Useful for II year B.Tech AGRI , II year B.E. Mech

1. ME3491 THEORY OF MACHINES
STUDY MATERIALS
Class: II year -III Semester/ B. Tech. Agricultural Engineering
Name of Faculty member: J.Edison Thangaraj

2. UNIT I
KINEMATICS OF MECHANISMS
Mechanisms – Terminology and definitions.
Kinematics inversions of 4 bar and slide crank chain
Kinematics analysis in simple mechanisms
Velocity and acceleration polygons
Analytical methods
Computer approach
Cams – Classifications – Displacement Diagrams
Layout of Plate Cam Profiles – Derivatives of Followers Motion
Circular Arc and Tangent Cams
Two marks Q&A
1. Define Kinematic Link & Mention its characteristics.
Each part of a machine which moves relative to some other parts is called a kinematic link
or element. It need not be a rigid body, but it must be a resistant body.
Characteristics:
 It must be a resistant body.
 It should have relative motion.
2. Define Structure.
It is an assemblage of number of rigid resistant bodies having no relative motion between
them and meant for carrying loads having straining action. There is no energy
transformation within the structure and members are just for supporting and carrying loads.
eg: a railway bridge, a roof truss, frames, etc.,
3. Distinguish between a Machine and a Structure.
MACHINE STRUCTURE
1. The parts of the machine move relative to
one another.
1. The members of a structure do not move
relative to one another.
2. A machine transforms the available energy
into some useful work.
2. In a structure, no energy is transformed into
useful work.
The links of a machine may transmit both
power and motion.
Eg: shaper, planar, scooter, car, etc.,
3. The members of a structure transmit forces
only.
Eg: bridge, truss, lathe bed, etc.,

3. 4. Distinguish between a Machine and a Mechanism.
MACHINE MECHANISM
1. Machine transmits forces and couples . 1. Mechanism transmits and modifies motion.
2. Machine may have many mechanisms for
transmitting power or mechanical work.
2. A mechanism is the skeleton outline of the
machine to produce definite motion between
various links.
Eg: Lathe, milling machine, shaper Eg: quick return mechanism, slider crank
mechanism, Geneva mechanism
5. Define Kinematic Pair.
Each part of a machine which moves relative to some other parts is called a kinematic
link. The two links or elements of a machine, when in contact with each other, are said to form a
pair
If the relative motion between them is completely or successfully constrained (i.e, in a
definite direction),the pair is known as kinematic pair. Kinematic pair can classified based on
the relative motion between the elements, type of contact between the elements and type of
closure.
6. Write a note on classification of kinematic pairs.
1. According to the type of relative motion between the elements.
 Sliding pair
 Turning pair
 Rolling pair
 Screw pair
 Spherical pair
2. According to the type of contact between the elements.
 Lower pair
 Higher pair
3. According to the type of closure
 Self closed pair
 Force closed pair
7. What are the three types of constrained motions?
 Completely constrained motion- Relative motion between two links takes place in
definite direction.
 Incompletely constrained motion- Relative motion between two links takes place
in more than one direction.
 Successfully constrained motion- Relative motion between two links takes place
in definite direction, not completely by itself but by some other means.

4. 8. Define Kinematic Chain.
The two links or elements of a machine, when in contact with each other, are said to form a pair.
When the kinematic pairs are coupled in such a way that the last link is joined to the first link to
transmit definite motion (i.e., completely or successfully constrained motion),it is called a
kinematic chain. Types of kinematic chain are four bar chain, single slider crank chain and
double slider crank chain.
9. What the different types of joints?
 Binary joint- Two links joined at the same point
 Ternary joint- Three links joined at the same point
 Quaternary joint- Four links joined at the same point
10. Define Mechanism.
When one of the links of a kinematic chain is fixed, the chin is known as mechanism. It may be
used for transmitting or transforming motion. A mechanism is the skeleton outline of the
machine to produce definite motion between various links by transmitting and modifying the
motion. Types of mechanisms are simple mechanisms and compound mechanisms. e.g: engine
indicators, typewriter, etc.,
11. What is meant by a Machine?
When a mechanism is required to transmit power or to do some particular type of work, then
that mechanism is called is called as machine. Machine is a device which converts the available
energy into some useful work by transmitting forces and couple.
eg: shaper, planar, lathe, scooter, car, etc.,
12. Define Degree of Freedom.
Degree of freedom is defined as the number of input parameters which must be independently
controlled in order to bring the mechanism into a particular position or useful engineering
purpose. It is also termed as mobility. Degree of freedom of simple mechanisms can be found out
by using the Kutzbach criterion.
n=3(l-1)-2j-h (Kutzbach equation is used to find the number of degree of freedom)
13. What is difference between Kutzbach equation and Grubler’s equation?
Kutzbach equation is used to find out number of degree of freedom in simple
mechanisms. Kutzbach criterion is given by
n = 3(l-1)-2j-h , where l- no.of links; j- no. of joints; h- no. of higher pair
Grubler’s equation is applicable to the mechanisms which have only one degree of
freedom and consists of only lower pair i.e, h = 0

5. n=3(l-1)-2j-h
1=3(l-1)-2j-0
3l-2j-4=0 (Grubler’s equation)
14. What is meant by inversion of mechanism?
If one link in a kinematic chain is fixed, then it is termed as mechanism. The method of
obtaining different mechanisms by fixing different links in a kinematic chain is known as
inversion of mechanism or Kinematic inversion.
Eg: Four bar chain – beam engine, locomotive engine, etc.,
Single slider crank chain- oscillating cylinder engine, rotary engine, etc.,
Double slider crank chain- oscillating cylinder engine, rotary engine, etc.,
15. State Grashoff’s law.
Grashoff’s law is applicable for four bar mechanisms. It states that, “The sum of the
shortest and longest link lengths should not be greater than the sum of the remaining two lengths
if there is to be continuous relative motion between the two links, in a four bar mechanism”
16. What are the different types of Kinematic Chain?
 Four bar chain or quadric cyclic chain- It consists of fixed link, crank, connecting rod and
output link or lever.
 Single slider crank chain- It consists of fixed link, crank, connecting rod and a slider.
 Double slider crank chain- It consists of fixed link, connecting links and two sliders
forming two sliding pairs and two turning pairs.
17. Differentiate Completely Constrained Motion and Incompletely
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. e.g,.
The motion of a piston is limited to a definite direction such that it will only reciprocate
irrespective of the direction of motion of the crank.
When the motion between a pair can take place in more than one direction, then the
motion is called an incompletely constrained motion. e.g., a circular bar or shaft either rotate or
slide in a circular hole.
18. Define 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, the motion is said to be successfully
constrained motion. Eg: a shaft in a foot step bearing in which the shaft may rotate in a bearing
or it may move upwards.

6. 19. Explain the types of kinematic pair according to the type of relative
motion between the elements
Sliding pair- When the two elements of a pair are connected in such a way that one can
only slide relative to another, the pair is known as a sliding pair. It has a completely constrained
motion. Eg: piston and cylinder, cross-head and guides of a reciprocating steam engine
Turning pair- When two elements of a pair connected in such a way that one can only
turn or revolve about a fixed axis of another link, the pair is known as the turning pair. It also has
a completely constrained motion. Eg: the crankshaft in a journal bearing in an engine
Rolling pair- When the two elements in a pair are connected in such a way that one link
rolls over the fixed link, the pair is known as rolling pair. Eg: ball bearing and roller bearing
Screw pair- When the two elements in a pair are connected in such a way that one
element can turn about the other by screw threads, the pair is known as screw pair. Eg: the lead
screw of a lathe with nut.
Spherical pair- When the two elements in a pair are connected in such a way that one
element turns or swivels about the other fixed element, the pair formed is called a spherical pair.
Eg: ball and socket joint, pen stand, etc.,
20. Classify lower pair and higher 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. Eg: sliding pairs, turning pairs, screw pairs
When the two elements of a pair have a line or point of 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. Eg: a pair of friction discs, toothed gearing, cam and follower.
21. What are the types of kinematic pair based on the type of closure
Self closed pair- When the two elements of a pair are connected together mechanically in
such a way that only required kind of relative motion occurs, it is then known as self closed pair.
Eg: lower pairs.
Force closed pair- When the two elements of a pair are not connected mechanically but
are kept in contact by the action of external forces, the pair is said to be a force closed pair. Eg:
cam and the follower as it is kept in contact by the forces exerted by spring and gravity.
22. What is the use of Oldham’s coupling?
Oldham’s coupling is the inversion of double slider crank chain. It is used for connecting
two parallel shafts whose axes are at a small distance apart (not coaxial). The shafts are coupled
in such a way that if one shaft rotates, the other shaft also rotates at the same speed by the

7. flanges attached at each ends and intermediate piece.
23. Define the terms mechanical advantage and transmission angle.
Mechanical advantage of a mechanism is the ratio of output force or torque to the input
force or torque at any instant. Transmission angle is defined as the angle between the coupler or
connecting rod and output link or follower of a four bar mechanism.
25. What do you meant by pantograph?
Pantograph is an instrument used for reproducing a displacement or to make a drawing
exactly to an enlarged or reduced scale and as exactly as possible. It is an inversion of four bar
chain mechanism. This mechanism is generally used in copying devices like engraving or profile
machine, minting, milling, sculpture duplication, etc.,
26. What is straight line generators and write its types.
Mechanisms which are used to obtain straight line are termed as straight line
mechanisms. A particular point in the mechanism traces out the straight line for specific purpose
find its application in machine design. It was extensively used in classical machines like steam
engines.
Types of straight line generators are the Scott Russel mechanisms, Peaucillier mechanisms,
Hart’s mechanisms and Watt’s indicator mechanisms.
27. Describe the terms (i) Universal joint or Hooke’s joint and (ii) Toggle
mechanisms.
A Hooke’s joint is used to connect two shafts which are intersecting at small angle. The
end of each shaft is forked to U-type and each fork provides two bearings. Applications of
Hooke’s joint are transmission from gear box to differential and knee joint in milling machines,
etc.,
Toggle mechanism- In slider crank mechanism, as the crank approaches one of its dead centre
position, the slider approaches zero. At the instant, the ratio of the crank movement to the slider
movement approaching infinity is proportional to the mechanical advantage. This principle is
used in toggle mechanisms.
UNIT -1 B (second Half)
1. Define Analysis of Mechanism.
Each link is subjected to definite displacement ,velocity and acceleration. Analyzing
various links in a mechanism to determine the velocity and acceleration is known as analysis of

