kinematic synthesis

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Kinematic Synthesis of Planar Mechanisms Yatin Kumar Singh Page 1 16.1 Introduction Kinematic Analysis is the process of determination of velocity and acceleration of the various links of an existing mechanism. Kinematic Synthesis deals with the determination of the lengths and orientation of the various lengths of the links so that a mechanism could be evolved to satisfy certain conditions. Kinematic synthesis of a mechanism requires the determination of lengths of various links that satisfy the requirements of motion of the mechanism. The usual requirements are related to specified positions of the input and output links. It is easy to design a planar mechanism, when the positions of input and output links are known at fewer positions as compared to large number of positions. 16.2 Movability (or Mobility) or Number Synthesis Movability of a mechanism means the number of degrees of freedom, which is equal to the number of independent coordinates required to specify its configurations in order to define its motion. This concept is also known as Number Synthesis. The Gruebler’s (or Kutzbach) criterion for degrees of freedom of planar mechanisms is given by: = ሺ − ሻ − − ሺ. ሻ Where w = + + + ⋯ = number of simple joints or lower pairs having one degree of freedom n2 = number of binary links n3 = number of ternary links n4 = number of quaternary links, and so on h = number of higher pairs having two degrees of freedom n = n2 + n3 + n4 + ··· + ni = total number of links. If h = 0, then = ሺ − ሻ − ሺ. ሻ = [ሺ + + +⋯ ሻ − ] − ሺ + + + ⋯ Simplifying and re-arranging the equation, we get = ሺ + ሻ + [ + + + ⋯ + ሺ − ሻ ] ሺ. ሻ For a fully constrained mechanism, F = 1. Thus = + [ + + + ⋯ + ሺ − ሻ ] ሺ. ሻ From Eq. (16.4), it is quite evident that the minimum number of binary links is equal to four. Therefore, the four-bar kinematic chain is the simplest mechanism. For F = 1, n2 η Ͷ F = 2, n2 η ͷ F = 3, n2 η ············

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Page 1: Kinematic Synthesis

Kinematic Synthesis of Planar Mechanisms

Yatin Kumar Singh Page 1

16.1 Introduction

Kinematic Analysis is the process of determination of velocity and acceleration of the various links of an

existing mechanism. Kinematic Synthesis deals with the determination of the lengths and orientation of the

various lengths of the links so that a mechanism could be evolved to satisfy certain conditions.

Kinematic synthesis of a mechanism requires the determination of lengths of various links that satisfy the

requirements of motion of the mechanism. The usual requirements are related to specified positions of the

input and output links. It is easy to design a planar mechanism, when the positions of input and output links

are known at fewer positions as compared to large number of positions.

16.2 Movability (or Mobility) or Number Synthesis

Movability of a mechanism means the number of degrees of freedom, which is equal to the number of

independent coordinates required to specify its configurations in order to define its motion. This concept is

also known as Number Synthesis.

The Gruebler’s (or Kutzbach) criterion for degrees of freedom of planar mechanisms is given by: = − − � − .

Where w���� � = + + + ⋯

= number of simple joints or lower pairs having one degree of freedom

n2 = number of binary links

n3 = number of ternary links

n4 = number of quaternary links, and so on

h = number of higher pairs having two degrees of freedom

n = n2 + n3 + n4 + ··· + ni = total number of links.

If h = 0, then = − − � . = [ + + + ⋯ − ] − + + + ⋯

Simplifying and re-arranging the equation, we get = + + [ + + + ⋯ + − ] .

For a fully constrained mechanism, F = 1. Thus = + [ + + + ⋯ + − ] .

From Eq. (16.4), it is quite evident that the minimum number of binary links is equal to four. Therefore, the

four-bar kinematic chain is the simplest mechanism.

For

F = 1, n2

F = 2, n2

F = 3, n2

············

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= �, ≥ � +

For a fully constrained motion, F = 1, so that

= − − � o� � = − .

16.3 Transmission Angle

Transmission angle μ is the interior angle between the coupler and output link as shown in Fig.16.1. If link AB

is the input link, the force applied to the output link DC is transmitted through the coupler BC. For a particular

value of force in the coupler rod, the torque transmitted to the output link (about point D) is maximum when the transmission angle μ is °. )f links BC and DC become coincident, the transmission angle is zero and the mechanism would lock or jam. )f μ deviates significantly from °, the torque on the output link decreases.

Sometimes, it may not be sufficient to overcome the friction in the system and the mechanism may be locked or jammed. (ence, μ is generally kept more than °. The transmission angle is nearly ° for the best mechanism. To find the positions where the transmission angle is maximum or minimum, apply cosine law to Δs ABD and BCD.

Fig.16.1 Four-bar mechanism + − � = � and + − � = � From Eqs. (1) and (2), we have + − − − � + � = For maximum or minimum values of μ, �� =

o� � − � × �� = �� = �� =

o� � = Since a and d are not zero, so sin θ = or θ = ° or °

Thus, transmission angle is maximum when θ = °, and minimum when θ = °.

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16.3.1 Transmission Angle in Slider–Crank Mechanism

The slider-crank mechanism ABC is shown in Fig. . where μ is the transmission angle. To find the positions of maximum or minimum values of μ, we have = − = � −

Fig.16.2 Slider-crank mechanism From Δ BEC, we have

= ° − � = � = � = �

From eq.1 and 2 � − = � Differentiating Eq. w.r.t. θ and equalizing it to zero, we get

� = − � × �� �� = − �� =

o� � = As a is not equal to zero, so cos θ = Thus θ = ° or ° gives the maximum or minimum values of transmission angle μ. 16.4 Limit Positions and Dead Centres of a Four-Bar Mechanism

For the design of four-link mechanism, limit positions and dead centres are essential.

Limit Position: It is the position in which the interior angle between its coupler and input link is either 180°

or 360°. In the limit position of the mechanism, the pivot points A, D and C lie on a straight line as shown in

Fig.16.3(b). In this case, the angle between the input link AD and coupler DC is 180°. A four-link mechanism

can have maximum two limit positions. The second limit position is shown in Fig.16.3(c) where, the angle

between the coupler DC and input link AD is 360°.

The angle of oscillation of output-link BC is given as,

Δϕ = ϕ − ϕ

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where ϕ and ϕ are the two extreme positions of output link, BC.

Fig.16.3 Limit positions of four-bar mechanism

Dead Centre Position: It is the position in which the interior angle between the coupler and the output link

(of a four-link mechanism) is either 180° or 360° (refer to Fig.16.4). In dead centre positions, the pivot points

B, C, and D lie in a straight line. There can be two (maximum) dead centre positions. A crank-crank four-bar

link mechanism does not have either a limit position or dead centre positions because both the cranks rotate

through 360°.

Fig.16.4 Dead centre positions of four-bar mechanism

16.5 Dimensional Synthesis

It deals with the determination of actual dimensions of the mechanism to satisfy the specified motion

characteristics. The actual dimensions could be the lengths of the links between adjacent hinge pairs, angles

between the arms of a bell-crank lever, or cam contour dimensions, etc. Synthesis of mechanisms can be

carried out by graphical methods or analytical methods.

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16.6 Graphical Method

16.6.1 Pole

Consider a four-bar mechanism O2 ABO4 in two positions O2A1B1O2 and O2A2B2O4 as shown in Fig.16.6. The

coupler link AB has moved from the position A1B1 to A2B2.

Fig.16.6 Pole of a four-bar mechanism

The input link O2A and output link O4B have moved through angles θ12 and ϕ12 respectively in the clockwise

direction. For the motion of the coupler AB from A1B1 to A2B2, P12 is its centre of rotation with respect to the

fixed link. P12 lies at the point of intersection of the perpendicular bisectors of the coupler link AB in its two

positions A1B1 and A2B2. Hence P12 is called the pole. P12 also lies at the point of intersection of the mid-

normals a12 and b12 of the chords A1 A2 and B1 B2 respectively.

Properties of Pole Point

Since AB = A1B1 = A2B2, the perpendicular bisectors of A1 A2 and B1 B2 pass through fixed centres O2 and O4.

Since coupler AB rotates about P12 from position A1B1 to A2 B2, and therefore, ΔA1 B1 P12 = ΔA2 B2 P12,

∠2 + ∠3 + ∠1 = ∠1 + ∠4 + ∠5

or angle subtended by A1 B1 at P12 = angle subtended by A2 B2 at P12.

Since ∠2 + ∠3 + ∠1 = ∠1 + ∠4 + ∠5

∠2 + ∠3 = ∠4 + ∠5

Thus A1 A2 and B1 B2 subtend equal angle at P12. Since P12 lies on the perpendicular bisectors of A1 A2 and B1

B2,

∠2 = ∠3 and ∠4 = ∠5

Since ∠2 + ∠3 = ∠4 + ∠5

And ∠2 = ∠3 and ∠4 = ∠5

∠2 = ∠4 and ∠3 = ∠5

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Therefore, input and output links subtend equal angles at pole P12 as they move from one position to

another.

Since ∠2 + ∠3 + ∠1 = ∠1 + ∠4 + ∠5 = ∠1 + ∠4 + ∠3 ( ∠5 = 3)

i.e., angle subtended by the coupler link AB = angle subtended by the fixed link O2O4.

The triangle A1B1P12 moves as one link about P12 to the position A2 B2 P12. Angular displacement of coupler A1

B1 = angular displacement of P12 B i.e., β12 = ∠4 + ∠5.

16.6.2 Relative Pole

The pole of a moving link is the centre of its rotation with respect to a fixed link. However, if the rotation of

the link is considered relative to another moving link, the pole is called as the relative pole. To determine the

relative pole, fix the link of reference and observe the motion of the other link in the reverse direction.

(a) Determination of Relative Pole for Four-Bar Chain

Consider the four-bar chain O2ABO4 in its two positions O2 AB1O4 and O2 AB2O as shown in Fig.16.7 with links

O2A fixed. The relative pole of AB relative to O2 A is at A. The relative pole of O4B relative to O2A is at R12, that

can be determined as follows:

Let θ12 = angle of rotation of O2A (clockwise) ϕ12 = angle of rotation of O4B (clockwise)

Fig.16.7 Determination of relative pole of four-bar chain

1. Assume O2 and A as the fixed pivots and rotate O2O4 about O2 through angle θ12 in the counter-clockwise

direction (opposite to the direction of rotation of O2A). Let od4 be the new position after the rotation of O2od4.

