👋 Introduction

A beam-column carries axial compression and bending at the same time, and the two make each other worse: the bending bows the member, and the compression acts on that bow.

This work example runs the full check of a hot-rolled I-section 610x305x238 UB in S355 under N_Ed = 800 kN, M_y,Ed = 800 kNm and M_z,Ed = 100 kNm over an 8.0 m buckling length, in the order a calculation report runs it:

Classify the section
Cross-section checks — compression, bending about both axes, and the biaxial + axial check
Member buckling checks — flexural buckling (χ_y, χ_z) and lateral-torsional buckling (χ_LT)
Interaction check — Eq. 6.61 and 6.62 with the Annex B interaction factors, the governing step

The method behind each step is explained in the companion article: Combined Compression and Bending in Eurocode 3 - A Simple Calculation Guide for Beam-Columns.

📐 Design Data

Below is the design information of the I-section (610x305x238), which we will check under compression and biaxial bending.

Steel Section
	Type	I-section
	Standard	610x305x238
	Depth, h	635.8mm
	Breadth, b	311.4mm
	Flange Thickness, t_f	31.4mm
	Web Thickness, t_w	18.4mm
	Root Radius, r	16.5mm

Material
Steel Grade		S355
Yield Strength, f_y		355 N/mm²
Young's Modulus, E		210,000 N/mm²
Shear Modulus, G		81,000 N/mm²
Partial Factor, γ_M0		1.0
Partial Factor, γ_M1		1.0

Design Actions and Restraints
Design Axial Force (Compression), N_Ed		800 kN
Design Bending Moment (Major Axis y), M_y,Ed		800 kNm
Design Bending Moment (Minor Axis z), M_z,Ed		100 kNm
Buckling Length, both axes, L_cr		8.0 m

🔢 Calculation Steps

*Unless specified, Below uses Eurocode 3 symbols EN 1993-1-1.

1.Section Properties of I section (610x305x238)

Area $A$	=	$3.03 \times 10^{4} mm^2$
Moment of Inertia, Major $I_{y}$	=	$2.09 \times 10^{5} cm^4$
Moment of Inertia, Minor $I_{z}$	=	$1.58 \times 10^{4} cm^4$
Plastic Modulus, Major $W_{pl,y}$	=	$7\,486 cm^3$
Plastic Modulus, Minor $W_{pl,z}$	=	$1\,571 cm^3$
Warping Constant $I_{w}$	=	$1.45 \times 10^{13} mm^6$
Torsional Constant $I_{t}$	=	$785 cm^4$

2.Section Classification Table 5.2

Material Factor $ε$	=	$\sqrt{235/f_{y}} = 0.81$
Outstand Flange $\frac{c_{f}}{t_{f}}$	=	$\frac{(b-t_{w}-2 \times r)/2}{t_{f}} = 4.14 ≤ 9\epsilon = 7.32, Flange \hspace{0.1cm} is \hspace{0.1cm} Class \hspace{0.1cm} 1$
Part Subject to Compression
Web, subject to Compression $\frac{c_{w}}{t_{w}}$	=	$\frac{h-2 \times t_{f}-2 \times r}{t_{w}} = 29.35 ≤ 38\epsilon = 30.92, Web \hspace{0.1cm} is \hspace{0.1cm} Class \hspace{0.1cm} 2$
Part Subject to Bending
Web, subject to Bending $\frac{c_{w}}{t_{w}}$	=	$\frac{h-2 \times t_{f}-2 \times r}{t_{w}} = 29.35 ≤ 72\epsilon = 58.58, Web \hspace{0.1cm} is \hspace{0.1cm} Class \hspace{0.1cm} 1$
Part Subject to Bending and Compression
Web compression ratio $\alpha$	=	$\frac{1}{2} \times (1 + \frac{N_{Ed}}{f_{y} \times c_{w} \times t_{w}}) = \frac{1}{2} \times (1 + \frac{800\,000}{355 \times 540 \times 18.4}) = 0.61$
Web, subject to bending and compression $\frac{c_{w}}{t_{w}}$	=	$\frac{h-2 \times t_{f}-2 \times r}{t_{w}} = 29.35 ≤ \frac{396 \epsilon}{13\alpha - 1} = 46.2, Web \hspace{0.1cm} is \hspace{0.1cm} Class \hspace{0.1cm} 1$

