Learn: Composite/internal-force moment sum
A composite-section resistance model may contain several force couples or resultants acting at different lever arms. This lesson adds two signed force-times-distance contributions about one common reference.
What the formula is saying
Each contribution is Fi zi. Compute each moment separately so the force and its own lever arm remain paired, then add them algebraically with a consistent sign convention.
Read the symbols in plain language
- F₁
- Internal force 1
Internal force 1. Keep the first force paired with its own perpendicular lever arm.
kNKilonewtons measure force; 1 kN = 1000 N.
- z₁
- Lever arm 1
Lever arm 1. Keep the first force paired with its own perpendicular lever arm.
mMetres measure length; 1 m = 1000 mm.
- F₂
- Internal force 2
Internal force 2. Use the same reference and sign convention for the second force contribution.
kNKilonewtons measure force; 1 kN = 1000 N.
- z₂
- Lever arm 2
Lever arm 2. Use the same reference and sign convention for the second force contribution.
mMetres measure length; 1 m = 1000 mm.
- MRd
- Result to find
Composite/internal-force moment sum. Compatible signed moment contributions can be added about their common reference.
kN·m
Sort out the units first
Forces are kN and lever arms are m, so each term and the total are kN·m. Lever arm means perpendicular distance to the force line of action, not an arbitrary sloping distance.
Assumptions before calculating
The force resultants and their lever arms have already been derived from an admissible section model and use the same reference. Signed force or signed lever arm may encode the moment direction, but do not encode it twice.
This is a first-generation Eurocode teaching relationship or an explicitly simplified coefficient calculation. The numbers supplied here are exercise data, not a recommendation for any country. Check the adopted edition, relevant clause, National Annex, applicability conditions and all other limit states before any real design.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
Read the supplied values as one complete study case. Find MRd and explain the result in the stated output unit.
- F₁ · Internal force 1
- 1000 kN
- z₁ · Lever arm 1
- 0.4 m
- F₂ · Internal force 2
- 500 kN
- z₂ · Lever arm 2
- 0.2 m
Calculate the first moment contribution
Keep the first force paired with its own perpendicular lever arm.
(1000) × (0.4) = 400 kN·mCalculate the second moment contribution
Use the same reference and sign convention for the second force contribution.
(500) × (0.2) = 100 kN·mSum the signed moments
Compatible signed moment contributions can be added about their common reference.
(400) + (100) = 500 kN·m
Does this worked answer make sense?
A force whose lever arm is zero contributes no moment about the selected reference. Equal and opposite moment contributions cancel even when the forces themselves are nonzero.
A second worked example — different values
A second case uses different data. Predict which way the answer will change, then calculate it without reusing the first answer.
- F₁ · Internal force 1
- 800 kN
- z₁ · Lever arm 1
- 0.35 m
- F₂ · Internal force 2
- -200 kN
- z₂ · Lever arm 2
- 0.2 m
Calculate the first moment contribution
Keep the first force paired with its own perpendicular lever arm.
(800) × (0.35) = 280 kN·mCalculate the second moment contribution
Use the same reference and sign convention for the second force contribution.
(-200) × (0.2) = -40 kN·mSum the signed moments
Compatible signed moment contributions can be added about their common reference.
(280) + (-40) = 240 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
Solve this separate case yourself. Use only the values below; the two worked examples use different data. Give the requested result in the selected unit.
- F₁ · Internal force 1
- 900 kN
- z₁ · Lever arm 1
- 0.4 m
- F₂ · Internal force 2
- 300 kN
- z₂ · Lever arm 2
- 0.15 m
Find: Learn: Composite/internal-force moment sum
A hint, not the answer
Each contribution is Fi zi. Compute each moment separately so the force and its own lever arm remain paired, then add them algebraically with a consistent sign convention.
Forces are kN and lever arms are m, so each term and the total are kN·m. Lever arm means perpendicular distance to the force line of action, not an arbitrary sloping distance.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Calculate the first moment contribution
Keep the first force paired with its own perpendicular lever arm.
(900) × (0.4) = 360 kN·mCalculate the second moment contribution
Use the same reference and sign convention for the second force contribution.
(300) × (0.15) = 45 kN·mSum the signed moments
Compatible signed moment contributions can be added about their common reference.
(360) + (45) = 405 kN·m
Avoid the common trap
Do not cross-pair F1 with z2. Do not add force magnitudes when the moments oppose, or sum moments taken about different reference points without transformation.
When this method applies — and when it does not
This is moment summation, not automatic composite design. Neutral axis, material stress limits, force equilibrium, shear connection and section classification are not calculated.
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.
Reading focus: Composite/internal-force moment sum. First-generation EN 1993/EN 1994 teaching: cross-section resistance, stability, connections or composite action as relevant. Member classification and other limit states remain separate checks.
Lesson updated: · Both examples and the separate practice case are checked against an independent high-precision numerical implementation. This verifies arithmetic for the stated model, not engineering certification.