Australian Beam Capacity Calculator — AS 4100

Launch the Beam Capacity Calculator →

Design per AS 4100:2020

This calculator checks steel beam flexural and shear capacity per AS 4100-2020 (Steel Structures Standard) for Australian UB and UC sections. Select "AS 4100" from the code selector at the top of the calculator to use Australian capacity factors and design rules.

Section Moment Capacity — Clause 5.2

For compact sections: φMsx = φ × Zex × fy (φ = 0.9)

The section moment capacity depends on the section's compactness classification per AS 4100 Table 5.1. Compact sections with λ ≤ λs can develop the full plastic moment. If the section is non-compact or slender, Zex is replaced by the effective section modulus Ze.

Member Moment Capacity — Clause 5.6

For laterally unrestrained beams: φMbx = φ × ÃƒÂŽÃ‚±m × ÃƒÂŽÃ‚±s × Zex × fy

• αm = moment modification factor (similar to AISC Cb)
• αs = slenderness reduction factor based on Mo (elastic lateral buckling moment)
• When the beam is fully laterally restrained, φMbx = φMsx

Shear Capacity — Clause 5.10

φVv = φ × 0.6 × fy × Aw (φ = 0.9)

For stiffened or slender webs, the shear buckling capacity Vb per Clause 5.11 may govern. Aw = d × tw for hot-rolled sections.

Worked Example

Problem: Check a 530UB92.4 beam, Grade 300, with full lateral restraint.

Span: 8.0 m | Load: w* = 45 kN/m (factored) | Steel: Grade 300 (fy = 300 MPa)

Solution:

1. Factored moment: M* = 45 × 8²/8 = 360 kNm
2. Section properties: Zex = 2480 × 10³ mm³ (from section table)
3. Section capacity: φMsx = 0.9 × 2480 × 10³ × 300 × 10⁻⁶ = 669.6 kNm
4. Check: 360 ≤ 669.6 → OK (usage: 54%)
5. Shear: V* = 45 × 8 / 2 = 180 kN; φVv = 0.9 × 0.6 × 300 × 9.9 × 528 × 10⁻³ = 847 kN → OK

Result: 530UB92.4 is adequate for both flexure and shear.

Second Worked Example — Laterally Unrestrained Beam

Problem: Check a 460UB67.1 beam, Grade 300, spanning 6.0 m between lateral restraints, supporting a factored UDL of 25 kN/m. No intermediate lateral restraint.

Span: 6.0 m | Unbraced segment: 6.0 m | Grade 300 (fy = 300 MPa, fu = 440 MPa)

Solution:

1. Section properties: Zex = 1300 × 10³ mm³, Iy = 13.7 × 10⁶ mm⁴, J = 368 × 10³ mm⁴, Iw = 457 × 10⁹ mm⁶
2. Factored moment: M* = 25 × 6.0²/8 = 112.5 kNm
3. Section capacity: φMsx = 0.9 × 1300 × 10³ × 300 × 10⁻⁶ = 351 kNm
4. Elastic LTB moment Mo (Clause 5.6.1.2): Mo = √[(π²EIy / L²) × (GJ + π²EIw / L²)] = √[(π² × 200000 × 13.7 × 10⁶ / 6000²) × (80000 × 368 × 10³ + π² × 200000 × 457 × 10⁹ / 6000²)] × 10⁻⁶ = 245 kNm
5. Slenderness reduction factor: αs = 0.6 × [√((Ms/Mo)² + 3) - (Ms/Mo)] per Clause 5.6.1.1 Ms/Mo = 351/245 = 1.43 → αs = 0.6 × [√(1.43² + 3) - 1.43] = 0.6 × [√5.045 - 1.43] = 0.6 × 0.816 = 0.490
6. Moment modification factor: αm = 1.13 (UDL on simply-supported beam, load applied to top flange)
7. Member capacity: φMbx = 0.9 × 1.13 × 0.490 × 1300 × 10³ × 300 × 10⁻⁶ = 194.3 kNm
8. Check: 112.5 ≤ 194.3 → OK (usage: 58%)

Result: 460UB67.1 is adequate for the 6.0 m unbraced span. Member capacity is governed by lateral-torsional buckling at 58% utilisation. For longer unbraced lengths, consider adding fly bracing or a larger section.

AS 4100 Clause Reference Summary

Serviceability and Deflection

Beyond strength checks, AS 4100 Appendix B specifies deflection limits for steel beams:

• Total deflection: Δmax ≤ L/250 (typical for floors supporting non-structural elements)
• Incremental live load deflection: ΔLL ≤ L/500 (for floors with brittle finishes)
• Camber: Pre-camber beams to L/300 to offset dead load deflection
• Vibration: Check natural frequency fn ≥ 3 Hz for office floors per AS 4100 Supplement 1

For the 530UB92.4 worked example: Δmax = 5wL⁴/(384EI) = 5 × 45 × 8000⁴ / (384 × 200000 × 554 × 10⁶) = 21.6 mm < 8000/250 = 32 mm → OK.

