ESE 2019 - Civil Engineering ESE Subjectwise Conventional Solved Paper 1

CIVIL ENGINEERING ESE SUBJECTWISE CONVENTIONAL SOLVED PAPER-I

1995-2018

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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@iesmasterpublications.com, info@iesmaster.org Web : iesmasterpublications.com, iesmaster.org

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Copyright Š 2018, by IES MASTER Publications. 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, New Delhi. Violates are liable to be legally prosecuted.

First Edition

:

2016

Second Edition

:

2017

Third Edition

:

2018

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

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

6.

BUILDING MATERIAL .......................................................................................... 784 – 866

...................................................................................................... 706 – 783

UNIT

Strength of Material

1

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.

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 would be the value of  at failure when computed by the different theories of failure?

[15 Marks]

Sol.

2 = + 

Given data : 1   1.5  ;

2    ;

1 = + (1.5)

3 = –  0.5   * 3 = (–0.5)

Elastic limit in tension (f y) = 210 MPa.  = 0.3

(Macroscopic View of a Point)

Determine :  at failure when computed by different theories of failure.

(1)

Maximum Principle Stress Theory : As per this theory for no failure maximum principal Stress should be less than yield stress under uniaxial loading.

So,  

(2)

 1  1.5 

 fy

1.5  fy



 

fy 1.5



 210    140.00     1.5 

  140 MPa

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

2

|

ESE Subjectwise Conventional Solved Paper-I 1995-2018

Among ( x ,  y ,  z ),  x will be maximum because 1 is maximum 1.5    0.5   1.5 – 0.3  0.15  1.35 –    E E E  E  E 1.35 210 210     1.35    155.55 MPa. E E

x =





(3)

Maximum shear stress theory : For no failure, maximum shear stress should be less than or equal to maximum shear stress under uniaxial loading.

Since we have 3–D case,  1 – 3 1 – 2 2 – 3  , ,  Maximum shear stress = maximum   2 2 2  fy Maximum shear stress under uniaxial loading : y = 2  From this theory   –   –   – 3   fy Maximum  1 3 , 1 2 , 2  2  2 2 2  1.5  –  –0.5  210 1 – 3     105 2 2 2 2  210



(4)

Maximum strain energy theorem : For no failure, maximum strain energy absorbed at a point should be less than or equal to total strain energy per unit volume under uniaxial loading, when material is subjected to stress upto elastic limit.

Total strain energy =

12  22  32 – 2  12  23  3 1  2E

Total strain energy per unit volume under uniaxial loading = 



According to this theory,

12

2

fy

2

2E

2

  2  3 – 2  1 2   2 3  3   fy 2

1.5 2    2   –0.5 2 – 2  0.3 1.5   –  0.5 – 1.5 0.5    2102

2102    114.73 MPa  3.35 Maximum distortion energy theory :- For no failure, maximum shear strain energy in a body should be less than maximum shear strain energy due to uniaxial loading.

1  1 – 2 2   2 – 3 2   3 – 1 2   fy2  2 1 1.5 –  2     0.5 2   –0.5 –1.5 2   2102  2 2



(5)

  105 MPa.



  



0.252  2.252  42  2 × 2102 

2



2  2102 6.5

  

  116.487 MPa

2  2102 6.5

|

Strength of Material

3

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.

2.

[10 Marks ]

Sol. y x –6

90 = 200 × 10

–6

45 = 500 × 10 90° 45° x –6

0 = 600 × 10

We know that

and

x = max

min

=

 x  y 2

x   y 2



x   y 2

cos 2  2

 xy 2

  x   y    xy        2   2 

sin 2

... (i)

2

... (ii)

Thus to determine the principal strain, we need normal strain in two mutually perpendicular direction and shear strain (xy) associated with these directions.

From (i) 45 = 500 × 10–6 =

 xy 0  90 0  90  cos (2  45)  sin (2  45) 2 2 2

 cos 90  0     sin 90  1 

 600  200 600  200  10 6   106  cos90 + xy 2 2 2

 xy  200  10 6

 From (ii)

2



[ xy   0  90 ]

max

min

=

0  90    90    xy    0    2  2   2 

2

2

 600  200   600  200   200  12 12  106     =    10   10  2   2   2  2 2 6 = [400  (200)  (100) ]  10

= [400  223.607]  10 6 

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 2P =

 xy / 2 x  y / 2



200 200 1   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 strain transformation eq.  xy 0  90 0  90  cos 2P  sin 2P   x = 2 2 2 200  600  200 600  200   cos 2 (13.282)  sin 2 (13.482)  10 6 =   2 2 2 

x = 623.607 × 10–6



P = 13.282° is associated with major principal strain



i.e., direction of major principal strain is at 13.282° in anticlockwise direction from 0 strain direction and hence direction of minor principal strain is at 103.282° in anticlockwise direction from 0 strain direction.

Calculation of principal stresses:

 

max (min )  = max E E min  max  = min E E max – 0.3 min = 623.607 ×10–6 × 2 × 105 N/mm2

max– 0.3 min = 124.72 N/mm2 min – 0.3 max = 176.393



×10–6

... (i) × 2×

105

N/mm2

... (ii)

0.3 min – 0.09 max = 35.279 × 0.3

... (iii)

From (i) + (iii) (10.09)max = 124.7 + 35.279 × 0.3 

max  148.685 N/mm2



min  79.885 N/mm2

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 strain By analytical approach we have found that xy is (+)ve. This implies that it is associated with (+)ve shear stress as shown below. y x

|

Strength of Material

5

Hence strains are shown as y xy

2

xy x 1

direction of y 6   6 200  10  200  10 ,   2 

 xy    y ,   i.e.  2 

 xy 2

–3

(623.607 × 10 , 0)

(176.393 × 10–6 , 0)

direction of minor principal strain

–6

(400 × 10 , 0)

2P

direction of major principal  strain

 xy    200  106   x ,  i.e,  600  106 ,   2   2 

direction of x

max = (400 × 10–6) + R min = (400 × 10–6) –R

[R = radius of circle]

   200  0    106 R =  (600  400)2   = 223.607 × 10–6   2    max = 623.607 × 10–6 2



min = 176.393 × 10–6 100 1  tan2P = 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 13.28  3.

180 = 103.28° in anticlockwise direction from the direction of x.

2

Draw bending moment & shear force diagrams for the beam loaded as shown in fig.

10 kN

3kN/m

1m P 2m

1m

1m

2m

2m

[ 15 Marks ]