C I V I L E N G I N E E R I N G · load P at C. The stiffness of spring at A is 2K and that of B is K. A B C P D a a The ratio of forces of spring at A and that of spring at B is

C I V I L E N G I N E E R I N GESE TOPICWISE OBJECTIVE SOLVED

PAPER–I

FROM 1995-2018

UPSC Engineering Services Examination,

State Engineering Service Examination & Public Sector Examination.

Regd. office : Phone : F-126, (Lower Basement), Katwaria Sarai, New Delhi-110016 011-26522064Mobile : E-mail:

Web : 8010009955, 9711853908 [email protected], [email protected]

iesmasterpublications.com, iesmaster.org

Typeset at : IES Master Publication, New Delhi-110016

© No part of this booklet may be reproduced, or distributed in anyform or by any means, electronic, mechanical, photocopying, orotherwise or stored in a database or retrieval system without theprior permission of IES MASTER PUBLICATION, New Delhi.Violaters are liable to be legally prosecuted.

First Edition : 2016Second Edition : 2017Third Edition : 2018

ISBN :

IES MASTER PublicationF-126, (Lower Basement), Katwaria Sarai, New Delhi-110016Phone : 011-26522064, Mobile : 8010009955, 9711853908E-mail : [email protected], [email protected] : iesmasterpublications.com, iesmaster.org

Preface

It is an immense pleasure to present topic wise previous years solved paper ofEngineering Services Exam. This booklet has come out after long observation anddetailed interaction with the students preparing for Engineering Services Exam andincludes detailed explanation to all questions. The approach has been to provideexplanation in such a way that just by going through the solutions, students will beable to understand the basic concepts and will apply these concepts in solving otherquestions that might be asked in future exams.

Engineering Services Exam is a gateway to an immensly satisfying and high exposurejob in engineering sector. The exposure to challenges and opportunities of leading thediverse field of engineering has been the main reason for students opting for thisservice as compared to others. To facilitate selection into these services, availabilityof arithmetic solution to previous year paper is the need of the day. Towards this endthis book becomes indispensable.

Mr. Kanchan Kumar ThakurDirector–IES Master

CONTENTS

1. Strength of Material ----------------------------------------------------------------------------- 001 – 214

2. Structure Analysis ------------------------------------------------------------------------------- 215 – 340

3. Steel Structure ----------------------------------------------------------------------------------- 341 – 458

4. RCC and Prestressed Concrete ------------------------------------------------------------ 459 – 586

5. PERT CPM --------------------------------------------------------------------------------------- 587 – 680

6. Building Material --------------------------------------------------------------------------------- 681 – 791

Contents

1. Strength of Materials ---------------------------------------------------------------------------------- 01 – 36

2. Shear Force and Bending Moment ----------------------------------------------------------------- 37 – 72

3. Deflection of Beams ----------------------------------------------------------------------------------- 73 – 96

4. Transformation of Stress and Strain --------------------------------------------------------------- 97 – 123

5. Combined Stresses--------------------------------------------------------------------------------- 124 – 136

6. Bending Stress in Beams ------------------------------------------------------------------------- 137 – 159

7. Shear Stress in Beams ---------------------------------------------------------------------------- 160 – 170

8. Torsion of Circular Shafts -------------------------------------------------------------------------- 171 – 187

9. Columns ---------------------------------------------------------------------------------------------- 188 – 201

10. Springs ------------------------------------------------------------------------------------------------ 202 – 206

11. Thick and Thin Cylinders/Spheres --------------------------------------------------------------- 207 – 212

12. Moment of Inertia ------------------------------------------------------------------------------------ 213 – 214

Elastic constants, stress, plane stress, Mohr’s circle of stress, strains, plane strain, Mohr’scircle of strain, combined stress, Elastic theories of failure; Simple bending, shear; Torsionof circular and rectangular sections and simple members.

