RAGHU ENGINEERING COLLEGE (AUTONOMOUS) DEPARTMENT OF

RAGHU ENGINEERING COLLEGE

(AUTONOMOUS)

DEPARTMENT OF MECHANICAL ENGINEERING

Mechanics of solids

UNIT – 1 – Simple Stresses & Strains

Prepared by

Dr Rugada Vaikunta Rao

Professor

1Dr R Vaikunta Rao Mechanics of Solids Mechanical Engineering REC

Contents

Introduction

Types of Materials

Simple Stress and Strain

Types of Stresses

Stresses in Composite Bars

Bars with varying cross-sections

Thermal Stresses

Principal Stresses

Mohr’s Circle

Strain EnergyDr R Vaikunta Rao Mechanics of Solids Mechanical Engineering REC 2

References

1. Strength of materials / GH Ryder / Mc Millan publishers

India Ltd

2. Solid Mechanics, by Popov

3. Mechanics of Materials / Gere and Timoshenko, CBS

4. Strength of Materials -By Jindal, Umesh Publications.

5. Analysis of structures by Vazirani and Ratwani.

Dr R Vaikunta Rao Mechanics of Solids Mechanical Engineering REC 3

Introduction

Mechanics of solids, also called strength of materials, is a

subject which deals with the behaviour of solid objects subject to

loads.

Strength is the ability to resist deformation.

Strength of materials is to estimate material behaviour when it

is subjected to a load or combination of different loads.


SOM / SM: Strength of Materials

MOS: Mechanics of Solids

Types of Materials

Elastic Materials: A material undergoes a deformation when

subjected to external loading and deformation disappears on

the removal of loading.

Plastic Materials: A material undergoes a continuous

deformation when subjected to external loading and deformation

will not disappears on the removal of loading.

Rigid Materials: A rigid material does not undergo any

deformation when subjected to external loading.


Simple Stress and Strain

Stress : The force of internal resistance offered by a body against

deformation is called stress.

Strain : The ratio of change of dimension of the body to the

original dimension is known as strain. Strain is dimensionless.

, /Strain e dl l

, /Stress

Tensile Stress : When a body is subjected to two equal and

opposite axial pulls, then the stress induced at any section of the

body is known as tensile stress

Tensile Strain : The ratio of the increase in length to the original

length is known as tensile strain.

Types of stresses and strains

Compressive Stress : When a body is subjected to two equal and

opposite axial pushes, then the stress induced at any section of the

body is known as compressive stress.

Compressive Strain : The ratio of the decrease in length to the

original length Is known as compressive strain.


Shear Stress and Shear Strain : When a body is subjected to two

equal and opposite forces acting tangentially across the resisting

section, then the stress induced is called shear stress (τ), The

corresponding strain is known as shear strain (φ).


According to Hooke's law when a material is loaded within

elastic limit, the stress is directly proportional to strain.

σ ∝ ε or σ = E ×ε

where E=young’s modulus

Young’s Modulus


It has been found experimentally that within the elastic limit,

the shear stress is directly proportional to shear strain.

Mathematically τ ∝ φ or τ = G . φ or τ / φ = G

G = shear modulus or modulus of rigidity

Shear Modulus


Stress-strain Diagram



Dr R Vaikunta Rao Mechanics of Solids Mechanical Engineering REC

Factor of Safety is the ratio of ultimate tensile stress to the

working (or permissible) stress.

The strain in the direction of applied load is known as

lngitudinal strain.

The strain at right angles to the direction of applied load is

known as lateral strain.

/ ( ) /LateralStrain b b or d d

/LongitudinalStrain l l

15

Example Problems: Stress and Strain


Problem1

Example Problems : Stress and Strain


Problem 2



Problem 3



Problem 4.

A circular rod of diameter 16 mm and 500 mm long is subjected to a tensile

force 40 kN. The modulus of elasticity for steel may be taken as 200 kN/mm2.

Find stress, strain and elongation of the bar due to applied load.

