guru nanak institutions technical campus (autonomous)
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III YEAR B.Tech MECH - I SEMESTER L T P C 3 0 0 3
DYNAMICS OF MACHINERY
Pre-requisite: Kinematics of Machinery
UNIT–I: Precession: Gyroscopes – effect of precession – motion on the stability of moving vehicles such as motorcycle – motorcar – aeroplanes and ships.
Static and Dynamic Force Analysis: Static force analysis of planar mechanisms – Analytical Method – Dynamic Force Analysis – D’Alembert’s principle, Dynamic Analysis of 4-link mechanism, Slider Crank Mechanism.
UNIT–II: Turning Moment Diagram and Flywheels: Engine Force Analysis – Piston Effort, Crank Effort, etc., Inertia Force in Reciprocating Engine – Graphical Method - Turning moment diagram –fluctuation of energy – flywheels and their design - Inertia of connecting rod- inertia force in reciprocating engines – crank effort and torque diagrams.
UNIT–III: Friction: Pivots and collars – uniform pressure, uniform wear – friction circle and friction axis: lubricated surfaces – boundary friction – film lubrication. Clutches – Types – Single plate, multi-plate and cone clutches.
Brakes and Dynamometers: Types of brakes: Simple block brake, band and block brake-internal expanding shoe brake-effect of braking of a vehicle. Dynamometers – absorption and transmission types. General description and methods of operation.
UNIT–IV: Governors: Types of governors - Watt, Porter and Proell governors. Spring loaded governors – Hartnell and Hartung with auxiliary springs. Sensitiveness, isochronism and hunting – stability – effort and power of the governors.
Balancing: Balancing of rotating masses- Primary, Secondary, and higher balancing of reciprocating masses. Analytical and graphical methods. Unbalanced forces and couples. Examination of “V” and multi cylinder in-line and radial engines for primary and secondary balancing- locomotive balancing – Hammer blow – Swaying couple – variation of tractive effort.
UNIT–V: Vibrations: Free Vibration of mass attached to vertical spring – Transverse loads – vibrations of beams with concentrated and distributed loads. Dunkerly’s method –
Guru Nanak Institutions Technical Campus
(Autonomous)
Raleigh’s method. Whirling of shafts – critical speed – torsional vibrations – one, two and three rotor systems.
TEXT BOOKS: 1. Theory of Machines, 2009, 3rd Edition / S. S. Rattan, Tata Mcgraw Hill Education Private Limited; 2. Kinematics & Dynamics Of machinery,1st edition, 2009 / Norton, McGraw Hill Education;
REFERENCE BOOKS: 1. Theory of Machines,2005, 3rd edition / Thomas Bevan, CBS; 2. Theory of Machines: Kinematics and Dynamics,2011, 3rd edition / Sadhu Singh, Pearson Education India. 3. Theory of Machines and Mechanisms,2009, 3rd edition / John J. Uicker Jr , Gordon R. Pennock , Joseph E. Shigley, Oxford; 4. Mechanism and Machine Theory, 1992, 2nd edition / JS Rao and RV Duggipati , New Age Publishers; 5. A Textbook of Theory of Machines, 2016,5th Revised Edition / Dr. R.K. Bansal, Dr. J.S. Brar, Laxmi Publications (P) Ltd; 6. Theory of Machines, 2005, 14th edition / R.S. Khurmi, J.K. Gupta, S Chand & Co Ltd; 7. Theory of Machines: Kinematics and Dynamics, 1st Edition, 2010 / B V R Gupta, I K International Publishing House;
Dr. G. Sankaranarayanan Dr. B. Anjaneya Prasad Dr. K. Vijaya kumar Reddy
(Chairman, BOS) (JNTUH Nominee) (Academic Council Nominee)
Dr. L. Siva Rama Krishna Dr. M. Ramalinga Reddy Dr. M. Harinatha Reddy (Academic Council Nominee) (Member) (Member)
Dr. J. Krishnaiah Dr. A. Rajkumar Mr. G. Kiran Reddy (Industry Representative) (Member) (Alumni)
UNIT WISE
QUESTION BANK
2 Marks Questions (UNIT-1)
(1) Explain briefly about spin, precession and gyroscopic planes? (R15 MAY
2018) (BTL2, CO1)
Ans:
Spin Plane: The plane in which the wheel is intended to rotate is called spin
plane. In the above figure, spin axis is along Y axis and hence spin plane is the
X-Z plane.
Precession axis: The axis about which the spin axis itself is made to turn is
called precession axis. In the above figure, precession axis is along X axis and
hence precession plane is the Y-Z plane.
Gyroscopic plane: Wheel gives an unintentional rotation when it is made to
precess. This effect is called gyroscopic effect. The axis about which this
rotation occurs is called the gyroscopic axis and the plane in which this effect is
observed is called gyroscopic plane. In the above figure, gyroscopic axis is along
Z axis and hence gyroscopic plane is the X-Y plane.
(2) Explain the influence of gyroscopic couple on the stability of four-
wheeler when it is taking a turn? (R13 NOV/DEC 2018) (BTL2, CO1)
Ans: Gyroscopic couple due to four wheels, 𝐶𝑊 = 4 𝐼𝑊 𝜔𝑊 𝜔𝑃
Gyroscopic couple due to rotating parts of the engine, 𝐶𝑒 = 𝐼𝑒 𝜔𝑒 𝜔𝑃
Total gyroscopic couple, C = 𝐶𝑊 ± 𝐶𝑒
+ sign is taken when the wheel and engine rotating parts rotate in the same
direction.
The four-wheel vehicle tends to turn outwards when taking a turn due to the
effects of gyroscopic couple and the centrifugal force.
(3) Explain the gyroscopic effect on aero-planes, when it is steered right-
hand side on horizontal plane. (R16 APR 2018) (BTL2, CO1)
Ans: a) If the engine rotates clockwise when viewing from the rear end, the
reactive gyroscopic couple tends to raise the tail and dip the nose of aero plane.
b) If the engine rotates counter-clockwise when viewing from the rear end, the
reactive gyroscopic couple tends to raise the nose and dip the tail of aero plane.
(4) Explain the effect of gyroscopic couple on ships when there is pitching
and rolling. (R13 NOV/DEC 2018) (BTL2, CO1)
Ans: Pitching: It is defined as the limited angular motion of the ship about the
transverse axis.
When the pitching is upward, the effect of the reactive gyroscopic couple will
try to move the ship toward star-board.
When the pitching is downward, the effect of the reactive gyroscopic couple is
to turn the ship towards port side.
Rolling: It is defined as the limited angular motion of the ship about the
longitudinal axis.
In case of rolling, the axis of precession (i.e. longitudinal axis) is always parallel
to the axis of spin for all positions. Hence, there is no effect of the gyroscopic
couple acting on the body of a ship.
(5) Explain the application of gyroscopic principles to aircrafts. (R16 DEC
2018) (BTL2, CO1)
Ans: When a single-engine aero plane turns to the left, the nose tends to dip;
when the turn is to the right, the gyroscopic effect tends to raise the nose. When
a twin-engine plane, with propellers turning in opposite directions, alters
course, the leading edge of one wing tends to dip and the leading edge of the
other tends to rise, so that additional stresses on the structure are introduced.
(6) Give the applications of gyroscopic principle? (BTL2, CO1)
Ans: It is used: 1) In instrument or troy known as gyroscope,
2) In ships in order to minimize the rolling and pitching effects of waves, and 3)
In aero plane, monorail cars, gyrocompasses, etc.
(7) Describe the effect of the gyroscopic couple on a disc fixed at a certain
angle to a rotating shaft? (R13 MAR 2017) (BTL2, CO1)
Ans:
If 𝐼𝑂𝑃 is the mass mass moment of inertia of the disc about the polar axis OP,
then the gyroscopic couple acting on the disc,
CP = IOP ω cos Ɵ ω sin Ɵ = IOP ω2 sin 2Ɵ
If 𝐼𝐷 is the mass mass moment of inertia of the disc about the polar axis OD,
then the gyroscopic couple acting on the disc, CD = IOD ω2 sin 2Ɵ
The resultant gyroscopic couple acting on the disc,
C = CP - CD = ω2 sin2Ɵ ( IOP - IOD )
But = and =
C = ω2 sin2Ɵ [ - ] = ω2 sin2Ɵ [ - ] = sin 2Ɵ
This resultant gyroscopic couple will act in the anticlockwise direction as seen
from the top.
(8) What is the difference between applied and constrained forces? Explain.
(R13 NOV/DEC 2016) (BTL1, CO2)
Ans: Applied forces: Forces acting from outside on a system of bodies are called
applied forces. Usually, these forces are applied through direct physical or
mechanical contact. However, forces like electric, magnetic and gravitational
are applied without actual physical contact.
Constraint forces: A Pair of action and reaction forces acting on a body are
called constrained forces. As the constraint forces at a mechanical contact
occur in pairs, they have no net force effect on the system of bodies. However,
for an individual body isolated from the system, only one of each pair of
constraint forces has to be considered.
(9) Distinguish between the applied torque and reaction torque in gyroscopic
motion. (R13 MAY/JUN 2019) (BTL1, CO1)
Ans: The torque required to cause the axis of spin to precess in a plane is known
as the applied torque. The reaction torque tends to rotate the axis of spin in the
opposite direction.
(10) With reference to Naval ships, explain the terms: Bow, Stern, Port,
Starboard, Steering, Pitching and Rolling. (R16 DEC 2018) (BTL1, CO2)
Ans: Bow is the fore end of the ship. Stern is the rear end of the ship.
Starboard is the right-hand side of the ship while looking in the direction of
motion. Port is the left-hand side of the ship while looking in the direction of
motion. Steering is the turning of a complete ship in a curve towards left or
right, while it moves forward.
Pitching is the movement of a complete ship up and down in a vertical plane
about transverse axis.
Rolling is the movement of a ship in a linear fashion.
