Question

A solid circular shaft carries a torque of 50 Nm. If the allowable shear stress of the material is 140 MPa, assuming factor of safety 2, the minimum diameter required for the shaft is

• A. 8 mm
• B. 32 mm
• C. 16 mm
• D. 24 mm

Answer 1

Answer:

C. 16 mm

Given: T = 50 Nm, $\tau = 140MPa$ FOS = 2,,,

$\frac{\tau}{Fos}$=$\frac{16D}{\pi d^3}$$\Rightarrow \frac{140}{2}$=$\frac{16\times50\times10^3}{\pi d^3}\Rightarrow d=15.38 mm or 16 mm$