2. Locate the point B2 by drawing arc with centres A and od4 and radii equal to AB1 and O4B1 respectively. Then

O2 AB2od4 is inversion of O2AB1od4.

3. Draw mid-normals of O2od4 and B1 B2 passing through O2 and A respectively to intersect at R12, which is the

required relative pole. ϕ12 − θ12 = angle of rotation of the output link O4B relative to the input link.

The angle will be negative if O4 B > O2A and positive if O4B < O2A.

Angular displacement of R12O4 = angular displacement of O4B1. ∠O4R12O’4 = (∠ϕ12 − ∠θ12) assuming O4B > O2A

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2 ∠ = − ∠ϕ12 − ∠θ12) ∠ = ∠ϕ − ∠θ )n ΔO2 R12 O4, ∠4 = ∠1 + ∠2 ∠θ = ∠ϕ − ∠θ + ∠ ∠ = ∠

= ∠ϕ + ∠θ + ∠

∠ = ∠ϕ

Procedure:

1. Join O2O4 and extend it further. Rotate O2O4 about O2 through an angle ½ θ12 in a direction opposite to that of

O2A.

2. Again rotate O2O4 about O4 through an angle ½ ϕ12 in a direction opposite to that of O4B.

3. The point of intersection of these two positions of O2O4 after rotation about O2 and O4, is the required relative

pole R12.

4. The angles subtended by O4 od4 and B1 B2 at R12 are the same.

i.e., ∠O4 R12 od4 = ∠B1 R12 B2

or 2 ∠O4 R12 O2 = 2 ∠B1 R12 A

or ∠O4 R12 O2 = ∠B1 R12A

Angle subtended by the fixed pivots (O2 and O4) at the relative pole = Angle subtended by the coupler AB.

(b) Determination of Relative Pole for Slider-Crank Mechanism

Consider the slider-crank mechanism shown in Fig.16.8. The point B on the slider reciprocates through a

horizontal distance x. Its centre of rotation will lie at infinity on a vertical line where the point O4 can also be

assumed to lie. Then O2 O4 will also be a vertical line through O2. Rotate O2 O4 about O4 through / θ12 in the

counter-clockwise direction. Rotating O2O4 about O4 through / θ12 would mean a vertical line towards left

of O2, at a distance of x/2. The intersections of the two lines locate R12. The procedure to locate the relative

pole of a slider-crank mechanism is as follows:

Fig.16.8 Determination of relative pole of slider-crank mechanism

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1. Draw two parallel lines l1 and l2 at a distance equal to the eccentricity e (if there is any eccentricity).

2. Select a line segment O2C = x/2 on line l1 such that C is measured in a direction opposite to the motion of the

slider.

3. Draw perpendicular lines O2P1 and C P2 to line l1.

4. Make ∠1/2 θ12 at point O2 with the line O2P1 in a direction opposite to the rotation of the input link.

5. The intersection of this line with the line CP2 extended locates the relative pole R12.

16.7 Design of Mechanisms by Relative Pole Method

16.7.1 Four-Bar Mechanism

(a) Two-Position Synthesis

Let for a four-bar mechanism, length of the fixed link O2O4 along with the angular displacement θ12 (between

position 1 and 2) of the input link O2A and angular displacement ϕ12 (between positions 1 and 2) of the

output link O4B, are known.

To design the mechanism (Fig.16.9), first locate the relative pole R12 � = � � �

= ∠θ − ∠ϕ Assu��n� O B ∠O A = ∠�

Procedure:

1. Construct an angle ψ12 at an arbitrary position R12. Join any two points on the two arms of the angle to obtain

the coupler link AB of the mechanism. Join A with O2 to get the input and output links respectively.

2. Locate point B arbitrarily so that BO4 is the output link. Construct ∠BR12 Z = ψ12. Take any suitable point A or

R12 Z. Join AB and O2A.

3. Instead of locating point B as above, locate point A arbitrarily so that O2A is the input link. Construct ∠AR12 Y = ψ12. Select any point B on R12 Y. Join A and B to get AB. Join B with O4 to get output link O4B.

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Fig.16.9 Two-position method for four-bar mechanism

(b) Three Position Synthesis Let three positions of input link θ1, θ2 and θ3, and three positions of output link ϕ1, ϕ2 and ϕ3 are known. The

relative poles R12 and R13 can be determined considering, θ12 = θ2 − θ1 and ϕ12 = ϕ2 − ϕ1 to locate R12

and θ13 = θ3 − θ1 and ϕ13 = ϕ3 − ϕ1 to locate R13 An��� subt�nd�d at R , ∠� = ∠θ − ∠ϕ

An��� subt�nd�d at R , ∠� = ∠θ − ∠ϕ

Fig.16.10 Three-position method for four-bar mechanism Construct the angles ψ12 and ψ13 at the points R12 and R13 respectively at arbitrary positions such that the

arms of the angles intersect at A and B. Join A and B to get the coupler AB. Join A with O2 and B with O4 to get

input and output links respectively, as shown in Fig.16.10.

16.7.2 Slider–Crank Mechanism

(a) Two Position Synthesis:

Let θ12 = angular displacement of input link O2 A angle between θ1 and θ2) in clockwise direction.

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x = linear displacement of slider to the right

e = eccentricity

Fig.16.11 Two position method for slider-crank mechanism

Procedure:

1. Draw two parallel lines l1 and l2 at a distance equal to eccentricity, e.

2. Construct an angle equal to θ12/2 at point R12, chosen arbitrarily and at a convenient position. The

intersection of an arm N of this angle with line l2 gives the position of the slider at point B.

3. Select point A arbitrarily on the other arm M of the angle. Join A with O2 to get the input link (crank) O2A. Join

A and B to get the coupler (connecting rod).

(b) Three-Position Synthesis Three positions of input link θ1, θ2, and θ3 are known, along with the corresponding slider positions x1, x2 and

x3. Find R12 and R13 as shown in Fig.16.12.

Fig.16.12 Three position method for slider-crank mechanism

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Here θ12/2= angle made by the fixed link at R12. θ13/2= angle made by the fixed link at R13.

Procedure:

1. Construct angle θ12/2= ∠M1 R12 N1 at R12 in an arbitrary position with arm N1 locating point B.

2. Draw angle θ13/2= ∠M2 R13 N2 at R13 with an arm N2 along R13 B.

3. Intersection of the two arms M1 at R12 and M2 at R13 (not through B) of the two angles locates the point A. Join

A with B to get the coupler AB.

16.8 Errors in Kinematic Synthesis of Mechanisms

There are three types of errors present in the design of linkages for function generation. These errors are:

1. Structural,

2. Mechanical, and

3. Graphical.

In function generation, there is correlation between the motion of input and output links. If the motions of

input link is represented as x0, x1, x2, …, xn+1 and the corresponding motions (which is dependent on input

variables x0, x1, …, xn+1) of the output link is represented by y0, y1, y2, …, yn+1, these can be shown on the graph

as indicated in Fig.16.13. We observe that at certain points (P1, P2, and so on) the desired function and

generated function agree well. These points are called Precision Points. The number of such points (from 3

to 6 generally) is equal to the number of design parameters. Except these points the generated function curve

and desired function curve do not agree and are deviating by certain amount of error which is known as

Structural Error. Structural error is the difference between the generated function and the desired

function for a certain value of input variable. So, the precision points are spaced in such a way as to minimize

the structural error of the linkage.

Fig.16.13 Errors in function generation of linkages

Mechanical Errors are caused because of mechanical defects such as improper machining, casting of

components of the linkage, clearance in the components because of rubbing, overloading of linkages, etc.

Graphical Error is caused because of inaccuracy in drawing of perpendicular or parallel lines. It may occur

because of wrong graphical construction and wrong choice of scale. Also, there may be human errors in

drawing work.

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16.9 Analytical Method

16.9.1 Function Generation

In function generation, the motion of input (or driver) link is correlated to the motion of output (or follower) link. Let θ and ϕ be the angles of rotation of input and output links respectively. Let y = f(x) be the function to be generated. The angle of rotation θ of the input link O2A represents the independent variable x and the

angle of rotation ϕ of the output link O4B represents the dependent variable y, as shown in Fig.16.14. The relation between x and θ and that between y and ϕ is generally assumed to be linear. Let θi and ϕi be the initial values of θ and ϕ representing xi and yi respectively. � − �− = � = . = � − �− .

And ϕ − ϕ− = � = . = ϕ − ϕ− .

where the constants rx and ry are called Scale Factors. The subscripts i and f denote the initial and final

values.

Fig.16.14 Function generation method

. . Chebyshev’s Spacing for Precision Points

Let xi and xj be the initial and final values of variable x respectively. A function f(x) is desired to be generated

in the interval xi x xf. Let the generated function be F(x, R1, R2, …, Rn), where R1, R2, …, Rn are the design

parameters. The difference E(x) between the desired function and generated function can be represented by, = − , � , � … . � .

At precision points, say for x = x1, x2, …, xn, the desired and generated functions agree and E(x) = 0. At other

points E(x) will have some value, called the structural error. It is desirable that E(x) should be minimum. Therefore, the spacing of precision points is very important. The precision points, according to Chebyshev s spacing, are given by:

= + [ − �] , = , , .

w���� = + ; = − ; = nu�b�� o� p��c�s�on po�nts

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16.9.3 Graphical Method to Locate Precision Points

The Precision points can be obtained by the graphical method from the following steps:

1. Draw a circle of radius b and centre on the x-axis at a distance a from point O. 2. Inscribe a regular polygon of side 2n in this circle such that the two sides are perpendicular to the x-axis.

3. Determine the locations of n accuracy points by projecting the vertices on x-axis as shown in Fig.16.15. It is

sufficient to draw semi-circles only showing inscribed polygon to get the values of precision points.

Fig.16.15 Graphical method to determine precision points

Example: Derive Freudenstein s equation for a four bar linkage. Solution:

Consider a four-bar mechanism as shown in Fig.16.16 in equilibrium. The magnitudes of AB, BC, CD and DA are a, b, c and d respectively. θ, β and ϕ are the angles of AB, BC and DC respectively with the x-axis. AD is the

fixed link. AB is the input link and DC the output link.

