How is shear stud capacity (Qn) calculated per AISC 360?

Shear stud nominal strength Qn per AISC 360-22 Equation I8-3 is the lesser of two limit states: (1) Concrete crushing/failure: Qn = 0.5 × Asc × √(f'c × Ec). (2) Steel stud fracture: Qn = Rg × Rp × Asc × Fu. Where: Asc = stud cross-sectional area (in²) = πd²/4. f'c = concrete compressive strength (psi). Ec = concrete modulus of elasticity = w_c^1.5 × √f'c (for normal weight: Ec = 145^1.5 × √f'c = 1,744 × √f'c ≈ 57,000√f'c psi). Fu = stud minimum tensile strength = 65 ksi per ASTM A108. Rg = geometry factor: 1.0 for studs in solid slab or with deck ribs perpendicular, 0.85 for deck ribs parallel with 2 studs per rib, 0.75 for deck ribs parallel with 1 stud per rib (per AISC Table I8-2). Rp = position factor: 0.75 for studs welded directly to flange, 0.60 for studs in weak position in deck rib, 1.0 for strong position. For 3/4 in stud (Asc = 0.442 in²), solid slab, normal weight concrete, no deck (Rg = Rp = 1.0): For f'c = 3,000 psi: concrete limit = 0.5 × 0.442 × √(3,000 × 3,320 × 10³) = 0.5 × 0.442 × √(9.96 × 10⁹) = 0.5 × 0.442 × 99,800 = 22.1 kips. Steel limit = 1.0 × 1.0 × 0.442 × 65 = 28.7 kips. Concrete controls → Qn = 22.1 kips nominal. For f'c = 4,000 psi: concrete limit ≈ 26.1 kips, steel limit = 28.7 kips → concrete controls. For f'c = 5,000 psi: concrete limit ≈ 29.3 kips, steel limit = 28.7 kips → steel stud controls. LRFD design strength: φQn = 0.65 × Qn = 14.4 kips per stud for f'c = 3,000 psi (steel limit with φ = 0.65). Note: the concrete limit uses Ec in psi × 10⁻⁶ in the square root term — careful unit handling is essential for correct Qn values.

How many shear studs are needed for full composite action?

The number of shear studs required for full composite action depends on the total horizontal shear force that must be transferred between the steel beam and concrete slab. Total horizontal shear: Vh = min(0.85 × f'c × Ac, Fy × As) where Ac = effective concrete area = be × tc (effective width × slab thickness above deck), and As = steel beam area. Number of studs per half-span: N = Vh / Qn (between point of zero moment and point of maximum moment). For a uniformly loaded simply-supported beam, studs are placed symmetrically: N per half-span from each support to midspan. Example: W21×44 (A = 13.0 in², Fy = 50 ksi) with 4 in slab (f'c = 4 ksi) on 2 in deck (tc = 2 in above deck), be = 90 in (L/8 for 30 ft span at 8 ft spacing). Ac = 90 × 2 = 180 in². Concrete limit: 0.85 × 4 × 180 = 612 kips. Steel limit: 50 × 13.0 = 650 kips. Vh = min(612, 650) = 612 kips. For 3/4 in studs, Qn ≈ 21.2 kips (f'c = 4 ksi), φQn = 13.8 kips. N = 612 / 13.8 = 44 studs per half-span = 88 total. Typical stud layout: pairs of studs at 12 in spacing = 30 pairs along the 30 ft beam. If fewer studs are provided, the composite action is partial: the composite moment capacity is reduced based on the actual shear transfer V'h = N_actual × Qn. Minimum composite action: per AISC I3.2c, the number of studs must provide at least 25% of full composite shear transfer. Below 25%, the beam is conservatively designed as non-composite. For typical floor beams at 8-10 ft spacing: expect 30-60 studs per half-span for full composite action.

How are shear studs through metal deck designed differently?