Solution: Load P = 40 kN = 40 × 1000 N

E = 200 kN/mm2 = 200 × 103 N/mm2

L = 500 mm Diameter of the rod d = 16 mm

Therefore, sectional area A = πd 2/ 4 = 201.06 mm2

Stress p =P/A= 40 (1000) / 201.06 = 198.94 N/mm2

Strain e = pE=(198.94) 200 (10)3 = 0.0009947

Elongation = PL/AE = (40 X 1000) (500) /201.06 200 103 = 0.497 mm



Problem 5.A surveyor’s steel tape 30 m long has a cross section of 15 mm × 0.75 mm. With this, line AB is measured as 150 m. If the force applied during measurement is 120 N more than the force applied at the time of calibration, what is the actual length of the line? Take modulus of elasticity for steel as 200 kN/mm2



Problem 6.A tensile test was conducted on a mild steel bar. The following data was obtained for the test:a) Diameter of the steel bar = 3 cmb) Gauge length of the bar = 20 cmc) Load at elastic limit = 250 kN.d) Extension at a load of 150 kN = 0.21 mme) Maximum Load = 380 kNf) Total extension = 60 mmg) Diameter of the rod at the failure = 2.25 cm.

Determine1) Young’s Modulus 2) The stress at elastic limit3) The percentage of elongation 4)The percentage decrease in area

Solution:Area of the rod = 32 = 7.0685 cm2= 7.0685 X 10-4 m2

4

1) Determination of Young’s Modulus:To find Young’s Modulus, Calculate the value of stress and strain with in elastic limit.

The load at elastic limit is given as 250 kN.Extension at a load of 150 kN = 0.21 mmSo these values are used for stress and strain within the elastic limit (Contd…)



Problem 6.

Stress = load / Area = (150 x 1000)/ 7.0685 x 10-4 = 21220.9 x 104 N/m2

Strain = extension/ Original length = 0.21/ (20 x10)= 0.00105Young’s Modulus E= (Stress/ Strain)

= 21220.9 x 104/ 0.00105 = 20209523 x 104 N/m2

Young’s Modulus E = 202.095 x 109 N/m2= 202.095 GN/m2.

2) Determination of stress at the elastic limit:Stress = 250 x 1000 /7.0685 x 10-4

= 35368 x 104 N/M2 =353.68 x 106 N/M2 = 353.68 MN/M2

3) The percentage elongation = (Total elongation/ original length) X 100= [(60 mm)/ (20 X 10 mm)] X 100 = 30%

4) The percentage decrease in area = {(Original area – Area at failure)/ Original area } X 100

(Contd…)


Factor of Safety is the ratio of ultimate tensile stress to the

working (or permissible) stress.

The strain in the direction of applied load is known as

lngitudinal strain.

The strain at right angles to the direction of applied load is

known as lateral strain.

/ ( ) /LateralStrain b b or d d

/LongitudinalStrain l l

23


Poisson's ratio: The ratio of lateral strain to the longitudinal

strain is a constant for a given material, when the material is

stressed within the elastic limit, is called Poisson's ratio.

Poisson's ratio µ= Lateral strain / Longitudinal strain

Volumetric Strain : The ratio of the change in volume to the

original volume is called the volumetric strain.

ev=Change in volume /Original volume

24


Bulk Modulus : The ratio of direct stress to the corresponding

volumetric strain is known as bulk modulus

Shear Modulus : The ratio of shear stress to the corresponding

shear strain within the elastic limit is known as Modulus of

Rigidity or Shear Modulus. This is denoted by C or G or N.

G

25

Relation Between Bulk Modulus and Young’s Modulus

Relation Between Young’s Modulus and Modulus of Rigidity


Example Problems: Volumetric Strain


Problem 7.

