the radii of the two base circles. The radii of the base circles is given by: O1 N 1 = O1 P cos φ

and

O2 N 2 = O2 P cos φ

( 4.6)

Also the centre distance between the base circles: O1O2 = O1 P + O2 P =

O1 N 1 O2 N 2 O1 N 1 + O2 N 2 + = cos φ cos φ cos φ

( 4.7 )

where φ is the pressure angle or the angle of obliquity. It is the angle which the common normal to the base circles make with the common tangent to the pitch circles. 4.2.2 Constant Velocity Ratio For a constant velocity ratio, the position of P should remain unchanged. In this case, the motion transmission between two gears is equivalent to the motion transmission between two imagined slip-less cylinders with radius R1 and R2 or diameter D1 and D2. We can get two circles whose centers are at O1 and O2, and through pitch point P. These two circles are termed pitch circles. The velocity ratio is equal to the inverse ratio of the diameters of pitch circles. This is the fundamental law of gear-tooth action. The fundamental law of gear-tooth action may now also be stated as follow (for gears with fixed center distance) A common normal (the line of action) to the tooth profiles at their point of contact must, in all positions of the contacting teeth, pass through a fixed point on the line-of-centers called the pitch point Any two curves or profiles engaging each other and satisfying the law of gearing are conjugate curves, and the relative rotation speed of the gears will be constant(constant velocity ratio).

77

4.2.3 Conjugate Profiles To obtain the expected velocity ratio of two tooth profiles, the normal line of their profiles must pass through the corresponding pitch point, which is decided by the velocity ratio. The two profiles which satisfy this requirement are called conjugate profiles. Sometimes, we simply termed the tooth profiles which satisfy the fundamental law of gear-tooth action the conjugate profiles. Although many tooth shapes are possible for which a mating tooth could be designed to satisfy the fundamental law, only two are in general use: the cycloidal and involute profiles. The involute has important advantages; it is easy to manufacture and the center distance between a pair of involute gears can be varied without changing the velocity ratio. Thus close tolerances between shaft locations are not required when using the involute profile. The most commonly used conjugate tooth curve is the involute curve. (Erdman & Sandor). conjugate action : It is essential for correctly meshing gears, the size of the teeth ( the module ) must be the same for both the gears. Another requirement - the shape of teeth necessary for the speed ratio to remain constant during an increment of rotation; this behavior of the contacting surfaces (ie. the teeth flanks) is known as conjugate action. 4.3 Involute Curve The following examples are involute spur gears. We use the word involute because the contour of gear teeth curves inward. Gears have many terminologies, parameters and principles. One of the important concepts is the velocity ratio, which is the ratio of the rotary velocity of the driver gear to that of the driven gears.

4.1 Generation of the Involute Curve The curve most commonly used for gear-tooth profiles is the involute of a circle. This involute curve is the path traced by a point on a line as the line rolls without slipping on the circumference of a circle. It may also be defined 78 Figure 4.3 Involute curve

as a path traced by the end of a string, which is originally wrapped on a circle when the string is unwrapped from the circle. The circle from which the involute is derived is called the base circle. 4.2 Properties of Involute Curves 1. The line rolls without slipping on the circle. 2. For any instant, the instantaneous center of the motion of the line is its point of tangent with the circle. Note: We have not defined the term instantaneous center previously. The instantaneous center or instant center is defined in two ways. 1. When two bodies have planar relative motion, the instant center is a point on one body about which the other rotates at the instant considered. 2. When two bodies have planar relative motion, the instant center is the point at which the bodies are relatively at rest at the instant considered. 3. The normal at any point of an involute is tangent to the base circle. Because of the property (2) of the involute curve, the motion of the point that is tracing the involute is perpendicular to the line at any instant, and hence the curve traced will also be perpendicular to the line at any instant. There is no involute curve within the base circle. Cycloidal profile:

Epicycliodal Profile:

79

Hypocycliodal Profile:

The involute profile of gears has important advantages; •

It is easy to manufacture and the center distance between a pair of involute gears can be varied without changing the velocity ratio. Thus close tolerances between shaft locations are not required. The most commonly used conjugate tooth curve is the involute curve. (Erdman & Sandor).

2. In involute gears, the pressure angle, remains constant between the point of tooth engagement and disengagement. It is necessary for smooth running and less wear of gears. But in cycloidal gears, the pressure angle is maximum at the beginning of engagement, reduces to zero at pitch point, starts increasing and again becomes maximum at the end of engagement. This results in less smooth running of gears. 3. The face and flank of involute teeth are generated by a single curve where as in cycloidal gears, double curves (i.e. epi-cycloid and hypo-cycloid) are required for the face and flank respectively. Thus the involute teeth are easy to manufacture than cycloidal teeth. In involute system, the basic rack has straight teeth and the same can be cut with simple tools. Advantages of Cycloidal gear teeth: 1. Since the cycloidal teeth have wider flanks, therefore the cycloidal gears are stronger than the involute gears, for the same pitch. Due to this reason, the cycloidal teeth are preferred specially for cast teeth. 2. In cycloidal gears, the contact takes place between a convex flank and a concave surface, where as in involute gears the convex surfaces are in contact. This condition results in less wear in cycloidal gears as compared to involute gears. However the difference in wear is negligible

80

3. In cycloidal gears, the interference does not occur at all. Though there are advantages of cycloidal gears but they are outweighed by the greater simplicity and flexibility of the involute gears. Properties of involute teeth: 1. A normal drawn to an involute at pitch point is a tangent to the base circle. 2. Pressure angle remains constant during the mesh of an involute gears. 3. The involute tooth form of gears is insensitive to the centre distance and depends only on the dimensions of the base circle. 4. The radius of curvature of an involute is equal to the length of tangent to the base circle. 5. Basic rack for involute tooth profile has straight line form. 6. The common tangent drawn from the pitch point to the base circle of the two involutes is the line of action and also the path of contact of the involutes. 7. When two involutes gears are in mesh and rotating, they exhibit constant angular velocity ratio and is inversely proportional to the size of base circles. (Law of Gearing or conjugate action) 8. Manufacturing of gears is easy due to single curvature of profile.

System of Gear Teeth The following four systems of gear teeth are commonly used in practice: 1. 14 ½O Composite system 2. 14 ½O Full depth involute system 3. 20O Full depth involute system 4. 20O Stub involute system The 14½O composite system is used for general purpose gears. It is stronger but has no interchangeability. The tooth profile of this system has cycloidal curves at the top and bottom and involute curve at the middle portion. The teeth are produced by formed milling cutters or hobs.

81

The tooth profile of the 14½O full depth involute system was developed using gear hobs for spur and helical gears. The tooth profile of the 20o full depth involute system may be cut by hobs. The increase of the pressure angle from 14½o to 20o results in a stronger tooth, because the tooth acting as a beam is wider at the base. The 20o stub involute system has a strong tooth to take heavy loads. Involutometry Addendum Circle Pitch Circle C B

A

r

ra

E F

Gear

Base Circle

O The study of the geometry of the involute profile for gear teeth is called involumetry. Consider an involute of base circle radius ra and two points B and C on the involute as shown in figure. Draw normal to the involute from the points B and C. The normal BE and CF are tangents to the Base circle. Let ra= base circle radius of gear rb= radius of point B on the involute rc= radius of point C on the involute and Φb= pressure angle for the point B Φc= pressure angle for the point C 82

tb= tooth thickness along the arc at B tc= tooth thickness along the arc at C From ∆OBE and ∆OCF ra = rb × cos φb (1)

ra = rc × cos φc ( 2) Therefore rb × cos φb = rc × cos φc From the properties of the Involute: Arc AE = Length BE and Arc AF = Length CF ArcAE BE = = tan φb OE OE ∠ AOB = ∠AOE − φb = tan φb − φb ∠ AOE =

∴ Inv.φb = tan φb − φb

 Expression ( tan φb − φb ) is  called involute function   

Similarly: ArcAF BE = = tan φc OF OF ∠ AOC = ∠AOF − φc = tan φC − φC ∠ AOF =

∴ Inv.φc = tan φc − φc At the po int B ∠AOD = ∠AOB +

tb 2rb

= tan φb − φb +

tb 2rb

At the po int C ∠AOD = ∠AOC +

tc 2rb

= tan φc − φc +

tc 2rc

83

Equating the above equations : t t tan φb − φb + b = tan φc − φc + c 2rb 2rc inv. φb +

tb t = inv.φc + c 2rb 2rc

 t  t c =  inv.φb − inv. φc + b 2rc 2rb   = tooth thickness at C Using this equation and knowing tooth thickness at any point on the tooth, it is possible to calculate the thickness of the tooth at any point Path of contact: O2 RA

Wheel

φ R

φ

N L P

K

Base Circle

Pitch Circle Addendum Circles

M r

φ

ra O1

Pinion

Pitch Circle Base Circle

Consider a pinion driving wheel as shown in figure. When the pinion rotates in clockwise, the contact between a pair of involute teeth begins at K (on the near the base circle of pinion or the outer end of the tooth face on the wheel) and ends at L (outer end of the tooth face on the pinion or on the flank near the base circle of wheel). MN is the common normal at the point of contacts and the common tangent to the base circles. The point K is the intersection of the addendum circle of wheel and the common tangent. The point L is the intersection of the addendum circle of pinion and common tangent. The length of path of contact is the length of common normal cut-off by the addendum circles of the wheel and the pinion. Thus the length of part of contact is KL which is the sum

84

of the parts of path of contacts KP and PL. Contact length KP is called as path of approach and contact length PL is called as path of recess. ra = O1L = Radius of addendum circle of pinion, and R A = O2K = Radius of addendum circle of wheel r = O1P = Radius of pitch circle of pinion, and R = O2P = Radius of pitch circle of wheel. Radius of the base circle of pinion = O1M = O1P cosφ = r cosφ and radius of the base circle of wheel = O2N = O2P cos φ = R cosφ From right angle triangle O2KN KN =

( O2 K ) 2 − ( O2 N ) 2

= ( RA ) − R 2 cos 2 φ PN = O2 P sin φ = R sin φ 2

Path of approach: KP KP = KN − PN = ( RA ) − R 2 cos 2 φ − R sin φ Similarly from right angle triangle O1ML 2

ML =

( O1L ) 2 − ( O1M ) 2

=

( ra ) 2 − r 2 cos 2 φ

MP = O1 P sin φ = r sin φ Path of recess: PL PL = ML − MP = ( r ) − r 2 cos 2 φ − r sin φ Length of path of contact = KL a 2

KL = KP + PL =

( RA ) 2 − R 2 cos 2 φ + ( ra ) 2 − r 2 cos 2 φ − ( R + r ) sin φ

85

Arc of contact: Arc of contact is the path traced by a point on the pitch circle from the beginning to the end of engagement of a given pair of teeth. In Figure, the arc of contact is EPF or GPH. O2 RA

Wheel

φ R L

E

Gear Profile

φ

N F

P

G M

Pitch Circle

r

φ

ra O1

Considering the arc of contact GPH.

Addendum Circles

H

K

Pinion

Pitch Circle

Base Circle

The arc GP is known as arc of approach and the arc PH is called arc of recess. The angles subtended by these arcs at O1 are called angle of approach and angle of recess respectively. Length of arc of approach = arc GP =

Length of arc of recess = arc PH =

Lenght of path of approach KP = cos φ cos φ

Lenght of path of recess PL = cos φ cos φ

Length of arc contact = arc GPH = arc GP + arc PH =

KP PL KL Length of path of contact + = = cos φ cos φ cos φ cos φ

Contact Ratio (or Number of Pairs of Teeth in Contact) The contact ratio or the number of pairs of teeth in contact is defined as the ratio of the length of the arc of contact to the circular pitch. Mathematically,

Contat ratio =

Length of the arc of contact PC

86

Where:

PC = Circular pitch = π × m

and

m = Module.

87

Number of Pairs of Teeth in Contact Continuous motion transfer requires two pairs of teeth in contact at the ends of the path of contact, though there is only one pair in contact in the middle of the path, as in Figure.

R

r

The average number of teeth in contact is an important parameter - if it is too low due to the use of inappropriate profile shifts or to an excessive centre distance.The manufacturing inaccuracies may lead to loss of kinematic continuity - that is to impact, vibration and noise. The average number of teeth in contact is also a guide to load sharing between teeth; it is termed the contact ratio Length of path of contact for Rack and Pinion: PITCH LINE Pc h

a b

T c

RACK

Base Circle

PINION

φ φ PITCH LINE

RACK

φ° c

88

Let r

= Pitch circle radius of the pinion = O1P

Φ = Pressure angle ra. = Addendu m radius of the pinion a = Addendum of rack EF = Length of path of contact EF = Path of approach EP + Path of recess PF sin φ =

AP a = EP EP

(1)

a sin φ Path of recess = PF = NF − NP From triangle O1 NP : Path of approach = EP =

From triangle O1NF:

(2) (3)

NP = O1 P sin φ = r sin φ O1 N = O1 P cos φ = r cos φ

(

NF = O1 F 2 − O1 N 2

) = (r 1 2

2 a

− r 2 cos 2 φ

)

1 2

Substituting NP and NF values in the equation (3)

(

Path of racess = PF = ra2 − r 2 cos 2 φ

)

1 2

− r sin φ

∴ Path of length of contact = EF = EP + PF = Exercise problems refer presentation slides Interference in Involute Gears

O2 RA

(

a + ra2 − r 2 cos 2 φ sin φ

)

1 2

− r sin φ

Wheel Base Circle

φ R L

φ

N

P

Pitch Circle Addendum Circles

K M r

φ

ra O1

Pinion

Pitch Circle Base Circle

89

Figure shows a pinion and a gear in mesh with their center as O1andO2 respectively. MN is the common tangent to the basic circles and KL is the path of contact between the two mating teeth. Consider, the radius of the addendum circle of pinion is increased to O1N, the point of contact L will moves from L to N. If this radius is further increased, the point of contact L will be inside of base circle of wheel and not on the involute profile of the pinion. The tooth tip of the pinion will then undercut the tooth on the wheel at the root and damages part of the involute profile. This effect is known as interference, and occurs when the teeth are being cut and weakens the tooth at its root. In general, the phenomenon, when the tip of tooth undercuts the root on its mating gear is known as interference.

