The maximum shear stress (q_max) in a rectangular beam is:
Try it first: pick an option, then read the explanation.
Correct answer: B. 1.50 times the average
Asked in SSC JE 24/01/2018 (Evening)
Detailed Explanation
Put y = 0 (the neutral axis) in the parabolic shear stress expression for a rectangle and compare with the average V/(bd).
\[ \tau_{max} = \frac{6V}{bd^3}\cdot\frac{d^2}{4} = \frac{3}{2}\cdot\frac{V}{bd} = 1.5\,\tau_{avg} \]
Why “1.50 times the average” Is Correct
The peak at the neutral axis is 1.50 times the average shear stress.
| 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. 1.25 times the average: 1.25 is not the ratio for any standard section.
- C. 1.75 times the average: 1.75 is too high for a rectangle.
- D. 2.0 times the average: 2.0 is the ratio for a thin-walled circular tube.
Key Point
Rectangle τmax/τavg = 1.5; circle 4/3; thin tube 2.
Key Concepts for Students
- τavg = V/A.
- τ = VAȳ/(Ib) gives the whole distribution.
- The extra 50 % is why shear checks use τmax, not τavg.
Practice more: this is question 18 in Bending and Shear Stress MCQs – Part 2. Try the whole set and check every answer instantly.