(11) Explain the term, gyroscopic couple? (BTL1, CO1)
Ans: If a body having moment of inertia I and rotating about its own axis at ω
rad/sec is also caused to turn at ωP rad/sec about an axis perpendicular to axis
of spin, then it experiences a gyroscopic couple of magnitude (I.ω.ωP) in an axis
which is perpendicular to both the axis of spin and axis of precession.
(12) State the D’Alembert’s principle for rectilinear motion and angular
motion. (R16 APR 2018) (BTL1, CO2)
Ans: For rectilinear motion, the D’Alembert’s principle states that inertia forces
and external forces acting on a body taken together give static equilibrium. (or)
𝐹𝑖 + (-m. 𝑓𝑔 ) = 0, where 𝐹𝑖 = Inertia force, m = mass of the body and 𝑓𝑔 =
acceleration of c entre of mass of the body.
For angular motion, the D’Alembert’s principle states that the inertia couples
and external torques applied to a body keep it in static equilibrium. (or) 𝐶𝑖 + (-
𝐼𝑔.α) = 0, where 𝐶𝑖 = Inertia couple, 𝐼𝑔 = Moment of inertia of the body about an
axis passing through centre of gravity and perpendicular to plane of rotation
and α = angular acceleration of the body.
(13) State the principle of superposition. (BTL1, CO2)
Ans: This principle states that if number of forces act on a system, the net effect
is equal to the sum of the individual effects of the forces taken one at a time.
(14) Explain about dynamically equivalent system. (R13 MAY/JUN 2019)
(BTL2, CO2)
Ans: A dynamically equivalent system means that the rigid link is replaced by a
link with two-point masses in such a way that it has the same motion as the
rigid link when subjected to the same force, i.e. the centre of mass of the
equivalent link has the same linear acceleration and the link has the same
angular acceleration.
(15) Explain about equivalent inertia force. (R13 NOV 2015) (BTL2, CO2)
Ans: Equivalent offset inertia force accounts for both inertia force and inertia
couple. This is obtained by displacing the line of action of the inertia force from
the centre of mass. The perpendicular displacement h of the force from the
centre of mass is such that the torque so produced is equal to the inertia couple
acting on the body.
i.e 𝑇𝑖 = 𝐶𝑖 or 𝐹𝑖 x h = 𝐶𝑖
or h = 𝐶𝑖
𝐹𝑖
= − 𝐼𝑔 𝛼
− 𝑚 𝑓𝑔 = m 𝑘2 α
𝑚 𝑓𝑔
= 𝑘2 𝛼
𝑓𝑔
h is taken in such a way that the force produces a moment about the centre of
mass, which is opposite in sense to the angular acceleration α.
Multiple-choice questions (UNIT-1)
(1) The magnitude of the gyroscopic couple applied to a disc of moment of
inertia I, spinning with an angular velocity ω and having angular velocity of
precession 𝜔𝑝 is
(a) 𝐼2ω𝜔𝑝 (b) 𝐼 𝜔2𝜔𝑝 (c) I ω 𝜔𝑝2 (d) I ω 𝜔𝑃
(2) The gyroscopic acceleration is given by
(a) 𝛿𝜔
𝛿𝑡 (b) ω 𝛿𝜃
𝛿𝑡 (c) r 𝛿𝜃
𝛿𝑡 (d) r 𝛿𝜔
𝛿𝑡
(3) If the air screw of an aero plane rotates clockwise when viewed from the
rear and the aero plane takes a right turn, the gyroscopic effect will be
(a) Tend to raise the tail and depress the nose
(b) Tend to raise the nose and depress the tail
(c) Tilt the aero plane about spin axis
(d) None of the above
(4) The axis of spin, the axis of precession and the axis of gyroscopic torque
are in
(a) Two parallel planes (b) two perpendicular planes
(c) three perpendicular planes (d) three parallel planes
(5) The effect of gyroscopic torque on the naval ship when it is rolling and the
rotor is spinning about the longitudinal axis is
(a) To raise the bow and lower the stern
(b) To lower the bow and raise the stern
(c) To turn the ship to one side
(d) None of the above
(6) In static equilibrium, the vector sum of all the forces acting on the body
and all the moments about _____ point is zero.
(a) A fixed point (b) a particular
c) any arbitrary (d) a permanent
(7) The forces generally considered in the design of mechanisms are:
(a) applied forces (b) inertia forces
(c) frictional forces (d) all of them
(8) If the resultant of forces acting on a body does not pass through the
centre of mass, then the inertia force and inertia couple is replaced
(a) by equivalent inertia force (b) equivalent inertia couple
(c) equivalent offset inertia force (d) equivalent offset inertia couple
(9) According to D' Alembert's principle, the body is in equilibrium
position if
(a) inertia force is applied in the direction
opposite to the resultant force
(b) inertia force is applied in the same direction of the resultant force
(c) both a. and b.
(d) none of the above
(10) Which of the following conditions should be satisfied for a two
point mass statistically equivalent system?
1. m = m1 + m2
2. m1 l2 = m2 l1
3. m1 l1 = m2 l2
4. IG = IG1 + IG2
a. Condition 1 and condition 2
b. Condition 1 and condition 3
c. Condition 2 and condition 4
d. All the above conditions should be satisfied
Answers:
1. d 2. b 3. a 4. c 5. D 6. c 7. d 8. c 9. a 10. b
Fill in the blanks (UNIT-1)
(1) The axis of precession is ____________ to the plane in which the axis of spin
is going to rotate.
(2) When the pitching of a ship is upward, the effect of gyroscopic couple acting
on it will be to move the ship towards __________.
(3) Up and down motion of bow and stern along transverse axis is called
________ of the ship.
(4) Gyroscopic effect is not observed in ____________actions performed by the
ships?
(5) Degree of freedom for gyroscope rotor is ______________.
(6) In a dynamically equivalent system, a uniformly distributed mass is
divided into __________ point masses.
(7) A pair of action and reaction forces acting on a body are known as
__________ forces.
(8) If the lines of action of three or more forces intersect at a point, it is
known as the ____________ point.
(9) If the resultant of forces acting on a body does not pass through the
centre of mass, then the inertia force and inertia couple is replaced by
_______________.
(10) “If number of forces act on a system, the net effect is equal to the sum of
the individual effects of the forces taken one at a time.’ This statement is called
principle of _____________.
Answers:
1. Perpendicular, 2. star-board 3. Pitching, 4. Rolling, 5. 3, 6. Two
7. Constraint forces, 8. Concurrency, 9. Equivalent offset inertia force.
10. Superposition
10 Marks Questions (UNIT-1)
(1) a. Develop the expression for gyroscopic couple. (6 M) (BTL3, CO1)
b. A uniform disk of diameter 300 mm and mass 5 kg is mounted on one end
of arm of length 600 mm. The other end of the arm is free to rotate in a
universal bearing. If the disc rotates about the arm with the speed of 300 rpm
clockwise, looking from the front, calculate its precession about the vertical
axis? (R16 DEC 2018) (BTL4, CO1)
(2) The turbine rotor of a ship has a mass of 2000 kg and rotates at a speed of
3000 rpm clockwise when viewed from stern. The rotor has radius of gyration
of 0.5 m.
a. Determine the gyroscopic couple and its effect when the ship steers to the
right in a curve of 100 m radius at a speed of 16.1 knots (1 knot = 1855 m/h)
(BTL4, CO1)
b. Calculate the torque and its effects when ship pitches simple harmonic
motion, the bow falling with its maximum velocity, the period of pitching is 50
seconds and the total angular displacement between two extreme positions of
pitching is 12 degrees. Find the maximum acceleration during the pitching
motion. (R16 APR 2018) (BTL4, CO1)
(3) a. An aeroplane makes a half circle of 100 m radius towards when flying at
400 kmph. The engine and propeller of plane weigh 500 kg, and have a radius
of gyration of 30 cm. The engine rotates at 3000 rpm ccw, when viewed from
the front end. Find the gyroscopic couple. (BTL4, CO1)
b. Develop equation for the limiting value of the Angle of heel (θ) to avoid
skidding of two-wheeled vehicle. (R15 MAY 2018) (BTL3, CO1)
(4) A motor cycle along with the rider weighs 2 kN, the centre of gravity of the
machine and rider combined being 60 cm above the ground, with the machine
in vertical position. The moment of inertia of each road wheel is 1030 N/mm2,
and the rolling diameter is 60 cm. The engine rotates at 6 times of the road
wheels and in the same sense. The moment of inertia of rotating parts of the
engine is 165 N/mm2. Determine the angle of heel necessary if the unit is
speeding at 62.5 km/h round a curve of 30.4 m. (R13 MAY/JUN 2019) (BTL4,
CO1)
(5) A rear engine automobile is travelling along a track of 100 m mean radius.
Each of the four road wheels has a moment of inertia of 2.5 kg-m2 and an
effective diameter of 0.6 m. The rotating parts of the engine have a moment of
inertia of 1.2 kg-m2. The engine axis is parallel to the rear axle and the crank
shaft rotates in the same sense as the road wheels. The ratio of engine speed
to back axle speed is 3:1. The automobile has a mass of 1600 kg and has its
centre of gravity 0.5 m above road level. The width of the track of the vehicle is
1.5 m. Determine the limiting speed of the vehicle around the curve for all four
wheels to maintain contact with the road surface. Assume that the road surface
is not cambered and centre of gravity of the automobile lies centrally with
respect to the four wheels. (R13 Mar 2017) (BTL4, CO1)
6) In a four-link mechanism shown in figure, torque 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. (R15 Nov/Dec 2017) (BTL4,
CO2)
7) a. Explain about the dynamically equivalent system? (BTL2, CO2)
b. Calculate the torque required for the static equilibrium of an in-line slider
crank mechanism in the position when crank angle is 450 from the inner dead
centre. The dimensions are, crank length = 300 mm, connecting rod length =
700 mm and the piston force = 40 N. Also find the torque required assuming
that the coefficient for all bearing is 0.1. The three journal bearings all have
radii of 10 mm, and the crank is rotating in the clockwise. Assume no friction in
the bearing. (R09 Nov 2015) (BTL4, CO2)
8) Referring to the figure below, O2A = 10 cm, AB = 15 cm, O4B = 15 cm, O4Q = 8
cm and AP = 7 cm.
a. Determine the force acting perpendicular to link 2 and passing through its
mid-point, for static equilibrium. What are the pin forces? (BTL4, CO2)
b. Instead of a force on link 2 , if link 2 is a driving crank, determine the couple
C2 necessary for static equilibrium. Determine also the pin forces. (R15 MAY
2018) (BTL4, CO2)
9) In a four-link mechanism ABCD, AB = 350 mm, BC = 500 mm, CD = 400 mm,
AD = 700 mm, DE = 150 mm, angle DAB = 60° (AD is the fixed link). A force of
35 N acts at E on link DC as shown in the figure. Determine the force on the
link AB required at the midpoint in the direction shown in the diagram for the
static equilibrium of the mechanism. The coefficient of friction is 0.4 for each
revolving pair. Assume impending motion of AB to be counter-clockwise.