Wheel

Undercut

Pinion

Similarly, if the radius of the addendum circles of the wheel increases beyond O 2M, then the tip of tooth on wheel will cause interference with the tooth on pinion. The points M and N are called interference points. Interference may be avoided if the path of the contact does not extend beyond interference points. The limiting value of the radius of the addendum circle of the pinion is O1N and of the wheel is O2M. The interference may only be prevented, if the point of contact between the two teeth is always on the involute profiles and if the addendum circles of the two mating gears cut the common tangent to the base circles at the points of tangency. When interference is just prevented, the maximum length of path of contact is MN. Maximum path of approach = MP = r sin φ Maximum path of recess = PN = R sin φ Maximum length of path of contact = MN MN = MP + PN = ( r + R ) sin φ ( r + R ) sin φ = ( r + R ) tan φ Maximum length of arc of contact = cos φ

90

Methods to avoid Interference 1. Height of the teeth may be reduced. 2. Under cut of the radial flank of the pinion. 3. Centre distance may be increased. It leads to increase in pressure angle. 4. By tooth correction, the pressure angle, centre distance and base circles remain unchanged, but tooth thickness of gear will be greater than the pinion tooth thickness. Minimum number of teeth on the pinion avoid Interference The pinion turns clockwise and drives the gear as shown in Figure. Points M and N are called interference points. i.e., if the contact takes place beyond M and N, interference will occur. The limiting value of addendum circle radius of pinion is O1N and the limiting value of addendum circle radius of gear is O2M. Considering the critical addendum circle radius of gear, the limiting number of teeth on gear can be calculated. Let Ф = pressure angle R = pitch circle radius of gear = ½mT r = pitch circle radius of pinion = ½mt T & t = number of teeth on gear & pinion m = module

O2 RA

Wheel

φ R L

Base Circle

φ

N

P

Pitch Circle Max. Addendum Circles

K M r

φ

ra O1

Pinion

Pitch Circle Base Circle

91

aw = Addendum constant of gear (or) wheel ap = Addendum constant of pinion aw. m = Addendum of gear

ap. m = Addendum of pinion

G = Gear ratio = T/t From triangle O1NP, Applying cosine rule

O1 N 2 = O1P 2 + NP 2 − 2 × O1 P × PN cos O1 PN

= r 2 + R 2 sin 2 φ − 2 r R sin φ cos( 90 + φ ) = r 2 + R 2 sin 2 φ + 2 r R sin 2 φ

(

 R 2 sin 2 φ 2 R sin 2 φ  RR  2  2 = r 2 1 + +  = r 1 +  + 2  sin φ  2 r r   rr    PN = O2 P sin φ = R sin φ )

Limiting radius of the pinion addendum circle: 1

1

 RR  2 mt  T  T   2 2 O1 N = r 1 +  + 2  sin 2 φ  = 1 + + 2   sin φ  2  t  t    rr  

Addendum of the pinion = O1N - O1P

1

mt  T  T  2  2 mt apm = 1 + + 2   sin φ  − 2  t  t 2    mt  T = 1 + 2  t 

1   T   2 2   + 2  sin φ  − 1  t   

Addendum of the pinion = O1N - O1P 1   t  T  T  2 2  a p = 1 +  + 2  sin φ  − 1  2  tt     2a p t= 1   2 2 −1 ( ) 1 + G G + 2 sin φ    

(

)

92

The equation gives minimum number of teeth required on the pinion to avoid interference. If the number of teeth on pinion and gear is same: G=1

2a=p 1. 14 ½O Composite system t = O 2. 14 ½ Full depth involute system =  2 3. 20O Full depth involute system  1 + 3 sin =φ 4. 20O Stub involute system = 

(

)

1 2

12 32 −18 1 14

Minimum number of teeth on the wheel avoid Interference From triangle O2MP, applying cosine rule and simplifying, The limiting radius of wheel addendum circle: 1

 rr 2  O2 M = R 1 +  + 2  sin 2 φ    RR  1

mT = 2

 tt  2 2 1 + + 2   sin φ   T T    

Addendum of the pinion = O2 M- O2P

O2 RA

Wheel

φ R L

Base Circle

φ

N

P

Pitch Circle Max. Addendum Circles

K M r

φ

ra O1

Pinion

Pitch Circle Base Circle

93

1   mT  tt  2 2  aw m = 1 +  + 2  sin φ  − 1  2  T  T     1    T  t t  2 2  aw = 1 +  + 2  sin φ  − 1  2  T  T     2aW T = 1   2   1 1   1 +  + 2  sin 2 φ  − 1  G  G     

The equation gives minimum number of teeth required on the wheel to avoid interference.

Minimum number of teeth on the pinion for involute rack to avoid Interference The rack is part of toothed wheel of infinite diameter. The base circle diameter and profile of the involute teeth are straight lines.

φ φ Pc

PITCH LINE

RACK

h

a

φ°

b

T c

φ

PINION

PITCH LINE

M P

L

φ°

H

K

RACK

c

94

Let t = Minimum number of teeth on the pinion r = Pitch circle radius of the pinion = ½ mt

Φ

= Pressure angle

AR.m = Addendum of rack The straight profiles of the rack are tangential to the pinion profiles at the point of contact and perpendicular to the tangent PM. Point L is the limit of interference. Addendum of the rack:

AR × m = LH = PL sin φ

= ( OP sin φ ) sin φ = OP sin 2 φ

= r sin 2 φ mt 2 = sin φ 2 2 AR ∴To avoid interference : t = sin 2 φ Backlash: The gap between the non-drive face of the pinion tooth and the adjacent wheel tooth is known as backlash. If the rotational sense of the pinion were to reverse, then a period of unrestrained pinion motion would take place until the backlash gap closed and contact with the wheel tooth reestablished impulsively.

Backlash is the error in motion that occurs when gears change direction. The term "backlash" can also be used to refer to the size of the gap, not just the phenomenon it causes; thus, one could speak of a pair of gears as having, for example, "0.1 mm of backlash." A pair of gears could be designed to have zero backlash, but this would presuppose perfection in manufacturing, uniform thermal expansion characteristics throughout the system, and no lubricant. Therefore, gear pairs are designed to have some backlash. It is usually provided by reducing the tooth thickness of each gear by half the desired gap distance. 95

In the case of a large gear and a small pinion, however, the backlash is usually taken entirely off the gear and the pinion is given full sized teeth. Backlash can also be provided by moving the gears farther apart. For situations, such as instrumentation and control, where precision is important, backlash can be minimised through one of several techniques. Standard (cutting) Pitch Circle

O2

O2 RA

φ R

Wheel N

R'

Wheel Base Circle

R N'

φ c

P

φ'

P

c'

Operating Pitch Circle

M r

φ Standard (cutting) Pitch Circle

ra

M'

Pinion

φ'

r

O1

Figure a

Standard (cutting) Pitch Circle

∆c

O1

r'

Pinion

Standard (cutting) Pitch Circle Base Circle

Figure b

Let r = standard pitch circle radius of pinion R = standard pitch circle radius of wheel c = standard centre distance = r +R r’ = operating pitch circle radius of pinion R’ = operating pitch circle radius of wheel c’ = operating centre distance = r’ + R’ Ф = Standard pressure angle Ф’ = operating pressure angle h = tooth thickness of pinion on standard pitch circle= p/2

96

h’ = tooth thickness of pinion on operating pitch circle Let H = tooth thickness of gear on standard pitch circle H1 = tooth thickness of gear on operating pitch circle p = standard circular pitch = 2п r/ t = 2пR/T p’ = operating circular pitch = 2п r1/t = 2пR1/T ∆C = change in centre distance B = Backlash t

= number of teeth on pinion

T = number of teeth on gear. Involute gears have the invaluable ability of providing conjugate action when the gears' centre distance is varied either deliberately or involuntarily due to manufacturing and/or mounting errors.

r R c = = r ' R ' c' c'× cos φ ' = c × cos φ ∴ c' = c

cos φ cos φ '

Now

∆c = c'−c = c

 cos φ  cos φ − c = c − 1 cos φ ' cos φ '  

On the operating pitch circle: Operating pitch = sum of tooth thickness + Backlash p' = h'+ H '+ B (1)

By involutometry : h  h' = 2r ' inv.φ − inv.φ '+  2r   h   H ' = 2 R' inv.φ − inv.φ '+ 2 R  

97

Substituting h’ and H’ in the equation (1): h h    p ' = 2r ' inv.φ − inv.φ '+  + 2 R' inv.φ − inv.φ '+ +B 2r  2 R     r ' R'  p ' = h +  + 2 inv.φ ( r '+ R ') − 2 inv.φ ' ( r '+ R ') + B r R  c' c'  p ' = h +  + 2 c' inv.φ − 2 c ' inv.φ '+ B c c c' ∴ B = p − 2h + 2c ' [ inv.φ '−inv.φ ] c 2πr ' 2πr c' B= −2 + 2c' [ inv.φ '−inv.φ ] t 2t c 2π  c'  B=  r '− r  + 2c' [ inv.φ '−inv.φ ] t  c 2π  r'   r '− r  + 2c ' [ inv.φ '−inv.φ ] t  r Backlash = B = 2c ' [ inv.φ '−inv.φ ] B=

There is an infinite number of possible centre distances for a given pair of profile shifted gears, however we consider only the particular case known as the extended centre distance. Non Standard Gears: The important reason for using non standard gears are to eliminate undercutting, to prevent interference and to maintain a reasonable contact ratio. The two main non- standard gear systems: (1) Long and short Addendum system and (2) Extended centre distance system. Long and Short Addendum System: The addendum of the wheel and the addendum of the pinion are generally made of equal lengths. Here the profile/rack cutter is advanced to a certain increment towards the gear blank and the same quantity of increment will be withdrawn from the pinion blank.

98

Therefore an increased addendum for the pinion and a decreased addendum for the gear is obtained. The amount of increase in the addendum of the pinion should be exactly equal to the addendum of the wheel is reduced. The effect is to move the contact region from the pinion centre towards the gear centre, thus reducing approach length and increasing the recess length. In this method there is no change in pressure angle and the centre distance remains standard. Extended centre distance system: Reduction in interference with constant contact ratio can be obtained by increasing the centre distance. The effect of changing the centre distance is simply in increasing the pressure angle. In this method when the pinion is being cut, the profile cutter is withdrawn a certain amount from the centre of the pinion so the addendum line of the cutter passes through the interference point of pinion. The result is increase in tooth thickness and decrease in tooth space. Now If the pinion is meshed with the gear, it will be found that the centre distance has been increased because of the decreased tooth space. Increased centre distance will have two undesirable effects. NOTE: Please refer presentation slides also for more figure, photos and exercise problems References: 12. Theory of Machines and Mechanisms by Joseph Edward Shigley and John Joseph Uicker,Jr. McGraw-Hill International Editions. 13. Kinematics and Dynamics of Machines by George H.Martin. McGraw-Hill Publications. 14. Mechanisms and Dynamics of Machinery by Hamilton H. Mabie and Fred W. Ocvirk. John Wiley and Sons. 15. Theory of Machines by V.P.Singh. Dhanpat Rai and Co. 16. The Theory of Machines through solved problems by J.S.Rao. New age international publishers. 17. A text book of Theory of Machines by Dr.R.K.Bansal. Laxmi Publications (P) Ltd. 18. Internet: Many Web based e notes

99

Chapter 5: Gears Trains A gear train is two or more gear working together by meshing their teeth and turning each other in a system to generate power and speed. It reduces speed and increases torque. To create large gear ratio, gears are connected together to form gear trains. They often consist of multiple gears in the train. The most common of the gear train is the gear pair connecting parallel shafts. The teeth of this type can be spur, helical or herringbone. The angular velocity is simply the reverse of the tooth ratio. Any combination of gear wheels employed to transmit motion from one shaft to the other is called a gear train. The meshing of two gears may be idealized as two smooth discs with their edges touching and no slip between them. This ideal diameter is called the Pitch Circle Diameter (PCD) of the gear. Simple Gear Trains The typical spur gears as shown in diagram. The direction of rotation is reversed from one gear to another. It has no affect on the gear ratio. The teeth on the gears must all be the same size so if gear A advances one tooth, so does B and C. t = number of teeth on the gear, D = Pitch circle diameter, N = speed in rpm D m = module = t and module must be the same for all gears otherwise they would not mesh. m=

v v

ωA

ωB

GEAR 'A'

GEAR 'B' (Idler gear)

ωC

DA DB D = = C tA tB tC

D A = m t A;

DB = m t B

and

DC = m tC

ω = angular velocity. D v = linear velocity on the circle. v = ω = ω r 2

GEAR 'C'

The velocity v of any point on the circle must be the same for all the gears, otherwise they would be DC DA DB slipping.

v = ωA

= ωB = ωC 2 2 2 ω A DA = ω B DB = ωC DC

ω A m t A = ω B m t B = ωC m t C ω A t A = ω B t B = ωC t C or in terms of rev / min N A t A = N B t B = N C tC 100

Application: a) to connect gears where a large center distance is required b) to obtain desired direction of motion of the driven gear ( CW or CCW) c) to obtain high speed ratio Torque & Efficiency The power transmitted by a torque T N-m applied to a shaft rotating at N rev/min is given by: 2π N T 60 In an ideal gear box, the input and output powers are the same so; P=

2π N1 T1 2π N 2 T2 = 60 60 T2 N N1 T1 = N 2 T2 ⇒ = 1 = GR T1 N2

P=

It follows that if the speed is reduced, the torque is increased and vice versa. In a real gear box, power is lost through friction and the power output is smaller than the power input. The efficiency is defined as: Power out 2π × N 2 T2 × 60 N T η= = = 2 2 Power In 2π × N1 T1 × 60 N1 T1 Because the torque in and out is different, a gear box has to be clamped in order to stop the case or body rotating. A holding torque T3 must be applied to the body through the clamps. The total torque must add up to zero. T1 + T2 + T3 = 0 If we use a convention that anti-clockwise is positive and clockwise is negative we can determine the holding torque. The direction of rotation of the output shaft depends on the design of the gear box. Compound Gear train Compound gears are simply a chain of simple gear trains with the input of the second being the output of the first. A chain of two pairs is shown below. Gear B is the output of the first pair and gear C is the input of the second pair. Gears B and C are locked to the same shaft and revolve at the same speed. For large velocities ratios, compound gear train arrangement is preferred. The velocity of each tooth on A and B are the same so: ωA tA = ωB tB -as they are simple gears. Likewise for C and D, ωC tC = ωD tD.

Input B A

D Output

C Compound Gears GEAR 'B'

GEAR 'A'

GEAR 'D' GEAR 'C'

101

ωC ω D ω A ωB = and = tB tA tD tC t × ωB t × ωD ωA = B and ωC = D tA TC t × ωB t D × ωD ω A × ωC = B × tA tC ω A × ωC t B t D = × ω B × ω D t A tC

Since gear B and C are on the same shaft ω B = ωC

ωA tB tD = × = GR ω D t A tC Since ω = 2 × π × N The gear ratio may be written as : N ( In ) t B t D = × = GR N ( Out ) t A t C Reverted Gear train The driver and driven axes lies on the same line. These are used in speed reducers, clocks and machine tools. N t ×t GR = A = B D N D t A × tC If R and T=Pitch circle radius & number of teeth of the gear RA + RB = RC + RD and

tA + tB = tC + tD

102

Epicyclic gear train: Epicyclic means one gear revolving upon and around another. The design involves planet and sun gears as one orbits the other like a planet around the sun. Here is a picture of a typical gear box. This design can produce large gear ratios in a small space and are used on a wide range of applications from marine gearboxes to electric screwdrivers. Basic Theory The diagram shows a gear B on the end of an arm. Gear B meshes with gear C and revolves around it when the arm is rotated. B is called the planet gear and C the sun.

B

Planet wheel B

Arm

Arm 'A'

First consider what happens when the planet gear orbits the sun gear. C C Sun wheel

103

Observe point p and you will see that gear B also revolves once on its own axis. Any object orbiting around a center must rotate once. Now consider that B is free to rotate on its shaft and meshes with C. Suppose the arm is held stationary and gear C is rotated once. B spins about its own center and the tC number of revolutions it makes is the ratio . B will rotate by this number for every complete tB revolution of C. tC Now consider that C is unable to rotate and the arm A is revolved once. Gear B will revolve 1 + tB because of the orbit. It is this extra rotation that causes confusion. One way to get round this is to imagine that the whole system is revolved once. Then identify the gear that is fixed and revolve it back one revolution. Work out the revolutions of the other gears and add them up. The following tabular method makes it easy. Suppose gear C is fixed and the arm A makes one revolution. Determine how many revolutions the planet gear B makes. Step 1 is to revolve everything once about the center. Step 2 identify that C should be fixed and rotate it backwards one revolution keeping the arm fixed as it should only do one revolution in total. Work out the revolutions of B. Step 3 is simply add them up and we find the total revs of C is zero and for the arm is 1. Step 1 2 3

Action Revolve all once Revolve C by –1 revolution, keeping the arm fixed

A 1

Add

1

0

B 1 t + C tB 1+

tC tB

C 1 -1 0

 tC  The number of revolutions made by B is 1 +  Note that if C revolves -1, then the direction of B tB   tC is opposite so + . tB Example: A simple epicyclic gear has a fixed sun gear with 100 teeth and a planet gear with 50 teeth. If the arm is revolved once, how many times does the planet gear revolve? Solution: Step 1 2 3

Action Revolve all once Revolve C by –1 revolution, keeping the arm fixed Add

A 1 0 1

B 1 100 + 50 3

C 1 -1 0

104

Gear B makes 3 revolutions for every one of the arm. The design so far considered has no identifiable input and output. We need a design that puts an input and output shaft on the same axis. This can be done several ways.

