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ESE - 19Mechanical Engineering

(Volume - II)(Strength of Materials, Engineering Materials, Theory of Machines, Machine Design,

Production Engineering, Industrial Engineering & Operations Research, Engineering Mechanics, Mechatronics & Robotics, Maintenance Engineering )

Previous years Objective Questions with Solutions, Subject wise & Chapter wise(1992 - 2018)

ACEEngineering Publications

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Authors : Subject experts of ACE Engineering Academy, Hyderabad

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ForewordUPSC Engineering Services in MECHANICAL ENGINEERING

Volume - II Objective Questions: From 1992 – 2018

Currently, the Stage-I (Prelims) of ESE (Mechanical Engineering) consists of two objective

papers. Paper - I is for General Studies & Engineering Aptitude, while Paper-II is of

Mechanical Engineering for 300 Marks and for 3 hours duration. In stage-II (Mains)

Mechanical Engineering, the technical syllabus is divided into two papers which contain

Conventional Questions.

The Objective Questions included in this volume are for the following subjects only.

1. Strength of Materials 2. Engineering Materials

3. Theory of Machines 4. Machine Design

5. Production Engineering 6. Industrial Engineering & Operations Research

7. Engineering Mechanics 8. Mechatronics & Robotics 9. Maintenance Engineering

Based on the new pattern for Prelims, Volume - II is redesigned using the previous questions from 1992

onwards for the above mentioned subjects.

The style, quality and content of the Solutions for previous ESE Questions of Mechanical Engineering, will

encourage the reader, especially the student whether above average, average or below average to learn the

concept and answer the question in the subject without any tension. However, it is the reader who should

confirm this and any comments and suggestions regarding this would be warmly received by the Academy.

It is observed that majority of ESE objective Questions are being asked as it is, in many PSUs, state service

commissions, state electricity boards and even in GATE exam. Hence, we strongly recommend all students who

are competing for various competitive exams to use this book according to the syllabus of the exam concerned.

This book can also be used by fresh Teachers in engineering colleges to improve their Concepts.

We proudly say that questions and solutions right from 1992 onwards are given in this book. The questions

which appeared in early 90’s are most conceptual oriented and these are being repeated in the recent exams

in a different way. Hence, we advise the students to practice these questions compulsorily. The student is also

advised to analyze why only a particular option is correct and why others are not. Evaluate yourself, in which

case, these other options are correct. With this approach you yourself can develop four questions out of one

question. The student is advised to solve the problems without referring to the solutions. The student has to

analyze the given question carefully, identify the concept on which it is framed, recall the relevant equations,

find out the desired answer, verify the answer with the final key such as (a), (b), (c), (d), then go through the

hints to clarify his/her answer. The student is advised to have a standard text book ready for reference to

strengthen the related concepts, if necessary. The student is further advised not to write the solution steps in

the space around the question. By doing so, one loses an opportunity of effective revision.

With best wishes to all those who wish to go through the following pages.

Y.V. Gopala Krishna Murthy,

M Tech. MIE,

Chairman & Managing Director,

ACE Engineering Academy,

ACE Engineering Publications.

Weightage of Subjects in Engineering Services Examination

MECHANICAL ENGINEERING

ESE – 2017 and 2018 (Prelims, Subjectwise Weightage)

SUBJECT 2017 2018

Thermal Engineering 45 53

Heat Transfer 07 07

Fluid Mechanics & Turbomachinery 17 19

Renewable Sources of Energy 03 01

Engineering Mechanics 05 04

Strength of Materials 11 08

Engineering Materials 08 11

Theory of Machines 23 16

Machine Design 06 03

Production Technology 10 07

Industrial Engineering & Operations Research 07 04

Maintenance Engineering 01 -

Mechatronics and Robotics 07 17

Total No. Marks 150 150

Previous years Questions with Solutions, Subject wise & Chapter wise MECHANICAL ENGINEERING (1992 – 2018)

MAIN INDEX

S.No. Name of the Subject Page No.

1 Strength of Materials 01 − 180

2 Engineering Materials 181 − 294

3 Theory of Machines 295 − 492

4 Machine Design 493 − 598

5 Production Engineering 599 − 798

6 Industrial Engineering & Operations Research 799 − 910

7 Engineering Mechanics 911 − 917

8 Mechatronics and Robotics 918 − 925

9 Maintenance Engineering 926 − 926

PageNo.01

UPSC Engineering Service (ESE)

SYLLABUS

STRENGTHOFMATERIALS

Stresses and Strains-Compound Stresses and Strains, Bending Moment and Shear

Force Diagrams, Theory of Bending Stresses-Slope and deflection-Torsion, Thin

and thick Cylinders, Spheres.

ChapterNo.

NameoftheChapter QuestionsPageNo.

SolutionsPageNo.

01 Simple Stress 03 – 16 17 – 35

02 Complex Stress 36 – 45 46 – 59

03 SFD & BMD 60 – 70 71 – 82

04 Pure Bending 83 – 87 88 – 93

05 Shear Stress in Beams 94 – 96 97 – 101

06 Torsion 102 – 109 110 – 119

07 Slopes and Deflections 120 – 123 124 – 129

08 Springs 130 – 136 137 – 143

09 Thin & Thick Cylinders 144 – 151 152 – 159

10 Struts & Column 160 – 166 167 – 174

11 Strain Energy 175 – 177 178 – 180

CONTENTS

PageNo.02

ACE Engineering Publications Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag |Tirupati | Kukatpally| Kolkata

SimpleStress 1Chapter

01. Which of the following is true ( =

Poisson’s ratio) (ESE-92)

(a) 0 < < 1/2 (b) 1 < < 0

(c) 1 < < –1 (d) – < <

02. Select the proper sequence (ESE-92)

1. Proportional Limit

2. Elastic limit

3. Yielding

4. Failure

(a) 2, 3, 1, 4 (b) 2, 1, 3, 4

(c) 1, 3, 2, 4 (d) 1, 2, 3, 4

03. A vertical hanging bar of length L and

weighing w N/unit length carries a load W

at the bottom. The tensile force in the bar at

a distance y from the support will be given

by (ESE-92)

(a) W + wL (b) W + w(L y)

(c) (W + w)y/L (d) W +w

W(L y)

04. The temperature stress is a function of

(ESE-92)

1. Coefficient of linear expansion

2. Temperature rise

3. Modulus of elasticity

(a) 1 and 2 only (b) 1 and 3 only

(c) 2 and 3 only (d) 1, 2 and 3

05. A steel rod of 1 sq. cm. cross sectional area

is 100 cm long and has a Young’s modulus

of elasticity 2× 106 kgf/cm2. It is subjected

to an axial pull of 2000 kgf. The elongation

of the rod will be (ESE-93)

(a) 0.05 cm (b) 0.1 cm

(c) 0.15 cm (d) 0.20 cm

06. Match List – I with List-II and select the

correct answer using the codes given below

the lists: (ESE-93)

List – I (Material Properties)

A. Ductility

B. Toughness

C. Endurance limit

D. Resistance to penetration

List – II (Tests)

1. Impact

2. Fatigue test

3. Tension

4. Hardness test

Codes: A B C D

(a) 3 2 1 4

(b) 4 2 1 3

(c) 3 1 2 4

(d) 4 1 2 3

` : 4 : Strength of Materials


07. If the value of Poisson’s ratio is zero, then it

means that (ESE-94)

