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1995-2018
Office: F-126, (Lower Basement), Katwaria Sarai, New Delhi-110 016
Phone: 011-2652 2064 Mobile: 81309 09220, 97118 53908
Email: [email protected], [email protected]
Web: iesmasterpublications.com, iesmaster.org
CIVIL ENGINEERINGESE SUBJECTWISE
CONVENTIONAL SOLVED PAPER-I
First Edition : 2016
Second Edition : 2017
Third Edition : 2018
Typeset at : IES Master Publication, New Delhi-110016
IES MASTER PUBLICATIONF-126, (Lower Basement), Katwaria Sarai, New Delhi-110016
Phone : 011-26522064, Mobile : 8130909220, 9711853908
E-mail : [email protected], [email protected]
Web : iesmasterpublications.com, iesmaster.org
All rights reserved.
Copyright © 2018, by IES MASTER Publications. No part of this booklet may bereproduced, or distributed in any form or by any means, electronic, mechanical,photocopying, recording, or otherwise or stored in a database or retrieval system withoutthe prior permission of IES MASTER, New Delhi. Violates are liable to be legallyprosecuted.
Engineering Services Exam (ESE) is one of most coveted exams written by engineering students aspiringfor reputed posts in the various departments of the Government of India. ESE is conducted by the UnionPublic Services Commission (UPSC), and therefore the standards to clear this exam too are very high. Toclear the ESE, a candidate needs to clear three stages – ESE Prelims, ESE Mains and Personality Test.
It is not mere hard work that helps a student succeed in an examination like ESE that witnesses lakhsof aspirants competing neck to neck to move one step closer to their dream job. It is hard work along withsmart work that allows an ESE aspirant to fulfil his dream.
After detailed interaction with students preparing for ESE, IES Master has come up with this book whichis a one-stop solution for engineering students aspiring to crack this most prestigious engineering exam.The book includes previous years’ solved conventional questions segregated subject-wise along with detailedexplanation. This book will also help ESE aspirants get an idea about the pattern and weightage ofquestions asked in ESE.
IES Master feels immense pride in bringing out this book with utmost care to build upon the exampreparedness of a student up to the UPSC standards. The credit for flawless preparation of this book goesto the entire team of IES Master Publication. Teachers, students, and professional engineers are welcometo share their suggestions to make this book more valuable.
Mr. Kanchan Kumar ThakurDirector–IES Master
PREFACE
CONTENTS
1. STRENGTH OF MATERIAL .................................................................................. 001 – 203
2. STRUCTURE ANALYSIS ...................................................................................... 204 – 427
3. STEEL STRUCTURE ............................................................................................ 428 – 557
4. RCC AND PRESTRESSED CONCRETE ................................................................ 558 – 705
5. PERT CPM .................................................................................................... .. 706 – 783
6. BUILDING MATERIAL .......................................................................................... 784 – 866
IES –1995
1. The principal stresses at a point in an elastic material are 1.5 (tensile), (tensile)and 0.5 (compressive). The elastic limit in tension is 210 MPa and = 0.3. What wouldbe the value of at failure when computed by the different theories of failure?
[15 Marks]
Sol. Given data :
1 1.5 ; 2 ; 3 = – 0.5 *
2 = +
1 = + (1.5 )
3 = (–0.5 )
(Macroscopic View of a Point)
Elastic limit in tension (fy) = 210 MPa.
= 0.3
Determine : at failure when computed by different theories of failure.
(1) Maximum Principle Stress Theory : As per this theory for no failure maximum principal Stressshould be less than yield stress under uniaxial loading.
So, 1 1.5 yf
1.5 yf yf1.5
210 140.001.5
140 MPa(2) Maximum principal strain theory : As per this theory, for no failure maximum principal strain
should be less than yield strain under uniaxial loading
i.e., max y
E
SYLLABUS
Basics of strength of materials, Types of stresses and strains, Bending moments and shear force,concept of bending and shear stresses;
Elastic constants, Stress, Plane stress, Strains, Plane strain, Mohr’s circle of stress and strain.Elastic theories of failure. Principal Stresses, Bending, Shear and Torsion.
Strength of MaterialUNIT
1
2 | ESE Subjectwise Conventional Solved Paper-I 1995-2018
Among x y z x( , , ), will be maximum because 1 is maximum
x = 1.5 0.5 1.5 – 0.3 0.15 1.35–
E E E E E
1.35
E
210E
2101.35 155.55 MPa.
(3) Maximum shear stress theory : For no failure, maximum shear stress should be less thanor equal to maximum shear stress under uniaxial loading.
Since we have 3–D case,
Maximum shear stress = 1 3 2 31 2– ––maximum , ,
2 2 2
Maximum shear stress under uniaxial loading : y = yf2
From this theory
1 3 2 31 2– ––Maximum , ,
2 2 2
yf2
1 3–2
2102
1.5 – –0.5
2
105
2 210 105 MPa.
