CIVIL ENGINEERING
ESE TOPICWISE OBJECTIVE SOLVED
PAPER-I
1995-2024
Office: F-126, (Lower Basement), Katwaria Sarai, New Delhi-110 016
Phone: 011-26522064 Mobile: 81309 09220, 97118 53908
Email: info.publications@iesmaster.org, info@iesmaster.org
Web: iesmasterpublications.com, iesmaster.org
IES MASTER PUBLICATION
F-126, (Lower Basement), Katwaria Sarai, New Delhi-110016
Phone : 011-41013406, Mobile : 8130909220, 9711853908
E-mail : info.publications@iesmaster.org
Web : iesmasterpublications.com
All rights reserved.
Copyright © 2024, by IES MASTER Publication. No part of this booklet may be
reproduced, or distributed in any form or by any means, electronic, mechanical,
photocopying, recording, or otherwise or stored in a database or retrieval system without
the prior permission of IES MASTER Publication, New Delhi. Violates are liable to be
legally prosecuted.
First Edition : 2016
Second Edition : 2017
Third Edition : 2018
Fourth Edition : 2019
Fifth Edition : 2020
Sixth Edition : 2021
Seventh Edition : 2022
Eighth Edition : 2023
Ninth Edition : 2024
Typeset at : IES Master Publication, New Delhi-110016
PREFACE
Engineering Services Examination (ESE) is the gateway to an immensely satisfying job in the engineering
sector of India that offers multi-faceted exposure. The exposure to challenges and opportunities of
leading the diverse field of engineering has been the main reason behind engineering students opting
for Engineering Services as compared to other career options. To facilitate selection into these services,
availability of numerical solution to previous years’ paper is the need of the day.
It is an immense pleasure to present previous years’ topic-wise objective solved papers of ESE. The
revised and updated edition of this book is an outcome of regular and detailed interaction with the
students preparing for ESE every year. The book includes solutions along with detailed explanation to
all the questions. The prime objective of bringing out this book is to provide explanation to each and
every question in such a manner that just by going through the solutions, ESE aspirants will be able to
understand the basic concepts, and have the capability to apply these concepts in solving other questions
that might be asked in future exams. Towards this end, this book becomes indispensable for every ESE
aspiring candidate.
IES Master Publication
New Delhi
CONTENT
1. Strength of Material ...................................................................................... 001 – 220
2. Structure Analysis ......................................................................................... 221 – 342
3. Steel Structure ......................................................................................... 343 – 467
4. RCC and Prestressed Concrete ................................................................... 468 – 599
5. PERT CPM ......................................................................................... 600 – 700
6. Building Material ......................................................................................... 701 – 818
Civil Engineering
2 ESE Topicwise Objective Solved Paper-I 1995-2024
1 STRENGTH OF MATERIALS
4. The stress-strain curve for an ideally plastic
IES-1995 material is
1. Given that f or an element in a body of
homogeneous isotropic material subjected to plane
Stress
Stress
stress; x , y and z are normal strains in x, y,, (a) (b)
z directions respectively and is the Poisson’s
Strain
ratio, the magnitude of unit volume change of the Strain
element is given by
(a) x y z (b) x ( y z )
Stress
Stress
(c) (d)
(c) ( x y z ) (d) 1/ε x 1/ε y 1/ε z
Strain Strain
2. A solid metal bar of uniform diameter D and length
L is hung vertically from a ceiling. If the density 5. A steel cube of volume 8000 cc is subjected to
of the material of the bar is and the modulus an all round stress of 1330 kg/sq. cm. The bulk
of elasticity is E, then the total elongation of the modulus of the material is 1.33 × 106 kg/sq. cm.
bar due to its own weight is The volumetric change is
(a) L / 2E (b) L2 / 2E (a) 8 cc (b) 6 cc
(c) 0.8 cc (d) 10–3 cc
E
(c) E / 2L (d) 6. In terms of bulk modulus (K) and modulus of
2L2
rigidity (G), the Poisson’s ratio can be expressed
3. A rigid beam ABCD is hinged at D and supported as
by two springs at A and B as shown in the given
figure. The beam carries a vertical load P at C. (a) (3K – 4G)/(6K+4G) (b) (3K+4G)/(6K– 4G)
The stiffness of spring at A is 2K and that of B (c) (3K – 2G)/(6K+ 2G) (d) (3K+2G)/(6K – 4G)
is K.
