A simply supported beam of length a carries a linearly varying load from zero at one end to w at the other end. Determine the maximum bending moment in terms of w and a (do not change symbols).

Introduction / Context:
The beam is simply supported and subjected to a triangular (linearly varying) load that is zero at one end and reaches w at the other end over span a. This classic load case appears in civil/mechanical design when wind pressure, soil pressure, or linearly distributed weights act on members. The task is to compute the location and magnitude of the maximum bending moment (M_max).

Given Data / Assumptions:

• Support conditions: simply supported at both ends.
• Load intensity: q(x) varies linearly from 0 to w over length a.
• Elastic, small-deflection theory; prismatic beam.
• We use x measured from the zero-load end.

Concept / Approach:
For a triangular load q(x) = (w/a)*x. The total load is W_total = (1/2)*w*a, acting at 2a/3 from the zero-load end. Compute reactions, write shear V(x) = R_A − ∫0^x q(s) ds, set V = 0 to find where bending moment is extremal, then integrate V to get M(x) and evaluate at the critical x.

Step-by-Step Solution:
1) Resultant load: W_total = (1/2)*w*a at x̄ = 2a/3 from the zero-load end.2) Reactions: R_A = W_total*(distance to right support)/a = (1/2*w*a)*(a/3)/a = w*a/6; R_B = W_total − R_A = w*a/3.3) Shear: V(x) = R_A − ∫0^x (w/a)*s ds = w*a/6 − (w/(2a))*x^2.4) Critical point: set V = 0 → w cancels → a/6 = x^2/(2a) → x = a/√3 from the zero-load end.5) Moment from V: M(x) = ∫ V dx = (w*a/6)*x − (w/(2a))*x^3/3 = (w*a*x)/6 − (w*x^3)/(6a).6) Evaluate at x = a/√3: M_max = (w*a^2)/(6√3) − (w*a^2)/(18√3) = w*a^2/(9√3).

Verification / Alternative check:
Units: w has units of force/length, multiplying by a^2 yields force*length, consistent with bending moment. Also, symmetry of the triangular load about the heavier end justifies the maximum occurring closer to that end at x = a/√3 from zero-load end.

Why Other Options Are Wrong:

• w a^2/8 and w a^2/12: Typical of uniform loads or different patterns, not this triangular case.
• w a^2/(6√3): This is the intermediate term before subtracting the cubic contribution; it overestimates M_max.

Common Pitfalls:

• Placing the resultant at the centroid wrongly (a/3 instead of 2a/3) from the zero-load end.
• Using uniform load formulae for a non-uniform (triangular) load.

Final Answer:
w a^2 / (9√3).