A circular beam section is subjected to a shear force of 40π kN. The maximum shear stress the material can take is 6 MPa. With a factor of safety of 2, the safe diameter of the section is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 15/11/2022 (Shift I)
Detailed Explanation
For a solid circular section the maximum shear stress, at the neutral axis, is 4/3 of the average V/A. With a factor of safety of 2, the working stress is half the material's maximum (the same as doubling the shear force).
\[ \tau_{allow} = \frac{6}{2} = 3\ \text{N/mm}^2,\qquad \tau_{max} = \frac{4}{3}\cdot\frac{V}{\pi D^2/4} = \frac{16V}{3\pi D^2} \]
\[ 3 = \frac{16\times 40\pi\times 10^3}{3\pi D^2} \Rightarrow D^2 = \frac{640\times 10^3}{9} = 71\,111 \Rightarrow D = 266.66\ \text{mm} \]
Why “266.66 mm” Is Correct
The safe diameter is 266.66 mm. The π in the shear force cancels the π in the area, which is why the numbers come out neatly.
Why the Other Options Are Wrong
- A. cannot be predicted using the given data: All the data needed (V, τ, FOS) are given.
- B. 133.33 mm: 133.33 mm is half the answer; it gives four times the allowable stress.
- C. 533.33 mm: 533.33 mm is twice the required diameter — safe but not the answer.
Key Point
Circle: τmax = (4/3)·V/(πD²/4); apply the factor of safety to the stress (or the load).
Key Concepts for Students
- Rectangle 1.5, circle 4/3, thin tube 2 — ratios of τmax to τavg.
- Dividing the stress by FOS or multiplying the load by FOS gives the same size.
- Shear stress varies as the area ratio, so D changes as the square root.
Practice more: this is question 8 in Bending and Shear Stress MCQs – Part 1. Try the whole set and check every answer instantly.