Discussion
A simply supported horizontal beam is subjected to a distributed transverse load varying linearly from \(q_{0}\) at A to zero at B, as shown in the figure. Which one of the following options is correct?
Answer: A
The reaction at A is larger than that at B.
The triangular load has total magnitude \(\frac{1}{2}q_{0}l\), acting at its centroid, \(l/3\) from A, the heavy end. Taking moments about A and then summing vertical forces:
\[ \begin{aligned} R_{B}\,l &= \tfrac{1}{2}q_{0}l \cdot \tfrac{l}{3} \Rightarrow R_{B} = \tfrac{1}{6}q_{0}l \\ R_{A} &= \tfrac{1}{2}q_{0}l - \tfrac{1}{6}q_{0}l = \tfrac{1}{3}q_{0}l \end{aligned} \]
So \(R_{A} = 2R_{B}\). The reactions would be equal only for a load symmetric about mid-span. A simply supported beam has two unknown vertical reactions and two equilibrium equations, so the reactions are statically determinate.
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