AS 4100 Composite Beam Design — AS 2327 Worked Example (310UB40.4 + 120mm Slab)

Complete Australian composite beam design walkthrough per AS 4100:2020 and AS 2327.1:2017. This worked example designs a 310UB40.4 Grade 300PLUS steel beam with a 120 mm reinforced concrete slab on trapezoidal Bondek decking, spanning 9.0 m at 3.6 m spacing. Shear stud capacity, full and partial shear connection, construction-stage checks, and long-term deflection are covered.

The Steel Calculator WASM engine performs composite beam checks automatically for AS 2327 (Australia), AISC 360 Chapter I (USA), EN 1994-1-1 (Europe), and CSA S16 Clause 17 (Canada).

PRELIMINARY — NOT FOR CONSTRUCTION. All results are for educational and reference use only. Must be independently verified by a Chartered Professional Engineer (CPEng) or RPEQ before use in any project.

Design Problem Definition

An office floor beam in a Brisbane commercial building spans 9.0 m between columns at 3.6 m centres. The slab is a 120 mm reinforced concrete topping on 0.75 mm Bondek II trapezoidal steel decking with 54 mm rib height. The beam is propped during construction to eliminate construction-stage stresses in the bare steel.

Design Data:

• Span: L = 9.0 m, simply supported
• Beam spacing: 3.6 m centres
• Section: 310UB40.4 Grade 300PLUS (AS/NZS 3679.1, fy = 300 MPa)
• Slab: 120 mm topping + 54 mm Bondek rib, total depth 174 mm
• Concrete: f'c = 32 MPa normal-weight, density 2,400 kg/m^3
• Shear studs: 19 mm diameter, 100 mm as-welded height, Grade 450 (AS 1554.6)
• Deck orientation: ribs perpendicular to beam, one stud per rib
• Propping: propped at mid-span, props removed after concrete reaches 75% f'c
• Reinforcement: SL82 mesh (8 mm at 200 mm each way)

Loads:

• Dead load (slab + deck + services): 3.2 kPa
• Superimposed dead (ceiling + partitions): 1.0 kPa
• Live load (office): 3.0 kPa per AS 1170.1 Table 3.1

Line Loads on Beam:

• G_construction = 3.2 kPa x 3.6 m = 11.5 kN/m (carried by composite section — propped)
• G_superimposed = 1.0 kPa x 3.6 m = 3.6 kN/m
• Q = 3.0 kPa x 3.6 m = 10.8 kN/m
• ULS: w* = 1.2(11.5 + 3.6) + 1.5(10.8) = 18.1 + 16.2 = 34.3 kN/m
• SLS: w_sls = 11.5 + 3.6 + 0.7 x 10.8 = 22.7 kN/m

Design Actions:

• M* = 34.3 x 9.0^2 / 8 = 347.3 kN.m
• V* = 34.3 x 9.0 / 2 = 154.4 kN

Section Properties — 310UB40.4 Grade 300PLUS

Step 1 — Effective Flange Width (AS 2327 Clause 3.1.2)

The effective concrete flange width on each side of the beam centreline is the lesser of:

• L/8 = 9,000 / 8 = 1,125 mm per side
• s/2 = 3,600 / 2 = 1,800 mm per side
• Distance to slab edge: not applicable (interior beam)

b_ef per side = 1,125 mm. Total effective flange width: b_ef = 2 x 1,125 = 2,250 mm.

Step 2 — Concrete and Steel Force Capacities

Concrete Compression Capacity (AS 2327 Clause 4.2)

The compressive force the slab can develop, assuming the plastic neutral axis lies within the slab:

F_cc = 0.85 x f'_c x b_ef x t_c = 0.85 x 32 x 2,250 x 120 = 7,344 x 10^3 N = 7,344 kN

Where t_c = 120 mm is the concrete topping thickness above the deck ribs. The rib concrete is ignored in compression per AS 2327 Clause 3.1.4 because the ribs run perpendicular to the beam.

Steel Tensile Capacity

F_st = A_g x f_y = 5,150 x 300 = 1,545 x 10^3 N = 1,545 kN

Since F_st (1,545 kN) < F_cc (7,344 kN), the steel section governs. The plastic neutral axis (PNA) lies within the concrete slab — a favourable condition that maximises the lever arm.

