Elasticity 1

AKASH MULTIMEDIA 5

PHYSICS - I C ELASTICITYELASTICITYInter atomic forces

Elastic Modulii

Behaviour of wire under stress

Elastic energy

*

*

*

*

1.1 INTRODUCTIONElasticity deals with property of a material, its

strength and ability to withstand against externalforces which are acting on it. While selecting asuitable material for a project, an engineer is alwaysinterested to know its strength. The strength of amaterial may be defined as an ability to resist itsfailure under the action of external forces. As a matterof fact the properties of a material under the actionof external forces are very essential, for an engineer,to enalbe him, in designing him all types of structuresand machines.The properties of matter like elasticity,surface tension, viscocity, can be studied well withthe help of interatomic and intermolecular forces.

The forces acting among charged particles inan atom are responsible for structure of atom. Theelectromagnetic forces acting among atoms areresponsible for the structure of molecules. Theelectromagnetic forces acting among the moleculesare responsible for the structure of matter and theirelastic behaviour.

1.2 INTERATOMIC FORCESThe forces acting between the atoms due to

electrostatic interaction between the charges ofthe atoms are called interatomic forces. Thusinteratomic forces are electrical in nature. Theinteratomic forces are active if the distance betweenthe two atoms is of the order of atomic size 1010 m .

During interaction between the two atoms, thefollowing electrostatic forces will be active(i)Attractive forces between the nucleus of one atom

and electrons of the other atom. These attractiveforces tend to decrease the potential energy of thepair of atoms.

(ii) The repuslive forces between the nucleusof one atom with the nucleus of another atom andelectrons of one atom with the electrons of the otheratom. These repulsive forces tend to increase thepotential energy of pair of atoms.

The potential energy U is related with the force

F by the relations. dUFdr

. The variation ofpotential energy U(r) and interatomic force F(r) withseparation r between two atoms have been shown ingraphs (fig (1), fig (2))

repulsion

attraction

F

rO r0

x

fig.(2)From graphs the following points are observed.

(i)At large distances, the potential energy isnegative and becomes more negative as r decrease.It implies that interatomic force in this region is

1

ELASTICITY

AKASH MULTIMEDIA 6

PHYSICS - I C

attractive. For a particular value of r denoted by x infig.(2), the attractive interatomic force becomesmaximum. After this distance x, the attractive forcestarts decreasing rapidly with the further decrease inthe value of r

(ii) At a distance r0 the potential energy attainsminimum value (maximum negative value). At thisstage, the two atoms will be in a state of equilibrium.The distance r0 is called as normal or equilibriumdistance. At this distance, the attractive force betweentwo atoms will become zero.

(iii) As the distance is further decreased belowr0, the potential energy starts increasing, becomeszero for a particular value of r and after this becomespositive. In this region, the interatomic force isrepulsive, the repulsive force increases very rapidlyas the distance between the two atoms decreases.So, the two atoms cannot be fused together easily.therefore, atoms are regarded as hard elastic spheres.

1.3 INTER MOLECULAR FORCES

The force between the molecules due toelectrostatic interaction between the charges ofthe molecules are called intermolecular forces.Thus intermolecular forces are also electrical inorigin. These forces are active if the separationbetween two molecules is of the order of molecular

size 910 m

The variation of intermolecular forces withdistance is shown in fig.

repulsion

attraction

F

rO r0

x

fig.(2)

(i) For large distnace r, the intermolecular force isneglisibly small. As the distance decreases, the forceof attraction increases. At a particular distance x,

the force of attraction becomes maximum. After thisdistance, the force of attraction decreases andbecomes zero at a distance r0. It is found that foce ofattraction between the molecules varies inversely asthe seventh power of intermolecular distance r, i.e.,

a a7 7

1 aF orFr r

The negative sign indicates that

the force is attractive in nature.

(i i) When the distance between the moleculesbecomes less than r0, the force becomes repulsive innature. The repulsive force increases very rapidlywith decrease in intermolecular distance. It is foundthat repuslive force varies inversely as the ninth

power of r, ie. r r9 91 bF or Fr r

1.4 : Some impor tant defimitions(i)Deforming Force : When an external force isapplied on a body which is not free to move, themolecules of the body are forced to undergo achange in their relative positions. Due to thischange, the body may suffer a change in length(or) volume (or) shape. Such a body is said to bedeformed. The applied force is called deformingforce.(ii) Restor ing force : The force developed withinthe body on account of r elat ive moleculardisplacement is called internal force (or) elasticforce (or) restor ing force.

At equilibrium the restoring force developed ina body is equal and opposite to the deforming forceapplied on the body.(iii) Rigid body : A body is said to be r igid if therelative posit ions of its constituent par ticlesremain unchanged inspite of any amount ofdeforming force. There is no perfect rigid body. Thenearest approach to a rigid body is diamond.(iv) Elasticity : I t is the proper ty of mater ial of abody by vir tue of which the body regains itsor iginal length, volume and shape after thedeforming forces have been removed.

AKASH MULTIMEDIA 7

PHYSICS - I C ELASTICITY

If a body regains its original length, volume andshape completely when the deforming forces areremoved, then the body is said to be a perfectly elasticbody. There is no perfectly elastic body in nature.The nearest approach to a perfectly elastic body isQuartz fiber.(v) Reason for elasticity : In a solid, atoms andmolecules are arranged in such a way that eachmolecule is acted upon by the forces due toneighbouring molecules. These forces are knownas intermolecular forces. The two molecules in theirequilibrium positions are at certain separation (r =r0) called inter molecular seperation. At thisseparation the potential energy is minimum. Onapplying the deforming forces, the molecules eithercome closer or go far apart from each other. In boththe cases potential energy of molecules is greaterthan the minimum. Since every system tends toremain in the state of minimum potential energy, themolecules has a tendency to come back to its originalposition. Tendency of the body to recover its originalconfiguration can be interpreted as due to thepresence of some forces known as restoring forcesacting in a direction opposite to that of deformingforces. This gives rise to the property of elasticity.When the deforming forces are removed, theserestoring forces bring the molecules of solid to theirrespective equilibrium positions (r = r0) and hencethe body regains its original form.(vi) Plasticity : The property of material of a bodyby vir tue of which it does not regain its or iginalshape and size (i.e it remains in the deformedstate) even after the removal of deforming forceis called plasticity.

If a body does not have any tendency to recoverits original configuration on the removal of deformingforce, then the body is said to be a perfectly plasticbody. There is no perfectly plastic body in nature,the nearest approach to a perfect plastic body is putty.

Note 1.1 : Most of the bodies are neitherperfectly elastic nor perfectly plastic. They arepartially elastic.

1.5 Stress : When a deforming force is applied on abody, there will be relative displacement of theparticles. Due to property of elasticity an internalrestoring force is developed which tends to restorethe body to its original state.Definition :The internal restoring force acting perunit area of crosssection of the deformed bodyis called stress.

Resec

storing force FStressArea of cross tion A

At equilibrium, as the restoring force is equal inmagnitude and opposite to external deforming force,stress can also be equal to external deforming forceper unit area on a body.

If F is the external deforming force appliedon the Area A of the body then

( )( )

Deforming force FStressArea A

* Stress is a tensor quantity* SI unit of stress is Pascal (N/m2)* C.G.S. unit of stress is dyne cm2.* Dimensional formula of stress is (ML1T2)* The units and dimensions of stress are sameas that of pressure.Stress developed in a body depends upon how

the external forces are applied over it. On this basisthere are two types of stresses. They are (a) Normalstress (b) Tangential ( or) shearing stress.(a) Normal Stress : I f the stress is normal to thesur face, it is called normal stress. The stress isalways normal in the case of change in length of awire (or) volume of a body.

The normal stress can further be compressive(or) tensile depending upon whether it produces adecrease (or) increase in length or volume.(i) Longitudinal stress : When a normal stresschanges the length of a body then it is calledlongitudinal stress. (or)

ELASTICITY

AKASH MULTIMEDIA 8

PHYSICS - I C

When a force is applied normal to the cross-sectional area of the body such that its lengthchanges then the restor ing force developed perunit cross-sectional area is called longitudinalstress.

Longitudinal stress = Deforming force (F)

Area of cross - section (A)It can be further divided into two types, they

are tensile stress and compressive stress.1) If a rod is stretched by two equal forces

applied normal to its crosssectional area,therestoring force developed per unit area in this case iscalled tensile stress.

F F

Body subjected to tensile force

A B C

F F

Tensile stress

A B B C

Ex : A string fixed at one end and stretched atthe other end experience tensile stress.

2) If a rod is compressed under the action ofapplied forces, the restoring force per unit area iscalled compressive stress.

F F

Body subjected to compressive force

A B C

F F

Compressive stress

A B B C

Ex: The pillars of a building experiencecompressive stress.(ii) Volume (or) Bulk stress : When a normal stresschanges the volume of a body then it is calledvolume stress.

When forces of equal magnitude act on a bodynormally from all directions, the volume of the bodychanges. The body develops internal restoring forcesthat are equal and opposite to the forces applied. Theinternal restoring force per unit area in this case isknown as volume stress.

