The proportional region within the elastic limit of the stress-strain curve (region OA which is the linear region where hook’s law is obeyed) 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.

Young’s Modulus

Experimental observation shows that for a given material, the magnitude of the strain produced is the same whether the stress is tensile or compressive.

The ratio of tensile (or compressive) stress \((\sigma)\) to the longitudinal (tensile) strain \((\varepsilon)\) is defined as Young’s modulus and is denoted by the symbol \(Y\).

\(
Y=\frac{\sigma}{\varepsilon}=\frac{\text { Tensile stress }}{\text { Tensile strain }} \dots(9.7)
\)

\(
\begin{aligned}
Y & =(F / A) /(\Delta L / L) \\
& =(F \times L) /(A \times \Delta L) \dots(9.8)
\end{aligned}
\)
Since strain is a dimensionless quantity, the unit of Young’s modulus is the same as that of stress t.e., \(\mathrm{N} \mathrm{m}^{-2}\) or Pascal (Pa). The table below gives the values of Young’s moduli and yield strengths of some materials.

If the material has a circular cross-section, then \(A=\pi r^2\)
\(
\therefore \quad \mathrm{Y}=\frac{\mathrm{F} L}{\left(\pi \mathrm{r}^2\right) \Delta \mathrm{L}}
\)

Example 1: A load of \(4.0 \mathrm{~kg}\) is suspended from a ceiling through a steel wire of length \(20 \mathrm{~m}\) and radius \(2.0 \mathrm{~mm}\). It is found that the length of the wire increases by \(0.031 \mathrm{~mm}\) as equilibrium is achieved. Find Young modulus of steel. Take \(g=3.1 \pi \mathrm{m} \mathrm{s}^{-2}\).

Solution:

\(
\begin{aligned}
& \text {The longitudinal stress }=\frac{(4 \cdot 0 \mathrm{~kg})\left(3 \cdot 1 \pi \mathrm{m} \mathrm{s}^{-2}\right)}{\pi\left(2 \cdot 0 \times 10^{-3} \mathrm{~m}\right)^2} \\
& =3.1 \times 10^6 \mathrm{~N} \mathrm{~m}^{-2} \text {. } \\
& \text { The longitudinal strain }=\frac{0.031 \times 10^{-3} \mathrm{~m}}{2.0 \mathrm{~m}} \\
& =0.0155 \times 10^{-3} \\
&
\end{aligned}
\)
\(
\text { Thus } Y=\frac{3.1 \times 10^6 \mathrm{~N} \mathrm{~m}^{-2}}{0.0155 \times 10^{-3}}=2.0 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2} \text {. }
\)

Example 2: A structural steel rod has a radius of \(10 \mathrm{~mm}\) and a length of \(1.0 \mathrm{~m}\). A \(100 \mathrm{kN}\) force stretches it along its length. Calculate (a) stress, (b) elongation, and (c) strain on the rod. Young’s modulus, of structural steel, is \(2.0 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}\).

Solution:

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
\(
\begin{aligned}
\text { Stress } & =\frac{F}{A}=\frac{F}{\pi r^2} \\
& =\frac{100 \times 10^3 \mathrm{~N}}{3.14 \times\left(10^{-2} \mathrm{~m}\right)^2} \\
& =3.18 \times 10^8 \mathrm{~N} \mathrm{~m}{ }^{-2}
\end{aligned}
\)
The elongation,
\(
\begin{aligned}
\Delta L & =\frac{(F / A) L}{Y} \\
& =\frac{\left(3.18 \times 10^8 \mathrm{~N} \mathrm{~m}^{-2}\right)(1 \mathrm{~m})}{2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}} \\
& =1.59 \times 10^{-3} \mathrm{~m} \\
& =1.59 \mathrm{~mm}
\end{aligned}
\)
The strain is given by
\(
\begin{aligned}
\text { Strain } & =\Delta L / L \\
& =\left(1.59 \times 10^{-3} \mathrm{~m}\right) /(1 \mathrm{~m}) \\
& =1.59 \times 10^{-3} \\
& =0.16 \%
\end{aligned}
\)

Example 3: A copper wire of length \(2.2\) \(\mathrm{m}\) and a steel wire of length \(1.6 \mathrm{~m}\), both of diameter \(3.0 \mathrm{~mm}\), are connected end to end. When stretched by a load, the net elongation is found to be \(0.70 \mathrm{~mm}\). Obtain the load applied.

