Max Bending Moment — Centre Point Load
Also known as PL/4
Worked example: 10 kN at midspan of 6 m → 15 kN·m — press Try an example to run it live, then adjust anything.
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Max Bending Moment — Centre Point Load explained
Drop a single load P at the middle of a beam that simply rests on two supports and each support takes P/2. Walk out to midspan and the moment there is (P/2)(L/2) = PL/4 — the largest anywhere on the beam, and it falls off linearly to zero at each support. A 10 kN hoist hung at the centre of a 6 m span produces M = 10 000 × 6 ÷ 4 = 15 000 N·m = 15 kN·m.
The lesson buried in the L is that span is expensive: doubling the span doubles the moment for the same load, and the beam you need grows much faster than that once deflection is added to the picture. The trap is the word simply supported. This result assumes the ends are free to rotate — a beam of the same span with fixed ends carries only PL/8 at midspan, and a cantilever of length L with the load at its tip carries PL, four times as much. Check the end conditions before reaching for the coefficient, and remember to add the beam's own weight, which contributes its own wL²/8.
Max Bending Moment — Centre Point Load formula
- = Maximum bending moment (N·m)
- = Point load (N)
- = Span (m)
Missing one of these? Work it out first, then come back
- Maximum bending moment — Max Bending Moment — Uniform Load, Max Moment — Simple Beam, Off-Centre Point Load
- Point load — Beam Deflection — Simply Supported, Centre Load, Max Moment — Simple Beam, Off-Centre Point Load
- Span — Max Bending Moment — Uniform Load, Beam Deflection — Simply Supported, Centre Load