Introduction
Elasticity is the property of material body by virtue of which its
regain its original configuration, when external force is removed is
called elasticity.
The property of the material body by virtue of which it does not
regain its original configuration when the external force is removed is
called plasticity.
Cause of Elasticity
In a solid atoms and molecules are arranged in such a way that each
molecule is acted upon by the forces due to the neighbouring molecules.
When no deforming force is applied on the body, each molecule of the
solid is in its equilibrium position and the inter molecular forces of
the solid are maximum. On applying deforming force, the molecules are
displaced from their equilibrium position. Inter molecular force gets
changed and restoring forces are developed. It is explained by using
spring- ball model. Deforming force is removed, these restoring force
bring the molecule to its equilibrium positions. Thus the body regains
its original shape and size.
Spring-ball model for the illustration of elastic behaviour of
solids.
The restoring mechanism can be visualised by taking a model of
spring-ball system shown above. Here the balls represent atoms and
springs represent interatomic forces.
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. 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
froce.
Stress( \(\sigma\)
)
When deforming force is applied on the body then the equal restoring
force in opposite direction is developed inside the body. The restoring
force per unit area is called stress.
\[Stress(\sigma) =
\frac{\text{~}\text{restoring force}\text{~}}{\text{~}\text{Area of
cross section of the body}\text{~}}\]
Stress can be tensile or compressive as given below-
| Tensile |
\[F \leftarrow F \leftarrow O \rightarrow
F \rightarrow F\] |
| Compressive \(F \rightarrow \square
\rightarrow \frac{F}{A}\) as \(F
\propto A\) |
|
Strain
Suppose we stretch a wire by applying tensile forces of magnitude F
to each end. The length of the wire increases from L to \(L + \Delta L\). The fractional length
change is called the strain. It is a dimensionless quantity.
\[\text{~}\text{strain}\text{~} =
\frac{\Delta L}{\text{ }L}\]
Hooke's law for tensile and compressive forces
Suppose we had wires of the same composition and length but different
thicknesses. It would require larger tensile forces to stretch the
thicker wire the same amount as the thinner one. We conclude that the
tensile force required is proportional to the cross-sectional area of
the wire ( \(F \propto A\) ). Thus, the
same applied force per unit area produces the same deformation on wires
of the same length and composition.
Hooke's Law
stress \(\propto\) strain
\[\frac{F}{\text{ }A} =
Y\frac{\Delta\text{ }L}{\text{ }L}\]
equation still says that the length change ( \(\Delta L\) ) is proportional to the
magnitude of the deforming forces (F). Stress and strain account for the
effects of length and cross-sectional area ; the proportionality
constant Y depends only on the inherent stiffness of the material from
which the object is composed ; it is independent of the length and
cross-sectional area.
Comparing equation \(F = k\Delta L\)
and \(\frac{F}{A} = Y\frac{\Delta L}{L},F =
Y\frac{\Delta L}{L}\) A. Y is called the elastic modulus or
Young's modulus, Y has the same units as those of stress ( Pa or \(N/m^{2}\) ) since strain is
dimensionless.
Young's modulus can be thought of as the inherent stiffness of a
material ; it measures the resistance of the material to elongation or
compression. Material that is flexible and stretches easily (for
example, rubber) has a low Young's modulus. A stiff material (such as
steel) has a high Young's modulus. It takes a larger stress to produce
the same strain.
Hooke's law holds up to a maximum stress called the proportional limit.
For many materials, Young's modulus has the same value for tension and
compression. Some composite materials, such as bone and concrete, have
significantly different Young's moduli for tension and compression. The
different properties of these two substances lead to different values of
Young's modulus for tensile and compressive stress.
Illustration :
A light wire of length \(4m\) is
suspended to the ceiling by one of its ends. If its crossectional area
is \(19.6{\text{ }mm}^{2}\), what is
its extension under a load of 10 kg . Young's modulus of steel \(= 2 \times 10^{ll}Pa\).
Sol. Given quantities - original length \(L =
4\text{ }m\); force \(F = 10 \times 9.8
= 98\text{ }N\); and \(Y = 2 \times
10^{II}{Nm}^{- 2}l =\) ?
