Clearly, the hollow shaft and solid shaft will be subjected to equal and opposite torques, say, T. Also, if \theta_{ h } \text { and } \theta_{ s } be the angles of twist of hollow and solid shafts at right end, respectively; then from compatibility of deformation, we can write

\frac{T L}{G J_{ h }}+\frac{T L}{G J_{ s }}=\frac{\pi}{2}

or            \left\lgroup \frac{T L}{G} \right\rgroup\left[\frac{J_{ s }+J_{ h }}{J_{ s } J_{ h }}\right]=\frac{\pi}{2}

or            T=\left\lgroup\frac{\pi}{2} \right\rgroup\left\lgroup\frac{G}{L} \right\rgroup\left\lgroup\frac{J_{ s } J_{ h }}{J_{ s }+J_{ h }} \right\rgroup            (1)

Now, total strain energy of the entire system will be

U=U_{ h }+U_{ s }

U=\frac{T^2 L}{2 G J_{ h }}+\frac{T^2 L}{2 G J_{ s }}=\left\lgroup \frac{T^2 L}{2 G} \right\rgroup\left\lgroup \frac{J_{ h }+J_{ s }}{J_{ s } J_{ h }} \right\rgroup

Replacing T from Eq. (1),

U=\frac{1}{2}\left\lgroup \frac{\pi}{2} \right\rgroup^2\left\lgroup \frac{G}{L} \right\rgroup\left\lgroup \frac{J_{ h } J_{ s }}{J_{ h }+J_{ s }} \right\rgroup                  (2)

Now,

J_{ s }=\frac{\pi}{32} d_{ s }^4 \quad \text { and } \quad J_{ h }=\frac{\pi}{3}\left(d_0^4-d_i^4\right)

Therefore,

J_s=\frac{\pi}{32}(6)^4  mm ^4=127.235  mm ^4

and        J_{ h }=\frac{\pi}{32}\left(9^4-6^4\right)  mm ^4=516.89  mm ^4

From Eq. (2), putting the values we get

U=\left\lgroup \frac{\pi^2}{8} \right\rgroup\left\lgroup \frac{82 \times 10^9}{2.5} \right\rgroup \frac{(127.235)(516.89)\left(10^{-12}\right)}{(127.235+516.89)}  J

or              U = 4.13 J

The total strain energy of the system is 4.13 J.