Solution:

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 \(A\). we have stress = strain \(\times\) Young’s modulus. Therefore
\(
W / A=Y_c \times\left(\Delta L_c / L_c\right)=Y_s \times\left(\Delta L_s / L_s\right)
\)
where the subscripts \(c\) and \(s\) refer to copper and stainless steel respectively. Or,
\(
\Delta L_c / \Delta L_s=\left(Y_s / Y_c\right) \times\left(L_c / L_s\right)
\)
Given \(L_c=2.2 \mathrm{~m}, L_s=1.6 \mathrm{~m}\),
From Table \(Y_c=1.1 \times 10^{11} \mathrm{~N}_{\mathrm{s}} \cdot \mathrm{m}^{-2}\), and \(Y_s=2.0 \times 10^{11} \mathrm{~N} . \mathrm{m}^{-2}\).
\(\Delta L_c / \Delta L_s=\left(2.0 \times 10^{11} / 1.1 \times 10^{11}\right) \times(2.2 / 1.6)=2.5\).
The total elongation is given to be
\(
\Delta L_c+\Delta L_s=7.0 \times 10^{-4} \mathrm{~m}
\)
Solving the above equations, \(\Delta L_c=5.0 \times 10^{-4} \mathrm{~m}\), and \(\Delta L_s=2.0 \times 10^{-4} \mathrm{~m}\). Therefore
\(
\begin{aligned}
W & =\left(A \times Y_c \times \Delta L_c\right) / L_c \\
& =\pi\left(1.5 \times 10^{-3}\right)^2 \times\left[\left(5.0 \times 10^{-4} \times 1.1 \times 10^{11}\right) / 2.2\right] \\
& =1.8 \times 10^2 \mathrm{~N}
\end{aligned}
\)

Determination of Young’s Modulus of the Material of a Wire

Let’s make an experimental arrangement as shown in the figure below to determine the value of Young’s modulus of a material of wire under tension. Take two identical straight wires (same length and equal radius) A and B. Now fix its end from a fixed, rigid support. The wire A is the reference wire, and it carries a millimeter main scale M and a pan to place weight. The wire B is the experimental wire. It also carries a pan in which known weights are placed. At the bottom of the wire, B attaches a vernier scale V.

Now, after putting the weight in the pan connected to B, it exerts a downward force. In the influence of this downward force (tensile Stress), wire B gets stretched. This elongation (increase in length) of wire B is measured by the vernier scale. The reference wire A is used to compensate for any change in length that may occur due to change in room temperature. Initially, give a small load to both the wires A and B so that both be straight and take the Vernier reading. Now increase the load gradually in wire B and note the vernier reading. The difference between these two vernier readings gives the change in length produced in the wire.

The ratio of tensile (or compressive) stress \((\sigma)\) to the longitudinal (tensile) strain \((\varepsilon)\) is defined as Young’s modulus and is denoted by the symbol \(Y\).

