Deflection of a simply supported steel beam – midspan point load A simply supported steel beam of span L carries a concentrated load W at its centre. What is the central deflection in terms of W, L, E, and I (second moment of area)?

Difficulty: Easy

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

Explanation:


Introduction / Context:
Deflection control is a key serviceability criterion in beam design. The classical elastic curve solution for a simply supported beam under a central point load is a staple result, used to check deflection limits against code-prescribed span ratios.



Given Data / Assumptions:

  • Beam is prismatic with constant E and I.
  • Boundary conditions: simple supports at both ends (no end fixity).
  • Loading: a single concentrated load W at midspan.
  • Small deflections and linear elastic behavior.


Concept / Approach:
Use the standard integration of the Euler–Bernoulli beam equation: E * I * d^2y/dx^2 = M(x). For a central point load, the bending moment diagram is triangular, peaking at W * L / 4 at midspan. Integrating twice with appropriate boundary conditions yields the closed-form deflection at midspan.



Step-by-Step Solution:

Write maximum bending moment at midspan: Mmax = W * L / 4.Integrate beam equation over half-span, apply symmetry (slope zero at midspan, deflection zero at supports).Obtain central deflection: δ = (W * L^3) / (48 * E * I).


Verification / Alternative check:
Compare with tabulated deflection coefficients: for central point load on simply supported beam, coefficient = 1/48, matching the derived expression.



Why Other Options Are Wrong:
1/24 overestimates deflection by a factor of 2; (W * L^2)/(8 * E * I) has wrong dimensions; 5/384 corresponds to uniformly distributed load, not point load; (W * L)/(4 * E * I) is dimensionally inconsistent.



Common Pitfalls:
Mixing up the coefficient with that for uniformly distributed loads; forgetting to use consistent units for E and I.



Final Answer:
δ = (W * L^3) / (48 * E * I)

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