Question 12.12: Redo the previous problem by replacing P with the torsional ...
Redo the previous problem by replacing P with the torsional moment, T at point B (Figure 12.23).
Let us draw the free-body diagram of the segment BC, defined by the angle \phi(0 \leq \phi \leq \pi) as shown in Figure 12.23, of the bent shaft in Figure 12.24:
From the foregoing free-body diagram shown, we observe at point C, the reaction moment is T which has components \left(M_{ b }\right)_{ C } \text { about } OC \text { and }\left(M_{ t }\right)_{ C } about EF, where EF is tangent at C. Clearly, EF and OC are perpendicular to each other at C. Therefore,
\left(M_{ b }\right)_{ C }=T \sin \phi
and torsional moment at C is
\left(M_{ t }\right)_{ C }=T \cos \phi
The normal stress is
\sigma_n=\frac{32\left(M_{ b }\right)_{ C }}{\pi d^3}=\left\lgroup \frac{32 T}{\pi d^3} \right\rgroup \sin \phi
and shear stress
\tau=\frac{16\left(M_{ t }\right)_{ C }}{\pi d^3}=\left\lgroup \frac{16 T}{\pi d^3} \right\rgroup \cos \phi
Thus, the principal stress at C is
\sigma_1=\frac{\sigma_n}{2}+\sqrt{\left\lgroup \frac{\sigma_n}{2} \right\rgroup^2+\tau^2}=\left\lgroup \frac{16 T}{\pi d^3} \right\rgroup\left[\sin \phi+\sqrt{\sin ^2 \phi+\cos ^2 \phi}\right]
or \sigma_1=\left\lgroup\frac{16 T}{\pi d^3} \right\rgroup (1+\sin \phi)
Clearly, \sigma_1 will be maximum when (1+\sin \phi) is maximum which implies when \sin \phi becomes maximum, that is, when \phi=90^{\circ} . Therefore, \sigma_1 is maximum when \phi=90^{\circ} .
