The distribution of shear stress intensity over a rectangular section of a beam follows:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 22/01/2018 (Evening)
Detailed Explanation
Use τ = VAȳ/(Ib) for a layer at distance y from the neutral axis. In a rectangle, b is constant, so τ follows the first moment of area Aȳ above the layer.
\[ A\bar y = b\left(\frac{d}{2}-y\right)\cdot\frac{1}{2}\left(\frac{d}{2}+y\right) = \frac{b}{2}\left(\frac{d^2}{4}-y^2\right) \]
\[ \tau = \frac{V}{2I}\left(\frac{d^2}{4}-y^2\right) = \frac{6V}{bd^3}\left(\frac{d^2}{4}-y^2\right) \]
Why “a parabolic curve” Is Correct
τ depends on y², so the distribution is a parabolic curve, with τ = 0 at y = ±d/2 and τmax = 1.5V/(bd) at the neutral axis.
Why the Other Options Are Wrong
- A. a circular curve: No circular arc arises from the expression.
- B. a straight line: A straight line describes bending stress, not shear stress.
- D. an elliptical curve: An ellipse would need τ² in the relation; τ itself varies with y².
Key Point
Rectangle: shear stress is parabolic, τmax = 1.5τavg at the N.A.
Key Concepts for Students
- Bending stress is linear in y; shear stress is parabolic in y.
- Shear stress is zero at free surfaces (top and bottom).
- Maximum shear and maximum bending stress occur at different fibres.
Practice more: this is question 10 in Bending and Shear Stress MCQs – Part 1. Try the whole set and check every answer instantly.