GATE - 19 - 2019 Mechanical... · Previous GATE Questions & Solutions To Strength of Materials Chapter No. Name of the Chapter Questions Page No. Solutions Page No. 01 Simple Stress

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Mechanical EngineeringPrevious GATE Questions with Solutions, Subject Wise & Chapter wise

(1987 - 2018)

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ForewordGATE in Mechanical Engineering

Starting year : 1987 From 1992– 2002 : Objective and Conventional Questions From 2003 – 2013 : Objective Questions with multiple choices 2014 onwards : Objective Questions with multiple choices & Numerical Answer Type Questions

The style, quality and content of the Solutions for previous years’ GATE 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 questions in the subject without any tension. However, it is the reader who should confirm this and any comments and suggestions would be warmly received by the Academy.

The student should not miss to go through the solutions for conventional questions asked prior to 2003, as more concepts are brought out in them that will facilitate to answer the numerical answer type questions, Common data and Linked answer questions , if any, easily.

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 the question 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 answer. This will help to face numerical answer type questions, better. The student is advised to have a standard text book ready for reference to strengthen the related concepts, if necessary. The student is advised not to write the solution steps in the space around the question. By doing so, he loses an opportunity of effective revision.

As observed in the GATE – 16 and thereafter, number of sets may be possible, being online exams. Hence, don’t skip any subject. All are the equally important.It is believed that this book is a Valuable aid to the students appearing for competitive exams like ESE, JTO, DRDO, ISRO and Other PSUs. This book can also be used by fresh lecturers in Engineering in improving their Concepts.

Mathematics & General Aptitude Previous Questions & Solutions of GATE of all branches are available in separate booklets.

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.

Applied Mechanics and Design

Engineering Mechanics: Free-body diagrams and equilibrium; trusses and frames; virtual work; kinematics and dynamics of particles and of rigid bodies in plane motion; impulse and momentum (linear and angular) and energy formulations, collisions.

Mechanics of Materials: Stress and strain, elastic constants, Poisson’s ratio; Mohr’s circle for plane stress and plane strain; thin cylinders; shear force and bending mo-ment diagrams; bending and shear stresses; deflection of beams; torsion of circular shafts; Euler’s theory of columns; energy methods; thermal stresses; strain gauges and rosettes; testing of materials with universal testing machine; testing of hardness and impact strength.

Theory of Machines: Displacement, velocity and acceleration analysis of plane mech-anisms; dynamic analysis of linkages; cams; gears and gear trains; flywheels and gov-ernors; balancing of reciprocating and rotating masses; gyroscope. Vibrations: Free and forced vibration of single degree of freedom systems, effect of damping; vibration isolation; resonance; critical speeds of shafts.

Machine Design: Design for static and dynamic loading; failure theories; fatigue strength and the S-N diagram; principles of the design of machine elements such as bolted, riveted and welded joints; shafts, gears, rolling and sliding contact bearings, brakes and clutches, springs.

Fluid Mechanics and Thermal Sciences

Fluid Mechanics: Fluid properties; fluid statics, manometry, buoyancy, forces on sub-merged bodies, stability of floating bodies; control-volume analysis of mass, momen-tum and energy; fluid acceleration; differential equations of continuity and momen-tum; Bernoulli’s equation; dimensional analysis; viscous flow of incompressible fluids, boundary layer, elementary turbulent flow, flow through pipes, head losses in pipes, bends and fittings.

Heat-Transfer: Modes of heat transfer; one dimensional heat conduction, resistance concept and electrical analogy, heat transfer through fins; unsteady heat conduction, lumped parameter system, Heisler’s charts; thermal boundary layer, dimensionless parameters in free and forced convective heat transfer, heat transfer correlations for flow over flat plates and through pipes, effect of turbulence; heat exchanger perfor-mance, LMTD and NTU methods; radiative heat transfer, Stefan-Boltzmann law, Wien’s displacement law, black and grey surfaces, view factors, radiation network analysis.

