CHAPTER NINE

MECHANICAL PROPERTIES OF SOLIDS

9.1 INTRODUCTION

In Chapter 7, we studied the rotation of the bodies and then

realised that the motion of a body depends on how mass is

distributed within the body. We restricted ourselves to simpler

situations of rigid bodies. A rigid body generally means a

hard solid object having a definite shape and size. But in

reality, bodies can be stretched, compressed and bent. Even

the appreciably rigid steel bar can be deformed when a

sufficiently large external force is applied on it. This means

that solid bodies are not perfectly rigid.

A solid has definite shape and size. In order to change (or

deform) the shape or size of a body, a force is required. If

you stretch a helical spring by gently pulling its ends, the

length of the spring increases slightly. When you leave the

ends of the spring, it regains its original size and shape. The

property of a body, by virtue of which it tends to regain its

original size and shape when the applied force is removed, is

known as elasticity and the deformation caused is known

as elastic deformation. However, if you apply force to a lump

of putty or mud, they have no gross tendency to regain their

previous shape, and they get permanently deformed. Such

substances are called plastic and this property is called

plasticity. Putty and mud are close to ideal plastics.

The elastic behaviour of materials plays an important role

in engineering design. For example, while designing a

building, knowledge of elastic properties of materials like steel,

concrete etc. is essential. The same is true in the design of

bridges, automobiles, ropeways etc. One could also ask —

Can we design an aeroplane which is very light but

sufficiently strong? Can we design an artificial limb which

is lighter but stronger? Why does a railway track have a

particular shape like I? Why is glass brittle while brass is

not? Answers to such questions begin with the study of how

relatively simple kinds of loads or forces act to deform

different solids bodies. In this chapter, we shall study the

9.1 Introduction

9.2 Elastic behaviour of solids

9.3 Stress and strain

9.4 Hooke’s law

9.5 Stress-strain curve

9.6 Elastic moduli

9.7 Applications of elastic

behaviour of materials

Summary

Points to ponder

Exercises

Additional exercises

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236 PHYSICS

elastic behaviour and mechanical properties of

solids which would answer many such

questions.

9.2 ELASTIC BEHAVIOUR OF SOLIDS

We know that in a solid, each atom or molecule

is surrounded by neighbouring atoms or

molecules. These are bonded together by

interatomic or intermolecular forces and stay

in a stable equilibrium position. When a solid is

deformed, the atoms or molecules are displaced

from their equilibrium positions causing a

change in the interatomic (or intermolecular)

distances. When the deforming force is removed,

the interatomic forces tend to drive them back

to their original positions. Thus the body regains

its original shape and size. The restoring

mechanism can be visualised by taking a model

of spring-ball system shown in the Fig. 9.1. Here

the balls represent atoms and springs represent

interatomic forces.

Fig. 9.1 Spring-ball model for the illustration of elastic

behaviour of solids.

If you try to displace any ball from its

equilibrium position, the spring system tries to

restore the ball back to its original position. Thus

elastic behaviour of solids can be explained in

terms of microscopic nature of the solid. Robert

Hooke, an English physicist (1635 - 1703 A.D)

performed experiments on springs and found

that the elongation (change in the length)

produced in a body is proportional to the applied

force or load. In 1676, he presented his law of

elasticity, now called Hooke’s law. We shall

study about it in Section 9.4. This law, like

Boyle’s law, is one of the earliest quantitative

relationships in science. It is very important to

know the behaviour of the materials under

various kinds of load fr

om the context of

engineering design.

9.3 STRESS AND STRAIN

When forces are applied on a body in such a

manner that the body is still in static equilibrium,

it is deformed to a small or large extent depending

upon the nature of the material of the body and

the magnitude of the deforming force. The

deformation may not be noticeable visually in

many materials but it is there. When a body is

subjected to a deforming force, a restoring force

is developed in the body. This restoring force is

equal in magnitude but opposite in direction to

the applied force. The restoring force per unit area

is known as stress. If F is the force applied normal

to the cross–section and A is the area of cross

section of the body,

Magnitude of the stress = F/A (9.1)

The SI unit of stress is N m

–2

or pascal (Pa)

and its dimensional formula is [ ML

–1

–2

There are three ways in which a solid may

change its dimensions when an external force

acts on it. These are shown in Fig. 9.2. In

Fig.9.2(a), a cylinder is stretched by two equal

forces applied normal to its cross-sectional area.

The restoring force per unit area in this case

is called tensile stress. If the cylinder is

compressed under the action of applied forces,

the restoring force per unit area is known as

compressive stress. Tensile or compressive

stress can also be termed as longitudinal stress.

In both the cases, there is a change in the

length of the cylinder. The change in the length

∆L to the original length L of the body (cylinder

in this case) is known as longitudinal strain.