As shown in figure, a small solid sphere is placedin a fluid such that it is compressed uniformly on allsides. The force applied by the fluid acts inperpendicular direction at each point of the surfaceand the body is said to be under hydrauliccompression. This leads to decrease in its volumewithout any change of its geometrical shape. Theinternal restoring force per unit area in this case isknown as bulk stress and is equal to the hydraulicpressure in magnitude (applied force per unit area)

V

V V

FF

F

F

F

F

F

F

Volume stress(or)

Bulk stress

Forcesurface area

FA==

= Pressure (P)(b) Shear ing stress : When the stress is tangentialto the sur face due to the application of forcesparallel to the sur face, then the stress is calledtangential (or) shear ing stress.

Shearing stress = Force

surface areaFA=

If two equal and opposite deforming forces areapplied parallel to the two surfaces of the cube asshown in the figure, there is relative displacementbetween the opposite faces of the cube. The restoringforce per unit area developed due to the appliedtangential force is known as tangential (or) shearingstress.

F

F

fixed

A

A

Note 1.2 : If deforming force is applied on a bodysuch that normal stress is developed in a body, thenthe length or volume of the body may change.

AKASH MULTIMEDIA 9


Note 1.3 : If deforming force is applied tangential tothe surface, such that tangential stress is developedin a body, then the shape of the body may change.Note 1.4 :

q

nA

F

e rF

F

When a force F acts at an angle ' 'q withoutward normal n to the area A as shown in figure.In this case, the stress will have the normal andtangential components.

To find the linear (or) longitudinal stress, takethe component of the force perpendicular to the planeof a given area A, then divide this component ( erF )by the area A.

Longitudinal stress = cos q erF F

A ATo find the shearing stress, take the component

of force parallel to the plane of the given area andthen devide elF by the area A.

Shearing stress = sin qel

F FA A

The total stress = longitudinal stress + shearing stressBut not F/A.

* Problem 1.1A steel wire of 2mm in diameter is stretched

by applying a force of 72N. Find the stress in thewire.Solution : r 31 10 m; F=72N

The stress = 22 372

1 10

F FA rp p

= 7 2672 2.292 1010

Nmp

.

1.6 Strain :When the forces (or) a torque acting upon a body

causes relative displacements of its particles, achange in length (or) volume (or) shape is produced.The body is then said to be strained.Def :The r at io of change pr oduced in thedimensions of a body by a system of forces orcouples in equilibr ium to its or iginal dimensionsis called strain.

Being the ratio of two similar quantities, strainis a dimensionless quantity and has no unit. Likestress, strain is a tensor.

Strain is classified into three types dependingupon the change produced in a body, they are

(i) Longitudinal strain (ii) Volume strain(iii) Shearing strain

(i) Longitudinal strain : It is the ratio of the changein length of a body to its or iginal length.

Consider a wire of length ' ' and is suspendedfrom a rigid support. Let a stretching force F beapplied normally to its face. Let the wire suffer achange in its length.

F

F

F

Longitudinal strain =

change in lengthoriginal length

I f the length increases due to tensile stress,the corresponding strain is called tensile strain.I f the length decreases due to compressive stress,the strain is called compressive strain.(ii) Volume strain : I t is the ratio of change involume of body to its or iginal volume.

Let v be the volume of a given body. Underthe action of a normal stress, let the change in volumeof the body be V .Then

V- V

F V

F

F

F

Volume strain = change in volume V

original volume V

ELASTICITY

AKASH MULTIMEDIA 10

PHYSICS - I C

(iii) Shear ing strain : I t is defined as angle q inradians through which a plane perpendicular tothe fixed sur face of the cubical body gets turnedunder the effect of tangential force .

I t is also the ratio of the displacement of a

layer to its distance from the fixed layer. xq

Consider a cube of material fixed at its lower

face and acted upon by a tangential force F at itsupper surface as shown in figure (a). The uppersurface is displaced relative to lower surface by xas show in the figure (b). The perpendicular distancebetween upper and lower fixed layer is ' ' . Then

F

A B

G

E H

C D

F

FixedSurface

F

fig (a)

A A BBx

F

F C DFixed

q

fig (b)

q

Shearing strain = xq

* Problem 1.2A copper wire of length 1m is stretched by

1cm. Find the strain on the wire

Solution : The strain 21 10 0.01

1eL

* Problem 1.3If a platinum wire is stretched by 0.5% what

is the strain on the wire?Solution :

Fractional increase in the lengtheL

= strain

The strain =0.5% =0.5 0.005100

.

1.7 Elastic limit : The maximum stress withinwhich a body can regain its or iginal size andshape after the removal of the deforming force iscalled elastic limit. If the stress developed in a bodyexcedes this limit, then it will not get the initial sizeand shape completely, even after the removal ofdeforming force.1.8 Hookes law: Hookes law states that withinelastic limit, the stress is directly propor tional tothe strain

stress strain (for small deformations)stress = E x (strain)

stress Estrain

Where E is proportionality constant and it is alsocalled modulus of elasticity.Def : Modulus of elasticity of the mater ial of abody is the stress in the body to produce unitstrain (within the elastic limit).E depends on the nature of the material, temperatureand impurities. It is independent of dimentions ofthe body.

SI unit of E is Pascal (Pa) 21pa 1N / m C.G.S. unit is dyne / cm2

Dimensional formula is 1 2ML T 1.9 Factors effecting Elasticity:(i) ANNEALING : The processes responsible formaking uniform structures from a given samplereduce the elasticity of a material while thoseresponsible for generating smaller regular units insidethe sample increase the elasticity. Hence annealingdecreases elasticity while hammering and rollingincreases it.(i i)IMPURITIES: The impurity having higherelasticity than the sample to which it is addedincreases the elasticity while the impurity with smallerelasticity decreases the elasticity of the sample.

(iii)TEMPERATURE: Normally, elasticity of thematerial gets decreased with rise in temperature.

AKASH MULTIMEDIA 11


However, INVAR STEEL is a material whose elasticbehaviour is not affected by rise in temperature.1.10 Types of modulii

Depending on the type of stress developedand the resulting strain, we have the followingthree modulii of elasticity.

(i) Youngs modulus(ii) Bulk modulus(iii) Modulus of Rigidity

1.11 Youngs modulus (Y) : I t is the ratio oflongitudinal stress to the longitudinal strainwithin elastic limit.

i.e., longitudinal stessYlongitudinal strain

F

F

Consider a wire of length ' ' and cross -sectional area A. One end of the wire fixed to rigidsupport and a stretching force F is applied normallyto its face as shown in the figure. Due to thestretching force, the length of the wire changes by . Then at equilibrium.

Longitudinal stress = FA

Longitudinal strain =

Y = longitudinal stresslongitudinal strain =

FFA

A

If a force is applied on a wire of radius rby hanging a load of mass M as shown in figure

then 2

2

MgMgrY Yr

pp

M

Problem 1.4Show that stress required to double the

length of wire (or) to produce 100% longitudinalstrain is numer ically equal to Youngs modulus.Sol : 1 , 2 2

2 , FAY

But as 1 , Hence

F YA

Problem 1.5A load (M) suspended from a wire produces

an elongation (e) in the wire then find the r ise intemperature required to produce same elongationin the same wire.

sol :

FAY

ta

....... 1 FAY ....... 2ta

But from given data

aa

F Ft tAY AY

Note 1.5 : If two wires having lengths 1 2, ; cross-sectional areas A1, A2 and Youngs modulii Y1, Y2are stretched by forces F1, F2 then

As 1 1 1 2 2

2 2 2 2 1

F Y F A eYAe Y F A e

Note 1.6 : In terms of volume V of a wire.

As 2 2F F F FY Y y

Ae A e Ve Ve

Where V is the volume of the wire

as m md VV d

2F dy

me

ELASTICITY

AKASH MULTIMEDIA 12

PHYSICS - I C

m mass of the wired density of material of the wire

In the above formula If V, F, Y are same for two

wires then2

2 1 12

2 2

eee

a

In the above formula if Y, d, F are same for two wires

then 22

1 1 2

2 2 1

e mem e m

a

Note 1.7 : As

2 2

F AF FVYAe A e A e

2 2 4

FV FVY YA e r ep

as m md VV d

2 4p Fm FmY Y

dA e d r eIn the above formula if F, V, Y are same for

two wires then 1eA

a (or) 41er

a

Problem 1.6When a body of mass m, density dB is

suspended from a wire, its elongation is e whenthe body is in air. I f the body is completelyimmersed in a non viscous liquid of density dthen its elongation is

In air

In liquid

F = Wa = mg 1 1

B

dF W mgd

as Fe AYe Fa (as , A, Y are same in both cases)

1 1mg

e Fe F

1B

dd

mg

1 1

B

de ed

Problem 1.7Two wires of same length and radius are

j oined end to end and loaded. The Youngsmodulii of the mater ials of the two wires are Y1and Y2. I f the combination behaves as a singlewire then its Youngs modulus is

W

e1

e2

1Y A

2Y A

W

e2AY

1 2e e e

but

1 21 2

2, ,

eq

F F Fe e eAY AY AY

1 2

2

eq

F F FAY AY AY

1 2

2 1 1

eqY Y Y

1 2

1 2

2eq

Y YYY Y

Problem 1.8Two wires of same length and radius are

j oined in par allel and loaded. The Youngsmodulii of the mater ial of the wires are Y1 & Y2.I f the combination is taken as a single wire thenits Youngs modulus is

W

1Y

A A2y

W2A

eqY

F = F1 + F2 1 22Y A e Y Ae Y Ae

1 22 eqY Y Y

1 2

2eqY YY

AKASH MULTIMEDIA 13


* Problem 1.9The length of a metal wire is 1 when the

tension in it is T1 and is 2 when the tension isT2. Then the actual length of the wire isSol : Let its original length be ' ' .

