Cantilever beam of length l carrying a uniformly distributed load w per unit length: Does the maximum bending moment occur at the middle of the span?

Difficulty: Easy

Correct Answer: False

Explanation:


Introduction / Context:
For a cantilever with UDL, locating the critical section is key to safe design. The support (fixed end) must resist the largest bending moment and shear.



Given Data / Assumptions:

  • Cantilever of span l, fixed at one end and free at the other.
  • Uniformly distributed load w along the entire span.
  • Linear elastic behaviour; self-weight can be included in w.


Concept / Approach:
Bending moment in a cantilever is largest where the lever arm of the resultant load to the support is greatest. For UDL, the resultant equals w * l acting at midspan, but the resisting moment is taken at the fixed end.



Step-by-Step Solution:

Resultant load: W = w * l at l/2 from the fixed end.Reaction and fixing moment at the wall balance W.Maximum bending moment magnitude at the fixed end: M_max = w * l * (l/2) = w * l^2 / 2.Bending moment reduces linearly to zero at the free end; thus the middle is not critical.


Verification / Alternative check:
From differential relations, dM/dx = V; for UDL, V varies linearly and M is quadratic, peaking at the fixed boundary.



Why Other Options Are Wrong:
Claiming the midspan is maximum ignores boundary conditions of a cantilever.Other conditional statements do not apply to this canonical case.



Common Pitfalls:
Transferring simply supported intuition to cantilevers; forgetting that internal moments are highest at the restraint.



Final Answer:

False

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