Composite Beam Calculator

Composite steel-concrete beam design per AISC 360. Stud shear connector layout, effective slab width, partial composite ratio, and deflection. Educational use only.

This page documents the scope, inputs, outputs, and computational approach of the Composite Beam Calculator on steelcalculator.app. The interactive calculator runs in your browser; this documentation ensures the page is useful even without JavaScript.

What this tool is for

What this tool is not for

Key concepts this page covers

Inputs and outputs

Typical inputs: steel beam section, span and spacing, concrete slab thickness, deck profile (rib height, width, spacing), concrete strength f'c, stud diameter, and applied dead and live loads.

Typical outputs: full and partial composite moment capacity, required number of studs, stud spacing, effective slab width, lower-bound moment of inertia, live load deflection, and demand-to-capacity ratio.

Computation approach

The calculator determines the effective slab width per AISC I3.1a, computes the stud capacity Qn per AISC I8.2a (including deck rib reduction factors per I8.2c), then determines the plastic neutral axis location by force equilibrium for the desired composite ratio. The nominal moment capacity is computed from the plastic stress distribution. Deflection is checked using the lower-bound moment of inertia that accounts for slip in partial composite beams.

AISC 360 Chapter I — Key Formulas

Effective slab width (AISC I3.1a)

be = 2 × min(L/8, s/2, distance to edge)

For interior beams:
  be per side = min(L/8, s/2)

Where:
  L = beam span
  s = spacing to adjacent beam

Shear stud capacity (AISC I8.2a)

Qn = 0.5 × Asc × √(f'c × Ec) ≤ 0.5 × Asc × Fu

Where:
  Asc = cross-sectional area of stud shank (in²)
  f'c = concrete compressive strength (ksi)
  Ec = concrete modulus of elasticity = w^1.5 × 33 × √f'c (ksi, w in pcf)
  Fu = stud tensile strength (ksi, typically 65 ksi)

For 3/4" studs: Asc = 0.442 in²
For 5/8" studs: Asc = 0.307 in²

Deck rib reduction factors (AISC I8.2c)

Qn_reduced = Qn × Rg × Rp

Rg (group factor):
  1 stud per rib: Rg = 1.0
  2 studs per rib: Rg = 0.85
  3+ studs per rib: Rg = 0.70

Rp (position factor, perpendicular deck):
  Strong position, wr/hr ≥ 1.5: Rp = 1.0
  Strong position, wr/hr < 1.5: Rp = 0.85
  Weak position: Rp = 0.60

Where wr = average rib width, hr = rib height

Plastic neutral axis (PNA) — force equilibrium

C = min(0.85 × f'c × be × ts, ΣQn, As × Fy)

If a = C / (0.85 × f'c × be) ≤ ts:
  PNA is in the slab
  φMn = 0.90 × C × (d/2 + ts - a/2)

If a > ts:
  PNA is in the steel beam
  Locate PNA by equating tension and compression forces in the steel section
  φMn = 0.90 × Σ(Ci × di) or Σ(Ti × di) about the PNA

Lower-bound moment of inertia (AISC Commentary I3.2)

ILB = Is + √(ΣQn / C) × (Icomp - Is)

Where:
  Is = moment of inertia of bare steel section
  Icomp = moment of inertia of fully composite section
  ΣQn = total shear stud capacity
  C = full composite force = min(0.85f'c × Ac, As × Fy)
  ΣQn/C = composite ratio

For deflection under service loads: Δ = 5wL⁴ / (384 × E × ILB)

Worked Example — W16x26 Composite Beam

Problem: A W16x26 (A992) spans 25 ft at 6 ft on center. The floor slab is 1.5" composite deck with 3.5" of lightweight concrete (110 pcf, f'c = 4 ksi). Service superimposed dead = 20 psf, live load = 80 psf. Determine studs and check capacity.

Step 1 — Effective width and concrete properties

be per side = min(25×12/8, 6×12/2) = min(37.5, 36) = 36 in
be = 72 in total

Slab total thickness: ts = 1.5 + 3.5 = 5.0 in above top of beam flange
Ec = (110)^1.5 × 33 × √4 = 1,151 × 33 × 2 = 2,630 ksi (LW concrete)

Step 2 — Stud capacity

3/4" studs (Asc = 0.442 in²)
Qn = 0.5 × 0.442 × √(4 × 2630) = 0.221 × 102.6 = 22.7 kips
Check: 0.5 × 0.442 × 65 = 14.4 kips → Qn = 14.4 kips (stud tensile governs)