The overall cross-section is:
- Class 2 subjected to Compression
- Class 1 subjected to Bending
- Class 1 subjected to Bending and Compression

3.Section Check with Design Action

3.1Compression check - Class 2 section Cl.6.2.4

Compression Resistance(Cl. 6.2.4(2), Eq. 6.10)
$N_{c,Rd}$	=	$\frac{A \times f_{y}}{\gamma_{M0}}$
	=	$\frac{3.03 \times 10^{4} \times 355}{1.0}$
	=	$\text{10 757 kN}$

Pure Compression Check(Cl. 6.2.4(1), Eq. 6.9)
$\frac{N_{Ed}}{N_{c,Rd}}$	=	$\frac{N_{Ed}}{N_{c,Rd}} \le 1$
	=	$\frac{800}{10\,757}$
	=	$0.07 \le 1,\ \text{OK!}$

∴ The section is sufficient for Pure Compression

3.2Bending check about y - Class 1, plastic modulus Cl.6.2.5

Design Resistance for Bending at y(Cl. 6.2.5(2), Eq. 6.13) ^ⓘ
$M_{c,y,Rd}$	=	$\frac{W_{pl,y} \times f_{y}}{\gamma_{M0}}$
	=	$\frac{7\,486 \times 355}{1.0}$
	=	$\text{2 658 kNm}$

Bending Check at major axis(Cl. 6.2.5, Eq. 6.12)
$\frac{M_{Ed}}{M_{c,y,Rd}}$	=	$\frac{M_{Ed}}{M_{c,y,Rd}} \le 1$
	=	$\frac{800}{2\,658}$
	=	$0.30 \le 1,\ \text{OK!}$

∴ The section is sufficient for Pure Bending about y

3.3Bending check about z Cl.6.2.5

Design Resistance for Bending at z(Cl. 6.2.5(2), Eq. 6.13)
$M_{c,z,Rd}$	=	$\frac{W_{pl,z} \times f_{y}}{\gamma_{M0}}$
	=	$\frac{1\,571 \times 355}{1.0}$
	=	$\text{558 kNm}$

Bending Check at minor axis(Cl. 6.2.5, Eq. 6.12)
$\frac{M_{Ed}}{M_{c,z,Rd}}$	=	$\frac{M_{Ed}}{M_{c,z,Rd}} \le 1$
	=	$\frac{100}{558}$
	=	$0.18 \le 1,\ \text{OK!}$

∴ The section is sufficient for Pure Bending about z

3.4Combined axial and biaxial bending check Cl.6.2.9

3.4.1Combined axial and bending at major axis check

For major axis of I section,reduction effect from axial force need not be made, if it satisfied below two criterias:
Criteria 1: $0.25 N_{pl,Rd}$	=	$0.25 \times 1.08 \times 10^{4} = 2692 kN ≥ 800 kN, Pass$
Criteria 2: $\frac{0.5 \times h_{w} \times t_{w} \times f_{y}}{\gamma_{M0}}$	=	$\frac{0.5 \times 573 \times 18 \times 355}{1.0} = 1871 kN ≥ 800 kN, Pass$
∴ Current Section pass both certeria, no further reduction factor apply to y axis from axial force.

3.4.2Combined axial and bending at minor axis check

For minor axis of I section,reduction effect from axial force need not be made, if it satisfied below two criterias:
Criteria 1: $\frac{h_{w} \times t_{w} \times f_{y}}{\gamma_{M0}}$	=	$\frac{573 \times 18 \times 355}{1.0} = 3743 kN ≥ 800 kN, Pass$
∴ Current Section pass the certeria, no further reduction factor apply to z axis from axial force.