Step 3 — Shear Stud Design (AS 2327 Clause 3.2)

Single Stud Capacity

For a 19 mm diameter headed stud, Grade 450, in 32 MPa normal-weight concrete:

Stud cross-sectional area: A_sc = pi x 19^2 / 4 = 283.5 mm^2

Stud failure mode: f_vs1 = 0.8 x f_uc x A_sc = 0.8 x 450 x 283.5 = 102.1 kN

Concrete cone failure mode (AS 2327 Eq. 3.2.3(2)): E_c = 0.043 x rho^1.5 x sqrt(f'_c) = 0.043 x 2,400^1.5 x sqrt(32) = 5.66 x 30,100 = 30,100 MPa (approximately)

f_vs2 = 0.29 x alpha x d_sc^2 x sqrt(f'_c x E_c)

For a stud height-to-diameter ratio h_sc / d_sc = 100 / 19 = 5.26 > 4, alpha = 1.0.

f_vs2 = 0.29 x 1.0 x 19^2 x sqrt(32 x 30,100) = 0.29 x 361 x sqrt(963,200) = 0.29 x 361 x 981.4 = 102.7 kN

The stud failure mode governs at f_vs = 102.1 kN.

Design Shear Capacity Per Stud

phi_sc = 0.85 (AS 2327 Table 3.1)

phi x f_vs = 0.85 x 102.1 = 86.8 kN per stud

Deck Reduction Factor (AS 2327 Clause 3.2.4)

For trapezoidal decking with ribs perpendicular to the beam and one stud per rib:

k_deck = 0.6 x (b_0 / h_p) x (h_sc / h_p - 1.0) <= 1.0

Where b_0 = average rib width = 150 mm (Bondek II), h_p = rib height = 54 mm, h_sc = 100 mm.

k_deck = 0.6 x (150 / 54) x (100 / 54 - 1.0) = 0.6 x 2.78 x 0.852 = 1.42

But k_deck is capped at 1.0 per Clause 3.2.4(a), so k_deck = 1.0. For Bondek II with 19 mm studs at 100 mm height, the deck geometry does not reduce stud capacity.

Design capacity per stud after deck reduction: phi x f_vs_eff = 86.8 x 1.0 = 86.8 kN.

Step 4 — Number of Studs Required

Full Shear Connection

For full shear connection, the total horizontal shear force to be transferred is:

V_h = F_st = 1,545 kN (steel governs)

Number of studs for half-span (zero moment at support to maximum moment at mid-span):

n_sc_half = V_h / (phi x f_vs_eff) = 1,545 / 86.8 = 17.8, use 18 studs per half-span.

Total studs for full shear connection: 36 studs.

At one stud per rib and Bondek II rib spacing of 300 mm, the maximum number of studs available in one half-span is:

n_available = L / (2 x rib_spacing) = 9,000 / (2 x 300) = 15 studs per half-span.

Only 15 studs can physically fit in each half-span — full shear connection cannot be achieved with single studs at every rib. Either double studs per rib or partial shear connection must be used.

Partial Shear Connection (AS 2327 Clause 4.3)

Using 15 studs per half-span (one per rib, the maximum available):

Degree of shear connection: eta = n_sc_actual / n_sc_full = 15 / 18 = 0.833 (83.3%).

Check minimum: eta_min = 0.4 for L < 20 m. 0.833 > 0.4 — acceptable.

Horizontal shear force transferred: V_h_actual = 15 x 86.8 = 1,302 kN.

Step 5 — Composite Moment Capacity (AS 2327 Clause 4.2)

For partial shear connection with the PNA in the slab (steel governs):

Stress Block Depth in Slab

a = V_h_actual / (0.85 x f'_c x b_ef) = 1,302 x 10^3 / (0.85 x 32 x 2,250) = 1,302,000 / 61,200 = 21.3 mm below the top of the slab.

The PNA at 21.3 mm depth lies entirely within the 120 mm concrete topping — the PNA is in the slab as assumed.

Lever Arm

Distance from the concrete compression centroid to the steel tension centroid:

d_sc = d/2 + h_p + t_c - a/2 = 304/2 + 54 + 120 - 21.3/2 = 152 + 54 + 120 - 10.7 = 315.3 mm

Composite Moment Capacity

M_s_composite = V_h_actual x d_sc = 1,302 x 0.3153 = 410.6 kN.m

Check against bare steel moment capacity: M_s_bare = S_x x f_y = 640 x 10^3 x 300 = 192.0 kN.m

Composite action increases moment capacity by 410.6 / 192.0 = 2.14 times the bare steel capacity.

phi = 0.90 (AS 4100 Table 3.4, bending)

phi x M_s_composite = 0.90 x 410.6 = 369.5 kN.m

Utilisation Check

M* / (phi x M_s_composite) = 347.3 / 369.5 = 0.940 — design is adequate with 6% reserve.

