Beam deflection under central load: A simply supported beam A of length l, breadth b, and depth d carries a central load W. Another identical beam B carries a central load of 2W. How does the midspan deflection of beam B compare with that of beam A?

Introduction / Context:
For linearly elastic beams, midspan deflection under a central point load is directly proportional to the magnitude of that load when all geometric and material properties remain the same. This question checks recall of proportionality in classic beam-bending relations for a simply supported beam.

Given Data / Assumptions:

• Beam A and Beam B are identical: same length l, breadth b, depth d, modulus of elasticity E, and moment of inertia I.
• Both beams are simply supported with a central point load.
• Loads: W on A, and 2W on B.
• Small deflection theory, linear elasticity, and no shear deformation effects.

Concept / Approach:
For a simply supported beam with a central point load P, the maximum deflection at midspan is given by
delta_max = (P * l^3) / (48 * E * I)
Since E, I, and l are unchanged between the two beams, the deflection ratio equals the load ratio.

Step-by-Step Solution:

Verification / Alternative check:
A quick dimensional check confirms deflection dependence on load P linearly, so doubling P doubles the deflection when geometry and material remain constant.

Why Other Options Are Wrong:

• One-fourth / one-half / four times / one and a half times: These contradict the direct proportionality of deflection to load for this case.

Common Pitfalls:
Confusing this with uniformly distributed load or cantilever formulae; mixing up how deflection scales with depth (via I) versus with load P.

Final Answer:
double