3.4.3Combined axial and biaxial bending check

Ratio of axial force $n$	=	$\frac{N_{Ed}}{N_{pl,Rd}} = \frac{800}{1.08 \times 10^{4}} = 0.07$
Exponent, major axis (I- and H-sections) $\alpha$	=	$2.0$
Exponent, minor axis $\beta$	=	$5 \times n = 0.37 < 1.0, \text{ take } \beta = 1.0$

Biaxial Bending and Axial Check(Cl. 6.2.9.1(6))
$\left(\frac{M_{y,Ed}}{M_{N,y,Rd}}\right)^{\alpha} + \left(\frac{M_{z,Ed}}{M_{N,z,Rd}}\right)^{\beta}$	=	$\left(\frac{M_{y,Ed}}{M_{N,y,Rd}}\right)^{\alpha} + \left(\frac{M_{z,Ed}}{M_{N,z,Rd}}\right)^{\beta} \le 1$
	=	$\left(\frac{800}{2\,658}\right)^{2} + \left(\frac{100}{558}\right)^{1}$
	=	$0.09 + 0.18 = 0.27 \le 1,\ \text{OK!}$

∴ The section is sufficient for Biaxial Bending and Axial.

4.Buckling Check

4.1Flexural buckling - both axes Cl.6.3.1

4.1.1Elastic critical buckling force Cl.6.3.1.2

Elastic Critical Force, y axis(Cl. 6.3.1.2(1))
$N_{cr,y}$	=	$\frac{\pi^2 \times E \times I_{y}}{(L_{cr})^2}$
	=	$\frac{\pi^2 \times 210\,000 \times 2.095 \times 10^{9}}{(8\,000)^2}$
	=	$\text{67 800 kN}$

Elastic Critical Force, z axis(Cl. 6.3.1.2(1))
$N_{cr,z}$	=	$\frac{\pi^2 \times E \times I_{z}}{(L_{cr})^2}$
	=	$\frac{\pi^2 \times 210\,000 \times 1.584 \times 10^{8}}{(8\,000)^2}$
	=	$\text{5 130 kN}$

4.1.2Non-dimensional slenderness λ̄ Cl.6.3.1.2

Slenderness, y axis(Cl. 6.3.1.2(1))
$\bar\lambda_{y}$	=	$\sqrt{\frac{A \times f_{y}}{N_{cr,y}}}$
	=	$\sqrt{\frac{10\,757}{67\,800}}$
	=	$\text{0.40}$

Slenderness, z axis(Cl. 6.3.1.2(1))
$\bar\lambda_{z}$	=	$\sqrt{\frac{A \times f_{y}}{N_{cr,z}}}$
	=	$\sqrt{\frac{10\,757}{5\,130}}$
	=	$\text{1.45}$

4.1.3Buckling curves Table 6.2

For this hot-rolled I-section:
$h/b$	=	$635.8/311.4 = 2.05 > 1.2$
$t_{f}$	=	$31.4 mm ≤ 40 mm$
From Table 6.2 gives:
- about y-y: buckling curve a, with imperfection factor α = 0.21 from Table 6.1
- about z-z: buckling curve b, with imperfection factor α = 0.34 from Table 6.1

4.1.4Reduction factors Φ and χ Cl.6.3.1.2

Intermediate Factor, y axis(Cl. 6.3.1.2(1))
$\phi_{y}$	=	$0.5[1 + \alpha(\bar\lambda_{y} - 0.2) + (\bar\lambda_{y})^2]$
	=	$0.5[1 + 0.21(0.40 - 0.2) + 0.40^2]$
	=	$\text{0.60}$

Reduction Factor, y axis(Cl. 6.3.1.2(1))
$\chi_{y}$	=	$\frac{1}{\phi_{y} + \sqrt{\phi_{y}^2 - (\bar\lambda_{y})^2}}$
	=	$\frac{1}{0.60 + \sqrt{0.60^2 - 0.40^2}}$
	=	$\text{0.95}$

Intermediate Factor, z axis(Cl. 6.3.1.2(1))
$\phi_{z}$	=	$0.5[1 + \alpha(\bar\lambda_{z} - 0.2) + (\bar\lambda_{z})^2]$
	=	$0.5[1 + 0.34(1.45 - 0.2) + 1.45^2]$
	=	$\text{1.76}$

Reduction Factor, z axis(Cl. 6.3.1.2(1))
$\chi_{z}$	=	$\frac{1}{\phi_{z} + \sqrt{\phi_{z}^2 - (\bar\lambda_{z})^2}}$
	=	$\frac{1}{1.76 + \sqrt{1.76^2 - 1.45^2}}$
	=	$\text{0.36}$

