Difficulty: Easy
Correct Answer: True
Explanation:
Introduction / Context:Serviceability design limits deflection to protect finishes and function. For common loading cases, knowing where the maximum deflection occurs avoids unnecessary full integrations.
Given Data / Assumptions:
Concept / Approach:The classical solution for a simply supported beam with full-span UDL gives maximum deflection at the midspan (x = l/2) with value:delta_max = 5 w l^4 / (384 E I). Symmetry of loading and supports ensures the extremum is at the center.
Step-by-Step Solution:
Write bending moment: M(x) = (w/2)(l x − x^2).Curvature: d^2y/dx^2 = M(x)/(E I).Integrate twice, apply boundary conditions y(0) = 0, y(l) = 0.Evaluate deflection at x = l/2 → delta_max = 5 w l^4 / (384 E I).Verification / Alternative check:Use standard beam tables or Castigliano’s theorem; both return the same midspan maximum and value.
Why Other Options Are Wrong:“False” contradicts the symmetric solution.Linear varying w changes the position; but here w is uniform.End fixity is a different support condition; not applicable.Cantilevers have different deflection profiles; the maximum is at the free end, not midspan.
Common Pitfalls:Mixing formulas for point loads and UDLs; forgetting the 5/384 factor; confusing simply supported with fixed-ended results.
Final Answer:
True
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