Question

Maximum shear stress developed on the surface of a solid circular shaft under pure torsion is 240 MPa. If the shaft diameter is doubled then the maximum shear stress developed corresponding to the same torque will be

• A. 120 MPa
• B. 60 MPa
• C. 30 MPa
• D. 15 MPa

Answer 1

The correct option is C 30 MPa

${\tau }_{max}=\frac{16T}{\pi (d{)}^3}=240$

When shaft diameter is doubled then

${\tau }_{max}^{\prime }=\frac{16T}{\pi (2d{)}^3}=\frac{16T}{8\pi d^3}$

$=\frac{{\tau }_{max}}{8}=\frac{240}{8}=30MPa$