(a) the material is rigid

(b) there is no longitudinal strain in the

material

(c) the material is perfectly plastic

(d) the longitudinal strain in the material is

infinite

08. The stretch in a steel rod of circular section,

having a length ‘L’ subjected to a tensile

load ‘P’ and tapering uniformly from a

diameter d1 at one end to a diameter d2 at

the other end, is given (ESE-95)

(a) P.L/4Ed1.d2 (b) PL./E.d1d2

(c) PL/4E(d1.d2) (d) 4P.L/.E.d1.d2

09. The total extension of the bar loaded as

shown in the figure is (ESE-95)

A = Area of cross-section

E = Modulus of elasticity

(a) AE

3010 (b)

AE

1026

(c) AE

309 (d)

AE

2230

10. If Poisson’s ratio for a material is 0.5, then

the elastic modulus for the material is

(ESE-95)

(a) three times its shear modulus

(b) four times its shear modulus

(c) equal to its shear modulus

(d) indeterminate

11. A bar of uniform cross-section of one sq.cm

is subjected to a set of five forces as shown

in the given figure, resulting in its

equilibrium. The maximum tensile stress (in

kgf/cm2) produced in the bar is

(ESE-97)

(a) 1 (b) 2 (c) 10 (d) 11

12. Match List – I (Elastic properties of an

isotropic elastic material) with List – II

(Nature of strain produced) and select the

correct answer using the codes given below

the Lists: (ESE-97)

List – I

A. Young’s modulus

B. Modulus of rigidity

C. Bulk modulus

D. Poisson’s ratio

A B C D E

1 2 3 4

1 2 3 4

11 kgf 2 kgf 1 kgf 5 kgf 5 kgf

10T 9T

10 mm

3T

2T

10 mm 10 mm

: 5 : Simple Stress


List – II

1. Shear strain

2. Normal strain

3. Transverse strain

4. Volumetric strain

Codes: A B C D

(a) 1 2 3 4

(b) 2 1 3 4

(c) 2 1 4 3

(d) 1 2 4 3

13. The relationship between the Lame’s

constant ‘’, Young’s modulus ‘E’ and the

Poisson’s ratio ‘’ is (ESE-97)

(a) = )21)(1(

E

(b) )1)(21(

E

(c) = )1(

E

(d) = )1(

E

14. A 10 cm long and 5 cm diameter steel rod

fits snugly between two rigid walls 10 cm

apart at room temperature. Young’s

modulus of elasticity and coefficient of

linear expansion of steel are 2 106 kgf/cm2

and 12 106/0C respectively. The stress

developed in rod due to a 100oC rise in

temperature will be (ESE-97)

(a) 61010 kgf/cm2

(b) 6 109 kgf/cm2

(c) 2.4 103 kgf/cm2

(d) 2.4 104 kgf/cm2 15. For a composite bar consisting of a bar

enclosed inside a tube of another material

and when compressed under a load ‘W’ as a

whole through rigid collars at the end of the

bar. The equation of compatibility is given

by (suffixes 1 and 2 refer to bar and tube

respectively) (ESE-98)

(a) W1+W2 = W (b) W1+W2 = constant

(c) 22

2

11

1EA

W

EA

W (d)

12

2

21

1EA

W

EA

W

16. A tapering bar (diameters of end sections

being d1 and d2) and a bar of uniform cross-

section ‘d’ have the same length and are

subjected to the same axial pull. Both the

bars will have the same extension if ‘d’ is

equal to (ESE-98)

(a) 2

dd 21 (b) 21dd

(c) 2

dd 21 (d) 2

dd 21

17. The number of independent elastic

constants required to express the stress-

strain relationship for a linearly elastic

isotropic material is (ESE-98)

(a) One (b) Two

(c) Three (d) Four



18. The deformation of a bar under its own

weight as compared to that when subjected

to a direct axial load equal to its own weight

will be (ESE-98)

(a) The same (b) One-fourth

(c) Half (d) Double

19. A slender bar of 100 mm2 cross-section is

subjected to loading as shown in the below

figure. If the modulus of elasticity is taken

as 200 109 Pa, then the elongation

produced in the bar will be (ESE-98)

(a) 10 mm (b) 5 mm

(c) 1 mm (d) Nil

20. If the rod is fitted snugly between the

supports as shown in the figure, is heated,

the stress induced in it due to 20oC rise in

temperature will be (ESE-99)

(=12.5106/oC, E=200 GPa)

(a) 0.07945 MPa (b) 0.07945 MPa

(c) 0.03972 MPa (d) 0.03972 MPa

21. The number of elastic constants for a

completely anisotropic elastic material is

(ESE-99)

(a) 3 (b) 4 (c) 21 (d) 25

22. A rod of material with E=200 103 MPa and

=10-3 mm/mmC is fixed at both the ends.

It is uniformly heated such that the increase

in temperature is 30oC. The stress

developed in the rod is

(ESE-00)

(a) 6000 N/mm2 (tensile)

(b) 6000 N/mm2 (compressive)

(c) 2000 N/mm2 (tensile)

(d) 2000 N/mm2 (compressive)

23. Assertion (A): Poisson’s ratio of a

material is a measure of change in

dimension in one direction due to loading in

the perpendicular direction.

Reason (R): The nature of lateral strain in a

uniaxially loaded member is opposite to that

of linear strain. (ESE-00)

(a) Both A and R are true and R is the

correct explanation of A

(b) Both A and R are true but R is NOT

the correct explanation of A

(c) A is true but R is false

(d) A is false but R is true

0.5m 1.0m 0.5m

200 kN 100 kN 100 kN 200 kN

10 mm

0.5 m k = 50kN/m

: 7 : Simple Stress


C

P = 120 N

B

2l l

A

24. The Poisson’s ratio of a material which has

Young’s modulus of 120 GPa and shear

modulus of 50 GPa, is (ESE-01)

(a) 0.1 (b) 0.2 (c) 0.3 (d) 0.4

25. Consider the following statements:

Thermal stress is induced in a component in

general, when

1. A temperature gradient exists in the

component.

2. The component is free from any

restraint.

3. It is restrained to expand or contract

freely.

Which of the above statements are correct?

(ESE-02)

(a) 1 and 2 (b) 2 and 3

(c) 3 alone (d) 2 alone

26. A straight bar is fixed at edges A and B. Its

elastic modulus is E and cross-section is A.

There is a load P = 120 N acting at C.

Determine the reactions at the ends.

(ESE-02)

(a) 60 N at A, 60 N at B

(b) 30 N at A, 90 N at B

(c) 40 N at A, 80 N at B

(d) 80 N at A, 40 N at B

27. For a given material, the modulus of rigidity

is 100 GPa and Poisson’s ratio is 0.25. The

value of modulus of elasticity in GPa is

(ESE-02)

(a) 125 (b) 150

(c) 200 (d) 250

28. A rigid beam of negligible weight is

supported in a horizontal position by two

rods of steel and Aluminium, 2 m and 1 m

long having values of cross-sectional areas

1 cm2 and 2 cm2 and E of 200 GPa and 100

GPa respectively. A load P is applied as

shown in the figure. (ESE-02)

If the rigid beam is to remain horizontal

then

(a) The forces on both sides should be equal

(b) The force on Aluminium rod should be

twice the force on steel

(c) The force on the steel rod should be

twice the force on Aluminium

(d) The force P must be applied at the

centre of the beam

P

1 m Aluminium

2 m Steel

Rigid Beam



29. A cube having each side of length a, is

constrained in all directions and is heated

uniformly so that the temperature is raised

to ToC. If is the thermal coefficient of

expansion of the cube material and E is the

modulus of elasticity, the stress developed

in the cube is (ESE-03)

