Question

When a simply-supported elastic beam of span $L$ and flexural rigidity $EI$ is loaded with a uniformly distributed load $w$ per unit length, the deflection at the mid-span is \[ \Delta_0=\frac{5}{384}\,\frac{wL^4}{EI}. \] If the load on one half of the span is now removed, the mid-span deflection \underline{\hspace{1.5cm}.}

• A. reduces to $\Delta_0/2$
• B. reduces to a value less than $\Delta_0/2$
• C. reduces to a value greater than $\Delta_0/2$
• D. remains unchanged at $\Delta_0$

Hint:

For linear elastic beams, break complicated loads into simpler parts and use superposition. If a full-span UDL is the sum of two symmetric half-UDLs, the mid-span deflection under a single half-UDL must be $\frac{1}{2}$ of the full-UDL value.

Answer 1

The Correct Option is A

Step 1: Use superposition.
Write the full UDL as the sum of two half-span UDLs: one on $[0,L/2]$ and one on $[L/2,L]$.
Let $\delta_{\text{half}}$ be the mid-span deflection when only one half is loaded (either left or right).

Step 2: Exploit symmetry at mid-span.
The mid-span deflection from loading the left half equals that from loading the right half (by symmetry).
Hence, the full UDL mid-span deflection is \[ \Delta_0=\delta_{\text{half}}+\delta_{\text{half}}=2\,\delta_{\text{half}}. \]

Step 3: Conclude for the new loading.
With only one half loaded, the mid-span deflection is \[ \delta_{\text{half}}=\frac{\Delta_0}{2} =\frac{1}{2}\left(\frac{5}{384}\frac{wL^4}{EI}\right) =\frac{5}{768}\frac{wL^4}{EI}. \] Therefore, it reduces to $\Delta_0/2$.
\[ \boxed{\delta_{\text{mid (half UDL)}}=\Delta_0/2} \]