8. mechanism. In kinematic analysis, a particular given mechanism is investigated based on the
mechanism geometry plus other known characteristics (such as input angular velocity, angular
acceleration, etc.).
2. What are the different methods of analyzing a mechanism?
 Reactive velocity & reactive acceleration,
 Instantaneous centre method,
 Klein’s construction method,
 Complex algebra method,
 Vector algebra method
 Analytical method
3. Define Rubbing Velocity.
Two links of a mechanism having turning point will be connected by pins. Velocity of rubbing of
pins depends on the angular velocity of links relative to each other as well as direction. The
rubbing velocity is defined as the algebraic sum between the angular velocity of two links which
are connected by pin, multiplied by the radius of the pin.
Rubbing Velocity = (ω1±ω2) * r
Where, r is the radius of the pin
ω1 and ω2 are angular velocities of link 1 and 2 connected by the pin.
4. Define Relative velocity
Relative velocity is a measurement of velocity between two objects moving in different
frames of reference. It is the vector difference between the velocities of two bodies : the velocity
of a body with respect to another regarded as being at rest. Consider two bodies A and B moving
along the parallel lines in the same direction with the absolute velocities VA and VB , then the
relative velocity of A with respect to B is given by
VAB = VB -VA
5.Differentiate between translation and rotation.
Translation: A slide or translation takes place in a body moves in one direction from one place
to another place. It appears like sliding motion.
Rotation: It is a circular movement of an object around a center of rotation. A rotation is a rigid
body movement, which unlike a translation keeps a point fixed.
6.What is a configuration diagram?
It is a skeleton or a line diagram which represents a machine or a mechanism, to study the

9. velocity and acceleration of any mechanisms. It is also called as space diagram. It is a line sketch
of a given mechanism drawn to a suitable scale & it is the basis for the construction of both
velocity and acceleration diagram.
7.What are the steps to be followed during the graphical method of
analysis of mechanisms?
Steps to be followed during the graphical method of analysis of mechanisms are:
(i) Draw the configuration diagram to the suitable scale.
(ii) Locate all fixed points in a mechanism as a common point in velocity diagram.
(iii) Choose a suitable scale for the velocity vector diagram.
(iv) Draw the velocity vector of each link perpendicular to the link.
(v) Measure the velocity vectors of each link and convert the same to the suitable scale
for respective velocity values.
8.Differentiate kinematic analysis and kinematic synthesis.
Kinematic analysis: the dimensions of the links are known and the output characteristics like
displacement, velocity and acceleration of a mechanism are determined.
Kinematic synthesis: It is the design or creation of a mechanism to obtain a set of required output
characteristics. Synthesis is opposite to the analysis.
UNIT I- C ( Third half of Unit I)
1. What is a cam? What are the different basic types of cam?
A cam is a rotating machine element that transmits reciprocating or oscillating motion to
another element, a follower. A cam may be defined as a machine element having a curved
outline or a curved groove, which, by its oscillation, rotation or reciprocating motion, gives a
predetermined specified motion to another element called the follower.
Types: Radial or Disc cam, Cylindrical Cam, Wedge Cam, Spiral Cam, Tangent Cam, Circular
Arc Cam
2. What are the important components of a cam and follower mechanism?
Any cam and follower mechanism essentially consists of three components;
(i) Cam - the driving member
(ii) Follower - the driven member and,
(iii) Guide (or) Frame - supporting and guiding member.
A cam may be defined as a machine element gives a predetermined specified motion to another
element called the follower.

10. 3. What are the applications of a cam and follower mechanism? Give
their advantages and disadvantages.
Cam and follower mechanisms find their place in many machines and mechanisms and
the following are a few examples: (1) Internal Combustion (IC) Engines (to actuate the inlet and
outlet valves). (ii) Textile Machinery (iii) Machine Tools like Lathe, Shaping Machine, Planar
Machine, Slotting Machine (in feed mechanisms) (iv) Paper Cutting Machines (v) Printing
Machines (vi) Clocks
Advantages of Cams: ✓ Very long life ✓ Quiet Operations ✓ Low shock and acceleration ✓
High load carrying capacity Disadvantages of cams: ✓ Rotary input is required ✓ Proper
manufacturing of the cam must be more essential
4. Why roller follower is preferable over a knife-edge follower? Explain in
brief.
During the operation of the cam mechanism, due to the relative movement between the
pointed knife-edge of the follower and the cam, excessive friction and wear occurs in the knife-
edge follower. As the roller rotates about its own axis also during the cam rotation, the friction
and wear is greatly reduced in a roller follower. In view of the above, the roller follower is
preferable over a knife-edge follower.
i) Due to the sliding motion between the contact surfaces of the knife edge and the cam,
the small area of the contacting surfaces of roller and cam, the small area of the contacting
surface results in excessive wear.
ii) Whereas in roller follower the rolling motion takes place between the contact surfaces
of roller and cam. As the rate of wear is greatly reduced, roller follower is preferred over a knife-
edge follower.
5. Compare the knife edge, roller and mushroom follower of cam
Knife edge follower: used in low speed cams. The wear rate is high at the point of contact.
Roller follower : used where more space is available. The operation is smooth with less
vibration. Used in gas engine valves, oil engine valves, aircraft engine valves.
Mushroom follower : used where the space is limited. Used in automobile inlet & exhaust
valves.
6. Briefly explain important radial cam nomenclature:
Base circle : Smallest circle that can be drawn to the cam profile

11. Cam profile : The surface of the cam that comes in contact with the follower
Trace point: Is the reference point on the follower to trace the cam profile. In case of a knife
edge follower, the sharp end point is the trace point and in case of a roller follower, the centre of
the roller is the trace point
Pitch curve: Is the path of the tracing point Prime circle: Is the smallest circle drawn tangent to
the pitch curve Pitch point: Is the point on the pitch curve at which the pressure angle is the
maximum
Pitch circle: Is the circle passes through the pitch point & concentric with the base circle
Lift or stroke or throw: Is the maximum displacement of the follower from the base circle of
the cam
7 .What are the types of followers?
i) According to the contact of surface:
a) Knife edge follower
b) Roller follower
c) Flat faced/mushroom follower
d) Spherical faced follower
ii) According to the type of motion:
a) Reciprocating or translating follower
b) Rotating or oscillating follower
iii) According to the path of motion:
a) Radial follower
b) Offset follower
8. What is dwell period in cam mechanism? and Define angle of dwell.
If there is zero movement or zero displacement of the follower during the motion of cam,
it is called as dwell.
Angle of dwell is the angle through which the cam rotates while the follower remains
stationary at the lower or higher positions.
9. List the various cams based on their shapes. (May 2013)
Translation or flat cam
Radial or disc cam
Cylindrical or drum cam

12. Globoidal cam
Conjugate or double disc cam
10. What are positive drive and preloaded spring cam?
Positive drive cams do not require any external force to maintain the contact between the
cam and follower. Eg. : Cylindrical cam.
Preloaded spring cam require helical compression spring in the compressed condition to
maintain the contact. Eg. : Radial cam.
11. Define trace point in cams.
It is a reference point on the follower which is used to generate the pitch curve. In case
of roller follower, the centre of roller represents the trace point whereas in case of knife edge
follower, the knife edge represents the trace point and the corresponding pitch curve represents
the cam profile.
12. Define pressure angle. (May 2011, Dec. 2013)
It is the angle between the direction of follower motion and a normal to the pitch curve. It
is very important while designing a cam profile. If its value is too large, it will jam a
reciprocating follower in its bearings.
13. What are the different types of follower motion used in cam-follower
mechanism?[Dec. 2010]
During the operation, the cam rotates at a uniform angular velocity and the follower may have
one of the following motions:
Uniform velocity motion (Uniform motion)
Simple harmonic motion
Uniform acceleration and retardation motion
Cycloidal motion
14. Which motion of follower is used for low and moderate cam speed
applications ?Why ?
i) Uniform acceleration and retardation motion of follower is used for low and medium
speed Operations.
ii) Due to infinite jerk, there will be shock loads produced which causes vibration and

13. high stresses. Therefore, used for above applications.
15. State the types of cams with specified contours.
Following are the various types of symmetrical or specified contour cams:
Tangent cam: When the flanks of the cam are straight and tangential to the base circle and nose
circle, then the cam is known as tangent cam.
Circular arc cam: When the flanks of the cam connecting the nose and base circles are of
convex circular arc, such cams are referred as circular arc cams.
Eccentric cam: An eccentric cam is a disc with its centre of rotation positioned 'off centre'.
16. Explain the effects of change in pressure angle in cam profile. (May
2011, Dec. 2013)
i) For lower value of pressure angle, the component of force along the motion of follower
can be kept low which reduces the friction between follower and its guides.
ii) Similarly, if the value of pressure angle is high then the friction between follower and
guides is more. This may cause the follower to jam in its guides
iii) Hence the maximum value of pressure angle is up to 30 ° to 35 °
17. Define undercutting in a cam mechanism. (Dec.2010)
If at any point on the cam surface the curvature is too sharp, then the follower may not
follow the prescribed path of pitch curve. This causes very high amount of contact stresses On
the surface of cam and sometimes remove some material from the cam profile which is called as
undercutting.
Conditions to avoid undercutting:
By using a smaller roller diameter
By increasing the size of cam
By using internal cams.
By decreasing the follower lift
PART B QUESTIONS
UNIT I - Kinematics of mechanisms
1. Explain with a neat sketch, how an offset slider crank mechanism can be used as quick -
return motion mechanism. Derive an expression to find the quick –return ratio.
2. Explain Elliptical Trammel

14. 3. Explain Coupling rod of locomotive.
4. Explain the constrained motion with neat sketch.
5. What is kinematic pair and Briefly Explain about its classification.
6. Explain beam engine
7. Explain double lever mechanism
8. Explain pendulum pump and oscillating cylinder engine.
9. Explain Whitworth quick -return motion mechanism
10. Explain Scotch yoke mechanism & Oldham’s coupling.
11. A crank and slotted lever mechanism used in a shaper has a centre distance of 300 mm
between the centre of oscillation of the slotted lever and the centre of rotation of the
crank. The radius of the crank is 120 mm. Find the ratio of time of cutting stroke to time
of return stroke
12. In a crank and slotted lever quick return mechanism, the distance between the two fixed
centres is 240 mm and the length of the driving crank is 120 mm. Find the inclination of
the slotted bar with the vertical in the extreme position and the time ratio of cutting stroke
to the return 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.
UNIT I - B
1. The crank and connecting rod of the theoretical steam engine are 0.5 m and 2 m respectively. The
crank makes 180 rpm in the clockwise direction. When, it is turned 450
from the inner dead centre
position. Determine 1. Velocity of the piston, 2. Angular velocity of connecting rod, 3. Velocity of
the point E on the connecting rod 1.5 m from, the gudgeon pin.
2. In the fig., the angular velocity of the crank OA is 600 rpm. Determine the linear velocity of slider
D and the angular velocity of link BD, when the angle is inclined at an angle of 750
to vertical. The
dimensions of the various links are OA= 28 mm, AB= 44 mm, BC= 49 mm and BD =46 mm. The
centre distance between the centre of rotation O and C is 65 mm. The path of travel of the slider is
11 mm below the fixed point C. The slider moves along a horizontal path and OC is vertical.

