Question 10.2: A prismatic steel rod of length L and cross-sectional area A...
A prismatic steel rod of length L and cross-sectional area A hangs vertically under its own weight. Calculate the strain energy stored in the rod. Assume γ is the specific weight (i.e., weight per unit volume) of the rod and that the material is following Hooke’s law.
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)-ν\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}+\tau_{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=\iiint_{\sout{V}} \tilde{u} d \sout{V}=\iiint_{\sout{V}}\left\{\int_0^\in \sigma d \in\right\} d \sout{V}
From Eq. (2),
U=\iiint_{\sout{V}}\left\lgroup \frac{\gamma^2 x^2}{2 E}\right\rgroup d \sout{V}
but d\sout{V}=A \cdot d x 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}
