regulation 2017 academic year : 2018-2019 me8492 ... · elliptical trammels: it is an instrument...

REGULATION : 2017 ACADEMIC YEAR : 2018-2019

JIT-JEPPIAAR/MECH /Mr.D.Raguraman and Mr.D.Arunkumar /IIrd Yr/SEM 04 /ME8492/KINEMATICS OF

MACHINERY/UNIT 1-5/QB+Keys/Ver1.0

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ME8492 KINEMATICS OF MACHINERY L T P C

4 0 0 4 OBJECTIVES:

• To understand the basic components and layout of linkages in the assembly of a system machine.

• To understand the principles in analyzing the assembly with respect to the displacement, velocity, and

acceleration at any point in a link of a mechanism.

• To understand the motion resulting from a specified set of linkages, design few linkage mechanisms and

cam mechanisms for specified output motions.

• To understand the basic concepts of toothed gearing and kinematics of gear trains and the effects of friction

in motion transmission and in machine components.

UNIT I BASICS OF MECHANISMS 9

Classification of mechanisms – Basic kinematic concepts and definitions – Degree of freedom, Mobility – Kutzbach

criterion, Gruebler’s criterion – Grashof’s Law – Kinematic inversions of four-bar chain and slider crank chains –

Limit positions – Mechanical advantage – Transmission Angle – Description of some common mechanisms – Quick

return mechanisms, Straight line generators, Universal Joint – rocker mechanisms.

UNIT II KINEMATICS OF LINKAGE MECHANISMS 9

Displacement, velocity and acceleration analysis of simple mechanisms – Graphical method– Velocity and

acceleration polygons – Velocity analysis using instantaneous centres – kinematic analysis of simple mechanisms –

Coincident points – Coriolis component of Acceleration – Introduction to linkage synthesis problem.

UNIT III KINEMATICS OF CAM MECHANISMS 9

Classification of cams and followers – Terminology and definitions – Displacement diagrams –Uniform velocity,

parabolic, simple harmonic and cycloidal motions – Derivatives of follower motions – Layout of plate cam profiles

– Specified contour cams – Circular arc and tangent cams – Pressure angle and undercutting – sizing of cams.

UNIT IV GEARS AND GEAR TRAINS 9

Law of toothed gearing – Involutes and cycloidal tooth profiles –Spur Gear terminology and definitions –Gear tooth

action – contact ratio – Interference and undercutting. Helical, Bevel, Worm, Rack and Pinion gears [Basics only].

Gear trains – Speed ratio, train value – Parallel axis gear trains – Epicyclic Gear Trains.

UNIT V FRICTION IN MACHINE ELEMENTS 9

Surface contacts – Sliding and Rolling friction – Friction drives – Friction in screw threads –Bearings and

lubrication – Friction clutches – Belt and rope drives – Friction in brakes- Band and Block brakes.

TOTAL: 45 PERIODS

TEXT BOOKS:

1. F.B. Sayyad, “Kinematics of Machinery”, MacMillan Publishers Pvt Ltd., Tech-max Educational resources, 2011.

2. Rattan, S.S, “Theory of Machines”, 4th Edition, Tata McGraw-Hill, 2014.

3. Uicker, J.J., Pennock G.R and Shigley, J.E., “Theory of Machines and Mechanisms”, 4th Edition, Oxford

University Press, 2014.

REFERENCES:

1. Allen S. Hall Jr., “Kinematics and Linkage Design”, Prentice Hall, 1961

2. Cleghorn. W. L, “Mechanisms of Machines”, Oxford University Press, 2014

3. Ghosh. A and Mallick, A.K., “Theory of Mechanisms and Machines", 3rd Edition Affiliated East-West Pvt. Ltd.,

New Delhi, 2006.

4. John Hannah and Stephens R.C., "Mechanics of Machines", Viva Low-Prices Student Edition, 1999.

5. Thomas Bevan, "Theory of Machines", 3rd Edition, CBS Publishers and Distributors, 2005.




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Subject Code: ME8492 Year/Semester: II /04

Subject Name: Kinematics of Machinery

Subject Handler: Mr.D.Raguraman & Mr.D.Arunkumar

UNIT I BASICS OF MECHANISMS

Classification of mechanisms – Basic kinematic concepts and definitions – Degree of freedom, Mobility

– Kutzbach criterion, Gruebler’s criterion – Grashof’s Law – Kinematic inversions of four-bar chain and

slider crank chains – Limit positions – Mechanical advantage – Transmission Angle – Description of

some common mechanisms – Quick return mechanisms, Straight line generators, Universal Joint –

rocker mechanisms.

PART * A

Q.No. Questions

1 Define the Grubler’s criterion for plane mechanism with mathematical expression. (Nov.

2017, Nov.2015) BTL1

The Grubler’s criterion applies to mechanisms with only single degree of freedom joints where

the overall movability of the mechanism is unity. Substituting n = 1 and h = 0 in Kutzbach

equation, we have

1 = 3 (l – 1) – 2 j or 3l – 2j – 4 = 0

This equation is known as the Grubler's criterion for plane mechanisms with constrained

motion.

A little consideration will show that a plane mechanism with a movability of 1 and only

single degree of freedom joints can not have odd number of links. The simplest possible

mechanisms of this type are a four bar mechanism and a slider-crank mechanism in which l = 4

and j = 4.

2 Name any two inversions of single slider crank chain. (Nov.2017) BTL1

• Pendulum pump or bull engine

• Oscillating cylinder engine

• Rotary internal combustion engine or gnome engine

• Crank and slotted lever quick return motion mechanism

• Whitworth quick return mechanism

3 What type of kinematic pair exists between human shoulder and arm based on nature of

contact and nature of relative motion? ( May 2017) BTL1

Higher pair will exist between human shoulder and arm based on nature of contact and nature of

relative motion.

4 Differentiate rigid and flexible links. (Nov. 2016) BTL3




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

motion. Strictly speaking, rigid links do not exist. However, as the deformation of a connecting

rod, crank etc. of a reciprocating steam engine is not appreciable; they can be considered as rigid

links.

Flexible link: A flexible link is one which is partly deformed in a manner not to affect the

transmission of motion. For example, belts, ropes, chains and wires are flexible links and transmit

tensile forces only.

5 Define transmission angle of a four bar mechanism. What are the maximum and minimum

values of transmission angle? Sketch them. (May 2017, Nov.2016, May 2012, Nov 2011, Nov

2003) BTL2

The angle between the coupler link and the driven link (or follower) is known as transmission

angle. It is denoted by “γ”.

Tan γ = Force component tending to move the driven link

Force component tending to apply pressure on driven link bearings

The minimum and the maximum value of the transmission angle for the four-bar mechanism will

be given by:

cos γminmax

=a4

2 + a32 − a1

2 − a22

2a3a4±

a1a2

a3a4

ϴ = Crank angle

Worst value is less than 450

6 Classify kinematic pairs based on nature of contact. Give examples. (May 2016) BTL2

The kinematic pairs according to the type of contact between the elements may be classified as

discussed below :

(a) Lower pair. When the two elements of a pair have a surface contact when relative motion

takes place and the surface of one element slides over the surface of the other, the pair

formed is known as lower pair.

Examples: It will be seen that sliding pairs, turning pairs and screw pairs form lower

pairs.

(b) Higher 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, then the pair is known as higher pair.




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Examples: Pair of friction discs, toothed gearing, belt and rope drives, ball and roller

bearings and cam and follower are the examples of higher pairs.

7 When a linkage becomes mechanism? (May 2016) BTL1

When one of the links of a kinematic chain is fixed, the chain is known as mechanism. It may be

used for transmitting or transforming motion e.g. engine indicators, typewriter etc.

8 What are the three conditions to obtain a four bar crank rocker mechanism? (Nov.2015)

BTL1

The three conditions to obtain a four bar crank rocker mechanism:

• One link rotates (Crank)

• One link is fixed and

• Other two links oscillates

9 Define inversion of kinematic chain. (Nov.2015) BTL1

When one of links is fixed in a kinematic chain, it is called a mechanism. So it can be able to

obtain as many mechanisms as the number of links in a kinematic chain by fixing, in turn,

different links in a kinematic chain. This method of obtaining different mechanisms by fixing

different links in a kinematic chain is known as inversion of the mechanism.

10 Name any two inversion of the four bar chain. (Nov.2015) BTL2

• Beam engine (crank and lever mechanism)

• Coupling rod of a locomotive (Double crank mechanism)

• Watt’s indicator mechanism (Double lever mechanism)

11 Determine the degree of freedom of the mechanism shown in the figure below: (May 2015)

BTL5

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

Here, n = 11, l =9 and h = 0.

F = 3(11-1)-2(9) = 12

12 Differentiate between rigid and resistant bodies. (Nov.2014) BTL3

Rigid body means a body with no deformations when the required force is transmitted.

A body is said to be resistant if it is capable of transmitting the required force with negligible

deformation.

13 The ratio between the width of the front axle and that of wheel base of a steering

mechanism is 0.44. At the instant when the front inner wheel is turned by 180, what should

be the angle turned by the outer front wheel for perfect steering? (Nov.2014) BTL4

Given : c/b = 0.44; ϴ = 18° w.k.t. fundamental equation for correct steering,

cot ∅ − cot θ =c

b cot ∅ − cot 18° = 0.44 cot ∅ = 3.518




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∅ = 𝟏𝟓. 𝟗° Answer

14 Differentiate machine and structure. (May 2014) (May 2013) (Nov 2010) (May 2007)(May

2005) (Nov 2002) BTL3

S.NO. Machine Structure

1 Relative motion exists between its parts No relative motion exists between its

members

2 It transforms available energy in to useful

work

It does not convert the available energy in

to work

15 Classify the constrained motion. (May 2014) BTL2

Constrained motions are classified in to:

• Completely constrained motion

• Incompletely constrained motion

• Successfully constrained motion

16 Define sliding connectors. (Nov.2013) BTL1

Sliding connectors are used when one slider (the input) is to drive another slider (the output).

Usually the two sliders operate in the same plane but in different directions.

17 What is meant by kinematic pair? (May 2013) (May 2010) (Nov 2006) BTL1

When any two links are connected in such a way that their relative motion is completely or

successfully constrained, they form a kinematic pair.

18 State the difference between mechanism and structure. ( May 2013) BTL2 S.NO. Mechanisms Structure

1 Relative motion exists between its parts No relative motion exists between its

members. Also it does not convert the

available energy in to work

19 State Grashof’s law for a four-bar linkage. (Nov 2012)( May 2009)( May 2008)( Nov 2017)

BTL1

Grashof’s law states that for a planar four bar mechanism, the sum of the shortest and longest

links must be less than or equal to the sum of the lengths of two other links, if there is to be

continuous relative rotation between two members.

20 What is low degree of complexity? (Nov.2013) BTL1

In a complex mechanism, if only one radius of path curvature of one motion transfer point is not

known, such a mechanism is called a mechanism with low degree of complexity.

PART * B

1 Write in detail with neat sketch, any three inversions of double slider Crank chain. (13M)

(Nov.2017) BTL3

Answer: Page 113 - R.S.KHURMI

Elliptical trammels: It is an instrument used for drawing ellipses. This inversion is obtained by

fixing the slotted plate (link 4), as shown in Fig. 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, as shown in Fig. (a). A little consideration will show




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that AP and BP are the semi-major axis and semi-minor axis of the ellipse respectively.

(6 M)

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 Fig., 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.

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 Fig. 5.36 (a). The shafts to be connected have two flanges (link 1 and link 3) rigidly fastened at

their ends by forging.

(7 M)

2 Describe with neat sketch, the mechanisms obtained by the inversions of four-bar chain.

(13 M) (Nov.2017, May 2014) BTL3


(4 M)

Beam engine (crank and lever mechanism):

Parts of the mechanism of a beam engine (also known as crank and lever mechanism) which

consists of four links, is shown in Fig. In this mechanism, when the crank rotates about the fixed

centre A, the lever oscillates about a fixed centre D. The end E of the lever CDE is connected to

a piston rod which reciprocates due to the rotation of the crank. In other words, the purpose of

this mechanism is to convert rotary motion into reciprocating motion.

(3M)

Coupling rod of a locomotive (Double crank mechanism):

The mechanism of a coupling rod of a locomotive (also known as double crank mechanism)

which consists of four links is shown in Fig. In this mechanism, the links AD and BC (having




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equal length) act as cranks and are connected to the respective wheels. The link CD acts as a

coupling rod and the link AB is fixed in order to maintain a constant centre to centre distance

between them. This mechanism is meant for transmitting rotary motion from one wheel to the

other wheel.

(3 M)

Watt’s indicator mechanism (Double lever mechanism):

A Watt’s indicator mechanism (also known as Watt's straight line mechanism or double lever

mechanism) which consists of four links, is shown in Fig.. The four links are fixed link at A, link

AC, link CE and link BFD. It may be noted that BF and FD form one link because these two parts

have no relative motion between them. The links CE and BFD act as levers. The displacement of

the link BFD is directly proportional to the pressure of gas or steam which acts on the indicator

plunger. On any small displacement of the mechanism, the tracing point E at the end of the link

CE traces out approximately a straight line.

The initial position of the mechanism is shown in Fig. by full lines whereas the dotted lines show

the position of the mechanism when the gas or steam pressure acts on the indicator plunger.

(3 M)

3 What is a kinematic inversion? Discuss any three applications of inversions of slider crank

mechanism with suitable sketches. (13 M) (May 2017) BTL3


When one of links is fixed in a kinematic chain, it is called a mechanism. So we can obtain as

many mechanisms as the number of links in a kinematic chain by fixing, in turn, different links in

a kinematic chain. This method of obtaining different mechanisms by fixing different links in a

kinematic chain, is known as inversion of the mechanism.