With rib reduction (perpendicular deck, 1 stud per rib, wr/hr ≈ 1.5):
Rg = 1.0, Rp = 1.0 → Qn = 14.4 kips per stud

Step 3 — Full composite stud count

W16x26: As = 7.68 in², d = 15.69 in

C = 0.85 × 4 × 72 × 5.0 = 1,224 kips (concrete)
T = 7.68 × 50 = 384 kips (steel) → GOVERNS

N_full = 384 / 14.4 = 26.7 → 27 studs per half beam (54 total)

Step 4 — 50% partial composite (more economical)

ΣQn_50% = 0.50 × T = 0.50 × 384 = 192 kips
N_50% = 192 / 14.4 = 13.3 → 14 studs per half beam (28 total)

a = ΣQn / (0.85 × f'c × be) = 192 / (0.85 × 4 × 72) = 0.78 in < 5.0 in
PNA in slab ✓

φMn = 0.90 × 192 × (15.69/2 + 5.0 - 0.78/2)
    = 0.90 × 192 × (7.85 + 5.0 - 0.39)
    = 0.90 × 192 × 12.46
    = 2,153 kip-in = 179 kip-ft

Step 5 — Demand vs capacity

Slab weight: 5/12 × 110 = 45.8 psf + deck: 1.2 psf + beam: 26/6 = 4.3 psf
Total dead: 45.8 + 1.2 + 4.3 + 20 = 71.3 psf
Live: 80 psf

wu = 1.2 × 71.3/1000 × 6 + 1.6 × 80/1000 × 6 = 0.513 + 0.768 = 1.281 kip/ft
Mu = 1.281 × 25² / 8 = 100.1 kip-ft

φMn / Mu = 179 / 100.1 = 1.79 → OK ✓

Step 6 — Deflection at 50% composite

ILB = Is + √(0.50) × (Icomp - Is)
Is = 301 in⁴ (W16x26)
Icomp ≈ 1,100 in⁴ (estimated)

ILB = 301 + 0.707 × (1,100 - 301) = 301 + 565 = 866 in⁴

Live load deflection:
Δ = 5 × (80/1000 × 6) × 25⁴ × 1728 / (384 × 29000 × 866)
Δ = 5 × 0.480 × 390,625 × 1728 / 9,616,896,000
Δ = 0.34 in

L/360 = 300/360 = 0.83 in → 0.34 < 0.83 ✓

The 50% partial composite design with 28 studs provides adequate strength and stiffness.

Composite Beam Selection Guide

Span (ft) Spacing (ft) Recommended Section Composite Ratio Approx. phi-Mn (kip-ft)
20 5 W12x19 25-50% 90-120
25 6 W16x26 50% 170-200
30 5 W18x35 50-75% 280-380
30 8 W21x44 50-75% 400-520
35 6 W21x44 75% 450-550
40 6 W24x55 75-100% 620-780
45 6 W27x84 100% 1,000+

Values are approximate for A992 steel with 4 ksi NW concrete slab. Use the calculator for exact values.

Worked Example — Stud Layout with Perpendicular Metal Deck Ribs

Problem: A W21x44 (A992) spans 30 ft at 6 ft on center. The floor uses 2-inch composite metal deck (e.g., Vulcraft 2VLI) with 3.5 inches of normal-weight concrete cover above the ribs (total slab thickness = 5.5 inches, f'c = 4 ksi). The deck ribs run perpendicular to the beam. Rib dimensions: wr (average rib width) = 3.75 in, hr (rib height) = 2.0 in, rib spacing = 12 in. Design a stud layout for 50% partial composite action using 3/4-inch studs.

Step 1 — Determine stud capacity with rib reduction

3/4" studs: Asc = 0.442 in²
Ec = 145^1.5 × 33 × √4 = 1,746 × 33 × 2 = 115,236 × 2 ≈ 4,031 ksi (NW concrete)

Qn_base = 0.5 × 0.442 × √(4 × 4031) = 0.221 × 127.0 = 28.1 kips
Check upper bound: 0.5 × 0.442 × 65 = 14.4 kips → stud tensile governs
Qn_base = 14.4 kips

Rib reduction factors (perpendicular deck):
  wr/hr = 3.75/2.0 = 1.875 ≥ 1.5 → strong position

  1 stud per rib: Rg = 1.0, Rp = 1.0
  2 studs per rib: Rg = 0.85, Rp = 1.0

Qn (1 per rib) = 14.4 × 1.0 × 1.0 = 14.4 kips
Qn (2 per rib) = 14.4 × 0.85 × 1.0 = 12.2 kips per stud

