A simply supported beam AB of span L carries a UDL w over the entire span. The bending moment at a distance L/4 from end A is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 10/10/2023 (5 pm–7 pm)
Detailed Explanation
The bending moment at a section is the sum of the moments, about that section, of all forces on one side. On the left of x = L/4 there is the reaction at A and a UDL of length L/4 whose resultant acts at its middle.
\[ R_A = R_B = \frac{wL}{2} \]
\[ M_{L/4} = \frac{wL}{2}\cdot\frac{L}{4} - w\cdot\frac{L}{4}\cdot\frac{L}{8} = \frac{wL^2}{8} - \frac{wL^2}{32} = \frac{3wL^2}{32} \]
Check with the general formula \( M = \frac{wx}{2}(L - x) = \frac{w}{2}\cdot\frac{L}{4}\cdot\frac{3L}{4} = \frac{3wL^2}{32} \).
Why “3wL²/32” Is Correct
The moment at quarter span is 3wL²/32 — option (d) — which is three-quarters of the mid-span maximum wL²/8.
Why the Other Options Are Wrong
- A. wL²/32: wL²/32 is only the UDL term; the reaction term wL²/8 was left out.
- B. Zero: The moment is zero only at the supports.
- C. wL²/8: wL²/8 is the maximum, at mid-span, not at L/4.
Key Point
SS beam with UDL: M(x) = (wx/2)(L − x).
Key Concepts for Students
- At x = L/4, M = 0.75 × M_max.
- The resultant of a UDL acts at the middle of its loaded length.
- The BMD under a UDL is a quadratic parabola.
Practice more: this is question 10 in Bending Moment and Shear Force MCQs – Part 1. Try the whole set and check every answer instantly.