7. Two bars one of material A and the other of material
a a a B of same length are tightly secured between two
unyielding walls. Coefficient of thermal expansion
of bar A is more than that of B. When temperature
D rises the stresses induced are
A B C
(a) tension in both materials
P
(b) tension in material A and compression in
The ratio of forces of spring at A and that of material B
spring at B is (c) compression in material A and tension in
(a) 1 (b) 2 material B
(c) 3 (d) 4 (d) compression in both materials
Civil Engineering
STRENGTH OF MATERIALS 3
8. A column of height ‘H’ and area at top ‘A’ has the
IES-1996
same strength throughout its length, under its
own weight and applied stress ‘P0’ at the top. 14. A bar of circular cross-section varies uniformly
Density of column material is ‘ ’. To satisfy the from a cross-section 2D to D. If extension of the
above condition, the area of the column at the bar is calculated treating it as a bar of average
bottom should be. diameter, then the percentage error will be
(a) 10 (b) 25
HP0 gH
(a) (b) P (c) 33.33 (d) 50
Ae g Ae 0
15. The length, coefficient of thermal expansion and
gH H
(c)
P
(d)
gP
Young’s modulus of bar ‘A’ are twice that of bar
Ae 0 Ae 0 ‘B’. If the temperature of both bars is increased
9. A bar of diameter 30 mm is subjected to a tensile by the same amount while preventing any
load such that the measured extension on a gauge expansion, then the ratio of stress developed in
length of 200 mm is 0.09 mm and the change is bar A to that in bar B will be
diameter is 0.0045 mm. The Poisson’s ratio will (a) 2 (b) 4
be
(c) 8 (d) 16
(a) 1/4 (b) 1/3
16. The lists given below refer to a bar of length L,
(c) 1/4.5 (d) 1/2 cross sectional area A, Young’s modulus E,
10. When a mild-steel specimen fails in a torsion- Poisson’s ratio and subjected to axial stress
test, the fracture looks like ‘p’. Match List-I with List-II and select the correct
answer using the codes given below the lists:
(a)
List-I List-II
(b) A. Volumetric strain 1. 2(1 + )
B. Strain energy per unit volume 2. 3(1 – 2 )
(c)
p
C. Ratio of Young’s modulus to 3. (1 2)
E
(d) bulk modulus
11. A 2 m long bar of uniform section 50 mm2 extends p2
D. Ratio of Young’s modulus to 4.
2 mm under a limiting axial stress of 200 N/ 2E
modulus of rigidity
mm2. What is the modulus of resilience for the
bar? 5. 2(1 – )
(a) 0.10 units (b) 0.20 units Codes:
(c) 10000 units (d) 200000 units A B C D
12. The stress level, below which a material has a (a) 3 4 2 1
high probability of not failing under reversal of
stress, is known as (b) 5 4 1 2
(a) elastic limit (b) endurance limit (c) 5 4 2 1
(c) proportional limit (d) tolerance limit (d) 2 3 1 5
13. If E = 2.06 × 105 N/mm 2, an axial pull of 17. If all dimensions of prismatic bar of square cross-
60 kN suddenly applied to a steel rod 50 mm in section suspended freely from the ceiling of a
diameter and 4 m long, causes an instantaneous roof are doubled then the total elongation produced
elongation of the order of by its own weight will increase
(a) 1.19 mm (b) 2.19 mm (a) eight times (b) four times
(c) 3.19 mm (d) 11.9 mm (c) three times (d) two times
IES MASTER Publication
Civil Engineering
22 ESE Topicwise Objective Solved Paper-I 1995-2024
EXPLANATIONS
1. (a) Unit volume change, FA 2kB A 2 A 2 1.5 B
= = = 3
V Final volume – Initial volume FB kB B B B
=
V Initial volume
4. (c) An ideal plastic material experiences no elastic
V (1 x ) (1 y ) (1 z ) 1
= deformation.
V 1
= 1 + x +y + z + xy + yz + zx + xy z–1 P
5. (a) Bulk modulus =
V / V
product of strain terms are very small, so ne-
glecting them 1330
1.33 106 = V / 8000
V
hence = x y z V = – 8 cc
V
(–) ve sign indicates reduction in volume if stress
2. (b) Elongation in length, dx is d
is compressive in nature.
Pdx
d = for a force of P on element (dx)
AE 6. (c) We know,
L Ax dx E = 2G (1 + ) ... (i)
d = 0 AE E = 3K (1 – 2 ) ... (ii)
2
L x
dx dx (where is poisson’s ratio)
= 0
E 2 Equation (i) (ii)
x
L2 Ax 2 G 1
= 1=
2E 3 K (1 2)
L
x2 L2 3K 6K = 2G 2G
= 2E 2E 3K 2G
0
=
6K 2G
Alternative
7. (d) As the temperature rises, both the bars will
The elongation of bar due to its own weight
have tendency to expand but they are fixed
(w) is between two unyielding walls so they will not
WL ( AL)·L be allowed to expand. Hence in both the bars
= =
2AE 2AE compressive stress will develop.
L2
= P0
2E
8. (c) A
3. (c) Given, KA = 2 KB
Force carried by spring at A ax
x
= FA = k A A 2k BA H dx
Force carried by spring at B
ax + dax
= FB = kBB
a 2a
A B
As we move down weight of column will add
up to produce stresses. Since the column has
B same strength, so to satisfy the condition, the
A X-sectional area must increase as we move
Deflected shape
down
Let area at distance x be ax and in length dx
From similar triangles wt, added = gax dx
A But stress has to remain constant
= B A = 1.5 B
3a 2a
