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?
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Aone-fourth
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Bone-half
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Cdouble
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Dfour times
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Eone and a half times
Answer
Correct Answer: double
Explanation
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 bydelta_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:
For Beam A: delta_A = (W * l^3) / (48 * E * I)For Beam B: delta_B = (2W * l^3) / (48 * E * I)Take the ratio: delta_B / delta_A = (2W) / W = 2Therefore, delta_B = 2 * delta_A, i.e., double.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