A homogeneous simply supported prismatic beam of width B, depth D and span L carries a concentrated load of 200 kN, which can be placed anywhere on the span. The maximum flexural stress developed in the beam is:
Try it first: pick an option, then read the explanation.
Correct answer: C. 300L/BD²
Asked in SSC JE 11/12/2020 (Evening)
Detailed Explanation
A point load W at distance a from one support of a simple span gives M = Wa(L − a)/L under the load. This is largest when a = L/2, so the worst position is mid-span.
\[ M_{max} = \frac{WL}{4} = \frac{200L}{4} = 50L,\qquad Z = \frac{BD^2}{6} \]
\[ \sigma_{max} = \frac{M}{Z} = \frac{50L\times 6}{BD^2} = \frac{300L}{BD^2} \]
Why “300L/BD²” Is Correct
With the load at mid-span, the maximum flexural (bending) stress is 300L/BD² (in the units of the load).
Why the Other Options Are Wrong
- A. 600L/BD²: 600L/BD² uses M = WL/2, the moment of a cantilever, not of a simple span.
- B. 75L/BD³: BD³ is the wrong dimension; Z for a rectangle is BD²/6.
- D. 75L/2BD²: 75L/2BD² is far too small; it does not use Z = BD²/6.
Key Point
Single moving load on a simple span: worst at mid-span, M = WL/4; σ = 6M/(BD²).
Key Concepts for Students
- For a moving load, place it to maximise M (influence line idea).
- σ = M/Z is the flexure formula at the extreme fibre.
- Doubling depth cuts the stress to one quarter.
Practice more: this is question 11 in Bending and Shear Stress MCQs – Part 1. Try the whole set and check every answer instantly.