Radius of each journal is 50 mm. Also, find the torque on AB for its impending
clockwise motion. (R13 NOV/DEC 2018) (BTL4, CO2)
10) a. State the conditions for equilibrium of a body under the action of (i) three
forces, and (ii) two forces and a torque. (BTL1, CO2)
b. A slider-crank mechanism shown in figure is subjected to a piston load of 1
kN. The link lengths are: AB = 250 mm; BC = 600 mm; Determine the input
torque to the link AB for static equilibrium of the mechanism. (R09 Dec 2014)
(BTL4, CO2)
2 Marks Questions (UNIT-2)
(1) Write the formula for velocity of the piston in reciprocating engine and
explain how does velocity is related to crank angle θ. (R15 NOV/DEC 2017)
(BTL1, CO2)
Ans: Velocity of piston, VP = ω r [sin Ɵ + 𝑆𝑖𝑛 2𝜃
2𝑛]
where ω = Angular velocity of the crank, rad/sec
r = radius of crank
Ɵ = angle turned by the crank
n = ratio of length of connecting rod to the radius of the crank = l / r
When Ɵ = 0°, velocity of piston = 0,
Ɵ =180°, velocity of piston = 0
Ɵ =90°, velocity of piston = ω r
(2) Explain the term coefficient of fluctuation of energy. What are the
parameters required to calculate coefficient of fluctuation of energy? (R15
NOV/DEC 2017) (BTL1, CO2)
Ans: Coefficient of fluctuation of energy: It is the ratio of maximum fluctuation
of energy to the work done per cycle.
𝐶𝐸 = 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑓𝑙𝑢𝑐𝑡𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 (𝛥𝐸)
𝑊𝑜𝑟𝑘𝑑𝑜𝑛𝑒 𝑝𝑒𝑟 𝑐𝑦𝑐𝑙𝑒
Work done per cycle = Tm. Ɵ
Tm = Mean torque, and Ɵ = Crank angle
Parameters required to calculate coefficient of fluctuation of energy are, Mean
angular speed of the flywheel and crank angle.
(3) When and why is the correction couple applied while considering the
inertia of the connecting rod of a reciprocating engine. (R13 NOV 2015)
(BTL1, CO2)
Ans: The difference of the torques is known as correction couple. This couple
must be applied, when the masses are placed arbitrarily to make the system
dynamical equivalent.
(4) List out the few machines in which flywheel is used. (BTL1, CO2)
Ans: 1. Punching machines,
2. Shearing machines,
3. Riveting machines, and
4. Crushing machines.
(5) What types of stresses are set up in the flywheel rims? (BTL1, CO2)
Ans: 1. Tensile stress due to the centrifugal force.
2. Tensile bending stress due to restraint of the arms.
3. Shrinkage stresses due to the unequal rate of cooling of casting.
(6) Explain the terms fluctuation of energy and fluctuation of speed as
applied to flywheels. (R13 MAR 2017) (BTL2, CO2)
Ans: Fluctuation of energy: The variations of energy above and below the mean
resisting torque line are called fluctuations of energy.
Fluctuation of speed: The variations of speed above and below the mean
resisting torque line are called fluctuations of speed.
(7) Draw the turning moment diagram of a single cylinder double acting
steam engine. (R16 APR 2018) (BTL1, CO2)
Ans:
Single cylinder double acting steam engine
(8) Explain the turning moment diagram of a four-stroke cycle internal
combustion engine. (BTL2, CO2)
Ans: In a four-stroke cycle internal combustion engine, there is one working
stroke after the crank has turned through two revolutions, i.e 4π radians.
Since the pressure inside the engine cylinder is less than the atmospheric
pressure during the suction stroke, therefore a negative loop is formed. During
the compression stroke, the work is done on the gases, therefore a higher
negative loop is obtained. During the expansion stroke, the fuel burns and the
gas expand, therefore a large positive loop is obtained. In this stroke the work
is done by the gases. During the exhaust stroke, the work is done on the gases,
therefore a negative loop is formed.
(9) Discuss the turning moment diagram of a multi-cylinder engine. (BTL2,
CO2)
Ans: The turning moment diagram for a compound steam engine having three
cylinders is shown in the following figure.
The resultant turning moment diagram is the sum of the turning moment
diagrams for the three cylinders. The first cylinder is the high-pressure
cylinder, second cylinder is the intermediate cylinder and the third cylinder is
the low-pressure cylinder. The cranks, in case of three cylinders are usually
placed at 120° to each other.
(10) Explain the terms: Piston effort, Crank effort. (R13 MAY/JUN 2019)
(BTL1, CO2)
Ans: Piston effort: It is the net force applied on the piston.
Crank effort: It is the net force applied to the crank pin perpendicular to the
crank which gives the required turning moment on the crank shaft.
(11) Write the formula for energy stored in a flywheel when it has
fluctuation in its speed. (R13 NOV/DEC 2018) (BTL1, CO2) (or)
Prove that the maximum fluctuation of energy, ΔE = E x 2 𝐂𝐒, where E = Mean
kinetic energy of the flywheel and 𝐂𝐒 = Coefficient of fluctuation of speed.
Ans: Energy stored in flywheel is given by,
E = ½ I 𝜔2 = ½ m 𝑘2 𝜔2
As the speed of the flywheel changes from ω1 to ω2, the maximum fluctuation
of energy,
ΔE = Maximum K.E – Minimum K.E = ½. I. ω12 – ½. I. 𝜔2
2
= ½ I (ω1 + ω2)(ω1 − ω2) = I. ω(ω1 − ω2)
= I. ω2 [ω1− ω2
ω] = I. ω2. CS = 2. E. CS (in N-m or joules)
(12) What is the function of flywheel? How does it differ from that of a
governor? (BTL2, CO2)
Ans: The function of a flywheel is:
a. To absorb energy when demand of energy id less than the supply
b. To give out energy when demand of energy is more than the supply.
Governor Flywheel
The function of a governor is to
regulate the mean speed of an
engine, when there are variations
in the load.
The function of a flywheel is to
reduce the fluctuations of speed
caused by the engine turning
moment during each cycle of
operation.
It is provided or, prime movers
such as engines and turbines.
It is provided on engine and
fabricating machines viz., rolling
mills, punching machines, shear
machines, presses, etc.
It works intermittently, i.e., only
when there is change in load.
It works continuously from cycle
to cycle.
It has no influence over cyclic
speed fluctuation
It has no influence on mean speed
of the prime mover
(13) Define the terms coefficient of fluctuation of energy and coefficient of
fluctuation of speed in case of flywheels. (R16 DEC 2018) (BTL1, CO2)
Ans: Coefficient of fluctuation of energy: It is the ratio of maximum fluctuation
of energy to the work done per cycle.
𝐶𝐸 = 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑓𝑙𝑢𝑐𝑡𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 (𝛥𝐸)
𝑊𝑜𝑟𝑘𝑑𝑜𝑛𝑒 𝑝𝑒𝑟 𝑐𝑦𝑐𝑙𝑒
Coefficient of fluctuation of speed: The ratio of the maximum fluctuation of
speed to the mean speed is called the coefficient of fluctuation of speed.
𝐶𝑆 = 𝑁1− 𝑁2
𝑁
where N1 = Maximum speed
N2 = Minimum speed and
N = Mean speed = 𝑁1+ 𝑁2
2
(14) What are turning moment diagrams? Explain the uses of turning
moment diagram of reciprocating engines. (BTL2, CO2)
Ans: Turning moment diagram is the graphical representation of the turning
moment or crank effort for various position of the crank.
In turning moment diagram, the turning moment is taken as the ordinate (Y-
axis) and crank angle as abscissa (X-axis).
Uses of turning moment diagrams: 1) The area under the turning moment
diagram represents work done per cycle. This area multiplied by number of
cycles per second gives the power developed by the engine.
2) Dividing the area of the turning moment diagram with the length of the base
gives the mean turning moment. This enables to find out the fluctuation of
energy,
3) The maximum ordinate of the turning moment diagram gives the maximum
torque to which the crankshaft is subjected to. This enables to find the diameter
of the crank shaft.
(15) Explain the term maximum fluctuation of energy and maximum
fluctuation of speed in flywheel? (BTL1, CO2)
Ans: The different between the maximum and the minimum energies is known
as maximum fluctuation of energy.
∆E = Maximum energy – Minimum energy.
The difference between the maximum and the minimum speeds is known as
maximum fluctuation of speed.
Multiple-choice questions (UNIT-2)
(1) Crank effort is the net force applied at the crankpin ________ to the crank
which gives the required turning moment on the crank shaft.
(a) Parallel (b) Perpendicular (c) at 45° (d) 135°
(2) In an engine, the work done by inertia forces in a cycle is
(a) Positive (b) Zero (c) Negative (d) None
(3) When the crank is at the inner dead centre, in a horizontal reciprocating
steam engine, then the velocity of the piston will be
(a) Zero (b) Minimum (c) Maximum (d) None
(4) Acceleration of piston of a reciprocating engine is ___________.