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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.

Oscillating cylinder engine:

The arrangement of oscillating cylinder engine mechanism, as shown in Fig. 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.

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., while the crank (link 2) is fixed. In this mechanism, when the

connecting rod (link 4) rotates, the piston (link 3) reciprocates inside the cylinders forming link 1.

4 (i) Find the degrees of freedom for the mechanisms shown in Fig (i). (7 M) BTL5

Answer: Page 1.28 - JAYAKUMAR




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

Here, n = 5, n = 5, l = 5 and h = 0.

F = 3(5-1)-2(5) = 2

I.e., two inputs to any two links are required to yield definite motions in all the links.

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

Here, n = 4, n =2, n = 6, l = 7 and h = 0.

F = 3(6-1)-2(7) = 1

I.e., one input to any one link will result in definite motion of all the links.

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

Here, n = 11, l = 15 (two lower pairs at the intersection of 3, 4, 6; 2, 4, 5; 5, 7, 8; 8, 10, 11)

and h = 0.

F = 3(11-1)-2(15) = 0

(ii) Explain mechanical advantage and transmission angle related to four bar

mechanism. (6 M) (May 2017) BTL3


Mechanical Advantage

It is defined as the ratio of the load to the effort. In a four bar mechanism, as shown in Fig., the

link DA is called the driving link and the link CB as the driven link. The force FA acting at A is

the effort and the force FB at B will be the load or the resistance to overcome. We know from the

principle of conservation of energy, neglecting effect of friction,

If we consider the effect of friction, less resistance will be overcome with the given effort.

Therefore the actual mechanical advantage will be less.

Let η = Efficiency of the mechanism.

∴ Actual mechanical advantage,

The mechanical advantage may also be defined as the ratio of output torque to the input torque.

Let TA = Driving torque,

TB = Resisting torque,

ωA and ωB = Angular velocity of the driving and driven links respectively.




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∴ Ideal mechanical advantage,

. . . (Neglecting effect of friction)

and actual mechanical advantage,

. . . (Considering the effect of friction)

(4 M)

Transmission angle: The angle between the coupler link and the driven link (or follower) is

known as transmission angle. It is denoted by “γ”.

Tan γ = Force component tending to move the driven link

Force component tending to apply pressure on driven link bearings

(2 M)

5 (i) Explain different types of constrained motion with suitable examples. (6 M) BTL3

(ii) Describe the working of Peaucellier mechanism and Offset slider mechanism. (7 M)

(Dec 2016) BTL3 Answer: Page 96 & 234 - R.S.KHURMI

(i) Types of constrained motions:

Completely constrained motion:

When the motion between a pair is limited to a definite direction irrespective of the direction of

force applied, then the motion is said to be a completely constrained motion. For example, the

piston and cylinder (in a steam engine) form a pair and the motion of the piston is limited to a

definite direction (i.e. it will only reciprocate) relative to the cylinder irrespective of the direction

of motion of the crank.

The motion of a square bar in a square hole, as shown in Fig., and the motion of a shaft with

collars at each end in a circular hole, as shown in Fig., are examples of completely constrained

motion.

(2 M)

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. The change in the direction of impressed force may

alter the direction of relative motion between the pair. A circular bar or shaft in a circular hole, as

shown in Fig., is an example of an incompletely constrained motion as it may either rotate or

slide in a hole. These both motions have no relationship with the other.




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(2 M)

Successfully 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, then the motion is said to be successfully

constrained motion. Consider a shaft in a foot-step bearing as shown in Fig.. The shaft may rotate

in a bearing or it may move upwards. This is a case of incompletely constrained motion. But if

the load is placed on the shaft to prevent axial upward movement of the shaft, then the motion of

the pair is said to be successfully constrained motion. The motion of an I.C. engine valve (these

are kept on their seat by a spring) and the piston reciprocating inside an engine cylinder are also

the examples of successfully constrained motion.

(2 M)

Peaucellier mechanism:

It consists of a fixed link OO1 and the other straight links O1A, OC, OD, AD, DB, BC and CA

are connected by turning pairs at their intersections, as shown in Fig.. The pin at A is constrained

to move along the circumference of a circle with the fixed diameter OP, by means of the link

O1A.

(4 M)

Offset slider mechanism:

If the eccentricity, c (or a1), is zero ( c = 0) the slider crank mechanism is called an in-line slider-

crank and the stroke is twice the crank length (s = 2a2). If the eccentricity is not zero ( c ¹0), it is

usually called an offset slider-crank mechanism.




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(3 M)

6 Describe Kutzbach criterion with neat sketches and explain the concept of plane

mechanisms. (13 M) (May 2012) BTL4

Answer: Page 1.36 – JAYAKUMAR

Kutzbach criterion for determining the number of degrees of freedom or movability (n) of a plane

mechanism is

n = 3(l-1) -2j – h

(3 M)

The number of degrees of freedom or movability (n) for some simple mechanisms having no

higher pair (i.e. h = 0), as shown in Fig. 5.16, are determined as follows :

1. The mechanism, as shown in Fig. 5.16 (a), has three links and three binary joints, i.e.

l = 3 and j = 3.

∴ n = 3 (3 – 1) – 2 × 3 = 0

(1 M)

2. The mechanism, as shown in Fig. 5.16 (b), has four links and four binary joints, i.e.

l = 4 and j = 4.

∴ n = 3 (4 – 1) – 2 × 4 = 1

(1 M)

3. The mechanism, as shown in Fig. 5.16 (c), has five links and five binary joints, i.e.

l = 5, and j = 5.

∴ n = 3 (5 – 1) – 2 × 5 = 2

(1 M)

4. The mechanism, as shown in Fig. 5.16 (d), has five links and six equivalent binary joints

(because there are two binary joints at B and D, and two ternary joints at A and C), i.e.

l = 5 and j = 6.

∴ n = 3 (5 – 1) – 2 × 6 = 0

(1 M)

5. The mechanism, as shown in Fig. 5.16 (e), has six links and eight equivalent binary joints

(because there are four ternary joints at A, B, C and D), i.e. l = 6 and j = 8.

∴ n = 3 (6 – 1) – 2 × 8 = – 1

(1 M)

It may be noted that

(a) When n = 0, then the mechanism forms a structure and no relative motion between the

links is possible, as shown in Fig. 5.16 (a) and (d).

(b) When n = 1, then the mechanism can be driven by a single input motion, as shown in Fig.

5.16 (b).

(c) When n = 2, then two separate input motions are necessary to produce constrained




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motion for the mechanism, as shown in Fig. 5.16 (c).

(d) When n = – 1 or less, then there are redundant constraints in the chain and it forms a

statically indeterminate structure, as shown in Fig. 5.16 (e).

(2 M)

Fig. Plane mechanism

(3 M)

7 In a Whitworth quick return motion mechanism, as shown in Fig., the distance between the

fixed centers is 50 mm and the length of the driving crank is 75 mm. The length of the

slotted lever is 150 mm and the length of the connecting rod is 135 mm. Find the ratio of the

time of cutting stroke to the time of return stroke and also the effective stroke. (13 M)(May

2012) BTL5

Answer: Page 112 - R.S.Khurmi

Given : CD = 50 mm ; CA = 75 mm ; PA = 150 mm ; PR = 135 mm

The extreme positions of the driving crank are shown in Fig.. From the geometry of the figure,

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

We know that

Length of effective stroke

In order to find the length of effective stroke (i.e. R1R2), draw the space diagram of the

mechanism to some suitable scale, as shown in Fig. 5.33. Mark P1R2 = P2R2 = PR. Therefore by

measurement we find that,

Length of effective stroke = R1R2 = 87.5 mm Ans.

PART * C

1 In a crank and slotted lever quick return motion mechanism, the distance between the fixed

centres is 240 mm and the length of the driving crank is 120 mm. Find the inclination of the




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slotted bar with the vertical in the extreme position and the time ratio of cutting stroke to

extreme stroke.

If the length of the slotted bar is 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. (15) (Dec. 2015) BTL5


Given : AC = 240 mm ; CB1 = 120 mm ; AP1 = 450 mm

Inclination of the slotted bar with the vertical

Let ∠CAB1 = Inclination of the slotted bar with the vertical.

(5 M)

The extreme positions of the crank are shown in Fig.. We know that

(5 M)

Time ratio of cutting stroke to the return stroke

We know that

(5 M)

2 Sketch and describe the working of two different types of quick return mechanisms. Give

examples of their applications. Derive an expression for the ratio of times taken in forward

and return stroke for one of these mechanisms. (15 M) (May 2015, May 2014) BTL4

Answer: Page - R.S.KHURMI 111

Crank and slotted lever quick return motion mechanism. This mechanism is mostly used in

shaping machines, slotting machines and in rotary internal combustion engines.

this mechanism, the link AC (i.e. link 3) forming the turning pair is fixed, as shown in Fig. The

link 3 corresponds to the connecting rod of a reciprocating steam engine. The driving crank CB




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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 (i.e. R1R2) 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 (or through an angle β) in the clockwise direction. The return stroke occurs when the

crank rotates from the position CB2 to CB1 (or through angle α) in the clockwise direction. Since

the crank has uniform angular speed, therefore,

Since the tool travels a distance of R1 R2 during cutting and return stroke, therefore travel of the

tool or length of stroke

= R1R2 = P1P2 = 2P1Q = 2AP1 sin ∠ P1 AQ

(8 M)

Whitworth quick return motion mechanism: This mechanism is mostly used in shaping and

slotting machines. In this mechanism, the link CD (link 2) forming the turning pair is fixed, as

shown in Fig. The link 2 corresponds to a crank in a reciprocating steam engine. The driving

crank CA (link 3) rotates at a uniform angular speed. The slider (link 4) attached to the crank pin

at A slides along the slotted bar PA (link 1) which oscillates at a pivoted point D. The connecting

rod PR carries the ram at R to which a cutting tool is fixed. The motion of the tool is constrained

along the line RD produced, i.e. along a line passing through D and perpendicular to CD.

(4 M)

Since the crank link CA rotates at uniform angular velocity therefore time taken during the

cutting stroke (or forward stroke) is more than the time taken during the return stroke. In other

words, the mean speed of the ram during cutting stroke is less than the mean speed during the

return stroke.

The ratio between the time taken during the cutting and return strokes is given by




2-16

(3 M)

3 A Hooke's joint connects two shafts whose axes intersect at 18°. The driving shaft rotates at

a uniform speed of 210 r.p.m. The driven shaft with attached masses has a mass of 60 kg

and the radius of gyration of 120 mm. Determine the torque required at the driving shaft. If

a steady torque of 180 Nm resists rotation of the driving shaft and the angle of rotation is

45° and angle between the shafts at which the total fluctuation of speed of the driven shaft is

limited to 18 r.p.m. (15 M) (Dec.2012) BTL5

Answer: Page SQ.92 - JAYAKUMAR

Given: N = 210 r.p.m or ω = 2 π × 210/60 = 21.9975 rad/s ; α = 18° ; m = 60 kg ;

k = 120 mm = 0.12 m ; Steady torque = 180 N-m ; θ = 45°

(2 M)

(i) Torque required at the driving shaft (T1):

We know that the max. angular acceleration of the driven shaft,

α2 =−ω1

2. cos α. sin 2θ . sin2α

(1 − cos2θ. sin2α)2

= - 48.46 rad/s2

(3 M)

Accelerating torque required on the driven shaft = Iα = mk2 α2 = -41.87 N-m

(2 M)

Total torque required on the driven shaft is given by

T2 = Steady torque + Accelerating torque = 138.13 N.m

(2 M)

We know that P = T1ω1 = T2ω2

ω2

ω1=

COS α

1−COS2θ .Sin2α = 0.998

T1 =137.95 N.m

(2 M)

(ii) Angle between the shafts at which maximum fluctuation of speed of the driven

shaft is 18 rpm:

We know that the max. fluctuation of speed of the driven shaft

(ω2)max − (ω2)min = ω1 (Sin2α

Cos α)

On simplification of quadratic equation: cos2α + 0.1 cos α − 1 = 0

α = 18.19° Ans.

(4 M)




2-17



Subject Handler: Mr. D.Raguraman & D.Arunkumar

UNIT II KINEMATICS OF LINKAGE MECHANISMS

Displacement, velocity and acceleration analysis of simple mechanisms – Graphical method– Velocity

and acceleration polygons – Velocity analysis using instantaneous centres – kinematic analysis of simple

mechanisms – Coincident points – Coriolis component of Acceleration – Introduction to linkage

synthesis problem.

PART * A

Q.No. Questions

1 Define Coriolis component of acceleration. (Nov.2017, Nov.2015, May2014, Nov.2011,

Nov.2010, Nov.2009, Nov.2007, May 2007) BTL1

When the distance between two points is not fixed, (i.e., the distance varies), then the total

acceleration will have one more additional acceleration along with radial and tangential

acceleration known as “Coriolis component of acceleration”. Coriolis component of acceleration

occurs when a point on one link is sliding along another rotating link, such as quick return

mechanism.

2 State the Arnold Kennedy theorem. (Nov.2017) BTL2

Arnold Kennedy theorem states that if three bodies have relative motion with each other, then

their relative instantaneous centers lie on a straight line.