We know 1 1e T T .... (1) 2 2 T ...... (2)

From (1) & (2)1 1

2 1 2 1 2 12 2

T T T T TT

2 1 1 2 2 1T T T T 2 1 1 22 1

T TT T

Problem 1.10The length of a rubber cord is 1 metres

when the tension in it is 4N and 2 metres whenthe tension is 5N. Then the length in meters whenthe tension is 9 N is

sol : Let ' ' be the original length and 3 bethe length of the wire when the tension is 9N.

We know e F 1 4....... 1 , 2 5....... 2

3 9....... 3

From (1) & (2) 1

1 22

4 5 5 4 45

1 25 4 ....... 4

From (1) & (3)

1 1 21

3 3 1 2

5 44 49 5 4 9

1 1 2

3 1 2

5 4 45 4 9

2 1

3 1 2

4 4 45 4 9

2 1 3 1 236 36 4 20 16 2 1 3 3 2 120 16 4 5 4

* Problem 1.11A steel wire of 1mm diameter and of length

1m is stretched by applying force of 10N. I f theincrease in length is 0.064mm, find (i) the stress,(ii) the strain and (iii) the Youngs modulus of thewire.

Solution : r = 0.5 x 103m; L=1mF = 10N ; e=0.064 x 103m

i) Stress = 2F FA rp = 2

10

p =

71.273 10 N m-2

(ii) Strain =30.064 10

1

eL

= 0.064 x 103

(iii) Y= 7

3

1.273 100.064 10

StressStrain

= 1.989x1011Nm2

* Problem 1.12A steel wire of diameter 1 mm and length

2m is stretched by applying a force of 2kg wt.Calculate (i) the increase in length of the wire,(i i ) the st r ain and (i i i ) t he st r ess.(g = 9.8 ms2, Y = 2 x 1011 N m2)

Sol: 31r 10 m; L 2m;2 F = 2kg wt = 2 x 9.8N; Y = 2 x 1011 Nm2

i) 2F LYr e

223 11

FL 2 9.8 2er Y 1 10 2 10

2

= 2.495x104m

ii) The strain=4e 2.495 10

L 2

= 41.248 10iii) The stress = Y x strain

= 4

11 2.495 102 102

= 7 22.495 10 Nm

ELASTICITY

AKASH MULTIMEDIA 14

PHYSICS - I C

*Problem 1.13What mass must suspended from the free

end of a steel wire of length 2m and diameter1mm to stretch it by 1mm? (Y=2x1011Nm2)Sol: 31r 10 m; L 2m;2 2

Mg LYr e

2Y r e

MgL

211 3 312 10 10 1 10

29.8 2

210 100 8.015kg

9.8 4 39.2

*Problem 1.14A brass wire of length 5m and cross section

1mm2 is hung from a r igid support, with a brassweight of volume 1000 cm3 hanging from theother end. Find the decrease in the length of thewir e, when the br ass weight is completelyimmersed in water.(Ybrass=10

11 Nm2; g=9.8ms2; 3water 1gcm )

Sol: When a weight is hung in air from the otherend of a wire, F = Mg. The increase in length of thewire, e = ?

Young's modulus, F LYA e

MgLeAY

.

When weight hung in a liquid,Weight of the body in the liquid = Mg - V gwhere V is the volume of the bodyThis is the force, F acting on the wire i.e.,F = Mg V gIncrease in length of the wire,

/ Mg v g LeAY

which is less than the increase in length of thewire when the weight is in air.

Decrease in length = e in air - e| in liquid

Mg V g LMgLAY AY

V gLAY

Here, V = 1000cm3 = 1000 x 10-6m3 =1gcm3=1x103kgm3;g=9.8ms2;L=5m

2 6 2 11 2A 1mm 1 10 m ;Y 1 10 Nm The decrease in length

= 6 3

6 11

1000 10 1 10 9.8 51 10 1 10

= 49 x 10-5 m = 0.49 mm

*Problem 1.15A copper wire and a steel wire of radii in

the ratio 1:2, lengths in the ratio 2:1 are stretchedby the same force. I f the Young's modulus ofcopper = 1.1 x 1011Nm2 find the ratio of theirextensions(youngsmodulus of steel = 2 x 1011 N/m2).Sol: we know 2

FLer Y

2

1 1 2 2

2 2 1 1

e L r Y Fe L r Y F

Here r1:r2= 1:2, L1:L2 = 2:1 Y1=1.1x1011

Nm2; Y2 = 2.0x1011Nm2

2 111

112

e 2 2 2.0 10 16 160e 1 1 1.1 10 1.1 11

e1:e2 = 160:11

*Problem 1.16An aluminium wire and a steel wire of the

same length and cross-section are joined end toend. The composite wire is hung from a r igidsupport and a load is suspended from the freeend. I f the increase in the length of the compositewire is 2.7mm, find the increase in the length ofeach wire.(YA1=2x1011Nm2, Ysteel=7x1011Nm2)Sol: Total increase in length, e = e1 + e2.

e1 + e2 = 2.7 mm

AKASH MULTIMEDIA 15


we know FLeAY

As F, A, L are same for both the wires. So, 1eY

1 2

2 1

e Ye Y

= 11

11

2 10 207 10 7 , 1 2

20e e7

substituting in e1 + e2 = 2.7 mm

1 220 e e 2.7mm7

227e 2.7mm7

e2 = 0.7 mm

1 220 20e e 0.7 2.0mm7 7

* Problem 1.17A block of mass 1 kg is fastended to one end

of a wire of cross - sectional area 2 mm2 and isrotated in a ver tical circle of radius 20 cm. Thespeed of the block at the bottom of the circle is3.5 m s1. Find the elongation of the wire whenthe block is at the bottom.Sol: i) Tension at the bottom of the circle,

2mvT mgr

21 3.5

1 9.80.2

= 61.25 + 9.8 = 71.05 NThis tension in the string is equal to the force,F i.e. F = 71.05N, L =r = 0.2m.The increase in length ,

6 11

71.05 0.22 10 2 10

FLeAY

= 3.553 x 105mii) Tension at the top of the circle,

T = Tension at the bottom 6 mg

= 71.05 6 x1x9.8 = 71.05 58.8 = 12.25 N.

F = 12.25 N; L = 0.2 m.

The increase in length

6 11

12.25 0.22 10 2 10

FLeAY

= 50.6125 10 m

* Problem 1.18A steel wire of length 2 m and cross sectional

area 2 mm2 is fixed at one end and stretched bysuspending a block of mass 2 kg on the sur faceof the moon. I f the Ypungs modulus of steel is 2x 1011 N m2 find the increase in the length of thesteel wire.

(g on the moon 16

of the g on the earth)Sol: L = 2 m ; A = 2 mm2 = 2 x 106 m2;

M = 2 kg ; g on the moon = 29.8

6 6g ms

1122 10 ; ?

NY em

The increase in length ,

6 11

2 9.8 26 2 10 2 10

FL MgLeAY AY

= 1.633 x 105m

Problem 1.19One end of a uniform wire of length L and

mass M is attached r igidly to a point in the roofand a load of mass m is suspended from its lowerend. I f A is the area of cross - section of the wirethen find the stress in the wire at height x fromits lower end (x < L)

L

(L-x)

x

m

P

Tension in the string at point P is

T = wt of load + wt of wire of length x

MT mg xgL

Stress at P = T/A = mg MxgA AL

Problem 1.20A metal r ing of radius r and cross - sectional

area A is to be fitted on to a wooden circulardisc of radius R (R > r ). I f the Youngs modulusof the mater ial of the r ing is Y the force withwhich the metal r ing expands is

(R > r)rx

R

ELASTICITY

AKASH MULTIMEDIA 16

PHYSICS - I C

Initial length of wire 2 rpfor it to be fitted onto a wodden disc, its final

length must be 1 2 Rp

1 2e R rp , FY Ae

2 p

YAYAeF

2 pR r YA R r

rr YA R rF

r Problem 1.21

I f two wires are arranged as shown in thefigure. What are the elongations of upper andlower wiresSol : for lower wire F = m2g

2 2 2

2 2

F m ge eAy Ay

m1

m2

1

22y

1yA

AFor upper wire F = (m1 +m2)g

1 2 111 1

m m gFe eAy Ay

Problem 1.22As shown in adjucent figure if a load of mass

(m) is attached at lower end of lower wire. Thenfind the displacements of the points B, C, D are

Sol : As shown in figureelongation of first wire

11

1

mge

Ay

1

22y

1y

AD

A

m

AB

C33y

elongation of 2nd wire 2

22

mge

Ay

elongation of 3rd wire 3

33

mge

Ay

displacement of B is e1displacement of C is e1 + e2displacement of D is e1 + e2 + e3

Problem 1.23A copper wire of negligible mass, length ,

cross - sectional area (A) is kept on a smoothhorizontal table with one end fixed, a ball of massm is attached at other end. The wire and theball are rotated with angular velocity w . If wireelongates by then find Youngs modulus ofwire. I f on increasing the angular velocity fromw to w 1 the wire breakdown, obtain breakingstress Sol : a) r

F = T = mrw2

2w m as in small , 2w F m

FyAe

2w

m

yA

b) We know Breaking stress

= secBreaking force

Area of cross tion

= 21wm

A Problem 1.24

A stone of mass (m) is attached to one end ofa small wire of length and cross - sectionalarea (A) suspended ver tically. The stone is nowrotated in hor izontal plane such that the wiremakes an angle ' 'q wi th ver t ical . Find theincrease in length of wire if its Youngs modulusis Y.Sol : From fig. cos cos

qq

mgT mg T

2sin q wT mR

F TeAY AY

mg

cosT q

sinT qR

L T

S

q

q

cosmge

AY q

AKASH MULTIMEDIA 17


Note : If ' 'q was not given but R, w and m wasgiven then in such case

222 2 2 2 2cos sinT T mg mRwq q

22 2T mg mRw

and use TeAy

Problem 1.25A light rod of length 2 m is suspended from

the ceiling hor izontally by means of two ver ticalwires of equal length tied to its ends. One of thewires is made of steel and is of crosssection 103m2 and the other is of brass of cross section 2 x103m2. Find out the position along the rod atwhich a weight may be hung to produce; i) equalstress in both wires

ii) equal strains in both wiresYoungs modulus of brass = 1 x 1011 N/m2

Youngs modulus of steel = 2 x 1011 N/m2

(2 x)x

W

A B

T1 T2

Brasswire

Steelwire

C

Sol : Suppose a1 and a2 are the cross - sectionalareas, and Y1 and Y2 are the Youngs moduli of steeland brass wire respectively. Let T1 and T2 aretensions in the steel and brass wires respectively.

