Question

Derive an expression for the elastic energy stored per unit volume of a wire.

Answer 1

When a body is stretched, work is done against the restoring force (internal force). This work done is stored in the body in the form of elastic energy.

Consider a wire whose un-stretch length is L and the area of cross-section is A. Let a force produce an extension 1 and further assume that the elastic limit of the wire has not been exceeded and there is no loss in energy. Then, the work done by the force F is equal to the energy gained by the wire.

The work done in stretching the wire by dl, dW = Fdl

The total work done in stretching the wire from 0 to l is

W = ${\displaystyle {\int }_0^{\text{l}}\text{Fdl}}$ .............(1)

From Young’s modulus of elasticity,

Y = ${\displaystyle \frac{\text{F}}{\text{A}}\times \frac{\text{L}}{\text{l}}\Rightarrow F=\frac{\text{YAl}}{\text{L}}}$ .........(2)

Substituting equation (2) in equation (1), we get

W = ${\displaystyle {\int }_0^{\text{l}}\frac{\text{YAl}}{\text{L}}\text{dl}}$

Since l is the dummy variable in the integration, we can change l to l’ (not in limits), therefore

W = ${\displaystyle {\int }_0^{\text{l}}\frac{\text{YAl’}}{\text{L}}\text{dl’}=\frac{\text{YA}}{\text{L}}{\left(\frac{{\text{l’}}^2}{2}\right)}_0^{\text{l}}=\frac{\text{YA}}{\text{L}}\frac{{\text{l}}^2}{2}=\frac{1}{2}\left(\frac{\text{YAl}}{\text{L}}\right)\text{l}=\frac{1}{2}\text{Fl}}$

W = ${\displaystyle \frac{1}{2}}$ Fl = Elastic potential energy

Energy per unit volume is called energy density

u = ${\displaystyle \frac{\text{Elastic potential energy}}{\text{Volume}}=\frac{\frac{1}{2}\text{Fl}}{\text{AL}}}$

= ${\displaystyle \frac{1}{2}\frac{\text{F}}{\text{A}}\frac{\text{l}}{\text{L}}=\frac{1}{2}\times \text{Stress}\times \text{Strain}}$