Discussion
A solid circular shaft needs to be designed to transmit a torque of 50 N.m. If the allowable shear stress of the material is 140 MPa, assuming a factor of safety of 2, the minimum allowable design diameter in mm is
Answer: B
The minimum design diameter is 16 mm.
The design shear stress is 140/2 = 70 MPa. For a solid shaft in torsion:
\[ \begin{aligned} \tau &= \frac{16T}{\pi d^{3}} \\ d^{3} &= \frac{16 \times 50\,000}{\pi \times 70} = 3638\ \text{mm}^{3} \\ d &= 15.4\ \text{mm} \end{aligned} \]
The next listed size up is 16 mm, so that is the minimum allowable design diameter. 8 mm is far too small, and without the factor of safety the diameter would be 12.2 mm, which still rounds up to 16 among these options. 24 and 32 mm are safe but not the minimum.
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