Question

A hollow circular shaft has an outer diameter of 100 mm and inner diameter of 50 mm. If the allowable shear stress is 125 MPa, the maximum torque(in kN-m) that the shaft can resist is .

• A. 23

Answer 1

The correct option is A 23
${\tau }_{allowable}=125\frac{N}{m{m}^2}$



$\frac{T_{max}}{J}=\frac{{\tau }_{allowable}}{R}$
$\Rightarrow T_{max.}=\frac{125}{50}\times (\frac{\pi }{32}\times ({100}^4-{50}^4))$
$=23009711.82$ N-mm
$=23.00\times {10}^{+6}$N-mm
$=23$ kN-m