A simply supported beam of length a carries a load varying from zero at one end to w per unit length at the other. The maximum bending moment is:

Asked in SSC JE 22/01/2018 (Morning)

Detailed Explanation

Find the reactions, write the shear at a section x from the zero-load end, set it to zero to locate the maximum moment, then evaluate the moment there. (In the exam the four expressions were numbered 1–4 and the answer was “4 only”.)

Total load \( = \tfrac{1}{2}wa \), acting at \( 2a/3 \) from A ⇒ \( R_A = \dfrac{wa}{6},\ R_B = \dfrac{wa}{3} \).

Intensity at x: \( \dfrac{wx}{a} \). Shear: \( V_x = \dfrac{wa}{6} - \dfrac{wx^2}{2a} = 0 \Rightarrow x = \dfrac{a}{\sqrt3} \)

\[ M_x = \frac{wa}{6}x - \frac{wx^3}{6a} \Rightarrow M_{max} = \frac{wa^2}{6\sqrt3} - \frac{wa^2}{18\sqrt3} = \frac{wa^2}{9\sqrt3} \approx 0.064\,wa^2 \]

Why “wa²/(9√3)” Is Correct

The maximum moment is wa²/(9√3) — option (d), occurring at 0.577a from the zero-load end.

Simply supported beam of span a with load rising from zero at A to w at B. Reactions: wa/6 at A and wa/3 at B. The SFD is a parabola that crosses zero at x = a/√3 ≈ 0.577a from A; there the BMD (a cubic curve) peaks at wa²/(9√3).

Why the Other Options Are Wrong

A. wa/27: wa/27 has the units of a force, not a moment.
B. wa²/27: wa²/27 ≈ 0.037wa² is too small.
C. w²a/√27: w²a/√27 has wrong units (w squared).

Key Point

Triangular load on SS beam: M_max = wL²/(9√3) at L/√3 from the zero end.

Key Concepts for Students

Reactions are in the ratio 1 : 2 (zero end : heavy end).
The SFD is a parabola, the BMD a cubic curve.
Compare: full UDL of the same peak w gives wL²/8 ≈ 0.125wL².

Practice more: this is question 36 in Bending Moment and Shear Force MCQs – Part 3. Try the whole set and check every answer instantly.