Question

Maximum shear stress developed on the surface of a solid circular shaft under pure torsion is 160 MPa. If the shaft diameter is doubled, then the maximum shear stress developed corresponding to the same torque will be:

• A. 10 MPa
• B. 30 MPa
• C. 40 MPa
• D. 20 MPa

Answer 1

Answer:

D. 20 MPa

Given: $\tau_{max} =160 MPa$; $d_{1} = d$ $d_{2} = 2d$ the maximum shear stress varies as, $\tau_{max} \propto \frac{1}{d^ 3} \Rightarrow \frac {\tau_{max_1}}{\tau_{max_2}}$ =$(\frac{ d_2}{ d_1})^ 3$ $\Rightarrow$$\frac{160}{\tau_{max_2}}$$=(\frac {2d}{d})^ 3 \Rightarrow \tau_{max_2} =20 MP a$