Learn: Simple ULS combination — study form
An ultimate-limit-state load combination applies factors to permanent and variable actions before adding their contributions. This lesson contains one permanent action and one leading variable action so the basic bookkeeping is easy to see.
What the formula is saying
Factor each action separately, then add the two design contributions. The partial factor belongs to its own action; multiplying the total characteristic load by one shared factor gives a different calculation.
Read the symbols in plain language
- γG
- Permanent-action factor
Permanent-action factor. Apply the permanent-action coefficient to Gk before combining it with the variable action.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- Gk
- Permanent action
Permanent action. Apply the permanent-action coefficient to Gk before combining it with the variable action.
kNKilonewtons measure force; 1 kN = 1000 N.
- γQ
- Variable-action factor
Variable-action factor. The variable action has its own coefficient and must be treated separately.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- Qk
- Variable action
Variable action. The variable action has its own coefficient and must be treated separately.
kNKilonewtons measure force; 1 kN = 1000 N.
- Ed
- Result to find
Simple ULS combination — study form. These are compatible force contributions in the same stated study combination.
kN
Sort out the units first
Gk and Qk are forces in kN in this exercise; γG and γQ have no unit. The answer is kN. For a real structure, action effects must be combined consistently rather than adding unlike quantities.
Assumptions before calculating
All coefficients are supplied exercise data. The designer must choose them from the relevant adopted standard, design situation and National Annex; the calculator does not select a country or verify that selection.
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 Ed and explain the result in the stated output unit.
- γG · Permanent-action factor
- 1.35
- Gk · Permanent action
- 100 kN
- γQ · Variable-action factor
- 1.5
- Qk · Variable action
- 50 kN
Factor the permanent action
Apply the permanent-action coefficient to Gk before combining it with the variable action.
(1.35) × (100) = 135 kNFactor the leading variable action
The variable action has its own coefficient and must be treated separately.
(1.5) × (50) = 75 kNAdd the two design contributions
These are compatible force contributions in the same stated study combination.
(135) + (75) = 210 kN
Does this worked answer make sense?
If Qk is zero, only the permanent contribution remains. Increasing one characteristic action changes only its own factored contribution.
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.
- γG · Permanent-action factor
- 1.35
- Gk · Permanent action
- 180 kN
- γQ · Variable-action factor
- 1.5
- Qk · Variable action
- 60 kN
Factor the permanent action
Apply the permanent-action coefficient to Gk before combining it with the variable action.
(1.35) × (180) = 243 kNFactor the leading variable action
The variable action has its own coefficient and must be treated separately.
(1.5) × (60) = 90 kNAdd the two design contributions
These are compatible force contributions in the same stated study combination.
(243) + (90) = 333 kN
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.
- γG · Permanent-action factor
- 1.35
- Gk · Permanent action
- 120 kN
- γQ · Variable-action factor
- 1.5
- Qk · Variable action
- 80 kN
Find: Learn: Simple ULS combination — study form
A hint, not the answer
Factor each action separately, then add the two design contributions. The partial factor belongs to its own action; multiplying the total characteristic load by one shared factor gives a different calculation.
Gk and Qk are forces in kN in this exercise; γG and γQ have no unit. The answer is kN. For a real structure, action effects must be combined consistently rather than adding unlike quantities.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Factor the permanent action
Apply the permanent-action coefficient to Gk before combining it with the variable action.
(1.35) × (120) = 162 kNFactor the leading variable action
The variable action has its own coefficient and must be treated separately.
(1.5) × (80) = 120 kNAdd the two design contributions
These are compatible force contributions in the same stated study combination.
(162) + (120) = 282 kN
Avoid the common trap
Do not factor a load that has already been factored. Do not use the example factors as universal national values, and do not add another variable action at full value without checking its combination role.
When this method applies — and when it does not
This is a deliberately limited two-action persistent/transient study combination. It omits accompanying variable actions, alternative expressions, favourable/unfavourable distinctions, accidental and seismic combinations, pattern loading and action-effect analysis.
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: Simple ULS combination — study form. Basis-of-design reading: partial factors, design values, and combinations of actions. The displayed coefficients are supplied exercise data.
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.