ME 2203 – KINEMATICS OF...

Chettinad College of Engineering & Technology

ME 2203 Kinematics of Machinery 1

ME 2203 – KINEMATICS OF MACHINERY

III SEMESTER B.E DEGREE

MECHANICAL ENGINEERING

2012 - 2013

COMPILED BY

S. RAJKUMAR., M.E., MBA., MSW., (Ph.D).,

Assistant Professor,

Department of Mechanical Engineering,

Chettinad College of Engineering & Technology,

Karur – 639114.

ME 2203 KINEMATICS OF MACHINERY UNIT I BASICS OF MECHANISMS



Definitions – Link, Kinematic pair, Kinematic chain, Mechanism, and Machine. –Degree

of Freedom – Mobility - Kutzbach criterion (Gruebler’s equation) -Grashoff's law-

Kinematic Inversions of four-bar chain and slider crank chain - Mechanical Advantage -

Transmission angle.

Description of common Mechanisms - Offset slider mechanism as quick return

mechanisms, Pantograph, Straight line generators (Peaucellier and Watt mechanisms),

Steering gear for automobile, Hooke’s joint, Toggle mechanism, Ratchets and

escapements - Indexing Mechanisms.

UNIT II KINEMATIC ANALYSIS

Analysis of simple mechanisms (Single slider crank mechanism and four bar

mechanism) - Graphical Methods for displacement, velocity and acceleration; Shaping

machine mechanism - Coincident points – Coriolis acceleration - Analytical method of

analysis of slider crank mechanism and four bar mechanism. Approximate analytical

expression for displacement, velocity and acceleration of piston of reciprocating engine

mechanism.

UNIT III KINEMATICS OF CAMS Classifications - Displacement diagrams - Parabolic, Simple harmonic and Cycloidal

motions – Graphical construction of displacement diagrams and layout of plate cam

profiles - circular arc and tangent cams - Pressure angle and undercutting.

UNIT IV GEARS Classification of gears – Gear tooth terminology - Fundamental Law of toothed gearing

and involute gearing – Length of path of contact and contact ratio - Interference and

undercutting - Gear trains – Simple, compound and Epicyclic gear trains - Differentials.

UNIT V FRICTION Dry friction – Friction in screw jack – Pivot and collar friction - Plate clutches - Belt and

rope drives - Block brakes, band brakes.

TEXT BOOKS

1. Ambekar A. G., Mechanism and Machine Theory, Prentice Hall of India,

New Delhi, 2007.

2. Uicker J.J.,Pennock G.R., Shigley J.E., “Theory of Machines and Mechanisms”

(Indian Edition), Oxford University Press, 2003.



REFERENCE BOOKS 1. Thomas Bevan, “Theory of Machines”, CBS Publishers and Distributors, 1984.

2. Ramamurti,V.,’ Mechanism and Machine Theory”, Second Edition, Narosa

Publishing House, 2005

3. Ghosh A and A.K.Mallick, “Theory of Mechanisms and Machines”, Affiliated East-

West Pvt. Ltd., New Delhi, 1998.

4. Rao J.S and Dukkipati R.V, “Mechanism and Machine Theory”, Wiley-Eastern

Ltd., New Delhi, 199 2.

5. John Hannah and Stephens R.C, “Mechanics of Machines”, Viva Low-Prices

Student Edition, 1999.

BIS Codes of Practice/Useful Websites

1. IS 2458 : 2001, Vocabulary of Gear Terms – Definitions Related to Geometry

2. IS 2467 : 2002 (ISO 701: 1998), International Gear Notation – Symbols for

Geometric Data.

3. IS 5267 : 2002 Vocabulary of Gear Terms – Definitions Related to Worm Gear

Geometry.

4. IS 5037 : Part 1 : 2004, Straight Bevel Gears for General

Engineering and Heavy Engineering - Part 1: Basic Rack.

5. IS 5037 : Part 2 : 2004, Straight Bevel Gears for General

Engineering and Heavy Engineering - Part 2: Module and Diametral Pitches.

WEBSITE: www.howstuffworks.com

UNIT I BASICS OF MECHANISMS

Definitions – Link, Kinematic pair, Kinematic chain, Mechanism, and Machine. –Degree of Freedom – Mobility - Kutzbach criterion (Gruebler’s equation) -Grashoff's law- Kinematic Inversions of four-bar chain and slider crank chain - Mechanical Advantage -Transmission angle. Description of common Mechanisms - Offset slider mechanism as quick return mechanisms, Pantograph, Straight line generators (Peaucellier and Watt mechanisms), Steering gear for automobile, Hooke’s joint, Toggle mechanism, Ratchets and escapements - Indexing Mechanisms.

PART- A (2 Marks)

1. What is meant by spatial mechanism?



Spatial mechanism have special geometric characteristics in that all revolute axes are parallel and perpendicular to the plane of motion and all prism axes lie in the plane of motion.

2. Define a kinematic link A part of a machine which moves relative to some other party is known as a kinematic link. 3. What is resistant body?

A body is said to be a resistant body if it is capable of transmitting the required forces with negligible deformation.

4. What are the characteristic of a link? A link has a following characteristic

1. It should have relative motion and 2. It must be a resistant body 5. What is the deference between a rigid link and a flexible link?

A rigid link is one which does not under go any deformation while transmitting motion. A flexible link is one which is partly deformed in a manner not to affect the transmission of motion

6. What is fluid link?

A fluid link is one which is formed by a fluid in a system and the motion is transmitted through the fluid by pressure or compression only.

7. Define structure with application It is an assemblage of a number of resistant bodies having no relative motion between them and meant for carrying loads having straining action.

E.g. Railway Bridge, Roof truss and Machine frame. 8. Differentiate machine and structure.

1. The party of a machine move relative to one another whereas the member of a structure do not move relative to one another

2. A machine transforms the available energy into some useful work, whereas in a structure no energy is transformed into useful work .

3. links of a machine transmit power and motion ,but members of the structure transmit forces only

9. Define kinematic link or element. (May 2010)

Each part of a machine which moves relative to some other part is known as kinematic link or element.

10. List out the types of links. The types of links are:

1. Rigid link 2. Flexible link 3. Fluid link

11. Define rigid link, flexible link and fluid link.

Rigid link: A rigid link is one which does not undergo any deformation while transmitting motion.



Flexible link: It is one which is partly deformed in a manner not to affect the transmission motion. Eg. Belt, rope. Fluid link: It is one which is formed by having a fluid in a closed and the motion is transmitted through the fluid by pressure.

12. Define pair, kinematic pair and list out the classification of kinematic pair. The two links or elements of a machine, when in contact with each other are said to form a pair. If relative motion between them is completely or successfully constrained, the pair is known as kinematics pair Kinematic pairs may be classified:

1. According to the type of relative motion between the elements. 2. According to the type of contact between the elements. 3. According to the type of closure.

13. List out the type of kinematic pairs according to the type of relative motion between the elements.

1. Sliding pair 2. Turning pair 3. Rolling pair 4. Screw pair 5. Spherical pair

14. List out the type of kinematic pairs according to the type of contact between the elements.

1. Lower pair 2. Higher pair

15. List out the type of kinematic pairs according to the type of closure. 1. Self closed pair 2. Force-closed pair

16. Define sliding pair, turning pair, rolling pair, screw pair and spherical pair

When the two elements of a pair are connected in such a way that one only slide relative to the other, the pair is known as sliding pair.

When the two elements of a pair are connected in such a way that one can revolve about a fixed axis of another link, the pair is known as turning pair.

When the two elements of a pair are connected in such a way that one roll over another fixed link, the pair is known as rolling pair. When the two elements of a pair are connected in such a way that one element can turn about the other by screw threads, the pair is known as screw pair. When the two elements of a pair are connected in such a way that one element turns about the other fixed element, the pair is known as spherical pair.

17. Define Lower pair and Higher pair. (May 2010)



When the two elements of a pair have a surface contact when the relative motion take place and the surface of one element slides over the surface of other, the pair formed is known as lower pair.

When the two elements of a pair have a line or point contact when relative motion takes place and the motion between the two elements is partly turning and partly sliding, the pair is known as higher pair.

18. Define self closed pair and force closed pair. When the two elements of a pair are connected together mechanically in such

a way that only required kind of relative motion occurs, it is known as self closed pair. Example: Piston and cylinder

When the two elements of a pair are not connected mechanically but are kept in contact by the action of external forces, the pair is said to be a force closed pair. Example: Cam and follower

19. List out the types of constrained motions. The types of constrained motion are: 1. Completely constrained motion. 2. Successfully constrained motion and 3. Incompletely constrained motion.

20. Define completely, successfully and incompletely constrained motions When the motion between a pair is limited to a definite direction irrespective of the direction of force applied, the motion is said to be a completely constrained motion. When the motion between the elements, forming a pair is such that the constrained motion is not completed by itself, but by some other means is called as successfully constrained motion. When the motion between a pair can takes place in more than one direction, the motion is said to be in incompletely constrained motion.

21. Define mechanism and list out its types. When one of the links of a kinematic chain is fixed, the chain is called a mechanism.

The types are: 1. Simple mechanism 2. Compound mechanism 3. Complex mechanism 4. Planer mechanism 5. Spatial mechanism

22. Define a kinematic chain (May 2010) When the kinematic pairs are coupled in such a way that the last link is joined to the first link to transmit definite motion, it is called a kinematic chain.

23. Writ the equation for the number of links and number of joints as applied to kinematic chains with lower pair. (May 2010)

Number of links, 42 −= pl

Where, l - Number of link p – Number of pair



Number of joints 22

3−= lj

Where, j - Number of joints

24. What are the different types of joint? Binary joint, Ternary joint, Quaternary joint are the various types of joint in a chain

25. Explain the different types of joint

If the two links are joined at the same connection, the joint is known as a binary joint. When three links are joined at the same connection, the joint is known as a ternary joint

When four links are joined at the same connection, the joint is called as quaternary joint .

26. List out the degrees of freedom of some system. The degrees of freedom of some system are:

1. A rigid body has 6 degrees of freedom. 2. A ball and a socket joint have three degrees of freedom. 3. A circular shaft rotating in a whole and also translating parallel to its axis

has two degrees of freedom. 4. The position of a crank of a slider crank mechanism has one degree of

freedom.

27. Define ‘Degree of freedom’ (or) what is meant by mobility? (May 2010) The ‘Mobility of a mechanism’ or ‘Degree of freedom’ is defined as the number of input parameters (usually pair variables) which must be controlled independently in order to bring the device into a particular position. The number of degree of freedom of a mechanism or kutzbach criterion (n) is given by,

hjln −−−= 2)1(3

l= Number of links, j= Number of joints and

h= Number of higher pairs.

28. Find out the value of mobility for a simple four bar mechanism without any higher pair using Kutzbach criterion .

For Four bar mechanism without any higher pair h=0, l=4, j=4 Apply Kutzbach criterion

hjln −−−= 2)1(3

n = 3(4-1)-2(4)-0 n = 9-8 n = 1

29. Give the expression for grubler’s criterion to determine the number of

degrees of freedom of planer mechanism.

212)1(3 PPNF −−−=

Where F= degrees of freedom N= total number of links in a mechanism.



P1=number of pairs having one degree of freedom (lower pairs) P2=number of pairs having two degrees of freedom (higher pairs)

30. List out the four types of joints commonly found in planar mechanism.

The four types of joints are: 1. Revolute 2. Prismatic 3. Rolling contact 4. Cam or Gear joint

31. What is mean by mechanism?

When one of the links of a kinematics is fixed, the chain is known as a mechanism Example: Typewriter, Minidrafter

32. What is an equivalent mechanism? Give some examples The new mechanism thus obtained has the same number of degrees of freedom as the original mechanism is called equivalent mechanism.

1. A turning pair replaced by sliding pair. 2. A spring can be replaced by two binary links and 3. A cam pair can be replaced by one binary link with two turning pairs at

each end.

33. State the grashoff’s law. (May 2010) Grashoff’s law states that “for a four bar mechanism the sum of the lengths of the

largest and shortest links should be less than the sum of the length of other links”.

34. What is meant by inversion of the mechanism? The method of obtaining different mechanisms by fixing different links in a

kinematic chain is known as inversion of the mechanism 35. list some of the inversions of single slider crank chain

The inversions of single slider crank chain are. 1) Pendulum pump (or) bull engine 2) Oscillating cylinder engine 3) Rotary IC engine 4) Crank and slotted lever quick return motion mechanism 5) With worth quick return motion mechanism

36. list some of the inversions of double slider crank chain

The inversions of double slider crank chain are. 1) Oldham’s coupling 2) Scotch yoke mechanism 3) Elliptical trammels

37. Name four kinds of kinematic chains employing four lower pairs.

1. Four bar chain or quadric cyclic chain 2. Single slider crank chain 3. Double slider crank chain and 4. Compound chain.



38. Mention the application of Oldham’s coupling. The application of Oldham’s coupling is to connect two parallel shafts, when the

distance between two shafts is small.

39. Differentiate single slider crank chain and double slider crank chain.

Single slider crank chain Double slider crank chain

1. It is a four link mechanism having three turning pairs and one sliding pair.

1. It is a four link kinematic chain having two turning pair and two sliding pair.

2. It converts rotary motion to reciprocating motion and vice versa.

2. It converts rotary motion to translatory motion and vice versa.

40. Define time ratio(Q)

Time ratio is the ratio of time of quicker stroke to that time of slower stroke.

Q= (time of quicker stroke/ time of cutting stroke) =β

β

−

+0

0

180

180

41. Define Toggle position. If the driver and coupler lie in the same straight line at this point mechanical

advantage is maximum. Under this condition the mechanism is known as toggle position.

42. Define pantograph.

A pantograph is an instrument used to reproduce to an enlarged or a reduced scale and as exactly as possible the path described by a given point.

PART- B (12 Marks)

1. A crank and slotted lever mechanism used in a shaper has a center distance

of 300mm between the center of oscillation of the slotted lever and the

center of rotation of the crank. The radius of the crank is 120mm. find the

ratio of the time of cutting to the time of return stroke.

Given

AB = 300mm AE = 120mm

To find Ratio of the time of cutting of the time of return stroke.


ME 2203 Kinematics of Machinery

10

Solution

From the figure,

−=∠

290sinsin 0 β

ABE

4.0300

120===

AB

AE

( )4.0sin2

90 10 −=−=∠β

ABE

=23.60

000 4.666.23902

=−=β

08.132=β

We know that,

72.18.132

8.1323603600

000

=−

=−

==β

β

β

α

rnstrokeTimeofretu

ingstrokeTimeofcutt

2. In a crank and slotted lever mechanism, the length of the fixed link is

300mm and that of the driving crank is 150mm. Determine the maximum

angle the slotted lever will make with the fixed link. Also determine the ratio

of the time of cutting and the return strokes. If the length of the slotted lever

is 700mm, what would be the length of the stroke, assuming that the line of

the stroke passes through the extreme positions of the free end of the

slotted lever? (May 2010)

Given

AB = 300mm = 0.3m AE = 150mm = 0.15m BP1 = 700mm = 0.7m

Solution



11

1. Inclination of the slotted bar with the fixed link:

Let =∠ABE inclination of the slotted bar with the vertical. Figure shows the

extreme positions of the crank.

We know that,

−=∠

290sinsin 0 β

ABE

5.03.0

15.0===

AB

AE

( )5.0sin2

90 10 −=−=∠β

ABE

030=β

2. Time ratio of cutting stroke to the return stroke: We know that,

00 302

90 =−β

0120=β

0

000

120

120360360 −=

−==

β

β

β

α

rnstrokeTimeofretu

ingstrokeTimeofcutt

=2. 3. Length of the stroke: Length of the stroke = P1P2 = 2(P1P)



12

( )

−=

290sin2 0

1

βBP

( )00 6090sin7.02 −= x

L = 450mm

3. In a Whitworth quick return mechanism, as shown in the figure, the distance

between the fixed centers 50mm and the length of the driving cranks is 7mm.

Find the ratio of the time of cutting stroke to the time of return stroke.

Given: AB = 50mm = 0.05m AE = 75mm =0.075m CE = 150mm = 0.15m ED = 135mm = 0.135m

Solution: The extreme position of the driving crank is shown in the figure. From the geometry of the figure,

667.0075.0

05.0

2cos

2

===AC

ABβ [ ]2ACAE =Q

00 4.962.482

== ββ

or

Ratio of time of cutting stroke to the time of return stroke:

0

000

4.96

4.96360360 −=

−==

β

β

β

α

rnstrokeTimeofretu

ingstrokeTimeofcutt

4. The distance between two parallel shafts is 18mm and they are connected

by an Oldham’s coupling. The driving shaft revolves at 160r.p.m. What will be

the maximum speed of the tongue of the intermediate piece along its groove?

Given: X= 18mm = 0.018m



13

N= 160r.p.m.

To find: Maximum velocity of sliding

sradN

/75.1660

160*2

60

2===

ππω

We know that maximum velocity of sliding in the Oldham’s coupling

ω*xv = 75.16*018.0=v smv /302.0=

5. Prove that the mechanism shown in the figure is a constrained kinematic

chain.

Solution: From the figure, l=6, j=7, and h= 0, p= 5 ∴Degree of freedom, n=3(l-1)-2j-h =3(6-1)-2*7-0 = 1 Thus it is a single degree of freedom system.

Two conditions to be satisfied are:

l= 2p-4 (I)

And j= 22

3−l (II)

Where l= Number of links =6

p= Number of pairs =5

j= Number of joints =7

From equation (I), 6=2x5-4 =6

So equation (I) is satisfied.

From equation (II), 7262

37 =−= x

∴L.H.S = R.H.S



14

So equation (II) is also satisfied.

Hence, the given mechanism is a constrained kinematic chain.

6. Determine the mobility (number of degree of freedom) of all the linkages

shown in figure,

Solution:

Number of degrees of freedom,

hjln −−−= 2)1(3

Where, l= Number of links

j= Number of joints

h= Number of higher pairs

From the figure (a),

l=5, h=0, It has two binary joints and two ternary joints.

So j=2+2x2=6

( )0

06213

2)1(3

=

−−−=

−−−=

n

xln

hjln

Therefore linkage forms structure.

For figure (b),

l=6, h=0, It has seven binary joints, so j=7



15

1

072)16(3

2)1(3

=

−−−=

−−−=

n

xn

hjln

Therefore it has single degree of freedom.

For figure (c),

l=3, j=2, h=1 (ie, roller constitutes a higher pair)

1

122)13(3

2)1(3

=

−−−=

−−−=

n

xn

hjln

Therefore the linkage has single degree of freedom.

For figure (d),

l=7 h=0, It has eight binary joints, so j=8

2

082)17(3

2)1(3

=

−−−=

−−−=

n

xn

hjln

Therefore it has two degrees of freedom.

For figure (e),

l=4, h=1 (ie, roller constitutes a higher pair)

j=3 (ie, it has three binary joints)

2

132)14(3

2)1(3

=

−−−=

−−−=

n

xn

hjln

Therefore the linkage has two degrees of freedom.

7. What are the different types of constraints? Explain each.

Types of constrained motions:

a. Completely constrained motion

b. Incompletely constrained motion

c. Successfully constrained motion.

a. Completely constrained motion:

When the motion between a pair is limited to a definite direction, then the motion

is

said to be a completely constrained motion.

Examples:



16

1. Square bar moving in a square hole (a)

2. A shaft with collars at its ends moving in a round hole (b).

b. Incompletely constrained motion:

When the motion between a pair can take place in more than one direction, then

the motion is called an incompletely constrained motion.

Examples:

A circular shaft moving in a circular hole has two types of motion; (a) rotary and (b)

reciprocating (sliding). So the motion is incompletely constrained and both the

motions are independent of each other.

c. Successfully constrained motion: When the motion between the elements is not completed by self, but by some

other means, then the motion is said to be successfully constrained motion.

Examples:

1. The motion of a piston inside the engine cylinder is not completed by itself but

due to the rotation of crank. Similarly the engine valves are not having their own

motion but they are operated by rocker arms.



17

2. A shaft in a foot-step bearing is also an example for successfully constrained

motion.

8. What is meant by degrees of freedom of a mechanism? Explain kutzbach

criterion for determining degree of freedom for mechanisms.

Degrees of freedom (Mobility):

The mobility of a mechanism is defined as the number of input parameters

(usually pair variables) which must be controlled independently in order to bring the

device into a particular position.

It is possible to express the number of degrees of freedom of a mechanism

interms of the number of links and the number of pair connections of a given type.

This is known as number synthesis.

Distinction between movability and mobility:

Movability includes the six degrees of freedom of the device as the whole, as

through the ground link were not fixed, and thus applies to a kinematic chain.

Mobility neglects these and considers only the internal relative motions, thus

applying to a mechanism.

Description of mobility: A link is supposed to have n degrees of freedom if it has n independent variables

associated with its position in the plane. These are the number of independent

parameters required to specify the position of all links of the mechanism with respect

to the fixed link.



18

Let there are two links 1 and 2, in which link 1 is fixed, as shown in figure. The link

2 has a point A over it and translated by coordinates XA and YA. It can be written as A

(XA, YA). Let us say the line joining points A and B makes an angle θ with the fixed

link 1 (OX). Thus link 2 is completely specified by three variables (XA, YA, θ). In other

words, it can be stated that an unconstrained rigid link in the plane has three degrees

of freedom.

Let number of links = l

In a mechanism, one of the links is to be fixed.

So, number of movable links= (l-1)

And Total number of degrees of freedom before they are connected to any other link=

3(l-1)

If j= number of binary joints or lower pairs and,

h= number of higher pairs.

Then the number of degrees of freedom of a mechanism (n) is given by

n=3(l-1)-2j-h

This equation is called kutzbach criterion for the movability of a mechanism having

plane mechanism.