How to Use

1. Select "AS 4100" from the code selector
2. Choose an Australian UB or UC section from the database
3. Select Grade 300 or Grade 350 steel
4. Enter the unbraced length (0 for fully restrained)
5. Enter the applied factored moment and shear
6. Review φMs, φMb, and φVv results

Moment Modification Factor alpha_m — AS 4100 Table 5.6.1

The alpha_m factor per AS 4100 Clause 5.6.1.1 adjusts the member moment capacity for non-uniform moment distribution along the unbraced segment. Values for common Australian beam configurations:

Important Australian practice note: The alpha_m values in AS 4100 Table 5.6.1 assume the segment is restrained against lateral deflection at both ends but is free to rotate on plan. For segments where both ends are fully fixed against lateral rotation, an additional 1.2 multiplier may be justified by rational analysis (per Commentary C5.6.1.1).

Slenderness Reduction Factor alpha_s — Detailed Calculation

The alpha_s factor per AS 4100 Clause 5.6.1.1 is determined from:

alpha_s = 0.6 x [sqrt((Ms/Moa)^2 + 3) - (Ms/Moa)]

Where Moa = alpha_m x Mo (Mo is the elastic LTB moment). The ratio Ms/Moa represents the degree of slenderness:

For efficient Australian beam design, target Ms/Moa <= 1.0 (alpha_s >= 0.60). This keeps the LTB reduction manageable. When Ms/Moa > 2.0 (alpha_s < 0.28), the beam is dominated by elastic buckling and adding more steel is inefficient — provide lateral restraint instead.

Web Bearing Check — AS 4100 Clause 5.13

For concentrated loads at supports (the most common case), the bearing capacity phi_Rb depends on the stiff bearing length b_s:

phi_Rb = phi x 1.25 x b_s x tw x fy (end bearing, yield line pattern)

For a 530UB92.4 beam (tw = 9.9 mm, fy = 300 MPa) supported on a 100 mm bearing plate: phi_Rb = 0.9 x 1.25 x 100 x 9.9 x 300 / 1000 = 334 kN > V* = 180 kN. OK.

If the bearing length is insufficient, provide a bearing stiffener (full-depth plate welded to the web and flanges) per Clause 5.14. Bearing stiffeners are also required at interior point loads exceeding the web bearing capacity.

Australian UB vs UC vs PFC Sections — Selection Guide

For beams with moderate axial load (N*/phiNs > 0.2), always check combined actions per AS 4100 Clause 8.3. A UC section may be lighter than a UB section when the axial component governs the weak-axis interaction.

Related Australian Resources

• Australian Beam Sizes (UB/UC/PFC)
• Australian Steel Grades
• AS 4100 Beam Design Example
• Australian Column Capacity Calculator
• All Australian References

FAQ

What is the difference between φMs and φMb? φMs is the section moment capacity assuming full lateral restraint (no LTB). φMb is the member moment capacity accounting for lateral-torsional buckling. For beams with continuous lateral restraint, φMb = φMs. For unrestrained beams, φMb ≤ φMs per AS 4100 Clause 5.6.

Can I check laterally unrestrained beams? Yes. Enter the segment length between points of lateral restraint, and the calculator computes φMb using the elastic lateral buckling moment Mo per AS 4100 Clause 5.6.1.2.

What steel grades are available for AS 4100? Grade 300 (fy = 300 MPa, fu = 440 MPa) for standard sections and Grade 350 (fy = 350 MPa, fu = 480 MPa) for higher-strength applications per AS/NZS 3679.1.

What is the αm moment modification factor? The moment modification factor αm per AS 4100 Clause 5.6.1.1 accounts for the shape of the bending moment diagram between lateral restraints. For a uniform moment (double curvature), αm = 1.0. For a simply-supported beam with UDL, αm = 1.13. For a cantilever with tip load, αm = 2.25. Higher αm values reduce the effective slenderness and increase φMb. Clause 5.6.1.1 Table provides αm for common moment patterns.

How do I check deflection and serviceability? Per AS 4100 Appendix B, compute the deflection under service loads (not factored). Δ = 5wL⁴/(384EI) for UDL on a simply-supported beam. Check against span/250 for total deflection and span/500 for incremental live load. If deflection governs, consider pre-cambering the beam by L/300 or selecting a deeper section (which increases I faster than it adds weight).

What about web bearing and stiffeners? AS 4100 Clause 5.13 checks concentrated load transfer at supports and point load locations. The bearing capacity φRb depends on the stiff bearing length bs and whether the load is applied at the beam end (yield line pattern) or interior. If φRb is insufficient, add web stiffener plates per Clause 5.14.

Disclaimer: This content is for educational purposes only. Results must be verified by a licensed professional engineer. Steel Calculator provides preliminary design tools — NOT a substitute for professional engineering judgment.