Simply supported beam with a triangularly varying load: Load varies linearly from zero at support B to w per unit length at support A over span l. What is the shear force at B (i.e., reaction at B)?

Introduction:Triangularly distributed loads produce resultant forces equal to the triangle area and act at one-third from the larger-intensity end. This problem checks equilibrium and load-resultant concepts.Given Data / Assumptions:

• Simply supported beam; span l.
• w at A, zero at B (linear variation).
• Statics; small deflection effects ignored.

Concept / Approach:Replace the triangular load with its single resultant W and its line of action. Use moment equilibrium about a support to compute reactions.Step-by-Step Solution:

Verification / Alternative check:Sum of vertical forces: R_A + R_B = W = wl/2; with R_B = wl/6 ⇒ R_A = wl/3 (consistent).Why Other Options Are Wrong:

• wl/3 or 2wl/3 or wl/2 or wl: do not satisfy both force and moment equilibrium for this load pattern.

Common Pitfalls:Placing the resultant at midspan; forgetting it lies at one-third from the larger load end; mixing up which end has intensity w.

Final Answer: