CSA S16 Composite Beam — Shear Stud Design Reference

Complete reference for CSA S16:19 Clause 17 composite steel-concrete beam design. Covers shear stud connector strength in solid slabs and metal deck, effective flange width, moment capacity for partial and full composite action, degree of partial shear connection, longitudinal shear reinforcement in the slab, and rib orientation effects for profiled steel decking. Includes a worked example for a composite beam with headed studs.

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CSA S16 Composite Beam Design — Overview

Composite beams combine a structural steel beam (typically a W-shape) with a reinforced concrete slab through mechanical shear connectors. The concrete slab acts as the compression flange of the composite section, significantly increasing both the stiffness and the flexural strength compared to the steel beam alone.

CSA S16:19 Clause 17 governs composite beam design in Canada. The key principles are:

Composite Action Benefits

Shear Stud Connectors (CSA S16 Clause 17.5)

Headed stud connectors are welded to the top flange of the steel beam and embedded in the concrete slab. The stud transfers horizontal shear between the steel and concrete, enabling composite action.

Stud Strength in Solid Slab

qr = φsc × 0.50 × Asc × √(f'c × Ec) ≤ φsc × Asc × Fu

where:
  qr  = factored shear resistance of one stud (N)
  φsc = 0.80 (resistance factor for shear connectors)
  Asc = cross-sectional area of stud shank (mm²)
  f'c = concrete compressive strength (MPa)
  Ec  = concrete modulus of elasticity (MPa)
       = 4,500 × √(f'c) for normal-weight concrete (CSA A23.3)
  Fu  = specified minimum tensile strength of stud (MPa)
       = 415 MPa for ASTM A108 studs (standard)

The two-part formula ensures the stud resistance is governed either by the concrete bearing strength (the √(f'c × Ec) term) or by the stud tensile strength (the Asc × Fu term).

Stud Strength in Metal Deck

When the slab is cast on profiled steel deck (most Canadian composite floor construction), the stud strength is reduced based on the rib geometry and orientation:

qr = φsc × 0.50 × Asc × √(f'c × Ec) × (0.80 / √Nr) × (hr / hs - 0.30) ≤ reduction

where:
  Nr  = number of studs per rib (≤ 3)
  hr  = nominal rib height (mm)
  hs  = overall stud height (mm)

Additional reductions apply when:

• Ribs are parallel to the beam: multiply by 0.60 for any number of studs per rib
• Rib spacing > 600 mm: multiply by 0.85
• Deck thickness < 1.2 mm: special consideration required

Standard Stud Sizes

Values shown for f'c = 30 MPa solid slab, Ec = 24,650 MPa, Nr = 1 stud per rib.

Effective Flange Width (CSA S16 Clause 17.4)

The effective width of the concrete slab acting as the compression flange of the composite beam is:

be = min(L/4, s)

where:
  be = effective flange width (mm)
  L  = beam span (mm)
  s  = centre-to-centre spacing of beams (mm)

For edge beams (beam on one side only):

be = min(L/8, actual overhang)

For beams with metal deck ribs parallel to the beam, the effective width is further limited by the rib geometry. The effective flange width must also account for any haunch between the beam top flange and the slab bottom.

Degree of Partial Shear Connection (CSA S16 Clause 17.6)

Full composite action requires enough shear connectors to develop the full plastic moment of the composite section. Partial composite action uses fewer connectors, reducing the moment capacity but potentially achieving a more economical design.

Full Shear Connection

The number of studs required for full composite action:

Nf = C / qr

where:
  C = compressive force in concrete slab at full composite action
    = min(Asteel × Fy, 0.85 × f'c × be × tc)
  qr = factored shear resistance per stud (N)

Partial Shear Connection

The degree of partial shear connection η = N/Nf determines the reduced moment capacity:

Mrc = φ × [T × y + Σ(qr × y_i)] / η

where:
  η = N / Nf ≤ 1.0 (degree of shear connection)
  T = total tension force in steel section at full composite (N)
  y = lever arm between tension and compression resultants (mm)

CSA S16 Clause 17.6.3 requires minimum partial shear connection:

• η ≥ 0.40 for beams with uniformly distributed loads (span ≤ 12 m)
• η ≥ 0.50 for beams with concentrated loads or spans > 12 m
• η ≥ 0.75 for beams with unshored construction (steel alone supports fresh concrete)

Longitudinal Shear Reinforcement (CSA S16 Clause 17.8)

The horizontal shear force at the steel-concrete interface must be transferred to the slab in a manner that prevents longitudinal splitting. The longitudinal shear per unit length:

V_l = N_s × qr / L_shear

where:
  N_s    = number of studs between critical sections
  L_shear = distance between points of maximum and zero moment (mm)

The required longitudinal shear reinforcement:

Av_req = V_l × γ_v / (φ_s × Fy_r)

where:
  γ_v = 0.50 (shear plane factor per CSA S16 Cl. 17.8.2)
  φ_s = 0.85 (reinforcing steel resistance factor)
  Fy_r = yield strength of transverse reinforcement (MPa)

Transverse reinforcement (typically welded wire mesh or rebar) must be provided within the slab to resist the longitudinal shear forces. Minimum reinforcement: 0.002 × slab cross-sectional area per CSA A23.3.

