Beam theory – Bending moment at mid-span for UDL\n\nA simply supported beam of span l carries a uniformly distributed load of intensity w per unit length over the entire span. What is the bending moment at the center of the beam?

Introduction / Context:
For basic loading cases, memorizing key formulas helps in quick checks and preliminary sizing. A simply supported beam under a full-span uniformly distributed load (UDL) is among the most common scenarios.

Given Data / Assumptions:

• Span l, UDL = w (force per unit length).
• Supports are simple (no end moments), beam is prismatic.
• Linear elastic behavior.

Concept / Approach:
By symmetry, reactions at supports are R_A = R_B = w l / 2. The bending moment at a distance x from the left is M(x) = R_A x − w x^2 / 2. The maximum occurs where shear V(x) = 0, i.e., at x = l/2.

Step-by-Step Solution:
Compute reactions: R_A = R_B = w l / 2.Moment at mid-span: M(l/2) = (w l / 2)(l/2) − w(l/2)^2 / 2.Simplify: M(l/2) = w l^2 / 4 − w l^2 / 8 = w l^2 / 8.Hence, the center bending moment equals w l^2 / 8.

Verification / Alternative check:
The parabolic bending moment diagram for full UDL peaks at the center with the known value w l^2 / 8; area under shear diagram also corroborates this result.

Why Other Options Are Wrong:
Zero is inconsistent with UDL; w l^2 / 2 and /4 are too high; /12 is not the standard value for this case.

Common Pitfalls:
Forgetting the 1/2 factor in the distributed load moment term; mixing with the point-load mid-span case P L / 4.

Final Answer:
w l^2 / 8