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:

• χLT = reduction factor for LTB (≤ 1.0)
• Wy = Wpl,y for Class 1 or 2 sections; Wel,y for Class 3; Weff,y for Class 4
• γM1 = 1.00 (UK NA value)

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)]

LTB Buckling Curves (Table 6.4, UK NA)

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

• UK Steel Grades Reference — EN 10025-2 grade selection for UK projects
• UK Steel Mechanical Properties — fy, fu, and elongation tables
• UK Universal Beam and Column Sizes — UB/UC section dimensions and properties
• UK Bolt Capacity Tables — Class 8.8 and 10.9 bolt resistance
• UK Beam Design Guide — EN 1993-1-1 flexure, shear, and LTB
• UK Connection Design Guide — EN 1993-1-8 bolted and welded joints
• All UK Steel Design References — complete library

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.

Related Pages

• EN 1993 Steel Design Overview
• European Steel Properties
• EN 1993 Beam Design Guide
• EN 1993 Column Buckling
• EN 1990 Load Combinations
• UK Steel Chemical Composition
• UK Steel Charpy Values

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.