👋 Introduction

Steel beams carry bending, but an unrestrained beam can fail in a way the cross-section check never sees: the compression flange moves sideways and the section twists long before the steel yields.

This work example runs the lateral-torsional buckling check of a hot-rolled I-section 610x305x238 UB in S355 carrying a design bending moment M_Ed = 1,000 kNm about the major axis over a laterally unrestrained length of 8.0 m, in the same order a calculation report runs it:

Classify the section — the class decides which modulus the bending resistance is built from
Cross-section bending check M_c,y,Rd
LTB check M_b,Rd — elastic critical moment M_cr, slenderness λ̄_LT, buckling curve and the reduction factor χ_LT, with the buckling resistance governing

The concept behind each step is explained step by step in the companion article: Lateral Torsional Buckling in Eurocode 3 - A Simple Calculation Guide for Beams.

📐 Design Data

Below is the design information of the I-section (610x305x238), which we will check under major-axis bending and lateral-torsional buckling.

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 Bending Moment (Major Axis y), M_Ed		1,000 kNm
Unrestrained Length between Lateral Restraints, 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 $I_{z}$	=	$1.58 \times 10^{4} cm^4$
Elastic Modulus, Major $W_{el,y}$	=	$6\,589 cm^3$
Plastic Modulus, Major $W_{pl,y}$	=	$7\,486 cm^3$
Warping Constant $I_{w}$	=	$1.45 \times 10^{13} mm^6$
Torsional Constant $I_{t}$	=	$785 cm^4$

**Compared with the column check, two properties are new here: the warping constant I_w and the torsional constant I_t. They measure how well the section resists twisting, and together with the minor-axis stiffness I_z they control the elastic critical moment below.

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 $\frac{c_{w}}{t_{w}}$	=	$\frac{h-2 \times t_{f}-2 \times r}{t_{w}} = 29.35 ≤ 72\epsilon = 58.58, Web \hspace{0.1cm} is \hspace{0.1cm} Class \hspace{0.1cm} 1$

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

**The same web that is Class 2 in compression is Class 1 in bending, and for a beam bent about the major axis it is the bending case that enters the check: the full plastic modulus W_pl,y applies (an elastic modulus would only enter if the section were Class 3).

3.Section Check with Design Action

3.1Bending check - Class 1 section 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)
$\frac{M_{Ed}}{M_{c,y,Rd}}$	=	$\frac{M_{Ed}}{M_{c,y,Rd}} \le 1$
	=	$\frac{1\,000}{2\,658}$
	=	$0.38 \le 1,\ \text{OK!}$

∴ The section is sufficient for Pure Bending

3.2Lateral-torsional buckling check Cl.6.3.2

Following the check order of the LTB guide: M_cr → λ_LT → buckling curve → χ_LT → M_b,Rd.

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 $C_{1}$	=	$1.0 (conservative)$

3.2.1Elastic critical moment M_cr

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 $\sqrt{\frac{I_{w}}{I_{z}} + \frac{(L_{cr})^2 \times G \times I_{t}}{\pi^2 \times E \times I_{z}}}$	=	$\sqrt{91\,557 + 123\,960} = 464 mm$

Eurocode 3 does not give a formula for M_cr. the M_cr above comes from the standard expression for a doubly symmetric I-beam with the load at the shear centre (NCCI SN003).

3.2.2Non-dimensional slenderness λ̄_LT Cl.6.3.2.2

Slenderness, Lateral-Torsional(Cl. 6.3.2.2(1), Eq. 6.56 definition)
$\bar\lambda_{LT}$	=	$\sqrt{\frac{W_{pl,y} \times f_{y}}{M_{cr}}}$
	=	$\sqrt{\frac{2\,658}{2\,381}}$
	=	$\text{1.06}$

3.2.3Buckling curve Table 6.5

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

3.2.4Reduction factors Φ_LT and χ_LT Cl.6.3.2.3

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)
$\chi_{LT}$	=	$\frac{1}{\phi_{LT} + \sqrt{\phi_{LT}^2 - \beta \times (\bar\lambda_{LT})^2}}$
	=	$\frac{1}{1.08 + \sqrt{1.08^2 - 0.75 \times 1.06^2}}$
	=	$\text{0.61}$

