Question

The maximum shear stress produced in a shaft in $5N/m{m}^2$. The shaft is of 40 mm diameter. What is the approximate value of twisting moment?

• A. 628 N-m
• B. 63 N-m
• C. 126 N-m
• D. 251 N-m

Answer 1

The correct option is B 63 N-m

Twisting moment, $T={\tau }_{max}J$

Torsional section modulus,

$J=\frac{\pi d^3}{16}=\frac{\pi \times (40{)}^3}{16}=12566m{m}^3$

$\therefore T=5\times 12566=62832N-mm\underset{–}{\sim }63N-m$