Syllabus

1

1. Given that for an element in a body ofhomogeneous isotropic material subjected to

plane stress; x y, and z are normal strains

in x, y, z directions respectively and is thePoisson’s ratio, the magnitude of unit volumechange of the element is given by

(a) x y z (b) x y z( )

(c) x y z( ) (d) x y z1/ε 1/ε 1/ε

2. A solid metal bar of uniform diameter D andlength L is hung vertically from a ceiling. If thedensity of the material of the bar is and themodulus of elasticity is E, then the totalelongation of the bar due to its own weight is

(a) L / 2E (b) 2L / 2E

(c) E / 2L (d) 2E

2L

3. A rigid beam ABCD is hinged at D andsupported by two springs at A and B as shownin the given figure. The beam carries a verticalload P at C. The stiffness of spring at A is2K and that of B is K.

A B C

P

D

a a a

The ratio of forces of spring at A and that ofspring at B is(a) 1 (b) 2(c) 3 (d) 4

4. The stress-strain curve for an ideally plasticmaterial is

(a)

Strain

Stre

ss

(b)

Strain

Stre

ss(c)

Strain

Stre

ss

(d)

Strain

Stre

ss

5. A steel cube of volume 8000 cc is subjectedto an all round stress of 1330 kg/sq. cm. Thebulk modulus of the material is 1.33 × 106 kg/sq. cm. The volumetric change is

(a) 8 cc (b) 6 cc(c) 0.8 cc (d) 10–3 cc

6. In terms of bulk modulus (K) and modulus ofrigidity (G), the Poisson’s ratio can beexpressed as

(a) (3K – 4G)/(6K+4G) (b) (3K+4G)/(6K– 4G)(c) (3K – 2G)/(6K+ 2G) (d) (3K+2G)/(6K – 4G)

7. Two bars one of material A and the other ofmaterial B of same length are tightly securedbetween two unyielding walls. Coefficient ofthermal expansion of bar A is more than thatof B. When temperature rises the stressesinduced are

Strength of Materials

CHAPTER

Strength of Materials  | 3

(a) tension in both materials(b) tension in material A and compression in

material B

(c) compression in material A and tension inmaterial B

(d) compression in both materials8. A column of height ‘H’ and area at top ‘A’ has

the same strength throughout its length, underits own weight and applied stress ‘P0’ at thetop. Density of column material is ‘ ’. Toosatisfy the above condition, the area of thecolumn at the bottom should be.

(a)HP0

gAe

(b)

gH

0PAe

(c) 0

gHPAe

(d) 0

HgPAe

9. A bar of diameter 30 mm is subjected to atensile load such that the measured extensionon a gauge length of 200 mm is 0.09 mm andthe change is diameter is 0.0045 mm. ThePoisson’s ratio will be(a) 1/4 (b) 1/3(c) 1/4.5 (d) 1/2

10. When a mild-steel specimen fails in a torsion-test, the fracture looks like

(a)

(b)

(c)

(d)

11. A 2 m long bar of uniform section 50 mm2

extends 2 mm under a limiting axial stressof 200 N/mm2. What is the modulus ofresilience for the bar?(a) 0.10 units (b) 0.20 units

(c) 10000 units (d) 200000 units12. The stress level, below which a material has

a high probability of not failing under reversalof stress, is known as

(a) elastic limit (b) endurance limit(c) proportional limit (d) tolerance limit

13. If E = 2.06 × 105 N/mm2, an axial pull of60 kN suddenly applied to a steel rod 50 mmin diameter and 4 m long, causes aninstantaneous elongation of the order of(a) 1.19 mm (b) 2.19 mm(c) 3.19 mm (d) 11.9 mm

14. A bar of circular cross-section varies uniformlyfrom a cross-section 2D to D. If extension ofthe bar is calculated treating it as a bar ofaverage diameter, then the percentage errorwill be

(a) 10 (b) 25(c) 33.33 (d) 50

15. The length, coefficient of thermal expansionand Young’s modulus of bar ‘A’ are twice thatof bar ‘B’. If the temperature of both bars isincreased by the same amount while preventingany expansion, then the ratio of stressdeveloped in bar A to that in bar B will be(a) 2 (b) 4(c) 8 (d) 16