(a) TE /2 (b) TE/(12)

(c) TE/2 (d) TE/(1+2)

30. The modulus of elasticity for a material is

200 GN/m2 and Poisson’s ratio is 0.25.

What is the modulus of rigidity? (ESE-04)

(a) 80 GN/m2 (b) 125 GN/m2

(c) 250 GN/m2 (d) 320 GN/m2

31. Which one of the following is correct in

respect of Poisson’s ratio () limits for an

isotropic elastic solid? (ESE-04)

(a) (b) 1/4 1/3

(c) –1 1/2 (d) –1/2 1/2

32. A bar of length L tapers uniformly from

diameter 1.1 D at one end to 0.9 D at the

other end. The elongation due to axial pull

is computed using mean diameter D. What

is the approximate error in computed

elongation? (ESE-04)

(a) 10% (b) 5%

(c) 1% (d) 0.5%

33. A bar of copper and steel form a composite

system. They are heated to a temperature of

400C. What type of stress is induced in the

copper bar? (ESE-04)

(a) Tensile

(b) Compressive

(c) Both tensile and compressive

(d) Shear 34. A cube with a side length of 1 cm is heated

uniformly 10C above the room temperature

and all the sides are free to expand. What

will be the increase in volume of the cube?

(Given coefficient of thermal expansion is

‘’ per 0C) (ESE-04)

(a) 3 cm3 (b) 2 cm3

(c) cm3 (d) zero 35. If E, G and K denote Young’s modulus,

Modulus of rigidity and bulk modulus,

respectively for an elastic material, then

which one of the following can be possibly

true? (ESE-05)

(a) G = 2K (b) G = E

(c) K = E (d) G = K = E 36. Consider the following statements:

1. Strength of steel increases with Carbon

content.

2. Young’s modulus of steel increases with

carbon content.

3. Young’s Modulus of steel remains

unchanged with variation of carbon

content.

: 9 : Simple Stress


Which of the statements given above is/are

correct? (ESE-05)

(a) 1 only (b) 2 only

(c) 1 and 2 (d) 1 and 3

37. A solid uniform metal bar of diameter D

and length ‘L’ is hanging vertically from its

upper end. The elongation of the bar due to

self weight is: (ESE-05)

(a) Proportional to L and inversely

proportional to D2

(b) Proportional to L2 and inversely

proportional to D2

(c) Proportional of L but independent of D

(d) Proportional of L2 but independent of D

38. E, G, K and represent the elastic modulus,

shear modulus, bulk modulus and Poisson’s

ratio respectively of a linearly elastic,

isotropic and homogeneous material. To

express the stress-strain relations

completely for this material, at least

(ESE-06)

(a) E, G and must be known

(b) E, K and must be known

(c) Any two of the four must be known

(d) All the four must be known

39. A metal rod is rigidly fixed at its both ends.

The temperature of the rod is increased by

100C. If the coefficient of linear expansion

and elastic modulus of the metal rod are

10 106/C and 200 GPa respectively, then

what is the stress produced in the rod?

(ESE-06)

(a) 100 MPa (tensile)

(b) 200 MPa (tensile)

(c) 200 MPa (compressive)

(d) 100 MPa (compressive) 40. In a tensile test, near the elastic limit zone

(ESE-06)

(a) tensile stress increases at a faster rate

(b) tensile stress decreases at a faster rate

(c) tensile stress increases in linear

proportion to the stress

(d) tensile stress decreases in linear

proportion to the stress

41. Two tapering bars of the same material are

subjected to a tensile load P. The lengths of

both the bars are the same. The larger

diameter of each of the bar is D. The

diameter of the bar A at its smaller end is

D/2 and that of the bar B is D/3. What is the

ratio of elongation of the bar A to that of the

bar B? (ESE-06)

(a) 3:2 (b) 2:3 (c) 4:9 (d) 1:3

42. An orthotropic material, under plane stress

condition will have: (ESE-06)

(a) 15 independent elastic constants

(b) 4 independent elastic constants

(c) 5 independent elastic constants

(d) 9 independent elastic constants



43. Which one of the following expresses the

total elongation of a bar of length L with a

constant cross-section of A and modulus of

Elasticity E hanging vertically and subject

to its own weight W? (ESE-07)

(a) AE

WL (b)

AE2

WL

(c) AE

WL2 (d)

AE4

WL

44. If the ratio G/E (G = Rigidity modulus, E =

Young’s modulus of elasticity) is 0.4, then

what is the value of the Poisson ratio?

(ESE-07)

(a) 0.20 (b) 0.25 (c) 0.30 (d) 0.33

45. What is the phenomenon of progressive

extension of the material i.e., strain

increasing with the time at a constant load,

called? (ESE-07)

(a) Plasticity (b) Creeping

(c) Yielding (d) Breaking

46. Which one of the following statements is

correct? (ESE-07)

If a material expands freely due to heating,

it will develop

(a) thermal stress

(b) tensile stress

(c) compressive stress

(d) no stress

47. What is the relationship between the linear

elastic properties Young’s modulus (E),

rigidity modulus(G) and bulk modulus(K)?

(ESE-08)

(a) G

3

K

9

E

1 (b)

G

1

K

9

E

3

(c) G

1

K

3

E

9 (d)

G

3

K

1

E

9

48. A 100 mm 5 mm 5 mm steel bar free to

expand is heated from 150 to 400 C. What

shall be developed? (ESE-08)

(a) Tensile stress

(b) Compressive stress

(c) Shear stress

(d) No stress

49. What is the relationship between elastic

constants E, G and K? (ESE-09)

(a) E = GK9

KG

(b) E =

GK

KG9

(c) E = G3K

KG9

(d) E =

GK3

KG9

50. A bar produces a lateral strain of magnitude

60 × 105 m/m, when subjected to tensile

stress of magnitude 300 MPa along the axial

direction. Find the elastic modulus of the

material, if the Poisson’s ratio is 0.3.

(ESE-09)

(a) 100 GPa (b) 150 GPa

(c) 200 GPa (d) 400 GPa

: 11 : Simple Stress


51. Assertion (A): A cast-iron specimen shall

fail due to shear when subjected to a

compressive load.

Reason (R): Shear strength of cast-iron in

compression is more than half its

compressive strength (ESE-10)

52. A prismatic bar, as shown in figure is

supported between rigid supports. The

support reactions will be: (ESE-11)

(a) RA = 3

10kN and RB =

3

20kN

(b) RA = 3

20kN and RB =

3

10kN

(c) RA = 10 kN and RB = 10 kN

(d) RA = 5 kN and RB = 5 kN

53. If a piece of material neither expands nor

contracts in volume when subjected to

stresses, then the Poisson’s ratio must be:

(ESE-11)

(a) Zero (b) 0.25 (c) 0.33 (d) 0.5

54. If a rod expands freely due to heating, it will

develop: (ESE-11)

(a) Bending stress (b) Thermal stress

(c) No stress (d) Compressive stress

55. What are the materials which show

direction dependent properties, called?

