Australian Beam Design — AS 4100 Flexure and LTB Guide
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Comprehensive guide for beam flexural design per AS 4100:2020 Clause 5. Covers section moment capacity (φMs), member moment capacity including lateral-torsional buckling (φMb), shear capacity (φVu), deflection serviceability, and combined actions.
Design Workflow
| Step | Check | Formula | Clause |
|---|---|---|---|
| 1 | Section Classification | λ = b/tf, d1/tw per Table 5.2 | 5.2 |
| 2 | Section Moment Capacity | φMs = φ × fy × Sx (if compact) | 5.2.3 |
| 3 | Member Moment Capacity (LTB) | φMb = φ × αm × αs × Msx | 5.6 |
| 4 | Shear Capacity | φVu = φ × 0.6 × fy × Aw × kv | 5.9 |
| 5 | Deflection Serviceability | δ ≤ L/250 (live), L/125 (total) | 16.4 |
| 6 | Combined Bending + Shear | (M*/φM)² + (V*/φV)² ≤ 1.0 | 5.12 |
| 7 | Web Bearing / Buckling | R* ≤ φRb per Clause 5.13 | 5.13 |
Lateral-Torsional Buckling — Clause 5.6
φMb = φ × αm × αs × Msx ≤ φMsx
Slenderness reduction factor: αs = 0.6 × [√(λs⁴ + 1) − λs²]^(1/2) where λs = √(Msx/Mo)
Elastic buckling moment: Mo = √(π²EIyLe² × (GJ + π²EIw/Le²)) / Le
| Segment Length Le | αm | αs (410UB59.7) | φMb (kNm) |
|---|---|---|---|
| 2.0 m | 1.0 | 0.96 | 506 |
| 4.0 m | 1.13 | 0.85 | 547 |
| 6.0 m | 1.13 | 0.72 | 463 |
| 8.0 m | 1.13 | 0.60 | 386 |
| 10.0 m | 1.13 | 0.50 | 322 |
Note: αs decreases rapidly for longer spans.
Shear Capacity — Clause 5.9
Without tension field action: φVu = φ × 0.60 × fy × Aw × kv
Where kv = 1.0 (unstiffened web), Aw = d × tw
With stiffeners (tension field): Increase for d1/tw > 82/√(fy/250)
| Section | Aw (mm²) | φVu (unstiffened) | φVu (stiffened, 1.0m) |
|---|---|---|---|
| 310UB46.2 | 2,844 | 461 kN | 580 kN |
| 410UB59.7 | 3,176 | 515 kN | 680 kN |
| 610UB125 | 5,513 | 893 kN | 1,250 kN |
| 700UB173 | 7,843 | 1,271 kN | 1,800 kN |
Worked Example
Problem: Design 410UB59.7 beam, 8 m span, UDL w*=25 kN/m (factored). Grade 300. Restrained at supports and midspan.
Solution:
- M* = 25×64/8 = 200 kNm. V* = 25×8/2 = 100 kN
- Section class: flange λ=6.3 < 9 compact, web λ=51 < 80 compact
- φMsx = 0.90 × 280 × 1,142×10³ = 288 kNm > 200 OK
- LTB: Le=4.0m, αm=1.13 (UDL), φMb=0.90×1.13×0.85×288= 249 kNm > 200 OK
- Shear: φVu = 0.90 × 0.60 × 300 × 3,176 = 515 kN > 100 OK
- Deflection (service): δ = 5wL⁴/(384EI) → Check L/250 live load OK
Design Resources
- [[Australian Steel Grades|/reference/australian-steel-grades/]] | [[Australian Steel Properties|/reference/australian-steel-properties/]] | [[Australian Beam Sizes|/reference/au-beam-sizes/]] | [[Australian Bolt Capacity|/reference/australian-bolt-capacity/]] | [[AS 4100 Beam Design|/reference/as4100-beam-design-example/]] | [[All Australian References|/reference/]]
FAQ
When does LTB govern beam design? LTB governs for long spans with low lateral restraint. For 410UB59.7, LTB governs above Le=4 m. For Le=10 m, capacity drops to 322 kNm (vs 288 kNm section).
What is the segment length Le? The distance between lateral restraints to the compression flange. Each unbraced segment must be checked individually with its own αm and αs.
How is shear stiffened web designed? Transverse stiffeners at spacing ≤ d1 increase shear capacity via tension field action per Clause 5.9.4. Unstiffened webs must satisfy d1/tw < 82/√(fy/250).
Educational Use Only — This reference is for educational and preliminary design purposes only. All structural designs must be independently verified by a licensed Professional Engineer (PE) or Structural Engineer (SE) in accordance with AS 4100:2020 and all applicable Australian Standards. Results are not for construction.