Learn: SLS characteristic combination — study form
The characteristic serviceability combination represents a particular level of actions for checks such as movement, cracking or long-term response. It is not interchangeable with an ultimate-limit-state combination or the other two serviceability combinations.
What the formula is saying
Keep the leading variable action Q1 at its characteristic value and reduce the accompanying Q2 by ψ0. This expresses their different roles in the characteristic combination.
Read the symbols in plain language
- Gk
- Permanent action
Permanent action. Add the compatible force terms; permanent action is not multiplied by the variable-action factors.
kNKilonewtons measure force; 1 kN = 1000 N.
- Qk,1
- Leading variable action
Leading variable action. Keep the leading variable action Q1 at its characteristic value and reduce the accompanying Q2 by ψ0. This expresses their different roles in the characteristic combination.
kNKilonewtons measure force; 1 kN = 1000 N.
- ψ0,2
- Combination factor
ψ0 for accompanying Q2 in this characteristic combination.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- Qk,2
- Accompanying variable action
Accompanying variable action. Apply the factor appropriate to the second variable action, not to the permanent action.
kNKilonewtons measure force; 1 kN = 1000 N.
- Ed
- Result to find
SLS characteristic combination — study form. Add the compatible force terms; permanent action is not multiplied by the variable-action factors.
kN
Sort out the units first
G, Q1 and Q2 are compatible action contributions in kN in this exercise. All ψ factors are decimal ratios between 0 and 1. The result is a force combination in kN, not the resulting deflection or crack width.
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.
- Gk · Permanent action
- 100 kN
- Qk,1 · Leading variable action
- 50 kN
- ψ0,2 · Combination factor
- 0.7
- Qk,2 · Accompanying variable action
- 30 kN
Find the leading-action contribution
Keep the leading variable action Q1 at its characteristic value and reduce the accompanying Q2 by ψ0. This expresses their different roles in the characteristic combination.
(50) = 50 kNFind the second-action contribution
Apply the factor appropriate to the second variable action, not to the permanent action.
(0.7) × (30) = 21 kNAdd permanent and variable contributions
Add the compatible force terms; permanent action is not multiplied by the variable-action factors.
(100) + (50) + (21) = 171 kN
Does this worked answer make sense?
With positive actions and factors between 0 and 1, the result lies between G and G + Q1 + Q2. Setting a variable factor to zero removes only that variable 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.
- Gk · Permanent action
- 160 kN
- Qk,1 · Leading variable action
- 80 kN
- ψ0,2 · Combination factor
- 0.7
- Qk,2 · Accompanying variable action
- 40 kN
Find the leading-action contribution
Keep the leading variable action Q1 at its characteristic value and reduce the accompanying Q2 by ψ0. This expresses their different roles in the characteristic combination.
(80) = 80 kNFind the second-action contribution
Apply the factor appropriate to the second variable action, not to the permanent action.
(0.7) × (40) = 28 kNAdd permanent and variable contributions
Add the compatible force terms; permanent action is not multiplied by the variable-action factors.
(160) + (80) + (28) = 268 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.
- Gk · Permanent action
- 120 kN
- Qk,1 · Leading variable action
- 60 kN
- ψ0,2 · Combination factor
- 0.7
- Qk,2 · Accompanying variable action
- 20 kN
Find: Learn: SLS characteristic combination — study form
A hint, not the answer
Keep the leading variable action Q1 at its characteristic value and reduce the accompanying Q2 by ψ0. This expresses their different roles in the characteristic combination.
G, Q1 and Q2 are compatible action contributions in kN in this exercise. All ψ factors are decimal ratios between 0 and 1. The result is a force combination in kN, not the resulting deflection or crack width.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Find the leading-action contribution
Keep the leading variable action Q1 at its characteristic value and reduce the accompanying Q2 by ψ0. This expresses their different roles in the characteristic combination.
(60) = 60 kNFind the second-action contribution
Apply the factor appropriate to the second variable action, not to the permanent action.
(0.7) × (20) = 14 kNAdd permanent and variable contributions
Add the compatible force terms; permanent action is not multiplied by the variable-action factors.
(120) + (60) + (14) = 194 kN
Avoid the common trap
Do not add ULS partial factors to this teaching SLS expression. Do not apply one ψ factor to the entire sum. Read the factor role rather than assuming that the input name p1 always means ψ1.
When this method applies — and when it does not
The model contains one permanent and two variable contributions only. Choosing the relevant SLS combination, leading action, favourable effects, imposed-deformation treatment and acceptance limit remains a separate design task.
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: SLS characteristic 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.