Table Of Content

MADE ERSH Publications UPSC ESE 2018 MAIN EXAM 18 Years Previous Solved Pa Subjectwise Conventional Solved Questions Fully Solved with Thoroughly Revised Complete Explanations and Updated ITIRDE ERSU MADE EASY Publications Corporate Office: 44-A/4, Kalu Sarai (Near Hauz Khas Metro Station), New Delhi-110016 E-mail: [email protected] Contact: 011-45124660, 0-8860378007 Visit us at: www.madeeasypublications.org ESE-2018: Mains Examination Mechanical Engineering : Paper-II © Copyright, by MADE EASY Publications. All rights are reserved. No part of this publication may be reproduced, stored in or introduced into a retrieval system, or transmitted in any form or by any means (electronic, mechanical, photo-copying, recording or otherwise), without the prior written permission of the above mentioned publisher of this book. 1st Edition : 2011 2nd Edition : 2012 3rd Edition : 2013 4th Edition : 2014 5th Edition : 2015 6th Edition : 2016 7th Edition : 2017 MADE EASY PUBLICATIONS has taken due care in collecting the data and providing the solutions, before publishing this book. Inspite of this, if any inaccuracy or printing error occurs then MADE EASY PUBLICATIONS owes no responsibility. MADE EASY PUBLICATIONS will be grateful if you could point out any such error. Your suggestions will be appreciated. © All rights reserved by MADE EASY PUBLICATIONS. No part of this book may be reproduced or utilized in any form without the written permission from the publisher. ESE 2018 UPSC ENGINEERING SERVICES EXAMINATION Mains Examination Thoroughly Revised and Years Updated Previous SOlved Papers Mechanical Engineering Conventional Solved Questions Paper-II CONVENTIONAL SOLVED PAPERS of UPSC ENGINEERING SERVICES EXAMINATION Also useful for State Engineering Services Examinations www.madeeasypublications.org Director's Message During the last few decades of engineering academics, India has witnessed geometric growth in engineering graduates. It is noticeable that the level of engineering knowledge has degraded gradually, while on the other hand competition has increased in each competitive examination including GATE and UPSC examinations. Under such scenario higher level efforts are required to take an edge over other competitors. The objective of MADE EASY books is to introduce a simplified approach to the overall concepts of related stream B. Singh (Ex. IES) in a single book with specific presentation. The topic-wise presentation will help the readers to study & practice the concepts and questions simultaneously. The efforts have been made to provide close and illustrative solutions in lucid style to facilitate understanding and quick tricks are introduced to save time. Following tips during the study may increase efficiency and may help in order to achieve success. • Thorough coverage of syllabus of all subjects • Adopting right source of knowledge, i.e. standard reading text materials • Develop speed and accuracy in solving questions • Balanced preparation of Paper-I and Paper-II subjects with focus on key subjects • Practice online and offline modes of tests • Appear on self assessment tests • Good examination management • Maintain self motivation • Avoid jumbo and vague approach, which is time consuming in solving the questions • Good planning and time management of daily routine • Group study and discussions on a regular basis • Extra emphasis on solving the questions • Self introspection to find your weaknesses and strengths • Analyze the exam pattern to understand the level of questions • Apply shortcuts and learn standard results and formulae to save time B. Singh (Ex. IES) CMD, MADE EASY Group ESE 2018: Main Examination Mechanical Engineering: Paper-II Conventional Solved Questions of UPSC Engineering Services Examination Contents SI. Topic Pages 1. Mechanisms and Machines 1-50 2. Strength of Materials 51-95 3. Design of Machine Elements 96-139 