Question 11.CA.11: A torque T is applied at the end D of shaft BCD (Fig. 11.30)......
A torque T is applied at the end D of shaft BCD (Fig. 11.30). Knowing that both portions of the shaft are of the same material and length, but that the diameter of BC is twice the diameter of CD, determine the angle of twist for the entire shaft.
In Concept Application 11.4, the strain energy of a similar shaft was determined by breaking the shaft into its component parts BC and CD. Making n = 2 in Eq. (1) of Concept Application 11.4 gives
U=\frac{17}{32} \frac{T^2 L}{2 G J}
where G is the modulus of rigidity of the material and J is the polar moment of inertia of segment CD. Making U equal to the work of the torque as it is slowly applied to end D and recalling Eq. (11.40), write
U=\int_0^{\phi_1} T d \phi=\frac{1}{2} T_1 \phi_1 (11.40)
\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}