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)?
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Awl/6
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Bwl/3
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Cwl
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D2wl/3
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Ewl/2
Answer
Correct Answer: wl/6
Explanation
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:
Resultant magnitude: W = (1/2)wlLocation: from the higher-intensity end (A) at l/3Taking moments about A: R_Bl = W*(l/3) = (wl/2)(l/3) = wl^2/6Therefore R_B = wl/6Verification / 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:
w*l/6