The displacement along x-axis is, a cos θ + b cos β = d + c cos ϕ

Or b cos β = c cos ϕ − a cos θ + d � = � − � + = � + � + − � � − � + �

The displacement along y-axis is, a sin θ + b sin β = c sin ϕ

or b sin β = c sin ϕ − a sin θ � = � − � = � + � − � �

Adding Eqs. (1) and (2), we get

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Fig. . Deriving freudenstein’s equation for four-bar mechanism

b2 = c2 + a2 + d2 − ac sin θ sin ϕ + cos θ cos ϕ − ad cos θ + cd cos ϕ

or 2cd cos ϕ − ad cos θ + a2 − b2 + c2 + d2 = ac sin θ sin ϕ + cos θ cos ϕ)

Dividing throughout by 2ac, we get

� − � + − + + = � − � = � − �

� � + � � + � = � − � .

w���� � = ; � = − ; � = − + + Eq. (16.9) is known as Freudenstein Equation.

16.9.4 Freudenstein’s Equation for the Precision Points Freudenstein s equation helps to determine the length of links of a four-bar mechanism. The displacement

equation of a four-bar mechanism, shown in Fig.16.17 is given by:

2cd cos ϕ − ad cos θ + a2 − b2 + c2 + d2 = ac cos θ cos ϕ + sin θ sin ϕ)

Fig. . Freudenstein’s equation for the precision points

Dividing throughout by 2ac and rearranging, we get

� − � + − + + = � − �

��t � = ; � = − ; � = − + + ���n � � + � � + � = � − � . Eq. . is known as the Freudenstein s equation.

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Let the input and output are related by some function, such as, y = f(x).

For three specified positions, let θ1, θ2, θ3 = three positions of input link ϕ1, ϕ2, ϕ3 = three positions of output link

Then substituting these values in Eq. (16.10), we get

k1 cos ϕ1 + k2 cos θ1 + k3 = cos θ1 − ϕ1)

k1 cos ϕ2 + k2 cos θ2 + k3 = cos θ2 − ϕ2)

k1 cos ϕ3 + k2 cos θ3 + k3 = cos θ3 − ϕ3)

These equations can be written in the matrix form as:

| � � � � � � | { � � � } = { � − �� − �� − � }

These equations can be solved by any numerical technique. Using Cramer s rule, let

= | � � � � � � | ; = | � − � �� − � �� − � � | = | � � − � � � − � � � − � | ; = | � � � − � � � � − � � � � − � |

���n � = ; � = ; � = Knowing k1, k2 and k3, the values of a, b, c, and d can be calculated. Value of either a or d can be assumed to be unity to obtain the proportionate values of other parameters.

Example: Design a four-bar mechanism to coordinate three positions of the input and output links given by: θ1 = 25, ϕ1 = °; θ2 = 35°, ϕ2 = °; θ3 = 50°, ϕ3 = 60°

Solution: cos θ1 = cos ° = . , cos θ2 = cos ° = . , cos θ3 = cos 50° = 0.6428

cos ϕ1 = cos 30° = 0.8660, cos ϕ2 = cos 40° = 0.7660, cos ϕ3 = cos 60° = 0.5000 cos θ1 − ϕ1 = cos ° − 30°) = 0.9962 cos θ2 − ϕ2 = cos ° − ° = . cos θ2 − ϕ2 = cos ° − ° = .

� = | . .. .. . | = . × −

� = | . .. .. . | = − . × −

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� = | . .. .. . | = − . × −

� = | . . .. . .. . . | = . × −

� = �� = − . × −. × − = . =

Let d = 1 unit, then a = 5.598 units

� = �� = − . × −. × − = − . = − , = . un�ts

� = �� = . × −. × − = . = − + + = . − + . +× . × .

= .

The mechanism is shown in Fig.16.18.

Fig.16.18 Four-bar mechanism developed by three position precision points

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Unit-2

Yatin Kumar Singh Page 1

Introduction

Kinematics deals with the study of relative motion between the various links of a machine ignoring the forces

involved in producing such motion. Thus, kinematics is the study to determine the displacement velocity and

acceleration of the various links of the mechanism. A machine is a mechanism or a combination of

mechanisms that not only imparts definite motion to the various links but also transmits and modifies the

available mechanical energy into some kind of useful energy.

2.2 Velocity Diagrams

Displacement: The displacement of a body is its change of position with reference to a certain fixed point.

Velocity: Velocity is the state of change of displacement of a body with respect to time. It is a vector quantity.

Linear Velocity: It is the rate of change of velocity of a body along a straight line with respect to time. Its

units are m/s.

Angular Velocity: It is the rate of change of angular position of a body with respect to time. Its units are

rad/s. The relationship between velocity v and angular velocity ω is: = � .

where r = distance of point undergoing displacement from the centre of rotation.

Relative Velocity: The relative velocity of a body A with respect to a body B is obtained by adding to the

velocity of A the reversed velocity of B. If va > vb, then = ⃗⃗⃗⃗ − ⃗⃗⃗⃗ o� = −

Similarly, = ⃗⃗⃗⃗ − ⃗⃗⃗⃗ o� = −

Fig.2.1 Relative velocity of a point

Consider two points A and B on a rigid link rotating clockwise about A as shown in Fig.2.1 (a). There can be no

relative motion between A and B as long as the distance between A and B remains the same. Therefore, the

relative motion of B with respect to A must be perpendicular to AB. Hence, the direction of relative velocity of

two points in a rigid link is always along the perpendicular to the straight line joining these points. Let

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relative velocity of B with respect to A be represented by vba = ω·AB, then ab is drawn perpendicular to AB to

a convenient scale, as shown in Fig.2.1 (b). Similarly, the linear velocity of any other point C on AB with

respect to A is vca = ω·AC and is represented by vector ac. Hence,

= �.�. =

= =

Hence the point C divides the vector ab in the same ratio as the point C divides the link AB.

2.3 Determination of Link Velocities

There are two methods to determine the velocities of links of mechanisms.

Relative velocity method

Instantaneous centre method

2.3.1 Relative Velocity Method

Consider a rigid link AB, as shown in Fig.2.2 (a), such that the velocity of A (va) is vertical and velocity of B (vb)

is horizontal. To construct the velocity diagram, take a point o. Draw oa representing the magnitude and direction of velocity of A. Draw ob along the direction of vb. From point a draw a line ab perpendicular to AB, meeting ob in b. Then oab is the velocity triangle, as shown in Fig.2.2 (b). Ob = vb; ab = vba, i.e. the velocity of B

with respect to A. Vector ab is called the velocity image of link AB. The velocity of any point C in AB with

respect to A is given by,

= = . ( )

= + ( )

= + . = + =

Hence vector oc represents the velocity of point C.

Fig.2.2 Relative velocity of points in a kinematic link

Consider a link A1B1 first undergoing rotation by an amount Δθ to A1B 1 and then undergoing translation by an amount ΔsA to occupy the new position A2B2, as shown in Fig.2.3 (a). Then ∆ = ∆ + ∆

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Now let the link A1B1 first undergo linear translation ΔsA to A2B”2 and then angular rotation of Δθ to A2B2, as

shown in Fig.2.3 (b). Then ∆ = ∆ + ∆ Eqs. a and b are same. Dividing by Δt, we get

∆∆� = ∆∆� + ∆∆�

Or = +

Therefore, the velocity of point B is obtained by adding vectorially the relative velocity of point B w.r.t. point

A to the velocity of point A. Now ∆ = . ∆ o� ∆∆� = . ∆∆�

= .�

Also ∠ϕ = 90°

Fig.2.3 Relative velocity of points in a kinematic link

The following conclusions may be drawn from the above analysis:

The velocity of any point on the kinematic link is given by the vector sum of the velocity of some other point

in the link and the velocity of the first point relative to the other.

The magnitude of the velocity of any point on the kinematic link relative to the other point in the kinematic

link is the product of the angular velocity of the link and distance between the two points under

consideration.

The direction of the velocity of any point on a link relative to any other point on the link is perpendicular to

the line joining the two points.

2.3.3 Relative Angular Velocities Consider two links OA and OB connected by a pin joint at O, as shown in Fig. . . Let ω1 and ω2 be the angular

velocities of the links OA and OB, respectively. Relative angular velocity of OA with respect to OB is,

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� = � − �

and relative angular velocity of OB with respect to OA is,

� = � − � = −�

If r = radius of the pin at joint O, then rubbing (or sliding) velocity at the pin joint O, = � − � �,���n ��� ��nk� �o��� �n ��� �a�� d���c��on . a = � + � �,���n ��� ��nk� �o��� �n ��� o��o���� d���c��on . b

Fig.2.4 Relative angular velocities

2.3.4 Relative Velocity of Points on the Same Link

Consider a ternary link ABC, as shown in Fig.2.5(a), such that C is any point on the link. Let va and vb be the

velocities of points A and B, respectively. Then = +

� = =

= + = +

Fig.2.5 Relative velocity of points on the same link

The velocity diagram is shown in Fig.2.5 (b).

Angular velocity of link ABC, � = = = = = = .

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2.3.5 Forces in a Mechanism

Consider a link AB subjected to the action of forces and velocities, as shown in Fig.2.6. Let A be the driving end

and B the driven end. When the direction of the forces and velocities is the same, then

Energy at A = Energy at B × = ×

= .

Fig.2.6 Force and velocity diagram

Considering the effect of friction, the efficiency of transmission,

= ����������� =

Or = .

When the forces are not in the direction of the velocities, then their components along the velocities should

be taken.

2.3.6 Mechanical Advantage

��������� ��������, = ��� ���������� ������ =

�o� a ��c�an���, = ������ ����������� ������ = = �� .

Considering the effect of friction, = �� .

2.3.7 Four-Bar Mechanism

(a) Consider the four-bar mechanism, as shown in Fig.2.7 (a), in which the crank O1A is rotating clockwise

with uniform angular speed ω. The linear velocity of point A is va = ω × O1A and it is perpendicular to O1A.

Therefore, draw o1a ⊥ O1A to a convenient scale in Fig.2.7 (b). The velocity of point B is perpendicular to O2B.

Therefore, at point o1, draw o1b ⊥ O2B. The relative velocity of B with respect to A is perpendicular to AB.

Therefore, draw ab ⊥ AB meeting the line drawn perpendicular to O2B at b. Then vb = o1b, and vba = ab. To find

the velocity of joint C, draw ac ⊥ AC and bc ⊥ BC to meet at c. Join o1c. Then vc = o1c.

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Now = + = + = = + = + = = + = + =

To find the velocity of any point D in AB, we have = o� = ( ) .

Fig.2.7 Four-bar mechanism with ternary link

Thus, locate point d in ab and join o1d. Then o1d = vd. To find the relative velocity of C w. r. t. D, join cd. Then

vcd = dc. The velocity image of link ABC is abc.