(ESE-11)

(a) Homogeneous materials

(b) Viscoelastic materials

(c) Isotropic materials

(d) Anisotropic materials

56. An elastic material of Young’s modulus E

and Poisson’s ratio is subjected to a

compressive stress of 1 in the longitudinal

direction. Suitable lateral compressive

stress 2 are also applied along the other

two lateral directions to limit the net strain

in each of lateral directions to half of the

magnitude that would be under 1 acting

alone. The magnitude of 2 is (ESE-12)

(a) 1)1(2

(b) 1)1(2

(c) 1)1(

(d) 1)1(

57. A copper rod 400 mm long is pulled in

tension to a length of 401.2 mm by applying

a tensile load of 330 MPa. If the

deformation is entirely elastic, the Young’s

modulus of copper is (ESE-12)

(a) 110 GPa (b) 110 MPa

(c) 11 GPa (d) 11 MPa

C

1 m 2 m

10 kN B A



58. A rod of length L tapers uniformly from a

diameter D at one end to a diameter d at the

other. The Young’s modulus of the material

is E. The extension caused by an axial load

P is (ESE-12)

(a) E)dD(

PL422

(b) E)dD(

PL422

(c) DdE

PL4

(d)

DdE

PL2

59. A bar of copper and steel form a composite

system which is heated through a

temperature of 400C. The stress induced in

the copper bar is (ESE-12)

(a) Tensile

(b) Compressive

(c) Both tensile and compressive

(d) Shear

60. Elastic limit of cast iron as compared to its

ultimate breaking strength is (ESE-12)

(a) Half

(b) Double

(c) Approximately same

(d) None of the above

61. In the arrangement as shown in the figure,

the stepped steel bar ABC is loaded by a

load P. The material has Young’s modulus

E = 200 GPa and the two portions AB and

BC have area of cross section 1 cm2 and 2

cm2 respectively. The magnitude of load P

required to fill up the gap of 0.75 mm is :

(ESE-13)

(a) 10 kN (b) 15 kN

(c) 20 kN (d) 25 kN

62. Consider the following statements:

Modulus of rigidity and bulk modulus of a

material are found to be 60 GPa and 140

GPa respectively. Then: (ESE-13)

1. Elasticity modulus is nearly 200 GPa

2. Poisson’s ratio is nearly 0.3

3. Elasticity modulus is nearly 158 GPa

4. Poisson’s ratio is nearly 0.25

Which of these statements are correct ?

(a) 1 and 3 (b) 2 and 4

(c) 1 and 4 (d) 2 and 3

63. A 16 mm diameter bar elongates by 0.04%

under a tensile force of 16 kN. The average

decrease in diameter is found to be 0.01%.

Then: (ESE-13)

1. E = 210 GPa and G = 77 GPa

2. E = 199 GPa and = 0.25

3. E = 199 GPa and = 0.30

4. E = 199 GPa and G = 80 GPa

Which of these values are correct ?

(a) 3 and 4 (b) 2 and 4

(c) 1 and 3 (d) 1 and 4

Gap 0.75 mm 1m

B P

C A

1m



64. In a body, thermal stress is induced because

of the existence of : (ESE-13)

(a) Latent heat

(b) Total heat

(c) Temperature gradient

(d) Specific heat

65. A hole of diameter d is to be punched in a

plate of thickness t. For the plate material,

the maximum crushing stress is 4 times the

maximum allowable shearing stress. For

punching the biggest hole, the ratio of

diameter of hole to plate thickness should

be equal to: (ESE-13)

(a) 1/4 (b) 1/2 (c) 1 (d) 2

66. The modulus of rigidity and the bulk

modulus of a material are found as 70 GPa

and 150 GPa respectively. Then (ESE-14)

1. Elasticity modulus is 200 GPa

2. Poisson’s ratio is 0.22

3. Elasticity modulus is 182 GPa

4. Poisson’s ratio is 0.3

Which of the above statements are correct?

(a) 1 and 2 (b) 1 and 4

(c) 2 and 3 (d) 3 and 4

67. A steel rod, 2 m long, is held between two

walls and heated from 20C to 60C.

Young’s modulus and coefficient of linear

expansion of the rod material are 200 103

MPa and 10 10–6/C respectively. The

stress induced in the rod, if walls yield by

0.2 mm, is (ESE-14)

(a) 60 MPa tensile

(b) 80 MPa tensile

(c) 80 MPa compressive

(d) 60 MPa compressive

68. A tension member of square cross-section

of side 10 mm and Young’s modulus E is to

be replaced by another member of square

cross-section of same length but Young’s

modulus E/2. The side of the new square

cross-section, required to maintain the same

elongation under the same load, is nearly.

(ESE-14)

(a) 14 mm (b) 17 mm

(c) 8 mm (d) 5 mm

69. An aluminum bar of 8 m length and a steel

bar of 5 mm longer in length are kept at

300C. If the ambient temperature is raised

gradually, at what temperature the

aluminum bar will elongate 5 mm longer

than the steel bar (the linear expansion

coefficients for steel and aluminum are

1210–6/C and 2310–6/C respectively)?

(ESE-14)

(a) 50.74C (b) 69.0C

(c) 143.7C (d) 33.7C



70. A rod of length l tapers uniformly from a

diameter D at one end to a diameter D/2 at

the other end and is subjected to an axial

load P. A second rod of length l and of

uniform diameter D is subjected to an axial

load P. Both the rods are of same material

with Young’s modulus of elasticity E. The

ratio of extension of the first rod to that of

the second rod is (ESE-14)

(a) 4 (b) 3 (c) 2 (d) 1

71. A hole of diameter 35 mm is to be punched

in a sheet metal of thickness t and ultimate

shear strength 400 MPa, using punching

force of 44 kN. The maximum value of t is

(ESE-14)

(a) 0.5 mm (b) 10 mm

(c) 1 mm (d) 2 mm

72. A copper rod of 2cm diameter is completely

encased in a steel tube of inner diameter 2

cm and outer diameter 4 cm. Under an axial

load, the stress in the steel tube is 100

N/mm2. If Es = 2Ec, then the stress in the

copper rod is (ESE – 15)

(a) 50 N/mm2 (b) 33.33 N/mm2

(c) 100 N/mm2 (d) 300 N/mm2

73. The figure shows a steel piece of diameter

20 mm at A and C, and 10 mm at B. The

lengths of three sections A, B and C are

each equal to 20 mm. The piece is held

between two rigid surfaces X and Y.

= 1.2105/C and Young’s modulus

E = 2 105 MPa for steel :

When the temperature of this piece

increases by 50C, the stresses in A and B

are , respectively, (ESE – 15)

(a) 120 MPa and 480 MPa

(b) 60 MPa and 240 MPa

(c) 120 MPa and 120 MPa

(d) 60 MPa and 120 MPa 74. For a material following Hooke’s law, the

value of elastic and shear Moduli are 3105

MPa and 1.2105 MPa, respectively. The

value for bulk modulus is (ESE – 15)

(a) 1.5105 MPa (b) 2105 MPa

(c) 2.5105 MPa (d) 3105 MPa

75. At a point in body, 1 = 0.0004 and

2 = 0.00012. If E = 2105 MPa and

= 0.3, the smallest normal stress and the

largest shearing stress are (ESE – 15)

(a) 40 MPa and 40 MPa

(b) 0 MPa and 40 MPa

(c) 80 MPa and 0 MPa

(d) 0 MPa and 80 MPa

A B

C

20 20 10

20 20 20 Y X



76. A steel rod of cross-sectional area 10 mm2

is subjected to loads at points P,Q.R and S

as shown in the figure below:

If Esteel=200 GPa, the total change in length

of the rod due to loading is (ESE – 16)

(a) –5 m (b) –10 m

(c) –20 m (d) –25 m

77. A circular steel rod of 20 cm2 cross-

sectional area and 10 m length is heated

through 50oC with ends clamped before

heating. Given, E = 200 GPa and coefficient

of thermal expansion, = 10 10-6/oC, the

thrust thereby generated on the clamp is

(a) 100 kN (b) 150 kN (ESE – 16)

(c) 200 kN (d) 250 kN

78. Two steel rods of identical length and

material properties are subjected to equal

axial loads. The first rod is solid with

diameter d and the second is a hollow one

with external diameter D and internal

diameter 50 % of D. If the two rods

experience equal extensions, the ratio of

D

dis (ESE – 16)

(a) 4

3 (b)

2

3 (c)

2

1 (d)

4

1

79. A steel rod 10 m long is at a temperature of

20oC. The rod is heated to a temperature of

60oC. What is the stress induced in the rod

if it is allowed to expand by 4 mm, when E

= 200 GPa and = 12 10-6/oC? (ESE – 16)

(a) 64 MPa (b) 48 MPa

(c) 32 MPa (d) 16 MPa

80. An isotropic elastic material is characterized

by (ESE – 16)

(a) two independent moduli of elasticity

along two mutually perpendicular

directions.

