Learn: Steel buckling resistance
A compressed steel member may buckle before reaching gross-section yield. This equation applies the supplied buckling reduction factor to the reference yielding force and then introduces the member partial factor.
What the formula is saying
Start with A fy, multiply by χ and divide by γM1. The factor χ belongs to the buckling mode being assessed; it is not an arbitrary safety margin to choose after seeing the answer.
Read the symbols in plain language
- χ
- Buckling reduction factor
Buckling reduction factor. The selected reduction factor accounts for the modelled member instability effect.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- A
- Area
Area. Multiply the compatible gross section area by its characteristic yield strength.
mm²Square millimetres measure area; 1 mm² = 10⁻⁶ m².
- fy
- Yield strength
Specified yield stress of the relevant steel grade and thickness, before the material partial factor unless explicitly stated otherwise.
N/mm²One N/mm² equals one MPa.
- γM1
- Partial factor
Supplied material/resistance partial factor in the divisor. Select its code clause and National Annex before real design.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- Nb,Rd
- Result to find
Steel buckling resistance. Divide by the member partial factor and convert the design force from N to kN.
kN
Sort out the units first
Area is mm² and yield strength is N/mm², so the force is initially N. χ and γm are dimensionless; division by 1000 changes the final force to kN.
Assumptions before calculating
Use the stated first-generation teaching equation with compatible section properties, material strengths and supplied partial factors. The required section class, buckling curve, National Annex values and design situation must be established separately.
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 Nb,Rd and explain the result in the stated output unit.
- χ · Buckling reduction factor
- 0.7
- A · Area
- 5000 mm²
- fy · Yield strength
- 355 N/mm²
- γM1 · Partial factor
- 1
Find the reference yielding force
Multiply the compatible gross section area by its characteristic yield strength.
(5000) × (355) = 1775000 NApply the buckling reduction
The selected reduction factor accounts for the modelled member instability effect.
(0.7) × (1775000) = 1242500 NApply the member factor and convert
Divide by the member partial factor and convert the design force from N to kN.
(1242500) ÷ (1) ÷ 1000 = 1242.5 kN
Does this worked answer make sense?
At χ = 1 the buckling reduction disappears in this equation. At fixed area, strength and partial factor, halving χ halves the reported resistance.
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.
- χ · Buckling reduction factor
- 0.55
- A · Area
- 4000 mm²
- fy · Yield strength
- 275 N/mm²
- γM1 · Partial factor
- 1
Find the reference yielding force
Multiply the compatible gross section area by its characteristic yield strength.
(4000) × (275) = 1100000 NApply the buckling reduction
The selected reduction factor accounts for the modelled member instability effect.
(0.55) × (1100000) = 605000 NApply the member factor and convert
Divide by the member partial factor and convert the design force from N to kN.
(605000) ÷ (1) ÷ 1000 = 605 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.
- χ · Buckling reduction factor
- 0.8
- A · Area
- 6000 mm²
- fy · Yield strength
- 355 N/mm²
- γM1 · Partial factor
- 1.1
Find: Learn: Steel buckling resistance
A hint, not the answer
Start with A fy, multiply by χ and divide by γM1. The factor χ belongs to the buckling mode being assessed; it is not an arbitrary safety margin to choose after seeing the answer.
Area is mm² and yield strength is N/mm², so the force is initially N. χ and γm are dimensionless; division by 1000 changes the final force to kN.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Find the reference yielding force
Multiply the compatible gross section area by its characteristic yield strength.
(6000) × (355) = 2130000 NApply the buckling reduction
The selected reduction factor accounts for the modelled member instability effect.
(0.8) × (2130000) = 1704000 NApply the member factor and convert
Divide by the member partial factor and convert the design force from N to kN.
(1704000) ÷ (1.1) ÷ 1000 ≈ 1549.090909 kN
Avoid the common trap
Do not use χ from the wrong buckling axis or apply it twice. Do not replace γM1 with a different partial factor merely because a different resistance calculation used that value.
When this method applies — and when it does not
This gross-area form assumes the applicable non-Class-4 member model and a valid reduction factor for the governing mode. Effective sections, combined bending and compression, torsional modes and connection effects need their separate checks.
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: Steel buckling resistance. 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.