When shear studs are welded through corrugated metal deck (the standard composite floor construction method), the stud capacity Qn is reduced by geometry (Rg) and position (Rp) factors per AISC Table I8-2: Deck orientation and reduction: Strong position (stud in the 'strong' part of the deck rib, nearer the beam web) — Rg × Rp = 1.0 (no reduction). Weak position (stud in the 'weak' part, nearer the deck edge) — Rg × Rp = 0.75 for 1 stud per rib, 0.60 for 2+ studs per rib with deck perpendicular and e_mid-ht ≥ 2 in. Deck ribs perpendicular to beam: most common configuration. Studs welded through deck into beam flange. Rg = 1.0, Rp = 0.75 (weak position) or 1.0 (strong position). Net reduction factor = 0.75-1.0. Deck ribs parallel to beam: less common, requires wider beam flange. Rg = 0.85 for 2 studs per rib, 0.75 for 1 stud per rib. Deck geometry requirements per AISC I8.1: (1) maximum rib height hr ≤ 3 in for studs ≤ 3/4 in diameter, (2) average rib width wr ≥ 2 in (for 3/4 in studs), (3) stud height after welding H ≥ hr + 1.5 in (minimum embedment into slab above deck), (4) stud diameter ≤ 3/4 in through steel deck (unless Qn is verified by testing), (5) concrete cover: minimum 1/2 in of concrete above the top of the stud. Example: 3/4 in stud through 2 in deck (hr = 2 in), weak position, deck perpendicular: φQn = 0.65 × 1.0 × 0.75 × 0.442 × 65 = 14.0 kips nominal × 0.65 = 9.1 kips LRFD (vs. 14.4 kips for solid slab — 37% reduction). For the W21×44 example with through-deck studs: N = 612 / 9.1 = 67 studs per half-span (52% more studs than solid slab). This is why deck profile selection (rib geometry) significantly impacts stud count and total cost.

What are the spacing and layout requirements for shear studs?

Shear stud spacing and layout requirements per AISC 360 Section I8.2d and I8.2e: Along the beam (longitudinal): minimum spacing = 6 × stud diameter = 4.5 in for 3/4 in studs, 3.75 in for 5/8 in studs. This prevents interaction between adjacent stud welds during installation. Maximum spacing = 8 × slab thickness (total slab depth, not just above deck) or 36 in, whichever is smaller. For a 4 in slab: max spacing = 32 in (8 × 4 = 32 < 36). Transverse (across beam flange): minimum spacing = 4 × stud diameter center-to-center = 3.0 in for 3/4 in studs. This ensures adequate concrete between studs for shear transfer. Edge distance: stud center to edge of flange ≥ stud diameter + cover (typically 1.5-2 in from flange edge). Studs per deck rib: typically 1 or 2 studs per rib for deck perpendicular to beam. The rib width must accommodate the stud(s) with adequate clearance. For 2 studs per rib: the rib must be ≥ 4 in wide for 3/4 in studs. Stud pattern: uniform spacing is typical — constant spacing along the beam length. This is conservative because shear demand varies (maximum at supports, zero at midspan for uniform load), but the simplicity justifies the extra studs. For optimization, stud spacing can taper from closer at supports to wider at midspan. Minimum embedment: total stud height after welding must protrude at least 1.5 in above the top of the deck ribs into the concrete slab. Common stud lengths: 3 in (for 1.5 in deck + 1.5 in above), 4 in (for 2 in deck + 2 in above), 5 in (for 3 in deck + 2 in above). The stud must also have minimum 1/2 in concrete cover above its top. Detailing: specify stud diameter, length after weld, spacing, and number per rib on the structural drawings. The fabricator welds studs in the shop using a stud welding gun — this is NOT a field operation for typical building construction.

How much extra capacity does composite action provide compared to non-composite?

Composite action (steel beam + concrete slab acting together via shear studs) significantly increases both strength and stiffness compared to the steel beam alone: Strength increase: for a W21×44 beam (Mn_steel = φ × Zx × Fy = 0.90 × 84.0 × 50/12 = 315 kip-ft): Full composite (4 in slab, f'c = 4 ksi, 90 in effective width): the plastic neutral axis typically lies in the slab, giving a compression block depth a = Fy×As/(0.85×f'c×be) = 50×13.0/(0.85×4×90) = 2.12 in, with moment arm = d/2 + tc − a/2 = 21.06/2 + 2 − 2.12/2 = 10.53 + 2 − 1.06 = 11.47 in. Mn_composite = 0.85 × Fy × As × moment arm = 0.85 × 50 × 13.0 × 11.47/12 = 527 kip-ft. Ratio = 527/315 = 1.67 — approximately 67% more moment capacity than non-composite. Stiffness increase: the effective moment of inertia for composite section I_eff ≈ 1.5-3.0 × I_steel, depending on the degree of composite action. This reduces deflection by 30-60% for the same load, often allowing shallower beams for the same span. Practical benefits: (1) shallower beams (W18 composite vs W24 non-composite for same span), (2) longer spans (40+ ft composite vs 30 ft non-composite for same depth), (3) reduced floor-to-floor height saving on cladding, stairs, and MEP costs, (4) vibration performance improves with composite action due to higher stiffness. The capacity gain is highest when the plastic neutral axis falls in the slab (full composite action). If the PNA is in the steel beam, the gain is smaller (typically 30-50%). For partial composite action (fewer studs than full composite): the capacity increase is roughly proportional to √(N_actual/N_full) — 50% of studs gives ~70% of the full composite capacity increase. Minimum 25% composite action required per AISC I3.2c.

