A fixed beam carries a uniformly distributed load over its entire span L. The points of contraflexure occur on either side of the centre at a distance of:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 03/03/2017 (Evening)
Detailed Explanation
Points of contraflexure are where the bending moment changes sign (M = 0). Write the moment at x from A with RA = wL/2 and the fixing moment wL²/12.
\[ \frac{wL}{2}x - \frac{wx^2}{2} - \frac{wL^2}{12} = 0 \;\Rightarrow\; x^2 - Lx + \frac{L^2}{6} = 0 \;\Rightarrow\; x = \frac{L}{2}\left(1 \pm \frac{1}{\sqrt3}\right) \]\[ \left|x - \frac L2\right| = \frac{L}{2\sqrt3} \approx 0.289L \quad (\text{i.e. } 0.211L \text{ from each support}) \]
Why “L/(2√3)” Is Correct
The points of contraflexure are L/(2√3) on either side of the centre.
Why the Other Options Are Wrong
- A. L/√3: L/√3 ≈ 0.577L from the centre lies beyond the supports.
- B. L/3: L/3 ≈ 0.333L from the centre would be only 0.167L from the supports.
- D. L/(4√3): L/(4√3) ≈ 0.144L is half the correct distance.
Key Point
Fixed beam, UDL: contraflexure 0.211L from each support = L/(2√3) from the centre.
Key Concepts for Students
- Central point load: contraflexure at L/4 from each end.
- Between the contraflexure points the beam acts like a simply supported span of 0.577L.
- Measure carefully — from the centre or from the supports.
Practice more: this is question 7 in Fixed Beam MCQs – Part 1. Try the whole set and check every answer instantly.