Longitudinal strain

∆

(9.2)

However, if two equal and opposite deforming

forces are applied parallel to the cross-sectional

area of the cylinder, as shown in Fig. 9.2(b),

there is relative displacement between the

opposite faces of the cylinder. The restoring force

per unit area developed due to the applied

tangential force is known as tangential or

shearing stress.

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MECHANICAL PROPERTIES OF SOLIDS 237

As a result of applied tangential force, there

is a relative displacement ∆x between opposite

faces of the cylinder as shown in the Fig. 9.2(b).

The strain so produced is known as shearing

strain and it is defined as the ratio of relative

displacement of the faces ∆x to the length of

the cylinder L.

Shearing strain

∆

= tan

(9.3)

where

is the angular displacement of the

cylinder from the vertical (original position of

the cylinder). Usually

is very small, tan

is nearly equal to angle

, (if

= 10°, for

example, there is only 1% difference between

and tan

It can also be visualised, when a book is

pressed with the hand and pushed horizontally,

as shown in Fig. 9.2 (c).

Thus, shearing strain = tan

≈

(9.4)

In Fig. 9.2 (d), a solid sphere placed in the

fluid under high pressure is compressed

uniformly on all sides. The force applied by the

fluid acts in perpendicular direction at each

point of the surface and the body is said to be

under hydraulic compression. This leads to

decrease in its volume without any change of

its geometrical shape.

The body develops internal restoring forces

that are equal and opposite to the forces applied

by the fluid (the body restores its original shape

and size when taken out from the fluid). The

internal restoring force per unit area in this case

Robert Hooke

(1635 – 1703 A.D.)

Robert Hooke was born on July 18, 1635 in Freshwater, Isle of Wight. He was

one of the most brilliant and versatile seventeenth century English scientists.

He attended Oxford University but never graduated. Yet he was an extremely

talented inventor, instrument-maker and building designer. He assisted Robert

Boyle in the construction of Boylean air pump. In 1662, he was appointed as

Curator of Experiments to the newly founded Royal Society. In 1665, he became

Professor of Geometry in Gresham College where he carried out his astronomi-

cal observations. He built a Gregorian reflecting telescope; discovered the fifth

star in the trapezium and an asterism in the constellation Orion; suggested that

Jupiter rotates on its axis; plotted detailed sketches of Mars which were later

used in the 19

century to determine the planet’s rate of rotation; stated the

inverse square law to describe planetary motion, which Newton modified later

etc. He was elected Fellow of Royal Society and also served as the Society’s

Secretary from 1667 to 1682. In his series of observations presented in Micrographia, he suggested

wave theory of light and first used the word ‘cell’ in a biological context as a result of his studies of cork.

Robert Hooke is best known to physicists for his discovery of law of elasticity: Ut tensio, sic vis (This

is a Latin expression and it means as the distortion, so the force). This law laid the basis for studies of

stress and strain and for understanding the elastic materials.

(a) (b) (c) (d)

Fig. 9.2 (a) A cylindrical body under tensile stress elongates by

∆

L (b) Shearing stress on a cylinder deforming it by

an angle

every point (hydraulic stress). The volumetric strain is

∆

V/V, but there is no change in shape.

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238 PHYSICS

is known as hydraulic stress and in magnitude

is equal to the hydraulic pressure (applied force

per unit area).

The strain produced by a hydraulic pressure

is called volume strain and is defined as the

ratio of change in volume (∆V) to the original

volume (V

Volume strain

∆

(9.5)

Since the strain is a ratio of change in

dimension to the original dimension, it has no

units or dimensional formula.

9.4 HOOKE’S LAW

Stress and strain take different forms in the

situations depicted in the Fig. (9.2). For small

deformations the stress and strain are

proportional to each other. This is known as

Hooke’s law.

Thus,

stress ∝ strain

stress = k × strain (9.6)

where k is the proportionality constant and is

known as modulus of elasticity.

Hooke’s law is an empirical law and is found

to be valid for most materials. However, there

are some materials which do not exhibit this

linear relationship.

9.5 STRESS-STRAIN CURVE

The relation between the stress and the strain

for a given material under tensile stress can be

found experimentally. In a standard test of

tensile properties, a test cylinder or a wire is

stretched by an applied force. The fractional

change in length (the strain) and the applied

force needed to cause the strain are recorded.

The applied force is gradually increased in steps

and the change in length is noted. A graph is

plotted between the stress (which is equal in

magnitude to the applied force per unit area)

and the strain produced. A typical graph for a

metal is shown in Fig. 9.3. Analogous graphs

for compression and shear stress may also be

obtained. The stress-strain curves vary from

material to material. These curves help us to

understand how a given material deforms with

increasing loads. From the graph, we can see

that in the region between O to A, the curve is

linear. In this region, Hooke’s law is obeyed.

The body regains its original dimensions when

the applied force is removed. In this region, the

solid behaves as an elastic body.

