Canadian Lateral-Torsional Buckling — LTB per CSA S16-19 Clause 13.6

Complete reference for lateral-torsional buckling of beams per CSA S16-19 Clause 13.6. Covers M_u formulas for the three LTB regimes (plastic, inelastic, elastic), Lp/Lr slenderness limits, omega_2 moment modification factor, and a step-by-step worked example for a W610x125 beam.

Quick access: CSA S16 beam design → | Cb/omega_2 factor → | Compact section limits →

CSA S16 LTB Framework

Per CSA S16-19 Clause 13.6, the factored moment resistance of a laterally unsupported beam segment is:

Mr = phi × Mu where phi = 0.90

The nominal moment resistance Mu depends on the unbraced length Lb relative to the limiting lengths Lp and Lr:

Full plastic (Lb ≤ Lp): Mu = Mp = Zx × Fy

Inelastic LTB (Lp < Lb ≤ Lr): Mu = omega_2 × [Mp - (Mp - My)(Lb - Lp)/(Lr - Lp)] ≤ Mp

Elastic LTB (Lb > Lr): Mu = omega_2 × Mcr ≤ Mp

Where My = Sx × Fy (yield moment) and Mcr is the elastic critical moment.

Lp and Lr Calculations for Doubly Symmetric Sections

Lp (Full Plastic Moment Length)

Per CSA S16 Clause 13.6.1:

Lp = 1.76 × ry × sqrt(E/Fy)

For 350W steel: Lp = 1.76 × ry × sqrt(200000/350) = 1.76 × ry × 23.9 = 42.1 × ry

Lr (Inelastic Limit Length)

For doubly symmetric sections, the elastic LTB limit Lr is determined where Mcr = My:

Lr = 1.76 × ry × sqrt(E/Fy) × sqrt(1 + (Ly/Lz)^2/2) (simplified form)

Where Ly relates to the weak-axis flexural stiffness and Lz to the torsional stiffness. For practical design, Lr values are tabulated in the CISC Handbook of Steel Construction.

Lp and Lr for Common W-Shapes (350W)

Sections with larger ry values have longer Lp and Lr, meaning they maintain full plastic capacity over longer unbraced lengths. Deep sections (W760, W920) have excellent LTB resistance.

Elastic Critical Moment (Mcr)

For a doubly symmetric section in uniform moment (omega_2 = 1.0):

Mcr = (pi/Lb) × sqrt(E × Iy × G × J + (pi × E / Lb)^2 × Iy × Cw)

Where:

• E = 200,000 MPa (elastic modulus)
• G = 77,000 MPa (shear modulus)
• Iy = weak-axis moment of inertia (mm^4)
• J = torsional constant (mm^4)
• Cw = warping constant (mm^6)

For practical design, Mcr is most easily obtained from CISC Handbook tables or the beam capacity calculator.

Mcr for Common Sections at Lb = Lr (350W)

By definition, at L = Lr, Mcr = My, providing continuity between the inelastic and elastic LTB formulas.

Omega_2 (Equivalent Moment Factor)

The omega_2 factor accounts for non-uniform moment diagrams across the unbraced segment, analogous to AISC's Cb factor.

Omega_2 Values for Common Load Cases

A detailed omega_2 table for multiple load cases is available on the Cb Factor reference page.

LTB for Non-Uniform Sections

For tapered beams, castellated beams, and cellular beams, the LTB calculation must account for:

1. Varying cross-section properties: Iy, J, and Cw vary along the length
2. Modified omega_2 factors: Account for both moment gradient and section property variation
3. Reduced Mp at the minimum section: The plastic moment at the most heavily stressed cross-section

For these sections, a finite element LTB analysis or conservative treatment as a uniform section with minimum properties is recommended. The CISC Handbook does not provide pre-computed values for tapered sections.

Worked Example — W610x125 LTB Check

Given: W610×125, Grade 350W. Span = 12.0 m, unbraced segment Lb = 6.0 m (braced at third points). Uniformly distributed dead load = 12 kN/m, live load = 18 kN/m. Two concentrated live loads of 40 kN at third points.

