ESE 2021 - Civil Engineering ESE Topicwise Conventional Solved paper - 1 by IES Master Publication

CIVIL ENGINEERING ESE TOPICWISE CONVENTIONAL SOLVED PAPER-I

1995-2019

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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

STRENGTH OF MATERIALS

01 – 202

STRUCTURE ANALYSIS

203 – 406

STRUCTURAL DYNAMICS

407 – 409

STEEL STRUCTURE

410 – 532

RCC AND PRESTRESSED CONCRETE

533 – 690

PERT CPM

691 – 767

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

Strength of Materials ................................................................................. 02 – 23

Shear Force and Bending Moment ........................................................... 24 – 81

Deflection of Beam ................................................................................. 82 – 107

Transformation of Stress and Strain ...................................................... 108 – 137

Combined Stress .................................................................................. 138 – 156

Bending Stress in Beam ....................................................................... 157 – 170

Shear Stress in Beams ........................................................................ 171 – 177

Torsion of Circular shaft ........................................................................ 178 – 197

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  104 E 2.04  106

y = 

x  0.3  4.878  104  1.4635  10 4 E

z = 

x  1.4635  104 E

 v =  x   y  2   1.951  10 4

[(–) because comp.]

Civil Engineering

ESE Topicwise Conventional Solved Paper-I

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

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 



 0

Pdx Pdx  AE 0  D  kx 2 E   4 1

4Pdx

4P  2   D1  kx  E E

 D 0

1  kx

2

 4P  1    E  k D1  kx  

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

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

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



50 kN

Normal force diagram

Normal force diagram will be constant equal to 50 kN (tension) (+) 50 kN C B Normal Force Diagram



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 +

250N/mm +

Normal stress Diagram