In the region from A to B, stress and strain

are not proportional. Nevertheless, the body still

returns to its original dimension when the load

is removed. The point B in the curve is known

as yield point (also known as elastic limit) and

the corresponding stress is known as yield

strength (

) of the material.

If the load is increased further, the stress

developed exceeds the yield strength and strain

increases rapidly even for a small change in the

stress. The portion of the curve between B and

D shows this. When the load is removed, say at

some point C between B and D, the body does

not regain its original dimension. In this case,

even when the stress is zero, the strain is not

zero. The material is said to have a permanent

set. The deformation is said to be plastic

deformation. The point D on the graph is the

ultimate tensile strength (

) of the material.

Beyond this point, additional strain is produced

even by a reduced applied force and fracture

occurs at point E. If the ultimate strength and

fracture points D and E are close, the material

is said to be brittle. If they are far apart, the

material is said to be ductile.

As stated earlier, the stress-strain behaviour

varies from material to material. For example,

rubber can be pulled to several times its original

length and still returns to its original shape.

Fig. 9.4 shows stress-strain curve for the elastic

tissue of aorta, present in the heart. Note that

although elastic region is very large, the material

Fig. 9.3 A typical stress-strain curve for a metal.

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MECHANICAL PROPERTIES OF SOLIDS 239

does not obey Hooke’s law over most of the

region. Secondly, there is no well defined plastic

region. Substances like tissue of aorta, rubber

etc. which can be stretched to cause large strains

are called elastomers.

9.6 ELASTIC MODULI

The proportional region within the elastic limit

of the stress-strain curve (region OA in Fig. 9.3)

is of great importance for structural and

manufacturing engineering designs. The ratio

of stress and strain, called modulus of elasticity,

is found to be a characteristic of the material.

9.6.1 Young’s Modulus

Experimental observation show that for a given

material, the magnitude of the strain produced

is same whether the stress is tensile or

compressive. The ratio of tensile (or compressive)

stress (

) to the longitudinal strain (

) is defined as

Young’s modulus and is denoted by the symbol Y.

Y =

(9.7)

From Eqs. (9.1) and (9.2), we have

Y = (F/A)/(∆L/L)

= (F × L) /(A × ∆L) (9.8)

Since strain is a dimensionless quantity, the

unit of Young’s modulus is the same as that of

stress i.e., N m

–2

or Pascal (Pa). Table 9.1 gives

the values of Young’s moduli and yield strengths

of some material.

From the data given in Table 9.1, it is noticed

that for metals Young’s moduli are large.

Therefore, these materials require a large force

to produce small change in length. To increase

the length of a thin steel wire of 0.1 cm

cross-

sectional area by 0.1%, a force of 2000 N is

required. The force required to produce the same

strain in aluminium, brass and copper wires

having the same cross-sectional area are 690 N,

900 N and 1100 N respectively. It means that

steel is more elastic than copper, brass and

aluminium. It is for this reason that steel is

Fig. 9.4 Stress-strain curve for the elastic tissue of

Aorta, the large tube (vessel) carrying blood

from the heart.

Table 9.1 Young’s moduli and yield strenghs of some material

# Substance tested under compression

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240 PHYSICS

preferred in heavy-duty machines and in

structural designs. Wood, bone, concrete and

glass have rather small Young’s moduli.

Example 9.1 A structural steel rod has a

radius of 10 mm and a length of 1.0 m. A

100 kN force stretches it along its length.

Calculate (a) stress, (b) elongation, and (c)

strain on the rod. Young’s modulus, of

structural steel is 2.0 × 10

N m

-2

Answer We assume that the rod is held by a

clamp at one end, and the force F is applied at

the other end, parallel to the length of the rod.

Then the stress on the rod is given by

Stress

( )

−

100 10 N

3.14 10 m

= 3.18 × 10

N m

–2

The elongation,

(

)

F/A L

∆ =

(

)

( )

8 –2

11 –2

3.18 10 N m

2 10 N m

= 1.59 × 10

–3

= 1.59 mm

The strain is given by

Strain = ∆L/L

= (1.59 × 10

–3

m)/(1m)

= 1.59 × 10

–3

= 0.16 % t

Example 9.2 A copper wire of length 2.2

m and a steel wire of length 1.6 m, both of

diameter 3.0 mm, are connected end to end.

When stretched by a load, the net

elongation is found to be 0.70 mm. Obtain

the load applied.

Answer The copper and steel wires are under

a tensile stress because they have the same

tension (equal to the load W) and the same area

of cross-section A. From Eq. (9.7) we have stress

= strain × Young’s modulus. Therefore

W/A = Y

× (∆L

) = Y

× (∆L

)

where the subscripts c and s refer to copper

and stainless steel respectively. Or,

∆L

/∆L

= (Y

) × (L

)

Given L

= 2.2 m, L

= 1.6 m,

From Table 9.1 Y

= 1.1 × 10

N.m

–2

, and

= 2.0 × 10

N.m

–2

∆L

/∆L

= (2.0 × 10

/1.1 × 10

) × (2.2/1.6) = 2.5.

