MECHANICAL ENGINEERING ESE TOPICWISE OBJECTIVE SOLVED PAPER-II
FROM (1995-2017)
UPSC Engineering Services Examination State Engineering Service Examination & Public Sector Examination.
IES MASTER PUBLICATION Office : F-126, (Lower Basement), Katwaria Sarai, New Delhi-110016 Phone : 011-26522064, Mobile : 8130909220, 9711853908 E-mail : info@iesmaster.org Web : www.iesmasterpublications.com
IES MASTER PUBLICATION F-126, (Lower Basement), Katwaria Sarai, New Delhi-110016 Phone : 011-26522064, Mobile : 8130909220, 9711853908 E-mail : info@iesmaster.org Web : www.iesmasterpublication.org
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First Edition : 2017
Typeset at : IES Master Publication, New Delhi - 110016
PREFACE
It is an immense pleasure to present topic wise previous years solved paper of Engineering Services Exam. This booklet has come out after long observation and detailed interaction with the students preparing for Engineering Services Exam and includes detailed explanation to all questions. The approach has been to provide explanation in such a way that just by going through the solutions, students will be able to understand the basic concepts and will apply these 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 in engineering sector. The exposure to challenges and opportunities of leading the diverse field of engineering has been the main reason for students opting for this service as compared to others. To facilitate selection into these services, availability of arithmetic solution to previous year paper is the need of the day. Towards this end this book becomes indispensable.
Mr. Kanchan Kumar Thakur Director–IES Master
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
1 S YL L A B U S Kinematic 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 vibration analysis of mechanical systems. Critical speeds and whirling of shafts. Automatic Controls.
Contents 1.
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 IES – 2017 1.
IES – 2016
In the 4-bar mechanism as shown, the link PQ measures 30 cm and rotates uniformly at 100 rev/min. The velocity of point Q on link PQ is nearly
5.
Statement (II) : Two links in this mechanism oscillate with one sliding relative to the other.
R 50 cm
6.
60 cm
Q P
2.
70 cm
4.
Consider the following motions : 1.
Piston reciprocating inside an engine cylinder
2.
Motion of a shaft between foot-step bearings
S
(a) 2.54 m/s
(b) 3.14 m/s
(c) 4.60 m/s
(d) 5.80 m/s
Which of the following mechanisms are examples of forced closed kinematic pairs?
Which of the above can rightly be considered as successfully constrained motion?
1. Cam and roller mechanism
(a) 1 only
(b) 2 only
2. Door-closing mechanism
(c) Both 1 and 2
(d) Neither 1 nor 2
3. Slider-crank mechanism
3.
Statement (I) : In quick return motion mechanism, Coriolis acceleration exists.
7.
Coriolis component of acceleration depends on
Select the correct answer using the code given below.
1.
angular velocity of the link
(a) 1 and 2 only
(b) 1 and 3 only
2.
acceleration of the slider
(c) 2 and 3 only
(d) 1, 2 and 3
3.
angular acceleration of the link
A planer mechanism has 10 links and 12 rotary joints. Using Grubler’s criterion, the number of degrees of freedom of the mechanism is
Which of the above is/are correct? (a) 1 only
(b) 2 only
(a) 1
(b) 3
(c) 1 and 3
(d) 2 and 3
(c) 2
(d) 4
The number of instantaneous centres of rotation in a slider-crank quick return mechanism is (a) 10
(b) 8
(c) 6
(d) 4
8.
In a circular arc cam with a roller follower, acceleration of the follower depends on 1.
cam speed and location of centre of circular arc
2.
roller diameter and radius of circular arc
Theory of Machines
3
Which of the above is/are correct?
9.
(a) 1 only
(b) 2 only
4. Mathematically not accurate except in three positions
(c) Both 1 and 2
(d) Neither 1 nor 2
5. Has only turning pairs
In a crank and slotted lever quick return motion mechanism, the distance between the fixed centres is 200 mm. The lengths of the driving crank and the slotted bar are 100 mm and 500 mm, respectively. The length of the cutting stroke is (a) 100 mm
(b) 300 mm
(c) 500 mm
(d) 700 mm
6. Controls movement of two front wheels
13.