Note:

If there are no higher pairs (ie, two degree of freedom pairs) then h=0, then

kutzbach criterion, n=3(l-1)-2j

9. Sketch and describe the working of two different types of quick return

mechanisms. Give examples of their application. Derive an expression for the

ratio of times taken in forward and return stroke for the same.

Crank and slotted lever quick return motion mechanism:

This mechanism is mostly used in shaping machines, slotting machines and in

rotary internal combustion engines.



19

In this mechanism, the link AC (i.e. link3) forming the turning pair is fixed, as

shown in the figure. The link3 corresponds to the connecting rod of a reciprocating

steam engine. The driving crank CB revolves with uniform angular speed about the

fixed centre C. A sliding block attached to the crank pin at B slides along the slotted

bar AP and thus causes AP to oscillate about the pivoted point A. A short link PR

transmits the motion from AP to the ram, which carries the tool and reciprocates

along the line of stroke R1R2. The line of stroke of the ram is perpendicular to AC

produced.

In the extreme positions, AP1 and AP2 are tangential to the circle and the cutting

tool is at the end of the stroke. The forward or cutting stroke occurs when the crank

rotates from the position CB1 to CB2 in the clockwise direction. The return stroke

occurs when the crank rotates from the position CB2 to CB1 in the clockwise direction.

Since the crank has uniform angular speed, therefore,

α

α

β

β

α

β −

−==

0

0

360

360or

rnstrokeTimeofretu

ingstrokeTimeofcutt

Since the tool travels a distance of R1R2 during the cutting and return stroke,

therefore travel of the tool or length of stroke

AQPAPQPPPRR 1112121 sin22 ∠====

2cos2

290sin2 0

1

ααAPAP =

−= APAP =1Q

AC

CBAPx 12=

AC

CB1

2cos =

αQ



20

AC

CBAPx 12= CBCB =1Q

Whitworth quick return motion mechanism:

It is a mechanism used in workshops i.e. in shaping and slotting machines to cut

metals. In this mechanism, link 2 (i.e. crank) is fixed, link 3 rotates, link 4 reciprocates

and link 1 oscillates as shown in figure.

Initially, let the slider 4 be at B1 point C2, then the tool will be in its extreme left

position. When the crank further rotates and the slider 4 reaches to point C1, the tool will

be in its extreme right position. The distance between extreme left and right positions is

the stroke length.

When the link AE rotates from the position AB1 to AB2, then the ram moves from

left to right and corresponding movement of the crank link will be from BC1 to BC2. When

the link AE further rotates, link AE moves from AB2 to AB1, then the ram moves from

right to left and the crank BC correspondingly moves from BC2 to BC1.

When crank moves from BC2 to BC1, it is the backward stroke. The time taken

during the left to right motion of the ram will be equal to the time taken by the crank from

BC1 position to BC2. The time taken during the right to left motion of the ram will be equal

to the time taken by the crank from BC2 position BC1. When the tool is in its left extreme

position, it is the return or backward stroke.

Let =α obtuse angle C1BC1 at B

=β Acute angle C1BC2 at B

α

α

β

β

β

α

−

−==

0

0

360

360or

rnstrokeTimeofretu

ingstrokeTimeofcutt

10. Sketch and explain inversions of double slider crank chain.



21

Double Slider Crank Chain:

A kinematic chain, which consists of two turning pairs and two sliding pairs, is

known as double slider crank chain. The three inversions of a double slider crank

chain are

a. Elliptical trammels

b. Scotch yoke mechanism

c. Oldham’s coupling.

a. Elliptical trammels:

It is an instrument used for drawing ellipses. This inversion is obtained by fixing

the slotted plate (link 4), as shown in the figure. The fixed plate or link 4 has two

straight grooves cut in it, at right angles to each other. The link 1 and link 3 are known

as sliders and form sliding pairs with link 4. The link AB (link 2) is a bar, which forms

turning pair with links 1 and 3.

When the links 1 and 3 slide along their respective grooves, any point on the link 2

such as P traces out an ellipse on the surface of link 4. A little consideration will show

that AP and BP are the semi-major axis of the ellipse respectively. This can be

proved as follows:

Fig. (a) Elliptical trammels

Fig. (b)



22

Let us take OX and OY as horizontal and vertical axes and let the link BA is inclined at an angle θ with the horizontal as shown in fig (b). Now the co-ordinates of the point P on the link BA will be

x=PQ=APcosθ; and y= PR=BPsinθ

Or θθ sin;cos ==BP

yand

AP

x

Squaring and adding,

( ) ( )

1sincos 22

2

2

2

2

=+=+ θθAP

y

AP

x

This is the equation of an ellipse. Hence, the path traced by point P is an ellipse

whose semi-major axis is AP and semi-minor axis is BP.

b. Scotch yoke mechanism:

This mechanism is used for converting rotary motion into a reciprocating motion.

The inversion is obtained by fixing either the link 1 or link 3. In the figure, link 1 is fixed. In

this mechanism, when the link 2 (which corresponds to crank) rotates about B as centre,

the link 4 (which corresponds to a frame) reciprocates. The fixed link 1 guides the frame.

c. Oldham’s coupling:

An Oldham’s coupling is used for connecting two parallel shafts whose axes are at

a small distance apart. The shafts are coupled in such a way that if one shaft rotates, the

other shaft also rotates at the same speed. This inversion is obtained by fixing the link 2

as shown in the figure. The shafts to be connected have two flanges (link 1 and link 3)

rigidly fastened at their ends by forging.



23

When the driving shaft A is rotated, the flange C (link 1) causes the intermediate

piece (link 4) to rotate at the same angle through which the flange has rotated, and it

further rotates the flange D (link3) at the same angle and thus the shaft B rotates. Hence,

links 1, 3 and 4 have the same angular velocity at every instant. A little consideration will

show that there is a sliding motion between the link 4 and each of the other links 1 and 3.

If the distance between the axes of the shafts is constant, the centre of

intermediate piece will describe a circle of radius equal to the distance between the axes

of the two shafts. Therefore, the maximum sliding speed of each tongue along its slot is

equal to the peripheral velocity of the centre of the disc along its circular path.

Let =ω Angular velocity of each shaft in rad/s, and

r= Distance between the axes of the shafts in metres.

∴Maximum sliding speed of each tongue (m/s)

rv ω=

11. Sketch and explain the different types of Ratchets and

Escapements.

Ratchets and escapements are used in locks, jacks, clockwork and other

applications requiring some form of intermittent motion. There are many different forms

of ratchets and escapements for different applications.

a. Ratchets and Escapements Mechanism 1:



24

The Ratchet as shown in figure, allows only one direction of wheel, pawl is

held against the wheel by gravity or a spring. This type of arrangement is used for lifting

jacks.

b. Ratchets and Escapements Mechanism 2:

An escapement used for rotary adjustment is shown in the figure. c. Ratchets and Escapements Mechanism 3: (Graham’s Escapement)

Graham’s escapement is used to regulate the movement of clockwork, as shown

in figure. Anchor drives a pendulum whose oscillating motion is caused by the two clicks

of the escapement wheel. One is a push click, the other a pull click. The lifting and

engaging of each click is caused by oscillation of the pendulum.



25

12. Sketch and explain the various inversions of a four bar chain.

The four bar chain is also known as quadric cycle chain. It has four links and four

pairs, which are turning in nature. Links are of different lengths.

One of the rotating links is known as the crank or driver and the other link as

follower or rocker. The member connecting the crank and the follower is known as

connecting rod and the fixed link is the frame. Some important inversions of this chain

are described here.

a. Coupled wheels of locomotive:

The mechanism of coupled locomotive is shown in fig. It consists of four links. The

opposite links are equal in length. Links 1 and 3 work as cranks so the mechanism is

known as double crank mechanism.



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b. Beam Engine:

It is also known as crank and lever mechanism. This mechanism is used to

convert the rotary motion into reciprocating one.

c. Pantograph:

Pantograph is advice used to reproduce a displacement in a reduced or an

enlarged scale. It is used for duplicating the drawings maps, plans, etc. It is a quadratic

cycle in the form of a parallelogram as shown in figure. All the four pairs are turning in

nature.

According to the geometry of the figure

''

AE

AE

AC

AC=

d. Watt mechanism:

This mechanism was invented by watt for his steam engine to guide the

piston rod. Links OA and BC are parallel in the mean position of the mechanism. They



27

are connected with a link AB. OA and BC ,D will trace an approximate straight line where

point D is located on AB such that OA

BC

DB

AD=

13. Write down the Grashof’s law for a four bar mechanism. What is

the

significance of Grashof’s law?

Grashof’s law states that the sum of the shortest and longest links cannot be

greater than the sum of the remaining two links lengths, if there is to be continuous

relative motion between two members.

The significance of Grashof’s law is,

1. Grashof’s law specifies the order in which the links are connected in a

kinematic chain.

2. Grashof’s law specifies which link of the four bar chain is fixed.

3. ( ) ( )qpls +≤+ Should be satisfied, if not, no link will make a complete

revolution relative to another.

Where,

s= Length of the shortest link

l= Length of the longest link, and

p & q are the lengths of other two links.

UNIT I

BASICS OF MECHANISMS

Assignment Topics/Questions

For Batch - A (01-33) PART A

1. Define Degrees of Freedom (Mobility).



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2. Write down the Kutzbach criterion for plane mechanism. 3. What is meant by spatial mechanism? 4. Define Transmission Angle and Mechanical Advantage

PART B

1. a) Explain any two inversion of four bar chain. b) Explain the first inversion of Single Slider Crank Chain

2. (i) Define and explain inversion of mechanisms. (ii) Sketch and explain oldham’s coupling. (iii) Design a four-bar crank rocker quick return mechanism to give a time ratio of 1.25 with rocker swing angle as 75° clockwise. Assume the output link (rocker) length as 50 mm and in the left extreme position it is vertical.

UNIT I

BASICS OF MECHANISMS


For Batch - A (34-66)

PART A

1 Define Kinematic Chain. 2. What are the essential design features of high speed cams? 3. Define (a) Module (b) Diametral Pitch of gears 4. Define Grashof’s Law.

PART B

1. (i) Distance between two parallel shafts connected by oldham’s coupling is 25 mm. Determine maximum speed of sliding of tongue of intermediate piece in the slot in

the flange if driving shaft is run at 250 rpm. (4) (ii) Discuss the application of Grashoff’s law in identifying the input and output motions of four-bar mechanism.

2. With the help of a neat sketch explain the working of Whitworth quick return mechanism..


Analysis of simple mechanisms (Single slider crank mechanism and four bar

mechanism) - Graphical Methods for displacement, velocity and acceleration; Shaping

machine mechanism - Coincident points – Coriolis acceleration - Analytical method of

analysis of slider crank mechanism and four bar mechanism. Approximate analytical

expression for displacement, velocity and acceleration of piston of reciprocating engine

mechanism.



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PART- A (2 Marks)

1. Define displacement analysis, velocity analysis and acceleration analysis.

Displacement analysis is defined as the analysis involved in determining the position of all links in a mechanism as the driven link is displaced. Velocity analysis is defined as the analysis which involves determining how fast certain points on the links of a mechanism are traveling. Acceleration analysis is defined as the analysis which involves determining the manner in which certain points on the links of a mechanism either speeding up or slowing down.

2. State the relative velocity method of analysis.

Relative velocity method of analysis states that “Two points that reside on the same

link can only have a relative velocity that is in a direction perpendicular to the line that connects the two points.

3. Define velocity image and acceleration image

The triangle in the shape of velocity polygon is termed as velocity image of the link.

The triangle in the shape of acceleration polygon is termed as acceleration image. 4. List out any two properties of acceleration image.

1. The acceleration image of each rigid link is a scale reproduction of the shape on the link in the acceleration polygon.

2. The point o in the acceleration polygon is the image of all points with zero absolute acceleration. It is the acceleration image of the fixed link.

5. Define apparent angular velocity and apparent angular acceleration

When two rigid bodies rotate with different angular velocities, the vector difference between the two bodies is called apparent angular velocity.

When two rigid bodies rotate with different angular acceleration, the vector difference between the two bodies is called apparent angular acceleration.

6. Define rubbing velocity. Rubbing velocity is defined as the algebric sum between the angular velocities of the two links which are connected by pin joints, multiplied by the radius of the pin. Rubbing velocity at the pin joint = rω Where ω =Angular velocity of turning member r=Radius of the pin.

7. List out the methods for determining the velocity of point on a link.

The methods are: 1. Instantaneous centre method 2. Relative velocity method

8. Define Instantaneous centre of rotation.

Any displacement of a rigid link having motion in one plane, can be considered as pure rotational motion of a rigid link as a whole about the same centre and that centre is called instantaneous centre of rotation.



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9. Define instantaneous centre of a moving body.

Instantaneous centre of a moving body may be defined as that centre which goes on

changing from one instant to another.

10. What is an instantaneous axis A line drawn through an instantaneous centre and perpendicular to the plane of motion is called instantaneous axis.

11. List out the properties of instantaneous centre and define number of instantaneous centre. (May 2010)

1. A rigid link rotates instantaneously relative to another link at the instantaneous centre 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.

The number of instantaneous centre is a mechanism which is equal to the number of

possible combinations of two links. Number of instantaneous centre:

2

)1( −=

nnN Where n= number of links.

12. State the Kennedy’s three centre inline theorem. (May 2010)

Kennedy’s theorem states that “If three bodies move relative to each other, they have three instantaneous centre and lie on a straight line”.

13. Define angular velocity ratio theorem. Angular velocity theorem states that “Ratio of the angular velocities of any two bodies moving in a constrained system is inversely proportional to the ratio of the distances of their common instantaneous centre from their centre of rotation”.

14. When the angular velocity is positive? The angular velocity ratio is positive only when the common instant centre falls outside the other two centres.

15. When the angular velocity ratio does become negative? The angular velocity ratio becomes negative only when the common instant centre falls in between the other two centres.

16. Define collineation axis The lines connecting instant centre are called collineation axis.

17. State the Freudenstein’s theorem Freudenstein’s theorem states that “At an extreme of the output to input angular velocity ratio of a four-bar linkage, the collineation axis is perpendicular to the coupler link”.

18. State the inverse of Freudenstein’s theorem. The inversion theorem states that “At an extreme value of the velocity ratio of a

four bar linkage occurs when the collineation axis is perpendicular to the follower”.



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19. List out the advantages of complex algebra methods The advantages of complex-algebra methods are

1. Increased accuracy 2. It gives solution by digital computer at a large number of positions once the

program is written. 3. The solutions of a loop-closure equation lead to tedious algebric

manipulation. 4. The solution given by complex algebra method to velocity analysis is quite

straight forward.

20. Define coupler curve and coupler point The path generated by a point on the coupler link is known as coupler curve. The generating point on the coupler link is called as coupler point or tracer point.

21. Define the term synthesis. Synthesis is the term given to describe the process of designing a mechanism that produces a desired output motion for a given input motion.

22. Define structural error and mechanical error.

The difference between desired motion and the actual motion produced is known as

structural error. The error resulting from tolerances in the length of links and bearing clearances is known as mechanical error.

23. When the Coriolis component of acceleration is calculated?

When a point on one link is sliding along another rotating link, such as in quick return motion mechanism, then the Coriolis component of acceleration is calculated.

24. Define Coriolis component of acceleration. (May 2010) If the distance between the two points does not remain fixed and the second point slides, the total acceleration will contain an additional component of acceleration known as Coriolis acceleration. Coriolis component of acceleration= ωv2 Where v= sliding velocity, ω = angular velocity.

25. What are coincident points?

When a point in one link is sliding along another rotating link, then the point is known as coincident point.

I.e., coincident point is the point on slider such that the slider slides along the another rotating link.

PART- B (12 Marks)

1. The mechanism shown has a crank 50 mm radius which rotates at 2000

rev/min. Determine the velocity of the piston for the position shown. Also determine the angular velocity of link AB about A.



32

SOLUTION

Note the diagrams are not drawn to scale. The student should do this using a

suitable scale for example 1 cm = 1 m/s.

This is important so that the direction at 90o to the link AB can be transferred to

the velocity diagram.

Angular speed of the crank ω = 2πN/60 = 2π x 2000/60 = 209.4 rad/s

(vA)O = ω x radius = 209.4 x 0.05 = 10.47 m/s.

First draw vector oa. (Diagram a)

Next add a line in the direction ab (diagram b)

Finally add the line in the direction of ob to find point b and measure ob to get

the velocity.

The velocity of B relative to O is 7 m/s.

The tangential velocity of B relative to A is the vector ab and this gives 9.2 m/s.

The angular velocity of B about A is found by dividing by the radius (length of AB).

ω for AB is then 9.2/0.09 = 102.2 rad/s. (note this is relative to A and not an

absolute angular velocity)

2. Find the angular velocity of the output link when the input rotates at a constant

speed of 500 rev/min. The diagram is not to scale.



33

SOLUTION First calculate ω1.

ω1 = 2π x 500/60 = 52.36 rad/s.

Next calculate the velocity of point B relative to A.

(VB)A = ω1 x AB = 52.36 x 1 = 52.36 m/s.

Draw this as a vector to an appropriate scale. Next draw the direction of velocity C

relative to B at right angles to the link BC passing through point b on the velocity

diagram.

Next draw the direction of the velocity of C relative to D at right angles to link

DC passing through point a (which is the same as point d). Point c is where the two lines

intersect,

Determine velocity cd by measurement or any other method. The velocity of point C

relative to D and is 43.5 m/s.

Convert this into angular velocity by dividing the length of the link DC into it. ω2 =

43.5/0.7 = 62 rad/s.

3. A piston, connecting rod and crank mechanism is shown in the diagram. The

crank



34

rotates at a constant velocity of 300 rad/s. Find the acceleration of the piston and

the

angular acceleration of the link BC. The diagram is not drawn to scale.

SOLUTION:

First calculate the tangential velocity of B relative to A.

(vB)A = ω x radius = 300 x 0.05 = 15 m/s.

Next draw the velocity diagram and determine the velocity of C relative to B.

From the

velocity diagram (vC)B = 7.8 m/s

Next calculate all accelerations possible and construct the acceleration diagram to

find the acceleration of the piston.

The tangential acceleration of B relative to A is zero in this case since the link has no

angular acceleration (α = 0).

The centripetal acceleration of B relative to A

aR = ω2x AB = 3002 x 0.05 = 4500 m/s2. The tangential acceleration of C relative to B is

unknown.

The centripetal acceleration of C to B

aR = v2/BC = 7.82 /0.17 = 357.9 m/s2.



35

The stage by stage construction of the acceleration diagram is as follows.

First

draw the centripetal acceleration of link AB (Fig.a). There is no tangential

acceleration so designate it ab. Note the direction is the same as the direction of

the link towards the centre of rotation but is starts at a and ends at b.

Next add the centripetal acceleration of link BC (Figure b). Since there are two

accelerations for point C designate the point c1. Note the direction is the same

as the direction of the link towards the centre of rotation.

Next add the tangential acceleration of point C relative to B (Figure c). Designate

it c1 c. Note the direction is at right angles to the previous vector and the length is

unknown. Call the line a c line.

Next draw the acceleration of the piston (figure d) which is constrained to be in

the horizontal direction. This vector starts at a and must intersect the c line.

Designate this point c.

The acceleration of the piston is vector ac so (aC) B = 1505 m/s2. The tangential

acceleration of C relative to B is c1 c = 4000 m/s2.

At the position shown the connecting rod has an angular velocity and acceleration

about its end even though the crank moves at constant speed.

The angular acceleration of BC is the tangential acceleration divided by the length

BC.

ω(BC) = 4000 / 0.17 = 23529 rad/s2.

4. The diagrams shows a “rocking lever” mechanism in which steady rotation of

the

wheel produces an oscillating motion of the lever OA. Both the wheel and the lever

are mounted in fixed centers. The wheel rotates clockwise at a uniform angular

velocity (�) of 100 rad/s. For the configuration shown, determine the following.



36

(i) The angular velocity of the link AB and the absolute velocity of point A.

(ii) The centrifugal accelerations of BC, AB and OA.

(iii)The magnitude and direction of the acceleration of point A.

The lengths of the links are as follows.

BC = 25 mm AB = 100 mm OA = 50 mm OC = 90 mm

SOLUTION The solution is best done graphically. First draw a line diagram of the mechanism to

scale. It should look like this.

Next

calculate the velocity of point B relative to C and construct the velocity diagram.

(vB)C = ω x

radius = 100 x 0.025 = 2.5 m/s



37

Scale the following velocities from the diagram.

(vA)O = 1.85 m/s {answer (i)} (vA)B = 3.75 m/s

Angular velocity = tangential velocity/radius

For link AB,ω= 3.75/0.1 = 37.5 rad/s. {answer (i)} Next calculate all the accelerations

possible.Radial acceleration of BC = 2 x BC = 1002 x 0.025 = 250 m/s2.

Radial acceleration of AB = v2/AB = 3.752/0.1 = 140.6 m/ s2.

Check same answer from 2 x AB = 37.52 x 0.1 = 140.6 m/ s2.

Radial Acceleration of OA is v2/OA = 1.852/0.05 = 68.45 m/ s2.

Construction of the acceleration diagram gives the result shown.

The acceleration of

point A is the vector o- a shown as a dotted line. Scaling this

we get 560 m/s2.

5. Find the angular acceleration of the link CD for the case shown.

SOLUTION:

First calculate or scale the length CB and find it to be 136 mm.

Next find the velocities and construct the velocity diagram. Start with link AB as this

has a known constant angular velocity.

(vB)A = ω x radius = 480 x 0.08 = 38.4 m/s



38

1. Next calculate all the accelerations possible.

2. The centripetal acceleration of B to A is 38.42/0.08 = 18 432 m/s2

3. The centripetal acceleration of C to D is 152/0.16 = 1406 m/s2

4. The centripetal acceleration of C to B is 312/0.136 = 7066 m/s2.

5. We cannot calculate any tangential acceleration at this stage.

6. The stage by stage construction of the acceleration diagram follows.

7. First draw the centripetal acceleration of B to A (Figure a). There is no tangential to

add on).