Steel Shear Stud Design — Composite Beam Capacity

Q: How much extra capacity does composite action provide compared to non-composite?

Composite action (steel beam + concrete slab acting together via shear studs) significantly increases both strength and stiffness compared to the steel beam alone: Strength increase: for a W21×44 beam (Mn_steel = φ × Zx × Fy = 0.90 × 84.0 × 50/12 = 315 kip-ft): Full composite (4 in slab, f'c = 4 ksi, 90 in effective width): the plastic neutral axis typically lies in the slab, giving a compression block depth a = Fy×As/(0.85×f'c×be) = 50×13.0/(0.85×4×90) = 2.12 in, with moment arm = d/2 + tc − a/2 = 21.06/2 + 2 − 2.12/2 = 10.53 + 2 − 1.06 = 11.47 in. Mn_composite = 0.85 × Fy × As × moment arm = 0.85 × 50 × 13.0 × 11.47/12 = 527 kip-ft. Ratio = 527/315 = 1.67 — approximately 67% more moment capacity than non-composite. Stiffness increase: the effective moment of inertia for composite section I_eff ≈ 1.5-3.0 × I_steel, depending on the degree of composite action. This reduces deflection by 30-60% for the same load, often allowing shallower beams for the same span. Practical benefits: (1) shallower beams (W18 composite vs W24 non-composite for same span), (2) longer spans (40+ ft composite vs 30 ft non-composite for same depth), (3) reduced floor-to-floor height saving on cladding, stairs, and MEP costs, (4) vibration performance improves with composite action due to higher stiffness. The capacity gain is highest when the plastic neutral axis falls in the slab (full composite action). If the PNA is in the steel beam, the gain is smaller (typically 30-50%). For partial composite action (fewer studs than full composite): the capacity increase is roughly proportional to √(N_actual/N_full) — 50% of studs gives ~70% of the full composite capacity increase. Minimum 25% composite action required per AISC I3.2c.

Shear studs (headed steel anchors) transfer horizontal shear between the steel beam and concrete slab in composite construction. Without adequate studs, the steel and concrete act independently. With properly designed studs, they act as a single composite section with significantly greater strength and stiffness. This page covers shear stud design per AISC 360-22 Chapter I.

How Shear Studs Work

In a composite beam:

The concrete slab is in compression (top)
The steel beam is in tension (bottom) and partially in compression
Horizontal shear at the steel-concrete interface must be transferred by the studs
Each stud acts as a small shear connector, resisting the horizontal shear force

Without studs, the beam capacity is that of the steel alone. With full composite action, the capacity can increase by 50-100%.

Shear Stud Specifications

Per ASTM A108 and AWS D1.1:

Property	1/2 in Stud	5/8 in Stud	3/4 in Stud	7/8 in Stud
Diameter (in)	0.500	0.625	0.750	0.875
Cross-section area (in²)	0.196	0.307	0.442	0.601
Head diameter (in)	1.0	1.25	1.25	1.50
Head thickness (in)	5/16	5/16	3/8	3/8
Min length after weld	2.0	2.5	3.0	3.5
Common length (in)	3, 4	3, 4, 5	3, 4, 5, 6	4, 5, 6
Fy (ksi)	51	51	51	51
Fu (ksi)	65	65	65	65

The 3/4 inch diameter stud is by far the most commonly used in building construction.

Stud Nominal Strength

Per AISC Equation I3-3:

Qn = min(0.5 × Asc × √(f'c × Ec), Rc × Asc × Fu)

where:

Asc = stud cross-sectional area (in²)
f'c = concrete compressive strength (psi)
Ec = modulus of elasticity of concrete (psi)
Fu = stud minimum tensile strength (65 ksi)
Rc = reduction factor (1.0 for solid slab, reduced for metal deck)

Simplified Stud Capacity (3/4 in, Solid Slab)

f'c (psi)	Ec (ksi)	Qn per stud (kips)	Controlling Mode
3,000	3,320	17.7	Concrete
3,500	3,590	19.5	Concrete
4,000	3,830	21.2	Concrete
4,500	4,070	22.8	Concrete
5,000	4,290	23.5	Steel stud

Note: 3/4 in stud, Asc = 0.442 in², Fu = 65 ksi. Steel limit = 0.5 × 0.442 × 65 = 14.4 kips (with Ω=2.0 for ASD) or Qn = 23.5 kips (nominal, for LRFD: φQn = 17.6 kips).