Problem 1: In an ecicyclic gear train shown in figure, the arm A is fixed to the shaft S. The wheel B having 100 teeth rotates freely on the shaft S. The wheel F having 150 teeth driven separately. If the arm rotates at 200 rpm and wheel F at 100 rpm in the same direction; find (a) number of teeth on the gear C and (b) speed of wheel B.

C 100 rpm

F150 S B100 B C

Arm A

200 rpm

Solution: TB=100;

TF=150;

NA=200rpm; NF=100rpm:

Since the mod ule is same for all gears : the number of teeth on the gears is proportional to the pitch cirlce : ∴ rF = rB + 2rC ⇒

TF = TB + 2TC 150 = 100 + 2 × TC TC = 25 → Number of teeth on gears C

The gear B and gear F rotates in the opposite directions: 105

∴ Train value = − also

TV =

∴

−

TB TF

N L − N Arm N F − N A = N F − N Arm N B − N A

( general exp ression for epicyclic gear train)

TB N F − N A = TF N B − N A

100 100 − 200 = ⇒ N E = 350 150 N B − 200 The Gear B rotates at 350 rpm in the same direction of gears F and Arm A. −

Problem 2: In a compound epicyclic gear train as shown in the figure, has gears A and an annular gears D & E free to rotate on the axis P. B and C is a compound gear rotate about axis Q. Gear A rotates at 90 rpm CCW and gear D rotates at 450 rpm CW. Find the speed and direction of rotation of arm F and gear E. Gears A,B and C are having 18, 45 and 21 teeth respectively. All gears having same module and pitch.

D E

P

A B Q

C

Arm F

Solution: TA=18 ;

TB=45;

TC=21;NA = -90rpm;

ND=450rpm:

Since the module and pitch are same for all gears : the number of teeth on the gears is proportional to the pitch cirlce : ∴ rD = rA + rB + rC ⇒

TD = T A + TB + TC TD = 18 + 45 + 21 = 84 teeth on gear D

106

Gears A and D rotates in the opposite directions: T T ∴ Train value = − A × C TB TD

N L − N Arm N D − N F = N F − N Arm N A − N F

also

TV =

∴

−

T A TC N D − N F × = TB TD N A − N F

−

18 × 21 450 − N F = 45 × 84 − 90 − N F

⇒

N F = Speed of Arm = 400.9 rpm − CW

Now consider gears A, B and E: rE = rA + 2rB ⇒

TE = T A + 2T B TE = 18 + 2 × 45 TE = 108 → Number of teeth on gear E

Gears A and E rotates in the opposite directions:

∴ Train value = −

TA TE

NE − NF NA − NF

also

TV =

∴

−

⇒

N − 400.9 18 = E 108 − 90 − 400.9 N E = Speed of gear E = 482.72 rpm − CW

TA N E − N F = TE N A − N F

−

Problem 3: In an epicyclic gear of sun and planet type shown inAnnular figure 3, 'A' the pitch circle diameter of the annular wheel A is to be nearly 216mm and module 4mm. When the annular ring is stationary, the spider that carries three planet wheels P of equal size to make one revolution for every five revolution of the driving spindle carrying the sun wheel. Spider 'L' Determine the number of teeth for all the wheels and the exact pitch circle diameter of the annular Sunthe Wheel 'S' determine the wheel. If an input torque of 20 N-m is applied to the spindle carrying sun wheel, fixed torque on the annular wheel. Planet Wheel 'P'

107

Solution: Module being the same for all the meshing gears: TA = TS + 2TP PCD of A 216 TA = = = 54 teeth m 4 Spider arm L

Operation Arm L is fixed & Sun wheel S is given +1 revolution Multiply by m (S rotates through m revolution) Add n revolutions to all elements

Sun Wheel S

Planet wheel P

TS

Annular wheel A

TP −

0

+1

0

m

−

n

m+n

n−

TS TP

TS m TP TS m TP

TA = 54 −

TS TP T × =− S TP T A TA −

TS m TA

n−

TS m TA

If L rotates +1 revolution: ∴ n=1 (1) The sun wheel S to rotate +5 revolutions correspondingly: ∴ n+m=5 (2) From (1) and (2) m=4 When A is fixed: n−

TS m=0 TA

∴

TS =

⇒

T A = 4 TS

54 = 13.5 teeth 4

But fractional teeth are not possible; therefore TS should be either 13 or 14 and TA correspondingly 52 and 56. Trial 1:

Let

TA = 52

and

TS = 13

108

∴ ⇒

T A − TS 52 − 13 = = 19.5 teeth 2 4 TA = 56 and TS = 14 T −T 56 − 14 ∴ TP = A S = = 21teeth 2 4 TA = 56, TS = 14 and TP = 21 PCD of A = 56 × 4 = 224 mm

∴

Torque on L × ωL = Torque on S × ωS 5 Torque on L × ωL = 20 × = 100 N − m 1 Fixing torque on A = (TL – TS) = 100 – 20 = 80 N-m

∴ Trial 2:

Let

Also

TP =

This is impracticable

This is practicable

] D

Problem 4: The gear train shown in figure 4 A is used in an indexing mechanism of a milling machine. The drive is from gear wheels A and B to the bevel gear wheel D through the gear train. The following table gives the number of teeth on each gear.

C

Arm B H G

Gear A B C D E F E Number of F 72 72 60 30 28 24 teeth Diametral 08 08 12 12 08 08 pitch in mm How many revolutions does D makes for one Figure 4 revolution of A under the following situations: a. If A and B are having the same speed and same direction b. If A and B are having the same speed and opposite direction c. If A is making 72 rpm and B is at rest d. If A is making 72 rpm and B 36 rpm in the same direction Solution: Gear D is external to the epicyclic train and thus C and D constitute an ordinary train. Operation

Arm C (60)

E (28)

F (24)

A (72)

B (72)

G (28)

H (24)

109

Arm or C is fixed & wheel A is given +1 revolution Multiply by m (A rotates through m revolution) Add n revolutions to all elements (i)

−

28 7 =− 24 6

+1

-1

+1

28 7 = 24 6

7 m 6

+m

-m

+m

7 m 6

n+m

n-m

n+m

0

-1

0

-m

−

n

n-m

n−

7 m 6

For one revolution of A: n+m=1 (1) For A and B for same speed and direction: From (1) and (2): n = 1 and m = 0 ∴

(ii)

(iii)

(iv)

n+m = n–m

n+

7 m 6

(2)

If C or arm makes one revolution, then revolution made by D is given by: N D T C 60 = = =2 NC TD 30

∴ N D = 2 NC A and B same speed, opposite direction: (n + m) = - (n – m) (3) n = 0; m = 1 ∴ When C is fixed and A makes one revolution, D does not make any revolution. A is making 72 rpm: (n + m) = 72 B at rest (n – m) = 0 ⇒

n = m = 36 rpm 60 ∴ C makes 36 rpm and D makes 36 × = 72 rpm 30 A is making 72 rpm and B making 36 rpm (n + m) = 72 rpm and (n – m) = 36 rpm (n + (n – m)) = 72; ⇒ n = 54 60 ∴ D makes 54 × =108 rpm A2 30 P2

Problem 5: Figure 5 shows a compound epicyclic gear train, gears S1 and S2 being rigidly attached to the shaft Q. If the shaft P rotates at 1000 rpm clockwise, while the annular A2 is driven in counter clockwise direction at 500 rpm, determine the speed and direction of rotation of shaft Q. The number of teeth in the wheels are S1 = 24; S2 = 40; A1 = 100; A2 = 120.

A1 P1 P

Q S2 S1

Figure 5

Solution: Consider the gear train P A1 S1:

110

Arm P

Operation

S1 (24) +

Arm P is fixed & wheel A1 is given +1 revolution

0

Multiply by m (A1 rotates through m revolution)

0

Add n revolutions to all elements If A1 is fixed:

A1 (100)

n

+1

Operation

P 100 ×− 1 P1 24

OR

25 =− 6

Arm P is fixed & wheel A1 is given -1 revolution

25 m 6

+m

−

n+ m

n−

Add +1 revolutions to all elements

25 m 6

Arm P

A1 (100 )

S1 (24) −

0

-1

A1 P ×− 1 P1 S1

=+

A1 S1

0

-1

100 25 = 24 6

+1

0

25 31 + 1= 6 6

n+ m; gives n = - m

NP n 1 6 = = = N S1 n + 256 n 316 31 ∴

NP =

6 N S1 31

Now consider whole gear train: Operation A1 is fixed & wheel A2 is given +1 revolution Multiply by m (A1 rotates through m revolution) Add n revolutions to all elements

A1 (100) 0

A2 (120) +1

S1 (24), S2 (40) Arm P and Q P 120 + ×− 2 P2 40 =−3

6 31 18 =− 31 − 3×

18 m 31

0

+m

− 3m

−

n

n+ m

n − 3m

n−

When P makes 1000 rpm: and A2 makes – 500 rpm:

18 m = 1000 31 n+ m = -500 n−

18 m 31

(1) (2)

111

18 m = 1000 31 ( 31 × 1000) + ( 500 × 31) = − 49 m − 500 − m −

from (1) and (2):

∴

and n = 949 − 500 = 449 rpm NQ = n – 3 m = 449 – (3 × -949) = 3296 rpm

∴ Problem 6. An internal wheel B with 80 teeth is keyed to a shaft F. A fixed internal wheel C with 82 teeth is concentric with B. A Compound gears D-E meshed with the two internal wheels. D has 28 teeth and meshes with internal gear C while E meshes with B. The compound wheels revolve freely on pin which projects from a arm keyed to a shaft A co-axial with F. if the wheels have the same pitch and the shaft A makes 800 rpm, what is the speed of the shaft F? Sketch the arrangement. Data:

tB = 80;

m = − 949 rpm

C Arm

B

C E

E

D

D

F

B

A A B80 C82 D28 N A=800rpm

tC = 82;

D = 28;

NA = 800 rpm

Solution: The pitch circle radius is proportional to the number of teeth:

rC − rD = rB − rE tC − t D = t B − t E Operation

Compound Gear wheel

82 Arm − 28 B= (80) 80 − t E

Arm is fixed & Bt E = 26

E(26)

Multiply by m (B rotates through m revolution) Add n revolutions to all elements

D (28)

80 80 + + of teeth 26 on gear 26 E

0

+1

0

+m

n

40 40 m+n m+n m+n 13 13

is given ONE − number revolution (CW)

+

40 m 13

C (82)

+

40 m 13

+ +

80 28 × 26 82

40 14 × m 13 41

40 14 × m+n 13 41

112

Since the wheel C is fixed and the arm (shaft) A makes 800 rpm,

⇒ n = 800rpm 40 14 × m+n =0 13 41 40 14 × m + 800 = 0 13 41 m = −761.42rpm Speed of gear B = m + n = − 761.42 + 800 = 38.58rpm Speed of gear B = Speed of shaft F = 38.58rpm Problem 7: The fig shows an Epicyclic gear train. Wheel E is fixed and wheels C and D are integrally cast and mounted on the same pin. If arm A makes one revolution per sec (Counter clockwise) determine the speed and direction of rotation of the wheels B and F. Arm

D15

B20

E20

F30

C35

113

Solution: Data:

tB = 20;

tC = 35;

Operation

tD = 15;

tE = 20; tF = 30 NA = 1rps-(CCW) Compound Gear wheel

Arm B (20)

D (15) Arm is fixed & B is given ONE revolution (CW) Multiply by m (B rotates through m revolution) Add n revolutions to all elements

0

0

n

F (30)

C (35)

4 35 ×− 3 20 7 =+ 3 −

20 − 15

+1

20 − 15

+m

4 4 − m − m 3 3

m+n

E (20)

7 m 3

7 20 ×− 3 30

−

14 m 9

7 4 4 14 n− m n− m m+n n− m 3 3 3 9

Since the wheel E is fixed and the arm A makes 1 rps-CCW

⇒

n = −1rps

and

7 m −1 = 0 3

⇒

7 m+n = 0 3 3 m = = 0.429 7

Speed of gear B = m + n = 0.429 − 1 = −0.571rps (CCW ) Speed of gear F = n −

14 14 m = −1 − 0.429 = −1.667 (CCW ) 9 9

Problem 7: In the gear train shown, the wheel C is fixed, the gear B, is keyed to the input shaft and the gear F is keyed to the output shaft.

D60

Output Shaft

A

F32

C80

E30

B20

Input Shaft

114

The arm A, carrying the compound wheels D and E turns freely on the out put shaft. If the input speed is 1000 rpm (ccw) when seen from the right, determine the speed of the output shaft. The number of teeth on each gear is indicated in the figures. Find the output torque to keep the wheel C fixed if the input power is 7.5 kW. Solution: Data : tB = 20; tC = 80; tD = 60; tE = 30; tF = 32; NB = 1000 rpm (ccw) (input speed); P = 7.5 kW

Operation

Arm

B (20) Input

Compound Gear wheel D (60)

0

+1

20 1 = 60 3

Multiply by m (B rotates through m revolution)

0

m

1 m 3

Add n revolutions to all elements

n

m+n

Arm is fixed & B is given +1 revolution

C (80)

F (32)

1 30 ×− 3 32 5 − 16

E (30)

1 3

1 60 ×− 3 80 1 =− 4

1 m 3

1 − m 4

−

5 m 16

1 1 1 5 m+n m+n n− m n− m 4 3 3 16

Input shaft speed = 1000 rpm (ccw) i.e., gear B rotates – 1000 rpm

m + n = −1000 Gear C is fixed ;

1 n− m = 0 4

− 1000 − m − 0.25m = 0 1000 = −800 1.25 n = − 1000 + 800 = − 200 m=−

115

Speed of F = n −

5 m 16

5 = 50 16 Speed of the output shaft F = +50rpm (CW ) = − 200 + 800

2 × π × N BTB 60 2 × π × −1000 × TB 7.5 × 1000 = 60 7500 × 60 TB = − = −71.59 Nm 2 × π ×1000

Input power = P =

From the energy equation; TB N B + TF N F + TC N C = 0 Since C is fixed : N C = 0 TB N B + TF N F = 0 − 71.59 × 1000 + TF × 50 = 0 TF = +1431.8 Nm From the torque equation : TB + TF + TC = 0 − 71.59 + 1431.8 + TC = 0 ∴

TC = −1360.21Nm

A1

The Torque required to hold the wheel C = 1360.21 Nm in the same direction of wheel A2 Problem 8: Find the velocity ratio of two co- P1 axial shafts of the epicyclic gear train as shown in figure 6. S1 is the driver. The number of teeth on the gears are S1 = 40, A1 = 120, S2 = 30, A2 = 100 and the sun wheel S2 is fixed. Determine also the magnitude and direction of the torque required to fix S2, if a torque of 300 N-m is applied in a clockwise direction to S1

P2

S1

S2

116

Figure 6

Solution: Consider first the gear train S1, A1 and A2 for which A2 is the arm, in order to find the speed ratio of S1 to A2, when A1 is fixed. (a) Consider gear train S1, A1 and A2: Operation

A2 (100)

A1 (120)

0

+1

0

+m

− 3m

n

n+ m

n − 3m

A2 is fixed & wheel A1 is given +1 revolution Multiply by m (A1 rotates through m revolution) Add n revolutions to all elements

S1 (40) −

120 = −3 40

m = −n A1 is fixed: N S1 n + 3n = =4 N A2 n ∴ N S1 = 4 N A2 (b) Consider complete gear train: Operation

A1 (120)

A1 is fixed & wheel S2 is given +1 revolution Multiply by m (A1 rotates through m revolution) Add n revolutions to all elements S2 is fixed

⇒

0 0 n

A2 (100) −

30 3 =− 100 10 3 =− m 10 3 n− m 10

−

S1 (40)

S2 (30)

6 3 ×4=− 10 5

+1

−

6 m 5

+m

n−

6 m 5

n+ m

m=-n

6 n+ n N S1 5 = 11 × 10 = 22 = 3 N A2 5 13 13 n+ n 10 Input torque on S1 = TS1 = 300 N-m, in the direction of rotation.