A steel bar 300 mm long, 50 mm wide and 40 mm thick is subjected to a pull of 300kN in the direction of its length. Determine the change in Volume. Take E = 2 X 105N/mm2 and µ=0.25

Solution:Given length = L = 300 mm width = b =50 mmThickness = t = 40 mm pull = 300 kN = 300 X 103 NOriginal volume = L b t = 600000 mm3

The longitudinal strain = the strain in the direction of load = dL /L= stress in the direction of load / E

But stress in the direction of load = (P/area) = {p/(b x t)}= 150 N/mm2

dL/L = 150 / (2x 105) = 0.00075Now volumetric strain is given by ev =

L

dL(1- 2 µ)

Substituting the values ev =0.000375Change in volume = dV= 0.000375 x 600000 =225 mm3



Problem 8.

A metallic bar 300 mm X 100 mm x 40 mm is subjected to a force of 5kN(tensile), 6 kN(tensile) and 4 kN(tensile) along x, y and z directions respectively. Determine thechange in the volume of the block.

Solution :Dimensions of the bar = 300 mm x 100 mm x 40 mmVolume = V = (x) X (y) X (z)= 300 X 100 X 40 = 1200000 mm3

Load in the x direction = 5 kN = 5000 N Load in the y direction = 6 kN = 6000 NLoad in the z direction = 4 kN = 4000 NValue of E = 2 x 105 N/mm2

Poisson’s ratio = µ =0.25∴ Stresses in the x-direction,



Problem 9.A steel rod 5m long and 30 mm in diameter is subjected to an axial tensile load of50 kN. Determine the change in length, diameter and volume of the rod.

Take E= 2 x 105 N/mm2 and Poisson's ratio =0.25

Solution: Given Length = 5 m = 5 x 103 mm; Diameter = d = 30 mm

Example Problems: Elastic Constants


Problem 10.For a material, Young’s Modulus is given as 1.2 x 105 N/mm2 and Poisson’s ratio ¼.Calculate the Bulk Modulus.

Solution:Using equation K = E/3(1- 2 µ) we can find the value of Bulk modulus.

K= (1.2 x 105)/ 3(1-2(¼))

K= (2.4 x 105)/3

Ans 0.8 x 105 N/mm2

Example Problems: Elastic Constants


A bar of 30 mm diameter is subjected to a pull of 60 kN. The measured extension ongauge length of 200 mm is 0.1 mm and change in diameter is 0.004 mm. Calculate

i) Young’s modulus ii) Poisson’s ratio iii) Bulk ModulusSolution: Given diameter of the bar, d = 30 mm

Problem 11.

Stresses in Composite Bars

A composite bar is made up of two or more different materials,

joined together, in such a manner that the system extends or

contracts as one unit, when subjected to tension or compression.


P1 = Load carried by bar 1,

A1 = Cross-sectional area of bar 1,

σ1 = Stress produced in bar 1,

E1 = Young's modulus of bar 1,

P2, A2, σ2, E2 = Corresponding values of bar 2,

P = Total load on the composite bar,

l = Length of the composite bar, and

δl = Elongation of the composite bar.

We know that P = P1 + P2

Stress in bar 1,

Strain in bar 1,


δl1 =δl2

P = P1 + P2 = σ1.A1 +σ2.A2

Bars with varying cross-sections

A typical bar with cross-sections varying in steps and subjected

to axial load

length of three portions L1, L2 and L3 and the respective cross-

sectional areas are A1, A2, A3

E = Young’s modulus of the material

P = applied axial load.


Stress, strain and extension of each of these portions are

29

Portion Stress Strain Extension

1 σ1 = P/ A1 e1 =σ1 / E δ1 = P L1 / A1E

2 σ2 = P/A2 e2 =σ2 / E δ2 = P L2 / A2E

3 σ3 = P/A3 e3 =σ3 / E δ3 = P L3 / A3E


Example Problems: Bars of Varying Sections


Problem 12 the Fig



Problem 13

the Figure

(Contd…)

Example Problems


Problem 13 (Contd…)



Problem 14

the fig.

(Contd…)

Example Problems


Problem 14 (Contd…)



Problem 15.

(Contd…)

Example Problems


Problem 15.