Section Properties (W610×125):

• Zx = 3430 × 10^3 mm^3, Sx = 3050 × 10^3 mm^3
• Iy = 50.8 × 10^6 mm^4, ry = 52.1 mm
• J = 982 × 10^3 mm^4, Cw = 2.12 × 10^12 mm^6

Step 1 — Moment Envelope: Factored load: wf = 1.25 × 12 + 1.5 × 18 = 42 kN/m Factored point loads: Pf = 1.5 × 40 = 60 kN each Mf at midspan = 42 × 12^2 / 8 + 60 × 4 = 756 + 240 = 996 kN·m

Step 2 — Lp and Lr: Lp = 42.1 × 52.1 = 2.19 m Lr = 74.2 × 52.1 = 3.87 m (for this section) Lb = 6.0 m > Lr = 3.87 m → Elastic LTB

Step 3 — Mcr: Mcr = (pi/6000) × sqrt(200000 × 50.8e6 × 77000 × 982e3 + (pi × 200000/6000)^2 × 50.8e6 × 2.12e12) Mcr = 524 × 10^-6 × sqrt(7.68e20 + 3.67e21) = 524 × 10^-6 × 2.11e11 = 110,200 kN·mm = 1102 kN·m

Step 4 — Omega_2: For UDL + two concentrated loads at third points, the moment diagram has three segments of equal length. The centre segment has the highest moment. For the centre segment with M2/M1 ≈ 1.0 (nearly uniform moment in the centre segment due to two symmetric loads): omega_2 ≈ 1.0 (conservative — could use 1.12 for distributed load)

Step 5 — Mu and Mr: Mu = omega_2 × Mcr = 1.0 × 1102 = 1102 kN·m ≤ Mp = 0.90 × 3430 × 10^3 × 350 / 10^6 = 1080 kN·m Mu = 1080 kN·m (governed by Mp) Mr = phi × Mu = 0.90 × 1080 = 972 kN·m

Step 6 — Check: Mf = 996 kN·m ≤ Mr = 972 kN·m — Marginal. Increase section to W610×140 or reduce unbraced length to 4.0 m.

Result: W610×125 is inadequate by 2.5%. Increase to W610×140 or add intermediate bracing.

Design Resources

For efficient design, use the CISC Handbook of Steel Construction which provides pre-computed Mr values for all standard W-shapes at various unbraced lengths. The handbook tables account for:

• All three LTB regimes
• Section class effects
• Omega_2 factors incorporated for the critical segment
• 350W and 350WT steel grades

Frequently Asked Questions

What is the difference between Lp and Lr in CSA S16 LTB design? Lp is the maximum unbraced length for which the beam can reach its full plastic moment Mp. For Lb ≤ Lp, Mu = Mp. Lr is the length at which elastic LTB begins — the transition point where Mcr = My. For Lb between Lp and Lr, inelastic LTB governs with a linear transition from Mp to My. For Lb > Lr, elastic LTB governs and Mu = omega_2 × Mcr ≤ Mp.

How does the omega_2 (Cb) factor affect LTB resistance in CSA S16? Omega_2 increases the LTB resistance for non-uniform moment diagrams. For a simply supported beam with UDL, omega_2 = 1.13 gives 13% higher Mu than uniform moment. For double curvature bending, omega_2 up to 2.50 provides significant benefit. The factor applies in all three regimes but is capped so Mu ≤ Mp.

What is the most effective way to increase LTB capacity of a steel beam? The most effective way is reducing the unbraced length Lb by adding lateral bracing (intermediate purlins, bridging, or cross-frames). Halving Lb can increase Mu by 2-4x depending on the slenderness range. The second most effective approach is increasing the weak-axis radius of gyration ry: deeper sections with wider flanges have larger ry and longer Lp/Lr limits.

How do you calculate Mcr for a doubly symmetric W-shape? Mcr = (pi/Lb) × sqrt(E × Iy × G × J + (pi × E / Lb)^2 × Iy × Cw). For W-shapes, the warping term (containing Cw) typically dominates over the St. Venant torsion term (containing J). The CISC Handbook provides pre-computed Mcr values as part of the beam selection tables for all standard sections and common unbraced lengths.

Related Pages

• CSA S16 Beam Design — Flexural Design Guide
• Cb Factor — Omega_2 Values Table
• Canadian Compact Section Limits
• Canadian Beam Sizes — W-Shape Table
• CSA S16 Column Design
• Beam Capacity Calculator
• All Canadian References

This page is for educational reference. LTB design per CSA S16-19 Clause 13.6. Verify section properties against CISC Handbook. Results are PRELIMINARY — NOT FOR CONSTRUCTION without independent PE/SE verification.