GATE Syllabus for Mechanical Engineering

Thermodynamics: Thermodynamic systems and processes; properties of pure sub-stances, behaviour of ideal and real gases; zeroth and first laws of thermodynamics, calculation of work and heat in various processes; second law of thermodynamics; thermodynamic property charts and tables, availability and irreversibility; thermody-namic relations.

Applications: Power Engineering: Air and gas compressors; vapour and gas power cycles, concepts of regeneration and reheat. I.C. Engines: Air-standard Otto, Diesel and dual cycles. Refrigeration and air-conditioning: Vapour and gas refrigeration and heat pump cycles; properties of moist air, psychrometric chart, basic psychrometric processes. Turbomachinery: Impulse and reaction principles, velocity diagrams, Pelton-wheel, Francis and Kaplan turbines.

Materials, Manufacturing and Industrial Engineering

Engineering Materials: Structure and properties of engineering materials, phase dia-grams, heat treatment, stress-strain diagrams for engineering materials.

Casting, Forming and Joining Processes: Different types of castings, design of pat-terns, moulds and cores; solidification and cooling; riser and gating design. Plas-tic deformation and yield criteria; fundamentals of hot and cold working processes; load estimation for bulk (forging, rolling, extrusion, drawing) and sheet (shearing, deep drawing, bending) metal forming processes; principles of powder metallurgy. Principles of welding, brazing, soldering and adhesive bonding.

Machining and Machine Tool Operations: Mechanics of machining; basic machine tools; single and multi-point cutting tools, tool geometry and materials, tool life and wear; economics of machining; principles of non-traditional machining processes; prin-ciples of work holding, design of jigs and fixtures. Metrology and Inspection: Limits, fits and tolerances; linear and angular measurements; comparators; gauge design; in-terferometry; form and finish measurement; alignment and testing methods; tolerance analysis in manufacturing and assembly.

Computer Integrated Manufacturing: Basic concepts of CAD/CAM and their integra-tion tools.

Production Planning and Control: Forecasting models, aggregate production plan-ning, scheduling, materials requirement planning.

Inventory Control: Deterministic models; safety stock inventory control systems. Operations Research: Linear programming, simplex method, transportation, assign-ment, network flow models, simple queuing models, PERT and CPM.

Previous GATE Questions with Solutions

Subjectwise and Chapterwise (1987 - 2018)

S.No. Name of the Subject Page No.

1 Engineering Mechanics 01 – 44

2 Strength of Materials 45 – 132

3 Theory of Machines 133 – 222

4 Machine Design 223 – 280

5 FM & Turbo Machinery 281 – 396

6 Heat Transfer 397 – 486

7 Thermodynamics 487 – 658

8 Production 659 – 1010

9 IM & OR 1011 – 1122

10 Engineering Materials 1123 – 1148

PageNo.45

PreviousGATEQuestions&Solutions

To StrengthofMaterials

ChapterNo. NameoftheChapter Questions

PageNo.SolutionsPageNo.

01 Simple Stress 47 – 54 55 – 62

02 Complex Stress 63 – 67 68 – 72

03 SFD & BMD 73 – 76 77 – 80

04 Centroids and Moment of Inertia 81 – 82 83 – 83

05 Pure Bending 84 – 87 88 – 91

06 Shear Stress in Beams 92 – 92 93 – 93

07 Springs 94 – 95 96 – 97

08 Torsion 98 – 102 103 – 108

09 Slopes and Deflections 109 – 112 113 – 117

10 Thin Cylinders 118 – 120 121 – 123

11 Column & Struts 124 – 125 126 – 128

12 Propped and Fixed Beams 129 – 129 130 – 130

13 Strain Energy 131 – 131 132 – 132

CONTENTS

PageNo.46

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10 cm 5 cm 95 MPa

One Mark Questions

Simplestresses 1Chapter

01. A large uniform plate containing a rivet-hole

is subjected to uniform uniaxial tension of

95 MPa. The maximum stress in the plate is

(GATE-ME-92)

(a) 100 MPa (b) 285 MPa

(c) 190 MPa (d) Indeterminate

02. A free bar of length l m is uniformly heated

from 0C to a temperature of tC. is the

coefficient of linear expansion and E the

modulus of elasticity. The stress in the bar is

(GATE-ME-95)

