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Elasticity - NetBadi.com ExclusiveElasticity - NetBadi.com ExclusiveElasticity - NetBadi.com ExclusiveElasticity - NetBadi.com ExclusiveIntroduction:

In all engineering construction the component parts of a structure must be assigned

definite physical sizes. Such parts must be properly proportioned to resist the actual or probable

forces that may be imposed upon them. When an external force acts on a body, the body tends to

undergo some deformation. Due to chesion between the molecules, the body resists deformation.

This resistance by which material of the body posses the deformation is known as strength of

material or mechanics of solid. A plastic material undergoes a continuous deformation during

the period of loading and the deformation is permanent and the material does not regain its

original dimensions on the removal of the loading.

A rigid material not undergo any deformation any deformation when subjected to an external

loading. The behaviour of member subjected to forces depends not only on the fundamental laws

of Newtoninan mechanics that govern the equilibrium of the forces but also on the physical

characteristics of the material of which the member is fabricated.

Mechanics of solids of strength of materials is a blended science of experiment and Newtonian

postulates of analytical mechanics. Complete, carefully drawn diagrammatic sketches of problems

to be solved will pay large dividends in a quicker and more complete mastery of this subject.

1.2 Elasticity1.2 Elasticity1.2 Elasticity1.2 ElasticityIt is the property of the material which enables it to regain its original shape after deformation

within the elastic limit:

(Or)

Elasticity is the property of a material by virtue of which deformation caused by applied load

disappears upon removal of the load.

Elasticity is a tensile property of the material.

Examples: Steel, rubber

Steel is said to be more elastic than rubber.

1.3 Plasticity1.3 Plasticity1.3 Plasticity1.3 Plasticity

It is the property of the material which enables the formation of permanent deformation.

(Or)

The plasticity of a material is its ability to undergo some degree of permanent deformation without

rupture of failure.

Examples: Clay, lead

Note: Plastic deformation will takes place only after the elastic range.

Applications: Forming, shaping and extruding operations.

Deforming force: An external force applied to a body which change its size or shape or both is

called deforming force.

Perfectly Elastic body: A body is said to be perfectly elastic if it completely remains its original

form when the deforming force is removed. Since no material can regain completely its original

form so the concept of perfectly elastic body is only an ideal concept. A quartz fiber is the nearest

approach to the perfectly elastic body.

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Perfectly Plastic body: A body is said to be perfectly if it does not regain its original form even

slightly when the deforming force is removed. Since every material partially regain its original

form on the removal of deforming force, so the concept of perfectly plastic body is only an ideal

concept. Paraffin wax, wet clay are the nearest approach to a perfectly plastic bodies.

Cause of Elasticity: In a solid, atoms and molecules are arranged in such a way that each

molecule is acted upon by the forces due to the neighbouring molecules. These forces are known

as intermolecular forces. When no deforming force is applied on the body, each molecule of the

solid (i.e body) is in its equilibrium position and the inter molecular forces between the molecules

of the solid are maximum. On applying the deforming force on the body, the molecules either

come closer or go far apart from eact other. As a result of this, the molecules are displaced from

their equilibrium position. In other words intermolecular forces of get changed and restoring

forces are developed on the molecules. When the deforming force is removed, these restoring

forces bring the molecules of the solid of their respective equilibrium —————————————

————————— regains its original form.

DEFINITION OF STRESS:

The force of resistance per unit area, offered by a body against deformation is known as stress.

(Or)

When a body is acted upon by external force or load, internal resisting forces are set up and it is

then said to be in a state of stress.

The intensity of the force perpendicular or normal to the section is called the normal stress at a

point.

The other component of the intensity of force acts parallel to the plane of the elementary area as

in figure. 1.2. This component of the intensity of force is called the shearing stress.

In mathematically,

The normal stress 2

, /0

Lim F d FN m

A A d A�

�� �

� � �

And Shearing stress 2

, /0

Lim F dUN m

A A d A�

�� �

� � �

Where F = force acting normal to the cut.

A = corresponding area

V = component of the force parallel to the cut.

Note: Normal stresses result from force components perpendicular to the plane of the cut, while

shearing stresses result from components parallel to the plane of the cut.

The unit of stress is the 2/N m , also designated a Pascal (Pa).

1.5 TYPES OF STRESS COMPRESSIVE, TENSILE AND SHEARING STRESS:1.5 TYPES OF STRESS COMPRESSIVE, TENSILE AND SHEARING STRESS:1.5 TYPES OF STRESS COMPRESSIVE, TENSILE AND SHEARING STRESS:1.5 TYPES OF STRESS COMPRESSIVE, TENSILE AND SHEARING STRESS:

The important types of simple stresses are:

1. Compressive stress 2. Tensile stress, and 3. Shear stress.

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1.5 COMPRESSIVE STRESS:1.5 COMPRESSIVE STRESS:1.5 COMPRESSIVE STRESS:1.5 COMPRESSIVE STRESS:

15.1 The stress induced in a body, when subjected to two equal and opposite pushes as shown in

Fig 1.3 as a result of which there is a decrease in length of the body, is known stress is given by

2Re/

( )p p

Push Psisting Force Por N m

Area A Area A A� �� �

1.52 The stress induced in a body, when subjected to two equal and opposite pulls in Fig 14. as

a result of which there is an increase in length, is known as tensile stress .

Tensile Stress: Let: P = Pull (or force) acting on the body.

A = cross – sectional area of the body.

L = original length of the body.

dL = increase in length due to pull P acting on the body

� = stress induced in the body and

� = strain (i.e tensile strain).

An equilibrium, P = Resisting force as shown in Fig. 1.4.

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This resisting force per unit area is known as stress or intensity of stress.

�Tensile stress Re

sect

sisting force R

Cross tional area� �

� or 2Re

/t

sisting load P PN m

A A� � � .

1.5..3 SHEAR STRESS :1.5..3 SHEAR STRESS :1.5..3 SHEAR STRESS :1.5..3 SHEAR STRESS :

The stress induced in a body, when subjected to two equal opposite forces which are acting

tangentially across the resisting section as shown in Fig 1.5 as a result of which the body finds

to shear off across the section, is known as shear stress.

� Shear Stress tanShear resis ce

Shear area� �

= R P

A A�

1

P P

L width L� � �

Where L = Length of block

Width = unity

3. Bulk stress or All around stress or Pressure: When force is acting all along the surface

normal to the area, then force acting per unit area is known s pressure. The effect of pressure is

to produce volume change. The shape of the body may or may not change depending upon the

homogeneity of body.

Illustration 1

Find out longitudinal stress and tangential stress on a fixed block

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Sol. Longitudinal or normal stress

02

1

100 sin305 /

5 2N m� � �

Tangential stress0

2

1

100 cos305 3 /

5 2N m� � �

Illustration 2

Find out Bulk stress on the spherical object of radius 10

� cm if area and mass of piston is 50 cm2

and 50 kg respectively for a cylinder filled with gas.

Sol. 5 5 2

4

50 101 10 2 10 /

50 10gas a

mgP P N m

A �

� � � � �

Bulk stress = P

gas = 2 × 105 N/m2.

1.19 BARS OF UNIFORM STRENGTH :1.19 BARS OF UNIFORM STRENGTH :1.19 BARS OF UNIFORM STRENGTH :1.19 BARS OF UNIFORM STRENGTH :

Fig. 1.46 shows a bar subjected to an external tensile load P. Let us find the shape of bar of

which self weight of the bar is considered and is having uniform stress on all sections when

subjected to an axial load P. In which the area of the bar increases from the lower end to upper

end.

Let A1 = Area of upper and

A2 = Area of lower and

= weight per unit volume of the bar.

� = uniform stress on the bar.

Let the area of the section be A at a distance x from the lower end. Let the area be (A + dA) at a

distance (x + dx) from the lower end. Consider the equilibrium of the strip ABCD. Total force

acting upwards = Total force acting downwards.

A dA A Adx

or A dA A Adx

� �

� � �

�� � � �

� � � �

1.6.3 VOLUMETRIC STRAIN:1.6.3 VOLUMETRIC STRAIN:1.6.3 VOLUMETRIC STRAIN:1.6.3 VOLUMETRIC STRAIN:

The ratio of change of volume of the body to the original volume is known as volumetric strain.

� the volumetric strain is given by

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v

change in volume dV

original volume V� � � .

1.6.4 SHEAR STRAIN:1.6.4 SHEAR STRAIN:1.6.4 SHEAR STRAIN:1.6.4 SHEAR STRAIN:As shown in Fig. 1.7, the bottom face of the block is fixed height h, length L and width unity. The

face ABCD will distorted to ABC1D

1 through an angle � as a result of force P.

� Shear strain � is given by

tantan

Transverse displacement

Dis ce AD� ��

Or 1

DD dL

AD h� � �

Note: Shearing strain is independent of the individual angles made with the coordinate axes.

1.7 ELEASTIC LIMIT:1.7 ELEASTIC LIMIT:1.7 ELEASTIC LIMIT:1.7 ELEASTIC LIMIT:

The body will regain its previous shape and size only when the deformation caused by the external

force, is within a certain limit. Thus there is a limiting value of force up to and within which, the

deformation completely disappears on the removal of the force. The value of stress corresponding

to this limiting force is knows as the elastic limit of the material.

Or d A Adx� �

Or .dA

dxA

�

�

Integrating the above equation, we get 1ln A x C

�

� �

Where C1 = constant of integration

At x = 0, A = A2.

� 1n A2 = C

1

� 1n A = �

x + 1n A2

Or 1n A – 1n A2 =

�

.x

Or 2

ln .A

xA

�

� ��� �

� �

Or / .

2

x

e

A

A

�� Or / .

2.

x

eA A ��

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The above equation gives the area at a distance x from lower end

At x = L, A = A1

� / .

1 2.

x L

eA A ��

1.11 DEFINITION OF STRAIN, TYPES OF STRAINS:

When a body is subjected to some external force, there is some change of dimension of the body.

The ratio of changed of dimension of the body to the original dimension is known as strain. It is

a dimensionless quantity, but it is convenient to refer to it as having the dimension of meter per

meter, i.e. m/m. Sometimes strain is given in percent. It is represented as � (epsilon).

Strain may be

1. Tensile strain

2. Compressive strain.

3. Volumetric strain and

4. Shear strain.

1.11.1 TENSILE STRAIN:

If there is some increase in length of a body due to external force, then the ratio of increase of

length to the original length of the body is known as tensile strain.

� Tensile strain is given by

1

Increase in length d L

original length L� � �

1.11.2 COMPRESSIVE STRAIN:

If there is some decrease in length of the body, then the ratio of decrease of the length of the body

to the original length is known as compressive strain.

� Compressive strain is given by

1

Increase in length d L

original length L� � �

1.8 HOOKE’S LAW:

It is observed that when a material is loaded such that the intensity of stress is within a certain

limit, the intensity of stress is proportional to the strain produced by the stress. The ratio of the

intensity of stress to the corresponding strain is a constant within the elastic limit. This constant

is known as modulus of elasticity or young’s modulus. Thus stress � Strain.

1.9 YOUNG’S MODULUS OR MODULUS OF ELASTICITY

The ratio of tensile or compressive stress to the corresponding strain is a constant. This constant,

knows as the modulus of elasticity and it is denoted by E. Thus.

StressE or

Strain��

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� or E = Tensile or compressive stress

Tensile or compressive strain

��

�

� or E = ��

. Now P d Land

A L� � ��

PPL PLAor or dL or

dL AE AL

��

� �

� Change in length dL or � L

sec ' mod

PL Load length of specimen

AE cross tional area young s ulus

� �

� .

1.11 TENSILE TEST ON MILD STEEL SPECIMEN:

If a ductile material is subjected to the tensile test, it passes through the following four important

stages after it finally fractures:

1. Limit of proportionally 2. Elastic limit.

3. Yield point 4. Maximum load point 5. Breaking point.

In the tensile test, a ductile material, in the form of a circular rod, is subjected to a gradually

increasing tension. The strain both longitudinal and lateral increases at first proportionally to

the stress. The limit of proportionally is the stage up to which the material perfectly obeys Hook’s

law.

The elastic limit is a state of loading at and before which the strain disappears completely on the

removal of the load, although Hook’s law doesn’t hold good between the limit of proportionality

and the elastic limit.

On further loading, the resultant strain begins to increase more quickly than the corresponding

stress and continuous to grow more as the load increase till the yield point reaches. The stress

at which this sudden stretch occurs is called the yield point of the material.

At a certain value of load, the extension continuous at a slow rate without any further loading. In

this state the material is plastic and the phenomenon of slow extension, increasing with the time

at a constant load is known as creeping. The cross – sectional area decreases in the same proportion

as the length increases during the ductile elongation. The volume of the bar under test remains

constant. The cross – sectional area reduces almost uniformly along the bar (Fig. 19.)

A – Limit of proportionality.

B – Elastic limit.

C – Yield point.

D – Maximum stress point.

E – Breaking or fracture point.

On continuously increasing the load after the creeping has taken place, a maximum value of load

is reached at which the material stretches locally over a short length forming a neck. The load

reduction in area is such that the load necessary to break the bar at the neck is considerably less

than the maximum load on the bar before the local extension takes place.

The breaking load divided by the reduced area of section shows that the actual stress intensity is

greater than at any previous load. If the load the divided by the original area of cross – section,

the result is the normal stress intensity, which is less in such a ductile material like soft material,

at the breaking load than at the maximum load.

As the extension continues, cross – sectional area becomes smaller and smaller there by requiring

lesser and lesser load to continue extension till fracture occurs.

Elastic after effect:

We know that some material bodies takes some time to regain their original configuration

when the deforming force is removed. The dealy in regaining the original configuration by the

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configuration by the bodies on the removal of deforming force is called elastic after effect. The

elastic after effect is negligibly small for quartz fiber and phosphor bronze. For this reason, the

suspensions made from quartz and phosphor – bronze are used in galvanometers and

electrometers.

For glass fiber elastic after effect is very large. It takes hours for glass fiber to return to its

original state on removal of deforming force.

Elastic Fatigue:

The loss of strength of the material due to repeated strains on the material is called

elastic fatigue. That is why bridges are declared unsafe after a longtime of their use.

1.12 BRAS OF VAYING SECTION:

Figure.

A bar of different lengths and of different diameters is shown in Fig. 1.10. Let this bar is subjected

to an axial load P. Though each section is subjected to the same axial load P, yet the stresses,

strains and change in lengths will be different. The total change in length will be obtained by

adding the changes in length of individual section.

Let P = Axial load acting on the bar,

L1 = Length of section (1)

A1 = Cross – sectional are of section (1)

L2, A

2 = Length and cross – sectional area of section (2)

L2, A

3 = Length and cross – sectional area of section 3, and E = Young’s modulus for the bar.

Section (1) Section (2) Section (3)

Stress 1sec

Load

Area of tion� � Stress 2

sec

Load

Area of tion� �

3

3

P

A� �

1

1

P

A� �

2

2

P

A� �

And Strain 1

1

1

P

E A E

�� � � Strain 2

2

P

A E� � Strain 2

3

P

A E� �

Or 1

1

1 1

dL P

L A E� � � Or 2

2

P

A E� � Or

3

3

3 3

dL P

L A E� � �

Or 1

1

1

PLdL

A E� Or

1

2

2

PLdL

A E� Or

3

3

3

PLdL

A E�

Or change in length of section Or change in length of section Or change in length of

section.