(b) Now consider the four-bar mechanism, as shown in Fig.2.8 (a), in which the crank AB is rotating at angular velocity ω in the anticlockwise direction. The absolute linear velocity of B is ω · AB and is perpendicular to AB. Draw ab ⊥ AB to a convenient scale to represent vba, as shown in Fig.2.8 (b). From b,

draw a line perpendicular to BC and from a another line perpendicular to CD to meet each other at point c. Then, ac = vca and cb = vbc. To find the velocity of any point E in BC, we have

= o� = ( ) . Thus, locate point e in cb and join ae. Then ae = vea.

Fig.2.8 Four-bar mechanism

The rubbing velocities at pins of joints are:

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: � . ; : � ± � . ; : � ± � . ; : � .

where r is the radius of the pin.

Use the + ve sign when angular velocities are in the opposite directions.

� = and � = 2.3.8 Slider–Crank Mechanism

Consider the slider–crank mechanism, as shown in Fig.2.9 (a), in which the crank OC is rotating clockwise

with angular speed ω. PC is the connecting rod and P is the slider or piston. The linear velocity of C, vc = OC ·

To draw the velocity diagram, draw a line oc = vc from point o, as shown in Fig.2.9 (b), representing the

velocity of point C to a convenient scale. From point c draw a line perpendicular to CP. The velocity of slider P

is horizontal. Therefore, from point o draw a line parallel to OP to intersect the line drawn perpendicular to

CP at p. Then the velocity of the piston, vp = op and the velocity of piston P relative to crank pin C is vpc = cp.

To find the velocity of any point E in CP, we have

Fig.2.9 Slider-crank mechanism

=

= ( ) .

Thus locate point e in cp, join oe. Then ve = oe

Rubbing Velocities are: : � . ; : � . ; : (� + � ). 2.3.9 Crank and Slotted Lever Mechanism

Consider the crank and slotted lever mechanism, as shown in Fig.2.10 (a). The crank OB is rotating at a uniform angular speed ω. Let OB = r, AC = l, and OA = d. Linear velocity of B, vb = rω. Draw ob = vb and

perpendicular to OB to a convenient scale, as shown in Fig.2.10 (b). vb is the velocity of point B on the crank

OB. The velocity of the slotted lever is perpendicular to AC. The velocity of the slider is along the slotted lever.

Hence draw a line from b parallel to AC to meet the line perpendicular to AC at p. Now, ap = vpa, and

= = ( ) .

From point c, draw a line perpendicular to CD and from point o draw a line parallel to the tool motion, to

intersect at point d.

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�������� �� ������� ����, =

The component of the velocity of the crank perpendicular to the slotted lever is zero at positions B1 and B2.

Thus for these positions of the crank, the slotted lever reverses its direction of motion.

Fig.2.10 Crank and slotted lever mechanism ���� o� cu���n� ���ok����� o� ���u�n ���ok� = α° − α

At positions B3 and B4, the component of velocity along the lever is zero, i.e. the velocity of the slider at the

crank pin is zero. Thus the velocity of the lever at crankpin is equal to the velocity of crankpin, i.e. rω. The velocities of lever and tool at these points are minimum. The maximum cutting velocity occurs at B3 and

maximum return velocity occurs at B4.

�a���u� cu���n� ���oc��y = .� ( ) = � ( + ) .

�a���u� ���u�n ���oc��y = .� ( ) = � ( − ) .

�a���u� cu���n� ���oc��y�a���u� ���u�n ���oc��y = ( −+ ) .

2.3.10 Drag Mechanism

The drag mechanism is shown in Fig.2.11 a . Link rotates at constant angular speed ω. Link rotates at a non-uniform velocity. Ram 6 will move with nearly constant velocity over most of the upward stroke to give a

slow upward stroke and a quick downward stroke when link 2 rotates clockwise.

To determine the velocity diagram, draw o1a = ω · O1A, perpendicular to O1A, as shown in Fig.2.11 (b). From a draw ab ⊥ AB and from o1 draw o1b ⊥ O2B to intersect at b. Then, o1b = vb and ab = vba. At b draw bc ⊥ BC

and at o1 draw o1c ⊥ O2C, to intersect at c. Then o1c = vc.

Now at o2 draw o2d || O2D and from c draw a line perpendicular to CD to intersect at d. Then o2d = vd, the

velocity of ram 6. o2bc is the velocity image of O2BC.

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Fig.2.11 Drag mechanism

2.3.11 Whitworth Quick-Return Motion Mechanism

In the Whitworth mechanism, as shown in Fig.2.12 a , the crank O A rotates with constant angular speed ω. The link PB oscillates about pin O2. The ram C reciprocates on the guides.

Fig.2.12 Whitworth quick-return motion mechanism

The velocity diagram has been drawn in Fig.2.12 (b). va = o1a = ω · O1A. Draw o1a ⊥ O1A to a convenient scale.

At o2, draw a line perpendicular to PB and at a draw a line parallel to O2P to intersect at p. Then ap = vpa

represents the velocity of slider at P and o2p = vpo2 the velocity of lever PB at P. Produce po2 to b so that

=

Draw bc ⊥ BC and o2c || O2C to intersect at c. Then vc = o2c is the velocity of the ram.

Example: In the mechanism shown in Fig.2.14 (a), the crank O1A rotates at a uniform speed of 650 rpm.

Determine the linear velocity of the slider C and the angular speed of the link BC. O1A = 30 mm, AB = 45 mm,

BC = 50 mm, O2B = 65 mm, and O1O2 = 70 mm.

Solution

Given: N = 650 rpm, lengths of links.

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Procedure:

1. Draw the configuration diagram to a convenient scale as shown in Fig.2.14 (a).

2. Calculate the angular velocity of crank O1A.

� = � × = . �ad/�

Fig.2.14

3. Calculate the linear velocity of point A,

va = ω·O1A = 68.07 × 0.03 = 2.04 m/s

4. Draw the velocity diagram as shown in Fig.2.14 (b) to a scale of 1 cm = 0.5 m/s.

5. Draw va = o1a ⊥ O1A such that o1a = 4.08 cm.

6. Draw a line from o2 perpendicular to O2B and another line from a perpendicular to AB to intersect at b. Then

ab = vba and o2b = vb

7. From b, draw a line perpendicular to BC and another line from o2 parallel to the path of motion of the slider to

intersect at c. Then

Velocity of the slider C, vc = o2c

By measurement, we get vc = o2c = 1.4 cm = 0.7 m/s ; vcb = bc = 1.3 cm = 0.65 m/s

An�u�a� ���oc��y o� ��nk BC = ��� = .. = �ad �⁄ c�ock���� abou� B

2.4 Instantaneous Centre Method

A link as a whole may be considered to be rotating about an imaginary centre or about a given centre at a

given instant. Such a centre has zero velocity, i.e. the link is at rest at this point. This is known as the

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Instantaneous Centre or Centre of Rotation. This centre varies from instant to instant for different

positions of the link. The locus of these centres is termed the Centrode.

2.4.1 Velocity of a Point on a Link

Consider two points A and B on a rigid link, having velocities va and vb, respectively, as shown in Fig.2.50 (a).

From A and B draw lines perpendicular to the directions of motion and let them meet at I. Then I is the

instantaneous centre of rotation of the link AB for its give position. )f ω is the instantaneous angular velocity of the link AB, then va = ω · )A and vb = ω · )B. Thus

= ( ) . .

The velocity diagram for the link AB has been drawn in Fig.2.50 (b). Triangles IAB and oab are similar. Hence

Fig.2.50 Concept of instantaneous centre

= = .

= = = = � where C is any point on the link AB.

2.4.2 Properties of Instantaneous Centre

The properties of the instantaneous centre are as follows:

1. At the instantaneous centre of rotation, one rigid link rotates instantaneously relative to another for the

configuration of the mechanism considered.

2. The two rigid links have no linear velocities relative to each other at the instantaneous centre.

3. The two rigid links have the same linear velocity relative to the third rigid link, or any other link.

2.4.3 Number of Instantaneous Centres

The number of instantaneous centres in a mechanism is equal to the number of possible combinations of two

links. The number of instantaneous centres,

= − .

where n = number of links.

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2.4.4 Types of Instantaneous Centres

The instantaneous centres for a mechanism are of the following types:

1. Fixed Instantaneous Centres.

2. Permanent Instantaneous Centres.

3. Neither Fixed nor Permanent Instantaneous Centres.

Fig.2.51 Instantaneous centres in a four-bar mechanism

Consider a four-bar mechanism shown in Fig.2.51. For this mechanism, n = 4.

Hence N = − / =

The instantaneous centres are: 12, 13, 14, 23, 24, 34.

The instantaneous centres 12 and 14 remain at the same place for the configuration of the mechanism, and

are therefore called fixed instantaneous centres. The instantaneous centres 23 and 34 move when the

mechanism moves. But the joints are permanent; therefore, they are called permanent instantaneous centres.

The instantaneous centres 13 and 24 vary with the configuration of the mechanism and are neither fixed nor

permanent instantaneous centres.

2.4.5 Location of Instantaneous Centres

The following observations are quite helpful in locating the instantaneous centres:

1. For a pivoted or pin joint, the instantaneous centre for the two links lies on the centre of the pin (Fig.2.52 (a)).

2. In a pure rolling contact of the two links, the instantaneous centre lies at their point of contact (Fig.2.52 (b)).

This is because the relative velocity between the two links at the point of contact is zero.

3. In a sliding motion, the instantaneous centre lies at infinity in a direction perpendicular to the path of motion

of the slider. This is because the sliding motion is equivalent to a rotary motion of the links with radius of

curvature equal to infinity (Fig.2.52(c)). If the slider (link 2) moves on a curved surface (link 1), then the

instantaneous centre lies at the centre of curvature of the curved surface (Fig.2.52 (d) and (e)).

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Fig.2.52 Locating instantaneous centres

2.4.6 Arnold–Kennedy Theorem

This theorem states that if three-plane bodies have relative motion among themselves their three

instantaneous centres must lie on a straight line.

Consider three rigid links 1, 2, and 3; 1 being a fixed link. 12 and 13 are the instantaneous centres of links 1, 2

and 1, 3 respectively. Let 23 be the instantaneous centre of links 2, 3, lying outside the line joining 12 and 13,

as shown in Fig.2.53. The links 2 and 3 are moving relative to link 1. Therefore, the motion of their

instantaneous centre 23 is to be the same whether it is considered in body 2 or 3. If the point 23 is considered

on link 2, then its velocity v2 is perpendicular to the line joining 12 and 23. If the point 23 lies on link 3, then

its velocity v3 must be perpendicular to the line joining 13 and 23. The velocities v2 and v3 of point 23 are in

different directions, which is not possible. The velocities v2 and v3 of instantaneous centre 23 will be equal

only if it lies on the line joining 12 and 13. Hence all the three instantaneous centres 12, 13 and 23 must lie on

a straight line.

