Lateral-Torsional Buckling (LTB) — AISC 360 Chapter F Reference

Lateral-torsional buckling is the governing failure mode for most unbraced steel beams. When a beam is loaded in bending, the compression flange acts like a column and can buckle laterally while the beam simultaneously twists. The unbraced length Lb -- the distance between points of lateral support for the compression flange -- determines whether the beam reaches its full plastic moment or is limited by LTB.

Three zones of flexural behavior

AISC 360-22 Chapter F divides flexural capacity into three zones based on Lb relative to Lp and Lr:

Zone 1 — Plastic (Lb <= Lp): Full plastic moment available. Mn = Mp = FyZx. Lp = 1.76rysqrt(E/Fy). For Fy = 50 ksi: Lp = 42.4ry (inches).

Zone 2 — Inelastic LTB (Lp < Lb <= Lr): Capacity decreases linearly from Mp to 0.7FySx. Mn = Cb*[Mp - (Mp - 0.7FySx)*(Lb - Lp)/(Lr - Lp)], capped at Mp.

Zone 3 — Elastic LTB (Lb > Lr): Capacity governed by elastic critical moment. Mn = FcrSx, capped at Mp. Fcr = (Cbpi^2E/(Lb/rts)^2)sqrt(1 + 0.078(Jc/(Sxho))(Lb/rts)^2).

Zone boundary equations

Lp = 1.76 * ry * sqrt(E / Fy)                        [Eq. F2-5]
Lr = 1.95 * rts * sqrt(E / (0.7*Fy)) * sqrt((J*c)/(Sx*ho) + sqrt((J*c/(Sx*ho))^2 + 6.76*(0.7*Fy/E)^2))   [Eq. F2-6]

Where rts = sqrt(sqrt(Iy*Cw)/Sx), c = 1.0 for doubly symmetric I-shapes, ho = distance between flange centroids.

The Cb factor — moment gradient modifier

Cb accounts for non-uniform moment distribution. Uniform moment (Cb = 1.0) is the worst case.

Cb = 12.5*Mmax / (2.5*Mmax + 3*MA + 4*MB + 3*MC)

All values are absolute moments at quarter, mid, and three-quarter points of the unbraced segment.

Computing Cb from moment values

Given a simply supported beam with uniform load, Lb = 20 ft, wu = 2.5 kip/ft:

• MA (at L/4 = 5 ft): M = 2.5*5*(20-5)/2 = 93.8 kip-ft
• Mmax (at L/2 = 10 ft): M = 2.5*10*(20-10)/2 = 125 kip-ft
• MB (at L/2): same as Mmax = 125 kip-ft
• MC (at 3L/4 = 15 ft): same as MA = 93.8 kip-ft

Cb = 12.5*125 / (2.5*125 + 3*93.8 + 4*125 + 3*93.8) = 1562.5 / (312.5 + 281.4 + 500 + 281.4) = 1562.5 / 1375.3 = 1.14 (matches table).

Lp and Lr values for common W-shapes (Fy = 50 ksi)

Lp is typically 4-7 feet for common sizes. Metal deck at 2-3 ft spacing prevents LTB entirely.

Lp and Lr for different Fy values (W18x50)

Higher Fy reduces Lp and Lr, meaning the beam transitions to LTB-governed behavior at shorter unbraced lengths. This is a key tradeoff: higher strength steel is more susceptible to LTB.

phiMn vs Lb — W18x50 (Fy = 50 ksi)

At Lb = 20 ft with Cb = 1.14, the W18x50 retains 40% of its full plastic capacity.

Worked example — W18x50, Lb = 12 ft

Given: W18x50, A992, uniform load, Lb = 12 ft. Zx = 101 in^3, Sx = 88.9 in^3, ry = 1.65 in.

Step 1: Lp = 5.83 ft, Lr = 16.4 ft. Since 5.83 < 12 < 16.4, inelastic LTB (Zone 2).

Step 2: Cb = 1.14 (simply supported, uniform load).

Step 3: Mp = 50*101/12 = 420.8 kip-ft. 0.7FySx = 0.7*50*88.9/12 = 259.3 kip-ft.

Mn = 1.14*[420.8 - (420.8-259.3)*(12-5.83)/(16.4-5.83)] = 1.14*[420.8-94.2] = 1.14*326.6 = 372.3 kip-ft. Check: 372.3 < 420.8 OK.

phiMn = 0.90*372.3 = 335 kip-ft (vs full phiMp = 379 kip-ft, a 12% reduction from the 12-ft unbraced length).

Worked example — W24x76, Lb = 8 ft, Cb = 1.32

Given: W24x76, A992, midpoint load, Lb = 8 ft. Zx = 176 in^3, Sx = 155 in^3, ry = 1.92 in.

Step 1: Lp = 6.78 ft, Lr = 18.8 ft. Since 6.78 < 8 < 18.8, Zone 2.

Step 2: Cb = 1.32 (simply supported, midpoint load).

Step 3: Mp = 50*176/12 = 733 kip-ft. 0.7FySx = 0.7*50*155/12 = 452 kip-ft.