(Contd…)

(Contd…)

Example Problems


Problem 15. (Contd…)

Example Problems : Bars of Composite Sections


Problem 16.

(Contd…)

Example Problems


Problem 16.

(Contd…)

(Contd…)

Example Problems



Example Problems: Bars of Composite Sections


Problem 17.

A compound tube consists of a steel tube 140 mm internal diameter and 160 mmexternal diameter and an outer brass tube 160 mm internal diameter and 180 mmexternal diameter. The two tubes are of the same length. The compound tube carriesan axial load of 900 kN. Find the stresses and the load carried by each tube and theamount it shortens. Length of each tube is 140mm. Take E for steel as Es=2 x 105

N/mm2 and E for brass as Eb =1x 105 N/mm2.

Solution:

Internal dia of steel tube = 140 mmExternal dia of steel tube = 160 mmInternal dia of brass tube = 160 mmExternal dia of brass tube = 180 mm

(Contd…)

Example Problems: Bars of Composite Sections


Problem 17.

Thermal Stresses

Stresses due to Change in Temperature.

Whenever there is some increase or decrease in the

temperature of a body, it causes the body to expand or contract.

If the body is allowed to expand or contract freely, with the rise

or fall of the temperature, no stresses are induced in the body.

But, if the deformation of the body is prevented, some stresses

are induced in the body. Such stresses are known as thermal

stresses.


If the ends of the body are fixed to rigid supports

• so that its expansion is prevented, then compressive strain

induced in the body,

• Increase or decrease in length, δl = l × α × t

∴ Thermal stress,


If the free expansion is prevented fully

• Since support is not permitting it, the support force P develops

to keep it at the original position.

• Magnitude of this force is such that contraction is equal to free

expansion.

δl = l ×α ×t

34Dr R Vaikunta Rao Mechanics of Solids Mechanical Engineering REC 52

If free expansion is prevented partially

Expansion prevented Δ = α tL – δ


Example Problems: Thermal Stresses


Problem 19.A rod is 2 m long at a temperature of 100c. Find the expansion of the rod, when the temperature is raised to 80oc. If this expansion is prevented, find the stress induced in the material of the rod. Take E = 1.0 x 105 MN/m2 and α = 0.000012 per degree centigrade.



Problem 20.

3cm

5m

(Contd…)



Problem 20.

3cm

5m

Uniformly Tapering Circular Rod







Example Problems: Tapered Circular Sections

Problem 21.

Uniformly Tapering Rectangular Plate



Total Extension dL =Total Extension dL =

Remember Point

States of stresses


Uni-axial stress: only one non-zero principal stress, Figure represents Uni-axial state of stress.


BIAXIAL STRESS SYSTEMS


A biaxial stress system has a stress state in two directions and a shear stress typically showing in Fig..

Element of a structure showing a biaxial stress system

When a Biaxial Stress state occurs in a thin metal, all the stresses are in the plane of the material. Such a stress system is called PLANE STRESS. We can see plane stress in pressure vessels, aircraft skins, car bodies, and many other structures.

Principal planes and stresses


The planes, which have no shear stress, are known as PRINCIPAL PLANES. • Hence principal planes are the planes of zero shear stress. • These planes carry only normal stresses. • The normal stresses, acting on a principal plane, are known as principal

stresses.

Methods for determining principal planes and stresses 1. Analytical method 2. Graphical method

1. Analytical method on oblique section The following are the two cases considered

• A member subjected to a direct stress in one plane • A member subjected to like direct stresses in two mutually perpendicular

directions.


Member Subjected to Direct stress in one plane

σ𝑡

σ𝑡

σ𝑡


LetP= Axial force on the member & A= Area of cross section which is perpendicular to the line of action of force P

Hence the member is subjected to stress along X axis

Consider cross section EF which is perpendicular to line of action of force P


P P

θ

E

F

G

P P

P

P

90(90- θ)

θθ(90-θ)

t

n

E G

F


=

(a) (b)

This Stress on the section FG, is parallel to axis of the member (i.e. along x-axis)

This Stress may be resolved into two components.