(b) two independent moduli of elasticity

along two mutually perpendicular

directions and Poission’s ratio.

(c) a modulus of elasticity, a modulus of

rigidity and Poission’s ratio.

(d) any two out of a modulus of elasticity, a

modulus of rigidity and Poisson’s ratio.

81. Measured mechanical properties of material

are same in a particular direction at each

point. This property of the material is

known as (ESE – 16)

(a) Isotropy (b) Homogeneity

(c) Orthotropy (d) Anisotropy

82. The modulus of rigidity of an elastic

material is found to be 38.5% of the value

of its Young’s modulus. The Poisson’s ratio

of the material is nearly (ESE – 17)

(a) 0.28 (b) 0.30 (c) 0.33 (d) 0.35

200 N P Q 400 N

300 N R S 100 N

1000 mm 500 mm 500 mm



83. A bar produces a lateral strain of

magnitudes 60 10–5 m/m when subjected

to a tensile stress of magnitude 300 MPa

along the axial direction. What is the elastic

modulus of the material if the Poisson’s

ratio is 0.3? (ESE – 17)

(a) 200 GPa (b) 150 GPa

(c) 125 GPa (d) 100 GPa

84. A 10 mm diameter bar of mild steel of

elastic modulus 200 109 Pa is subjected to

a tensile load of 50000 N, taking it just

beyond its yield point. The elastic recovery

of strain that would occur upon removal of

tensile load will be (ESE – 17)

(a) 1.3810–3 (b) 2.6810–3

(c) 3.1810–3 (d) 4.6210–3

85. A rigid beam of negligible weight is

supported in a horizontal position by two

rods of steel and aluminium, 2 m and 1 m

long, having values of cross sectional areas

100 mm2 and 200 mm2 and Young’s

modulus of 200 GPa and 100 GPa,

respectively. A load P is applied as shown

in the figure below :

If the rigid beam is to remain horizontal,

then (ESE – 18)

(a) the force P must be applied at the

centre of the beam.

(b) the force on the steel rod should be

twice the force on the aluminium rod

(c) the force on the aluminium rod should

be twice the force on the steel rod

(d) the forces on both the rods should be

equal.

Rigid beam P

1 m Aluminium 2 m Steel



SOLUTIONS

KEY SHEET

01. (a) 02. (d) 03.(b) 04.(d) 05. (b) 06. (c) 07. (d) 08. (d) 09. (b) 10 (a)

11. (d) 12. (c) 13. (a) 14. (c) 15. (c) 16. (b) 17. (b) 18. (c) 19. (d) 20.(b)

21. (c) 22. (b) 23. (b) 24. (b) 25.(c) 26. (d) 27. (d) 28. (b) 29. (b) 30.(a)

31. (c) 32. (c) 33. (b) 34. (a) 35. (c) 36.(d) 37. (d) 38. (c) 39. (c) 40. (b)

41. (b) 42. (d) 43.(b) 44. (b) 45.(b) 46.(d) 47.(d) 48.(d) 49.(d) 50. (b)

51.(c) 52. (b) 53.(d) 54. (c) 55.(d) 56. (b) 57. (a) 58. (c) 59. (b) 60.(c)

61. (b) 62. (d) 63.(b) 64.(c) 65.(c) 66. (d) 67. (d) 68.(a) 69.(c) 70.(c)

71. (c) 72. (a) 73. (b) 74. (b) 75. (b) 76.(d) 77.(c) 78.(b) 79.(d) 80.(d)

81.(b) 82.(b) 83.(b) 84.(c) 85.(c)

01. Ans: (a)

Sol: Elastic constants of a material are always

positive.

We know that,

E = G (1+ )

0G

E1

1 + = 0

> –1

Also E = 3K(1 – 2)

0K3

E21

2

1

2

1

Thus from above, it is concluded that,

–2

11

From the given options, option (a) is most

appropriate.

: 18 : Strength of Materials


A B

C

D

E

F

X X

y

L y

W

L

02. Ans: (d)

Sol: Stress-strain diagram for ductile material is

shown in the figure below:

A = Proportional limit

B = Elastic limit

C = Upper yield point

D = Lower yield point

E = Ultimate stress

F = Failure stress

From above it is concluded that correct

sequence for a given properties is

proportionality limit, elastic limit, yielding

and failure.

03. Ans: (b)

Sol: For a bar, section X-X is shown in the

figure below:

Force at section X-X,

= load carried by the bar at bottom(W) +

self weight of the bar of length (L – y)

= W + w(L-y)

04. Ans: (d)

Sol: Stress due to temperature =E(T)

where, E = Young’s modulus,

= Co-efficient of linear expansion,

T = Change in temperature

Thus, temperature is a function of all three

variables mentioned above.

05. Ans: (b)

Sol: Given data:

A = 1 cm2, L = 100 cm,

E = 2 106 kgf/cm2, P = 2000 kgf

The elongation due to axial pull is given by,

cm1.01021

1002000

AE

PL6

06. Ans: (c)

Sol:

Ductility can be tested by tension test.

Toughness is determined by impact

testing.

Endurance limit can be determined by

fatigue test.

Resistance to penetration is for hardness

test.



d1

d2

P P

x

L

x

d1

07. Ans: (d)

Sol: Poisson's ratio is defined as,

strain)axial(alLongitudin

strainLaterial

If = 0 then either lateral strain is zero or

longitudinal strain is infinite. Thus, option

(d) is correct.

Note :

If material is rigid then slope of stress-strain

diagram is infinite which indicates modulus

of elasticity is infinite.

For a perfectly plastic material, no volume

change occurs.

Thus,

21V

V

= 0.5 (∵V = 0)

08. Ans: (d)

Sol:

Let, D1 = Diameter at the larger end

d2 = diameter at the smaller end

L = Length of the bar

E = Young’s modulus of the bar

Consider a very small length x at a

distance x from the small end.