Using the relation,
Young's modulus \((Y) =
\frac{\text{~}\text{longitudinal
stress}\text{~}}{\text{~}\text{Iongitudinal}\text{
strain}\text{~}}\)
We have
\[\begin{matrix}
& \ Y = \frac{F/A}{l/L}\ \Rightarrow \ l = \frac{FL}{YA} \\
& \ \therefore\ l = \frac{98 \times 4}{2 \times 10^{ll} \times 19.6
\times 10^{- 6}} = 1 \times 10^{- 4}\text{ }m \\
& \ = 0.1\text{ }mm
\end{matrix}\]
Illustration :
Two vertical rods of equal lengths, one of steel and the other of
copper, are suspended from the ceiling, at a distance l apart and are
connected rigidly to a rigid horizontal light bar at their lower
ends.
If \(A_{S}\) and \(A_{C}\) be their respective cross sectional
areas, and \(Y_{S}\) and \(Y_{C}\) their respective Young's modulii of
elasticities, find where should a vertical force \(F\) be applied to the horizontal bar, in
order that the bar remains horizontal. (Fig.)
Sol. Let the force \(F\) be applied at
a distance \(x\) from the steel bar,
measured along the horizontal bar.
Let \(F_{S}\) and \(F_{C}\) be the loads on steel and copper
rods respectively, so
\[\begin{array}{r}
F_{S} + F_{C} = F\#(i)
\end{array}\]
Since the rigid horizontal bar remains horizontal so, the extensions
produced in the two rods and hence strains remains same.
\[\begin{array}{r}
\text{~}\text{i.e.,}\text{~}\ \frac{F_{S}}{A_{S}Y_{S}} =
\frac{F_{C}}{A_{C}Y_{C}}\#(ii)
\end{array}\]
Solving (i) and (ii) \(F_{S} =
\frac{FA_{S}Y_{S}}{A_{S}Y_{S} + A_{C}Y_{C}}\)
and
\[F_{C} = \frac{FA_{C}Y_{C}}{A_{S}Y_{S} +
A_{C}Y_{C}}\]
Now, taking moments about the steel bar.
\[F_{C}l = Fx\ \Rightarrow x =
\frac{F_{C}}{F}l \Rightarrow \frac{A_{C}Y_{C}l}{A_{S}Y_{S} +
A_{C}Y_{C}}\]
or
\[x = l/\left\lbrack 1 + \left(
\frac{A_{s}}{A_{c}} \right)\left( \frac{Y_{s}}{Y_{c}} \right)
\right\rbrack\]
Elastic potential energy
It is the potential energy stored inside the body due to change their
configuration. If F force is applied on a body as shown below.
For differential change in length dx the work done by restoring force F
is dw
\[\begin{matrix}
& dw = - Fdx\ \therefore\left( \text{ }F = \frac{AY}{\text{ }L}x
\right) \\
& dw = - \frac{AY}{\text{ }L}xdx \\
& {\text{ }W}_{\text{elastic}\text{~}} = - \frac{AY}{\text{
}L}\int_{0}^{l}\mspace{2mu}\mspace{2mu} xdx \\
& \Delta U = - W = \frac{AYl^{2}}{2\text{ }L}\ = \frac{1}{2}\left(
\frac{Yl}{\text{ }L} \right)\left( \frac{l}{\text{ }L} \right)(AL)F \\
& \ \therefore U_{i} = 0,U_{f} = U
\end{matrix}\]
Elastic potential energy \((U) =
\frac{1}{2}\) (stress) (strain) (volume)
Elastic potential energy per unit volume
\[= \frac{1}{2} \times
\text{~}\text{stress}\text{~} \times
\text{~}\text{strain}\text{~}\]
The above formula holds good for any type of strain. Change in
equilibrium, restoring force \(=\)
external force F
Then \(U = \frac{1}{2}\left( \frac{YA}{L}
\right)l^{2}\)
\[= \frac{1}{2}\left( \frac{Y}{\text{ }L}l
\right)Al = \frac{1}{2}\text{ }Fl\]
Illustration :
A uniform heavy rod of weight \(W\),
cross- sectional area \(A\) and length
\(L\) is hanging from a fixed support.
Young's modulus of the material of the rod is \(Y\). Neglect the lateral contraction. Find
the elongation of the rod.