4. Manufacturing Engineering 140-246 5. Industrial Engineering 247-292 6. ESE 2017 : Conventional Solved Paper (Paper-II) 293-330 0000 Revised Syllabus of ESE : Types of Kinematics Pair, Mobility, Inversions, Kinematic Analysis, Velocity and Acceleration Analysis of Planar Mechanisms, CAMs with uniform acceleration and retardation, cycloidal motion, oscillating followers; Vibrations -Free and forced vibration of undamped and damped SDOF systems, Transmissibility Ratio, Vibration Isolation, Critical Speed of Shafts. Gears - Geometry of tooth profiles, Law of gearing, Involute profile, Interference, Helical, Spiral and Worm Gears, Gear Trains- Simple, compound and Epicyclic; Dynamic Analysis - Slider - crank mechanisms, turning moment computations, balancing of Revolving & Reciprocating masses, Gyroscopes -Effect of Gyroscopic couple on automobiles, ships and aircrafts, Governors. Machine Vibration, Data acquisition, Fault Detection, Vibration Monitoring, Field Balancing of Rotors, Noise Monitoring, Flywheels. Q.1 A single cylinder two stroke vertical engine has a bore of 30 cm and a stroke of 40 cm with a connecting rod of 80 cm long. The mass of the reciprocating parts is 120 kg. When the piston is at quarter stroke and moving down, the pressure on it is 70 N/cm2. If the speed of the engine crank shaft is 250 rpm clockwise, find the turning moment on the crank shaft. Neglect the mass and inertia effects on connecting rods and crank. [15 marks : 2001] Solution: Length of stroke = 40 cm Length of crank = 20 cm 40 In quarter stroke the distance travelled = — = 10 cm 4 8 cp 0 x = r(1 - cos 0) 0 C) 3 mg Given that when piston is at quarter stroke means x = Stroke/4 = 2r/4 = r/2 r/2 = r(1 - cos 0) 1/2 = (1 - cos 0) = 60° 1 I 2 1 I cos 13 = n2 -sin u = 4.v16 -sin2 60° = 0.976 = 12.5° Now piston effort F = F - F + (mg) p g I 2 ESE-Mains Mechanical Engineering • Paper-II Solved Papers where Fg = gas pressure force = p1A1 F = 70 x x (30)2 4 Fg = 49.455 kN = Inertia force 2 ( cos20 cos20) mr03 cos° 120 x 0.2 x (211x 25° 2(cos0 + = 60 ) n = 6.168 kN mg = weight of piston = 120 x 9.81 1.172 kN Put all these values in equation (i), we get Fp = 49.455 - 6.168 + 1.192 Fp = 44.46 kN Turning moment M = FT X r, where, FT = Fc sin (0 +(3) where Fc= connecting rod force = Fp/cosr3 Fp sin(0 + p) 44.46 x 103 x sin(60 +12.5)x 0.20 M= x r cos i3 cos12.5° = 8.68 kN-m Q.2 The turning moment of an engine is given by T = 10000 + 2000 sin 20 - 1800 cos 20, 0-crank angle from TDC N = 250 rpm, / = 80 cm, r = 20 cm, mr = 120 kg. Assuming load is constant. Find (i) Power of the engine. (ii) Moment of inertia of the wheel. (If the speed variation from the mean speed is not exceed ±0.25%) (iii) Angular acceleration of the wheel at 0 = 45°. [15 marks : 2001] Solution : 27EN Power of engine = 26.18 rad/sec Tmeanx w = — 60 Power of engine = 10000 x 26.18 = 261.8 kW T= 10000 + 2000 sin 20 -1800 cos 20 (T - Tm) = AT = 2000 sin 20 - 1800 cos 20 T AT =- 0 1800 tan 20 = = 0.9 2000 0= ET-c+21°, n=0,1, 2 2 Max. flucluation of energy 1.111° 21° 111° 201° 291° 360° AE = AT • c19 J21° 111° = 521. (2000 sin20 1800 cos29)de 1800 = 1000[-cos 20[1211:1 [sin 2011110 = 1486.3 + 900 x 1.3383 2 21° 2690.735 N-m MIME EFISH Mechanisms and Machines 3 AE = 2E • Cs AE = Ico2 • Cs 2 0.5 2690.735 = / x 26.18 X 100 I = 785.166 kg-m2 (iii) At 0 = 45° AT = 2000 sin 90° - 1800 cos 90° = 2000 N-m AT= Ia AT 2000 a - / 785.166 a = 2.547 rad/sec2 Q.3 Can the above engine be completely balanced? If yes, explain the method of balancing. If not, what is the technique adopted for minimisation of unbalance. Find out the mass required at a radius of 30 cm from crank shaft for such balancing. Also find the resultant unbalance force at the crank position, as given in part (a). [10 marks : 2001] Solution: No the above engine cannot be completely balanced. To minimize unbalance we introduce a fraction of unbalanced mass at the crank