Question

Find the elastic deformation energy (in J) of a steel rod whose one end is fixed and the other is twisted through an angle $\phi ={6.0}^{\circ }$. The length of the rod is equal to $l=1.0m$, and the radius to $r=10mm$. (G=77.2 GPa)

Answer 1

When the rod is twisted through an angle $\theta$, a couple $N\left(\theta \right)=\frac{\pi r^{4}G}{2l}\theta$ appears to resist this.
Work done in twisting the rod by an angle $\phi$ is then ${\int }_0^{\phi }N\left(\theta \right)d\theta =\frac{\pi r^{4}G}{4l}{\phi }^2=7J$ on putting the values.