Chapter 9 Mechanical Properties Of Solids
CHAPTER NO.9 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 ourselvea
to simpler
stiuations of rigid bodies. A rigid body generally
means a hard solid object. having a definite shape and size. But tn
reality, bodica can be stretched, compressed and
bent. Even the appreciably rigid steel bar can be deformed when a sufficiently
large external force ts applied on tt. This means that solid bodies are not
perfectly rigid.
Asolid has defintte 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 apring by gently pulling tts
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, ifyou apply force to a lump of putty or mud, they have no groas
tendency to regain their
previous shape, and they get permanently deformed.
Such substances are called plastic and this property 1s called
Plasticity. Putty and muid are close to ideal
plastics.
The elastic behaviour of materials plays an
tmportant role in engineering design. For example, while designing a
building, knowledge of clastic properties of materials
Ifke steel,
concrete etc. is essential. The same ts true in the
design of bridges, automobiles, ropeways cic. One could alsa 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 raflway track have
a particular shape like I? Why {a glass brittle while braas is not? Answers to
such questiona begin with the study of how relatively sfmple kinds of loads or
forces act to deform
different. solids bodies. In this chapter, we shall
study the elastic behaviour and mechanical properties of
solids which would anawer many such
questions.
9.2 ELASTIC BEHAVIOUR OF SOLIDS
We know that in a solid, each atom or molecule ie
surrounded by neighbouring atoms or molecules. These are bonded together by
interatomic or intermolecular forces and stay
in a stable equilforiun position. When a solid ts
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.
If you try to displace any ball from its
equilfbrium postition, the spring system trica to
restore the ball back to ite original poattion. Thus elastic behaviour of
solids can be explained in
terms of microscopic nature of the solid. Robert
Hooke, an English physicist (1635 - 1703 AD)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 ft in Section 9.4. This law, like Boyle's law, is one of the carliest
quantitattve relationships in science. It is very important to
know the behaviour of the materials under various
kinds of load from the context of engineering design.
8.3 STRESS AND STRAIN
When forces are applied on a bedy 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 bedy. This
restoring force is equal in magnitude but opposite in direction to the applied
force. The restoring force per unit arca is known as strese.
If Fis the force applied and A is the area of cross
section of the body,
Magnitude of the stress = F/A 8.1)
The SI unit of stress is Nm? or pascal (Pa)and its
dimensional formula is | MLOT? J.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 ig.9.2(a), a cytinder is stretched by two equal
forces applied nannal 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 mit area is known as
compressive stress. Tensile or compressive stress
can alse be termed as longttudinal stress.In both the cases, there is a change
in the length of the cylinder. The change in the length
AL te 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 stresa.
Robert Hooke
(1635 - 1703 AD.)
Robert Hooke was born on July 18, 1635 in
Freshwater, Isle of Wight. He waa
one of the most brilliant and versatile seventeenth
century Engiah scientists. Ea cy 5 He attended Oxford University but never
graduated. Yet he was an extremely eee.talented inventor, instrument-maker and
building designer. He assisted Robert aA Boyle in the construction of Boylean
air pump. In 1862, he was appointed aa ad a
Curator of Expertments to the newly founded Royal
Society. In 1865, he became 5 = Profeascr of Geometry in Gresham College where
he carried out his astronomi- a Ree.cal observations. He built a Gregorian
reflecting telescope; diacovered the fifth ra » & star in the trapezium and
an aateriam in the constellation Orion; suggested that * H a Jupiter rotates on
ita axis; plotted detailed sketches of Mara which were later ra a ‘ay used in the
19 century to determine the planet’s rate of rotation; stated the ;inverse
square law to deacribe planetary motion, which Newton modified later etc. He
was elected Fellow of Royal Society and also served aa the Society's Secretary
from 1667 to 1682. In his series of observations presented in Micrographia, he
suggested
wave theory of Hght and first used the word ‘cell’
in a biological context as a result of his studies of cork.Robert Hooke is best
known toc physicists for his diacovery of law of elasticity: Ut teusio, sic vis
(This is a Latin expreasion and it means as the distortion, ao the force). Thin
law laid the basin for studies of streas and strain and for understanding the
elastic materials.As a result of appHed tangential force, there is a relative
displacement Ax between opposite
faces of the cylinder as shown in the Fig. 9.2(b).
The strain so produced is known as shearing etrain
and it is defined as the ratio of relative displacement of the faces Ax to the
length of the cylinder L.
AX
Shearing strain “FT = tan @ (9.3)
where @ is the angular displacement of the cylinder
from the vertical (original position of the cylinder}. Usually @ 1s very small,
tan @ is nearly equal to angle @, (if @= 10°, for example, there is only 1%
difference between 6
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 8 = @ (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 appHed 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 is known as hydraulic stress
and in magnitude is equal to the hydraulic preasure (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 (AV) to
the original volume (V).
AV Volume strain = vy (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 x strain (9.6)where k is the
proportionality constant and is
known as moduhus 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 roportional.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 (c,) of the material.
If the load is increased further, the stress
developed exceeds the yield strength and strain increases rapidly even for a
amall 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 (c,) 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.4s 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 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
ofstress 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 (0) to the longitudinal strain (@ is defined
as Young's modulus and is denoted by the symbol Y.0 Y= : @.7) From Eqs. (9.1)
and {9.2), we have Y=(F/A/(AL/D = (Fx D /(Ax AD) (2.8)
Since strain is a dimensionless quantity, the unit
of Young's modulus is the same as that of stress i.e., N nr® or Pascal (Pa).
Table 9.1 gives the values of Young's moduli and yield strengths of some
materials.
From the data given in Table 9.1, it ia 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 ie more elastic than copper, brass and aluminium. It is for this reason that steel is
preferred in heavy-duty machines and in structural designs. Wood,
bone, concrete and
glass have rather small Young's moduli.Example 9.1 Astructural steel rod has a
radius of 10 mm and a length of 1.0m. A 100 EN force stretches ft along its length.Calculate (a) stress, (b) elongation, and (c)strain on the rod. Young’s modulus, of structural steel is 2.0 x 1071 N m7.
Answer We assume that the rod ts 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
: FF
Sttess=— = Zz
A
~ 100x10°N
3.14x( 107 m}-
=3.18 x 10° N m?
The elongation,
apa ADL
Y
( 3.18x10°N m™? dm)
* 2x10'' Nm?
=1.59x 105m
= 1.59 mm
The strain is given by
Strain = AL/L
= (1.59 x 10° m)/{1m)
= 1.59x 105
= 0.16% <
Example 9.2 A copper wire of length 2.2
m and a steel wire of length 1.6m, 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 x Young's
modulus. Therefore W/A= Y,x (AL,/L) = Y, x (AL,/L)
where the subscripts c and s refer to copper and
stainless steel respectively. Or,AL JAL, = (Y,/Y) x W/L)Given L,= 2.2 m, L,=
1.6m,From Table 9.1 Y, = 1.1 x 10"! N.m®, and Y¥,=2.0x 10" N.m*.
AL/AL, = (2.0 x 10"./1.1 x 10") x
(2.2/1.6) = 2.5.The total elongation is given to be AL, + AL, = 7.0 x 10+m
Solving the above equations,AL,=5.0 x10¢m, and
AL,=2.0 x 104m.
Therefore Ws (Ax Y, x AL}/L,=n (1.5x 1057 x (6.0 x
104 x 1,1 x 10%) /2.2]
=1.8x10?N 4
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
tabks, 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.
Answer Total maas 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 x 9.8 N = 2156 N.
Weight supported by each thighbone of the performer
= %4 (2156) N = 1078 N.
From Table 9.1, the Young's modulus for bone is
given by Y = 9.4x 10°Nm?,
Length of each thighbone L = 0.5 m
the radius of thighbone = 2.0 cm
Thus the cross-sectional area of the thighbone A =x
(2x 107)? m? = 1.26 x 10° m?.Using Eq. (9.8), the compression in each thighbone
(AJ) can be computed as
AL = [FxD/(x Al
= ((1078x0.5)/(9.4.x 10° x 1.26 x 10%]
= 4.55 x 105 m or 4.55 x 10° cm.
This is a very small change! The fractional decrease
in the thighbone is AL/L = 0.000091 or 0.0091%. <
9.6.2 Determination of Young's Modulus of the
Material of a Wire
Atypical experimental arrangement to determine the
Young's modulus ofa 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 streas. 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.)
Both the reference and experimental wires are given
an initial amall 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 rand L
be the initial radius and length of the
experimental wire, respectively. Then the areca of
cross-section of the wire would be ar*. Let M be the mass that produced an
elongation AL in the wire. Thus the applied force is equal to Mg,
where gis the acceleration due to gravity. From Eq.
(9.8), the Young’s modulus of the material of the experimental wire is given by
o Mg L
Veg Ter AL
= Mgx L/lar* x AD (9.9)
0.6.3 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 (c,) /shearing strain
G = (F/A)/(Ax/1)
= (Fx D/{Ax AX (9.10)
Similarly, from Eq. (9.4)
G =(F/A/é
= F/(Ax 4 (9.11)
The shearing stress 6, can also be expressed as
= Gxe@ (9.12)
SI unit of shear modulus is N nr 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
matorials
Exampie 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 x
104 N. The lower edge is riveted to the floor.
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 =50cmx 10cm
=0.5mx0.1m
= 0.05 m*
Therefore, the stress applied ts
= (9.4 x 10* N/0.05 m’)= 1.80 x 10° N.m?
