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2 MECHNICL ENGINEERING ESE TOPICWISE OBJECTIVE SOLVED PPER-II FROM ( ) UPSC Engineering Services Examination State Engineering Service Examination & Public Sector Examination. IES MSTER PUBLICTION Office : F-16, (Lower Basement), Katwaria Sarai, New Delhi Phone : , Mobile : , info@iesmaster.org Web :
3 IES MSTER PUBLICTION F-16, (Lower Basement), Katwaria Sarai, New Delhi Phone : , Mobile : , info@iesmaster.org Web : No part of this booklet may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying or otherwise or stored in a database or retrieval system without the prior permission of IES MSTER PUBLICTION, New Delhi. Violaters are liable to be legally prosecuted. First Edition : 017 Typeset at : IES Master Publication, New Delhi
4 PREFCE 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
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6 CONTENT 1. Theory of Machines Machine Design Strength of Materials Engineering Materials Production Engineering Industrial Engineering Miscellaneous
7 1 S YL L BU 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. utomatic Controls. Contents 1. nalysis of Planer Mechanism Governors and Cam Slider Crank Mechanism and Flywheel Gear and Gear Trains Balancing Linear Vibration nalysis utomatic Control and Whirling of Shafts
8 1 IES 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 P Q 50 cm R 70 cm 60 cm (a).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? 1. Cam and roller mechanism. Door-closing mechanism 3. Slider-crank mechanism Select the correct answer using the code given below. (a) 1 and only (b) 1 and 3 only (c) and 3 only (d) 1, and 3 3. planer mechanism has 10 links and 1 rotary joints. Using Grubler s criterion, the number of degrees of freedom of the mechanism is (a) 1 (b) 3 (c) (d) 4 4. The number of instantaneous centres of rotation in a slider-crank quick return mechanism is (a) 10 (b) 8 (c) 6 (d) 4 S IES Statement (I) : In quick return motion mechanism, Coriolis acceleration exists. Statement (II) : Two links in this mechanism oscillate with one sliding relative to the other. 6. Consider the following motions : 1. Piston reciprocating inside an engine cylinder. Motion of a shaft between foot-step bearings Which of the above can rightly be considered as successfully constrained motion? (a) 1 only (b) only (c) Both 1 and (d) Neither 1 nor 7. Coriolis component of acceleration depends on 1. angular velocity of the link. acceleration of the slider 3. angular acceleration of the link Which of the above is/are correct? (a) 1 only (b) only (c) 1 and 3 (d) 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 of circular arc. roller diameter and radius of circular arc
9 Theory of Machines 3 Which of the above is/are correct? (a) 1 only (b) only (c) Both 1 and (d) Neither 1 nor 9. In a crank and slotted lever quick return motion mechanism, the distance between the fixed centres is 00 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 IES Statement (I) : Hooke s joint connects two non-parallel non-intersecting shafts to transmit motion with a constant velocity ratio. Statement (II) : Hooke s joint connects two shafts the axes of which do not remain in alignment while in motion. 11. In a crank and slotted lever type quick return mechanism, the link moves with an angular velocity of 0 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 (a) 30 m/s and direction is such as to rotate slider velocity in the same sense as the angular velocity (b) 30 m/s and direction is such as to rortate slider velocity in the opposite sense as the angular velocity (c) 60 m/s and direction is such as to rotate slider velocity in the same sense as the angular velocity (d) 60 m/s and direction is such as to rotate slider velocity in the opposite sense as the angular velocity 1. Which of the following are associated with ckerman steering mechanism used in automobiles? 1. Has both sliding and turning pairs. Less friction and hence long life 3. Mechanically correct in all positions 4. Mathematically not accurate except in three positions 5. Has only turning pairs 6. Controls movement of two front wheels (a), 4, 5 and 6 (b) 1,, 3 and 6 (c), 3, 5 and 6 (d) 1,, 3 and four-bar mechanism is as shown in the figure bleow. t the instant shown, B is shorter than CD by 30 cm. B is rotating at 5 rad/s and CD is ratating at rad/s: B The length of B is (a) 10 cm (c) 30 cm D C IES 014 (b) 0 cm (d) 40 cm 14. 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).0 (c). (d) Consider the following statements: In a slider-crank mechanism, the slider is at its dead centre position when the 1. slider velocity is zero. 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) and 3 (d) and Which one of the following mechanisms is an inversion of double slider-crank chain?