(4) Maximum strain energy theorem : For no failure, maximum strain energy absorbed at a pointshould be less than or equal to total strain energy per unit volume under uniaxial loading, whenmaterial is subjected to stress upto elastic limit.
Total strain energy = 2 2 21 2 3 1 2 2 3 3 1– 2
2E
Total strain energy per unit volume under uniaxial loading = 2
yf2E
According to this theory, 2 2 2 21 2 3 1 2 2 3 3 y– 2 f
2 2 21.5 –0.5 – 2 0.3 1.5 – 0.5 –1.5 0.5 2102
2 2210
3.35 114.73 MPa
(5) Maximum distortion energy theory :- For no failure, maximum shear strain energy in a bodyshould be less than maximum shear strain energy due to uniaxial loading.
2 2 21 2 2 3 3 1
1 – – –2 2
yf
2 2 21 1.5 – 0.5 –0.5 –1.52 2102
2 2 20.25 2.25 4 2 × 2102
2 22 210
6.5
22 210
6.5
116.487 MPa
Strength of Material | 3
2. The strain measurements from a rectangular strain rosette were e0 = 600 × 10–6, e45 =500 × 10—6 and e90 = 200 × 10–6. Find the magnitude and direction of principal strains.If E = 2 × 105 N/mm2 and = 0.3 find the principal stresses.
[10 Marks ]
Sol.
90°45°
90 = 200 × 10–6
45 = 500 × 10–6
0 = 600 × 10–6
y
x
x
We know that x = x y x y xycos 2 sin 22 2 2
... (i)
and max min =2 2
x y x y xy
2 2 2
... (ii)
Thus to determine the principal strain, we need normal strain in two mutually perpendicular directionand shear strain (xy) associated with these directions.
From (i)
45 = xy0 90 0 90 cos (2 45 ) sin (2 45 )2 2 2
xy 0 90[ ]
500 × 10–6 = 6 6600 200 600 20010 10 cos902 2
+ xy
2 cos 90 0
sin 90 1
6xy 200 10
From (ii)
max min =22
xy0 90 0 902 2 2
=2
6 12 12600 200 600 200 20010 10 102 2 2
=2 2 6[400 (200) (100) ] 10
= 6[400 223.607] 10
max = 623.607 × 10–6 = major principal strain
min = 176.393 × 10–6 = minor principal strain
4 | ESE Subjectwise Conventional Solved Paper-I 1995-2018
Also, we know that,
tan 2P = xy
x y
/ 2 200 200 1/ 2 600 200 400 2
P = 13.282° or 103.282°One of these angles will be associated with major principal strain and other with minor principal strain.To determine which of the angle is associated with major principal strain, let us put the value of P in straintransformation eq.
x =xy0 90 0 90
P Pcos 2 sin 22 2 2
= 6600 200 600 200 200cos 2 (13.282 ) sin 2 (13.482) 102 2 2
x = 623.607 × 10–6
P = 13.282° is associated with major principal straini.e., direction of major principal strain is at 13.282° in anticlockwise direction from 0 strain direction andhence direction of minor principal strain is at 103.282° in anticlockwise direction from 0 strain direction.
Calculation of principal stresses:
max min( )E E
= max
maxminE E
= min
max – 0.3 min = 623.607 ×10–6 × 2 × 105 N/mm2
max– 0.3 min = 124.72 N/mm2 ... (i)min – 0.3 max = 176.393 ×10–6 × 2× 105 N/mm2 ... (ii)
0.3 min – 0.09 max = 35.279 × 0.3 ... (iii)From (i) + (iii)
max(10.09) = 124.7 + 35.279 × 0.3
2max 148.685 N/mm
2min 79.885 N/mm
Alternative approach (Mohr circle approach):
If we use Mohr transformation, we will not have to check which of the two angles 13.28° and 103.28°corresponds to major principal strainBy analytical approach we have found that xy is (+)ve. This implies that it is associated with (+)ve shearstress as shown below.
y
x
Strength of Material | 5
Hence strains are shown as
1
2
x
xy
xy
y
6xy 6
y200 10, i.e. 200 10 ,
2 2
xy
2
(176.393 × 10 , 0)–6
(400 × 10 , 0)–6 2P
(623.607 × 10 , 0)–3 direction of major principal
straindirection of
minor principal strain
xyx ,
2
66 200 10i.e, 600 10 ,
2
direction of x
direction of y
max = (400 × 10–6) + Rmin = (400 × 10–6) –R [R = radius of circle]
R =2
62 200 0 10(600 400)2
= 223.607 × 10–6
max = 623.607 × 10–6
min = 176.393 × 10–6
tan2P =100 1
600 400 2
P = 13.28°
Major principal strain is at 13.28° in anticlockwise direction from the direction of x and minor
principal strain is 18013.28
2 = 103.28° in anticlockwise direction from the direction of x.
3. Draw bending moment & shear force diagrams for the beam loaded as shown in fig.
10 kN 3kN/m1m
P
2 m 2 m2 m 1 m 1 m
[ 15 Marks ]
