Deflection sensitivity: A simply supported rectangular beam carries a central point load. If the beam's width is doubled (depth unchanged), by what factor does the midspan deflection change?

Introduction / Context:
Understanding how geometric changes affect beam deflection is central to structural stiffness tuning. For a simply supported beam with a central point load, deflection depends on span, material (E), and the section's second moment of area (I). This question isolates the effect of width, keeping depth constant.

Given Data / Assumptions:

• Beam is simply supported, rectangular cross-section.
• Central concentrated load P acts at midspan.
• Material modulus E and depth d remain constant.
• Only width b is doubled.

Concept / Approach:
For a simply supported beam with a midspan point load, the maximum deflection is δ = (P * L^3) / (48 * E * I). For a rectangle, I = b * d^3 / 12 about the neutral axis. Thus, δ ∝ 1/I ∝ 1/b when depth d is unchanged. Doubling b halves δ.

Step-by-Step Solution:
Start: δ1 = (P * L^3) / (48 * E * (b * d^3 / 12)) = (P * L^3) / (4 * E * b * d^3).Change: b → 2b, new I = 2b * d^3 / 12, so δ2 = (P * L^3) / (4 * E * (2b) * d^3) = (1/2) * δ1.Therefore, the deflection reduces to half: factor = 1/2.

Verification / Alternative check:
Dimensional consistency holds. Also, from stiffness k_bending ∝ E * I / L^3, doubling b doubles I and hence doubles stiffness, which halves deflection for the same load.

Why Other Options Are Wrong:
'1/8': would require I to increase eightfold (e.g., doubling depth, not width).'2', '4', '8': imply deflection increases, which contradicts δ ∝ 1/b for fixed depth.

Common Pitfalls:
Confusing effects of width and depth; using I ∝ d^3 incorrectly with b; forgetting the 1/I dependence of deflection.

Final Answer:
1/2