The total elongation is given to be

∆L

+ ∆L

= 7.0 × 10

-4

Solving the above equations,

∆L

= 5.0 × 10

-4

m, and ∆L

= 2.0 × 10

-4

Therefore

W = (A × Y

× ∆L

)/L

= π (1.5 × 10

-3

)

× [(5.0 × 10

-4

× 1.1 × 10

)/2.2]

= 1.8 × 10

N t

Example 9.3 In a human pyramid in a

circus, the entire weight of the balanced

group is supported by the legs of a

performer who is lying on his back (as

shown in Fig. 9.5). The combined mass of

all the persons performing the act, and the

tables, plaques etc. involved is 280 kg. The

mass of the performer lying on his back at

the bottom of the pyramid is 60 kg. Each

thighbone (femur) of this performer has a

length of 50 cm and an effective radius of

2.0 cm. Determine the amount by which

each thighbone gets compressed under the

extra load.

Fig. 9.5 Human pyramid in a circus.

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MECHANICAL PROPERTIES OF SOLIDS 241

Answer Total mass of all the performers, tables,

plaques etc. = 280 kg

Mass of the performer = 60 kg

Mass supported by the legs of the performer

at the bottom of the pyramid

= 280 – 60 = 220 kg

Weight of this supported mass

= 220 kg wt. = 220 × 9.8 N = 2156 N.

Weight supported by each thighbone of the

performer = ½ (2156) N = 1078 N.

From Table 9.1, the Young’s modulus for bone

is given by

Y = 9.4 × 10

N m

–2

Length of each thighbone L = 0.5 m

the radius of thighbone = 2.0 cm

Thus the cross-sectional area of the thighbone

A = π × (2 × 10

-2

)

= 1.26 × 10

-3

Using Eq. (9.8), the compression in each

thighbone (∆L) can be computed as

∆L = [(F × L)/(Y × A)]

= [(1078 × 0.5)/(9.4 × 10

× 1.26 × 10

-3

)]

= 4.55

× 10

-5

m or 4.55 × 10

-3

cm.

This is a very small change! The fractional

decrease in the thighbone is ∆L/L = 0.000091

or 0.0091%. t

9.6.2 Determination of Young’s Modulus of

the Material of a Wire

A typical experimental arrangement to determine

the Young’s modulus of a material of wire under

tension is shown in Fig. 9.6. It consists of two

long straight wires of same length and equal

radius suspended side by side from a fixed rigid

support. The wire A (called the reference wire)

carries a millimetre main scale M and a pan to

place a weight. The wire B (called the

experimental wire) of uniform area of cross-

section also carries a pan in which known

weights can be placed. A vernier scale V is

attached to a pointer at the bottom of the

experimental wire B, and the main scale M is

fixed to the reference wire A. The weights placed

in the pan exert a downward force and stretch

the experimental wire under a tensile stress. The

elongation of the wire (increase in length) is

measured by the vernier arrangement. The

reference wire is used to compensate for any

change in length that may occur due to change

in room temperature, since any change in length

of the reference wire due to temperature change

will be accompanied by an equal change in

experimental wire. (We shall study these

temperature effects in detail in Chapter 11.)

Fig. 9.6 An arrangement for the determination of

Young’s modulus of the material of a wire.

Both the reference and experimental wires are

given an initial small load to keep the wires

straight and the vernier reading is noted. Now

the experimental wire is gradually loaded with

more weights to bring it under a tensile stress

and the vernier reading is noted again. The

difference between two vernier readings gives

the elongation produced in the wire. Let r and L

be the initial radius and length of the

experimental wire, respectively. Then the area

of cross-section of the wire would be πr

. Let M

be the mass that produced an elongation ∆L in

the wire. Thus the applied force is equal to Mg,

where g is the acceleration due to gravity. From

Eq. (9.8), the Young’s modulus of the material

of the experimental wire is given by

Mg L

∆

= Mg × L/(πr

× ∆L) (9.9)

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242 PHYSICS

9.6.3 Shear Modulus

The ratio of shearing stress to the corresponding

shearing strain is called the shear modulus of

the material and is represented by G. It is also

called the modulus of rigidity

G = shearing stress (σ

)/shearing strain

G = (F/A)/(∆x/L)

= (F × L)/(A × ∆x) (9.10)

Similarly, from Eq. (9.4)

G = (F/A)/

= F/(A ×

) (9.11)

The shearing stress σ

can also be expressed as

= G ×

(9.12)

SI unit of shear modulus is N m

–2

or Pa. The

shear moduli of a few common materials are

given in Table 9.2. It can be seen that shear

modulus (or modulus of rigidity) is generally less

than Young’s modulus (from Table 9.1). For most

materials G ≈ Y/3.

