AS 4100 Beam Design Worked Example — 310UB40.4

Complete AS 4100:2020 beam design walkthrough: section moment capacity (Ms), shear capacity (Vv), lateral torsional buckling (Mb), web bearing, and deflection. The worked example designs a 310UB40.4 Grade 300 steel beam spanning 7.2m at 3.6m tributary spacing. All code references are to AS 4100:2020.

This page covers the full beam design workflow under the Australian limit state standard. The Steel Calculator WASM engine performs these checks automatically for AS 4100, AISC 360, EN 1993, CSA S16, and IS 800.

PRELIMINARY — NOT FOR CONSTRUCTION. All results are for educational and reference use only and must be independently verified by a Chartered Professional Engineer registered with Engineers Australia before use in any project.

Design Problem Definition

A floor beam in a Melbourne office building spans 7.2m between columns with a 3.6m tributary width. The beam is simply supported with lateral restraint provided by the concrete slab at 1.8m centres (purlin spacing). The slab is composite with shear studs, but for this example we treat the beam as a bare steel member carrying construction-stage dead load plus the imposed live load from the composite slab.

Design Data:

Line Loads:

Design Actions:

Section Properties — 310UB40.4 Grade 300

Australian universal beams are manufactured to AS/NZS 3679.1 in Grade 300PLUS by InfraBuild. The 310UB40.4 is a medium-weight beam suitable for floor framing in multi-storey construction.

Geometric Properties

Property Symbol Value Units
Depth d 304 mm
Flange width b_f 165 mm
Flange thickness t_f 10.2 mm
Web thickness t_w 6.1 mm
Root radius r 11.4 mm
Depth between flanges d_1 283.6 mm
Area A_g 5,150 mm^2
Second moment of area (major) I_x 85.0 x 10^6 mm^4
Elastic section modulus Z_x 559 x 10^3 mm^3
Plastic section modulus S_x 640 x 10^3 mm^3
Torsion constant J 160 x 10^3 mm^4
Warping constant I_w 79.7 x 10^9 mm^6
Radius of gyration (minor) r_y 33.2 mm
Mass per metre — 40.4 kg/m

Step 1 — Section Classification (AS 4100 Clause 5.2)

Section classification determines which moment capacity formula applies and whether the section can develop its plastic moment. AS 4100 Table 5.2 gives the limiting width-to-thickness ratios for Class 1 (Plastic), Class 2 (Compact), Class 3 (Non-compact), and Class 4 (Slender).

Flange Slenderness

The outstand flange slenderness parameter:

lambda_e = (b_f - t_w) / (2 x t_f) x sqrt(f_y / 250)

lambda_e = (165 - 6.1) / (2 x 10.2) x sqrt(300 / 250) = 7.79 x 1.095 = 8.53

Class 1 limit: lambda_ep = 9 —> 8.53 < 9, flange is Class 1 (Plastic).

Class 2 limit: lambda_ey = 16 —> 8.53 < 16, confirmed.

Web Slenderness

For a web in pure bending:

lambda_e = (d_1 / t_w) x sqrt(f_y / 250)

lambda_e = (283.6 / 6.1) x 1.095 = 46.49 x 1.095 = 50.91

Class 1 limit: lambda_ep = 45 —> 50.91 > 45, web is NOT Class 1.

Class 2 limit: lambda_ey = 82 —> 50.91 < 82, web is Class 2 (Compact).

Class 3 limit: lambda_ey = 115 —> 50.91 < 115, confirmed.

Final Classification

Flange is Class 1, web is Class 2. The section is Class 2 (Compact): the plastic moment can be reached but with limited rotation capacity due to the web. For Class 2 sections:

Z_e = min(S, 1.5 x Z) = min(640, 1.5 x 559) = min(640, 838.5) = 640 x 10^3 mm^3

The effective section modulus equals the plastic section modulus for strength design.

