Derive an expression for strain energy and show that the strain energy per unit volume of a stretched wire is proportional to the square of the strain.

Explanation

Let a uniform wire of length L and area of cross section A, suspended from a rigid support, be stretched by applying a load Mg at the other end. The internal restoring force gradually and uniformly increases from 0 to F=Mg. During the time the wire stretches at some instant, let the magnitude of the applied force be f and the elongation produced be x.When fully stretched, x becomes `DeltaL`.
Young's modulus,
`Y=("longitudinal stress")/("longitudinal strain")=(f//A)/(x//L)=(fL)/(Ax)`
`:. f=(YAx)/(L)" "`......(1)
The work done by the force f, against the restoring force, in stretching the wire through a further infinitesimal length dx is
W=force `xx`displacement
=fdx
`=(YAx)/(L)dx`
The total work done in producing the elongation `DeltaL` is
`W=int dW=underset(0)overset(DeltaL) int (YAx)/(L) dx=(YA)/(L) underset(0)overset(DeltaL)int x dx`
`=(YA)/(L)[(x^(2))/(2)]_(0)^(2DeltaL)`
`=(YADeltaL^(2))/(2L)=(1)/(2)((YADeltaL)/(L))DeltaL" "`......(3)
Now, `Y=(F//A)/(DeltaL//L)=(FL)/(ADeltaL)" "`....(4)
`:. F=(YADeltaL)/L)" "`......(5)
Hence, `W=(1)/(2)FDeltaL" "`....(6)
This work done is stored in the strained wire as potential energy and is called the strain energy.
`:.` Strain energy per unit volume of the strained wire
`=(W)/AL)=(Y)/(2)A((DeltaL)^(2))/(L)xx(1)/(AL)`
`=(Y)/(2)((DeltaL)/(L))^(2)`
`=(Y)/(2)("tensile strain")^(2)" "`.......(7)
It is proportional to the square of the strain as Y is constant.