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MECHANICAL ENGINEERINGESE TOPICWISE OBJECTIVE SOLVED
PAPER-II
UPSC Engineering Services Examination
State Engineering Service Examination & Public Sector Examination.
FROM (1995-2017)
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Typeset at : IES Master Publication, New Delhi - 110016
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First Edition : 2017
Mr. Kanchan Kumar ThakurDirector–IES Master
It is an immense pleasure to present topic wise previous years solved paper of EngineeringServices Exam. This booklet has come out after long observation and detailed interaction withthe students preparing for Engineering Services Exam and includes detailed explanation to allquestions. The approach has been to provide explanation in such a way that just by goingthrough the solutions, students will be able to understand the basic concepts and will applythese concepts in solving other questions that might be asked in future exams.
Engineering Services Exam is a gateway to a immensly satisfying and high exposure job inengineering sector. The exposure to challenges and opportunities of leading the diverse fieldof engineering has been the main reason for students opting for this service as compared toothers. To facilitate selection into these services, availability of arithmetic solution to previousyear paper is the need of the day. Towards this end this book becomes indispensable.
PREFACE
CONTENT
1. Theory of Machines ................................................................................. 01–144
2. Machine Design...................................................................................... 145–254
3. Strength of Materials .............................................................................. 255–396
4. Engineering Materials ............................................................................ 397–462
5. Production Engineering ........................................................................ 463–593
6. Industrial Engineering ........................................................................... 594–704
7. Miscellaneous ......................................................................................... 705–710
S YL L A BU SKinematic and dynamic analysis of planner mechanisms. Cam. Gears and gear trains. Flywheels, Governors,Balancing of rigid rotor and field balancing. Balanacing of single and multicylinder engines. Linear vibrationanalysis of mechanical systems. Critical speeds and whirling of shafts. Automatic Controls.
Contents1. Analysis of Planer Mechanism ---------------------------------------------------------------- 01–34
2. Governors and Cam ------------------------------------------------------------------------------ 35–54
3. Slider Crank Mechanism and Flywheel ----------------------------------------------------- 55–67
4. Gear and Gear Trains --------------------------------------------------------------------------- 68–94
5. Balancing ------------------------------------------------------------------------------------------ 95–109
6. Linear Vibration Analysis --------------------------------------------------------------------- 110–136
7. Automatic Control and Whirling of Shafts ----------------------------------------------- 137–144
1
1
IES – 20171. In the 4-bar mechanism as shown, the link PQ
measures 30 cm and rotates uniformly at 100rev/min. The velocity of point Q on link PQ isnearly
50 cm
R
60 cm
SP
Q70 cm
(a) 2.54 m/s (b) 3.14 m/s(c) 4.60 m/s (d) 5.80 m/s
2. Which of the following mechanisms areexamples of forced closed kinematic pairs?1. Cam and roller mechanism2. Door-closing mechanism3. Slider-crank mechanismSelect the correct answer using the code givenbelow.(a) 1 and 2 only (b) 1 and 3 only(c) 2 and 3 only (d) 1, 2 and 3
3. A planer mechanism has 10 links and 12 rotaryjoints. Using Grubler’s criterion, the number ofdegrees of freedom of the mechanism is(a) 1 (b) 3(c) 2 (d) 4
4. The number of instantaneous centres of rotationin a slider-crank quick return mechanism is(a) 10 (b) 8
(c) 6 (d) 4
IES – 2016
5. Statement (I) : In quick return motionmechanism, Coriolis acceleration exists.
Statement (II) : Two links in this mechanismoscillate with one sliding relative to the other.
6. Consider the following motions :
1. Piston reciprocating inside an enginecylinder
2. Motion of a shaft between foot-stepbearings
Which of the above can rightly be consideredas successfully constrained motion?
(a) 1 only (b) 2 only
(c) Both 1 and 2 (d) Neither 1 nor 2
7. Coriolis component of acceleration depends on
1. angular velocity of the link
2. acceleration of the slider
3. angular acceleration of the link
Which of the above is/are correct?