IES – 2015 10.
(c) 2, 3, 5 and 6
(d) 1, 2, 3 and 5
A four-bar mechanism is as shown in the figure bleow. At the instant shown, AB is shorter than CD by 30 cm. AB is rotating at 5 rad/s and CD is ratating at 2 rad/s:
Statement (I) : Hooke’s joint connects two non-parallel non-intersecting shafts to transmit motion with a constant velocity ratio.
In a crank and slotted lever type quick return mechanism, the link moves with an angular velocity of 20 rad/s, while the slider moves with a linear velocity of 1.5 m/s. The magnitude and direction of Coriolis component of acceleration with respect to angular velocity are
(b) 30 m/s2 and direction is such as to rortate slider velocity in the opposite sense as the angular velocity
C
A D
The length of AB is (a) 10 cm
(b) 20 cm
(c) 30 cm
(d) 40 cm
IES – 2014 14.
(a) 30 m/s2 and direction is such as to rotate slider velocity in the same sense as the angular velocity
15.
m/s2
(c) 60 and direction is such as to rotate slider velocity in the same sense as the angular velocity
In a crank and slotted lever quick-return motion, the distance between the fixed centres is 150 mm and the length of the driving crank is 75 mm. The ratio of the time taken on the cutting and return strokes is (a) 1.5
(b) 2.0
(c) 2.2
(d) 2.93
Consider the following statements: In a slider-crank mechanism, the slider is at its dead centre position when the 1. slider velocity is zero
(d) 60 m/s2 and direction is such as to rotate slider velocity in the opposite sense as the angular velocity 12.
(b) 1, 2, 3 and 6
B
Statement (II) : Hooke’s joint connects two shafts the axes of which do not remain in alignment while in motion. 11.
(a) 2, 4, 5 and 6
2. slider velocity is maximum 3. slider acceleration is zero
Which of the following are associated with Ackerman steering mechanism used in automobiles?
4. slider acceleration is maximum (a) 1 and 4
(b) 1 and 3
1. Has both sliding and turning pairs
(c) 2 and 3
(d) 2 and 4
2. Less friction and hence long life 3. Mechanically correct in all positions
Which of the above statements are correct?
16.
Which one of the following mechanisms is an inversion of double slider-crank chain?
IES Topic Wise Objective Solved Paper-II 1995-2017
4 (a) Elliptic trammels
Which of these statements are correct?
(b) Beam engine
(a) 1, 2 and 3
(b) 1 and 3 only
(c) Oscillating cylinder engine
(c) 1 and 2 only
(d) 2 and 3 only
(d) Coupling rod of a locomotive 17.
18.
22.
The number of instantaneous centres of rotation for a 10-link kinematic chain is (a) 36
(b) 90
(c) 120
(d) 45
A slider moves with uniform velocity v on a revolving link of length r with angular velocity . The Coriolis acceleration component of a point on the slider relative to a coincident point on the link is equal to
Universal or Hooke’s joint is used to connect two non-coaxial rotating shafts transmitting power and intersecting at a small angle transmitting power. For a constant rotating speed of the driving shaft, the maximum fluctuation of instantaneous speed of the driven shaft : (a) Varies as the square of the angle (in radian) between the two shafts
(a) v parallel to the link
(b) Varies as the angle (in radian) between the two shafts
(b) 2v perpendicular to the link (c) v perpendicular to the link
(c) Varies as the inverse of the angle (in radian) between the two shafts
(d) 2v parallel to the link
(d) is nil
IES – 2012
IES – 2013
19.
20.
21.
The minimum number of links in a constrained planer mechanism involving revolute pairs and two higher pairs is: (a) 3
(b) 4
(c) 5
(d) 6
A man is climbing up a ladder which is resting against a vertical wall. When he was exactly half way up, the ladder started slipping. The path traced by the man is: (a) Parabola
(b) Hyperbola
(c) Ellipse
(d) Circle
23.