P P

P

P

90(90- θ)

θθ(90-θ)

t

n

E G

F

P P

θ

E

F

G



Resolved into two components (i) Normal to section FG & Tangential to section FG,

P P

P

P

90(90- θ)

θθ(90-θ)

t

n

E G

F

P P

θ

E

F

G



P P

P

P

90(90- θ)

θθ(90-θ)

t

n

E G

F

σ𝑡

σ𝑛



Direct stress in one plane: E

F

P P

E

F

135o

P P

45o


Remember Point

When a member is subjected to a direct stress (𝝈) in one plane, then the stresses on an oblique plane (which is incline at an angle θ with the normal cross-section) are given by:

Example Problems: Principal Stresses


. Area of cross-section normal to x-axis = 3 x 0.5 = 1.5 cm2 Area of cross-section normal to y-axis = 4 x 0.5 = 2 cm2

Problem 21.



Problem 22.


Member subjected to direct stresses in two mutually perpendicular directions

These are Important….


Principle Planes• Principle Planes are the planes on which shear stress is zero. • To locate the position of principle planes


When a member is subjected to two direct stress in two mutually perpendicular directions, then the stresses on an oblique plane at an angle

θ with the axis of the minor stress (or with the plane of major stress)

Remember Point



Problem 23.




Problem 24.A small block is 4 cm long, 3 cm high and 0.5 cm thick. It is subjected to uniformly distributed tensile forces of resultants 1200 N and 500 N as shown in Fig. below. Compute the normal and shear stresses developed along the diagonal AB.

Area of cross-section normal to x-axis = 3 x 0.5 = 1.5 cm2

Area of cross-section normal to y-axis = 4 x 0.5 = 2 cm2



Members subjected to direct stresses in two mutually perpendicular directions accompanied by simple shear stress

When a member is subjected to a simple shear stress (τ), then the stresses on an oblique plane are given as


When a Members subjected to two direct stresses (𝝈𝟏 , 𝝈𝟐) in two mutually perpendicular directions accompanied by simple shear stress (τ) , then the stresses, on

an oblique plane inclined at an angle θ with the axis of minor stress are given by

Remember Point




Problem 25.A rectangular block of material is subjected to a tensile stress of 110 N/mm2 on one plane and a tensile stress of 47 N/mm2 on the plane at right angles to the former. Each of the above stresses is accompanied by a shear stress of 63 N/mm2 and that associated with the former tensile stress tends to rotate the block anticlockwise. Find: (i) The direction and magnitude of each of the principal stress and (ii) Magnitude of the greatest shear stress

Given σ1 = 110 N/mm2; σ2 = 47 N/mm2 ; τ = 63 N/mm2 ; θ = 45°

(Contd…)





Principal stresses

Principal strains

2 21, 2

1[( ) ( ) 4

2x y x y xy

1[ ( )]

2

1[ ( )]

2

1[ ( )]

2

x x y z

y y x z

z z x y

e

e

e

84

Finally…

Pull

Consider a Structure and pull it

σ𝑥 σ𝑥

σ𝑦

σ𝑦

τ𝑥𝑦

τ𝑦𝑥

τ𝑥𝑦

τ𝑦𝑥

Shear loading leading to shear stresses

Loading is tensile but inclined planes feel shear.

Principal Stresses and Planes

Mohr’s circle


It is a graphical method of finding normal, tangential and

resultant stresses on an oblique plane.

Mohr's circle can be drawn for following cases (I, II, III):

I. A body subjected to two mutually perpendicular principal

tensile stress of unequal intensities

II. A body subjected to two mutually perpendicular principal

stresses which are unequal and unlike (i.e., one is tensile and

other is compressive)

III. A body subjected to two mutually perpendicular principal

tensile stresses accompanied by a simple shear stresses


𝐸𝐷 = σ𝑡

A

B

σ1

σ2

C O

2θ

E

D

𝐴𝐷 = σ𝑛

θ

R

Mohr’s circle: Type I

Mohr's Circle of a body subjected to two mutually perpendicular

Principal Tensile stressesof unequal intensities

Steps:1. Take any point A and draw a

horizontal line through capital A.

