Question 11.11: A torque T is applied at the end D of shaft BCD (Fig. 11.32)...

The strain energy of a similar shaft was determined in Example 11.04 by breaking the shaft into its component parts $B C$ and $C D$. Making $n=2$ in Eq. (11.23),

$U=\frac{1+n^4}{2n^4} \frac{T^{2} L}{2 G J}$      (11.23)

we have

$U=\frac{17}{32} \frac{T^{2} L}{2 G J}$

where $G$ is the modulus of rigidity of the material and $J$ the polar moment of inertia of portion $C D$ of the shaft. Setting $U$ equal to the work of the torque as it is slowly applied to end $D$, and recalling Eq. (11.49),

$U=\int_{0}^{\phi_{1}}T\,d\phi={\textstyle\frac{1}{2}}T_{1}\phi_{1}$     (11.49)

we write

$\frac{17}{32} \frac{T^{2} L}{2 G J}=\frac{1}{2} T \phi_{D / B}$

and, solving for the angle of twist $\phi_{D / B}$,

$\phi_{D / B}=\frac{17 T L}{32 G J}$