Maximum shear stress developed on the surface of a solid circular shaft under pure torsion is 240 MPa. If the shaft diameter is doubled, the maximum shear stress developed corresponding to the same torque will be : - (5) Question not attempted

The maximum shear stress ($\tau_{max}$) in a solid circular shaft under pure torsion is given by the formula: $\tau_{max} = \frac{16T}{\pi D^3}$, where T is the torque and D is the diameter of the shaft. Given: $\tau_{max1} = 240$ MPa. When the shaft diameter is doubled, the new diameter $D_2 = 2D_1$. The new maximum shear stress $ \tau_{max2} = \frac{16T}{\pi (D_2)^3} = \frac{16T}{\pi (2D_1)^3} = \frac{16T}{\pi 8D_1^3} = \frac{1}{8} \left(\frac{16T}{\pi D_1^3}\right)$. Since $\tau_{max1} = \frac{16T} {\pi D_1^3}$, we have $\tau_{max2} = \frac{1} {8} \tau_{max1}$. $\tau_{max2} = \frac{1}{8} \times 240$ MPa $= 30$ MPa.