(a) tE (b) tE/2

(c) zero (d) none of the above

03. Two identical circular rods of same diameter

and same length are subjected to same

magnitude of axial tensile force. One of the

rod is made of mild steel having the

modulus of elasticity of 206 GPa. The other

rod is made of cast iron having the modulus

of elasticity of 100 GPa. Assume both the

materials to be homogeneous and isotropic

and the axial force causes the same amount

of uniform stress in both the rods. The

stresses developed are with in the

proportional limit of the respective

materials. Which of the following

observations is correct? (GATE-ME-03)

(a) Both rods elongate by the same amount

(b) Mild steel rod elongates more than the

cast iron

(c) Cast iron rod elongates more than the

mild steel rod

(d) As the stresses are equal strains are

also equal in both the rods

04. In terms of Poisson’s ratio () the ratio of

Young’s Modulus (E) to Shear Modulus (G)

of elastic materials is (GATE-ME-04)

(a) 2(1 + ) (b) 2 (1 – )

(c) )1(2

1 (d) )1(2

1

05. A uniform, slender cylindrical rod is made

of a homogeneous and isotropic material.

The rod rests on a frictionless surface. The

rod is heated uniformly. If the radial and

longitudinal thermal stresses are represented

by r and z respectively, then

(GATE-ME-04)

: 48 : Strength of Materials


E1 E2

P

S L/2

P

S

L

(a) r = 0 ,z = 0 (b) r 0, z = 0

(c) r = 0, z 0 (d) r 0, z 0

06. A steel rod of length L and diameter D,

fixed at both ends, is uniformly heated to a

temperature rise of T. The Young’s

modulus is E and the coefficient of linear

expansion is ‘’. The thermal stress in the

rod is (GATE-ME-07)

(a) 0 (b) T

(c) E T (d) E TL

07. A rod of length L having uniform cross-

sectional area A is subjected to a tensile

force P as shown in the figure below. If the

Young’s modulus of the material varies

linearly from E1 to E2 along the length of the

rod, the normal stress developed at the

section – SS is (GATE-ME-13)

(a) A

P (b)

21

21

EEA

EEP

(c) 1

2

AE

PE (d)

2

1

AE

PE

08. A circular rod of length ‘L’ and area of

cross-section ‘A’ has a modulus of elasticity

‘E’ and coefficient of thermal expansion ‘’.

One end of the rod is fixed and other end is

free. If the temperature of the rod is

increased by T, then

(GATE-ME-14-SET-1)

(a) stress developed in the rod is ET and

strain developed in the rod is T

(b) both stress and strain developed in the

rod are zero

(c) stress developed in the rod is zero and

strain developed in the rod is T

(d) stress developed in the rod is ET and

strain developed in the rod is zero

09. A metallic rod of 500mm length and 50mm

diameter, when subjected to a tensile force

of 100kN at the ends, experiences an

increase in its length by 0.5mm and a

reduction in its diameter by 0.015mm. The

Poisson’s ratio of the rod material is ____

(GATE-ME-14-SET-1)

10. A steel cube, with all faces to deform, has

Young’s modulus, E, Poisson’s ratio, , and

coefficient of thermal expansion, . The

pressure (hydrostatic stress) developed

within the cube, when it is subjected to a

uniform increase in temperature, T, is

given by (GATE-ME-14-SET-2)

: 49 : Simple Stresses


(a) 0 (b)

21

ET

(c)

21

ET (d)

213

ET

11. The stress-strain curve for mild steel is

shown in the figure given below. Choose the

correct option referring to both figure and

table (GATE-14-SET-3)