Let x is distance of the position of the hangingweight from the steel wire.

i) First case : For equal stress in both wires, wehave

1 2

1 2

T Ta a

(or) 1 23 310 2 10T T

(or) 2 12T T ......... (i)As the whole system is in equilibrium, so

t = 0. Taking moment of all the forces acting onthe rod about C, we have

1 2 2 0T x T x ......... (ii)Solving equations (i) and (ii), we get

43

x mii) Second case : For equal strains in both the wires

e1 = e2

1 2

1 1 2 2

T Ta Y a Y

(or) 1 23 11 3 1110 2 10 2 10 10T T

(or) T1 = T2 ........... (iii)From equations (ii) and (iii) , we getx = 1 m

Problem 1.26A steel wire of area of cross-section A and

length 2Lis clamped firmly between two pointsseperated by a distance 2' L' . A body is hung fromthe middle point of the wire such that the middlepoint sags by a distance x. Calculate the mass ofthe body and the angle made by the str ing withthe horizontal

LL

T T

Mg

x

Since ' ' is small xsin tan

L

F Ly .A e

1 22 2 /YAe YAF L x L

L L

ELASTICITY

AKASH MULTIMEDIA 18

PHYSICS - I C

21pa 1N / m 2 2

21 22

YA x YA xF L L L LL L LL

2

22 YAxF

L

2

2

222

22

T sin mgT mg for small angles sin

F mgYAx. mg

L2

2

22

YAx x. mgLL

,3

3YAXM

L g3

3 x Mg

YAL,

1 3

/x MgL YA ,

xTanL

1 3 1 31

/ /Mg MgTan TanYA YA

Problem 1.27A sphere of radius 0.1 m and mass 8p kg is

attached to the lower end of a steel wire of length5.0 m and diameter 103m. The wire is suspendedfrom 5.22 m high ceiling of a room. When thesphere is made to swing as a simple pendulum, itjust grazes the floor at its lowest point. Calculatethe velocity of the sphere at the lowest position.Y for steel = 1.994 x 1011 N/m2.

5.22m

0.2m

T

2mvCFFr

mg

Sol : As the length of the wire is 5m and diameter2 x 0.1 = 0.2 m and at lowest point it grazes the floorwhich is at a distance 5.22 m from the roof, theincrease in length of the wire at lowest point

5.22 5 0.2L = 0.02 m

So tension in the wire (due to elasticity)

211 41.994 10 5 10 0.02199.4

5YAT L NL

pp

and as equation of circular motion of a mass m tiedto a string in a vertical plane is

2 / cosmv r T mg q So at lowest point

2 /mv r T mg [as 0q ]But here r = 5 + 0.02 + 0.1 = 5.12 m

So 28 / 5.12 1.99.4 8 9.8p p p vi.e., 2 121 5.12 / 8 77.44v , so v = 8.8 m/s.1.12 Elongation of wire due to its own weigt:

Consider a wire of length ' ' and cross -sectional area A. If density of its material is d then

weight of the wire W A dg

dx

x

Let the wire is hanging from the rigid support.The wire extends due to self weightLet us consider an element of thickness dx at

a distance x from the free end.The weight of the wire of length x is w1 = (Ax)dgThe extension of the element due to this weight is

1w dx xAdg dx dgde xdxAY AY Y

The total extension 0 0

dge de x dxY

= 0

dg x dxY

, 2

2dge

Y also 2

WeAY

AKASH MULTIMEDIA 19


* Note 1.8 : The above formula can also derived byconsidering total weight at center of mass and usingeffective length / 2 .

/ 2FY

Ae

2 2

mgFeAY AY

/ 2wt

cm

A

2

dg

A

Y

, 2

2dgeY

Note 1.9 : In the above case, if a force F is appliedat the lower end in addition to its weight then thetotal elongation is

2

2dg FeY AY

1.13 Thermal Stress:

When a rod whose ends are rigidly fixed such that itis prevented from expansion or contractionundergoes a change in temperature, due to thermalexpansion or contraction a compressive or tensilestress is developed init. Due to this thermal stress therod will exert a large force on the supports. If thechange in temperature of a rod of length ' ' is oCq .Then

Thermal strain = a q

as aq

Thermal stress = Y (thermal strain)

Thermal stress = a qYForce a q F YA

Problem 1.28Two rods of different metals, having the same

area of cross - section A, are placed end to endbetween two massive walls as shown in fig. If thetemperature of both the rods are now raised by

0t C then

a) Find the force with which the rods acts oneach other at heigher temperature.b) Find the lengths of the rods at the highertemperature.

Assume that there is no change in the cross-sectional area of the rods and the rods do notbend. There is no deformation of walls.

211 1aY 2 2a YA B

Sol : a) Due to heating the increases in length of thecomposite rod will be

1 1 2 2 1 1 2 2I t t ta a a a ... (1)due to compressive force F from the walls,

due to elasticity, the decrease in length will be

1 2 1 21 2 1 2

D

F F FAY AY A Y Y

.... (2)

As the length of the composite rod remainsunchanged the increase in length due to heating mustbe equal to decrease in length due to compression.

1 21 1 2 21 2

a a Ft

A Y Y

1 1 2 21 2

1 2

a a

A tF

Y Y

b) As initially the length of one rod in L1 and due

to heating its length increases by 1 1 1H ta ,while due to compression its length decreases by

111

CFAY

so its final length

11 1 1 1H C

= 1

1 1 11

a FtAY

ELASTICITY

AKASH MULTIMEDIA 20

PHYSICS - I C

for the other rod 12 2 2 2H C

1 22 2 2 2

2

a FtAY

Note 1.10: In the above problem length of compositerod remains unchanged, i.e. 1 11 2 1 2 . butthat of individual rods changes i.e. 11 1 and

12 2 .

Note 1.11 : In the above problem if the displacementof junction point was asked,

This displacement is equal to change in lengthof any one of the rod

11 1 1

1

FtAY

a where

1 1 2 21 2

1 2

tFA

Y Y

a a

If 1 2 then 1 1 2 2

1 2

Y Y tY Y

a a

* Problem 1.29A steel wire, 2mm in diameter, is j ust

st r etched between two f i xed point s at atemperature of 300C. Determine its tension whenthe temperature falls to 200C. (Coefficient oflinear expansion of steel = 0.000011/0C; Young'smodulus for steel = 2.0 x 1011 Nm2)Sol: Thermal stress =

F Y tA

Tension in the wire 2 1Y A t t .Here,Y=2.0x1011Nm2;

6 011 10 / C;t2 = 30

0C; t1 = 200C; radius=1mm=1x103m;

22 3 6 2A r 1 10 10 m The tension in the wire

11 6 62.0 10 11 10 10 30 20 = 69.14N

1.14 Analogy of rod as a spr ing :

We know stress FY Ystrain A

(or)

AYF

= F

K

AY constant, depends on type of material

and geomentry of rod.

F k (or) (F = kx)

Where AYk is the equivalent spring

constant.

Problem 1.30A mass m is attached with rod as shown in

figure. This mass is slightly stretched and releasedthen find the time per iod. (Y is Youngs modulusof rod, A is cross sectional area of rod, ' ' is itslength).

A

m

Y

m

AYk

2 2 m mT Tk AY

p p

1.15 Bulk modulus (K) :I t is defined as the ratio of the volume stress

(normal stress) to the volume strain within theelastic limitvolume stressK

volume strain

V

V V

FF

F

F

F

F

F

F

AKASH MULTIMEDIA 21


when a solid (or) fluid is subjected to a changein pressure, its volume changes but the shape remainsunchanged. The force per unit area, applied normallyand uniformaly to the surface of the body i.e pres-sure gives the stress and the change in volume perunit volume gives the volume strain.

