EN 1993-1-1 Column Buckling — Curves a0-d, Imperfection Factors, Nb,Rd

Quick Reference: The EN 1993 column buckling check uses the Perry-Robertson formulation with five buckling curves (a0, a, b, c, d). The design buckling resistance Nb,Rd = chi _ A _ fy / gamma_M1, where chi is the reduction factor. Use the free column capacity calculator for instant EN 1993 checks.

The Buckling Reduction Factor chi (Cl. 6.3.1)

The design buckling resistance of a compression member is:

Nb,Rd = chi _ A _ fy / gamma_M1 (for Class 1, 2, 3 cross-sections)

For Class 4 cross-sections, use Aeff instead of A and fy instead of fy.

The reduction factor chi accounts for both the slenderness of the member and the imperfection sensitivity of the cross-section type:

chi = 1 / (Phi + sqrt(Phi^2 - lambda_bar^2)) ... but chi <= 1.0

where:

• Phi = 0.5 [1 + alpha * (lambda_bar - 0.2) + lambda_bar^2]
• lambda_bar = sqrt(A * fy / Ncr) for Class 1-3 sections
• Ncr = pi^2 _ E _ I / Lcr^2 (Euler buckling load)
• alpha is the imperfection factor from Table 6.1

Buckling Curves and Imperfection Factors

EN 1993-1-1 Table 6.2 defines five buckling curves. The higher the curve letter, the higher the imperfection factor (more conservative):

For a typical UKB in S355 buckling about the strong axis: use curve a (alpha = 0.21) if h/b > 1.2 and tf <= 40mm, or curve b (alpha = 0.34) if h/b <= 1.2.

For buckling about the weak axis (z-z), hot-rolled I-sections use curve b (alpha = 0.34). Welded I-sections use curve c (alpha = 0.49) for both axes.

Alpha Values at a Glance

A column with lambda_bar = 1.0 designed with curve 'a' achieves about 80% of the squash load, while the same column designed with curve 'd' achieves only about 46%. This is why section selection matters for compression members — a hot-rolled section may have twice the buckling capacity of an equivalent welded section.

Worked Example: UKC Column

Given: 254x254x89 UKC in S355, Lcr = 5.0 m, pinned-pinned (effective length factor = 1.0), buckling about weak axis (z-z).

Section properties: A = 113 cm^2, iz = 65.5 mm, h/b = 260.3/256.3 = 1.02 <= 1.2, tf = 17.3 mm <= 40mm

Step 1 — Buckling curve selection: h/b <= 1.2, tf <= 40mm, hot-rolled, about z-z → buckling curve b, alpha = 0.34

Step 2 — Non-dimensional slenderness: Ncr = pi^2 _ 210,000 _ (A _ iz^2/1e6) / (5000)^2 = pi^2 _ 210,000 _ (0.0113 _ 0.0655^2) / 25,000,000 Ncr = pi^2 _ 210,000 _ 4.85e-5 / 25e6 _ 1e6 = ... lambda_bar = sqrt(A _ fy / Ncr) = sqrt(11300 * 355 / Ncr)

For this section, lambda_bar ~ 1.1 (use software for exact value).

Step 3 — Reduction factor chi: Phi = 0.5 [1 + 0.34*(1.1 - 0.2) + 1.1^2] = 0.5 [1 + 0.306 + 1.21] = 1.258 chi = 1 / (1.258 + sqrt(1.258^2 - 1.1^2)) = 1 / (1.258 + 0.610) = 0.535

Step 4 — Buckling resistance: Nb,Rd = 0.535 _ 11300 _ 355 / 1.00 = 2,147,000 N = 2,147 kN

The squash load (chi = 1.0) would be A * fy = 4,012 kN. Buckling reduces capacity by ~47% for this slenderness.

Effective Length and Buckling Length Lcr

For a column in a frame, the buckling length Lcr = k * L where k is the effective length factor. EN 1993-1-1 provides two approaches:

1. Sway mode method: Lcr = L * max( sqrt( (1-0.2(eta1+eta2) - 0.12eta1eta2) / (1-0.8(eta1+eta2) + 0.6*eta1*eta2) ), ... ) for non-sway; similar for sway
2. Simplified: k = 1.0 for pinned-pinned, k = 0.7 for fixed-pinned, k = 0.5 for fixed-fixed, k = 2.0 for cantilever

Cross-Section Classification for Compression

For columns, classification follows Cl. 5.5, but the stress distribution differs from beams — pure compression means the full cross-section is in uniform compression. This is more onerous than the bending classification.

For a 254x254x89 UKC in S355 in pure compression:

Flange (internal part in uniform compression): c/tf limits per Table 5.2 (Sheet 1):

• Class 1: 33 * epsilon = 26.7
• Class 2: 38 * epsilon = 30.8
• Class 3: 42 * epsilon = 34.0

c = (b - tw - 2r) / 2 = (256.3 - 10.3 - 2*12.7) / 2 = 110.3 mm. c/tf = 110.3/17.3 = 6.38 < 26.7 → Class 1.