Rib Orientation Effects for Metal Deck

Profiled steel deck ribs can be oriented either parallel or perpendicular to the beam span, significantly affecting the composite behaviour:

Ribs Perpendicular to Beam (Preferred)

• Studs are welded through the deck directly to the beam flange
• Stud strength reduction: moderate (0.80/√Nr factor)
• Headed studs work primarily in shear (ideal orientation)
• Deck ribs provide transverse stiffness
• Most common in Canadian composite floor construction

Ribs Parallel to Beam

• Studs are welded through individual deck sheets or through pre-punched holes
• Stud strength reduction: significant (0.60 multiplier)
• Studs may be offset from the beam centreline if deck ribs are narrow
• Requires generous stud height to ensure full slab embedment
• Avoided in Canadian practice unless required by framing geometry

Deck Rib Geometry — Typical Values

Worked Example — Composite Beam with Headed Studs

Problem: Design a composite beam for a Canadian office building. Span = 10.0 m, beam spacing = 3.0 m. Slab: 150 mm total depth, 75 mm concrete on 75 mm metal deck. Concrete: f'c = 30 MPa (normal weight, 1,850 kg/m³). Steel beam: W460x74, Grade 350W. Studs: 19 mm × 125 mm headed studs, ASTM A108. Deck ribs perpendicular to beam, 1 stud per rib. Construction loads: unshored.

Loads

Dead load (slab + deck): 0.15 × 24 = 3.60 kPa = 10.8 kN/m
Superimposed dead: 1.0 kPa (mechanical, ceiling, finishes) = 3.0 kN/m
Live load (office, NBCC Table 4.1.5.3): 2.4 kPa = 7.2 kN/m
Beam self-weight: 0.74 kN/m

Total unfactored: w = (10.8 + 3.0 + 0.74) + 7.2 = 21.74 kN/m
Factored (ULS): wf = 1.25 × 14.54 + 1.5 × 7.2 = 18.18 + 10.80 = 28.98 kN/m

Step 1 — Steel Section Alone (Construction Check)

Mf_construction = 1.25 × (10.8 + 0.74) × 10² / 8 = 1.25 × 11.54 × 100 / 8 = 180.3 kN·m

W460x74: Mr_steel = φ × Zx × Fy = 0.90 × 1,460 × 10³ × 350 = 459.9 kN·m > 180.3 kN·m OK

Step 2 — Composite Section Properties

Effective flange width:

be = min(L/4, s) = min(10,000/4, 3,000) = min(2,500, 3,000) = 2,500 mm

Step 3 — Full Composite Moment Capacity

Compressive force in concrete:

C = 0.85 × f'c × be × tc = 0.85 × 30 × 2,500 × 75 = 4,781,250 N = 4,781 kN

Tension force in steel:

T = Asteel × Fy = 9,480 × 350 = 3,318,000 N = 3,318 kN

Since T < C, the neutral axis is in the concrete slab (full composite):

a = T / (0.85 × f'c × be) = 3,318,000 / (0.85 × 30 × 2,500) = 52.1 mm

Lever arm = d/2 + tc + hr - a/2 = 460/2 + 75 + 75 - 52.1/2
         = 230 + 75 + 75 - 26.1 = 353.9 mm

Mrc_full = φ × T × lever arm = 0.90 × 3,318,000 × 353.9 = 1,057 × 10⁶ N·mm = 1,057 kN·m

Step 4 — Required Number of Studs

Stud strength in metal deck (75 mm rib, 1 stud per rib, perpendicular):

Asc = 284 mm² (19 mm stud)
Ec = 4,500 × √30 = 24,650 MPa

qr = 0.80 × 0.50 × 284 × √(30 × 24,650)
qr = 0.80 × 0.50 × 284 × √739,500
qr = 0.80 × 0.50 × 284 × 860
qr = 0.80 × 122,120
qr = 97,696 N = 97.7 kN

Check stud shank limit:
qr_max = 0.80 × 284 × 415 = 94,288 N = 94.3 kN

Since 97.7 kN > 94.3 kN, use qr = 94.3 kN per stud (stud tensile strength governs)

Number of studs for full composite action (per half-span):

Nf = T / qr = 3,318 / 94.3 = 35.2 → 36 studs per half-span
Total studs per span = 72

Step 5 — Partial Composite Action

For unshored construction, η_min = 0.75 per CSA S16 Clause 17.6.3. Try η = 0.80:

N = η × Nf = 0.80 × 36 = 28.8 → 28 studs per half-span
Total = 56 studs

C_actual = N × qr = 28 × 94,285 = 2,640,000 N

a = C_actual / (0.85 × f'c × be) = 2,640,000 / (0.85 × 30 × 2,500) = 41.4 mm
y = d/2 + tc + hr - a/2 = 230 + 75 + 75 - 20.7 = 359.3 mm

Mrc_partial = 0.90 × 2,640,000 × 359.3 = 854 × 10⁶ N·mm = 854 kN·m

Mf = 28.98 × 10² / 8 = 362.3 kN·m
D/C = 362.3 / 854 = 0.42 OK

Step 6 — Stud Layout

56 studs per span, placed in pairs at each rib (28 ribs × 2 studs per rib = 56):

