Australian Shear Stud Connectors — AS 2327-2017 Composite Beam Design

Comprehensive reference for headed shear stud connectors in composite steel-concrete beams per AS 2327-2017 (Composite structures -- Composite steel-concrete construction in buildings). Covers stud capacity, dimensions, placement requirements, degree of shear connection, and transverse reinforcement requirements for typical Australian composite floor systems.

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Composite Beam Principle

A composite steel-concrete beam consists of a steel beam (typically UB or WB section) acting compositely with a concrete slab through the action of shear connectors (headed studs) welded to the top flange. The shear connectors transfer longitudinal shear between the slab and the steel beam, enabling the two components to act as a single structural unit.

The composite action provides:

• Increased flexural stiffness (30-50% greater than the steel beam alone)
• Increased moment capacity (the concrete slab carries compression while the steel beam carries tension)
• Reduced steel depth for the same span (shallower floor zones)
• Longer spans (up to 15 m for typical UB sections with composite slabs)

Shear Stud Capacity — AS 2327 Clause 4.2

The design shear capacity of a headed stud connector is the lesser of:

1. Stud shank failure: phi_v x f_us = phi_v x f_us x A_shank

2. Concrete crushing/pull-out: phi_v x f_cc = phi_v x 0.29 x alpha x d^2 x sqrt(f'_c x E_c)

where:

• phi_v = 0.85 (capacity factor for shear connectors)
• f_us = tensile strength of the stud material (typically 415 MPa for standard SD studs)
• A_shank = pi x d^2 / 4 (cross-sectional area of the stud shank)
• d = stud shank diameter (mm)
• f'_c = characteristic compressive strength of the concrete (MPa)
• E_c = modulus of elasticity of concrete (MPa) = 0.043 x rho^1.5 x sqrt(f'_c) for normal weight concrete
• alpha = 1.0 for studs in solid slabs (standard composite slab); 0.85 for studs in deck ribs perpendicular to beam; 0.60 for studs in deck ribs parallel to beam

Standard Shear Stud Dimensions (Australian Supply)

The 19 mm diameter stud is the most common size in Australian composite construction. After welding, the stud height must be not less than the specified height minus 2 mm, and the weld collar must be full 360 degrees around the stud base.

Design Shear Capacity — 19 mm Stud in Standard Concrete Grades

For 19 mm studs in Grade 32 concrete (standard for Australian suspended slabs), the concrete crushing/pull-out governs at 100 kN. For higher-strength concretes, the stud shank failure governs at 100 kN.

Stud Placement Requirements — AS 2327 Clause 4.4

Longitudinal Spacing

The maximum spacing ensures adequate composite action along the beam. At the maximum spacing, the concrete between studs can still transfer the shear flow to the studs through diagonal compression struts without distress.

Transverse Spacing

Studs Through Profiled Steel Decking

For composite slabs using profiled steel decking (Bondek, Condeck, KingFlor, etc.):

• Studs welded through the deck (through-deck welding) are standard practice
• The deck rib orientation relative to the steel beam affects the stud capacity
• Deck ribs perpendicular to the beam: reduction factor alpha = 0.85 (typical)
• Deck ribs parallel to the beam: reduction factor alpha = 0.60 (greater reduction due to rib splitting)
• Stud height after welding: minimum 75 mm above the top of the steel deck

Degree of Shear Connection — AS 2327 Clause 4.5

The degree of shear connection (beta) is the ratio of the total longitudinal shear force provided by the studs to the horizontal shear force required for full composite action:

beta = sum(phi P_stud) / min(F_c, F_s)

where:

• F_c = 0.85 x f'_c x A_c (concrete compression capacity of the slab)
• F_s = A_s x f_y (steel tension capacity of the beam)
• sum(phi P_stud) = total capacity of all studs in the shear span

Partial Shear Connection

AS 2327 permits partial shear connection (beta < 1.0) for simply supported beams, provided that:

• beta >= 0.40 (minimum degree of shear connection)
• The reduced moment capacity M_beta is calculated using the linear interpolation: M_beta = M_s + beta x (M_c - M_s)
• M_s = steel section moment capacity alone
• M_c = full composite moment capacity
• The studs are ductile (stud elongation at fracture > 6 mm)

Partial shear connection is commonly used to reduce the number of studs required when the full composite moment capacity is not needed for the design loads.