Step 2 — Required number of studs for 50% composite

W21x44: As = 13.0 in², Fy = 50 ksi
T = 13.0 × 50 = 650 kips

50% composite: ΣQn = 0.50 × 650 = 325 kips

Option A — 1 stud per rib (strong position):
  N = 325 / 14.4 = 22.6 → 23 studs per half beam, 46 total
  Beam length = 30 ft = 360 in
  Rib spacing = 12 in → 30 ribs along the span
  Available ribs per half beam = 15 ribs (at 12 in on center)
  Need 23 studs but only 15 ribs → cannot fit with 1 per rib

Option B — 2 studs per rib:
  N = 325 / 12.2 = 26.6 → 27 studs per half beam
  Available ribs per half beam = 15
  2 studs × 15 ribs = 30 studs per half beam > 27 needed ✓

Select: 2 studs per rib at 12 in on center = 30 studs per half (60 total)
Actual ΣQn = 60 × 12.2 = 732 kips > 325 kips
Actual composite ratio = 732 / 650 = 1.13 → full composite achieved

For economy, reduce to achieve closer to 50%:
  Use 2 studs in every other rib = 15 studs per half (30 total)
  ΣQn = 30 × 12.2 = 366 kips
  Composite ratio = 366 / 650 = 0.56 → acceptable ✓

Step 3 — Verify stud placement rules

AISC I8.2c placement requirements:
  Minimum stud spacing along beam = 6 × stud diameter = 6 × 0.75 = 4.5 in
  Our spacing = 24 in (every other rib at 12 in) → OK ✓

  Minimum transverse spacing (2 studs in one rib):
    = 4 × stud diameter = 4 × 0.75 = 3.0 in
  Place studs at 3.5 in apart transversely within the rib → OK ✓

  Stud height above deck:
    Minimum = 1.5 in above top of rib after welding
    Use 5-inch studs: height above rib = 5 - 2 (deck) - 0.125 (burnoff) = 2.875 in > 1.5 in ✓

  Stud diameter ≤ 2.5 × flange thickness:
    W21x44 flange tf = 0.525 in → 2.5 × 0.525 = 1.31 in > 0.75 in ✓

Step 4 — Capacity check at 56% composite

be per side = min(30×12/8, 6×12/2) = min(45, 36) = 36 in → be = 72 in

C = ΣQn = 366 kips
a = 366 / (0.85 × 4 × 72) = 1.49 in < 5.5 in → PNA in slab ✓

φMn = 0.90 × 366 × (20.7/2 + 5.5 - 1.49/2)
    = 0.90 × 366 × (10.35 + 5.5 - 0.75)
    = 0.90 × 366 × 15.10
    = 4,971 kip-in = 414 kip-ft

This stud layout (2 studs in every other rib, 30 total) provides 56% composite action with 414 kip-ft capacity.

Frequently Asked Questions

How do deck ribs affect shear stud capacity? When studs are placed inside the ribs of a metal deck, the confined concrete volume limits the stud capacity. AISC applies reduction factors Rp and Rg that depend on the number of studs per rib, the rib width-to-height ratio, and whether the ribs run perpendicular or parallel to the beam. These reductions can decrease individual stud capacity by 25-50% compared to studs in a flat soffit slab.

What is the minimum composite ratio allowed by AISC? AISC 360 allows a minimum of 25% composite action (the sum of stud capacities is at least 25% of the smaller of AsFy or 0.85f'cAc). Below 25%, the connection is too flexible to reliably transfer horizontal shear, and the deflection prediction becomes unreliable. Most practical designs use 50-75% composite action, which provides good economy while keeping deflections within limits.

How does composite action reduce beam weight? A composite beam uses the concrete slab as a compression flange, which dramatically increases the effective section modulus compared to the bare steel section alone. This means a lighter steel section can carry the same load. Typical weight savings are 20-30% compared to non-composite design, which offsets the cost of the shear studs and makes composite construction the standard approach for steel floor framing.

What is the construction sequence for composite floor beams? The typical sequence is: (1) erect steel beam, (2) place metal deck and connect to beam flange, (3) weld shear studs through the deck, (4) place reinforcing mesh, (5) pour concrete slab, (6) remove temporary shoring if shored. During steps 1-5, the bare steel beam carries all loads including wet concrete. After step 5, the beam acts compositely for all subsequent loads. The unshored beam must be checked for construction-stage deflection (limit L/180 or 3/4 inch).