Fig.2.53 Proving three-centres theorem

2.4.7 Method of Locating Instantaneous Centres

The following procedure may be adopted to locate the instantaneous centres:

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1. Determine the number of instantaneous centres.

2. Make a list of all the instantaneous centres by writing the link numbers in the first row and instantaneous

centres in ascending order in columns. For example, for a four-bar chain shown in Fig.2.54 (a), we have

3. Locate the fixed and permanent instantaneous centres by inspection. Tick mark (tick) these instantaneous

centres.

4. Locate the remaining neither fixed nor permanent instantaneous centres (circled) by using Arnold–Kennedy s theorem. This is done by a circle diagram, as shown in Fig.2.54 (b). Mark points on a circle equal to the

number of links in the mechanism. Join the points by solid lines for which instantaneous centres are known

by inspection. Now join the points forming the other instantaneous centres by dotted lines. The instantaneous

centre shall lie at the intersection of the lines joining the instantaneous centres of the two adjacent triangles

of the dotted line. For example, in Fig.2.54 (b), the centre 13 is located at the intersection of lines (produced)

joining the instantaneous centres 12, 23, and 14, 34. Similarly, the centre 24 is located at the intersection of

the lines (produced) joining the centres 23, 34 and 12, 14.

Fig.2.54 Four-bar mechanism

2.4.8 Determination of Angular Velocity of a Link

The angular velocities of two links vary inversely as the distances from their common instantaneous centre to

their respective centres of rotation relative to the frame. For example, for the four-bar mechanism shown in Fig. . a , if ω2 is the angular velocity of link , then angular velocity ω4 of link 4 will be given by the

following relationship: �� = −−

If the respective centres of rotation are on the same side of the common instantaneous centre, then the

direction of angular velocities will be same. However, if the respective centres of rotation are on opposite

sides, then the direction of angular velocities will be opposite.

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Similarly, �� = −−

Example: Locate the instantaneous centres of the slider crank mechanism shown in Fig.2.56(a). Find the velocity of the slider. OA = mm, AB = mm, and OB = mm, ω2 = 12 rad/s cw.

Fig.2.56 Slides-crank mechanism

Solution:

Given: n = , ω2 = 12 rad/s cw, lengths of links.

Number of instantaneous centres = 6

1. Draw the configuration diagram to a convenient scale as shown in Fig.2.56 (a).

2. Locate fixed centres 12 and 34.

3. Locate permanent centre 23. Instantaneous centre 14 is at infinity.

4. Locate neither fixed nor permanent centres 13 and 24 by using Arnold–Kennedy s three centres theorem. 13: 12, 23; 14, 34

24: 12, 14; 23, 34

By measurements, we have − = mm − = 390 mm � = � . �� = × . = . � �⁄ = � . −

Hence, � = � [ −− ] = × = . � �⁄ When ω3 is the angular velocity of link 3 about 13.

Velocity of slider,

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� = � − = � [ −− ] − = � .�� [ −− ] = × . × = . � �⁄ 3.1 Introduction

The acceleration may be defined as the rate of change of velocity of a body with respect to time. The

acceleration can be linear or angular. Linear acceleration is the rate of change of linear velocity of a body with

respect to time. Angular acceleration is the rate of change of angular velocity of a body with respect to time.

Acceleration Diagram: It is the graphical representation of the accelerations of the various links of a

mechanism drawn on a suitable scale. It helps us to determine the acceleration of various links of the

mechanism.

The determination of acceleration of various links is important from the point of view of calculating the forces

and torques in the various links to carry out the dynamic analysis.

3.2 Acceleration of a Body Moving in a Circular Path Consider a body moving in a circular path of radius r with angular speed ω, as shown in Fig.3.1(a). The body is initially at point A and in time δt moves to point B. Let the velocity change from v to v + δv at B and the angle covered be δθ.

Fig.3.1 Acceleration for a link

The change in velocity can be determined by drawing the velocity diagram as shown in Fig.3.1(b). In this diagram, oa = v, ob = v + δv, and ab = change in velocity during time δt. The vector ab is resolved into two components ax and xb, parallel and perpendicular to oa, respectively.

Now = − = ��� � − = + � ��� − = ��� � = + � ��� �

The rate of change of velocity is defined as the acceleration. It has two components: tangential and normal.

The tangential component of acceleration ft is the acceleration in the tangential direction. It is defined by,

= � = + � ��� −� = + � −� B�cau�� �o� ��a�� an��� co� � ≈

= �� = = � = ( �) = .

Where α = angular acceleration.

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The normal component of acceleration fn is the acceleration in the direction normal to the tangent at that

instant. This component is directed towards the centre of the circular path. It is also called the radial or

centripetal acceleration. It is defined by,

= � = + � ���� = + � �� B�cau�� �o� ��a�� an��� ��n � ≈ �

= �� + � �� ≈ �� �����c��n� ��cond ����, b��n� ��a��

= = � = � = .

Two cases arise regarding the motion of the body.

1. When the body is rotating with uniform angular velocity, then dω/dt = 0, so that ft = 0. The body will have

only normal acceleration, fn = r ω2.

2. When the body is moving on a straight path, r will be infinitely large and v2/r will tend to zero, so that fn = 0.

The body will have only tangential acceleration, ft = r . 3.3 Acceleration Diagrams

3.3.1 Total Acceleration of a Link

Consider two points A and B on a rigid link, as shown in Fig.3.2 (a), such that the point B moves relative to point A with an angular velocity ω and angular acceleration . Centripetal (or normal or radial) acceleration of point B with respect to point A is,

= � . = .

Tangential acceleration of point B with respect to point A, = . .

Total acceleration of B w.r.t. A, = + ������ ��� = � × + × = � + . .

��� = = � .

The acceleration diagram has been represented in Fig.3.2 (b).

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Fig.3.2 Total acceleration of a link

3.3.2 Acceleration of a Point on a Link

The accelerations of any point X on the rigid link w.r.t. A as shown in Fig.3.2 (a) are: = � . � . = . � . ����� ������������, = +

Therefore, fxa Denoted by ax in the acceleration diagram shown in Fig.3.2 (b) is inclined to XA at the same angle . Triangles abx and ABX are similar. Thus, point x can be fixed on the acceleration image,

corresponding to point X on the link. ����� ������������ �� � �������� �� , = ����� ������������ �� � �������� �� , =

Acceleration Image: Acceleration images are helpful to find the accelerations of offset points of the links.

The acceleration image is obtained in the same manner as a velocity image. An easier method of making Δabx similar to ΔABX is by making AB on AB equal to ab and drawing a line

parallel to BX, meeting AX in X . AB X is the exact size of the triangle to be made on ab.

Take ax = AX and bx = B X

Thus the point x is located.

The method is illustrated in Fig.3.2(c).

3.3.3 Absolute Acceleration for a Link Consider the rigid link AB such that point B is rotating about A with angular velocity ω and angular acceleration , as shown in Fig. . (a). The point A itself has acceleration fa. The acceleration diagram is

shown in Fig.3.3 (b). ������� ����������� �� , = + + = + � . + . . Similarly for any other point X, ������� ����������� �� �, = + + = + � . + . .

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Fig.3.3 Absolute acceleration of a rigid link

3.3.5 Acceleration Diagram for Four-Bar Mechanism

The four-bar mechanism is shown in Fig.3.5 (a). The velocity of point B, vb = ω · AB. The velocity diagram is

shown in Fig.3.5 (b), In this diagram,

Fig.3.5 Acceleration diagram for four-bar mechanism

vb = ab ⊥ AB; dc ⊥ DC and bc ⊥ BC ���� �������� �� �������� �� , = = �������� �� �������� �� , =

Now we calculate the accelerations of various points and links.

= = ; = = ′ ; = = ′′ = + = ′ + ′ = + = ′′ + ′′

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The acceleration diagram is shown in Fig.3.5(c) to a suitable scale. To construct the acceleration diagram,

proceed as follows:

1. Draw ab = fb parallel to AB, which is known in magnitude and direction.

2. Draw bc = fncb parallel to BC, which is known in magnitude and direction.

3. Draw cc , representing fncb perpendicular to bc , which is known in direction only.

4. Now draw dc = fncd parallel to CD, which is known in magnitude and direction.

5. Draw c c, representing ftcd, perpendicular to dc to intersect c c at point c.

6. Join dc and bc. Then Acc����a��on o� C ���a���� �o B, = Acc����a��on o� C ���a���� �o D, = =

3.3.7 Acceleration Diagram for Slider-Crank Mechanism

(i) Velocity Diagram

For the slider crank mechanism as shown in Fig.3.7 (a). va = ω·OA. The velocity diagram is shown in Fig. .

(b) In this diagram,

Fig.3.7 Acceleration diagram of slider-crank mechanism

va = oa ⊥ OA

ab ⊥ AB

ob || OB

Then

Velocity of slider, vb = ob

Velocity of connecting rod, vba = ab

(ii) Acceleration Diagram

The acceleration diagram is shown in Fig.3.7(c), in which = � × = =

= = ′ The following steps may be adopted to draw the acceleration diagram:

1. Draw oa = fa parallel to OA on a convenient scale to represent the normal acceleration of the crank OA.

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2. Draw ab = fnba parallel to AB to represent the normal acceleration of the connecting rod, which is known in

magnitude and direction.

3. Draw a line perpendicular to AB at b to represent the tangential acceleration ftba of AB, which is unknown in

magnitude.

4. Draw a line parallel to OB at o to represent the acceleration fb of the slider B. Let these two lines intersect at b.

Join ab. Then

Linear acceleration of slider B, fb = ob

5. To find the acceleration of any point C in AB, we have

=

Locate point c in ab. Join oc. Then

Acceleration of point C, fc = oc

Example: A four-bar mechanism with ternary link is shown in Fig.3.9 (a). The lengths of various links is given

as below: O1O2 = 600 mm, O1A = 300 mm, AB = 400 mm, O2B = 450 mm, AC = 300 mm, BC = 250 mm, AD =

100 mm, and ∠AO1O2 = 75°.

