Problem Statement

A floor beam spans 10.0 m simply supported in a commercial office building. The beam supports a composite slab (Bondek II, 150 mm total thickness) that provides continuous lateral restraint to the top flange at 3.0 m centres (the secondary beam spacing). The construction stage (bare steel only) must also be checked.

Loads (AS 1170.1):

• Dead load: 5.0 kPa (slab, Bondek, services, ceiling, partitions allowance 1.0 kPa per AS 1170.1 Table 3.1)
• Live load: 3.0 kPa (office general, AS 1170.1 Table 3.1)
• Beam spacing: 3.0 m centres

Service loads: w_G = 5.0 x 3.0 = 15.0 kN/m (dead) w_Q = 3.0 x 3.0 = 9.0 kN/m (live)

Materials: AS/NZS 3679.1 Grade 300 (fy = 300 MPa for t <= 11 mm, fy = 280 MPa for 11 < t <= 17 mm; fu = 440 MPa). E = 200,000 MPa, G = 80,000 MPa.

Step 1 — Factored Load Combinations (AS 1170.0 Clause 4.2.2)

For strength limit state:

• 1.35G = 1.35 x 15.0 = 20.25 kN/m
• 1.2G + 1.5Q = 1.2 x 15.0 + 1.5 x 9.0 = 18.0 + 13.5 = 31.5 kN/m (governs)

Factored design load: w* = 31.5 kN/m

Step 2 — Design Actions

M* = w* x L^2 / 8 = 31.5 x 10.0^2 / 8 = 393.8 kNm V* = w* x L / 2 = 31.5 x 10.0 / 2 = 157.5 kN

Step 3 — Section Properties (310UB40.4)

From OneSteel/ASI Hot Rolled and Structural Steel Products, 7th Edition:

Step 4 — Section Classification (AS 4100 Table 5.2)

Flange slenderness (hot-rolled UB): lambda_e = (b_f - t_w) / (2 x t_f) x sqrt(fy_f / 250) = (165 - 6.1) / (2 x 10.2) x sqrt(300/250) = 158.9 / 20.4 x 1.095 = 7.79 x 1.095 = 8.53

Flange limits (Table 5.2, flat element, one supported edge, HR): lambda_ep (compact) = 9 (for fy = 300 MPa, limit = 9 x sqrt(250/300) = 8.22) lambda_ey (non-compact) = 16 (limit = 16 x sqrt(250/300) = 14.6)

8.53 > 8.22 but < 14.6: Flange is non-compact.

Web slenderness (hot-rolled UB, both edges supported): lambda_e = (d - 2 x t_f) / t_w x sqrt(fy_w / 250) = (304 - 20.4) / 6.1 x sqrt(300/250) = 283.6 / 6.1 x 1.095 = 46.5 x 1.095 = 50.9

Web limits (Table 5.2, flat element, both edges supported, bending): lambda_ep (compact) = 82 (limit = 82 x sqrt(250/300) = 74.9)

50.9 < 74.9: Web is compact.

Section classification: Non-compact (governed by flange). Section has compact web, non-compact flange.

Step 5 — Section Moment Capacity (Clause 5.2.3)

For non-compact sections, the section capacity is interpolated between the yield moment and plastic moment:

Effective section modulus (Clause 5.2.4): Z_e = Zx + (lambda_ey - lambda_e) / (lambda_ey - lambda_ep) x (Sx - Zx) = 570 + (14.6 - 8.53) / (14.6 - 8.22) x (641 - 570) = 570 + (6.07 / 6.38) x 71 = 570 + 0.951 x 71 = 570 + 67.6 = 637.6 x 10^3 mm^3

Section moment capacity: phi_Msx = phi x fy_f x Z_e = 0.90 x 300 x 637.6 x 10^3 / 10^6 = 172.2 kNm

Check: M* = 393.8 kNm >> 172.2 kNm — section capacity grossly inadequate for this span!

The 310UB40.4 is clearly too small for a 10 m span at this loading. For illustration, let's continue with the LTB check and then size up the beam.

Step 6 — LTB Member Capacity (Clause 5.6)

Assume the construction stage without slab restraint: L_e = 3.0 m (secondary beams provide lateral restraint at 3.0 m centres).

Elastic buckling moment Mo: Mo = sqrt[(pi^2 x E x Iy / Le^2) x (G x J + pi^2 x E x Iw / Le^2)]

Iy = 6.56 x 10^6 mm^4 (from section properties) Mo = sqrt[(pi^2 x 200000 x 6.56 x 10^6 / 3000^2) x (80000 x 150 x 10^3 + pi^2 x 200000 x 72.5 x 10^9 / 3000^2)] = sqrt[(1,439,000) x (12.0 x 10^9 + 15.9 x 10^9)] = sqrt(1,439,000 x 27.9 x 10^9) = sqrt(40.1 x 10^15) = 200.5 x 10^6 Nmm = 200.5 kNm

Slenderness reduction factor: Msx = Sx x fy_f = 641 x 10^3 x 300 / 10^6 = 192.3 kNm (plastic section moment) lambda_s = sqrt(Msx / Mo) = sqrt(192.3 / 200.5) = sqrt(0.959) = 0.979

alpha_s (Clause 5.6.1.1): alpha_s = 0.6 x [sqrt(lambda_s^4 + 3) - lambda_s^2]^0.5 = 0.6 x [sqrt(0.979^4 + 3) - 0.979^2]^0.5 = 0.6 x [sqrt(0.919 + 3) - 0.959]^0.5 = 0.6 x [sqrt(3.919) - 0.959]^0.5 = 0.6 x [1.979 - 0.959]^0.5 = 0.6 x sqrt(1.020) = 0.6 x 1.010 = 0.606

alpha_m (moment modification factor, Clause 5.6.1.2): For a simply supported UDL segment, alpha_m = 1.13 (Table 5.6.1, segment with end restraints restrained laterally).