Thus if the volume 'V' of the body decreasesby an amount V when the pressure on its surfaceis increased uniformaly by p , then in equilibrium.

volume stress = p

volume strain V

V

Bulk modulus P PK K V

V VV

The negative sign shows that with increase in

pressure, the volume decreses.Note 1.12 : All the states of matter possess bulkmodulus

solids liquids gasesK K K Note 1.13 : Gases have two bulk modulli, they are1.16 Isothermal Bulk modulus of elasticity Eq :

The bulk modulus of a gas in isothermal con-dition is defined as isothermal Bulk modulus ofeleasticity. We know for isothermal process

PV = constantDifferentiating both sides

pdV + Vdp = 0

PdV = V dp

dpP

dVV

dpP E PdVV

q

Hence isothermal elasticity is equal topressure.

1.17 Adiabatic Bulk modulus of elasticity Ef :The bulk modulus of a gas in adiabatic conditionis defined as adiabatic Bulk modulus of elasticity.

For adiabatic process PV = constantdifferentiating both sides

1 0P V dV V dpg gg 1P V dV V dpg gg 1P V V dV V dpg gg

1p dV dpV

g

dp dpp pdV dVV V

g g E pf g

Hence, adiabatic Bulk modulus of elasticity isequal to g times pressureNote 1.14 : Ratio of adiabatic to isothermal Bulkmodulus of elasticity

E PE P

f

q

g g g > 1 E Ef q

adiabatic bulk modulus of elasticity is g times

to the isothermal bulk modulus of elasticity.

1.18 Compressibility :The reciprocal of bulk modulus is called com-

pressibility. i.e

Compressibility = 1

modBulk ulus

1 VV P

O.F = M1LT2S.I unit of compressibility is N1 m2

Note 1.15 : A rigid body and an ideal liquid are in-compressible i.e., compressibility is zero implies bulkmodulus is infinite*1.19 Density of compressed liquid :

If a liquid of density ' r ', volume V and bulkmodulus 'K' is compressed, then its density increases

density mVr

ELASTICITY

AKASH MULTIMEDIA 22

PHYSICS - I C

1 VV V

rrr -------- (1)

But by definition of bulk modulus

V P V PK

V V K ----------------(2)

from (1) and (2) P

Kr

r

1

1P PK K

r r r r rr

1 1r r

PK

Also 1 1 C Pr r where C is the com-pressibility.

* Problem 1.31A volume of 103 m3 is subjected to a

pressure of 10 atmosphere. The change in volumeis 10-6 m3. Find the bulk modulus of water.(Atmospheric pressure= 51 10 N m2 )Sol : 3 3 6 310 ; 10 ; 10V m V m P atm

= 5 6 210 10 1 10 Nm

K = P VV

V= 6

3 9 26

1 10 10 1 1010

Nm

* Problem 1.32Determine the pressure required to reduce

the given volume of water by 1 %. Bulk modu-lus of water is 92 10 N m-2

Sol : 1 ,

100V

V K= 92 10 N m-2 P = ?

PK VV

, VP K

V

= 9 712 10 2 10

100 N m -2

Problem 1.33A solid sphere of radius 'R' made of a mate-

r ial of bulk modulus B is surrounded by a liquidin a cylindr ical container. A massless piston ofarea 'A' floats on the sur face of the liquid. Findthe fractional change in the radius of the sphere

dRR

, when a mass M is placed on the piston tocompress the liquid.Sol : As for a spherical body

343

V Rp , V R3V R

Now by definition of bulk modulus

.

P V P Mg MgB V i e as PV V B AB A

13 3 dR V dR Mg

R V R AB

Problem 1.34A uniform pressure 'P' is exer ted on all sides

of a solid cube at temperature t0C . By whatamount should the temperature of the cube beraised in order to br ing its volume back to thevolume it had before the pressure was applied,if the bulk modulus and coefficient of volumeexpansion of the mater ial are B and g respec-tively.Sol : As by definition of bulk modulus

PB VV

, with increase in pressure

decrease in volume of the cube will be given

by VPVB

, (as P P )Now with rise in temperature due to thermal

expansion, volume increases so if ' q ' is the rise in

temperature then VV V as

Vg q g

q

As the volume of the cube remains constant

VP PVB B

g q qg

AKASH MULTIMEDIA 23


Problem 1.35When a rubber ball of volume V, bulk modu-

lus 'K ' is taken to a depth 'h' in water, then de-crease in its volume is

PKV

V

P VV

K

h

m

v

Pa

p h gr

m

v1aP ph g V

Kr

So decrease in volume of r h g VballK

1.20 Modulus of r igidity h (or) shear modulusWithin elastic limit, the ratio of shear ing

stress to the shear ing strain is called modulus ofr igidity of the mater ial of the body.

shearing stressshearing strain

h

F

F

fixed

A

A

L

A A1 C C1

F B

F

DFixed

xq q

consider a cube of material fixed at its lowerface and acted upon by a tangential force 'F' at itsupper surface having area A as shown in the figure.

// elF FShearing stress AAAs shown in above figure, the shearing force

'F' causes the consecutive horizontal layers of the

cube to be slightly displaced or sheared relative toone another, each line such as AB or CD in the cubeis rotated through an angle 'q ' by this shear..

The shearing strain is the angle q in radiansthrough which a line normal to a fixed surface hasturned. For small values of angle

1AA xshearing strainAB L

q

/shear stress F A Fshear strain A

hq q

* In this case shape of a body changes but its vol-ume remains unchanged.* Only solids can exhibit a shearing as these havedefinite shape.

* Problem 1.36A 5.0 cm cube of substance has its upper face

displaced by 0.65 cm, by a tangential force of 0.25N. Calculate the modulus of r igidi ty of thesubstance.

Sol : h =FLA , A = L

2, 2FL FL L

h

Here, 25.0 10L m 20.65 10 m ; F=0.25 N.

2 2

0.255.0 10 0.65 10

h =40.25 10

3.25

= 769.2 N m2

* Problem 1.37A tangential force of 2100 N is applied on a

sur face of area 63 10 m 2 which is 0.1 m froma fixed face. The force produces a shift of 7mmof upper sur face wi th r espect to bot tom.Calculate the modulus of r igidity of the material.Sol:

F = 2100 N ; A = 63 10 m2 ;

L =0.1m; 37 10 m.

6 32100 0.1

3 10 7 10FLA

h =1x10

10 Nm2

ELASTICITY

AKASH MULTIMEDIA 24

PHYSICS - I C

* Problem 1.38A steel plate of face area 2 cm2 and the

thickness 1.0cm is fixed r igidly at the lower face.A tangential force of 10 N is applied on the uppersur face . Find the lateral displacement of theupper sur face with respect to the lower sur face.Rigidity modulus of steel = 108.4 10 Nm -2Sol : 2 4 22 2 10A cm m ;

L=1.0 cm= 21 10 m; F=10 N. 108.4 10h N m-2 ; ?

FLA

h

The lateral displacement of the upper face with

respect to the lower face is FLA

h

=2

4 10

10 1 102 10 8.4 10

= 71 10

16.8 m = 95.952 10 m.

Problem 1.39Calculate the force F needed to punch a

1.46 cm diameter hole in a steel plate 1.27 cm thick(as shown in fig). The ultimate shear strength ofsteel is 345 M N/m2

Sol : As in punching, shear elasticity is involved, thehole will be punched it

11FA

ultimate shear stress

F

F11 > (shear stress) X AreaF11 min = (3.45 X 10

8) 2 rLp 2A rLp= 8 2 23.45 10 2 3.14 0.73 10 1.27 10 200KN 1.21 Some important points on modulus ofelasticity

1) Young's modulus (Y) and rigidity modulus( h ) exist only for solids but not for liquids. This isbecause liquids and gases cannot be deformed alongone dimension only and also cannot sustain ( shearstrain). Bulk modulus (K) exists for all states of mater(solids, liquids and gases)

2) Gases being most compressible are least elas-tic while solids are most elastic.

solid liquid gasE E E

Type of stress

Stress StrainChange in

shape volumeElastic

modulusName of modulus

State ofMatter

Tensileor

compressive

Two equal andopposite forcesperpendicular

to opposite faces

Elongation orcompression

parallel to forcedirection

(longitudinal strain)

Young'smodulus

Solid

/F As /L L

F LY

A L

Shearing

Two equal andopposite forces

parallel to oppositesurfaces [forces

in each case suchthat total force andtotal torque on the

body vanishes]

Pure shear, Shearmodulus Solidq F

GA

qYes No

BulkForces perpendicular

everywhere to thesurface, force per

unit area (pressure)same everywhere

Volume change(compression or

elongation)Bulk

modulusSolid, liquid

and gasNo Yes /pB

V V

/V V

YesNo

AKASH MULTIMEDIA 25


3) For a perfectly rigid body as ,L V or0f , So Y, K or h will be 'a ' i.e elasticity of a

rigid body is infinite.4) Greater the value of modulli of elasticity, more

elastic is the material.

But as 1 1,a a Y KL V and

1h af for a

constant stress, smaller change of shape or size for agiven stress corresponds to greater elasticity.ex :(1) For same load, more elongation is producedin rubber wire than in steel wire of same cross-sec-tion hence steel is more elastic than rubber.ex : (2) Water is more elastic than air as volumechange in water is less for same applied pressure

5) The value of moduli of elasticity is indepen-dent of the magnitude of the stress and strain. It de-pends on the nature of the material of the body.

6) For a given material there can be differentmoduli of elasticity depending on the type of stressapplied and the strain produced.

7) In a suspension br idge as there is a stretchin the ropes by the load of the bridge, the elasticityinvolved is linear or tensile.

8) In an automobile tyre as air is compressedthe elasticity involved is volume, i.e., bulk.