1

1

1

PLdL

A E� 2

2

2

PLdL

A E� 3

3

3

PLdL

A E�

� Total change in the length of the bar.

dL = dL1 + dL

2 + dL

3

31 2

1 2 3

PLPL PLdL

A E A E A E� � �

Or 31 2

1 2 3

LL LPdL

E A A A

� �� � �� �

� �……………….. (1.1)

Equation (1.1) in used when the young’s modulus of different sections is same. If the young’s

modulus of different sections is different, then total change in length of the bar is given by

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31 2

1 1 2 2 3 3

LL LdL P

A E A E A E

� �� � �� �

� �

Principle of Superposition:

When a number of loads are acting on a body the resulting strain, according to the

principle of superposition, will be the algebraic sum of the strains caused by the individual

forces.

Free body diagram. If an elastic body is subjected to a number of direct forces (tensile or

compressive) at different sections along the length of the body, the deformation of individual

sections can be very easily found if the free body diagrams are drawn for the individual sections.

The total deformation of the body will then be equal to the algebraic sum of deformation of the

individual sections.

Thus, in Fig 1.13 (a), the bar AB is acted upon by several forces. The free body diagrams for AC,

AD and DB have been drawn in Fig 1.13 (b), (c) and (d) respectively.

Figure 10(b)

3

1

To draw the free body diagrams for AC, consider all the forces to the left of the normal section at

A. The net force is 10 kN tensile. Considering all the force to the right of the normal section at C,

the net force acting is equal to (9 + 3 – 2) = 10 kN. Thus the section AC of length L1 is subjected

to a tensile force of 10 kN.

� The extension of the section will be1

1

1

10L

A E� � , where A

1 is the area of cross – section of the

section AC. Similarly, the extension of the section CD will be 2

2

2

7L

A E� � and the extension of the

section DB will be 3

3

3

9 L

A E� �

�The total extension � of the bar AB will be given by 1 2 3� � � �� � �

Note:

1. The positive sign of the answer indicates that the rod elongates, as a + ve sing is associated

with tensile forces.

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2. Negative sign of the answer indicates the rod shortened, as a – ve sing in associated

with compressive forces.

1.13 EXTENSION OF A TAPERING ROD:

A bar uniformly tapering from a diameter d1 at one end to a diameter d

2 at the other end is shown

in Fig. 1.21.

dx

Let P = Axial tensile load on the bar

L = Total length of the bar

E = Young’s modulus

Consider a small element of length dx of the bar at a distance x from the left end. Let the

diameter of the bar be Dx at a distance x from the left end.

1 21 .x

d dD d x

L

�� �� � � � �

� �

Dx = d

1 – qx where

1 2d dq

L

��

Area of cross – section of the bar at a distance x from the left end,

2 21[ . ]

4 4x xA D d q x

� �� �

Then the stress at a distance x from the left end is given by

2 2

1 1

4

/ 4 . .x

x

Load P P

A d q x d q x�

� �� � �

� �

And the strain 21

4

[ . ]

xx

stress P

E E E d q x

�

�� � � �

� . Extension of the small elemental length dx.

dL = strain. dx x

dL dXor dL dX

L X

� ��� � � � �� �

�

dL = x� . dx

21

4.

[ ]

PdL dx

E d q x��

�

Total extension of the bar is obtained by integrating the above equation between the limits 0 and

L

� Total extension,

2

1210 0

4 . 4.

[ ]

L LP dX P

dL d q x dxEE d q x ��

�� � �

�

2

1

0

4.

Ld q x qP

dxE q�

�� �

��

[Multiplying and dividing by (- q) ]

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1

1

1 00

4 4 1

1 .

LL

d q xP P

E q E q d q x� �

�� �� � �� �� � � �� �� � � �� �

01 1

4 1 1P

E q d q L d q�

� �� �� �

�� �� �

1 1

4 1 1P

E q d q L d�

� �� �� �

�� �� �

1 2 1 2 11

4 1 1

.

PdL

d d d d dE d L

L L�

� �� �� �� �� �� �� � � �

� � � �� �� � � �� �

dL = 1 2

1 2 1 2 1 2

4 4

.

d dPL PL

E d d d d E d d� �

� ��� �� �

� � �

� Total extension = 1 2

4

.

PL

E d d�

Note: If the rod is uniform diameter, then d1 = d

2 = d.

� Total extension, dL = 1 2

4

. / 4 .

PL PL PL

AEE d d E� �� �

2.10 UNIFORMLY TAPERING RECTAGULAR BARS:

Fig shows a uniformly tapering bar of rectangular cross – section, length L and thickness t. The

width of the bar at one end is b1 and the width at the other end is b

2, where b

2 > b

1. The bar is

subjected to an axial force P.

Consider a very short section XX of length x� and width bx, situated at distance x from end A.

Width bx = b

1 +

2 11

b bx b kx

L

�� � , where

2 1b bk

L

�� .

� Extension of the strip

1

.

.

xP

b kx t E

��� � �

� . Hence the extension of the whole length of the rod is.

1 010 0

1. log

Lx LL

e

x

Pd x Pb k x

b kx t E t E k�

�

�

� �� � � � � �� ��!

Or 1 2

1 1

log loge e

b k L bP P

k t E b k t E b

�� � �

Or 2

2 1 1

logebP L

b b b� �

�

2.12 ELONGATION OF BAR OF UNIFORMLY TAPERING SECTION:

Fig shows a bar of uniformly tapering section of length L, hanging freely under its own weight.

Consider an elementary section of length x� , at a distance x from the free end. Let Ax be the area

of cross section of the elementary section. The extension of this elementary section is given by

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.

.

x

x

W x

A E

�� � �

Where Wx = weight of the portion below the section =

1

3 A

x. x. "

(where " is the specific weight or unit weight of the material)

. .1 .

3 . 3

x

x

A x x x x

A E E

" � " ��� � � �

Hence total extension of the bar, 0 0

3

Lx L

x

x x

E

" ��

�

�

� � �!

Which gives

2 2

6 6

L g L

E E

" �� � � (2.20)

If d is the diameter of bar at its uppermost section (i.e at the support) total weight of the bar

21

3 4W d L

�"� �

� � � �� �

or 2

12WW

d L�� . Substituting this value of " in Eq. 2.20, we get

2

2

2

W L W L

AEd E�� � �

Where A = area of cross – section of the bar at its support.

2.13 ELONGATION OF TRUNCATED CONE SHAPED BAR:

Fig. shows a solid bar in the form of a truncated cone, having a diameter d1 at the free end

diameter d2 at the support. The length of the bar is L and it is hanging freely.

Prolong AD and BC to meet at O. Let l’ be the length of cone DCO and l be the length of the

full cone ABO. From Eq. 2.20, the extension of the cone ABO under is own weight is given by

2

6c

l

E

"� � ………….. (1).

Similarly, the extension of cone DCO, under its own weight …………… (2).

2'

6

l

E

"� �

The weight of cone DCO = W’ = 2 21 1

1' '

3 4 12d l d l

� �" "� �

�� �� �

Again, from Eq. 2.16, the extension of a tapering circular bar of length L, under an axial pull W’

is given by

21

1 2 1 2

4 ' 4' '

12tc

W L Ld l

E d d E d d

�"

� �� �

� � � �� � or

1

2

'' .3

tc

dLl

E d

"� �

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Now extension of ABCD = Extension of ABO – extension of DCO – extension of ABCD under

weight W’ or ' 'tc c c tc� � � � � � � or

2 21

2

' '.

6 6 3tc

dl l Ll

E E E d

" " "� � � .

In the above expression, l and l’ are not directly known and hence these are to be eliminated.

From the geometry of the cone, 2

1

'1

' ' '

d l L l L

d l l l

�� � � �

Hence 2 1 1

1 2 1

, ''

d d d LLfrom which l

l d d d

�� �

� ……. (i)

and 2 2 1 2

1 1 2 1 2 1

'd d d L d L

l ld d d d d d

� � �� � ……. (ii)

Substituting these values in (4), we get

2 2 2 22 1 1

2 1 2 1 2 2 16 6 3tc

d L d L L d

E d d E d d E d d d

" " "� � � �� � � �� � � �

� � �� � � �

Or

3 2 22 1 2 1 2 1

2

2 2 16

tc

d d d d d d

E d d d

" � �� � �� �� �� ��� �

.

Or

3 3 222 1 1 2

2

2 2 1

2 3

6tc

d d d dL

E a d d

" � �� �� �� �� ��� �

……….. (2.22).

This is the required expression. When d1 = 0 d and d

2 = d (i.e case of 2.12, figure. 2.40), we get

2 3 2

36 6

L d L

E Ed

" "� �� � �� �

� �� � which is the same as Eq. 2.20.

ELONGATION OF ROD UNDER IT’S SELF WEIGHT:

Let rod is having self weight ‘W’, area of cross – section ‘A” and length ‘L’. Considering on element

at a distance ‘x’ from bottom, then T = W

L x elongation ‘dx’ element =

.T dx

Ay.

Total elongation s = 0 0

.

2

L LT dx W x d y W L

Ay L A y Ay

�� � .

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Note: One can do directly by considering total weight at C.M. and using effective length l/2.

Illustration:

A thin uniform copper rod length l and mass m rotates uniformly with angular velocity on a horizontal plane about a vertical axis passing through one of its ends. Determine the tension

of the rod as a function of distance r from the axis and find its elongation.

Solution: Consider a small portion of length dr at a distance r from the axis of rotation.

Tension = 2

0

,

imdr r

l

=

2 2

2

l

r

m r

l

� �� �� �� �

22 2

2

ml r

l

� �� �� �

2 22

21

2

m rl

l l

� �� �� �

� �� �

2 2

21

2

m l r

l

� �� �� �

� �� � extension =

stresslength

Y = 2

tensiondr

a Y�

2 2

2 21

2

m l r dr

l a Y

�

� �� �� �

� �� �, where a = radius of

Where

2 2

2 21 .

2

m l rdr

a Y l

�

� �� �� �

� �� �

� total extension

2 2

2 2

0

1 .2

lm l r

dra Y l

�

� �� �� �� �

� �

2 3

2 2

02 3

l

m l rr

a Y l

�

� �� �� �

� �� �

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2

2 32

m l ll

a Y

�� �� �� �� �

2

2

2.32

m l l

a Y

��

2 2

23

m l

a Y

�� .

But, m 2

.a l� #2 2 2 2 3

2

.

33

a l l lTotal extension

Ya Y

� # #

�� � � .

Illustration: Find out the elongation in block. If mass, area of cross – section and young modulus

of block are m, A and y respectively.

Sol.

Acceleration, a = ' 'F mthen T m a where m

m e�

m F FxT

l m l� � Elongation in element ‘dx’ =

T dx

Ay

Total elongation, 2o o

T dx F x dx Fd

Ay A y Ay� � � �

� �

�

�

Note: - Try this problem, if friction is given between block and surface ( $ = friction coefficient(,

and Case: (I) F < $ mg

(II) F > $ mg

Ans: In both cases answer will be 2

F

Ay

�

Illustration: In a ring having linear charge density ' '" , made up of wire cross – section area ‘a’,

young modulus y, a charge Q0 is placed at it’s center. If initial radius is ‘R’, then find out change

in radius.

Sol. Considering an element of angular width ' 2 '� - dq = '.2R" � 2 sinF T � � .

0 0

2

. ' 2 .2 sin 2 sin

'2'

kdqQ k R QT T

RR

" �� �

�

If � is small, then the sin � = � .

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Further, 0

'

k QT

R

"� But,

2 ' '.

2

R Rstress R Ry strain

strain R R

�

�

� �� � � .

0'

' '

k QTy R R

a R R R ay

"� � �

�

2 0' ' 0

k QR RR

ay

"� �

� �� � � �� �

� �

24

'2

R RR

�% �� [(-) is only when charge is opposite nature]

Then

1/2

2 2

4 2' 1 1 ' 1 12 2

R RR R

R R

� �� �� � � �� � � � � �� �� � � �� � � �� �� �

[neglecting higher term].

'R RR

�� � Change in radius

0 0

0

'4

k Q QR R R

R ayR ayR

" "��

� � � � � ��

1.14 STRESSES IN BARS OF COMPOSITE SECTION:

A bar, made up of two or more bars of equal lengths but of different materials rigidly fixed with

each other and behaving as one unit for extension or compression when subjected to an axial

tensile or compressive loads, is called a composite bar. For the composite bar the following two

points are important.

P

1 2

1. The extension or compression in each bar is equal. Hence deformation per unit length,

i.e., strain in each bar is equal i.e.,

1 2

1 2

1 2

PL LL L l

E E AE E

� � �� � � � � �� �� �

�

2. The total external load on the composite bar is equal to the sum of the loads carries by

each different material. i.e., 1 1 2 2

P A A� �� � . Form equations (1) and (2) the stresses

1 2and� � may be computed.

Modular Ratio: The ratio of 1

2

E

E is called the modular ratio of the first material to the second.

Illustration: A light rod of length 2m is suspended from the ceiling horizontally by means of two

vertical wires of equal length tied to its ends. One of the wire is made of steel and is of cross –

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section 0.1 cm2 and the other is made of brass of cross – section 0.2 cm2. Find out the position

along the rod at which a weight may be hung to produce.

(a) Equal stress in both wire (b) equal strain in both wire.

Young’s modulus of elasticity of brass and steel are 10 × 1011 dyne cm-2 and 20 × 1011 dyne cm-2

respectively.

Solution: Let the rod BD be suspended by the wires AB and CD as shown in figure. If W is the

weight suspended at O at a distance x from the steel wire AB, the force acting at point O is the

vertically downward direction producing tensions T1 and T

2 in the two strings.

a. Let A1

a. Let A1 and A

2 be the areas of cross – section of the respective wires, i.e,

A1 = 0.1 cm2 and A

2 = 0.2 cm2.

Stress in the steel wire = 1

0.1

T

Stress in the brass wire = 2

0.2

T. For the stresses to be equal, or

1 2

0.1 0.2

T T�

or 1

2

1

2

T

T� . ………………. (1).

For the system to be in equilibrium, the moments of the forces T1 and T

2 about O must be equal.

Hence x T1 = (200 - x)T

2………………. (2).

Using Eq. (1) we have 1

2

200 1

2

T x

T x

�� � or 400 – 2 x = x x =

400

3 = 133.3 cm.

b. Young’s modulus Stress

YStrain

� ; Stress = T

A ; Strain =

T

AY .

The strains in the two wires will be equal if 1 2

1 1 2 2

T T

A Y A Y� or

1 1 1

2 2 2

T A Y

T A Y� .

Now Y1 = 20 × 1011 dyne cm-2 and Y

2 = 10 × 1011 dyne cm-2.

Therefore

11

1

112

0.1 20 101

0.2 10 10

T

T

� �

or T

1 = T

2.

Using Eq. (2) we have x = (200 – x) or x = 100 cm.

Thus the weight W must be hung from the middle of the rod to produce equal strains in the two

wires.

Example: A steel rod of cross – sectional area 2000 mm2 and two brass rods each of cross – sectional

area of 1200 mm2 together support a load of 60 kN, as shown in Fig. 2.45. Find the stresses in the

rods. Take E for steel = 2 × 105 N/mm2 and E for brass = 1 × 103N/mm2.

Solution: Let us use suffix 1 for steel rod and suffix 2 for brass rods.