TA 2

TS 2

∴ Resisting torque on A2;

ωS 1 TS 1

ωA 2

117

22 = 507.7 N − m 13 → opposite to directiojn of rotation

T A2 = 300 ×

∴ Referring to the figure: TS 2 = 507.7 − 300 = 207.7 N − m

(CW )

118

Subject: KINEMATICS OF MACHINES Topic: VELOCITY AND ACCELERATION Session – I •

Introduction

Kinematics deals with study of relative motion between the various parts of the machines. Kinematics does not involve study of forces. Thus motion leads study of displacement, velocity and acceleration of a part of the machine. Study of Motions of various parts of a machine is important for determining their velocities and accelerations at different moments. As dynamic forces are a function of acceleration and acceleration is a function of velocities, study of velocity and acceleration will be useful in the design of mechanism of a machine. The mechanism will be represented by a line diagram which is known as configuration diagram. The analysis can be carried out both by graphical method as well as analytical method. •

Some important Definitions

Displacement: All particles of a body move in parallel planes and travel by same distance is known, linear displacement and is denoted by ‘x’. A body rotating about a fired point in such a way that all particular move in circular path angular displacement and is denoted by ‘θ’. Velocity: Rate of change of displacement is velocity. Velocity can be linear velocity of angular velocity. Linear velocity is Rate of change of linear displacement= V =

dx dt

Angular velocity is Rate of change of angular displacement = ω =

dθ dt

Relation between linear velocity and angular velocity. x = rθ dx dθ = r dt dt

119

V = rω ω=

dθ dt

Acceleration: Rate of change of velocity f= Thirdly α =

dv d 2 x = 2 Linear Acceleration (Rate of change of linear velocity) dt dt dω d 2θ = 2 Angular Acceleration (Rate of change of angular velocity) dt dt

We also have, Absolute velocity: Velocity of a point with respect to a fixed point (zero velocity point).

A ω2 O2 Va = ω 2 x r Va = ω2 x O2 A Ex: Vao2 is absolute velocity. Relative velocity: Velocity of a point with respect to another point ‘x’ B A

3 4

2 O2

O4

Ex: Vba  Velocity of point B with respect to A Note: Capital letters are used for configuration diagram. Small letters are used for velocity vector diagram.

120

This is absolute velocity Velocity of point A with respect to O2 fixed point, zero velocity point. B A

3

Vba = or Vab Vba = or Vab Equal in magnitude but opposite in direction. B

O4

Vb  Absolute velocity is velocity of B with respect to O 4 (fixed point, zero velocity point) b

Vba

Vb O2, O4 Vab a Velocity vector diagram Vector O 2 a = Va= Absolute velocity Vector ab = Vab ba = Va

Relative velocity

Vab is equal magnitude with Vba but is apposite in direction.

121

Vector O 4 b = Vb absolute velocity. To illustrate the difference between absolute velocity and relative velocity. Let, us consider a simple situation. A link AB moving in a vertical plane such that the link is inclined at 30 o to the horizontal with point A is moving horizontally at 4 m/s and point B moving vertically upwards. Find velocity of B. Va = 4 m/s ab Absolute velocity

Horizontal direction (known in magnitude and directors)

Vb = ?

Absolute velocity

ab

Vertical direction (known in directors only)

A

30o

Va

O

4 m/s

a Vab

Vb

C

B Vba

Velocity of B with respect to A is equal in magnitude to velocity of A with respect to B but opposite in direction. •

Relative Velocity Equation y

Rigid body

A ya

R O

O4

θ xA

x

122

Fig. 1 Point O is fixed and End A is a point on rigid body.

Rotation of a rigid link about a fixed centre. Consider rigid link rotating about a fixed centre O, as shown in figure. The distance between O and A is R and OA makes and angle ‘θ’ with x-axis next link xA = R cos θ, yA = R sin θ. Differentiating xA with respect to time gives velocity. d xA dθ = R ( − sin θ) dt dt = - Rω sin θ

Similarly,

dy A dθ = R ( − cos θ ) dt dt = - Rω cos θ

Let,

d yA

d xA = VAx dt ω=

dt

= VAy

dθ = angular velocity of OA dt

∴ VAx = - Rω sin θ VAy = - Rω cos θ ∴ Total velocity of point A is given by VA =

( − Rω sin θ ) 2 + ( − Rω cos θ ) 2

VA = Rω •

Relative Velocity Equation of Two Points on a Rigid link 123

Rigid body

B

yB

R sin θ A

yA

R cos θ x

xB

xA

Fig. 2 Points A and B are located on rigid body From Fig. 2 xB = xA + R cos θ

yB = yA + R sin θ

Differentiating xB and yB with respect to time we get, d xB d dθ = VBx = xA + R ( − sin θ) dt dt dt =

Similarly,

d yB dt =

Similarly,

= VBy = d yA dt

d xA + Rω sin θ = VAx − Rω sin θ dt d yA dt

+ R ( cos θ )

dθ dt

+ Rω cos θ = VAy − Rω cos θ

VA = VAx

VAy = Total velocity of point A

VB = VBx

VBy = Total velocity of point B

= VAx

(Rω sin θ)

VAy

Rω cos θ

124

(Rω sin θ + R ωcos θ)

= ( VAx

VAy )

= ( VAx

VAy ) VA Similarly, ( R ωsin θ + Rω cos θ) = Rω

∴VB = VA

Rω = VA

VBA

∴VBA = VB – VA Velocity analysis of any mechanism can be carried out by various methods. 1. By graphical method 2. By relative velocity method 3. By instantaneous method •

By Graphical Method The following points are to be considered while solving problems by this method. 1. Draw the configuration design to a suitable scale. 2. Locate all fixed point in a mechanism as a common point in velocity diagram. 3. Choose a suitable scale for the vector diagram velocity. 4. The velocity vector of each rotating link is ⊥r to the link. 5. Velocity of each link in mechanism has both magnitude and direction. Start from a point whose magnitude and direction is known. 6. The points of the velocity diagram are indicated by small letters.

To explain the method let us take a few specific examples. 1. Four – Bar Mechanism: In a four bar chain ABCD link AD is fixed and in 15 cm long. The crank AB is 4 cm long rotates at 180 rpmω (cw) while link CD rotates about D is 8 cm long BC = AD and | BAD = 60o. Find angular velocity of link CD. C 15 cm B 8 cm wBA 60o A

15 cm

D

125

Configuration Diagram Velocity vector diagram Vb = ωr = ωba x AB =

2πx 120 x 4 = 50.24 cm/sec 60

Choose a suitable scale 1 cm = 20 m/s = ab ⊥r to CD

c

a, d

Vcb

⊥r to BC ⊥r to AB b

Vcb = bc Vc = dc = 38 cm/s = Vcd We know that V =ω R Vcd = ωCD x CD WcD =

Vcd 38 = = 4.75 rad/s (cw) CD 8

2. Slider Crank Mechanism: In a crank and slotted lover mechanism crank rotates of 300 rpm in a counter clockwise direction. Find (i)

Angular velocity of connecting rod and

(ii)

Velocity of slider.

126

A 60 mm

150 mm

45o

B Configuration diagram

Step 1: Determine the magnitude and velocity of point A with respect to 0, VA = ωO1A x O2A =

2π x 300 x 60 60

= 600 π mm/sec Step 2: Choose a suitable scale to draw velocity vector diagram. a

Va

⊥r to AB

⊥r to OA

b

O

Along sides B

Velocity vector diagram Vab = ab =1300mm/sec ωba =

Vba 1300 = = 8.66 rad/sec BA 150

Vb = ob velocity of slider Note: Velocity of slider is along the line of sliding. 3. Shaper Mechanism: In a crank and slotted lever mechanisms crank O2A rotates at ω rad/sec in CCW direction. Determine the velocity of slider.

127

6

D

Scale 1 cm = ……x…. m

5

C ω O2

3 B

2

4

O1

Configuration diagram Scale 1 cm = ……x…. m/s

a VBA

c

VAO2 = VA

b VBO1 VDC d

O1O2 Velocity vector diagram

Va = ω2 x O2A O1 b O 1 c = O1 B O1C To locate point C O C ∴ O1c = O1 b  1   O1 B  To Determine Velocity of Rubbing

128

Two links of a mechanism having turning point will be connected by pins. When the links are motion they rub against pin surface. The velocity of rubbing of pins depends on the angular velocity of links relative to each other as well as direction. For example: In a four bar mechanism we have pins at points A, B, C and D. ∴ Vra = ωab x ratios of pin A (rpa) + sign is used ωab is CW and Wbc is CCW i.e. when angular velocities are in opposite directions use + sign when angular velocities are in some directions use - ve sign. VrC = (ωbc + ωcd) radius r VrD = ω cd rpd Problems on velocity by velocity vector method (Graphical solutions) Problem 1: In a four bar mechanism, the dimensions of the links are as given below: AB = 50 mm, CD = 56 mm

BC = 66 mm and

AD = 100 mm

o At a given instant when | DAB = 60 the angular velocity of link AB is 10.5 rad/sec in CCW direction.

Determine, i)

Velocity of point C

ii)

Velocity of point E on link BC when BE = 40 mm

iii)

The angular velocity of link BC and CD

iv)

The velocity of an offset point F on link BC, if BF = 45 mm, CF = 30 mm and BCF is read clockwise.

v)

The velocity of an offset point G on link CD, if CG = 24 mm, DG = 44 mm and DCG is read clockwise.

vi)

The velocity of rubbing of pins A, B, C and D. The ratio of the pins are 30 mm, 40 mm, 25 mm and 35 mm respectively.

129

Solution: Step -1: Construct the configuration diagram selecting a suitable scale. Scale: 1 cm = 20 mm C G

B F

60o A

D

Step – 2: Given the angular velocity of link AB and its direction of rotation determine velocity of point with respect to A (A is fixed hence, it is zero velocity point). Vba = ωBA x BA = 10.5 x 0.05 = 0.525 m/s Step – 3: To draw velocity vector diagram choose a suitable scale, say 1 cm = 0.2 m/s. •

First locate zero velocity points.

•

Draw a line ⊥r to link AB in the direction of rotation of link AB (CCW) equal to 0.525 m/s. b Vba = 0.525 m/s e, g a, d

f C

Ved

•

From b draw a line ⊥r to BC and from d. Draw d line ⊥r to CD to interest at C.

•

Vcb is given vector bc

Vbc = 0.44 m/s

•

Vcd is given vector dc

Vcd = 0.39 m/s

Step – 4: To determine velocity of point E (Absolute velocity) on link BC, first locate the position of point E on velocity vector diagram. This can be done by taking corresponding ratios of lengths of links to vector distance i.e.

130

be BE = bc BC

∴ be =

BE 0.04 x Vcb = x 0.44 = 0.24 m/s BC 0.066

Join e on velocity vector diagram to zero velocity points a, d / vector de = Ve = 0.415 m/s. Step 5: To determine angular velocity of links BC and CD, we know Vbc and Vcd. ∴ Vbc =ω BC x BC ∴ WBC = Similarly,

Vbc 0.44 = = 6.6 r / s . (cw) BC 0.066

Vcd = WCD x CD ∴ WCD =

Vcd 0.39 = = 6.96 r / s (CCW) CD 0.056

Step – 6: To determine velocity of an offset point F •

Draw a line ⊥r to CF from C on velocity vector diagram.

•

Draw a line ⊥r to BF from b on velocity vector diagram to intersect the previously drawn line at ‘f’.

•

From the point f to zero velocity point a, d and measure vector fa/fd to get V f = 0.495 m/s.

Step – 7: To determine velocity of an offset point. •

Draw a line ⊥r to GC from C on velocity vector diagram.

•

Draw a line ⊥r to DG from d on velocity vector diagram to intersect previously drawn line at g.

•

Measure vector dg to get velocity of point G. Vg = dg = 0.305 m / s

Step – 8: To determine rubbing velocity at pins •

Rubbing velocity at pin A will be

131

Vpa = ωab x r of pin A Vpa = 10.5 x 0.03 = 0.315 m/s •

Rubbing velocity at pin B will be Vpb = (ωab + ωcb) x rad of point at B. [ωab CCW and ωcbCW] Vpb = (10.5 + 6.6) x 0.04 = 0.684 m/s.

•

Rubbing velocity at point C will be = 6.96 x 0.035 = 0.244 m/s

Problem 2: In a slider crank mechanism the crank is 200 mm long and rotates at 40 rad/sec in a CCW direction. The length of the connecting rod is 800 mm. When the crank turns through 60o from Inner-dead centre. Determine, i)

The velocity of the slider

ii)

Velocity of point E located at a distance of 200 mm on the connecting rod extended.

iii)

The position and velocity of point F on the connecting rod having the least absolute velocity.

iv)

The angular velocity of connecting rod.

v)

The velocity of rubbing of pins of crank shaft, crank and cross head having pins diameters 80,60 and 100 mm respectively.

Solution: Step 1: Draw the configuration diagram by selecting a suitable scale. E

A F 45o O

B G

Va = Woa x OA

132

Va = 40 x 0.2 Va = 8 m/s Step 2: Choose a suitable scale for velocity vector diagram and draw the velocity vector diagram. •

Mark zero velocity point o, g.

•

Draw oa ⊥r to link OA equal to 8 m/s e a

Scale: 1 cm = 2 m/s

f b

o, g

•

From a draw a line ⊥r to AB and from o, g draw a horizontal line (representing the line of motion of slider B) to Xseet the previously drawn line at b.

•

ab give Vba=4.8 m/sec

BE x ab AB mark the point e on extension of vector ba. Join e to o, g. ge will give velocity of point E. Step – 3: To mark point ‘e’ since ‘E’ is on the extension of link AB drawn be =

Ve = ge =8.4 m/sec Step 4: To mark point F on link AB such that this has least velocity (absolute). Draw a line ⊥r to ab passing through o, g to cut the vector ab at f. From f to o, g. gf will have the least absolute velocity. •

To mark the position of F on link AB. Find BF by using the relation. fb ab = BF AB

133

BF =

fb x AB =200mm ab

Step – 5: To determine the angular velocity of connecting rod. We know that Vab = ωab x AB ∴ ω ab =

Vab = 6 rad/sec AB

Step – 6: To determine velocity of rubbing of pins. •

Vpcrankshaft = ωao x radius of crankshaft pin = 8 x 0.08 = 0.64 m/s

•

•

VPcrank pin = (ωab + ωoa) rcrank pin= (6 +8)0.06 =0.84 m/sec

•

VP cross head = ωab x rcross head = 6 x 0.1 = 0.6 m/sec

Problem 3: A quick return mechanism of crank and slotted lever type shaping machine is shown in Fig. the dimensions of various links are as follows. O1O2 = 800 mm, O1B = 300 mm, O2D = 1300 mm and DR = 400 mm

The crank O1B makes an angle of 45o with the vertical and relates at 40 rpm in the CCW direction. Find:

•

i)

Velocity of the Ram R, velocity of cutting tool, and

ii)

Angular velocity of link O2D.

Solution:

Step 1: Draw the configuration diagram.

134

R Tool

R

200

D

D

B on orank, O, B O1

C on O2D

B 2

45o

C

O1

O2 O2

Step 2: Determine velocity of point B. Vb = ωO1B x O1B ωO1B =

2πN O1B 2π x 40 = = 4.18 rad / sec 60 60

Vb = 4.18 x 0.3 = 1.254 m/sec Step 3: Draw velocity vector diagram. Choose a suitable scale 1 cm = 0.3 m/sec b d c r

O1O2

o Draw O1b ⊥r to link O1B equal to 1.254 m/s.