(a) 1 only (b) 2 only
(c) 1 and 3 (d) 2 and 3
8. In a circular arc cam with a roller follower,acceleration of the follower depends on
1. cam speed and location of centre ofcircular arc
2. roller diameter and radius of circular arc
Theory of Machines 3
Which of the above is/are correct?(a) 1 only (b) 2 only(c) Both 1 and 2 (d) Neither 1 nor 2
9. In a crank and slotted lever quick return motionmechanism, the distance between the fixedcentres is 200 mm. The lengths of the drivingcrank and the slotted bar are 100 mm and 500mm, respectively. The length of the cuttingstroke is(a) 100 mm (b) 300 mm(c) 500 mm (d) 700 mm
IES – 201510. Statement (I) : Hooke’s joint connects two
non-parallel non-intersecting shafts to transmitmotion with a constant velocity ratio.Statement (II) : Hooke’s joint connects twoshafts the axes of which do not remain inalignment while in motion.
11. In a crank and slotted lever type quick returnmechanism, the link moves with an angularvelocity of 20 rad/s, while the slider moveswith a linear velocity of 1.5 m/s. The magnitudeand direction of Coriolis component ofacceleration with respect to angular velocityare(a) 30 m/s2 and direction is such as to rotate
slider velocity in the same sense as theangular velocity
(b) 30 m/s2 and direction is such as to rortateslider velocity in the opposite sense asthe angular velocity
(c) 60 m/s2 and direction is such as to rotateslider velocity in the same sense as theangular velocity
(d) 60 m/s2 and direction is such as to rotateslider velocity in the opposite sense asthe angular velocity
12. Which of the following are associated withAckerman steering mechanism used inautomobiles?
1. Has both sliding and turning pairs
2. Less friction and hence long life
3. Mechanically correct in all positions
4. Mathematically not accurate except inthree positions
5. Has only turning pairs
6. Controls movement of two front wheels
(a) 2, 4, 5 and 6 (b) 1, 2, 3 and 6
(c) 2, 3, 5 and 6 (d) 1, 2, 3 and 5
13. A four-bar mechanism is as shown in the figurebleow. At the instant shown, AB is shorterthan CD by 30 cm. AB is rotating at 5 rad/sand CD is ratating at 2 rad/s:
B C
A
D
The length of AB is
(a) 10 cm (b) 20 cm
(c) 30 cm (d) 40 cm
IES – 201414. In a crank and slotted lever quick-return motion,
the distance between the fixed centres is 150mm and the length of the driving crank is 75mm. The ratio of the time taken on the cuttingand return strokes is
(a) 1.5 (b) 2.0
(c) 2.2 (d) 2.93
15. Consider the following statements:
In a slider-crank mechanism, the slider is atits dead centre position when the
1. slider velocity is zero
2. slider velocity is maximum
3. slider acceleration is zero
4. slider acceleration is maximum
Which of the above statements are correct?
(a) 1 and 4 (b) 1 and 3
(c) 2 and 3 (d) 2 and 4
16. Which one of the following mechanisms is aninversion of double slider-crank chain?
4 IES Topic Wise Objective Solved Paper-II 1995-2017
(a) Elliptic trammels(b) Beam engine(c) Oscillating cylinder engine(d) Coupling rod of a locomotive
17. The number of instantaneous centres of rotationfor a 10-link kinematic chain is(a) 36 (b) 90(c) 120 (d) 45
18. A slider moves with uniform velocity v on arevolving link of length r with angular velocity. The Coriolis acceleration component of apoint on the slider relative to a coincident pointon the link is equal to(a) v parallel to the link(b) v2 perpendicular to the link(c) v perpendicular to the link(d) v2 parallel to the link
IES – 201319. The minimum number of links in a constrained
planer mechanism involving revolute pairs andtwo higher pairs is:(a) 3 (b) 4(c) 5 (d) 6
20. A man is climbing up a ladder which is restingagainst a vertical wall. When he was exactlyhalf way up, the ladder started slipping. Thepath traced by the man is:(a) Parabola (b) Hyperbola(c) Ellipse (d) Circle
21. Consider the following statements regardingmotions of various points of the same link ofa mechanism:1. The magnitude of the relative velocity of a
point A with respect to a point B on rotationof link is given by multiplication of angularvelocity with distance between the points.
2. The direction of the relative velocity of apoint A with respect to a point B on arotating link is perpendicular to AB.