Point on

the graph

Description of the point

P 1. Upper yield point

Q 2. Ultimate Tensile Strength

R 3. Proportionality Limit

S 4. Elastic Limit

T 5. Lower Yield Point

U 6. Failure

(a) P-1, Q-2, R-3, S-4, T-5, U-6

(b) P-3, Q-1, R-4, S-2, T-6, U-5

(c) P-3, Q-4, R-1, S-5, T-2, U-6

(d) P-4, Q-1, R-5, S-2, T-3, U-6

12. If the Poisson’s ratio of an elastic material is

0.4, the ratio of modulus of rigidity to

Young’s modulus is ____ (GATE-14-SET-4)

13. The number of independent elastic constants

required to define the stress-strain

relationship for an isotropic elastic solid is

________ (GATE-14-SET-4)

14. A rod is subjected to an uniaxial load within

linear elastic limit. When the change in the

stress is 200 MPa, the change in the strain is

0.001. If the Poisson’s ratio of the rod is 0.3

the modulus of rigidity ( in GPa) is______

(GATE –15 –Set 2)

15. The room-temperature stress () - strain()

curves of four materials P,Q,R and S are

shown in the figure below. The material that

behaves as a perfectly plastic material is

(GATE - PI –15)

(a) P (b) Q (c) R (d) S

16. Consider the following statements:

(P) Hardness is the resistance of a material

to indentation.

(Q) Elastic modulus is a measure of

ductility.

(R) Deflection depends on stiffness.

U

P

T

R

S

Q

Str

ess (

N/m

m2 )

Stress e (%)

Q

R S

P



(S) The total area under the stress-strain

curve is a measure of resilience.

Among the above statements, the correct

ones are (GATE - PI-16)

(a) P and Q only. (b) Q and S only.

(c) P and R only. (d) R and S only.

17. The Poisson’s ratio for a perfectly

incompressible linear elastic material is

(GATE – 17 – SET – 1)

(a) 1 (b) 0.5 (c) 0 (d) infinity

18. In the engineering stress-strain curve for

mild steel, the Ultimate Tensile Strength

(UTS) refers to (GATE – 17 – SET – 1)

(a) Yield stress

(b) Proportional limit

(c) Maximum stress

(d) Fracture stress

19. A steel bar is held by two fixed supports as

shown in the figure and is subjected to an

increase of temperature T = 100oC. If the

coefficient of thermal expansion and

Young’s modulus of elasticity of steel are

1110–6/oC and 200GPa, respectively, the

magnitude of thermal stress (in MPa)

induced in the bar is_______

(GATE – 17 – SET – 2)

20. If E is the modulus of elasticity in GPa, G is

the shear modulus in GPa and v is the

Poisson’s ratio of a linear elastic and

isotropic material, the three terms are

related as (GATE – PI – 17)

(a) E = G (1 – 2v) (b) E = 2G (1 – v)

(c) E = G (1 + 2v) (d) E = 2G (1 + v)

21. In a linearly hardening plastic material, the

true stress beyond initial yielding

(GATE – 18 – SET – 1)

(a) increases linearly with the true strain

(b) decreases linearly with the true strain

(c) first increases linearly and then

decreases linearly with the true strain

(d) remains constant

22. Length, width and thickness of a plate are

400 mm, 400 mm and 30 mm, respectively.

For the material of the plate, Young’s

modulus of elasticity is 70 GPa, yield stress

is 80 MPa and Poisson’s ratio is 0.33. When

the plate is subjected to a longitudinal

tensile stress of 70 MPa, the increase in the

volume (in mm3) of the plate is ________

(PI_GATE – 18)



1 m

40 mm diameter

Two Marks Questions

T2 T1 b

a

P l l

A

M N

50N

400mm 500mm

250N 200N

100N

1700mm

K L

01. Determine the temperature rise necessary to

induce buckling in a 1m long circular rod of

diameter 40 mm shown in the figure below.