135

o From b draw a line along the line of O2B and from O1O2 draw a line ⊥r to O2B. This intersects at c bc will measure velocity of sliding of slider and O 2 C will measure the velocity of C on link O2C. o Since point D is on the extension of link O2C measure O 2 d such that O 2d = O 2C

O2D . O 2 d will give velocity of point D. O 2C

o From d draw a line ⊥r to link DR and from O1O2. Draw a line along the line of stroke of Ram R (horizontal), These two lines will intersect at point r O 2 r will give the velocity of Ram R. o To determine the angular velocity of link O2D determine Vd = O 2 d . We know that Vd = ωO2D x O2D.

∴ ωO2d =

O 2d r/s O 2D

•

Problem 4: Figure below shows a toggle mechanisms in which the crank OA rotates at 120 rpm. Find the velocity and acceleration of the slider D.

•

Solution:

136

120

All the dimensions in mm A

45o 40 190

100 135

B

120

D

Configuration Diagram Step 1: Draw the configuration diagram choosing a suitable scal. Step 2: Determine velocity of point A with respect to O. Vao = ωOA x OA Vao =

2π x 120 = 0.4 = 5.024 m / s 60

Step 3: Draw the velocity vector diagram. o Choose a suitable scale o Mark zero velocity points O,q o Draw vector oa ⊥r to link OA and magnitude = 5.024 m/s. a

b

O,q

D

Velocity vector diagram o From a draw a line ⊥r to AB and from q draw a line ⊥r to QB to intersect at b.

137

ab = Vba and qb = Vbq . o Draw a line ⊥r to BD from b from q draw a line along the slide to intersect at d. dq = Vd (slider velocity) •

Problem 5: A whitworth quick return mechanism shown in figure has the following dimensions of the links.

The crank rotates at an angular velocity of 2.5 r/s at the moment when crank makes an angle of 45o with vertical. Calculate OP (crank) = 240 mm a) the velocity of the Ram S OA = 150 mm b) the velocity of slider P on the slotted level AR = 165 mm c) the angular velocity of the link RS. RS = 430 mm •

Solution:

Step 1: To draw configuration diagram to a suitable scale. R

S

A

O o

45

P on slider Q on BA B

Configuration Diagram Step 2: To determine the absolute velocity of point P. VP = ωOP x OP

138

Vao =

2π x 240 x 0.24 = 0.6 m / s 60

Step 3: Draw the velocity vector diagram by choosing a suitable scale. P 0.6 m

q

S O, a, g r Velocity vector diagram o Draw op ⊥r link OP = 0.6 m. o From O, a, g draw a line ⊥r to AP/AQ and from P draw a line along AP to intersect previously draw, line at q. Pq = Velocity of sliding. aq = Velocity of Q with respect to A. Vqa = aq = o Angular velocity of link RS = ω RS = sr rad/sec SR •

Problem 6: A toggle mechanism is shown in figure along with the diagrams of the links in mm. find the velocities of the points B and C and the angular velocities of links AB, BQ and BC. The crank rotates at 50 rpm in the clockwise direction.

139

C Q

100

B 140 OA = 30 AB = 80

A

BQ = 100

50 rpm

BC = 100 All dimensions are in mm •

O

Solution

Step 1: Draw the configuration diagram to a suitable scale. Step 2: Calculate the magnitude of velocity of A with respect to O. Va = ωOA x OA  2π x 50   x 0.03 = 0.05 π m / s = 0.1507 m / s Va =   60  b

a

c

O, q

Vector velocity diagram Step 3: Draw the velocity vector diagram by choosing a suitable scale. o Draw Oa ⊥r to link OA = 0.15 m/s o From a draw a link ⊥r to AB and from O, q draw a link ⊥r to BQ to intersect at b.

140

ab = Vba = ωab =

and qb = Vb = 0.13 m / s

ab qb = 0.74 r / s (ccw ) ωbq =1.3 r / s (ccw ) AB aB

o From b draw a line ⊥r to Be and from O, q these two lines intersect at C. OC = VC = 0.106 m / s bC = VCb = ω BC = •

bc = 1.33 r / s (ccw ) BC

Problem 7: The mechanism of a stone crusher has the dimensions as shown in figure in mm. If crank rotates at 120 rpm CW. Find the velocity of point K when crank OA is inclined at 30o to the horizontal. What will be the torque required at the crank to overcome a horizontal force of 40 kN at K. 500 60o

A 100

M hz

h2 100

200 400

K

600

600

320

360 B

200

D

C

Configuration diagram •

Solution:

Step 1: Draw the configuration diagram to a suitable scale. Step 2: Given speed of crank OA determine velocity of A with respect to ‘o’.  2π x 120   x 0.1 = 1.26 m / s Va = ωOA x OA =  60  

141

Vk(hz)

d

o, q, m

k

b

a

c Velocity vector diagram

Step 3: Draw the velocity vector diagram by selecting a suitable scale. o Draw Oa ⊥r to link OA = 1.26 m/s o From a draw a link ⊥r to AB and from q draw a link ⊥r to BQ to intersect at b. o From b draw a line ⊥r to BC and from a, draw a line ⊥r to AC to intersect at c. o From c draw a line ⊥r to CD and from m draw a line ⊥r to MD to intersect at d. o From d draw a line ⊥r to KD and from m draw a line ⊥r to KM to x intersect the previously drawn line at k. o Since we have to determine the torque required at OA to overcome a horizontal force of 40 kN at K. Draw a the horizontal line from o, q, m and c line ⊥r to this line from k. ∴ ( ωT ) I = ( ωT ) O P

V = ωR

P

T=FxP

F=

T r

∴ ωOA TOA = Fk Vk horizontal ∴ TOA = TOA =

Fk Vk ( hz ) ω OA 40000 X 0.45 = 12.6

N-m

142

•

Problem 8: In the mechanism shown in figure link OA = 320 mm, AC = 680 mm and OQ = 650 mm.

Determine, i)

The angular velocity of the cylinder

ii)

The sliding velocity of the plunger

iii)

The absolute velocity of the plunger When the crank OA rotates at 20 rad/sec clockwise.

•

Solution:

Step 1: Draw the configuration diagram. A B on AR (point on AR below Q) C

60o O

R

Step 2: Draw the velocity vector diagram o Determine velocity of point A with respect to O. Va = ωOA x OA = 20 x 0.32 = 6.4 m/s o Select a suitable scale to draw the velocity vector diagram. o Mark the zero velocity point. Draw vector oa ⊥r to link OA equal to 6.4 m/s. a

O, q

c b

o From a draw a line ⊥r to AB and from o, q, draw a line perpendicular to AB.

143

o To mark point c on ab We know that ∴ ac =

ab ac

=

AB AC

ab x AC = AB

o Mark point c on ab and joint this to zero velocity point. o Angular velocity of cylinder will be. ωab =

Vab = 5.61 rad/sec (cω) AB

o Studying velocity of player will be qb = 4.1 m/s

o Absolute velocity of plunger =

•

OC qc

= 4.22 m/s

Problem 9: In a swiveling joint mechanism shown in figure link AB is the driving crank which rotates at 300 rpm clockwise. The length of the various links are: AB = 650 mm

Determine, i)

The velocity of slider block S

AB = 100 mm

ii)

The angular velocity of link EF

BC = 800 mm

iii)

The velocity of link EF in the swivel block.

DC = 250 mm BE = CF EF = 400 mm

•

Solution:

Step 1: Draw the configuration diagram.

OF = 240 mm FS = 400 mm

144

F

400

S O

P G

400 300

E

B 45o A

D

Step 2: Determine the velocity of point B with respect to A. Vb = ωBA x BA Vb =

2π x 300 x 0.1 = 3.14 m/s 60

Step 3: Draw the velocity vector diagram choosing a suitable scale. o Mark zero velocity point a, d, o, g. f

b P

S

a, d, o, g c Velocity vector diagram

o From ‘a’ draw a line ⊥r to AB and equal to 3.14 m/s. o From ‘b’ draw a line ⊥r to DC to intersect at C. 145

o Mark a point ‘e’ on vector bc such that be = bc x

BE BC

o From ‘e’ draw a line ⊥r to PE and from ‘a,d’ draw a line along PE to intersect at P. o Extend the vector ep to ef such that ef = ef x EF EP

o From ‘f’ draw a line ⊥r to Sf and from zero velocity point draw a line along the slider ‘S’ to intersect the previously drawn line at S. o

Velocity of slider gS = 2.6 m / s . Angular Velocity of link EF.

o Velocity of link F in the swivel block = OP = 1.85 m / s . •

Problem 10: Figure shows two wheels 2 and 4 which rolls on a fixed link 1. The angular uniform velocity of wheel is 2 is 10 rod/sec. Determine the angular velocity of links 3 and 4, and also the relative velocity of point D with respect to point E. φ 50 mm φ 40 mm B

3 60 mm

o

30

ω2

A 2 G

•

C

D 4 F

Solution:

Step 1: Draw the configuration diagram. Step 2: Given ω2 = 10 rad/sec. Calculate velocity of B with respect to G. Vb = ω2 x BG Vb = 10 x 43 = 430 mm/sec.

146

Step 3: Draw the velocity vector diagram by choosing a suitable scale. B C

30o E

D

F

G 50 mm

Redrawn configuration diagram •

Velocity vector diagram c b

e

d

g, f

o Draw gb = 0.43 m/s ⊥r to BG. o From b draw a line ⊥r to BC and from ‘f’ draw a line ⊥r to CF to intersect at C. o From b draw a line ⊥r to BE and from g, f draw a line ⊥r to GE to intersect at e. o From c draw a line ⊥r to CD and from f draw a line ⊥r to FD to intersect at d. •

Problem 11: For the mechanism shown in figure link 2 rotates at constant angular velocity of 1 rad/sec construct the velocity polygon and determine. i)

Velocity of point D.

ii)

Angular velocity of link BD.

147

iii) •

Velocity of slider C.

Solution:

Step 1: Draw configuration diagram. D

6

O2 = 50.8 mm AB = 102 mm

O6

BD = 102 mm DO6 = 102 mm

5 102 mm

AC = 203 mm

A 3

B

45o

C

4

O2 Step 2: Determine velocity of A with respect to O2. Vb = ω2 x O2A Vb = 1 x 50.8 = 50.8 mm/sec. Step 3: Draw the velocity vector diagram, locate zero velocity points O2O6. d Vd

a Udb b

O2O6

C

o From O2, O6 draw a line ⊥r to O2A in the direction of rotation equal to 50.8 mm/sec.

148

o From a draw a line ⊥r to Ac and from O2, O6 draw a line along the line of stocks of c to intersect the previously drawn line at c. o Mark point b on vector ac such that ab = ab x AB AC

o From b draw a line ⊥r to BD and from O2, O6 draw a line ⊥r to O6D to intersect at d. Step 4:

Vd = O 6 d = 32 mm/sec ωbd =

bd = BD

Vc = O 2 C = ADDITIONAL PROBLEMS FOR PRACTICE •

Problem 1: In a slider crank mechanism shown in offset by a perpendicular distance of 50 mm from the centre C. AB and BC are 750 mm and 200 mm long respectively crank BC is rotating eω at a uniform speed of 200 rpm. Draw the velocity vector diagram and determine velocity of slider A and angular velocity of link AB. B A 50 mm C

•

Problem 2: For the mechanism shown in figure determine the velocities at points C, E and F and the angular velocities of links, BC, CDE and EF.

149

C

All dimensions are in mm

120

E

B 100

100 60

120

150

o

rpm A

•

F

120

D

50

The crank op of a crank and slotted lever mechanism shown in figure rotates at 100 rpm in the CCW direction. Various lengths of the links are OP = 90 mm, OA = 300 mm, AR = 480 mm and RS = 330 mm. The slider moves along an axis perpendicular to ⊥r AO and in 120 mm from O. Determine the velocity of the slider when | AOP is 135o and also mention the maximum velocity of slider.

45o

B

O C

D A

•

Problem 4: Find the velocity of link 4 of the scotch yoke mechanism shown in figure. The angular speed of link 2 is 200 rad/sec CCW, link O2P = 40 mm.

4

P 3 2 Q on link 4 45o

150

•

Problem 5: In the mechanism shown in figure link AB rotates uniformly in Cω direction at 240 rpm. Determine the linear velocity of B and angular velocity of EF. E

C

B

AB = 160 mm BC = 160 mm CD = 100 mm

A

o

45

AD = 200 mm EF = 200 mm

100 mm

F

CE = 40 mm

151

II Method • Instantaneous Method To explain instantaneous centre let us consider a plane body P having a non-linear motion relative to another body q consider two points A and B on body P having velocities as Va and Vb respectively in the direction shown.

Va B

A I

P

Vb q

Fig. 1 If a line is drawn ⊥r to Va, at A the body can be imagined to rotate about some point on the line. Thirdly, centre of rotation of the body also lies on a line ⊥r to the direction of Vb at B. If the intersection of the two lines is at I, the body P will be rotating about I at that instant. The point I is known as the instantaneous centre of rotation for the body P. The position of instantaneous centre changes with the motion of the body.

Va

A

B

P Vb q

I

Fig. 2 In case of the ⊥r lines drawn from A and B meet outside the body P as shown in Fig 2.

A

Va

B

Vb

I at ∞ Fig. 3 If the direction of Va and Vb are parallel to the ⊥r at A and B met at ∞. This is the case when the body has linear motion.

152

153

Subject: KINEMATICS OF MACHINES Topic: VELOCITY AND ACCELERATION Session – I •

Introduction

Kinematics deals with study of relative motion between the various parts of the machines. Kinematics does not involve study of forces. Thus motion leads study of displacement, velocity and acceleration of a part of the machine. Study of Motions of various parts of a machine is important for determining their velocities and accelerations at different moments. As dynamic forces are a function of acceleration and acceleration is a function of velocities, study of velocity and acceleration will be useful in the design of mechanism of a machine. The mechanism will be represented by a line diagram which is known as configuration diagram. The analysis can be carried out both by graphical method as well as analytical method. •

Some important Definitions

Displacement: All particles of a body move in parallel planes and travel by same distance is known, linear displacement and is denoted by ‘x’. A body rotating about a fired point in such a way that all particular move in circular path angular displacement and is denoted by ‘θ’. Velocity: Rate of change of displacement is velocity. Velocity can be linear velocity of angular velocity. Linear velocity is Rate of change of linear displacement= V =

dx dt

Angular velocity is Rate of change of angular displacement = ω =

dθ dt

Relation between linear velocity and angular velocity. x = rθ dx dθ = r dt dt V = rω

154

ω=

dθ dt

Acceleration: Rate of change of velocity f= Thirdly α =

dv d 2 x = 2 Linear Acceleration (Rate of change of linear velocity) dt dt dω d 2θ = 2 Angular Acceleration (Rate of change of angular velocity) dt dt

We also have, Absolute velocity: Velocity of a point with respect to a fixed point (zero velocity point).

A ω2 O2 Va = ω 2 x r Va = ω2 x O2 A Ex: Vao2 is absolute velocity. Relative velocity: Velocity of a point with respect to another point ‘x’ B A

3 4

2 O2

O4

Ex: Vba  Velocity of point B with respect to A Note: Capital letters are used for configuration diagram. Small letters are used for velocity vector diagram. This is absolute velocity 155

Velocity of point A with respect to O2 fixed point, zero velocity point. B A

3

Vba = or Vab Vba = or Vab Equal in magnitude but opposite in direction. B

O4

Vb  Absolute velocity is velocity of B with respect to O 4 (fixed point, zero velocity point) b

Vba

Vb O2, O4 Vab a Velocity vector diagram Vector O 2 a = Va= Absolute velocity Vector ab = Vab ba = Va

Relative velocity

Vab is equal magnitude with Vba but is apposite in direction. Vector O 4 b = Vb absolute velocity.

156

To illustrate the difference between absolute velocity and relative velocity. Let, us consider a simple situation. A link AB moving in a vertical plane such that the link is inclined at 30 o to the horizontal with point A is moving horizontally at 4 m/s and point B moving vertically upwards. Find velocity of B.