8. Next add the centripetal acceleration of C to B (figure b)

9. Next draw the direction of the tangential acceleration of C to B of unknown

length at right angles to the previous vector (figure c). Designate it as a c line.

10. We cannot proceed from this point unless we realize that points a and d are the

same (there is no velocity or acceleration of D relative to A). Add the

centripetal acceleration of C to D (figure d). This is 1406 m/s2 in the direction of link CD.

Designate it d c2.

Finally draw the tangential acceleration of C to D at right angles to the previous

vector to intersect the c line . From the diagram determine c2 c to be 24 000 m/s2. This

is the tangential acceleration of C to D. The angular acceleration of the link DC is then: α

(CD) = 24000/0.16 = 150 000 rad/s2 in a clockwise direction.

Note that although the link AB rotates at constant speed, the link CD has angular

acceleration.

6. A horizontal single cylinder reciprocating engine has a crank OC of radius

40 mm and a connecting rod PC 140 mm long as shown. The crank rotates at 3000 rev/min clockwise. For the configuration shown, determine the velocity and acceleration of the piston.

The sliding piston has a mass of 0.5 kg and a diameter of 80 mm. The gas

pressure acting on it is 1.2 MPa at the moment shown. Calculate the effective

turning moment acting on the crank. Assume that the connecting rod and crank

has negligible inertia and friction.



39

SOLUTION

Draw the space diagram to scale:

The moment arm should be scaled and found to be 34 mm (measured at right angles

to the connecting rod PC.

Calculate the velocity of C relative to O.

ω = 2πN/60 = 2π x 3000/60 = 314.16 rad/s

(VC)O = ω x radius = 314.16 x 0.04 = 12.57 m/s

Draw the velocity diagram.

From the velocity diagram we find the velocity of the piston is 11 m/s.

Next calculate all the accelerations possible. Point C only has a radial acceleration

towards O.

Radial acceleration of C is v2/radius = 12.572/0.04 = 314.16 m/s2

Point P has radial and tangential acceleration relative to C. Tangential acceleration is

unknown. Radial acceleration = (vP)C2/CP = 92/0.14 = 578.57 m/s2. Now draw the

acceleration diagram and it comes out like this.



40

The acceleration of the piston is 770 m/s2.

Now we can solve the forces.

Pressure force = p x area = 1.2 x 106 x π x 0.082/4 = 6032 N and this acts left to right.

Inertia force acting on the piston = M a = 0.5 x 770 = 385 N and this must be provided

by the pressure force so the difference is the force exerted on the connecting rod. Net

Force = 6032 – 385 = 5647 N.

The connecting rod makes an angle of 11o to the line of the force (angle scaled from

space diagram). This must be resolved to find the force acting along the line of the

connecting rod.

The force in the connecting rod is 5637 cos 11o = 5543 N.

This acts at a radius of 34 mm from the centre of the crank so the torque provided by the

crank is T = 5545 x 0.034 = 188.5 N


Assignment Topics/Questions For Batch - A (01-33)



41

PART A 1. Define Coincident points

2. Define Coriolis Component

3. How will you determine the magnitude and direction of coriolis component of

acceleration?

3. Define rubbing velocity at a pin joint.

PART B

1. (i) A four bar mechanism DABC has the following dimensions :

DA = 300 mm; CB = AB = 360 mm; DC = 600 mm. Link DC is fixed and angle ADC is

Driving link DA turns clockwise at 100 rpm. Constant driving torque is 50 N-m.

Determine the following: (1) Velocity of point B (2) Angular velocity of driven link CB

(3) Mechanical advantage of mechanism in this position (4) Resisting torque. (12)

(ii) Sketch a four-bar crank rocker mechanism in (1) Maximum transmission angle

position and (2) toggle position where mechanical advantage is infinity.

2. In a slider crank mechanism, the length of the crank and the connecting rod are

150

mm and 600 mm respectively./ The crank [position is 60° from IDC, the crank shaft

speed is 450 r.p.m. clockwise. Using analytical method Determine (1) Velocity and

acceleration of the slider, and (2) Angular velocity and angular acceleration of the

connecting rod.

UNIT II

KINEMATIC ANALYSIS


For Batch - B (34-66)

PART A

1. What are the types of instantaneous Centres?



42

2. Define Kennedy’s theorem.

3. Define Instantaneous centre.

4. Define Angular Velocity ratio theorem.

PART B

1. The Crank of a slider crank mechanisms rotates clockwise at a Constant speed of

300 r.p.m. The crank is 150 mm and connecting rod is 600 mm long. Determine 1. Linear

velocity and acceleration of the mid Point of the connecting rod, and 2. Angular velocity

and angular acceleration of the connecting rod, at a crank angle of 45° from inner dead

centre position.

2. (i) Find the number of instantaneous centres for a six link mechanism. State the

use of instantaneous centre method in kinematic analysis.

(ii) Crank of a slider crank mechanism rotates clockwise at a constant speed of 300 rpm,

crank and connecting rod are of lengths 150 mm and 600 respectively. Determine the

following; at a crank angle of 45° from inner dead centre position (1) Linear velocity and

acceleration of the midpoint of connecting rod (2) Angular velocity and angular

acceleration of the connecting rod.

UNIT III KINEMATICS OF CAMS Classifications - Displacement diagrams - Parabolic, Simple harmonic and Cycloidal

motions – Graphical construction of displacement diagrams and layout of plate cam

profiles - circular arc and tangent cams - Pressure angle and undercutting.

PART- A (2 Marks)

1. Define cam. (May 2010) A cam is a rotating machine, which gives reciprocating or oscillating motion to another element known as follower.

2. List out the uses of cam.



43

Cams are used in: 1. Automatic machines 2. Internal combustion engines 3. Machine tools 4. Printing control mechanisms 5. Spinning and weaving textile machineries.

3. List out the types of cam. Cams are classified according to

1. Shape 2. Follower movement and 3. Manner of constraint of the follower.

4. Give the classification of cam according to shape.

1. Wedge and flat cams 2. Radial or disc cams 3. Spiral cams 4. Cylindrical cams 5. Conjugate cams 6. Globoidal cams 7. Spherical cams.

5. Give the classification of cams according to manner of constraint of the follower.

1. Pre-loaded spring cam 2. Positive drive cam 3. Gravity cam.

6. List out the types of followers. (May 2010) Cam followers are classified according to

1. Shape 2. Movement 3. Location of line of movement.

7. Give the classification of cam follower according to shape. 1. Knife edge follower 2. Roller follower 3. Mushroom follower.

8. Give the classification of cam follower according to movement.

1. Reciprocating follower 2. Oscillating follower

9. Give the classification of cam follower according to Location of line of

movement. 1. Radial follower 2. Offset follower.

10. Define displacement diagram.

During the rotation of cam through one cycle of input motion, the follower executes

a series of events. These events are represented in a graphical form known as displacement diagram.



44

11. Define base circle, pitch circle and prime circle (May 2010) Base circle is the smallest circle that can be drawn to the cam profile. Pitch circle is a circle drawn from the centre of the cam through the pitch points. Prime circle is the smallest circle that can be drawn from the centre of the cam

and tangent to the pitch curve.

12. Define trace point and pitch point. Trace point is a reference point on the follower and is used to generate pitch

curve. Pitch point is a point on the pitch curve having the maximum pressure angle.

13. Define Pitch curve. It is the curve generated by the trace point as the follower moves relative to the

cam.

14. Define lift or stroke. It is the maximum travel of the follower from the lowest position to the top most position. It is also called throw or stroke.

15. List out the types of follower motions. (May 2010) The follower has the following motions:

1. Uniform velocity 2. Simple harmonic motion 3. Uniform acceleration and deceleration 4. Cycloidal motion.

16. Define dwell period. (May 2010) The period during which the follower remains at rest are known as dwell periods.

17. Define return. Return is the period in which the motion of the follower is towards the cam centre.

18. Define cycloid.

Cycloid is a curve traced by a point on a circle when the circle rolls without slipping

on a straight line.

19. What principle is used in constructing the cam profile?

In constructing the cam profile, the principle of kinematic inversion is used. (i.e.) the cam is imagined to be stationary and the follower is allowed to rotate in the opposite direction to the cam’s rotation.

20. Define tangent cam and circular arc cam.

When the flanks of the cam are straight and tangential to the base circle and nose circle, then the cam is known as tangent cam. When the flanks of the cam connecting the base circle and nose are of convex circular arcs, then the cam is known as circular arc cam.

21. List out the parameters by which cam size is defined. The cam size is defined by the following parameters.

1. Pressure angle 2. Radius of curvature of cam profile



45

3. Hub size.

22. List out the methods used to reduce the pressure angle. The methods used to reduce the pressure angle are

1. Increase the diameter of base circle 2. Increase the angle of rotation of the cam, thereby lengthening the pitch

curve for the specified follower displacement. The cam profile becomes flatter and the pressure angle becomes smaller.

3. Select the motion curve for a smaller pressure angle. 4. By changing the offset of the follower.

23. Define pressure angle.

The pressure angle is defined as the angle between the axis of the follower stem and

the line of the force exerted by the cam onto the roller follower, the normal to the pitch curve through the trace point.

24. Define undercutting.

Undercutting is defined as a method to limit addendum the driven gear so that it passes through the interference point thus giving a new beginning of contact.

25.Write the maximum velocity of follower moving in SHM

Vmax at outstroke π ws/2θ0 Vmax at return stroke = π ws/2θr 26.Write the maximum acceleration of follower moving in SHM.

a0= π²w²s/2(θ0 )2

aR = π² w²s/2(θR)²

27.Write the maximum velocity of follower in cycloidal motion.

VR = 2 w s/θR ; Vo = 2 w s/θ o

28.Maximum acceleration of follower moving in cycloidal motion. ao = 2π w² s/(θo )²; a r =2π w²s/(θR )²

29.What is prime circle? It is the smallest circle that can be drawn from center of cam shaft and tangent to pitch curve.

For knife-edge follower the prime circle & base circle are same. But in roller follower prime Circle is larger than base circle by radius of follower.

PART- B (12 Marks)

1. Design a cam for operating the exhaust valve of an oil engine. It is required to

give equal uniform acceleration and retardation during opening and closing of the



46

valve each of corresponds to 60 deg of cam rotation. The valve must remain in the

fully open position for 20 deg of cam rotation.

The lift of the valve is 37.5 mm and the least radius of the cam is 40mm.The

follower is provided with a roller of radius 20mm and its line of stroke passes through the

axis of the cam.

Solution

2. Draw the profile of a cam operating with a knife edged follower having a lift of

30mm. The cam raises the follower with SHM for 1500 of its rotation followed by a

period of dwell for 600. The follower descends for the next 1000 rotation of the cam

with uniform velocity, again followed by a dwell period. The cam rotates at a



47

uniform velocity of 120 rpm and has the least radius of 20mm. What will be the

maximum velocity and acceleration of the follower during the lift?

S= 30mm=0.3m; ө0=150*Л/180=2.618 rad; өR =100*Л/180=1.745 rad; N=120 rpm.

Maximum velocity and acceleration of the follower during the lift:

Angular velocity of the cam shaft,

ω = 2 Л N / 60 = 2 Л 120 / 60 =12.57 rad/s;

Maximum velocity of the follower during lift,

v0= Л ω S / 2 ө0 = 0.226 m/s;

Maximum Acceleration of the follower during lift,

ao= Л2 ω 2 S /2(ө0 )= 3.413 m/s2.

3. It is required to set out the profile of a cam to give the following motion to the

reciprocating follower with a flat mushroom contact face:

a. Follower to have a stroke of 20mm during 1200 of cam rotation.

b. Follower to dwell for 300 of cam rotation.

c. Follower to return to its initial position during 1200 of cam rotation and,

d. Follower to dwell for remaining 900 of cam rotation.



48

The minimum radius of the cam is 25mm. The outstroke of the follower is performed with

simple harmonic motion and the return stroke with equal uniform acceleration and

retardation.

4. Design a cam to raise a valve with SHM 50mm in 1/3 of revolution , keep it fully

through 1/12 revolution and to lower it with harmonic motion in 1/6 revolution . The

value remains closed during the rest of the revolution. The diameter of the roller is

20 mm and minimum radius of the cam is 25mm. the diameter of cam shaft is

25mm.the axis of the valve rod passes through the axis of the cam shaft. if the

cam shaft rotates at uniform speed of 100 rpm, find the maximum velocity and

acceleration of a valve during raising and lowering.



49

S= 50mm=0.5m; ө0=360*1/3=120 deg =2.094 rad; өR=360*1/6=60 deg= 1.047 rad;

N=100 rpm.

Diameter of the roller = 20 mm;

Radius of cam=25 mm;

Diameter of the cam shaft=25 mm; Dwell=1/12*360=30 deg.

Maximum velocity of the follower during its ascent and descent:


ω = 2 Л N / 60 = 2 Л 100 / 60 =10.47 rad/s;

Maximum velocity of the follower during ascent,

v0= Л ω S / 2 ө0 = 0.39 m/s;

Maximum Acceleration of the follower during ascent,

ao= Л2 ω 2 S /2(ө0)2= 6.17 m/s2.

Maximum velocity of the follower during descent,

v0= Л ω S / 2 өR = 0.78 m/s;

Maximum Acceleration of the follower during descent,

ao= Л2 ω 2 S /2(өR)2= 24.67 m/s2.



50

5. The following data are for a disc cam mechanism with roller follower:

Minimum radius of the cam=35mm;

Lift of the follower =40mm;

Offset of the follower =10mm;

Roller Diameter = 15mm;

Cam rotation angles are as mention below: During ascent-120 deg:

Dwell=80 deg;

During descent=80 deg.

Cam rotates in clockwise direction and the follower motion is simple harmonic

during both ascent and descent.

(i) Draw the Displacement diagram,

(ii) Draw the Cam profile.



51

6.A disc cam used for moving a knife edge follower with simple harmonic motion

during lift and uniform acceleration and retardation motion during rotates in

clockwise direction at 300 rpm.The line of motion of the follower has an offset

10mm to the right of the camshaft axis. The minimum radius of the cam is

30mm.The lift of the follower is 40 mm.the cam rotation angles are lift 60 deg;

dwell 90 deg; return 120 deg and remaining angle for dwell. Draw the cam profile

and determine the maximum velocity and acceleration during the lift and return.



52

S= 40mm=0.4m; ө0=60deg =1.047 rad; өR=120 deg= 2.094 rad; N=300 rpm.

Maximum velocity of the follower during its lift and return:


ω = 2 Л N / 60 = 2 Л 300 / 60 =31.416 rad/s;


v0= Л ω S / 2 ө0 = 1.8853 m/s;


ao= Л2 ω 2 S /2(ө0)2= 177.72 m/s2.

Maximum velocity of the follower during return,

v0= Л ω S / 2 өR = 1.2 m/s;

Maximum Acceleration of the follower during return,

ao= 4 ω 2 S /(өR)2= 36 m/s2.



53

7. The following particulars relate to a symmetrical circular cam operating a flat

faced follower: Least radius = 37.5 mm;

Nose radius =5mm

Lift of the follower = 18.75mm

Angle of lift= 75°

Speed= 600 r.p.m

Determine (1) the following dimensions of the cam and

(2) The acceleration of the follower

a) At the beginning of the lift

b) At the end of contact with the circular flank,

c) At the beginning of contact with nose and

d) At the apex of the nose

Given data: r1 =OD = 37.5 mm



54

r2 = QE = QK= 5 mm

x = JK = 18.75 mm

α = 75 °

N = 600 r.p.m

Solution: ` ω = 2πN / 60= 2π * 600 / 60 = 62.84 rad/s

1) Principal dimensions of the cam: a symmetrical circular cam operating a flat faced follower is shown in figure φ = angle of contact on the circular flank,

R = Radius of circular flank = PD, r = OQ = distance between cam centre and nose center, We know that the displacement or lift of the follower(x), 18.75 = r + r2 - r1 = r + 5 – 37.5 = r-32.5 r = OQ = 18.75 + 32.5=51.25 mm From the geometry of figure PQ = PE – EQ = PD – EQ = OD –EQ = OP + r1 - r2

= OP + 37.5 – 5 = (OP +32.5) mm Now from a triangle OPQ

(PQ) 2 = (OP) 2 + (OQ) 2- 2 *OP*OQ COS β

(OP +32.5)2 = (OP)2 +(51.25)2 -2 *51.25 COS (180° –75° )

After simplification, we get OP= 40.82 mm Radius of the circular flanks = PD = OP+ OD= OP+ r1

R= 40.82 + 37.5 + 78.32 mm PQ= OP+ 32.5= 40.82+32.5=70.32 mm From the triangle OPQ, the angle φ is given by

φsin

OQ=

βsin

pq

Sinφ = PQ

OQ βsin*= 51.25 * sin (180°-75°) / 73.22 = 0.675



55

φ = 42.47°

Acceleration of the follower: a) At the beginning of the lift(θ=00)

a= ω 2(R – r1)cos θ = (62.84)2(78.32-37.5)cos 00=161.19 m/s2

b) At the end of contact with the circular flank,( θ=φ )

a= ω 2(R – r1)cos φ

=(62.84)2(78..32-37.5)cos 42.47o=118 m/s2

c) At the beginning of contact with nose, (θ=φ )

a = -ω 2 * OQ cos(α -θ)= -=(62.84)2 * 51.25*cos(65 o-75 o)

= 170.63 m/s2

d) At the apex of nose(α -θ0 =00)

= -ω 2 * OQ cos(α -θ)= ω 2r= 202.38 m/ s2.

UNIT - III

KINEMATICS OF CAMS



PART A 1. Sketch the velocity and acceleration profile of a follower which moves with SHM.



56

2. Sketch the velocity and acceleration profile of a follower which moves with

uniform velocity.

3. What are the different motions of the follower?

4. Define pitch curve of the cam.

PART B

1. A cam drivers a flat reciprocating follower manner:

i. Follower moves outwards through a distance of 20 mm with simple

harmonic motion during the first 120o rotation of the cam

ii. Follower dwells during next 30oof cam rotation.

iii. Follower moves inward with simple harmonic motion for the next 1200

of cam rotation

iv. the follower dwells for the remaining period

Draw the profile of the cam, when the minimum radius of the cam is 25 mm. Also

calculate the maximum velocity and acceleration during outward and inward motion of

follower when the cam rotates with 200 rpm

2. A cam with minimum radius of 25mm,rotatig clockwise direction with a uniform

speed of 100 rpm is to be designed to give the motion for a roller follower as follows

1. The raise through 50mm during 120o rotation of cam with SHM

2. Fully raised through next 30o

3. To lower during 600 with UAUR

4. Dwell for the remaining period

Draw the profile of the cam when the line of stroke of the follower is offset by 15mm

from the axis of the cam shaft

UNIT III

KINEMATICS OF CAMS



PART - A

1. Define dwell period.

2. Explain offset follower.

3. List the classifications of cam followers based on shape.

4. What are the essential design features of high speed cams?



57

PART - B 1. Construct a tangent cam and mention the important terminologies on it. Also

derive the expressions for displacement, velocity, acceleration of a reciprocating

roller follower when the roller has contact with the nose.

2. A cam is to give the following motion to a knife edged follower:

(a) Outstroke during 60° of cam rotation

(b) Dwell for the next 30° of cam rotation

(c) Return stroke during next 60° of cam rotation and

(d) Dwell for the remaining 210° of cam rotation

The stroke of the follower is 40 mm and the minimum radius of the cam is 50 mm.

The follower moves with uniform velocity during both the outstroke and return

strokes. Draw the profile of the cam when (a) the axis of the follower passes

through the axis of the cam shaft, and (b) the axis of the follower is offset by 20

mm from the axis of the cam shaft.

UNIT IV GEARS Classification of gears – Gear tooth terminology - Fundamental Law of toothed gearing

and involute gearing – Length of path of contact and contact ratio - Interference and

undercutting - Gear trains – Simple, compound and Epicyclic gear trains - Differentials.

PART- A

1. Define spur gear. A spur gear is a cylindrical gear whose tooth traces are straight line generation of the reference cylinder. They are used to transmit rotary motion between parallel shafts.

2. Define addendum and dedendum

Addendum is the radial distance of a tooth from the pitch circle to the top of the tooth. Dedendum is the radial distance of a tooth from the pitch circle to the bottom of the tooth.

3. Define addendum and dedendum circle.



58

Addendum circle is the circle drawn through the top of the teeth and is concentric with pitch circle. Dedendum circle is the circle drawn through the bottom of the teeth. It is also called root circle.

4. Define circular pitch.

It is the distance measured on the circumference of the pitch circle from a point of one tooth to the corresponding point on the next tooth. It is denoted by Pc.

DTPc

π=

Where D = diameter of pitch circle. T= Number of teeth on the wheel.

5. Define diameter pitch. (May 2010)

It is the ratio of number of teeth to the pitch circle diameter. It is denoted as Pd.

PcD

TPd

π==

Where T= Number of teeth D= Pitch circle diameter.

6. Define module. It is the ratio of pitch circle diameter to the number of teeth.

Module,T

Dm =

7. Define clearance and clearance circle. Clearance is the radial distance from the top of the tooth to the bottom of the tooth in a messing gear. A circle passing through the top of the meshing gear is known as clearance circle.

8. Define total depth and working depth.

Total depth is the radial distance between the addendum and dedendum of a gear.

Working depth is the radial distance from the addendum circle to the clearance circle.

9. State the law of gearing (May 2010)

Law of gearing states that, “The common normal at the point of contact between a pair of teeth must always pass through the pitch point”.

10. Define velocity of sliding. (May 2010)

Velocity of sliding is the velocity of one tooth relative to its mating tooth along the common tangent at the point of contact.

