Free Tool · EN 1993-1-1 §6.3.1 · Flexural Buckling · Steel Columns

Column Buckling Capacity (Nb,Rd)

Flexural buckling resistance of steel compression members per EN 1993-1-1 §6.3.1. Select section and steel grade, enter effective buckling lengths, get χ reduction factors and N_b,Rd for both axes with full step-by-step working.

EN DE FR ES IT NL PL SV DA

Presets:

Section

Material & Loads

Steel grade

γM1 — partial factor

γM1 = 1.0 per EN 1993-1-1 §6.1(1). NL/DE/BE national annexes all confirm 1.0.

NEd — design compression (kN)

Leave 0 to skip utilization check.

Effective Buckling Lengths

Lcr,y — major axis (m)

Lcr,z — minor axis (m)

Lcr = K·L where K = 1.0 pin-pin, 0.7 fixed-pin, 0.5 fixed-fixed.

Buckling Results

PASS

3580.9 Nb,Rd

69.8% Utilization η

Z Governing axis

Utilization η: 69.8%

Buckling Results — Y / Z

Y-axis (major)

Curve (Table 6.2)B

α (Table 6.1)0.34

Ncr (kN)25761.9

λ̄0.4533

Φ0.6458

χ0.9043

Nb,Rd this axis (kN)4786.8

Z-axis (minor) ★

Curve (Table 6.2)C

α (Table 6.1)0.49

Ncr (kN)8764.4

λ̄0.7771

Φ0.9434

χ0.6765

Nb,Rd this axis (kN)3580.9

Step-by-step — EN 1993-1-1 §6.3.1

Step	Value	Reference
N_pl,Rd = A·f_y/γM1	5293.1 kN	§6.2.4 Eq 6.7
N_cr,gov = π²·E·I / L_cr²	8764.4 kN	§6.3.1.2
λ̄ = √(N_pl,Rd / N_cr)	0.7771	Eq 6.50
Buckling curve (Table 6.2)	C	Table 6.2
α — imperfection factor	0.49	Table 6.1
Φ = 0.5[1+α(λ̄−0.2)+λ̄²]	0.9434	Eq 6.49
χ = 1/(Φ+√(Φ²−λ̄²)) ≤ 1.0	0.6765	Eq 6.49
N_b,Rd = χ · N_pl,Rd	3580.9 kN	Eq 6.47

χ vs λ̄ — Buckling Curves

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FAQ

How does buckling curve selection work?

EN 1993-1-1 Table 6.2 assigns curves based on section type, h/b ratio, flange thickness, and fabrication. Rolled IPE/HEA with h/b > 1.2 and tf ≤ 40 mm: y-axis → curve a, z-axis → curve b. Rolled H sections with h/b ≤ 1.2 and tf ≤ 40 mm: y → b, z → c. Welded I sections tf ≤ 40 mm: y → b, z → c; tf > 40 mm: y → c, z → d. Each curve has an imperfection factor α from Table 6.1: a0 = 0.13, a = 0.21, b = 0.34, c = 0.49, d = 0.76.

What is non-dimensional slenderness λ̄?

λ̄ = √(A·fy / Ncr) where Ncr = π²·E·I / Lcr². For Class 1–3 sections the gross area A is used. When λ̄ ≤ 0.2 (or N_Ed/N_cr ≤ 0.04), buckling effects may be ignored per §6.3.1.2(4) and χ = 1.0.

How is χ calculated?

χ = 1 / (Φ + √(Φ² − λ̄²)) ≤ 1.0 where Φ = 0.5·[1 + α·(λ̄ − 0.2) + λ̄²]. The buckling resistance is then Nb,Rd = χ·A·fy / γM1.

What effective length Lcr should I use?

Lcr = K·L where K depends on end conditions. K = 1.0 for pin-pin (non-sway), K = 0.7 for fixed-pin, K = 0.5 for fixed-fixed. In sway frames K may exceed 1.0 — consult EN 1993-1-1 §5.2.2. Different values often apply to each axis when lateral restraint differs.

Does this tool cover combined axial + bending?

No. This tool covers pure compression per Eq 6.49. For beam-columns under N + M, use the interaction equations §6.3.3 (Eqs 6.61 + 6.62) with k-factors from Annex A or B.

Is γM1 = 1.0 correct for all national annexes?

Yes. NL (NEN-EN), DE (DIN EN), BE, and FR national annexes all confirm γM1 = 1.0 for §6.3.1. The recommended value in the base EN 1993-1-1 is also 1.0. Some legacy designs used 1.1 but this is not EN-compliant.