Assume the rod to be pinned at its ends and

the coefficient of thermal expansion as

2010-6/C. Assume uniform heating of the

bar. (GATE-ME-93)

02. Below Fig. shows a rigid bar hinged at A

and supported in a horizontal position by

two vertical identical steel wires. Neglect

the weight of the beam. The tension T1 and

T2 induced in these wires by a vertical load

P applied as shown are (GATE-ME-94)

(a) 1 2

PT T

2

(b) 222221 ba

PbT,

ba

PaT

ll

(c) 222221 ba

PaT,

ba

PbT

ll

(d) 222221 ba2

PbT,

ba2

PaT

ll

03. A 200 × 100 × 50 mm steel block is

subjected to a hydrostatic pressure of 15

MPa. The Young’s modulus and Poisson’s

ratio of the material are 200 GPa and 0.3

respectively. The change in the volume

of the block in mm3 is (GATE-ME-03)

(a) 85 (b) 90 (c) 100 (d) 110

04. The figure below shows a steel rod of 25

mm2 cross sectional area. It is loaded at four

points. K, L, M and N. Assume Esteel = 200

GPa. The total change in length of the rod

due to loading is (GATE-ME-04)

(a) 1 m (b) – 10 m

(c) 16 m (d) – 20 m

05. A steel bar of 40 mm × 40 mm square cross-

section is subjected to an axial compressive

load of 200 kN. If the length of the bar is 2

m and E = 200 GPa. The decrement in

length of the bar will be (GATE-ME-06)

(a) 1.25 mm (b) 2.70 mm

(c) 4.05 mm (d) 5.40 mm



L L

E F 3E

A B P

C

0, 0

0.2, 100

0.6, 140

0.8, 130

0 20 40 60 80 100 120 140 160

0 0.2 0.4 0.6 0.8 1 Engg. Strain (%)

Eng

g. S

tres

s (M

Pa)

P

Q

R S

63kN 35kN 49kN

21kN

D C B A

06. A bar having a cross-sectional area of 700

mm2 is subjected to axial loads at the

positions indicated. The value of stress in

the segment BC is (GATE-ME-06)

(a) 40 MPa (b) 50 MPa

(c) 70 MPa (d) 120 MPa

07. A solid steel cube constrained on all six

faces is heated so that the temperature rises

uniformly by T. If the thermal coefficient

of the material is , young’s modulus is E

and the Poisson’s ratio is , the thermal

stress developed in the cube due to heating

is (GATE-ME-12)

(a)

21

ET (b)

21

ET2

(c)

21

ET3 (d)

213

ET

08. A 200 mm long, stress free rod at room

temperature is held between two immovable

rigid walls. The temperature of the rod is

uniformly raised by 250C. If the Young’s

modulus and coefficient of thermal

expansion are 200 GPa and 110–5/C,

respectively, the magnitude of the

longitudinal stress (in MPa) developed in the

rod is _____ (GATE-ME-12)

09. A metallic bar of uniform cross-section with

specific weight of 100 kN/m3 is hung

vertically down. The length and Young’s

modulus of the bar are 100 m and 200 GPa,

respectively. The elongation of the bar, in

mm, due to its own weight is _________.

(GATE - PI –15) 10. A horizontal bar with a constant cross-

section is subjected to loading as shown in

the figure. The Young’s modules for the

sections AB and BC are 3E and E,

respectively.

For the deflection at C to be zero, the ratio

P/F is ____ (GATE – 16 – SET – 1)

11. A hypothetical engineering stress-strain

curve shown in the figure has three straight

lines PQ, QR, RS with coordinates P(0,0), Q

(0.2,100), R(0.6, 140) and S(0.8, 130). ‘Q’ is

the yield point, R is the UTS point and S

the fracture point.



45o

45o

L

L

B

A

P

C

0.2 mm 250 mm

The toughness of the material (in MJ/m3) is

_______ (GATE – 16 – SET – 1)

12. In the figure, the load P = 1 N, length L = 1

m, Young’s modulus E = 70 GPa, and the

cross-section of the links is a square with

dimension 10 mm 10 mm. All joints are

pin joints.

The stress (in Pa) in the link AB is _____

(Indicate compressive stress by a negative

sign and tensile stress by a positive sign)

(GATE – 16 – SET – 2)

13. A circular metallic rod of length 250 mm is

placed between two rigid immovable walls

as shown in the figure. The rod is in perfect

contact with the wall on the left side and

there is a gap of 0.2 mm between the rod

and the wall on the right side. If the

temperature of the rod is increased by

200oC, the axial stress developed in the rod

is ______MPa.

