Max Bending Moment — Uniform Load
Also known as wL²/8
Worked example: 5 kN/m over 4 m → 10 kN·m — press Try an example to run it live, then adjust anything.
Enter your known values, leave one input blank, and solves for the missing one. Tap a variable’s symbol to see what it means, with a typical value. Try different units for next level excitement!
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UniversityMechanics of Materials
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Max Bending Moment — Uniform Load explained
wL²/8 is the most-used number in structural engineering. Spread a load evenly along a simply supported beam — its own weight, a floor, snow on a roof, a run of water-filled pipe — and the moment diagram is a parabola peaking at midspan with the value wL²/8. A 5 kN/m load on a 4 m span gives M = 5000 × 16 ÷ 8 = 10 000 N·m. The square on L is the important part: stretch the span by 50% and the moment goes up by 125%.
Enter w as force per unit length — this calculator borrows the N/m unit family, which also offers kN/m, lbf/ft and lbf/in. Two traps. First, converting an area load to a line load requires the tributary width: 2.4 kPa of floor load on joists at 400 mm centres is 2.4 × 0.4 = 0.96 kN/m per joist, and forgetting the tributary width is the most common error in the whole calculation. Second, this is the simply supported case; a fixed-fixed beam peaks at wL²/12 over the supports and only wL²/24 at midspan, and a propped cantilever gives wL²/8 at the fixed end. Continuous multi-span beams are somewhere in between, which is why they use less steel.
Max Bending Moment — Uniform Load formula
- = Maximum bending moment (N·m)
- = Uniform load per unit length (N/m)
- = Span (m)
Missing one of these? Work it out first, then come back
- Maximum bending moment — Max Bending Moment — Centre Point Load, Max Moment — Simple Beam, Off-Centre Point Load
- Uniform load per unit length — Beam Deflection — Simply Supported, Uniform Load, Cantilever Deflection — Uniform Load
- Span — Max Bending Moment — Centre Point Load, Beam Deflection — Simply Supported, Centre Load