15. 3. The crank of a slider crank mechanism rotates clockwise at a constant speed of 300 rpm. The crank
is 150 mm and the connecting rod is 600 mm long. Determine (i).linear velocity and acceleration of
midpoint of the connecting rod and angular velocity (ii).angular acceleration of the connecting rod
at crank angle of 45° from inner dead centre position
4. The engine mechanism shown in Fig. has crank OB = 50 mm and length of connecting rod AB
= 225 mm. The centre of gravity of the rod is at G which is 75 mm from B. The engine speed is
200 r.p.m. For the position shown, in which OB is turned 45° from OA, Find 1. the velocity of
G and the angular velocity of AB, and 2. the acceleration of G and angular acceleration of AB.
5. In a pin jointed four bar mechanism ABCD, the lengths of various links are as follows: AB = 25
mm ; BC = 87.5 mm ; CD = 50 mm and AD = 80 mm. The link AD is fixed and the angle BAD
= 135°. If the velocity of B is 1.8 m/s in the clockwise direction, find 1. velocity and
acceleration of the mid point of BC, and 2. angular velocity and 3. angular acceleration of link
CB and CD.
6. In a four bar chain ABCD , link AD is fixed and the crank AB rotates at 10 radians per second
clockwise. Lengths of the links are AB = 60 mm ; BC = CD = 70 mm ; DA = 120 mm. When
angle DAB = 60° and both B and C lie on the same side of AD, find 1. angular velocities
(magnitude and 1. direction) of BC and CD ; and 2. angular acceleration of BC and CD.
7. In the toggle mechanism, as shown in Fig., D is constrained to move on a horizontal path. The
dimensions of various links are : AB = 200 mm; BC = 300 mm ; OC = 150 mm; and BD = 450
mm. The crank OC is rotating in a counter clockwise direction at a speed of 180 r.p.m.,
increasing at the rate of 50 rad/s2
. Find, for the given configuration 1. velocity and acceleration
of D, and 2. angular velocity and angular acceleration of BD.

16. 8. A mechanism as shown in Fig., the link AB rotates with a uniform angular velocity of 30 rad/s.
The lengths of various links are : AB = 100 mm ; BC = 300 mm ; BD = 150 mm ; DE = 250
mm ; EF = 200 mm ; DG = 165 mm. Determine the velocity and acceleration of G for the
given configuration.
9. In a mechanism as shown in Fig, the crank OA is 100 mm long and rotates in a clockwise
direction at a speed of 100 r.p.m. The straight rod BCD rocks on a fixed point at C. The links
BC and CD are each 200 mm long and the link AB is 300 mm long. The slider E, which is
driven by the rod DE is 250 mm long. Find the velocity and acceleration of E.

17. UNIT I-C
1. Draw a cam profile to drive an oscillating roller follower to the specification given below
 Follower to move outwards through an angular displacement of 20° during the first
120° rotation of the cam
 Follower to return to its initial position during next 120° rotation of the cam
 Follower to dwell during next 120° of cam rotation.
The distance between the pivot centre and roller centre is 120 mm and distance between pivot
centre and cam axis is 130 mm; minimum radius of cam = 40 mm, radius of roller is 10 mm;
inward and outward strokes takes place with simple harmonic motion.
2. Draw the profile of a cam to give following motion to a follower with a flat face: (i) Follower
to have a stroke of 20 mm during 120° of cam rotation. (ii) Follower to dwell for 30° of cam
rotation. (iii) Follower to return to its initial position during 120° of cam rotation.
(iv)Follower to dwell for remaining 90° of cam rotation. Minimum radius of cam = 25 mm.
Out stroke and return stroke of the follower are simple harmonic.
3. Draw the profile of a cam to give following motion to a reciprocating follower with a flat
face:
 Follower to have a stroke of 20 mm during 120° of cam rotation.
 Follower to dwell for 30° of cam rotation.
 Follower to return to its initial position during 120° of cam rotation.
 Follower to dwell for remaining 90° of cam rotation.
Minimum radius of cam =25 mm.
Out stroke and return stroke of the follower as simple harmonic motion and uniform
acceleration and retardation motion respectively.
4. A cam with a minimum radius of 25 mm, rotating in clockwise direction with a uniform
speed of 100 rpm is to be designed to give the motion for a roller follower as follows.
 To rise through 50 mm during 120° rotation of cam with UAUR.
 Fully raised through next 30°.
 To lower during next 60° with SHM.
 Dwell for the remaining period.
Draw the profile of the cam when the line of stroke of the follower is offset by 15 mm
from the axis of the camshaft.
5. Draw a cam profile to drive an oscillating roller follower to the specification given below
 Follower to move outwards through an angular displacement of 20° during the
first 120° rotation of the cam
 Follower to return to its initial position during next 120° rotation of the cam
 Follower to dwell during next 120° of cam rotation.
The distance between the pivot centre and roller centre is 120 mm and distance between
pivot centre and cam axis is 130 mm; minimum radius of cam = 40 mm, radius of roller
is 10 mm; inward and outward strokes takes place with simple harmonic motion.
6. Construct the profile of a cam to suit the following specification.

18. Cam shaft diameter 40 mm least radius of cam is 25 mm diameter of the roller 10 mm dia
and angle of lift 120° angle of fall 150° lift of the follower 40 mm number of pauses are two
equal intervals between the motions. During the lift the motion is SHM and for falling
Uniform acceleration and retardation the speed of the cam shaft is uniform. The line of stroke
of the follower is off -set by 12.5 mm from the centre of the cam.
7. A cam is designed for knife edge follower with the following data cam lift 40 mm during 90°
of cam rotation with SHM, dwell for next 30°, during the next 60° of cam rotation the
followers returns to its original position with SHM , dwell during the remaining 180°. Draw
the profile of the cam when the line of stroke is off set 20 mm from the axis of cam shaft. The
radius of the base circle of the cam is 40 mm.
8. Draw the profile of the cam for operating the exhaust valve of an oil engine. It is required to
give equal uniform acceleration and retardation during opening and closing of the valve each
of which corresponds to 60° of cam rotation. The valve must remain in the fully open
position for 20° of cam rotation. The valve is 37.5 mm and least radius of the cam 40 mm. the
follower is provided with roller of radius 20 mm and the line of stroke passes through the
axis of the cam.
9. A disc cam used for moving a knife edge follower with SHM during lift and uniform
acceleration and retardation motion during return. Cam rotates at 300 rpm clockwise
direction. The line of motion of the follower has an offset 10 mm to the right angle of the
cam shaft axis. The minimum radius if the cam is 30 mm. the lift of the follower is 40 mm.
the cam rotates an angle are lift 60°, dwell 90°, return 120 °and remaining angle for dwell.
Draw the profile of the cam and find maximum velocity and acceleration during the lift and
return.
10. It is required to set out the profile of a cam to give the following motion to the reciprocating
motion follower with flat mushroom contact face.
Follower to have a stroke 20 mm during 120 ° of cam rotation
Follower to dwell for next 30° of cam rotation
Follower to return to its original position during 120 ° of cam rotation
Follower to dwell for the remaining period.
The minimum radius of the cam is 25 mm the outstroke of the Follower is performed with
SHM and return outstroke of the Follower is uniform acceleration and retardation.
11. Draw the cam profile for the following data:
Base circle radius of cam is 50 mm, lift is 40 mm, angle of ascent with cyclodal motion is
60°, angle of dwell is 90°, angle of descent with uniform acceleration and retardation 90°
remaining dwell period. Speed of the cam 300 rpm. Follower off set by 10 mm. type of
follower knife edge.
12. Draw the cam profile for the following data:
Base circle radius of cam is 50 mm, lift is 40 mm, angle of ascent with SHM motion is 90°,
and angle of dwell is 90°, angle of descent with uniform acceleration and retardation 90°

19. remaining dwell period. Speed of the cam 300 rpm. type of follower roller with 10 mm
radius.
13. A cam with 30 mm is minimum diameter is rotating clock wise at uniform speed of 1200 rpm
and has to give the following motion to roller 10 mm in diameter:
The follower to complete stroke of 25 mm during 120° of cam rotation with uniform
acceleration and retardation
Follower is dwell for next 60°
Follower to return to its original position during 900 ° of cam rotation with uniform
acceleration and retardation
Follower is dwell for next 90 °
Draw the profile of cam if the axis of the roller follower is passes through the axis of the cam.
14. A cam drives a flat reciprocating follower in the following manner:
The follower moves outwards through a distance of 20 mm with SHM during the first 120° of
cam rotation
Follower is dwell for next 30°
The follower moves inwards with SHM during the next 120° of cam rotation.
Follower is dwell for remaining period.
Draw the profile of the cam when the minimum radius of the cam is 25 mm. and also
calculate the maximum velocity and acceleration during outstroke and inward motion of the
follower when the cam rotates with 200 rpm.
15. From the following data draw the profile of a cam in which the follower moves with SHM
during ascent while it moves with uniformly accelerated and decelerated motion during
descent.
Least radius of the cam = 50 mm
Angle of ascent = 48°
Angle of dwell = 42°
Angle of descent = 60°
Lift of the follower = 40 mm
Diameter of the roller = 30 mm
If the cam rotates at 360 rpm anticlockwise find the maximum velocity and acceleration of
the follower during descent.

20. UNIT-II GEARS AND GEAR TRAINS
Spur gear – law of toothed gearing
Involute gearing – Interchangeable gears
Interchangeable gears
Gear tooth action interference and undercutting
Nonstandard teeth
Gear trains
Parallel axis gears trains
Epicyclic gear trains
Automotive transmission gear trains
Part A questions
1. Define gear.
Gears are defined as toothed wheels which can transmit power and motion from one shaft
to another shaft by means or successive engagement of teeth.
2. Give advantages of gear drive.
 It can transmit large power with low velocity.
 It has high efficiency.
 It is compact in size.
 It is more reliable than other drives.
3. Give disadvantages of gear drive.
 Manufacturing cost of gear is high.
 Manufacturing process of gears is complicated.
 It requires precise alignment of shafts.
 It requires lubrication system.
4. What are the types of gears ?
1) Parallel shaft axes gears
2) Intersecting shaft axes gears
3) Non-intersecting and perpendicular shaft axes gears
4) Non-intersecting and non-perpendicular shaft axes gears.

21. 5. Give types of parallel shaft axes gears.
1) Spur gears
2) Helical gears
3) Herringbone gears
4) Rack and pinion gears
5) Internal gears
6. Define addendum.
Addendum is defined as the radial distance of a gear tooth from the pitch circle to the top
of gear tooth. It is also defined as the radial height of the gear tooth above the pitch circle.
7. What is backlash ?
Backlash is the difference between the tooth space and the tooth thickness which is
measured along the pitch circle. Theoretically, it should be zero but practically some backlash
must be allowed to prevent jamming of the teeth due to tooth errors and thermal expansion
during manufacturing.
8. Explain circular pitch and dedendum
The circular pitch is defined as the distance measured along the circumference of the
pitch circle from a point on one tooth to the corresponding point on the next tooth.
πD
PC = = π ∙ m
T

22. Dedendum is the radial distance of a gear tooth from the pitch circle to the bottom of the
gear tooth. It is also defined as the radial depth of the gear tooth below the pitch circle.
9. What is conjugate action?
When a pair of mating gear teeth act against each other, rotary motion is produced which
is transmitted from the driving gear to the driven gear. These gears have tooth profiles which are
so arranged that a constant angular velocity ratio is produced and maintained while meshing.
These gears are said to have conjugate action and the corresponding tooth profiles are said to
have conjugate profiles.
10. What is conjugate teeth?
If any arbitrary shape of the tooth is selected for tooth profile of one of the two gears in
mesh, the profile of the other gear may be found to satisfy the law of gearing. Such teeth are
called as conjugate teeth.
11. Why conjugate teeth are not used generally?
The conjugate teeth profile will transmit the desired motion but objection to such
arbitrarily selected profiles is the difficulty of manufacturing, production, Cost and
standardization. Hence conjugate teeth are not used generally.
12. Differentiate between involute tooth profile and cycloidal tooth profile.
Sl.
No.
Involute tooth profile Cycloidal toooth profile
1.
The profile of involute gears is
the single curvature
The profile of cycloidal gears is the double
curvature i.e., combination of epicycloid and
hypo-cycloid.
2.
Due to single curvature profile
its manufacturing is easy.
Due to double curvature profile its manufacturing
is difficult.
13. Define path of contact.
Path of contact is defined as the path traced by the point of contact of two teeth from the
beginning to the end of engagement
14. What is length of path of contact?
The length of path of contact is the length of the common normal cut-off by the
addendum circles of the two gears.