Stud Capacity Through Metal Deck

When studs are welded through corrugated metal deck, the capacity is reduced:

Deck Orientation	Weak Position	Strong Position
Strong (ribs perpendicular)	0.70 × Qn	1.0 × Qn
Weak (ribs parallel)	0.90 × Qn	1.0 × Qn

Additional reductions apply when the rib height exceeds the stud length or when the deck flute is narrow. See AISC Table I3-2 for the complete reduction factors.

Number of Studs Required

Full Composite Action

The total horizontal shear force to be transferred equals the lesser of:

Vh = min(0.85 × f'c × Ac, Fy × As)

where Ac = concrete area within effective width, As = steel beam area.

Number of studs for full composite: N = Vh / Qn (per shear span)

Partial Composite Action

When fewer studs are provided, the composite section has reduced capacity:

V'h = N × Qn

The composite moment capacity is calculated based on the actual shear transfer V'h.

Minimum degree of composite action: AISC requires the composite moment capacity to exceed the non-composite moment capacity by at least some amount. In practice, a minimum of 25-50% composite action is typical.

Stud Layout

Spacing Requirements

Requirement	Limit
Minimum spacing (along beam)	6 × stud diameter (4.5 in for 3/4 in stud)
Maximum spacing	8 × slab thickness, or 36 in
Minimum transverse spacing	4 × stud diameter (3 in for 3/4 in stud)
Maximum transverse spacing	Not specified (practical: within slab effective width)
Studs per rib (metal deck)	1 or 2 per rib typical
Edge distance (stud to slab edge)	Not less than stud head diameter

Typical Layout Patterns

For a W21x44 beam spanning 30 ft with 3/4 in studs:

Composite Level	Studs per half-span	Total studs	Approx. capacity increase
Non-composite	0	0	0%
25%	8	16	~25%
50%	15	30	~50%
75%	23	46	~75%
100%	30	60	~85-100%

Effective Concrete Width

Per AISC Section I3.1a:

be = min(b/2, L/8) on each side of beam centerline

where b = beam spacing, L = beam span.

Beam Spacing (ft)	Span (ft)	Effective Width (in)
4	20	48 (controlled by b/2)
5	25	60 (controlled by b/2)
6	30	72 (controlled by b/2)
8	30	90 (controlled by L/8)
10	30	90 (controlled by L/8)

Frequently Asked Questions

What size shear studs are used in composite beams? 3/4 inch diameter is the most common. 5/8 inch is used for lighter beams and deck profiles. 7/8 inch is used for heavy bridge girders. Studs are typically 3-5 inches long.

How many shear studs do I need? For full composite action, the number is determined by dividing the total horizontal shear (Vh) by the capacity per stud (Qn). For a W21x44 spanning 30 ft with 4 ksi concrete, approximately 30 studs per half-span (60 total) are needed for full composite action.

Can I have too many studs? Yes. Beyond full composite action, additional studs do not increase capacity. They add cost and weld time without benefit. The number should be limited to that required for full composite action, or a specified partial composite level.

Do I need studs on both sides of the beam? Studs are typically placed on both sides of the beam top flange, but this is for symmetry and constructability, not a code requirement. Single-row studs are acceptable if the capacity works out.

How are studs installed? Studs are welded to the top flange using a stud welding gun (arc welding with a ferrule ceramic). The process takes 1-2 seconds per stud. Studs can be welded through metal deck (with proper ferrules and procedure).

Composite Beam Design — Composite beam analysis
Beam Capacity Calculator — Flexural and shear checks
Steel Deck Types — Corrugated deck profiles
Beam Sizes — W-shape section properties
Steel Floor Systems — Framing and deck selection

Disclaimer

This is a calculation tool, not a substitute for professional engineering certification. All results must be independently verified by a licensed Professional Engineer (PE) or Structural Engineer (SE) before use in construction, fabrication, or permit documents. The user is responsible for the accuracy of all inputs and the verification of all outputs.