Simply supported beam with symmetric, zero-end, maximum-at-centre triangular load\nFor a simply supported beam carrying a distributed load that varies from zero at both ends to w per metre at the centre (symmetric triangular), the location of maximum bending moment is:

Introduction / Context:
In beams with non-uniform distributed loads, the maximum bending moment occurs where the shear force crosses zero. For a symmetric triangular load peaking at midspan and vanishing at the supports, determining the location is key to design.

Given Data / Assumptions:

• Simply supported beam, symmetric load w(x) increasing to a maximum at midspan and decreasing to zero at both ends.
• Linear elastic behavior; standard sign conventions.

Concept / Approach:
Shear force V(x) is the integral of −w(x), and bending moment M(x) is the integral of V(x). For a symmetric loading about midspan, reactions are equal, and the shear diagram is antisymmetric about the centre. The location where V(x) = 0 under symmetric loading is the centre.

Step-by-Step Reasoning:
Because loading is symmetric, support reactions are equal.The shear force changes sign only once and, by symmetry, this occurs at the midspan.Therefore, M(x) reaches an extremum at midspan, which is the maximum bending moment for this case.

Verification / Alternative check:
Constructing the shear and moment diagrams analytically for a symmetric triangular load confirms V = 0 at midspan and M maximum there.

Why Other Options Are Wrong:
Quarter-span locations are typical for certain point or trapezoidal loadings, not this symmetric zero-end triangular case; at supports, moment is zero for simply supported beams.

Common Pitfalls:
Confusing the zero-shear point for asymmetric loads; assuming uniform load behavior (where maximum also occurs at midspan, but for different reasons).

Final Answer:
At the centre of the span