23. 15. Define arc of contact. [Dec. 2012]
Arc of contact is defined as the path traced by the point on a pitch circle from start to the
end of engagement of a given pair of teeth.
16. What is interference?
The phenomenon when the tip of tooth undercuts the root on its matting gear is known as
interference. It will occurs on mating of two non-involute or non-conjugate tooth profiles.
17. What is undercutting?
If the gear of pinion is cut with a standard tool like gear hob which will also interfere
with the portion of the tooth below the base circle and will cut away the interfering material, then
this process is called as undercutting.
18. What are the methods to avoid interference and undercutting ?[May
2014]
1) Modified involute profile of tooth.
2) Modified addendum of the gear and pinion
3) Increased centre distance.
19. What are the types of standard tooth profile?
1) 14. 5 ° composite system.
2) 14. S° full depth involute system
3) 20 ° full depth involute system
4) 20° stab involute system
20. What is the significance of contact ratio in gears? [Dec. 2010]
The ratio of the length of arc of' contact to the circular pitch is known as contact ratio.
For smooth and continuous operation contact ratio must be as high as possible. If the contact
ratio is 1.5 it means there are alternately one pair and two pairs of teeth in contact and on a time
basis average is 1. 5.
21. Define the following terms used in gear. [Dec. 2011, May
2012]
a) Pressure angle b) Module
a) Pressure angle:
It is the angle between the common normal drawn at the point of contact of the mating
gears and the common tangent at the pitch point. The standard values of pressure angles are 14.
5° and 20°.

24. b) Module:
It is the ratio of pitch circle diameter in mm to the number of teeth of gear.
m =
PCD of gear
Number or teeth on gear
= D mm
T

25. 22 . List down the common forms of teeth. [Dec. 2012, May 2013]
The common forms of teeth are :
1) Cycloidal profile teeth
2) Involute profile teeth
23.What is the condition stated by law of gearing? [ May 2013, May 2014]
The condition stated by law of gearing is "The common normal at the point of contact
between a pair of teeth must always pass through the pitch point which is the fundamental
condition that must be satisfied while designing the profiles for the teeth of gear".
24. Define gear ratio. [ Dec. 2013]
The gear ratio is defined as the ratio of pinion speed to the gear speed. It is also defined
as the ratio of number of teeth on gear to the number of teeth on pinion or it is the ratio of
diameter of gear to the diameter of pinion.
25. What is a gear train? & applications of gear train?
Gear train is combination of two or more gears which is used to transmit motion from
one shaft to another shaft.
Applications of gear train
1) Automobiles 2) Clocks 3) Strips
26. What are the types of gear train?
1) Simple gear train 2) Compound gear train
3) Reverted gear train 4) Epicyclic gear train
27.What is the advantage of compound gear train over simple gear train?
The main advantage of compound gear train over simple gear train is that a higher speed
reduction from the first shaft to last shaft can be obtained by using small gears. Hence compound
gear train requires less space as compared to simple gear train for the given speed ratio.
28. What is epicyclic gear train?.& advantage

26. A gear train having relative motion of axes is called a planetary or an epicyclic gear train.
In an epicyclic gear train, the axis of at least one of the gears also moves relative to the frame.
Example-hoists.
1) It can be used to transmit high velocity ratio
2) They are compact.
29. What is the use of differential used in automobile?
Use of differential in automobile
1) To compensate for the difference in distance that the outer wheel travels while the
vehicle is taking a turn.
2) To avoid skidding.
30. What are the role of' idlers in gear trains?
1) To connect gears where a large centre distance is required idlers are used.
2) Idlers are used to obtain the desired direction or motion of the driven gear
31. Write short note on differentials.
Differentials is the application of the epicyclic gear train with bevel gear. The function of
a differential gear of an automobile is to transmit the motion from engine to rear wheels and
rotate the rear wheels at different speeds while the automobile is taking a turn.
PART B QUESTIONS
1. Two mating gears have 20 and 40 involutes 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 wheel, length of path of contact, arc of contact and contact ratio.
2. A pairs of gears having 40 and 20 teeth respectively are rotating in mesh the speed of the
smaller being 2000 rpm. Determine the velocity of sliding between the gear teeth face at the
point of engagement, at the point of pitch and the point of disengagement if the smaller gear
is a driver. Assume that the gear teeth are 20° involute form, addendum length is 5 mm and
module is 5 mm
3. A pinion having 18 teeth engages with an internal gear of 72 teeth. If the gear having involute
profile teeth with 20° pressure angle, module of 4 mm and the addendam on pinion and gear
are 8.5 mm and 6.5 mm respectively , find the length of path of contact. arc of contact and
contact ratio.
4. Two spur gears of 24 and 36 teeth of 8 mm module and 20° pressure angle are in mesh.
Addendum of each gear is 7.5 mm. the teeth of involutes form. Determine (i).the angle

27. through which the pinion turns while any pair of teeth are in contact. (ii). the velocity of
sliding between the teeth when the contact on the pinion is at a radius of 102 mm. the speed
of the pinion is 450 rpm.
5. The following data relate to pair of 20° pressure angle is in mesh,
Module is 6 mm, number of teeth on pinion and gear are 17 and 49 respectively and
addendum on gear and pinion is one module.
Determine
 The number of pairs of teeth in contact
 The angle through which the gear turns while any pair of teeth are in contact.
 The ratio of sliding velocity to rolling velocity at the point of engagement.
6. Two involutes gears of 20° pressure angle are in mesh. The number of teeth on pinion is 20
and gear ratio is 2. If the pitch expressed in module is 5 mm and pitch line speed is 1.2 m/s
assuming addendum as standard and equal to one module. Find
 The angle turned through by pinion when one pair of teeth is in mesh
 The maximum velocity of sliding.
7. Two gear wheels mesh externally and are give the velocity ratio of 3 to 1. The teeth are
involute form; module = 6 mm, addendum = one module, pressure angle =20°. The pinion
rotates 90 rpm . determine
 The number of teeth on pinion to avoid the interference on it and the corresponding
number of teeth on gear wheel
 The length of path and arc of contact
 The number of pairs of teeth in contact
 The maximum velocity of sliding
8. A single reduction gear of 120kw with a pinion 250mm pitch circle diameter & speed
650 rpm is supported in bearing on either side. Calculate the total load due to power
transmitted, the pressure angle being
9. The number of teeth on each of the two equal spur gears in mesh are 40 .The teeth have
20° involute profile & the module is 6mm . If the arc of contact is 1.75 times the circular
pitch, find the addendum.
10. A pinion having 30 teeth drives a gear having 80 teeth.The profile of the gears is involute
with 20° pressure angle , 12mm module & 10 mm addendum. Find the length of path of
contact , arc of contact & the contact ratio.

28. 11. Two involute gears of 20° pressure angle are in mesh. The number of teeth on pinion is
20 & the gear ratio is 2. If the pitch expressed in module is 5mm & the pitch line speed is
1.2m/s, assuming addendum as standard & equal to one module, find:
(1) The angle turned through by pinion when one pair of teeth is in mesh;&
(2). The maximum velocity of sliding
12. A pair of gears, having 40 & 20 teeth respectively, arc rotating in mesh, the speed of the
smaller being 2000 rpm. Determine the velocity of sliding between the gear teeth faces at
the point of engagement, at the pitch point, and at the point of disengagement if the
smaller gear is the driver. Assume that the gear teeth are 20° involute form, addendum
length is 5 mm & the module is 5mm. Also find the angle through which the pinion turns
while any pairs of teeth are in contact.
13. The following data relate to pair of 20° involute gears in mesh; module =6mm, 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 in contact, and (3). The ratio of
sliding to rolling motion when the tip of a tooth on the larger wheel (a)is just making
contact, (b) is just leaving contact with its mating tooth and (c) is at the pitch point.
14. A pinion having 18 teeth engages with an internal gear having 72 teeth. If the gears have
involute profiled teeth with 20° pressure angle, module of 4mm & the addenda on pinion
and gear are 8.5 mm & 3.5 mm respectively, find the length of path of contact.
15. 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

29. 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, &
contact ratio.
16. Determine the minimum number of teeth required on a pinion, in order to avoid
interference which is to gear with, (1). A wheel to give a gear ratio of 3 to 1 ;& (2).an
equal wheel..The pressure angle is 20° and a standard addendum of 1 module for the
wheel may be assumed.
17. A pair of spur gears with involute teeth is to give a gear ratio of 4:1. The arc of approach
is not to be less than the circular pitch and smaller wheel is the driver. The angle of
pressure is 14.5°. find(1).the least no of teeth that can be used on each wheel, and (2).the
addendum of the wheel in terms of the circular pitch?
18. A pair of involute spur gears with 16° pressure angle and pitch of the module 6mm is in
mesh. The number of teeth on pinion is 16 & its rotational speed is 240 rpm. When the
gear ratio is 1.75, find in order that the interference is just avoided;(1). The addenda on
pinion and gear wheel;(2)the length of path of contact ;and (3).the maximum velocity of
sliding of teeth on either side of the pitch point.
19. A pair of 20° full depth involute spur gears having 30 and 50 teeth respictively of module
4mm are in mesh. The smaller gear rotates at 1000rpm. Determine (1).sliding velocities
at engagement and at disengagement of pair a teeth,and (2).contact ratio.
20. Two gears wheels mesh externally and are to give a velocity ratio of 3 to 1.the teeth are
of involute form; module=6mm, addendum=one module, pressure angle =20°.the pinion
rotates at 90 rpm. 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.
21. A pinion of 20 involute teeth and 125 mm pitch circle diameter drives a rack. The
addendum of both pinion and rack is 6.25mm. what is the least pressure angle which can
be used to avoid interference? With this pressure angle,find the length of the arc of
contact and the minimum number of teeth in contact at a time
22. A pair of spiral gears is required to connect two shafts 175mm apart, the shaft angle