9) In transmitting power an automobile shaftis sheared as it rotates, so the elasticity involved isshear, i.e., rigidity.

10) When a coiled spr ing is stretched, the de-formation of the wire of the spring is in the form of atwisting strain so the elasticity involved is shear, i.e.,rigidity.

11) In a water lift pump as the water is com-pressed, the elasticity involved is volume, i.e., bulk

12) The shape of rubber heels changes understress, the elasticity involved is shear, or rigidity.1.22 Poisson's ratio :

When a wire is stretched by a force along itslength, then its length increases and the radius (or)diameter decreases as shown in the figure.

The ratio of change in radius (or) diameterto the or iginal radius (or) diameter is called lat-eral strain

L (D)r

r r D D

F

DLateral strainD

(or) r

r

The ratio of change in length to the originallength is called longitudinal strain

Longitudinal strain

Lateral strain is directly proportional to the lon-gitudinal strain

lateral strain a (longitudinal strain)Lateral strain = s (longitudinal strain)Where 's ' is poisson's ratio. It depends on the

nature of the material.Poisson's ratio ( s ) is defined as the ratio of lateralstrain to longitudinal strain.

' s Lateral strainpoisson s ratioLongitudinal strain

DDs

negative sign indicates that the radius or diam-eter of the wire decreases when it is stretched.

Poissons's ratio has no units and dimensions asit is ratio of two strains.

The theroetical limits of poisson's ratio are from 1 to + 0.5. But its practical limits are from 0 to 0.5and generally between 0.2 and 0.4.

1.23 Relation among volume strain, Lateral strainand poisson's ratio :

Consider a wire of length ' ' and radius 'r', thenits volume 2 (1)V rp

2V rV r

ELASTICITY

AKASH MULTIMEDIA 26

PHYSICS - I C

But we know

rrs

rr

s

2VV

s

1 2VV

s

Note 1.16 : If a material has 0.5s then 0V

V ,

0V , there in no change in the volume ofthe body and the material is said to be incompressible.

* Problem 1.40A 3 cm long copper wire is stretched to in-

crease its length by 0.3 cm. find the lateral strainin the wire, if the Poissons ratio for copper is0.26.Sol : L = 3cm ; 0.3 ; 0.26.L cm s

Longitudinal strain 0.3 0.13

LL

LateralStrainLongitudinalStrain

s

The Lateral Strain = sLongitudinalStrain= 0.26 0.1= 0.026.

* Problem 1.41Find the fractional increase in volume of a

wire of circular cross section if its longitudinalstrain is 1% . 0.30s Sol: we know that

1 2dV dLV L

s

Here, s 11% 0.01; 0.30.100dLL

The fractional increase in volume, VV 0.01(12x0.30)=0.01x0.40=0.004.

1.24 BEHAVIOUR OF A METAL WIREUNDER INCREASING LOAD

Consider a metal wire having its upper end fixedto a rigid support and loaded at the lower end byattaching a weight hanger. Let the load be increasedgradually. To study the behaviour of the metal wireunder increasing load, a graph is plotted betweenthe stress on the Y axis and the strain on the X axis. In general, the curve shown in figure is obtainedfor ductile materials that can be drawn into wires.

1. From O to P the graph is a straight line show-ing that stress is proportional to strain i.e., the wireobeys Hooke's law upto the point P. So, P is calledthe proportionality limit of the wire.2. From P to E as the graph is slightly curved, thestress is not proportional to strain. If the load isremoved at any point between O and E it will regainits natural length. The point E is called the elasticlimit. In case of some materials, the wire may obeyHooke's law upto E coinciding E with P i.e., P willbe the elastic limit for such materials.3. On increasing the load beyond elastic limit, thegraph is more curved upto the point Y called yieldpoint. From E to Y, the wire does not obey Hooke'slaw indicating that for a small increase in load thereis greater increase in length. If the load applied onthe wire is removed between E and Y, the wire doesnot regain its natural length completely. It will havea permanent increase in length. This behaviour of

AKASH MULTIMEDIA 27


the wire is shown by the dashed line which is astraight line that cuts the xaxis not at O but at O1.OO1 is the permanent set.

Strain |OO

Permanent increase in length |OO 4. When the wire crosses the point Y strainincreases rapidly without any increase in the load.So, yield point is defined as the point beyond whichstrain increases rapidly without any increase in load.Beyond the point Y as the wire becomes thin andthe stress for the same load becomes larger and largerincreasing the strain further and further. If the loadis not removed the strain increases continuously tillthe wire reaches a point T. The stress correspondingto T is called the tensile strength of the given material.The tensile strength is ratio of maximum load towhich the wire may be subjected by slowlyincreasing the load to the original area of cross-section of the wire.5. Beyond the point T, the thinning of the wire isno longer uniform and the wire shows necks.Immediately, as this occurs, the stress decreasesautomatically and the part TB is obtained. At B thewire ultimately breaks. B is called breaking point.6. If large deformation occurs between the elasticlimit and the breaking point, the material is ductile.Ex : copper, silver, gold etc.,7. If the deformation between the elastic limit andthe breaking point is very small or if the wire breaksimmediately after crossing the elastic limit, thematerial is brittle. Ex : glass, ceramic etc.1.25 BREAKING STRESS :i) The breaking stress of a wire is the maximumstress at which the wire breaks.

ii) Breaking stress = BreakingForceinitialareaofcrosssection

iii) Breaking force = Breaking stress x area of crosssection.iv) Breaking stress a) depends only on the natureof material of the wire b) is independent of the lengthand area of cross-section of the wire.v) Breaking force a) is independent of length ofthe wire b) depends on the area of cross-section andnature of material of the wire.

vi) Breaking force is proportional to area of cross-section.vii) If we cut a cable that can support a maximum load ofW into two equal parts, then each part can support amaximum load of W.viii) A very long wire suspended vertically maybreak due to its own weight

The maximum length of a wire that can hungwithout breaking under its own weight is

breaking stress = mg A g gA A

Breaking Stressg

r

Note 1.17 : A metal rope of density b has breakingstress (B:S). This rope is used to measure the depthof the sea. Then the depth of the sea that can bemeasured without breaking is

B. S. =

b

b bb

mg 1 A g 1g

A A

r rrr r

r r

bB.stress

g r r

(Whoe PL is the density of sea water)

*Problem 1.42

Find the greatest length of the wire made ofmater ial of breaking stress 8x108Nm2 anddensity 8x103kgm3 that can be suspended froma rigid support without breaking. (g = 10 ms2)Sol:

Greatest length of the wire without breaking.

Braking StressLgr

Here, breaking stress = 8 x 108 Nm2;

r = 8 x 103 kg m3; g = 10 ms2.

84

3

8 10 1 10 108 10 10

L m km

ELASTICITY

AKASH MULTIMEDIA 28

PHYSICS - I C

* Problem 1.43A block of mass 1 kg is fastended to one end

of a copper wire of cross- sectional area 1 mm2and is rotated in a ver tical circle of radius 20 cm.I f the breaking stress of copper is 5 x 108 Nm2,find the maximum number of revolutions theblock make in the minute without the str ingbreaking.Sol : Maximum tension on the string = Breakingstress x Area of cross - section

= 5x108 x1x10 6= 500 N.When a body revolves in a horizontal circle,Tension on the string = Centripetal force

500 = mr 2wWhere m = 1 kg, r = 0.20m, w = ?

500 = 1 x 0.2 x 2w2 500 2500

0.2w

Maximum angular speed, 150 rad sw .t= 60 s,n= ?

2 ntpw , 250 60

npThe maximum number of revolut ions ,

50 60 1500 477.4 .2

n rpmp p

1.26 Elastic Fatigue : When a body is subjected toa repeated stress, even within the elastic limit, itbecomes weak since it loses its elastic property tosome extent temporarily. If greater stress is appliedon the body without knowing present state, cracksdevelop within the body and it breaks. This occurseven for a stress lesser than that of breaking stress.Ex : When a metal wire is bent once, it may notbreak. But, it breaks when it is bent repeatedly at thesame point. This weakness or the state of temporaryloss of elastic nature of the body when subjected torepeated stress is called elastic fatigue.

If the material is given some rest, i.e., kept inunstrained state for some time, it regains its originalelastic nature.

1.27 ELASTIC AFTER EFFECT:When stress is removed the strain does not

reduce to zero at once. It takes some time for thestrain to become zero after the removal of stress, thedelay in recovering back to the original condition onremoval of deforming force is called elastic after-effect. This effect is very much dominant in glasswhile it is totally absent in quartz, phosphor bronze,silver and gold.1.28 Strain energy : When a wire has natural length,the potential energy corresponding to the atomic andmolecular forces is minimum. When the wire isdeformed, internal forces called restoring forces areset up and work is to be done against these forces toproduce the deformation. This work done is storedin the wire as potential energy which is called strainenergy.Strain Energy is the energy stored in a body dueto its deformation :1.29 Expression for strain energy or work donein stretching a wire : Consider a metal wire oflength L and crosssectional area A fixed at one endand is stretched by an external force applied at theother end. The force is so adjusted that the wire isonly slowly stretched. This ensures that at any timeduring the extension the external force is equal tothe tension in the wire. When the extension is e, thewire is under a longitudinal stress F/A, where F isthe tension at that instant of time. The strain is x/L.

Let the force acting on a wire suspended from arigid support be F. The work done in increasing itslength by de is

dW = FdeThe total work done in increasing the length of

wire by e is obtained by integrating the aboveexpression between the limits 0 and e.

