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Simply supported beam of span l under uniformly distributed load w (per unit length): The location of maximum deflection and its value—does the maximum occur at midspan?

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:

  • Straight, prismatic beam; simply supported ends.
  • Uniformly distributed load w along the whole span.
  • Elastic analysis; constant E and I.


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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