Question

Find how the volume density of the elastic deformation energy is distributed in a steel rod depending on the distance $r$ from its axis. The length of the rod is equal to $l$, the torsion angle to $\phi$.

• A. $u=\frac{1}{2}\frac{G{\phi }^2{r}^2}{l^2}$
• B. $u=\frac{3}{2}\frac{G{\phi }^2{r}^2}{l^2}$
• C. $u=\frac{5}{2}\frac{G{\phi }^4{r}^2}{l^2}$
• D. None of these

Answer 1

The correct option is A $u=\frac{1}{2}\frac{G{\phi }^2{r}^2}{l^2}$

We know that $E=\frac{G\pi r^4{\phi }^2}{4l}$

$\frac{dE}{dr}=\frac{4G\pi r^3{\phi }^2}{4l}$

$E=\frac{4G\pi r^3{\phi }^{2}dr}{4l}$

Also $dV=2\pi lrdr$ , where $V$ is volume.

$\frac{dE}{dV}=\frac{G{r}^2{\phi }^2}{2l^2}$