Question

A steel wire of diameter $1 \times {10^{ - 3}}{\text{m}}$ is stretched by a force of $20{\text{N}}$ . Calculate the strain energy per unit volume. ( ${Y_{steel}} = 2 \times {10^{11}}{\text{N}}{{\text{m}}^{ - 2}}$ ).

Answer 1

The strain energy refers to the amount of energy stored in the wire because of the stretching. It depends on the stress of the wire and Young’s modulus of the material used to make the wire which in our case is steel. The strain energy per unit volume is also referred to as the strain energy density of the steel wire.

Formulas used:
The area of cross-section of a wire is given by, $A = \pi {r^2}$ where $r$ is the radius of the wire.
The strain energy of a body is given by, $U = \dfrac{{{\sigma ^2}}}{{2YV}}$ where $\sigma$ is the stress applied, $Y$ is Young’s modulus of the material and $V$ is the volume of the body.
The stress of a body is given by, $\sigma = \dfrac{F}{A}$ where $F$ is the applied force and $A$ is the area of the body.

Complete step by step answer.
Step 1: List the parameters mentioned in the question.
The diameter of the steel wire that gets stretched is given to be $d = 1 \times {10^{ - 3}}{\text{m}}$ .
Then its radius will be $r = \dfrac{d}{2} = \dfrac{{1 \times {{10}^{ - 3}}}}{2} = 0 \cdot 5 \times {10^{ - 3}}{\text{m}}$ .
The force applied to stretch the wire is given to be $F = 20{\text{N}}$ .
Young’s modulus of steel is given to be ${Y_{steel}} = 2 \times {10^{11}}{\text{N}}{{\text{m}}^{ - 2}}$ .
Step 2: Express the relation for the area of the wire.
The area of cross-section of the wire will be $A = \pi {r^2}$
Substituting for $r = 0 \cdot 5 \times {10^{ - 3}}{\text{m}}$ in the above expression for the cross-sectional area of the wire we get, $A = \pi {\left( {0 \cdot 5 \times {{10}^{ - 3}}} \right)^2} = \pi \times 2 \cdot 5 \times {10^{ - 7}}{{\text{m}}^2}$ .
Thus the area of cross-section of the wire is $A = \pi \times 2 \cdot 5 \times {10^{ - 7}}{{\text{m}}^2}$
Step 3: Express the relation for the strain energy of the wire to find the strain energy per unit volume.
The strain energy of the steel wire is given by, $U = \dfrac{{{\sigma ^2}}}{{2{Y_{steel}}V}}$ -------- (1) where $\sigma$ is the stress applied, ${Y_{steel}}$ is Young’s modulus of steel and $V$ is the volume of the wire.
Since we can express the stress of the given wire as $\sigma = \dfrac{F}{A}$
Equation (1) becomes, $U = \dfrac{{{F^2}}}{{2{Y_{steel}}{A^2}V}}$ ------- (2)
Substituting for $F = 20{\text{N}}$ , ${Y_{steel}} = 2 \times {10^{11}}{\text{N}}{{\text{m}}^{ - 2}}$ and $A = \pi \times 2 \cdot 5 \times {10^{ - 7}}{{\text{m}}^2}$ in equation (2) we get, $U = \dfrac{{{{20}^2}}}{{2 \times 2 \times {{10}^{11}} \times {{\left( {\pi \times 2 \cdot 5 \times {{10}^{ - 7}}} \right)}^2}V}} = \dfrac{{1621 \cdot 1{\text{J}}}}{V}$
Then we have the strain energy per unit volume as $U = 1621 \cdot 1{\text{J}}{{\text{m}}^{ - 3}}$ .

Note: Here the stretching of the wire will cause the length of the wire to change. As the wire is stretched a restoring force will develop in it. The restoring force acting on a unit area of the wire is referred to as stress of the given wire. The energy stored in the wire will be released when the wire is unstretched so as to return it to its initial state.