3 Find the resultant acceleration of an 80mm radius crank rotating at a constant angular

velocity of 10rad/s, at the crank-pin position. (May 2017) BTL3

aOA =v2

length of link;

vOA = ωOA × OA = 10 x 0.08 = 0.8 m/s

aOA =0.82

0.08= 8 m/s2

4 Illustrate the instantaneous centers of a typical four bar mechanism. (May 2017) BTL3

Number of instantaneous centre, N = n(n−1)

2

Where, N =Number of links,

In four bar chain mechanism, n=4

N = 4 (4−1)

2= 6




2-18

5 Define rubbing velocity at pin joint. What will be the rubbing velocity at pin joint when

two links move in the same and opposite directions? (Nov.2016, Nov.2015, May 2013,

Nov.2012, May 2012, May 2010, May 2007, Nov.2005, Nov.2004, Nov.2002) BTL1

Rubbing velocity is defined as the algebraic 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 pin joint when the two links move in opposite direction = (ω1 + ω2) r

Rubbing velocity at pin joint when the two links move in same direction = (ω1 − ω2) r

6 State and prove Kennedy’s three centre theorem. (Nov.2016) BTL2

Kennedy’s three centre theorem: If three bodies have relative motion with each other, then

their relative instantaneous centres must lie on a straight line.

Proof: The instantaneous centre is same whether it is regarded as a point on the first link or as a

point on the second link. Therefore, the velocity of the point Ibc cannot be perpendicular to both

lines Iab Ibc and Iac. Ibc unless the point Ibc lies on the line joining the points Iab and Iac. Thus the

three instantaneous centres (Iab, Iac and Ibc) must lie on the same straight line.

7 What is relative pole with respect to velocity analysis? (May 2016) BTL1

In slider crank mechanism, the relative pole is the centre of rotation of the connecting rod relative

to the crank rotation and the corresponding slider displacement.

8 What are the different methods used for finding the velocity? (May 2016) BTL1

The different methods used for finding the velocity:

• Relative velocity method

• Instantaneous centre method

9 Give the relation to find the magnitude of Coriolis component of acceleration. ( May 2014,

May2008) BTL2

ac = 2vsω

Where,

ω - Angular velocity and vs - Velocity of sliding.

10 In a revolving stage with a speed of 3 r.p.m., a person is walking with a speed of 0.5 m/s

along a radial path. Determine the magnitude of the Coriolis component of acceleration in

this motion. BTL3

Given data: N = 3 r.p.m. ; 𝛚 =𝟐𝛑(𝟑)

𝟔𝟎= 𝟎. 𝟑𝟏𝟒 𝐫𝐚𝐝 𝐬⁄ ; 𝐯𝐬 = 𝟎. 𝟓 m/s

Coriolis component of acceleration, ac = 2vsω = 2 x 0.5 x 0.314 = 0.314 m/s2 Ans.

11 Name the mechanism in which Coriolis component of acceleration is taken into account.

(Nov.2015, Nov.2014, May 2010) BTL2

In the mechanism such as crank and slotted lever mechanism, Whitworth quick return

mechanism and oscillating cylinder mechanism, Coriolis acceleration is taken into account.




2-19

In the mechanisms such as four-bar chain, slider-crank mechanism and toggle mechanisms,

Coriolis acceleration is not taken into account.

12 Write the relation between the number of instantaneous centres and the number of links in

a mechanism. (Nov.2015, May 2015) BTL2


2

Where, N =Number of links

13 Depict all the directions of Coriolis component of acceleration that arise in a completed

cycle of quick return motion of the crank mechanism. (May 2015) BTL3

Since the direction of ac depends on the direction of vs and ω, therefore there are four possible

cases of direction of ac, as shown in Fig.

14 What is the need of finding acceleration of linkage in a mechanism? (Nov.2014) BTL1

Since the dynamic forces are functions of accelerations ( F = ma), therefore the determination of

acceleration of various links become very important in the design of any mechanism.

15 State the relationship between crank angle (ϴ) and connecting rod angle () of single slider

mechanism. (Nov.2011) BTL2

𝐬𝐢𝐧 ∅ = 𝐫

𝐥 𝐬𝐢𝐧 𝛉 =

𝐬𝐢𝐧 𝛉

𝐧

16 Define instantaneous centre. (May 2014, Nov.2009, Nov.2003, Nov.2002 ) BTL1

The combined motion of rotation and translation of the link may be assumed to be a motion of

pure rotation about some centre known as virtual centre or instantaneous centre.

17 Differentiate rotation and translation. (Nov.2013) BTL3

Translation is defined as a state of motion of body for which the displacement difference

between any two points is zero.

Rotation is a state of motion of the body for which different points of the body are equal.

18 Write down the expression for finding the number of instantaneous centres in a mechanism.

(Nov.2103, May 2013) BTL2


2

Where, N =Number of links

19 What is a configuration diagram? What is its use? (May 2012) BTL1

• Configuration diagram is a line sketch of a given mechanism drawn to a suitable scale.

• The configuration diagram forms the basis for the construction of both velocity and

acceleration diagrams.

20 State the Freudenstein’s equation for a four-bar mechanism. (Nov.2007) BTL2

Freudenstein’s equation is given by

k1 cos ∅ + k2 cos θ = cos(θ − ∅)

Where, k1 =d

a ; k2 =

−d

c; and k3 =

a2−b2+c2+d2

2 a c

a,b,c and d are magnitudes of four links.

ϴ and are the angles made with horizontal by the input and follower links respectively.

PART * B

1 In a four-link mechanism, the dimensions of the links are as under:




2-20

AB = 50mm; BC= 66 mm; CD = 56 mm;_AD = 100 mm.

At an instant when ∠DAB = 600, the link AB has an angular velocity of 10.5 rad./s in

counter clock wise direction; Determine:

(i) The velocity of point C;

(ii) The velocity of point E on the link BC when BE = 40 mm;

(iii) The angular velocity of the links BC and CD;

(iv) The velocity of an offset point F on the link BC, if BF = 45 mm, CF = 30 mm and

BCF is read clockwise;

(v) The velocity of an offset point G on the link CD if CG = 24 mm, DG = 44 mm and

DCG is read clockwise; and

(vi) The velocities of rubbing at pins A,B,C and D. The radii of the pins are 30,40, 25 and

35 mm respectively. (13 M) (May 2013) BTL4


The problem can be solved using relative velocity method.

Step-1 : Configuration diagram: Take 1 cm =10 mm

Step-2 : Velocity of input link: ωBA = 10.5 rad/s (given)

VBA = ωBA X AB = 0.525 m/s

(2 M)

Step-3: Velocity diagram (1 cm = 0.1 m/s)

1. Draw the velocity triangle abc

2. Locate the offset point F

3. Locate the offset point G

4. Locate E on the velocity diagram

VBE

VBC=

BE

BC

VBE = 0.206 m/s

(2 M)

Step-4 : Velocity of various links:

VCA = Vector ca = 0.393 m/s Ans

VEB = Vector eb = 0.206 m/s Ans

VFA = Vector fa = 0.504 m/s Ans




2-21

ωBC =VBC

BC = 5.17 rad/s (Counter clockwise about B)

ωCD =VCD

CD = 7.01 rad/s (Counter clockwise about D)

(2 M)

(5 M)

Rubbing Velocitites at pins A,B,C and D:

Velocity of rubbing at pin B = (ωAB + ωBC) × rB = 0.6268 m/s

Velocity of rubbing at pin C = (ωCD + ωBC) × rC = 0.3045 m/s

Velocity of rubbing at pin A = (ωAD + ωAB) × rA = 0.315 m/s

Velocity of rubbing at pin D = (ωAD + ωCD) × rD = 0.2454 m/s

(2 M)

2 The crank of a slider-crank mechanism is 15 cm and the connecting rod is 60 cm long. The

crank makes 300 r.p.m. in the clockwise direction. When it has turned 450 from the inner

dead centre position, determine:

(i) Velocity of slider,

(ii) Angular velocity of connecting rod,

(iii)Linear velocity of mid-point of the connecting rod.

(13 M) (Nov.2015, May 2014) BTL4


Step 1. Configuration diagram (1 cm = 75 mm)

Step 2. Velocity of input link

NOA = 300 rpm (Given)

ωOA =2πNOA

60 = 31.42 rad/s

VOA = ωOA × OA = 4.713 m/s

(3 M)




2-22

(5 M)

Step 3. Velocity diagram (1 c.m. = 1.5 m/s)

Step 4. Velocity of various links:

(i) Velocity of slider:

From measurement from velocity diagram

vB = vob= vector ob = 4 m/s

(ii) Angular velocity of connecting rod AB:


velocity of connecting rod AB: vector ba = 3.35 m/s

Angular velocity of connecting rod

ωAB =vAB

AB = 5.58 rad/s (Counter clockwise about A)

(iii)Linear velocity of the mid-point of the connecting rod:


vC = voc= vector oc = 4.1 m/s

(5 M)

3 ABCD is a four bar chain with link AB fixed. The lengths of the links are AB = 62.5 mm;

BC = 175 mm; CD = 112.5 mm; and AD = 200mm. The crank AB rotates at 10 rad./s

clockwise. Draw the velocity and acceleration diagram when angle BAD = 600 and B and C

lie on the same side of AD. Find the angular velocity and angular acceleration of links BC

and CD. (13 M) (Nov.2015, Nov.2009, Nov.2004, Nov.2002) BTL 5




VBA = ωBA × AB = 0.625 m/s

(2 M)





2-23

(5 M)



VCB = Vector bc = 0.35 m/s Ans

VCD = Vector dc = 0.44 m/s Ans

ωBC =VBC

BC = 2 rad/s (Counter clockwise about B)

ωCD =VCD

CD = 3.91 rad/s (Clockwise about D)

(3 M)

Step 5. Acceleration diagram: (1 c.m. = 1.5 m/s2)

Radial and tangential components of acceleration of various links:

Step 6. Acceleration of various links:

From measurement from acceleration diagram,

aCB′ = vector xc′ = 24 m/s2 (Clockwise about B) Ans.

aCD′ = vector yc′ = 48.89 m/s2 (Counter clockwise about D) Ans.

(3 M)

4 In a small steam engine running at 600 rad/min clockwise, length of crank is 80 mm and

ratio of connecting rod length to crank radius is 3. For the position when crank makes

450 to horizontal to horizontal, determine:

(i) The velocity and acceleration of the piston;

(ii) The angular velocity and angular acceleration of the connecting rod; and

(iii) The linear velocity and acceleration of appoint X on connecting rod 80 mm

from crank pin (13 M) (May 2017, Nov. 2006) BTL5

Answer: Page 2.42 JAYAKUMAR





2-24


VOA = ωOA × OA = 8 m/s

(2 M)

Step 3. Velocity diagram (1 c.m. = 2 m/s)



VB = Vector ob = 7 m/s Ans

Velocity of connecting rod, VBA = Vector ab = 6 m/s Ans

Velocity of point X on AB, Vx = Vector x = 7 m/s Ans

(2 M)

Step 5. Acceleration diagram: (1 c.m. = 1.5 m/s2)

Radial and tangential components of acceleration of various links:

(2 M)



Acceleration of piston B, aB = Vector o’b’ = 600 m/s2 Ans

Tangential component of acceleration of connecting rod AB, a’AB = Vector o’b’ = 560 m/s2 Ans

Acceleration of point X, ax = Vector o’x’ = 700 m/s2 Ans




2-25

(5 M)

αAB =aAB

′

AB = 2.33 rad/s2 (Clockwise about B) Ans.

(2 M)

5 In the mechanism, as shown in Fig, the crank OA rotates at 20 rpm anticlockwise and gives

motion to the sliding blocks B and D. The dimensions of the various links are OA = 300 mm;

AB =1200 mm; BC = 450 mm and CD = 450 mm.

For the given configuration, determine: (i) velocities of sliding at B and D; (ii) angular

velocity of CD; (iii) linear acceleration D; and (iv) angular acceleration of CD. (13 M)

(Nov.2007) BTL5



(2 M)


Angular velocity of crank OA, ωOA =2πNOA

60 = 2.09 rad/s (CCW about O)

Velocity of crank OA, VOA = ωOA × OA = 0.627 m/s

(3 M)


(3 M)






2-26

Velocity of slider B, VB = Vector ob = 0.42 m/s Ans

Velocity of slider D, VD = Vector od = 0.27 m/s Ans

ωCD =VCD

CD = 0.82 rad/s (Counter Clockwise about D)

Linear acceleration of slider D, a’DC = 1.31 m/s2 Ans

αCD =aDC

′

CD = 2.91 rad/s2 (Clockwise about B) Ans.

(5 M)

6 Locate all the instantaneous centres of the slider crank mechanism shown in Fig., Find:

(i) velocity of piston, and

(ii) angular velocity of connecting rod. (13 M) (Nov 2017, May 2017, Nov. 2013) BTL 5


Step 1. Configuration diagram (1 cm = 0.2 m)

(2 M)


Velocity of input link, crank OA, VOA = ωOA × OA = 6 m/s

(2 M)

Step 3. Locate the required instantaneous centres:

N = n(n-1)/2 = 6

(2 M)




2-27


Velocity of piston, vB = ω2RI12 I24

= 5.16 m/s

Velocity of link OA, vAO = ω3RI13 I23

; ωAB = ω3 = 2.11 rad/s

(3 M)

(4 M)

7 In a reciprocating engine mechanism, the lengths of the crank and connecting rod are I50

mm and 600 mm respectively. The crank position is 60° from inner dead centre. The crank

– shaft speed is 450 rpm (clockwise). Using analytical method, determine:

(1) the velocity of the piston;

(ii) the acceleration of the piston, and

(m) the crank angle for maximum velocity of the piston and the corresponding velocity.

(13 M) (Nov.2012) BTL5


(i) Velocity of piston (vp):

vp = rω [sin θ +sin 2θ

2n] = 6.89 m/s

(4 M)

(ii) Acceleration of the piston (ap):

ap = ω2r [cos θ +cos 2θ

n] = 124.89 m/s2 Ans.