(a) r𝜔2 (sin 𝜃 + sin 2𝜃
𝑛) (b) rω (cos 𝜃 +
cos 2𝜃
𝑛)
(b) r𝜔2 (cos 𝜃 + cos 2𝜃
4𝑛) (d) r𝜔2 (cos 𝜃 +
cos 2𝜃
𝑛)
(5) The essential condition of placing the two masses, so that the system
becomes dynamically equivalent is
(a) L1 . l2 = 𝑘𝐺2 (b) l1 . l2 = 𝑘𝐺 (c) l1 = 𝑘𝐺 (d) l2 = 𝑘𝐺
where l1 and l2 = Distance of two masses from the centre of gravity of the
body, and 𝑘𝐺 = Radius of gyration of the body
(6) The ratio the maximum fluctuation of energy to the ______ is called
coefficient of fluctuation of energy.
(a) Minimum fluctuation of energy (b) Work done per cycle
(c) Torque (d) None
(7) The maximum fluctuation of energy is the
(a) Sum of maximum and minimum energies
(b) Difference between the maximum and minimum energies
(c) Ratio of the maximum energy and minimum energy
(d) Ratio of the mean resisting torque to the work done per cycle
(8) The amount of energy absorbrd by a flywheel is found from
(a) Speed-energy diagram
(b) Velocity-crank angle diagram
(c) Acceleration-crank angle diagram
(d) Torque-crank angle diagram
(9) In a turning moment diagram, the variations of energy above and below the
mean resisting torque line is called
(a) Fluctuation of energy
(b) Maximum fluctuation of energy
(c) Coefficient of fluctuation of energy
(d) None of the above
(10) The ratio of maximum fluctuation of speed to the mean speed is called
(a) Fluctuation of speed
(b) Maximum fluctuation of speed
(c) Coefficient of fluctuation of speed
(d) None of these
Answers:
1. b 2. a 3. a 4. b 5. a 6. b 7. b 8. d 9. a 10. c
Fill in the blanks (UNIT-2)
(1) The _____________ is the net force applied on the piston.
(2) The _____________ force arises due to the mass and acceleration of the
reciprocating parts.
(3) _______________ Principle is used to rduce a dynamic problem into as
equivalent static problem.
(4) When the crank is at the inner dead centre in a horizontal reciprocating
steam engine, then the velocity of the piston is _______,
(5) If a rigid link in a system is replaced by a link with two point masses in such
a way that it has the same motion as the rigid link when subjected to the
same force, then the system is called ______________.
(6) A plot of T vs. θ is known as the____________________.
(7) A ___________ is used to control the variations in speed during each cycle
of an engine.
(8) The area under the turning moment diagram represents the ___ per cycle.
(9) If mean speed of the prime mover is increased then the coefficient of
fluctuation of speed will _____________.
(10) The difference between the greatest speed and the least speed is known
as the ________________.
Answers:
1. Piston effort. 2. Inertia. 3. D-Alembert’s principle. 4. Zero. 5. A dynamically
equivalent system. 6. Turning moment diagram. 7. Flywheel. 8. Work done. 9.
Remains same. 10. The maximum fluctuation of speed.
10 Marks Questions (UNIT-2)
(1) In a vertical petrol engine, the crank radius is 6 cm, and the connecting rod
is 22 cm long. The piston weighs 9.8 N. The connecting rod may be regarded as
being equivalent to a mass of 0.5 kg at the piston together with a mass of 1 kg
at the crank pin. Find the amount and the direction of the force exerted on the
crank pin when the crank has moved 30° from the top dead centre. The engine
speed is 2000 rpm, and in this position the force on the piston due to gas
pressure is 7.35 N. (R13 MAY/JUN 2019) (BTL4, CO2)
(2) The connecting rod of a gas engine weighs 700 N, and has a radius of
gyration of 400 mm about an axis through the center of gravity. The length of
the rod between centers is 1 m and the C.G. is 350 mm from the crank pin
center. If the crank is 250 mm long, and revolves at a uniform speed of 300 rpm,
find the magnitude and direction of the inertia force on the rod, and of the
corresponding torque on the crankshaft when the inclination of the crank to the
IDC is 1350. (R13 NOV/DEC 2016) (BTL4, CO2)
(3) The length of connecting rod of a gas engine is 500 mm, and its C.G. lies at
165 mm from the crank pin centre. The rod has a mass of 80 kg and a radius of
gyration of 180 mm about an axis passing through the centre of mass. The
stroke of piston is 225 mm, and the crank speed is 300 rpm. Determine the
inertia force on the crankshaft when the crank has turned through 125° from
the inner dead centre. (R13 NOV 2015) (BTL4, CO2)
(4) In a vertical double-acting steam engine, the connecting rod is 4.5 times the
crank. The weight of the reciprocating parts is 120 kg and the stroke of the
piston is 440 mm. The engine runs at 250 rpm. If the net load on the piston due
to steam pressure is 25 kN when the crank has turned through an angle of 120°
from the top dead centre, determine the (i) thrust in the connecting rod (ii)
pressure on slide bars (iii) tangential force on the crank pin (iv) thrust on the
bearings (v) turning moment on the crank shaft. (BTL4, CO2)
(5) By using graphical method, find the inertia force for the following data of an
I.C. engine.
Bore = 175 mm, stroke = 200 mm, engine speed = 500 rpm, length of
connecting rod = 400 mm, crank angle = 60° from T.D.C. and mass of
reciprocating parts = 180 kg. (BTL4, CO2)
(6) The turning moment diagram for a multi-cylinder engine has been drawn to
a scale of 1 mm = 600 N-m vertically and 1mm = 3° horizontally. The areas
above and below the mean torque line are +52, -124, +92, -140, +85, -72, and
+107 mm2, when engine is running at a speed of 600 rpm. If the total fluctuation
of speed is not to exceed ±1.5% of the mean speed. Find the necessary mass of
the flywheel of radius 05 m. (R16 DEC 2018) (BTL4, CO2)
(7) The turning moment diagram for a petrol engine is drawn to the following
scales: Turning moment, 1 mm = 5 N-m; crank angle, 1 mm = 1°. The turning
moment diagram repeats itself at every half revolution of the engine and the
areas above and below the mean turning moment line taken in order are 295,
685, 40, 340, 960, 270 mm2. The rotating parts are equivalent to a mass of 36
kg at a radius of gyration of 150 mm. Determine the coefficient of fluctuation
of speed when the engine runs at 1800 rpm. (R16 APR 2018) (BTL4, CO2)
(8) The turning moment diagram for a multi-cylinder has been drawn to a scale
of 1 mm = 325 Nm vertically and 1 mm = 30 horizontally. The areas above and
below the mean torque line are -26, +378, -256, +306, -302, +244, -380, +261
and -225 sq.mm. The engine is running at a mean speed of 600 rpm. The total
fluctuation of speed is not to exceed ±1.8% of the mean speed. If the radius of
flywheel is 0.7 m, find the mass of the flywheel. (R15 NOV/DEC 2017) (BTL4,
CO2)
(9) In a turning moment diagram, the areas above and below the mean torque
line taken in order are: 5.81, 3.23, 3.87, 5.16, 1.94, 3.87, 2.58 and 1.94 cm2
respectively. The scales of the diagram are: Turning moment. 1 cm = 7 kN-m;
crank, 1 cm = 60°. The mean speed of the engine is 120 rpm and the variation
of speed must not exceed ±3 % of the mean speed. Assuming the radius of
gyration of the flywheel to be 106.67 cm, find the weight of the flywheel to
keep the speed within the given limits. (R15 MAY 2018) (BTL4, CO2)
(10) With reference to a reciprocating engine mechanism, derive the relations
for: a) The angular velocity and angular acceleration of the connecting rod, and
b) Turning moment on the crank shaft. (R13 MAY/JUN 2019) (BTL3, CO2)
2 Marks Questions (UNIT-3)
(1) What do you understand by self-locking brake and self-energized brake?
(R15 MAY 2018) (BTL2, CO3)
Ans: In self-locking brake, no external force is required to apply the brake. In
self-energizing brake, frictional force helps in applying the brake.
(2) What is principle of clutches, list out different types of clutches. (R15
NOV/DEC 2017) (BTL1, CO3)
Ans: Principle of clutch: It operates on the principle of friction. When two
surfaces are brought in contact and are held against each other due to friction
between them, they can be used to transmit power. If one is rotated, then other
also rotates. One surface is connected to engine and other to the transmission
system of automobile. Thus, clutch is nothing but a combination of two friction
surfaces.
Types of clutches: Some types of clutches used in vehicles are given below.
(a) Friction Clutch: It may be (i) single plate clutch, (ii) multi-plate clutch, or (iii)
cone clutch. Multi-plate clutch can be either wet or dry. A wet clutch is
operated in an oil batch whereas a dry clutch does not use oil.(b) Centrifugal
clutch. (c) Semi-centrifugal clutch. (d) Hydraulic clutch. (e) Positive clutch.
(f) Vacuum clutch. (g) Electromagnetic clutch.
(b)
(3) How does differential band brake work? Write the condition for self-
locking when drum rotates clockwise. (R15 NOV/DEC 2017) (BTL1, CO3)
Ans:
In a differential band brake, neither end of the band is attached to the fulcrum
of the lever. The two ends of band are attached to the two points on opposite
side of the fulcrum as in figure.
Condition for self-locking: T1 a = T2 b
(4) Discuss the various types of the brakes. (BTL1, CO3)
Ans: Brakes can be classified as follows:
1) Block or shoe brakes. 2) Band brakes.
3) Band and block brakes. 4) Internal expanding shoe brakes.
(5) What is meant by the expression friction circle? (R13 MAR 2017) (BTL2,
CO3)
Ans: Consider a shaft rotating inside a bearing in clockwise direction. The
lubricant between the journal and bearing forms a thin layer which gives rise to
a greasy friction. Therefore, the reaction R doesn’t act vertically upward, but
acts at another point of pressure B. This is due to the fact that when shaft
rotates, a frictional force F = µ RN acts at the circumference of the shaft which
has a tendency to rotate the shaft in opposite direction of motion and this shifts
the point A to point B.