Given: A1 = 2000 mm2 and L

1 = 400 mm

A2 = 1200 mm2 and L

2 = 300 mm.

From the equilibrium of the system. p1A

1 + 2p

2A

2 = p

Or 2000 p1 + 2400 p

2 = 500 × 103. ……………….. (i).

Also,1 2

� � �

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�1 1 2 2

1 2

p L p L

E E�

Or1 2

5 5

400 300

2 10 1 10

p p �

� p

1 = 1.5 p

2. ……………….. (ii).

Hence from (i), we get 200 (1.5 p2) + 2400 p

2 = 500 × 103

From which p2 = 92.59 N/mm2

� p1 = 1.5 × 92.59 � 138.89 N/mm2.

Analogy of Rod as a spring:

stress Fy y

strain A� �

��

�

Or Ay

F � ���

Ay

� = constant, depends on type of material and geometry of rod. F = �� .

Where Ay

� = equivalent spring constant.

A ,y

(a)

A ,y (b)

(c)

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For the system of rods shown in figure (a), the replaced spring system is shown in figure (b) two

spring in series. Figure (c) represents equivalent spring system. Figure (d) represents another

combination of rods and their replaced spring system.

A ,y ,

(d)

A ,y ,

A ,y ,

FF

Illustration: A mass ‘m’ is attached with rods as shown in figure. This mass is slightly stretched

and released whether the motion of mass is S.H.M, if yes then find out the time period.

1 11

2 22

1

2

Sol. 1 21 2

1 2 1 2

2 2eqeq

m k kk k mk T

k k k k k� �

�� � �

�

1 1 2 21 2

1 2

A y A ywhere k and k� �

� �.

ELASTIC POTENTIAL ENERGY STORED IN A STRETCHED WIRE OR IN A ROD:

Strain energy stored in equivalent spring

21

2U kx�

Where x =

2 2 2

2 2

1 1,

2 2

F Ay Ay F Fk U

Ay AyA y� � �

� � �

� �.

Equation can be re – arranged

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21

2

2

1

2

F AU

yA�

�[ � A = volume of rod, F/A = stress].

21

.2

stressU volume

y�

Again, 1

2

F FU A

A Ay�

�� y

1

2U � stress × strain × volume.

1

2U � y (strain)2 × volume.

Strain energy density =

221 1 1

.2 2 2

stressstrain energyy strain stress strain

volume y� � �

Illustration: Higher is mass – less A ball of mass ‘m’ drop from a height ‘h’, which sticks to

hanger after striking. Neglect over turning, find out the maximum extension in rod.

A ,y

m

h

A ,y

Hanger

Sol. Applying energy consecration

21 2

1 2

1

2

k kmg h x x

k k� �

�

Where 1 1 2 2

1 2

1 2

A y A yk k� �

� �

& 1 2 1 2

1 2 1 2 1 1

eq

A A y yk

A y A y�

�� �

Keq

x2 – 2 mgx – 2mgh = 0.

2 22 4 8

2

eq

eq

mg m g mgh kx

k

% �� ;

2 2

max 2

2

eq eqeq

mg m g mghx

k kk� � �

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1.15 TEMPERA TURE STRESSES OR THERMAL STRESSES:

It was possible to disregard the deformations caused by temperature in statically determinate

systems since in such situations the members are free to expand or contract.

The free deformations caused by a change in temperature must be known for the determination

of stresses caused by temperature.

Thermal stresses are set up in a body, when the temperature of the body is raised or lowered of

the body is not allowed to expand or contract freely. But if the body is allowed to expand or

contract freely, no stresses will be set up in the body.

Consider a body which is heated to a certain temperature.

Let L = Original length of the body

Y = Rise in temperature

E = Young’s modulus

� - co – coefficient of thermal expansion per 0C.

dL = Extension of rod due to rise of temperature.

� dL = � TL.

And compressive strain = Decrease in length

original length

= TL

TL

���

But Stress

Estrain

�

� Stress = Strain × E = � . T. E

And Load or thrust on the rod Stress × Area = � . T. E × A.

If the ends of the body are fixed to rigid supports, so that it expansion is prevented, then

compressive stress and strain will be set up in the rod. These stresses and strains are knows as

temperature stresses and temperature strain.

� Temperature strain,Expansion prevented

orginal length��

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23

= .dL TL

TL L

��� �

And Temperature, � = temp. strain × E.

� = � TE.

2.19 TEMPERATURE STRESSES IN BARS OF TAPERING SECTION:

Consider a bar of uniformly tapering section, shown in the fig. rigidly fixed between two

end supports. When the temperature is raised by t, compressive force P will be induced since the

bar is not free to expand. This force is the same for all cross – sections, and hence maximum

stress will be induced at section AA where diameter is d1.

If the bar were free to expand, we have t L t�� � ……(i)

The force induced In the bar will be a compressive force P which is required to prevent free

expansion of � given by (i). Now, for an element of length dx, the deformation due to P is

.

x

P dx

A E�� �

� Total deformation 0

.L

x

P dx

A E� �

But

222 1 2 1

1 14 4

x

d d d dA d x d kx where x

L L

� �� �� �� � � � �� �

� �

21 010

4 4 1LL

P dx P

E k E d kxd bx� �� �

�� � � � � ��� � �

Or 21 2 1 1 22 1

4 1 4PL PL

d d d E d dE d d ��

� �� � � �� �� �� � �

…………. (ii)

Equating (i) and (ii), we get, t� � �

Or1 2

4PLL t

E d d�

�� …………. (ii)

From which 1 2 .4

EP d d t

�� ……… (2.29)

Maximum stress induced at section AA is given by

2 2,max 1 2 1

14 4t

dEP d d t d E t

d

� �� �� � � �� & �� � � �

� � � � ……..(2.30).

If d1 = d

2, P

t = E t� , which is the same as Eq. 2.27.

Again, if the support yields by an amount a, the amount of expansion checked will be t a� � .

Hence

t a� � � �

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Or 1 2

4PLL t a

E d d�

�� � . Form which 1 2

4

E L t aP d d

L

� � �� �� � �� �

.

Illustration: When composile rod is free, then composite length increases to 2.002 m for temp

200 C to 1200 C. When composite rod is fixed between the support, there is no change in component

length find y and � of steel, if 13 2 5 01.5 10 / 1.6 10 /cu cuy N m C� �� � .

0.5m

Sol. s s c cT T� �� � � � �� � �

002 = 5[1.5 0.5 1.6 10 ] 100s� ��

56 01.2 10

8 10 /1.5

s C��

�� � .

s s�

c c�

There is no change in component length for steel.

0ss s

s

Fx T

AY�� � � �

��

s

s

FT

AY�� � ………… (A)

For copper 0cc c

c

Flx T

Ay�� � � ��

c

FlT

Ay�� � ………… (B)

B/A s c

c s

y

y

��

�

13 5

6

1.5 10 16 10

8 10

cs c

s

y y��

�

�

� � �

13 23 10 /sy N m�

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1.51.1 TEMPERATURE STRESSES IN COMPOSITE BARS:

Fig. 1.31 (a) shows a composite bar consisting of two members a bar of gun metal and another of

steel.

Let the composite bar be heated through some temperature. If the members are free to expand

then no stress will be induced I the members. But the two members are rigidly fixed and hence

the composite bar as a whole will expand by the same amount.

Gum

metalSteel

(a)

Gum

metalSteel

(b)

Gum

metalSteel

(c)

As the coefficient of linear expansion of gun metal is more than of the steel, the gun metal will

expand more than the steel. Hence the free expansion of gun metal will be more than that of

steel. The expansion of the composite bar, as a whole, will be less than that of the gun metal, but

more than that of the steel.

Hence that of the induced in the gun metal will be compressive where as the stress in steel will

be tensile as shown in figure. 1.30 (c). Therefore the load or force on the gun metal will be

compressive where as on the steel the load will be tensile.

Let Ag = Area of cross – section of gun metal bar.

g� = stress in gun metal

g� = strain in gun metal

g� = linear – coefficient of expansion for gun metal /0C.

Eg = Young’s modulus for gun metal

A,. � ., � ., and � , corresponding values of area, stress, strain and linear coefficient of expansion

for steel

E, = Young’s modulus for steel.

Now load on the gun metal = stress in gun metal × Area of gun metal

= g gA�

Ann load on the steel = s sA�

Compression in gun metal = Tension in the steel

Or load on the gun metal = load on the steel.

i.e g gA� = s sA� …………….. (1)

Also we know that

Actual expansion of gun metal = Actual expansion of steel …………….. (2)

But

Actual expansion of gun metal = Free expansion of gun metal – Contraction.

= due to compressive steel induced in gun metal.

= . . . .g

gg

T L LE

�� �

And actual expansion of steel = Free expansion of steel + expansion due to tensile stress in steel.

= . . . .sss

T L LE

�� �

Substituting these values in equation (2), we have

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. . . . . . . .g s

g sg s

T L L T L LE E

� �� �� � � .

APPLICATIONS OF ELASTICITY:

1. Some of the important applications of the elasticity of the materials are discussed as

follows. The material used in bridges lose its elastic strength with time bridges are declared

unsafe after long sue.

2. The estimate the maximum height of a mountain: The pressure at the base of the

mountain = h # g = stress. The elastic limit of a typical rock is 3 × 108 N m-2 . The stress

must be less than the elastic l imits, otherwise the rock begins to flow.

8 84 3 3 2

3

3 10 3 1010 3 10 ; 10

3 10 10h m kg m g ms

g#

#� �

' ' ' �

� or h == 10 km It may be noted

that the height of Mount Everest is nearly 9 km.

INTER – ATOMIC FORCE – CONSTNAT:

The inter atomic force F’ developed between the atoms is directly proportional to the change in

distance � r between them, F’ = k � r. ………………… (i)

Where k is inter atomic force – constant. Let L be the length of a wire and r0 the normal distance

between the atoms of the wire. Suppose, when a force F is applied, the length of the wire increases

by and the distance between its atoms increases from r0 to r

0 + � r.

Then longitudinal strain = 0

r

L r

��

�

………………… (ii)

If A be the area of cross – section of the wire, it will have A/ 20r chains of atoms. Therefore, the

inter atomic force (force acting on chain) is

20

20

' ./

F rexternal force FF

number of chains AA r� � �

Inter atomic force – constant is –

20 0

0

/' F r A rF Fk r

r r A r� � �

� � � ——— by equation (i)

But r0/ r� = L/, k = 0 0

/

/

F L F Ar r

A L�

� �—— by equation (ii).

But, by definition, /

/

F A

L� = Y (Young’s modulus of the material of the wire) k = Yr

0.

Thus, inter – atomic force – constant k is equal to the product of Young’s modulus of the materials

of the wire and the normal distance r0 between the atoms of the wire.

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Illustration : For steel rod young modulus = 2 × 1011 N/m2. inter atomic spacing = 2.0 A0 Find (a)

inter atomic force constant (N/A0) (b) increase in inter atomic spacing for a stress of 2 × 109 N/

m2.

Sol. (a) k = yr0 = 2 × 1011 × 2 × 10-10 = 40 N/m =

9 0

10 0

404 10 /

10N A

A

�� .

(b) F

rk

� � where F = stress × 20r = 2 × 109 × (3 × 10-10)2

= 18 × 10-11 N

11

9

18 10

4 10r

�

�

� �

= 4.5 × 10-2 A0 = .045 A0 Ans.

LATERAL STRAIN, POISSON’S RATIO AND VOLUMETRIC STRAIN:

When a body is subjected to an axial tensile load, there is an increase in the length of the body,

but at the same time there is a decrease in other dimensions of the body at right angles to the

line of action of the applied load.

Longitudinal strain: When a body is subjected to an axial tensile or compressive load, there is

an axial deformation in the length of the body.

The ratio of axial deformation to the original length of the body is known as longitudinal (or

linear) strain.

1.16.1 LATERAL STRAIN

The strain at right angles to the direction of applied load is known as lateral strain. Let a rectangular

bar of length L, breadth b and depth d is subjected to an axial tensile load P as shown in figure.

1.35. The length of bar will increase while the breadth and depth will decrease.

Let � L = Increase in length

� b = decrease in breadth, and

� d = decrease in depth

Then Longitudinal strain = L

L

� and Lateral strain

b dor

b d

��.

1.16.2 POISSION’S RTIO:

It is defined as 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, it is known as

Poisson’s Ratio. It is represented as $ or 1.

m theoretical value – 1 to 0.5 practical value 0 to 0.5.

Mathematically, Poisson’s Ratio =

b dor

lateral strain b dLlongitudinal strain

L

$

� �

� �� .

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Note: The value of Poisson’s ratio varies from 0.25 to 0.30, for rubber, its value ranges from 0.45

to 0.50. For structural steel, experiments indicate that approximately $ = 0.3, $ = 0.1, for concrete

and $ = for cork.

1.16.3 VOLUMETRIC STRAIN

It is defined as the ratio of change in volume to the original volume of a body is called volumetric

strain. It is denoted by v�

Mathematically, Volumetric strain, v

V Changein volume

V original volume

�� � � .

1.16.3.1 Volumetric Strain of Rectangular Bar:

Let a rectangular bar of l units long, b units wide and d units deep undergo small changes by

,l b and b� � � respectively in length, width and depth.

� Original volume (V) = b × d × l

Final volume = (b + � b) (d + � d) (l + � l)

= lbd + bd � l + lb � d + dl � b

= original volume + change in volume

[� ignoring products of small quantities]

� change in volume V bd l lb b b�� � � � � �

i.e, v l d b� �� � � � �

1.16.3.2 Volumetric Strain of Cylindrical Rod:

Let l = length of the rod in m

D = diameter of the rod.

Let the length and diameter change by l and d� � respectively.

� Original volume (V) = 2

4d L

�

= 2

4d d l l

�� � � �

= 2 2

2 .4d l d l ld d

��� � �

= 2 2

2 .4 4d l d l ld d

� �� � � � .

[� Ignoring products and higher powers of small quantities]

� Change in volume = 2

24

V d l ld d�

� � � � �

� Volumetric strain,

2

2

2 .v

change in volume V d l ld d

original volume V d l

� � � �� � � � .

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� 2. 2.v l d

l d

l d

� �� � � �� � �

� Volumetric strain = strain of the length + twice the strain of the diameter.

1.16.3.3 Volumetric Strain of a Sphere:

Let d = diameter of solid sphere.

d + � d = increase in diameter. Original volume of the sphere

(V) =

3

6

d�

Final volume of the sphere, = 3

6d d

�� �

= 3 23

6d d d

�� � [� ignoring higher powers of � d]

� Change in volume = 2

36

V d d�

� � �

� Volumetric strain, V

change in volume V

original volume V

�� � �

2

3

33.V d

d d d

dd

� �� �� � � � �� �

� �

� Volumetric strain = 3 times the strain of the diameter.

1.16.3.4 Volumetric Strain of a Rectangular Bar Subjected

To Three Forces which are Mutually Perpendicular

Consider a rectangular block of dimension x, y, z subjected to three direct tensile stresses along

three mutually perpendicular axis as shown in figure. 1.37.

y( )

x( )

z( ) y( )

x

z

Then volume of block V = x y z

Taking logarithm to both sides, we have

1n = ln x + ln y + ln z

Differentiating the above equation, both sides we get

1 1 1 1

2dV d x d y d z

V x y� � � or

dV dx dy dz

V x y z� � � . ……………. (i).