Mn = 1.32*[733 - (733-452)*(8-6.78)/(18.8-6.78)] = 1.32*[733 - 26.2] = 1.32*706.8 = 933 kip-ft. Capped at Mp = 733 kip-ft.

phiMn = 0.90*733 = 660 kip-ft. The Cb = 1.32 factor restored full plastic capacity even with Lb = 8 ft > Lp = 6.78 ft.

Bracing requirements

Effective lateral bracing must have both strength and stiffness:

• Strength: Brace must resist 0.8*Mmax/Cd (AISC Appendix 6), typically 2-4% of the compression flange force.
• Stiffness: Brace stiffness must be at least beta_brace = 2*Pbracing*N/(L*theta_0), where theta_0 is the initial imperfection.

Types of lateral bracing

Multi-code comparison

AS 4100-2020: Uses phi*Mb = phi*alpha_m*alpha_s*Ze*fy, where alpha_m = moment modification factor (like Cb) and alpha_s = slenderness reduction factor based on reference buckling moment Mo. The Mo calculation uses the full elastic critical moment including warping: Mo = sqrt((pi^2*E*Iy/Lb^2)*(G*J + pi^2*E*Cw/Lb^2)).

EN 1993-1-1: Uses chi_LT reduction factor: Mb,Rd = chi_LT*Wy*fy/gamma_M1, determined from non-dimensional slenderness lambda_LT and buckling curves (a-d). The method accounts for section type through the imperfection factor alpha_LT.

CSA S16-19: Uses Mr = phi*Mu with inelastic transition similar to AISC. Mu = (Mp*Mu2) / (Mp + (Mu2 - Mp)*(L/Lp)^(1/3)) for the inelastic range, which differs from AISC's linear interpolation.

LTB capacity comparison: W18x50, Fy = 50 ksi, Lb = 12 ft, Cb = 1.14

Common mistakes

1. Confusing bracing point with support point. A brace point is any location where the compression flange is restrained laterally, including intermediate braces, not just supports.

2. Ignoring Cb when it helps. Using Cb = 1.0 for all cases can be 14-30% overly conservative for typical loading. Always compute Cb for the actual moment diagram.

3. Using Cb > 1.0 for cantilevers. The AISC Cb equation does not capture cantilever LTB behavior. Use Cb = 1.0 for cantilevers unless specifically justified.

4. Bracing the tension flange instead of the compression flange. LTB involves the compression flange. Bracing the tension flange does not prevent LTB unless it provides torsional restraint.

5. Not checking negative moment regions. Bottom flange is in compression at supports of continuous beams. Ensure bottom flange is braced in these regions.

6. Assuming metal deck always braces. Metal deck provides bracing only if it is through-fastened or welded to the beam. Standing seam roof deck may not provide adequate bracing.

7. Using ry from the wrong section. For non-symmetric sections (channels, angles), the principal axis may not align with the geometric axis.

Frequently asked questions

What is lateral-torsional buckling? LTB is a failure mode where a beam's compression flange displaces laterally and the cross-section twists simultaneously. It is analogous to column buckling but involves both lateral translation and rotation.

How do I prevent LTB? Provide lateral bracing to the compression flange at intervals no greater than Lp. Metal deck or concrete slab with shear studs provides continuous bracing to the top flange.

What is the unbraced length? Lb is the distance between points where the compression flange is restrained against lateral displacement. A 30-ft beam with bracing at third points has Lb = 10 ft.

Does LTB apply to HSS sections? HSS (round, square, and rectangular) are much less susceptible to LTB than I-shapes because they have high torsional stiffness (J). Round HSS and square HSS do not experience LTB. Rectangular HSS may require LTB checks for high aspect ratios (b/t > 4) per AISC Table F4-1.

When does Cb restore full plastic capacity? When Cb * Mn_Zone2 >= Mp. This occurs when the moment gradient is steep enough that the Cb factor compensates for the LTB reduction. For common loading, this happens when Lb is only slightly above Lp.

What about continuous beams? Each unbraced segment of a continuous beam must be checked independently with its own Cb factor. Negative moment regions (at supports) have the bottom flange in compression and need separate bracing consideration.

How do I check a beam with non-uniform moment? Compute the moments at A (quarter point), B (midpoint), C (three-quarter point), and Mmax, then use the Cb equation. For beams with point loads, use the actual moment diagram rather than tabulated Cb values.

AISC F2 LTB Detailed Procedure

AISC 360-22 Chapter F, Section F2 provides the step-by-step procedure for determining the flexural strength of doubly symmetric I-shaped members loaded in the plane of their web. The procedure depends on the unbraced length Lb relative to the limiting laterally unbraced lengths Lp and Lr.