T = W. r tan ɸ = µ W r
If a circle is drawn with centre O and radius OC = r sin ɸ, then the circle is called
friction circle of a bearing.
(6) Which of the two assumptions, uniform intensity of pressure or uniform
rate of wear, would you make use of in designing friction clutch and why?
(BTL2, CO3)
Ans: The uniform pressure theory gives a higher frictional torque than the
uniform wear theory. Therefore, in case of friction clutches, uniform wear
should be considered, unless otherwise stated.
(7) What is a brake? What is the difference between a brake and a clutch?
(R13 MAY/JUN 2019) (BTL1, CO3)
Ans: Brakes are used to stop a moving member or to control its speed. Clutches
are used to keep the driving and driven member moving together.
(8) Do you recommend the uniform pressure theory or uniform wear theory
to find the friction torque of a bearing. Justify your answer. (R13 NOV 2015)
(BTL2, CO3)
Ans: Though the actual power loss will be a little less than the calculated,
uniform pressure theory is recommended to find the friction torque of a
bearing.
(9) List the assumptions made in estimating the HP absorbed by friction in a
footstep bearing. (R13 MAY 2018) (BTL1, CO3)
Ans: The HP absorbed by friction in a footstep bearing is estimated on the basis
of two assumptions.
1) The intensity of pressure on the bearing surface is constant. Pressure is
assumed to be uniform over the surface are and the intensity of pressure is
given by, Pressure = 𝑎𝑥𝑖𝑎𝑙 𝑓𝑜𝑟𝑐𝑒
𝑐𝑟𝑜𝑠𝑠−𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎
2) Uniform wearing of the bearing surface. The intensity of pressure should
vary inversely proportional to the elemental areas. P.r = constant
(10) Distinguish between brakes and dynamometers. (R16 DEC 2018) (BTL2,
CO3)
Ans:
Brakes Dynamometers
It absorbs energy and stops motion. It measures power.
Used to retard or stop vehicle. Used to measure speed, torque or
power of the rotating shaft.
No torque is measured. Torque measurement is basic
function.
(11) What is the difference between absorption and transmission
dynamometers? (BTL2, CO3)
Ans: In the absorption dynamometers, the entire energy or power produced by
the engine is absorbed by the friction resistances of the brake and is
transformed into heat, during the process of measurement.
In the transmission dynamometers, the energy is not wasted in friction but is
used for doing work. The energy or power produced by the engine is
transmitted through the dynamometer to some other machines where the
power developed is suitably measured.
(12) Though cone clutches provide high frictional torque, yet they have
become obsolete. Why? (R13 MAY/JUN 2019) (BTL2, CO3)
Ans: Cone clutches became obsolete due to following reason:
1) Small cone angle. 2) Exposure to dirt and dust tends to bind two cones and
it becomes difficult to engage them.
3) This is also mainly because of the frictional contact surface of the cone
clutch. In this case of clutch there are high chances of unequal wear of friction
lining. Once this happens, the clutch does not engage with the flywheel
properly. As a result, their efficiency drops and they need to be repaired all the
time.
(13) Define the following terms: Friction, Limiting force of friction, co-
efficient of friction and angle of friction. (R13 NOV/DEC 2016) (BTL1, CO3)
Ans: Friction: Friction is the resistance to motion of one object moving relative
to another.
Limiting force of friction: The maximum value of frictional force, which comes
in to play, when a body just begins to slide over the surface of the other body,
is known as limiting force of friction.
Co-efficient of friction: It is defined as the ratio of the limiting friction (F) to the
normal reaction (𝑅𝑁) between the two bodies.
Angle of friction: Angle made by the resultant of normal reaction and limiting
frictional force with the normal reaction is called angle of friction.
(14) Explain about friction between lubricated surfaces. (BTL2, CO3)
Ans: When lubricant is applied between two surfaces in contact then the friction
may be classified into the following two types depending upon the thickness of
layer of a lubricant.
1) Boundary friction: It is the friction, experienced between the rubbing
surfaces when the surfaces have a very thin layer of lubricant. It is also called
greasy friction or non-viscous friction.
2) Film friction: It is the friction experienced between the rubbing surfaces when
the surfaces have a thick layer of the lubricant. It is also called fluid friction or
viscous friction.
(15) Explain the characteristics of the materials for brake lining. (BTL2, CO3)
Ans: The material used for the brake lining should have the following
characteristics.
1) It should have high coefficient of friction with minimum fading.
2) It should have low wear rate.
3) It should have high heat resistance.
4) It should have high heat dissipation capacity.
5) It should have adequate mechanical strength.
6) It should not be affected by moisture and oil.
Multiple-choice questions (UNIT-3)
(1) In a flat pivot bearing, the moment due to friction with uniform pressure is
assumed to act at
(a) r (b) r/2 (c) r/3 (d) 2r/3
(2) The radius of a friction circle drawn for a journal rotating in a bearing
depends on the coefficient of friction and _______________.
(a) Angular velocity of the journal
(b) Radius of the journal
(c) Clearance between the journal and the bearing
(d) Weight of the shaft
(3) The frictional torque transmitted by a disc or plate clutch is same that of
(a) Flat pivot bearing (b) Flat collar bearing
(c) Conical pivot bearing (d) Trapezoidal pivot bearing
(4) For a self-locking screw, efficiency should be
(a) <50% (b) >50% (c) =50% (d) None
(5) The frictional torque transmitted in a conical pivot bearing considering
uniform wear is
(a) ½ μ W R cosec α (b) 23
μ W R cosec α
(c) 34
μ W R cosec α (d) μ W R cosec α
where R = Radius of the shaft, and α = Semi-angle of the cone
(6) The brakes commonly used in railway trains is
(a) Shoe brake (b) band brake
(c) band and block brake (d) internal expanding brake
(7) Which type of brake is commonly used in cars
(a) Band brake (b) Shoe brake
(c) Band and block brake (d) Internal expanding shoe brake
(8) Which of the following is an absorption type dynamometer?
(a) Prony brake dynamometer (b) Rope brake dynamometer
(c) Froude’s hydraulic dynamometer (d) All of the above
(9) In an internal expanding shoe brake, more than 50% of the total braking
torque is supplied by
(a) Leading shoe (b) Trailing shoe (c) Both (d) None
(10) Lubricant is used in a rope brake dynamometer is
(a) Oil (b) Water (c) Grease (d) None
Answers:
1. d 2. b 3. b 4. a 5. a 6. a 7. d 8. d 9. a 10. b
Fill in the blanks (UNIT-3)
(1) _________energy is absorbed by the brakes of an elevator during braking
process?
-
(2) ____________theory is used to determine safe design of bearings?
(3) Condition for uniform wear theory is ______________.
(4) In case of an old clutch, _____________ theory is more appropriate to find
out frictional torque.
(5) in a self-locking brake, the force required to apply the brake is __________.
(6) When the frictional force helps the applied force in applying the brake, the
brake is called __________________.
(7) The frictional torque for flat pivot by using uniform wear theory is given by
______________________,
(8) A device that is used to measure the frictional resistance is __________.
(9) A _____________ dynamometer is used for measuring large powers.
(10) Self-locking brake is used in ___________ applications.
Answers:
1. Potential 2. Uniform pressure 3. P.r = Constant 4. Uniform wear
5. zero. 6. Self-energizing. 7. ½ μWR
8. Dynamometer. 9. Torsion 10. Back-stop.
10 Marks Questions (UNIT-3) (1) a. Write a note on uniform pressure. (3M) (R16 DEC 2018) (BTL2, CO3)
b. A conical pivot with an angle of cone as 120° supports a load of 20 kN. The
external diameter is 2 times the internal diameter. The shaft rotates at 200
rpm. If the intensity of pressure is to be 0.3 N/mm2 and coefficient of friction as
0.1, find the power lost in working against the friction. Assuming uniform
pressure. (7M) (R16 DEC 2018) (BTL4, CO3)
(2) a. Discuss the various types of the brakes with neat sketch. (5M) (R16 DEC
2018) (BTL2 CO3)
b. Describe the construction and operation of a Prony brake dynamometer. (5M)
(R16 DEC 2018) (BTL3, CO3)
(3) With a neat sketch, describe the principle and working of an internal
expanding shoe brake. Derive the expression for the force exerted by the cam,
(a) on the leading shoe, (b) on the trailing shoe. (R16 APR 2018) (BTL3, CO3)
(4) a. What are the leading and trailing shoes of an internal expanding shoe
brake? Explain. (5 M) (R16 APR 2018) (BTL2, CO3)
b. Describe the construction and operation of rope brake absorption
dynamometer. (5M) (R16 APR 2018) (BTL3, CO3)
(5) Figure shows a differential band brake of drum diameter 400 mm. The ends
of the band are fixed to the points on the opposite side of fulcrum of the lever
at a distance of 50 mm and 160 mm from the fulcrum as shown in the figure.