\(Y=\frac{\sigma}{\varepsilon}=\frac{\text { Tensile stress }}{\text { Tensile strain }}\)

Let \(r\) and \(L\) be the initial radius and length of the experimental wire, respectively.
Then the area of the cross-section of the wire would be \(\pi r^2\).
Let \(M\) be the mass that produced an elongation \(\Delta L\) in the wire.
Thus the applied force is equal to \({Mg}\), where \(g\) is the acceleration due to gravity.
Young’s modulus of the material of the experimental wire is given by
\(
\begin{aligned}
Y & =(F / A) /(\Delta L / L) \\
& =(F \times L) /(A \times \Delta L) \\
& =\frac{M g}{\pi r^2} \cdot \frac{L}{\Delta L} \\
& = M g \times L /\left(\pi r^2 \times \Delta L\right) \dots(9.9)
\end{aligned}
\)

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 \(\left(\sigma_{\mathrm{s}}\right) /\) shearing strain
\(G=(F / A) /(\Delta x / L)\)
\(=(F \times L) /(A \times \Delta x) \dots(9.10)\)
We know that shearing strain = \(\text { Shearing } \operatorname{strain}=\frac{\Delta x}{L}=\tan \theta\approx \theta\)
\(
\begin{aligned}
G & =(F / A) / \theta \\
& =F /(A \times \theta) \dots(9.11)
\end{aligned}
\)
The shearing stress \(\sigma_{\mathrm{s}}\) can also be expressed as \(\sigma_{\mathrm{s}}=G \times \theta \quad \dots(9.12)\)

SI unit of shear modulus is \(\mathrm{N} \mathrm{m}^{-2}\) or \(\mathrm{Pa}\).
Dimensional formula is \(\mathrm{ML}^{-1} \mathrm{~T}^{-2}\)
For most of the materials \(\mathrm{G} \approx \frac{\mathrm{Y}}{3}\)

The shear moduli of a few common materials are given in the Table below.

Example 4: A block of unknown material kept on a table(The square face is placed on the table.), is under a shearing force. The following data is given, calculate the shear modulus of the material.
Dimensions of the block \(=60 \mathrm{~mm} \times 60 \mathrm{~mm} \times 20 \mathrm{~mm}\)
Shearing Force \(=0.245 \mathrm{~N}\)
Displacement \(=5 \mathrm{~mm}\)

Solution:

Substituting the values in the formula we get

Shearing stress

\(
\begin{aligned}
& =\frac{F}{A}=\frac{0.245}{60 \times 60 \times 10^{-6}} \\
& =\frac{2450}{36} \mathrm{~N} / \mathrm{m}^2
\end{aligned}
\)

Shearing strain

\(
=\frac{\Delta x}{l}=\frac{5}{20}=\frac{1}{4}
\)
Thus, shear modulus,
\(
\begin{aligned}
& G=\frac{\text { Shearing stress }}{\text { Shearing strain }} \\
& =\frac{2450 \times 4}{36}=\frac{2450}{9} \\
& =272.22 \mathrm{~N} / \mathrm{m}^2 .
\end{aligned}
\)

Example 5: A square lead slab of side 50 \(\mathrm{cm}\) and thickness \(10 \mathrm{~cm}\) is subject to a shearing force (on its narrow face) of \(9.0 \times\) \(10^4 \mathrm{~N}\). The lower edge is riveted to the floor. How much will the upper edge be displaced?

Solution:

The lead slab is fixed and the force is applied parallel to the narrow face as shown in Figure above. The area of the face parallel to which this force is applied is
\(
\begin{aligned}
A & =50 \mathrm{~cm} \times 10 \mathrm{~cm} \\
& =0.5 \mathrm{~m} \times 0.1 \mathrm{~m} \\
& =0.05 \mathrm{~m}^2
\end{aligned}
\)
Therefore, the stress applied is
\(
\begin{aligned}
& =\left(9.4 \times 10^4 \mathrm{~N} / 0.05 \mathrm{~m}^2\right) \\
& =1.80 \times 10^6 \mathrm{~N}^{-\mathrm{m}^{-2}}
\end{aligned}
\)
We know that shearing strain \(=(\Delta x / L)=\) Stress \(/ G\).
Therefore the displacement \(\Delta x=(\) Stress \(\times L) / G\)
\(=\left(1.8 \times 10^6 \mathrm{~N} \mathrm{~m}^{-2} \times 0.5 \mathrm{~m}\right) /\left(5.6 \times 10^9 \mathrm{~N} \mathrm{~m}^{-2}\right)\)
\(=1.6 \times 10^{-4} \mathrm{~m}=0.16 \mathrm{~mm}\)

