Question

A shaft subjected to pure torsion develops the maximum shear stress of 80 MPa. If the shaft diameter is doubled, then the maximum shear stress developed in the shaft corresponding to the same torque is

• A. 10 MPa
• B. 20 MPa
• C. 40 MPa
• D. 80 MPa

Hint:

When the diameter of a shaft is doubled, the shear stress induced by the same torque is reduced by a factor of 16.

Answer 1

The Correct Option is A

Step 1: Formula for shear stress in torsion.
The shear stress \(\tau\) induced by a torque \(T\) in a shaft is given by: \[ \tau = \frac{T \cdot r}{J} \] where:
\(r\) is the radius of the shaft,
\(J\) is the polar moment of inertia, which for a solid circular shaft is given by:
\[ J = \frac{\pi d^4}{32} \] where \(d\) is the diameter of the shaft.
Step 2: Effect of doubling the diameter.
When the diameter of the shaft is doubled, the radius \(r\) increases by a factor of 2, and the polar moment of inertia \(J\) increases by a factor of \(2^4 = 16\). Thus, when the diameter doubles, the new shear stress \(\tau_{\text{new}}\) will be: \[ \tau_{\text{new}} = \frac{\tau_{\text{old}}}{16} \]
Step 3: Substituting the known values.
The initial shear stress is 80 MPa. So, the new shear stress will be: \[ \tau_{\text{new}} = \frac{80}{16} = 10 \, \text{MPa} \] Thus, the correct answer is 10 MPa.