The brake is to sustain a torque of 300 Nm. The co-efficient of friction between
band and the brake is 0.2. The angle of contact is 210° and the length of lever
from the fulcrum is 600 mm. Determine a) The force required at the end of the
lever for the clockwise and anti-clockwise rotation of the drum, b) Value of OB
for the brake to be self-locking for clockwise rotation. (R13 NOV/DEC 2018)
(BTL4 CO3)
(6) 100 kW is transmitted at 3000 rpm by a multiple disc friction clutch. The
plates are having friction surface with a coefficient of friction 0.07, and the axial
intensity of pressure is not to exceed 1.5 bar. External radius is 1.25 times the
internal radius, and the external radius is 12.5 cm. Determine the number of
plates needed to transmit the required torque. Assume uniform wear. (R15 MAY
2018) (BTL4, CO3)
(7) a. A torsion dynamometer is fitted on a turbine shaft to measure the angle
of twist. It is observed that the shaft twists 2° in a length of 5 m at 600 rpm. The
shaft is solid and has a diameter of 250 mm. If the modulus of rigidity is 84 GPa,
find the power transmitted by the turbine. (5M) (R15 MAY 2018) (BTL4, CO3)
b. Which of the two assumptions, uniform intensity of pressure or uniform rate
of wear, will you make use of in designing a friction clutch and why? Give
reason. (5M) (R15 MAY 2018) (BTL3, CO3)
(8) A thrust bearing has contact surfaces of 40 cm and 30 cm external and
internal diameters respectively. Calculate the number of collars required for an
end thrust of 16 tonnes. The coefficient of friction is 0.04 and the maximum
intensity of allowable pressure is 0.35 MPa. Calculate the HP lost in friction at a
speed of 120 rpm? (R13 MAY/JUN 2019) (BTL4, CO3)
(9) A rotor is driven by a co-axial motor through a single plate clutch, both sides
of the plate being effective. The external and internal diameters of the plate
are respectively 220 mm and 160 mm and the total spring load pressing the
plates together is 570 N. The motor armature and shaft have a mass of 800 kg
with an effective radius of gyration of 200 mm. The rotor has a mass of 1300
kg with an effective radius of gyration of 180 mm. The coefficient of friction for
the clutch is 0.35. The driving motor is brought up to a speed of1250 rpm when
the current is switched off and the clutch suddenly engaged. Determine,
a) The final speed of motor and rotor.
b) The time to reach this speed, and
c) The kinetic energy lost during the period of slipping.
How long would slipping continue if it is assumed that a constant resisting
torque of 60 Nm were present? If instead of a resisting torque, it is assumed
that a constant driving torque of 60 Nm is maintained on the armature shaft,
what would then be slipping time? (R13 MAR 2017) (BTL4, CO3)
(10) The below figure shows a brake shoe applied to a drum by a lever AB which
is pivoted at a fixed-point A and rigidly fixed to the shoe. The radius of the drum
is 160 mm. The coefficient of friction at the brake lining is 0.3. If the drum
rotates clockwise, find the braking torque due to the horizontal force of 600 N
at B. (R13 MAR 2017) (BTL4, CO3)
2 Marks Questions (UNIT-4)
(1) Define the terms relating to governors: Isochronism and Hunting. (R16
APR 2018) (BTL1, CO1)
Ans: Isochronism: A governor is said to be isochronous, when the equilibrium
speed is constant for all radii of rotation of the balls, within the working range.
An isochronous governor will be infinitely sensitive.
Hunting: It is a condition in which the speed of the engine controlled by the
governor fluctuates continuously above and below the mean speed. It is caused
by a governor which is too sensitive.
(2) Explain about secondary balancing in multi-cylinder in-line engine? (R16
APR 2018) (BTL2, CO4)
Ans: In multi-cylinder in-line engines, each imaginary secondary crank with a
mass attached to the crank pin is inclined to the line of stroke at twice the angle
of the actual crank. The values of the secondary forces and couples may be
obtained by considering the revolving mass. This is done in the similar way as
discussed for primary forces. The following two conditions must be satisfied in
order to give a complete secondary balance of an engine:
1.The algebraic sum of the secondary forces must be equal to zero. In other
words, the secondary force polygon must close, and
2.The algebraic sum of the couples about any point in the plane of the
secondary forces must be equal to zero. In other words, the secondary couple
polygon must close.
(3) Distinguish between primary and secondary balancing. (R16 DEC 2018)
(BTL2, CO4)
Ans: 1) The primary unbalanced force is maximum twice in one revolution of the
crank. The secondary force is maximum four times in one revolution of the
crank.
2) Maximum secondary unbalanced force is 1/n times the maximum primary
unbalanced force.
3) In case of moderate speeds, the secondary unbalanced force is so small that
it may me neglected as compared to primary unbalanced force.
(4) Define the term sensitiveness in governors. (R15 NOV/DEC 2017) (BTL1,
CO1)
Ans: The sensitiveness is defined as the ratio of the mean speed to the
difference between the maximum and minimum speeds. A governor is said to
be sensitive, when it really responds to a small change of speed.
(5) How do you balance V-engines? (R16 DEC 2018) (BTL2, CO4)
Ans: Consider a symmetrical two-cylinder V-engine, as shown in Figure. The
common crank OC is driven by two connecting rods PC and QC. The lines of
stroke OP and OQ are inclined to the vertical OY, at an angle α.
Inertia force due to the reciprocating parts of cylinder 1, along the line of stroke
Inertia force due to the reciprocating parts of cylinder 2, along the line of stroke
(6) Define the terms ‘Variation in tractive force’, ‘Swaying couple’ and
‘Hammer blow’ for an uncoupled two-cylinder locomotive engine? (R13
MAY/JUN 2019) (BTL1, CO4)
Ans: Variation in tractive force: A variation in tractive force of an engine is
caused by the unbalanced portion of the primary force which acts along the line
of stroke of a locomotive engine.
Swaying couple: Swaying couple tends to make the leading wheels sway from
side to side due to unbalanced primary forces along the lines of stroke.
Hammer blow: Hammer blow is the maximum vertical unbalanced force caused
by the mass to balance the reciprocating masses. Its value is mr𝜔2.
(7) What is meant by static and dynamic unbalance in machinery? How can
the balancing be done? (R15 MAY 2018) (BTL2, CO4)
Ans: Static balance refers to the ability of a stationary on object to its balance.
Static balancing definition refers to the ability of a stationary object to its
balance. This occurs when a parts centre of gravity is on the axis of rotation.
However, the dynamic balance definition is the ability of an object to balance
whilst in motion or when switching between positions.
Dynamic balance is the ability of an object to balance whilst in motion or when
switching between positions. Dynamic balancing is when the rotating system
doesn’t yield any other force or couple. Other than the force that is needed the
system will rotate without the need for any additional external force or
pressure to be applied.
(8) What are primary and secondary forces in the reciprocating engine? (R15
NOV/DEC 2017) (BTL1, CO4)
Ans: Primary unbalanced force, FP = m. 𝜔2. r cos θ
Secondary unbalanced force, FS = m. 𝜔2. r cos 2𝜃
𝑛
where m = Mass of the reciprocating parts
ω = Angular speed of the crank
r = Radius of the crank
θ = Angle of inclination of the crank with the line of stroke
n = Ratio of length of the connecting rod to the crank radius = l/r
(9) What is the difference between the Porter and Proell governors? (R15
MAY 2018) (BTL2, CO1)
Ans: 1) For a given mass of Governor ball and center load the speed of Proell
governor lower than that of porter governor.
3) In order to have the same equilibrium speed for the given values, balls of
smaller masses are used in the Proell governor than in the Portel governor.
(10) What are the effects of friction and of adding a central weight to the
sleeve of a Watt governor? (R13 MAR 2017) (BTL2 CO1)
Ans: 1) When the loaded sleeve moves up and down the spindle, the frictional
force acts on if in a direction opposite to that of motion sleeve.
2) When a central load attached to the sleeve, the load moves up down the
central spindle. This additional downward force increases the speed of
revolution required to enable the balls to rise to any to any pre-determined
level. Height of the governor increases.
(11) Explain the term partial balancing of primary force. (R13 NOV/DEC
2016) (BTL2, CO4)
Ans: The primary force acts from O to P along the line of stroke. Hence
balancing of primary force is considered as equivalent to the balancing of mass
m rotating at the crank radius r. This is balanced by having a mass B at a radius
b, placed diametrically opposite to the crank pin C. 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.
(12) What is the difference between centrifugal governors and inertia
governors? (R13 NOV/DEC 2016) (BTL2, CO1)
Ans:
Centrifugal governor Inertia governor
The operation of a centrifugal
governor depends on the change in
In addition to centrifugal force, the
position of governor ball and the
speed and centrifugal force on the
governor balls.
operation of governor controlled by
force of angular acceleration and
retardation of the spindle.
Only centrifugal force is in
controlling action
In inertia governor, both centrifugal
force and inertia force are in action.
The sensitiveness is less when
compared to inertia governor Highly sensitive to varying load.
The response is slower than that of
inertia governor.
The reaction of inertia governor is
faster than that of the centrifugal
governor (quick response).
Easy to balance the revolving parts. Hard to balance revolving parts.
This type of governor more
frequently used. They are not popular.
(13) What are the limitations of a Watt governor? (BTL1, CO1)
Ans: 1) Watt governors are limited to in vertical position applications.
2) Watt governor is used in very slow speed engine. At higher speed, the
sensitivity will decrease.
(14) What do you understand by inside cylinder locomotives and outside
cylinder locomotives? (R13 NOV/DEC 2016) (BTL2, C04)
Ans: Inside cylinder locomotives: The front-mounted cylinders are placed
inside (between the frames). Example: Planet locomotive. Inside cylinders give
a more stable ride with less yaw or nosing but access for maintenance is more
difficult. Some designers used inside cylinders for aesthetic reasons.
Outside cylinder locomotives: The front-mounted cylinders are placed
outside. Examples: GNR Sterling 4-2-2
(15) Explain effort and power of a governor. (BTL2, CO1)
Ans: Effort of a governor: The effort of a governor is the mean force exerted at
the sleeve for a given percentage change of speed.
Power of a governor: The power of a governor is the work done at the sleeve
for a given percentage change of speed.
Power = Mean effort x lift of sleeve
Multiple-choice questions (UNIT-4)
(1) A Hartnell governor is a
(a) Pendulum type governor (b) Spring loaded governor
(c) Dead weight governor (d) Inertia governor
(2) The height of a Watt’s governor (in metres) is equal to
(a) 8.95/N2 (b) 89.5/N2
(c) 895/N2 (d) 8950/N2
where N = Speed of the arm and ball about the spindle axis.
(3) For two governors A and B, the lift of sleeve of governor A is more than that
of governor B, for a given frictional change in speed. It indicates that
(a) governor A is more sensitive than governor B
(b) governor B is more sensitive than governor A
(c) both governors A and B are equally sensitive
(d) none of the above
(4) Which of the following governor is used to drive a gramophone?