16. The lists given below refer to a bar of lengthL, cross sectional area A, Young’s modulusE, Poisson’s ratio and subjected to axialstress ‘p’. Match List-I with List-II and selectthe correct answer using the codes given belowthe lists:

List-I List-II

A. Volumetric strain 1. 2(1 + )

B. Strain energy per unit volume 2. 3(1 – 2 )

C. Ratio of Young’s modulus to 3.p (1 2 )E

bulk modulus

D. Ratio of Young’s modulus to 4.2p

2Emodulus of rigidity

5. 2(1 – )

  4 | ESE Topicwise Objective Solved Paper-I 1995-2018

Codes:A B C D

(a) 3 4 2 1

(b) 5 4 1 2

(c) 5 4 2 1

(d) 2 3 1 5

17. If all dimensions of prismatic bar of squarecross-section suspended freely from the ceilingof a roof are doubled then the total elongationproduced by its own weight will increase(a) eight times (b) four times(c) three times (d) two times

18. The stress at which a material fractures underlarge number of reversals of stress is called.(a) endurance limit (b) creep(c) ultimate strength (d) residual stress

Directions: The following items consist of twostatements, one labelled the ‘Assertion A’ and theother labelled the ‘Reason R’ you are to examinethese two statements carefully and decide if theAssertion A and the Reason R are individually trueand if so, whether the Reason is a correct explanationof the Assertion. Select your answers to these itemsusing the codes given below and mark your answersheet accordingly.

Codes:(a) Both A and R are true and R is correct

explanation of A(b) Both A and R are true but R is not a

correct explanation of A(c) A is true but R is false(d) A is false but R is true

19. Assertion (A) : Strain is a fundamentalbehaviour of the material, while the stress isa derived conceptReason (R) : Strain does not have a unitwhile the stress has a unit.

20. Assertion (A) : The amount of elasticdeformation at a certain point, which an elasticbody undergoes, under given stresses is thesame irrespective of the stresses being tensile

or compressive.

Reason (R) : The modulus of elasticity andPoisson’s ratio are assumed to be the samein tension as well as compression.

21. Assertion (A) : A mild steel tension specimenhas a cup and cone fracture at failure.

Reason (R): Mild steel is weak in shear andfailure of the specimen in shear takes placeat 45° to the direction of the applied tensileforce.

22. A round steel bar of overall length 40 cmconsists of two equal portions of 20 cm eachhaving diameters of 10 cm and 8 cmrespectively. If the rod is subjected to a tensileload of 10 tones, the elongation will be givenby

6 2E = 2 × 10 kg / cm

(a)1 1 1

10 25 16

cm (b)2 1 1

10 25 16

cm

(c)3 1 1

10 25 16

cm (d)4 1 1

10 25 16

cm

23. A copper bar of 25 cm length is fixed by meansof supports at its ends. Supports can yield (total)by 0.01 cm. If the temperature of the bar israised by 100°C, then the stress induced in the

bar for 6c 20 10 /°C & 6 2

cE =1 10 kg/cm

will be

(a) 2 × 102 kg/cm2 (b) 4 × 102 kg/cm2

(c) 8 × 102 kg/cm2 (d) 16 × 102 kg/cm2

24. A given material has Young’s modulus E,modulus of rigidity G and Poisson’s ratio 0.25.The ratio of Young’s modulus to modulus ofrigidity of this material is

(a) 3.75 (b) 3

(c) 2.5 (d) 1.5

25. A prismatic bar of uniform cross-sectional areaof 5 cm2 is subjected to axial loads as shownin the given figure.

Strength of Materials  | 19

1. (a)

2. (b)

3. (c)

4. (c)

5. (a)

6. (c)

7. (d)

8. (c)

9. (b)

10. (a)

11. (a)

12. (b)

13. (a)

14. (a)

15. (b)

16. (a)

