Question

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

• A. 240 MPa
• B. 120 MPa
• C. 60 MPa
• D. 30 MPa

Hint:

Remember that for a solid circular shaft under constant torque, the maximum shear stress is inversely proportional to the cube of the diameter (\( \tau_{max} \propto \frac{1}{d^3} \)).

Answer 1

The Correct Option is D

Step 1: Recall the torsion formula relating maximum shear stress and torque.
The maximum shear stress \( \tau_{max} \) on the surface of a solid circular shaft subjected to a torque \( T \) is given by: \[ \tau_{max} = \frac{TR}{J}, \] where \( R \) is the radius of the shaft and \( J \) is the polar moment of inertia of the circular cross-section. For a solid circular shaft with diameter \( d \), the radius \( R = \frac{d}{2} \) and the polar moment of inertia \( J = \frac{\pi d^4}{32} \). Substituting these into the formula: \[ \tau_{max} = \frac{T \left( \frac{d}{2} \right)}{\frac{\pi d^4}{32}} = \frac{Td}{2} \times \frac{32}{\pi d^4} = \frac{16T}{\pi d^3}. \] Step 2: Analyze the relationship between maximum shear stress and diameter for a constant torque.
From the formula \( \tau_{max} = \frac{16T}{\pi d^3} \), for a constant torque \( T \), the maximum shear stress \( \tau_{max} \) is inversely proportional to the cube of the diameter \( d \): \[ \tau_{max} \propto \frac{1}{d^3}. \] Step 3: Determine the new maximum shear stress when the diameter is doubled.
Let the initial diameter be \( d_1 \) and the initial maximum shear stress be \( \tau_{max,1} = 240 \, \text{MPa} \). When the diameter is doubled, the new diameter is \( d_2 = 2d_1 \). Let the new maximum shear stress be \( \tau_{max,2} \). Since the torque remains the same, we can write the ratio of the shear stresses as: \[ \frac{\tau_{max,2}}{\tau_{max,1}} = \frac{\frac{1}{d_2^3}}{\frac{1}{d_1^3}} = \left( \frac{d_1}{d_2} \right)^3. \] Substituting \( d_2 = 2d_1 \): \[ \frac{\tau_{max,2}}{\tau_{max,1}} = \left( \frac{d_1}{2d_1} \right)^3 = \left( \frac{1}{2} \right)^3 = \frac{1}{8}. \] Step 4: Calculate the new maximum shear stress \( \tau_{max,2} \).
\[ \tau_{max,2} = \frac{1}{8} \tau_{max,1} = \frac{1}{8} \times 240 \, \text{MPa} = 30 \, \text{MPa}. \] Step 5: Select the correct answer.
The maximum shear stress developed corresponding to the same torque when the shaft diameter is doubled will be 30 MPa, which corresponds to option 4.