Maximum bending moment for a simply supported beam under UDL A simply supported beam of span L carries a uniformly distributed load of intensity w (force per unit length) over its entire length. What is the maximum bending moment (in consistent units)?

Introduction / Context:
Uniformly distributed loads (UDL) are common in slabs, bridge decks, and purlins. Determining the peak bending moment quickly and correctly is foundational to sizing members and reinforcement and to checking deflections and crack control.

Given Data / Assumptions:

• Beam is simply supported with clear span L.
• Load is uniformly distributed along the full length with intensity w (force/length).
• Linear elastic, small-deflection theory; self-weight may be included in w.

Concept / Approach:
For a simply supported beam with UDL over the entire span, reactions are equal and the bending moment diagram is a parabola. The maximum bending moment occurs at midspan and equals w L^2 / 8. Using W = w L (total load), the same result can be written as (W * L) / 8.

Step-by-Step Solution:
Support reactions: R_A = R_B = w L / 2.Moment at a section x from the left: M(x) = R_A * x − w x^2 / 2.Differentiate: dM/dx = R_A − w x = 0 → x = L / 2 (midspan).Maximum moment: M_max = (w L / 2) * (L / 2) − w (L / 2)^2 / 2 = w L^2 / 8.

Verification / Alternative check:
Area method: the shear diagram is a straight line from +wL/2 to −wL/2; the area under shear up to midspan equals w L^2 / 8, which is the peak moment.

Why Other Options Are Wrong:
w L / 8 has wrong dimensions (moment must be forcelength). W L / 8 is correct only if W = w L is clearly defined; option C is a note, not a fundamental expression. w L^2 / 12 and w L^2 / 4 are standard values for other boundary conditions or checks but not for this case.

Common Pitfalls:
Confusing w (intensity) with W (total load). Dimensional analysis is a quick sanity check: bending moment must carry units of forcelength.

Final Answer:
w L^2 / 8