Angular velocity of crank O1A = 20 rad/s

Angular acceleration of crank O1A = 100 rad/s2

Determine (a) acceleration of coupler AB, (b) acceleration of lever O2B, (c) acceleration of points C and D, and

(d) angular acceleration of ternary link.

Solution:

Linear velocity of A, va = ω × O1A = 20 × 0.3 = 6 m/s

1. Draw the configuration diagram as shown in Fig.3.9 (a) to a scale of 1 cm = 100 mm.

2. Draw the velocity diagram as shown in Fig.3.9 (b) to a scale of 1 cm = 1 m/s

3. Measure ab = vba = 3.5 cm = 3.5 m/s; o2b = vb = 4.7 cm = 4.7m/s; o1c = vc = 3.5 cm = 3.5 m/s, vca = ac = 2.7 cm =

2.7 m/s; vcb = bc = 2.1 cm = 2.1 m/s

4. Now ���� = ����

�� = . × = . c�

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5. Locate point d in ab and join o1d. Then vd = o1d = 5.5 cm = 5.5 m/s.

Fig.3.9 Four-bar mechanism with ternary link

6. Calculate the accelerations of various points as follows:

� = �� � = . = �/�

�� = � × � � = × . = �/�

� = ��� = .. = . �/�

� = �� � = .. = . �/�

� = ��� = .. = . �/�

� = ��� = .. = . �/�

7. Draw the acceleration diagram on a scale of 1 cm = 10 m/s2, as shown in Fig.3.9(c)

8. Measure the various required lengths on the acceleration diagram.

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Acceleration of coupler AB, fba = ab = 7.8 cm = 78 m/s2

Acceleration of lever O2B, fb = o2b = 8.6 cm = 86 m/s2

Acceleration of point C, fc = o1c = 7.9 cm = 79 m/s2

9. Locate point d in ab from the relation, �� = �� × ���� = . c�

Acceleration of point D, fd = o1d = 11 cm = 110 m/s2.

Tangential acceleration of AB, ftba = b b = 7.2 cm = 72 m/s2 An�u�a� acc����a��on o� ��, � = ���� = .. = �ad/� c�

3.4 Coriolis Acceleration

The total acceleration of a point with respect to another point in a rigid link is the vector sum of its normal

and tangential components. This holds true when the distance between the two points is fixed and the

relative acceleration of the two points on a moving rigid link has been considered. If the distance between the

two points varies, i.e., the second point which was considered stationary, now slides, the total acceleration

will contain one additional component, called Coriolis Component of Acceleration. Consider a slider B on a link OA such that when the link OA is rotating clockwise with angular velocity ω the slider B moves outward with linear velocity v, as shown in Fig. . . Let in time δt the angle turned through by link OA be δθ to occupy the new position OA and the slider moves to position E. The slider can be considered

to move from B to E as follows:

Fig.3.12 Concept of coriolis acceleration

1. From B to C due to angular velocity ω of link OA. 2. C to D due to outward velocity v of the slider.

3. D to E due to acceleration perpendicular to the rod, i.e., due to Coriolis acceleration. Now arc DE = arc EF − arc FD = arc EF − arc BC = OF · dθ − OB · dθ

= OF − OB dθ = BF · dθ = CD · dθ

Now

CD = v · dt

and

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dθ = ω · dt

Hence arc

DE = (v · dt) · (ω · dt)

Now arc DE = chord DE, as dθ is very small. ������o�� = �

Bu� = .

Where fcr is the constant Coriolis acceleration of the particle. Hence from (1) and (2), we get = � .

The direction of the Coriolis acceleration component is such so as to rotate the sliding velocity vector in the

same sense as the angular velocity of OB. This is achieved by turning the sliding velocity vector through 90° in

a manner that the velocity of this vector is the same as that of angular velocity of OA. The method of finding

the direction of Coriolis acceleration is illustrated in Fig.3.13.

Fig.3.13 Finding direction of Coriolis acceleration

Example: In the crank and slotted lever type quick return motion mechanism shown in Fig.3.14 (a), the crank

AB rotates at 120 rpm. Determine (a) velocity of ram at D, (b) magnitude of Coriolis acceleration component,

and (c) acceleration of ram at D · AB = 200 mm, OC = 800 mm, CD = 600 mm, OA = 300 mm.

Solution:

Velocity diagram

� = � × = . �ad/�

vb = vba = ω·AB = . × . = . m/s

1. Draw the configuration diagram to a scale of 1 cm = 100 mm, as shown in Fig.3.14 (a). By measurement, OP =

450 mm.

2. Draw the velocity diagram as shown in Fig.3.14 (b) to a scale of 1 cm = 0.5 m/s, by adopting the following

steps.

3. Draw ba ⊥ AB = 5.02 cm to represent the linear velocity of point B on link AB.

4. Let the coincident point of B on the slotted lever OC be P Draw a line at o perpendicular to OC and another

line at b parallel to OC to meet at point p.

5. By measurement, op = 4.2 cm. Then

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�� = �� × ���� = . × = . c�

6. Produce op to oc.

7. Draw a line parallel to XX, the line of stroke of the ram D, and a line at c perpendicular to CD to meet the

previous line at d. Then

Velocity of the slider D, vd = od = 7.8 cm = 3.9 m/s

Velocity of point C, vc = oc = 7.47 cm = 3.74 m/s An�u�a� ���oc��y o� ��o��d ����� �C, � = ��� = .. = . �ad �⁄ c�

vpo = op = 4.2 cm = 2.1 m/s

vbp = pb = 3 cm = 1.5 m/s

vcd = cd = 3.2 cm = 1.6 m/s

Acceleration Diagram

1. Calculate the accelerations of various points.

2. Coriolis acceleration of B relative to P, � � = � × � = × . × . = . �/�

The direction of coriolis component of acceleration is shown in Fig.3.14 (d), which is perpendicular to OC. � = ��� = . . = . �/�

� = ��� = .. = . �/�

� = ��� = .. = . �/�

3. Draw the acceleration diagram as shown in Fig.3.14(c) to a scale of 1 cm = 2 m/s2, by adopting the following

steps.

4. Draw ab = fnba = 15.75 cm parallel to AB.

5. Draw p b = fcrbp = 7.0 cm perpendicular to OC at point b, and another line perpendicular to OC representing

fsbp, the sliding acceleration of point B relative to point P. fbp = pb is the total acceleration of B relative to P.

6. Draw op = fnpo = 4.9 cm parallel to OP and another line p p perpendicular to it to meet the previous line at

point p.

7. Join bp and op. By measurement, op = 5.3 cm.

�� = �� × ���� = . × = . c�

8. Extend op to point c so that oc = 9.4 cm.

9. Draw cd = fndc = 2.13 cm parallel to CD and draw a line perpendicular to CD at d to represent the tangential

acceleration of CD.

10. Draw a line at o parallel to the line of action of the ram at D meeting previous line at d. Join cd. Then

Acceleration of the ram at D, fd = od = 2.5 cm = 5 m/s2

Velocity of ram = 3.9 m/s

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Coriolis acceleration = 14.025 m/s2

Acceleration of ram = 2.5 m/s2

Fig.3.14 Crank and slotted lever mechanism

. Klein’s Construction Klein s construction is used to draw the velocity and acceleration diagrams for a single slider-crank

mechanism on its configuration diagram. The line that represents the crank in the configuration diagram also

represents the velocity and the acceleration of its moving end to some scale. The following steps may be

adopted to draw the velocity and acceleration diagrams, as shown in Fig.3.24:

1. Draw the configuration diagram OAB of the slider-crank mechanism to a convenient scale.

2. Draw OI ⊥ OB and produce BA to meet O) at C. Then OAC represents the velocity diagram. OA = ωr to some scale from which the scale of velocity diagram is determined. OC = vbp, CA = vba, OA = vao.

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3. With AC as the radius and A as the centre, draw a circle.

4. Locate the mid-point D of AB. With D as centre and DA as radius, draw the circle to intersect the previously

drawn circle at E and F. Join EF intersecting AB at G.

5. EF meets OB at H. If it does not meet OB then produce EF to meet OB at H. Join AH.

6. Then OAGH is the acceleration diagram. = ; = Ac����a��on o� ����on ���d�� B , = � . = � . ; = � .

Fig. . Klein’s construction

7. To find the acceleration of any point X in AB, draw a line XX || OB to intersect HA at X. Join OX. Then = � . �

= = � .

4.1 Introduction

The lower pairs are those which have surface contact between them. We study, in this chapter, the various

mechanisms for the generation of straight line motion, both exact and approximate. These mechanisms have a vital role to play in generating the configurations for machines. The steering gears and (ooke s joint are other lower pairs that are discussed.

4.2 Pantograph

It is a mechanism to produce the path traced out by a point on enlarged or reduced scale. Fig.4.1 (a) shows

the line diagram of a pantograph in which AB = CD, BC = AD and ABCD is always a parallelogram. OQP is a

straight line. P describes a path similar to that described by Q. It is used as a copying mechanism.

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Fig.4.1 Pantograph

Proof: To prove that the path described by P is similar to that described by Q, consider Δs OAQ and OBP which

are similar, because ∠BOP is common.

∠AQO = ∠BPO being corresponding angles as AQ || BP.

��nc�, = = .

In the displaced position shown in Fig.4.1 (b), as all links are rigid,

B1O = BO, D1A1 = DA, A1O = AO

and P1B1 = PB, B1A1 = BA, A1Q1 = AQ

��nc�, = As A1B1C1D1 is a parallelogram, A1D1 || B1C1, i.e., A1Q1 || B1P1.

OQ1P1 is again a straight line so that Δs QA1Q1 and OB1P1 are similar.

= .

From Eqs. (4.1) and (4.2), we get

=

because OA = OA1 and OB = OB1.

Hence QQ1 is similar to PP1 or they are parallel. Thus path traced by P is similar to that of Q.

The pantograph is used in geometrical instruments, manufacture of irregular objects, to guide cutting

tools, and as indicator rig for cross head.

4.3 Straight Line Motion Mechanisms

Straight line motion can be generated by either sliding pairs or turning pairs. Sliding pairs are bulky and gets

worn out rapidly. Therefore, turning pairs are preferred over sliding pairs for generating straight line motion.

Straight line motion can be generated either accurately or approximately.

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4.3.1 Accurate Straight Line Motion Mechanisms

The mechanisms for the generation of accurate straight line motion are as follows:

1. Peaucellier Mechanism

2. Hart Mechanism

3. Scott-Russel Mechanism.

1. Peaucellier Mechanism: A line diagram of the Peaucellier mechanism is shown in Fig.4.2.

OR = OS and OO1 = O1Q.