Table 9.2 Shear moduli (G) of some common

materials

Material G (10

–2

or GPa)

Aluminium 25

Brass 36

Copper 42

Glass 23

Iron 70

Lead 5.6

Nickel 77

Steel 84

Tungsten 150

Wood 10

Example 9.4 A square lead slab of side 50

cm and thickness 10 cm is subject to a

shearing force (on its narrow face) of 9.0 ×

N. The lower edge is riveted to the floor.

How much will the upper edge be displaced?

Answer The lead slab is fixed and the force is

applied parallel to the narrow face as shown in

Fig. 9.7. The area of the face parallel to which

this force is applied is

A = 50 cm × 10 cm

= 0.5 m × 0.1 m

= 0.05 m

Therefore, the stress applied is

= (9.4 × 10

N/0.05 m

)

= 1.80 × 10

N.m

–2

Fig. 9.7

We know that shearing strain = (∆x/L)= Stress /G.

Therefore the displacement ∆x = (Stress × L)/G

= (1.8 × 10

N m

–2

× 0.5m)/(5.6 × 10

N m

–2

)

= 1.6 × 10

–4

m = 0.16 mm t

9.6.4 Bulk Modulus

In Section (9.3), we have seen that when a body

is submerged in a fluid, it undergoes a hydraulic

stress (equal in magnitude to the hydraulic

pressure). This leads to the decrease in the

volume of the body thus producing a strain called

volume strain [Eq. (9.5)]. The ratio of hydraulic

stress to the corresponding hydraulic strain is

called bulk modulus. It is denoted by symbol B.

B = – p/(∆V/V) (9.13)

The negative sign indicates the fact that with

an increase in pressure, a decrease in volume

occurs. That is, if p is positive, ∆V is negative.

Thus for a system in equilibrium, the value of

bulk modulus B is always positive. SI unit of

bulk modulus is the same as that of pressure

i.e., N m

–2

or Pa. The bulk moduli of a few

common materials are given in Table 9.3.

The reciprocal of the bulk modulus is called

compressibility and is denoted by k. It is defined

as the fractional change in volume per unit

increase in pressure.

k = (1/B) = – (1/∆p) × (∆V/V) (9.14)

It can be seen from the data given in Table

9.3 that the bulk moduli for solids are much

larger than for liquids, which are again much

larger than the bulk modulus for gases (air).

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MECHANICAL PROPERTIES OF SOLIDS 243

Gases have large compressibilities, which vary

with pressure and temperature. The

incompressibility of the solids is primarily due

to the tight coupling between the neighbouring

atoms. The molecules in liquids are also bound

with their neighbours but not as strong as in

solids. Molecules in gases are very poorly

coupled to their neighbours.

Table 9.4 shows the various types of stress,

strain, elastic moduli, and the applicable state

of matter at a glance.

Example 9.5 The average depth of Indian

Ocean is about 3000 m. Calculate the

fractional compression, ∆V/V, of water at

the bottom of the ocean, given that the bulk

modulus of water is 2.2

××

× 10

N m

–2

. (Take

g = 10 m s

–2

)

Answer The pressure exerted by a 3000 m

column of water on the bottom layer

p = h

g = 3000 m × 1000 kg m

–3

× 10 m s

–2

= 3 × 10

kg m

–1

-2

= 3 × 10

N m

–2

Fractional compression ∆V/V, is

∆V/V = stress/B = (3 × 10

N m

-2

)/(2.2 × 10

N m

–2

)

= 1.36 × 10

-2

or 1.36 % t

Table 9.3 Bulk moduli (B) of some common

Materials

Material B (10

N m

–2

or GPa)

Solids

Aluminium 72

Brass 61

Copper 140

Glass 37

Iron 100

Nickel 260

Steel 160

Liquids

Water 2.2

Ethanol 0.9

Carbon disulphide 1.56

Glycerine 4.76

Mercury 25

Gases

Air (at STP) 1.0 × 10

–4

Thus, solids are the least compressible, whereas,

gases are the most compressible. Gases are about

a million times more compressible than solids!

Table 9.4 Stress, strain and various elastic moduli

Type of Stress Strain Change in Elastic Name of State of

stress shape volume Modulus Modulus Matter

Tensile Two equal and Elongation or Yes No Y = (F×L)/ Young’s Solid

or opposite forces compression (A×∆L) modulus

compressive perpendicular to parallel to force

(σ = F/A) opposite faces direction (∆L/L)

(longitudinal strain)

Shearing Two equal and Pure shear,

Yes No G = F/(A×

) Shear Solid

(σ

= F/A) opposite forces modulus

parallel to oppoiste or modulus

surfaces forces of rigidity

in each case such

that total force and

total torque on the

body vanishes

Hydraulic Forces perpendicular Volume change No Yes B = –p/(∆V/V) Bulk Solid, liquid

everywhere to the (compression or modulus and gas

surface, force per elongation)

unit area (pressure) (∆V/V)

same everywhere.