157

Va = 4 m/s ab Absolute velocity

Horizontal direction (known in magnitude and directors)

Vb = ?

Absolute velocity

ab

Vertical direction (known in directors only)

A

o

Va

O

4 m/s

a Vab

30

Vb

C

B Vba

Velocity of B with respect to A is equal in magnitude to velocity of A with respect to B but opposite in direction. •

Relative Velocity Equation y

Rigid body

A ya

R O

O4

θ xA

x

Fig. 1 Point O is fixed and End A is a point on rigid body.

Rotation of a rigid link about a fixed centre. Consider rigid link rotating about a fixed centre O, as shown in figure. The distance between O and A is R and OA makes and angle ‘θ’ with x-axis next link xA = R cos θ, yA = R sin θ. 158

Differentiating xA with respect to time gives velocity. d xA dθ = R ( − sin θ) dt dt = - Rω sin θ

Similarly,

dy A dθ = R ( − cos θ ) dt dt = - Rω cos θ

Let,

d yA

d xA = VAx dt ω=

dt

= VAy

dθ = angular velocity of OA dt

∴ VAx = - Rω sin θ VAy = - Rω cos θ ∴ Total velocity of point A is given by VA =

( − Rω sin θ ) 2 + ( − Rω cos θ ) 2

VA = Rω •

Relative Velocity Equation of Two Points on a Rigid link

159

Rigid body

B

yB

R sin θ A

yA

R cos θ x

xB

xA

Fig. 2 Points A and B are located on rigid body From Fig. 2 xB = xA + R cos θ

yB = yA + R sin θ

Differentiating xB and yB with respect to time we get, d xB d dθ = VBx = xA + R ( − sin θ) dt dt dt =

Similarly,

d yB dt =

Similarly,

= VBy = d yA dt

d xA + Rω sin θ = VAx − Rω sin θ dt d yA dt

+ R ( cos θ )

dθ dt

+ Rω cos θ = VAy − Rω cos θ

VA = VAx

VAy = Total velocity of point A

VB = VBx

VBy = Total velocity of point B

= VAx

(Rω sin θ)

VAy

Rω cos θ

160

(Rω sin θ + R ωcos θ)

= ( VAx

VAy )

= ( VAx

VAy ) VA Similarly, ( R ωsin θ + Rω cos θ) = Rω

∴VB = VA

Rω = VA

VBA

∴VBA = VB – VA Velocity analysis of any mechanism can be carried out by various methods. 4. By graphical method 5. By relative velocity method 6. By instantaneous method

•

By Graphical Method The following points are to be considered while solving problems by this method. 7. Draw the configuration design to a suitable scale. 8. Locate all fixed point in a mechanism as a common point in velocity diagram. 9. Choose a suitable scale for the vector diagram velocity. 10. The velocity vector of each rotating link is ⊥r to the link. 11. Velocity of each link in mechanism has both magnitude and direction. Start from a point whose magnitude and direction is known. 12. The points of the velocity diagram are indicated by small letters.

To explain the method let us take a few specific examples. 4. Four – Bar Mechanism: In a four bar chain ABCD link AD is fixed and in 15 cm long. The crank AB is 4 cm long rotates at 180 rpm (cw) while link CD rotates about D is 8 cm long BC = AD and | BAD = 60o. Find angular velocity of link CD.

161

C 15 cm B 8 cm ωBA 60o A

D

15 cm Configuration Diagram

Velocity vector diagram Vb = ωr = ωba x AB =

2πx 120 x 4 = 50.24 cm/sec 60

Choose a suitable scale 1 cm = 20 m/s = ab ⊥r to CD

c

a, d

Vcb

⊥r to BC ⊥r to AB b

Vcb = bc Vc = dc = 38 cm/sec = Vcd We know that V =ω R Vcd = ωCD x CD ωcD =

Vcd 38 = = 4.75 rad/sec (cw) CD 8

5. Slider Crank Mechanism: 162

In a crank and slotted lever mechanism crank rotates of 300 rpm in a counter clockwise direction. Find (iii)

Angular velocity of connecting rod and

(iv)

Velocity of slider. A 60 mm

150 mm

45o

B Configuration diagram

Step 1: Determine the magnitude and velocity of point A with respect to 0, VA = ωO1A x O2A =

2π x 300 x 60 60

= 600 π mm/sec Step 2: Choose a suitable scale to draw velocity vector diagram. a

Va

⊥r to AB

⊥r to OA

b

O

Along sides B

Velocity vector diagram Vab = ab =1300mm/sec ωba =

Vba 1300 = = 8.66 rad/sec BA 150

Vb = ob velocity of slider Note: Velocity of slider is along the line of sliding.

163

6. Shaper Mechanism: In a crank and slotted lever mechanisms crank O2A rotates at ω rad/sec in CCW direction. Determine the velocity of slider. 6

D

Scale 1 cm = ……x…. m

5

C ω O2

3 B

2

4

O1

Configuration diagram Scale 1 cm = ……x…. m/s

a VBA

c

VAO2 = VA

b VBO1 VDC d

O1O2 Velocity vector diagram

Va = ω2 x O2A O1 b O 1 c = O1 B O1C To locate point C O C ∴ O1c = O1 b  1   O1 B  164

To Determine Velocity of Rubbing Two links of a mechanism having turning point will be connected by pins. When the links are motion they rub against pin surface. The velocity of rubbing of pins depends on the angular velocity of links relative to each other as well as direction. For example: In a four bar mechanism we have pins at points A, B, C and D. ∴ Vra = ωab x ratios of pin A (rpa) + sign is used ωab is CW and Wbc is CCW i.e. when angular velocities are in opposite directions use + sign when angular velocities are in some directions use - ve sign. Vrb = (ω ab + ω bc) radius rpb VrC = (ω bc + ω cd) radius rpc VrD = ω cd rpd

Problems on velocity by velocity vector method (Graphical solutions) Problem 1: In a four bar mechanism, the dimensions of the links are as given below: AB = 50 mm, CD = 56 mm

BC = 66 mm and

AD = 100 mm

o At a given instant when | DAB = 60 the angular velocity of link AB is 10.5 rad/sec in CCW direction.

Determine, i)

Velocity of point C

ii)

Velocity of point E on link BC when BE = 40 mm

iii)

The angular velocity of link BC and CD

iv)

The velocity of an offset point F on link BC, if BF = 45 mm, CF = 30 mm and BCF is read clockwise.

165

v)

The velocity of an offset point G on link CD, if CG = 24 mm, DG = 44 mm and DCG is read clockwise.

vi)

The velocity of rubbing of pins A, B, C and D. The ratio of the pins are 30 mm, 40 mm, 25 mm and 35 mm respectively.

Solution: Step -1: Construct the configuration diagram selecting a suitable scale. Scale: 1 cm = 20 mm C G

B F

60o A

D

Step – 2: Given the angular velocity of link AB and its direction of rotation determine velocity of point with respect to A (A is fixed hence, it is zero velocity point). Vba = ωBA x BA = 10.5 x 0.05 = 0.525 m/s Step – 3: To draw velocity vector diagram choose a suitable scale, say 1 cm = 0.2 m/s. •

First locate zero velocity points.

•

Draw a line ⊥r to link AB in the direction of rotation of link AB (CCW) equal to 0.525 m/s. b Vba = 0.525 m/s e, g a, d

f C

Ved

•

From b draw a line ⊥r to BC and from d. Draw d line ⊥r to CD to interest at C.

•

Vcb is given vector bc

Vbc = 0.44 m/s

166

•

Vcd is given vector dc

Vcd = 0.39 m/s

Step – 4: To determine velocity of point E (Absolute velocity) on link BC, first locate the position of point E on velocity vector diagram. This can be done by taking corresponding ratios of lengths of links to vector distance i.e. be BE = bc BC

∴ be =

BE 0.04 x Vcb = x 0.44 = 0.24 m/s BC 0.066

Join e on velocity vector diagram to zero velocity points a, d / vector de = Ve = 0.415 m/s. Step 5: To determine angular velocity of links BC and CD, we know Vbc and Vcd. ∴ Vbc =ω BC x BC ∴ ωBC = Similarly,

Vbc 0.44 = = 6.6 r / s . (cw) BC 0.066

Vcd = ωCD x CD ∴ ωCD =

Vcd 0.39 = = 6.96 r / s (CCW) CD 0.056

Step – 6: To determine velocity of an offset point F •

Draw a line ⊥r to CF from C on velocity vector diagram.

•

Draw a line ⊥r to BF from b on velocity vector diagram to intersect the previously drawn line at ‘f’.

•

From the point f to zero velocity point a, d and measure vector fa to get Vf = 0.495 m/s.

Step – 7: To determine velocity of an offset point. •

Draw a line ⊥r to GC from C on velocity vector diagram.

•

Draw a line ⊥r to DG from d on velocity vector diagram to intersect previously drawn line at g.

•

Measure vector dg to get velocity of point G. 167

Vg = dg = 0.305 m / s Step – 8: To determine rubbing velocity at pins •

Rubbing velocity at pin A will be Vpa = ωab x r of pin A Vpa = 10.5 x 0.03 = 0.315 m/s

•

Rubbing velocity at pin B will be Vpb = (ωab + ωcb) x rpb of point at B. [ωab CCW and ωcbCW] Vpb = (10.5 + 6.6) x 0.04 = 0.684 m/s.

•

Rubbing velocity at point C will be = 6.96 x 0.035 = 0.244 m/s

Problem 2: In a slider crank mechanism the crank is 200 mm long and rotates at 40 rad/sec in a CCW direction. The length of the connecting rod is 800 mm. When the crank turns through 60o from Inner-dead centre. Determine, vi)

The velocity of the slider

vii)

Velocity of point E located at a distance of 200 mm on the connecting rod extended.

viii)

The position and velocity of point F on the connecting rod having the least absolute velocity.

ix)

The angular velocity of connecting rod.

x)

The velocity of rubbing of pins of crank shaft, crank and cross head having pins diameters 80,60 and 100 mm respectively.

Solution: Step 1: Draw the configuration diagram by selecting a suitable scale.

168

E

A F 45o

B

O

G

Va = Woa x OA Va = 40 x 0.2 Va = 8 m/s Step 2: Choose a suitable scale for velocity vector diagram and draw the velocity vector diagram. •

Mark zero velocity point o, g.

•

Draw oa ⊥r to link OA equal to 8 m/s e a

Scale: 1 cm = 2 m/s

f b

o, g

•

From a draw a line ⊥r to AB and from o, g draw a horizontal line (representing the line of motion of slider B) to intersect the previously drawn line at b.

•

ab give Vba=4.8 m/sec

BE x ab AB mark the point e on extension of vector ba. Join e to o, g. ge will give velocity of point E. Step – 3: To mark point ‘e’ since ‘E’ is on the extension of link AB drawn be =

Ve = ge =8.4 m/sec Step 4: To mark point F on link AB such that this has least velocity (absolute).

169

Draw a line ⊥r to ab passing through o, g to cut the vector ab at f. From f to o, g. gf will have the least absolute velocity. •

To mark the position of F on link AB. Find BF by using the relation. fb ab = BF AB BF =

fb x AB =200mm ab

Step – 5: To determine the angular velocity of connecting rod. We know that Vab = ωab x AB ∴ ω ab =

Vab = 6 rad/sec AB

Step – 6: To determine velocity of rubbing of pins. •

Vpcrankshaft = ωao x radius of crankshaft pin = 8 x 0.08 = 0.64 m/s

•

VPcrank pin = (ωab + ωoa) rcrank pin= (6 +8)0.06 =0.84 m/sec

•

VP cross head = ωab x rcross head = 6 x 0.1 = 0.6 m/sec

170

•

Problem 3: A quick return mechanism of crank and slotted lever type shaping machine is shown in Fig. the dimensions of various links are as follows. O1O2 = 800 mm, O1B = 300 mm, O2D = 1300 mm and DR = 400 mm

The crank O1B makes an angle of 45o with the vertical and relates at 40 rpm in the CCW direction. Find:

•

iii)

Velocity of the Ram R, velocity of cutting tool, and

iv)

Angular velocity of link O2D.

Solution:

Step 1: Draw the configuration diagram.

R Tool

R

D

200 D

B on orank, O, B C on O2D

O1

B 2

45o

C

O1

O2 O2

Step 2: Determine velocity of point B. Vb = ωO1B x O1B ωO1B =

2πN O1B 2π x 40 = = 4.18 rad / sec 60 60

Vb = 4.18 x 0.3 = 1.254 m/sec 171

172

Step 3: Draw velocity vector diagram. Choose a suitable scale 1 cm = 0.3 m/sec b d c O1O2

r o Draw O1b ⊥r to link O1B equal to 1.254 m/s.

o From b draw a line along the line of O2B and from O1O2 draw a line ⊥r to O2B. This intersects at c bc will measure velocity of sliding of slider and O 2 C will measure the velocity of C on link O2C. o Since point D is on the extension of link O2C measure O 2 d such that O 2d = O 2C

O2D . O 2 d will give velocity of point D. O 2C

o From d draw a line ⊥r to link DR and from O1O2. Draw a line along the line of stroke of Ram R (horizontal), These two lines will intersect at point r O 2 r will give the velocity of Ram R. o To determine the angular velocity of link O2D determine Vd = O 2 d . We know that Vd = ωO2D x O2D.

∴ ωO2d =

O 2d r/s O 2D

173

•

Problem 4: Figure below shows a toggle mechanisms in which the crank OA rotates at 120 rpm. Find the velocity and acceleration of the slider D.

•

Solution: 120

All the dimensions in mm A

45o 40 190

100 135

B

120

D

Configuration Diagram Step 1: Draw the configuration diagram choosing a suitable scal. Step 2: Determine velocity of point A with respect to O. Vao = ωOA x OA Vao =

2π x 120 = 0.4 = 5.024 m / s 60

Step 3: Draw the velocity vector diagram. o Choose a suitable scale o Mark zero velocity points O,q o Draw vector oa ⊥r to link OA and magnitude = 5.024 m/s.

174

a

b

O,q

D

Velocity vector diagram o From a draw a line ⊥r to AB and from q draw a line ⊥r to QB to intersect at b. ab = Vba and qb = Vbq . o Draw a line ⊥r to BD from b from q draw a line along the slide to intersect at d. dq = Vd (slider velocity) •

Problem 5: A whitworth quick return mechanism shown in figure has the following dimensions of the links.

The crank rotates at an angular velocity of 2.5 r/s at the moment when crank makes an angle of 45o with vertical. Calculate OP (crank) = 240 mm d) the velocity of the Ram S OA = 150 mm e) the velocity of slider P on the slotted level AR = 165 mm f) the angular velocity of the link RS. RS = 430 mm •

Solution:

Step 1: To draw configuration diagram to a suitable scale.

175

R

S

A

O o

45

P on slider Q on BA B

Configuration Diagram

176

Step 2: To determine the absolute velocity of point P. VP = ωOP x OP Vao =

2π x 240 x 0.24 = 0.6 m / s 60

Step 3: Draw the velocity vector diagram by choosing a suitable scale. P 0.6 m

q

S O, a, g r Velocity vector diagram o Draw op ⊥r link OP = 0.6 m. o From O, a, g draw a line ⊥r to AP/AQ and from P draw a line along AP to intersect previously draw, line at q. Pq = Velocity of sliding. aq = Velocity of Q with respect to A. Vqa = aq = o Angular velocity of link RS = ω RS = sr rad/sec SR

177

•

Problem 6: A toggle mechanism is shown in figure along with the diagrams of the links in mm. find the velocities of the points B and C and the angular velocities of links AB, BQ and BC. The crank rotates at 50 rpm in the clockwise direction. C Q

100

B 140 OA = 30 AB = 80 BQ = 100

A

50 rpm

BC = 100 All dimensions are in mm

•

O

Solution

Step 1: Draw the configuration diagram to a suitable scale. Step 2: Calculate the magnitude of velocity of A with respect to O. Va = ωOA x OA  2π x 50   x 0.03 = 0.05 π m / s = 0.1507 m / s Va =   60 

178

b

a

c

O, q

Vector velocity diagram Step 3: Draw the velocity vector diagram by choosing a suitable scale. o Draw Oa ⊥r to link OA = 0.15 m/s o From a draw a link ⊥r to AB and from O, q draw a link ⊥r to BQ to intersect at b. ab = Vba = ωab =

and qb = Vb = 0.13 m / s

ab qb = 0.74 r / s (ccw ) ωbq =1.3 r / s (ccw ) AB aB

o From b draw a line ⊥r to Be and from O, q these two lines intersect at C. OC = VC = 0.106 m / s bC = VCb = ω BC = •

bc = 1.33 r / s (ccw ) BC

Problem 7: The mechanism of a stone crusher has the dimensions as shown in figure in mm. If crank rotates at 120 rpm CW. Find the velocity of point K when crank OA is inclined at 30o to the horizontal. What will be the torque required at the crank to overcome a horizontal force of 40 kN at K.