∴ The critical reduction factor is χ_z = 0.36 — minor-axis buckling governs

4.1.5Design buckling resistance and the check Cl.6.3.1.1

Buckling Resistance, z axis(Cl. 6.3.1.1, Eq. 6.47) ^ⓘ
$N_{b,Rd}$	=	$\chi \times \frac{A \times f_{y}}{\gamma_{M1}}$
	=	$0.36 \times 10\,757$
	=	$\text{3 870 kN}$

Buckling Check under axial compression(Cl. 6.3.1.1, Eq. 6.46)
$\frac{N_{Ed}}{N_{b,Rd}}$	=	$\frac{N_{Ed}}{N_{b,Rd}} \le 1$
	=	$\frac{800}{3\,870}$
	=	$0.21 \le 1,\ \text{OK!}$

∴ The section is sufficient for Flexural Buckling

4.2Lateral-torsional buckling Cl.6.3.2.3

According to Cl. 6.3.2.3:
Plateau slenderness $\bar\lambda_{LT,0}$	=	$0.4 (maximum value)$
Curve shape factor $\beta$	=	$0.75 (minimum value)$
Curve shape factor $k_{c}$	=	$1 (minimum value)$
For Calculation of M_cr:
Moment diagram factor $C_{1}$	=	$1.0 (conservative)$

Elastic Critical Moment for Lateral-Torsional Buckling(NCCI SN003)
$M_{cr}$	=	$C_{1} \times \frac{\pi^2 \times E \times I_{z}}{(L_{cr})^2} \times \sqrt{\frac{I_{w}}{I_{z}} + \frac{(L_{cr})^2 \times G \times I_{t}}{\pi^2 \times E \times I_{z}}}$
	=	$1.0 \times 5\,129 \text{ kN} \times 464 \text{ mm}$
	=	$\text{2 381 kNm}$

where
Lateral Euler term $\frac{\pi^2 \times E \times I_{z}}{(L_{cr})^2}$	=	$\frac{\pi^2 \times 210\,000 \times 1.584 \times 10^{8}}{(8\,000)^2} = 5\,129 kN$
Torsional term $\sqrt{\frac{I_{w}}{I_{z}} + \frac{(L_{cr})^2 \times G \times I_{t}}{\pi^2 \times E \times I_{z}}}$	=	$\sqrt{91\,557 + 123\,960} = 464 mm$

Slenderness, Lateral-Torsional(Cl. 6.3.2.2(1))
$\bar\lambda_{LT}$	=	$\sqrt{\frac{W_{pl,y} \times f_{y}}{M_{cr}}}$
	=	$\sqrt{\frac{2\,658}{2\,381}}$
	=	$\text{1.06}$

For this hot-rolled I-section, the special method selects:
$h/b$	=	$635.8/311.4 = 2.05 > 1.2$
From Table 6.5 gives:
- buckling curve c, with imperfection factor α_LT = 0.49 from Table 6.3

Intermediate Factor, lateral-torsional(Cl. 6.3.2.3(1))
$\phi_{LT}$	=	$0.5[1 + \alpha_{LT} \times (\bar\lambda_{LT} - \bar\lambda_{LT,0}) + \beta \times (\bar\lambda_{LT})^2]$
	=	$0.5[1 + 0.49 \times (1.06 - 0.4) + 0.75 \times 1.06^2]$
	=	$\text{1.08}$

Reduction Factor, lateral-torsional(Cl. 6.3.2.3(1), Eq. 6.57)
$\chi_{LT}$	=	$\frac{1}{\phi_{LT} + \sqrt{\phi_{LT}^2 - \beta \times (\bar\lambda_{LT})^2}}$
	=	$\frac{1}{1.08 + \sqrt{1.08^2 - 0.75 \times 1.06^2}}$
	=	$\text{0.61}$

Modification Factor for moment distribution(Cl. 6.3.2.3(2))
$f$	=	$1 - 0.5 \times (1 - k_{c}) \times [1 - 2.0 \times (\bar\lambda_{LT} - 0.8)^2]$
	=	$1 - 0.5 \times (1 - 1.0) \times [1 - 2.0 \times (1.06 - 0.8)^2]$
	=	$\text{1.0}$