Step 2 — Section Moment Capacity Ms (AS 4100 Clause 5.2)

For a Class 2 section with full lateral restraint:

M_s = Z_e x f_y = 640 x 10^3 x 300 = 192.0 x 10^6 N.mm = 192.0 kN.m

Design Moment Capacity

phi = 0.90 (AS 4100 Table 3.4, bending)

phi x M_s = 0.90 x 192.0 = 172.8 kN.m

Flexural Utilisation (full restraint assumption)

M* / (phi x M_s) = 189.0 / 172.8 = 1.094 — this slightly exceeds unity at full moment.

This tells us the 310UB40.4 is borderline under the maximum design moment if we assume full lateral restraint. However, the actual design moment reduces near the supports — a more refined approach or adding intermediate restraints will be needed. We will check lateral torsional buckling next.

Step 3 — Shear Capacity Vv (AS 4100 Clause 5.11)

The nominal shear capacity depends on the web slenderness. For unstiffened webs, the shear buckling check is:

d_p / t_w = (d - 2 x t_f) / t_w = (304 - 20.4) / 6.1 = 283.6 / 6.1 = 46.49

The shear yield limit per AS 4100 Clause 5.11.2:

Limit = 82 / sqrt(f_y / 250) = 82 / sqrt(300 / 250) = 82 / 1.095 = 74.84

Since 46.49 < 74.84, the web is stocky enough that shear yielding (not buckling) governs.

Nominal Shear Capacity

For a web where shear yielding controls:

V_w = 0.6 x f_y x A_w

where A_w = d x t_w = 304 x 6.1 = 1,854 mm^2

V_w = 0.6 x 300 x 1,854 = 333,720 N = 333.7 kN

Design Shear Capacity

phi = 0.90 (shear)

phi x V_v = 0.90 x 333.7 = 300.4 kN

Shear Utilisation

V* / (phi x V_v) = 105.1 / 300.4 = 0.350 — shear is not governing.

Since V* < 0.6 x (phi x V_v) = 180.2 kN, no shear-bending interaction check is required per Clause 5.12.

Step 4 — Lateral Torsional Buckling Mb (AS 4100 Clause 5.6)

When the compression flange is not continuously laterally restrained, the member moment capacity Mb accounts for lateral-torsional buckling (LTB). The lateral restraint spacing is 1.8m (composite deck purlins at 1.8m centres).

Step 4a — Modified Slenderness lambda_n

lambda_n = (L_e / r_y) x sqrt(f_y / 250)

Using the reduced effective length approach: L_e = k_t x k_l x k_r x L_s

For a simply supported beam with load applied to the top flange and segment with lateral restraint at both ends:

L_e = 1.0 x 1.0 x 1.0 x 1,800 = 1,800 mm

lambda_n = (1,800 / 33.2) x sqrt(300 / 250) = 54.22 x 1.095 = 59.4

Step 4b — Slenderness Reduction Factor alpha_s

AS 4100 Table 5.6.1 gives alpha_s as a function of lambda_n. For lambda_n = 59.4:

For a rolled I-section with k_f = 1.0, the slenderness reduction factor alpha_s is obtained from AS 4100 Table 5.6.1(1), which provides tabulated values as a function of the modified slenderness lambda_n and the section type (rolled or welded).

Interpolating from Table 5.6.1(1) for lambda_n = 59.4:

alpha_s = 0.835 (rolled I-section, compact section table).

Step 4c — Moment Modification Factor alpha_m

alpha_m accounts for the shape of the bending moment diagram between lateral restraints. For a simply supported beam under uniformly distributed load, from AS 4100 Table 5.6.1:

For a UDL on a simply supported span with end restraints: alpha_m = 1.13 (bending moment gradient factor)

Step 4d — Member Moment Capacity Mb

M_b = alpha_m x alpha_s x M_s <= M_s

M_b = 1.13 x 0.835 x 192.0 = 181.2 kN.m

Design LTB Capacity

phi x M_b = 0.90 x 181.2 = 163.1 kN.m

LTB Utilisation

M* / (phi x M_b) = 189.0 / 163.1 = 1.159 — LTB would exceed capacity.

This result shows that at 1.8m restraint spacing, the 310UB40.4 is marginally below the design requirement for the full span moment. Two options:

  1. Reduce lateral restraint spacing — adding a bridging member or purlin at mid-span of the 1.8m segment would reduce L_e and increase alpha_s.
  2. Increase section — a 310UB46.2 would provide additional capacity with minimal weight increase.

