The ratio of maximum shear stress to average shear stress in a circular beam is:
Try it first: pick an option, then read the explanation.
Correct answer: D. 4/3
Asked in SSC JE 22/01/2018 (Morning)
Detailed Explanation
The book lists four values and asks "1 only … 4 only"; the values are shown here directly. At the neutral axis of a circle, Aȳ = d³/12 and b = d.
\[ \tau_{max} = \frac{V\,(d^3/12)}{(\pi d^4/64)\,d} = \frac{16V}{3\pi d^2} = \frac{4}{3}\cdot\frac{V}{\pi d^2/4} \]
Why “4/3” Is Correct
The ratio is 4/3.
| Section | τmax / τavg | Where τmax occurs |
|---|---|---|
| Rectangle | 3/2 = 1.5 | Neutral axis |
| Solid circle | 4/3 ≈ 1.33 | Neutral axis |
| Triangle | 3/2 = 1.5 | Mid-height (h/2); at the N.A. it is 4/3 τavg |
| Square with a diagonal vertical (diamond) | 9/8 = 1.125 | d/8 above and below the N.A. (d = diagonal) |
| Thin-walled circular tube | 2 | Neutral axis |
| I-section | Much higher than τavg | Neutral axis, in the web |
Why the Other Options Are Wrong
- A. 2/3: 2/3 is less than 1 — the peak can never be below the average.
- B. 3/2: 3/2 is for a rectangle.
- C. 3/4: 3/4 is also less than 1.
Key Point
Circle 4/3; rectangle 3/2.
Key Concepts for Students
- τmax always ≥ τavg.
- Thin tube: 2.
- Diamond: 9/8.
Practice more: this is question 55 in Bending and Shear Stress MCQs – Part 4. Try the whole set and check every answer instantly.