Step 6 — Shear Capacity (AS 4100 Clause 5.11)

With the concrete slab providing continuous lateral restraint, the bare steel shear capacity governs. The shear check for the steel section alone:

V_w = 0.6 x f_y x d x t_w = 0.6 x 300 x 304 x 6.1 = 333.7 kN

phi x V_v = 0.90 x 333.7 = 300.4 kN

V* = 154.4 kN < 300.4 kN — shear is satisfactory.

Since V* < 0.6 x phi x V_v = 180.2 kN, no shear-bending interaction per AS 4100 Clause 5.12 is required.

Step 7 — Deflection Checks (AS 2327 Clause 5.2)

Short-Term Deflection (Composite Section)

The composite second moment of area I_comp is calculated using the transformed section method with modular ratio n = E_s / E_c.

E_s = 200,000 MPa. For 32 MPa concrete: E_c = 0.043 x 2,400^1.5 x sqrt(32) = 30,100 MPa.

n = 200,000 / 30,100 = 6.64 (use 7.0 per AS 2327 simplified approach).

Transformed slab width: b_tr = b_ef / n = 2,250 / 7.0 = 321 mm.

The composite neutral axis depth from the top of the slab, measured from first principles using the transformed area method:

Area of transformed slab: A_tr = b_tr x t_c = 321 x 120 = 38,520 mm^2 Area of steel: A_s = 5,150 mm^2

Centroid of transformed slab from top of slab: y_slab = 60 mm Centroid of steel from top of slab: y_steel = 120 + 54 + 304/2 = 326 mm

Composite centroid from top of slab: y_bar = (A_tr x 60 + A_s x 326) / (38,520 + 5,150) = (2,311,200 + 1,678,900) / 43,670 = 91.4 mm

I_comp = I_s + A_s x (326 - 91.4)^2 + (1/12) x b_tr x t_c^3 + A_tr x (91.4 - 60)^2 = 85.0x10^6 + 5,150 x (234.6)^2 + (1/12) x 321 x 120^3 + 38,520 x (31.4)^2 = 85.0x10^6 + 283.5x10^6 + 46.2x10^6 + 38.0x10^6 = 452.7 x 10^6 mm^4

This is 5.3 times the bare steel I_x (85.0 x 10^6 mm^4).

Short-Term Deflection (SLS load, composite section)

delta_short = 5 x w_sls x L^4 / (384 x E_s x I_comp) = 5 x 22.7 x (9,000)^4 / (384 x 200,000 x 452.7 x 10^6) = 5 x 22.7 x 6.561 x 10^15 / (3.477 x 10^16) = 7.448 x 10^17 / 3.477 x 10^16 = 21.4 mm

L / 250 = 9,000 / 250 = 36.0 mm — total deflection limit is satisfied.

Long-Term Deflection (Creep + Shrinkage)

Per AS 2327 Clause 5.2.2, long-term effects are accounted for by doubling the modular ratio for sustained loads (use 2n for creep). For the superimposed dead load portion (G_super = 3.6 kN/m):

Transformed slab width for long-term: b_tr_lt = 2,250 / 14 = 161 mm.

Recalculating I_comp_lt with n = 14 yields approximately 320 x 10^6 mm^4 — a 29% reduction from the short-term value.

Creep deflection increment from superimposed dead load: delta_creep = 5 x 3.6 x (9,000)^4 / (384 x 200,000 x 320 x 10^6) - short-term SDL contribution = 20.3 - 4.8 = 15.5 mm (approximate creep component)

Live load deflection (n = 7): delta_live = 5 x (0.7 x 10.8) x (9,000)^4 / (384 x 200,000 x 452.7 x 10^6) = 7.1 mm

L / 360 = 9,000 / 360 = 25.0 mm — live load deflection is satisfactory.

Total long-term deflection: delta_total = delta_short + delta_creep = 21.4 + 15.5 = 36.9 mm. This marginally exceeds L/250 = 36.0 mm. In practice, the slab reinforcement and continuity over supports (if continuous) would reduce this. For a simply supported beam, pre-camber of 10-15 mm would address the excess.

