EN 1993-1-1 Column Design Worked Example — HEA 200 (Clause 6.3, Buckling Curves a₀-d)

Quick Reference: This worked example demonstrates a complete EN 1993-1-1 column buckling design check for an axially loaded HEA 200 column in S275 steel with an effective length of 4.5 m. The verification covers cross-section classification (Cl. 5.5), the Perry-Robertson formulation for the reduction factor χ, buckling curve selection per Table 6.2, flexural buckling about both axes (Cl. 6.3.1), torsional buckling check (Cl. 6.3.1.4), and the design buckling resistance Nb,Rd. Use the free column capacity calculator for instant EN 1993 checks.

PRELIMINARY — NOT FOR CONSTRUCTION. All calculations are illustrative educational examples. Results must be verified by a licensed Professional Engineer before use in any design project.

Design Data

HEA 200 Section Properties

Step 1: Section Classification for Compression (Cl. 5.5)

Epsilon for S275: ε = √(235/275) = 0.924

Web Classification (uniform compression)

cw = h − 2tf − 2r = 190 − 2 × 10 − 2 × 18 = 134 mm

c/tw = 134 / 6.5 = 20.62

Class 1 limit for internal part in pure compression: 33ε = 33 × 0.924 = 30.49

20.62 < 30.49 → Web is Class 1 ✓

Flange Classification (uniform compression)

cf = (b − tw − 2r) / 2 = (200 − 6.5 − 2 × 18) / 2 = 78.75 mm

c/tf = 78.75 / 10.0 = 7.88

Class 1 limit for outstand flange in pure compression: 9ε = 9 × 0.924 = 8.32

7.88 < 8.32 → Flange is Class 1 ✓

Conclusion: HEA 200 in S275 is Class 1 for pure compression.

Step 2: Buckling Curve Selection (Table 6.2)

For hot-rolled H-sections with tf ≤ 40 mm in S235–S420 steel:

Major-axis buckling (y-y): α = 0.34 Minor-axis buckling (z-z): α = 0.49

The complete set of EN 1993-1-1 buckling curves:

Step 3: Flexural Buckling About y-y Axis (Major Axis)

Non-Dimensional Slenderness

Reference slenderness: λ₁ = 93.9ε = 93.9 × 0.924 = 86.8

λ̄,y = (Lcr,y / iy) / λ₁ = (4500 / 82.8) / 86.8 = 54.35 / 86.8 = 0.626

Alternative computation via Euler load:

Ncr,y = π²EIy / Lcr,y² = π² × 210,000 × 36.92 × 10⁶ / 4500²

Ncr,y = 3,793,000 N = 3793 kN

λ̄,y = √(A × fy / Ncr,y) = √(5380 × 275 / 3,793,000) = √(1,479,500 / 3,793,000) = √(0.390) = 0.624

(Slight difference due to rounding; use λ̄,y = 0.626 from the slenderness ratio method.)

Reduction Factor χy (Perry-Robertson, Curve b, α=0.34)

Φy = 0.5 [1 + α(λ̄,y − 0.2) + λ̄,y²]

Φy = 0.5 [1 + 0.34 × (0.626 − 0.2) + 0.626²]

Φy = 0.5 [1 + 0.34 × 0.426 + 0.392]

Φy = 0.5 [1 + 0.145 + 0.392] = 0.5 × 1.537 = 0.7685

χy = 1 / [Φy + √(Φy² − λ̄,y²)]

χy = 1 / [0.7685 + √(0.7685² − 0.626²)]

χy = 1 / [0.7685 + √(0.5906 − 0.3919)]

χy = 1 / [0.7685 + √(0.1987)] = 1 / [0.7685 + 0.4457] = 1 / 1.2142

χy = 0.824 ≤ 1.0 ✓

Buckling Resistance y-y

Nb,Rd,y = χy × A × fy / γM1 = 0.824 × 5380 × 275 / 10⁶

Nb,Rd,y = 0.824 × 1479.5 = 1219.1 kN

Unity check: NEd / Nb,Rd,y = 850 / 1219.1 = 0.697 → OK ✓

Step 4: Flexural Buckling About z-z Axis (Minor Axis)

Non-Dimensional Slenderness

λ̄,z = (Lcr,z / iz) / λ₁ = (4500 / 49.8) / 86.8 = 90.36 / 86.8 = 1.041

Reduction Factor χz (Curve c, α=0.49)

Φz = 0.5 [1 + α(λ̄,z − 0.2) + λ̄,z²]

Φz = 0.5 [1 + 0.49 × (1.041 − 0.2) + 1.041²]

Φz = 0.5 [1 + 0.49 × 0.841 + 1.084]

Φz = 0.5 [1 + 0.412 + 1.084] = 0.5 × 2.496 = 1.248

χz = 1 / [Φz + √(Φz² − λ̄,z²)]

χz = 1 / [1.248 + √(1.248² − 1.041²)]

χz = 1 / [1.248 + √(1.557 − 1.084)]

