Q.Derive an expression for strain energy per unit volume of the material of a wire.

Step 1. Consider a wire of original length $L$, cross-sectional area $A$, stretched by a force $F$ producing total elongation $l$. Since stress and strain increase proportionately throughout the stretching, $Y = \dfrac{F/A}{l/L}$, so the force needed at any intermediate extension $x$ is $f = \dfrac{YAx}{L}$.

Step 2. For a further small extension $dx$ at that stage, the work done is $dW = f\,dx = \dfrac{YAx}{L}\,dx$.

Step 3. Integrating from $x=0$ to $x=l$: $W = \displaystyle\int_0^l \dfrac{YAx}{L}\,dx = \dfrac{YA}{L}\left[\dfrac{x^2}{2}\right]_0^l = \dfrac{YAl^2}{2L}$.

Step 4. Using $Y = FL/(Al)$ to simplify: $W = \dfrac{1}{2}\cdot\dfrac{YAl}{L}\cdot l = \dfrac{1}{2}Fl$, i.e. $\text{Strain energy} = \dfrac{1}{2}(\text{load})(\text{extension})$. …

Integrate the incremental work f dx = (YAx/L) dx from x = 0 to l, then divide by volum …

Forgetting the factor of ½ that comes from integrating a linearly increasing force — using W = (load)(extension …