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244 PHYSICS

9.6.5 POISSON’S RATIO

Careful observations with the Young’s modulus

experiment (explained in section 9.6.2), show

that there is also a slight reduction in the cross-

section (or in the diameter) of the wire. The strain

perpendicular to the applied force is called

lateral strain. Simon Poisson pointed out that

within the elastic limit, lateral strain is directly

proportional to the longitudinal strain. The ratio

of the lateral strain to the longitudinal strain in

a stretched wire is called Poisson’s ratio. If the

original diameter of the wire is d and the

contraction of the diameter under stress is ∆d,

the lateral strain is ∆d/d. If the original length

of the wire is L and the elongation under stress

is ∆L, the longitudinal strain is ∆L/L. Poisson’s

ratio is then (∆d/d)/(∆L/L) or (∆d/∆L) × (L/d).

Poisson’s ratio is a ratio of two strains; it is a

pure number and has no dimensions or units.

Its value depends only on the nature of material.

For steels the value is between 0.28 and 0.30,

and for aluminium alloys it is about 0.33.

9.6.6 Elastic Potential Energy

in a Stretched Wire

When a wire is put under a tensile stress, work

is done against the inter-atomic forces. This

work is stored in the wire in the form of elastic

potential energy. When a wire of original length

L and area of cross-section A is subjected to a

deforming force F along the length of the wire,

let the length of the wire be elongated by l. Then

from Eq. (9.8), we have F = YA × (l/L). Here Y is

the Young’s modulus of the material of the wire.

Now for a further elongation of infinitesimal

small length dl, work done dW is F ×

dl or YAldl/

L. Therefore, the amount of work done (W) in

increasing the length of the wire from L to L + l,

that is from l = 0 to l = l is

W =

= ×

∫

YAl YA l

L L

W =

 

× × ×

 

 

Y AL

Young’s modulus × strain

volume of the wire

stress × strain × volume of the

wire

This work is stored in the wire in the form of

elastic potential energy (U). Therefore the elastic

potential energy per unit volume of the wire (u) is

u =

σ ε

(9.15)

9.7 APPLICATIONS OF ELASTIC

BEHAVIOUR OF MATERIALS

The elastic behaviour of materials plays an

important role in everyday life. All engineering

designs require precise knowledge of the elastic

behaviour of materials. For example while

designing a building, the structural design of

the columns, beams and supports require

knowledge of strength of materials used. Have

you ever thought why the beams used in

construction of bridges, as supports etc. have

a cross-section of the type I? Why does a heap

of sand or a hill have a pyramidal shape?

Answers to these questions can be obtained

from the study of structural engineering which

is based on concepts developed here.

Cranes used for lifting and moving heavy

loads from one place to another have a thick

metal rope to which the load is attached. The

rope is pulled up using pulleys and motors.

Suppose we want to make a crane, which has

a lifting capacity of 10 tonnes or metric tons (1

metric ton = 1000 kg). How thick should the

steel rope be? We obviously want that the load

does not deform the rope permanently.

Therefore, the extension should not exceed the

elastic limit. From Table 9.1, we find that mild

steel has a yield strength (

) of about 300 ×

N m

–2

. Thus, the area of cross-section (A)

of the rope should at least be

A ≥ W/

= Mg/

(9.16)

= (10

kg × 9.8 m s

-2

)/(300 × 10

N m

-2

)

= 3.3 × 10

-4

corresponding to a radius of about 1 cm for

a rope of circular cross-section. Generally

a large margin of safety (of about a factor of

ten in the load) is provided. Thus a thicker

rope of radius about 3 cm is recommended.

A single wire of this radius would practically

be a rigid rod. So the ropes are always made

of a number of thin wires braided together,

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MECHANICAL PROPERTIES OF SOLIDS 245

like in pigtails, for ease in manufacture,

flexibility and strength.

A bridge has to be designed such that it can

withstand the load of the flowing traffic, the force

of winds and its own weight. Similarly, in the

design of buildings the use of beams and columns

is very common. In both the cases, the

overcoming of the problem of bending of beam

under a load is of prime importance. The beam

should not bend too much or break. Let us

consider the case of a beam loaded at the centre

and supported near its ends as shown in

Fig. 9.8. A bar of length l, breadth b, and depth d

when loaded at the centre by a load W sags by

an amount given by

= W l

/(4bd

Y) (9.17)

Fig. 9.8 A beam supported at the ends and loaded

at the centre.

