Beam under tent-shaped load: A simply supported beam carries a load that varies gradually from zero at both supports to w per metre at midspan (symmetric). Is the statement “the bending moment at the centre is w l / 4” correct?

Introduction / Context:
Different load distributions produce different shear and bending moment diagrams. Dimensional consistency and known reference cases help validate formulas. The given statement resembles the classic point-load case at midspan rather than a distributed tent-shaped load.

Given Data / Assumptions:

• Simply supported beam of span l.
• Load intensity is zero at both supports and increases linearly to w per metre at midspan, then symmetrically decreases.
• w denotes load intensity (force per unit length).

Concept / Approach:
For a central point load W on a simply supported beam, M_max at midspan equals W * l / 4 (units of forcelength). For a uniformly distributed load q, M_max = q * l^2 / 8 (forcelength). For a tent-shaped distributed load peaking at midspan, the correct midspan moment involves w * l^2 multiplied by a coefficient distinct from 1/8 and certainly not w * l / 4, which is dimensionally inconsistent if w is per metre.

Step-by-Step Solution:

Verification / Alternative check:
Detailed integration of the symmetric triangular intensity gives M_center proportional to w * l^2 with a coefficient less than that for UDL. Regardless, w * l / 4 is invalid dimensionally and numerically.

Why Other Options Are Wrong:

• “Yes” contradicts both dimensional analysis and known beam results.
• “Only for very small spans” or “Depends on end fixity” are irrelevant; the form remains wrong.
• “Only if w is total load” would redefine symbols, but the stem clearly defines w as intensity.

Common Pitfalls:
Memorizing W l / 4 (point load) and misapplying it to distributed loads; ignoring unit consistency; mixing up W (total load) with w (load per unit length).

Final Answer:
No