11. Define conjugate action. When the tooth profiles are so shaped so as to produce a constant angular velocity ratio during meshing, then the surfaces are said to be conjugate.

12. When a set of gears is interchangeable? What are the conditions to be satisfied for interchangeability of all gears?



59

A set of gears is interchangeable when any two gears selected from set will mesh and satisfy the fundamental law of gearing. For interchangeability of all gears, the set must have the same circular pitch, module, diameter pitch, pressure angle, addendum and dedendum and tooth thickness must be one half of the circular pitch.

13. What are the standard pressure angle values? Give the advantages of lower and larger pressure angle. The standard pressure angle values are 14.50, 200.

The advantages of lower pressure angle are 1. High contact ratio 2. Reduced wear 3. Reduced tooth load 4. Reduced bearing loads

The advantages of larger pressure angle are 1. Broader teeth at the base 2. Stronger in bending 3. It is used on the pinion without undercutting the teeth.

14. Define interference.

The phenomenon when the tip of tooth cuts under the root on its mating gear is known as interference.

15. List out the ways to avoid interference. Interference can be avoided by:

1. Undercutting 2. Making sub teeth 3. Increasing the pressure angle. 4. Cutting the gears with long and short addendum gear teeth.

16. What is the principle reason for employing non standard gears? The principle reason foe employing non standard gears are

1. To eliminate undercutting 2. To prevent interference 3. To maintain reasonable contact ratio.

17. What is the advantage of non standard gears?

The advantage of non standard gears is the tendency to maintain a better balance of strength between pinion and gear.

18. Define gear.

A gear may be defined as any toothed member designed to transmit or receive motion from another member by successively engaging tooth.

19. What are the advantages and disadvantages of gear drive? (May 2010)

Advantages 1. It transmits exact velocity ratio 2. It is used to transmit large power 3. It has high efficiency.

Disadvantages 1. The manufacture of gears requires special tool and equipment. 2. The error in cutting teeth may cause vibrations and noise during operation.



60

20. Define helix angle.

It is the angle between a line drawn through one of the teeth and the centre line of the shaft on which the gear is maintained. It is denoted by β

21. List out the classification of toothed wheels.

The gears or toothed wheels may be classified as 1. According to the position of axes of the shafts

a. parallel b. intersecting c. non-intersecting and non-parallel

2. According to the peripheral velocity of gears a. low velocity

b. medium velocity c. high velocity 3. According to type of gearing

a. external gearing b. internal gearing c. rack and pinion

4. According to position of teeth on the gear surface a. straight b. inclined c. curved

22. What is maximum length of arc of contact for two mating gears to avoid interference?

(r+R) sinФ/2 23. What are simple gear train?

The arrangement of gear in parallel shafts, and gears are serially in mesh gears

(idle gears) have no effect on speed ratio. But the No. of idle gears/ decides the direction of rotation of driven gear.

24. Explain speed ratio & train value.

Speed ratio = speed of driven/speed of follower N1 /N2 = T2 / T1 Train value = speed of follower /speed of driver N2 /N1 = T1 / T2

25. What are compound gear train

When there are more than one gear on a shaft, its called compound tram’s of gear. Here the intermediate gears have role in fixing the speed ration. Speed ratio = speed of driven/speed of driven =

product of No. of teeth of driven/product of No.of teeth of driven train value = 1/speed ratio 26. What is Epicyclic gear train?

In a gear train, when axes of the shaft over which the gears are mounted, move relative to the fixed axis, its called Epicyclic gear train.

27. What type of gearing is used in watches? Reverted gear train.



61

PART- B (12 Marks)

1. A pinion having 30 teeth drives a gear having 80 teeth. The profile of the gear

in involute with 20o pressure angle, 12mm module and addendum 10mm. Find the

length of path of contact, arc of contact and contact ratio.

Given data:

Tp = 30 ; Ta = 80 ; φ = 30o m = 12 mm; Addendum = 10 mm To find:

i. Length of path of contact ii. Arc of contact and iii. Contact ratio

Solution: i. Length of path of contact;

We know that pitch circle radius of gear wheel, 2

amTR =

mm4802

8012=

×=

and pitch circle radius of pinion, mmmT

rp

1802

3012

2=

×==

Addendum radius of pinion, rA = r + addendum = 180 + 10 = 190 mm. Length of path of approach (KP),

φφ sin222RCosRRKP A −−=

( ) ( ) ooCos 20sin48020480480

222−−=

= 27.28 mm. and the length of path of recess (PL),



62

φφ sin222rCosrrPL

A−−=

( ) ( ) ooCos 20sin18020180190

222−−=

Length of path of contact (KL) = Length of path of approach (KP)

+ Length of path of recess (PL) = 27.28 + 24.98 = 52.26 mm

ii. Length of arc of contact: Length of arc of contact =

oCps 20

26.52=

= 55.61 mm iii. Contact ratio: Contact ratio = =

12

61.55

×=

π

= 1.47 say 2

2. A pinion having 20 teeth engages with an internal gear having 80 teeth. If the

gears have involute profiled teeth with 20o pressure angle, module of 10 mm and

addendum of 10 mm, find the path of contact, arc of contact and the contact ratio.

Given Tp = 20 Ta = 80 φ = 20o

m = 10 mm Addendum = 10 mm To find,

i. Length of path of contact, ii. Length of arc of contact, and iii. Contact ratio

Length of path of contact

Cos φ

Length of arc of contact

Cosφ

Cos φ


πm



63

Solution:

We know that pitch circle radii of gear wheel and pinion are,

.4002

8010

2mm

mTR a =

×==

.1002

2010

2mm

mTr a =

×==

Radius of addendum circle of pinion, rA = r + addendum = 100 + 10 = 110 mm. Radius of addendum circle of gear wheel (Since it is an internal gear) RA = R- addendum on wheel = 400 - 10 = 390 mm. For internal gearing, length of path of approach (KP),

φφ 222sin CosRRRKP A −−=

( ) ( )2222040039020sin400 oo Cos−−×=

= 32.8 mm

Length of path of recess, φφ rsomCosrrPL A −−= 222

( ) ( ) ooCosPL 20sin10020100110222

−−=

= 23 mm. Length of path of contact, KL = KP +PL = 32.8 + 23 = 55.8 mm ii. Length of arc of contact: Length of arc of contact =

10

38.59

×=

π

= 1.89 say 2


Cos φ



64

3. A pair of gears, having 40 and 30 teeth respectively are of 25o involute form.

The addendum length is 55mm and module pitch is 2.5 mm. If the smallest wheel

is driver and rotates at 1500 rpm. Find the velocity of sliding at the point of

engagement and at the point of disengagement.

Given, TG = 40 TP: = 30

φ = 25o

m = 2.5 mm A/p = 1500 rpm Addendum = 5 mm To find;

i. Velocity of sliding at the point of engagement ii. Velocity of sliding at the point of disengagement

Solution:

Angular velocity of pinion, sec/08.15760

15002

60

25rad

N P

P =×

==π

ω

Gear ratio,

P

a

a

p

T

T=

ω

ω (or)

G

P

PaT

T×= ωω

sec/81.11740

3008.157 rad=×=

Let r, rA = Pitch circle and addendum circle radii of pinion R, RA = Pitch circle and addendum circle radii of gear

.5.372

305.2

2mm

mTr P =

×==

.502

405.2

2mm

mTR a =

×==

rA = r + addendum = 37.5 +5 = 42.5 mm RA = R + addendum = 50 +5 = 55mm

Length of path of approach, φφ sin222RCosRRKP A −−=



65

( ) ( ) ooCosKP 25sin50255055 222−−=

= 10.04 mm

Length of path of recess, φφ sin222rCosrrPL A −−=

( ) ( ) ooCos 25sin5.37255.375.42 222−−=

= 9.67 mm

i. Velocity of sliding at the point of engagement

( )21 ωω +=Vs Length of path of approach

( )KP21 ωω +=

( ) 04.1081.11708.157 ×+=

smmVs /89.2759= = 2.75 m/s

ii. Velocity of sliding at the point of disengagement

( )21 ωω +=Vs Length of path of recess

( ) 67.981.11708.157 +=

smm /2658= = 2658 mm/s Vs = 2.66 m/s

4. Two gear wheels of module pitch 4.5 mm having 24 and 33 teeth respectively.

The; pressure angle is 20o and each wheel has a standard addendum of 1 module.

Find the length of arc of contact and maximum velocity of sliding it the speed of

smaller wheel is 120 rpm.

Given, m = 4.5 mm TP: = 24 TG = 33,

φ = 20o

Np = 120 rpm Addendum = 1 module = 4.5 mm; To find;

iii. Velocity of sliding at the point of engagement iv. Velocity of sliding at the point of disengagement

Solution:


1202

60

2rad

N P

P =×

==ππ

ω

Gear ratio,

P

a

a

p

T

T=

ω

ω (or)

G

P

PaT

T×= ωω



66

sec/14.933

2457.12 rada =×=ω

Pitch circle radii of pinion and gear are given as,

mmmT

r P 542

245.4

2=

×==

mmmT

R G 25.742

335.4

2=

×==

Addendum radius of pinion, rA = r + addendum = 54 + 4.5 = 58.5 mm Addendum radius of gear wheel, RA = R+ addendum = 74.25 + 4.5 = 78.75 mm i. Length of arc of contact:


( ) ( ) ooCos 20sin25.742025.745.36 222−−=

= 11.1 mm


( ) ( ) ooCos 20sin5420545.58 222−−=

= 10.6 mm Length of path of contact, KL = KP + PL = 11.1 + 10.6 = 21.7 mm Length of arc of contact =

mmCos

o23

20

7.21==

ii. Maximum velocity of sliding

( )KPVs 21 ωω += as KP > PL

( ) 1.1114.957.12 ×+=

= 241 mm/s = 0.241 m/s

5. The following data refers to two matching involute gears of 20o pressure angle.

Number of teeth on pinion : 20 Gear ratio : 2 Speed of pinion : 250 rpm Module : 12 mm


Cosφ



67

If the addendum of each wheel is such that the path of approach and path of

recess on each side are half the maximum possible length. Find

i). addendum for both the wheels ii). The length of arc of contact

iv. Contact ratio & The maximum sliding velocity v. Angle turned through by the pinion and vi. Angle turned through by the gear wheel when one pair of teeth is in

contact. Given:

TP: = 20 ; Gear ratio, G = 2 =P

a

T

T;

Np = 250 rpm m = 12mm Solution:


2502

60

2rad

N P

P =×

==ππ

ω

Gear ratio,

2==P

a

T

TG

402022 =×=×= pG TT

2===p

G

G

P

T

TG

ω

ω

sec/08.132

16.26

2radP

n ===⇒ω

ω

Pitch circle radii of pinion and gear wheel is given by,

mmmT

r P 1202

2012

2=

×==

mmmT

R G 2402

4012

2=

×==

i. Addendum for both the wheels; Given that addendum on pinion and gear is such that the path of approach and recess are half of their maximum possible values, so,



68

2

sinsin222 φ

φφr

RCosRRA =−− (1)

and2

sinsin222 φ

φφR

rCosrrA =−− (2)

Substituting the values of R and r in (1) we get

2

20sin12020sin24020240 222

ooo

A CosR =−−

RA = 247.77 mm

∴ Addendum of gear wheel = RA – R = 247.77 – 240 = 7.77 mm

Substituting the values of R and r in (2) we get,

2

sinsin222 φ

φφR

rCosrrA =−−

2

20sin24020sin12020120 222

ooo

A Cosr =−−

rA = 139.5 mm

∴ Addendum of pinion = rA – r = 139.5 – 120 = 19.5 mm ii. Length of arc of contact;


( ) ( ) ooCos 20sin2402024077.247 222−−=

= 20.52 mm

φφ sin222rCosrrPL A −−=

( ) ooCos 20sin120201205.139 222−−=

= 41.08 mm. Length of path of contact, KL = KP+ PL = 20.52 + 41.08 = 61.6 mm

∴ Length of arc of contact = mmCosCos

KLo

51.6520

56.61==

φ

iii. Contact ratio: Contact ratio =

2738.112

51.65say=

×π


π m



69

iv. Maximum sliding velocity: ( ) PLVs ×+= 21 ωω as PL > KP

( ) smsmm /612.1/97.161108.4108.1316.26 ==×+=

v. Angle turned through by the pinion: Angle turned by pinion =

= oo

28.311202

36051.65=

×

×

π

vi. Angle turned through by the gear wheel; Angle turned by gear =

2402

36051.65

×

×=

π

o

= 15.64o

6. The following data relate to a pair of 20o involute gear in module = 6 mm.

Number of teeth on pinion = 17, Number of teeth on gear = 49. Addenda (or) pinion

and gear wheel = 1 module.

Find 1. The number of pair of teeth in contact, 2. The angle turned through by the pinion and the gear wheel when one

pair of teeth is in contact and 3. The ratio of sliding to rolling motion when the tip of a tooth on the large

wheel i. just making contact ii. just leaving contact with its mating tooth and iii. At the pitch point.

Given:

φ: = 20o Tp = 17 rpm m = 6mm TG = 49 Addenda on pinion and gear wheel = 1 module = 6 mm Solution:

i. Number of pairs of teeth in contact, Pitch circle radii of the pinion and the gear wheel is given by

mmmT

r P 512

176

2=

×==

mmmT

R G 1472

496

2=

×==

Length of arc of contact Circumference of pinion

o360×


Circumference of gear

o360×



70

Addendum radius of pinion, rA + addendum = 51 + 6 = 57 mm. Addendum radius of gear wheel, RA = RA + addendum = 147 + 6 = 153 mm. Length of path of approach,


( ) ( ) ooCos 20sin14720147153 222−−=

= 15.5 mm

Length of path of recess φφ sin222rCosrrP AL −−=

( ) ooCos 20sin51205157 222−−=

= 13.41 mm. Length of path of contact PLKPKL += mm91.2841.135.15 =+=

Length of arc of contact =

o

Cos20

91.28=

= 30.8mm

∴Number of pairs of teeth in contact i.e Contact ratio =

26.16

8.30

8

8.30say

m=

×==

π

2. Angle turned through by the pinion and gear wheel when one pair of teeth is in contact

Angle turned through by the pinion =

oo 6.34360512

8.30=×

×=

π

Angle turned through by the pinion =

oo 123601472

8.30=×

×=

π

3. Ratio of sliding to rolling motion:


Cosφ


Circular pitch (PL)


Circumference of pinion


Circumference of gear wheel

x 360o

x 360o



71

Gear ratio,

)(orT

T

p

G

G

P ==ω

ω

PP

G

PPG

T

Tωωωω 34.0

49

17=×=×=

and rolling velocity, rVpR.ω=

sec/5151. mmR PPG ωωω =×=

i. At the instant when the tip of a tooth on the larger wheel is just making contact

with its mating teeth.

The sliding velocity = (ωp+ωG) Length of path of approach.

= ( ) PGP Kωω +

( ) smmVs pP /88.205.15347.0 ωωω =+=

Ratio of sliding velocity to rolling velocity.

41.051

88.20==

p

p

R

S

V

V

ω

ω

ii. At the instant when the tip of a tooth on the larger wheel is just leaving contact

with its mating teeth.

The sliding velocity, ( )×+= GPsV ωω length of path of recess

( ) PLGP ×+= ωω

( ) sec/1.1841.13347.0 mmV pPS =×+= ωω

∴ Ratio of siding velocity to rolling velocity,

355.051

1.18==

p

p

R

S

V

V

ω

ω

iii. At the instant when the tip of a tooth on the larger wheel is at the pitch point.

At the pitch point the length of path of contact is zero. So the sliding velocity will

be zero. ∴ The ratio of sliding velocity to rolling velocity is zero.

7. Two spur wheel shave 30 teeth each of involute shape. The circular pitch is 25

mm and pressure angle is 20o. Determine the addendum of the wheels if the arc of

contact is twice e the circular pitch.



72

Given TP = TG = 30 PC = 25mm φ = 20o Arc of contact = 2 x Pc

To find: Addendum of the wheels.

Solution: Length of arc of contact = 2 x circular pitch Length of path of contact Cos φ

Length of path of contact = mmCos o 472050 =×

Module, π

Pcm =

Since number of teeth on pinion and gear wheel are same, pitch circle radii of pinion and gear wheel are also same.

20

30

2×===

πcPmTp

rR

3.1191525

=×

=π

Addendum radii of pinion and gear are equal

rA = RA


−− φφ sin2 222

RCosRRA

( ) 4720sin3.119203.1192 222=

−− oo

A RCosR

125655.41342

+=AR

mmRA 2.129=

Addendum on gear wheel = RA- R = 129.2 - 119.3 = 9.9 mm

Addendum on pinion = rA = RA = 9.9 mm

8. A pair of spur gear wheels with 14 and 21 teeth are of involute profile angle 16o.

Find maximum addend on the pinion and gear wheel to avoid interference if the

side of the pitch point of pinion runs at 300 rpm.

Given TP = 14, TG = 21

= 2 x 25



73

φ = 16o

m = 6 mm Np = 300 rpm

Solution:

Angular velocity of pinion , 60

2 p

p

Nπω =

srad /4.3160

3002=

×=

π

Gear ratio, p

G

G

p

T

T=

ω

ω

(or) sradT

T

p

G

pG /93.2021

144.31 =×=×= ωω

1. Maximum addenda on the pinion to avoid interference:

−

++= 121

2

2φSinT

T

T

TmTA

P

G

P

Gp

p

−

++

×= 1162

14

21

14

211

2

146 2 oSin

= 7.68 mm 2 Maximum addenda on gear wheel to avoid interference:

−

++= 121

2

2φSinT

T

T

TmTA

G

P

G

pG

w

−

++

×= 1162

21

14

21

141

2

216 2 oSin

= 4.12 3 Maximum velocity of sliding pitch circle radii of pinion and gear wheel are given as,

mmmT

rp

422

146

2=

×==

mmmT

R G 632

216

2=

×==

Addendum radius of pinion, rA = r + Addendum of pinion = 42 + 7.68 = 49.68 mm Addendum radius of gear wheel, RA = R + Addendum of gear wheel = 63+ 4.12 = 67.12 mm

Length of path of approach, φφ RSinCosRRKP A −−= 222



74

( ) oo RSinCos 16166312.67 222−−=

= 11.58 mm.

Length of path of recess, φφ rSinCosrrPL A −−= 222

( ) oo SinCos 1642164268.49 222−−=

= 16.22 mm. Maximum velocity of sliding of engagement,

( ) ppp Kωω +=

( ) smm /98.60558.1193.204.31 =×+=

= 0.606 m/s

a. Maximum velocity of sliding at disengagement, ( ) LGp Pωω +=

( ) smm /8.84822.1693.204.31 =×+=

= 0.849 m/s

9. A pair of involute spur gears with 142

1pressure angle and having 6 mm module

is in mesh. The number of teeth on the pinion is 16 and it rotational speed is 300

rpm. When the gear ratio is 2, Find the addendum on the pinion in order to avoid

interference.

Given,

φ = 142

1 o,

m = 6 mm Tp = 16

Np = 300 rpm

G = 2=p

G

T

T

Solution: Addendum on the pinion in order to avoid interference,

−

++= 121

2

2φSinT

T

T

TmTA

P

G

P

Gp

p

( )

−++

×= 1

2

1142221

20

166 2Sin

Ap = 18.78mm



75

10. Determine the minimum number of teeth required on a pinion, in order to

avoid interference which is to gear with i. a wheel to give a gear ratio of 4 to 1; and

ii. An equal wheel. The pressure angle is 20o and a standard addendum of 1

module for the wheel may be assumed.

Given,

,4==P

G

T

TG φ = 20o

AW = 1 module,

Solution;

Minimum number of teeth for a gear ratio of 4;1

( )[ ]12/1/11

2

2(min)

−++=

φSinGGG

ATP

−

++

×=

124

1

4

114

12

2φSin

1492.131436.0

2say==

2. Minimum number of teeth for equal wheel, Put G = 1 , is the above Tp(min) equation,

131

2

2(min)

−+=

φ

ω

Sin

ATP

12031

12

2 −+

×=

oSin

162.0

2=

= 12.34 say 13



76

11. Two 20o involute spur gears have a module of 10mm. The addendum is one

module. The larger gear has 50o teeth and the pinion 13 teeth. Does the

interference occur? If it occurs, to what value should the pressure angle be

changed to eliminate interference?

Given,

φ = 20o, m = 10mm, Addendum = 1 module = 10 mm TG = 50 TP = 13

Solution:

i. To check whether interference occurs or not, Pitch circle radii of pinion and gear wheel is given by,

mmmT

r P 652

1310

2=

×==

mmmT

R G 2502

5010

2=

×==

Addendum radius of pinion, rA = r + addendum, = 65+10 = 75 mm Addendum radius of gearwheel, RA = R + addendum, = 250+10 = 260 mm To avoid interference, maximum addendum radius RA (max) is given by

( ) ( )22

(max) sincos φφφ rRSinRRA ++=

( ) ( )22

20sin652025020cos250 ooo Sin ++=

= 258.45 mm Actual addendum radius of wheel, RA = 260 mm. As the actual addendum radius RA is more than the maximum value RA (max) therefore, interference occurs.

ii. New value of pressure angle to eliminate interference; Equating RA(max) to RA, we will get new value of pressure angle just to eliminate interference. i.e). RA (max) = A



77

( ) ( ) ArRSinR =++22

sincos φφφ

( ) ( ) 260sin65250cos250222

=++ φφφ Sin

2222 260sin315cos250 =+ φφ

861.0cos2 =φ

928.0cos =φ

o88.21=⇒ φ (or) 21o 521

Thus if the pressure angle is increased to 21o 521 the interference is avoid.