‘We kmow that shearing strain = (Ax/L)= Stress
/G.Therefore the displacement Ax= (Stress x D/G = (1.8 x 10° N mr x 0.5m)/(5.6
x 10° N m4)=1.6x 104+ m=0.16 mm <
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/(AV/V) (9.13)
The negative sign indicates the fact that with an
increase in preasure, a decrease in volume occurs. That is, if p is positive,
AV 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 te., N m? 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/Ap) x (AV/Y) (9.14)
It can be seen from the data given in Table 9.3 that
the bulk moduli for aolids are much larger than for liquids, which are again
much larger than the bulk modulus for gases (air).
Thus solids are least compressible whereas gases are
most compressible. Gases are about a million times more compressible than
solids! Gases have
large compressthilities, which vary with pressure
and temperature. The incompreasibility 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, AV/V, of water at the bottom
of the ocean, given that the bulk modulus of water is 2.2 x 10° N m~. (Take g
=10ms”)
Answer The pressure exerted bya 3000 m
column of water on the bottom layer
P=hpg =3000m x 1000 kgm*x 1lOms?
=3 x10’ kgm's =3 x10’Nm?
Fractional compression AV/Y, is
AV/V =stress/B =(3x10’Nur9)/Q.2x10°Nor4
=1.36x10%or 1.36% <4
9.7 APPLICATIONS OF ELASTIC BEHAVIOUR
OF MATERIALS
The elastic behaviour of materials plays an important
role in everyday life. All engineering deaigns require precise knowledge of the
elastic behaviour of materials. For example while designing a buflding. 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 limft. From Table 9.1, we find that mild steel has a
yield strength (S,) of about 300 x 16° N m®. Thus, the area of
cross-section (A) of the rope should at least be
A2W/S, = Mo/S, (9.15)
= (104 kgx 10 ms /(800 x 10°N m4)
= 3.3 x 104 m?
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 thie
radius would practically be a rigid rod. So the ropes are always made of a
number of thin wires braided together, 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 trafiic, the force of winds and its own
weight. Similarly, in the design of buildings 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 ita ends as
shown in
Fig. 9.8. A bar of length 1, breadth b, and depth d
when loaded at the centre by a load W sagas by an
amount given by
&= WI? /(4bd*y) (9.16)
This relation can be derived using what you have
already learnt and a little calculus. From Eq. (8.16), 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 bis more efiective in
reducing the bending, since 6 is proportional to d~* and only to b (of course
the length i of the span should be as amall 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. 8.9(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.
Use of pillars or columns is also very common.in
buikiings 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 fimction, the cost and long period,
reliability of usable
materials etc.
The answer to the question why the maximum height of
a mountain on earth is ~10 kan 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 fpg where p is the density
of the material of the mountain and g is the acceleration 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 ia not a case
of pressure or buik compression. There is a shear component,approximately hpg
itself. Now the elastic limit
for a typical rock is 30 x 10’ N m®, Equating this
to hpg, with p = 3 x 10° kg mm gives
hpg =30x 10’ Nm. Or
nh = 30x10 Nm?/8x 10 kgm*>x10ms4
=10kn
which is more than the height of Mt. Evereatt
SUMMARY
1. Streseis the restoring force per unit area and
strain is the fractional change in dimension.In general there are three types
of streasea (a) tenaile strese — longitudinal atreas (associated with
stretching) or compressive streas (associated with compression),(b) shearing
atresa, and (c) hydraulic stress.
2. For amall deformations, streaa is directly
proportional to the strain for many materials.This is known as Hooke's law. The
constant of proportionality is called modulus of dasticity. Three elastic
moduli viz., Young's modulus, shear modulus and bulk modulus
are used to describe the elastic behaviour of
objects as they respond te deforming forces that act on them.
Aclase of solids called elastomers does not obey
Hooke’a law.
3. When an object is under tension or compreseion,
the Hooke's law takes the form F/A = YAL/L where AL/Lis the tenaile or
compressive strain of the object, F ia the magnitude of the
applied force causing the strain, A is the cross-sectional
area over which F is applied (perpendicular to A) and Yis the Young's modulus
for the object. The stress is F/A.