30. being 70°. The velocity ratio is to be 1.5 to 1, the faster wheel having 80 teeth &a pitch
circle diameter of 100 mm. find the spiral angles for each wheel. If the torque on the
faster wheel is 75N-m; find the axial thrust on each shaft neglecting friction.
23. In a spiral gear drive connecting two shafts, the approximate center distance is 400 mm
and the speed ratio=3. The angle between the two shafts is 50° and the normal pitch is
18mm. the spiral angle for the driving and driven wheels are equal. Find (1).no of teeth
on each wheel, (2).exact center distance, and (3). Eficiency of the drive,if friction angle
=6°
24. A drive on a machine tool is to be made by two spiral gear wheels, the spirals of which
are of the same hand and has normal pitch of 12.5mm. the wheels are of equal diameter
and the center distance between the axes of the shafts is approximately 134 mm. the angle
between the shafts is 80° and the speed ratio 1.25. determine(1). The spiral angle of each
wheel, (2).the number of teeth on each wheel (3). The efficiency of the drive, if the
friction angle is 6°, and (4).the maximum efficiency..
25. The pitch circle diameter of the smaller of the two spur wheels which mesh extrenally
and have involute teeth is 100 mm. The number of teeth are 16 and 32. The pressure
angle is 20° and the addendum is 0.32 of the circular pitch. Find the length of the path of
contact of pair of teeth.
26. A pair of gears, having 40 and 30 teeth respectively are of 25° Involute form. The
addendum length is5mm and the module pitch is 2.5mm. If the smaller wheel is the
driver and rotates at 1500 rpm, find the velocity of sliding at the point of engagement and
at the point of disengagement.
27. Two gears of module 4mm have 24 and 33 teeth. The pressure angle is 20° and each gear
has a standard addendum of one module. Find the length of arc of contact and the
maximum velocity of sliding if the pinion rotates at 120rpm
28. The number of teeth in gears 1 and 2 are 60 and 40; module=3mm,
pressure angle = 20°,and addendum=0.318 of the circular pitch . determine the velocity
of the sliding when the contact is at the tip of the teeth of gear 2 and the gear 2 rotates at
800 rpm.
29. Two spur gears of 24 teeth and 36 teeth of 8mm module and 20°pressure angle are in
mesh .addendum of each gear is 7.5 mm. The teeth are of involute form . determine

31. (1).the angle through which the pinion turns while any pair of teeth are in contact,and (2)
the velocity of sliding between the teeth when the contact on the pinion is at a radius of
102 mm. the speed of the pinion is 450 rpm.
30. A pinion having 20 involute teeth of nodule pitch 6mm rotates at 200 rpm. And transmits
1.5kw to a gear wheel having 50 teeth . The addendum on both wheel is ¼ of the circular
pitch . the angle of obliquity is 20°. Find (1) the length of path of approach;(2) the length
of arc of contact,(3). The normal force between the teeth at an instant where is only pair
of teeth in contact
31. Two mating involute spur gears of 20° pressure angle have a gear ratio of 2 . The number
of teeth on the pinion is 20 and its speed is 250 rpm . The module pitch of the teeth is
12mm. if the addendum on each wheel is such that the path of approach and the path of
recess on each side are half the maximum possible length , find (1). The addendum for
pinion and gear wheel ;(2).the length of arc of contact ; (3)the maximum velocity of
sliding during approach and recess.
32. Two mating gears have 20 and 40 involute teeth of module 10mm and 20° pressure
angle. If the addendum on each wheel is such that the path of contact is maximum and
interference is just avoided, find the addendum for each gear wheel , path of contact , arc
of contact and contact ratio.
33. A 20° involute pinion with 20 teeth drives a gear having 60 teeth. Module is 8 mm and
addendum of each gear is 10 mm.(1)state whether interference occurs or not, (2).find the
length of path of approach and arc of approach if pinion is the driver.
34. A pair of spur wheel with involve teeth is to give a gear ratio of 3 to 1.The arc of
approach is not to be less than the circular pitch and the smaller wheel is the driver . The
pressure angle is 20°.What is the least number of teeth that can be used on each wheel?
What is the addendum of the wheel in terms of the circular pitch?
35. Two gears wheels mesh externally and are to give a velocity ratio of 3 . the teeth are of
involute form of module 6. The standard addendum is 1 module. If the pressure angle is
18° and pinion rotates at 90 rpm. Find (1). The number of teeth on each wheel so that the
interference is just avoided,(2). The length of the path of contact and (3) the maximum
velocity of sliding between the teeth.

32. 36. A pinion with 24 involute teeth of 150 mm of pitch circle diameter drives a rack . The
addendum of the pinion and rack is 6 mm. Find the least pressure angle which can be
used if under cutting of the teeth is to be avoided. Using this pressure angle, find the
length of the arc of contact and the minimum number of teeth in contact at one time.
37. In a epicyclic gear train, an arm carries two gears A & B having 36 and 45 teeth
respectively. If the arm rotates at 150 rpm in anticlockwise direction about the center of
the gear A which is fixed , determine the speed of the gear B . if the gear A instead of
being fixed, makes 300 rpm in the clockwise direction, what will be the speed of gear B?
38. In a reverted epicyclic gear train, the arm A carries two gears B and C and a compound
gear D-E. The gear B meshes with gear E and the gear C meshes with gear D. the number
of teeth on gears B,C and D are 75,30,& 90 respectively. Find the speed and direction of
gear C when gear B is fixed and the arm A makes 100 rpm clockwise.
39. In an epicyclic gear train, the internal wheels A&B and compound wheels C&D rotate
independently about axis O. the wheels E&F rotate on pins fixed to the arm G .E gears
with A& C and F gears with B&D. all the wheels have the same module and the number
of teeth are : Tc =28: Td =26; Te= Tf =18. (1) sketch the arrangement;(2) find the no of
teeth on A& B;(3) if the arm G makes 100 rpm clockwise and A is fixed find the speed of
B(4) if the arm G makes 100 rpm clockwise ,and wheel A makes 10 rpm counter
clockwise ; find the speed of wheel B.
40. Two shafts A&B are co axial. A gear C (50 teeth) is rigidly mounted on the shaft A. a
compound gear D-E gears with an internal gear G .D has 20 teeth and gears with C and E
has 35 teeth and gears with an internal gear G. The gear G is fixed and is concentric with
the shaft axis . the compound gear D-E is mounted on a pin which projects from an arm
keyed to the shaft B. sketch the arrangement and find the number of teeth on internal gear
G assuming that all gears have the same module. If the shaft A rotates at 110 rpm. Find
the speed of shaft B.
41. An internal wheel B with 80 teeth is keyed to a shaft F. a fixed internal wheel C with 82
teeth is concentric with B . a compound wheel D-Egears with the two internal wheels ; D
has 28 teeth and gears with C while E gears with B. the compound wheels revolve freely
on a pin which projects from a disc keyed to a shaft A cooxial with F. If the wheels have
the same pitch and the shaft A makes the same pitch and the shaft A makes 800 rpm.,
what is the speed of the shaft F? sketch the arrangement.
42. A compound epicyclic gear train in fig. the gears A, D and E are free to rotate on the axis
P. the compound gear B & C rotate together on the axis Q at the end of the arm F. all gear
s are equal pitch. The number of external gear teeth on the gears A, B and C are 18, 45

33. and 21 respectively. The gears D and E are annular gears. The gear A rotates at 100 rpm in
CCW and D rotates at 450 rpm CW. Find the speed and direction of the arm and gear E.
43. In an epicyclic gear train of Sun and Planet type as show in fig. the pitch circle diameter of
internal toothed ring D is to be 216 mm and module 4 mm. when the ring D is stationary. The
spider A which carries three planet wheels C of equal in size is to be make one revolution in
the same sense as a sun wheel B for every five revolutions of the driving spindle carrying the
sun wheel B. Det. The suitable number of teeth for all wheels.
UNIT-III FRICTION IN MACHINE ELEMENTS
TWO MARKS Q&A

34. 1. What is drive?
The mechanism used to transmit power and speed from the prime-mover to the driven machine is
called as drive.
2. What are the types of friction?
According to the nature of surface
a) Static friction b) Dynamic friction
According to the condition of surface
a) Dry friction b) Greasy friction c) Film friction.
3. What is frictional devices ? Give examples.
There are devices in which the friction is desirable and efforts are made maximize it, is called as
frictional devices. Example-1) Belts 2) Ropes 3) Brakes 4) Clutches
4. What is dry friction?
When there is relative motion between two completely un lubricated surfaces, then the friction
between them is called as dry friction.
Types of dry friction: i) Sliding friction (ii) Rolling friction.
5. Define film friction & Define greasy friction
When the two surfaces in contact are completely separated by a thick layer of lubricant and the
friction occurs due to the resistance to relative motion between the surfaces and lubricant, then the
friction is called as film or viscous or fluid friction.
When a very thin layer of lubricant is interposed between two contacting surfaces, then the
friction between them is known as greasy or skin or boundary friction.
6. Define sliding friction & rolling friction
If the two surfaces have sliding motion with respect to each other, the friction between them is
called as sliding or solid friction. Example-Nut and bolt.
If the two surfaces have rolling motion with respect to each other, the friction between them is
called as rolling friction. Example-Ball and roller bearings.
7. Define helix angle.
Helix angle is defined as angle made by helix of the thread with a plane perpendicular to the
axis of the screw.
8. What are the functions of the clutch?
1) when clutch is engaged, the clutch transmits maximum power from engine crankshaft
gearbox input shaft.
2) When clutch is engaging (clutch pedal position-moving up), the clutch accommodates
minor slippages and hence provides smooth drive transmission without jerks.
9. Why shall self locking screw have lesser efficiency?
The self locking screws require friction in between the thread surface the nut. Thus it requires
more effort to lift the body. For this reason the self locking screw have lesser efficiency.

35. 10. What is meant by friction clutches?
Friction clutches are the clutches that work on the friction principle that when two independent
disc have relative motion between them, friction is caused.
11. Give advantages & disadvantages of single plate clutch.
Advantages of single plate clutch are
 Simple in design, construction and working.
 Better heat dissipation from single plate.
 Gear changing with single plate clutch is easier.
 It has better torsional vibration absorbing capacity.
Disadvantages of single plate clutch are
 For higher power transmission, the surface area of clutch plate increases and
thereby increasing the overall size of clutch.
 Clutch pedal force required is high.
12. Give advantages & disadvantage of multi-plate clutch.
Advantages :
1)The overall size of clutch is smaller.
2) It has higher torque transmitting capacity.
3) Drive transmission smoother.
4) Wear and tear of clutch is lower.
Disadvantages
1) The design or clutch-plate set is complicated.
2) It is difficult to service.
3) The cost or rnulti-plate clutch is higher.
13. Give applications of multi-plate clutch.
It is used where high torque transmission is required such as racing car's etc.
It is also used where overall space is constrained such as scooters, motorbikes, etc.
14. What are the significance of friction with regard to power
transmission devices like clutches and bearings?
The power transmission devices like clutches and bearings work on the principle of friction.
When two friction surfaces are brought in contact with each other and pressed, they are united due to
the friction between them.
15. Explain positive clutches.
These clutches are used when positive drive is required. The simplest type of positive clutch is
Jaw clutch which permits one shaft to drive another shaft through a direct contact between interlocking
jaws. These type of clutches are used in sprocket wheels, gears, pulleys, etc.
16. What are the desirable properties of belt materials?
 High coefficient of friction to transmit the power from one pulley to another.
 High tensile strength to avoid tearing.
 High wear resistance and durability.

