Question

derive an expression for elastic potential energy stored per unit volume

Answer 1

if wire is slowly streched by dL on applying a streching force F then workdone during streching is ---
W=1/2k(dL)^2
also we know F = -k.dL
use this In workdone ,
W= -1/2.F . dL

negative of workdone is known as potential energy .
hence,
elastic potential energy (U)=-W =1/2.F . dL
now ,
elastic potential energy per unit volume
U/V=1/2.F.dL/V
but we know ,
V=A.L
A is area and L is the length .

U/V=1/2 (F/A). (dL/L)

we also know ,
F/A= stress and dL/L = strain
use this

U/V= 1/2 .stress .strain.
=1/2.young's modulus x (strain)^2

Answer 2

restoration force   F = - k x
work done in stretching a spring from x = x to x + dx  (by dx) is 
   = dW = - F * dx = k x dx

W = energy stored in the spring as elastic P E
     = integral of   k x  dx     from x = 0 to A
     = 1/2 k x²    from x = 0 to A 
    = 1/2 k A²

=================
If the question is about the energy stored in a wire (of cross section area A and length L) when it is pulled with a force F :

Let the wire be stretched by an amount x. Now let it be stretched by a further dx amount.

Y = σ/ε = F (L+x) / (A x)
So  F = Y A x /(L + x)

Work done (energy stored as PE) in stretching from L+x to L+x+dx :
dW = F dx  = Y A x dx /( L+x)
W = YA * integral x = 0 to x,  { x dx/(L+x) }
     = YA * integral x =0 to x,   { 1 - L/(L+x) } dx
     = YA * [ x -  L * Ln (L+x)/L ]
 W = U = YA * [x - L * Ln (1 + ε) ]     =  YA [ x -  L (ε - ε²/2) ]
U    = Y A L ε² /2

PE per volume    = W /AL = 1/2 *Y  * ε²
U        = 1/2 * σ * ε