EN 1993-1-1 Column Design Worked Example — HEB240 Axial Compression

Full step-by-step worked example for EN 1993-1-1:2005 steel column design using a European HEB240 section in S355 steel. This example covers cross-section classification (Clause 5.5), axial compression resistance Nc,Rd for Class 1 cross-sections (Clause 6.2.4), flexural buckling resistance Nb,Rd (Clause 6.3.1), buckling curve selection (Table 6.2), the reduction factor chi calculation using the Perry-Robertson formula (Clause 6.3.1.2), and buckling about both the strong (y-y) and weak (z-z) axes. All calculations follow the UK National Annex values.

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Design Problem Statement

Problem: Verify a HEB240 column in S355 steel, pin-ended about both axes, with a system length L = 5.0 m. The column is part of a braced multi-storey frame and carries axial compression only (no significant moments — less than 5% of Mc,Rd, which may be neglected for the axial check). The column is in a heated internal environment (corrosion category C1, no reduction for temperature).

Design axial load: N_Ed = 800 kN (compression, factored ULS)

Design to EN 1993-1-1 with UK National Annex:

• gamma_M0 = 1.00
• gamma_M1 = 1.00 (UK NA)
• Steel grade: S355JR (EN 10025-2), fy = 355 MPa for tf <= 40 mm

Step 1 — HEB240 Section Properties

From the European section tables (ArcelorMittal Sections Handbook):

Material: S355, fy = 355 N/mm^2 for tf = 17.0 mm <= 40 mm, E = 210,000 N/mm^2, nu = 0.3

The HEB240 has h/b = 240/240 = 1.0, making it a stocky H-section with nearly equal strong-axis and weak-axis depth. This gives relatively balanced buckling resistance about both axes.

Step 2 — Cross-Section Classification (Clause 5.5)

Flange Classification

Outstand width c = (b - tw - 2*r) / 2 = (240 - 10.0 - 42.0) / 2 = 188.0 / 2 = 94.0 mm

epsilon = sqrt(235/fy) = sqrt(235/355) = sqrt(0.662) = 0.814

c/tf = 94.0 / 17.0 = 5.53

Limits for flange outstand in pure compression (Table 5.2, Sheet 2):

• Class 1: c/tf <= 9*epsilon = 9 * 0.814 = 7.33

• Class 2: c/tf <= 10*epsilon = 10 * 0.814 = 8.14

• Class 3: c/tf <= 14*epsilon = 14 * 0.814 = 11.39

  5.53 <= 7.33 → Flange is Class 1.

Web Classification

Web depth between fillets: cw = h - 2tf - 2r = 240 - 34.0 - 42.0 = 164.0 mm

cw/tw = 164.0 / 10.0 = 16.40

For a web in pure compression (axial load only), the limiting value for Class 1 (Table 5.2, Sheet 1): cw/tw <= 33*epsilon = 33 * 0.814 = 26.85

16.40 <= 26.85 → Web is Class 1.

Result: HEB240 in S355 is Class 1 for pure compression. The full cross-sectional area may be used without reduction for local buckling.

Step 3 — Cross-Section Compression Resistance (Clause 6.2.4)

For Class 1, 2, and 3 cross-sections in compression:

Nc,Rd = A * fy / gamma_M0
Nc,Rd = 10,600 * 355 / 1.00 = 3,763,000 N = 3,763 kN

This is the squash load — the maximum axial load the section can carry if buckling is prevented. For a stocky column with very low slenderness, this would be the governing limit.

Cross-section utilisation: N_Ed / Nc,Rd = 800 / 3,763 = 0.213 — the section has substantial reserve if buckling does not govern.

Step 4 — Flexural Buckling Resistance (Clause 6.3.1)

Buckling Length

For a pin-ended column in a braced frame:

Lcr,y = Lcr,z = 1.0 * L = 5,000 mm (both axes)

Non-Dimensional Slenderness lambda_bar

The non-dimensional slenderness for flexural buckling:

lambda_bar = sqrt(A * fy / Ncr)

Where Ncr is the elastic critical buckling load (Euler load):

Ncr = pi^2 * E * I / Lcr^2

Strong axis (y-y) buckling:

Ncr,y = pi^2 * 210,000 * 11,260*10^4 / (5,000^2)
      = 9.8696 * 210,000 * 112.6*10^6 / 25*10^6
      = 9.8696 * 210,000 * 112.6 / 25 * 10^6
      = 9.8696 * 945,840 * 10^6
      = 9,335 * 10^6 N
      = 9,335 kN

lambda_bar_y = sqrt(A * fy / Ncr,y)
             = sqrt(10,600 * 355 / 9,335*10^3)
             = sqrt(3,763,000 / 9,335,000)
             = sqrt(0.403)
             = 0.635

Alternatively, using the radius of gyration:

lambda_bar = (Lcr / i) / (93.9 * epsilon)
           = (5,000 / 103.1) / (93.9 * 0.814)
           = 48.5 / 76.4
           = 0.635  (agrees)

Weak axis (z-z) buckling:

Ncr,z = pi^2 * 210,000 * 3,923*10^4 / (5,000^2)
      = 9.8696 * 210,000 * 39.23*10^6 / 25*10^6
      = 9.8696 * 329.5 * 10^6
      = 3,252 * 10^6 N
      = 3,252 kN

lambda_bar_z = sqrt(A * fy / Ncr,z)
             = sqrt(10,600 * 355 / 3,252*10^3)
             = sqrt(3,763,000 / 3,252,000)
             = sqrt(1.157)
             = 1.076

Weak-axis buckling governs (larger lambda_bar). Further checks focus on weak-axis buckling.

Step 5 — Buckling Curve Selection (Table 6.2)

For flexural buckling, the buckling curve depends on:

• Section type: hot-rolled I-section
• Axis of buckling
• Steel grade
• h/b ratio

Strong axis (y-y):

• h/b = 240/240 = 1.0 <= 1.2
• tf = 17.0 mm <= 100 mm
• → Buckling curve b (alpha = 0.34)

Weak axis (z-z):

• h/b = 1.0 <= 1.2
• tf = 17.0 mm <= 100 mm
• S355 steel (fy > 235 MPa)
• → Buckling curve c (alpha = 0.49)

The weak-axis buckling curve is less favourable (higher imperfection factor), reflecting that weak-axis buckling is more sensitive to residual stresses and geometric imperfections in rolled H-sections.

Step 6 — Reduction Factor chi (Clause 6.3.1.2)

The reduction factor chi accounts for the effects of initial imperfections and residual stresses on the column buckling strength. It is derived from the Perry-Robertson formulation.

For weak-axis buckling (z-z, curve c, alpha = 0.49):

lambda_bar_z = 1.076

Phi = 0.5 * [1 + alpha * (lambda_bar - 0.2) + lambda_bar^2]
    = 0.5 * [1 + 0.49 * (1.076 - 0.2) + 1.076^2]
    = 0.5 * [1 + 0.49 * 0.876 + 1.158]
    = 0.5 * [1 + 0.429 + 1.158]
    = 0.5 * 2.587
    = 1.294

chi_z = 1 / (Phi + sqrt(Phi^2 - lambda_bar^2))
      = 1 / (1.294 + sqrt(1.294^2 - 1.076^2))
      = 1 / (1.294 + sqrt(1.674 - 1.158))
      = 1 / (1.294 + sqrt(0.516))
      = 1 / (1.294 + 0.718)
      = 1 / 2.012
      = 0.497

For strong-axis buckling (y-y, curve b, alpha = 0.34):

lambda_bar_y = 0.635

Phi = 0.5 * [1 + 0.34 * (0.635 - 0.2) + 0.635^2]
    = 0.5 * [1 + 0.34 * 0.435 + 0.403]
    = 0.5 * [1 + 0.148 + 0.403]
    = 0.5 * 1.551
    = 0.776

chi_y = 1 / (0.776 + sqrt(0.776^2 - 0.635^2))
      = 1 / (0.776 + sqrt(0.602 - 0.403))
      = 1 / (0.776 + sqrt(0.199))
      = 1 / (0.776 + 0.446)
      = 1 / 1.222
      = 0.818

Weak-axis buckling gives chi_z = 0.497, governing over strong-axis chi_y = 0.818.

Step 7 — Buckling Resistance Nb,Rd

The design buckling resistance of the compression member:

Nb,Rd = chi * A * fy / gamma_M1

Governed by weak-axis buckling:

Nb,Rd = chi_z * A * fy / gamma_M1
      = 0.497 * 10,600 * 355 / 1.00
      = 0.497 * 3,763,000
      = 1,870,000 N
      = 1,870 kN

Utilisation check:

N_Ed / Nb,Rd = 800 / 1,870 = 0.428 — OK (43% utilised)

The column is adequate. The buckling check utilises 43% of the capacity, leaving significant reserve for incidental moments or future load increases.