36.  High flexibility
17. What is meant by angle of contact (Lap angle)?
It is the angle made by a common normal drawn to the tangent line at the point of engagement
and at the point of disengagement of the belt on a pulley, at its centre
18. Give advantages of flat belt.
 Flat belts are easy to produce, hence have low cost.
 Flat belt are simple to design and manufacture.
 They have higher efficiency.
 Can be operated in dusty and abrasive atmosphere.
19. Give disadvantages of flat belt.
 Flat belts have limited frictional contact and hence have lower power transmitting
capacity.
 Used for limited speed reduction up to 4: 1.
 They are restricted to single belt usage
20. What are the disadvantages of V-belt drive over flat belt?
 V-belt cannot be used for 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
timing devices.
 It is a costlier system
21. What is a brake?
A brake is a device with the help of which artificial frictional resistance is applied to a moving
machine or member, in order to stop or retard the motion of a body.
22. What is the difference between clutch and brake?
The main difference between clutch and brake is that, clutch is used to keep the driving and
driven member moving together whereas brake is used to stop or to control the speed of moving
member.
23. List out commonly used breaks.
 Hydraulic brakes, eg. Pumps.
 Electric brakes, eg. Eddy current brakes.
 Mechanical brakes, eg. Axial brakes.
24. Give desirable characteristics of brake lining material.
 A high and uniform co-efficient of friction.
 Should withstand high temperatures.
 Should have high resistance to wear.
 Adequate mechanical and thermal strengths.
26. Mention the types of cluthes
Types of friction clutches.:
a) Disc Or plate clutches b) Cone clutches
c) Semi-centrifugal clutches d)Centrifugal clutches

37. UNIT-III FRICTION IN MACHINE ELEMENTS
PART B QUESTIONS
1. A single dry plate clutch transmits 7.5 kW power at 900 rpm. The axial pressure limited to
0.07 N/mm2
if the co efficient of friction is 0.25. Find the mean radius and face width of friction
lining assuming the ratio of mean radius to face width as 4 and outer and inner radii of the clutch
plate
2. A single plate clutch with both sides effective has outer and inner radius 150 mm and 100 mm
respectively. The maximum pressure intensity at the any point is not exceeds to 0.1 N/mm2
. The co
efficient of friction is 0.3. Det. The power transmitted by the clutch at a speed of 2500 rpm for both
uniform wear and uniform pressure theory.
3. A friction clutch multi plate is meant for transmitting a power of 55 kW at 1600 rpm. The co
efficient of friction is 0.1. Actual intensity pressure not exceeds to 0.16 N/mm2
. The external radius
is 125mm and is 1.25 times of internal radius. Det. the number of plate needed to transmits the
required torque.
4. The pitch of 50 mm mean diameter threaded screw of screw jack is 12.5 mm. The co efficient of
friction for screw is 0.13. Det. torque required on the screw to raise a load of 25 kN. Assuming load
rotates with screw. Det. the ratio of torque required to raise and lowered the load. And efficiency of
the machine
5. A load of 10 kN is raised by means of screw jack, having a square threaded screw of 12 mm pitch
and mean diameter 50 mm If the force of 100 N is applied at the end of the lever to rise the load
what should be the length of the lever used? The co efficient of friction is 0.15. What is the
mechanical advantage obtained?.
6. The following data is required to a screw jack the pitch of the thread screw is 8 mm and diameter is
40 mm. The co efficient of friction is 0.1 (screw and nut) and load 20 kN. Assuming load rotates
with screw. Det. the ratio of torque required to rise and lower and efficiency of the machine
7. A screw jack of square thread of mean dia 60 mm and pitch 8 mm. the co efficient of friction at
screw thread 0.09. a load of 3 kN is used to lift 120 mm. Determine a torque required at work done
in lifting load through 120 mm. find also efficiency of screw jack.
8. The mean diameter of screw jack having pith 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 screw jack when
Load rotes with screw jack and
The load rest on the loose head with does not rotate with the screw.
The external and internal diameter of bearing surface are 60 mm and 10 mm respectively. The co
efficient of friction for screw as well as bearing surface may be taken as0.08.
9. A plate clutch has three discs on the driving side and two on driven side. The outside and inside
diameter is 240 mm and 120 mm respectively. Assuming uniform pressure and µ is
0.3. Find the total spring load passing plate together to transmit 285 k W at 1575 rpm.
If there are 6 springs each of stiffness 13 k N/m and each of the contact surface has worn away by
1.25 mm, find the maximum power that can be transmitted assuming uniform wea
UNIT IV
FORCE ANALYSIS

38. Two marks
questions
1. State D’Alembert’s principle. (NOV 2003, NOV 2006 , MAY 2016, NOV 16 , 15 )
D’Alembert’s principle states that the inertia forces and torques, and the external forces and
torquesacting together result in static equilibrium
2. Define Piston effort & crank effort? (APR/MAY 17, MAY/JUN 2016, NOV/DEC
2012)
 Piston effort:Piston effort is defined as the net or effective force applied on the piston, along
the lineof stroke. It is also known as effective driving force (or) net load on the gudgeon pin.
 Crank effort:Crank effort is the net effort ( force ) applied at the crank pin perpendicular to
the crank
, which gives the required turning moment on the crankshaft.
3. What is meant by crank pin effort? (APR/MAY 18, NOV/DEC 2018)
The component of force on the connecting rod perpendicular to the crank is known as crank pin
effort.
FT = sin( + ∅)
cos ∅
4. Define coefficient of fluctuation of energy(MAY / JUNE 2006 , NOV / DEC
2006, NOV / DEC 2003 ,NOV/DEC 2016 ,15 , MAY/JUNE 16)
It is the ratio between max fluctuation of energy and the work done per cycle.
CE =
Maximum Fluctuation of energy
Work done per cycle
5. What are the conditions for a body to be in static and dynamic equilibrium?
(APR 2004 , NOV / DEC2005 , MAY 2006 , NOV 2007)
 Necessary and sufficient conditions for static and dynamic equilibrium are
1. Vector sum of all forces acting on a body is zero.
2. The vector sum of the moments of all forces acting about any arbitrary point or axis is zero.
 First condition is the sufficient condition for static equilibrium together with second condition is
necessary for dynamic equilibrium.
6.Define inertia and inertia force. (NOV/DEC 2016)
 Inertia:
Applied and Constrained Forces
Free body diagrams
Static Equilibrium conditions
Two, Three and four members
Static Force analysis in simple machine members
Dynamic Force Analysis
Inertia Forces and Inertia Torque
D’Alembert’s principle
Superposition principle – dynamic Force Analysis in simple machine
members

39. Thepropertyofmatterofferingresistancetoanychangeofitsstateofrestorof
uniformmotioninastraightlineis knownasinertia.
 Inertia force:The inertia force is an imaginary force, which when acts upon a rigid body,
brings it inan equilibrium position.
Inertia force = - Accelerating force = - ma
7.What is the purpose of flywheel used in an engine?(NOV/DEC 2016 ,14)
The purpose of flywheel is to reduce the fluctuations of speed caused by the fluctuation of
the engineturning moment during each cycle of operation.
In both forging and pressing operations, flywheels are required to control the variations in speed
duringeach cycle of an engine.
9 . What is engine shaking force ?( MAY / JUNE 2009, NOV/DEC 13 )
The force produces in an engine due to the mass of the piston , and mass of the connecting rod
is calledengine shaking force.
10. Define the significance of inertia force analysis ?(APR / MAY 2018,NOV /
DEC 2009 )
Inertia force analysis reduces the dynamic analysis problem into an equivalent static analysis
problemby determining the required torque and the direction,
11.Define Inertia force & Inertia torque ? ( NOV / DEC 2010 , MAY/JUNE 16 )
 Inertia force:It is an imaginary force, which when acts upon a rigid body brings
it in anequilibrium position.
Inertia force = - Accelerating force = - m.a
 Inertia torque:It is an imaginary torque, which when applied upon the rigid body,
brings it inequilibrium position. It is equal to the accelerating couple in magnitude but
opposite in direction.
12.In a two force planar member , specify the conditions for static equilibrium?
(NOV / DEC 2010)
i. The forces are of the same magnitude
ii. The forces act along the same line
iii. The forces are in opposite directions
13 State the principle of super position. (NOV / DEC 2010)
The principle of super position states that for linear systems the individual responses to several
disturbances or driving functions can be superposed on each other to obtain the total response of the
system
14. Differentiate the functions of flywheel and governor. ( NOV / DEC 2012 )
S. No Flywheels Governors
1 The function of flywheel is to reduce the Its function is to control the mean speed over
fluctuations of speed during a cycle above a period for output load variations
and below the mean value for constant
load from the prime mover.
2 It works continuously from cycle to cycle. Its works intermittently i.e. only when there is
change in the load.
3 It has no influence on mean speed of the It has no influence over cyclic speed
prime mover. fluctuations.

40. 16. Define coefficient of fluctuation of speed.( MAY / JUNE 2006, NOV/DEC 2018)
The ratio of maximum fluctuation of speed to the mean speed is called as coefficient of
maximumfluctuation of speed.
CS = 1− 2
=
2( 1− 2)
1+ 2
Where,N1– Max Speed
N2 -Min Speed
17.List the uses of turning moment diagram (APR/MAY 17)
 To determine the work done per cycle and power developed
 To determine the mean torque and the fluctuation of energy
 To find the diameter of the crankshaft
 To design the flywheel
18.Statethe principle of virtual work. (AUA/M2015)
The principle of virtual work can be stated as the work done during a virtual displacement
from theequilibrium is equal to zero.
18. Define maximum fluctuation of energy. (MAY/JUNE 16 ,NOV/DEC 16)
The difference between maximum and minimum energies during a cycle is called as
maximumfluctuation of speed.
ΔE = Max energy – Min Energy.
19. What is freebodydiagram? (AU2005, AUM/J2009)
A free body diagram is a sketch of the isolated or free body which shows all the Pertinent
weight forces, the externally applied loads ,and there action from its supports and connections acting
upon it by the removed elements.
20.Distinguish between static force and inertia force(MAY/JUNE 16)
 While analyzing the mechanism, if mass of the body and inertia force are not
considered, then itis called static force.
 Inertia force is a fictitious force which when acts upon a rigid body, brings it in equilibrium.
21. Distinguish between radial cam and cylindrical cam.(APR/MAY 10)
 In radial cams, the follower reciprocates or oscillates in a direction perpendicular to
the camaxis.
 In cylindrical cams, a cylinder which has a circumferential contour cut in the surface,
rotates about its axis. The follower reciprocates or oscillates in a direction perpendicular
to the cam axis.
22.Explain surge and windup.(NOVDEC 08)
 Spring surge: Spring surge means vibration of the retaining spring.
 Windup: Twisting effect produced in the camshaft during the raise of heavy load
follower iscalled as windup.
23.Why multi cylinder needs small flywheel?(APR/MAY 08)
In a multi cylinder engine there are one or more number of power strokes per revolution of
the crankshaft so lesser energy is stored in the flywheel. Hence a smaller flywheel is sufficient.
24.Define applied force and constraint force. ( NOV / DEC 2004 , NOV / DEC
2005)
 The external force acting on a system of body from outside the system is called applied
force. Theapplied forces are classified as active and reactive force
 The constraint forces are the forces existing internally within the body

41. 25. List out few machines in which flywheel are used.
Punching Machines B) Shearing Machines C) Rivetting Machines D) Crushing Machines.
PART B QUESTIONS
1. In a four link mechanism as in Fig, torques T3 and T4 have magnitudes of 30 Nm and 20 Nm
respectively. The link lengths are AD = 800 mm, AB = 300 mm, BC = 700 mm and CD = 400 mm. For
the static equilibrium of the mechanism, determine the required input torque T2.(NOV/DEC 17)

42. 2.Refer Fig. Determine the couple on crank2 to be applied for equilibrium of the system, when a force
of 500Nacts on the connecting rod at point C as shown. Also determine the resultant of forces exerted of
forces exerted on the frame of the engine.
(AUA/M2015)
FORCE ANALYSIS RECIPROCATING ENGINE
1. The crank pin radius of a horizontal engine is 300mm. The mass of the reciprocating parts is 250kg.
When the crank has travelled 60⁰ from I.D.C., the difference between the driving and the back pressures is
0.35N/mm2
. The connecting rod length between centers is 1.2m and the cylinder bore is 0.5m. If the
engine runs at 250 r.p.m., and if the effect of piton rod is neglected, calculate the pressure on the slide
bars, the thrust in the connecting rod, the tangential force on the crank pin and the turning moment on the
crank shaft. (APR/MAY 17)(APR/MAY 18) case study question.

