The shear stress distribution over the depth d of a beam of rectangular cross-section, subjected to transverse loading, is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 03/03/2017 (Morning)
Detailed Explanation
The book shows four sketches; they are described here in words. Shear stress is zero at the free top and bottom surfaces and grows towards the neutral axis as more area lies above the layer.
\[ \tau(y) = \frac{6V}{bd^3}\left(\frac{d^2}{4}-y^2\right),\qquad \tau(0) = \frac{3V}{2bd} \]
Why “a single symmetric parabola, maximum at mid-depth” Is Correct
The expression is a parabola in y, symmetric about the neutral axis and peaking there. So the distribution is a single symmetric parabola with its maximum at mid-depth.
Why the Other Options Are Wrong
- A. two triangles meeting at zero at mid-depth: Two triangles meeting at zero describes the bending stress diagram (if one triangle is flipped), not shear.
- B. two parabolas pinched to zero at mid-depth: Shear stress is maximum, not zero, at the neutral axis.
- C. a single triangle, maximum at one edge: A single triangle would mean τ varies linearly and is maximum at an edge, where it is actually zero.
Key Point
Rectangle: parabolic shear stress, zero at the edges, maximum 1.5 τavg at the N.A.
Key Concepts for Students
- Shear and bending stress diagrams are complementary: one peaks where the other is zero.
- τavg = V/(bd).
- Shear stress at a free surface must be zero (complementary shear).
Practice more: this is question 14 in Bending and Shear Stress MCQs – Part 1. Try the whole set and check every answer instantly.