If we reduce the effective segment length to L_s = 0.9 m (adding an intermediate brace):

lambda_n = (900 / 33.2) x 1.095 = 29.7 alpha_s (from Table 5.6.1) approximately 0.956 M_b = 1.13 x 0.956 x 192.0 = 207.5 kN.m phi x M_b = 0.90 x 207.5 = 186.8 kN.m M* = 189.0 / 186.8 = 1.012 — still borderline.

Recommendation: Upgrade to 310UB46.2 for this application, or verify that the slab provides continuous torsional restraint (which would eliminate LTB entirely).

Step 5 — Deflection Check (AS 4100 Appendix B)

Serviceability limits per AS 1170.0 and AS 4100 Appendix B Table B1 for floor beams:

Total Deflection (G + 0.7 x Q at SLS)

delta_total = 5 x w_sls x L^4 / (384 x E x I_x)

delta_total = 5 x 18.4 x (7,200)^4 / (384 x 200,000 x 85.0 x 10^6)

= 5 x 18.4 x 2.688 x 10^15 / (384 x 200,000 x 85.0 x 10^6)

= 2.473 x 10^17 / 6.528 x 10^15

= 37.9 mm

This exceeds the L/250 limit of 28.8 mm. Total deflection governs — the 310UB40.4 is inadequate for serviceability at this span.

Live Load Deflection

delta_live = 5 x (0.7 x Q_lineload) x L^4 / (384 x E x I_x)

d_live = 5 x (0.7 x 10.8) x 2.688 x 10^15 / 6.528 x 10^15

= 5 x 7.56 x 2.688 x 10^15 / 6.528 x 10^15

= 15.6 mm

This is within the L/360 = 20.0 mm limit for live load deflection alone. The problem is the total deflection including dead load.

Step 6 — Web Bearing at Supports (AS 4100 Clause 5.13)

Web bearing capacity at the support must be checked against the concentrated support reaction. Assume a 150 mm bearing length at the support.

Design Bearing Capacity

phi x R_by = phi x 1.25 x b_bf x t_w x f_y (when the bearing length extends at least 1:1 from the end)

Where b_bf = bearing length + 2.5 x (t_f + r) = 150 + 2.5 x (10.2 + 11.4) = 150 + 54 = 204 mm

phi x R_by = 0.90 x 1.25 x 204 x 6.1 x 300 = 0.90 x 466,650 = 420.0 kN

R* = 105.1 kN < 420.0 kN — bearing capacity at the support is adequate by a large margin.

Web Bearing Buckling

phi x R_bb = phi x alpha_c x A_b x f_y

The effective compression member for web bearing buckling is a strip of web with an effective length:

L_e_b = 2.5 x d_1 = 2.5 x 283.6 = 709 mm

A_b = t_w x b_bf_eff = 6.1 x 204 = 1,244 mm^2

r_w = t_w / sqrt(12) = 6.1 / 3.464 = 1.76 mm

lambda_n_b = (L_e_b / r_w) x sqrt(f_y / 250) = (709 / 1.76) x 1.095 = 403 x 1.095 = 441

This is extremely slender — web buckling at the support would govern unless bearing stiffeners are provided. For a standard office beam with light support reactions in a composite slab, web stiffeners are typically not required if the reaction R* is below about 50 kN. However, our reaction of 105.1 kN exceeds this, so:

Provide 2 x 80 x 10 mm bearing stiffeners at each support.

Summary of Design Checks — 310UB40.4

Limit State Clause Capacity (phi-R) Design Action D/C Ratio Status
Section moment Cl. 5.2 172.8 kN.m 189.0 kN.m 1.094 FAIL
Lateral-torsional buckle Cl. 5.6 163.1 kN.m 189.0 kN.m 1.159 FAIL
Shear (web yielding) Cl. 5.11 300.4 kN 105.1 kN 0.350 PASS
Web bearing Cl. 5.13 420.0 kN 105.1 kN 0.250 PASS
Total deflection App. B 28.8 mm 37.9 mm 1.316 FAIL
Live load deflection App. B 20.0 mm 15.6 mm 0.780 PASS

The 310UB40.4 fails section moment capacity marginally and total deflection significantly at a 7.2 m span. For production design, the 310UB46.2 (I_x = 101 x 10^6 mm^4, Z_e = 790 x 10^3 mm^3) would be the minimum recommended section.