The diameter at a distance x from the small

end = d2 + xL

dd 21

The extension of a small length x is given

by,

ExL

ddd

4

x.P

E.A

x.Pd

221

2x

Extension of the whole rod is given by,

=

L

02

212

dx

ExL

ddd

P4

=

L

0

2

212 dxx

L

ddd

E

P4

=

L

0

212

21

L

x)dd(d

1

dd

L

E

P4

=

21

21

21 dd

dd

ddE

PL4

= 21ddE

PL4

09. Ans: (b)

Sol: Free body diagrams of each section are

drawn below:

10 T 10 T

7T 7T

9T 9T 1

2

3



Total extension of the bar is given by

321

=AE

P

AE

P

AE

P 321

321 PPPAE

= 9710AE

10 =

EA

1026

10 Ans: (a)

Sol: Given data:

= 0.5;

We know that,

E = 2 G (1 + )

E = 2 G (1 + 0.5) = 3 G

E = 3 G

where, E = Elastic modulus

G = Shear modulus

11. Ans: (d)

Sol: A free body diagrams of each section are

drawn below:

Maximum tensile force, F1 = 11 kgf

Maximum tensile stress,

A

F1max

11

11

= 11 kgf/cm2

12. Ans: (c)

Sol: The Young's modulus (E) is the slope of the

stress-strain diagram in the linearly elastic

region. Thus, it is related to normal stress

() and normal strain () by following

relationship:

= E. (Hook’s law)

Modulus of rigidity (also called shear

modulus of elasticity) is based on shear

stress () and shear strain () as per the

following relationship:

= G.

Poisson's ratio is defined as,

strainaxialalLongitudin

straintransverseLateral

Thus, it is related to transverse strain.

13. Ans: (a)

Sol: The Lame constant,

211

E

where, E = Young's modulus

= Poisson's ratio

14. Ans (c)

Sol: Given data: T = 100 oC

E = 2106 kgf/cm2 , = 12 10–6 /oC

11 kgf 11 kgf

9 kgf 9 kgf

10 kgf 10 kgf

5 kgf 5 kgf

1

2

3

4

D = 5 cm

L = 10 cm



W W1 W2

Collar

Inner bar(1)

Outer tube(2)

= E (T)

= 12 10-6 2 106 100

= 2400 kgf/cm2 = 2.4 103 kgf/cm2

15. Ans: (c)

Sol: The composite bar is shown in the figure

below as mentioned in the problem:

When the composite bar is compressed

under a load 'W' as a whole through rigid

collars, the equation of compatibility is

obtained from the fact that inner bar (1) and

outer tube (2) shorten the same amount.

Thus, 1 = 2

22

2

11

1

EA

W

EA

W

16. Ans: (b)

Sol: Extension of tapered bar is given by,

21

Taper dEd

PL4L

Extension of a uniform cross-section bar is

given by,

2uniform Ed

PL4L

As, (L)Taper = (L)Uniform

d2 = d1d2 21ddd

17. Ans: (b)

Sol: The number of independent elastic

constants required to express the stress-

strain relationship for different types of

materials are given below.

Type of

Material

Number of Independent

Elastic Constants Required

Isotropic 2

Orthotropic 9

Anisotropic 21

18. Ans: (c)

Sol: Self weight deformation bar is given by,

(L)own weight = AE2

WL

Deformation of a bar due to axial pull (W)

is given by,

(L)Axial load =AE

WL

Thus, loadaxialwerightown L2

1L

19. Ans: (d)

Sol: The free body diagrams of each section are

shown below:

100kN 100kN 100kN 100kN

0.5 m 0.5 m

1 m

100kN 100kN



Total elongation of bar is given by,

AE

PL

AE

PL

AE

PLL 321

Total

= 5.010011005.0100AE

1

= 0

20. Ans: (b)

Sol: Given data:

L = 0.5m

T = 20C

= 12.510–6/C

E = 200 GPa

k = 50kN/m, D = 10 mm

Free expansion of the bar due to

temperature rise is given by,

Expansion of rod = L(T)

= 0.512.510-620

= 0.12510-3 m

Force due to spring stiffness is given by;

Fk = k

= 0.12510-350103 = 6.25 N

Stress induced in a rod due to spring force

(compression) is given by,

2

k

104

25.6

A

F

MPa07945.0 (Compression)

21. Ans: (c)

Sol: Refer to the solution of Question No.17.

22. Ans: (b)

Sol: Given data: E = 200 GPa,

= 10–3/oC, T = 30oC

Thermal stress () = E T

= 20010310330

= 6000 N/mm2 (compression)

As the temperature of the bar rises, the bar

tries to expand. But the bar is restricted to

expand, thus, compressive stress is induced.

23. Ans: (b)

Sol:

Poisson's ratio is defined as,

strainalLongitudin

strainLateral

When a prismatic bar is loaded in tension as

shown in the figure below, the axial

elongation is accompanied by lateral

contraction (normal to the direction of the

applied load).

From above it is concluded that Poisson's

ratio is a measure of change in dimension in

one direction due to loading in the

perpendicular direction.

Also the lateral strain and longitudinal strain

are opposite in nature to each other.

Thus, both the statements are true but

reason is not the correct explanation of

assertion.

P P



24. Ans: (b)

Sol: Given data:

E = 120 GPa,

G = 50 GPa

We know that,

E = 2 G (l + )

120 = 2 50(l + )

= 0.2

25. Ans: (c)

Sol: Thermal strain due to change in temperature

is given by,

= (T)

here, = Co-efficient of thermal expansion

T = Change in temperature of a

component

When the above strain is restrained, then

stress induces in the component which is

called thermal stress.

Thus, only statement (3) is correct.

26. Ans: (d)

Sol: The free body diagram of each section is

shown below.

Assume that reactions RA and RB are tensile

in nature.

Given that both the ends are fixed, then

compatibility condition,

(l)AB + (l)BC = 0

0AE

2R

AE

R BA

RA + 2RB = 0 ------ (1)

Also, RB + 120 = RA (∵ P = 120 N)

RA – RB = 120 ------(2)

From equation (1) and (2)

–3RB = 120

RB = – 40 N

(RB is in opposite direction to our

assumption)

And, RA = 80 N

27. Ans: (d)

Sol: Given data:

G = 100 GPa,

= 0.25

We know that,

E = 2G (1 + )

= (2 100) (1 + 0.25)

= 250 GPa

28. Ans: (b)

Sol: For the rigid beam to be horizontal

(l)S = (l)Al

AS AE

P

AE

P

3

1A3

s

101002

)1(P

102001

)2(P

2PS = PAl

RA RA

A B l

RB

B C 2l

RB



29. Ans: (b)

Sol: As the cube is constrained in all directions,

x = y = z = 0

0EE

.E

T

021E

T

21

TE

Alternate approach:

Since the cube is constrained on all sides, it

undergoes hydrostatic stress. Therefore we

can use v

vK

.

where, K = Bulk modulus,

v = Volumetric strain

v = K V

3)21(3

E

)T(3)21(3

E

)21(

ET

30. Ans: (a)

Sol: Given data:

E = 200 GPa,

= 0.25

We know that,

E = 2 G (1 + )

200 = 2 G (1 + 0.25)

100 = G (1.25)

G = GPa8025.1

100

31. Ans: (c)

Sol: Refer to the solution of Question No. 01.

32. Ans: (c)

Sol: Given data:

d1 = 1.1 D, d2 = 0.9 D, d = D

Extension of a uniform cross-section bar is

given by,

22Uniform D.E.

PL4

Ed

PL4L

Extension of tapered bar is given by,

21

Taper dEd

PL4L

2D99.0E

PL4

D9.0D1.1E

PL4

Percentage error,

100L

LL

Taper

UniformTaper

%1100

99.0

1

199.0

1

33. Ans: (b)

Sol: Coefficient of thermal expansions,

For steel is, s = 12 10–6 /C

For copper is, Cu = 17 10–6 /C

Copper

Steel

[li(T)]S

P

P

[li(T)]Cu

(PL/AE)S

(PL/AE)Cu

li

lf



li = Initial length of composite bar

lp = Final length of composite bar

Free expansion due to increase in

temperature:

(s)T = liS(T)

(c)T = liCu(T)

Here, Cu > S, thus difference in increase

in length = li (c – s)(T) is eliminated by

compressing the copper bar by force P and

pulling out the steel bar by an equal tensile

force P to satisfy the compatibility condition

of final length (lf) of the composite bar.