17. (b)

18. (a)

19. (b)

20. (a)

21. (a)

22. (a)

23. (d)

47. (b)

48. (b)

49. (a)

50. (a)

51. (a)

52. (b)

53. (a)

54. (b)

55. (b)

56. (a)

57. (c)

58. (c)

59. (c)

60. (c)

61. (b)

62. (d)

63. (b)

64. (c)

65. (d)

66. (c)

67. (c)

68. (b)

69. (c)

93. (c)

94. (a)

95. (b)

96. (b)

97. (d)

98. (d)

99. (c)

100. (a)

101. (d)

102. (a)

103. (a)

104. (d)

105. (d)

106. (c)

107. (c)

108. (c)

109. (a)

110. (a)

111. (b)

112. (b)

113. (d)

114. (d)

115. (a)

70. (c)

71. (a)

72. (c)

73. (c)

74. (a)

75. (a)

76. (d)

77. (c)

78. (a)

79. (b)

80. (b)

81. (c)

82. (a)

83. (a)

84. (a)

85. (d)

86. (c)

87. (b)

88. (b)

89. (d)

90. (d)

91. (b)

92. (c)

24. (c)

25. (c)

26. (d)

27. (a)

28. (d)

29. (a)

30. (b)

31. (a)

32. (c)

33. (d)

34. (c)

35. (a)

36. (a)

37. (d)

38. (a)

39. (d)

40. (b)

41. (a)

42. (a)

43. (d)

44. (a)

45. (d)

46. (b)

116. (b)

117. (c)

118. (a)

119. (b)

120. (d)

121. (d)

122. (b)

123. (c)

124. (c)

125. (b)

126. (d)

127. (a)

128. (c)

129. (a)

130. (b)

131. (b)

132. (a)

133. (*)

134. (b)

136. (c)

136. (c)

137. (b)

138. (a)

  2 0 | ESE Topicwise Objective Solved Paper-I 1995-2018

1. (a) Unit volume change,VV

=Final volume – Initial volume

Initial volumeVV

= x y z(1 ) (1 ) (1 ) 11

= 1 + x +y + z + xy + yz + zx + xy z–1product of strain terms are very small, soneglecting them

henceVV

= x y z

2. (b) Elongation in length, dx is d

d = PdxAE

for a force of P on element (dx)

dxdx

xAx

d =L

0

Ax dxAE

=2L

0

xE 2

=2L

2E

=L 2 2

0

x L2E 2E

AlternativeThe elongation of bar due to its own weight(w) is

=WL2AE

= ( AL)·L

2AE

=2L

2E

3. (c) Given, KA = 2 KBForce carried by spring at A

= FA = A Ak B A2k Force carried by spring at B

= FB = B Bk

Deflected shape

A

A

B

Ba 2a

From similar trianglesA

3a=

B

2a A = B1.5

A

B

FF =

B A

B B

2kk

=

A

B

2 =

B

B

2 1.5 3

4. (c) An ideal plastic material experiences nowork (non strain) hardening during plasticdeformation.

5. (a) Bulk modulus =P

V / V

61.33 10 =

1330V / 8000

V = – 8 cc(–) ve sign indicates reduction in volume ifstress is compressive in nature.

6. (c) We know,E = 2G (1 + ) ... (i)E = 3K (1 – 2 ) ... (ii)

(where is poisson’s ratio)Equation (i) (ii)

1 = 23

Gk

1(1 2 )

3K 6k = 2G 2G

=

3k 2G6k 2G

7. (d) As the temperature rises, both the bars willhave tendency to expand but they are fixedbetween two unyielding walls so they willnot be allowed to expand. Hence in boththe bars compressive stress will develop.