This relation can be derived using what you

have already learnt and a little calculus. From

Eq. (9.16), we see that to reduce the bending

for a given load, one should use a material with

a large Young’s modulus Y. For a given material,

increasing the depth d rather than the breadth

b is more effective in reducing the bending, since

is proportional to d

-3

and only to b

-1

(of course

the length l of the span should be as small as

possible). But on increasing the depth, unless

the load is exactly at the right place (difficult to

arrange in a bridge with moving traffic), the

deep bar may bend as shown in Fig. 9.9(b). This

is called buckling. To avoid this, a common

compromise is the cross-sectional shape shown

in Fig. 9.9(c). This section provides a large load-

bearing surface and enough depth to prevent

bending. This shape reduces the weight of the

beam without sacrificing the strength and

hence reduces the cost.

(a) (b) (c)

Fig. 9.9 Different cross-sectional shapes of a beam.

(a) Rectangular section of a bar;

(b) A thin bar and how it can buckle;

bearing bar.

The use of pillars or columns is also very

common in buildings and bridges. A pillar with

rounded ends as shown in Fig. 9.10(a) supports

less load than that with a distributed shape at

the ends [Fig. 9.10(b)]. The precise design of a

bridge or a building has to take into account

the conditions under which it will function, the

cost and long period, reliability of usable

material, etc.

(a) (b)

Fig. 9.10 Pillars or columns: (a) a pillar with rounded

ends, (b) Pillar with distributed ends.

The answer to the question why the maximum

height of a mountain on earth is ~10 km can

also be provided by considering the elastic

properties of rocks. A mountain base is not under

uniform compression and this provides some

shearing stress to the rocks under which they

can flow. The stress due to all the material on

the top should be less than the critical shearing

stress at which the rocks flow.

At the bottom of a mountain of height h, the

force per unit area due to the weight of the

mountain is h

g where

is the density of the

material of the mountain and g is the acceleration

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246 PHYSICS

SUMMARY

1. Stress is the restoring force per unit area and strain is the fractional change in dimension.

In general there are three types of stresses (a) tensile stress — longitudinal stress

(associated with stretching) or compressive stress (associated with compression),

(b) shearing stress, and (c) hydraulic stress.

2. For small deformations, stress is directly proportional to the strain for many materials.

This is known as Hooke’s law. The constant of proportionality is called modulus of

elasticity. Three elastic moduli viz., Young’s modulus, shear modulus and bulk modulus

are used to describe the elastic behaviour of objects as they respond to deforming forces

that act on them.

A class of solids called elastomers does not obey Hooke’s law.

3. When an object is under tension or compression, the Hooke’s law takes the form

F/A = Y∆L/L

where ∆L/L is the tensile or compressive strain of the object, F is the magnitude of the

applied force causing the strain, A is the cross-sectional area over which F is applied

(perpendicular to A) and Y is the Young’s modulus for the object. The stress is F/A.

4. A pair of forces when applied parallel to the upper and lower faces, the solid deforms so

that the upper face moves sideways with respect to the lower. The horizontal displacement

∆L of the upper face is perpendicular to the vertical height L.

This type of deformation is

called shear and the corresponding stress is the shearing stress. This type of stress is

possible only in solids.

In this kind of deformation the Hooke’s law takes the form

F/A = G × ∆L/L

where ∆L is the displacement of one end of object in the direction of the applied force F,

and G is the shear modulus.

5. When an object undergoes hydraulic compression due to a stress exerted by a surrounding

fluid, the Hooke’s law takes the form

p = B (∆V/V),

where p is the pressure (hydraulic stress) on the object due to the fluid, ∆V/V (the

volume strain) is the absolute fractional change in the object’s volume due to that

pressure and B

is the bulk modulus

of the object.

POINTS TO PONDER

1. In the case of a wire, suspended from celing and stretched under the action of a weight (F)

suspended from its other end, the force exerted by the ceiling on it is equal and opposite

to the weight. However, the tension at any cross-section A of the wire is just F and not

2F. Hence, tensile stress which is equal to the tension per unit area is equal to F/A.

2. Hooke’s law is valid only in the linear part of stress-strain curve.

3. The Young’s modulus and shear modulus are relevant only for solids since only solids

have lengths and shapes.

4. Bulk modulus is relevant for solids, liquid and gases. It refers to the change in volume

when every part of the body is under the uniform stress so that the shape of the body

remains unchanged.

due to gravity. The material at the bottom

experiences this force in the vertical direction,

and the sides of the mountain are free. Therefore,

this is not a case of pressure or bulk compression.

There is a shear component, approximately h

itself. Now the elastic limit for a typical rock is

30 × 10

N m

-2

. Equating this to h

g, with

= 3 × 10

kg m

-3

gives

g = 30 × 10

N m

-2

h = 30 × 10

N m

-2

/(3 × 10

kg m

-3

× 10 m s

-2

)

= 10 km

which is more than the height of Mt. Everest!

