A cantilever of length l and uniform strength carries a UDL w. Keeping the depth d constant, the width varies from zero at the free end to ______ at the fixed end.
Try it first: pick an option, then read the explanation.
Correct answer: B. (3w/σd²) × l²
Asked in SSC JE 23/01/2018 (Morning)
Detailed Explanation
In a beam of uniform strength the extreme-fibre stress is the same σ at every section. With the depth fixed, the width must follow the bending moment.
\[ \frac{w x^2}{2} = \sigma\,\frac{b_x d^2}{6} \Rightarrow b_x = \frac{3 w x^2}{\sigma d^2} \]
\[ x = l:\quad b = \frac{3w}{\sigma d^2}\,l^2 \]
Why “(3w/σd²) × l²” Is Correct
The width at the fixed end is (3w/σd²) × l²; in plan the beam is a parabola, zero width at the tip.
Why the Other Options Are Wrong
- A. (2w/σd²) × l²: The factor 2 comes from forgetting the 6 in Z = bd²/6.
- C. (3w/σd) × l²: d must be squared (Z ∝ d²).
- D. (5w/σd) × l²: Both the factor and the power of d are wrong.
Key Point
Uniform strength, constant depth: b ∝ M; cantilever + UDL: b = 3wl²/(σd²) at the fixed end.
Key Concepts for Students
- Constant width instead: d ∝ √M.
- Leaf springs approximate beams of uniform strength.
- Near the free end a minimum width is still needed for shear.
Practice more: this is question 54 in Bending and Shear Stress MCQs – Part 4. Try the whole set and check every answer instantly.