Question

Shear stress developed at point A on the surface of the shaft of radius is 50 mm is

• A. 25.46 MPa
• B. 50.93 MPa
• C. 38.20 MPa
• D. 76.39 MPa

Answer 1

The correct option is B 50.93 MPa

Torque at A = $10\times 1$ = 10 kNm
$\therefore$ Shear stress , $\tau =\frac{T\times r}{I_p}=\frac{10\times {10}^6\times 50}{\frac{\pi }{32}\times {100}^4}=50.93MPa$