The ratio of the maximum to the average shear stress on a rectangular section is:
Try it first: pick an option, then read the explanation.
Correct answer: C. 1.5
Asked in SSC JE 22/01/2018 (Evening), 04/03/2017 (Evening), CWC & MES 2011
Detailed Explanation
At the neutral axis the area above is b·d/2 with its centroid at d/4, so Aȳ = bd²/8.
\[ q_{max} = \frac{F\cdot (bd^2/8)}{(bd^3/12)\,b} = \frac{3}{2}\cdot\frac{F}{bd} \Rightarrow \frac{q_{max}}{q_{avg}} = 1.5 \]
Why “1.5” Is Correct
The ratio is 1.5: the peak is 50 % above the average.
| 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. 1: A ratio of 1 would mean uniform shear, which is impossible since τ = 0 at the free surfaces.
- B. 1.25: 1.25 does not correspond to the rectangle.
- D. 2: 2 is for a thin circular tube.
Key Point
Rectangle q_max/q_avg = 1.5.
Key Concepts for Students
- Circle 4/3, diamond 9/8, thin tube 2.
- q_avg = F/A.
- The derivation uses τ = FAȳ/(Ib).
Practice more: this is question 31 in Bending and Shear Stress MCQs – Part 3. Try the whole set and check every answer instantly.