Sol. Consider a small length \(dx\) of
tne rod at a distance \(x\) from the
fixed end. The part below this small element has length \(L - x\). The tension T of the rod at the
element equals the weight of the rod below it.
\[T = (L - x)\frac{W}{L}\]
Elongation in the element is given by
\[\begin{matrix}
& \text{~}\text{elongation}\text{~} = \text{~}\text{original
length}\text{~} \times \text{~}\text{stress}\text{~}/Y \\
& \ = \frac{Tdx}{AY} = \frac{(L - x)Wdx}{LAY}
\end{matrix}\]
The total elongation \(=
\int_{0}^{L}\mspace{2mu}\frac{(L - x)Wdx}{LAY}\)
\[= \frac{W}{LAY}\left( Lx -
\frac{x^{2}}{2} \right)_{0}^{L} = \frac{WL}{2AY}\]
Illustration:
A wire having a length \(l = 2m\),
and cross sectional area \(A = 5{\text{
}mm}^{2}\) is suspended at one of its ends from a ceiling. What
will be its strain energy due to its own weight, if the density and
Young's modulus of the material of the wire be \(d = 9\text{ }g/{cm}^{3}\) and \(Y = 1.5 \times 10^{11}{Nm}^{- 2}\) ?
Sol. Consider an elemental length of the wire of length \(dx\), at a distance \(x\) from the lower end.
Clearly, this length is acted upon by the external force equal to the
weight of the portion of wire below it \(=
x\) Adg. In equilibrium, the restoring force \(f = x\) Adg.
\(\therefore\ \) stress \(= \frac{f}{A} = xdg\).
Now, elastic potential energy stored in the elemental length will be
\[\begin{matrix}
& dU = \frac{1}{2} \times \text{~}\text{stress}\text{~} \times
\text{~}\text{strain}\text{~} \times \text{~}\text{volume}\text{~} \\
& \ = \frac{1}{2} \times
\frac{(\text{~}\text{stress}\text{~})^{2}}{Y} \times
\text{~}\text{volume}\text{~}
\end{matrix}\]
\[= \frac{1}{2} \times \frac{(xdg)^{2}}{Y}
\times Adx = \frac{1}{2}\frac{d^{2}g^{2}A}{Y}x^{2}dx\]
\(\therefore\ \) Total elastic
potential energy \(U = \int_{}^{}\
dU\)
\[\begin{matrix}
& \ =
\int_{0}^{l}\mspace{2mu}\mspace{2mu}\frac{I}{2}\frac{d^{2}g^{2}A}{Y}x^{2}dx
\\
& \ = \frac{I}{6}d^{2}g^{2}\frac{\text{ }Al^{3}}{Y}
\end{matrix}\]
Substituting the values,
\[\begin{matrix}
& U = \frac{1}{6} \times \frac{\left( 9 \times 10^{3} \right)(9
\cdot 8)^{2} \times 5 \times 10^{- 6} \times 2^{3}}{1.5 \times 10^{ll}}
\\
& \ = 3.46 \times 10^{- 7}\text{ }J
\end{matrix}\]
Illustration :
Find out the shift in point \(B,C\) and
\(D\).
\[\begin{matrix}
& Y_{AB} = 2.5 \times 10^{10}\text{ }N/m^{2} \\
& Y_{BC} = 4 \times 10^{10}\text{ }N/m^{2} \\
& Y_{CD} = 1 \times 10^{10}\text{ }N/m^{2} \\
& A = 10^{- 7}{\text{ }m}^{2}
\end{matrix}\]
Sol. \(\ Y = \frac{F/A}{\Delta L/L} =
\frac{FL}{A\Delta L}\)
\[\Rightarrow \ \Delta L = \frac{FL}{AY} =
\frac{MgL}{AY}\]
Shift of point \(B\left( \Delta L_{B}
\right) = \Delta L_{AB} = \frac{10 \times 10 \times 0.1}{10^{- 7} \times
2.5 \times 10^{10}} = 4 \times 10^{- 3}\text{ }m = 4\text{
}mm\)
Shift of point \(C\left( \Delta L_{C}
\right) = \Delta L_{B} + \Delta L_{BC} = 4 \times 10^{- 3} + \frac{100
\times 0.2}{10^{- 7} \times 4 \times 10^{10}}\)
\[= 9 \times 10^{- 3}\text{ }m = 9\text{
}mm\]
Shift of point \(D\left( \Delta L_{D}
\right) = \Delta L_{C} + \Delta L_{CD} = 9 \times 10^{- 3} + \frac{100
\times 0.5}{10^{- 7} \times 1 \times 10^{10}}\)
\[= 9 \times 10^{- 3} + 15 \times 10^{- 3}
= 24\text{ }mm\]
Illustration :
A load of 10 KN is supported from a pulley which in turn is supported
by a rope of cross-sectional area \(1 \times
10^{3}{\text{ }mm}^{2}\) and modulus of elasticity \(10^{3}\text{ }N/{mm}^{2}\), as shown in
figure. Neglecting the friction at the pulley determine the deflection
of load.
