Discussion
Calculate the maximum torque for a solid circular shaft having a polar moment of inertia of \(30 \times 10^{6}\) mm\(^4\) that can transmit, if the maximum angle of twist is 1° in a length of 1.5 m.
(Take the modulus of rigidity equal to 90 GPa)
Answer: B
The torsion equation gives \(T/J = G\theta/L\), so \(T = GJ\theta/L\). With \(G = 90 \times 10^3\) N/mm\(^2\), \(J = 30 \times 10^6\) mm\(^4\), \(\theta = 1^\circ = \pi/180\) rad and \(L = 1500\) mm:
\[ \begin{aligned} T \\ &= \frac{90\times 10^3 \times 30 \times 10^6 \times \pi/180}{1500} \\ &= 10\pi \times 10^6\ \text{N-mm} \end{aligned} \]That is \(10\pi\) kN-m. The options 5, 15 and 20 pi kN-m would result from a different angle or length (for example, 20 pi if the length were taken as 0.75 m).
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