Maximum beam moment from a uniform load
Use this lesson for a simply supported beam carrying a uniform load over its entire span. The largest sagging moment is at the middle because the loading is symmetric.
What the formula is saying
At midspan the left reaction contributes (wL/2)(L/2). The load on the left half contributes an opposite moment (wL/2)(L/4). Subtracting gives wL²/8.
Read the symbols in plain language
- w
- UDLN/m
- L
- Spanm
Sort out the units first
Use w in N/m and L in m. Because L is squared, (N/m) × m² becomes N·m. The example uses 10,000 N/m, equal to 10 kN/m.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
- w · UDL
- 10000 N/m
- L · Span
- 4 m
Square the span
A double span does not merely double the moment under fixed w: it makes it four times larger.
(4)^2 = 16 m²Multiply by the load intensity
Keep w as force per metre, not as the total load.
(10000) × (16) = 160000 N·mDivide by eight
The divisor belongs to this support and loading arrangement.
(160000) ÷ 8 = 20000 N·m
Does this worked answer make sense?
For the worked case, RA = 20,000 N. At midspan: 20,000 × 2 − 20,000 × 1 = 20,000 N·m.
Now try your own values
Change a value or its unit. The same method will show your calculation, step by step.
Results update only when you calculate. The lesson example above stays unchanged.
Your turn — check your understanding
Use these new values. Work it out first, then check your answer.
- w · UDL
- 6000 N/m
- L · Span
- 6 m
Find: Maximum beam moment from a uniform load
A hint, not the answer
At midspan the left reaction contributes (wL/2)(L/2). The load on the left half contributes an opposite moment (wL/2)(L/4). Subtracting gives wL²/8.
Use w in N/m and L in m. Because L is squared, (N/m) × m² becomes N·m. The example uses 10,000 N/m, equal to 10 kN/m.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Square the span
A double span does not merely double the moment under fixed w: it makes it four times larger.
(6)^2 = 36 m²Multiply by the load intensity
Keep w as force per metre, not as the total load.
(6000) × (36) = 216000 N·mDivide by eight
The divisor belongs to this support and loading arrangement.
(216000) ÷ 8 = 27000 N·m
Avoid the common trap
Do not enter wL into the w field: it would count the span twice. A load only on part of the beam needs another model.
When this method applies — and when it does not
An ideal straight beam with the supports and load stated here. Loads are magnitudes; self-weight is omitted unless already included. This is an equilibrium result, not a check of strength, deflection or stability. Enter the nonnegative magnitude of the downward load; sign conventions for internal moments are explained separately.
For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.
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Sources & further reading
References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.
Lesson updated: · Worked examples checked against the implemented formula; not an independent engineering certification.