3.2.5Moment distribution factor f and χ_LT,mod Cl.6.3.2.3

The moment shape also enters through the factor f, with the correction factor k_c from Table 6.6 (k_c = 1.0 for a uniform moment, ψ = 1):

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)
$\chi_{LT,mod}$	=	$\frac{\chi_{LT}}{f}$
	=	$\frac{0.61}{1.0}$
	=	$\text{0.61}$

3.2.6Design buckling resistance moment and the check Cl.6.3.2.1

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)
$\frac{M_{Ed}}{M_{b,Rd}}$	=	$\frac{M_{Ed}}{M_{b,Rd}} \le 1$
	=	$\frac{1\,000}{1\,621}$
	=	$0.62 \le 1,\ \text{OK!}$

∴ The section is sufficient for Lateral-Torsional Buckling — the buckling check governs.

📋 Summary

The I Section (610x305x238) is sufficient under Moment at Major Axis (1,000 kNm), it :
✓ Bending Check at major axis (util: 38%)
✓ Lateral-Torsional Buckling Check (util: 62%) - governing

Quantity	Cross-section bending (Cl. 6.2.5)	Lateral-torsional buckling (Cl. 6.3.2)
Slenderness	—	λ̄_LT = 1.06
Buckling curve (α_LT)	—	c (0.49)
Resistance	M_c,y,Rd = 2 658 kNm	M_b,Rd ≈ 1 621 kNm
Utilisation	0.38	0.62 — governing

🔧 Improvement of Lateral Torsional Buckling Capacity

If the compression flange were restrained at mid-span:

L_cr would halve to 4.0 m, M_cr would rise from 2 381 kNm to about 7 180 kNm,
λ̄_LT would fall to 0.61 and χ_LT would rise to about 0.88, taking the utilisation down from 62% to roughly 43%.

∴ Adding lateral restraints is the most effective fix for LTB.

🙋 Frequently asked questions

1. What is lateral torsional buckling in an I-section beam?

When a slender beam bends about its major axis, the compression flange behaves like a slender column: it cannot bow in the plane of the web, so it moves sideways — and because the tension flange resists that movement, the whole section twists. The beam can therefore fail well below its cross-section capacity (2 658 kNm here). The guide article explains the mechanism with diagrams.

2. Which Eurocode 3 clause covers the LTB check?

Cl. 6.3.2 of EN 1993-1-1. It offers the general case (Cl. 6.3.2.2) and the special method for rolled or equivalent welded sections (Cl. 6.3.2.3), which has a longer plateau (λ̄_LT,0 = 0.4), a flatter curve (β = 0.75) and the optional factor f. This example uses the special method, as most tools do for rolled beams — the National Annex may restrict the choice.

3. What is the elastic critical moment M_cr?

The theoretical moment at which a perfect, elastic beam would buckle sideways — the beam version of the Euler critical load. It collects the sideways bending stiffness (π²EI_z/L²) and the torsional stiffness (G·I_t, I_w), modified by the moment shape through C₁. Eurocode 3 does not give the formula — it comes from guidance such as NCCI SN003 or from structural software.

4. When can lateral torsional buckling be ignored?

When the compression flange is fully restrained and twist is prevented (e.g. by a concrete slab or closely spaced bracing), for sections not susceptible to LTB such as CHS/SHS, and when the beam is stocky enough: λ̄_LT ≤ 0.4 or M_Ed/M_cr ≤ 0.16 (Cl. 6.3.2.2(4) with the recommended values). Here λ̄_LT = 1.06, so the full check is required.

5. How does the bending moment diagram affect the result?

A uniform moment is the worst case: it gives C₁ = 1.0 and k_c = 1.0. A varying diagram (≈ 1.13 for a UDL, ≈ 1.35 for a central point load) gives a higher M_cr and, in the special method, a moment-gradient benefit through the factor f. Keep C₁ and k_c consistent — both describe the same moment diagram.

6. Does the position of the load matter?

Yes. A load applied on the top flange is destabilising and lowers M_cr, while a load on the bottom flange is stabilising. The simplified M_cr formula used here assumes loading at the shear centre; for top-flange loading use the full expression from NCCI SN003 or structural software.

📄 Need a Beam Lateral Torsional Buckling Check in Your Design Report?

If you want a ready-to-submit design report with the lateral-torsional buckling check, 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.

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#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.