Web (internal part in uniform compression): c/tw limits:

• Class 1: 33 * epsilon = 26.7
• Class 2: 38 * epsilon = 30.8
• Class 3: 42 * epsilon = 34.0

c = d = 260.3 - 217.3 - 212.7 = 200.3 mm. c/tw = 200.3/10.3 = 19.45 < 26.7 → Class 1.

The section is Class 1 for pure compression. Use A (gross area) in the buckling resistance formula.

The Perry-Robertson Formulation

The EN 1993 buckling curves are based on the Perry-Robertson equation, which models the behaviour of a geometrically imperfect column with an initial out-of-straightness e0:

**sigma_cr = (fy + (1+eta)sigma_Euler) / 2 - sqrt( ((fy + (1+eta)sigma_Euler)/2)^2 - fy*sigma_Euler )

where eta = alpha _ (lambda_bar - 0.2) and sigma_Euler = pi^2 _ E / (Lcr/i)^2.

The imperfection factor alpha is calibrated from test data for each buckling curve (a0 through d). The 0.2 offset in the lambda_bar term (lambda_bar - 0.2) creates a plateau where chi = 1.0 for low slenderness — columns with lambda_bar <= 0.2 fail by squashing, not buckling.

The five curves reflect different levels of geometric imperfection and residual stress:

• a0 (alpha=0.13): Hot-rolled S460+ hollow sections — highest quality, lowest imperfection
• a (alpha=0.21): Hot-rolled I-sections with h/b > 1.2 — good straightness
• b (alpha=0.34): Hot-rolled I-sections with h/b <= 1.2 — stockier sections
• c (alpha=0.49): Welded sections, hot-rolled I about z-z — moderate imperfections
• d (alpha=0.76): Thick welded plates (tf > 40mm) — largest imperfections

Complete Buckling Curve Selection (Table 6.2)

Worked Example 1: 203x203x60 UKC in S355 — Strong Axis

Given: Lcr = 4.0 m, columns pinned both ends, buckling about y-y (strong axis).

Section: A = 76.6 cm^2, iy = 8.96 cm, h/b = 209.6/205.8 = 1.02 <= 1.2, tf = 14.2 mm <= 40 mm.

Step 1 — Curve selection: h/b <= 1.2, tf <= 40mm, hot-rolled, about y-y → curve b, alpha = 0.34.

Step 2 — Slenderness: Ncr = pi^2 _ E _ Iy / Lcr^2 = pi^2 _ 210,000 _ (76.6 _ 8.96^2) / 4,000^2 = 9.8696 _ 210,000 _ 6,147 / 16,000,000 = 9.8696 _ 210 * 6,147 / 16,000 (converting to kN) = 795 kN

Wait — using Iy = A _ iy^2 = 76.6 _ 8.96^2 = 6,147 cm^4: Ncr = 9.8696 _ 210,000 N/mm^2 _ 6,14710^4 mm^4 / (4,000 mm)^2 = 9.8696 * 210,000 * 6.14710^7 / 16*10^6 = 9.8696 * 210,000 * 3.842 / 1 = 7,960,000 N = 7,960 kN

lambda*bar = sqrt(A * fy / Ncr) = sqrt(7,660 _ 355 / 7,960,000) = sqrt(2,719,300 / 7,960,000) = sqrt(0.3416) = 0.584

Step 3 — Reduction factor: Phi = 0.5 _ [1 + 0.34_(0.584 - 0.2) + 0.584^2] = 0.5 * [1 + 0.131 + 0.341] = 0.736 chi = 1 / (0.736 + sqrt(0.736^2 - 0.584^2)) = 1 / (0.736 + sqrt(0.2007)) = 1 / (0.736 + 0.448) = 1 / 1.184 = 0.845

Step 4 — Buckling resistance: Nb,Rd = 0.845 _ 7,660 _ 355 / 1.0 = 2,298,000 N = 2,298 kN.

Squash load = 7,660 * 355 / 1.0 = 2,719 kN. Reduction = 15.5%.

Worked Example 2: 203x203x60 UKC — Weak Axis

Same column, buckling about z-z (weak axis): iz = 5.19 cm, curve b (alpha = 0.34).

Ncr,z = pi^2 _ 210,000 _ (76.6 _ 5.19^2) _ 10^4 / 4,000^2 = 9.8696 _ 210,000 _ 2,06310^4 / 1610^6 = 2,671,000 N = 2,671 kN

lambda_bar,z = sqrt(7,660 * 355 / 2,671,000) = sqrt(2,719,300 / 2,671,000) = sqrt(1.018) = 1.009

Phi = 0.5 _ [1 + 0.34_(1.009 - 0.2) + 1.009^2] = 0.5 * [1 + 0.275 + 1.018] = 1.147 chi = 1 / (1.147 + sqrt(1.147^2 - 1.009^2)) = 1 / (1.147 + sqrt(0.2978)) = 1 / (1.147 + 0.546) = 0.591

Nb,Rd,z = 0.591 _ 7,660 _ 355 / 1.0 = 1,607 kN (41% reduction).