Young’s modulus of the material of the rod

is 200 GPa and the coefficient of thermal

expansion is 10–5 per oC.

(GATE – 16 – SET – 2) 14. A square plate of dimension L L is

subjected to a uniform pressure load P =

250 MPa on its edges as shown in the

figure. Assume plane stress conditions. The

Young’s modulus E = 200 GPa.

The deformed shape is a square of

dimension L – 2. If L = 2 m and = 0.001

m, the Poisson’s ratio of the plate material

is ________. (GATE – 16 – SET – 3)

15. A horizontal bar, fixed at one end (x = 0),

has a length of 1 m, and cross-sectional area

of 100 mm2. Its elastic modulus varies

along its length as given by E(x) = 100 e–x

GPa, where x is the length coordinate (in m)

along the axis of the bar. An axial tensile

load of 10 kN is applied at the free end

(x = 1). The axial displacement of the free

end is _______ mm. (GATE – 17 – S– 1)

P

P

P

P L



L = 1m

H = 10 mm

m = 100 kg

g

B = 100

A = 500

(MPa)

A

B

B 0.5

dam

ped

L/2

Aluminium

L/2

P

Perfectly bonded interface dam

ped

end

Steel

16. A point mass of 100 kg is dropped onto a

massless elastic bar (cross-sectional area

= 100 mm2, length = 1 m, Young’s modulus

= 100 GPa) from a height H of 10 mm as

shown (Figure is not to scale). If g = 10

m/s2, the maximum compression of the

elastic bar is _______ mm.

(GATE – 17 – SET – 1)

17. The true stress () - true strain () diagram

of a strain hardening material is shown in

figure. First, there is loading up to point A,

i.e up to stress of 500 MPa and strain of 0.5.

Then from point A, there is unloading up to

point B, i.e , to stress of 100MPa. Given

that the Young’s modulus E = 200 GPa, the

natural strain at point B (B) is _____

(correct to three decimal places).

(GATE – 18 – SET – 1)

18. A bimetallic cylindrical bar of cross

sectional area 1 m2 is made by bonding

Steel (Young's modulus = 210 GPa) and

Aluminium (Young's modulus = 70 GPa) as

shown in the figure. To maintain tensile

axial strain of magnitude 10-6 in Steel bar

and compressive axial strain of magnitude

10-6 in Aluminium bar, the magnitude of the

required force P (in kN) along the indicated

direction is

(GATE – 18 – SET – 2)

(a) 70 (b) 140

(c) 210 (d) 280



One Mark Solutions

Key Sheet

One Mark Questions

01. (c) 02. (c) 03. (c) 04. (a) 05. (a) 06. (c) 07. (a)

08. (c) 09. (0.3) 10. (a) 11. (c) 12. (0.36) 13. (2) 14. (76.92)

15. (d) 16. (c) 17. (b) 18. (c) 19. (220) 20. (d) 21. (a)

22. (1632)

Two Marks Questions

01. 49.34C 02. (b) 03. (b) 04. (b) 05. (a) 06. (a) 07. (a)

08. (500) 09. (2.5) 10. (4) 11. (0.85) 12. (0) 13. (240) 14. (0.2)

15. (1.718) 16. (1.5177) 17. (0.498) 18. (d)

01. Ans: (c)

Sol: Maximum stress in plate develops across

rivet hole, let P be the axial tensile force.

P = 95 t 10cm = t (10 – 5)cm

Maximum stress, MPa1905

1095

02. Ans: (c)

Sol: As the bar is free to expand, there will not

be any thermal stress induced in the bar.

03. Ans: (c)

Sol: AE

PLδ l

E

1δ l

As E of cast iron is less than the mildsteel

then the change in length is more in cast iron

04. Ans: (a)

Sol: We know that,

1G2E

12G/E



E1 E2

P

L/2

P

S

L/2 P Q

R

S

T

U

05. Ans: (a)

Sol: Free to expand means no stresses will be

developed.