But V

dV change in volume

V original volume� ��

dx

x = Strain in the x – direction = x�

dx

y = Strain in the y – direction = y�

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dx

z = Stain in the z – direction = z� .

Substituting these values in equation (i), we get

Let x� = Tensile stress in x – x direction

y� = Tensile stress in y – y direction

z� = Tensile stress in z – z direction

E = Young’s modulus,

$ = Poisson’s ratio.

Now x� will produce a tensile strain equal to x

E

� in the direction of x, and a compressive strain

equal to .x

E

�$

� �� �� �

in the direction of y and z. Similarly y will produce a tensile strain equal to y

E

�

in the direction of y and a compressive equal to .y

E

�$

� �� �� �� �

in the direction of x and z . Similarly z�

will produce a tensile strain equal to z

E

�� �� �� �

in the direction of ‘z’ and a compressive strain equal

to .z

E

�$� �

� �� �

in the direction of x and y. Hence y zand� � will produce compressive strains equal to

. .y zandE E

� �$ $

� � � �� � � �� � � �� �

in the direction of x.

� Note tensile strain along x – direction is given by

. .yx z

xE E E

�� �$ $� � � �

Similarly . .y x z

yE E E

� � �$ $� � � �

. .yxz

zE E E

���$ $� � � � .

Adding all the strains, we get

1 2.

11 2

x y z x y z x y z

x y z

E E

E

� � � $ � � �

� � � $

� �� � � � � � � � � � �� �

� �� � � �� �

But x y z� � � � � = volumetric strain.

= dV

V

1 2x y z

v

dV

V E

� � �$

� �� �� � � .

Note:

1. If any of the stresses is compressive, it may be regarded as negative and the above equation

will hold good.

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2. If the value of dV

V is positive, it represents increase in volumes whereas the negative

value of dV

V represents a decrease in volume.

The minus sign is included in the definition of K because an increase of pressure on the surface

of the body always causes a decrease in its volume. That is, if #� is positive, v� is negative. By

including a minus sign in its definition, we make the bulk modulus itself a positive quantity.

Compressibility. The reciprocal of the bulk modulus is called the ‘compressibility’.

From its definition,

Compressibility = 1 / 1v v v

K p v p

� �� � � �

� �

The units of bulk modulus are the same as those of pressure and the units of compressibility are

those of a reciprocal pressure.

1.10 SHEAR MODULUS OR MODULUS OF RIGIDITY OR RIGIDITY MODULUS:

It is defined as the ratio of shear stress to the corresponding shear strain within the elastic

limits, is known. The is denoted by C or G or N.

TauShear stressG

Shear strain Theta

�

�� � �

Illustration

A rubber cube of side 5 cm has one side fixed while a tangential force equal to 1800 N is

applied to opposite face find the shearing strain and the lateral displacement of the strained face.

Modulus of rigidity for rubber is 2.4 × 106 N/m2.

Sol.25 10

F XL m

A L(�� �

Strain 4 8

1800 180 30.3

25 24 1025 10 2.4 10

F

A�

( �� � � � �

2 20.3 0.3 5 10 1.5 10 1.5

xx m mm

L

� �� � � � .

If a cylinder is fixed at one end and the other end is subjected to a constant torque T, co – axial

with the cylinder, the latter end suffers an angular displacement about the axis of the cylinder.

The cylinder is said to be twisted. The torque is known as torsional torque and is given by

41( )

2

n naT J

l l

� � �� � .

Where n = modulus of rigidity of the material of the rod

A = radius of the cylinder; l = length of the cylinder; � = the twist in the cylinder.

And J = moment of inertia of cross – section of the cylinder about the cylinder axis = 41

2a� .

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Then

41

2

na

l

� = torsional torque per unit twist = torsional rigidity of the cylinder.

1.17 DEFINITON OF BULK MODULUS:

When a body is subjected to the mutually perpendicular like and equal direct stresses, “the ratio

of direct stress to the corresponding volumetric strain” is found to be constant for a given material

when the deformation is within a certain limit. This ratio is known as bulk modulus and it is

represented by K. Mathematically,

Bulk modulus (K) = - direct stress

volumetric train

= -

tensile or compressive

V

V

�

�� �� �� �

.

Illustration .

Find the depth of lake at which density of water is 1% greater than at the surface. Given

compressibily k = 50 × 10-6 /atm.

Sol.

VB

V V B

V

# #� � �� � �

��

We know atm h g# # #� � or m = # V = const.

d # . v + dv. # = 0 ;

d # V + dV. # = 0 d dV

V

##

� �

i.e, B

# ##

� ��

1

100

##

� � .

1

100

h g

B

#� [assuming # = const.]

1

100 100

Bh g

k# � �

5

6

1 1 10

100 50 10h g#

�

�

.

5 3

6

0 100 102

505000 10 1000 10g km

�

) � � �

Ans.

1.18.1 RELATION BETWEEN THE THREE ELASTIC CONSTANTS

RELATION BETWEEN THE MODULUS OF ELASTICITY AND THE BULK MODULUS.

Let L = length of the cube

� L = change in length of the cube.

E = Young’s modulus of the material of the cube.

� = Tensile stress acting on the faces

$ = Poisson’s ratio

Then volume of cube, V = L3

Let us consider the strain of one of the sides of the cube (say AB) under the action of the three

mutually perpendicular stresses. This side will suffer the following three strains:

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F

B

G

CD

E

H

A

1. Strain of AB due to stresses on the faces AEHD and BFGC. This strain is tensile and is

equal to E

�� �� �� �

.

2. Strain of AB due of stresses on the faces AEFB and DHGC. This is compressive lateral

strain and is equal to .E

�$� �

�� �� �

.

3. Strain of AB due to stresses on the faces ABCD and EFGH. This is also compressive

lateral strain and is equal to .E

�$� �

�� �� �

.

.

.

lateral strain L S

longiludinalstrain

E

Lateral strainE

$ $�

�$

� �� � �� �

� �� �� �� �

� �� �� �� �� �

�

.

The total strain of AB is given by

. .dL

L E E E

� � �$ $� � �

1 2dL

L E

�$� � . …………. (i)

Now original volume of cube, V = L3. …………. (ii)

If dL is the change in length, then dV is the change in volume. Differentiating equation (ii), with

respect to L dL L� �� .

dV = 3L2 × dL …………. (iii).

Dividing equation (iii) by equation (ii), we get 2

3

3dV L dL

V L

� = 3

dL

L

� �� �� �

.

Substituting the value dL

L

� �� �� �

from equation (i), in the above expression, we get.

31 2

dV

V E

�$� � . And bulk modules is K = V

V

��� �

� �� �

= 3 1 2E

��

$� .

= 3 1 2

E

$� .

Or E = 3 1 2$� .

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This is the expression between Young’s modulus and Bulk modulus and Poisson’s ratio.

Note: From the above expression, we have to calculate the Poisson’s ratio.

3.

6

K E

k$

�� �� � �

� �

RELATIONSHIP BETWEEN MODULUS OF EALSTICITY AND MODULUS OF RIGIDITY:

Consider a square block of unit thickness is subjected to a set of shear stresses of magnitude

� n the faces AB, CD and the faces AD and CB, then the diagonal strain due to shear stress �

is given by equation 1E

�$� � .

� Total tensile strain along diagonal 1BDE

�$� � . ……………. (i).

But also the total strain in diagonal 1

2BD shear strain�

1

2

shear strain

G� .

1

2 G

�� . ……………. (ii).

Equating the two tensile strain along diagonal BD, we get G�*

� ��� �

� ��

11

2E G

� �$� � or

1 1

2E G

$�� or E = 2G 1 $�

1.18.3: RELEATION BETWEEN YOUNG’ S MODULUS,

MODULUS OF REGIDITY AND BULK MODULUS.

We know, E = 2G 1 $� ………………… (i)

E = 3K ………………… (ii)

12

E

G$� � � ………………… (iii)

And 1 23

E

K$� � ………………… (iv).

Multiplying equation (iii) by 2 and adding equation (iv), we have

2 + 2 3

E

K$ �

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1 23

33

2 3 .

E

K

E K GE E

G K K G

$� �

�� � �

9. .

3

K GE

K G� �

�.

Important points :

1. If the length of a wire is doubled, the longitudinal strain will be 2 1

1

21

L LL L L L

L L L L

�� �� � � � .

But as Y = (stress/strain). Y = stress (as strain = 1].

2. As for a loaded wire: 2

2

FL F LL as Y and A r

A Lr Y�

�

� �� � � �� ��� �

. So if same stretching force is

applied to different wire of same material. 2

F LL

r� � [as F and Y = constant].

3. In case of compression of a fluid, as density = m V

soV V

##

#� �

� � � . But by definition of

bulk modulus, i.e. V p V p

B orV V B

� � �� � � �

� so

'[ ' ]

p por as

B B

# # ## # #

# #� � � �

� � � � �

Or ' [1 / ] [1 ]p B C p# # #� � � � � � [as 1/B = C].

4. In case of bending of a beam of length L, breadth b and thickness d, by a load Mg at the

middle, depression � is given by

3

34

Mg L

bd Y� � and for a beam of circular cross – section of

radius r and length L,

3

412

Mg L

r Y�

�� .

5. In case of twisting of a cylinder (or wire) of length L and radius r, elastic restoring couple

per unit twist per unit twist is given by

4

2

rC

L

� (� , when ( is modulus of rigidity of the of

the material of wire.

6. In case of a rod length L and radius r fixed at one end, angle of shear * is related to angle

of twist � by the relation. L r* ��

Elastic Constants Y, B, ( & �

1. Poisson discovered that within limit of proportionality the ratio of the lateral strain to

longitudinal strain is constant for a given material. This constant in his honour is called

Poisson’s ratio and is represented by � . It has no units and dimensions. It has been

established that theoretically – 1 < � (1/2) while practically no substance had been

found for which � is negative, i.e. practically o < � < (1/2).

2. The value of moduli o elasticity is independent of the magnitude of the stress and strain.

it depends only on the nature of the material of the body.

3. For a given material there can be different moduli of elasticity depending on the type of

stress applied and the strain resulting

4. The moduli of elasticity has same dimensional formula and units as that of stress since

strain is dimensions, i.e. the dimensional formula Y, B or ( is [ML -1T-2] while units dyne/

cm2 or Newton/m2.

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5. Greater the value of moduli of elasticity, more elastic is the material. But as Y � (1/

� L), B � (1/ � L) and 1/( � * for a constant stress, so smaller change of shape or size

for a given stress corresponds to greater elasticity*.

6. The moduli of elasticity Y and ( exist only for solids †as liquids and gases cannot be

deformed along one dimension only and also cannot sustain shear strain. However, B

exists for all states of matter, viz., solid, liquid or gas.

7. Gases being most compressible are least elastic while solids are most, i.e., the bulk modulus

of gases is very low while that for liquids and solids is very high, i.e., solid liquid gasE E E+ + .

8. Gases have two bulk – moduli, namely, isothermal elasticity E� and adiabatic elasticity

E* .It has been found that at a given pressure p, so that 1, . .,p

v

CEi e E E

E C

** �

�

�� � + +

i.e., adiabatic elasticity is greater than isothermal elasticity.

9. With rise in temperature the distance between atoms increases and so the elastic restoring

force will decrease. This is turn will decrease the elasticity, i.e., with rise in temperature

Y, B and ( decrease. Usually this temperature dependence is not taken into account

unless specified.

10. Moduli of elasticity are three, viz., Y, B and ( while elastic constants are four, viz., Y, B (and � . Poisson’s ratio � is not modulus of elasticity as it is the ratio of two strains and

not of strain. Elastic constants are found to depend on each other through the relations:

Y = 3B (1 - 2� ) and Y = 2( (1 + � ). Eliminating � or Y between these, we get

9 3 2

3 6

B BY and

B B

( (�

( (� �

� � .

* For a rigid body ,L V or *� � ; so Y, B or will be , , i.e., elasticity of a rigid body is infinite! † In

case

of solids usually Y = 3(

Concept Building Questions:

Question I. Is the following statement true of false: A metal rod (Young’s modulus Y) has a length

L and area of cross – section A. The work done in stretching the rod by and amount � L is

2

2

YA L

L

�.

Answer: True.

Work done in stretching a wire: In stretching a wire work is done against internal restoring

forces. This work is stored in the body as elastic potential energy or strain energy. To calculate it,

consider a wire of length L and cross – section A. If a force F acts along the length of the wire and

stretches it by x, then as /

,/

stress F A FL YAY F x

strai x L Ax L� � � � .

So work done for an additional small increase dx in length, dW = F dx = (YA/L)x dx.

Hence total work done in increasing the length by L� , 2

0

1

2

L YA YAW xdx L

L L

�� � �

Properties of Matter:

Question – II. In the language of Physics which is more elastic (a) rubber or steel (b) air or water?

Answer: In Physics elasticity is defined as the ratio of stress applied to resulting strain, i.e.,

stressE

strain� . So for a given stress E � 1/strain. i.e., a smaller change in shape or size under the

action of a same force corresponds to higher elasticity.

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So (a) as for identical steel and rubber bar under the action of same force, rubber stretches

more; it is less elastic than steel.

(b) as for equal volumes of water and air under the action of same pressure, water compresses

less, it is more elastic than air.

Question III. What kind of elasticity is sued in (a) suspension bridge (b) an automobile tyre (c) an

automobile drive shaft (d) a coil spring (e) a water lift pump (f) rubber heels.

Answer: (a) In a suspension bridge as there is a stretch in the ropes by the load of the bridge, the

elasticity involved is linear or tensile.

(b) In an automobile tyre a air is compressed the elasticity involved is volume, i.e., bulk.

(c) In transmitting power an automobile shaft is sheared as it rotates, so the elasticity involved is

shear, i.e., rigidity.

(d) When a coiled spring a is stretched, the deformation of the wire of the spring is in the form of

a twisting strain so the elasticity involved is shear, i.e., rigidity.

(e) As in a water lift pump, the water is compressed; the elasticity involved is volume, i.e, bulk.

(f) As the shape of rubber heels changes under stress, the elasticity involved is shear, or rigidity.

Concept Building Examples:

1. A submerged wreck is lifted from a dock basin by means of a crane to which is attached

a steel cable 10 m long, of cross – sectional area 5 cm2 and Young’s modulus 5 × 1010 N/

m2. The wreck has mass 104 kg and mean density 8000 kg/m3. Find the change in extension

of the cable as the load is lifted clear of the water. Assume that at all the times the tension

in the cable is the same throughout its length. (Water = 1000 kg/m2 )

Solution :

Given data,

Length of cable, L = 10 m

Area of cross – section, A = 5 cm2 = 5 × 10-4 m2

Young’s modulus Y = 5 × 10-4 N/m2

Mass of wreck, M = 104 kg

Mean density of wreck, # = 8000 kg/m3

Now, Volume of the wreck is,

4310 10

8000 8

MV m

#� � � .

� When the wreck is in water, extension of the cable is,

1

g BM F L

AY

�� �� ; Where F

B is buoyant force.

Here MB = 1

w gV L

AY

# #�� �� ………………….. (i)

When the wreck is out of the water, the extension of the cable is,

1

gV L

AY

#�� ………………….. (ii)

� Change in extension = 1 2�� �

= gw g w gV LV L V L

AY AY AY

## # #�� �

= 3

4 10

101000 9.8 10

84.9 10

5 10 5 10

��

� � � �

� � �

m.

