Eurocode 3 Steel Column Compression & Buckling Calculator

This professional-grade tool computes the axial compression buckling resistance of steel columns according to Eurocode 3 (EN 1993-1-1). It is designed for structural engineers and advanced students to verify design capacity, slenderness and utilization rapidly and accurately.

Calculator

Buckling axis
Total steel area in mm². Use design (gross) area of the member.
mm²
Second moment of area about y-axis in mm⁴. Required if buckling about y.
mm⁴
Second moment of area about z-axis in mm⁴. Required if buckling about z.
mm⁴
Characteristic yield strength in MPa (N/mm²). Typical S355: 355 MPa.
MPa
Use 210000 MPa for structural steel unless specified otherwise.
MPa
Clear length between restraints, in meters. We convert to mm internally.
m
Common effective length factors k: fixed–fixed 0.5, fixed–pinned 0.7, pinned–pinned 1.0, fixed–free 2.0. Choose the case that best represents your boundary conditions.
Per EN 1993-1-1 Table 6.1: a0=0.13, a=0.21, b=0.34, c=0.49, d=0.76. Select per section type/axis/NA.
Used for cross-section resistance Npl,Rd = A·fy / γM0. Recommended value: 1.0 (check NA).
Used for buckling resistance Nb,Rd = χ·A·fy / γM1. Recommended value: 1.0 (check NA).
Optional. Enter the design axial compression action in kN for utilization and pass/fail.
kN

Results

Effective length Le mm ( m)
Euler critical load Ncr
Non-dimensional slenderness λ̄
Imperfection α / φ / χ
Section plastic resistance Npl,Rd
Buckling resistance Nb,Rd
Utilization η = NEd / Nb,Rd
Status Awaiting input

Data Source and Methodology

Authoritative source: EN 1993-1-1:2005 + A1:2014 Eurocode 3 — Design of steel structures — Part 1-1: General rules and rules for buildings, Clause 6.3 (Member buckling in compression). Official overview: European Commission – JRC Eurocodes. Additional context: Eurocode Applied.

Tutti i calcoli si basano rigorosamente sulle formule e sui dati forniti da questa fonte.

The Formula Explained

$$ N_{b,Rd} = \chi \cdot \frac{A\, f_y}{\gamma_{M1}} $$ $$ \chi = \frac{1}{\varphi + \sqrt{\varphi^2 - \bar{\lambda}^2}} \quad \text{with} \quad \varphi = \tfrac{1}{2}\left[1 + \alpha\left(\bar{\lambda} - 0.2\right) + \bar{\lambda}^2 \right] $$ $$ \bar{\lambda} = \sqrt{\frac{A f_y}{N_{cr}}}, \qquad N_{cr} = \frac{\pi^2 E I}{L_e^2}, \qquad L_e = k \cdot L $$ $$ N_{pl,Rd} = \frac{A\, f_y}{\gamma_{M0}} $$ where: A [mm²], E [MPa], fy [MPa], I [mm⁴], L [mm], k [–], α per buckling curve.

Glossary of Variables

Definitions of inputs and outputs
SymbolNameUnitDescription
AAreamm²Gross cross-sectional area of the member.
Iy, IzSecond moment of areamm⁴About y or z axis. Choose the governing axis.
fyYield strengthMPaCharacteristic yield stress of steel.
EElastic modulusMPaModulus of elasticity of steel (≈210000 MPa).
LMember lengthmClear length between effective restraints.
kEffective length factorDepends on end restraints; Le = k·L.
LeEffective lengthmmUsed in Euler buckling check.
NcrEuler critical loadkNIdeal elastic buckling load.
λ̄Non-dimensional slendernessEC3 slenderness parameter, function of Ncr.
αImperfection factorSet by the chosen buckling curve.
φAuxiliary factorUsed to compute reduction factor χ.
χReduction factorReduces plastic strength for buckling.
Npl,RdPlastic section resistancekNA·fy/γM0.
Nb,RdBuckling resistancekNχ·A·fy/γM1.
NEdDesign axial loadkNApplied factored compression action.
ηUtilizationNEd / Nb,Rd.

How It Works: A Step-by-Step Example

Given: A = 6500 mm², Iy = 8.5e8 mm⁴, Iz = 2.4e8 mm⁴, fy = 355 MPa, E = 210000 MPa, L = 3.0 m, end condition pinned–pinned (k = 1.0), buckling about y-axis, curve b (α = 0.34), γM0 = γM1 = 1.0.

  1. Effective length: Le = k·L = 1.0 × 3.0 m = 3000 mm.
  2. Euler load about y: Ncr = π² E Iy / Le².
  3. Slenderness: λ̄ = sqrt(A·fy / Ncr).
  4. φ = 0.5 [1 + α(λ̄ − 0.2) + λ̄²] and χ = 1 / (φ + sqrt(φ² − λ̄²)).
  5. Npl,Rd = A·fy/γM0 and Nb,Rd = χ·A·fy/γM1. If NEd = 850 kN, utilization η = 850 / Nb,Rd.

The calculator performs these steps instantly and displays intermediate values so you can audit the calculation.

Frequently Asked Questions (FAQ)

What differentiates curves a0, a, b, c, d?

They represent different sensitivity to imperfections and residual stresses. Less favorable curves (higher α) produce lower χ at a given slenderness. Use the Eurocode’s Table 6.1 and your NA to select the correct curve.

Can I enter radius of gyration instead of I?

Yes indirectly: I = A·i². If you know i, compute I and enter it. Future versions will add a dedicated toggle.

Does the tool include shear or bending interaction?

No. This module covers pure axial compression buckling. For combined bending and axial force, use EC3 interaction checks per Clause 6.3.3.

Why do I see χ = 1.0 for very stocky members?

At very low slenderness, buckling is not governing and χ approaches 1.0 (limited slightly below 1.0 per EC3 expression).

What accuracy/rounding is used?

Internally the tool uses full precision JS floating-point and reports values rounded to sensible engineering units (e.g., kN with two decimals).

Tool developed by Ugo Candido. Content verified by CalcDomain Engineering Editorial.
Last reviewed for accuracy on: .