Let us consider Figure 10.18

If we consider the free-body diagram of the portion of the rod as shown in Figure 10.18, we can write

\sigma_A=\gamma A x \Rightarrow \sigma=\gamma x            (1)

Clearly, the strain energy density of small element of length dx as shown in Figure 10.18 is given from Eq. (10.12) for one-dimensional stress situation as

\begin{aligned} d \tilde{u}=& \frac{1}{E}\left[\left(\sigma_{x x} d \sigma_{x x}+\sigma_{y y} d \sigma_{y y}+\sigma_{z z} d \sigma_{z z}\right)-v\left\{ d \left(\sigma_{x x} \sigma_{y y}+\sigma_{y y} \sigma_{z z}+\sigma_{z z} \sigma_{x x}\right)\right\}\right] \\ &+\frac{1}{G}\left(\tau_{x y} d \tau_{x y}+\tau_{y z} d \tau_{y z}+\left.\tau\right|_{z x} d \tau_{z x}\right) \end{aligned}              (10.12)

d \tilde{u}=\sigma d \in=\frac{\sigma d \sigma}{E}=\frac{\gamma^2 x}{E} d x

Therefore, at any position x

\tilde{u}=\int_0^x \frac{\gamma^2 x}{E} d x=\frac{\gamma^2 x^2}{2 E} d x

\tilde{u}=\tilde{u}(x)=\frac{\gamma^2 x^2}{2 E}=\int_0^{\in} \sigma d \in                  (2)

Now, total strain energy stored in the rod is given by

U=\underset{\sout{V}}{\iiint} \tilde{u} d \sout{V} =\underset{\sout{V}}{\iiint}\left\{\int_0^∈ \sigma d \in\right\} d \sout{V}

From Eq. (2),

U=\underset{\sout{V}}{\iiint}\left\lgroup \frac{\gamma^2 x^2}{2 E} \right\rgroup d \sout{V}

but d\sout{V} = A⋅dx. So from the above equation, we get

U=\int_0^L \frac{\gamma^2 x^2}{2 E} A d x=\left\lgroup \frac{\gamma^2 A}{2 E}\right\rgroup \int_0^L x^2 d x

or          U=\frac{\gamma^2 A L^3}{6 E}

The total strain energy stored in the rod due to its self-weight is

\frac{\gamma^2 A L^3}{6 E}