Point P describes a straight line perpendicular to OO1 produced.

Fig.4.2 Peaucellier straight line mechanism

Proof: Δs ORQ and OSQ are congruent, because OR = OS, OR = QS and OQ is common.

∠ROQ = ∠SOQ. Also ∠OQR = ∠OQS and Δs PRO = Δs PSQ. ∠OQR + ∠RQP = ∠OQS + ∠SQP

But OQ is a straight line. ∠OQR + ∠RQP = ∠OQS + ∠SQP = 180°

Hence OQP is a straight line.

Now OR2 = OT2 + RT2

RP2 = RT2 + TP2

OR2 − RP2 = OT2 − TP2 = (OT + TP)(OT − TP) = OP · OQ

But OR and RP are always constant.

Hence OP · OQ = constant.

Draw PP1 ⊥ OO1 produced and join QQ1. Δs OQQ1 and OPP1 are similar, because ∠OQQ1 = ∠OP1P = 90° and ∠QOQ1 is common. Hence =

or OQ · OP = OQ1 · OP1 = constant

Now OQ1 = 2OO1 = constant

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Hence OP1 = constant

or point P moves in a straight line.

2. Hart Mechanism: The Hart mechanism is shown in Fig.4.3. OO1 is a fixed link and O1Q is the rotating link.

The point Q moves in a circle with centre O1 and radius O1Q. ABCD is a trapezium so that AB = CD, BC = AD and

BD || AC. = =

P describes a straight line perpendicular to OO1 produced as Q moves in a circle with centre O1.

Proof: �n � , = ��nc� ||

Fig.4.3 Hart Straight Line Mechanism

Δs ABD and AOP are similar, because ∠AOP = ∠ABD being corresponding angles and ∠DAB is common. ∴ = o� = . .

�����a��y a� =

Hence OQ || AC

Now AC || BD, OQ || BD, OP || BD Hence OP || OQ

Since O is a common point, therefore OQP is a straight line. Now Δs BOQ and BAC are similar. ∴ = o� = . .

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From Eqs. (4.3) and (4.4), we get

. = ( . ) . ( . ) = ( . ) . .

Draw DE ⊥ AC. = − = + . = − + = + − + = − + = − . From ΔAED,

AE2 = AD2 − DE2 From ΔGED,

EC2 = CD2 − DE2 − = − .

From Eqs. (4.6) and (4.7), we get . = − . . = ( . ) − = ��������

as AO, BO · AB, AD, and CD are fixed.

Hence P describes a straight line perpendicular to OO1 produced as point Q moves in a circle with centre O1.

3. Scott-Russel Mechanism: The Scott-Russel mechanism is shown in Fig.4.4. It consists of a sliding pair and

turning pairs. It can be used to generate approximate and accurate straight lines.

Fig.4.4 Scott-Russel straight line mechanism (CP = CQ)

1. When OC = CP = CQ, Q describes a straight line QO ⊥ OP provided P moves in a straight line along OP.

2. If CP ≠ CQ then Q describes an approximate straight line perpendicular to OP provided P moves along a

straight line OP such that

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=

Proof:

(a) As OC = CP = CQ, hence ∠POQ = 90° and OQ ⊥ OP. If P moves along OP then Q moves along a line

perpendicular to OP.

Hence OP = OC cos θ + CP cos θ = PQ cos θ ∠CPI = 90° − θ

Now ∠ICP = ∠COP + ∠CPO = 2θ

Fig.4.5 Scott-Russel mechanism when CP ≠ CQ

Then ∠CIP = ° − ∠CPI − ∠ICP = ° − ° + θ − θ = ° − θ ∠CIP = ∠CPI

Or CI = CP ∴ CP = CQ = CO = CI and ∠POQ = 90°

Hence OPIQ is a square. As the path of Q is ⊥ OI, I is the instantaneous centre. Q will move along a line

perpendicular to OP.

(b) When CP ≠ CQ, it will form an elliptical trammel, as shown in Fig.4.5. + =

Because x = CQ cos θ and y = PC sin θ

= ���� − ����� �������� − ����� ���� = �

As point C moves in a circle, for Q to move along an approximate straight line,

= �

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Limitations: When OC ⊥ OP, P coincides with O and OQ = 2OC. Here a small displacement of P shall cause a

large displacement of Q, requiring a relatively small displacement of P to give displacement to Q. This requires

highly accurate sliding surfaces.

4.3.2 Approximate Straight Line Motion Mechanisms

The mechanisms for the generation of approximate straight line motion are:

1. Grasshopper Mechanism

2. Watt Mechanism

3. Tchebicheff Mechanism

1. Grasshopper Mechanism: The Grasshopper mechanism is shown in Fig.4.6. The crank OC rotates about a

fixed point O · O1 is a fixed pivot for link O1P. For small angular displacements of O1P, point Q on link PCQ will

trace approximately a straight line path perpendicular to OP if = �

Fig.4.6 Grasshopper mechanism

2. Watt Mechanism: The Watt mechanism is shown in Fig.4.7 (a). Links OA and O1B oscillates about O and O1

respectively. AB is a connecting link. P will trace an approximate straight line if

=

In Fig.4.7 (b), θ and ϕ are the amplitudes of oscillation and I is the instantaneous centre of A′B′. P′ is the point

which lies on the approximate straight line described by P.

Proof: � = ′ and = ′

� = ′ . ( ′) .

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Fig.4.7 Watt mechanism

If ∠B′P′I = 90°, then sin θ = B’P’/B’) o� = ′ ′′ ′ ; � ≈ ′ ′′

� ≈ ′ ′′ . ′ ′′ ′ .

From Eqs. (4.9) and (4.10), we get ′ . ( ′) = ′ ′′ . ′ ′′ ′

�o� ′′ = ′ ′′

Δs OAA′ and IP′A are approximately similar, as well as Δs O1BB′ and B′P′I are approximately similar.

≈ ′ ′′ ′ ≈

Therefore, P divides the coupler AB in the ratio of the lengths of oscillating links. Hence P will describe an

approximate straight line for a certain position of its path.

3. Tchebicheff Mechanism: The Tchebicheff mechanism is shown in Fig.4.8. In this mechanism, OA = O1B

and AP = PB. P is the tracing point. Let AB = 1, OA = O1B = x, and OO1 = y. �o� = − − = − = + .

Draw AL ⊥ OO1, then OL = OO1 − O1L

Also O1L = AP1

and = ( − ) = ( − )

A�a�n = − = − − = +

Further OA2 = AL2 + OL2

= ( − ) + [ + ]

Fig.4.8 Tchebicheff mechanism

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o� = + + .

From Eqs. (4.11) and (4.12), we get + = + +

Or y = 2

Substituting in Eq. (4.11), we find that, x = 3.5

i.e., AB : OO1, : OA :: 1 : 2 : 3.5

P moves horizontally because instantaneous centre of AB will lie on point C which is the intersection of OA

and BO1. Thus C lies just below P.

4.4 Intermittent Motion Mechanisms

These mechanisms are used to convert continuous motion into intermittent motion. The mechanisms used

for this purpose are the Geneva wheel and the ratchet mechanism.

1. Geneva Wheel: The Geneva wheel as shown in Fig. 4.10 consists of a plate 1, which rotates continuously and

contains a driving pin P that engages in a slot in the driven member 3. Member 2 is turned ¼ th of a

revolution for each revolution of plate 1. The slot in member 2 must be tangential to the path of pin upon engagement in order to reduce shock. The angle is half the angle turned through by member 2 during the

indexing period. The locking plate serves to lock member 2 when it is not being indexed. Cut the locking plate

back to provide clearance for member 2 as it swings through the indexing angle. The clearance arc in the

locking plate will be equal to twice the angle α.

2. Ratchet Mechanism: This mechanism is used to produce intermittent circular motion from an oscillating or

reciprocating member. Fig.4.11 shows the details of a ratchet mechanism. Wheel 4 is given intermittent

circular motion by means of arm 2 and driving pawl 3. A second pawl 5 prevents 4 from turning backward

when 2 is rotated clockwise in preparation for another stroke. The line of action PN of the driving pawl and

tooth must pass between centers O and A in order to have the pawl 3 remain in contact with the tooth. This

mechanism is used particularly in counting devices.

Fig.4.10 Geneva wheel Fig.4.11 Ratchet mechanism

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4.5 Parallel Linkages

These are the four-bar linkages in which the opposite links are equal in length and always form a

parallelogram. There are three types of parallel linkages: parallel rules, universal drafting machine and lazy

tongs. They are used for producing parallel motion.

1. Parallel Rules: A parallel rule is shown in Fig.4.12, in which AB = CD = EF = GH = IJ and AC = BD, CE = DF, EG

= FH, GI = HJ. Here AB, C, EF, GH and IJ will always be parallel to each other.

2. Universal Drafting Machine: A universal drafting machine is shown in Fig.4.13, in which AB = CD; AC = BD;

EF = GH; and EG = FH. Position of points A and B are fixed. Similarly the positions of points E and F are fixed

with respect to C and D. The positions of scales I and II are fixed with respect to points G and H. Then ABDC is

a parallelogram. Line CD will always be parallel to AB so that the direction of CD is fixed. Therefore, the

direction of EF is fixed. Further EFHG is a parallelogram, so GH is always parallel to EF such that the direction

of GH is also fixed, whatever their actual position may be.

3. Lazy Tongs: Lazy tongs, as shown in Fig.4.14, consists of pin joint A attached to a fixed point. Point B moves

on a roller. All other joints are pin jointed. The vertical movement of the roller affects the movement of the

end C to move in a horizontal direction. It is used to support a telephone or a bulb at point C for horizontal

movements.

Fig.4.12 Parallel rule Fig.4.13 Universal drafting machine

Fig.4.14 Lazy tongs

4.7 Automobile Steering Gear Mechanisms

The mechanisms that are used for changing the direction of motion of an automobile are called steering gears.

Steering is done by changing the direction of motion of front wheels only as the rear wheels have a common

axis which is fixed and moves in a straight line only.

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4.7.1 Fundamental Equation for Correct Steering

When a vehicle takes a turn all the wheels should roll on the road smoothly preventing excessive tyre wear.

This is achieved by mounting the two front wheels on two short axles, called stub axles. The stub axles are

pin-jointed with the main front axle which is rigidly attached to the rear axle, on which the back wheels are

attached. Steering is done by front wheels only.