Change in = 4.9 mm.

2. A device to project a toy rocket vertically makes use of the energy stored in a stretched

rubber cord. Assuming that the cord obeys Hooke’s law, find the length by which the cord

must be extended if the rocket is to be projected to a height of 20 m.

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Mass of rocket = 0.3 kg

Unstretched length cord, = 0.2 m

Area of cross – section of cord, = 2.5 × 10-5 m2

Yrubber

= 8 × 108 N/m2.

Solution:

M = 0.3 kg; H = 20 m;

L = 0.2 m; A = 2.5 × 10-5 m2;

Y = 8 × 108 N/m2.

In this problem, when the rubber cord is stretched strain energy is stored in it. This strain

energy is then used to project the rocket.

Now, energy required to project a rocket of mass M to a height H is,

E = Mg H ……………… (i)

If l is the elongation of the rubber cord then, strain energy is stored in it is,

2

2

Y AU

L�

�

………………. (ii) equating (i) and (ii) we get,

2

2g

Y AM H

L �

�

2

8 5

2 2 0.3 9.8 20 0.2

8 10 2.5 10

gM H L

Y A �

� �

�

2 31.176 10

�� � �

2 31.176 10�� � � = 0.03429 m = 0.03429 m or 3.429 cm.

3. A highly rigid cubical block A of small mass M and side L is fixed rigidly into another

cubical block B of same dimensions and of low modulus of rigidity ( , such that the lower

face of A completely covers the upper face of B. The lower face of B is rigidly held on a

horizontal surface. A small force F is applied perpendicular to one of the side faces of A.

After the force is withdrawn will the block execute a SHM? If yes, then find the time

period of oscillations.

Solutions:

Due to the action of force F, upper face of block B gets displaced by x, say. The corresponding

shear is � . For small � , we can write. x

L� � . Now, modulus of rigidity is given by

F

A(

��

2x xF A L L x

L L( � ( (� � �

F L x(� � .

Now, the restoring force f is equal in magnitude but opposite in direction to the applied force F.

f L x(� � f x�� � .

(�

The block A performs a SHM.

Now, f = M 2

2

d xL x

dt(� �

22

2

d x Lx x

Mdt

( � �

� � � � �� �� �

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22

L M

M T L

( � �

( � � .

4. A steel wire of diameter 0.8 mm and length 1 m is clamped firmly at two points A and B

which are 1 m apart and in the same horizontal plane. A body is hung from the middle

point of the wire such that the middle point stays 1 cm lower from the original position.

Calculate the mass of the body. Given Y = 2 × 1011 N/m2.

Solution:

Diameter of wire, d = 0.8 mm 8 × 10-4 m, Length of wire, L = 1q m , Let, m be the mass of the body

hanging from the middle point O of the middle point gets shifted to C. Such that,

OC = 1 cm = 0.01 m

�

�

�

Let, T be the tension in the wire,

� From FBD of the mass, we get, 2T cos � = mg

� T = 2cos

mg

� ………………………. (i).

Let, l be the elongation produced in the left half (AO) of the wire, by the tension T.

� Now length of the left half is,

2 20.5 0.01

2

LAC � � � ��

2 20.5 0.01 0.5 � � �� .

Now, for the length AC, the Young’s modulus is,

2

2

2cos2

2

LT

mg LY

A d��

� �� �� �� �

� �� �� �

��

.

2

cosd Y

mg L

��� � �

New 2 2

0.01cos

0.5 0.01� �

�.

22 2

2 2

0.010.5 0.01 0.5

0.5 0.01

d Ym

g L

�� � � �

�

2 2

2 2

10 0.51

0.5 0.01

d Ym

g L

� � � �� � �� �

� ��� �.

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Here,

1/22

2 2 2

1 12 1 0.02

0.5 0.01 0.010.5 1

0.5

�� � �

� � �� � �

� �.

240.02

2 1 2 1 2 102

�� �- � � � � �� �

� �

2 2410

[ 0.5 2(1 2 10 )]d Y

mgL

� ��

� � � �

2 2410

2 10d Y

mgL

� ��

� �

4 11 48 10 2 10 2 10

9.8 1m

� � � � �

m� = 0.08207 kg or m = 82.07 gm.

5. A thin uniform copper rod of length and mass m rotates uniformly with an angular velocity

in a horizontal plane about a vertical axis passing through one of its end. Determine

the tension in the rod as a function of the distance r from the rotational axis. Find the

elongation of the rod.

Solution:

The situation is shown in figure.

Consider an element of thickness dr at a distance r from the rotational axis.

Mass of element dr = m

dr� �

� �� ��

.

Centrifugal force acting at a distance r

= 2 2m m

dr r r dr � � �� �

� �� �.

This centrifugal force provides the tension in the rod.

Hence, 2

0

T t

r

mdT r dr � � �

.

2 22 2

2

11

2 2r

m r rT m

� � � �� � �� �� � � �� �� � � �

�

�� �

……………………. (1).

For calculation of increase in length, let us again consider an element dr at a length. We known

that strain = (stress/Y)

/T AChange in length

Original length Y� �

d T T ror d

r AY AY� �

�� ……………………. (2).

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Now mass of the part dr = # A dr

And centrifugal force = m r 2

Or T = # A dr r 2 Substituting in equation (2), we get,

22 2Adr r r r dr

dAY Y

# # � �� . ……………………. (3).

The elongation �� of the rod can be obtained by integrating this expression. Thus

2 2 3 22 3

0 00

1

3 3

rd r dr Or

Y Y Y

# # # � � �� � � � � �� �

� �� �

�� �

� � � �

Example: The aluminium and steel pipes shown in figure 2.47 are fastened to rigid supports at one

of their ends and to a rigid plate C at the other ends. Derive expression for axial stresses in the two

pipes.

Solution:

From the inspection of figure 2.47, we observe that steel pipe will carry tensile stress (ps)

while aluminium pipe will carry compressive stress (pa). If R

s and R

a are the reactions at the two

ends, as marked in figure 2.47, we get from static equilibrium:

Rs + R

a = 2 P.

Since Rs causes tensile stress in steel pipe, R

s = p

s. A

s. Similarly, R

a causes compressive stress in

aluminium pipe and hence Ra = p

a A

a.

ps A

s + p

a A

a = 2 P …………………. (i)

Also, extension of steel pipe = contraction of aluminium pipe

Or2ss

s a

p Lp L

E E�

Or2. 2 .s s

s a a

a a

E ELp p p

L E E� � …………………. (ii)

Substituting in (i), we get 2 . 2sa s a a

a

Ep A p A PE

� � .

From which

2

2 .a

ss a

a

Pp

EA AE

��

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42

Hence

/ 44 .

2 2 .

s as

s as a s a

a s

E E Pp P

E EA A A AE E

� �� � .

Example :- A rigid bar ABC is supported by three rods of the same material and of equal diameter

as shown in figure 2.48. Calculate the forces in the bars due to an applied force P, if the bar ABC

remains horizontal after the load has been applied. Neglect the weight f the rigid bar.

Solution:

Let us used suffix 1 for outer bars and suffix 2 for the inner bar.

From statical equilibrium, 2 P1 + P

2 = P …………….. (i)

2L

LL

From compatibility, 1 2� � �

Or 212P LP L

AE AE� from which P

1 = 2 P

2…………….. (ii)

Substituting in (i), we get 2 (2 P2) + P

2 = P, from which P

2 = 0.2 P

P1 = 2 × 0.2 P = 0.4 P.

Example: - A bar of uniform section is fixed at both the ends and is loaded with a force P shown in

figure 2.61. Determine the reactions at both the ends and the extension of AC.

Solution:

Let R1 and R

2 be the two reactions, obviously in the directions shown.

From static’s, R1 + R

2 = P ………… (1)

Since the end are fixed in position,

Extension of AB = compression of BC

1 2R a R b

AE AE� � …………. (2)

Or a R1 = b R

2 substituting in (1) we get

2 1b

R Pa

� �� �� �

� �

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43

From which 2

Pa PaR

a b L� �

�

Hence 1 .b Pa Pa

Ra a b L

� ��

These are thus the values of reactions R1 and R

2

Now extension of AB 1

1

R a P ab

AE L AE� � �

Example: A rigid bar AB is supported by three equidistant rods in the same vertical plane, and

is loaded as shown in figure. All the three rods have the same area of cross – section. Determine

the forces in the three rods, if the bars 1 and 3 are of steel and bar 2 is of brass. Take Es/E

b = 2.

Solution:

From static’s, P1 + P

2 + P

3 = P ………. (1).

Also, taking moments about B, P3 (2a) + P

2 (a) = P × 1.5 a.

Or 2 P3 + P

2 = 1.5 P.

There are three unknown forces (i.e., P1, P

2, and P

3), while we have only two equations

obtained from statical equilibrium.

The third equation is obtained from the deformation pattern of the structure. The dotted

lines shows the deformed position of the structure, where 1 2 3, and� � � are the extensions of the

three rods. From compatibility, we obtain.

1 22

2

� � �� �

Or32 1

2 2 1 1 3 32 2

PP P

A E A E A E� �

Here, A1 = A

2 = A

3.

Also, E1 = E

3 = 2 E

2 = 2 E (say)

32 1

2 2 2 2

PP P

E E E� � �

Or 4P2 = P

1 + P

3……….. (3).

Substituting this value of (P1 + P

3) in (1), we get

4 P2 + P

2 = P from which 2 0.2

5

PP P� �

Hence from (2), 2P3 + 0.2 P = 1.5 P.

From which P3 = 0.65 P

Substituting these values of P2 and P

3 in (1), we get

P1 + 0.2 P + 0.65 P = P

From which P1 = 0.15 P.

Example: A load P is supported by a system of three rods shown in figure. The outer rods are

indentical and are of the same material while the inner rod is of different material. Determine the

forces in the rods. Hence if P = 30 kN and � = 300, determine value of these forces when the outer

bars are of steel and the central bar is of brass. Take the ratio of moduli of elasticity of steel and

brass as 2.

Solution:

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From static’s 2 P1 cos � + P

2 = P ……. (1).

The dotted lines show the deformed shape of the structure, in which point O the position O’.

Assuming that � does not change appreciably, we get.

1 2� � � = �

Or1 1 2 2

1 1 2 2

cosP L P L

A E A E�� ……. (2).

��

�

But, from geometry, L2/L

1 = cos �

Hence, from (2),

21 1 2 1 1 21 2 2

2 2 1 2 2 1

. . cos . . cosA E L A E L

P P PA E L A E L

� �� � …….. (a).

Substituting this value of P1 in (2), w3, we get

31 12 2

2 2

2 cosA E

P P PA E

� � �

From which 2

31 1

2 2

21 cos

PP

A E

A E�

�� ….. (i)

Hence

21 11

31 1 2 22 22

2 2 1 1

. cos2

1 cos 2coscos

A EP PP

A E A EA E

A E A E

�� �

�

� �� �� �� �� �� �� �� �

….. (ii)

Taking the values : P = 30 kN, � = 300, A1 = A

2, E

1/E

2 = 2 we get

2 3 00.278 8.34

1 2 2 cos 30

PP P kN� � �

�

And1

0

2 0

0.417 12.51

2cos302cos 30

PP kN� � �

�

Example: Three identical in - connected bars are arranged as shown in figure, and support a load

P. Find the axial are force in each bar and also the vertical displacement of the point of application

of load. Neglect any possibility of lateral buckling of the bar. All the bars are of the same length and

same areas of cross – section.

Solution:

Let the use suffix 1 for bars AO and BO and suffix 2 for CO. obviously, force P1 in bars AO

and BO will be tensile while force P2 in bar OC will be compressive.

From static’s, 2P1 cos 600 + P

2 = P

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From which P1 + P

2 = P …………. (i).

21

The dotted lines show the deformed position of the structure. Assuming that the angles between

the rods do not change appreciably, we get

01 2 2

1cos 60

2� � � � �

Or1 21

2

P L P L

AE AE�

From which P2 = 2P

1 …………. (ii)

Substituting in (i), we get P1 + 2P

1 = P

From which P1 =

2

3

P

Also, 2

2

2

2

P L P L

AE AE� � �

Example: A load of 10 kN is suspended by the ropes shown in figure. (a) and (b) In both the cases,

the cross sectional area of the ropes is 100 mm2 and the values of E is 200 kN/mm2. In case (a), the

rope ABC is continuous over a smooth pulley from which the load is suspended in such a way that

both the ropes are stretched by the same amount. Determine, for both the cases, the stresses in the

ropes and the deflections of the pulley and the block due to the load.

Solution: Let us used suffix 1 for rope AB and 2 for rope CB. Load = 10 kN; length of AB = li = 6

m; length of CB = l2 = 8m. Area of rope = 200 mm2.

Case (a): Since the load is applied on a smooth pulley, the external load (= 10 kN) is

shared equally by both the ropes.

� P in each rope 10

2� = 5 kN.

(Note that reactions at A and C, each are equal to 5 kN)

Hence stress 25 1000

25 /200

p N mm

� �

Also, 1 2 5 6000 8000

3.50100 200

P l lplmm

AE AE

� �� � � � �

Case (b): Load is suspended is such a way that:

Or 1 2� � �

Or2 2 22

1

1

8 4

6 3

lp p p p

l� � � …….. (1).

Also, from static’s p1 A

1 + p

2 A

2 = P

Or 100 (p1 + p

2) = 10 × 1000

Or p1 + p

2 = 100

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Substituting the value of p1 in (2), we get ……… (2).

2 24100

3p p� �

From which p2 = 42.86 N/mm2

Hence, p1 = 42.86 ×

2457.14 /

3N mm� .

Also,1 1 57.14 6000

1.71200 1000

p lmm

E

� � � �

.

Example: A rigid beam ABC is hinged at D and supported by two springs A and B as shown in

figure. The beam carries a vertical load W at a point C at a distance of b from the hinged end. The

flexibilities (deflection/unit load) of the springs at A and B are d1 and d

2 respectively. Determine the

forces in the springs and also the reaction at the hinged support.

Solution:

The dotted line in figure shows the deformed shape of rigid beam. Let PA and P

s be the

loads carried by springs A and B.

For spring A, extension 1.A AP d� � .

For spring B, extension 2.B BP d� �

Or 3

2A B� � �

Substituting the values of A Band� �

1 2

3. .

2A BP d P d� or

2

1

3

2A B

dP P

d�

Also, by taking moments about hinge D, we get

PA × 3 b + P

B × 2 b = W. b or 3 P

A + 2 P

B = W.

Substituting the value of PA from (1) in (2), we get

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47

2

1

33 22

B B

dP P W

d

� �� �� �

� �

From which 1

1 2

2

4 9B

dP W

d d�

� and 2

1 2

3

4 9A

dP W

d d�

� .

Example: A solid cylinder of steel is placed inside a copper tube. The assembly is compressed

between rigid plates by forces P. Find the value of increase in temperature so that all the load is

carried by the copper tube.

Solution:

Let the increase in temperature be t. Due to this increase, copper will expand more than

steel. Difference between the two expansions = 1 c sL � �� � � t. If the load P is to be borne entirely

by the copper tube, the shortening c� of the copper tube due to P should be equal to t� .

c t� � � �

Or c s

c c

P LL t

A E� �� � or

c c c s

Pt

A E � �� �

�

Example: A steel bar AB of length L is fixed at the ends and is subjected to an non – uniform

increase in temperature represented by the expression

2

2.x

xt t

L� , as shown in figure. Determine the

value of compressive stress set up in the bar.

