Question

Calculate the strain energy per unit volume in a brass wire of length $3m$ and are of cross-section $0.6m{m^2}$ when it is stretched by $3mm$ and a force of $6kgwt$ is applied to its free end.

Answer 1

Hint: Strain is defined as the ratio of change in configuration to the original configuration. Strain Energy is defined as the work done in changing the original configuration of the object to said configuration.

Complete step by step answer:
Whenever an external force is applied on a fixed object, that force tends to change the configuration of that object, rather than changing its state of motion. This force is known as the deforming force. This deforming force produces effects on the object like Stress and Strain.
Stress: Stress is defined as the ratio of the deforming force applied on the object to the cross-section area on which the force is being applied.
Strain: Strain is defined as the ratio of change in configuration to the original configuration. Since strain is a ratio of two similar quantities, it has no dimensions and units. Strain are of three types – Linear, Volumetric and Shear.
Linear Strain: When the deforming force produces a change in length alone, the strain produced in the object is known as the linear strain.
Volumetric Strain: When the deforming force produces a change in volume of the object, the strain produced in the object is known as volumetric strain.
Shear Strain: When the deforming force produces a change in shape of the object without changing its volume, the strain produced in that object is known as Shear Strain.
Now in the above case, we are concerned with Linear Strain only as there is a change in length. So,
$Stress = \dfrac{F}{A}$
Where, $F = $Deforming force$ = 6kgwt = 6 \times 9.8N = 58.8N$
     $A = $Cross-section area$ = 0.6m{m^2} = 6 \times {10^{ - 7}}{m^2}$
So, $Stress = \dfrac{{58.8}}{{6 \times {{10}^{ - 7}}}}$
$ \Rightarrow Stress = 9.8 \times {10^7}N{m^{ - 2}}$
And $Strain = \dfrac{{\Delta L}}{L}$
Where, $\Delta L = $Change in length$ = 3mm = 3 \times {10^{ - 3}}m$
     $L = $Original Length$ = 3m$
So, $Strain = \dfrac{{3 \times {{10}^{ - 3}}}}{3}$
$ \Rightarrow Strain = {10^{ - 3}}$
Now, for strain energy per unit volume,
${U_V} = \dfrac{1}{2} \times Stress \times Strain$
Where, ${U_V} = $Strain Energy per unit volume.
$ \Rightarrow {U_V} = \dfrac{1}{2} \times 9.8 \times {10^7} \times {10^{ - 3}}$
$ \Rightarrow {U_V} = 4.9 \times {10^4}J{m^{ - 3}}$
$\therefore {U_V} = 49000J/{m^3}$

And thus, the strain energy stored per unit volume in the wire is $49000J/{m^3}$.

Note: The strain energy per unit volume is actually the potential energy produced in the object because of the change in its configuration per unit volume. This energy is stored in the object because of the work done in performing that change in configuration. The strain energy can be calculated by other formulae as well, which are
${U_V} = \dfrac{1}{2} \times \dfrac{{F \times \Delta L}}{{A \times L}} = \dfrac{1}{2} \times Stress \times Strain = \dfrac{1}{2} \times Y \times {\left( {Strain} \right)^2} = \dfrac{1}{{2Y}} \times {\left( {Stress} \right)^2}$
Where, $Y = $Young’s modulus$ = \dfrac{{Stress}}{{Strain}}$.