12. In a spiral gear drive connecting two shafts, the approximate centre

distance is 400 mm and the speed ratio = 3. The angle between the two shafts is

50o and the normal pitch is 18 mm. The spiral angle for the driving and driven

wheels is equal. Find,

1. Number of teeth on each wheel,

2. Exact centre distance and

3. Efficiency of the drive, it friction angle = 6o

Given, C = 400 mm

θ = 50o 31

2 ==T

TG

φ = 6 o, PN = 18 mm Solution;

1. Number of teeth on each wheel,

21 ααθ +=

o252

21 ===θ

αα

Centre distance between two shafts C is given by

+=

21

1 1

2 ααπ Cos

G

Cos

TPC N



78

+

+

21

1 1

2 ααπ Cos

G

Cos

GTPN [ ]21 αα =Q

,64.1225

31

2

18400 1 T

Cos

To

=

+×=

π

(or)

32)(64.3164.12

4001 orT ==

96323. 12 =×== TGT

2 Exact centre distance,

+=

21

1

1

1

2 ααπ Cos

G

Cos

TPC N

+=

1

1 1

2 απ Cos

GTPN

+×=

oCos

C25

31

2

3218 11

π

= 404.5 mm 3. Efficiency of the drive,

( )( ) 21

12

αφα

αφαη

CosCos

CosCos

−

+=

( )( )φα

φα

−

+=

1

1

Cos

Cos [ ]21 αα =Q

( )( )oo

oo

Cos

Cos

625

625

−

+=

= 0.907 %7.90=η

13. A pair of spiral gears connects two shafts inclined at 80o. The velocity ratio

is 2 and driver has 25 teeth of normal pitch of 12 mm and spiral angle of 30o. Find

the centre distance between the shafts.

Given,



79

θ = 80o 21

2

2

1 ===T

TG

ω

ω

T1 = 25, PN = 12 mm α = 30o Solution;

Shaft angle, o8021 =+= ααθ

oothen 50,30 21 == αα

;22

2 ==T

G if T1 = 25 then T2 = 50

Circular pitch is given by, 1

1

1

1T

d

Cos

PP N

C

π

α==

mmCosCos

PTd

o

N 11030

1225

1

1

1 =×

×==

παπ

and,

2

2

2

2T

d

Cos

PP N

C

π

α==

mmCosCos

PTd

o

N 29750

1250

2

2

2 =×

×==

παπ

Centre distance between shafts,

mmdd

C 5.2032

297110

2

21 =+

=+

=

14. A spiral wheel reduction gear, of ratio 3 to 2, is to be used on a machine,

with the angle between the shafts 80o. The approximate centre distance between

the shafts is 125 mm. The normal pitch of the teeth is 10mm and wheel diameters

are equal. Find the number of teeth on each wheel, pitch circle diameter and spiral

angles. Find the efficiency of the drive if the friction angle is 5o.

Given,

;2

3

1

2 ==T

TG θ = 80o



80

C = 125 mm PN = 10 mm d1 = d2, φ = 5o Solution;

Spiral angles, ( )21 &αα

2

1

1

2

1

2

α

α

Cos

Cos

T

T

d

d=

2/32

1

1

2 ==ω

ω

T

T

2121 &80 ddo ==+= ααθ

( )1

1

1 802

31

α

α

−=

oCos

Cos

( ) 111 cos380802 dSinSinCosCos oo =+ αα

9696.1

6528.2tan 1 =−

α1 =- 53.4

α2 =- 80o - 53.4 = 26.6o Number of teeth m each wheel (T1 & T2)

Centre distance,

+=

21

1 1

2 ααπ Cos

G

Cos

TPC N

+

×=

ooCosCos 6.26

2/3

4.53

1

2

2110125

π

125 = 5.34 T1

⇒ T1 = 23.4 say 24 T2 = GT1

= 3/2 x 24 = 36 Pitch circle diameters (d1 & d2)

Circular pitch , 11

1

1ααπ

π

Cos

P

Cos

dP N

c ==

o

N

CosCos

Pd

4.53

2410

1

1παπ

×==



81

= 128 mm d2 = d1 = 128 mm Efficiency of the drive,

( )( )

21

12

αφα

αφαη

CosCos

CosCos

−

+=

( )( ) oo

ooo

CosCos

CosCos

6.2654.53

4.5356.26

−

+=

= 0.855 %5.85=η

15. A right angled drive on a machine is to be made by two spiral wheels. The

wheels are of equal diameters with a normal pitch of 10mm and centre distance is

approximately 150 mm. If the speed ratio is 2.5 to 1. Find

a. Spiral angles of teeth,

b. Number of teeth on each wheel,

c. the exact centre distance,

d. transmission efficiency if friction angle is 6o.

Given, θ = 90o

d1 = d2

PN = 10 mm C = 150 mm

;5.22

1 ==ω

ωG

φ = 6o, Solution;

a. Spiral angles of teeth,

2

1

1

2

1

2

α

α

Cos

Cos

T

T

d

d=

1221 90)(90 ddor oo −==+αα



82

( )1

1

905.21

α

α

−=

oCos

Cos

111 cos5.29090 ααα =+ SinSinCosCos oo

5.2tan)(5.2 1

1

1 == αα

αor

Cos

Sin

o2.681 =α

o8.212 =α

b. Number of teeth on each wheel,

+=

21

1 1

2 ααπ Cos

G

Cos

TPC N

+=

ooCosCos

T

8.21

5.2

2.68

1

2

10150 1

π

T1 = 17.5 say 18 T2 = GT1 = 2.5 x x18 = 45 b. Exact centre distance;

+=

21

1 1

2 ααπ Cos

G

Cos

TPC N

+

×=

ooCosCos 8.21

5.2

2.68

1

2

1810

π

= 154.2 mm c. Transmission efficiency

( )( ) 21

12

αφα

αφαη

CosCos

CosCos

−

+=

( )( ) ooo

ooo

CosCos

CosCos

8.2162.68

2.6868.21

−

+=

= 0.758



83

%8.75=η

UNIT IV

GEARS



PART A

1. What are the standard pressure angle values? Give the advantages of lower and

larger pressure angle

2. State the law of gearing

3. Define interference in Gears

4. Define (a) Module (b) Diametral Pitch of gears

PART B

1. Two involutes gears of 200 pressure angle are in mesh. The number of teeth on

pinion 20 and the gear ratio is 2. If the pitch expressed in module is 5 mm, and the pitch

line speed is 1.2m/s, determine the angle turned through by pinion, when one pair of

teeth of teeth is in mesh. Also calculate the maximum of sliding. Take addendum as one

module



84

2. An epicyclic train has a pinion A having 15 teeth, centrally located and rigidly

connected to shaft of driving motor. Another gear B having 20 teeth is gearing with A and

also with annular fixed wheel D. Gear C is integral with B and meshes with another

annular

wheel E which is keyed to the shaft of driven unit. The arm rotates about driving shaft

and

carries compound gear B, C. Sketch the arrangement and determine speed of machine

for a

motor speed of 1000 rpm. Also determine torque on machine shaft for a motor torque of

100 N-m

UNIT IV

GEARS



PART A

1. What is axial pitch of a helical gear?

2. List out the applications of epicyclic gear train.

3. What is reverted gear train?

4. Define undercutting in Gears.

PART B

1. An epicyclic gear train consist of a sun wheel S, a stationary internal gears E and

three identical planet wheels p carried on a star shaped planet carrier C. The size of the

different toothed wheels are such that the planet carrier C rotates 1/5 th of the speed of

the sun wheel S. The Minimum number of teeth on any wheel is 16. The driving torque

on the sun wheel is 100 N-m. Determine (1) Number of teeth on different Wheels of the

train and (2) Torque necessary to keep the internal gear Stationary.



85

2. A pair of involutes spur gears with 160 pressure angles and pitch of module 6mm

is in mesh. The number of teeth in pinion is 16 and its rotational speed is 240 rpm. The

gear ratio is 1.75.In order to avoid the interference, determine

i) Addenda on pinion and wheel

ii) Length of path contact

iii) Maximum velocity of sliding on either side of pitch point

UNIT V FRICTION

Dry friction – Friction in screw jack – Pivot and collar friction - Plate clutches - Belt and

rope drives - Block brakes, band brakes.

PART- A (2 Marks)

1. Define friction. When a body moves or tends to move on another body, the property of the two bodies by virtue of which a force is developed between the two bodies, which opposes the motion is called friction.

2. List out types of friction

The types of friction are 1. static friction 2. dynamic friction 3. sliding friction 4. rolling friction 5. pivot friction 6. dry (or solid) friction 7. boundary (or skin) friction 8. Fluid (or film) friction.

3. Give the classification of dynamic friction.

1. sliding friction 2. rolling friction 3. Pivot friction.

4. Define sliding friction and rolling friction.

Sliding friction is the friction experienced by a body when it slides over another body. Rolling friction is the friction experienced between the surfaces which have balls

or



86

rollers interposed between them.

5. Define coefficient of friction. Coefficient of friction is defined as the ratio of the limiting friction (F) to the normal

reaction (RN) between the bodies. It is denoted mathematically as NR

F=µ

6. Define limiting force of friction.

The maximum value of frictional force which comes into play when a body just begins to slide over the surface of the other body is known as limiting force of friction.

7. State the laws of dynamic friction.

1. The force of friction always acts in a direction, opposite to that in which the body is moving.

2. The magnitude of the kinetic friction bears a constant ratio the normal reaction between the two surfaces.

3. For moderate speeds, the force of friction remains constant. But it decreases slightly with increase of speed.

8. Define limiting angle of friction.

It is defined as the angle which the resultant reaction R makes with the normal reaction (RN).

NR

F=φtan

9. Define angle of repose The angle of the plane when motion of an object on the plane is impending is

called angle of repose.

10. What is screw jack and how its efficiency is calculated. (May 2010) Screw jack is a device for lifting heavy loads by comparatively smaller effort at its handle. The efficiency of screw jack is defined as the ratio between ideal effort to the

actual effort.

d)tan(

tan

φα

αη

+=

11. Define overhauling of screws. The condition in which the load will start moving downward without the application of any torque is known as overhauling of screws.

12. Define self locking screw. The condition in which the load will start moving downward with the application of any torque is known as self locking screws.

13. List out the types of clutches

5. Friction clutches 6. Centrifugal clutches 7. Semi centrifugal clutch



87

8. Conical clutch 9. Dog and spline clutch 10. Hydraulic clutch 11. Electromagnetic clutch 12. Vacuum clutch

14. Define clutch.

Clutch is a mechanical device used in automobiles to engage and disengage the driving and driven shaft instantaneously at the desire of the driver.

15. List out the types of friction clutches.

The types of frictional clutches 1. Single plate clutch 2. Multiple plate clutch 3. Cone clutch

16. List out the factors upon which selection of a belt drive depends. 1. Speed of driving and driven shafts 2. Speed reduction ratio 3. Power to be transmitted 4. Centre distance between shafts 5. Positive drive requirements 6. shaft layout 7. Space available 8. Service condition.

17. List out the types of belt

1. Flat belt 2. V-belt 3. Circular belt or rope

18. List out the classification of belts according to the material used. 1. Leather belts 2. Cotton fabric belts 3. Rubber belts 4. Balata belts

19. What is slip of a belt?

It’s improper relative movement of belt & pulleys. i) if the drive pulley rotates without carrying belt(S1) ii) If the belt rotates without carrying follower pulley(S2)

..velocity ratio considering both slip is N 1/ N1 = d1/ d2 ( 1 - s 1+s 2 /100)

20. Explain creep of belt. When belts moves from tight side to slack, a certain portion of belt contracts, again when belt moves to tight side, a portion expands this relative change of shape of belt due to continuous motion is termed as creep.

21. Write the condition for transmission of maximum power.

Power transmitted is maximum, when 1/3 of the maximum tension is absorbed as centrifugal tension .



88

T = 3Tc Velocity of belt for maximum power

V= √T/3m 22. Whether centrifugal tension has any effect on power transmitter?

Power transmitted p = (T1 –T2) .V If centrifugal tension is considered Then p = [(T1- Tc) – (T2+ Tc)]

= (T1- T2).V 23. Explain centrifugal tension & initial tension?

When belt runs over pulley continuously the centrifugal force created. Which increase tension in both sides of both sides of belt. This is known as centrifugal tension. To is considered usually above speed of 10m/s In order to increase the grip between pulley and belt, the belt is subjected to some tension even in stationary condition, this is known as initial tension. To = T 1 + T2 /2 (without Tc) Centrifugal tension Tc

To = T 1 + T2+ 2Tc /2 Considering Tc initial tension To 24. What are the advantage of V-belt over other belt?

1. Slip between belt & pulley is negligible. 2. V- belts provide compactness 3. Even at high velocity ratio (10m/s), the drive is smooth.

25. Define brakes

Brakes are the devices used to reduce the speed of moving machine components by absorbing energy. The absorbed energy is converted into heat and released into the atmosphere or absorbed in another medium.

26. List out the classification of brakes. (May 2010)

1. Block or shoe brake 2. Band brake 3. Band and block brake 4. Internal expanding shoe brake.

27. List out the classification of band brake

1. Simple band brake 2. Differential band brake

28. List out the classification of block or shoe brake

1. Single block or shoe brake 2. Pivoted block or shoe brake 3. Double block or shoe brake

PART- B (12 Marks)

BRAKES



89

1. A vehicle of mass 1200 Kg is moving down the hill at a slope of 1:5 at 72 km/h.

It is to be stopped in a distance of 50m If the diameter of the tyre is 600 mm,

determine the average braking torque to be applied to stop the vehicle, neglecting

all the frictional energy except for the brake. If the friction energy is momentarily

stored in a 20 kg cast iron brake drum, what is the average temperature rise of the

drum? The specific heat for cast iron maybe taken as 520 J/Kgo C

Determine the minimum coefficient of friction between the tyres and the load in

order that the wheel do not skid, assuming that the weight is equally distributed

among all the four wheels.

Given,

m = 1200 Kg, Slope = 1:5, V = 72 Km/h = 20m/sec.

h = 50m; d = 600 mm (or) r = 300 mm = 0.3 m; mb = 20 Kg;

C = 5205/KgoC. Average braking torque to be applied to stop the vehicle, Solution: We know that kinetic energy of the vehicle,

( ) mNvmEK .2400002012002/1.2/122 =××==

and potential energy, 5/15081.91200.. ×××=×= slopehgmE p

= 117,720 N.m ∴Total energy of the vehicle,

pk EEE +=

= 240000 + 117720 = 357720 N.m Since the vehicle stopped in a distance of 50 m, therefore tangential baking force,

NFt 4.715450

357720==

We know that average braking torque to be applied to stop the vehicle,

rFTtB×=

= 7154.4 x 0.3



90

TB = 2146.32 N.m Average temperature rise of the drum.

Let ∆t = Average temperature rise of the drum in oC.. We know that the heat absorbed by the brake drum, Hg = Energy absorbed by the brake drum, = 357720 N.m = 3,57,720J (Q1N.m = 1 J ) Heat absorbed by the brake drum (Hg) tCmHg b ∆××=

357720 = 20 x 520 x ∆t

Cto4.34

10400

357720==∆⇒

Minimum coefficient of friction b/w the tyre and road, µ = Minimum coefficient, RN = Normal force = Weight of the vehicle = m.g = 1200 x 9.81 = 11772N We know that, tangential braking force (Ft),

NRFt µ=

7154.4 = µ x 11772

11772

4.715=⇒ µ

6.0=µ

2. A single block brake is shown in figure. The diameter of the drum 250 mm and

the angle of contact is 90o. If the operating force of 700N is applied at the end of a

lever and the coefficient of friction between the drum and the lining is 0.35.

Determine the torque that may be transmitted by the block brake.

Given, d = 250 mm (or) r = 125 mm

2θ = 90o = π/2 rad; P = 700 N

µ = 0.35



91

Since the angle of contact is greater than 60o, therefore equivalent coefficient of friction,

385.0

902

45sin35.04

22

41 =

×

××=

+=

o

o

SinSin

Sin

πθθ

θµµ

Let RN = Normal force pressing the block to the brake drum,

Ft = Tangential braking force = µ1 RN

Taking moment about o,

( ) t

tt

Nt FFF

RF 520200385.0

200200502002507001

=×=×=×=×++µ

(or) 45070050520 ×=− tt FF

NFt 670470

450700=

×=

We know that torque transmitted by the block brake,

rFT tB ×=

= 670 x 125 = 83750 N-mm

3. A rope drum of an elevator having 650 mm diameter is fitted with a brake drum

of 1m diameter. The brake drum is provided with four cast iron brake shoes each

subtending an angle of 45o. The mass of the elevator when loaded as 2000 Kg and

mNTB .75.83=



92

moves with a speed of 2.5 m/s. The brake has a sufficient capacity to stop the

elevator in 2.75 meters. Assuming the coefficient of friction between the brake

drum and shoes as 0.2.

Find

1. Width of the shoes, it the allowable pressure on the brake shoe is limited to 0.3 N/mm2 and 2. Heat generated in stopping the vehicle. Solution: Given, de = 650 mm (or) re = 325 mm = 0.325 m. d = 1m (or) 0.5m = 500 mm; n = 4,

2θ = 45o (or) θ = 22.5o m = 2000 Kg, V = 2.5 m/s, h = 2.75m,

µ = 0.2; Pb = 0.3N/mm2 1. Width of the shoe; We know that,

ahuv 222 =−

( ) 75.2205.22

×=− a

2/136.1 sma =⇒ Accelerating force = mass x acceleration = 2000 x 1.136 = 2272 N. ∴ Total load acting on the rope while moving,

W = Load on the elevator in Newton’s + Accelerating force = 2000 x09.81 + 2272 = 21892 N

We know the, torque acting on the shaft, T = W x re = 21892 x 0.325 = 7115 N-m.

∴Tangential force acting on the drum,

.142305.0

7115N

r

T===



93

The brake drum is provided with four cast iron shoes, therefore, tangential force acting on each shoe, .5.35574/14230 NF

t==

Normal load on each shoe,

NFtRN 5.177872.0

5.3557/ === µ

Projected hearing area of each shoe,

( ) ( ) 27.3825.22sin5002sin2 WmmWrWA

o

b =×== θ

Bearing pressure on the shoe, (P0)

WWA

R

b

N 465

7.382

5.787,173.0 ===

mmW 1553.0

5.46==⇒

2. Heat generated in stopping the elevator, Heat generated = Total energy absorbed = Kinetic energy + potential energy = ½ mv2 + mgh

= ( ) 75.281.920002520002

1 2××+××

= 60,205 N-m. = 60,205 KN-m = 60.205 KJ

4. A double shoe block, as shown in figure is capable of absorbing a torque of

1400 N-m. The diameter of the brake drum is 350 mm and the angle of contact for

each shoe is 100o. If the confinement of friction between the brake drum and

lining is 0.4; find

1. The spring force necessary to set the brake; and

2. The width of the brake shoes, if the bearing pressure on the lining

material is not to exceed 0.3 N/mm2

Solution:

Given, TB = 1400 N.m = 1400 x 103 N.mm, D = 350 mm (or) r = 175 mm.

2θ = 100o = 100 x π/180= 1.75 rad.

µ = 0.4, Pb = 0.3 N/mm2



94

Since the angle of contact is greater than 60o, therefore equivalent coefficient of friction,

45.0100sin75.1

50sin4.04

22

41 =×

××=

+=

o

o

Sin

Sin

θθ

θµµ

Taking moment about θ1, we have,

( ) 13520045.0

40175200450 1

1

11 ×+×=−+×=× t

t

tN FF

FRS

= 579.4 Ft1 [ ]1

11 / µtN

FR =Q

4.579

4501 ×=∴ SFt

Ft1 = 0.7765 Taking moment about 02, We have,

( ) 2

2

22 4.44420045.0

20040175450 FtF

RFS t

Nt =×=×=−+× [ ]1

12 / µtN

FR =

4501354.444 22 ×=− SFF tt

SFt 4504.309 2 =

SsFt 454.1/4.309

4502 =×=

We know that, torque capacity of the brake (TB)

( )rFF tt 21

3101400 +=×

( ) sss 25.390175454.1776.0 =+=

NS 358725.390

1400==⇒



95

2 Width of the brake shoes, (b), Projected bearing area for one shoe,

( )θsin2rbAb =

( )ob 50sin1752 ××=

= 2686 mm2

Normal force on the right hand side of the shoe,

NsF

R t

N 618645.0

3587776.0

45.0

776.01

1

1 =×

=×

==µ

Normal force on the left hand side of the shoe,

NsF

R t

N 1159045.0

3587454.1

45.0

454.11

2

2 =×

=×

==µ

We shall design the shoe for max. Normal force ie RN2.

Bearing pressure, b

N

bA

RP 2=

b268

115903.0 =

3.0268

11590

×=⇒ b

b = 144.2 mm

5. A simple band brake operates on a drum of 600 m in diameter that is running at

200 r.p.m. The coefficient of friction is 0.25. The brake band has a contact of 270o;

one end is fastened to a fixed pin and the other end to the brake arm 125 mm from

the fixed pin. The straight brake arm is 750 mm long and placed perpendicular

to the diameter that bisects the angle of contact.

a. What is the pull necessary on the end of the brake arm to stop the wheel if

35KW is being absorbed? What is the direction of this maximum pull?

b. What width of steel band of 2.5 mm thick is required for this brake if the

maximum tensile stress is not to exceed 50Mpa?

Solution:

Given d = 600 mm (or) r = 300 mm, N = 200 rpm,

µ = 0.25

θ = 270o = 270 x π/180 = 4.1713 rad. P = 35 KW = 35 x 103 W t = 2.5mm tσ = 50 MPa = 50 N/mm2



96

a. Pull necessary on the end of the brake drum to stop the wheel.

