Cantilever moment from a uniform load
A uniform load covers a cantilever’s full length. Replace the many small forces with one equivalent force, then measure its distance from the fixed end.
What the formula is saying
The resultant is wL and acts at the centre of the loaded length, L/2 from the fixing. Its moment magnitude is (wL)(L/2) = wL²/2.
Read the symbols in plain language
- w
- UDLN/m
- L
- Lengthm
Sort out the units first
w must be force per metre. With N/m and m, the moment is N·m. Divide by 1,000 to express it in kN·m.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
- w · UDL
- 2000 N/m
- L · Length
- 3 m
Find the equivalent total force
The rectangle under the load diagram has area w × L.
(2000) × (3) = 6000 NLocate the centre of that rectangle
The resultant acts at half the loaded length.
(3) ÷ 2 = 1.5 mFind the fixed-end moment magnitude
Multiply total force by the distance from the fixing.
(6000) × (1.5) = 9000 N·m
Does this worked answer make sense?
For the same full-span w and L, this moment magnitude is four times the simply supported maximum. The worked result is 9 kN·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
- 4000 N/m
- L · Length
- 2 m
Find: Cantilever moment from a uniform load
A hint, not the answer
The resultant is wL and acts at the centre of the loaded length, L/2 from the fixing. Its moment magnitude is (wL)(L/2) = wL²/2.
w must be force per metre. With N/m and m, the moment is N·m. Divide by 1,000 to express it in kN·m.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Find the equivalent total force
The rectangle under the load diagram has area w × L.
(4000) × (2) = 8000 NLocate the centre of that rectangle
The resultant acts at half the loaded length.
(2) ÷ 2 = 1 mFind the fixed-end moment magnitude
Multiply total force by the distance from the fixing.
(8000) × (1) = 8000 N·m
Avoid the common trap
wL²/8 belongs to a simply supported beam, not this cantilever. Partial loading changes the resultant position.
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.