(a) Watt governor (b) Porter governor
(c) Pickering governor (d) Hartnell governor
(5) Isochronism in a governor is desirable when
(a) The engine operates at low speeds
(b) The engine operates at high speeds
(c) The engine operates at variable speeds
(d) One speed is desirable under one load
(6) Static balancing involves balancing of
(a) Forces (b) Couples
(c) Forces as well as couples (d) masses
(7) If a rotating system is dynamically balanced, it is statically
(a) Balanced (b) Unbalanced
(c) Partially balanced (d) None
(8) The primary unbalanced force is maximum when the angle of crank with the
line of stroke is _____________
(a) 45° (b) 90°
(c) 135° (d) 180°
(9) Which of the following statements are correct?
(a) If a rotor is statically balanced, it is also dynamically balanced.
(b) If a rotor is dynamically balanced, it is also statically balanced.
(c) If a rotor is dynamically balanced, it may or may not be statically balanced.
(d) None of the above.
(10) __________ can completely balance a system of several masses revolving
in different planes on a shaft.
(a) Two masses in any two planes
(b) Two equal masses in any two planes
(c) Two masses in any of the two planes of revolving masses only.
(d) A single mass in any plane
Answers:
1. b 2. c 3. a 4. c 5. d 6. a 7. a 8. d 9. b 10. a
Fill in the blanks (UNIT-4)
(1) The height of Watt’s governor is proportional to ________________.
(2) If the ball masses of a governor occupy a definite specified position for each
speed, it is said to be _____________.
(3) If the ball masses of a governor have same equilibrium speed for all the radii
of rotation, it is said to be __________.
(4) Governor power is defined s the product of governor effort and ________.
(5) If the controlling force of a spring-controlled governor is expressed as
(a.r+b), where r is the radius of rotation and a and b are constants, it is a/an
_________ governor.
(6) If the balance mass is to be placed in a plane parallel to the plane of the
unbalance mass then the minimum number of balance masses required are __.
(7) The frequency of secondary force is ________ to that of primary force.
(8) A _______ cylinder inline engine working on four-stroke cycle is completely
balanced inherently.
(9) The angle between the axes of a 2-cylinder V-engine to balance the primary
forces completely is ___________.
(10) Primary unbalance force due to inertia of reciprocating parts of mass m
at a radius r moving with an angular velocity ω is _______.
Answers:
1. 1/N2 2. Stable 3. Isochronous 4. Sleeve lift. 5. Unstable
6. Two 7. Twice 8. Six 9. 90° 10. mrω2 cos θ
10 Marks Questions (UNIT-4) (1) Four masses m1, m2, m3 and m4 are 200 kg, 300 kg, 240 kg and 260 kg
respectively. The corresponding radii of rotation are 0.2 m, 0.15 m, 0.25m and
0.3 m respectively and the angles between successive masses are 45°, 75° and
135°. Find the position and magnitude of the balance mass required, if its radius
of rotation is 0.2 m. (R16 DEC 2018) (BTL4, CO4)
(2) In a porter governor, the upper and lower arms are 200 mm and 250 mm
respectively and pivoted on the axis of rotation. The mass of the central load is
15 kg, the mass of each ball is 2 kg and friction of the sleeve together with the
resistance of the operating gear is equal to a load of 24 N at the sleeve. If the
limiting inclinations of the upper arms to the vertical are 30° and 40°. Find the
range of speed of governor taking friction into account. (R16 DEC 2018) (BTL4,
CO4)
(3) The following data refer to two-cylinder locomotive with cranks at 90°,
reciprocating mass per cylinder = 300 kg, crank radius = 0.3 m, driving wheel
diameter = 1.8 m, distance between cylinder centre lines = 0.65 m, distance
between the driving wheel central planes = 1.55 m. Determine a) the fraction
of the reciprocating masses to be balanced, if the hammer blow is not to exceed
46 kN at 96.5 kmph, b) the variation in tractive effort and c) the maximum
swaying couple. (R15 MAY 2018) (BTL4, CO4)
(4) a. Draw a neat sketch of Proell governor. Establish a relation among the
various forces acting on the bent link. (5M) (BTL3, CO1)
b. Determine the (i) maximum speed (ii) minimum speed (iii) range of speed of
a Watt governor of open arm type shown in figure in which the length of arm AE
= 400 mm, and length EF = 60 mm, when the angle θ changes from 40° to 30°.
(5M) (R15 MAY 2018) (BTL4, CO1)
(5) a. Explain the terms: Primary disturbing force and secondary disturbing
force. (5M) (BTL2, CO4)
b. Explain the procedures for balancing of V – engine. (5M) (R13 NOV/DEC 2018)
(BTL2, CO4)
(6) The upper and lower ends of the links of a Proell governor are pivoted on
the axis of rotation of the governor. Each of the upper and lower links are each
25 cm long between centers and the lower links carry extension arms each 10
cm long and parallel to the governor axis when the radius of the ball path is 15
cm, Determine the equilibrium speed of the governor for this configuration if
each ball weighs 60 N and the central load weighs 390 N. (R13 MAY/JUN 2019)
(BTL4, CO1)
(7) A shaft carries five masses A, B, C, D and E which revolve at the same radius
in equidistant planes. The masses in planes A, C and D weigh respectively 500,
400 and 800 N. The angle between A and C is 90° and that between C and D is
135°. Find the weights in plane B and E and their angular positions so that the
shaft may be completely balanced. (R13 MAY/JUN 2019) (BTL4, CO4)
(8) Deduce the governing equation of a Porter governor, taking into account the
friction at the sleeve. Also discuss the effect of friction on the functioning of the
governor. (R13 NOV 2015) (BTL3, CO1)
(9) A shaft carries four revolving masses A, B, C and D in that order along the
axis. The mass A may be assumed to be concentrated at a radius of 12 cm, B at
15 cm, C at 14 cm, and D at 18 cm. The weights of A, C and D are 150 N, 100 N
and 80 N respectively. The planes of revolution of A and B are 15 cm apart and
those of B and C are 19 cm apart. The angle between the masses A and C is 90°.
Determine (a) the angles between masses A, B and C (b) the distance between
the planes of revolution of C and D and (c) the weight of mass B, so that the
shaft may be completely balanced. (R13 MAY/JUN 2019) (BTL4, CO4)
(10) Prove that for a Hartnell governor:
a) The total lift, h = 𝑏
𝑎 (𝑟2 − 𝑟1), and
b) Stiffness of spring, s = 𝑎𝑏
[𝑆2− 𝑆1
𝑟2− 𝑟1
]
where a and b = lengths of bell-crank lever of ball-arm and sleeve-arm
respectively; and S1 and S2 spring forces at minimum and maximum radii r1 and
r2 respectively. (R13 NOV/DEC 2016) (BTL3, CO1)
2 Marks Questions (UNIT-5)
(1) Define critical speed of a shaft. Why is critical speed encountered? (R16
APR 2018) (BTL1, CO5)
Ans: The speed at which resonance occurs is called critical speed of the shaft.
In other words, the speed at which the shaft runs so that the additional
deflection of the shaft from the axis of rotation becomes infinite is known as
critical speed.
The critical speed may occur due to one or more of the following reasons:
1) Eccentric mountings like gears, flywheels, pulleys, etc.,
2) Bending of the shaft due to self-weight
3) Non-uniform distribution of rotor material, etc.
(2) What are the causes of vibrations? (R16 DEC 2018) (BTL2, CO5)
Ans: 1) Unbalanced forces in the machine: Produced within the machine itself.
2) Dry friction between the two-mating surface: Self-excited vibration
produced.
3) External excitations: These excitations may be periodic, random or the
nature of an impact produced external to the vibrating system.
4) Earthquakes: These are responsible for the failure of many buildings, dams
5) Winds: These are cause the vibration of transmission and telephone line
under certain conditions.
(3) Write about the terms: free vibrations, forced vibrations and damped
vibrations. (R16 DEC 2018) (BTL1, CO5)
Ans: Free or natural vibrations: When no external force acts on the body, after
giving it an initial displacement, then the body is said to be under free or natural
vibrations.
Forced vibrations: When the body vibrates under the influence of external
force, then the body is said to be under forced vibrations.
Damped vibrations: When there is a reduction in amplitude over every cycle of
vibration, then the motion is said to be damped vibration.
(4) Write the formula for natural frequency ‘ f ’ and static deflection ‘ δ ‘ of
the cantilever beam loaded at free end and shaft having negligible mass.
(R15 NOV/DEC 2017) (BTL1, CO5)
Ans: Natural frequency, 𝑓𝑛 = 1
2𝜋 √
𝑔
𝛿
; Static deflection, δ = 𝑊𝑙3
3𝐸𝐼
W = Load at the free end, in Newton
l = Length of the beam in meter
E = Young’s modulus for the material of the beam in N/m2
I = Moment of inertia of the beam in m4
(5) What is a torsionally equivalent shaft? (R15 MAY 2018) (BTL1, CO5)
Ans: A shaft having diameter for different lengths can be theoretically replaced
by an equivalent shaft of uniform diameter such that they have the same total
angle of twist when equal opposing torques are applied at their ends. Such a
theoretically replaced shaft is known as torsion ally equivalent shaft.
(6) Explain briefly, the longitudinal, transverse and torsional free
vibrations. (R15 MAY 2018) (BTL2, CO5)
Ans: Longitudinal vibrations: When the particles of the shaft or disc moves
parallel to the axis of the shaft, then the vibrations are known as longitudinal
vibrations. In this case, tensile and compressive stresses induced in the shaft.
Transverse vibrations: When the particles of the shaft or disc move
approximately perpendicular to the axis of the shaft, then the vibrations are
known as transverse vibrations. Bending stresses are induced in the shaft.
Torsional vibrations: When the particles of the shaft or disc move in a circle
about the axis of the shaft, then the vibrations are known as torsional
vibrations. Torsional shear stresses are induced in the shaft.