Solution:

Consider a short length dx, at a distance x from end A.

Temperature =

2

2.x

xt t

L�

� Increase in length of dx due to temperature. rise

2

2.tx x

xdx t t dx

L� �� � � �

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48

Hence total increase in length 1�

2 3

2 2

0

.3 3

L

tx

x t L t Lt dxL L

� ��� . � � � �

Since this increase in length is restricted by end supports, compressive stress will be set

up in the bar, the value of which is given by.

.3 3

c

E t L E E tp t

L L

� �� �� � � �� �

� �

Problem: Calculate the elongations of (a) a copper wire of 1.4 mm diameter and (b) an aluminium

wire of 1 mm diameter as shown in figure 1.19, if Young’s modulus for copper and aluminium are

11 × 1010 N/m2 and 7 × 1010 N/m2 respectively.

Solution: By definition of Young’s modulus, F L

LY A

� � .

(a) As tension in copper wire will be due to load of 7 kg.

i.e., TCu

= M1 g = 7 × g N

So 23 10

7 9.8 0.5

22 / 7 0.7 10 11 10Cu

L�

� �

i.e, 3 39.8 5

10 0.2 1022 11Cu

L m� �� � �

(b) However, the tension in the aluminium wire will be

1 2 2 1A CuT M g T M M g� � � �

So 3

1 3 10

4 7 9.8 11.96 10

22 / 7 0.5 10 7 10A

L m�

�

� � � �

Ans:

Example: Two rods of different metals, having the same area of cross – section A, are placed end to

end between two massive walls as shown in the figure. The first rod has a length l1, coefficient of

linear expansion 1� and Y2. The temperature of both corresponding quantities for second rod are l

2

2� and Y2. The temperature of both the rods is now raised by T degrees. (a) Find the force with

which the rods act on each other at the higher temperature in terms of the given quantities. (b) Also

find the length of the rods at the higher temperature. Assume that there is no charge in the cross –

sectional area of the rods and the rods do not bend. There is no deformation of walls.

Solution: (a) Due to heating the increase in length of the composite rod will be

1 2

1 2D

L L F F LL as L

Y Y A AY

� � � �� � � � �� � � �

� �� �

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49

As the length of the composite rod remains unchanged the increase in length due of heating

must be equal to decrease in length due to compression, i.e.,

1 1 2 2

1 1 2 2[( / ) ( / )]

A L L TF

L Y L Y

� ���

� …………. (1)

LL

Y Y

(b) As initially the length of one rod is L1 and due to heating it increase by 1 1 1H

L L T�� � while

due to compression it decreases by 1 1 1/C

L F L AY� � , so its final length

1 1 1 1 1 1 1' [1 / ]H C

L L L L L T F AY�� � � � � � � �

Similarly for the other rod,

2 2 2 1 2 2 2' [1 / ]H C

L L L L L T F AY�� � � � � � � � .

Where F is given by Eqn. (1).

Problem: A uniform pressure p is exerted on all sides of a solid cube at temperature t0C. By what

amount of should the temperature of the cube be raised in order to bring is volume back to the

volume it had before the pressure was applied, if the bulk modulus and coefficient of volume

expansion of the material are B and � respectively.

Solution: As by definition of bulk modulus B = - V ( � p/ � V), with increase in pressure decrease

in volume of the cube will be given by,

[ ]p

V V as p pB

�� � � � ……………… (1).

As the volume of the cube remains constant,

, . .,p p

V V i eB B

� � ��

� � � � .

Problem: Calculate the force F needed to punch a 1.46 cm diameter hole in a steel plate 1.27 cm

thick. See the figure. The ultimate shear strength of steel is 345 MN/m2.

Solution: As in punching, shear elasticity is involved, the hole will be punched if 11F

A

� �� �� �

> Ultimate

shear stress

i.e, F11

> (Shear stress) × (Area).

So 8

11 min3.45 10 2F r L��

[as her A = 2 r L� ]

i.e., 8 2 2

11 min3.45 10 2 3.14 0.73 10 1.27 10 200F kN

� �� � . Ans:

Example: Show that work performed to make a hoop out of a steel band (Young’s modulus Y) of

length l, width b and thickness d is equal to

2 31

6

Y bd

l

�� �� �� �� �

.

Solution: While making a hoop out of a steel band, the central layer remains unstretched. The

layers above the central layer are extended in length while the lower layers are compressed in

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length. Let R be the radius of the central unstretched layer (shown dotted in figure) Thus l =

2 R� .

Consider a coaxial layer of radius x. The strain produced in this layer will be

2 21

2

x R x

R R

� ��� � �

� �� �� �

.

Hence, stress at this layer.

1x

YR

� �� �� �

� �.

Now the energy density (u) at this layer is given by

1

2u � × stress × strain

21 1

1 1 12 2

x x xY

R R R

/ 0� � / 0 � � � � � �1 2 1 2� � � �

� � 3 4 � �3 4.

b

d

b

R

The energy stored in an elementary volume at x,

21

1 {(2 ) }2

xdU u volume x dx b

R�� �

� � �� �� �

.

2

2

2

1. . 1 .2

2

dR

dR

xU Y dxb

R�

�

�

� �� � �� �

� �

222

2

dR

dR

Y bx R xdx

R

� �

�� �

3 2 222

2

2

dR

dR

Y bx R x R x dx

R

� �

�

� �� �� � �� �� �

3

12

Y bdd R

R

�� ''�

3

; 212 / 2

Y bdR

l

��

�� �

2 31

6

Y bd

l

�� �� � �� �

� �

Example: A thin cylindrical shell closed at both ends is subject to a uniform internal pressure P. The

wall thickness is h and the inner radius r. (a) Calculate the longitudinal and circumferential normal

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stresses existing in the walls due to internal pressure. Neglect any restraining effect of the cover

plates. (n) Calculate the increase in radius of the cylinder due to the internal pressure P.

Solution: (a) Considering length L of the cylindrical part of the shell, as shown in figure. From

symmetry the horizontal components of the radial pressures cancel. Consider a differential element

that subtends an angle � at the center. The vertical component of force due to pressure on this

element is (r d � ) sin � L.

For the equilibrium in the vertical direction. 0

2 sin 0v cF hL P r d L�

� � �. � � � � .

Integration, we get 0

2 cosc h L P r L�� �� �

Or2

c

P r

h� � . …………… (1).

Note that the resultant vertical force due to internal pressure can be determined by multiplying

the pressure by the horizontal projected area upon which it cats. To determine the longitudinal

stress l� , consider equilibrium in the horizontal direction.

22 0h lF P r r h� � �. � � � �

Or2

l

P r

h� �

The result shows that the circumferential stress is twice the longitudinal stress. Thus if the

water in a closed pipe freezes, the pipe will rupture along a line running longitudinally along the

cylinder.

(b) As Young’s modulus, Y�5

� , hence circumferential strain, Pr

cYh

5 � .

Note that c5 is unit strain, it acts over the circumference of the cylinder, 2� r. Hence the total

elongation of the circumference is

22 Pr

2cl rYh

�5 �� �

The final length of the circumference =

22 Pr

2 rYh

�� � � . Dividing this circumference by 2� , we get

the radius of the deformed cylinder to be r +

2Pr

Yh, so that the increase in radius is

2Pr

Yh. Longitudinal

strain due to longitudinal stress [eqn. (2)] is

2Pr

2l

Y h5 �

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The longitudinal strain induces a circumferential strain equal to - v l5 . Denoting it by c5 we

have 2

c

v P r

Y h5 � � which tends to decrease the radius of the cylinder as indicated by minus sign

Analogous t the procedure for increase of radius, the decrease in radius corresponding to strain c5

is given by

2

2

v P r

Y h. The st……. increase of radius due to internal pressure is thus

2 2P r

2

P r vrY h Y h

� �

2

12

Pr v

Y h

� �� �� �� �

.

1. A balloon ascends vertically slowly unreeling a long copper wire. Estimate the amount by

which the wire has stretched when 1 km of initially Unstretched wire has been unreeled.

The density of copper # is 9.0 × 103 kg/m3 and its Young’s modulus Y = 1.2 × 1011 N/m2.

1. Consider an element of wire of length dx at a height x from the ground as shown in figure.

Load on dx = (A x # )g. If e be the elongation dx, then

/

/

A x g AstressY

strain e dx

#� �

g x dxe

Y

#� �

Stretched length = dx + e = dx + g x dx

Y

#

= 1g x

dxY

#� ��� �

� �

Entire length = 0

1H g x

dxY

#� ��� �

� �

= H +

2 2

02 2

H

g x g HH

Y Y

# #� �� � � �

� �� �

= 3 6

3 3

11

9 10 9.8 1010 1 10

21.2 10H km m

� �� � �� ��

= 103 + 0.3675

� Increase in length = 0.3675 m.

2. A bar of cross section A is subjected to two equal and opposite tensile forces F at its ends.

Consider a plane through the bar making an angle � with a plane at right angle to the

bar:

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a. What is the tensile stress at this plane, in terms of F, A and � ?

b. What is the shearing stress at the plane, in terms of F,. A and � ?

c. For what value of � is the tensile stress a maximum?

d. For what value of � , the shearing stress is maximum?

2. The bar cross – sectional area A is shown in figure. Imagine a cut along the plane BB’.

Since this section is equilibrium, the bar on the left must have been exerting a horizontal

force F uniformly to the left. This force F has been resolved in two components. F cos �(normal to the plane BB’) and F sin � (along the plane BB’)

a. Tensile stress = sec

Normal force

area of corss tion

Here, Normal force = F cos �And Area of cross section = A/cos �

� Tensile stress =

2cos cos

/ cos

F F

A A

� ��

� .

b. Shearing stress = sec

Tangential force

area of corss tion

= cos sin cos sin 2

/ cos 2

F F F

A A A

� � � ��

� �

c. From part (A), ten tensile stress is given by, 2cosF

A

�

This will be maximum then cos2 � = 1, i.e., � = 0

d. Shearing stress sin 2

2

F

A

�.

This will be maximum when sin 2� = 1 or 2� = � /2, i.e., = � /4

3. A ring of radius R mad of end wire is rotated about a stationary vertical axis passing

through the its center and perpendicular to the plane of the ring. What is the number of

r. p. s at which the ring ruptures?

3. Consider an element of length dl on the periphery of the ring of the ring as shown in figure.

Mass per unit length of the ring = 2

m

R�

Mass of length dl = 2

mdl

R�� �� �� �

= 2

2

mdl R

R

�� �� �� �

…………….. (1).

Projection of tension T along center

2 sin 2T T is small� � �� � �

= T (2� ) = T (dl/R) ……………….. (2)

From equation (1) and (2), we get

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2

2

mdl R

R

�� �� �� �

= T (dl/R) ………...... (3)

Here, m = � r2 l #

Where, l is the length of wire and # is the density of its material and r is the cross –

sectional

radius.

Further, l = 2� R

2 2m r R� � #� �

Substituting the value of m in equation (3) and solving it, we get

2 2 2 2

2

TR or Stress S R

r# #

�� � �

max1 S

R

#� �

� � � �� �

for rupture

Number of r. p. s = max1

2 2

S

R

� � #

� �� � �

� �.

4. A steel cylindrical rod of length l and radius r is suspended by its end from the ceiling.

a. Find the elastic deformation energy U of the rod.

b. Define U in terms of tensile strain ��/l of the rod.

4. (a) Consider an element of length dx at a distance x from the fixed end. Let dU be the

energy

stored in the element of length dx. Then

1

2dU stress strain volume�

= 1

.2

m x g stressAdx

A Y

�

�

�

=

2

1

2

mg xAdx

Y A

� ��� �

� �� �

�

�

2 22 2

2

0

12

2

m gU x dx

Y A� � � �

�

� �

�

2 2 3 2 2 32 3

2 2.

32 6

m g m g

Y H Y A

� �� � � �� �

� �� �

� �� � �

� �

2 22 2 2 2 2 3

26 66

r gm g r g

Y A YY r

� # � #

�� � �

� �� �

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b. 1

2U stress strain volume�

= 1

2strain Y strain volume

= 21

2Y strain volume

2

21

2Y r�

�� �� �

� �

��

�

5. A sphere of radius 10 cm and mass 25 kg is attached to the lower and of steel wire which

is suspended from the ceiling of a room. The point of support is 521 cm above floor. When

the sphere is set swinging as a simple pendulum, its lowest point just grazes the floor.

Calculate the velocity of the ball at its lower position. [Young’s modulus of steel = 20 ×

1010 N m-2, Unstretched length of wire = 500 cm Radius of the steel wire = 0.05 cm]

5. The situation is shown in figure. Unstretched length of the wire = 5.0 m.

Stretched length of the wire = L + l

= 5.21 – 0.2 = 5.01 m.

5.01 5.0 0.01m m m� � � ��

Strain = 0.01/5

Let T be the tension in the wire.

Now, stress in wire = T/A

2

2 40.05 10

TStress N m

��

�� �

StressY

Strain�

Or 10

2 4

520 10

0.010.05 10

T

� � �

Solving we get, T = 341.1 N

Now, 2mv

T mgr

� �

Where, v is the velocity of the sphere at it crosses the equilibrium position and r is the distance

of

the center of the sphere from the ceiling.

R = 5.01 + 0.1 = 5.11 m

225314.1 25 9.8

5.11

v� � �

2 69.1 5.11

25v

� or v = 3.76 m/sec.

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6. A force of 106 N/m2 is required for breaking a material. If the density of material is 3 ×

103 kg/m3, then what should be the length of the wire made of material is so that it

breaks by its own weight?

6. Let L be the length of the wire required and A be its area of cross section.

� Weight of the wire, W = mg = AL # g, where # = its density

� Longitudinal stress developed at the point of support of the wire

WL g

A#�

But it is given that the wire will break if

6 610 , . ., 10W

i e LA

#6 6

Hence, length of the wire required, L, so that it breaks under its own weight is given by

610L g# �

6 6

3

10 1034.01

3 10 9.8L m

g#� � � �

.

7. A copper wire of negligible mass, 1 m length and cross – sectional area 10-6 m2 is kept on

a smooth horizontal table with one end fixed. A ball of mass 1 kg is attached to the other

end. The wire and the ball are rotating with an angular velocity of 20 rad/sec. If the

elongation in the wire is 10-3 m, obtain the Young’s modulus. If on increasing the angular

velocity to 100 rad/sec, the wire brakes down, obtain the breaking stress.

7. The ball of mass, m = 1kg is rotating in a horizontal plane in a circle of radius r = 1 + 10-

3 =

1.001 m where we have added 10-3 m to the original 1 m of the wire as the wire is extended

by

this amount.

� Centrifugal force acting on the wire,

2F m r � ………….. (i)

From Young’s modulus, Y A

FL

��

�………….. (ii)

Where A = cross – sectional area of the wire,

� L = increase in the length and

L = original length

From equations (i) and (ii), we have

2Y Am r

L

��

�

222

6

1 20 1.001 1/

10 0.001

m r LY N m

A

�

� �

� �

= 4.0 × 1011 N/m2.