Sol. longitudinal stress in the rope is
\[\sigma = \frac{T}{A} = \frac{5 \times
10^{3}}{10^{3}{\text{ }mm}^{2}} = 5\text{ }N/{mm}^{2}\]
Extension in the rope \(=
\frac{\text{~}\text{stress}\text{~}}{Y} \times L\)
\[\begin{matrix}
& \ = \frac{5\text{ }N/{mm}^{2}}{10^{3}\text{ }N/{mm}^{2}} \times
1500 \\
& \ = 7.5\text{ }mm
\end{matrix}\]
Deflection in the load \(=
\frac{7.5}{2}\)
\[= 3.75\text{ }mm\]
Stress-strain curve
The relation between the stress and the strain for a given material
under tensile stress can be found experimentally. The applied force is
gradually increased in steps and the change in length is noted. A graph
is plotted between the stress and the strain produced. 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.
Beyond Hooke's law
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 ( \(S_{y}\) ) of the
material. If the tensile or compressive stress exceeds the proportional
limit, the strain is no longer proportional to the stress. The solid
still returns to its original length when the stress is removed as long
as the stress does not exceed the elastic limit.
If the stress exceeds the elastic limit, the material is permanently
deformed. For still larger stresses, the solid factures when the stress
reaches the breaking point. The maximum stress that can be withstood
without breaking is called the ultimate strength. The ultimate strength
can be different for compression and tension ; then we refer to the
compressive strength or the tensile strength of the material. A ductile
material continues to stretch beyond its ultimate tensile strength
without breaking; the stress then decreases from the ultimate strength
(fig. (a) ). Examples of ductile solids are relatively soft metals, such
as gold, silver, copper, and lead. These metals can be pulled like
taffy, becoming thinner and thinner until finally reaching the breaking
point.
While as Brittle material can not stand beyond ultimate strength
(a)
(b)
Practice Exercise
Q. 1 A light rigid bar is suspended horizontally from two vertical
wires, one of steel and one of brass, as shown in figure. Each wire is
2.00 m long. The diameter of the steel wire is 0.60 mm and the length of
the bar \(AB\) is 0.20 m . When a mass
of 10 kg is suspended from the centre of AB bar remains
horizontal.
(i) What is the tension in each wire?
(ii) Calculate the extension of the steel wire and the energy stored
in it.
(iii) Calculate the diameter of the brass wire.
(iv) If the brass wire were replaced by another brass wire of diameter 1
mm , where should the mass be suspended so that AB would remain
horizontal? The Young modulus for steel \(=
2.0 \times 10^{11}\text{ }Pa\), the Young modulus for brass \(= 1.0 \times 10^{11}\text{ }Pa\).
Answers
Q. \(1\ \) (i) 50 N , (ii) 0.045 J ,
(iii) \(8.4 \times 10^{- 4}\text{ }m\),
(iv) \(x = 0.12\text{ }m\)
Shearing Stress
A cylinder subjected to shearing (tangential) stress deforms by an
angle \(\theta\).
However, if two equal and opposite deforming forces are applied parallel
to the cross-sectional area of the cylinder, as shown in fig, 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.
As a result of applied tangential force, there is a relative
displacement \(\Delta x\) between
opposite faces of the cylinder as shown in the fig. The strain so
produced is known as shearing strain and it is defined as the ratio of
relative displacement of the faces \(\Delta
x\) to the length of the cylinder \(L\).