179

500 60o

A 100

M hz

h2 100

200 400

K

600

600

320

360 B

200

D

C

Configuration diagram •

Solution:

Step 1: Draw the configuration diagram to a suitable scale.

180

Step 2: Given speed of crank OA determine velocity of A with respect to ‘o’.  2π x 120   x 0.1 = 1.26 m / s Va = ωOA x OA =  60   Vk(hz)

d

o, q, m

k

b

a

c Velocity vector diagram

Step 3: Draw the velocity vector diagram by selecting a suitable scale. o Draw Oa ⊥r to link OA = 1.26 m/s o From a draw a link ⊥r to AB and from q draw a link ⊥r to BQ to intersect at b. o From b draw a line ⊥r to BC and from a, draw a line ⊥r to AC to intersect at c. o From c draw a line ⊥r to CD and from m draw a line ⊥r to MD to intersect at d. o From d draw a line ⊥r to KD and from m draw a line ⊥r to KM to x intersect the previously drawn line at k. o Since we have to determine the torque required at OA to overcome a horizontal force of 40 kN at K. Draw a the horizontal line from o, q, m and c line ⊥r to this line from k. ∴ ( ωT ) I = ( ωT ) O P

V = ωR

P

T=FxP

F=

T r

∴ ωOA TOA = Fk Vk horizontal

181

∴ TOA = TOA = •

Fk Vk ( hz ) ω OA 40000 X 0.45 = 12.6

N-m

Problem 8: In the mechanism shown in figure link OA = 320 mm, AC = 680 mm and OQ = 650 mm.

Determine, iv)

The angular velocity of the cylinder

v)

The sliding velocity of the plunger

vi)

The absolute velocity of the plunger When the crank OA rotates at 20 rad/sec clockwise.

•

Solution:

Step 1: Draw the configuration diagram. A B on AR (point on AR below Q) C

60o O

R

Step 2: Draw the velocity vector diagram o Determine velocity of point A with respect to O. Va = ωOA x OA = 20 x 0.32 = 6.4 m/s o Select a suitable scale to draw the velocity vector diagram. o Mark the zero velocity point. Draw vector oa ⊥r to link OA equal to 6.4 m/s.

182

a

O, q

c b

o From a draw a line ⊥r to AB and from o, q, draw a line perpendicular to AB. o To mark point c on ab We know that ∴ ac =

ab ac

=

AB AC

ab x AC = AB

o Mark point c on ab and joint this to zero velocity point. o Angular velocity of cylinder will be. ωab =

Vab = 5.61 rad/sec (cω) AB

o Studying velocity of player will be qb = 4.1 m/s

o Absolute velocity of plunger =

•

OC qc

= 4.22 m/s

Problem 9: In a swiveling joint mechanism shown in figure link AB is the driving crank which rotates at 300 rpm clockwise. The length of the various links are: AB = 650 mm

Determine, iv)

The velocity of slider block S

AB = 100 mm

v)

The angular velocity of link EF

BC = 800 mm

vi)

The velocity of link EF in the swivel block.

DC = 250 mm BE = CF EF = 400 mm OF = 240 mm FS = 400 mm

183

•

Solution:

Step 1: Draw the configuration diagram. F

400

S O

P G

400 300

E

B 45o A

D

Step 2: Determine the velocity of point B with respect to A. Vb = ωBA x BA Vb =

2π x 300 x 0.1 = 3.14 m/s 60

Step 3: Draw the velocity vector diagram choosing a suitable scale. o Mark zero velocity point a, d, o, g. f

b P

S

a, d, o, g c

184

Velocity vector diagram o From ‘a’ draw a line ⊥r to AB and equal to 3.14 m/s. o From ‘b’ draw a line ⊥r to DC to intersect at C. o Mark a point ‘e’ on vector bc such that be = bc x

BE BC

o From ‘e’ draw a line ⊥r to PE and from ‘a,d’ draw a line along PE to intersect at P. o Extend the vector ep to ef such that ef = ef x EF EP

o From ‘f’ draw a line ⊥r to Sf and from zero velocity point draw a line along the slider ‘S’ to intersect the previously drawn line at S. o

Velocity of slider gS = 2.6 m / s . Angular Velocity of link EF.

o Velocity of link F in the swivel block = OP = 1.85 m / s . •

Problem 10: Figure shows two wheels 2 and 4 which rolls on a fixed link 1. The angular uniform velocity of wheel is 2 is 10 rod/sec. Determine the angular velocity of links 3 and 4, and also the relative velocity of point D with respect to point E. φ 50 mm φ 40 mm B

3 60 mm

o

30

ω2

A 2 G

•

C

D 4 F

Solution:

Step 1: Draw the configuration diagram.

185

Step 2: Given ω2 = 10 rad/sec. Calculate velocity of B with respect to G. Vb = ω2 x BG Vb = 10 x 43 = 430 mm/sec. Step 3: Draw the velocity vector diagram by choosing a suitable scale. B C

30o E

D

F

G 50 mm

Redrawn configuration diagram

186

•

Velocity vector diagram c b

e

d

g, f

o Draw gb = 0.43 m/s ⊥r to BG. o From b draw a line ⊥r to BC and from ‘f’ draw a line ⊥r to CF to intersect at C. o From b draw a line ⊥r to BE and from g, f draw a line ⊥r to GE to intersect at e. o From c draw a line ⊥r to CD and from f draw a line ⊥r to FD to intersect at d. •

•

Problem 11: For the mechanism shown in figure link 2 rotates at constant angular velocity of 1 rad/sec construct the velocity polygon and determine. iv)

Velocity of point D.

v)

Angular velocity of link BD.

vi)

Velocity of slider C.

Solution:

Step 1: Draw configuration diagram.

187

D

6

O2 = 50.8 mm AB = 102 mm

O6

BD = 102 mm DO6 = 102 mm

5 102 mm

AC = 203 mm

A 3

B

45o

C

4

O2 Step 2: Determine velocity of A with respect to O2. Vb = ω2 x O2A Vb = 1 x 50.8 = 50.8 mm/sec. Step 3: Draw the velocity vector diagram, locate zero velocity points O2O6. d Vd

a Udb b

O2O6

C

o From O2, O6 draw a line ⊥r to O2A in the direction of rotation equal to 50.8 mm/sec. o From a draw a line ⊥r to Ac and from O2, O6 draw a line along the line of stocks of c to intersect the previously drawn line at c. o Mark point b on vector ac such that ab = ab x AB AC

o From b draw a line ⊥r to BD and from O2, O6 draw a line ⊥r to O6D to intersect at d. 188

Step 4:

Vd = O 6 d = 32 mm/sec ωbd =

bd = BD

Vc = O 2 C = ADDITIONAL PROBLEMS FOR PRACTICE •

Problem 1: In a slider crank mechanism shown in offset by a perpendicular distance of 50 mm from the centre C. AB and BC are 750 mm and 200 mm long respectively crank BC is rotating eω at a uniform speed of 200 rpm. Draw the velocity vector diagram and determine velocity of slider A and angular velocity of link AB. B A 50 mm C

•

Problem 2: For the mechanism shown in figure determine the velocities at points C, E and F and the angular velocities of links, BC, CDE and EF. C

All dimensions are in mm

120

E

B 100

100 60

120

o

rpm A

•

150 F

120

50

D

The crank op of a crank and slotted lever mechanism shown in figure rotates at 100 rpm in the CCW direction. Various lengths of the links are OP = 90 mm, OA = 300 mm, AR = 480 mm and RS = 330 mm. The slider moves along an axis perpendicular

189

to ⊥r AO and in 120 mm from O. Determine the velocity of the slider when | AOP is 135o and also mention the maximum velocity of slider.

45o

B

O C

D A

190

•

Problem 4: Find the velocity of link 4 of the scotch yoke mechanism shown in figure. The angular speed of link 2 is 200 rad/sec CCW, link O2P = 40 mm.

4

P 3 2 Q on link 4 45o

•

Problem 5: In the mechanism shown in figure link AB rotates uniformly in Cω direction at 240 rpm. Determine the linear velocity of B and angular velocity of EF. E

C

B

AB = 160 mm BC = 160 mm CD = 100 mm

A

45o

AD = 200 mm EF = 200 mm

100 mm

F

CE = 40 mm

191

II Method • Instantaneous Method To explain instantaneous centre let us consider a plane body P having a non-linear motion relative to another body q consider two points A and B on body P having velocities as Va and Vb respectively in the direction shown.

Va B

A I

P

Vb q

Fig. 1 If a line is drawn ⊥r to Va, at A the body can be imagined to rotate about some point on the line. Thirdly, centre of rotation of the body also lies on a line ⊥r to the direction of Vb at B. If the intersection of the two lines is at I, the body P will be rotating about I at that instant. The point I is known as the instantaneous centre of rotation for the body P. The position of instantaneous centre changes with the motion of the body.

Va

A

B

P Vb q

I

Fig. 2 In case of the ⊥r lines drawn from A and B meet outside the body P as shown in Fig 2.

A

Va

B

Vb

I at ∞ Fig. 3 If the direction of Va and Vb are parallel to the ⊥r at A and B met at ∞. This is the case when the body has linear motion.

192

•

Number of Instantaneous Centers

The number of instantaneous centers in a mechanism depends upon number of links. If N is the number of instantaneous centers and n is the number of links. N=

•

n ( n − 1) 2

Types of Instantaneous Centers

There are three types of instantaneous centers namely fixed, permanent and neither fixed nor permanent. Example: Four bar mechanism. N=

n = 4.

n ( n − 1) 4( 4 − 1) =6 = 2 2 I13

I34 3 I23 4 2 I24 I12

1

I14

Fixed instantaneous center I12, I14 Permanent instantaneous center I23, I34 Neither fixed nor permanent instantaneous center I13, I24 •

Arnold Kennedy theorem of three centers:

193

Statement: If three bodies have motion relative to each other, their instantaneous centers should lie in a straight line.

194

Proof: 1

VA32

I12

VA22 I13

2 I23

3 A

Consider a three link mechanism with link 1 being fixed link 2 rotating about I12 and link 3 rotating about I13. Hence, I12 and I13 are the instantaneous centers for link 2 and link 3. Let us assume that instantaneous center of link 2 and 3 be at point A i.e. I 23. Point A is a coincident point on link 2 and link 3. Considering A on link 2, velocity of A with respect to I12 will be a vector VA2 ⊥r to link A I12. Similarly for point A on link 3, velocity of A with respect to I13 will be ⊥r to A I13. It is seen that velocity vector of VA2 and VA3 are in different directions which is impossible. Hence, the instantaneous center of the two links cannot be at the assumed position. It can be seen that when I23 lies on the line joining I12 and I13 the VA2 and VA3 will be same in magnitude and direction. Hence, for the three links to be in relative motion all the three centers should lie in a same straight line. Hence, the proof. Steps to locate instantaneous centers: Step 1: Draw the configuration diagram. Step 2: Identify the number of instantaneous centers by using the relation ( n − 1) n = . 2

N

Step 3: Identify the instantaneous centers by circle diagram. Step 4: Locate all the instantaneous centers by making use of Kennedy’s theorem. To illustrate the procedure let us consider an example. 195

196

A slider crank mechanism has lengths of crank and connecting rod equal to 200 mm and 200 mm respectively locate all the instantaneous centers of the mechanism for the position of the crank when it has turned through 30 o from IOC. Also find velocity of slider and angular velocity of connecting rod if crank rotates at 40 rad/sec. Step 1: Draw configuration diagram to a suitable scale. Step 2: Determine the number of links in the mechanism and find number of instantaneous centers.

( n − 1) n

N=

2

n = 4 links

N=

4( 4 − 1) =6 2 I13

I24 A 200 I12

2

3

I23

800

B

30o

O 1 I14 to ∞

1

4 I12

I14 to ∞

Step 3: Identify instantaneous centers. o Suit it is a 4-bar link the resulting figure will be a square.

197

I12

1

2 1

I24

2

I12 I41

I23

OR

I13

I23 I13

4 I34

I24 I14

I34

4

3

3

o Locate fixed and permanent instantaneous centers. To locate neither fixed nor permanent instantaneous centers use Kennedy’s three centers theorem. Step 4: Velocity of different points. Va = ω2 AI12 = 40 x 0.2 = 8 m/s also Va = ω2 x A13 ∴ ω3 =

Va AI13

Vb = ω3 x BI13 = Velocity of slider. •

Problem 2:

A four bar mechanisms has links AB = 300 mm, BC = CD = 360 mm and AD = o 600 mm. Angle | BAD = 60 . Crank AB rotates in Cω direction at a speed of 100 rpm. Locate all the instantaneous centers and determine the angular velocity of link BC. •

Solution:

Step 1: Draw the configuration diagram to a suitable scale. Step 2: Find the number of Instantaneous centers N=

( n − 1) n 2

=

4( 4 − 1) =6 2

Step 3: Identify the IC’s by circular method or book keeping method.

