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When a simply supported elastic beam of span $L$ and flexural rigidity $EI$ (where $E$ is the modulus of elasticity and $I$ is the moment of inertia of the section) is loaded with a uniformly distributed load $w$ per unit length, the deflection at the mid-span is $\Delta_0 = \frac{5wL^4}{384EI}$. If the load on one half of the span is now removed, the mid-span deflection ________.
- reduces to $\Delta_0/2$
- reduces to a value less than $\Delta_0/2$
- reduces to a value greater than $\Delta_0/2$
- remains unchanged at $\Delta_0$
Correct answer: reduces to a value greater than $\Delta_0/2$
Solution
The original deflection is due to UDL over the entire span. If the load on one half is removed, the mid-span deflection decreases, but not by half of $\Delta_0$ because the load distribution is no longer symmetric. By superposition, the reduction is less than the contribution of half the full-span load at the center, so the new deflection is greater than $\Delta_0/2$.
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