For comparison — strong-axis capacity: Nb,Rd,y = 0.818 * 3,763 = 3,078 kN (utilisation 800/3,078 = 0.260).

Step 8 — Alternative Column Size Checks

What if we used HEB220?

HEB220: A = 9,100 mm^2, iy = 9.41 cm, iz = 5.55 cm

lambdabar_z = (5,000/55.5) / (93.9 * 0.814) = 90.1 / 76.4 = 1.179 Phi = 0.5 _ [1 + 0.49(1.179-0.2) + 1.179^2] = 0.5 _ [1 + 0.480 + 1.390] = 1.435 chi*z = 1 / (1.435 + sqrt(1.435^2 - 1.179^2)) = 1 / (1.435 + sqrt(2.059 - 1.390)) = 1 / (1.435 + 0.818) = 0.444 Nb,Rd = 0.444 * 9,100 * 355 = 1,434 kN

Utilisation: 800/1,434 = 0.558 — OK (but less reserve). HEB220 would also work.

What if we used HEB200?

HEB200: A = 7,810 mm^2, iz = 5.06 cm

lambda*bar_z = (5,000/50.6) / 76.4 = 98.8 / 76.4 = 1.293 Phi = 0.5 * [1 + 0.49_(1.293-0.2) + 1.293^2] = 0.5 _ [1 + 0.536 + 1.672] = 1.604 chi_z = 1 / (1.604 + sqrt(1.604^2 - 1.293^2)) = 1 / (1.604 + sqrt(2.573 - 1.672)) = 1 / (1.604 + 0.949) = 0.392 Nb,Rd = 0.392 _ 7,810 * 355 = 1,087 kN

Utilisation: 800/1,087 = 0.736 — still OK. HEB200 is the most economical but leaves less reserve (26%).

Capacity Table — HEB Sections at 5.0 m Height (S355, pin-ended both axes)

Step 9 — Torsional and Torsional-Flexural Buckling (Clause 6.3.1.4)

For doubly-symmetric I-sections, torsional buckling may govern for columns with a short buckling length about the weak axis (Lcr,z small). Check:

The elastic critical load for torsional buckling:

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

Where i0^2 = iy^2 + iz^2 + y0^2 + z0^2 = 103.1^2 + 60.8^2 + 0 + 0 = 10,630 + 3,697 = 14,327 mm^2

For Lcr,T = 5,000 mm (same as flexural buckling length, conservative):

Ncr,T = (81,000 * 101.7*10^4 + pi^2 * 210,000 * 0.482*10^12 / 25*10^6) / 14,327
      = (82,377*10^6 + 40,020*10^6) / 14,327
      = 122,397*10^6 / 14,327
      = 8,543,000 N = 8,543 kN

lambda_bar_T = sqrt(A * fy / Ncr,T) = sqrt(3,763,000 / 8,543,000) = sqrt(0.440) = 0.664

Using buckling curve b for torsional buckling of rolled I-sections: Ncr,T > Ncr,z (8,543 > 3,252), so torsional buckling does not govern. The weak-axis flexural buckling check is the design case.

Step 10 — Summary of Results

Final Recommendation: HEB240 in S355 is adequate for N_Ed = 800 kN at 5.0 m pin-ended height, with 43% utilisation governed by weak-axis flexural buckling (curve c, chi_z = 0.497). A lighter HEB200 could achieve 74% utilisation, but the HEB240 provides desirable reserve for incidental eccentricity, connection fixity, and potential future load increases. For economy, HEB220 (utilisation 56%) represents a balanced design.

Frequently Asked Questions

Why is weak-axis buckling the governing limit state for H-section columns?

H-sections have significantly lower second moment of area about the weak axis (Iz) compared to the strong axis (Iy). For HEB240, Iz = 3,923 cm^4 vs Iy = 11,260 cm^4 — a ratio of 0.35. The radius of gyration iz = 6.08 cm vs iy = 10.31 cm. For the same unbraced length, the slenderness about z-z is inherently higher (lambda_bar_z = 1.076 vs lambda_bar_y = 0.635). Additionally, the buckling curve for weak-axis buckling (curve c, alpha = 0.49) is less favourable than for strong-axis buckling (curve b, alpha = 0.34), because weak-axis buckling is more affected by flange-tip residual stresses from the rolling process. The combination of higher geometric slenderness and more severe imperfection factor makes weak-axis buckling the design case for virtually all pin-ended H-section columns.