Stud Distribution

For simply supported beams with uniform loading, the required number of studs per unit length follows the shear flow diagram. The studs are commonly spaced uniformly (equal spacing) for simplicity, which is conservative because more studs are provided near the supports (where shear is highest) than the minimum required.

Transverse Reinforcement — AS 2327 Clause 4.6

Transverse reinforcement (perpendicular to the steel beam) is required in the concrete slab to resist the splitting forces generated by the stud shear transfer. The minimum transverse reinforcement area per unit length is:

A_t / s >= 0.0025 x b_f_slab

where b_f_slab is the effective width of the concrete flange in transverse bending.

For typical Australian composite slabs with SL82 mesh (8 mm bars at 200 mm c/c both ways), the transverse reinforcement area is: A_t / s = (pi x 8^2 / 4) / 200 = 50.3 / 200 = 0.251 mm^2/mm = 251 mm^2/m.

This satisfies the minimum requirement for effective widths up to about 4 m, which covers most commercial building beam spacings.

Worked Example: Stud Design for Composite Floor Beam

Problem: A 410UB59.7 Grade 300 steel beam at 3.0 m centres spans 10.0 m supporting a 120 mm thick composite slab (Grade 32 concrete) on Bondek II profiled steel decking (0.75 mm BMT, ribs perpendicular to beam). The beam is simply supported. Determine the number of 19 mm diameter shear studs required and their spacing.

Given:

• Beam: 410UB59.7, A_s = 7600 mm^2, f_y = 300 MPa
• Slab: 120 mm total thickness, 54 mm above deck ribs (Bondek II rib height = 54 mm, trough width at top = 150 mm)
• Concrete: f'_c = 32 MPa, normal weight
• Stud: 19 mm diameter, 125 mm height, f_us = 415 MPa
• Deck ribs: perpendicular to beam, alpha = 0.85
• Beam spacing: 3.0 m, effective slab width = b_eff approx. L/8 each side = 2 x 10000/8 = 1250 mm each side, limited to beam spacing / 2 = 1500 mm. b_eff = 1250 + 1250 + 178 = 2678 mm (including beam flange width)

Solution:

Step 1: Full shear connection force

Steel tension capacity: F_s = A_s x f_y = 7,600 x 300 x 10^(-3) = 2,280 kN

Concrete compression capacity: F_c = 0.85 x 32 x 2,678 x 54 x 10^(-3) = 0.85 x 32 x 144.6 = 3,934 kN

Note: for the slab above the deck ribs, only the concrete above the ribs (54 mm) is considered in compression, as the concrete in the ribs is below the neutral axis in hogging. Wait -- for simply supported beam, the slab is in compression, and the full slab depth is available.

Revised F_c: average slab depth = (120 x 3000 x 1000) ... let me use a simpler approach.

Compression area of slab: b_eff x t_s_above_deck = 2,678 x (120 - 54) = 2,678 x 66 = 176,748 mm^2

Plus the concrete in the ribs (3 ribs across the effective width at 300 c/c, each rib ~150mm wide at top): A_ribs_approx = 3 x 150 x 54 / 2 = 12,150 mm^2 (triangular approximation)

Total concrete area in compression: ~189,000 mm^2

F_c = 0.85 x 32 x 189,000 x 10^(-3) = 5,141 kN

F_s = 2,280 kN governs (steel yields before concrete crushes).

Step 2: Stud capacity

Stud shank failure: phi P_stud = 0.85 x 415 x pi x 19^2 / 4 x 10^(-3) = 0.85 x 415 x 284 x 10^(-3) = 100.0 kN

Concrete failure (through-deck, alpha = 0.85): phi P_cone = 0.85 x 0.29 x 0.85 x 19^2 x sqrt(32 x 30,100) x 10^(-3)

E_c = 0.043 x 2400^1.5 x sqrt(32) = 0.043 x 117,576 x 5.657 = 28,590 MPa ~ 30,100

phi P_cone = 0.85 x 0.29 x 0.85 x 361 x sqrt(963,200) x 10^(-3) = 0.85 x 0.29 x 0.85 x 361 x 981.4 x 10^(-3) = 74.4 kN

Concrete governs: design stud capacity = 74.4 kN per stud.