Shearing strain \(= \frac{\Delta x}{L} =
tan\theta\)
where \(\theta\) is the angular
displacement of the cylinder from the vertical ( \(\theta\) is very samll \(tan\theta \simeq \theta\) ).
Volume Deformation
Since the fluid presses inward on all sides of the object (figure),
the solid is compressed-its volume is reduced. The fluid pressure \(P\) is the force per unit surface area ; it
can be through of as the volume stress on the solid object. Pressure has
the same units as the other kinds of stress: N/m2 or Pa.
Fig. Forces on an object when submerged in a fluid
\[\text{~}\text{volume stress}\text{~} =
\text{~}\text{pressure}\text{~} = \frac{F}{\text{ }A} = P\]
The resulting deformation of the object is characterized by the
volume strain, which is the fractional change in volume :
\[\text{~}\text{volume strain}\text{~} =
\frac{\text{~}\text{change in volume}\text{~}}{\text{~}\text{original
volume}\text{~}} = \frac{\Delta V}{\text{ }V}\]
Bulk Modulus (B)
In fig., 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 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 ( \(\Delta V\) ) to the original volume ( \(V\) ).
\[\text{~}\text{Volume strain}\text{~} =
\frac{\Delta v}{v}\]
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.
\[\Delta P = - B\frac{\Delta V}{V}\
\text{~}\text{(Hooke's law for volume
deformation)}\text{~}\]
where \(V\) is the volume at
atmospheric pressure. The negative sign. equation \(\Delta P = - B\frac{\Delta V}{V}\) allows
the bulk modulus to be positive. The bulk moduli of liquids are
generally not much less than those of solids, since the atoms in liquids
are nearly as close together as those in solids.
Gases are much easier to compress than solids or liquids, so their bulk
moduli are much smaller. The bulk moduli of a few common materials are
given in Table
| Material Solids |
B ( \(10^{9}{Nm}^{- 2}\) or
GPa) |
| 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) |
\[1.0 \times 10^{- 4}\] |
Table: Bulk moduli \((B)\) of some
common Materials
Compressibility (k)
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.
\[k = \frac{1}{\text{ }B} = -
\frac{1}{\text{ }V}\left( \frac{\Delta\text{ }V}{\Delta P}
\right)\]
Poisson's ratio
When an elongation is produced by longitudinal stresses, a change is
produced in the lateral dimensions of the strained substance. Thus, when
a wire is stretched, its diameter diminishes ; and when the longitudinal
strain is small, the lateral strain is proportional to it. The ratio of
the lateral strain to the longitudinal strain is called Poisson's
ratio.
\(\left( l_{1},l_{2} \right.\ \),
and \(l_{3}\) are the dimentional when
no strain. \(\Delta l_{1},\Delta
l_{2}\), and
\(\Delta l_{3}\) are the change in
length of \(l_{1},l_{2}\), and \(l_{3}\) respectively)
\[\begin{matrix}
Y & \ = \frac{\frac{F}{\text{ }A}}{\frac{\Delta l_{1}}{l_{1}}} \\
\frac{\Delta l_{2}}{l_{2}} & \ = \frac{\Delta l_{3}}{l_{3}} = -
\sigma\frac{\Delta l_{1}}{l_{1}}
\end{matrix}\]
Illustration:
Sol. \(\ \) Longitudinal stress
\(= \frac{F}{A}\).
Longitudinal strain \(= \frac{F}{AY} =
\varepsilon_{l}\) (say)
Now, by definition of Poisson's ratio,
\[\sigma = \frac{- \text{~}\text{lateral
strain}\text{~}}{\text{~}\text{longitudinal strain}\text{~}} = \frac{-
\delta r/r}{\delta L/L}
\]or \(\ \delta r/r = - \sigma\delta
L/L\ \Rightarrow - \frac{\sigma F}{AY}\) [From eqn. (i)]
Since Volumetric strain \(=\) Strain in
length + Twice strain in radius.
\(\therefore\ \) Volumetric strain
\(= \frac{\delta L}{L} +
\frac{2\delta\delta}{r}\)
\[= \frac{F}{AY} + 2\left( - \frac{\sigma
F}{AY} \right)\ = \frac{F}{AY}(1 - 2\sigma)\]