43. 2.A horizontal steam engine running at120r.p.m.hasa bore of 250mm and a stroke of 400mm.The
connecting radius 0.6m and mass of the reciprocating partsis60 kg. When the crank has turned through an
angle of 45° from the inner dead centre ,the steam pressure on the cover end side is 550 kN/m2and that
on the crank end side is 70 kN/m2. Considering the diameter of the piston rod equal to50mm,
determine:1.Turning moment on the crank shaft,2.Thrust on the bearings, and 3. Acceleration of the
flywheel, if the power of the engine is20kW,mass of the flywheel 60kg and radius of gyration 0.6m.
(AUM/J2014) (APR/MAY 18) case study question.
3. The length of crank and connecting rod of a horizontal engine are 200mm and 1m respectively. The
crank is rotating at 400 rpm. When the crank has turned through 30° from the inner dead center, the
difference of pressure between cover and piston rod is 0.4 N/mm2
. If the mass of the reciprocating parts
is 100kg cylinder bore is 0.4m, then calculate, the inertia force, force on the piston, piston effort, thrust
on the sides of the cylinder walls, the thrust in the connecting rod, and the crank effort. (NOV/DEC 17)
4. The crank and connecting rod of a vertical single cylinder gas engine running at 1800 rpm are 60mm
and 240mm respectively. The diameter of the piston is 80mm and the mass of the reciprocating parts is
1.2kg. At a point during the power stroke when the piston has moved 20mm from the top dead center,
the pressure on the piston is 800 kN/m2
, Determine the (i) Net force on the piston (ii) Thrust in the
connecting rod (iii) Thrust on the sides of the cylinder walls (iv) The engine speed at which the above
values are zero. (NOV/DEC 16)
5. During a trail on steam engine, it is found that the acceleration of the piston is 36 m/s2
when the crank
has moved 30 from the inner dead center position. The net effective steam pressure on the piston is 0.5
MPa and the frictional resistance is equivalent to a force of 600 N. The diameter of the piston is 300mm
and the mass of the reciprocating parts is 180 kg. If the length of the crank is 30mm and the ratio of the
connecting rod length to the crank length is 4.5. Find (i) reaction on the guide bars (ii) thrust on the crank
shaft bearings and (iii) Turning moment on the crank shaft.(NOV/DEC 15)
6. The lengths of the crank and connecting rod of a reciprocating engine are 300mm and 1.5m
respectively. The crank is rotating clockwise at a speed of 120 r.p.m. The mass of connecting rod is
250kg and the distance of center of gravity of the rod from the crank pin center is 475mm. The radius of
gyration of the rod about center of gravity is 625mm. When the crank position is 40⁰ from the inner dead
center then find by graphical method and analytical method (i) Magnitude, position and direction of
inertia force due to the mass of the connecting rod (ii) Torque exerted on the crank shaft in magnitude
and direction. Take the mass of the reciprocating parts = 290kg. (MAY/JUNE 16)
7.The length of crank and connecting rod of a horizontal reciprocating engine are100mm and500 mm
respectively. The crank is rotating at400rpm.When the crank has turned30°from the IDC, find
analytically
1. Velocity of piston
2. Acceleration of piston
3. Angular velocity of connecting rod
4. Angular acceleration of connecting rod. (AUA/M2015)

44. 8.The lengths of crank and connecting rod of horizontal steam engine are 300mm and1.2m respectively
.When the crank has moved 30°from the inner dead center, the acceleration of piston is35m/s2.The
average frictional resistance to the motion of piston is equivalent to a force of 550N and net effective
steam pressure on piston is 500kN/m2.The diameter of piston is 0.3m and mass of reciprocating parts
is160kg.Determine(i)Reaction on the cross-head guides;(ii)Thrust on the crankshaft bearings; and (iii)
Torque on the crank shaft. (AUN/D2010)
9. A single cylinder vertical engine has a bore of 300mm andastrokeof400mm.Theconnecting rodis1m
long and the mass of the reciprocating parts is140kg.On the expansion stroke, with the crank at30°from
the top dead center, the gas pressure is0.7MPa.If the engine runs at250rpm, determine
(i)Net force acting on the piston
(ii) Resultant load on the gudgeon pin
(iii) Thruston the cylinder walls, and
(iv) The speed above which, other things remaining the same, the gudgeon pin load would be reversed in
direction. (AUN/D2006)
10.In a reciprocating engine mechanism, if the crank and connecting rod are 300mm and 1m long
respectively and the crank rotates at a constant speed of 200rpm. Determine analytically,
(i)The crank angle at which the maximum velocity occurs
(ii) Maximum velocity of the piston
(iii) Derive the relevant equations. (AUN/D2007)
11.A vertical double acting steam engine has a cylinder 300mm diameter and 450mm stroke and runs at
200 rpm. The reciprocating parts has a mass of 225kg and the piston rod is 50mm diameter. The
connecting rod is 1.2m long. When the crank has turned through 125° from the top dead center, the steam
pressure above the piston is 30kN/m2
and below the piston 1.5 kN/m2
. Calculate the effective turning
moment on the crank shaft. (NOV/DEC 18)
DYNAMICALLY EQUIVALENT SYSTEM
1. A connecting or is suspended from a point 25mm above the center of small end and 650mm above its
center of gravity, its mass being 37.5kg. When permitted to oscillate, the time period is found to be 1.87
seconds. Find the dynamical equivalent system constituted of two masses, one of which is located at the
small end center.(APR/MAY 18)
FLYWHEEL
1. A multi cylinder engine is to run at a speed of 600rpm. On drawing the turning moment diagram to a
scale of 1mm = 250Nm and 1mm = 3°, the areas above and below the mean torque line in mm2
are:
+160, -172, +168, -191, +197, -162. The speed is to be kept within ±1% of the mean speed of the engine.
Calculate the necessary moment of inertia of the flywheel. Determine the suitable dimensions of a
rectangular flywheel rim if the breadth is twice its thickness. The density of the cast iron is 7250 kg/m3
and its hoop stress is 6 MPa. Assume that the rim contributes 92% of the flywheel effect.(AP/MAY 18)

45. 2. The turning moment diagram of a four stroke engine is assumed to be represented by four triangles,
the areas of which from the line of zero pressure are: Suction stroke = 440mm2
, Compression stroke =
1600 mm2
, Expansion stroke = 7200 mm2
, Exhaust stroke = 660 mm2
. Each mm2
of area represents 3
Nm of energy. If the resisting torque is uniform, determine the mass of the rim of a flywheel to keep the
speed between 218rpm and 222rpm, when the mean radius of the rim is to be 1.25m. (NOV/DEC 17)
case study question.
3. Turning moment diagram for a multi cylinder engine has been drawn to a vertical scale of 1mm = 650
Nm and a horizontal scale of 1mm = 4.5⁰. The areas above and below the mean torque line are -28,
+380, -260, +310, -300,+242, -380, +265 and -229mm2
. The fluctuation of speed is limited to ±1.8% of
the mean speed which is 400 rpm. The density of the rim material is 7000kg/m3
and the width of the rim is
4.5 times the thickness. The centrifugal stress in the rim material is limited to 6 N/mm2
. Neglecting the
effect of the boss, and the arms, determine the diameter and the cross section of the flywheel rim.
(APR/MAY 17)
4. Turning moment diagram for a multi cylinder engine has been drawn to a scale of 1mm = 600 Nm
vertically and 1mm = 3⁰ horizontally. The intercepted areas between the output torque curve and the
mean resistance line, taken in order from one end, are as follows: +52, -124, +92, -140, +85, -72, and
+107 mm2, when the engine running at 600 r.p.m. If the total fluctuation of speed is not to exceed ±1.5%
of the mean, find the necessary mass of the flywheel of radius 0.5m. (NOV/DEC 16)
5. A single cylinder double acting steam engine develops 150kW at a mean speed of 80 rpm. The
coefficient of fluctuation of energy is 0.1 and the fluctuation of speed is ±2% of mean speed. If the mean
diameter of the flywheel rim is 2m and the hub and spokes provide 5% of the rotational inertia of the
flywheel, find the mass and cross sectional area of the flywheel rim. Assume the density of the flywheel
material (which is cast iron) as 7200 kg/m3
(NOV/DEC 15)
6. Turning moment diagram for a multi cylinder engine has been drawn to a scale of 1mm = 325 Nm
vertically and 1mm = 3⁰ horizontally. The areas above and below the mean torque line are -26, +378, -
256, +306, +244, -380, +261 and -225mm2
. The engine is running at a speed of 600 r.p.m. The total
fluctuation of speed is not to exceed ±1.8% of the mean speed. If the radius of the flywheel is 0.7m, find
the mass of the flywheel. (MAY/JUNE 16)
7.A certain machine requires a torque of (5000+500sinθ )N-m to drive it ,where θ is the angle of
rotation of shaft measured from certain datum .The machine is directly coupled to an engine which
produces a torque of(5000+600sin2θ)N-m. The flywheel and the other rotating parts attached to the
engine has a mass of 500kg at a radius of gyration of 0.4m.If the mean speed is150r.p.m., find:1.the
fluctuation of energy,2.the total percentage fluctuation of speed,and3.the maximum and minimumangular
acceleration of the flywheel and the corresponding shaft position. ( AUN/D2014)
8.The equation of the turning moment curve of a three crank engine is(5000+1500sin3θ)N-m, where θ is
the crank angle in radians .The moment of inertia of the flywheel is1000kg-m2and the mean speed
is300r.p.m.Calculate: 1.power of the engine ,and 2.the maximum fluctuation of the speed of the

