Keeping the breadth b constant, the depth of a cantilever of length l and uniform strength, carrying a UDL w, varies from zero at the free end to ______ at the fixed end.
Try it first: pick an option, then read the explanation.
Asked in SSC JE 25/01/2018 (Evening)
Detailed Explanation
The commission cancelled this question because the printed options were wrong (they had d in the denominator and l inside the root). The options here are rewritten so that the correct expression appears.
\[ \frac{w x^2}{2} = \sigma\,\frac{b\,d_x^2}{6} \Rightarrow d_x^2 = \frac{3 w x^2}{b\,\sigma} \Rightarrow d_x = x\sqrt{\frac{3w}{b\sigma}} \]
\[ x = l:\quad d = l\sqrt{\frac{3w}{b\sigma}} \]
Why “√(3w/(σb)) × l” Is Correct
The depth at the fixed end is √(3w/(σb)) × l, and the depth varies linearly from the free end.
Why the Other Options Are Wrong
- A. √(2w/(σb)) × l: The factor 2 forgets the 6 in Z = bd²/6.
- C. (3w/(σb)) × l: Without the square root the units do not give a length.
- D. √(3wl/(σb)): l must be outside the root because M ∝ l².
Key Point
Constant breadth, UDL: d = l√(3w/(bσ)) at the fixed end (linear variation).
Key Concepts for Students
- Constant depth instead: b = 3wl²/(σd²).
- Check units: w (N/mm)/(b σ) (mm·N/mm²) is dimensionless.
- Cancelled questions still test the method.
Practice more: this is question 61 in Bending and Shear Stress MCQs – Part 5. Try the whole set and check every answer instantly.