10 4 IES Topic Wise Objective Solved Paper-II (a) Elliptic trammels (b) Beam engine (c) Oscillating cylinder engine (d) Coupling rod of a locomotive 17. The number of instantaneous centres of rotation for a 10-link kinematic chain is (a) 36 (b) 90 (c) 10 (d) 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 (a) (b) (c) (d) v parallel to the link v perpendicular to the link v perpendicular to the link v parallel to the link IES 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 0. 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 (c) Ellipse (b) Hyperbola (d) Circle 1. 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 with respect to a point B on rotation of link is given by multiplication of angular velocity with distance between the points.. The direction of the relative velocity of a point with respect to a point B on a rotating link is perpendicular to B. 3. The resultant acceleration of a point with respect to a point B of the same link is perpendicular to B Which of these statements are correct? (a) 1, and 3 (c) 1 and only (b) 1 and 3 only (d) and 3 only. 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 (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 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 4. 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) 5. 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) 50 mm and 150 mm (b) 50 mm and 100 mm (c) 500 mm and 300 mm (d) 500 mm and 00 mm
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12 1 8 IES Topic Wise Objective Solved Paper-II Sol 1: Sol : Sol 3: Sol 4: Sol 5: Sol 6: Sol 7: (b) The angular velocity of link PQ, = N 100 rad/sec Velocity of point Q on link PQ, (a) 100 V = PQ = 3.14 m/sec Forced closed mechanism require external force to maintain mechanical contact between links. Slider-crank mechanism does not requires such force. (b) Degree of freedom using grubler criterion, DoF = 3 (n 1) J h F r = 3 (10 1) = 3 (c) No of instantaneous centre, (c) 4 3 C C 6 = n 4 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. (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, = V Sol 8: Sol 9: where V is sliding velocity of slider over oscillating link and is angular velocity of oscill ating link in qui ck return mechanism. (c) The expression for acceleration of follower involve following parameters (i) Radius of flank of cam surface, r f (ii) Least radius of cam profile or base circle radius r b. (iii) Radius of roller r r. (iv) Cam rotation speed ' ' (c) Sol 10: (d) Driving crank length O 1 = 100 mm Slotted Bar length O B = 500 mm In triangle O O, 00mm O 1 sin = C = 30 O O1 100 O O 00 1 B Length of cutting stroke, = BC OBsin 500 sin30 = = 500 mm 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
13 Theory of Machines 1 9 Sol 11: (c) The schematic of quick return mechanism C V V The Coriolis acceleration, a c = V = = 60 m/sec 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. C V V V Since nothing is mentioned about velocity of slider (away or toward centre) so assume it moves away from centre, the right. Sol 1: (a) The basic schematic of ckerman steering mechanism, w D B C ll pairs (, B, C, and D) are turning pair in four bar mechanism, 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 5 toward left and right C V Sol 13: (b) 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. The schematic of mechanism, B = 5rad/sec 1 C = rad/sec t the instant shown in figure, the linear velocity of point B and C will be same, V B = V CD B CD 1 5B = CD D (i) Since the difference between B and CD CD B = 30 From equation (i) CD = 5 B 5 B B 30 3 B 30 B = 30 = 0 cm 3 Sol 14: (b) The schematic of quick return mechanism- Distance between fixed centers- B = 150 mm B 150 mm 75m 90 C
14 0 IES Topic Wise Objective Solved Paper-II Length of driving crank, Sol 15: (a) BC = 75 mm In triangle BC, BC 75 1 cos = B 150 = 60 Since cutting of metal happens to be in forward stroke and angle rotated by crank BC during cutting = 360 = = 40 The angle turned by crank during return stroke. B = = 10 Ratio of cutting stroke time to return stroke time, x B = 40 = 10 = The distance travelled by slider B during rotation of crank O from dead position. x = r 1 cos n n sin r r O The velocity of slider B, V = d x dt V = sin r sin n sin t dead center, = 0 or velocity, V = 0 cceleration, a = a = cos r cos + n r r n Sol 16: (a) Sol 17: (d) Sol 18: (b) Sol 19: (b) = r 1 n 1 So acceleration is maximum. Hence at dead centres, the velocity is zero and acceleration is maximum. Elliptical trammel, skotch yoke mechanism and Oldham coupling are example of inversion of double slider crank mechanism. Number of I-centres of kinematic chain having n-links, = nn The schematic of mechanism having slider on rotating link is 0 The coriolis component of acceleration acting on slider, = V The direction of coriolis component of acceleration is always perpendicular to to rotating link. Gruebler s criterion for constrained planer mechanism i.e. one degree of freedom planer mechanism. higher pair, h = V F = 1 = 3 (n 1) j h 1 = 3n 3 j 3n = j + 6 j = 3n 6 n = number of links j = number of revolute/lower pair,
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16 Theory of Machines 1 Sol 0: (d) The RHS of equation should be positive and even number put, n = 3 Now put, j = = 3 odd number n = 4 j = = 6, even number So minimum number of link is 4. 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 O lternatively From figure y O x C Mid Point B C( x, y) y x B x = Ccos y = BC sin x y C BC x y C BC x = = 1 t mid point, C = BC = B/ x + y = B This is equation of circle. cos sin Sol 1. (c) Sol. (a) B V B Velocity of relative to B V B = B The direction of V B is perpendicular to B 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 one centripetal along B to and other tangential perpendicular to B. So resultant can never be perpendicular to B. The schematic of Hooke s joint N 1 Driving shaft N Driven shaft Maximum fluctuation of speed in universal or Hooke s joint, sin N N 1 = Ν cos for small value of Sol 3: (c) sin cos 1 N where is in radian 1 Driven shaft Driving shaft The velocity ratio in Hooke's joint, = cos 1 sin cos 1
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