Question 11.11: A torque T is applied at the end D of shaft BCD (Fig. 11.32)...
A torque \mathbf{T} is applied at the end D of shaft B C D (Fig. 11.32). Knowing that both portions of the shaft are of the same material and same length, but that the diameter of B C is twice the diameter of C D, determine the angle of twist for the entire shaft.
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}