8. (c)

ax

dx

x

A

P0

a + dax x

H

Strength of Materials  | 21

As we move down weight of column willadd up to produce stresses. Since thecolumn has same strength, so to satisfythe condition, the X-sectional area mustincrease as we move downLet area at distance x be ax and in lengthdx

wt, added = xga dxBut stress has to remain constant

P ax0

dw

P a + da )0 x x(

dx

dw = ax . dx g

0 x xP .a a . g (dx) = 0 x xP (a da )

So dax P0 = xga dx

x

x

daa

=0

g dxP

ln ax =0

g x CP

at x = 0, ax = AC = ln A

xan

Al =

0

gxP

at x = H, ln H

0

a gHA P

HaA

= 0gH/Pe

aH = 0gH/PAe

9. (b) Poisson’s ratio,

= – Lateral strainLongitudinal strain

= 0.0045 / 30

0.09 / 200

= 13

10. (a) In ductile material failure is due to shearwhich in case of torsion occurs at 90° tothe axis.

11. (a) Modulus of resilience = Energy storedupto elastic limit per unit volume

= 12

stress × stress

= 1 22002 2000 = 0.1 units

12. (b) Endurance limit is the stress level belowwhich even large no. of stress cycle cannotproduce fatigue failure.

13. (a) Energy stored in body = 2AL

2ELoad is applied suddenly

P

2AL

2E =

2AL P LP. L2E E

= 2PA

L =

2PLLE AE

=

3

2 5

2 60 10 4000 1.19 mm50 2.06 10

4Instantansis elongation is double that understatic loading.

14. (a) Diameter at a distance x

L

x

P2D

PD

dx

xD = xDDL

Extension of bar due to load P

d =2

1

d

2d

PdD .E4

x

x

N o t e : Extension of circular bar having varyingdiameter from d1 to d2 due to load P.

=

1 2

PLd d E4

When taking average diameter extension of bar is

av = PLAE

=

2PL

D 2D E4 2

= 2

16 PL9 D E

  2 2 | ESE Topicwise Objective Solved Paper-I 1995-2018

% error =

2 2

2

2PL 16 PLD E 9 D E

2PLD E

= 11.11%

15. (b) Stress developed in a bar due to temperature

=EL

(Deformation prevented)

= E L tL = E t

A

B

=

A

B

E tE t

= 2E 2 t

E t

= 4

16. (a) A = Volumetric strainVV

=

x y z 1 2E

= p 1 2E

= p 1 2E

B = Strain energy per unit volume

B = 1 stress strain2

= 2p

2E C : E = 3K 1 2

EK

= 3 1 2

D : E = 2G (1 )

17. (b) Elongation of bar due to it’s own weight

initial =wL

2AE =

(a a L) L ·2 (a a) E

=

2L2E

When all the dimensions are doubled

final =wL

2AE =

(2a 2a 2L) 2L ·2 (2a 2a) E

=

24L2E

= initial4

18. (a)

Stress

Endurancelimit

No. of cycles of loadingwhich causes fatiguefailure.

for ferrousmetals

19. (b) Assertion is correct because strain is thefundamental behaviour but stress is aderived concept because strain can bemeasured with some instrument and is afundamental quantity however stress canonly be derived, it cannot be measured.Reasoning is also correct but A does notfollow from R.

20. (a) If the material is homogeneous & isotropic,magnitude of deformation will be same if E& µ are same in all direction.

x = yx zE E E

... (i)

x = y zx

E E E

... (ii)

Magnitude of x in (i) as well as (ii) is same.

21. (a) Shear is maximum at 45° to the directionof the applied tensile force and mild steel isweak in shear so failure takes place in thedirection of maximum shear stress, in cupand cone shape fracture mode.

22. (a) Elongation due to tensile force P

=1 2

1 2

PL PLPLAE A E A E

= 2 26 6

10 1000 20 10 1000 2010 82 10 2 104 4

= 1 1 110 25 16

cm

23. (d) Stress induced in the bar due to rise intemperature

=E (Deflection prevented)L

=E (L t 0.01)L

=6

610 [25 20 10 100 0.01]25

= 1600 = 16×102 kg/cm2

24. (c) E = 2G (1+ )

EG = 2 (1+ ) = 2 × (1 + 0.25)

= 2.5