Actually, for the end segment of a beam with intermediate restraints at third points, the moment diagram is between 0 and approximately 0.89 x M_max at the first restraint. For this shape, alpha_m ~ 1.35 per ASI Design Capacity Tables.

Member moment capacity: phi_Mb = phi x alpha_m x alpha_s x Msx = 0.90 x 1.35 x 0.606 x 192.3 = 141.8 kNm

141.8 kNm << 393.8 kNm. Clearly undersized for the 10 m span.

Step 7 — Beam Resizing: 460UB82.1

Given the 10 m span and 31.5 kN/m factored load, a much larger section is needed. Try 460UB82.1 Grade 300:

460UB82.1 properties: d = 460 mm, b_f = 191 mm, t_f = 16.0 mm, t_w = 9.9 mm Sx = 1,830 x 10^3 mm^3, Zx = 1,610 x 10^3 mm^3, Ix = 371 x 10^6 mm^4, ry = 42.3 mm

Flange: t_f = 16.0 mm => fy_f = 280 MPa (thickness > 11 mm) Web: t_w = 9.9 mm => fy_w = 300 MPa

Classification (fy_f = 280 MPa): lambda_e_flange = (191 - 9.9) / (2 x 16.0) x sqrt(280/250) = 181.1/32.0 x 1.058 = 5.66 x 1.058 = 5.99 lambda_ep = 9 x sqrt(250/280) = 8.50

5.99 < 8.50: Compact flange.

lambda_e_web = (460 - 32.0) / 9.9 x sqrt(300/250) = 428/9.9 x 1.095 = 43.2 x 1.095 = 47.3 lambda_ep = 82 x sqrt(250/300) = 74.9

47.3 < 74.9: Compact web. Section is compact.

Section capacity: phi_Msx = 0.90 x 280 x 1,830 x 10^3 / 10^6 = 461.2 kNm > 393.8 kNm. OK (0.85 utilization).

Deflection check (live load serviceability): w_Q = 9.0 kN/m delta_Q = 5 x 9.0 x 10000^4 / (384 x 200000 x 371 x 10^6) = 15.8 mm L/500 = 20.0 mm (total deflection limit for floors with brittle finishes). 15.8 < 20.0. OK.

Total deflection (G + Q): delta_total = 5 x 24.0 x 10000^4 / (384 x 200000 x 371 x 10^6) = 42.1 mm L/250 = 40.0 mm. 42.1 > 40.0 — marginal. Camber 20 mm upward to compensate.

Final beam: 460UB82.1 Grade 300, camber 20 mm. Camber to be specified on shop drawings.

Step 8 — Shear Capacity (Clause 5.11)

Web shear (unstiffened web): d_p = d - 2 x t_f = 460 - 32 = 428 mm d_p / t_w = 428 / 9.9 = 43.2

For Grade 300 web (fy_w = 300 MPa): 82 x sqrt(250/300) = 74.9

43.2 < 74.9: shear buckling does not govern — web yields in shear.

Shear capacity: phi_Vv = phi x 0.6 x fy_w x Aw where Aw = d x t_w = 460 x 9.9 = 4,554 mm^2 phi_Vv = 0.90 x 0.6 x 300 x 4554 / 1000 = 737.7 kN >> 157.5 kN. OK.

Summary

Final specification: 460UB82.1, AS/NZS 3679.1 Grade 300, camber 20 mm upward. Bondek shear studs 19 mm dia. at 300 mm c/c for composite action.

Frequently Asked Questions

Why does the 310UB40.4 fail so badly at 10 m span? The 310UB40.4 is suitable for spans up to approximately 6-7 m at this loading. A 10 m span requires a beam roughly 3x stiffer (Ix scales with span^3 for deflection). The 460UB82.1 has Ix = 371 x 10^6 mm^4 vs 85.2 x 10^6 for the 310UB — a factor of 4.4, which provides the necessary stiffness. The beam depth also follows the span/22 rule: 10,000/22 = 455 mm, close to the 460 mm actual depth.

What is the difference between Zx and Sx in AS 4100? Zx is the elastic section modulus (yield moment = Zx x fy), while Sx is the plastic section modulus (plastic moment = Sx x fy). In AS 4100, "Z" is elastic and "S" is plastic — the opposite of North American notation where S is elastic and Z is plastic. This is a common source of confusion when working between Australian and US codes.

When can I use the ASI Design Capacity Tables instead of full calculation? The ASI Design Capacity Tables (ADCT) provide pre-computed phi_Msx, phi_Mbx for all standard UB/UC sections at various effective lengths. For standard hot-rolled sections in Grade 300, the tables are fully validated and accepted by all Australian building regulators. Use them for production design — hand calculation is for verification and non-standard cases only.

This page is for educational reference. Beam design per AS 4100:2020 Clause 5. Verify capacities against current ASI Design Capacity Tables (Blue Book). All structural designs must be independently verified by a licensed Professional Engineer or Structural Engineer. Results are PRELIMINARY — NOT FOR CONSTRUCTION.