Reduction Factor with moment distribution considered(Cl. 6.3.2.3(2), Eq. 6.58)
$\chi_{LT,mod}$	=	$\frac{\chi_{LT}}{f}$
	=	$\frac{0.61}{1.0}$
	=	$\text{0.61}$

Design Buckling Resistance Moment(Eq. 6.55)
$M_{b,Rd}$	=	$\chi_{LT,mod} \times \frac{W_{pl,y} \times f_{y}}{\gamma_{M1}}$
	=	$0.61 \times 2\,658$
	=	$\text{1 621 kNm}$

Buckling Check under bending(Cl. 6.3.2.1(1), Eq. 6.54)
$\frac{M_{Ed}}{M_{b,Rd}}$	=	$\frac{M_{Ed}}{M_{b,Rd}} \le 1$
	=	$\frac{800}{1\,621}$
	=	$0.49 \le 1,\ \text{OK!}$

∴ The section is sufficient for Lateral-Torsional Buckling

4.3Interaction check — bending and axial compression Cl.6.3.3

In below analysis, calculation of interactive factors k_yy, k_yz, k_zy, k_zz is using alternative method B as stated in Annex B of EC3.
For moment distrubtion factor from Table B3,
$C_{my}, C_{mz} and C_{mLT}$	=	$1$
For capacity of section subject to axial and bending from Table 6.7,
$N_{Rk}$	=	$A \times f_{y}$
$M_{Rk,y}$	=	$W_{pl,y} \times f_{y}$
$M_{Rk,z}$	=	$W_{pl,z} \times f_{y}$
$n_{y}$	=	$\frac{N_{Ed}}{\chi_{y} \times N_{Rk} / \gamma_{M1}} = 0.08$
$n_{z}$	=	$\frac{N_{Ed}}{\chi_{z} \times N_{Rk} / \gamma_{M1}} = 0.21$

Interaction Factor, kyy(Table B.1)
$k_{yy}$	=	$C_{my} \times (1 + (\bar\lambda_{y} - 0.2) \times n_{y}) ≤ C_{my} \times (1 + 0.8 \times n_{y})$
	=	$1.0 \times (1 + (0.40 - 0.2) \times 0.08) ≤ 1.0 \times (1 + 0.8 \times 0.08)$
	=	$1.02 < 1.06, take 1.02$

Interaction Factor, kzy(Table B.2)
$k_{zy}$	=	$1 - \frac{0.1 \times \bar\lambda_{z}}{C_{mLT} - 0.25} \times n_{z} ≥ 1 - \frac{0.1}{C_{mLT} - 0.25} \times n_{z}$
	=	$1 - 0.1933 \times 0.21 ≥ 1 - 0.1333 \times 0.21$
	=	$0.96 < 0.97, take 0.97$

Interaction Factor, kzz(Table B.1)
$k_{zz}$	=	$C_{mz} \times (1 + (2 \times \bar\lambda_{z} - 0.6) \times n_{z}) ≤ C_{mz} \times (1 + 1.4 \times n_{z})$
	=	$1.0 \times (1 + (2 \times 1.45 - 0.6) \times 0.21) ≤ 1.0 \times (1 + 1.4 \times 0.21)$
	=	$1.48 > 1.29, take 1.29$

Interaction Factor, kyz(Table B.1)
$k_{yz}$	=	$0.6 \times k_{zz}$
	=	$0.6 \times 1.29$
	=	$\text{0.77}$

Combined Buckling Check, major axis(Cl. 6.3.3, Eq. 6.61)
	=	$\frac{N_{Ed}}{\chi_{y} \times N_{Rk}/\gamma_{M1}} + k_{yy}\frac{M_{y,Ed}}{\chi_{LT,mod} \times M_{y,Rk}/\gamma_{M1}} + k_{yz}\frac{M_{z,Ed}}{M_{z,Rk}/\gamma_{M1}} \le 1$
	=	$\frac{800}{10\,220} + 1.02 \times \frac{800}{1\,621} + 0.77 \times \frac{100}{558}$
	=	$0.08 + 0.50 + 0.14 = 0.72 \le 1,\ \text{OK!}$