Bulk modulus (B)

The ratio of the volume stress (hydraulic stress) to the corresponding volume strain (hydraulic strain) is defined as bulk modulus. It is denoted \(\mathrm{by}{ }^{\prime} \mathrm{B}\) ‘.
\(
\begin{aligned}
\mathrm{B} & =\frac{\text { Volume Stress }}{\text { Volume Strain }} \\
& =\frac{\mathrm{F} / \mathrm{A}}{\Delta \mathrm{V} / \mathrm{V}} \\
& =\frac{\text { Pressure }}{\Delta \mathrm{V} / \mathrm{V}}=\frac{-{p} \mathrm{V}}{\Delta \mathrm{V}} \dots(9.13)
\end{aligned}
\)

The negative sign indicates the fact that with an increase in pressure, a decrease in volume occurs. That is if \(p\) is positive, \(\Delta V\) 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., \(\mathrm{N} \mathrm{m}^{-2}\) or \(\mathrm{Pa}\). The bulk moduli of a few common materials are given in the Table below.

Compressibility (K)

The reciprocal of bulk modulus is called compressibility and is denoted by \(\mathrm{K}\).
\(\mathrm{K}=\frac{1}{\mathrm{~B}}=\frac{-\Delta \mathrm{V}}{{p} \mathrm{V}} \dots(9.14)\)

SI unit : \(\mathrm{Pa}^{-1}\) or \(\mathrm{~N}^{-1}{m}^2\)
Dimensional formula: \(\mathrm{M}^{-1} \mathrm{LT}^2\)
The bulk modulus for solids is much larger than that for liquids, which is again larger than the bulk modulus for gases.

Example 6: The average depth of Indian Ocean is about \(3000 \mathrm{~m}\). Calculate the fractional compression, \(\Delta V / V\), of water at the bottom of the ocean, given that the bulk modulus of water is \(2.2 \times 10^9 \mathrm{~N} \mathrm{~m}^{-2}\). (Take \(g=10 \mathrm{~m} \mathrm{~s}^{-2}\) )

Solution:

The pressure exerted by a \(3000 \mathrm{~m}\) column of water on the bottom layer
\(
\begin{aligned}
p=h \rho g & =3000 \mathrm{~m} \times 1000 \mathrm{~kg} \mathrm{~m}^{-3} \times 10 \mathrm{~m} \mathrm{~s}^{-2} \\
& =3 \times 10^7 \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-2} \\
& =3 \times 10^7 \mathrm{~N} \mathrm{~m}^{-2}
\end{aligned}
\)
Fractional compression \(\Delta V / V\), is
\(
\begin{aligned}
\Delta V / V=\text { stress } / B & =\left(3 \times 10^7 \mathrm{~N} \mathrm{~m}^{-2}\right) /\left(2.2 \times 10^9 \mathrm{~N} \mathrm{~m}^{-2}\right) \\
& =1.36 \times 10^{-2} \text { or } 1.36 \%
\end{aligned}
\)

Poisson’s Ratio

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.

Poisson’s ratio = lateral strain / longitudinal strain

If the original diameter of the wire is \(d\) and the contraction of the diameter under stress is \(\Delta d\), the lateral strain is \(\Delta d / d\).

If the original length of the wire is \(L\) and the elongation under stress is \(\Delta L\), the longitudinal strain is \(\Delta L / L\).

Poisson’s ratio = \((\Delta d / d) /(\Delta L / L)\) or \((\Delta d / \Delta L) \times(L / d)\).

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 the material. For steels, the value is between \(0.28\) and \(0.30\), and for aluminium alloys it is about \(0.33\).