3. The resultant acceleration of a point A withrespect to a point B of the same link isperpendicular to AB
Which of these statements are correct?(a) 1, 2 and 3 (b) 1 and 3 only(c) 1 and 2 only (d) 2 and 3 only
22. Universal or Hooke’s joint is used to connecttwo non-coaxial rotating shafts transmittingpower and intersecting at a small angletransmitting power. For a constant rotatingspeed of the driving shaft, the maximumfluctuation of instantaneous speed of the drivenshaft :(a) Varies as the square of the angle (in radian)
between the two shafts(b) Varies as the angle (in radian) between
the two shafts(c) Varies as the inverse of the angle (in radian)
between the two shafts(d) is nil
IES – 201223. The driving and driven shaft connected by a
Hooke’s joint are inclined by an angle toeach other. The angle through which the drivingshaft turns is given by . The condition for thetwo shafts to have equal speeds is
(a) cos sin = (b) sin tan
(c) tan cos (d) cot cos =24. In a crank and slotted lever quick return motion
mechanism, the distance between the fixedcenters is 160 mm and the driving crank is 80mm long. The ratio of time taken by cuttingand return strokes is(a) 0.5 (b) 1(c) 1.5 (d) 2
25. In a elliptic trammel, the length of the linkconnecting the two sliders is 100 mm. Thetracing pen is placed on 150 mm extension ofthis link. The major and minor axes of theellipse traced by the mechanism would be(a) 250 mm and 150 mm(b) 250 mm and 100 mm(c) 500 mm and 300 mm(d) 500 mm and 200 mm
1 8 IES Topic Wise Objective Solved Paper-II 1995-2017
Sol–1: (b)The angular velocity of link PQ,
=2 N 2 100 rad/sec60 60
Velocity of point Q on link PQ,
V = 2 100PQ 0.3060
= 3.14 m/secSol–2: (a)
Forced closed mechanism require externalforce to maintain mechanical contactbetween links. Slider-crank mechanismdoes not requires such force.
Sol–3: (b)Degree of freedom using grubler criterion,
DoF = 3 (n – 1) – 2J – h – Fr
= 3 (10 –1)–2×12 – 0 – 0 = 3Sol–4: (c)
No of instantaneous centre,
= n 42 2
4 3C C 62
Sol–5: (c)In quick return mechanism, slider moveson oscilating/rotating link, so Coriolisacceleration exist. But in this mechanism,one link rotates and other oscillates andslider a third link slides.
Sol–6: (b)Successfully constraint motion means thatthe motion is constraint only in the presenceof external force. The piston inside cylinderexecutes constraint motion due to it designand gudgeon pin. While foot step bearingrequires a force to press the shaft in bearingto have constraint motion (rotation).
Sol–7: (a)The expression for Coriolis component ofacceleration,= 2 V
where V is sliding velocity of slider overoscillating link and ‘ ’ is angular velocityof osci l l at ing l ink in qui ck returnmechanism.
Sol–8: (c)The expression for acceleration of followerinvolve following parameters(i) Radius of flank of cam surface, ‘rf’(ii) Least radius of cam profile or base
circle radius ‘rb’.(iii) Radius of roller ‘rr’.(iv) Cam rotation speed ' '
Sol–9: (c)Driving crank length
O1A = 100 mmSlotted Bar length
O2B = 500 mm
In triangle O2O,A
sin = 1
1 2
O A 100O O 200
= 30°
O1
A
200m
m
O2
C B
Length of cutting stroke,
= 22BC 2O Bsin 2 500 sin30
= 2 × 500 × 0.5 = 500 mmSol–10: (d)
The Hooke’s joint connects two non-parallelshafts but intersecting. For constant velocityratio there are two Hooke’s joints inparticular torks orientation
Theory of Machines 1 9
Sol–11: (c)The schematic of quick return mechanism
V
C 2V
The Coriolis acceleration,
ac = 2V = 2 × 1.5 × 20= 60 m/sec2
Direction: The direction of Coriol iscomponent of acceleration depends uponvelocity of slider ‘C’. If slider moves awayfrom centre, the Coriolis acceleration willbe in the direction of link rotation. If theslider moves toward centre of rotation theCoriolis acceleration will be opposite to linkrotation as shown below.
V
C
A
C
A
2V2V
V
Since nothing is mentioned about velocityof slider (away or toward centre) so assumeit moves away from centre, the right.
Sol–12: (a)The basic schematic of Ackerman steeringmechanism,
A w D
B C• All pairs (A, B, C, and D) are turning
pair in four bar mechanism A, B, C andD.