How is the buckling curve selected for different column cross-sections in EN 1993-1-1?

EN 1993-1-1 Table 6.2 selects the buckling curve based on: cross-section shape, axis of buckling, steel grade, h/b ratio, and flange thickness. For hot-rolled I-sections: strong-axis buckling uses curve a (alpha = 0.21) for h/b > 1.2 and tf <= 40 mm, or curve b (alpha = 0.34) for h/b <= 1.2. Weak-axis buckling uses curve b for tf <= 40 mm or curve c (alpha = 0.49) for S355+ steel. For hollow sections: hot-finished uses curve a for S235-S355, cold-formed uses curve c. The UK NA may adjust these curves — designers must check the applicable National Annex.

What is the physical meaning of the chi factor in EN 1993 column design?

chi is the buckling reduction factor, derived from the Perry-Robertson equation that accounts for the combined effects of initial geometric imperfections (out-of-straightness) and residual stresses on column strength. chi = 1.0 means no reduction (the column fails by squashing, Nc,Rd). As lambda_bar increases (more slender column), chi decreases. The alpha (imperfection factor) in the Perry formula controls how rapidly chi drops off — a higher alpha (e.g., curve d, alpha = 0.76) gives a steeper reduction, reflecting the greater sensitivity of that cross-section type to imperfections. The Perry-Robertson formulation was developed from extensive full-scale column tests across European steel producers in the 1960s-70s (ECCS column curves).

How does the EN 1993-1-1 column design method compare to AISC 360?

Both codes use the same fundamental approach: non-dimensional slenderness parameter, a buckling curve/reduction factor, and the squash load multiplied by the reduction factor. Key differences: (1) EN 1993-1-1 uses 5 buckling curves (a0, a, b, c, d) with different imperfection factors, while AISC 360 uses a single column curve (Eq. E3-2/E3-3) but modifies it with a 0.877 factor for elastic buckling. (2) EN gamma_M1 = 1.00 (UK NA) while AISC phi_c = 0.90. (3) EN lambda_bar incorporates the section area A and yield stress directly, while AISC uses KL/r divided by the elastic limit. For the HEB240 example at 5.0 m: EN Nb,Rd = 1,870 kN. AISC (using equivalent W10x... section): phi*Pn ~ 1,550-1,750 kN. The difference mainly comes from gamma_M1 = 1.00 vs phi = 0.90. For practical design, both codes produce similar utilisation ratios for typical column designs.

Can the column design be optimised by providing different end restraints about each axis?

Yes, and this is a common optimisation. If the weak axis can be braced at mid-height by a secondary beam or tie, Lcr,z reduces to 2.5 m (half the system length). This reduces lambda_bar_z from 1.076 to (2,500/60.8) / 76.4 = 41.1 / 76.4 = 0.538. With chi_z ~ 0.83 (curve c), Nb,Rd increases to approximately 3,123 kN — nearly equal to the strong-axis capacity. This allows a lighter section (e.g., HEB200 or even HEB180) if two-axis symmetry of the framing provides this restraint. In multi-storey buildings, this is exactly what the floor beams provide — lateral restraint to the column weak axis at each floor level, making the inter-storey height the weak-axis buckling length.

Related Pages

• EN 1993 Column Buckling — Curves a0-d
• EN 1993 Beam Design — Worked Example IPE400
• EN 1993-1-1 Beam Design — Flexure, Shear & LTB
• EN 1993 Steel Grades — S235, S275, S355, S460
• EN 1993-1-8 Connection Design — Bolts & Welds
• Column Capacity Calculator — Free EN 1993 Tool
• Beam Capacity Calculator — Free Online Tool

Educational reference only. This worked example is for training and verification purposes. Verify all design values against the current EN 1993-1-1 and the applicable National Annex for your jurisdiction. Section properties should be taken from the manufacturer's published data or the SCI/BCSA Blue Book. Results are PRELIMINARY — NOT FOR CONSTRUCTION without independent verification by a qualified structural engineer.