(iii)Crank angle for maximum velocity of the piston the corresponding velocity:

Let θ = Crank angle from IDC at which the maximum velocity occurs

W.K.T., Velocity of the piston,

vp = rω [sin θ +sin 2θ

2n]

For max. velocity of the piston, 𝑑𝑣𝑝

𝑑𝜃 = 0 Or

𝑑

𝑑𝜃[rω [sin θ +

sin 2θ

2n]] = 0

(4 M)




2-28

θ = 77°

Substituting , θ = 77° in the vp, we get

Vp(max) = 7.27 m/s Ans.

(5 M)

PART * C

1 The following Fig. shows the mechanism of a radial valve gear. The crank OA turns

uniformly at I50 rpm and is planned at A to rod AB. The point C in the rod is guided

in the circular path with D as centre and DC as radius. The dimensions of various links

are: OA = I50 mm; AB = 550 mm; AC = 450 mm; DC = 500 mm; BE = 350 mm.

Determine velocity and acceleration of the ram E for the given position of the mechanism.

(15 M) (Nov.2006) BTL5


Step 1 : Cofiguration diagram:

Select suitable scale (Say 1 cm = 50 mm), to draw the configuration diagram.

(2 M)

Step 2: Velocity of input link:




(3 M)

Step 3. Velocity diagram:

Select suitable scale (Say 1 c.m. = 0.5 m/s), to draw the configuration diagram.




2-29

(3 M)

Step 4: Velocity of various links:

By measurement from the velocity diagram, we get

VBA = vector ab = 0.68 m/s

VCD = vector dc = 1.75 m/s

VEB = vector be = 1.96 m/s

VE = vector oe = 1.1 m/s

(2 M)

Step 5: Acceleration diagram:

Select suitable scale (Say 1cm = 1 m/s2)

(3 M)

Step 6: Acceleration of various links:

By measurement from the acceleration diagram, we get

aE = vector o′e′ = 3.22 m/s2

(2 M)

2 In the toggle mechanism shown in Fig., the slider D is constrained to move on a horizontal

path, The crank OA is rotating in the counter clockwise direction at a speed of 180 rpm

increasing at the rate of 50 rad/s2. The dimensions of the various links are as follows:




2-30

OA = 180 mm; CB = 240 mm; AB = 360 mm; and BD = 540 mm.

For the given configuration, determine:

(i) the velocity of the slider D;

(ii) the angular velocity and angular acceleration of links AB, BC and BD; and

(iii) the linear acceleration of the slider D. (15 M) (Nov.2007, May 2007) BTL5



(2 M)

Step 2: Velocity of input link:




(3 M)

Step 3. Velocity diagram:

Select suitable scale (Say 1 c.m. = 0.75 m/s), to draw the configuration diagram.

(3 M)

Step 4: Velocity of various links:

By measurement from the velocity diagram, we get

VBA = vector ab = 1 m/s

VBD = vector dc = 2.48 m/s

VBC = vector be = 2.92 m/s

VD = vector cd = 2.12 m/s

(2 M)

Step 5: Acceleration diagram:

Select suitable scale (Say 1cm = 10 m/s2)

(2 M)



𝐚𝐃 = 𝐯𝐞𝐜𝐭𝐨𝐫 𝐜′𝐝′ = 𝟏𝟑. 𝟕 𝐦/𝐬𝟐 Ans.

aBA′ = vector b′y′ = 37 m/s2

aBC′ = vector b′z′ = 29 m/s2




2-31

aDB′ = vector d′s′ = 38 m/s2

αAB =aBA

′

AB = 102.78 rad/s2 Ans.

αBC =aBC

′

BC = 120.83 rad/s2 Ans.

αBD =aBD

′

BD = 70.37 rad/s2 Ans.

(3 M)

3 The-mechanism of a warping machine, as shown in Fig, has dimensions as follows: O1A =

100 mm; AC = 700 mm; BC = 200 mm; BD = 150 mm; O2D=200 mm; O2E=400 mm,

O3C=200 mm.

The Crank O1A rotates at a uniform speed of 100 rad/s. For the given configuration,

determine:

(I) linear velocity of the point E on the bell crank lever,

(ii) acceleration of the points E and B, and

(m) angular acceleration of the bell crank lever.




2-32

(15 M) (Nov.2016, May 2010, Nov.2008) BTL5


(5 M)

(5 M)




2-33

(5 M)


2- 34




UNIT III KINEMATICS OF CAM MECHANISMS

Classification of cams and followers – Terminology and definitions – Displacement diagrams –Uniform

velocity, parabolic, simple harmonic and cycloidal motions – Derivatives of follower motions – Layout

of plate cam profiles – Specified contour cams – Circular arc and tangent cams – Pressure angle and

undercutting – sizing of cams.

PART * A

Q.No. Questions

1 Differentiate between radial cam and cylindrical cam. (Nov.2017) BTL3 Radial (Disk or plate) cam: The disk (or plate) cam has an irregular contour to impart a specific

motion to the follower. The follower moves in a plane perpendicular to the axis of rotation of the

camshaft and is held in contact with the cam by springs or gravity.

Cylindrical cam: The cylindrical cam has a groove cut along its cylindrical surface. The roller

follows the groove, and the follower moves in a plane parallel to the axis of rotation of the

cylinder.

2 Name the cam follower extensively used in air-craft engines. (Nov.2017) BTL1

Roller follower is extensively used in aircraft engines. Because the side thrust between the

follower and guide is lesser than knife edged follower. Also the wear rate is less.

3 Which type of cam follower motion is preferred for high speed engines? Why? (May 2017)

BTL3

The cycloidal motion is recommended for high speed engines as it has the lower vibration,

surface wear, contact stress, noise and shock.

4 Give any two applications of cam mechanism in IC engines. (May 2017) BTL2

The combination of cam and follower determines the timing of opening and closing the inlet and

exhaust valve of engine.

Mushroom and spherical faced followers are used in automobile engines.

The roller follower and cycloidal cam combination are applied in aircraft engines.

5 What is maximum velocity and acceleration of the follower on both the strokes of uniform

acceleration and retardation? (Nov.2016) BTL1

Maximum velocity during outward stroke at 𝜃 =𝜃0

2,

(𝑣𝑜)𝑚𝑎𝑥 =2 𝐿 𝜔

𝜃0; (𝑣𝑟)𝑚𝑎𝑥 =

2 𝐿 𝜔

𝜃𝑟

Maximum acceleration during outward stroke at 𝜃 =𝜃0

2,

(𝑎𝑜)𝑚𝑎𝑥 = ±4 𝐿 𝜔2

𝜃02 ; (𝑎𝑟)𝑚𝑎𝑥 = ±

4 𝐿 𝜔2

𝜃𝑟2

6 Classify cams based on their physical shape. (Nov.2016, May 2013) BTL2

Based on the physical shape cams are classified as:

• Radial (disc or plate) cams

• Wedge (or Flat) cams

• Cylindrical (or barrel) cams

• Conical cams

• Globoidal cams

• End (or face) cams


2- 35

7 Define trace point of a cam. (May 2016) BTL1

It is a reference point on the follower and is used to generate the pitch curve. In case of knife edge

follower the knife edge represents the trace point and the pitch curve corresponds to the cam

profile. In a roller follower the centre of the roller represents the trace point.

8 Define tangential cam. (May 2016) (Nov.2015) (May 2014) BTL1

When the flanks of the cam are straight and tangential to the base circle and nose circle, the cam

is known as tangential cam.

9 Why sometimes the axes of translating roller followers in cam follower mechanisms are

offset from the axis of cam rotating? (Nov.2015, Dec. 2012, Dec.2011) BTL3

An offset is usually provided on a side so as to decrease pressure angle at the point of maximum

velocity during outstroke in order to reduce the side thrust in guides of followers.

10 Define the following with respect to cam and follower mechanism

(a) Pressure angle

(b) Pitch circle (Nov.2015) BTL1

Pressure angle: It is the angle between the direction of the follower motion and a normal to the

pitch curve. This angle is very important in designing a cam profile. If the pressure angle is too

large, a reciprocating follower will jam in its bearings.

Pitch circle: It is a circle drawn from the centre of the cam through the pitch points.

11 Draw the displacement, velocity and acceleration diagrams for a follower when it moves

with simple harmonic motion. (May 2015) BTL3

12 Why a roller follower is preferred to that of a knife-edged follower? (May 2015, Nov-2009,

Nov.2006) BTL3

In roller followers, the wear rate is considerably reduced because of rolling motion between

contacting surfaces.

13 Write down the classification of cams according to the manner of constraint of the follower.

(Nov.2014) BTL2

• Gravity constraint cams

• Spring constraint cams


2- 36

• Positive mechanical constraint cams

14 Define the term Jump speed of a cam. (Nov.2014) BTL1

In a cam-follower system, the contact between the cam surface and the follower is maintained by

means of a retaining spring. Beyond a particular speed of a cam rotation, the follower may lose

contact with the cam, because of inertia force acting on the follower, and the speed is known as

“jump speed” of a cam.

15 What are the different motions of a follower? (May 2014, Nov.2010, May 2010, Nov.2008,

May 2006) BTL1

• Uniform velocity motion

• Simple Harmonic Motion (SHM)

• Uniform acceleration and retardation motion (UARM) (or parabolic motion)

• Cycloidal motion

16 Define pressure angle and state best value of the pressure angle. (Nov.2013) BTL1

Pressure angle: It is the angle between the direction of the follower motion and a normal to the

pitch curve. This angle is very important in designing a cam profile. If the pressure angle is too

large, a reciprocating follower will jam in its bearings.

The maximum pressure angle should be less than 300 for cams with reciprocating followers.

17 Write the procedure to draw the cam profile. (Nov.2013) BTL2

In order to draw the cam profile for a radial cam, first of all the displacement diagram for the

given motion of the follower is drawn. Then by constructing the follower in its proper position at

each angular position, the profile of the working surface of the cam is drawn.

18 What do you meant by undercutting in cams? (Nov.2010, Nov.2009, May 2006, May 2003)

BTL1

If the curvature of the pitch curve is too sharp, then the part of the cam shape would be lost and

thereafter the intended cam motion would not be achieved. Such a cam is said to be undercut.

19 Define Angle of dwell. (May 2013) BTL1

The period during which the follower remains at rest is called dwell period.

20 State the expressions for maximum velocity and acceleration of a follower moves with

cycloidal motion. (Nov.2012, Nov.2007, May 2007) BTL2

Maximum velocity during outward stroke at 𝜃 =𝜃0

2,

(𝑣𝑜)𝑚𝑎𝑥 =2 𝐿 𝜔

𝜃0; (𝑣𝑟)𝑚𝑎𝑥 =

2 𝐿 𝜔

𝜃𝑟

Maximum acceleration during outward stroke at 𝜃 =𝜃0

2,

(𝑎𝑜)𝑚𝑎𝑥 =2𝜋 𝐿 𝜔2

𝜃02 ; (𝑎𝑟)𝑚𝑎𝑥 =

2𝜋 𝐿 𝜔2

𝜃𝑟2

PART * B

1 A cam is to be designed for, a knife-edge follower with the following data:

Follower lift is 40 mm with SHM, during 90 °of cam rotation.

Dwell for the next 300

Follower return to its original position with SHM, during the next 60° of cam rotation.

Dwell for the remaining cam rotation.

The line of stroke of the follower passes through the axis of the cam shaft. Radius of the

. base circle of the cam is 40 mm.

(i) Draw the displacement diagram.

(ii) Draw the profile of the cam.


2- 37

(iii) Determine the maximum velocity and acceleration of the follower during forward and

return strokes, the cam rotates 200 rpm in clockwise direction. (13) (May 2014, May 2012)

BTL5


Construction of displacement diagram:

(4 M)

Cam profile:

(5 M)


2- 38

(2 M)

(2 M)

2 A cam is to be designed for a knife-edge follower with the following data:

(i) Cam lift = 40 mm during 900 of cam rotation with Simple harmonic motion.

(ii) Dwell for next 300

(iii) During the next 60'° of cam rotation, the follower returns to its original position with

simple harmonic motion..

(iv) Dwell for the remaining 1800

Draw the profile of the cam when the line of stroke is offset 20 mm from the axis of the

camshaft. The radius of the base circle of the cam is 40 mm. (13) (May 2009, Nov.2004)

BTL5




2- 39

(5 M)

Cam profile:

(8 M)

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:

(i) Follower to have a stroke of 20 mm during 1200 of cam rotation,

(ii) Follower to dwell for 300 of cam rotation,

(iii) Follower to return to its initial position during 1200 of cam rotation, and

(iv) Follower to dwell for remaining 900 of cam rotation.

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

with SHM and the return stroke with equal uniform acceleration and retardation. Draw

the profile of the cam. (13) ( May 2016, May 2010, Nov. 2008, Nov. 2006) BTL5



2- 40


(5 M)

Cam profile:

(8 M)

4 Draw the displacement, velocity and acceleration curves, when the follower moves with

SHM and derive the expression for maximum velocity and maximum acceleration. (13)

(May/June 2016) BTL5

Answer: Page 3.17 JAYAKUMAR


2- 41

(6 M)

(3 M)


2- 42

(2 M)

(2 M)

5 A cam drives a flat reciprocating follower in the following manner: During first 120°

rotation of the cam, follower moves outwards through a distance of 20mm with SHM. The

follower dwells during next 30° of cam rotation. During next 120° of cam rotation, the

follower moves inwards with SHM. The follower dwells for the next 90° of cam rotation.