Elastic Potential Energy in a Stretched Wire or body

When a wire is put under tensile stress, work is done against the inter-atomic forces. This work is stored in the wire in the form of elastic potential energy. Thus, the potential energy of the body is increased. This is called elastic potential energy. We shall derive an expression for the increase in elastic potential energy when a wire is stretched from its natural length.

Suppose a wire having natural length \(L\) and cross-sectional area \(A\) is fixed at one end and is stretched by an external force applied at the other end (figure below). The force is so adjusted that the wire is only slowly stretched. This ensures that at any time during the extension the external force equals the tension in the wire. When the extension is \(x\), the wire is under a longitudinal stress \(F / A\), where \(F\) is the tension at this time. The strain is \(x / L\).

If Young modulus is \(Y\), or,
\(
\begin{aligned}
\frac{F / A}{x / L} & =Y \\
F & =\frac{A Y}{L} x \dots(i)
\end{aligned}
\)
The work done by the external force in a further extension \(d x\) is, 
\(d W=F d x\),

Using (i),
\(
d W=\frac{A Y}{L} x d x
\)
The total work by the external force in an extension 0 to \(l\) is (let the length of the wire be elongated by \(l\))
\(
\begin{aligned}
W & =\int_0^l \frac{A Y}{L} x d x \\
& =\frac{A Y}{2 L} l^2 \\
& = \frac{1}{2} \times Y \times\left(\frac{l}{L}\right)^2 \times A L \\
& = \frac{1}{2} \times \text { Young’s modulus } \times \operatorname{strain}^2 \times \text { volume of the wire }
\end{aligned}
\)
This work is stored into the wire as its elastic potential energy.

Thus, the elastic potential energy of the stretched wire is,
\(
U=\frac{A Y}{2 L} l^2 \dots(ii)
\)
This may be written as
\(
\begin{aligned}
U & =\frac{1}{2}\left(A Y \frac{l}{L}\right) l \\
& =\frac{1}{2} \text { (maximum stretching force) (extension). }
\end{aligned}
\)
Equation (ii) may also be written as
\(
U=\frac{1}{2}\left(Y \frac{l}{L}\right) \frac{l}{L}(A L)
\)
\(
\text { or, Potential energy }=\frac{1}{2} \times \text { stress } \times \text { strain } \times \text { volume of wire. }
\)

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
\(
u=\frac{1}{2} \times \text { stress } \times \text { strain }=\frac{1}{2} \times \sigma \varepsilon \dots(9.15)
\)

Example 7: A steel wire of length \(2.0 \mathrm{~m}\) is stretched through \(2.0 \mathrm{~mm}\). The cross-sectional area of the wire is \(4.0 \mathrm{~mm}^2\). Calculate the elastic potential energy stored in the wire in the stretched condition. Young modulus of steel \(=2.0 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}\).

Solution:

The strain in the wire \(\frac{\Delta l}{l}=\frac{2.0 \mathrm{~mm}}{2.0 \mathrm{~m}}=10^{-a}\).
The stress in the wire \(=Y \times\) strain
\(
=2.0 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2} \times 10^{-3}=2.0 \times 10^8 \mathrm{~N} \mathrm{~m}^{-2} \text {. }
\)
The volume of the wire \(=\left(4 \times 10^{-6} \mathrm{~m}^2\right) \times(2.0 \mathrm{~m})\)
\(
=8.0 \times 10^{-6} \mathrm{~m}^3 \text {. }
\)
The elastic potential energy stored
\(
\begin{aligned}
& =\frac{1}{2} \times \text { stress } \times \text { strain } \times \text { volume } \\
& =\frac{1}{2} \times 2.0 \times 10^8 \mathrm{~N} \mathrm{~m}^{-2} \times 10^{-3} \times 8.0 \times 10^{-6} \mathrm{~m}^3 \\
& =0.8 \mathrm{~J} .
\end{aligned}
\)