Step 8 — Transverse Reinforcement (AS 2327 Clause 3.3)

The concrete flange must resist the longitudinal shear transferred by the studs. Per AS 2327 Clause 3.3.1, the minimum transverse reinforcement area per unit length is:

A_sf / s_f >= 0.002 x b_f x t_c = 0.002 x 2,250 x 120 = 540 mm^2/m in each direction.

SL82 mesh provides 415 mm^2/m per layer — two layers (top and bottom) provide 830 mm^2/m. This exceeds the minimum. Additionally, the longitudinal shear per shear plane must be checked:

v_L = V_h_actual / (2 x L/2) = 1,302,000 / 9,000 = 144.7 N/mm per shear plane.

The shear plane capacity per AS 2327 Clause 3.3.3 is 0.55 x f'_ctf x A_cv + A_sf x f_sy. At typical reinforcement ratios, the SL82 mesh provides adequate shear plane capacity for 144.7 N/mm longitudinal shear.

Summary of Design Checks — 310UB40.4 Composite Beam

The 310UB40.4 with 83.3% partial shear connection is adequate in strength but borderline in long-term deflection. Pre-camber of 12 mm is recommended.

AS 2327 vs AISC 360 Composite Beam Comparison

Frequently Asked Questions

How does AS 2327 calculate the design shear capacity of a stud connector?

AS 2327 Clause 3.2.3 calculates the nominal shear capacity of a headed stud as f_vs = 0.8 x f_uc x A_sc for stud failure and f_vs = 0.29 x alpha x d_sc^2 x sqrt(f'_c x E_c) for concrete cone failure, taking the lesser value. For a 19mm stud Grade 450 in 32 MPa concrete, the stud failure governs at approximately 102 kN. The capacity reduction factor phi_sc = 0.85 per AS 2327 Table 3.1, giving a design capacity of 86.7 kN per stud.

What is the minimum degree of shear connection required by AS 2327?

AS 2327 Clause 4.3.3 requires a minimum degree of shear connection of 0.4 for simply supported beams with spans up to 20 m. For continuous beams, the minimum is 0.4 at positive moment regions and 0.5 at negative moment regions if reinforcement is relied upon. The degree of shear connection eta = n_sc x f_vs / min(F_cc, F_st), where F_cc is the concrete slab compression capacity and F_st is the steel section yield capacity. Below eta = 0.4, the connection is classified as non-ductile and the full AS 2327 design model does not apply — the beam must be designed as non-composite or with supplementary mechanical interlock.

How does propping affect composite beam design per AS 2327?

Propping supports the steel beam during concrete placement and curing, eliminating construction-stage stresses in the bare steel. Per AS 2327 Clause 5.1, when the beam is propped at mid-span or third-points, the entire dead load (slab + finishes) is resisted by the composite section rather than the bare steel. This eliminates construction-stage deflection entirely and allows lighter steel sections. The downside is increased site labour and curing time before props can be removed. For unpropped construction, the bare steel must support wet concrete weight, and the composite section resists only superimposed dead and live loads — the steel section is typically 15-25% heavier than a propped alternative.

How does the effective flange width differ between AS 2327 and AISC 360?

AS 2327 Clause 3.1.2 defines the effective flange width as b_ef = min(L/8 per side, centre-to-centre beam spacing/2, distance to slab edge). This matches AISC 360 I3.1a exactly. However, AS 2327 also permits a refined effective width for edge beams equal to b_ef = L/16 + the actual overhang distance, capped at L/8. For interior beams at typical 8m x 8m grid spacing, both codes produce identical effective widths. AS 2327 additionally requires that the effective width at interior supports of continuous beams be taken as 70% of the span effective width to account for slab cracking under negative moment.

Try it now: Check your composite beam with our free Australian Beam Capacity calculator

Related Pages

• AS 4100 Beam Design — 310UB40.4 Worked Example — Flexure, shear, LTB
• AS 4100 Column Design — 200UC46 Worked Example — Axial and buckling
• AS 4100 Bolt Design Guide — Clause 9.3 — Bolted connections
• AS 4100 Fillet Weld Design — Clause 9.7 — Weld capacity
• AS 4100 Load Combinations — ULS/SLS/STB
• Australia AS 4100 Steel Design Guide — Full code reference
• Composite Beam Design — AISC 360 — US composite design
• Concrete on Steel Deck — Profile Guide — Deck profiles

This page is for educational reference. All resistance formulae are per AS 4100:2020 and AS 2327.1:2017. 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).