Question

A solid circular shaft of 40 mm diameter transmits a torque of 3200 N-m. The value of maximum stress developed is

• A. 400/π
• B. 800/π
• C. 1600/π
• D. 600/π

Answer 1

Answer:

B. 800/π

Given: T = 3200N - m $= 3200 \times 10 ^ 3 N - mm$ D = 40mm and $\tau=?$= ; $T=\frac {\pi}{16} \times \tau\times D^ 3 \Rightarrow 3200 \times 10 ^ 3$=$\frac {\pi}{16} \times \tau \times 40 ^ 3 \Rightarrow$ $\tau=\frac{800}{\pi}$