Combined Buckling Check, minor axis(Cl. 6.3.3, Eq. 6.62)
	=	$\frac{N_{Ed}}{\chi_{z} \times N_{Rk}/\gamma_{M1}} + k_{zy}\frac{M_{y,Ed}}{\chi_{LT,mod} \times M_{y,Rk}/\gamma_{M1}} + k_{zz}\frac{M_{z,Ed}}{M_{z,Rk}/\gamma_{M1}} \le 1$
	=	$\frac{800}{3\,870} + 0.97 \times \frac{800}{1\,621} + 1.29 \times \frac{100}{558}$
	=	$0.21 + 0.48 + 0.23 = 0.92 \le 1,\ \text{OK!}$

∴ The section is sufficient for Bending and Axial Compression Buckling — the governing check is Eq. 6.62 about the minor axis

📋 Summary

The I Section (610x305x238) is sufficient under Compression (800 kN), Moment at Major Axis (800 kNm) and Moment at Minor Axis (100 kNm):
Cross-Section Resistance
✓ Pure Compression Check (max util: 7%)
✓ Bending Check at major axis (max util: 30%)
✓ Bending Check at minor axis (max util: 18%)
✓ Bending and Axial at major axis is pass due to no reduction factor from axial
✓ Bending and Axial at minor axis is pass due to no reduction factor from axial.
✓ Biaxial Bending Moment and Axial (util: 27%)
Buckling Resistance
✓ Buckling Resistance under axial compression (util: 21%)
✓ Buckling Resistance under bending (util: 50%)
✓ Combined Axial & Bending Ratio (Eq. 6.61) (util: 72%)
✓ Combined Axial & Bending Ratio (Eq. 6.62) (util: 92%)

🙋 Frequently asked questions

1. What is the combined compression and bending check in Eurocode 3?

It is the stability check for a beam-column — a member carrying axial compression and bending at the same time. Clause 6.3.3 checks it with the two interaction equations 6.61 and 6.62, each adding an axial term and two bending terms weighted by the interaction factors k. The guide article explains every parameter.

2. Why can't compression and bending be checked separately?

Because compression and bending make each other worse: bending bows the member, and the compression acts on that bow, adding second-order moment. The interaction equations capture this amplification through the k factors — adding two separate utilisation ratios would miss it.

3. What are the interaction factors k_yy, k_yz, k_zy and k_zz?

They weight the bending terms in the interaction equations for second-order effects and the coupling between axes. This example uses Annex B (Method 2) — Tables B.1 and B.2 give ready-made formulae, and Table B.3 gives the moment-distribution factors (all C_m = 1.0 here, conservative).

4. Why are two interaction equations needed?

Because the member can buckle about either axis: Eq. 6.61 covers the major axis (y-y, with χ_y) and Eq. 6.62 the minor axis (z-z, with χ_z). Here the minor axis governs (0.92 against 0.72) — typical, because χ_z = 0.36 is far lower than χ_y = 0.95 for an I-column braced only at its ends.

5. What do α and β mean in the biaxial cross-section check?

They are the exponents of the two bending terms in the cross-section interaction of Cl. 6.2.9.1(6). For I- and H-sections: α = 2.0 and β = 5n, but not less than 1.0 (here 5n = 0.37, so β = 1.0). Check your code edition — some versions differ, and your National Annex may prefer the conservative α = β = 1.0.

📄 Need a Beam-Column Check in Your Design Report?

If you want a ready-to-submit design report with the combined compression and bending check, the buckling checks, section classification and the full cross-section check in Eurocode 3, feel free to try our application — what you need is just the design data values, and a ready-to-submit PDF report will be generated in a minute.

The ready-to-submit PDF Report is not generated by AI but programmed by chartered engineer, which is accurate, no hallucination and same input same output.

#steel design #work example #structural

References

Eurocode 3: Design of steel structures — Part 1-1: General rules and rules for buildings (EN 1993-1-1) —CEN (European Committee for Standardization) ·2005
NCCI: Elastic critical moment for lateral torsional buckling (SN003) —Access Steel

CivilSimple Team

The CivilSimple Team writes practical engineering guides for the profession and the curious. All articles are reviewed for technical accuracy before publication.