0

e

W dW Fde =

0

e YAe deL

FLYAe

2 1

2 2 2YA e YAe e FeL L

AKASH MULTIMEDIA 29


This is stored as strain energy in the wire.

Strain energy in the wire = 12

Fe

Strain energy per unit volume of the wire

1 12 2

Fe F eAL A .

12

Stress Strain

= 21

2Stress

YStressstrain

Y

= 212 strain YWhen the external force is withdrawn, the stress

disappears and the strain energy appears as heat. Theabove relation holds good for longitudinal, volumeand shearing strains.Note 1.18:

i) Work done in stretching a wire,

w = 12 x stretching force x extension.

ii) w = 12

F e = l

YAe21 2

= 2 2

2

1 1 F2 2 p

FAY r y

iii) w =12 x stress x strain x volume of the wireiv) Area under F-e graph gives the work done

or the strain energy stored in the wire.

Area = 12

F.e = WW

* Problem 1.44I f Youngs modulus of the mater ial of a wire

is 111.2 10 N m2, calculate the work done instretching the wire of length 3 m and cross- sec-tional area 4 mm2 when it is suspended ver ticallyand a load of 8 kg is attached to its lower end.Sol : Y = 111.2 10 Nm -2 : L =3 m; A = 4mm2

= 6 24 10 ;m M= 8 kg.

work done = 12stretching force increase in

length.But increase in length,

FLe

AY

The work done = 221 1

2 2 Mg LF L

AY AY

26 11

8 9.8 312 4 10 1.2 10

= 0.0192 J

Problem 1.45A metal wire of length and cross-sectional

area A has mass m. I t is stretched by an amounte by a load of mass M. I f the wire breaks at thepoint of suspension due to the load then find ther ise in temperature of the wire ?

Strain energy = 1 1F.e Mg e2 2As the wire breaks strain energy stored in the

wire appears in the form of heat.

1 Mg e ms t2

[s = specific heat of wire]

Mg et

2 ms

Problem 1.46A stone of mass m is projected from a

rubber catpult of length and cross-sectionalarea A stretched by an amount e. I f Y be theyoungs modulus of rubber then find the velocityof projection of stone ?Solution :

Strain energy = 1 F.e2

= 1 YAe e2

21 YAe2

As the stone is released, the strain energy in thecatpult appears in the form of kinetic energy of thestone.

221 YAe 1 mv

2 2

2

2 YAevm

,

2YAevm

ELASTICITY

AKASH MULTIMEDIA 30

PHYSICS - I C

1.30 Exper imental determination of Youngsmodulus (Y) Sear les apparatus

Youngs modulus of the material of a wire canbe experimentally determined by Searles method.

Descr iption : Two wires of the same material,length and area of cross section, suspended from arigid support carry at their lower ends, tworectangular metal frames as shown in figure. One ofthe wires is called experimental wire and the otherwire is called reference wire. The frame attached tothe reference wire carries a constant weight to keepthe wire stretched without any kinks. The frameattached to the experimental wire carriers a hanger,over which slotted weights can be slipped as required.A spirit level is hinged with one end to the frameattached to the reference wire and rests horizontallyon the tip of a micrometer screw which can beworked in the frame attached to the experimental wirealong a vertical scale marked in millimeter.

Working :1) A suitable load is kept on the hanger so that

the experimental wire is straight without kinks.2) The micrometer screw is adjusted so that the

air bubble in the spirit level comes in the centre. Thereading of the micrometer screw is noted.

3) Half a kilogram weight is then slipped intothe hanger. This elongates the experimental wire. Theframe attached to the experimental wire moves down

relative to the other frame and the air bubble shifts toone side. The micrometer screw is now adjusted totake back the air bubble to the centre and themicrometer screw reading is noted.

4) The experiment is repeated at least five timesevery time increasing the load by half a kilogramweight. Readings of the micrometer screw are notedwhile increasing and decreasing the load and meanreading is found.

5) The difference between the first and secondreadings gives the increase in length or extensionproduced in the experimental wire when the load isincreased by half a kilogram weight. The differencebetween the first and third readings gives theextension for a load of one kilogram weight.Similarly, the extensions for

11 ,2 2, .... kg wt are

found.6) A graph is plotted between the load and

extension. The graph is a straight line and gives theelongation e for a load Mg.

Load

Ext

ensi

on

7) The radius, r, of the experimental wire isfound by using a screw gauge and measuring thediameter at 6 or 7 places of the wire. The length, L,of the experimental wire is measured with the helpof a meter scale.

8) Substituting the values of r, L and Mge in

the formula of Young modulus.

F LYA e

= p Mg L

r e (F = Mg and A = = 2rp )

the Youngs modulus of the material iscalculated.Sources of error and their minimisation : Thereare two sources of error in the experiment.i) The support may yield when the load is attachedat the lower end of the experimental wire and the

AKASH MULTIMEDIA 31


measured value of increase in length may not becorrect. ii) While the experiment is carried out,temperature may change which causes some increasein length. The measured value of increase in lengthbecomes incorrect.

Both the errors are minimized by using thereference wire. The yield of support or the changeof temperature affects both the experimental andreference wires. The relative increase of theexperimental wire with respect to the reference wirewill give correct increase in length.

Long Answer Questions1. Define Hooke's law of elasticity. Descr ibe

Sear les method to determine the Young'smodulus of the mater ial of a wire.

Short Answer Questions1. Def ine Hooke' s law of elast ici t y,

propor tionality l imit, permanent set andBreaking stress.

2. Define modulus of elasticity, stress, strain thePoission's ratio.

3. Define deformation and deforming force.Mention the difference between elastic andplastic bodies.

4. Descr ibe the behaviour of a wire undergradually increasing load.

5. Define Young's modulus, Bulk modulus andRigidity modulus.

6. Define stress and explain the types of stress.7. Define strain and explain the types of strain.8. Define strain energy and derive the equation

for the same.9. Steel exhibits more elastic nature than

rubber Explain.10. I f a wire is bent continuously at a par ticular

point in opposi te dir ections, it br eaks.Explain.

Very Short Answer Questions1. Define Hooke's law of elasticity.2. State the units and dimensions of stress.3. State the units and dimensions of modulus

of elasticity.4. State the units and dimensions of Young's

modulus.5. State the units and dimensions of Rigidity

modulus.6. State the units and dimensions of Bulk

modulus.7. State the examples of near ly per fectly elastic

and plastic bodies .8. State the theoretical limits of Poisson's ratio.9. State the practical limits of Poisson's ratio.10. What is strain energy? State its expression

in terms of the applied force and extension.11. Express strain energy per unit volume in

terms of stress and strain, stress and Young'smodulus.

12. What is elastic fatigue?13. What are the sources of error in Sear le's

exper iment?14. How are the errors eliminated in Sear le's

exper iment?15. State the examples of ductile and br itt le

materials.Assess Yourself

1. Why is a spr ing made of steel but not ofcopper?

Ans. As Youngs modulus of steel is greater thanthat of copper, the strain produced is small for agiven stress in case of steel than copper. Hencesteel is prefered

2. A cable is cut to half its or iginal length. Caneach par t support the same maximum loadas the or iginal cable ?

Ans. Yes. Since the breaking stress is constant for agiven material and the breaking load = breakingstress x area of cross section; the maximum loadremains the same as the area of cross sectiondoes not change.

ELASTICITY

AKASH MULTIMEDIA 32

PHYSICS - I C

12. Is there any increase in temperature when awire breaks ?

Ans.Yes. Since the strain energy is converted intoheat energy.

Additional topic's for AIEEE

** (i) Elastic hysteresis : As a result of elastic aftereffect strain in a material lags behind the stress towhich it is subjected, this phenomenon of laggingbehind of strain with respect to stress is called elastichysteresis.

(ii) Hysteresis Curve : When a ductile materialis loaded and then unloaded, stress-strain graph ofthe material is as shown in Figure. It is seen that atthe time of unloading strain is larger than that at thetime of loading for the same stress. This lagging isknown as hysteresis. The area enclosed by hysteresiscurve represents the hysteresis loss during theprocess. This energy is lost as heat. The material isselected depending upon the type of use. Forexample when rubber is used as shock absorber thenwe want that large quantity of energy of mechanicalvibrations which are impressed upon it is dissipatedas heat. In this case, rubber having the stress-straincurve as shown in Fig . is choosen. For air crafttyres where wear and tear matters, rubber whosehysteresis curve has low area, is choosen (fig.)

.Load

Extension(iii) Bending of Beam :Beam is the structural member which can carrytransverse load. A simply supported beam issupported at its ends. A cantilever beam is fixed atone end.

3. When a spr ing is stretched, what type ofstrain is produced ?

Ans. When a spring is stretched there is neitherchange in the length of the wire forming thecoil nor change in its volume. The change takesplace in the shape of coil producing shear strain.

4. Can a liquid offer resistance to shear ingstress?

Ans. No. It cannot offer permanent resistance forcesto change its shape.

5. Can a liquid offer resistance to bulk stress?Ans. Yes. It can offer very great resistance forces

tending to decrease its volume.6. Can a gas offer resistance to shearing stress?Ans. No.7. Can a gas offer great resistance to bulk stress ?Ans. No. It can offer small resistance forces tending

to decrease its volume.8. What will be the modulus of elasticity of a

r igid body?Ans. Infinity.9. Can a shear strain be expressed in terms of

tensile strain and compressive strain ?Ans.Yes. If shear strain is equal to q , tensile and

compressive strains are each equal to 2q

.