opposite to unbalance Mw2 R = cm co2 r 2 MCOr If C is the fraction mass unbalanced MR = cmr Unbalance along line of stroke = m co2 r cos()- M cwt R cos0 = m co2 r cos() - cm co2 r cos° = m co2 r cosh (1 - c) 2 mw R unbalance along vertical = M co2 R sin 0 = cm vv2 r sin() V(mal202c2 sin2 + (m(0202 cos c) resultant t m(02_ 1_2 r \lc; sin2 0 + cos2 0(1— c)2 If C = - the forces are minimum 2 rn22r (2n 6x 250)2 x0.2 resultant x = 8224.6 N 0 mass required at a radius of 30 cm 1/2 x 120 x 20 M - - 40 kg 30 Q.4 The bearings of a shaft A and B are 5 m apart. The shaft carries three eccentric masses C, D and E. Which are 160 kg, 170 kg and 85 kg respectively. The respective eccentricity of each masses measured from axis of rotation is 0.5 cm, 0.3 cm 0.6 cm and distance from A is 1.3 m, 3 m and 4 m respectively. Determine angular position of each mass with respect to C so that no dynamic force is exerted at B and also find dynamic force at A for this arrangement when the shaft runs at 100 rpm. [15 marks : 2002] Solution : No dynamic force at B. Consider Rerefence Plane passing through A 4 ESE-Mains Mechanical Engineering • Paper-II Solved Papers Couple balancing mi m2 m3 160 kg 170 kg 85 kg C D E 0.5 cm 0.3 cm 0.6 cm [0— 1.3 m 1.7 m 1m 1 m A 01=00 x 100 — 2n 60 E = 10.47 rad/sec X-component mr1 cos° + m2r21 COSO + M r COSO + 0 = 0 111 1 2 2 3 3/3 3 160 x 0.5 x 1.3x 1 + 170 x 0.3 x 3 cos02 + 85 x 0.6 x 4 cos03 = 0 104 + 153 cos02 + 204 cos03 = 0 Y-component 0 + 153 sin02 + 204 sin03 = 0 153 sin02 + 204 sin03 = 0 Squaring and Adding equation (i) and equation (ii), 1532 + 2042 + 2 x 153 x 204 [cos02 cos03 + sin02 sin03] = 1042 cos (03— 02) = — 0.8684 03 — 02 = 150.27° 03 = 150.27° + 02 ...(iii) Putting in equation (ii), 153 sin02 + 204 sin (150.27 + 02) = 0 153 sin02 + 204 (0.496 cos02 — 8684 sin02) = 0 153 sin02 + 101.184 cos02— 177.15 sin02 = 0 24.15 sin02 = 101.184 cos02 101.184 tan02 = 02 = 76.6°, From equation (iii) 03 = 226.85° 24.15 , Let the dynamic force at A is F Static Balance X-component F • cos() + m1r1w12 cosOi + m2r2o.)22 cos02 + m3r3o332 cos03 + 0 = 0 Fcos0 + 160 x 0.005 x (10.47)2 cos0° + 170 x 0.003 x (10.47)2 cos 76.6° + 85 x (0.006) x (10.47)2 x cos (226.85°) Fcose = — 62.4178 Y-component Fsin 0 + 160 x 0.005 x (10.47)2 x sin 0° + 170 x 0.003 x 10.472 sin 76.6° + 85 x 0.006 x 10.47 x sin 226.85° Fsin0 = — 13.6 F = (62.4178)2 + (13.6)2 F = 63.88N MADE EASY Mechanisms and Machines 5 At angle (with C) -13.6 tan() = -62.4178 0 = 180 + 12.3 = 192.3° Q.5 The total sleeve movement in a Hartnell governor is 3 cm. The mass of rotating balls is 1.5 kg each. At the mid position of the sleeve arm, which is 6.5 cm long, is horizontal. The ball arm has a length of 7.5 cm. At the mid position of the sleeve, the balls rotate at a radius of 10.5 cm. Due to the mal adjustment of spring, the equilibrium speed of governor at top position is 415 rpm and in lowest position it is 430 rpm. Determine (i) Stiffness and initial compression of spring. (ii) Required initial compression in spring which gives equilibrium speed at the top most position which is 10 rpm more than the lowest position neglect bliquity. [15 marks : 2002] Solution : m = 1.5 kg Top position of sleeve In the top position of sleeve Lowest position x = 1.5 1 10.5 cm 7.5 6.5 x = 1.7308 cm Radius of rotation in top position r2 = 10.5+ 1.7308 = 12.23 cm r6.5 Radius of rotation in lowest position = 10.5 - 1.7308 = 8.77 cm cm h = 3 cm Equilibrium speed is w2 = 415 rpm = 43.46 rad/sec qiii (Fs2 + Mg) Fc2 x 7.5 - x6.5 2 (Fs + mg) 7.5 x 2 1.5 x 0.1223 x (43.46)2 x (Fs2 + Mg) 6.5 Fs2 + Mg = 804.133 In the lowest position (Fsi Fc1 x 7.5 = Mg)x6.5 2 (Fsi + Mg) mr1(1)12 X 7'5 = 2 x 6.5 x(7.5 5x 2) 1.5 x 0.08774-4312 (Fs1 + Mg) 3 6 . Fs1 + Mg = 615.55 Assuming M= 0 (i) Stiffness & Initial compression (Fs2 - Fs1) = kh - k - Fs1 Fs1 804.133 - 615.55 - 6286.1 N/m h 0.03 Initial compression Fs1 615.55 S 9.8 cm 1 k 6286.1