The general expression for the bending moment of a beam of length l is M = (wl/2)x − wx²/2. The beam carries:

Asked in SSC JE 24/01/2018 (Morning)

Detailed Explanation

Read the expression term by term. The first term is a reaction times its lever arm; the second is the moment of the load to the left of the section about the section.

\( R_A = \dfrac{wl}{2} \Rightarrow \) first term \( R_A x = \dfrac{wl}{2}x \)

Load on length x: \( wx \), acting at \( x/2 \Rightarrow \) moment \( wx \cdot \dfrac{x}{2} = \dfrac{wx^2}{2} \)

Check: \( \dfrac{d^2M}{dx^2} = -w \) (constant) ⇒ uniform load.

Why “a uniformly distributed load w per unit length” Is Correct

Both terms match a simply supported beam with a uniformly distributed load w — option (a). The expression peaks at x = l/2 with M = wl²/8.

Simply supported beam with UDL w over span L. Reactions wL/2. Shear varies linearly from +wL/2 to −wL/2 and is zero at mid-span; the BMD is a parabola with its peak wL²/8 at mid-span, where the shear is zero.

Why the Other Options Are Wrong

B. a load varying linearly from zero at one end to w at the other: A triangular load gives an x³ term in M.
C. an isolated load at mid-span: A central point load gives a linear (not quadratic) M on each half.
D. None of these: The expression fits a UDL exactly.

Key Point

d²M/dx² = −w: a quadratic M with constant second derivative means a UDL.

Key Concepts for Students

dM/dx = wl/2 − wx = V(x).
V = 0 at x = l/2 ⇒ M_max = wl²/8.
A triangular load gives a cubic M(x).

Practice more: this is question 34 in Bending Moment and Shear Force MCQs – Part 3. Try the whole set and check every answer instantly.