µθ=

2

1log3.2T

T

= 0.75 x 4.713 = 1.178

3.2/178.1log2

1 =

∴

T

T

= 0.5123

25.32

1 =T

T (1)

TB = braking Torque,

60

2 BNTP

π=

60

20021035 3 BT××

=×π

mmNNmTB .1016671667 3×==⇒

We also know that,

( ) ( )3002121 TTrTTTB −=−=

NTT 5557300

101667 3

21 =×

=− (2)

From (1) & (2) we get,



97

NTNT 2470&8027 21 ==

Now taking moments about θ,

25.62750 22 ×=×=× TODTP

= 2470 x 88.4 = 218 .348

NP 291750

348.218==

b. Width of the steel band: Maximum tension (T1)

tWtT ××= σ1

8027 = 50 x W x 2.5

5.250

8027

×=⇒W

W = 64.2 mm

6. A differential band brake as shown in fig, has a drum diameter of 600 mm and

the angle of contact is 240o. The brake band is 5mm thick and 100 mm wide. The

coefficient of friction between the band and the drum is 0.3. If the band is

subjected to a stress of 50 MPa. Find,

1. The least force required at the end of a 600 mm lever and

2. The torque applied to the brake drum shaft.

Solution: Given d = 600 mm, (or) r = 300 mm = 0.3 m.

θ = 240o = 240 x π/180 = 4.2 rad. t = 5 mm W = 100 mm

µ = 0.3 tσ = 50MPa = 50 N/mm2



98

1. Least force required at the end of lever,

26.1423.0log3.22

1 =×==

µθ

T

T

5478.03.2

26.1log

2

1 ==

T

T

53.32

1 =⇒T

T (1)

We know the maximum tension in the band,

15075600 21 ×=×+× TTP

NTT

P 1355600

75250001507082

600

75150 12 −=×−×

=×−×

=

N1355= (In magnitude)

Since P is negative, therefore the brake is self-locking 2. Torque applied to the drum shaft:

( )rTTTB 21 −=

( ) 3.0708225000−=

mNTB .5375=



99

7. A differential band brake has a force of 220N applied at the end of a lever as

shown in fig. The coefficient of friction between the band and the drum is 0.4. The

angle of lap is 180o. Find

1. The maximum and minimum force in the band, when a clockwise

torque of 450 N-m is applied to the drum; and

2. The maximum torque that the brake may sustain for counter

clockwise rotation of the drum.

Solution:

Given, P = 220 N, µ = 0.4,

θ = 180o = π rad. d = 150 mm; or r = 75 mm = 0.075 m.

1. Maximum and minimum force in the band We know that braking torque, (TB)

( )rTT 21450 −=

( ) 075.021 TT −=

NTT 6000075.0

45021 ==−

( )NTT 600021 +=

Now taking moment about the pivot 0, we have

10050200220 21 ×=×+× TT



100

( ) 1005060004400 22 ×=×+ TT

100300000504400 22 ×=++ TT

NT 68802 =⇒

And, 600021 += TT

= 6880 + 6000

.128801 NT =

2. Maximum torque that the brake can sustain.

257.14.0log3.2 0

2

1 =×==

πµ

T

T

5465.03.2

257.1log

2

1 ==

T

T

52.32

1 =T

T

Now taking moments about 0, we have,

2212 35210052.310050200220 TTTT =×=×=×+×

22 50.52.344000 TT −=

NT 146302

440002 ==⇒

NTT 51414652.352.3 21 =×==

We know that the maximum torque that the brake may sustain,

( )rTTTB 21 −=

( ) 075.0146514 ×−=

mNTB

−= 6.27

8. A band and block brake having 12 blocks, each of which sub bends an angle of

16o at the center, is applied to a rotating drum of diameter 600 mm. The blocks are

75 mm thick. The drum and the flywheel mounted on the same shaft have a mass

of 1800 Kg and have a combined radius of gyration of 600 mm. The tow ends of

the band are attached to pins on the opposite sides of the brake fulcrum at a



distance of 40mm and 150 mm from the fulcrum. It a force of 250N is applied at a

distance of 900 mm form the fulcrum, find;

i. The maximum braking torque,

ii. The angular retardation of the drum.

iii. The time taken by the system to be stationary from the rated speed of

300 rpm. Take coefficient of friction between the blocks and the

drum as 0.3.

Given, n = 10 2θ = 16o⇒θ = 8o d = 600 mm = 0.6 m r = 0.3m t = 75 mm = 0.075m m = 1800 Kg K = 600 mm = 0.6 m C = OD = 40 mm = 0.04 m b = 0A = 150 mm = 0.15 m f = 250 N; a = 0B = 900 mm = 0.9 m. Solution: 1. Maximum braking Torque: The braking torque will be maximum when the following conditions are satisfied.

a. b>c ; ie) OA>OD b. Brake drum rotates anti clockwise c. The applied force acts upwards.

Tension ratio,

n

T

T

−

+=

θµ

θµ

tan1

tan1

2

1

12

8tan3.01

8tan3.01

−

+=

o

o

= 2.752 ie) T1 = 2.752 T2 (1) Taking moment about o,

bTcTaF .. 21 =+

15.0.04.09.0250 21 ×=×+× TT



102

21 15.04.0225 TT =× (2)

Solving (1) and (2) we get, T2 = 5636 N T1 = 15511 N Maximum braking torque TB

( )

+−=

2

221

tdTTTB

( )

×+−=

2

075.026.0563615511BT

mNTB −= 3703

2 Angular retardation of the drum, (α) ;

αITB =

α2mKTB =

( ) α××=2

6.018003703

2

2/71.5

6.01800

3703Srad=

×=⇒α

3. Time taken by the system to come to rest from the speed of 300 rpm

Initial angular speed, sradN

/4.3160

3002

60

20 =

×==

ππω

Final angular speed, ω = 0,

tαωω −= 0

t×−= 71.54.310

71.5

4.31=⇒ t

t = 5.5 sec.

9. A vehicle moving on a length plane inclined at 12o with the horizontal at a

speed of 35 Km/hr has a wheel base 2.2m. The C.G of the vehicle is 1.1 m from the

rear wheels and 0.9 m above the inclined plane. Find the distance traveled by the

vehicle before coming to rest and the time taken to do so when,

1. The vehicle moves up the plane



103

2. The vehicle moves down the plane. The brakes are applied to all the

wheels and the coefficient of friction is 0.6

Solution: Given

α = 12o

µ = 35Km/hr = sm /722.93600

100035=

×

L = 2.2 m . x = 1.1 m

h = 0.9m ; µ = 0.9. 1. When the vehicle moves up the plane: Since the vehicle moves up, the retardation of the vehicle,

( )ααµ SinCosa += l

( )ooSinCos 12126.081.9 +=

= 7.797 m/s2 For uniform retardation,

( )

ma

S 061.677972

722.9

2

22

=×

==µ

From velocity of the vehicle due to retardation, V = u – at 0 = 10 – 7.797 x t

St 2825.1797.7

10==

2 When the vehicle moves down the plane;

( )ααµ sin−= Cosga

( )ooCos 12sin125.081.9 −=

= 3.718 m/s2 for uniform retardation,

( )

ma

S 71.12718.32

722.9

2

22

=×

==µ

Final velocity of the vehicle v



104

V = u – at 0 = 9.722 -3.718 x t

718.3

722.9=⇒ t

St 62.2=

CLUTCHES

1. Determine the maximum minimum and average pressure in a plate clutch when

the axial force is 5000 N. The outer and inner diameter of the friction surfaces are

200 mm and 100 mm respectively.

Given data:

W = 5000 N, d1 = 200 mm or r1 =100mm =1000x010-3 m d2 = 100 mm or r2 = 50 mm = 50 x 10-3 m.

Solutions (i) Maximum pressure: Since the intensity of the pressure is maximum at the inner radius of (r2) , therefore, Pmax x r2 = C or C = 50 x 10-3 P max. Axial force exerted on the contact surface (w) is given by, W = 2πc(r1 – r2)

5000 = 2π x 50 x 10-3 Pmax (100 x 10-3-50 x 10-3) = 0.0157 Pmax. Pmax = 31.83 x 104 N/m2. ii. Minimum pressure: Since the intensity of the pressure is minimum at the outer radius (r1), therefore, Pmax * r1 = C or C = 100 x 10-3 Pmin.

Minimum force exerted on the contact surface (ω) is given by,

min21 )(2 PrrCW −= π

( )3313 1050101001010025000 −−− ×−××××= π

= 0.0314 Pmin



105

Pmin = 15.92 x 104 N/m2 iii. Average pressure:

( )2

2

2

1 rr

W

−=

π

( ) ( )[ ]2323 105010100

5000

−− ×−×=

π

= 21.22 x 104 N/m2.

2. A car engine has its rated output of 12 KW. The maximum torque developed

is100 N-m. The clutch use is of single plate type having two active surfaces. The

axial pressure not to exceed 85 KN/m2.

The external diameter of the friction plate is 1.25 times the internal diameter.

Determine the dimensions of the friction plate and the axial force exerted by the

springs. Co-efficient of friction is 0.3.

Given data: P = 12 Kw = 12 x 103 W T = 100 N –m. N = 2 Pmax = 85 KN/m2 = 85 x w3 N/m2

D1 = 1.25 d2 or 25.12

1 =d

d or 25.1

2

1 =r

r

µ = 0.3 Solution: i. Dimensions of the friction plate Assume uniform wear. Since the intensity of the pressure is max at the inner radios (r2) therefore,

Total normal force on contact surfaces

Pav =

Cross sectional area on contact surfaces



106

mNrCorCXrP /1083 2

3

2max ××== and the axial force exerted by the springs,

)(2 21 rrCW −= π

3

10852 ××= πW ( )222 25.1 rrr −

= 1.335 x 105 Nr2

2

Torque transmitted,

+=

2.. 21 rrWnT µ

+∗×∗=

2

25.110335.13.02100 222

2

5 rrr

= 9.0124 x 104 (r2)3

mr 1035.02 = or 103.5 mm

r1 = 1.252 ⇒ 1.25 x 0.1035 = 0.1294 m or 129.4 mm ii. Axial force exerted on the springs

( ) ( )2

2

5

21 10335.12 rrrcW ×=−= π

= 1.335 x 105 x (0.1035)2 W = 1430 N

3. A friction clutch is used to rotate a m/c from a shaft rotating at an uniform

speed of 250r.p.m. The disk type clutch has both of its sides effective the

coefficient of the friction being 0.3. The outer of inner diameters of the plate are

200 mm and 120 mm respectively. Assume uniform wear of the clutch, the

intensity of the pressure is not to exceed 100 KN/m2. If the moment of inertia of

the rotating ports of the m/o is 6.5 Kg-m2. Determine time to attain the full speed

by the m/e and energy loss in slipping of the clutch. What will be the intensity of

the pressure, if the condition of uniform pressure of the clutch is considered, Also

the ration of the power transmitted with uniform wear to that with uniform

pressure.



107

Given data: N = 250 rpm. n =2

µ = 0.3 d1 = 200 mm (or) r1=100 mm = 0.1 m d2 = 120 mm or r2 = 60 mm = 60 x 10-3m. Pmax = 100 KN/m2 = 100 x 103 N/m2 I = 6.5 Kg –m2 Solution:

i. The time to attain the full speed by the m/c (with uniform wear)

Since the intensity of the pressure (p) is max at the inner radius r2, Therefore, Pmax, r2 = C C=100 x 103 x 60 x 10-3 = 6000 N/m Axial thrust exerted,

( )212 rrCW −= π

( )33 10601010060002 −− ×−×××= πW

= 1507.96 N. Torque transmitted,

+=

2.. 21 rrNnT µ

×+××××=

−−

2

10601010096.15073.02

33

T = T2.38 N-m We know that,

Power transmitted 60

2 NTP

π=

WP 189568

38.722502=

××=

π

Also, T = 1 α where α is angular acceleration.



108

α∗= 5.638.2T or 135.11=α rad/sec2

W.K.T

135.11==t

ωα

135.111

60

2=×

t

Nπ

135.1160

2502=

×

×

t

π

t = 2.353. ii. The energy loss in the slipping of the clutch,

Angle turned by the driving shaft, Q1 = 353.260

2502

60

2+

×=×=

ππω t

Nt

= 61.52 rad.

Angle turned by the driving shaft =2

022

1ttt αω +=

( )235.2135.11

2

10 ×∗+=

= 30.75 rad. Energy lost in friction ( )21 QQ −+=

= 72.38 (61.52 - 30.75) = 2226 N-m iii. Intensity of the pressure, if the condition is uniform pressure:

Intensity of the pressure, ( )2

2

2

1 rr

wP

−=

π

( ) ( )[ ]2323 106010100

96.1507

×−×=

−π

= 7500 N/m2 P = 75 KN/m2



109

iv. Ratio of the power with uniform wear to that with uniform pressure. Power transmitted with uniform wear = 1895 W

Torque transmitted with uniform pressure =

−

−2

2

2

1

3

2

3

1

3

2...

rr

rrWn µ

( ) ( )( ) ( )

−

−×××

32

33

06.01.0

06.01.0

3

296.15073.02

T = 73.89 N-m. Power transmitted with uniform pressure,

60

2NT

P π=

60

89.732502 ××=

π

= 1934 N

1934

1895=

= 0.98.

4. A multiple clutch consisting of 6 plates, each plate of external diameter 150

mm and internal diameter 100 mm, is to transmit 7.5 KW at 900 rpm. Assuming µµµµ =

0.1 determine the pressure on each effective pair of surfaces in contact.

Given data: Number of plate np = 6 d1 = 150 mm or r1 = 75 mm = 75 x 10-3 m d2 = 100 mm or r2 = 50 mm = 50 x 10-3 m P = 7.5 Kw = 7.5 x 103 W. N = 900 rpm. µ = 0.1 Solution:

Power transmitted with uniform wear

Power transmitted with uniform pressure



110

As given number of plates np = 6 Number of pairs of surfaces in contact N = np -1 = 6 -1 = 5


2 NTP

π=

60

9002105.7 3 ππ ××

=×

Total friction torque to be Transmitted, T = 79.57 N-m Since no assumption is given, We assume uniform wear,

+×=

2. 21 rrWT µ

×+×∗×=

−−

2

105010751.057.79

33

w

W = 12731 N ∴ Total Axial force or pressure on contact surfaces, W = 12731 N But N = 5 so pressure on each effective surface is,

N25465

12731=

5. A multi plate clutch has three discs on the driving shaft and two on the driven

shaft. Te outside diameter of the contact surfaces is 240 mm and inside diameter

is 120 mm. Assume c uniform wear coefficient of friction as 0.3 find the maximum

axial intensity of pressure between discs for transmitting 25 KW at 1575 r.p.m.

Given data, n1 = 3, n2 = 2; d1 = 240 mm or r1 = 120 mm = 0.12 m d2 = 120 mm or r2 = 60 mm = 60 x 10-3 m



111

µ = 0.3 P = 25 Kω = 25 x 103 W: N = 1775 rpm. Solution; Number of pairs of contact surfaces, 4123111 =−+=−+= nnn

Power transmitted , 60

2 NTP

π=

( )

60

51521025 3 T+

=×π

T = 151.6 N-m Torque transmitted for uniform wear is given by,

+=

2.. 21 rrWnT µ

+∗×=

2

06.012.01.046.151

2W

W = 1404 N But axial force exerted is given by,

( )212 rrW −= π

)(2 212max rrrPW −∗∗= π

∴ (C = Pmax - r2)

1404 = ( )16.012.006.02 max −∗∗ Pπ

23

max /1007.62 mNP ×=

2

max /62 mKNP =



112

6. A leather faced conical friction clutch has a cone angle of 30o. The intensity of

pressure between contact surface is not to exceed 6 x 104 N/m2 and the breath of

the conical surface is not to be greater than 1/3 of the mean radius if µµµµ = 0.20 and

the clutch transmit the 37 KN at 2000 r.p.m. Find the dimension of the contact

surface.

Given data:

α = 30o; Pn= = 6 x 104 N/m2

b = R/3;

µ = 0.2; P = 37 KW = 37 x 103 N = 2000 rpm. To find dimensions of contact surface r1 dr 2) Solution:


2 NTP

π=

60

200021037 3 T××

=×π

T = 176.66 N-m Torque transmitted is also given by,

bRPT n ..2 2πµ=

××××=

310462.0266.176 2 R

Rπ

= 25132.74 R3 R = 0.19155 m or 191.55 mm

3

19155.0

3==

Rb

= 0.06385 m or 63.85 mm.

We find that Sinxb

rr=

− 21



113

o

SinSinbrr 1506385.021 ×=−=− α

01653.021 =− rr (1)

Mean radius 19155.02

21 =+

=rr

R

3831.021 =+ rr (2)

Solving equations (1) & (2) we get, Outer radius of contact surface r1 = 0.1998 (or) 199.8 mm Inner radius of contact surface r2 = 0.18328 m (or) 183.28 mm

7. The semi-cone angle of a cone clutch is 12.5 o and the contact surfaces have a

mean diameter of 80 mm. Co-efficient of friction is 0.32. What is the minimum

torque required to produce slipping of the clutch for an axial force of 200N.

If the clutch is used for connect an electrical motor with a stationary fly wheel,

what is the time needed to attain the full speed and energy lost driving slopping?.

Motor speed is 900 rpm. And moment of inertia of the fly wheel is 0.4 Kg.m2 .

Given data:

α = 12.5 o D = 80 mm or R = 40 m = 40 x 10-3 m µ = 0.32 W = 200 N N = 900 r.p.m I = 0.4 Kg – m2 To find,

i. Minimum torque required to produce slipping of the clutch. ii. Time needed to attain the full speed, iii. Energy lost during speed.

Solution: i. Minimum Torque required to produce slipping W.K.T Torque required to produce slipping. T = µ .W.R Cos α

oCo 15sec104020032.0 3 ××××= − T = 11.828 N-m



114

ii. Time needed to attain the full speed,

W.K.T

T = Iα

11.828 = 0.4 x α

α = 29.54 rad/sec2 Angular speed of fly wheel

t.αω = or

==

α

π

α

ω 1

60

2 Nt

∗

×=

57.29

1

60

9002πt

t = 31.1873 Ans. iii. Energy lost during slipping,

Angle turned by driving shaft 187.360

90021 ∗

×==

πωtQ

.4.3001 radQ =

Angle turned by driving shaft 2

22

1ttQ o αω +=

( )2187.357.29

2

10 ×∗+=

Q2 150.2 rad. Energy lost in friction = T (Q1- Q2) = 11.828 (300.4 - 150.2) = 1777 N- m.

8. A centrifugal clutch is to transmit 15 KW at 900 r.p.m. The shoes are four

in number. The speed at which the engagement begins is ¾ th of the

running speed. The inside radius of the pulley rim is 150 mm and the

center of gravity of the shoe lies at 120 mm from the centre of the spider.



115

The shoes are lined with ferrode for which the coefficient of friction may

be taken as 0.25. Determine: 1 Mass of the shoes and 2 sizes of the

shoes, if angles subtended by the shoes at the centre of the spider is 60O

and the pressure exerted on the shoes is 0.1 N mm2.

Given

P = 15 KW = 15 x 103 W n = 4 r = 120 mm = 0.12 m

θ = 60o N = 900 rpm R = 150 mm = 0.15 m;

µ = 0.25; P = 0.1 N/mm2 = 0.1 x 106 N/m2 To find,

i. Mass of the shoes (m) and ii. Size of the shoes (b)

Solution; i. Mass of the shoes (m)

Angular speed, sec/26.9460

9002

60

2rad

N=

×==

ππω

Given that the speed at which the engagement beings (ω) is ¾ th of running

speed (ω) is 3/4th of running speed (ω) therefore.

sec/7.7026.944

3

4

31 rad=×== ωω

Power transmitted, 60

2 NTP

π=

T = 159 N-m Centrifugal force in each shoe,

( ) 12.026.9422 mrmFc == ω

= 1066m N.



116

Spring force in each shoes, ie, the centrifugal force at the engagement speed, ω1

( ) ( ) .60012.07.7022

1 NmrmFs =×== ω

Frictional force acting tangentially on each shoe

( ) ( ) mNmmFsFF c 5.116600106625.0 =−=−= µ

We know that the torque transmitted. T = n.F.B 159 = 4 x 116.5 m x 0.15 = 70 m m = 2.27 Kg. ii. Size of the shoes,

θ = 60o = rado

318060

ππ=×

Contact length of shoe = Angle subtended by the shoe x radius of shoe

.1571.015.03

. mRl =×==π

θ

We know that, Fc – Fs = l .b.p 1066 m – 600 m = 0.1571 x b x 0.1 x 106 466m = 466 x 2.27 = 0.1571 x b x 0.1 x 106 b = 0.0673 m = 67.3 mm.

UNIT V

FRICTION



117


For Batch - (01-33)

PART A

1. What is creep in the case of belt?

2 Which type of screw thread is preferable in power transmission?

3. What are timing belts?

4 List out any four desirable characteristics of brake lining material

PART B

1. i)Derive the expression for Frictional torque on cone clutch based on uniform

Pressure theory. (6)

(ii) The brake whose dimensions are shown in figure has a co-efficient of friction of

0.3 and is to have a maximum pressure of 1000 kPa against the friction material.

(1) Using an actuating force of 1750 N, determine the face width of the Shoes

(both shoes have same width) and (2) What torque will the brake absorb?

4. An open belt running over two pulleys of 1.5m and 1.0m diameters connects two

parallel shafts 4.80m a part .The initial tension in the belt in 3000N.the smaller

pulleys is rotating at 600 rpm. The mass of the belt is 0.673kg/m length .The

coefficient of friction between the belt and pulleys is 0.3.Find 1. The exact length

of the belt required, and 2.The power transmitted taking centrifugal tension into

account [May/June 2006 Part Time]

UNIT V

FRICTION


For Batch - (34-64)

PART A



118

1. What is the condition of maximum efficiency of a Screw jack?

2. Distinguish between sliding and rolling friction.

3. State the condition for transmission of maximum power in belt drives.

4. Write an expression for the ratio of tension between the tight and slack sides

of a band and block brake.