198

1

I12

2 I12

1

I14

I23

2

I12

OR

I23 I13

I13 4

3

4 I34

I24 I14

I34

3

Step 4: Locate all the visible IC’s and locate other IC’s by Kennedy’s theorem. I13

C

3

I34

B I23

4

2 I24

I12 A

Vb = ω2 x BI12 = Also

1

D

2π x 100 x 0.3 = m / sec 60

Vb = ω3 x BI13 ω3 =

•

I14

Vb = rad / sec BI13

For a mechanism in figure crank OA rotates at 100 rpm clockwise using I.C. method determine the linear velocities of points B, C, D and angular velocities of links AB, BC and CD. OA = 20 cm

AB = 150 cm

BC = 60 cm 199

CD = 50 cm

BE = 40 cm

OE = 135 cm C

A 2

6

3

30o

1

D

5 E

O

10 mm 4 B

Va = ωOA x OA Va =

2π x 100 x 0.2 = 2.1 m / s 60

n = 6 links N=

n ( n − 1) = 15 2

1

2 12

3

4

23 13

34 24

14

5 45

35 25

15

5 4 3 2 1 --15 ---

6 56

46 36

26 16

I16 @ ∞

I16 @ ∞

I16 @ ∞

I13 I45 I23 2 1

5 3

I14

6 I56

I12 I34

I15

200

I13

Link 3

A

3 B

Va = ω3AI13 ω3 =

Va = 2.5 rad / sec AI13

Vb = ω3 x BI13 = 2.675 m/s C

Link 4

I14 4 B Also

Vb = ω4 x BI14 ω4 =

Vb = 6.37 rad / sec BI14

VC = ω4 x CI14 = 1.273 m/s Link 5

C

5

D

I15

201

VC = ω5 x CI15

Answers Vb = 2.675 m/s VC = 1.273 m/s Vd = 0.826 m/s ωab = 2.5 rad/sec ωbc = 6.37 rad/sec ωcd = 1.72 rad/sec

VC =1.72 rad / sec ω5 = AI15 Vd = ω5 x DI15 = 0.826 m/s

•

In the toggle mechanism shown in figure the slider D is constrained to move in a horizontal path the crank OA is rotating in CCW direction at a speed of 180 rpm the dimensions of various links are as follows: OA = 180 mm

CB = 240 mm

AB = 360 mm

BD = 540 mm

Find, i)

Velocity of slider

ii)

Angular velocity of links AB, CB and BD. A

45o O

360

B 105

D C n = 6 links N=

n ( n − 1) = 15 2

202

1

2 12

3

4

23 13

34 24

14

5 45

35 25

15

5 4 3 2 1 --15 ---

6 56

46 36

26 16

I16 @ ∞

I16 @ ∞ I15

A

I23

2 I12

I46

I16 @ ∞

3 O

I13

I45 I34

4

B I35 5

C 6 I56

I24 Va = ω2 x AI12 = 3.4 m/s A

Link 3

ω3 I13 B Va = ω3 x AI13 ω3 =

Va = 2.44 rad / sec AI13

Vb = ω3 x BI13

203

B

Link 4 ω4 C

I14

Vb = ω4 x BI14 ω4 =

Vb =11.875 rad / sec AI14 I15

Link 5

B 5 D Vb = ω5 x BI15 Answers Vd = 2 m/s ωab = 2.44 rad/sec ωbc = 11.875 rad/sec Vd = ω5 x DI15 = 2 m/s ωcd = 4.37 rad/sec Figure shows a six link mechanism. What will be the velocity of cutting tool D and the angular velocities of links BC and CD if crank rotates at 10 rad/sec. Vb = 4.37 rad / sec ω5 = AI15

•

204

Q

25 90o

B All dimensions are in mm

15 15 C 45

45

60 A 15 o

30

O

D

205

I13

I14

I16 @ ∞

I34

I46

4 3 I45

I24 I23

2

I26

I16 @ ∞

I12

5 O

6

I56

I16 @ ∞

I15

Va = ω2 x AI12 = 10 x 0.015 Va = ω2 x AI12 = 0.15 m/s Link 3

I13

B ω3

A Va = ω3 x AI13 ω3 =

Va AI13

Vb = ω3 x BI13

206

Q I14

Link 4

B ω4

C Vb = ω4 x BI14 ω4 =

Vb = 4.25 rad / sec BI14

VC = ω4 x CI14 C

Link 5

5

D

•

I15

VC = ω5 x CI15

Answers

VC = 1.98 rad / sec ω5 = AI15

Vd = 1.66 m/s

Vd = ω5 x DI15 = 1.66 m/s

ωcd = 1.98 rad/sec

ωbc = 4.25 rad/sec

A whitworth quick return mechanism shown in figure has a fixed link OA and crank OP having length 200 mm and 350 mm respectively. Other lengths are AR = 200 mm and RS = 40 mm. Find the velocity of the rotation using IC method when crank makes an angle of 120o with fixed link and rotates at 10 rad/sec.

207

R 5 S 6

A 4 1

P O

3

2

B Locate the IC’s n = 6 links N=

n ( n − 1) = 15 2

1

2 12

3 23

13

4 34

24 14

5 45

35 25

15

26 16

56 46

36

6

5 4 3 2 1 --15 ---

208

I15 I16 @ ∞

I46

I45 6

5 I14

I56 1 I12

4 2

I23 3 I34

I24

VP = ω2 x OP = ……… m/s •

Acceleration Analysis

Rate of change of velocity is acceleration. A change in velocity requires any one of the following conditions to be fulfilled: o Change in magnitude only o Change in direction only o Change in both magnitude and direction When the velocity of a particle changes in magnitude and direction it has two component of acceleration. 1. Radial or centripetal acceleration fc = ω2r Acceleration is parallel to the link and acting towards centre.

209

A

r

Va A’ δθ

O1

Va’ cosδθ

ft oa

oa

δθ fcoa

O

Va’ sinδθ

Va’

f oa a1

Va’ = (ω + α δ t) r Velocity of A parallel to OA = 0 Velocity of A’ parallel to OA = Va’ sin δ θ Therefore change in velocity = Va’ sin δ θ – 0 Centripetal acceleration = fc =

( ω + αδt ) r sin δθ δt

as δt tends to Zero sin δ θ tends to δ θ ∴

( ωrδθ + αrδθδt ) δt

fc = ωr (dθ/ dt) =ω2r But V = ωr or ω = V/r Hence, fc =ω2r = V2/r 2. Tnagential Acceleration: Va’ = (ω + α δ t) r Velocity of A perpendicular to OA = Va Velocity of A’ perpendicular to OA = Va’ cos δ θ Therefore change in velocity = Va’ cos δ θ – Va Tnagnetial acceleration = ft =

( ω + αδt ) r cos δθ − ωr δt

210

as δt tends to Zero cos δ θ tends to 1 ∴

( ωr + αrδt ) − ωr δt

ft = αr Example: B

A

fCab = ω2AB Acts parallel to BA and acts from B to A. frab fab ftab

ft = αBA acts ⊥r to link. fBA = frBA + ftBA •

Problem 1: Four bar mechanism. For a 4-bar mechanism shown in figure draw velocity and acceleration diagram.

211

All dimensions are in mm

C 66 B 50

ω = 10.5 rad/sec

56

60o A

D 100

212

•

Solution:

Step 1: Draw configuration diagram to a scale. Step 2: Draw velocity vector diagram to a scale. Vb = ω2 x AB Vb = 10.5 x 0.05 Vb = 0.525 m/s Vc

C

a1d Vbc b Step 3: Prepare a table as shown below: Sl. No. 1.

Link AB

Magnitude fc = ω2ABr

Direction

Sense

Parallel to AB

A

Parallel to BC

B

fc = (10.5)2/0.525 fc = 5.51 m/s2 2.

BC

fc = ω2BCr fc = 1.75

3.

CD

ft = αr

⊥r to BC

fc = ω2CDr

Parallel to DC

– D

fc = 2.75 ft = ?

⊥r to DC

–

Step 4: Draw the acceleration diagram.

213

a1d1

11el to CD

11el to CD ⊥γ to BC

c1

c1′

fbc b1

11el to BC

b1

11el to AB

o Choose a suitable scale to draw acceleration diagram. o Mark the zero acceleration point a1d1. o Link AB has only centripetal acceleration. Therefore, draw a line parallel to AB and toward A from a1d1 equal to 5.51 m/s2 i.e. point b1. o From b1 draw a vector parallel to BC points towards B equal to 1.75 m/s2 (b11). o From b11 draw a line ⊥r to BC. The magnitude is not known. o From a1d1 draw a vector parallel to AD and pointing towards D equal to 2.72 m/s 2 i.e. point c1. o From c11 draw a line ⊥r to CD to intersect the line drawn ⊥r to BC at c1, d1c1 = fCD and b1c1 = fbc. To determine angular acceleration. t 1 αBC = f bc = c1b1 = 34.09 rad / sec (CCW ) BC BC t 1 αCD = f cd = c1c1 = 79.11 rad / sec (CCW ) CD CD

•

Problem 2: For the configuration of slider crank mechanism shown in figure below.

Calculate i)

Acceleration of slider B.

ii)

Acceleration of point E.

iii)

Angular acceleration of link AB. If crank OA rotates at 20 rad/sec CCW.

•

Solution:

214

E

450

All dimensions are mm

A 1600

480

B

60o

G

Step 1: Draw configuration diagram. Step 2: Find velocity of A with respect to O. Va = ωOA x OA Va = 20 x 0.48 Va = 9.6 m/s Step 4: Draw velocity vector diagram. e a 5.25 b

O1g

9.7

Step 4: Sl. No.

Link

Magnitude

Direction

Sense

1.

OA

fcaO = ω2OAr = 192

Parallel to OA

O

2.

AB

fcab = ω2abr = 17.2

Parallel to AB

A

ftab

–

⊥r to AB

–

–

Parallel to Slider

–

3.

Slider B

215

Step 5: Draw the acceleration diagram choosing a suitable scale. fb fab

o1g1

t

f ab

b11 fcab

192 172

a1 e1

ee1

216

o Mark o1g1 (zero acceleration point) o Draw o1g1 = C acceleration of OA towards ‘O’. o From a1 draw a1b11 = 17.2 m/s2 towards ‘A’ from b11 draw a line ⊥r to AB. o From o1g1 draw a line along the slider B to intersect previously drawn line at b1, a 1 b1 = f ab g 1 b1 = fb = 72 m/s2. o Extend a 1 b1 = a 1e1 such that

a 1 b1 A1 R 1 . = AB AE

o Join e1 to δ1g1, g 1e1 = fe = 236 m/s2. o αab =

f abt bb 167 = 104 rad/sec2 (CCW). = 1 1 = AB AB 1.6

Answers: fb = 72 m/sec2 fe = 236 m/sec2 αab = 104 rad/sec2 •

Problem 3: In a toggle mechanism shown in figure the crank OA rotates at 210 rpm CCW increasing at the rate of 60 rad/s2. •

Velocity of slider D and angular velocity of link BD.

•

Acceleration of slider D and angular acceleration of link BD.

217

A

150

200 45o

400

B

300

500

D

D

Q

G

Step 1 Draw the configuration diagram to a scale. Step 2 Find Va = ωOA x OA Va =

2π ( 210 ) x 0.2 = 4.4 m/s 60

Step 3: Draw the velocity vector diagram. a

b

o1,q,g

d Step 4: Sl. No. 1.

Link AO

Magnitude m/s2

Direction

fcaO = ω2r = 96.8

Parallel to OA

ftaO = αr = 12

⊥ to OA r

Sense O –

218

2.

AB

3.

BD

5.

Parallel to AB

f = αr =

⊥ to AB

fcbq = ω2r = 38.3

Parallel to BQ

ftbq = αr =

⊥r to BQ

–

fcbd = ω2r = 20

⊥r to BD

B

ftbd = αr =

⊥r to BD

Slider D

–

Parallel to slider motion

–

A

r

t ab

BQ

4.

fcab = ω2r = 5.93

–

– Q

Step 5: Draw the acceleration diagram choosing a suitable scale. o Mark zero acceleration point. fbd d1

fd

O1q1g1

q11 fab b11 a1

fcOA ftOA

d11 b1

a11

o Draw o1a11 = fcOA and a11a = ftOA ⊥r to OA from o

o1a1 = fa

o From a1 draw a 1 b1 = f c ab , from b11 draw a line ⊥r to AB. o From o1q1g1 draw o1q11 = fcbq and from q11 draw a line a line ⊥r to BQ to intersect the previously drawn line at b1 q 1 b1 = f bq

a 1 b1 = fab

o From b1 draw a line parallel to BD = fcbd such that b1d 11 = fcbd. o From d11 draw a line ⊥r to BD, from o1q1g1 draw a line along slider D to meet the previously drawn line at .

219

o

g 1d 1 = f d = 16.4 m/sec2.

o

b1d 1 = f bd = 5.46 m/sec2.

o αBD =

f bd 5.46 = 109.2 rad / sec 2 BD 0.5

Answers: Vd = 2.54 m/s ωbd = 6.32 rad/s Fd = 16.4 m/s2 αbd = 109.2 rad/s2

220

•

Coriolis Acceleration: It has been seen that the acceleration of a body may have two components. •

Centripetal acceleration and

•

Tangential acceleration.

However, in same cases there will be a third component called as corilis acceleration to illustrate this let us take an example of crank and slotted lever mechanisms. P1

P 2

Q

B1

B2

dθ

B on link 3

3 A on link 2

dθ

A1 ω2

O Assume link 2 having constant angular velocity ω2, in its motions from OP to OP1 in a small interval of time δt. During this time slider 3 moves outwards from position B to B2. Assume this motion also to have constant velocity VB/A. Consider the motion of slider from B to B2 in 3 stages. 1. B to A1 due to rotation of link 2. 2. A1 to B1 due to outward velocity of slider VB/A. 3. B1 to B2 due to acceleration ⊥r to link 2 this component in the coriolis component of acceleration. We have Arc B1B2 = Arc QB2 – Arc QB1 = Arc QB2 – Arc AA1 ∴ Arc B1B2 = OQ dθ - AO dθ

221

= A1B1 dθ = VB/A ω 2dt2 The tangential component of velocity is ⊥r to the link and is given by Vt = ωr. In this case ω has been assumed constant and the slider is moving on the link with constant velocity. Therefore, tangential velocity of any point B on the slider 3 will result in uniform increase in tangential velocity. The equation Vt = ωr remain same but r increases uniformly i.e. there is a constant acceleration ⊥r to rod. ∴ Displacement B1B2 = ½ at2 = ½ f (dt)2 ∴ ½ f (dt)2 = VB/A ω2 dt2 fcrB/A = 2ω 2 VB/A coriolis acceleration The direction of coriolis component is the direction of relative velocity vector for the two coincident points rotated at 90o in the direction of angular velocity of rotation of the link. Figure below shows the direction of coriolis acceleration in different situation. fcr ω2

ω2

ω2

fcr (a) Rotation CW slider moving up

(b) Rotation CW slider moving down

222

fcr

ω2

ω2

fcr (c) Rotation CCW slider moving up

(d) Rotation CCW slider moving down

A quick return mechanism of crank and slotted lever type shaping machine is shown in Fig. the dimensions of various links are as follows. O1O2 = 800 mm, O1B = 300 mm, O2D = 1300 mm and DR = 400 mm The crank O1B makes an angle of 45o with the vertical and rotates at 40 rpm in the CCW direction. Find: v)

Acceleration of the Ram R, velocity of cutting tool, and

vi)

Angular Acceleration of link AD. Tool

R

D

Solution:

R 200 D

B on orank, A B

Step 1: Draw the configuration diagram.

C on AD

O

B o

2

45

C

O

A A

223

Step 2: Determine velocity of point B. Vb = ω OB x OB ω OB =

2πNO1B 2π x 40 = = 4.18 rad / sec 60 60

Vb = 4.18 x 0.3 = 1.254 m/sec Step 3: Draw velocity vector diagram. Step 4: prepare table showing the acceleration components Choose a suitable scale 1 cm = 0.3 m/sec Sl. Link Magnitude m/s2 Direction Sense No. b Parallel to OB O fcob = ωd2r =5.24 1. OB – c c 2 Parallel to AB f ac = ω r A 2. AC r t – ⊥ to AB f ac = αr o.a r s Parallel to AB f bc =αr _ 3. BC r cc – ⊥ to AC f bc = 2vω = c 2 f bd = ω r = 20 Parallel to DR D 4. DR ftbd =α r r _ ⊥ to BD 5.

Slider R

ftbd = αr

Parallel to slider motion

– 224

r1 f

fr

o 1a 1

t dr

b1’’

fcob

fad ftab

d1 c

f dr

b 1’

r1 ’

fob b1

fsab fccbc

b1’’’

Acceleration of Ram = fr = o1 r Angular Acceleration of link AD α

bd

=

f bd BD

KLENIN’S Construction This method helps us to draw the velocity and acceleration diagrams on the construction diagram itself. The crank of the configuration diagram represents the velocity and acceleration line of the moving end (crank).

225

The procedure is given below for a slider crank mechanism.

A

ω

800

200

B

45º O

To draw the velocity vector diagram: Link OA represents the velocity vector of A with respect to O. Voa = oa = ω r = ω OA. b b

o a 200

ω 45º

800 a

o

Draw a line perpendicular at O, extend the line BA to meet this perpendicular line at b. oab is the velocity vector diagram rotated through 90º opposite to the rotation of the crank. Acceleration diagram: The line representing Crank OA represents the acceleration of A with respect to O. To draw the acceleration diagram follow the steps given below. • Draw a circle with OA as radius and A as centre. • Draw another circle with AB as diameter. • The two circles intersect each other at two points C and D.

226

•

Join C and D to meet OB at b1 and AB at E.

O1,a1,ba1and b1 is the required acceleration diagram rotated through 180º.

fb

b1 ftab

O1

fab

fa

ba1 fcab

a1

ω

ba1

200

800

45º O1

a

B b1

227