The weak-axis buckling capacity (1,607 kN) governs. Even for a UC section (nearly square), the difference in radius of gyration (89.6 mm vs 51.9 mm) means weak-axis buckling controls the design. This is why effective bracing to reduce Lcr,z is essential for columns.

Combined Bending and Axial Compression (Cl. 6.3.3)

Members subjected to combined compression and bending must satisfy two interaction formulae, which check in-plane buckling and out-of-plane buckling:

Eq. 6.61 (in-plane): NEd / (chi_y * NRk / gamma_M1) + k_yy * My,Ed / (chi_LT * My,Rk / gamma_M1) + k_yz * M_z,Ed / (M_z,Rk / gamma_M1) <= 1.0

Eq. 6.62 (out-of-plane): NEd / (chi_z * NRk / gamma_M1) + k_zy * My,Ed / (chi_LT * My,Rk / gamma_M1) + k_zz * M_z,Ed / (M_z,Rk / gamma_M1) <= 1.0

The interaction factors k_yy, k_yz, k_zy, k_zz are computed per Annex B (Method 1) or Annex A (Method 2 — simpler, often conservative). These factors depend on the equivalent uniform moment factors C_my, C_mz, C_mLT and the section type.

Simplified Check for Class 1/2 I-Sections (UK NA)

When M_z,Ed is negligible (minor axis moment from eccentricity < 5% of M_z,Rd), and chi_LT = 1.0 (fully restrained beam), the check simplifies to:

N_Ed/N_b,Rd + M_y,Ed/Mc,Rd <= 1.0 (this is the conservative UK industry approximation)

For a 254x254x89 UKC with N_Ed = 1,200 kN and M_y,Ed = 80 kN.m:

• N*b,Rd,z = chi_z * A _ fy = 0.535 _ 11,350 _ 355 / 1.0 = 2,156 kN
• Mc,Rd,y = Wpl,y * fy / gamma_M0 = 1,220*10^3 * 355 / 1.0 = 433 kN.m

Check: 1,200/2,156 + 80/433 = 0.557 + 0.185 = 0.742 < 1.0 → OK.

Torsional and Flexural-Torsional Buckling (Cl. 6.3.1.4)

For cross-sections with low torsional stiffness (angles, tees, cruciform sections, thin-walled open sections), the buckling mode may involve torsion. The elastic critical force for torsional buckling is:

Ncr,T = (1 / i0^2) _ ( G _ It + pi^2 _ E _ Iw / Lcr,T^2 )

where i0^2 = iy^2 + iz^2 + y0^2 + z0^2 (polar radius of gyration about the shear centre).

For standard hot-rolled I-sections, torsional buckling rarely governs because It is substantial for the flange thickness. However, for angles, channels, and fabricated plate girders, torsional buckling should be checked. Use the elastic critical force from torsional buckling to compute lambda_bar,T, then compute chi using buckling curve c or d as appropriate.

Effective Lengths in Framed Structures

EN 1993-1-1 Cl. 5.2.2(5) allows two approaches for effective length factors:

Simplified Method

Sway Classification

The frame must first be classified as sway or non-sway. A frame is non-sway (braced) if alpha_cr >= 10, where alpha_cr is the factor by which the design loads would have to be increased to cause elastic instability in a global sway mode.

For a non-sway column in a rigid-jointed frame, the effective length can be estimated from Annex E using the relative stiffness distribution factors eta1 and eta2 at the column ends.

UK National Annex Specifics

The UK NA to BS EN 1993-1-1 specifies:

• gamma_M1 = 1.00 for buckling resistance (same as gamma_M0 — no additional partial factor for instability)
• The plateau length lambda_bar,0 = 0.2 (standard EN 1993 value, confirmed in UK NA)
• For sway-sensitive frames, alpha_cr should be computed using the Horne method or second-order analysis per Cl. 5.2.1(3)

EN 1993 vs Other Codes — Column Buckling

The EN 1993 approach is the most refined, with 5 curves calibrated to specific section types. AISC uses a simpler single curve approach but compensates with the Q factor for slender elements. AS 4100 is functionally identical to EN 1993 (Perry-Robertson with 5 curves) but the alpha_b values differ — AS 4100 alpha_b corresponds roughly to the EN 1993 alpha numerically.

Related Pages

• EN 1993-1-1 Beam Design Guide — Section classification, Mc,Rd, LTB with full worked example
• EN 1993 Steel Grades — S235-S460 grade properties per BS EN 10025
• UK Steel Beam Sizes — UB, UC, PFC — Complete UK section tables
• Column Capacity Calculator — Free AISC/AS 4100/EN 1993 column check
• Understanding Eurocode 3 — Design Guide — EN 1993 overview

This page is for educational reference. Buckling design must comply with the applicable National Annex. Column buckling is a stability failure — the consequence of getting it wrong can be catastrophic. All results are PRELIMINARY — NOT FOR CONSTRUCTION without independent PE/SE verification.