PART B

1. Following data are for a screw jack. Screw pitch is 12.5 mm; mean diameter of

screw is 50 mm; co-efficient of friction is 0.13. i) Determine torques to raise and lower 20

kN load and efficiency of jack. (ii) Discuss the functions of clutches in automobiles.

(iii) Derive the expression to determine the ratio of tensions in a flat belt drive.

2. 100 kW is to be transmitted by a rope drive through a 160 cm diameter 45° grooved

pulley running at 200 rpm. Angle of overlap 140° and coefficient of friction between

pulley

and rope is 0.25. Mass of rope is 0.7 kg/m and it can withstand a tension of 800 N.

Considering centrifugal tension, find the following: (i) Number of ropes required, (ii) Initial

tension in the rope (May 2009)

B.E./B.Tech. Degree Examination, Nov/Dec, 2009

Third Semester

ME 2203 -- Kinematics of machinery

PART A

1. State Gruebler’s criterion for spatial mechanism.

The Grubler’s criterion applies to mechanisms with only single degree of freedom joints where the overall movability of the mechanism is unity. 3l – 2j – 4 = 0

where



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l – No. of links

j – No. of joints

2. Define ‘Mechanical Advantage’

If is defined as the ratio of load to the effort.

It may also be defined as output torque to input torque.

3.What is coriolis acceleration?

The acceleration of a moving point relative to a moving coordinate system may have two components of acceleration. One is tangential and the other one is perpendicular to the point. The perpendicular component of the point is known as coriolis component of acceleration.

4. What is meant by virtual centre?

Virtual centre (or) instantaneous centre of rotation of a moving body may be difined as the centre which goes on changing from one instant to another.

5. Why is roller follower preferred to knife edge follower?

In roller followers, the rolling motion takes place between the contacting surfaces, therefore the rate of wear is greatly reduced.

6. Define ‘Under cutting in cam’.

The process of removal of portion of cam which would have interfered with the follower is known as undercutting.

7. State the law of gearing.

The common normal at the point of contact between a pair of teeth always pass through the pitch point.

8. Write the mobility of differential mechanism.

The differential of an automobile requires that angular velocity two elements be fixed in order to know the velocity of remaining elements. The differential mechanism is said to have two degrees of freedom.

9. What do you mean by self-locking screws?

A screw is said to be self locking if the friction angle is greater than helix angle or coefficient of friction is greater than tangent of helix angle.

10. Give the effect of centrifugal tension in belf drives.



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The tension caused by centrifugal force is called centrifugal tension.

When the power transmitted is maximum, 1/3rd of maximum tension is absorbed as centrifugal tension.

PART B

11. (a) (i) Define transmission angle. Sketch a dray-link mechanism in maximum transmission angle and minimum transmission angle positions and explain.

Transmission angle

The angle between the output link and the coupler (connecting rod) is known as transmission angle (m)

The reason that mechanisms which appear to have three links are called four bar linkages is rooted in the theory of kinematics chains.

A chain of four links is the minimum basis for a useful mechanism. The classification of kinematics chains is based on a study of mechanism function as it is related to its form. Kinematics linkages that are identical except in the choice of the foundation link are called “inversions”. In describing the functional characteristics of inversions of four bar linkages it is usual to consider are pivoting link as the input or driver member.

In drag link mechanism, the shortest link DA is foundation link, and both driver and driven links AB and CD can make full revolutions.

(ii) Sketch and explain ‘whitworth’ quick return mechanism.

Whitworth quick return motion mechanism

This mechanism is also mostly used in shaping machines and slotting machines.

Here the link 2 is fixed. (This link 2 corresponds crank in a reciprocationg steam engine). The driving crank (link3) rotates with uniform speed. The slider (link 4) attached to the crank pin A. It slides along the slotted PA (link 1). So the slotted bar PA oscillates about a pivote point D. The connecting rod PR connects the slotted bar an ram which carries the tool and reciprocates.



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When the driving crank CA moves from CA¹ to CA2 (clockwise) through an angle ß, the tool is moving forward stroke (cutting stroke).

When the driving crank CA moves from CA2 to CA1 thorugh an angle ( œ ) (clockwise), the tool is moving back (return stroke).

Since ß is greater than œ and the driving crank CA rotates with uniform angular velocity, the return stroke is quicker than the cutting stroke. Thus it executes the quick return motion.

(or)

(b) (i) Describe various inversions of double slider rank mechanism with sketches.

Double slider crank chain

It consists of two turning pairs and two sliding pairs

Inversions of Double slider crank chain

Three important inversions

1. Elliptical trammel

2. Scotch yoke mechanism

3. Oldham’s coupling

1. Elliptical trammel

Elliptical trammel is an instrument used for drawing ellipses. Refer the fig.



122

This inversion is obtained by fixing the link 4 (slotten plate). This slotted plate has two straight grooves at righ angle to each other. (Just like + sign).

The slider (link 1) and slotted plate (link 4) form one sliding pair. The slider (link 3) and slotted plate (link 4) form another slider pair.

The connecting bar (link 2) and slider (link 1) form one turning pair; The connecting bar (link 2) and slider (link 3) form another turning pair. When the sliders slide along their respective grooves, any point on ‘connecting bar’ (say point P) traces out an ellipse.

Here AP is the half of major axis of the llipse and BP is the half of minor axis of the ellipse.

2. Scutch yoke mechanism

Scotch yoke mechanism is used to convert rotary motion into reciprocating motion. The inversion is obtained by fixing any one of the sliders link 1 or link 3.



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In this mechanism, link 1 is fixed. In such arrangement, the whole frame ie link 4 will reciprocate. When the crank (link 2) rotates about B as centre, the whole frame (link 4) reciprocates. The fixed link 1 guides the frame.

3. Oldham’s coupling

This coupling is used for connecting two parallel shafts whose axes are at a small distance apart. In this mechanism, when one shaft rotates, the other shaft also rotates at the same speed.

The link 1 (flange) is rigidly fastened to the end of the driving shaft by forging.

The link 3 (another flange) is rigidly fastened to the end of the driven shaft.



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The supporting frame (link 2) is fixed.

The link 1 and link 3 have diametrical slots cut in their inner faces.

In between to these two flanges, an intermediates pie (link 4) is sliding.

It has two tonges-two diametrical projections on bo faces – at right angles to each other.

The link 4 slides on the slots of the two links 1 and intermediate piece also rotates with the same speed. Hen the ling 3 also rotates with the same speed, i.e., links I, and 4 have the same angular velocity at every instant.

The distance between the axes of the shafts is consta and therefore the centre of the intermediate piece will follow the path of a circle with radius equal to the distance between the axes of the two shafts.

Therefore, maximum sliding speed of each tongue of the intermediate piece in the slot will be given by the periphery velocity of the centre of the disc along its circular path.

x distance between the axes of the shafts

V = w x r

(ii) Discuss about the straight line generators.

1. Peaucellier mechanism.

It consists of a fixed link OO1 and the other straight links O1, A, OC, OD, AD, DB, BC and CA are connected B turning pairs at their intersections as shown in fig. below.



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2. Hart’s mechanism

This mechanism requires only six links as compared with the eight links required by the peaucellier mechanism It consists of a fixed link OO1 and other straight links O, A, FC, CD, DE and EF are connected by turning pairs at their points of intersection as shown below.

12. (a) In a simple steam engint, the lengths of the crank and the connecting road are 100 mm and 400 mm respectively. The weight of the connecting rod is 50 kg and its centre of mass is 220 mm from the cross head centre. The radius of gyration about the centre of mass is 120 mm if the engine speed is 300 rpm and the crank has turned 45’ from IDC, determine

(i) The angular velocity and acceleration of the connecting rod.

(ii) Kinetic energy of the connecting rod.

Solution

NBO = 300 rpm



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W = 2 x 300

Since the crank length OB =0.1m,

ç linear velocity of B with respect to O

VBD =VB = WBO x OB =31.42 x0.1 =3.142 m/s



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(ii) Kinetic energy of connecting rod.

½ mu² =1/2 x 50 x 2.4² = 144 N.m



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(or)

(ii) PQRS is a four bar chain with a fixed link PS. The lengths of the links are: PQ = 62.5 mm, QR = 175 mm, RS = 112.5 mm and PS = 200 mm. The crank PQ rotates at 10 radls clockwise Draw the velocity and acceleration diagram when angle QPS = 60 º and find the angular velocity and angular acceleration of the links QR and Rs.

Solution

Procedure



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130



131



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13. (a) A disc cam used for moving a knife edge follower with SHM during life and uniform acceleration and retardation motion during return. Cam rotates at 300 rpm clockwise direction. The line of motion of the follower has an offset 10 mm to the right angle of cam shaft axis. The minimum radius of the cam is 30 mm. The lift of the follower is 40 mm. The cam rotation angles are: Lift 60 º, dwell. 90 º, return 120 º and remaining angle for dwell. Draw the cam profile and determine the maximum velocity and acceleration during the lift and return.

Solution

The displacement diagram

1. Draw a rectangle of length equal to 90 mm respectively 360 º.

2. Divide the rectangle in such a way outstroke 15 mm (60 º), dwell 22.5 mm

(90 º) , return 30 mm (120 º), dwell 22.5 mm (90 º)



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3. Here lift is with simple harmonic motion and return stroke is with uniform acceleration and retardation.

To draw profile of cam

1. Draw the base circle with radius equal to the minimum radius of cam (30 mm) with O as centre.

2. Draw a offset circle of radius 10 mm concentric with the base circle Now draw a

3. vertical axis line tangent to the offset circle. This vertical axis line is the axis of follower.

4. The axis of follower cuts the base circle at O.

5. Join OA. Mark angle AOS=60º to represent lift, angle SOT = 90º to represent

dwell and angle TOP = 120º to represent return stroke.

6. Divide the outstroke AOS into six equal parts as in displacement diagram and

mark 0, 1, 2, 3, 4, 5 and 6 similarly divide return stroke.

7. Now draw tangents from all the points.

8. Join all the points ABCDEFGHJKIMNPA by smooth curve. This curve is the

profile of cam.

Maximum velocity and acceleration dueling lift and return



134



135

(or)

(b) What is tangent cam? Derive the expressions for the velocity and acceleration of a roller follower in the tangent cam.

When the roller has contact with the nose

A roller having contact with the circular – nose at G is shown in Fig. The centre of roller lies at D. on the pitch curve. The displacement is usually measured from the top position of the roller, i.e. when the roller has contact at the apex of the nose (point H) and the centre of roller lies at J on the pitch curve.

Terminology



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Fig: tangent cm with reciprocating roller follower having contact with the nose



137



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14. (a) A Pair of spur gears with involute teeth is to give a gear ration of 3:1 The are of approach is not be less than the circular pitch and smaller wheel is the driver. The angle of pressure is 20º.

(i) What is the least number of teeth that can be used on each wheel? (ii) What is the addendum of the wheel in terms of circular pitch?

Solution

1. Least number of teeth on each wheel



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Let

t = Least number of teeth on smaller wheel

T=Least number of teeth on larger wheel

r= Pitch circle radius of smaller wheel

Maximum length of patch of approach.



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(or)

(b) In an epicylic gear trair, an annular wheel A having 54 teeth meshes with a planet wheel B which gears with a sun wheel C the wheels A and C being rotates about the axis of the wheels A and C. If the wheel A makes 20 rpm in a clockwise sense and the arm rotates at 100 rpm in the anticlockwise direction and the wheel C has 24 teeth, determine the speed and direction of rotation of wheel C.

From the geometry of the figure,



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Nc = x + y = 100 + 200 = 370 rpm



142

15. (a) (i), Derive the condition for maximum efficiency of screw. The efficiency of a screw jack may be defined as the ratio between ideal effort to actual effort.

(ii) A bicycle and rider and mass 100kg are tavelling at the rate of 16km/hr on a

level road. A brake is applied to the rear wheel which is 00.9 m in

diameter and this is the only resistance acting. How many turns will it

make before it comes to rest? The pressure applied to on the brake is

100 N and µ = 0.05.

Solution



143

( or)

(b) A rope drive is required to transmit 230 kw from a pulley of 1 m diameter

running at 450 rpm. The safe pull in each rope is 800 N and the mass of the rope is



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0.4 kg per meter length. The angle of lap and groove is 1600 and 450 respectively. If

µ = 0.3, Find the number of ropes required.

Solution:



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B.E./B.Tech. Degree Examination, Nov/Dec 2010

Third Semester

Mechanical Engineering

(Regulation 2004)



146

MA 2203 – Kinematics of Machinery

(Regulation 2008)

(Common to PTME 2203 – Kinematics of Machinery for B.E. (Part –Time) Third

Semester Mechanical Engineering (Regulation 2009)

Time: Three hours Maximum:

100 marks

Answer ALL questions

PART A - (10x2=20 marks)

1. Differentiate between a machine and a mechanism.

Machine Mechanism

The parts of a machine move relative

to one another

When one of links of a kinematic

chain is fixed, the chain is

mechanism.

A machine transforms the available

energy into some useful work.

It is used for transmitting or

transforming motion.

The links of a machine may transmit

both power and motion

When mechanism is required to

transmit power to do type of work it

them becomes machine.

Example : Lathe, Cycle Example : Typewriter, Engine

indicators.

2. Write and explain Gruebler’s equation.

The Grubler’s criterion applies to mechanism with only single degree of freedom

joints where overall movability of the mechanism is unity

i.e n = 1 &h = 0 in Kutzbach equation,

3J-2j- 4 = 0

This equation is known as the Goubler’s criterion.

3. Draw a sketch to explain how total acceleration of a link is obtained.

Acceleration of a link

4. What is Coriolis component of acceleration?

Coriolis component of acceleration

When a point on one link is sliding along another rotating link, such as in

quick return motion mechanism, then the coriolis component of

acceleration must be calculated.



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The tangential component of acceleration of the sliding element with

respect to the coincident point on the rotating element is known as coriolis

component of acceleration.

5. Define undercutting in a cam mechanism.

Undercutting

On the roller follower, the prime circle of a cam is proportioned to provide aa

satisfactory pressure angle, still the follower may not be completing the desired

motion. This can happen if the curvature of the pitch curve is too sharp. As it is

impossible to produce such a cam profile, the result is that the cam be undercut

and become pointed cam.

6. What are the different types of follower motions used in cam- follower

mechanisms?

Types of Follower

1. Knife edge follower

2. Roller follower

3. Flat faced follower

4. Spherical faced follower

7. What is the significance of contact ratio in gears?

Contact ratio

The contact ratio or the numbr of pairs of teeth in contact is defined as the ratio

the length of the are of contact to the circular pitch.

1. The contact ratio, usually, is not a whole number.

2. The theoretical minimum value for the contact ratio is one, that is there must

always be atleast one pair of teeth in contact ratio, more quietly the gears will

operate.

3. Larger the contact ratio, more quietly the gears will operate.

8. What are the advantages of epicyclic gar train?

Advantages of Epicyclic gear trains

1. The epicyclic gear trains are useful for transmitting high velocity ratios with

gears of moderate size in a comparatively lesser space.

2. Large Speed reductions are possible with epicyclic gears and if fixed wheel is

annular, amore compact unit could be obtained.

3. The epicyclic gear trains are used in the back gear of lathe, differential gears

of the automobiles, hoists, pulley blocks, wrist watches etc.

9. What is the role of friction in screw jack?

Screw Jack Friction.

The screw jack is a device, for lifting heavy loads, by applying a comparatively

smaller effort at its handle. The principle, on which a screw jack works, is similar

to that of a friction on inclined plane. A screw jack, which consists of a square

threaded rod (also called screw rod) which fits into the inner threads of the nut.

The load, to be raised or lowered, is placed on the application of an effort at the



148

end of the lever of the square threaded rod which is rotated by the application of

an effort at the lever for lifting or lowering the load. Since the principle on which a

screw jack works similar to that of an inclined plane, therefore the force apple on

the lever of a screw jack may be considered to horizontal. When the load is being

lifted, therefore the for of friction ( F= RN) will act downwards.

10. What is self-energizing brake?

The moment of frictional force (µ.RN.a) adds to the moment of force (P.I). In

words, the frictional force help to apply the brake. When the frictional force is great

enough to apply the brake with no external force, then the brake said to be self-

locking brake.

PART B - (5X16=80 marks)

11. (a)(i)Explain the working of toggle mechanism with a neat sketch.

Toggle mechanism

A mechanical linkage is a series of rigid link connected with joints to form a

closed chain, or a series o closed chains. Each link has two or more joints, and

the joints have various degrees of freedom to allow motion between the links. A

linkage is called a mechanism if two or more links are usually designed to take an

input and produce a different output, altering the motion, velocity, acceleration,

and applying mechanical advantage.

Many practical multilink mechanisms are made up of basic linkage combinations

such as the slider-crank and the four-bar mechanism. The toggle linkage shown in

Figure 1 is an example of a mechanism of this type; the toggle principle is applied

in ore crushers and in essentially static linkages that act as clamps. The linkage

analysis is made by considering the basic linkages separately.

Toggle mechanisms are used, where large resistances are to be overcome

through short distances. Here, effort applied will b small but acts over large

distance. In the mechanism shown in fig 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.

Tan α = F/2

P



149

F=2 P tan α

For small angles of a, F (effort) is much smaller than P (resistance).

This mechanism is used in rock crushers, presses, riveting machines etc.

An important feature of the toggle mechanism is its ability to produce high values

of force at the slider with relatively low torque input. While the study of

mechanisms is concerned primarily with motion, forces are of great importance to

the designer and are intimately reltd motion analysis. If a rigid mechanism has a

single input a a single output with negligible losses, the rate of energy in equals

the rat of energy output. Force ration are the invent of velocity ratios when inertia

effects are negligible.

11.(a) (ii) Explain the working of Pantograph or indexing mechanism with a

neat sketch.

Pantograph

A pantograph is an instrument used to reproduce to enlarged or a reduced

scale and as exactly as possible the path described by a given point. It consists of

a joints parallelogram ABCD as shown in Fig.2.

It is made up of bars connected by turning pairs. The bars BA and BC are

extended to O and E respectively, such that OA/OB = AD/BE Thus, for all relative

positions of the bars, the triangles OAD and OBE are similar and the points O,D

and E are in one straight line. It may be proved that point E traces out the same

point E traces out the same path as described by point D.

From similar triangles OAD and OBE. We find that OD/OE-AD/BE Let point

O be fixed and the points D and E move to some new positions D’ and E’ Then

OD/OE=OD/OE

A little consideration will show that the straight line DD’ is parallel to the

straight line EE’. Hence, if O is fixed to the frame of a machine by means of a

turning pair and D is attached to a point in the machine which has rectilinear

motion relative to the frame, them E will also trace out a straight line path.

Similarly, if E is constrained to move in a straight line, them D will trace out a

straight line parallel to the former.

A pantograph is mostly used for the reproduction of plane areas and

figures such as mps, plans etc., on enlarged or reduced scales. It is, sometimes,

used as an indicator rig in order to reproduce to a small scale the displacement of

the crosshead and therefore of the piston of a reciprocating steam engine. It is



150

also used to guide cutting tools. A modified form of pantograph is used to guide

cutting tools. A modified form of pantograph is used to collect power at the top of

an electric locomotive.

(or)

11.(b) Explain the working of any two inversions of a slider –crank chain

with neat sketches.

Single Slider Crank Chain

A single slider crank chain is a modification of the basic four bar chain. It

consist of one sliding pair and three turning pairs. It is, usually. Found in

reciprocating steam engine mechanism. This type of mechanism converts rotary

motion into reciprocating motion and vice versa. In a single slider crank chain, as

shown in Fig. 3, the links 1 and 2, links 2 and 3, and links 3 and 4 form three

turning pairs while the links 4 and 1 form a sliding pair.

The link 1 corresponds to the frame of the engine, which is fixed. The link 2

corresponds to the crank; link corresponds to the connecting rod and link 4

corresponds to cross-head. As the crank rotates, the cross-head reciprocates in

the guides and thus the piston reciprocates in the cylinder.

1. Pendulum pump or Bull engine

In this mechanism, the inversion is obtained by fixing the cylinder or link 4 (i.e

sliding pair), as shown in Fig. In this case, when the crank (link 2) rotates, the

connecting rod (link 3) oscillates about a pin pivoted to the fixed link 4 at A and

the piston attached to the piston rod ( link 1) reciprocates. The duplex pump

which is used to supply feed water to boilers have two pistons attached to link

1, as shown in Fig.4



151

2. Oscillating cylinder engine

The arrangement of oscillating cylinder engine mechanism, as shown in Fig.5,

is used to convert reciprocating motion into rotary motion. In this mechanism,

the link 3 forming the turning pair is fixed. The link 3 corresponds to the

connecting rod of a reciprocating steam engine mechanism. When the crank

(link 2) rotates, the piston attached to piston rod (link 1) reciprocates and the

cylinder (link 4) oscillates about a pin pivoted to the fixed link at A.

3. Rotary internal combustion engine or Gnome engine

Sometimes back, rotary internal combustion engines were used in aviation. But

now-a-days gas turbines are used in its place. It consists of seven cylinders in one

plane and all revolves about fixed centre D, as shown in Fig.6. while the crank

(link2) is fixed. In this mechanism, when the connecting rod ( link 4 ) rotates, the

piston (link 3) reciprocates inside the cylinders forming link 1.

12. (a) The lengths of various links of a mechanism shown in are as follows:

QA = 150 mm, CD=125 mm AC=600mm, BD=50 mm, CQ=QD=145 mm,



152

OQ=625mm The crank OA rotates at 60 rpm in the counter-clockwise

direction Determine the velocity of the slider B and the angular velocity of

the link BD when the crank has turned n angle of 450 with the vertical.