The wire will break if the centrifugal force acting on it exceeds the breaking force i.e., when

' 100 /rad s � � . Hence the wire snaps when F = m 2' r

= 1 × (100)2 × 1.001 = 104 N

� Breaking stress =

410 2

6

1010 /

sec 10

Breaking forceN m

corss tional area �� �

� .

8. A 0.5 m long cylindrical steel wire with radius 2 × 10-3 m is suspended vertically from a

rigid support & carries a bob of mass 100 kg at the other end. If the bob gets snapped.

Calculate the change in temperature of the wire ignoring Radiation loss (y = 2.1 × 1011 Pa,

Density = 7860 kg/m3, Specific heat = 420 J/kg K).

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8. Stored elastic potential energy = Heat generated

1

2U stress strain volume ms �� � � � �

Stress = ,M g M g

StrainA y A

� , Volume = LA.

2

. .2

M g LA L s

y A# �

� � �

2 2

2 23 11

100 10

2 2 7860 2 10 2.1 10 420.

M g

A y s�

# � �

� �� �� � � � � � � �� �

�� = 0.00457 K.

9. A solid copper cylinder of length L, is placed on a horizontal surface. A compressive force

F is applied to the cylinder in a vertically downward direction & is distributed uniformly

over the end face. Derive an expression for resulting change in volume assuming y to be

the Young’s modulus & r its radius. (Assume, � to be the Poisson’s Ratio).

9. If L� is considered to be the decrease in length of the cylinder due to application of force

‘F’.

2

/

/

F A L F Fy

L L L A y r y�

�� � � �

� ………………… (1)

Poisson’s Ratio /

/

Lateral strain r r

Longitudinal strain L L�

��� �

��

r L

r L�

� �� � �

Initial volume of cylinder V = 2r� L

Volume change 2V r L�� � �

= 2 2

.2 .r L r L r r L� � �� � � � �

= 2 22

2L r L L

r r LL r L L

� � �� � � �� � � �

� � � �� � � �� � � �

2 2

21 2 1 2 1 2

L E F LV r L r L

L yr y� � � � �

�

�� � � � � � � � .

10. Each of the three blocks P, Q & R shown has mass of 5 kgs. Each of the wires A & B has

the cross – sectional area of 0.005 cm2 & Young’s modulus of 2.5 × 1011 N/m2. Neglect

frictions and find the longitudinal strain in each of the wire Take g = 10 m/s2.

10. Block R will go down, blocks P and Q will move on frictionless horizontal table. If ‘a’ is the

common acceleration of the system and tension in wires A and B are TA & T

B.

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TA = 5 a ……………. (1)

TB - T

A = 5a ……………. (2)

5 × g - TB = 5a ……………. (3)

B

R

QA

P

a

T T

a

a

T

From (1) & (2), TB = 2 T

A

2 5 10BT a a� � �

� 5g – 10a = 5a

� 15 a = 5 g = 50 m/s2

� 2

10

3

ma

s�

10 505 5

3 3AT a N� � � � .

50 1002 2 .

3 3B AT T N� � �

' mod

Longitudinal stressLongitudinal strain

Young s ulus� �

22 2

11 211

2

50 50 50.005 10

3 3 1000

2.5 10 /2.5 10

N cm N m

Strain in wire AN N m

m

�� � �

4 3

73 4

11 2 11 11

50 5 10 50 10

10 103 310 5 10

32.5 10 / 2.5 10 2.5 10

N

N m

�

�

� � �

74

7 4

10 101.33 10

3 2.5 10 10

�� �

Strain in wire B =

42

23

11 2 11

100 5 10100 510

33 1000 10

2.5 10 / 2.5 10

m

N m

��

�

=

7 7

11 7 4

100 205 10 10

3 3

2.5 10 2.5 10 10

� � �

=4 4 420 2 8

10 10 2.66 10 .3 5 3

� � � � �

11. Show that work performed to make a loop out of a steel band (Young’s modulus) of length

l, width b & thickness d is equal to

2 31

6

y bd�� �� �� �� ��

.

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59

When hoop is made out of a steel band, then central layer of hoop remains Unstretched, the

layers above the central layer are extended in length while those below the central layer are

shortened in length. If R is the radius of central unstretched layer, then 2 .R��� .

Consider now a coaxial layer of radius x. The strain produced in this layer = 2 2

2 .

x R

R

� ���

1.x

R� �

1x

Stress at the layer y stress y strainR

� �� � � � � �

� ��

� u = (Energy density at this layer)

=

21 1 1

1 1 12 2 2

x x xstress strain y Y

R R R

� � � � � � � � � � �� � � � � �

� � � � � �

� dU = stored energy in an elementary volume at

21

1 2 .2

xx y x dx b

R�� �

� � � �� �

2 4 3 32

2 2

2

2

64 24 12

dR

dR

y b yb d d ybdU x R x dx R d R

RR R

� � ��

�

� �� � � � � � ''� �� �

� � �

3

2

122

y bdR

��

�

� �� �

� �� �

��

�

2 31

.6

ybd�� �� � �� �

� ��

12. A thin rod of negligible mass & area of cross section 4 × 10-6 m2, suspended vertically from

one end, has a length of 0.5 m at 1000 C. The rod is cooled to 00 C but prevented from

contracting by attaching a mass at its lower end. Find (i) This mass & (ii) The energy

stored in the rod. Y = 1011 N/m2, � = 10-5 K-1 & g = 10 m/s2.

12. (i) 6 21004 10 , 0.5A m L m�� �

Initial temperature = 1000 C

5 1 11 210 , 10 /k y N m� � �� �

g = 10 m/s2, final temperature = 00 C

100 0 100L L L T L�� � � � �

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� Longitudinal strain = 5 3

100

10 100 10L

TL

� � ��� � � �

� Thermal force generated = stress × C. S. Area

= 108 × A = 108 × 4 × 10-6 = 4× 102 N.

If m = mass attached at the end for ensuring that it does not contract.

22 4 10

4 10 40 .10

mg m Kg

� � � �

(ii) Stored energy = 1

2stress strain volume

8 3 6110 10 4 10 0.5 0.1 .

2Joule

� � �

13. A horizontal aluminium rod 5.0 cm in diameter, projects 6.0 cm from a wall. A 1000 kg

object is suspended from the end of the rod. Shear modulus of aluminium is 3.2 × 1010 N/

m2. If one neglects the mass of the rod find (a) Shear stress on the rod and (b) the vertical

deflection at the end of the rod.

13. (a) The shear stress is F/A.

F = 100 × 9.8 N = 9800 N

2A r��

2.50.025

100r m� �

�2 2

0.025 3.142 0.000625 0.0019637A m�� � �

6 2

2

98004.99 10 /

0.00196

Nstress N m

m� �

(b) Shear modulus /

/

F A

x L( �

�

6 2

10 2

/ 4.99 10 / 0.06

3.2 10 /

F A L N m mx

N m(

� � � �

= 4 4 54.99

0.06 10 0.093 10 0.93 10 .3.2

m m m� � � � �

Concept Building Objectives:

Elasticity:

Single Answer Correct:

1. If a metal wire is stretched a little beyond its elastic limit (or yield point), and released, it

will

a. Lose its elastic property completely.

b. Not contract.

c. Contract, but its final length will be greater than its initial length.

d. Contract only up to its length at the elastic limit.

Solution : (C)

2. The wires A and B shown in the figure are made of the same material and have radii rA

and rB respectively. The block between them has a mass m. When the force F is

3

mg, one

of the wires breaks. Which of the following is not true?

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61

a. A will break before B if rA = r

B

b. A will break before B if rA < 2r

B

c. Either A or B may break if rA = 2r

B

d. The lengths of A and B must be known to predict which wire will break.

Solution : (D)

Tension in B = TB =

3

mg

Tension in A = TA = T

B + mg =

4

3

mg

4A BT T� �

Stress 2

TS

r�� .

Wire breaks when S = breaking stress.

With, rA = r

B

We get, SA = 4S

B A breaks before ‘B’.

With rA < 2r

B, S

A > S

B A breaks before ‘B’.

With rA = 2r

B, S

A = S

B Either ‘A’ or ‘B’ may break.

Value of length of the wires is irrelevant here.

3. Two rods of equal cross – sections, one of copper and other of steel are joined to form a

composite rod of length 2 m. At 200 C the length of the copper rod is 0.5 m. When the

temperature is raised to 1200 C, the length of the composite rod increases to 2.002 m. If

the composite rod is fixed between two rigid walls and is thus not allowed to expand, it is

found that the length of the two component rods also does not change with the increase

in temperature. Calculate the coefficient of linear expansion

a. 0.8 × 10-6/0 C b. 0.8 × 10-5/0 C

c. 6 × 10-5/0 C d. 0.16 × 10-5/0 C.

Solution: (B)

With increase in temperature, due to thermal expansion, length of the rod changes i.e., L’ = L (1

+ � � t). So for composite rod, ' ' 1 1s c s s c cL L L t L t� �� � � � � � �

s c s s c cL L L L t� �� � � � �

According to problem,

2.002 = 2 [0.5 × 1.6 × 10-5 + 1.5 s� ] × 100.

Solving, we get, s� = 0.8 × 10-5/0 C.

4. The normal density of gold is # and its bulk modulus is K. the increase in density of a

lump of gold when a pressure P is applied uniformly on all sides is:

a. P

K

#b.

K

P

#c.

P

K# d. K

P#

Solution : (A)

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; 'M M

V V V# #� �

� �

1'

1 1V V V

V V V V

##

�� �� �

� � � � � �� �� � � �

/

P V PK or

V V V K

�� �

�

' '1 1

P Por

K K

# ## #

� � � � �

Or '

'P

orK K

# # # ## #

#�

� � �

5. A wire 3 m in length and 1 mm in diameter at 300 C is kept in a low temperature at - 1700

C and is stretched by hanging a weight of 10 kg at one end. The change in length of the

wire is [Y = 2 × 1011 N/m2, g = 10 m/s2 and � = 1.2 × 10-5/0 C]:

a. – 5.2 mm b. 2.5 mm c. 52 mm d. 25 mm.

Solution : (A)

The contraction in the length of the wire due to change in temperature

5 31.2 10 3 170 30 7.2 10L T m� � �� � � � � �

The expansion in the length of wire due to stretching force

3

11 6

10 10 32 10

2 10 0.75 10

F Lm

Y A

�

�

� � �

Resultant change in length = - 7.2 × 10-3 + 2 × 10-3 m = - 5.2 × 10-3 m = - 5.2 mm

Negative sign shows a contraction.

6. Two cylinders A and B of the same material have same length, their radii being in the ratio

of 1 : 2 respectively. The two are joined end to end as shown in the adjoining figure. The

upper end of A is rigidly fixed. The lower end of B is twisted through an angle � , the angle

of twist of the cylinder A is

a.15

16� b.

16

15� c.

16

17� d.

17

16�

Solution: (C)

For cylinder A :

4

'2

r� (� ��

�

For cylinder B :

42 '

2

r� ( � ��

��

�

4 42 16' '

2 2

r r� (� � ( � �� � �

� �

' 16 16 '� � �� �

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16'17

� �� � .

O1

O

O2

Previous IIT Subjectives:

Example:

A wire of density 9 g cm-3 is stretched between two clamps 100 cm apart, while subjected

to an extension of 0.05 cm. What is the lowest frequency of transverse vibrations in the wire,

assuming Young’s modulus of the material to be 9 × 1011 dyne cm-2

Solution:

Let F be stretching force and A be the area of cross – section of the wire.

By definition, Young’s modulus (Y) = linear stress

linear strain

/

/

F A F LY

L A� � �

� �

117 29 10 0.05

45 10100

F Ydyne cm

A L

� � � � �

�

The lowest frequency (fundamental frequency) of transverse vibration on the wire is given by

1

2

Fn

L m�

Where L = length of the stretched wire

F = stretching force

m = mass per unit length of wire

= (Area of cross – section of wire) × (density) = 9A

71 1 45 10

2 100.05 9 200.1 9

Fn

A

� � �

.

Example:

A steel wire of cross – sectional area 0.5 mm2 is held between two fixed supports. If the

tension in the wire is negligible and it is just taut at a temperature of 200 C, determine the

tension when the temperature falls it is 00 C. Young’s modulus of elasticity is 21 × 1011 dyne cm-

2 and the coefficient of linear expansion is 12 × 10-6 (C0)-1. Assume the distance between the

supports remains same.

Solution :

Let the original length = L cm

The change in length L�� �� t where � is the coefficient of linear expansion and t is the change

in temperature.

L tStrain t

L L

��� � � �

�

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64

6 512 10 20 24 10� � � The nature of the strain is tensile because the wire is kept taut and not allowed to contract.

� Tensile stress in the wire = Young’s modulus × strain

= 11 521 10 24 10

� = 5.04 × 108 dyne cm-2

The cross – sectional area of the wire = 0.5 mm2

= 0.005 cm2

� Tension (T) in the wire = tensile stress × area

= 5.04 × 108 × 0.005

= 25.2 × 105 dyne = 25.2 N

� The tension in the wire, when the temperature falls by 200 C = 25.2 N

� Strain = 0.01

5

Stress = Y. Strain

= 11 0.01

2 105

= 4 × 108

� Stretching force = area of cross section of wire x stress

2

80.054 10

100F � � �

� � �� �

= 100 � N

The force at the lowest point = 2mv

mgr

�

=

225

25 9.85.11

v � because it describes a circle of radius 5 + 0.01 + 0.1

= 5.11 m.

225

25 9.8 1005.11

vF �� � � �

225

314 2455.11

v� � �

= 59

2 59 5.11

25v

� �

V = 3.76 m/s.

Q. A sphere of radius 10 cm and mass 25 Kg is attached to the lower and of a steel wire which is

suspended from the ceiling of a room. The point of support is 5.21 m above the floor. When the

sphere is set swinging, as a simple pendulum, its lowest point just grazes the floor. Calculate the

velocity of the ball at the lowest position. (Young’s modulus of steel = 2 × 1011 N/m2; Unstretched

length of the wire 5m; radius of steel wire = 0.05 cm).

Solution:

Length of wire + diameter of sphere = 5 + 0.2 5.2 m

In the mean position, it just touches the floor. So, its elongation then is

5.2 – 5.2 = 0.01 m.

Q. A composite rod is made by joining a copper rod end to end with a second rod of different

material but of same cross – section. A 250C, the composite rod is 1 m in length, of which the

length of copper rod is 30 cm. At 12500 C, the length of the composite rod increases by 1.91 mm.

When the composite rod is not allowed to expand by holding it between two rigid walls, it is fund

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65

that the length of constituents do not change with rise in temperature. Find the Young’s modulus

and coefficient of linear expansion of the second rod. For copper, � = 1.7 × 10-5 C-1 and Y = 1.3 ×

1011 Nm-2.

Solution: Increase of length due to heating

= 1 1 2 2l t l t� ��

= 1.91 mm = 300 × 1.7 × 10-5 × 100 + 700 × 100 × 2�

21.91 0.51 70000mm �� � �

21.41 70000�� �

52 10 K� �� �

If length does not change thermal stresses on the rods must be equal and opposite.

112

1.7 1.310

2Y

� �

10 211.05 10 /N m�

� Young’s modulus of material of second rod

11 21.105 10 /N m� .

Problem : A solid sphere of radius R made of a material of bulk modulus B is surrounded by a

liquid in a cylindrical contaniner. A masselsess piston of area A floats on the surface of the

liquid. Find the fractional change in the radius of the sphere (dR/R) when a mass M is placed on

the piston to compress the liquid.

