The shear stress distribution over a beam of solid circular section is such that:
Try it first: pick an option, then read the explanation.
Correct answer: C. q_max = 1.33 q_mean
Asked in SSC JE 2010
Detailed Explanation
At the neutral axis of a circle the area above is a semicircle (πd²/8) whose centroid is 2d/(3π) above the axis.
\[ A\bar y = \frac{\pi d^2}{8}\cdot\frac{2d}{3\pi} = \frac{d^3}{12},\qquad q_{max} = \frac{F\,(d^3/12)}{(\pi d^4/64)\,d} = \frac{4}{3}\cdot\frac{F}{\pi d^2/4} \]
Why “q_max = 1.33 q_mean” Is Correct
The peak is 4/3 ≈ 1.33 times the mean shear stress.
Why the Other Options Are Wrong
- A. q_max = 2 q_mean: 2 is for a thin-walled circular tube.
- B. q_max = 1.5 q_mean: 1.5 is for a rectangle.
- D. q_max = 1.25 q_mean: 1.25 does not belong to any standard section.
Key Point
Solid circle: q_max = (4/3) q_mean.
Key Concepts for Students
- Semicircle centroid = 4r/(3π) = 2d/(3π) from the diameter.
- Shafts in bending + shear: check τ at the N.A.
- Thin tube: q_max = 2 q_mean.
Practice more: this is question 39 in Bending and Shear Stress MCQs – Part 3. Try the whole set and check every answer instantly.