Given data

N=60 rpm;

Solution

Displacement Diagram

Velocity Diagram



153

(or)

12.(b) For the slide-crank mechanism shown in Fig. Q.12 (b), determine (i)

the acceleration of slider B and (ii) acceleration of point C. The crank OA

rotates at 180 rpm. OA=500 mm, AB =1500 mm and AC =250 mm.

Given data :

OA=500mm; AB =1500 mm; AC=250 mm

N = 180 rpm

Solution


Velocity Diagram



154

b’c ‘= 146.55 mm

aB =127.4 m/sec 2

aC = 192.82 m/sec2

13. (a) Draw the cam profile for the following data: Bse circle radius of cam

=50mm, Lift = 40 mm, Angle of ascent with Cycloidal = 600, Angle of dwell =

900 Angle of descent with uniform velocity = 900,Speed of cam =300 rpm,

Follower offset = 10 mm, Type of follower Knife-edge.

Given data



155

Follower : Knife edge

Lift : 40 mm

Cam radius : 50 mm

Displacement : Cycloidal (Ascent); Uniform velocity (Descent)

Angle : 600,900,900

Offset : 10mm


1. Draw a base circle with radius equal to the minimum radius of the cam (i.e., 50

mm) with O as centre

2. Draw the axis of the follower at a distance of 10 mm from the axis of the cam,

which intersects the base circle at A.

3. Join AAO and draw an offset circle of radius 10 mm with centre O.

4. From OA, mark angle AOB = 600 to represent outstroke, angle BOC = 900 to

represent return stroke.



156

5. Divide the angular displacement during outstroke and return stroke (i.e., angle

AOB and angle BOC) into the same number of equal even parts as in

displacement diagram.

6. Now from the points 1,2,3… etc and 0’1’2’3’…… etc. on the base circle, draw

tangents to the offset circle and produce these tangents beyond the base circle

as shown in Fig.

7. Now set off 1’a’, 2’b’, 3’c’… etc. and 9j, 8h, 7g,…. Etc. from the displacement

diagram.

8. Join the points a’b’, c’, … d,c.b.a with a smooth curve. The curve AfiBA is the

complete profile of the cam.

(or)

13.(b) Draw the cam profile for the following data :Base circle rdius of cam =

50 mm, Lift = 40mm, Angle of ascent with SHM = 900 , Angle of descent with

uniform acceleration and deceleration = 900, Speed of cam = 300 rpm, Type

of follower =, Roller follower (with roller radius = 10mm).

Given data

Cam radius = 50mm

Lift = 40 mm

Angle = 900,900,900,

Displacement : SHM (Ascent ); UA&R (Descent)

Follower : Roller

Roller radius : 10mm



157

14.(a). The Pressure angle of two gears is 200 and has a module of 10 mm. The number of teeth on pinion is 24 is on gear 60. The addendum of pinion and gear is same and equal to one module. Determine (i) the number of paris of teeth in contact (ii) the angle of action of pinion and gear and the ratio of sliding to rolling velocity at the beginning of contact. Given Data:



158



159

(or)

14.(b). The pitch circle diameter of the angular gear in the epicyclic gear in Fig . is 425 mm and the module is 5 mm. when the annular gear 3 is stationary, the spindle A makes one revolution in the same sense as the sun gear 1 for every 6 revolutions of the driving spindle carrying the sun gear. All the planet gears are of the same size. Determine the number of teeth on all gears.

Given Data:

d3 = 425 mm ; module = 5 mm



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N3 = 0 ; NA = N1 / 6

15.(a). An open belt drive is used to connect two parallel shafts 4m apart. The diameter of bigger pulley is 1.5 m and that of the smaller pulley 0.5m. The mass of the belt is 1 kg/m length. The maximum tension is not exceed 1500 N. The coefficient of friction is 0.25. The bigger pullely which is the driver runs at 250 rpm. Due to slip, the speed of the driven pulley is 725 rpm. Calculate the power transmitted, and power lost in friction. Given Data:

solution:



161



162

Power Cost in Friction:

(or)

15. (b). A Vertical shaft 140 mm diameter rotating at 120 rpm rests on a flat end

footstep bearing. The shaft carries a vertical load of 30 KN. The co-efficient of

friction is 0.06. Estimate the power lost in friction assuming (1) uniform

pressure and (2) uniform wear.

Given Data : D = 140 mm; R = 70 mm; N = 120 rpm; W = 30 KN ; µ = 0.06 Solution



163



164

Question Paper Code : 13152

B.E./B.Tech. Degree Examination, Nov/Dec 2011

Third Semester

Mechanical Engineering

113302 – Kinematics of Machinery

(Regulation 2010)

Time: Three hours Maximum:

100 marks

Answer ALL questions

PART A - (10 x 2=20 marks)

1. Explain the term kinematic link. Give the classification of the kinematic link.

Each part of a machine, which moves relative to some other part, is known as a

kinematic link or element .

a. Rigid link

b. Flexiable link

c. Fluid link

2. Define Lower pair and Higher pair. When the two elements of a pair have a surface contact when the relative

motion take place and the surface of one element slides over the surface of other, the pair formed is known as lower pair.

When the two elements of a pair have a line or point contact when relative motion takes place and the motion between the two elements is partly turning and partly sliding, the pair is known as higher pair.

3. List out the type of kinematic pairs according to the type of closure. a. Self closed pair b. Force-closed pair

4. What is coriolis component of acceleration? When a point on one link is sliding along another rotating link, such as in quick return motion mechanism, then the Coriolis component of acceleration is calculated.

5. Define pressure angle of a cam mechanism and state the best value of the pressure angle.

The pressure angle is defined as the angle between the axis of the follower stem and the line of the force exerted by the cam onto the roller follower, the normal to the pitch curve through the trace point.

6. State the advantages of tangent cams. When the flanks of the cam are straight and tangential to the base circle and nose cirle, then the cam is known as tangent cam. These cams are usually symmetrical



165

about the centre line of the cam shaft. Such types of cams are used for operating the inlet and exhaust valves of internal combustion engines.

7. What is epicylic gear train? The epicyclic gear trains are useful for transmitting high velocity ratios with gears of moderate size in a comparatively lesser space.

8. What is the use of differential in automobile? In automobiles and other wheeled vehicles, the differential allows the driving road wheels to rotate at different speeds.

9. What is the difference between sliding friction and rolling friction? Give example. Sliding friction is the friction experienced by a body when it slides over another body. Rolling friction is the friction experienced between the surfaces which have balls or rollers interposed between them.

10. What are the advantages of V – Belt Drive? a. The V- belt drives gives compactness due to the small distance between

the centers of pulleys b. It provides longe life, 3 to 5 years. c. It can be easily installed and removed. d. The operation of the belt and pulley is quiet.

Part – B

11. (a). In crank and slotted lever quick return motion mechanism, the distance

between the fixed centre is 240 mm and the length of the driving crank is

120 mm. Find the inclination of the slotted bar with the vertical in the

extreme position and the time ratio of cutting stroke to be return stroke. If

the length of the slotted bar 450 mm, Find the length of the stroke if the line

of stroke passes through the extreme positions of the free end of the lever.

Given Data:

AC = 240 mm; CB1 = 120 mm; AP1 = 450 mm

Solution

Inclination of the slotted bar with the vertical



166



167

(or)

11.(b). What is Inversion of mechanism? With neat sketch and explain any two inversion of double slider crank chain mechanism. Solution Inversion of double slider crank chain mechanism 1. Elliptical Trammels

2. Scotch yoke mechanism



168

3. Oldham’s Coupling



169

12. a. The diagrams shows a “rocking lever” mechanism in which steady rotation

of the wheel produces an oscillating motion of the lever OA. Both the wheel and

the lever are mounted in fixed centers. The wheel rotates clockwise at a uniform

angular velocity (�) of 100 rad/s. For the configuration shown, determine the

following.

(i) The angular velocity of the link AB and the absolute velocity of point A.

(ii) The centrifugal accelerations of BC, AB and OA.

(iii)The magnitude and direction of the acceleration of point A.

The lengths of the links are as follows.

BC = 25 mm AB = 100 mm OA = 50 mm OC = 90 mm

SOLUTION The solution is best done graphically. First draw a line diagram of the mechanism to

scale. It should look like this.

Next

calculate the velocity of point B relative to C and construct the velocity diagram.



170

(vB)C = ω x

radius = 100 x 0.025 = 2.5 m/s

Scale the following velocities from the diagram.

(vA)O = 1.85 m/s {answer (i)} (vA)B = 3.75 m/s

Angular velocity = tangential velocity/radius

For link AB,ω= 3.75/0.1 = 37.5 rad/s. {answer (i)} Next calculate all the accelerations

possible.Radial acceleration of BC = 2 x BC = 1002 x 0.025 = 250 m/s2.

Radial acceleration of AB = v2/AB = 3.752/0.1 = 140.6 m/ s2.

Check same answer from 2 x AB = 37.52 x 0.1 = 140.6 m/ s2.

Radial Acceleration of OA is v2/OA = 1.852/0.05 = 68.45 m/ s2.

Construction of the acceleration diagram gives the result shown.

The acceleration of

point A is the vector o- a shown as a dotted line. Scaling this

we get 560 m/s2.

(or)



171

12. b. A horizontal single cylinder reciprocating engine has a crank OC of radius 40 mm and a connecting rod PC 140 mm long as shown. The crank rotates at 3000 rev/min clockwise. For the configuration shown, determine the velocity and acceleration of the piston.

The sliding piston has a mass of 0.5 kg and a diameter of 80 mm. The gas

pressure acting on it is 1.2 MPa at the moment shown. Calculate the effective

turning moment acting on the crank. Assume that the connecting rod and crank

has negligible inertia and friction.

SOLUTION

Draw the space diagram to scale:

The moment arm should be scaled and found to be 34 mm (measured at right angles

to the connecting rod PC.

Calculate the velocity of C relative to O.

ω = 2πN/60 = 2π x 3000/60 = 314.16 rad/s

(VC)O = ω x radius = 314.16 x 0.04 = 12.57 m/s

Draw the velocity diagram.

From the velocity diagram we find the velocity of the piston is 11 m/s.



172

Next calculate all the accelerations possible. Point C only has a radial acceleration

towards O.

Radial acceleration of C is v2/radius = 12.572/0.04 = 314.16 m/s2

Point P has radial and tangential acceleration relative to C. Tangential acceleration is

unknown. Radial acceleration = (vP)C2/CP = 92/0.14 = 578.57 m/s2. Now draw the

acceleration diagram and it comes out like this.

The acceleration of the piston is 770 m/s2.

Now we can solve the forces.

Pressure force = p x area = 1.2 x 106 x π x 0.082/4 = 6032 N and this acts left to right.

Inertia force acting on the piston = M a = 0.5 x 770 = 385 N and this must be provided

by the pressure force so the difference is the force exerted on the connecting rod. Net

Force = 6032 – 385 = 5647 N.

The connecting rod makes an angle of 11o to the line of the force (angle scaled from

space diagram). This must be resolved to find the force acting along the line of the

connecting rod.

The force in the connecting rod is 5637 cos 11o = 5543 N.

This acts at a radius of 34 mm from the centre of the crank so the torque provided by the

crank is T = 5545 x 0.034 = 188.5 N



173

13. a. Draw the profile of a cam operating with a knife edged follower having a lift

of 30mm. The cam raises the follower with SHM for 1500 of its rotation followed by

a period of dwell for 600. The follower descends for the next 1000 rotation of the

cam with uniform velocity, again followed by a dwell period. The cam rotates at a

uniform velocity of 120 rpm and has the least radius of 20mm. What will be the

maximum velocity and acceleration of the follower during the lift?

S= 30mm=0.3m; ө0=150*Л/180=2.618 rad; өR =100*Л/180=1.745 rad; N=120 rpm.

Maximum velocity and acceleration of the follower during the lift:


ω = 2 Л N / 60 = 2 Л 120 / 60 =12.57 rad/s;


v0= Л ω S / 2 ө0 = 0.226 m/s;


ao= Л2 ω 2 S /2(ө0 )= 3.413 m/s2.

(or)



174

13. b. Design a cam to raise a valve with SHM 50mm in 1/3 of revolution , keep it

fully through 1/12 revolution and to lower it with harmonic motion in 1/6 revolution

. The value remains closed during the rest of the revolution. The diameter of the

roller is 20 mm and minimum radius of the cam is 25mm. the diameter of cam shaft

is 25mm.the axis of the valve rod passes through the axis of the cam shaft. if the

cam shaft rotates at uniform speed of 100 rpm, find the maximum velocity and

acceleration of a valve during raising and lowering.

S= 50mm=0.5m; ө0=360*1/3=120 deg =2.094 rad; өR=360*1/6=60 deg= 1.047 rad;

N=100 rpm.

Diameter of the roller = 20 mm;

Radius of cam=25 mm;

Diameter of the cam shaft=25 mm; Dwell=1/12*360=30 deg.

Maximum velocity of the follower during its ascent and descent:


ω = 2 Л N / 60 = 2 Л 100 / 60 =10.47 rad/s;

Maximum velocity of the follower during ascent,

v0= Л ω S / 2 ө0 = 0.39 m/s;

Maximum Acceleration of the follower during ascent,

ao= Л2 ω 2 S /2(ө0)2= 6.17 m/s2.

Maximum velocity of the follower during descent,

v0= Л ω S / 2 өR = 0.78 m/s;

Maximum Acceleration of the follower during descent,

ao= Л2 ω 2 S /2(өR)2= 24.67 m/s2.



175

14. a. . A pair of gears, having 40 and 30 teeth respectively are of 25o involute

form. The addendum length is 55mm and module pitch is 2.5 mm. If the smallest

wheel is driver and rotates at 1500 rpm. Find the velocity of sliding at the point of

engagement and at the point of disengagement.

Given, TG = 40 TP: = 30

φ = 25o

m = 2.5 mm A/p = 1500 rpm Addendum = 5 mm To find;

vii. Velocity of sliding at the point of engagement viii. Velocity of sliding at the point of disengagement

Solution:


15002

60

25rad

N P

P =×

==π

ω

Gear ratio,

P

a

a

p

T

T=

ω

ω (or)



176

G

P

PaT

T×= ωω

sec/81.11740

3008.157 rad=×=

Let r, rA = Pitch circle and addendum circle radii of pinion R, RA = Pitch circle and addendum circle radii of gear

.5.372

305.2

2mm

mTr P =

×==

.502

405.2

2mm

mTR a =

×==

rA = r + addendum = 37.5 +5 = 42.5 mm RA = R + addendum = 50 +5 = 55mm


( ) ( ) ooCosKP 25sin50255055 222−−=

= 10.04 mm


( ) ( ) ooCos 25sin5.37255.375.42 222−−=

= 9.67 mm

i. Velocity of sliding at the point of engagement

( )21 ωω +=Vs Length of path of approach

( )KP21 ωω +=

( ) 04.1081.11708.157 ×+=

smmVs /89.2759= = 2.75 m/s

ii. Velocity of sliding at the point of disengagement

( )21 ωω +=Vs Length of path of recess

( ) 67.981.11708.157 +=

smm /2658=

= 2658 mm/s Vs = 2.66 m/s

(or)



177

14. b. The following data refers to two matching involute gears of 20o pressure

angle.

Number of teeth on pinion : 20 Gear ratio : 2 Speed of pinion : 250 rpm Module : 12 mm If the addendum of each wheel is such that the path of approach and path of

recess on each side are half the maximum possible length. Find

i). addendum for both the wheels ii). The length of arc of contact

v. Contact ratio & The maximum sliding velocity ix. Angle turned through by the pinion and x. Angle turned through by the gear wheel when one pair of teeth is in

contact. Given:

TP: = 20 ; Gear ratio, G = 2 =P

a

T

T;

Np = 250 rpm m = 12mm Solution:


2502

60

2rad

N P

P =×

==ππ

ω

Gear ratio,

2==P

a

T

TG

402022 =×=×= pG TT

2===p

G

G

P

T

TG

ω

ω

sec/08.132

16.26

2radP

n ===⇒ω

ω

Pitch circle radii of pinion and gear wheel is given by,

mmmT

r P 1202

2012

2=

×==



178

mmmT

R G 2402

4012

2=

×==

i. Addendum for both the wheels; Given that addendum on pinion and gear is such that the path of approach and recess are half of their maximum possible values, so,

2

sinsin222 φ

φφr

RCosRRA =−− (1)

and2

sinsin222 φ

φφR

rCosrrA =−− (2)

Substituting the values of R and r in (1) we get

2

20sin12020sin24020240 222

ooo

A CosR =−−

RA = 247.77 mm ∴ Addendum of gear wheel = RA – R = 247.77 – 240

= 7.77 mm Substituting the values of R and r in (2) we get,

2

sinsin222 φ

φφR

rCosrrA =−−

2

20sin24020sin12020120 222

ooo

A Cosr =−−

rA = 139.5 mm

∴ Addendum of pinion = rA – r = 139.5 – 120 = 19.5 mm ii. Length of arc of contact;


( ) ( ) ooCos 20sin2402024077.247 222−−=

= 20.52 mm

φφ sin222rCosrrPL A −−=

( ) ooCos 20sin120201205.139 222−−=

= 41.08 mm. Length of path of contact, KL = KP+ PL = 20.52 + 41.08 = 61.6 mm

∴ Length of arc of contact = mmCosCos

KLo

51.6520

56.61==

φ



179

iii. Contact ratio: Contact ratio =

2738.112

51.65say=

×π

iv. Maximum sliding velocity: ( ) PLVs ×+= 21 ωω as PL > KP

( ) smsmm /612.1/97.161108.4108.1316.26 ==×+=

v. Angle turned through by the pinion: Angle turned by pinion =

= oo

28.311202

36051.65=

×

×

π

vi. Angle turned through by the gear wheel; Angle turned by gear =

2402

36051.65

×

×=

π

o

= 15.64o

15.(a). The semi-cone angle of a cone clutch is 12.5 o and the contact surfaces

have a mean diameter of 80 mm. Co-efficient of friction is 0.32. What is the

minimum torque required to produce slipping of the clutch for an axial force of

200N.

If the clutch is used for connect an electrical motor with a stationary fly wheel,

what is the time needed to attain the full speed and energy lost driving slopping?.

Motor speed is 900 rpm. And moment of inertia of the fly wheel is 0.4 Kg.m2 .

Given data:

α = 12.5 o D = 80 mm or R = 40 m = 40 x 10-3 m

µ = 0.32 W = 200 N N = 900 r.p.m I = 0.4 Kg – m2 To find,

iv. Minimum torque required to produce slipping of the clutch. v. Time needed to attain the full speed,


π m


Circumference of pinion

o360×


Circumference of gear

o360×



180

vi. Energy lost during speed. Solution: i. Minimum Torque required to produce slipping W.K.T Torque required to produce slipping.

T = µ .W.R Cos α

oCo 15sec104020032.0 3 ××××= −

T = 11.828 N-m ii. Time needed to attain the full speed,

W.K.T

T = Iα

11.829 = 0.4 x α

α = 29.54 rad/sec2 Angular speed of fly wheel

t.αω = or

==

α

π

α

ω 1

60

2 Nt

∗

×=

57.29

1

60

9002πt

t = 31.1873 Ans. iii. Energy lost during slipping,

Angle turned by driving shaft 187.360

90021 ∗

×==

πωtQ

.4.3001 radQ =

Angle turned by driving shaft 2

22

1ttQ o αω +=

( )2187.357.29

2

10 ×∗+=

Q2 150.2 rad.



181

Energy lost in friction = T (Q1- Q2) = 11.828 (300.4 - 150.2) = 1777 N- m.

(or)

15.(b). A centrifugal clutch is to transmit 15 KW at 900 r.p.m. The shoes are

four in number. The speed at which the engagement begins is ¾ th of the

running speed. The inside radius of the pulley rim is 150 mm and the center of

gravity of the shoe lies at 120 mm from the centre of the spider. The shoes are

lined with ferrode for which the coefficient of friction may be taken as 0.25.

Determine: 1 Mass of the shoes and 2 sizes of the shoes, if angles subtended

by the shoes at the centre of the spider is 60O and the pressure exerted on the

shoes is 0.1 N mm2.

Given

P = 15 KW = 15 x 103 W n = 4 r = 120 mm = 0.12 m θ = 60o N = 900 rpm R = 150 mm = 0.15 m;

µ = 0.25; P = 0.1 N/mm2 = 0.1 x 106 N/m2 To find,

iii. Mass of the shoes (m) and iv. Size of the shoes (b)

Solution; i. Mass of the shoes (m)

Angular speed, sec/26.9460

9002

60

2rad

N=

×==

ππω

Given that the speed at which the engagement beings (ω) is ¾ th of running

speed (ω) is 3/4th of running speed (ω) therefore.

sec/7.7026.944

3

4

31 rad=×== ωω

Power transmitted, 60

2 NTP

π=

T = 159 N-m Centrifugal force in each shoe,



182

( ) 12.026.9422 mrmFc == ω

= 1066m N.

Spring force in each shoes, ie, the centrifugal force at the engagement speed, ω1

( ) ( ) .60012.07.7022

1 NmrmFs

=×== ω

Frictional force acting tangentially on each shoe

( ) ( ) mNmmFsFF c 5.116600106625.0 =−=−= µ

We know that the torque transmitted. T = n.F.B 159 = 4 x 116.5 m x 0.15 = 70 m m = 2.27 Kg. ii. Size of the shoes,

θ = 60o = rado

318060

ππ=×

Contact length of shoe = Angle subtended by the shoe x radius of shoe

.1571.015.03

. mRl =×==π

θ

We know that, Fc – Fs = l .b.p 1066 m – 600 m = 0.1571 x b x 0.1 x 106 466m = 466 x 2.27 = 0.1571 x b x 0.1 x 106 b = 0.0673 m = 67.3 mm.

ME 2203 – KINEMATICS OF...

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