The minimum radius of the cam is 25mm. draw the profile of the cam. (13)BTL5

Answer: Page 792 -R.S.KHURMI


(5 M)

Cam profile:


2- 43

(8 M)

6 The following data are for a disc cam mechanism with roller follower: Minimum radius of

the cam = 35 mm; Lift of the follower =40 mm; Offset of the follower = 10 mm right; Roller

diameter = 15 mm.

Cam rotation angles are as mentioned below:

During ascent = 1200; Dwell = 80°; During descent = 80°; Dwell = 800. Cam rotates in

clockwise direction and the follower motion is simple harmonic during both ascent and

descent.

(i) Draw the displacement diagram of the follower and indicate the relevant data.

(it) Draw the cam profile and indicate the relevant data. (13) (Nov. 2003) BTL5



(5 M)


2- 44

Cam profile:

(8 M)

7 In a cam with translating roller follower, the follower axis is offset to the right of the cam

hinge by12 mm. The roller radius is 10 mm and the cam rotates in the clockwise direction.

Layout the rise portion of the cam profile to meet the following specifications: Rise takes

place during I80° of cam rotation of which for the first 90° the rise is with constant

acceleration and the rest is with constant retardation. Take seven station points only. The

lift of the cam is 30 mm and the least radius of the cam is 25 mm. (13) BTL5



(5 M)

Cam profile:


2- 45

(8 M)

PART * C

1 From the following data, draw the profile of the cam in which the follower moves with

SHM during ascent while it moves with uniformly accelerated motion during descent:

Least radius of cam = 50 mm; Angle of ascent = 480; Angle of dwell between ascent and

descent = 420; Angle of descent = 600; Lift of follower = 40 mm; Diameter of follower 30

mm; Distance between the line of action of the follower and the axis of cam = 20 mm.

If cam rotates at 360 rpm clockwise, find the maximum velocity and acceleration of the

follower during descent. (15) (Nov.2013, May 2008) BTL5



(4 M)

Cam profile:


2- 46

(7 M)

(2 M)

(2 M)

2 A cam rotating clockwise at a uniform speed of 100 rpm is required to give motion to knife-

edge follower as below:

(i) Follower to move outwards through 25 mm during 1200 of cam rotation.

(ii) Follower to dwell for the next 600 of cam rotation.

(iii)Follower to return to its starting position during next 90°of cam rotation.

(iv) Follower to dwell for the remaining 90 °of cam rotation.

The minimum radius of cam is 50 mm; Draw the profile of the cam when the line of stroke

of the follower passes through the axis of the camshaft. IF the displacement of the follower

takes place with uniform and equal acceleration and retardation on both the outstroke and

return strokes, find the maximum velocity and acceleration during its outstroke and

return strokes. Also draw the displacement, velocity and acceleration diagrams for one

complete revolutions of the cam. (15) (May 2006, May 2005) BTL5




2- 47

(4 M) Cam profile:

(6 M)

Displacement, Velocity and acceleration diagrams:


2- 48

(5 M)

3 The following particulars refers to a symmetrical circular cam operating a flat faced

follower: Lease radius of cam = 16 mm; Nose radius = 3.2 mm; Distance between cam shaft

centre and nose centre = 25 mm, Angie of action of cam = 1500; Cam shaft speed = 600 rpm,

Assuming that there is a no dwell between ascent or descent, determine:

(i) the lift of the valve,


2- 49

(ii) the flank radius, and

(iii) the acceleration and retardation of the follower at a point where circular nose merges

into circular flank. (15) (Nov. 2006, May 2006, Nov.2005, Nov.2004) BTL5


(5 M)

(2 M)

(4 M)


2- 50

𝑎𝑚𝑖𝑛 = −𝜔2𝑟 𝑐𝑜𝑠 (𝛼 − ∅) = 69.02 m/s2 (retardation)

(4 M)




2- 51




UNIT IV GEARS AND GEAR TRAINS

Law of toothed gearing – Involutes and cycloidal tooth profiles –Spur Gear terminology and definitions –

Gear tooth action – contact ratio – Interference and undercutting. Helical, Bevel, Worm, Rack and Pinion

gears [Basics only]. Gear trains – Speed ratio, train value – Parallel axis gear trains – Epicyclic Gear

Trains.

PART * A

Q.No. Questions

1 Give the classification of gears based on position of teeth on the wheel. (Nov.2017) BTL2

The classification of gears based on position of teeth on the wheel:

• Straight gears

• inclined gears

• curved gears

2 Draw the compound gear train and write its speed ratio. (Nov.2017) BTL3

Speed ratio =Speed of the first driver

Speed of the last driven or follower

=Product of the number of teeth on the drivens

Product of the number of teeth on the drivers

3 What type of gear arrangement is used to traverse the carriage in lathe machine? (May

2017) BTL1

The reverted gear trains and epicyclic gear trains are used in back gear of lathe.

4 State law of gearing. (May 2017 ,Nov.2015, May 2014, May 2013, Nov.2009, Nov.2008, May

2007, Nov.2005, Nov 2004, Nov.2002) BTL2

The law of gearing states that for obtaining a constant velocity ratio, at any instant of teeth the

common normal at each point of contact should always pass through a pitch point, situated on the

line joining the centre of rotation of the pair of mating gears.

5 Define the terms: Module and Pitch Circle diameter of a spur gear. (Nov.2016) BTL1

Module: It is the ratio of the pitch circle diameter in millimetres to the number of teeth. It is




2- 52

usually denoted by m. Mathematically,

Module, m = D /T

6 What is meant by interference and undercutting of gears? (Nov.2016) BTL1

Interference:

The phenomenon when the tip of tooth undercuts the roots on its mating gear is known as

interference.

Undercutting:

A portion of its dedendum falls inside the base circle. The profile of the tool inside the base circle

is radial. A gear having its material removed in this manner is known as undercut and this process

is known as undercutting.

7 Define normal and axial pitch in helical gears. (May 2016) BTL1

Normal pitch: It is the distance between similar faces of adjacent teeth, along a helix on the pitch

cylinder normal to the teeth. It is denoted by PN.

Axial pitch: It is the distance measured parallel to the axis, between similar faces of adjacent

teeth. It is the same as circular pitch and is therefore denoted by pc. If α is the helix angle, then

circular pitch,

𝑃𝑐 =𝑝𝑁

cos 𝛼

8 What is the advantage when arc of recess is equal to arc of approach in meshing gears?

(May 2016) BTL1

When arc of recess equal to arc of approach, the work wasted by friction is minimum and

efficiency of drive is maximum.

9 What is interference in involute gear and how is it prevented? (Nov.2015-R2008, May 2015,

Nov.2014, May 2014, Nov.2004) BTL1

Interference:

The phenomenon when the tip of tooth undercuts the roots on its mating gear is known as

interference.

The interference may only be avoided, if the point of contact between the two teeth is always on

the involute profiles of both the teeth. In other words, interference may only be prevented, if the

addendum circles of the two mating gears cut the common tangent to the base circles between the

points of tangency.

10 How the epicyclic gear train works? (Nov.2015) BTL1

An epicylic gear is a planetary gear arrangement consisting of one or more planet(epicyclic) gears

(P) meshed and rotating round a central sun gear (S).

It's application can be in a automatic transmission in an automobile gear box.

Epicyclic gears consist of several components: sun, carrier, planets, and rings. The sun is the

center gear, meshing with the planets, while the carrier houses the planet gear shaft. As the carrier

rotates, planets rotate on planet gear shafts while orbiting the sun. Finally, the ring is the internal

gear that meshes with the planets.

11 What are the special advantages of epicyclic gear trains? (May 2015) BTL1

Advantages:

• It provides high power density in comparison to standard parallel axis gear trains.

• It provides a reduction volume, multiple kinematic combinations, purely torsional

reactions, and coaxial shafting.




2- 53

12 Differentiate between involute profile and cycloidal profile. (Nov.2014) BTL3

S.No Involute Tooth Profile Cycloidal Tooth Profile

1. Variation in centre distance does

not affect the velocity ratio.

The centre distance should not vary.

2. Pressure angle remains constant

throughout the teeth.

Pressure angle varies. It is zero at the

pitch point and maximum at the start

and end of engagement

3. Interference occurs. No interference occurs.

4. Weaker teeth. Stronger teeth.

13 Define gear ratio. (Nov.2013) BTL1

Gear ratio:

The ratio of the angular speed of the initial or driving member of a gear train or equivalent

mechanism to that of the final or driven member

14 Write short notes on differentials. (Nov.2013) BTL2

The function of a differential gear in an automobile is to:

• Transmit motion from engine to rear wheels

• Rotate the rear wheels at different speeds while the automobile is taking a turn

15 List down the common forms of teeth. (May 2013) BTL2

• Involute curve

• Cycloidal curve

16 Define the term ‘arc of contact’ in gears. (Nov.2012) BTL1

The arc of contact is the path traced by a point on the pitch circle from the beginning to the end of

engagement of two meshing teeth.

17 Name two applications of reverted gear train. (Nov.2012) BTL2

The reverted gear trains are used in automobile gear boxes, lathe back gears, clocks, etc.,

18 What are the advantages and disadvantages of involute gear tooth profile? (May 2009,

Nov.2006, Nov.2003) BTL2

Advantages: Variable centre distance; Constant pressure angle; Easy manufacturing

Disadvantages: Interference occurs; Weaker teeth; More wear and tear.

19 Explain any two methods of reducing or eliminating interference in gears. (May 2014,

Nov.2002) BTL2

• By modifying addendum of gear teeth

• By increasing the pressure angle

• By modifying tooth profile or profile shifting

• By increasing the centre distance

20 What are the roles of idlers in gear trains? [or - What are the uses of intermediate gears in a

gear train?] (May 2013) BTL1

Intermediate gears, also known as idler gears, are necessary:

• To change the direction of rotation of the driven gear without changing its angular

velocity, and




2- 54

• To bridge the gap between first and last gears, when the centre distance is large

PART * B

1 A pinion having 20 teeth engages with an internal gear having 80 teeth. If the gears have

involute profiled teeth with 200 pressure angle, module of 10 mm and addendum of 10 mm,

find the path of contact, arc of contact and the contact ratio. (13) (Nov.2017) BTL4

Answer: Page 4.41 - V.Jayakumar

Given: Tp = 20; TG =80; = 20; m =10; ap = aw = 10 mm

(i) Length of path of contact:

Pitch circle radii of pinion and gear wheel 𝑟 =𝑚𝑇𝑝

2 = 100 mm

𝑅 =𝑚𝑇𝐺

2 = 400 mm

𝑟𝑎 = 𝑟 + 𝑎𝑝= 110 mm

𝑅𝑎 = 𝑅 − 𝑎𝑊= 390 mm

(4 M)

Length of path of approach KP = 𝑅 𝑠𝑖𝑛∅ − √𝑅𝐴2 − 𝑅2𝑐𝑜𝑠2∅ = 32.8 mm

Length of path of recess, PL = √𝑟𝐴2 − 𝑟2𝑐𝑜𝑠2∅ - r 𝑠𝑖𝑛∅ = 23 mm

Length of path of contact, KL = KP + PL = 55.8 mm Ans.

(6 M)

(ii) Length of arc of contact:

𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑎𝑟𝑐 𝑜𝑓 𝑐𝑜𝑛𝑡𝑎𝑐𝑡 = 𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑝𝑎𝑡ℎ 𝑜𝑓 𝑐𝑜𝑛𝑡𝑎𝑐𝑡

cos ∅ = 59.38 mm

(2 M)

(iii) Contact ratio:

𝐶𝑜𝑛𝑡𝑎𝑐𝑡 𝑟𝑎𝑡𝑖𝑜 = 𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑎𝑟𝑐 𝑜𝑓 𝑐𝑜𝑛𝑡𝑎𝑐𝑡

𝜋 𝑚 = 1.89 pairs

(1 M)

2 An epicyclic gear consists of three gears A, B and C as shown in Fig. The gear A has 72

internal teeth and gear C has 32 external teeth. The gear B meshes with both A and C and is

carried on an arm EF which rotates about the centre of A at 18 r.p.m.. If the gear A is fixed,

determine the speed of gears B and C. (13) (May 2017) BTL4


Given : TA = 72 ; TC = 32 ; Speed of arm EF = 18 r.p.m.

Considering the relative motion of rotation as shown in Table

Table of motions

(5 M)




2- 55

Speed of gear C

We know that the speed of the arm is 18 r.p.m. therefore,

y = 18 r.p.m.

and the gear A is fixed, therefore

X = 40.5

So, Speed of gear C = x + y = 58.5 r.p.m. in the direction of arm

(4 M)

Speed of gear B

Let dA, dB and dC be the pitch circle diameters of gears A, B and C respectively. Therefore, from the geometry of Fig

(4 M)

3 A compound epicyclic gear is shown diagrammatically in Fig. The gears A, D and E are free

to rotate on the axis P. The compound gear B and C rotate together on the axis Q at the end

of arm F. All the gears have equal pitch. The number of external teeth on the gears A, B and

C are 18, 45 and 21 respectively. The gears D and E are annular gears. The gear A rotates at

100 r.p.m. in the anticlockwise direction and the gear D rotates at 450 r.p.m. clockwise. Find

the speed and direction of the arm and the gear E. (13) (Nov.2016) BTL4


From Fig, we can write that




2- 56

(4 M)




2- 57

(5 M)

(2 M)

(2 M)




2- 58

4 Explain gear nomenclature with neat diagram and define all salient terms pertaining to

the gear. (13) (May 2016) BTL4


GEAR NOMENCLATURE

1. Pitch circle. It is an imaginary circle which by pure rolling action, would give the same

motion as the actual gear.