06. Ans: (c)

Sol: Temperature stress = ET

07. Ans: (a)

Sol:

Normal stress = A

P

Normal stress for determinate structures is

independent of young’s modulus.

08. Ans: (c).

Sol: As the Circular rod is free to expand, stress

developed is zero. Strain developed is

(T).

09. Ans: 0.3

Sol: Given data:

Length, L = 500mm

Diameter, D = 50mm

Tensile force, P = 100 kN

Increase in length, L = 0.5mm

Decrease in diameter, D = 0.015mm

Poisson’s ratio, =

L

LD

D

linear

lateral

= )500/5.0(

)50/015.0( = 0.3.

10. Ans: (a)

Sol: A cube free to deform has no stresses.

11. Ans: (c)

Sol:

P – Proportionality limit

Q – Elastic limit

R – Upper yield point

S – Lower yield point

T – Ultimate tensile strength

U – Failure

12. Ans: 0.36

Sol: Poisson’s ratio, = 0.4

We know, E = 2G(1+ )

)1(2

1

E

G

= )4.01(2

1

= 0.36



13. Ans: (2)

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.

14. Ans: 76 to 78

Sol: Given data, 3.0

2mm/N200 , mm/mm001.0

Hooke’s law, E

23 mm/N10200001.0

200E

Also, 1G2E

MPa1092.7612

EG 3

= 76.92GPa

15. Ans: (d)

Sol: P Elastic – Plastic behavior

Q Elastic – Strain hardening behavior

R Elastic behavior

S Perfectly plastic behavior

16. Ans: (c)

Sol: Hardness of a material is measured by

indentation technique.

In a ductile material, plastic deformation

is more predominant that elastic one.

If a material undergo more strain

(deflection) in elastic region means

stiffness is low.

The total area under the stress-strain curve

is a measure of modulus of toughness.

Thus, only P and R are correct.

17. Ans: (b)

Sol: Poisson’s ratio of perfectly incompressible

(non – dilatant) material is 0.5 (maximum).

18. Ans: (c)

Sol: Refer to solution of Q. No. 11

19. Ans: 220

Sol: Given data:

Temperature increase, T = 100oC

Young’s modulus of elasticity,

E = 200103 MPa

Coefficient of thermal expansion,

= 1110–6/oC

Thermal stress, t = (T) E

= (1110–6)(100)(200103)

= 220 MPa

20. Ans: (d)

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

Also, E = 3K (1 - 2)



Two Marks Solutions

21. Ans: (a)

Sol: In a linearly hardening plastic material, the

true stress beyond initial yielding increases

because the cross sectional area decreases

continuously.

22. Ans: 1632

Sol: Original volume = 40040030

= 4800,000 mm3

Volumetric strain is given as

E

21321V

=

E

21001

1000

34.0

000,70

33.02170

V

V

48000001000

34.0V = 1632 mm3

01.

Sol: Euler’s buckling load

Pe = EIl 2

2---- (1)

Temperature thrust,

P = .AtE --- (2)

For equilibrium, equation l = 2 ;

AEtEI2

2

42

2

4064

E)1000(

26 404

Et1020

t = 49.340C

02. Ans: (b)

Sol: 0 AM

T2 (b) + T1(a) =P(l) ---- (1)

(a)δ(b)δ 21

AE

T(b)

AE

T 21 all

T1 = 2Tb

a

Sub P1 in equation (1)

T2 (b) + T2 )()( 2

lPb

a

T2 lPb

ab

22

T2 = )( 22 ba

blP

Similarly, T1 = )ba(

aP22

l

03. Ans: (b)

Sol: For hydrostatic pressure

zyX

EEEzyx

21E

ε3εεεV

δVε zyxV

501002000.32110200

153δV

3

3mm90Vδ



100 N 100 N K L

150 N 150 N L M

50 N 50 N M N

63 kN 63 kN

A B

28 kN 28 kN

B C

C

21kN 21 kN

D

04. Ans: (b)

Sol:

