Question 3.1: A solid steel bar of circular cross section (Fig. 3-11) has ...

(a) Maximum shear stress and angle of twist. Because the bar has a solid circular cross section, we can find the maximum shear stress from Eq. (3-12), as follows:

\tau _{max}=\frac{16T}{\pi d^{3} } =\frac{16\left(250  lb-ft\right)\left(12  in./ft\right) }{\pi \left(1.5  in.\right)^{3} } =4530 psi

In a similar manner, the angle of twist is obtained from Eq. (3-15) with the polar moment of inertia given by Eq. (3-10):

\phi =\frac{TL}{GI_{P} }                             (3-15)

I_{P}=\frac{\pi r^{4} }{2} =\frac{\pi d^{4} }{32}                      (3-10)

Thus, the analysis of the bar under the action of the given torque is completed.
(b) Maximum permissible torque. The maximum permissible torque is determined either by the allowable shear stress or by the allowable angle of twist.

Beginning with the shear stress, we rearrange Eq. (3-12) and calculate as follows:

\tau_{max} = \frac{16T}{\pi d^{3}}           (3-12)

T_{1}=\frac{\pi d^{3}\tau_{allow } }{16}=\frac{\pi }{16}\left(1.5  in.\right)^{3}\left(6000  psi\right)=3980  lb-in.  =331    lb-ft

Any torque larger than this value will result in a shear stress that exceeds the allowable stress of 6000 psi.

Using a rearranged Eq. (3-15), we now calculate the torque based upon the angle of twist:

T_{2} =\frac{GI_{P}\phi _{allow} }{L} =\frac{\left(11.5\times 10^{6}  psi \right)\left(0.4970  in.^{4} \right)\left(2.5^{\circ} \right)\left(\pi   rad/180^{\circ} \right) }{54  in.}=4618  lb-in.=385    lb-ft

Any torque larger than T_{2}  will result in the allowable angle of twist being exceeded. The maximum permissible torque is the smaller of  T_{1} and  T_{2}:

T_{max}=331 lb-ft

In this example, the allowable shear stress provides the limiting condition.