• This steering satisfy fundamental
equation of steering =
wcot cot
in three positions only namely in straightmotion 0 = and two positions 25 toward left and right
• Since very less sliding surfaces (turningpairs only) So longer life. Due to longerlife it is used generally despite it is notcorrect mathematically.
• Steering control is provided in frontwheels only in all steering mechanisms.
Sol–13: (b)The schematic of mechanism,
B C
A
D
1 5rad/sec=
2 2rad/sec=
At the instant shown in figure, the linearvelocity of point B and C will be same, VBA = VCD
1 2AB CD 5AB = 2CD …(i)
Since the difference between AB and CDCD – AB = 30
From equation (i)
CD = 5 AB2
5 AB AB 302
3 AB 302
AB = 2 30
3
= 20 cm
Sol–14: (b)The schematic of quick return mechanism-Distance between fixed centers-
AB = 150 mm
75m
C90°
A
B
150
mm
2 0 IES Topic Wise Objective Solved Paper-II 1995-2017
Length of driving crank,BC = 75 mm
In triangle ABC,
cos = BC 75 1AB 150 2
= 60°
Since cutting of metal happens to be inforward stroke and angle rotated by crankBC during cutting
= 360 2
= 360 – 120 = 240°The angle turned by crank during returnstroke.
B = 2 = 120° Ratio of cutting stroke time to return
stroke time,
=240 2120
= =
Sol–15: (a)
B
x r
r
OA
A
The distance travelled by slider B during rotation of crank OA from dead position A.
x = 2 2r 1 cos n n sin The velocity of slider B,
V = ddtx
V =
2 2
sin2r sin2 n sin
At dead center, = 0 or velocity, V = 0Acceleration,
a =
2 cos2r cos +n
a =
22 rr
n
= 2 1r 1n
So acceleration is maximum.Hence at dead centres, the velocity is zeroand acceleration is maximum.
Sol–16: (a)Elliptical trammel, skotch yoke mechanismand Oldham coupling are example ofinversion of double slider crank mechanism.
Sol–17: (d)Number of I-centres of kinematic chainhaving n-links,
= n n 1 10 9 45
2 2
Sol–18: (b)The schematic of mechanism having slideron rotating link is
0 V
The coriolis component of accelerationacting on slider,
= 2V
The direction of coriolis component ofacceleration is always perpendicular to torotating link.
Sol–19: (b)Gruebler’s criterion for constrained planermechanism i.e. one degree of freedom planermechanism.
F = 1 = 3 (n – 1) – 2j – h higher pair,
h = 2
1 = 3n – 3 – 2j – 23n = 2j + 6
2j = 3n – 6
n = number of linksj = number of revolute/lower pair,
Theory of Machines 2 1
The RHS of equation should be positive andeven numberput, n = 3
2j = 3 × 3 – 6 = 3 odd numberNow put,
n = 42j = 3 × 4 – 6 = 6, even number
So minimum number of link is 4.Sol–20: (d)
When the man is exactly mid way i.e. at C,the ladder starts slipping. The path traced byman will be circle. This is the case of ellipticaltrammel when the point of consideration ismid point of binary link. This point will trace acircle.
Mid PointC
B
A
y
O x
AlternativelyFrom figure
y
A
C( )x, y
O
x
B xy
x = ACcos
y = BC sin
2 2x y
AC BC
= 2 2cos sin
2 2x yAC BC
= 1
At mid point, AC = BC = AB/2
x2 + y2 =2AB
2
This is equation of circle.
Sol–21. (c)
VABA
B
Velocity of A relative to B
VAB = AB
The direction of VAB is perpendicular to AB
So statement one and two are correct.Thestatement three is wrong because there aretwo components of acceleration of point Bwith respect to A one centripetal along B toA and other tangential perpendicular to AB.So resultant can never be perpendicular toAB.
Sol–22. (a)The schematic of Hooke’s joint
N1
Driving shaft
N2Drive
n shaft
Maximum fluctuation of speed in universalor Hooke’s joint,
N2 – N1 =
2sinΝcos
for small value of
sin sin
cos 1
2N
where is in radianSol–23: (c)
12
Driving shaft
Driven sh
aft
The velocity ratio in Hooke's joint,
2
1
= 2 2
cos1 sin cos
