Question

The difference between two principal stresses is \(120 \text{ MPa}\), when a circular shaft is subjected to an axial force and shear force. What is the factor of safety based on maximum shear stress theory if elastic limit of the bar is \(300 \text{ MPa}\)?

• A. 1.03
• B. 2.5
• C. 3.25
• D. 5.0

Hint:

Under Tresca's maximum shear stress theory, for any two-dimensional stress case where the principal stresses have opposite signs or are standard planar outputs, the design relation simplifies directly to: \[ \text{F.O.S.} = \frac{\sigma_{\text{yield}}}{\sigma_1 - \sigma_2} \] This avoids calculating intermediate shear values.

Answer 1

The Correct Option is B

Step 1: Find the maximum shear stress from the given principal stresses.
On Mohr's circle, the radius of the circle gives the maximum shear stress, which is half the difference of the two principal stresses: \[ \tau_{max} = \frac{\sigma_1 - \sigma_2}{2} = \frac{120}{2} = 60 \text{ MPa} \]

Step 2: Find the allowable shear stress from the elastic limit.
By the maximum shear stress (Tresca) theory, yielding in shear happens at half the tensile elastic limit, so the shear yield stress is \(\tau_y = 300/2 = 150\) MPa.

Step 3: Take the ratio of the two shear stresses.
The factor of safety is simply how many times smaller the working shear stress is compared to the yield shear stress: \[ \text{F.O.S.} = \frac{\tau_y}{\tau_{max}} = \frac{150}{60} = 2.5 \]
This matches option (2).