CIVIL ENGINEERING ESE TOPICWISE CONVENTIONAL SOLVED PAPER-I
1995-2019
Office: F-126, (Lower Basement), Katwaria Sarai, New Delhi-110 016 Phone: 011-2652 2064
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IES MASTER PUBLICATION F-126, (Lower Basement), Katwaria Sarai, New Delhi-110016 Phone : 011-26522064, Mobile : 8130909220, 9711853908 E-mail : info.publications@iesmaster.org Web : iesmasterpublications.com
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First Edition
:
2016
Second Edition
:
2017
Third Edition
:
2018
Fourth Edition
:
2019
Typeset at : IES Master Publication, New Delhi-110016
PREFACE
Engineering Services Exam (ESE) is one of most coveted exams written by engineering students aspiring for reputed posts in the various departments of the Government of India. ESE is conducted by the Union Public Services Commission (UPSC), and therefore the standards to clear this exam too are very high. To clear 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 lakhs of aspirants competing neck to neck to move one step closer to their dream job. It is hard work along with smart 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 which is 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 detailed explanation. This book will also help ESE aspirants get an idea about the pattern and weightage of questions asked in ESE. IES Master feels immense pride in bringing out this book with utmost care to build upon the exam preparedness of a student up to the UPSC standards. The credit for flawless preparation of this book goes to the entire team of IES Master Publication. Teachers, students, and professional engineers are welcome to share their suggestions to make this book more valuable.
MR. KANCHAN KUMAR THAKUR DIRECTOR—IES MASTER
CONTENT
1.
STRENGTH OF MATERIALS
01 – 202
2.
STRUCTURE ANALYSIS
203 – 406
3.
STRUCTURAL DYNAMICS
407 – 409
4.
STEEL STRUCTURE
410 – 532
5.
RCC AND PRESTRESSED CONCRETE
533 – 690
6.
PERT CPM
691 – 767
7.
BUILDING MATERIAL
768 – 864
UNIT-1
STRENGTH OF MATERIALS
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.
CONTENTS
1.
Strength of Materials ................................................................................. 02 – 23
2.
Shear Force and Bending Moment ........................................................... 24 – 81
3.
Deflection of Beam ................................................................................. 82 – 107
4.
Transformation of Stress and Strain ...................................................... 108 – 137
5.
Combined Stress .................................................................................. 138 – 156
6.
Bending Stress in Beam ....................................................................... 157 – 170
7.
Shear Stress in Beams ........................................................................ 171 – 177
8.
Torsion of Circular shaft ........................................................................ 178 – 197
9.
Columns ............................................................................................... 198 – 198
10.
Thick and Thin Cylinder/Sphere ............................................................ 199 – 199
11.
Moment of Inertia ................................................................................. 200 – 202
CHAPTER 1 Q-1:
STRENGTH OF MATERIALS
A cylindrical piece of steel 80 mm dia and 120 mm long is subjected to an axial compressive force of 50,000 kg. Calculate the change in the volume of the piece if bulk modulus = 1.7 × 106 kg/cm2 and Poisons’ ratio = 0.3. [10 Marks, ESE-1997]
Sol:
Given: Axial compressive load = 50,000 kg Bulk modulus (k) = 1.7 × 10+6 kg/cm2
P = 50,000 kg
80 mm
Passion’s ratio () = 0.3 120 mm
Determine: We know that
V = V x y z V
x = y = Z = In our case,
x (y ) (z ) E E E y E
(z ) ( x ) E E
z ( x ) y E E E
y = 0 ; z = 0 50000 kg 995.2 kg/cm2 {(-ve) because its compressive} x = (8)2 cm2 4
Also, we know,
E = 3k (1 2) = 3 × 1.7 × 106 × (1 – 2× 0.3) = 2.04 × 106 kg/cm2
x =
x 995.2 4.878 4 104 E 2.04 106
y =
x 0.3 4.878 104 1.4635 10 4 E
z =
x 1.4635 104 E
v = x y 2 1.951 10 4
[(–) because comp.]
Civil Engineering
ESE Topicwise Conventional Solved Paper-I
3
V = –1.951 × 10–4 V
4 V = 1.951 10
(8)2 12 cm3 = –0.1176 cm3 4
Change in vol. is (–)ve Q-2:
There is volume reduction of 0.1176 cm3
A steel rod, circular in cross-section, tapers from 30 mm diameter to 15 mm diameter over a length of 600 mm. Find how much its length will increase under a pull of 20 kN if Young’s modulus of elasticity = 200 kN/mm2. Derive the formula used. [15 Marks, ESE-1998]
Sol:
D2 P
dx
Dx
P
D1
x L
For deriving the expression of elongation for tapered beam, we assume a tapered beam of Length = L, Small end Dia = D1, Larger end dia = D2
D – D1 Dx = D1 2 x , L
i.e.,
Dx = D1 + kx,
{where Dx is Dia at any distance x from smaller end}
D2 – D1 where k = L
Change in the length of element of length dx = d(L) d(L) =
Pdx AE L
So net elongation
d( L) =
0
L
L =
L
0
L
Pdx Pdx AE 0 D kx 2 E 4 1
4Pdx
4P 2 D1 kx E E
L
D 0
dx
1 kx
2
4P 1 E k D1 kx
L
0
4PL ED1D2
Values given, P = 20 kN = 20 ×103 N D1 = 15 mm,
D2 = 30 mm,
L = 600 mm, L =
E = 200 kN/mm2 = 200 × 103 N/mm2
4 20 103 600 15 30 200 103
0.1697mm
IES MASTER Publication
4 Q-3:
Civil Engineering
Strength of Materials
Draw the diagram of normal forces, stresses and displacements along the length of the stepped bar ABC shown in figure.
50 kN B
A
C
1.0 m
1.0 m
cross-sectional area over AB = 100 kN/mm2
mm2;
and area over BC = 200 mm2; modulus of elasticity = 200 [10 Marks, ESE-1998]
Sol:
50 kN B
A
C
1.0 m
1.0 m
Given: AAB = 100 mm2,
ABC = 200mm2,
E = 200 kN/mm2
Draw Diagram of normal forces normal stresses & displacement along the length
FBD for AB and BC, B A
B 50 kN
50 kN
C
50 kN
50 kN
Normal force diagram
Normal force diagram will be constant equal to 50 kN (tension) (+) 50 kN C B Normal Force Diagram
A
Normal stress diagram 50 1000 2 Stress, AB = 500N / mm 100
BC =
50 1000 250N / mm2 200
The normal stress diagram have discontinuity at interface as shown below 500N/mm +
2
250N/mm +
1m
1m
Normal stress Diagram
2