2. Take AB= σ1 and AC= σ2towards right from A to some suitable scale.

3. With BC as diameter describe a circle.

4. Let O is the centre of the circle.5. Now through O draw a line OE

marking an angle 2θ6. With OB.7. From E, Draw ED perpendicular

AB.8. Join AE. Then the normal end

tangential stresses on the oblique plane are give by AD and ED respectively.

9. The resultant stress on the oblique plane is given by AE.

AD= Normal stress on oblique planeED= Tangential stress on oblique planeAE= Resultant stress on Oblique plane

φ


𝐸𝐷 = σ𝑡

A

B

σ1

σ2

C O

2θ

E

D

𝐴𝐷 = σ𝑛

θ

R

Example Problem: Mohr’s circle: Type I

By MeasurementsAD= Normal stress on oblique plane = 10.50cmED= Tangential stress on oblique plane =2.60 cmAE= Resultant stress on Oblique plane = 10.82 cm

Problem 23.

ScaleLet 1 cm = 10 N/mm2

Thenσ1= 120/10 = 12cmsσ2= 60/10 = 6 cms

2 θ = 60°

GivenMajor Principal Stress σ1= 120 N/mm2

Minor Principal Stress σ2= 60 N/mm2

Then angle of oblique plane θ= 30°

FinallyNormal Stress = AD x Scale = 105 N/mm2

Tangential Stress = ED x Scale = 26 N/mm2

Resultant Stress = AE x Scale = 108.2 N/mm2

Solution:

φ


A BC

σ1 σ2

O

φ

E

2θ

D

𝐴𝐷 = σ𝑛

𝐸𝐷 = σ𝑡R

Mohr’s circle: Type II

Mohr's Circle of a body subjected to two mutually perpendicular

Principal stresses of which are unequal and un like (i.e., one tensile and one compressive)

Steps:1. Take any point A and draw a

horizontal line through A on both sides of A.

2. Take AB=σ1 (+ve) towards right of A and AC= σ2 (-ve) towards left of A to some suitable scale.

3. Bisect BC at O. with O as center and radius = CO or OB, draw a circle.

4. Through O draw a line OE

making an angle 2θ with OB.5. From E, Draw ED perpendicular

to AB. 6. Join AE and CE.7. Then Normal and Shear

stresses (i.e., Tangential stress) on the oblique plane are given by AD and ED.

8. Length AE= Resultant Stress

and φ is obliquity.

AD= Normal stress on oblique planeED= Tangential stress on oblique planeAE= Resultant stress on Oblique plane

(+ve) (-ve)

θ


Example Problem: Mohr’s circle: Type II

By MeasurementsAD= Normal stress on oblique plane = 6.25 cmED= Tangential stress on oblique plane =6.5 cmAE= Resultant stress on Oblique plane = 9 cm

Problem 26.


Thenσ1= 200/20 = 10cmsσ2= -100/20 = -5 cms

2 θ = 60°


Minor Principal Stress σ2= -100 N/mm2

Then angle of oblique plane θ= 60°90°-60°=30°



Resultant Stress = AE x Scale = 180 N/mm2

Solution:

A BC

σ1 σ2

O

φ

E

2θ

D

𝐴𝐷 = σ𝑛

𝐸𝐷 = σ𝑡R

Max Shear Stress = Radius of Mohr's Circle= (σ1+σ2)/2 = 150 N/mm2

θ

Mohr’s circle: Type IIIMohr's Circle of a body subjected to two mutually perpendicular

Principal Tensile Stresses accompined by a simple shear stress


A B

σ1

σ2

C O

E

2θ

D

G

F

R

𝐴𝐷 = σ𝑛

𝐸𝐷 = σ𝑡

φ

M

Steps:

1. Take AB=σ1 and AC=σ2towards right of A to some suitable scale.

2. Draw perpendiculars at B and C and Cut off BF and CG = Shear Stress (τ) to the same scale.

3. Bisect BC at O. Now with O as centre and Radius = OG or OF draw a circle.

4. Trough O, Draw a line OE making an Angle of 2θ with OF.

5. From E, draw ED perpendicular CB.

6. Join AE. Then AE= Resultant Stress and φ is obliquity.AD= Normal stress on oblique plane

ED= Tangential stress on oblique planeAE= Resultant stress on Oblique plane

α


Example Problem: Mohr’s circle: Type III

Solution:

σ𝑥 σ𝑥 = 65N/mm2

σ𝑦 = 35N/mm2

σ𝑦

τ𝑥𝑦= 25 N/mm2

τ𝑦𝑥

τ𝑥𝑦

τ𝑦𝑥

45o

Problem 27.

(Contd.)


Example Problem: Mohr’s circle: Type III

By MeasurementsAD= Normal stress on oblique plane = 7.5 cmED= Tangential stress on oblique plane =1.5 cmAE= Resultant stress on Oblique plane = ??? cm

Problem 27.


Thenσ1= 65/10 = 6.5 cmsσ2= 35/10 = 3.5 cmsτ = 25/10 = 2.5 cms

2 θ = 90°


Minor Principal Stress σ2= 35N/mm2

Shear Stress τ = 25N/mm2

Then angle of oblique plane θ= 45°



Resultant Stress = AE x Scale = ???? N/mm2

Solution:

A B

σ1

σ2

C O

E

2θ

D

G

F

R

𝐴𝐷 = σ𝑛

𝐸𝐷 = σ𝑡

φ

M

(Contd.)

α

Strain Energy


• The energy which is absorbed in the body due to straining effect is known as strainenergy.

• The straining of a body may be due to gradually applied load or suddenly or with animpact.

• The total strain energy stored in a body is commonly known as resilience.• Also resilience is also defined as the capacity of a strained body for doing work on

the removal of the straining force.

• The maximum strain energy stored in a body is known as proof resilience and itis the quantity of strain energy stored in a body when strained upto elastic limit.

Energy Stored in a Bow and Arrow

Derivation for an expression for strain energy stored

in a body when the load was applied gradually


Strain energy stored in a body is equal to the work done by the applied load in stretching the body.The work done by load P in stretching the body is stored in the body as strain energy in the bodyand this strain energy is recoverable after the load P is removed.LetP= Gradually applied load,X = extension of the body,A = Area of cross section,L = length of the body,V = volume of the body,E = young’s ModulusU = strain energy stored in the body and = stress induced in the bodyNow work done by the load = area of load extension curve( shaded area in the above fig)= Area of the triangle ONM = ½ x P x X

But load P = Stress x Area = x AExtension = e = (Stress / E ) x L = ( /E) x LWork done by load = ½ x A x ( /E) x L =½ (2/E) x A x L = (2/2E) x VBut work done by the load in stretching the body is equal to the strain energy stored in the body.Therefore energy stored in the body U = (2/2E) x V

GATE QuestionsQuestion 1. A 200 100 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-2007]

(A) 85 (B) 90 (C) 100 (D) 110

Hint . (Ans. B)

Here,

Now,


Where, 1/m=Poisson’s ratio

96

Question 2. A steel rod of length D, fixed at both ends, is uniformly

heated to a temperature rise of . The Young’s modulus is E and the

coefficient of linear expansion is . The thermal stress in the rod is

[GATE-ME-2007]

(A) 0 (B) (C) E (D) E

Hint . (Ans C)

Since rod is free to expand, therefore

Since rod is fixed at both ends, so thermal strain will be zero but there

will be thermal stresses.