Figure 4.22 shows an automobile taking right turn. When the vehicle takes a turn, the front wheels along with

stub axles turn about their respective pin-joints. The inner front wheel turns through a greater angle as

compared to that of outer front wheel. In order to avoid skidding, for correct steering, the two front wheels

must turn about the same instantaneous centre I, which lies on the axis of the rear wheels. Thus, the condition

for correct steering is that all the four wheels must turn about the same instantaneous centre.

Let AE = l = wheel base; CD = a = wheel track

AB = b = distance between the pivots of front axles

I = common instantaneous centre of all wheels

θ, ϕ = angles turned through by stub-axles of the inner and outer front wheels respectively

Fig.4.22 Automobile steering gear �o�, = − = ���� − ���

���� − ��� = .

Eq. (4.12) is called the Fundamental Equation for Correct Steering.

4.7.2 Steering Gears

A steering gear is a mechanism for automatically adjusting values of θ, ϕ for correct steering. The following

steering gears are commonly used in automobiles:

1. Davis Steering Gear

2. Ackermann Steering Gear

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1. Davis Steering Gear

Davis steering gear is shown in Fig.4.23. This type of gear has only sliding pairs. Two arms AG and BH are

fixed to the stub axles AC and BD respectively. CAG and DBH form two similar bell-crank levers pivoted at A

and B respectively. KL is a cross-link which is constrained to slide parallel to AB. The ends of the cross-link KL

are pin-jointed to two sliders S1 and S2 as shown. These sliders are free to slide on links AG and BH

respectively. The whole mechanism is in front of the front wheels.

During the straight motion of the vehicle, the gear is in the mid-position, with equal inclination of the arms AG

and BH with the verticals at A and B. The steering is achieved by moving cross-link KL to the right or left of

the mid-position. The steering gear for taking a right turn is shown in Fig.4.24. K′L′ shows the position of the

cross-link KL while taking a right turn.

Determination of Angle α:

Let x = distance moved by KL from mid-position = KK′ − LL′

y = horizontal distance of points K and L from A and B respectively

h = vertical distance between AB and KL

α = angle of inclination of track arms AG and BH with the vertical in mid-position

θ, ϕ = angles turned by stub axles

Fig.4.23 Davis steering gear for straight drive

From Fig.4.24, we have

y = AK sin α = BL sin α

��� − = − ��� − ���+ ��� ��� = −

�o� ��� =

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Fig.4.24 Davis steering gear taking a right turn

∴ − ���+ ��� = −

��� = − + .

�����a��y, ��� + � = +

�o ��a� ���� = + + .

From Eqs. (4.16) and (4.17), we get

���� − ��� = =

��� = =

Generally b/l =0.4 to 0.5, so that α = 11.3° to 14.1°. There will be friction and more wear due to sliding pairs

in the Davis steering gear. Therefore, it becomes inaccurate after some use. On account of these reasons, it is

not used in practice.

2. Ackermann Steering Gear

The Ackermann steering gear has only turning pairs. Fig.4.25 shows the gear for straight drive. The turning

pairs are: AK, KL, LB and AB. The two short arms AK and BL are of equal length and are connected by pin-

joints with front wheel axle AB at A and B respectively. AC and BD are the stub-axles so that CAK and DBL form

bell-crank levers. ABLK form a four - bar linkage. KL is the track rod. For correct steering, we have ���� − ��� =

��n��a��y = . �o . ≈ .

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Fig.4.25 Ackermann steering gear for straight drive

Determination of Angle α

1. Analytical Method: Consider the Ackermann steering gear, as shown in Fig.4.26 (a), taking a right turn.

The instantaneous centre I lies on a line parallel to the rear axis at a distance of approximately 0.3l above the

rear axis. It may be seen that the whole mechanism of the Ackermann steering gear is on the back of the front

wheels. From Fig.4.26 (b), we have

Projection of arc K’K on AB = Projection of arc L’L on AB o� ′ = ′

AK [sin α − sin α − ϕ) = BL [sin (α + θ − sin α]

Now AK = BL

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

or sin α − sin α cos ϕ + cos α sin ϕ = sin α cos θ + cos α sin θ − sin α

sin α − cos ϕ − cos θ) = cos α (sin θ − sin ϕ)

��� = ��� − ����− ���� − ��� .

The values of θ and ϕ are known for correct steering. Hence, the value of α can be determined from, Eq.

(4.18).

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Fig.4.26 Ackermann steering gear in displaced position

2. Graphical Method to Determine Angle α: Draw a horizontal line OX (Fig.4.27). Make ∠XOQ = θ and ∠XOP

= ϕ. With any radius, draw arc PRQ. Join PR and produce it so that PR = RM. Join MQ and produce. Draw ON ⊥

NQM. Then α = 90° − ∠XON.

Fig.4.27 Graphical method to determine angle α

Proof:

In Fig.4.28, let BL = x so that b = c + 2x sin α.

For correct steering,

b = l(cot ϕ − cot θ) = c cos ψ + x sin (α + θ) + x sin (α − ϕ)

If ψ is small, c cos ψ = c.

c + 2x sin α = c + x sin (α + θ) + x sin (α − ϕ)

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or 2 sin α = sin (α + θ) + sin (α + θ)

Fig.4.28 Graphical method to determine angle α

In Fig.4.29, let r = OP = OR = OQ. In right angled triangle OQN,

∠NOQ = 90° − α − θ = 90 − α − θ)

ON = r cos[90° − α + θ)] = r sin(α + θ)

Draw RL ⊥ ON then RL || SN

i.e. ∠SRQ = ∠α

Also ∠ORL = ∠α

OL = r sin α

Fig.4.29 Graphical method to determine angle α

Now NL = RS. )n ΔOPR,

���� = ��� ° − . �

= ������� . � = ��� . �

From triangle RSM in Fig.4.30, we have

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Fig.4.30 Graphical method to determine angle α ∠MRS = α − 0. ϕ ∠SMR = 90° − α − 0. ϕ)

A��o = ��� − . �

Now RM = PR

= ��� − . �

o�, = ��� . � ��� − . � = ��� − ��� − �

ON = OL + LN = OL + RS ��� + = ��� + ��� − ��� − � o�, ��� = ��� + + ��� − � .

o� ��� = ��� − ��� �− ��� − ���� .

Hence proved

4.8 (ooke’s Joint or Universal Coupling

It is a device to connect two shafts whose axes are neither coaxial nor parallel but intersect at a point. This is

used to transmit power from the engine to the rear axle of an automobile and similar other applications. It is

also called universal coupling. The (ooke s joint, as shown in Fig.4.31, consists of two forks connected by a centre piece, having the shape of

a cross or square carrying four trunnions. The ends of the two shafts to be connected together are fitted to the

forks.

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Fig.4.31 (ooke s joint

4.8.1 Velocities of Shafts

Let the driving shaft A and the driven shaft B be connected by a joint having two forks KL and MN. The two

forks are connected by a cross KLMN intersecting at O, as shown in Fig.4.32. Let the angle between the axes of

the shafts be α. Let fork KL move through an angle θ in a circle to the position K1L1 in the front view. The fork

MN will also move through the same angle θ. MN being not in the same plane shall move in an ellipse in the

front view to its new position M1N1. To find the true angle, project M1 to the top view, which cuts the

horizontal axis in R and fork MN in R1. Rotate R1 to R2 on the horizontal axis with centre O. Project R2 back in

the front view cutting the circle in M2. Join OM2 Measure angle MOM2, which is the true angle ϕ. Thus when

the driving shaft A revolves through an angle θ, the driven shaft B will revolve through an angle ϕ. Now ���� = ′′ ; ��� = ′′ ������� = ′′ × ′ ′

Bu� ′ = ′

∴ ������� = ′′ =

Now OR = OR1 cos α ������� = ���

��� = ��� ����

An�u�a� ���oc��y o� d����n ��a��, � =

An�u�a� ���oc��y o� d����n ��a��, � = �

Differentiating Eq. (4.20), we get

Fig.4.33 Speed polar diagram

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��c �. �� = co� �. ��c � �� �������� = co� �− ��n �. co� �

�� = ���− ��� . ��� .

Case 1: �� = co� � = − ��n �. co� �

��� = ±√ + ��� .

This condition occurs once in each quadrant, as shown in

Fig.4.33 by points 1, 2, 3 and 4.

Case 2: ωB/ωA is minimum when the denominator is maximum,

i.e., sin2 α cos2 θ must be minimum, or cos θ = 0°.

Thus ωB/ωA is minimum at θ = 90°, 270°, i.e. at points 6 and 8. Then �� = ��� .

Case 3: ωB/ωA is maximum when denominator is minimum,

i.e., when cos2 θ = 1, or cos θ = ±1, θ = 0° or 180°,

i.e. at points 5 and 7. Then �� = ��� .

Case 4: �a���u� ��uc�ua��on o� ���oc��y o� d����n ��a�� = � − ��

�o� � = �

�a���u� ����d ��uc�ua��on = ���� − � ���� = − ������ = ��� ��� .

For α to be small, tan α ≈ α and sin α ≈ α ������ ����� ����������� ≈ .

4.8.2 Angular Acceleration of Driven Shaft �� = ���− ��� . ��� � = � . . ��� [ ��� . ��� . ���− ��� . ��� ]

Fig.4.32 (ooke s coupling forks in displaced position

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An�u�a� acc����a��on o� d����n ��a��, = � ��� . ��� ���− ��� . ��� .

�o� acc����a��on �o b� �a���u� o� ��n��u�, = = � ��� . ��� [ ��� − ��� . ��� ��� ��� − − ��� . ��� . ���− ��� . ��� ]

= ��� + (��� − ) ��� − =

o�, ��� ≈ ���− ��� .

4.9 Double Hooke’s Joint The double (ooke s joint, as shown in Fig.4.34, is used to maintain the speed of driven shaft equal to the

driving shaft at every instant. To achieve this, the driving and the driven shafts should make equal angles with

the intermediate shaft and the forks of the intermediate shaft should lie in the same plane. Let γ be the angle

turned by the intermediate shaft while the angle turned by the driving shaft and the driven shaft be θ and ϕ

respectively. Then

tan θ = cos α tan γ and tan ϕ = cos α tan γ

Therefore, θ = ϕ

Fig.4.34 Double (ooke s joint

However, if the forks of the intermediate shafts lie in perpendicular planes to each other, then the variation of

speed of the driven shaft will be there.

(�� ) = ���

(�� ) = ���

(�� ) = ��� .

�����a��y (�� ) = ��� .