Step 3: Number of studs for full shear connection

Half-span shear force = min(F_s, F_c) = 2,280 kN (half of this goes to each end support)

Studs per half-span: n = F_s / (phi P_stud) = 2,280 / 74.4 = 30.6 -> 31 studs per half-span.

Total studs for full shear connection: 62 studs.

Step 4: Stud spacing

Beam length 10.0 m, studs in one row along each flange (two rows total): 31 studs per half-span per row = 62 total per row... no, that's 62 total.

Let me reconsider. With two studs per cross-section (one per flange row), the shear per stud is halved. So the number of stud positions per half-span = 31 / 2 = 15.5 -> 16 positions.

Spacing = 5,000 mm / 16 = 313 mm centre-to-centre.

Check: 313 mm > 5d = 95 mm (OK), < 600 mm (OK), < 4t_s = 4 x 120 = 480 mm (OK).

Step 5: Degree of shear connection

The design may use partial shear connection if the composite moment demand is less than the full composite capacity. With the minimum beta = 0.40, the required number of studs reduces to 0.40 x 62 = 25 studs total (13 per half-span). However, full shear connection is preferred for simply supported beams in high-usage applications.

Result: 62 x 19 mm shear studs required for full shear connection, at approximately 310 mm spacing in two rows along the beam. Concrete pull-out governs stud capacity at 74.4 kN per stud.

Frequently Asked Questions

What is the purpose of shear studs in composite beams?

Shear studs transfer longitudinal shear between the concrete slab and the steel beam, enabling composite action. Without shear connectors, the slab and beam would slide relative to each other and act as two independent elements -- the slab carrying compression through arching, the beam carrying tension and bending independently. The shear studs force the two to deform together, with the concrete carrying compression and the steel carrying tension, approximately doubling the flexural stiffness compared to the steel beam alone.

How are shear studs installed on Australian construction sites?

Shear studs are installed by through-deck stud welding: a ceramic ferrule is placed around the stud base, the stud is pressed against the steel flange through the deck, and an electric arc melts the stud tip and the flange surface. The stud is plunged into the molten pool, creating a full-penetration weld. The process takes approximately 0.5-1.0 second per stud. The deck is typically burned through during welding, which is acceptable provided a full 360-degree weld collar is visible after completion. Stud welding requires a minimum flange thickness of approximately 6 mm to avoid burn-through.

What is the minimum degree of shear connection per AS 2327?

AS 2327 permits partial shear connection with a minimum degree of shear connection beta = 0.40 for simply supported beams with ductile stud connectors. For continuous beams where moment redistribution depends on the composite capacity, beta >= 0.70 is recommended. Partial shear connection reduces the number of studs and therefore the fabrication cost, but the composite moment capacity is reduced by a roughly linear interpolation between the steel-only capacity (beta = 0) and the full composite capacity (beta = 1.0).

Why does the deck rib orientation affect shear stud capacity?

When profiled steel deck ribs are perpendicular to the steel beam, the stud is seated in a rib trough, and the concrete around the stud is confined by the adjacent rib walls. This produces a 45-degree concrete compression strut that transfers shear from the slab into the stud. When ribs are parallel to the beam, the stud is seated on a continuous flat surface of the deck, and the concrete confinement is reduced because the deck rib is a continuous channel that can split longitudinally under the stud bearing force. The reduction factor alpha = 0.60 (parallel) vs alpha = 0.85 (perpendicular) reflects this reduced confinement.

What reinforcement is required around shear studs in composite slabs?

Per AS 2327 Clause 4.6, transverse reinforcement must be provided across the effective width of the slab in the vicinity of the shear studs. The minimum transverse reinforcement ratio is 0.0025, which corresponds to SL82 mesh (8 mm at 200 mm c/c) for typical Australian composite slabs. For beams with high stud density or narrow effective widths, additional trimmer bars or U-bars may be required to resist the longitudinal shear splitting force. The transverse reinforcement should extend a distance of at least 300 mm beyond the outermost stud on each side of the beam.

Educational reference only. All design values must be verified against the current edition of AS 2327:2017 and the project specification. This information does not constitute professional engineering advice. Always consult a qualified structural engineer for design decisions.