Simply Supported Beam – Central Point Load A simply supported beam of span L and constant flexural rigidity E I carries a single concentrated load W at midspan. What is the maximum deflection at the centre in terms of W, L, E, and I?

Difficulty: Easy

Correct Answer: W L^3 / (48 E I)

Explanation:


Introduction / Context:
Deflection of simply supported beams under common loads is a staple result in structural analysis. The central point load case yields a widely used benchmark for serviceability checks and for verifying software and hand calculations.


Given Data / Assumptions:

  • Simply supported beam, span L.
  • Flexural rigidity E I is constant.
  • Single point load W at midspan.
  • Small deflections, linear elasticity, plane sections remain plane.


Concept / Approach:

By symmetry, reactions are W/2 at each support. Use the relation E I * d^2y/dx^2 = M(x) with a piecewise moment function or use standard integration for symmetric loading. Maximum deflection occurs at x = L/2.


Step-by-Step Solution:

Take origin at left support, 0 ≤ x ≤ L/2: M(x) = (W/2) xE I * d^2y/dx^2 = (W/2) xIntegrate twice and enforce y = 0 at x = 0 and continuity of slope and deflection at x = L/2Resulting centre deflection: delta_max = W L^3 / (48 E I)


Verification / Alternative check:

Matches tabulated formula and Castigliano theorem. Numerical check with w distribution equivalence confirms the magnitude.


Why Other Options Are Wrong:

W L^3 / (3 E I) is for a cantilever with end load. 5 W L^3 / (384 E I) is for a uniform load case with total load W. W L^2 / (8 E I) has incorrect dimension. W L^3 / (96 E I) is half the correct value.


Common Pitfalls:

Applying cantilever formulas, missing symmetry, or mixing total load with intensity.


Final Answer:

W L^3 / (48 E I)

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