AS 4100 vs International Beam Design — Comparison

Design Aspect AS 4100:2020 AISC 360-22 EN 1993-1-1
Moment capacity Ms = Ze x fy (Cl. 5.2) Mn = Mp or Fy x Zx (Ch F) Mc,Rd = Wpl x fy / gamma_M0
Capacity factor phi = 0.90 phi_b = 0.90 gamma_M0 = 1.00
LTB formulation alpha_m x alpha_s x Ms Cb x Mp (Ch F) chi_LT x My,Rd
Section classification 4 classes (Cl. 5.2) Compact/Non/Slender Class 1-4
Shear-bending interaction 0.6 x Vv threshold (Cl. 5.12) 0.6 x Vn threshold 0.5 x Vpl,Rd (Cl. 6.2.8)
Deflection limits (floor) L/250 total, L/360 live L/240 total, L/360 live L/200 total (UK NA)
Web bearing method Cl. 5.13 (yield + buckle) J10 (yield + crippling) Cl. 6.2.6 (shear only)

AS 4100 uses the same multi-curve approach as EN 1993 for LTB but with different notation. The alpha_s reduction factor in AS 4100 is structurally analogous to chi_LT in EN 1993, but the buckling curve imperfection parameters differ to reflect Australian fabrication tolerances.

Frequently Asked Questions

How does AS 4100 calculate the section moment capacity Ms for a steel beam?

AS 4100 Clause 5.2 calculates the nominal section moment capacity Ms = Ze x fy for sections with full lateral restraint. The effective section modulus Ze is the lesser of S (plastic modulus) or 1.5 x Z (elastic modulus) for compact sections. For a 310UB40.4 in Grade 300, Ms = 640x10^3 x 300 = 192 kN.m, with the capacity factor phi-Ms = 0.90 giving a design capacity of 173 kN.m.

When does lateral torsional buckling control AS 4100 beam design?

Lateral torsional buckling per AS 4100 Clause 5.6 controls when the beam's compression flange lacks continuous lateral restraint. The member moment capacity Mb = alpha-m x alpha-s x Ms, where alpha-s is the slenderness reduction factor. For a 310UB40.4 with 3.0m segment length, the modified slenderness lambda-n = 127 produces alpha-s = 0.461, reducing the design moment capacity from 173 kN.m to about 80 kN.m. Full lateral restraint (purlin at 1.5m spacing) can reduce lambda-n to 64 and increase alpha-s to 0.812.

How does AS 4100 handle shear and bending interaction in beam design?

AS 4100 Clause 5.12 requires a shear-bending interaction check when the design shear force V* exceeds 0.6 x phi-Vv. The interaction formula is M* <= phi-Mrv where Mrv is the reduced moment capacity accounting for the shear stress in the web. For a simply supported 310UB40.4 under uniform load, the maximum moment and maximum shear occur at different locations (mid-span and support), so interaction rarely governs. For a continuous beam over an internal support where both are high simultaneously, the interaction check becomes critical.

What is the alpha-m moment modification factor in AS 4100?

The alpha-m factor in AS 4100 Clause 5.6.1.1 accounts for the bending moment distribution between lateral restraints. For a simply supported beam under uniformly distributed load, alpha-m = 1.13. For a central point load on a simply supported span, alpha-m = 1.35. For a beam segment under a uniform moment (no gradient), alpha-m = 1.0. The factor amplifies the reference buckling moment and effectively increases the lateral torsional buckling capacity for beams with moment gradients.

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Related Pages

This page is for educational reference. All resistance formulae are per AS 4100:2020 with AS/NZS 3679.1 section properties. Verify the applicable edition of the National Construction Code for your project jurisdiction. Results are PRELIMINARY — NOT FOR CONSTRUCTION without independent review by a registered structural engineer (CPEng/RPEQ).