24.
Consider the following statements regarding motions of various points of the same link of a mechanism: 1. The magnitude of the relative velocity of a point A with respect to a point B on rotation of link is given by multiplication of angular velocity with distance between the points. 2. The direction of the relative velocity of a point A with respect to a point B on a rotating link is perpendicular to AB. 3. The resultant acceleration of a point A with respect to a point B of the same link is perpendicular to AB
25.
The driving and driven shaft connected by a Hooke’s joint are inclined by an angle to each other. The angle through which the driving shaft turns is given by . The condition for the two shafts to have equal speeds is (a) cos = sin
(b) sin tan
(c) tan cos
(d) cot = cos
In a crank and slotted lever quick return motion mechanism, the distance between the fixed centers is 160 mm and the driving crank is 80 mm long. The ratio of time taken by cutting and return strokes is (a) 0.5
(b) 1
(c) 1.5
(d) 2
In a elliptic trammel, the length of the link connecting the two sliders is 100 mm. The tracing pen is placed on 150 mm extension of this link. The major and minor axes of the ellipse 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
IES Topic Wise Objective Solved Paper-II 1995-2017
18
Sol–1:
where V is sliding velocity of slider over oscillating link and ‘ ’ is angular velocity of osci l l at i ng l i nk i n qui ck ret urn mechanism.
(b) The angular velocity of link PQ, 2N 2 100 rad/sec 60 60 Velocity of point Q on link PQ, =
Sol–8:
The expression for acceleration of follower involve following parameters
2 100 0.30 60 = 3.14 m/sec
V = PQ
Sol–2:
(a) Forced closed mechanism require external force to maintain mechanical contact between links. Slider-crank mechanism does not requires such force.
Sol–3:
(b)
(c)
(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) Sol–9:
Cam rotation speed ' '
(c) Driving crank length
Degree of freedom using grubler criterion, DoF = 3 (n – 1) – 2J – h – Fr
O1A = 100 mm Slotted Bar length
= 3 (10 –1)–2×12 – 0 – 0 = 3 Sol–4:
(c) No of instantaneous centre,
O2B = 500 mm
In triangle O2O,A sin =
43 6 = C2 C2 2 n
Sol–5:
4
= 30°
(c) C
In quick return mechanism, slider moves on oscilating/rotating link, so Coriolis acceleration exist. But in this mechanism, one link rotates and other oscillates and slider a third link slides.
Sol–7:
A
(b) Successfully constraint motion means that the motion is constraint only in the presence of external force. The piston inside cylinder executes constraint motion due to it design and gudgeon pin. While foot step bearing requires a force to press the shaft in bearing to have constraint motion (rotation). (a) The expression for Coriolis component of acceleration, = 2 V
B
O1 200mm
Sol–6:
O1A 100 O1O2 200
O2
Length of cutting stroke, = 2BC 2O2Bsin 2 500 sin30 = 2 × 500 × 0.5 = 500 mm Sol–10: (d) The Hooke’s joint connects two non-parallel shafts but intersecting. For constant velocity ratio there are two Hooke’s joints in particular torks orientation
Theory of Machines
19
Sol–11: (c)
•
The schematic of quick return mechanism V C
•
2V
Since very less sliding surfaces (turning pairs only) So longer life. Due to longer life it is used generally despite it is not correct mathematically. Steering control is provided in front wheels only in all steering mechanisms.
Sol–13: (b) The schematic of mechanism, B
C
The Coriolis acceleration, 1=5rad/sec
ac = 2V = 2 × 1.5 × 20 = 60 m/sec2 Direction: The direction of Coriolis component of acceleration depends upon velocity of slider ‘C’. If slider moves away from centre, the Coriolis acceleration will be in the direction of link rotation. If the slider moves toward centre of rotation the Coriolis acceleration will be opposite to link rotation as shown below.