Thus, compressive stress is induced in

copper bar and tensile stress is induced in

steel bar.

34. Ans: (a)

Sol: Volumetric strain, v = 3

= 3 L(T) = 3 (1) (1) = 3

Change in volume = v v

= 3(1 1 1) = 3 cm3

35. Ans: (c)

Sol: We know that,

E = 2G(1 + ) ----------- (1)

E = 3K(1 – 2) ----------- (2)

GK3

KG9E

----------- (3)

By putting G = 2K in equations (1) and (2),

= –0.1

By putting G = E in equation (1),

= –0.5

By putting K = E in equation (2),

= 0.33

Thus, most appropriate answer to have a

positive value of Poisson’s ratio is option

(c).

36. Ans: (d)

Sol: As shown in the figure below, the

proportional limit for a particular type of

steel alloy depends on its carbon content.

However, most grades of steel, from the

softest rolled steel to the hardest tool steel,

have about same modulus of elasticity.

37. Ans: (d)

Sol:

Deformation due to self- weight:

Let, L = Length of bar, = weight density

A = Area of cross section

E = Young’s Modulus of elasticity.

x

x

L

Soft steel (0.1 %C)

Structural steel (0.2 %C)

Machine steel (0.6 %C)

Hard steel (0.6 %C)

Spring steel (1 %C)



Consider a small section of length x at a

distance x from free end.

The deformation d is given by,

d = AE

xWx

where, Wx = weight of the portion below the

section

= A x

d = xAE

Ax

= E

xx

Total deformation of the rod,

L

0

L

0

dxE

xd =

E2

L2 =

E2

gL2

38. Ans: (c)

Sol: We know that, E = 2G (1 + )

Also, E = 3K (1 2 )

GK3

KG9E

From above it is concluded that, if any two

of the four constants (, E, G and K) is

known then we can find out the stress-strain

relationship.

39. Ans: (c)

Sol: Given data:

T = 100 C , = 10 10–6 /C

E = 200 GPa

Thermal stress = ET

= 200 103 10 106 100

= 200 MPa (Compressive)

As the temperature of the bar rises, the bar

tries to expand. But the bar is restricted to

expand, thus, compressive stress is induced.

40. Ans: (b)

Sol: After elastic limit, material yields where

stress decreases suddenly.

Also refer to the stress-strain diagram

shown in the solution of Question 02.

41. Ans: (b)

Sol:

Elongation of tapered bar, 21dEd

PL4

21dd

1

3

2

2

DD

3

DD

dd

dd

A21

B21

B

A

42. Ans: (d)


43. Ans: (b)

Sol: Self weight elongation = AE2

WL

Note: Option (a) is true when bar is

subjected to external load ‘W’ and if self

weight is ignored.

D A 2

D D B 3

D



44. Ans: (b)

Sol: Given data:

4.0E

G ,

We know that,

E = 2G (1+)

25.014.02

11

G2

E

45. Ans: (b)

Sol:

Creep is the permanent elongation of a

component under a static load maintained

for a period of time.

Plasticity is the property of the material to

undergo permanent deformation at constant

stress.

Yielding means a material begins to

deforms plastically.

The maximum stress a material can stand

before it breaks is called breaking stress or

breaking point.

46. Ans: (d)

Sol:

When the temperature of a material is

changed, its dimensions change. If this

change in dimension is prevented, then a

stress is set up in a material

When a temperature rises, the material is

prevented from expanding and, therefore,

compressive stress is induced. On the other

hand if the temperature decreases, tensile

stress is induced.

In a given problem, material expands freely,

thus no stress is induced.

47. Ans: (d)

Sol: We know that,

E = 2G(1 + ) and E = 3K (1 – 2)

1 + = G2

E

2 + 2 = G

E ……. (1)

and 1 – 2 = K3

E……. (2)

By adding equations (1) and (2),

3 =

K3

1

G

1E = )GK3(

KG3

E

KG

GK3

E

9

G

3

K

1

E

9

48. Ans: (d)


49. Ans: (d)

Sol: We know that,

E = 2G(1 + ) and E = 3K (1 – 2)

1 + = G2

E

2 + 2 = G

E ……. (1)



And 1 – 2 = K3

E……. (2)

By adding equations (1) and (2),

3 =

K3

1

G

1E = )GK3(

KG3

E

GK3

KG9E

50. Ans: (b)

Sol: Given data: = 0.3,

Lateral strain = –60 10–5,

t = 300 MPa

We know that,

strainalLongitudin

strainLateral

E

10603.0

t

5

E

3001060

3.05

5105.1E MPa = 150 GPa

51. Ans: (c)

Sol: A cast-iron (brittle material) is strong in

compression than shear. When it is

subjected to compressive load, it fails in an

oblique plane due to shear. Thus, assertion

is true.

When cast-iron is subjected to compressive

load, Mohr’s circle can be drawn as shown

below.

From Mohr’s circle, shear strength in

compression is given by,

2

cy

Thus, shear strength is half of its

compressive strength. Therefore, second

statement is wrong.

52. Ans: (b)


drawn below:

RA + RB = 10 kN …….. (1)

(l)AC + (l)BC = 0

(As both ends are fixed)

0EA

2)R(

E.A

)1(R BA

RA – 2RB = 0 …...… (2)

By solving equations (1) and (2),

RA = 3

20N , RB =

3

10N

RB RB

C B

RA RA

A C

1 m 2 m

c

y



2 2

2 2

1 1

53. Ans: (d)

Sol: If the material neither expands nor contracts

then it is incompressible (i.e. V = 0). For

incompressible material Poisson’s ratio is

0.5.

We know that,

21V

V

= 0.5 (∵V = 0)

54. Ans: (c)


55. Ans: (d)

Sol:

Homogeneous: In this, material properties

are same in one particular direction at any

point.

Viscoelastic: It is the property of material

that exhibit both viscous and elastic

characteristics when undergoing

deformation. They exhibit time dependent

strain.

Isotropic: In this, material properties are

same in all direction at one particular point

only.

Anisotropic materials show different

properties in different directions.

56. Ans: (b)

Sol:

When 1 is acting alone lateral strain

2 = E

1

When stresses act in all three directions,

then strain in lateral direction is given by,

EEE

122'2

Given that '2 should be 2

2 .

E2EEE

1212

(given)

E2EE2

1E

1112

2 = 112

57. Ans: (a)

Sol: Given data:

= 330 MPa

= 401.2 400 = 1.2 mm

We know that,

E = GPa1102.1

400330



58. Ans: (c)


59. Ans: (b)


60. Ans: (c)

Sol:

Materials that exhibit little or no yielding

before failure are referred as brittle

materials.

Thus for brittle material, yield stress and

failure stress are nearly same. Cast iron is

also brittle material.

61. Ans: (b)

Sol: Until the gap of 0.75 mm is filled, the load

P is taken by AB only.