2. Pitch circle diameter. It is the diameter of the pitch circle. The size of the gear is usually

specified by the pitch circle diameter. It is also known as pitch diameter.

3. Pitch point. It is a common point of contact between two pitch circles.

4. Pitch surface. It is the surface of the rolling discs which the meshing gears have replaced at

the pitch circle.

5. Pressure angle or angle of obliquity. It is the angle between the common normal to two gear

teeth at the point of contact and the common tangent at the pitch point. It is usually denoted by φ.

The standard pressure angles are 1

2 14 ° and 20°.

6. Addendum. It is the radial distance of a tooth from the pitch circle to the top of the tooth.

7. Dedendum. It is the radial distance of a tooth from the pitch circle to the bottom of the tooth.

8. Addendum circle. It is the circle drawn through the top of the teeth and is concentric with the

pitch circle.

9. Dedendum circle. It is the circle drawn through the bottom of the teeth. It is also called root

circle.

Note : Root circle diameter = Pitch circle diameter × cos φ, where φ is the pressure angle.

10. 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 usually denoted by pc.

Mathematically, Circular pitch, pc = π D/T

where D = Diameter of the pitch circle, and T = Number of teeth on the wheel.

11. Diametral pitch. It is the ratio of number of teeth to the pitch circle diameter in millimetres.

It is denoted by pd . Mathematically,




2- 59

Diametral pitch, d

where T = Number of teeth, and D = Pitch circle diameter.

12. Module. It is the ratio of the pitch circle diameter in millimeters to the number of teeth.

It is usually denoted by m. Mathematically,

Module, m = D /T

13. Clearance. It is the radial distance from the top of the tooth to the bottom of the tooth, in a

meshing gear. A circle passing through the top of the meshing gear is known as clearance circle.

14. Total depth. It is the radial distance between the addendum and the dedendum circles of a

gear. It is equal to the sum of the addendum and dedendum.

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

It is equal to the sum of the addendum of the two meshing gears.

16. Tooth thickness. It is the width of the tooth measured along the pitch circle.

17. Tooth space . It is the width of space between the two adjacent teeth measured along the pitch

circle.

18. Backlash. It is the difference between the tooth space and the tooth thickness, as measured

along the pitch circle. Theoretically, the backlash should be zero, but in actual practice some

backlash must be allowed to prevent jamming of the teeth due to tooth errors and thermal

expansion.

19. Face of tooth. It is the surface of the gear tooth above the pitch surface.

20. Flank of tooth. It is the surface of the gear tooth below the pitch surface.

21. Top land. It is the surface of the top of the tooth.

22. Face width. It is the width of the gear tooth measured parallel to its axis.

23. Profile. It is the curve formed by the face and flank of the tooth.

24. Fillet radius. It is the radius that connects the root circle to the profile of the tooth.

25. Path of contact. It is the path traced by the point of contact of two teeth from the beginning to

the end of engagement.

26. *Length of the path of contact. It is the length of the common normal cut-off by the

addendum circles of the wheel and pinion.

27. ** Arc of contact. It is the path traced by a point on the pitch circle from the beginning to the

end of engagement of a given pair of teeth. The arc of contact consists of two parts, i.e.

(a) Arc of approach. It is the portion of the path of contact from the beginning of the engagement

to the pitch point.

(b) Arc of recess. It is the portion of the path of contact from the pitch point to the end of the

engagement of a pair of teeth.

Sketch - (5 M)

Description - (8 M)

5 Fig. shows a differential gear used in a motor car. The pinion A on the propeller shaft has

12 teeth and gears with the crown gear B which has 60 teeth. The shafts P and Q form the

rear axles to which the road wheels are attached. If the propeller shaft rotates at 1000 r.p.m.

and the road wheel attached to axle Q has a speed of 210 r.p.m. while taking a turn, find

the speed of road wheel attached to axle P. (13) (May 2016) BTL4




2- 60


Given : TA = 12 ; TB = 60 ; NA = 1000 r.p.m. ; NQ = ND = 210 r.p.m.

𝑵𝑩 = 𝑵𝑨 ×𝑻𝑨

𝑻𝑩 = 200 rpm

(2 M)

(7 M)

Since the speed of gear B is 200 r.p.m., therefore from the fourth row of the table,

y = 200 ...(i)

Also, the speed of road wheel attached to axle Q or the speed of gear D is 210 r.p.m., therefore

from the fourth row of the table,

y – x = 210 or x = y – 210 = 200 – 210 = – 10

∴ Speed of road wheel attached to axle P

= Speed of gear C = x + y

= – 10 + 200 = 190 r.p.m. Ans.

(2 M)




2- 61

6 In an epicyclic gear train, an arm carries two gears A and B having 36 and 45 teeth

respectively. If the arm rotates at 150 r.p.m. in the anticlockwise direction about the centre

of the gear A which is fixed, determine the speed of gear B. If the gear A instead of being

fixed, makes 300 r.p.m. in the clockwise direction, what will be the speed of gear B ? (13)

(Nov. 2015) BTL4


Given : TA = 36 ; TB = 45 ; NC = 150 r.p.m. (anticlockwise)

(5 M)

Speed of gear B when gear A is fixed

Since the speed of arm is 150 r.p.m. anticlockwise, therefore from the fourth row of the table,

y = + 150 r.p.m.

Also the gear A is fixed, therefore

x + y = 0 or x = – y = – 150 r.p.m.

Speed of gear B when gear A makes 300 r.p.m. clockwise

Since the gear A makes 300 r.p.m.clockwise, therefore from the fourth row of the table,

x + y = – 300 or x = – 300 – y = – 300 – 150 = – 450 r.p.m.

(4 M)

2. Algebraic method

Let NA = Speed of gear A.

NB = Speed of gear B, and

NC = Speed of arm C.

Assuming the arm C to be fixed, speed of gear A relative to arm C

= NA – NC

and speed of gear B relative to arm C = NB – NC

Since the gears A and B revolve in opposite directions, therefore




2- 62

Speed of gear B when gear A is fixed

When gear A is fixed, the arm rotates at 150 r.p.m. in the anticlockwise direction, i.e.

NA = 0, and NC = + 150 r.p.m.

Speed of gear B when gear A makes 300 r.p.m. clockwise

Since the gear A makes 300 r.p.m. clockwise, therefore NA = – 300 r.p.m.

NB = – 450 × – 0.8 + 150 = 360 + 150 = 510 r.p.m. Ans.

(4 M)

7 Two gear wheels mesh externally and are to give a velocity ratio of 3 to 1. The teeth are of

involute form ; module = 6 mm, addendum = one module, pressure angle = 20°. The pinion

rotates at 90 r.p.m. Determine : 1. The number of teeth on the pinion to avoid interference

on it and the corresponding number of teeth on the wheel, 2. The length of path and arc of

contact, 3.The number of pairs of teeth in contact, and 4. The maximum velocity of sliding.

(13) (May 2014) BTL4


Given : G = T / t = 3 ; m = 6 mm ; AP = AW = 1 module = 6 mm ; = 20° ; N1 = 90 r.p.m. or ω1

= 2π × 90 / 60 = 9.43 rad/s

1. Number of teeth on the pinion to avoid interference on it and the corresponding number of teeth

on the wheel

We know that number of teeth on the pinion to avoid interference,

and corresponding number of teeth on the wheel,

T = G.t = 3 × 19 = 57 Ans.

(3 M)

2. Length of path and arc of contact

We know that pitch circle radius of pinion,

r = m.t / 2 = 6 × 19/2 = 57 mm

∴ Radius of addendum circle of pinion,

rA = r + Addendum on pinion (AP) = 57 + 6 = 63 mm

and pitch circle radius of wheel,

R = m.T / 2 = 6 × 57 / 2 = 171 mm

∴ Radius of addendum circle of wheel,

RA = R + Addendum on wheel (AW ) = 171 + 6 = 177 mm

We know that the path of approach (i.e. path of contact when engagement occurs),

= 15.7 mm and the path of recess (i.e. path of contact when disengagement occurs),




2- 63

=13.67 mm

Length of path of contact,

KL KP PL 15.7 13.67 29.37 mm Ans.

We know that length of arc of contact

= 31.25 mm

(4 M) 3. Number of pairs of teeth in contact

We know that circular pitch,

(3 M)

4. Maximum velocity of sliding

(3 M)

PART * C

1 The following data relate to a pair of 20° involute gears in mesh :

Module = 6 mm, Number of teeth on pinion = 17, Number of teeth on gear = 49 ; Addenda

on pinion and gear wheel = 1 module.

Find : 1. The number of pairs 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 larger wheel (i) is just making contact, (ii) is just

leaving contact with its mating tooth, and (iii) is at the pitch point. (15) (May 2017) BTL4


Given : = 20° ; m = 6 mm ; t = 17 ; T = 49 ; Addenda on pinion and gear wheel = 1 module = 6

mm

1. Number of pairs of teeth in contact

We know that pitch circle radius of pinion,

r = m.t / 2 = 6 × 17 / 2 = 51 mm

and pitch circle radius of gear,

r = m.T / 2 = 6 × 49 / 2 = 147 mm




2- 64

∴ Radius of addendum circle of pinion,

rA = r + Addendum = 51 + 6 = 57 mm

and radius of addendum circle of gear,

RA = R + Addendum = 147 + 6 = 153 mm

We know that the length of path of approach (i.e. the path of contact when engagement occurs),

=15.5 mm

and length of path of recess (i.e. the path of contact when disengagement occurs),

=13.41 mm

Length of path of contact, KL = KP + PL = 15.5 + 13.41 = 28.91 mm

(5 M)

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

We know that angle turned through by the pinion

(5 M)

3. Ratio of sliding to rolling motion




2- 65

(iii) Since at the pitch point, the sliding velocity is zero, therefore the ratio of sliding velocity

to rolling velocity is zero. Ans.

(5 M)

2 (i) Prove that the max. length of arc of contact between a pair of gear tooth to avoid

interference is (r+R) tan (6) BTL4


When interference is just avoided, the maximum length of path of contact is MN when the

maximum addendum circles for pinion and wheel pass through the points of tangency N and M

respectively as shown in Fig. In such a case,

Maximum length of path of approach,

MP = r sin




2- 66

and maximum length of path of recess,

PN = R sin

So, Maximum length of path of contact,

MN = MP + PN = r sin + R sin = (r + R) sin

and maximum length of arc of contact

=(r + R) sin ∅

cos ∅= (r + R) tan ∅

Sketch - (3 M)

Description - (3 M)

(ii) Two mating gears have 20 and 40 involute teeth of module 10 mm and 20°

pressure angle. The addendum on each wheel is to be made of such a length that

the line of contact on each side of the pitch point has half the maximum possible

length. Determine the addendum height for each gear wheel, length of the path of

contact, arc of contact and contact ratio. (9) (Nov. 2016) BTL4


Given : t = 20 ; T = 40 ; m = 10 mm ; φ = 20°

Addendum height for each gear wheel

We know that the pitch circle radius of the smaller gear wheel,

r = m.t / 2 = 10 × 20 / 2 = 100 mm

and pitch circle radius of the larger gear wheel,

R = m.T / 2 = 10 × 40 / 2 = 200 mm

Let RA = Radius of addendum circle for the larger gear wheel, and

rA = Radius of addendum circle for the smaller gear wheel.

RA = 206.5 mm

Addendum height for larger gear wheel

= RA - R = 6.5 mm Ans.

Now path of recess, ½ PL = PN

Addendum height for smaller gear wheel

= rA - r = 116.2 - 100 = 6.2 mm Ans.

(4 M)

Length of the path of contact

We know that length of the path of contact




2- 67

(2 M)

Length of the arc of contact

We know that length of the arc of contact

(1 M)

Contact ratio

We know that circular pitch,

(2 M)

3 Two involute gears of 20° pressure angle are in mesh. The number of teeth on pinion is 20

and the gear ratio is 2. If the pitch expressed in module is 5 mm and the pitch line speed is

1.2 m/s, assuming addendum as standard and equal to one module, find :

1. The angle turned through by pinion when one pair of teeth is in mesh ; and

2. The maximum velocity of sliding. (15) (Nov.2014) BTL4


(3 M)




2- 68

(3 M)

(3 M)

(3 M)

(3 M)




2 - 69




UNIT V FRICTION IN MACHINE ELEMENTS

Surface contacts – Sliding and Rolling friction – Friction drives – Friction in screw threads –Bearings

and lubrication – Friction clutches – Belt and rope drives – Friction in brakes- Band and Block brakes.

PART * A

Q.No. Questions

1 What is meant by crossed belt drive? (Nov.2017) BTL1

The crossed or twist belt drive is used with shafts arranged parallel and rotating in the opposite

direction.

• Cross belt is used where the direction of rotation of driven pulley is opposite to driving

pulley.

• Where we need more power transmission there we can use cross belt drive

2 Write the conditions for the maximum power transmission by a belt from one pulley to

another. (Nov.2017) BTL2

Power transmitted by belt P = (T1-T2) v

By substituting dP/dv =0, the conditions for maximum power transmission is obtained as

Tmax = 3 Tc

The power transmitted shall be maximum when the centrifugal tension (Tc) is one third of the

maximum belt tension (Tmax)

3 What kind of friction acts between the tyre and road in an automobile? (May 2017) BTL1

• Friction is the force between two objects as they move over one another such as a car's tire

and the surface which it is travelling on.

• There are two types of friction, only one which with we will be dealing with. These types

are known as static friction and kinetic friction. Static friction is the frictional force

required to start an object moving on another surface. Kinetic frictional force is the force

to keep the object in motion.