A 2 =2rad/sec D
At the instant shown in figure, the linear velocity of point B and C will be same,
1AB 2 CD 5AB = 2CD
V 2V 2V
C V
A
w
A
•
D
5 AB 2
5 AB AB 30 2 3 AB 30 2 2 30 AB = = 20 cm 3
Sol–14: (b) The schematic of quick return mechanism-
B
•
CD =
A
Since nothing is mentioned about velocity of slider (away or toward centre) so assume it moves away from centre, the right. Sol–12: (a) The basic schematic of Ackerman steering mechanism,
…(i)
Since the difference between AB and CD CD – AB = 30 From equation (i)
C
All pairs (A, B, C, and D) are turning pair in four bar mechanism A, B, C and D. This steering satisfy fundamental w equation of steering cot cot = in three positions only namely in straight motion = 0 and two positions 25 toward left and right
Distance between fixed centersAB = 150 mm B
150 mm
C
VBA = VCD
A
75
m
90°
C
IES Topic Wise Objective Solved Paper-II 1995-2017
20 Length of driving crank,
1 = r2 1 n So acceleration is maximum.
BC = 75 mm
In triangle ABC,
cos =
Hence at dead centres, the velocity is zero and acceleration is maximum.
BC 75 1 AB 150 2
Sol–16: (a)
= 60°
Since cutting of metal happens to be in forward stroke and angle rotated by crank BC during cutting
Sol–17: (d)
= 360 2
= 360 – 120 = 240° The angle turned by crank during return stroke. B = 2 = 120° Ratio of cutting stroke time to return stroke time,
=
Elliptical trammel, skotch yoke mechanism and Oldham coupling are example of inversion of double slider crank mechanism.
240 = =2 120
Number of I-centres of kinematic chain having n-links, =
n n 1 10 9 45 2 2
Sol–18: (b) The schematic of mechanism having slider on rotating link is 0
V
Sol–15: (a) A r B
A
x
O
r
The distance travelled by slider B during rotation of crank OA from dead position A.
x = r 1 cos n n2 sin2 The velocity of slider B, dx dt sin 2 V = r sin 2 2 2 n sin
V =
At dead center, = 0 or velocity, V = 0
The coriolis component of acceleration acting on slider, = 2V The direction of coriolis component of acceleration is always perpendicular to to rotating link. Sol–19: (b) Gruebler’s criterion for constrained planer mechanism i.e. one degree of freedom planer mechanism. F = 1 = 3 (n – 1) – 2j – h
higher pair, h = 2
Acceleration,
1 = 3n – 3 – 2j – 2 3n = 2j + 6
cos 2 a = r cos + n 2
a = r2
r2 n
2j = 3n – 6 n = number of links j = number of revolute/lower pair,
Theory of Machines
21
The RHS of equation should be positive and even number put,
Sol–21. (c) A
n = 3
VAB
2j = 3 × 3 – 6 = 3 odd number B
Now put, n = 4
Velocity of A relative to B
2j = 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 by man will be circle. This is the case of elliptical trammel when the point of consideration is mid point of binary link. This point will trace a circle. y
VAB = AB The direction of VAB is perpendicular to AB So statement one and two are correct.The statement three is wrong because there are two components of acceleration of point B with respect to A one centripetal along B to A and other tangential perpendicular to AB. So resultant can never be perpendicular to AB. Sol–22. (a) The schematic of Hooke’s joint
A C Mid Point
N2
x
O
B
Dr
ive
n
sh
af
t
N1
Alternatively
Driving shaft
From figure
Maximum fluctuation of speed in universal or Hooke’s joint,
y A
sin2 cos for small value of
N2 – N1 = Ν
C(x, y)
x
y
B
O
sin sin x
x = ACcos y = BC sin 2
x y AC BC 2
x y AC BC
2
= cos2 sin2
cos 1
N 2
where is in radian Sol–23: (c)
Dr
2
= 1
At mid point, AC = BC = AB/2 2
AB x 2 + y2 = 2 This is equation of circle.
ive
n
sh
af
t
2
1 Driving shaft
The velocity ratio in Hooke's joint, 2 cos = 1 1 sin2 cos2