10. I f a body is per fectly incompressible whatwill be its value of Poissons ratio ?

Ans. For a body to be perfectly incompressible,

0VV .We know that 1 2VV s

. Hence 1 2 0 0.5s s .11. Br idges are declared unsafe after long use.

Explain.Ans.After long use, the bridge loses its elastic

strength. It develops large strains correspondingto the same usual values of stress and the bridgemay collapse.

AKASH MULTIMEDIA 33


(iv) Deflection of beam : Deflection of beam at itscentre due to load placed as shown in Figure.

3

48W

YId for simply supported beam and

3

3W

YId for cantilever beam where I is called

geometric moment of area.

i) For rectangular cross - section 3

12bdI

b

d

ii) For circular cross - section 4

4rI p r

(v) Twisting of a shaft :Let us consider a shaft of length and radius r,,

whose one end is rigidly clamped and torque t isapplied at the free end. Because of this the free endis twisted by and angle q .

From the diagram, arc s rq f where q angle of twist and

f angle of shear

Torsional rigidity of shaft4

2rt ph

q

where h modulus of rigidity..Note 1.19: One end of the rod is fixed. The otherfree end is twisted through an angle 'q ' by applyinga torque 't ' then the work done on the rod (or) energystored in the rod is

1W2

where is in radians.

(vi) A hollow shaft is stronger than a solid shaftmade of same mater ial and of equal volume.

Consider a solid bar or shaft of radius r, length land made of material of modulus of rigidity . Itcan be proved that the torque required to produce

unit twist in the bar is given by 4r

2pht .....(1)

If the shaft is hollow with internal and externalradii r

1 and r

2 respectively, then torque required to

produce unit twist is given by

4 42 1r r2

pht

............(2)

Length of the hollow shaft is the same from eqs(1)and (2) we have

2 2 2 24 4 2 1 2 12 14 4

r r r rr r'r r

........(3)

As the two shafts are made from equal material,of same volume. Hence

2 2 22 1r r rp p or 2 2 22 1r r r .....(4)Substituting the value of 2r in eq. (3) we get

2 2 2 2 2 22 1 2 1 2 1

22 2 22 1

r r r r r r' 1rr r r

2 2 2 2 2 22 1 2 1r r r and r r r '

Therefore, torque required to twist a hollowcylinder is more than required to twist a solid cylinder.So, a hollow shaft is more stronger than a solid shaft.Due to this reason, electric poles are made hollow.(vii) RELATI ON BETWEEN Y, n, AND K :

(OPTI ONAL )Consider a unit cube with sides parallel to the

axes OX,OY and OZ.Let the forces P,Q and R areacting along X,Y and Z axes respectively. Since thearea of each face is unit, the force acting on eachface is equal to stress. Each force produces

ELASTICITY

AKASH MULTIMEDIA 34

PHYSICS - I C

elongation in its direction and contraction in the othertwo directions.

We know that,

Youngs modulus LongitudinalstressLongitudinalStrain

Longitudinal strain LongitudinalstressYoung's modulus

Extension along X -axis = P/YLateral strain = longitudinal strain.

LateralstrainPoisson s ratio

Longitudinalstrain'

PY

Compression along Y and Z axis PY

Elongation along each of Y and Z axes PY

Similarly, the elongations in other directions aretabulated as follows.

STRAIN PRODUCED ALONG

Stress X-axis Y-axis Z-axis

P along X-axis PY

PY

s

QY

sQY

RY

s RY

P Q RY Y

s Q P RY Ys R P Q

Y Ys

PY

s

QY

s

RY

s

Q along Y-axis

R along Z-axis

when P, Q and Ract simultaneously

(i) If P = Q = R, thenElongation produced in each side

P PPY Y Y

s s 2 1 2

Longitudinal strain

Pchangein length PYoriginal length Y

1 21 2

1

But, volume strain = 3 Longitudinal strain

P

Y3 1 2

Bulk modulus Normalstressvolumestrain

P YK P

Yss

3 3 1 21 2

Y K 3 1 2 1ii) Two forces, one elongative and the othercompressive force constitute shear,i.e., If Q = P and R = 0, then shear will be producedin the cube.Now the linear strain

P PPY Y Y

0 1

Shearing strain 2 linear strain

PY2 1

Rigidity modulus Tangential stressShearing strain

Pn PY2 1

Yn 1 2

2Fom equation (1)

YK

1 2 33

and from equation 2

Yn

2 2 4

On adding equations 3 and 4

Y YK n Y n K

3 1 133 3

Y n K

1 1 1 5

3 9

n KYK n

9 63

AKASH MULTIMEDIA 35


(viii) L imiting values of Poissons ratio :From equation (3)

YK

1 23

Y and K are positive, 1 2 must be positive

or 2 1 or 12

From equation (2) Yn

12

Y and n are positive, 1 must be positive1 0 or 1 so 11

2 . But can never be

negative.

Poissons ratio lies between 0 and 12 or

102

Note : From (1) and (2) K n3 1 2 2 1 K K n ns s 3 6 2 2

3 2 6 2s s K n K n K n K n3 2 2 3

K n

K n

3 2 72 3 .

Problem 1.47A steel rod of cross-sectional area 1m2 is

acted upon by forces shown in the fig. Determinethe total elongation of the bar.Take Y = 2.0 x 1011 N/m2.

1.5m 1m 2m

50 kN20 kN10 kN60 kN

A C DB

Sol: The action of forces on each part of rod is shown in fig.

60 kN

60 kN60 kN

10 kN

50 kN

20 kN 50 kN

50 kN

60 kN

60 kN70 kN

70 kN

50 kN50 kN

We know that the extension due to external force Fis given by

ABFAY 3 7

11

60 10 1.54.5 10

1 2 10 AB m

3 711

70 10 13.5 10

1 2 10 BC m

and 3 7

11

50 10 25.0 10

1 2 10 CD m

The total extension AB BC CD = 4.5 x 107 + 3.5 x 107 + 5.0 x 107

= 13 x 107m Problem 1.48

A uniform elastic plank moves over a smoothhor izontal plane due to a constant force F0distr ibuted uniformly over the end face. Thesur face area of the end face is equal to A andYoungs modulus of the mater ial is Y. Find thecompressive strain of the plank in the directionof acting force.

xm

dx

MF0

Sol : The force at any section is due to the inertiabehind the section. The stress therefore increases fromzero to maximum at the end where force is applied.

Consider a small element of length dx at adistance x from the free end. The force

Fx = ma

= 0 0F F xM x

L M L dx

Fx Fx

Elongation of the element

0

x

F xdxF dx Ld

AY AY

Total elongation

0

0

L F x dxALY

202F LALY

(or) 02 F LAY

Problem 1.49A slightly conical wire of length and radius

r1 and r 2 is stretched by two forces appliedparallel to length in opposite directions andnormal to end faces. I f Y denotes the Youngsmodulus, then find the elongation of the wire.

ELASTICITY

AKASH MULTIMEDIA 36

PHYSICS - I C

i.e., 2 2 2 21 .......(1)

2

L

r

T A rdr A L rr w r w So here

224 2 21 110 10 400

2 2T rp

6 218 10

4r Np

Problem 1.51

The tension in the rod will not be constantbut will vary from point to point. At the free end,i.e., r = L, it will be min = 0 while at the other endr = 0, it wil be max = 62 10 Np .(b) Now if dy is the elongation in the element oflength dr at position r where tension is T, by definitionof Youngs moulus,

dy T stressas straindr AY Y

Which in the light of Eqn. (1) gives

22 21

2dy L r dr

Yrw

so the elongation of the whole rod

2 2 3

2 2

0

12 3

L LL L r drY Y

rw rw

Here 24 33

11

10 400 0.51 1 103 2 10 3

L m

SYNOPSIS

1. Rigid body : A body whose shape and sizecannot be changed, however large the appliedforce may be, is called rigid body.

There is no perfectly rigid body in nature.

2. Deformation force : A force which changes thesize or shape or both of a body without moving itas a whole is called deformation force.

3. Restor ing force : The force which restores thesize and shape of the body when deformationforces are removed is called restoring force.Deformation force and restoring force are notaction reaction pair.

Sol :

dxx

r1r2

F F

r

dx

F F

r

Consider an element of length dx at distance xas shown in fig. The radius of the sect ion

2 11x

r rr r x

The extension of the element

x

F dxd

A Y 2

x

Fdxr Yp

Total extension

20 2 1

1p

Fdxr rr x Y

Problem 1.50A thin uniform metallic rod of length 0.5 m

and radius 0.1 m rotates with an angular velocity400 rad/s in a hor izontal plane about a ver ticalaxis passing through one of its ends. Calculatetension in the rod and the elongation of the rod.The density of mater ial of the rod is 104 kg/m3and the Youngs modulus is2 x 1011 N/m|2.Sol :(a) Consider an elementof length dr at a distance rfrom the axis of rotation asshown in fig. The centripetalforce acting on this elementwill be

2 2dT dmr Adr rw r w As this force is provided by tension in the rod

(due to elasticity), so the tension in the rod at adistance r from the axis of roataion will be due to thecentripetal force due to all elemetns between x = r tox = L.centripetal force due to all elemetns between x = r tox = L.

1 2

Fr r Yp

L

drr

w

Elasticity 1

Documents

Transcript of Elasticity 1