A bar of square section (area a²) is placed with one diagonal vertical. The maximum shear stress develops at a depth h below the top corner, where h is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 27/01/2018 (Morning)
Detailed Explanation
The commission cancelled this question: the printed options had lost the factor a (the book prints 2√3/4, 3√2/4, 2/3, √3/4). The options here are rewritten with the correct value. For a diamond, the width changes with depth, so the peak is not at the neutral axis.
Half-diagonal \(= a/\sqrt2\). At depth x from the top corner: width \(b_x = 2x\), area above \(A = x^2\), its centroid \(\bar y = \frac{a}{\sqrt2} - \frac{2x}{3}\) from the N.A.
\[ \tau_x = \frac{F\,x^2\left(\frac{a}{\sqrt2} - \frac{2x}{3}\right)}{\frac{a^4}{12}\cdot 2x} = \frac{6F}{a^4}\left(\frac{a\,x}{\sqrt2} - \frac{2x^2}{3}\right) \]
\[ \frac{d\tau}{dx} = 0 \Rightarrow x = \frac{3a}{4\sqrt2} = \frac{3\sqrt2}{8}a \approx 0.53a,\qquad \tau_{max} = \frac{9}{8}\,\tau_{avg} \]
Why “3√2a/8” Is Correct
The maximum shear stress is at 3√2a/8 below the top corner (and symmetrically above the bottom corner) — i.e. one-eighth of the diagonal from the neutral axis. Its value is 9/8 of the average, and at the neutral axis itself τ = τavg.
Why the Other Options Are Wrong
- A. a/√2 (at the neutral axis): At the neutral axis τ equals only the average; the peak is d/8 away from it.
- C. 2a/3: 2a/3 ≈ 0.667a is below the true location (0.530a).
- D. √3a/4: √3a/4 ≈ 0.433a is above the true location.
Key Point
Diamond (square, diagonal vertical): τmax = 9/8 τavg at d/8 from the N.A., i.e. 3√2a/8 from the corner.
Key Concepts for Students
- Triangle: τmax at h/2, = 1.5 τavg.
- Where width shrinks fast (near corners), τ need not peak at the N.A.
- Cancelled questions: learn the method, not the printed option.
Practice more: this is question 52 in Bending and Shear Stress MCQs – Part 4. Try the whole set and check every answer instantly.