χz = 1 / [1.248 + √(0.473)] = 1 / [1.248 + 0.688] = 1 / 1.936

χz = 0.517 ≤ 1.0 ✓

Buckling Resistance z-z

Nb,Rd,z = χz × A × fy / γM1 = 0.517 × 5380 × 275 / 10⁶

Nb,Rd,z = 0.517 × 1479.5 = 764.9 kN

Unity check: NEd / Nb,Rd,z = 850 / 764.9 = 1.111 → FAILS ✗

Step 5: Resolution — Increase Section Size

The HEA 200 fails minor-axis buckling. Options:

1. Reduce effective length with intermediate lateral bracing
2. Increase to HEA 220 (iz = 56.0 mm, A = 6430 mm²)
3. Increase to HEA 240 for margin

Let us verify HEA 220:

λ̄,z = (4500 / 56.0) / 86.8 = 80.36 / 86.8 = 0.926

Φz = 0.5 [1 + 0.49 × (0.926 − 0.2) + 0.926²] = 0.5 [1 + 0.356 + 0.857] = 1.107

χz = 1 / [1.107 + √(1.107² − 0.926²)] = 1 / [1.107 + √(1.225 − 0.857)] = 1 / [1.107 + 0.607] = 0.583

Nb,Rd,z = 0.583 × 6430 × 275 / 10⁶ = 1031 kN

NEd/Nb,Rd,z = 850/1031 = 0.824 → OK ✓

Check major axis for completeness:

λ̄,y = (4500 / 91.4) / 86.8 = 49.23 / 86.8 = 0.567

Φy = 0.5 [1 + 0.34 × (0.567 − 0.2) + 0.567²] = 0.5 [1 + 0.125 + 0.321] = 0.723

χy = 1 / [0.723 + √(0.723² − 0.567²)] = 1 / [0.723 + √(0.523 − 0.321)] = 1 / [0.723 + 0.449] = 0.853

Nb,Rd,y = 0.853 × 6430 × 275 / 10⁶ = 1508 kN > 850 kN ✓

Step 6: Torsional Buckling Check (Cl. 6.3.1.4)

For doubly symmetric H-sections, torsional buckling rarely governs, but EN 1993 requires verification.

Elastic torsional buckling load:

Ncr,T = (1/i₀²) × [G × It + (π² × E × Iw) / Lcr,T²]

where i₀² = iy² + iz² = 82.8² + 49.8² = 6856 + 2480 = 9336 mm² for HEA 200.

G = E / [2(1+ν)] = 210,000 / 2.6 = 80,769 MPa

Ncr,T = (1/9336) × [80,769 × 207,000 + (π² × 210,000 × 108 × 10⁹) / 4500²]

Ncr,T = (1/9336) × [16.72 × 10⁹ + 1.108 × 10¹¹] N

Ncr,T = (1/9336) × 127.5 × 10⁹ = 13,656,000 N = 13,656 kN

This is significantly larger than Ncr,z = 1368 kN, so torsional buckling does not govern as expected for H-sections.

The Perry-Robertson Formula — Theory and Derivation

The reduction factor χ in EN 1993-1-1 Cl. 6.3.1.2 is derived from the Ayrton-Perry formulation of the European buckling curves. The key relationship:

χ = 1 / [Φ + √(Φ² − λ̄²)] with χ ≤ 1.0

where Φ = 0.5 [1 + α(λ̄ − 0.2) + λ̄²]

This is mathematically equivalent to solving the quadratic equation for the Perry-Robertson imperfection:

(1 − χ)(1 − χ × λ̄²) = η × χ

where η = α(λ̄ − 0.2) is the generalized imperfection factor.

The term α(λ̄ − 0.2) in Φ represents the effect of geometric imperfections (initial out-of-straightness) and residual stresses. The plateau at λ̄ ≤ 0.2 reflects the observation that very stocky columns (low slenderness) reach yield without significant imperfection sensitivity. The λ̄² term in Φ captures the non-linear degradation of stiffness as buckling is approached.

Sensitivity Analysis: Effect of Buckling Curve Selection

For the HEA 200 column with λ̄,z = 1.041, the choice of buckling curve significantly affects capacity:

If the German NA permitted curve 'a' for minor-axis buckling (which it generally does not), the HEA 200 would just meet the requirement at a ratio of 0.994. Under the standard Eurocode curve assignment (curve c for z-z axis), it fails.

Key Takeaways

1. Minor-axis buckling governs for H-sections because iz is substantially smaller than iy. The design buckling resistance about z-z is typically 50-70% of the y-y axis resistance for the same effective length.
2. Buckling curve selection matters enormously. A shift from curve b to curve c reduces Nb,Rd by approximately 5-10% for intermediate slenderness. For slender columns (λ̄ > 1.5), the difference can exceed 15%.
3. The plateau λ̄ ≤ 0.2 means very short columns (Lcr/i < 17 for S275) achieve full squash load without buckling reduction. This is why column base regions and short bracing members often require only a cross-section check.
4. National Annexes modify buckling curve assignments. The UK NA, German NA, and French NA all apply subtle differences in curve selection for specific section types. Always verify against the applicable NA for the jurisdiction of design.

Related Pages

• EN 1993-1-1 Beam Design — IPE 300 Worked Example, χLT
• EN 1993 Column Buckling Guide — Curves a₀-d
• EN 1993-1-8 Base Plate Design — T-Stub, Concrete Bearing
• Column K-Factor Table — Effective Length Factors
• European Steel Grades (S235–S460)