Spacing = 10,000 / 28 = 357 mm → use 350 mm spacing

Check minimum spacing requirements:

Minimum longitudinal spacing: 6 × stud diameter = 6 × 19 = 114 mm → 350 mm OK
Maximum longitudinal spacing: 8 × slab thickness = 8 × 150 = 1,200 mm → 350 mm OK
Minimum transverse spacing: 4 × stud diameter = 4 × 19 = 76 mm → OK (pair at 100 mm typical)

Step 7 — Deflection Check

Composite moment of inertia (transformed section, cracked):

n = Es / Ec = 200,000 / 24,650 = 8.1

Transformed concrete area = be / n = 2,500 / 8.1 = 308.6 mm

Calculate neutral axis depth of transformed section:
ȳ = (Asteel × d/2 + Ac_transformed × (tc/2 + hr + d/2)) / (Asteel + Ac_transformed)

For the composite section, the cracked transformed moment of inertia:

I_comp ≈ 800 × 10⁶ mm⁴ (detailed calculation required for precise value)

Service load deflection (unfactored):
δ = 5 × w × L⁴ / (384 × E × I_comp)
δ = 5 × 21.74 × 10,000⁴ / (384 × 200,000 × 800 × 10⁶)
δ = 5 × 21.74 × 10¹⁶ / (384 × 200,000 × 800 × 10⁶)
δ = 1.087 × 10¹⁸ / 6.144 × 10¹³
δ = 17.7 mm

Span/deflection = 10,000 / 17.7 = 565 >> L/300 for total load OK
Live load deflection: 5 × 7.2 × 10⁴ / (384 × 200,000 × 800 × 10⁶) = 5.9 mm
L/300 = 33.3 mm → OK

Step 8 — Step 8 — Longitudinal Shear Reinforcement

Required transverse reinforcement per unit length (between critical section and mid-span):

V_l = 56 × 94,285 / 5,000 = 1,056 N/mm

Av_req = 1,056 × 0.50 / (0.85 × 400) = 1.55 mm²/mm

Provide 10M bars at 200 mm spacing (Av = 500 mm²/m = 0.50 mm²/mm) — Note: the calculated value is per unit of shear plane interface; typical detailing in Canadian practice uses welded wire mesh (WWF 152x152-MW18.7/MW18.7) per CSA A23.3 minimum requirements for the slab.

Step 9 — Summary

Frequently Asked Questions

What is the minimum degree of partial shear connection for composite beams? CSA S16 Clause 17.6.3 specifies minimum η values: η ≥ 0.40 for uniformly distributed loads on spans ≤ 12 m, η ≥ 0.50 for concentrated loads or spans > 12 m, and η ≥ 0.75 for unshored construction where the steel beam alone carries wet concrete. These minimums ensure adequate ductility and prevent sudden failure of the shear connection. For Canadian composite construction, η = 0.75 (from unshored construction) typically governs.

How does metal deck rib orientation affect shear stud strength? Ribs perpendicular to the beam are preferred and use the standard qr reduction factor (0.80/√Nr). Ribs parallel to the beam impose a further 0.60 multiplier on the stud strength, significantly reducing capacity. This is because studs in parallel ribs experience more complex stress states and less effective concrete confinement. Canadian practice strongly favours perpendicular deck orientation for composite beams.

What is longitudinal shear reinforcement and when is it required? Longitudinal shear reinforcement (CSA S16 Clause 17.8) prevents horizontal splitting of the concrete slab along the steel-concrete interface. It consists of transverse reinforcement across the shear plane, typically welded wire mesh or bent rebar. It is required in all composite beams with concentrated shear stud groups near supports. For uniformly loaded composite beams with studs evenly distributed, minimum reinforcement per CSA A23.3 (0.002 × slab area) typically suffices.

How do you calculate the effective width of a composite slab? Per CSA S16 Clause 17.4, the effective width is be = min(L/4, s) for interior beams — the smaller of one-quarter of the span or the centre-to-centre beam spacing. For edge beams, it is be = min(L/8, actual overhang). The effective width governs the compression capacity of the concrete slab and directly affects the composite moment resistance. Canadian design uses the plastic stress distribution method with the rectangular stress block (0.85 × f'c) across the effective width.

Related Pages

• Canada CSA S16 Guide — Full CSA S16:19 steel design reference
• Canadian Steel Beam Sizes — W, WWF, HSS sections per CISC
• Canadian Steel Grades — G40.21 300W to 480W
• CSA S16 Beam Design — Flexure, LTB, shear checks
• Load Combinations Calculator — Canadian load case calculator
• Beam Capacity Calculator — Multi-code beam capacity calculator
• Deflection Limits — Span/depth ratios per code

This page is for educational reference. CSA S16:19 composite beam design must comply with the current edition of CSA S16, CSA A23.3, and NBCC 2020. All results are PRELIMINARY — NOT FOR CONSTRUCTION without independent verification by a licensed Professional Engineer (P.Eng.).