4 State the functional difference between a clutch and a brake. (May 2017, Nov.2015, May

2015, May 2012, Nov.2011) BTL3

“The functional difference between a clutch and a brake is that a clutch connects two moving




2 - 70

members of a machine, whereas a brake connects moving member to a stationary member”.

Function of clutch Function of brake

A clutch is a mechanical device used to

connect or disconnect the driven shaft from

the driving shaft at the will of the operator

while power is transmitted from driving to

driven shaft.

A brake is also a mechanical device by means

of which motion of a body is retarded for

slowing down or to bring it to rest, by

applying artificial friction resistance.

5 Differentiate Self energizing and self locking brakes. (Nov.2016 , May 2016, Nov.2012) BTL3

Self energizing (or self actuating) brake:

When the moment of applied force and the moment of the frictional force are in the same

direction, then frictional force helps in applying the brake. This type of brake is known as self

energizing brake.

Self locking brake:

When the frictional force is sufficient enough to apply the brake with no external force, then the

brake is said to be self-locking brake.

6 What are the disadvantages of V-belt drive over flat belt? (Nov.2016) BTL1

• V belt cannot be used in large distance.

• It is not as durable as flat belt.

• Since the V belt subjected to certain amount of creep therefore it is not suitable for

constant speed applications such as synchronous machines, and timing devices.

• It is a costlier system

7 Why self locking screws have lesser efficiency? (May 2016, Nov.2012) BTL2

Self locking needs some friction on the thread surface of the screw and nut hence it needs higher

effort to lift a body and hence automatically the efficiency decreases.

8 What are the advantages of wire ropes over fabric ropes? (Nov.2015) BTL1

Advantages of wire ropes over cotton (fabric) ropes:

• These are lighter in weight,

• These offer silent operation,

• These can withstand shock loads,

• These are more reliable,




2 - 71

• They do not fail suddenly,

• These are more durable,

• The efficiency is high, and

• The cost is low.

9 Write the mathematical expression for the maximum efficiency of a screw jack. (Nov.2015,

May 2014) BTL2

𝜂𝑚𝑎𝑥 =1 − 𝑠𝑖𝑛𝜙

1 + 𝑠𝑖𝑛𝜙

Where, - Friction angle

10 Write mathematical expression for the length of the belt required for two pulleys of

diameters d1 and d2 and at distance x apart are connected by means of an open belt drive.

(Nov.2015) BTL2

𝐿 =𝜋

2(𝑑1 + 𝑑2) + 2𝑥 +

(𝑑1 − 𝑑2)2

4𝑥

11 What is centrifugal tension in a belt? How does it affect the power transmitted? (May

2015) BTL1

During operation, as the belt passes over a pulley the centrifugal effect due to its weight tends to

lift the belt from the pulley surface. This reduces the normal reaction and hence the frictional

resistance. The centrifugal force produces additional tension in the belt called “centrifugal

tension”.

The power transmitted shall be maximum when the centrifugal tension (Tc) is one third of the

maximum belt tension (Tmax)

12 What are the laws of solid dry friction? (Nov.2014) BTL1

• The frictional force is directly proportional to the normal reaction between the surfaces.

• The frictional force depends upon the nature of the surfaces in contact.

• The frictional force is independent of the area and the shape of the containing surfaces.

• The frictional force is independent of the velocity of sliding of one body relative to the

other body.

13 What is meant by crowning of pulleys in a flat belt drives? Also write its purpose.

(Nov.2014) BTL1

The rim of the pulley is given camber (i.e., taper) on the surface, so that the belt will not slip off




2 - 72

the pulley when it rotates. This process is known as “crowning” of pulley.

14 Define velocity ratio. (May 2014) BTL1

Velocity ratio is the ratio of speed of driving gear to the speed of the driven gear.

15 Define anti-friction bearing. (Nov.2013) BTL1

An anti friction bearing, also known as a rolling contact bearing, is used when very little friction

is needed for low differential surface speeds.

16 Differentiate multi plate clutch and cone clutch. (Nov.2013) BTL3

In disc (plate) clutches, friction lined flat plates are used.

In cone clutches, friction lined frustum of cone is used.

17 Distinguish between open and cross belt drive in terms of its application. (May 2013) BTL3

The open belt drive is used with shafts arranged parallel and rotating in same direction.

The crossed belt drive is used with shafts arranged parallel and rotating in the opposite direction.

18 Differentiate between self-locking and overhauling of screw. (May 2012) BTL3

We know that the torque required to lower the load by the screw,

𝑇 = 𝑃 ×𝐷

2= 𝑊 𝑡𝑎𝑛(∅ − 𝛼) ×

𝑑

2

If ∅ > 𝛼 , then torque required to lower the load will be positive, so an effort will be required to

lower the load. This type of screw is termed as “self-locking screw”.

We know that the torque required to lower the load by the screw,

𝑇 = 𝑃 ×𝐷

2= 𝑊 𝑡𝑎𝑛(∅ − 𝛼) ×

𝑑

2

If ∅ < 𝛼 , then torque required to lower the load will be negative. That means, the load will start

moving downward without the application of any torque. Such a condition is termed as

“overhauling of screw”.

19 What is limiting angle of friction? (Nov.2011) BTL2

The limiting angle of friction (∅) is defined as the angle at which the resultant reaction R makes

with the normal reaction RN.

20 What is the role of friction in screw jack? (Nov.2010) BTL2

Due to friction, the effort required to move the body up is increased.

When µ > tanα, the self locking in screws will be achieved.

PART * B

1 An electric motor driven power screw moves a nut in a horizontal plane against a force of




2 - 73

75 kN at a speed of 300 mm/min. The screw has a single square thread of 6 mm pitch on a

major diameter of 40 mm. The coefficient of friction at the screw threads is 0.1. Estimate

power of the motor. (13) (NOV.2017) BTL5


Given : W = 75 kN = 75 × 103 N ; v = 300 mm/min ; p = 6 mm ; do = 40 mm ; µ = tan = 0.1

We know that mean diameter of the screw,

d = d0 – p/2 = 0.037 m

tan 𝛼 =𝑝

𝜋𝑑= 0.0516

So, Force required at the circumference of the screw

𝑃 = 𝑤 tan(𝛼 + ∅) = 𝑤 [tan 𝛼+tan ∅

1−tan 𝛼 tan ∅] = 11.43 x 103 N (7 M)

and torque required to overcome friction,

T = P × d/2 = 211.45 N-m

T = P × d/2 = 11.43 × 103 × 0.037/2 = 211.45 N-m

We know that speed of the screw,

𝑁 =𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑛𝑢𝑡

𝑃𝑖𝑡𝑐ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑐𝑟𝑒𝑤 = 50 r.p.m.

And angular speed, ω = 2 π x 50 /60 = 5.24 rad/s

So, power of the motor = T ω = 1108 W = 1.108 kW Ans.

(6 M)

2 The following data relate to a screw jack:

Pitch of the threaded screw = 8 mm; Diameter of the threaded screw = 40 mm;

Coefficient of friction between screw and nut = 0.1; Load = 20 KN.

Assuming that the load rotates with the screw, determine:

(i)the ratio of torques required to raise and lower the load, and

(ii) the efficiency of the machine. (13) (May 2017, May2012) BTL4


Given: p= 8 mm = 0.008 m; d = 40 mm = 0.04 m; µ = tan = 0.1; W = 20 kN = 20 x 103 N

tan 𝛼 =𝑝

𝜋𝑑= 0.0637 ; 𝛼 = 3.64°

= tan -1 (0.1) = 5.71°

(3 M)

(i) Ratio of torques required to raise and lower the load:




2 - 74

Torque required to raise the load, T1 = 𝑤 tan(𝛼 + ∅)𝑑/2 = 65.861 N-m

Torque required to lower the load, T2 = 𝑤 tan(∅ − 𝛼)𝑑/2 = 14.457 N-m

So, ratio of torques, T1 / T2 = 4.556 Ans

(6 M)

(ii) Efficiency of the machine:

We know that,

η = tan α

tan(α+∅)= 0.386 or 38.6 % Ans.

(4 M)

3 A single plate clutch transmits 25 kW at 900 rpm. The maximum pressure intensity between

the plates is 85 kN/m2. The ratio of radii is 1.25. Both the sides of the plate are effective and

the coefficient of friction is 0.25. Determine (i) the inner diameter of the plate and (ii) the

axial force to engage the clutch. Assume theory of uniform wear.

(13) (May 2017) BTL5


Given: P = 25 kW = 25 x 103 W; N = 900 rpm; n = 2; pmax = 85 kN/m2 = 85 x 103 N/m2; µ = 0.25

(i) Inner diameter of the plate:

We know that, 𝑃 = 2𝜋𝑁𝑇

60

T = 265.26 N-m

Since the intensity of pressure is maximum at the inner radius (r2), therefore

Pmax r2 = C ; OR C = 85 x 103 r2 N/mm

and the axial thrust transmitted to the frictional surface,

W = 2 π C (r1 – r2) = 1.335 x 105 (r2)2

Mean radius for uniform wear is given by

R = (r1+r2)/2 = 1.125 r2

(7 M)




2 - 75

Torque transmitted, T = ½ n µ W (r1+r2)

r2 = 0.1523 m or 152.3 mm

r1 = 1.25 r2 = 190.375 mm Ans

(3 M)

(ii) Axial force to engage the clutch:

We know that the axial force to engage the clutch,

W = 2 π C (r1 – r2) = 3096.57 N Ans.

(3 M)

4 The mean diameter of the screw jack having pitch of 10 mm is 50 mm. A load of 20 kN is

lifted through a distance of 170 mm. Find the work done in lifting the load and efficiency of

the screw jack when

1. the load rotates with the screw, and

2. the load rests on the loose head which does not rotate with the screw.

The external and internal diameter of the bearing surface of the loose head are 60 mm and

10 mm respectively. The coefficient of friction for the screw as well as the bearing surface

may be taken as 0.08. (13) (Nov.2016) BTL5


Given : p = 10 mm ; d = 50 mm ; W = 20 kN = 20 × 103 N ; D2 = 60 mm or R2 = 30 mm ;

D1 = 10 mm or R1 = 5 mm ; µ = tan = µ1 = 0.08

tan α =p

πd= 0.0637


P = w tan(α + ∅) = w [tan α+tan ∅

1−tan α tan ∅] = 2890 N

T = P × d/2 = 72250 N-mm = 72.25 N-m

Since the load is lifted through a vertical distance of 170 mm and the distance moved by the

screw in one rotation is 10 mm (equal to pitch), therefore number of rotations made by the screw,

N = 170/10 = 17 (4 M)

1. When the load rotates with the screw

We know that work done in lifting the load




2 - 76

= T × 2πN = 7718 N-m Ans.

(2 M)

and efficiency of the screw jack,

η = tan α

tan(α+∅)=

tan α(1−tan α tan ∅)

tan α+ tan ∅ = 0.441 = 44.1 % Ans.

(2 M)

2. When the load does not rotate with the screw

We know that mean radius of the bearing surface,

R =R1+R2

2 = 17.5 mm

and torque required to overcome friction at the screw and the collar,

T = P ×d

2+ μ1WR = 100250 N-mm = 100.25 N-m

Work done by the torque in lifting the load

= T × 2πN = 10 710 N-m Ans.

(3 M)

We know that the torque required to lift the load, neglecting friction,

T0 = P0 ×d

2= w tan α ×

d

2 = 31850 N-mm = 31.85 N-m (∵ Po = W tan α)

Efficiency of the screw jack,

η = T0

T= 0.318 or 31.8 % Ans.

(2 M)

5 A leather faced conical clutch has a cone angle of 30º. If the intensity of pressure between

the contact surfaces is limited to 0.35 N/mm2 and the breadth of the conical surface is not to

exceed one-third of the mean radius, find the dimensions of the contact surfaces to transmit

22.5 kW at 2000 r.p.m. Assume uniform rate of wear and take coefficient of friction as 0.15.

(13) (Nov.2016) BTL5


Given : 2 α = 30º or α = 15º ; pn = 0.35 N/mm2; b = R/3 ; P = 22.5 kW = 22.5 × 103 W ;

N = 2000 r.p.m. or ω = 2 π × 2000/60 = 209.5 rad/s ; µ = 0.15

Let r1 = Outer radius of the contact surface in mm,

r2 = Inner radius of the contact surface in mm,

R = Mean radius of the the contact surface in mm,

b = Face width of the contact surface in mm = R/3, and




2 - 77

T = Torque transmitted by the clutch in N-m.

We know that power transmitted (P),

22.5 × 103 = T.ω = T × 209.5

∴ T = 22.5 × 103/209.5 = 107.4 N-m = 107.4 × 103 N-mm

(4 M)

We also know that torque transmitted (T ),

107.4 × 103 = 2π μ pn.R2. b = 2π × 0.15 × 0.35 × R2 × R/3 = 0.11 R3

∴ R3 = 107.4 × 103/0.11 = 976.4 × 103 or R = 99 mm Ans.

(4 M)

The dimensions of the contact surface are shown in Fig.

From Fig., we find that

𝑟1 − 𝑟2 = b sin α = R/3 x sin α = 8.54 mm ...(i)

and r1 + r2 = 2R = 198 mm ...(ii)

From equations (i) and (ii),

r1 = 103.27 mm, and r2 = 94.73 mm Ans.