(7) Write about Rayleigh’s method of finding the natural frequency of
transverse vibrations. (R13 N0V/DEC 2018) (BTL1 CO5)
Ans: Consider a shaft is loaded with point loads W1, W2, W3 and W4 etc. and y1,
y2, y3, y4 etc. be total deflection made under these loads.
According to Rayleigh’s method, the maximum potential energy is equal to
maximum kinetic energy.
½ Σ m g y = ½ 𝜔2 Σ m𝑦2 ; ω = √
𝑔 𝛴 𝑚𝑦
𝛴 𝑚 𝑦2
Natural frequency of transverse vibration, 𝑓𝑛 = 𝜔
2𝜋 = 1
2𝜋 √
𝑔 𝛴 𝑚𝑦
𝛴 𝑚 𝑦2
(8) What is node? Discuss. (BTL1, CO5)
Ans: The point or the section of the shaft whose amplitude of torsional vibration
is zero is known as node. At node, the shaft remains unaffected by the vibration.
(9) Discuss the methods to eliminate or reduce the undesirable vibrations.
(BTL2, CO5)
Ans: 1) Removing the causes of vibrations.
2) Putting the screens if noise is objectionable.
3) Placing the machinery on proper isolators.
4) Using shock absorbers.
5) Using dynamic vibration absorbers.
10) Explain the terms Periodic motion, Time period, Frequency, Amplitude
of motion, Resonance and Cycle (BTL1, CO5)
Ans: Periodic motion: A motion which repeats itself after equal interval of time.
Time period: Time taken to complete one cycle.
Frequency: Number of cycles per unit time.
Amplitude of motion: The maximum displacement of a vibrating body from the
mean position.
Resonance: The vibration of the system when the frequency of the external
force is equal to the natural frequency of the system.
Cycle: Motion completed during one time period.
11) Why is it important to find the natural frequency of a vibrating system?
(BTL2, CO5)
Ans: When the frequency of externally excited system equal to natural
frequency of vibration system it gets failure due to resonance. So, to avoid the
resonance at vibrating system natural frequency must be known.
12) Define degree of freedom of a vibrating system. (BTL1, CO5)
Ans: The minimum number of independent coordinates required to specify the
motion of a system at any instant is known as degrees of freedom of the
system.
13) Give examples of the bad and good effects of vibration. (BTL2, CO5)
Ans: Bad effects:
1. Excessive stresses. 2. Undesirable noise
3. Looseness of part and partial or complete failure of part.
Good effects:
1. Used in musical instruments. 2. Used in vibrating screens
3. Used in shakers. 4. Used in stress relieving
(14) What is Rayleigh’s Principle? (BTL1, CO5)
Ans: Rayleigh principle: It is stated that the distribution of the potential and
kinetic energies of conservation, elastic system in the fundamental mode of
vibration is such that the frequency is minimum.
(15) What is the basic assumption in deriving the Dunkerlay’s formula?
(BTL1, CO5)
Ans: 1) Dunkerlay’s formula is applicable to a uniform diameter shaft carrying
several loads.
2.This method can also account for self-weight of the shift.
Multiple-choice questions (UNIT-5) (1) When there is reduction in amplitude over every cycle of vibration, then the
body is said to have
(a) Free vibration (b) Forced vibration
(c) Damped vibration (d) None
(2) Longitudinal vibrations are said to occur when the particles of a body move
(a) Perpendicular to its axis (b) Parallel to its axis
(c) In a circle about its axis (d) None
(3) When a body subjected to transverse vibrations, the stress induced in a body
will be
(a) Shear stress (b) Tensile stress
(c) Compressive stress (d) None
(4) The factor which affects the critical speed of a shaft is
(a) Diameter of the disc (b) Span of the shaft
(c) Eccentricity (d) All of these
(5) The natural frequency (in Hz) of free longitudinal vibrations is equal to
(a) 1
2𝜋 √
𝑠
𝑚
(b) 1
2𝜋 √
𝑔
𝛿
(c) 0.4985
√𝛿 (d) Any one of these
where m = Mass of the body in kg; s = Stiffness of the body in N/m, and
δ = Static deflection of the body in meters
(6) The natural frequency of free torsional vibrations of a shaft is
(a) 2π √
𝑞
𝐼
(b) 2π √𝑞𝐼
(c) 1
2𝜋 √
𝑞
𝐼
(d) 1
2𝜋 √𝑞𝐼
where q = Torsional stiffness of the shaft, and
I = Mass moment of inertia of the disc attached at the end of the shaft
(7) At nodal point in a shaft, the amplitude of torsional vibration is
(a) Zero (b) Minimum
(c) Maximum (d) None
(8) A shaft carrying two rotors at its ends will have
(a) No node (b) One node
(c) Two nodes (d) Three nodes
(9) A shaft carrying three rotors will have
(a) No node (b) One node
(c) Two nodes (d) Three nodes
(10) In a spring mass system, if the mass is doubled and the spring stiffness is
halved, the natural frequency is
(a) Halved (b) Doubled
(c) Unchanged (d) Quadrupled
Answers:
1. c 2. b 3. b 4. d 5. d 6. c 7. a 8. b 9. c 10. a
Fill in the blanks (UNIT-5)
(1) The particles of a body move ______ an axis in torsional vibrations.
(2) Resonance is a phenomenon in which the frequency of the exciting force is
__________ to the natural frequency of the system.
(3) The frequency of damped vibrations is always _____ the natural frequency.
(4) The critical speed of rotating shaft with a mass at the centre is _____ the
natural frequency of transverse vibration of the system.
(5) A compressive coil spring is cut into two equal halves and each half is joined
in parallel, the ratio of spring stiffness will be now ____ times of the original.
(6) Free vibration is also termed as ______________.
(7) Forced vibration is also termed as ____________.
(8) The linear velocity at the centre of oscillation of a physical pendulum is
_________.
(9) The equivalent stiffness of two springs of equal stiffness in series becomes
_________.
(10) A torsional system having m rotors in a vibrating shaft has _____________
nodes.
Answers:
1. In a circle about. 2. Equal. 3. Less than. 4. Equal. 5. Four.
6. Transient vibration. 7. Steady-state vibration. 8. Zero 9. ½
10. (m – 1).
10 Marks Questions (UNIT-5) (1) a. Discuss briefly with neat sketches the longitudinal, transverse and
torsional free vibrations. (5M) (R16 DEC 2018) (BTL2, CO5)
b. Derive the length of torsionally equivalent shaft. (5M) (BTL3, CO5)
(2) A shaft 1.5 m long supported in flexible bearings at the ends carries two
wheels each of 50 kg mass. One wheel is situated at the centre of the shaft and
the other at distance of 375 mm from the centre towards left. The shaft is
hollow of external diameter 75 mm and internal diameter 40 mm. The density
of the material is 7700 kg/m3 and its modulus of elasticity is 200 GN/m2. Find
the lowest whirling speed of the shaft, taking into account the mass of the
shaft. (R16 APR 2018) (BTL4, CO5)
(3) a. In case of free torsional vibrations of two-rotor system, prove that the
node divides the length of the shaft in the inverse ratio of the moments of
inertia of the corresponding rotors. (5M) (BTL3, CO5)
b. In a spring-mass vibrating system, the natural frequency of vibration is 3.56
Hz. When the amount of suspended mass is increased by 5 kg, the natural
frequency is lowered to 2.9 Hz. Determine the original unknown mass and the
spring constant. (5M) (R15 MAY 2018) (BTL4, CO5)
(4) a. Find the frequency of transverse vibrations of a shaft which is simply
supported at the ends and is of 40 mm in diameter. The length of the shaft is 5
m. The shaft carries three point loads of masses 15 kg, 35 kg, and 22.5 kg at 1
m, 2 m and 3.4 m respectively from the left support. The Young’s modulus for
the material of the shaft is 200 GN/m2. The weight of the shaft is 18.394 N per
meter length. (5M) (R15 MAY 2018) (BTL4, CO5)
b. Describe with relevant sketches, the equilibrium method to find the natural
frequency of free longitudinal vibrations. (5M) (BTL3, CO5)
(5) a. A shaft is supported freely at its ends has a load of 1.2 kN placed at the
centre of the shaft. The diameter of shaft is 40 mm and its length is 700 mm.
Find the frequency of its natural transverse vibrations if E = 200 GN/m2. (5M)
(R13 MAY/JUN 2019) (BTL4, CO5)
b. Explain the Dunkerley’s method for finding the frequency of natural
transverse vibrations of a simply supported shaft carrying several
concentrated loads. (5M) (R13 MAY/JUN 2019) (BTL3, CO5)
(6) A shaft of length 1.25 m is 75 mm in diameter for the first 275 mm of its
length, 125 mm in diameter for the next 500 mm length, 87.5 mm in diameter
for the next 375 mm length and 175 mm in diameter for the remaining 100 mm
of its length. The shaft carries two rotors at two ends. The mass moment of
inertia of the first rotor is 75 kg m2 whereas of the second rotor is 50 kg m2.
Find the frequency of natural torsional vibrations of the system. The modulus
of the rigidity of shaft material may be taken as 80 GN/m2. (R13 NOV/DEC 2018)
(BTL4, CO5)
(7) A torsional system is shown in figure. Find the frequencies of torsional
vibrations and the positions of the nodes. Also find the amplitudes of
vibrations. G = 84 x 109 N m2. (BTL4, CO5)
(8) A flywheel is mounted on a vertical shaft as shown in figure. The both ends
of the shaft are fixed and its diameter is 50 mm. The flywheel has a mass of
500 kg. Find the natural frequencies of longitudinal and transverse vibrations.
Take E = 200 GN/m2. (BTL3, CO5)
(9) A vertical shaft of 5 mm diameter is 200 mm long and is supported in long
bearings at its ends. A disc of mass 50 kg is attached to the centre of the shaft.
Neglecting any increase in stiffness due to the attachment of the disc to the
shaft, find the critical speed of rotation and the maximum bending stress when
the shaft is rotating at 75% of the critical speed. The centre of the disc is 0.25
mm from the geometric axis of the shaft. E = 200 GN/m2. (BTL4, CO5)