Step 1: Determine Lp (plastic limit)
  Lp = 1.76 × ry × sqrt(E / Fy)    (AISC Eq. F2-5)

Step 2: Determine Lr (inelastic limit)
  Lr = 1.95 × rts × (E / (0.7 × Fy)) × sqrt(J × c / (Sx × ho)) × sqrt(1 + sqrt(1 + 6.76 × ((0.7 × Fy × Sx × ho) / (E × J × c))²))    (AISC Eq. F2-6)

where:
  rts = sqrt(sqrt(Iy × Cw) / Sx)
  c = 1.0 (doubly symmetric), or ho/2 × sqrt(Iy/Cw) (for monosymmetric)
  ho = distance between flange centroids
  J = torsional constant
  Sx = elastic section modulus

Step 3: Compare Lb to Lp and Lr
  If Lb ≤ Lp:  Mn = Mp = Fy × Zx (plastic moment, no LTB)
  If Lp < Lb ≤ Lr:  Mn = Cb × (Mp - (Mp - 0.7 × Fy × Sx) × ((Lb - Lp) / (Lr - Lp))) ≤ Mp
  If Lb > Lr:  Mn = Cb × (π² × E / (Lb/rts)²) × sqrt(1 + 0.078 × J × c / (Sx × ho) × (Lb/rts)²)  ≤ Mp

Lp and Lr Calculation Example

Consider a W24x62 beam with Fy = 50 ksi, E = 29,000 ksi.

From AISC Manual Table 1-1 for W24x62:
  Zx = 153 in³,  Sx = 131 in³
  ry = 1.26 in,   Iy = 18.9 in⁴
  J = 1.17 in⁴,   Cw = 4610 in⁶
  ho = 23.0 in,   rts = 1.47 in

Step 1: Lp = 1.76 × 1.26 × sqrt(29000/50)
           = 1.76 × 1.26 × 24.08
           = 53.4 in = 4.45 ft

Step 2: Lr calculation (using AISC Eq. F2-6):
  Lr = 1.95 × 1.47 × (29000/35) × sqrt(1.17 / (131 × 23))
       × sqrt(1 + sqrt(1 + 6.76 × ((35 × 131 × 23) / (29000 × 1.17))²))
  Lr ≈ 12.8 ft

Step 3: Results
  For Lb ≤ 4.45 ft:  φMn = 0.90 × 50 × 153 = 6885 kip-in = 574 kip-ft
  For Lb = 8.0 ft (between Lp and Lr):
    Mn = Cb × [Mp - (Mp - 0.7FySx)(Lb-Lp)/(Lr-Lp)]
       = Cb × [574 - (574 - 0.7×50×131/12) × (8-4.45)/(12.8-4.45)]
       ≈ Cb × 456 kip-ft  (for Cb = 1.0)

Cb Factor with Worked Example

The moment gradient coefficient Cb accounts for non-uniform moment diagrams. A higher Cb value means the beam is less susceptible to LTB because the maximum compressive stress occurs at only one point along the unbraced length.

Cb = 12.5 × Mmax / (2.5 × Mmax + 3 × MA + 4 × MB + 3 × MC)    (AISC Eq. F1-1)

where MA, MB, MC are moments at quarter-point, midpoint, and three-quarter point
of the unbraced segment, and Mmax is the maximum moment in the segment.

Worked example: A simply supported W18x35 beam spans 30 ft with a concentrated load at midspan. The beam is braced at the third points (Lb = 10 ft).

Moment diagram (symmetric, triangular):
  At left brace (10 ft):   M = 10/30 × P/2 × 30 × (1 - 10/30) = P × 6.67 ft-kips
  At quarter-point (7.5 ft): MA = P × 5.0 ft-kips
  At midpoint (15 ft):       MB = Mmax = P × 7.5 ft-kips
  At three-quarter (22.5 ft): MC = P × 5.0 ft-kips

Cb = 12.5 × 7.5P / (2.5 × 7.5P + 3 × 5.0P + 4 × 7.5P + 3 × 5.0P)
   = 12.5 × 7.5 / (18.75 + 15 + 30 + 15)
   = 93.75 / 78.75 = 1.19

This increases the LTB moment capacity by 19% compared to the Cb = 1.0
uniform moment (worst case) assumed in AISC Tables.

Lateral Bracing Types and Spacing

Effective lateral bracing must prevent both lateral translation and twist of the compression flange. AISC 360-22 Appendix 6 provides guidance on bracing strength and stiffness.

General rule: For typical building beams, bracing at the compression flange at intervals not exceeding Lp ensures the full plastic moment Mp is achieved. When Lb exceeds Lp, the moment capacity reduces per the AISC linear or elastic LTB equations.

LTB Resistance by W-Shape Depth

The table below shows the unbraced length limits Lp and Lr and the corresponding design moment strength for common W-shapes, assuming A992 Grade 50 steel (Fy = 50 ksi) and Cb = 1.0.

Note: φMr values shown are at Lb = Lr with Cb = 1.0. Values are approximate — verify with AISC Manual tables.

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Related references

• Cb Factor
• Compact Section Limits
• Beam Sizes
• Beam Formulas
• Deflection Limits
• Beam Design Guide
• Column Buckling Equations
• Stress-Strain Curve
• Steel Grades
• How to Verify Calculations

Disclaimer

This page is for educational and reference use only. It does not constitute professional engineering advice. All design values must be verified against AISC 360-22 Chapter F and the governing project specification. The site operator disclaims liability for any loss arising from the use of this information.

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