(5 M)

6 Following data is given for a rope pulley transmitting 24 kW :

Diameter of pulley = 400 mm ; Speed = 110 r.p.m.; angle of groove = 45° ; Angle of lap on

smaller pulley = 160° ; Coefficient of friction = 0.28 ; Number of ropes = 10 ; Mass in kg/m

length of ropes = 53 C2 ; and working tension is limited to 122 C2 kN, where C is girth of

rope in metres. Find initial tension and diameter of each rope. (13) (May 2016) BTL5


Given : PT = 24 kW ; d = 400 mm = 0.4 m ; N = 110 r.p.m. ; 2 β = 45° or β = 22.5°;

θ = 160° = 160 × π / 180 = 2.8 rad ; µ = 0.28 ; n = 10 ; m = 53 C2 kg/m ;

T = 122 C2 kN = 122 × 103 C2 N

Initial tension

We know that power transmitted per rope,




2 - 78

P =Total power transmitted

No of ropes=

PT

n = 2.4 kW = 2400 W

and velocity of rope, 𝑣 =𝜋𝑑𝑁

60=2.3 m/s

(4 M)

Let, T1 = Tension in the tight side of the rope, and

T2 = Tension in the slack side of the rope

We know that power transmitted per rope ( P )

2400 = (T1 – T2) v = (T1 – T2) 2.3

∴ T1 – T2 = 2400 / 2.3 = 1043.5 N ...(i)

We know that

2.3 log (T1

T2) = μθ cosec β = 2.05

log (T1

T2) = 0.8913

T1

T2= 7.786 …(Taking antilog of 0.8913) … (ii)

From equation (i) and (ii),

T1 = 1197.3 N, and T2 = 153.8 N

We know that initial tension in each rope,

T0 =T1+T2

2 = 675.55 N Ans.

(5 M)

Diameter of each rope

Let d1 = Diameter of each rope,

We know that centrifugal tension,

TC = m.v2 = 280.4 C2 N

and working tension (T),

122 × 103 C2 = T1 + TC

∴ C2 = 9.836 × 10–3 or C = 0.0992 m = 99.2 mm

We know that girth (i.e. circumference) of rope (C),

99.2 = π d1 or d1 = 31.57 mm Ans.

(4 M)

7 The cutter of a broaching machine is pulled by square threaded screw of 55mm external

diameter and 10 mm pitch. The operating nut takes the axial load of 400 N on a flat surface

of 60 mm internal diameter and 90 mm external diameter. If the coefficient of friction is




2 - 79

0.15 for all contact surfaces on the nut, determine the power required to rotate the

operating nut, when the cutting speed is 6 m/min. (13) (May 2016) BTL5


Given: do = 55 mm ; p = 10 mm = 0.01 m ; W = 400 N ; D2 = 60 mm or R2 = 30 mm ; D1 = 90

mm or R1 = 45 mm ; µ = tan = µ1 = 0.15

We know that mean diameter of the screw,

d = d0 – p/2 = 50 mm

tan α =p

πd= 0.0637

(4 M)


P = w tan(α + ∅) = w [tan α+tan ∅

1−tan α tan ∅] = 86.4 N

We know that mean radius of the flat surface,

R =R1 + R2

2= 37.5 mm

Total torque required,

T = P ×d

2+ μ1WR = 4410 N-mm = 4.41 N-m

(5 M)

Since the cutting speed is 6 m/min, therefore speed of the screw,

𝑁 =𝐶𝑢𝑡𝑡𝑖𝑛𝑔 𝑠𝑝𝑒𝑒𝑑

𝑃𝑖𝑡𝑐ℎ = 600 r.p.m.

And angular speed, ω = 2π x 600/60 = 62.84 rad/s

We know that power required to operate the nut

= T.ω = 4.41 × 62.84 = 277 W = 0.277 kW Ans.

(4 M)

PART * C

1 A compressor, requiring 90 kW is to run at about 250 r.p.m. The drive is by V-belts from an

electric motor running at 750 r.p.m. The diameter of the pulley on the compressor shaft

must not be greater than 1 metre while the centre distance between the pulleys is limited to

1.75 metre. The belt speed should not exceed 1600 m/min.

Determine the number of V-belts required to transmit the power if each belt has a cross

sectional area of 375 mm2, density 1000 kg/m3 and an allowable tensile stress of 2.5 MPa.

The groove angle of the pulley is 35°. The coefficient of friction between the belt and the




2 - 80

pulley is 0.25. Calculate also the length required of each belt. (15) (Nov 2016) BTL5


Given : P = 90 kW ; N2 = 250 r.p.m. ; N1 = 750 r.p.m. ; d2 = 1 m ; x = 1.75 m ;

v = 1600 m/min = 26.67 m/s ; a = 375 mm2 = 375 × 10–6 m2 ; ρ = 1000 kg/m3;

σ = 2.5 MPa = 2.5 × 106 N/m2 ; 2 β = 35° or β = 17.5° ; μ = 0.25

First of all, let us find the diameter of pulley on the motor shaft (d1). We know that

N2

N1=

d1

d2 or d1 =

N2d2

N1 = 0.33 m

We know that the mass of the belt per metre length,

m = Area × length × density = 0.375 kg

∴ Centrifugal tension, TC = m.v2 = 267 N

and maximum tension in the belt,

T = σ. a = 937.5 N

∴ Tension in the tight side of the belt,

T1 = T – TC = 670.5 N

(3 M)

Let T2 = Tension in the slack side of the belt.

For an open belt drive, as shown in Fig.,

sin α =O2M

O1O2=

r2−r1

x=

d2−d1

2x = 0.1914

∴ α= 11°

(3 M)

and angle of lap on smaller pulley (i.e. pulley on motor shaft),

θ = 180° – 2α = 180° – 2 × 11° = 158° = 158 × π / 180 = 2.76 rad

2.3 log (T1

T2) = μθ cosec β = 2.295

log (T1

T2) = 0.998




2 - 81

T1

T2= 9.954 …(Taking antilog of 0.998)

T2 = 67.36 N

(3 M)

Number of V-belts

We know that power transmitted per belt

= (T1 – T2) v = 16 086 W = 16.086 kW

∴ Number of V − belts = Total power transmitted

Power transmitted per belt = 5.6 or 6 Ans.

(3 M)

Length of each belt

We know that length of belt for an open belt drive,

𝐿 =𝜋

2(𝑑2 + 𝑑1) + 2𝑥 +

(𝑑2−𝑑1)2

4𝑥 = 5.664 m Ans.

(3 M)

2 2.5 kW of power is transmitted by an open belt drive. The linear velocity of the belt is 2.5

m/s. The angle of lap on the smaller pulley is 1650 . The coefficient of friction is 0.3.

Determine the effect on power transmission in the following cases:

(i) Initial tension in the belt is increased by 8%,

(ii) Initial tension in the belt is decreased by 8%,

(m) Angle of lap is increased by 8% by the use of an idler pulley, for the same speed and

the tension on the tight side, and

(iv) Coefficient of friction is increased by 8% by suitable dressing to the friction surface of

the belt.

Also state which of the above methods suggested could be more effective?

(15) (Nov. 2015) BTL5

Answer: Page 8.41 - JAYAKUMAR 8.41

Given: P =2.5 kW = 2500 W; v = 2.5 m/s; ϴ = 1650 = 1650 x (π/1800) = 2.88 rad; µ = 0.3

Power, P = (T1-T2) v

T1-T2 = 1000 N ... (i)

Tension ratio, 𝑇1

𝑇2= 𝑒𝜇𝜃= 2.37

T1 = 2.37 T2 ... (ii)

On solving equations (i) and (ii), we get

T1 = 1729.9N and T2 = 729.9 N

Initial tension, 𝑇0 =𝑇1+𝑇2

2 = 1229.9 n

(3 M)

(i) When Initial tension is increased by 8%:

New increased initial tension,




2 - 82

𝑇0′ =

108

100 × 𝑇0 = 1328.3 N

T1+T2 = 2656.6 N … (iii)

As µ and ϴ remain unchanged, 𝑒𝜇𝜃 or 𝑇1

𝑇2 is same

So, T1 = 2.37 T2

On solving equations (ii) and (iii), we get

T1 = 1868.3N and T2 = 788.3 N

Power, P = (T1- T2)v = 2700 W Or 2.7 kW

So, Increase in power = (2.7 – 2.5)/2.5 = 0.08 Or 8 % Ans.

(3 M)

(ii) When initial tension is decreased by 8%:

New decreased initial tension,

𝑇0′ =

92

100 × 𝑇0 = 1131.5 N

T1+T2 = 2263 N … (iv)

We know that, T1 = 2.37 T2

On solving equations (ii) and (iv), we get

T1 = 1591.5N and T2 = 671.5 N

Power, P = (T1- T2)v = 2300 W Or 2.3 kW

So, Decrease in power = (2.5 – 2.3)/2.5 = 0.08 Or 8 % Ans.

(3 M)

(iii) Angle of lap (θ) is increased by 8%:

New angle of lap, 𝜃′ =108

100 × 𝜃 = 3.1104 rad

New tension ratio is given by, 𝑇1

𝑇2= 𝑒𝜇𝜃

T1 = 2.54 T2 ... (v)

Tension on the tight side remains same as before,

T1 = 1729.9N

Then from equation (v), T2 = 680.5 N

Power, P = (T1- T2)v = 2624 W Or 2.624 kW

So, percentage increase in power = (2.624 – 2.5)/2.5 = 0.0496 Or 4.96 % Ans.

(3 M)

(iv) Coefficient of friction is increased by 8 %:

New co-efficient of friction, 𝜇′ =108

100 × 𝜇 = 0.27

T1 = 2.54 T2 ... (vi)

T1+T2 = 2459.8 N …(vii)

On solving equations (vi) and (vii), we get

T1 = 1764.9N and T2 = 694.9 N

Power, P = (T1- T2)v = 2675 W Or 2.675 kW

So, percentage increase in power = (2.675 – 2.5)/2.5 = 0.07 Or 7 % Ans.




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Conclusion: Since the power transmitted by increasing the initial tension by 8% is more,

therefore in order to increase the power transmitted, we shall adopt the method of increasing

the initial tension. Ans.

(3 M)

3 (i) The following data relate to a screw jack:

Pitch of the threaded screw = 8 mm; Diameter of the threaded screw = 40 mm; Coefficient

of friction between screw and nut = 0.1; Load = 20 KN.

Assuming that the load rotates with the screw, determine:

(a) the ratio of torques required to raise and lower the load, and

(b) the efficiency of the machine. (5) (Nov. 2105) BTL4


Given: p= 8 mm = 0.008 m; d = 40 mm = 0.04 m; µ = tan = 0.1; W = 20 kN = 20 x 103 N

tan 𝛼 =𝑝

𝜋𝑑= 0.0637 ; 𝛼 = 3.64°

= tan -1 (0.1) = 5.71°

(a) Ratio of torques required to raise and lower the load:

Torque required to raise the load, T1 = 𝑤 tan(𝛼 + ∅)𝑑/2 = 65.861 N-m

Torque required to lower the load, T2 = 𝑤 tan(∅ − 𝛼)𝑑/2 = 14.457 N-m

So, ratio of torques, T1 / T2 = 4.556 Ans

(3 M)

(b) Efficiency of the machine:

We know that,

η = tan α

tan(α+∅)= 0.386 or 38.6 % Ans.

(2 M)

(ii) A friction clutch is used to rotate a machine from a shaft rotating at a uniform

speed of 250 rpm. The disc type clutch has both of its sides effective, the

coefficient of friction being 0.3. The outer and inner diameters of the friction

plate are 200 mm and 120 mm respectively. Assuming uniform wear of the

clutch, the intensity of pressure is not to exceed 100 kN/m2. If the moment of

inertia of the rotating parts of the machine is 6.5 kg.m2, determine the time to

attain the full speed by the machine and the energy lost in slipping of the clutch.

What will be the intensity of pressure, if the condition of uniform pressure of the

clutch is considered? Also determine the ratio of power transmitted with uniform




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wear to that with uniform pressure. BTL5 (10) (Nov. 2015)


Given: 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 = 0.06 m; pmax = 100 kN/m2 = 100 x 103 N/m2; I = 6.5 kg.m2

(a) Time to attain the full speed by the machine (with uniform wear):

Since the intensity of pressure (p) is maximum at the inner radius r2, therefore

Pmax.r2 = C Or C = 6000 N/m

Axial thrust exerted, W = 2 π C (r1 – r2) = 1507.96 N

Torque transmitted, T =1

2n. μW(r1 + r2) = 72.38 N-m

We know that, P =2 π NT

60 = 1895 W

Also T = I α, Where α is the angular accleration

α = 11.135 rad/s2

We know that, α = ω/t = 11.135 t = 2.353 Ans.

(3 M)

Thus the full speed is attained by the machine in 2.35 seconds

(b) Energy lost in slipping of the clutch:

Angle turned by the driving shaft, 𝜃1 = 𝜔𝑡 =2𝜋𝑁

60 = 61.52 rad

Angle turned by the driven shaft, 𝜃2 = 𝜔0𝑡 +1

2𝛼𝑡2 = 30.75 rad

So, Energy lost in friction = T (𝜃1 − 𝜃2)= 2226 N.m

(2 M)

(c) Intensity of pressure, if the condition is uniform pressure:

Intensity of pressure, 𝑝 =𝑊

𝜋(𝑟12−𝑟2

2) = 75 kN/m2 Ans.

(2 M)

(d) Ratio of power transmitted with uniform wear to that with uniform pressure:

Power transmitted with uniform wear = 1895 W

Torque transmitted with uniform pressure = 2

3𝑛𝜇𝑊 [

𝑟13−𝑟2

3

𝑟12−𝑟2

2] = 73.89 N-m

Power transmitted with uniform pressure is given by P =2 π NT

60 = 1934 W




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So, Power transmitted with uniform wear

Power transmitted with uniform pressure = 0.98 Ans. (3 M)

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