Question

A solid circular shaft of 4 cm in diameter is subjected to a shear stress of 20 kN/cm, then the value of twisting moment (kN-cm) will be:

• A. 80 π
• B. 20 π
• C. 15 π
• D. 10 π

Answer 1

Answer:

A. 80 π

Given: D = 4 cm. $\tau_{max} =20 kN/cm^ 2 \Rightarrow T= \frac{\tau_{max} pi D^ 3}{16} \Rightarrow T = \frac {20\times \pi \times 4 ^ 3}{16} = 80\pi kNcm$