UK Lateral-Torsional Buckling Design — EN 1993-1-1 Clause 6.3.2 + UK NA

Complete guide to lateral-torsional buckling (LTB) of steel beams per EN 1993-1-1 Clause 6.3.2 with UK National Annex. Covers the χLT reduction factor, elastic critical moment Mcr calculation, LTB buckling curve selection, and a worked example using a UK Universal Beam (UB) section in S355.

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LTB Check per EN 1993-1-1 Clause 6.3.2

For a beam not fully laterally restrained, the design buckling resistance moment is:

Mb,Rd = χLT × Wy × fy / γM1

Where:

Non-Dimensional Slenderness for LTB

λLT,bar = √(Wy × fy / Mcr)

Where Mcr is the elastic critical moment for LTB.

Mcr — Elastic Critical Moment

For a simply supported beam with uniform moment (conservative) or general loading:

Mcr = C1 × π²EIz / (kzL)² × √[(kz/kw)² × Iw/Iz + (kzL)² × GIt / (π²EIz)]

Parameter Description
C1 Moment gradient factor (1.00 uniform moment, 1.13 UDL, 1.35 triangular)
Iz Weak-axis second moment of area
It Torsion constant
Iw Warping constant
kz End rotation factor (0.5 fixed, 0.7 partially fixed, 1.0 pinned)
kw End warping factor (0.5 fixed, 1.0 free)

LTB Buckling Curves (Table 6.4, UK NA)

Section Type Limits Curve αLT
Rolled I-sections h/b ≤ 2.0 b 0.34
Rolled I-sections h/b > 2.0 c 0.49
Welded I-sections h/b ≤ 2.0 c 0.49
Welded I-sections h/b > 2.0 d 0.76
Other sections all d 0.76

Most UK UB sections have h/b > 2.0 (e.g., 533UB: h/b = 2.51), so curve c typically applies. UK UC sections have h/b ≤ 2.0, so curve b applies.

χLT Reduction Factor

χLT = 1 / [ΦLT + √(ΦLT² − β × λLT,bar²)]

Using the simplified method (Clause 6.3.2.3 for rolled sections):

ΦLT = 0.5 × [1 + αLT × (λLT,bar − λLT,0) + β × λLT,bar²]

With recommended values: λLT,0 = 0.4, β = 0.75

Worked Example — 533UB in S355, 6.0 m Unbraced

Section: 533UB, S355 Span: L = 6.0 m, no intermediate lateral restraint Moment: MEd = 208 kN·m (from 46.2 kN/m UDL)

Section Properties

h = 529 mm, b = 211 mm, h/b = 2.51 > 2 → curve c (αLT = 0.49) Wpl,y = 3,040 cm³, Iy = 65,750 cm⁴, Iz = 2,450 cm⁴ It = 126 cm⁴, Iw = 1.24 dm⁶ fy = 355 MPa, E = 210 GPa, G = 81 GPa

Mcr Calculation (C1 = 1.13 for UDL)

λ1 = 93.9 × √(235/355) = 76.4

Mcr = 1.13 × π² × 210,000 × 2,450×10⁴ / 6,000² × √(1.24×10¹²/2,450×10⁴ + 6,000²×81,000×126×10⁴/(π²×210,000×2,450×10⁴))

= 440 kN·m (full calculation per worked example)

Slenderness and χLT

λLT,bar = √(Wpl,y fy / Mcr) = √(3,040×10³ × 355 / 440×10⁶) = √(1,079/440) = √2.452 = 1.566

ΦLT = 0.5 × [1 + 0.49 × (1.566 − 0.4) + 0.75 × 1.566²] = 0.5 × [1 + 0.571 + 1.839] = 1.705

χLT = 1 / [1.705 + √(1.705² − 0.75 × 2.452)] = 1 / [1.705 + √(2.907 − 1.839)] = 1 / [1.705 + 1.033] = 1 / 2.738 = 0.365

Buckling Resistance

Mb,Rd = 0.365 × 3,040×10³ × 355 / 1.00 = 394 kN·m

Utilisation: MEd / Mb,Rd = 208 / 394 = 0.528 — OK (53 % utilised)

With continuous lateral restraint from composite slab: Utilisation = 208 / 1,079 = 0.193 (19 %). The LTB check governs only when the beam is unrestrained.


Design Resources


Frequently Asked Questions

What is the difference between LTB buckling curves b and c for UK sections?

Curve b (αLT = 0.34) applies to rolled I-sections with h/b ≤ 2.0 (typically UK UC sections). Curve c (αLT = 0.49) applies to h/b > 2.0 (most UK UB sections). The higher αLT for curve c reflects the greater LTB sensitivity of deep, narrow UB sections compared to stocky UC sections. At λLT,bar = 1.0, curve b gives χLT ≈ 0.68 while curve c gives χLT ≈ 0.56.

How does the UK NA modify LTB design parameters?

The UK NA to BS EN 1993-1-1 adopts the recommended values for LTB: γM1 = 1.00, λLT,0 = 0.4, β = 0.75. The UK NA does not modify the buckling curve selection (Table 6.4) or the Mcr formula. The main UK-specific aspect is the use of UK UB/UC sections rather than European IPE/HEA sections.

When can LTB be ignored for a UK beam?

LTB can be ignored when the compression flange is continuously restrained (e.g., composite slab connected to the top flange with shear studs, or a concrete floor slab bearing directly on the top flange). EN 1993-1-1 Clause 6.3.2.1(2) states that LTB need not be checked if the compression flange has sufficient lateral restraint. LTB can also be ignored when λLT,bar ≤ λLT,0 = 0.4 (very stocky beams).

How do intermediate restraints affect LTB capacity?

Intermediate lateral restraints reduce the unbraced length L, which increases Mcr proportionally to 1/L² and reduces λLT,bar. For a 533UB at 6.0 m, Mcr = 440 kN·m (no restraint). With one midspan restraint (L = 3.0 m): Mcr ≈ 1,760 kN·m (4× higher), λLT,bar reduces to approximately 0.78, and χLT increases to approximately 0.68. This demonstrates the significant benefit of intermediate restraints for deep UB sections.


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Educational reference only. All design values are per BS EN 1993-1-1:2005 + UK National Annex and BS EN 10025-2:2019. Verify all values against the current editions of the standards and the applicable National Annex for your project jurisdiction. Designs must be independently verified by a Chartered Structural Engineer registered with the Institution of Structural Engineers (IStructE) or the Institution of Civil Engineers (ICE). Results are PRELIMINARY — NOT FOR CONSTRUCTION without independent professional verification.