MNLMKL δδδδ llll

33 1020025

800150

1020025

500100

31020025

40050

μm10m1010 6

05. Ans: (a)

Sol: AE

PLδ l

3

3

102004040

200010200δ

l =1.25 mm

06. Ans: (a)

Sol:

FBD of BC:

Area

Loadσ

MPaBC 40700

1028 3

07. Ans: (a)

Sol:

x = y = z =

Strain in X-direction due to temperature rise,

Tt

Strain in X-direction due to volumetric

stress

TEEEt

T12E

12

ET

21

TE

08. Ans: 500

Sol: Stress developed due to temperature rise

= tE

= 1 10-5 250 200 103 = 500 MPa



3E A B B C F (F–P) (F–P) (F)

E

450 450

450

FAB

FBC

P

09. Ans: 2.5

Sol: E2

2

mm5.2102002

1000100101009

23

10. Ans: 4 (range 3.9 to 4.1)

Sol:

AB + BC = 0

0

EA

LF

E3A

LPF

0F3

PF

F P + 3F = 0

4F = P 4F

P

11. Ans: 0.85 (range 0.849 to 0.851)

Sol: Toughness = Area under diagram

= 1 0.2 1 0.4100 100 140

2 100 2 100

+

1 0.2140 130

2 100

T = 0.1 + 0.48 + 0.27 = 0.85 MJ/m3

12. Ans: 0 (range -1 to 1)

Sol: Fx = 0

P sin45 = FB cos45

FBC = P

∑Fy =0

FBC sin45 – P cos45 + FAB = 0

ABF2

P

2

1P

FAB = 0

AB = 0

13. Ans: 240 (range 239 to 241)

Sol: 20010250t 5 = gapmm5.02

1

Deformation prevented

= 0.5 0.2 = 0.3 = AE

P

0.3 =

310200

250

= 240 MPa

14. Ans: 0.2 (range 0.18 to 0.22)

Sol: v = x + y + z

v E E E E E

Due to plane stress condition, z = 0

v 2 1E

2 2

2

2 2 0.001 2 P2 1

2 E

33

2502 10 2 1

200 10

200

1250

0.8 = 1 –

= 0.2



L

dx x

x

P

Strip

15. Ans: (1.718)

Sol:

Change in length of small strip,

xxxx

xx

EA

dxP

Total change in length of bar,

L

0xxxx

xxL

0total EA

dxP

Px-x = P = Constant

Ax-x = A = Constant

Ex-x = 100 e-x GPa

L

0

1

0 x9xtotal e

dx

A10010

P

e100

dx

A

P

10x

96

3

e1010010100

1010

m107182.1 3

total = 1.718 mm

16. Ans: 1.5177

Sol: m = 100 kg ; H = 10 mm ;

L = 1 ; m =1000 mm ;

E = 100 GPa ; g = 10 m/sec2

W = 100 × 10 = 1000 N

factorImpact

WL

EAh211

AE

WL

=

10001000

1010010100211

10100100

10001000 3

3

= 2001110

1 = 1.5177 mm

17. Ans: 0.498

Sol:

The curve traces a near straight line path

from the point of unloading (N.A), and its

slop is virtually identical to the modulus of

elasticity or parallel to the initial elastic

curve.

BA

BAE

B5.0

100500

= 200 103 MPa

3B 10200

4005.0

0.5 – B = 210–3

B = 0.5 – 210–3 = 0.498

B = 100

A = 500

(MPa)

A

B

B 0.5



18. Ans: (d)

Sol:

As = Aa= A = 1 m2 = 106 mm2

Step (1) :

s = Es s = 210103 10–6

a = Ea a = 20 103 10–6

Step (2) :

Ps = s As = 210103 10–6

Pa = a Aa = 70 103 10–6

Ps = 210 kN and Pa = 70 kN

P = Ps + Pa = 210 + 70 = 280 kN

L/2

Aluminium

L/2

P Steel

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GATE - 19 - 2019 Mechanical... · Previous GATE Questions & Solutions To Strength of Materials Chapter No. Name of the Chapter Questions Page No. Solutions Page No. 01 Simple Stress