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MECHANICAL PROPERTIES OF SOLIDS 247

5. Metals have larger values of Young’s modulus than alloys and elastomers. A material

with large value of Young’s modulus requires a large force to produce small changes in

its length.

6. In daily life, we feel that a material which stretches more is more elastic, but it a is

misnomer. In fact material which stretches to a lesser extent for a given load is considered

to be mor

e elastic.

7. In general, a deforming force in one direction can produce strains in other directions

also. The proportionality between stress and strain in such situations cannot be described

by just one elastic constant. For example, for a wire under longitudinal strain, the

lateral dimensions (radius of cross section) will undergo a small change, which is described

by another elastic constant of the material (called Poisson ratio).

8. Stress is not a vector quantity since, unlike a force, the stress cannot be assigned a

specific direction. Force acting on the portion of a body on a specified side of a section

has a definite direction.

EXERCISES

9.1 A steel wire of length 4.7 m and cross-sectional area 3.0 × 10

-5

stretches by the same

amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10

–5

under

a given load. What is the ratio of the Young’s modulus of steel to that of copper?

9.2 Figure 9.11 shows the strain-stress curve for a given material. What are (a) Young’s

modulus and (b) approximate yield strength for this material?

Fig. 9.11

9.3 The stress-strain graphs for materials A and B are shown in Fig. 9.12.

Fig. 9.12

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248 PHYSICS

The graphs are drawn to the same scale.

(a) Which of the materials has the greater Young’s modulus?

(b) Which of the two is the stronger material?

9.4 Read the following two statements below carefully and state, with reasons, if it is true

or false.

(a) The Young’s modulus of rubber is greater than that of steel;

(b) The stretching of a coil is determined by its shear modulus.

9.5 Two wires of diameter 0.25 cm, one made of steel and the other made of brass are

loaded as shown in Fig. 9.13. The unloaded length of steel wire is 1.5 m and that of

brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.

Fig. 9.13

9.6 The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a

vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The

shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?

9.7 Four identical hollow cylindrical columns of mild steel support a big structure of mass

50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively.

Assuming the load distribution to be uniform, calculate the compressional strain of

each column.

9.8 A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in

tension with 44,500 N force, producing only elastic deformation. Calculate the resulting

strain?

9.9 A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum

stress is not to exceed 10

N m

–2

, what is the maximum load the cable can support ?

9.10 A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long.

Those at each end are of copper and the middle one is of iron. Determine the ratios of

their diameters if each is to have the same tension.

9.11 A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is

whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle.

The cross-sectional area of the wire is 0.065 cm

. Calculate the elongation of the wire

when the mass is at the lowest point of its path.

9.12 Compute the bulk modulus of water from the following data: Initial volume = 100.0

litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 10

Pa), Final volume = 100.5

litre. Compare the bulk modulus of water with that of air (at constant temperature).

Explain in simple terms why the ratio is so large.

9.13 What is the density of water at a depth wher

e pressure is 80.0 atm, given that its

density at the surface is 1.03 × 103 kg m

–3

9.14 Compute the fractional change in volume of a glass slab, when subjected to a hydraulic

pressure of 10 atm.

9.15 Determine the volume contraction of a solid copper cube, 10 cm on an edge, when

subjected to a hydraulic pressure of 7.0 × 10

Pa.

9.16 How much should the pressure on a litre of water be changed to compress it by 0.10%?

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MECHANICAL PROPERTIES OF SOLIDS 249

Additional Exercises

9.17 Anvils made of single crystals of diamond, with the shape as shown in

Fig. 9.14, are used to investigate behaviour of materials under very high pressures. Flat

faces at the narrow end of the anvil have a diameter of 0.50 mm, and the wide ends are

subjected to a compressional force of 50,000 N. What is the pressure at the tip of the anvil?

Fig. 9.14

9.18 A rod of length 1.05 m having negligible mass is supported at its ends by two wires of

steel (wire A) and aluminium (wire B) of equal lengths as shown in

Fig. 9.15. The cross-sectional areas of wires A and B are 1.0 mm

and 2.0 mm

respectively. At what point along the rod should a mass m be suspended in order to

produce (a) equal stresses and (b) equal strains in both steel and aluminium wires.

Fig. 9.15

9.19 A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10

-2

stretched, well within its elastic limit, horizontally between two pillars. A mass of 100

g is suspended from the mid-point of the wire. Calculate the depression at the mid-

point.

9.20 Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0

mm. What is the maximum tension that can be exerted by the riveted strip if the

shearing stress on the rivet is not to exceed 6.9 × 10

Pa? Assume that each rivet is to

carry one quarter of the load.

9.21 The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven

km beneath the surface of water. The water pressure at the bottom of the trench is

about 1.1 × 10

Pa. A steel ball of initial volume 0.32 m

is dropped into the ocean and

falls to the bottom of the trench. What is the change in the volume of the ball when it

reaches to the bottom?

2020-21