How do I select the right stud diameter for composite beams? AISC permits 1/2", 5/8", 3/4", and 7/8" diameter headed studs. The 3/4" diameter is the most common because it balances capacity with weldability through metal deck. For beams lighter than W16, 5/8" studs may be preferred to avoid web crippling during welding. For heavy beams (W24 and above) with thick slabs, 7/8" studs reduce the total number needed. The stud length must extend at least 1-1/2" above the deck rib top after welding, with a minimum total length of 3× the diameter.

Can composite beams be used for roof framing? Yes, but it is less common than for floor framing because roof live loads are typically lower and the weight savings are smaller. Composite roof beams are used for long-span applications (40+ ft) where deflection controls the design. In this case, the composite section reduces live-load deflection significantly, allowing a lighter steel section. The deck is typically a deeper profile (2" or 3") with lightweight concrete fill for fire rating.

How does construction sequence loading affect composite beam design? During construction, the bare steel beam must support the wet concrete, metal deck, and construction loads before the slab gains strength and composite action develops. AISC Specification I3.2b requires the steel beam alone to resist all construction-stage loads (beam self-weight, deck, wet concrete, and a minimum construction live load of 20 psf per ASCE 37). The unshored beam is checked for both strength (using the bare steel section modulus) and deflection (typically limited to L/180 or 3/4 inch, whichever is less, to avoid ponding effects where wet concrete pools in the deflected shape). If the bare steel beam is inadequate, the designer has three options: (1) use a larger steel section, (2) use temporary shoring to reduce the span during construction, or (3) limit the placement area so that only a portion of the span is loaded at once. Ponding is particularly important for long-span beams where the additional concrete volume from deflection can increase the dead load by 10-15%.

What is the difference between shored and unshored composite design? In shored construction, temporary props support the beam during the concrete pour, meaning the bare steel beam carries no load before the slab cures and composite action is achieved. After shoring is removed, the composite section resists all dead and live loads. In unshored construction (far more common in practice), the bare steel beam resists its own weight, the deck, and the wet concrete. After curing, the composite section handles superimposed dead and live loads. Both approaches yield the same ultimate composite moment capacity because the plastic stress distribution depends only on the final force equilibrium, not the load history. However, the service stress distribution differs: shored beams have higher stresses in the concrete slab (which may crack earlier) and lower steel stresses, while unshored beams have higher steel stresses from the construction dead load locked into the bare steel section. Deflection calculations must account for the construction sequence in unshored design — the construction dead load deflection is computed using the bare steel moment of inertia, while subsequent live load deflection uses the composite lower-bound moment of inertia.

Can large openings be placed in the web of a composite beam? Yes, but the design requires careful analysis. Web openings in composite beams are commonly used to route mechanical ducts, plumbing, and electrical conduit through the floor framing, reducing the overall building floor-to-floor height. Openings can be rectangular or circular and are typically located in the web of the steel beam between the top and bottom flanges. For partial composite beams, the reduced web area affects the shear capacity and the moment capacity at the opening location. The design must consider: (1) Vierendeel bending at the corners of rectangular openings, where the top and bottom steel tees resist local moments from the transfer of shear across the opening, (2) global moment capacity reduction at the opening section, since the steel web contribution to the plastic neutral axis calculation is interrupted, (3) the concrete slab above the opening provides additional shear capacity through its depth, acting as a compression strut that helps transfer forces across the opening, and (4) local buckling of the steel tee sections above and below the opening. Reinforcing plates or stiffeners are often welded around large openings to restore capacity. As a rule of thumb, unreinforced rectangular openings should not exceed 60-70% of the beam depth in length or 60% of the beam depth in height.

How do I design continuous composite beams? Continuous composite beams develop negative moment (tension in the slab) at interior supports, which changes the design approach significantly compared to simply-supported spans. At interior supports, the concrete slab is cracked under negative bending, so only the steel beam and the longitudinal reinforcing steel within the effective slab width contribute to the composite section. This means the negative moment capacity depends on the amount of slab reinforcement, which is typically 0.25-0.50% of the concrete cross-sectional area. The positive moment regions (midspan) behave similarly to simple span composite beams with the slab in compression. AISC 360 permits moment redistribution for continuous composite beams: up to 20% redistribution from the elastic negative moment is allowed when the negative moment connection has sufficient rotation capacity. Continuous composite beams are more economical than simple spans because they reduce the positive moment demand, allowing lighter steel sections. However, the design must check: (1) negative moment capacity at supports, (2) lateral-torsional buckling of the bottom flange at supports (the bottom flange is in compression under negative moment), (3) stud layout must provide enough connectors in both positive and negative moment regions, and (4) column web stiffeners may be required at the interior joints to transfer the concentrated flange forces.

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