Question 3. A rod of length L and diameter D is subjected to a tensile

load P. which of the following is sufficient to calculate the resulting

change in diameter? [GATE-ME-2008]

(A) Young’s Modulus (B) Shear modulus

(C) Poisson’s ratio (D) Both Young’s modulus and the shear modulus

Hint .(Ans D)

Both Poison’s ratio and Young’s Modulus are required to calculate the

lateral strain in rod Now m, E and N are related by

So E and G are required if is given


Question 4. A solid steel cube constrained on all six face is heated so

that the temperature rises uniformly by . If the thermal coefficient of

the material is , young’s modulus is E and the Poisson’s ratio is v, the

thermal stress developed in the cube due to heating is

[GATE-ME-2012]

(A) (B) (C) (D)

Hint . (Ans B)


Question 5. A rod length L having uniform cross-sectional area A is

subjected to a tensile strength force P as shown in the figure below. If

the Young’s modulus of the material varies linearly from along the

length of the rod, the normal stress developed at the section SS is

[GATE-ME-2013]

(A)P/A (B) (C) (D)

Hint . (Ans A)

Do the force balance. Force at section P.


Question 6. The stress-strain behaviour of a material is shown in

figure. Its resilience and toughness, in Nm/m3, are respectively

[GATE-ME-2000]

(A)

(B)

(C)

(D)

Hint . (Ans. C)

Resilience=

Toughness=


Question 7. A 1.5 mm thick sheet is subject to unequal biaxial

stretching and the true strains in the directions of stretching are 0.05

and 0.09. The final thickness of the sheet in mm is [GATE-ME-2000]

(A) 1.414 (B) 1.304 (C) 1.362 (D) 289

Hint . (Ans A)

Question 8. The relationship between Young’s modulus(E), Bulk

Modulus (K), and Poisson’s ratio ( ) is given by [GATE-ME-2002]

(A) E=3K(1-2 )

(B) K=3E(1-2 )

(C) E=3K(1- )

(D) K=3E(1- )

Hint . (Ans A)


Question 9. The total area under the stress-strain curve of a mild steel

specimen tested up to failure under tension is a measure of

[GATE-ME-2002]

(A) Ductility (B) Ultimate strength

(C) Stiffness (D) Toughness

Hint . (Ans D)

Question 10. In a linearly hardening plastic material, the true stress

beyond initial yielding [GATE-ME-2018]

(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

Hint . (A) increases linearly with the true strain


Question 11. Two identical circular rods of same diameter and same length are

subjected to same magnitude of axial tensile force. One of the rods is made out of mild

steel having the modulus of elasticity of 206 GPa. The other rod is made out of cast

iron having the modulus of elasticity 100 GPa. Assume both the materials to be

homogeneous and isotropic and the axial forces 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 observation is correct?

[GATE-ME-2003]

(A) Both rods elongate with the same amount

(B) Mils steel rod elongate more than the cast iron rod

(C) Cast iron rod elongate more than the mild steel rod.

(D) As stresses are equal strains are also equal in both the rods.

Hint 5. (Ans C)

Emild steel=206 Gpa Ecast iron =100 GPa

Now, elongation in cast iron,

Elongation in mild steel,


Question 12. 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 the

length of the rod due to loading is [GATE-ME-2004]

(A) 1 m

(B) -10 m

(C) 16 m

(D) -20 m

Hint 6. (Ans. B)

Total change in length


Question 13. In terms of Poisson’s ratio(v) the ratio of Young’s

Modulus (E) to shear Modulus (G) of elastic material is

[GATE-ME-2004]

(A) 2(1+ ) (B) 2(1- ) (C) ½ (1+ ) (D) ½ (1- )

Hint 7. (Ans. A)

We know E=2G(1+ )

Question 14. A uniform slender cylinder 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 and , respectively, then [GATE-ME-2004]

(A) (B) (C) (D)

Hint 8. (Ans A)

Rod is not restrained but completely free; hence no stress will be

induced.Dr R Vaikunta Rao Mechanics of Solids Mechanical Engineering REC 10

6

Question 15. 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 2m and E=200 GPa, the elongation of the bar will be:

[GATE-ME-2006]

(A) 1.25 mm (B) 2.70 mm (C) 4.05 mm (D) 5.40 mm

Hint . (Ans A)

We know


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