P required to cause elongation of 0.75 mm

in AB is calculated by,

AE

P

0.75 = 32 10200)101(

1000P

P = 15 kN

62. Ans: (d)

Sol: Given data:

G = 60 GPa

K = 140 GPa

We know that,

E = .GPa5.157601403

601409

GK3

KG9

Also, E = 2G (1 + )

157.5 = (2 60) (1+ )

= 0.3

63. Ans: (b)

Sol: Given data:

.0004.0100

04.0%04.0

P = 16 103 N,

d = 16 mm

0001.0100

01.0%01.0

D

Dh

Poisson’s ration,

25.00004.0

0001.0h

Tensile stress,

MPa8016

4

1016

A

P

2

3

Young’s Modulus,

E =

MPa1020004.0

80 5

= 200 GPa

We know that,

E = 2G (1 + )

200 = 2G (1 + 0.25)

G = 80 GPa



64. Ans: (c)

Sol: Temperature gradient across the cross-

section causes thermal stress. Also, refer to

the solution of Question no. (46).

65. Ans: (c)

Sol: Crushing force on punching head

= Shear resistance to plate.

t.d.d4

2

Given that, 4

1t

d

66. Ans: (d)

Sol: Given data:

G = 70 GPa, K = 150 GPa

We know that,

E = 701503

701509

GK3

KG9

= 182 GPa

Also, E = 2G (1 + )

182 = 2 70 (1 + )

= 0.3

67. Ans: (d)

Sol: Given data:

l = 2 m,

Increase in temperature, t = 40C

E = 200 103 MPa ,

= 1010–6

Yielding of supports = 0.2 mm

Free expansion of bar = lt

= 2000 10 10–6 40

= 0.8 mm > yielding

To find stress develops in bar,

lt – = E

l

(0.8 – 0.2) = 310200

2000

= 60 MPa (Compression)

68. Ans: (a)

Sol: Given data:

x1 = 10 mm

Given that, l1 = l2

)2/E)(A(

))(P(

)E)(1010(

))(P( ll

A = 10 10 2 = 200 = x22

Side(x2) = 200 = 14 mm

69. Ans: (c)

Sol: Given data:

Aluminum length, lAl = 8000 mm

Steel length, lS = 8000 + 5

From the given compatibility condition,

(lt + 10)S = (lt)Al

(8005 12 10–6 t) + 10

= 8000 23 10–6 t

(0.096 t) + 5 = 0.184 t

t = 113.713C T2 – T1 = 113.713

T2 – 30 = 113.713

T2 = 143.71C



Steel

Copper

Steel

1cm

2cm 4cm

ls = lc

70. Ans: (c)

Sol: Case (I) : Tapering bar:

1 =

2

DD.E

PL4 =

2ED

PL8

Case (II) : Uniform cross-section bar:

2 = 2

2 ED

PL4

ED4

PL

22

1

71. Ans: (c)

Sol: Given data:

Punching force, P = 44 kN

Diameter of pole, D = 35 mm

Shear strength of plate, = 400 MPa

For the sheet to be punched,

Punching force ≥ Shear resistance

44 103 ≥ (.D.t)

44103 ≥ (.35.t)(400)

t 1 mm

Thus, maximum value of ‘t’ is limited to

1 mm.

72. Ans: (a)

Sol: Given data:

dC = 2 cm

(ds)i = 2 cm, (ds)o = 4 cm

s = 100 N/mm2

Es = 2 Ec

From compatibility condition,

( )s = ( )c

cs AE

P

AE

P

c

cc

s

ss

EE

2

c

c

s

csc mm/N50

E2

E100

E

E

73. Ans: (b)

Sol: When the temperature of the bar ABC

increases, the bar ABC will try to expand.

But this expansion is prevented, due to

which compressive stress is induces in the

bar.

D D/2

L

D

L

PA PA PC PC A C

PB PB B



Here, the bar is fixed at both the ends. Thus,

net deflection of the bar should be zero.

( t )A + ( t )B + ( t )C – AE

– BE

– CE

= 0

3( t ) – (A + B + C)E

= 0

3t = E

CBA

3 1.2 10–5 50 = 5

BA

102

2

2A + B = 360 MPa ----- (1)

CA

Also, PA = PB

AAA = BAB

A 202 = B 102

B = 4A ----- (2)

From equations (1) and (2),

6A = 360

A = 60 MPa

B = 240 MPa

74. Ans: (b)

Sol: Given data:

E = 3105 MPa

G = 1.2105 MPa

We know that,

E = GK3

KG9

3105 = 5

5

102.1K3

102.1K9

9 510 K + 3.61010 = 10.8105 K

3.61010 = 1.8105 K

K = 2105 MPa

75. Ans: (b)

Sol: Given data:

1 = 0.0004, 2 = – 0.00012

E = 2105 MPa, = 0.3

We know that,

1 = EE

21

0.0004 = )3.0(102

1215

1 – 0.32 = 80

1 = 80 + 0.32 ---------(1)

Also, 2 = EE

12

–0.00012 = )3.0(102

1125

2–0.31 = – 24

2 = 0.31 – 24 --------(2)

From equations (1) and (2),

2 = 0.3(80 + 0.32) – 24

2 = 24 + 0.9 2 – 24

2 = 0

From equation (1),

1 = 80 MPa

Now, max = 2

21 =

2

080= 40 MPa



Case (II) Extension = 2

P P 2

D D

E

P d P

E

Case (I) Extension = 1

76. Ans: (d)


shown below:

Total deflection of rod is given by,

(L)Total = (L)AB + (L)BC + (L)CD

333 1020010

1000200

1020010

500100

1020010

500200

100020050010050020031020010

1

= –0.025 mm

77. Ans: (c)

Sol: Given data:

Ac = 20 cm2 = 2000 mm2,

l = 10000 mm,

T = 50C,

E = 200 103

F = AE(T)

= 2000 200 103 10 10–6 50

= 200 kN

78. Ans: (b)

Sol:

Given that, 1 = 2

21 AE

PL

AE

PL

2

22

2

D

4D

4d

4

4

DDd

222

4

DDd

222

4

DD4 22

4

D3d

22

D4

3d

2

3

4

3

D

d

79. Ans: (d)

Sol: Given data:

L = 10000 mm

T1 = 20C , T2 = 60C ,

T = 40C

T = ? , = 4 mm

Thermal stress is given by,

T = E Strain restrained

L

TLE

10000

41000040101210200

63

= 16 MPa

P Q

Q R

R S

200 N 200 N

200 N 300 N

100 N 100 N

500 mm 500 mm

1000 mm



80. Ans: (d)

Sol: For an isotropic material, the number of

independent elastic constants is two only.

Relationships between elastic constants are

given below:

E = 2G (1 + )

E = 3K (1 – 2)

where, E = Modulus of elasticity

G = Modulus of rigidity

K = Bulk modulus

From above two equations, we can say that

if we know any of the two values (of E, G,

K or ), other two can be calculated.

81. Ans: (b)


82. Ans: (b)

Sol: Given data:

G = 0.385 E

We know that,

E = 2G (1+)

E = 2 (0.385E) (1+)

= 0.3

83. Ans: (b)

Sol: Given data:

x = 300 MPa,

y = –60 10–5 m/m,

y = 0, = 0.3

Using generalized Hook’s law,

xyy E

1

51060

3003.00E

E = 150 GPa

84. Ans: (c)

Sol: Given data:

E = 200 109 Pa,

F = 50000 N,

d = 10 mm

E

e

MPa6.63610

4

000,50

A

F

2

E = 200 109Pa = 200 103MPa

003183.010200

6.6363

e

= 3.183 10–3

85. Ans: (c)

Sol: Let, Pa = Force in Aluminium

Ps = Force in steel

From the given condition that the rigid

beam to remain horizontal,

s = a

AlSteel AE

PL

AE

PL

aa

aa

ss

ss

EA

LP

EA

LP

(Here, Ls = 2La, As = Aa/2, Es = 2Ea)

Pa = 2Ps

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