4. Apalr of forces when applied parallel to the
upper and lower faces, the solid deforms so thet the upper face moves sideways
with respect to the lower. The horizontal displacement AL of the upper face is
perpendicular to the vertical height L, This type of deformation is called
shear and the corresponding stress ie the shearing stress. This
type of atrese is poeaible only in solids,In this
kind of deformation the Hooke's law takea the form F/A = Gx AL/L where AL is
the displacement of one end of object in the direction of the applied force
F,and Gis the ahear 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(av/v,where p is the pressure (hydraulic stress) on the object due to the
fhiid, AV/V {the
volume strain) is the absolute fractional change in
the object's volume due to that pressure and B 1s the bulk moduhus 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 fta other end, the
force exerted by the ceiling on tt ia equal and opposite to the weight However,
the tension at arry cross-section A of the wire is just F
and not 2F. Hence, tenaile streas which is equal to
the tension per unit arca ia equal to F/A.
2. Hooke's law ia valid only in the Hncar part of
stress-strain curve.
3. The Young’s modulus and shear moduhis are
relevant only for solids since only solids have lengths and shapes.
4. Bulk modulus is relevant for solids, hquid and
gases. It refers to the change in vohime when every part of the body ie under
the uniform streas so that the ahape of the body remains unchanged.
5. Metals have larger valucs of Young's modulus than
alloys and dastomers. A material with lerge value of Young's mnodulus requires
a large force to produce small changes in ita length.
6. In daily life, we feel that a material which
stretches more is more elaatic, but it a ia misnomer. In fact material which
stretches to a leaser extent for a given load is considered
to be more clastic.
7. In general. a deforming force in one direction
can produce strains in other directions also. The proportionality between
stress and strain tn 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 amall change, which ia described by another elastic
constant of the materia) {called Poisson mito).
8. Stress is not a vector quantity simce, unlike a
force, the streas cannot be assigned a spectfic direction. Force acting on the
portion of a body on a specified aide of a section haa a defintts direction.
EXERCISES
9.1 A steel wire of length 4.7 m and crosa-sectional
area 3.0 x 10% m? stretches by the same amount as a copper wire of length 3.5 m
and crosa-sectional area of 4.0 x 10% m 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-strese curve for a
given material. What are (a) Young's modulus and {b) approximate yield strength
for this material?
9.3 The stresa-strain graphs for materials A and B
are ahown in Fig. 9.12.‘The graphe are drawn to the same scale.
(a) Which of the materials has the greater Young's
modulus?
(b} Which of the two fs 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 cofl is determined by its
shear modulus.
9.8 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 brags wire is 1.0 m. Compute the
elongationa of the steel and the brass wires.
9.8 The edge of an ahunintum 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?
8.7 Four identical hollow cylindrical columns of
mild steel support a big structure of mase 50,000 kg. The inner and outer radti
of each column are 30 and 60 an respectively.Assuming the load distribution to
be uniform, calculate the compressional strain of each column.
9.8 Apiece of copper having a rectangular
crose-section of 15.2 mm x 19.1 mm js pulled in tension with 44,500 N force,
producing only elastic deformation. Calculate the resulting
strain?
9.8 Asteel cable with a radius of 1.5 cm supports a
chairlift at a ski area. If the maximum atrese is not to exceed 10° N mm, what
is the maximum load the cable can support 7
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 masa, 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 maas is at the lowest point of ite path.
9.12 Compute the bulk modulus of water from the
following data: Initial vohime = 100.0 litre, Pressure increase = 100.0 atm (1
atm = 1.013 x 10° Pa), Final volume = 100.5 litre. Compare the bulk modulus of
water with that of air (at constant temperature).Explain in simple teams why
the ratio ia so large.
9.13 What is the density of water at a depth
where pressure is 80.0 atm, given that
ite density at the surface ia 1.03 x 103 kg m™°?
9.14 Compute the fractional change in volume of a
glass alab, when subjected to a hydraulic pressure of 10 atm.
9.18 Determine the volume contraction of a solid
copper cube, 10 am on an edge, when subjected to a hydraulic pressure of 7.0 x
10° Pa.
9.18 How much should the pressure on a litre of
water be changed to compress it by 0.10%?
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?
9.18 A rod of length 1.05 m having negligible mass
is supported at its ends by two wircs of steel (wire A) and aluminium (wire B)
of equal lengthe 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 stresecs and (b) cqual
strains in both steel and aluminium wires.
9.19 A mild steel wire of length 1.0 m and
cross-sectional area 0.50 x 10% cm? is stretched, well within its elastic
limit, horizontally between two pillars. A maes 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, cach of diameter 6.0 mm. What is the maximum tension
that can be exerted by the riveted strip if the shearing etrees on the rivet is
not to exceed 6.9 x 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 ie nearly eleven kan beneath the surface of water.
The water pressure at the bottom of the trench is about 1.1 x 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 ie the change in the volume of the ball when it
reaches to the bottom?