46. flywheel in percentage when(i)the resisting torque is constant, and(ii)the resisting torque is(5000+600sinθ) N-m.
( AUN/D2014)
9.A shaft fitted with a flywheel rotates at250r.p.m.and drives a machine. The torque of machine varies ina cyclic
manner over a period of 3 revolutions. The torque rises from750N-m to3000 N- m uniformly during ½ revolution
and remains constant for the following revolution .It then falls uniformly to750N-mduring the next1/2 revolution
and remains constant for one revolution, the cycle being repeated there after. Determine the power required to
drive the machine and percentage fluctuation in speed, if the driving torque applied on the shaft is constant and the
mass of the flywheel is 500 kg with radius of gyration of 600mm.
10.A single cylinder ,single acting, four stroke gas engine develops 25kW at320 rpm. The workdone bythe gases
during the expansion stroke is three times the workdone on the gases during the compression stroke, the
workdone during the suction and exhaust strokes being negligible. If the total fluctuation of speed is not to
exceed ±2%ofthe mean speed and the turning moment diagram during compression and expansion is assumed to
be triangular in shape, find the weight of the flywheel if its radius of gyration is0.5m. (AUN/D2014)
11.Thetorquedelivered by a two stroke engine is represented by T=(1000+300sin2θ-500cos2θ)N-m where θ is
the angle turned by the crank from the IDC. The engine speed is250rpm.The mass of the flywheel is 400kg and
radius of gyration 400mm. Determine,
(i)The power developed
(ii) The total percentage fluctuation of speed
(iii)Theangularaccelerationofflywheelwhenthecrankhasrotatedthroughanangleof60° From
the IDC.
(iv)The maximum angular acceleration and retardation of the flywheel. (AU 2010)
12.The variation of crankshaft torque of a four cylinder petrol engine may be approximately represented by
taking the torque as zero for crank angles 0° and 180° and as 260Nm for crank angles 20°and 45°,the
intermediate portions of the torque graph being straight lines. The cycle is being repeated in every half
revolution. The average speed is 600 rpm. Supposing that the engine drives a machine requiring constant torque
,determine the mass of the fly wheel of radius of gyration 250mm, which must be provided so that the total
variation of speed shall be one percent.(AUN/D2006)
13. an electric motor drives a punching machine. A flywheel fitted to the press has a radius of gyration of 0.5m
and runs at 250rpm. The press is capable to punch 800 holes per hour with each punching operation taking 1.5
seconds and requiring 12000Nm of work. Determine the rating of the machine in kW and the mass of the
flywheel if the speed of the flywheel does not drop below 230rpm.(NOV/DEC 18)

47. UNIT V
BALANCING AND VIBRATION
Static and Dynamic balancing
Balancing of revolving and reciprocating masses
Balancing machines
free vibrations – Equations of motion – natural Frequency
Damped Vibration
bending critical speed of simple shaft
Torsional vibration
Forced vibration
Harmonic Forcing
Vibration isolation.(gyroscopic principles)
TWO MARKS Q&A
1. Why balancing is necessary? (APR/MAY 17 MAY/JUNE 16)
 Forces caused by unbalance, disruptive vibration and noise are removed by balancing.
 This involves improving the mass distribution of a rotor so that smaller centrifugal force acts in its
bearings.
 So balancing is necessary.
2. Define static balancing of shaft? (NOV/DEC 15)
 Static balance occurs when the center of gravity of an object is on the axis or rotation.
 The object can therefore remain stationary, with the axis horizontal, without the application of ay
braking force.
 It has no tendency to rotate due to the force of gravity.
3. State the reason for choosing multi cylinder engine in comparison with that of
the single cylinderengine. (NOV/DEC 15)
 Able to produce more power output
 Ability to neutralize imbalances
 Capable of produce higher revolution per minute
 Reduced peak torque
4.What do you mean by partial balancing of `single cylinder engine? (NOV/DEC 15)
 Balancing is a process of eliminating the unbalanced shaking force and shaking couple.
 In most of the mechanisms, we can reduce the shaking force and shaking couple by adding appropriate
balancing mass, but it is usually not practical to eliminate them completely.
 It is called partial balancing of single cylinder engine.
5.In case of balancing of rotary masses in different planes how many planes in
which balancing masseswill be kept? (NOV/DEC 15)
In case of balancing of rotary masses in different planes, there are two planes in which balancing masses

48. will be kept.
6.Differentiate between the unbalanced force due to a reciprocating mass and
that due to revolvingmasses. ( NOV / DEC 2006 )
(i)Complete balancing of revolving mass can be possible. But fraction of reciprocating mass only
balanced.
(ii)The unbalanced force due to reciprocating mass varies in magnitude but constant in direction. But in
the case of revolving masses, the unbalanced force is constant in magnitude but varies in direction
7. Why complete balancing is not possible in reciprocating engine? (NOV / DEC
2008)
Balancing of reciprocating masses is done by introducing the balancing mass opposite to the crank. The vertical
component of the dynamic force of this balancing mass gives rise to “Hammer blow”. In order to reduce the
Hammer blow, a part of the reciprocating mass is balanced. Hence complete balancing is not possible in
reciprocating engines
8. Differentiate static and dynamic balancing ( MAY / JUNE 2016, APR/MAY 18)
S. No Static Balancing Dynamic Balancing
1 The dynamic forces as a result of the The arrangement made in static balancing
unbalanced masses are balanced by gives rise to a couple which tends to rock the
introducing balancing masses in the plane of shaft in the bearing.
rotation or diff planes Dynamic balancing considers the net couple
The net dynamic force acting on the shaft is as well as net dynamic force to do complete
made zero. balancing.
2 It deals with only balancing of dynamic It deals with balancing of dynamic force and
forces. balancing couple due to dynamic force.
9. Why single cylinder engines are not fully balanced ?(NOV / DEC 2004)
 A single cylinder engines cannot be fully balanced
 Because the unbalanced forces due to reciprocating masses remains constant in direction but varies in
magnitude
10. Why are the cranks of a locomotive, with two cylinders, placed at 90o to each
other? (NOV / DEC 06)
 In order to facilitate the starting of locomotive in any position the cranks of a locomotive are generally at
90o
to one another.
11. List down the planes of considering for uncoupled and coupled locomotives? (NOV
/ DEC 2007)
 In coupled locomotives , the two or three pairs of wheels are coupled by connecting their crank pins
with a connecting rod
12. What are primary and secondary unbalanced forces?
The primary unbalanced force may be considered as the component of the centrifugal force produced by
a rotating mass m placed at the crank radius r.
When the connecting rod is not too long (i.e when the obliquity of the connecting rod is considered) then
the secondary unbalanced mass due to reciprocating mass arises.

49. Primary unbalanced force = mrω2
cosθ
Secondary unbalanced force = mrω2cos 2
13. Explain the term primary balancing and secondary balancing? (MAY / JUNE 2009
MAY/JUNE 16)
 primary balancing is the balancing a fraction of primary unbalanced forces of an engine
 secondary balancing is the balancing a fraction of secondary unbalanced forces of an engine
14. Define tractive force. (APR / MAY 2010)
The resultant unbalanced force due to the two cylinders along the line of stroke is known as
tractive force.
15. What is Hammer Blow? (NOV / DEC 2013 APR/MAY 17 NOV/DEC 16)
The maximum magnitude of the unbalanced force along the perpendicular to the line of stroke is known as
hammer-blow
16. What is swaying couple? (APR / MAY 2005 NOV/DEC 14, NOV/DEC 2018)
The unbalanced force acting at a distance between the line of stroke of two cylinders, constitute a couple in the
horizontal direction. This couple is known as swaying couple
17. Why rotating masses are to be dynamically balanced? (APR/ MAY 2005, NOV/DEC
2018)
 If the rotating masses are not dynamically balanced, the unbalanced dynamic forces will cause worse
effects such as wear and tear on bearings and excessive vibrations on machines. It is very common in
cam shafts, steam turbine rotors, centrifugal pumps, etc.
18. State the condition for static balancing. (NOV / DEC 2005)
The net dynamic force acting on the shaft is equal to zero. This requires that the line of action of their
centrifugal forces must be same.
19.State the condition for dynamic balancing. (NOV / DEC 2017)
i. The net dynamic force action on the shaft is zero.
ii. The net couple due to dynamic forces acting on the shaft is zero.
20. What are the balancing machines are used? (MAY / JUNE 2006)
 Static balancing machines
 Dynamic balancing machines
 Universal balancing machines
21. What is meant by balancing of rotating masses? (Or) What is meant by
balancing of by singlerotating mass? (MAY / JUNE 2016, APR/MAY 18)
 The process of providing the second mass in order to counteract the effect of the centrifugal force of the
unbalanced first mass is called balancing of rotating masses.
22.Why is only a part of the unbalanced force due to reciprocating masses balanced
by revolving mass ? (NOV / DEC 2006 )

50.  Balancing of reciprocating masses is done by introducing the balancing mass opposite to the crank. The
vertical component of the dynamic force of this balancing mass gives rise to hammer blow in order to
reduce the hammer blow , a part of the reciprocating mass is balanced. Hence complete balancing is not
possible in reciprocating engines.
23. What are the different types of balancing machines? ( APR / MAY 2003 )
Different types of balancing machines are
 Pivot-Cradle Balancing Machine
 Nodal – Point method of balancing
 Micro Dynamic Balancing
 Mechanical Compensation method
24. What is dynamic balancing. ( APR / MAY 2003 )
A system of rotating masses is in dynamic balance when there does not exist any resultant
centrifugal force as well as resultant couple.
25. state the condition for complete balance of several masses revolving in
different planes of a shaft ? (NOV / DEC 2003)
1. The resultant centrifugal force must be zero.
2. The resultant couple must be zero.
26. List the effects of partial balancing of locomotives? ( NOV / DEC 2003)
1. Variation in tractive force along the line of stroke
2. Swaying couple
3. Hammer blow
27. What is the effect of hammer blow and what is the cause it? ( APR / MAY 2004)
The effect of hammer blow is to cause the variation in pressure between the wheel and the rail, such
that vehicle vibrates vigorously. Hammer blow is caused due to the effect of unbalanced primary force
acting perpendicular to the line of stroke.
UNIT V VIBRATION (second half)
1. Define Period and cycle of vibration. ( APR / MAY 2003)
Period is the time interval after which the motion is repeated itself. Cycle is defined as the
motion completed during one time period.
2.Define logarithmic decrement. ( May / June 2015,( APR / MAY 2017)
Logarithmic decrement is defined as the natural logarithm of the amplitude reduction factor. The
amplitude reduction factor is the ratio of any two successive amplitudes on the same side of the mean position.
3. Define resonance.( APR / MAY 2003)
When the frequency of external force is equal to the natural frequency of a vibrating body, the
amplitude of vibration becomes excessively large. This phenomenon is known as resonance.
4. What do you meant by degree of freedom in a vibrating system?( APR / MAY 2003, NOV 2007)
The number of independent coordinates required to completely define the motion of a system is known
as degree of freedom of the system
5. What are the various types of damping?( APR / MAY 2004 & NOV/DEC-2016/17)
 Viscous damping
 Coloumb damping