Solution: As for a spherical body,

34 1,

3 3

R VV R

R V�

� �� � …………….. (1)

Now by definition of bulk modulus,

. .,V Mg Mg

B V i e asV V B AB A

# ##

� � � � �� � � � � �� �� � �

.

So 3

dR Mg

R AB� .

Problem A thin rod of negligible mass and area of cross – seciotn 4 × 10-6 m2, suspended vertically

from one end, has a length of 0.5 m at 1000 C. The rod is closed to 00 C but prevented from

contracting by attacing a mass at the lower end. Find (i) this mass, and (ii) the energy stored in

the rod.

Given for the rod, Young’s modulus = 1011 N/m2, Coefficient of linear expansion = 10-5 K-

1 and g = 10 m/s2.

Solution: (i) Given that 6 2 0 5 1100 14 10 , 0.5 , 100 , 10A m L m C k� �� � �� � � �

11 2 2 0210 / , 10 / , 0Y N m g m s C�� � �

100 0 100L L L T L�� � � �

Or, Longitudinal strain 5 3 2

100

10 100 10 /L

T N mL

� � ��� � � � ��

Hence longitudinal stress = Y × Long. Strain = 1011 × 10-3 = 108 N/m2.

� Longitudinal thermal force = Stress × C.S. area = 108 × A = 108 × 4 × 10-6 = 4 × 102 N.

If m be mass attached at the lower end of rod to prevent it from contracting, then

22 4 10

4 10 40 .10

mg or m kg

� � �

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66

(ii) Energy stored in the rod U = (1/2) × stress × strain × volume = (1/2) × 108 × 10-3 (4 × 10-6 ×

0.5) = 0.1 Joule.

Problem: A 5 m long cylindrical steel wire with radius 2 × 10-3 m is suspended vertically from a

rigid support and carries a bob of mass 100 kg at the other end. If the bob gets snapped, calculate

the change in temperature of the wire ignoring radiation losses.

(For the steel wire: Young’s Modulus = 2.1 × 1011Pa; Density = 7860 kg/m3; Specific heat

420 J/kg – K)

Solution : Heat generated = Stored Elastic Potential Eenergy

1/ 2 , ,Mg Mg

ms U stress strain volume Stress Strain volume LAA Y A

�� �

� � � � � � �� �� �

.

2

. .2

Mg LA L s

Y A# �� � [Neglecting the strain due to the weight of wire]

22 2

23 3 3 3 11 3

100 10/ 2

2 7860 2 10 2 10 10 2.110 10 10 420

Mg A Y s� #� � � � � �

� �� �� � � � �

� �� � = 0.00457 K.

Q. A force of 106 Nm-2 is required for breaking a material. If the density of the material is 3 × 103

kg/m3, than what should be the length of the wire made of material so that its breaks by its own

weight?

Solution: Weight of wire = 2r l g� #

Stress =

26

210

r l gl g

r

� ##

�� �

6 3

3

10 1034

29.43 10 9.8l m� � � �

Problem A rod AD consisting of three segements AB, BC and CD joined together is hanging

vertically from a fixed support at A. The lengths of the segements are respectively 0.1m, 0.2 m

and 0.15 m. The cross – section of the rod is uniformly 10-4 m2. A weight of 10 kg is hung from D.

Calculate the displacement of points. B, C and D if YAB

= 2.5 × 1010 N/m2, YBC

= 4 × 1010 N/m2 and

YCD

= 1 × 1010 N/m2. (Neglect the weight of the rod).

Solution : By definition of Young’s modulus,

[ ]FL MgL

L as F MgAY AY

� � � �

So for rod AB 6

1 10 4

10 9.8 0.13.92 10

2.5 10 10L m

��

� � �

And for rod BC 6

2 10 4

10 9.8 0.214.7 10

4 10 10L m�

�

� � �

And for rod CD 6

3 10 4

10 9.8 0.1514.7 10

1 10 10L m

��

� � �

So displacement of B = 61 3.92 10L m

�� �

Displacement of C = 61 2 8.82 10L L m

�� � � �

And displacement D = 61 2 3 23.5 10L L L m�� � � � � �

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67

Problem: Two rods of equal cross – seciotns, one of copper and the other of steel, are joined to

form a composite rod of length 2.0 m. At 200 C the length of the copper and is 0.5 m. When the

temperature is raised to 1200 C the length of the walls and is, thus, not allowed to expand, it is

found that the lengths of the two componenet rods also do not change with the increase in

temperature. Calculate the Young’s modulus and the coefficient of linear expansion of steel.

Given

13 2 5 01.3 10 / 1.6 10 /Cu CuY N m and C� �� �

Solution: (a) As with increase in temperature due to thermal expansion the length of a rod

changes,

i.e., ' 1L L � �� � � .

So for composite rod ' ' 1 1s C s s C CL L L L� � � �� � � � � � �

' 's C s C s s C CL L L L L L� � �� � � � � �

According to given problem,

2.002 = 2 + [0.5 × 1.6 × 10-5 + 1.5 s� ] × 100

Or 5 5 5 01.5 2 10 0.8 10 , . , 0.8 10 /s si e C� �� � �� � �

(b) When the rods are fixed between the walls and its length remains unchanged.

s C s CH CL L L L� � � � � � �

However, as here length of individual rods also remains unchaged,

s s C CH C H CL L and L L� � � � � �

Now as /H C

L L and L FL Ay� �� � � � �

So / / .s s s s C C C CL FL AY and L FL AY� � � �� � � �

Dividing one by the other / /c s s cY Y� � �

So

513 13 2

5

1.6 101.3 10 2.6 10 /

0.8 10

Cs c

s

Y Y N m��

�

�

� � �

Problem: A copper wire of negligible mass, 1 m, length and cross – sectional area 10-6 m2 is kept on

a smooth horizontal table with one end fixed. A ball of mass 1 kg is attached to the other end. The

wire and the ball are rotating with an angular velocity of 20 rad/s. If the elongation in the wire is 10-

3m, obtain the Young’s modulus. If on increasing the angular velocity to 100 rad/s, the wire breaks

down, obtain the breaking stress:

Solution: According to given problem, for vertical equilibrium of ball,

R = mg ……………… (1).

And for motion of mass m in a circle of radius r at angular frequency in a horizontal plane, the

centripetal force required F = mr 2 ………….. (2).

Here this force is provided by the elasticity of the wire,

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68

i.e.,Y A

F T LL

� � � ……………. (3).

So equating F from Eqns. (2) and (3),

2 2 2

[ ]mr L m L

F as r L LA L A L

� � � � �

� ��

So

2 2

11 2

6 3

1 1 204 10 /

10 10Y N m

� �

� �

Further as wire breaks at max = 100 rad/s,

Breaking force = mr22 4

max 1 1 100 10 N � �

And as cross – section of wire is 10-6 m2,

Breaking stress =

410 2

6

1010 /

10

Breaking forceN m

Area �� � Ans:.

Problem A stone of 0.5 kg mass is attached to one end of a 0.8 m long aluminium wire of 0.7

diameter and suspended vertically. The stone is now rotated in a horizontal plane at a rate such

that the wire makes an angle of 850 with the vertical. Find the increase in the length of the wire.

[Young’s modulus of aluminum = 7 × 1010 N/m2; sin 850 = 0.9962 and cos 850 = 0.0872]

Solution: According to given problem, for vertical equilibrium of stone,

T = cos � = mg ……………….. (1).

And for circular motion of stone in a horizontal plane,

T = sin � = mR 2 ……………….. (2).

Now, as tension in the wire is provided by its elasticity.

2

T L T LY so L

A L r Y�� � �

� ……………….. (3).

So substituting the value of T from Eqn. (1) in (3),

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69

2cos

LL mg

r� �� �

23 10

0.5 9.8 0.8

3.14 0.35 10 7 10 0.0872�

�

i.e., L� = 1.668 × 10-3 m.

Q. A load of 31.4 kg is suspended forma wire of radius 10-3 m and density 9 × 103 m-3. Calculate

the change in temperature of the wire, if 75% of the work done is converted into heat. The

Young’s modulus and heat capacity of the a material of the wire are 9.8 × 1010 Nm-2 and 490 J kg-

1 K-1 respectively.

Solution: Work done in elongation = 1

2. Stretching force × elongation

= 131.4 9.8

2

F l

AY

=

2

10

131.4 9.8 31.4

2

9.8 10

l

A

= 10

15.7 31.4 9.8

10

l

A

= Heat produced

= mst = Al # × 490 t.

Where t is rise of temperature.

3

10

15.7 31.4 9.89 10 490

10

lAl t

A

�

10 3 6 6

15.7 31.4 9.8

490 10 9 10 3.14 10 3.14 10t

� �

� �

Since 2A r��

1

90t K�

Since only 75% of energy is converted into heat. Rise of temperature.

1 3

90 4�

= 1

120K .

Problem: A body of mass 3.14 kg is changing from one end of a wire of length 10.0 m. The radius

of the wire is changing uniformly from 9.8 × 10-4 m at one end to 5.0 × 10-4 m at the other end. Find

the change in the length of the wire. What will be the change in length if the ends are interchanged?

Young’s modulus of the material of the wire is 2 × 1011 N-m2.

Solution: As radius of wire is changing uniformly hence gradient of radius will be constant over

the whole length of the wire i.e.,

2 1r rdrK

dx L

�� � ………………. (1)

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70

Consider a small length element of the wire having initial length dx and radius r. Let dl is the

extension in the length of this element due to application of force.

Now, 2

M gY

r dl��

Or dl = 2 2

M g M g drY dx

Kr Y r Y� �� �

� � � �� �

[From eq. (1)]

L=5m

5.22m

L

Hence total extension in the length of wire

22

11

2 1

22 1

1r

r

rr

r rM g dr M g M gl dl

Y K Y K r Y K r rr� � �

�� �� � � � �� �� �

But according to eq. (1), K = (r2 – r

1)/L

Hence 2 1

2 1 2 1 2 1

r rM g L M g Ll

Y r r r r Y r r� �

�� �

� �

3

11 4 4

3.14 9.8 1010

3.14 2 10 9.8 10 5 10

�� �

� �

meter.

On interchanging the ends of the wire, extension in the length remains same.

Problem: A sphere of radius 0.1 m and mass 8� is attached to the lower end of a steel wire of

length 5.0 m and diameter 10-3. The wire is suspended from 5.22 m high ceiling of a room. When the

sphere is made to swing as a simple pendulum, it just grazer the floor at the lowest point. Calculate

the velocity of the sphere at the lowest position. Young modulus of steel is 1.994 × 1011N/m2.

Solution: Mass of sphere M = 8� kg, radius of sphere r = 0.1m, Initial length of wire = L = 5m.

Extension of wire at mean position when oscillating

L� = 5.22 – (L + 2r) = 5.22 – (5 + 2 × 0.1)

= 5.22 – 5.2 = 0.02 m

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71

If T be the tension in wire at mean position when oscillating, then

/

/

T AY

L L�

�

20. Y r LY A L

TL L

� ��� � �

11 41.994 10 5 10 0.021.99.4

5

��

� � � ……………. (1)

But in mean position, T – Mg = 2Mv

R……………. (2)

Where R = radius of circular path of oscillating sphere = 5.22 – 0.1 = 5.12 m

Hence from eq. (2), we get, 199.4� - 8� × 9.8 = 8� v2/5.12 or 199.4 – 78.4 = 8v2/5.12

Or 121 = 8v2/5.12 or v2 = (5.12 × 121)/8 = 0.64 × 121

0.64 121 0.8 11 8.8 /v m s� � � �

Problem: A wire of cross – sectional area 4 × 10-4 m2, modulus of elasticity 2 × 1011 N/m-2and length

1 m is stretched between two vertical rigid poles. A mass of 1 kg is suspended at its middle.

Calculate the angle it makes with horizontal.

Solution: Given that A = 4 × 10-4 m2; 2L = 1 m,

Y = 2 × 1011 Nm-2 Nm-2, m = 1kg.

From the figure, it is clear 2T sin� = mg ………. (1)

If CD = x, then sin � = x/L,

�7

��

�7 �7

According to definition of Young’s modulus: ………. (2)

22 2 1/2

2[( ) ]

2

Y A Y A Y A xT L L x L

L L L� � � � � � ………. (3)

1/32 3

2 32 ,

2

Y A x x x mg x mgmg or or

L Y A L Y AL L

� �� � � � �

� �

Or,

1/3 1/3

11 4

1 10sin

2 10 4 10

x mg

L Y A� �

�

� � � �� � � �� � � �

� �� �

2310

5 102

��� radian

Exam Try Questions

1. A copper ring with a radius of r = 100 cm and a cross – sectional area of A = 4 mm2 is fitted

onto a steel rod with a radius R = 100.125 cm. With what force F will be ring be expanded

if the modulus of elasticity f copper E = 12,000 kgf/mm2? Disregard the deformation of

the rod.

2. What work can be performed by a steel rod with a length l and a cross – sectional area A

when heated by t digress.

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72

3. A wire with a length of 2l is stretched between two posts. A lantern with a mass M is

suspended from the middle of the wire. The cross – sectional are of the wire is A and its

modulus of elasticity E. Determine the angle � of sagging of the wire, considering it to be

small.

4. A steel rod with a cross section A = 1 cm2 is tightly fitted between two stationary absolutely

rigid walls. What force F will the rod act with on the walls if it is heated by t� = 50 C? The

coefficient of linear thermal expansion of steel � = 1.1 × 10-5 deg-1and its modulus of

elasticity E – 20,000 kgf/mm2.

5. Two rods made of different materials are placed between massive walls. The cross section

of the rods is A and their respective lengths l1 and l

2. The rods are heated by t degrees.

Find the force with the rods act on each other if their coefficients of linear thermal expansion

1 2and� � and moduli of elasticity of their materials E1 and E

2 are known. Disregard the

deformation of the walls.

6. A homogeneous block with a mass m = 100 kg on three vertical wires of equal length

arranged symmetrically. Find the tension of the wire if the middle wire is steel and the

other two are of copper. All the wire have the same cross section. Consider the modulus

of elasticity of steel to be double that of copper.

7. A reinforced – concrete column is subjected to compression by a certain load. Assuming

that the modulus of elasticity of concrete Ec is one – tenth of that of iron E

i and the cross

– sectional area of the iron is one – twentieth of that of concrete, find the portion of the

load acting on the concrete.

8. A steel bolt is inserted into a copper tube as shown in figure. Find the induced in the bolt

and in the tube when the nut turned through one revolution if the length of the tube is l,

the pitch of the bolt thread is h and the cross – sectional areas of the bolt and the tube

area Ab and A

t respectively.

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73

9. A copper plate is soldered to two steel plates as shown in figure. What tensions will arise

in the plates if the temperature is increased by t0 C? All three plates have same cross

sections.

10. Find the maximum permissible linear velocity of a rotating thin lead ring if the ultimate

strength of lead u� = 200 kgf/cm2 and its density # = 11.3 g/cm3.

11. An iron AB has both ends fixed. Hook H is fastened with two nuts in a hole in the middle

of the block. The block is clamped by the nuts with a force F0. What forces will act on the

upper and lower nuts from the side of the block if the hook carried a load whose weight

can change from zero to G =2F0? Disregard sagging of the block and the weight of the

hook.

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