AISC 360-22 Composite Beam Design — W18x35 + 4.5 in. Slab Full Worked Example

Complete step-by-step composite beam design following AISC 360-22 Chapter I provisions (LRFD). This worked example covers a W18x35 steel beam acting compositely with a 4.5 in. normal-weight concrete slab on metal deck: effective width determination (Section I3.1a), composite section properties, plastic moment capacity of the composite section (Section I3.2), shear stud design (Section I8.2), and serviceability deflection checks. Every intermediate calculation is shown with actual numbers and code clause references.

Problem Statement

PRELIMINARY — NOT FOR CONSTRUCTION. All results presented here are for educational and reference use only. Values must be independently verified by a licensed Professional Engineer (PE) or Structural Engineer (SE) before use in any design or construction.

Design a composite floor beam for a commercial office building. The beam is simply supported with a 30 ft span, spaced at 10 ft on center. The floor consists of 4.5 in. normal-weight concrete (f'c = 4 ksi, unit weight = 145 pcf) on 2 in. 20-gage composite metal deck with ribs perpendicular to the beam. The steel beam is ASTM A992 W18x35. The beam is unshored during construction.

Design parameters:

• Steel beam: W18x35, ASTM A992 (Fy = 50 ksi, Fu = 65 ksi)
• Concrete slab: Normal weight, f'c = 4 ksi, t = 4.5 in.
• Metal deck: 2 in. deep, 20-gage, ribs perpendicular to the beam
• Span: 30 ft, simply supported
• Beam spacing: 10 ft center-to-center
• Construction: Unshored (steel beam alone carries wet concrete)

Loads:

• Dead load (superimposed): 25 psf (flooring, MEP, ceiling)
• Live load: 50 psf (office occupancy per ASCE 7-22 Table 4.3-1)
• Construction dead load: Concrete slab weight = (4.5/12) × 145 = 54.4 psf + deck = 3 psf = 57.4 psf

Section Properties (W18x35)

Step 1: Effective Slab Width (AISC I3.1a)

The effective width of the concrete slab on each side of the beam centerline is the minimum of three limits:

Limit 1 — One-eighth of the beam span:

be1 = span / 8 = (30 ft × 12 in./ft) / 8 = 360 / 8 = 45 in. per side

Limit 2 — One-half the distance to the adjacent beam web:

be2 = beam_spacing / 2 = (10 ft × 12) / 2 = 120 / 2 = 60 in. per side

Limit 3 — Distance to the slab edge (not applicable for interior beams):

be3 = infinity (interior beam, no slab edge within the floor)

Effective width per side: be_side = min(45, 60, ∞) = 45 in.

Total effective width:

be = 2 × 45 = 90 in. = 7.5 ft

The effective width of 90 in. means the concrete slab contributing to composite action extends 45 in. on each side of the beam centerline. For a 10 ft beam spacing with beams on both sides, the full 5 ft of slab on each side is available; however, the AISC span/8 rule caps the contribution at 45 in. per side.

Step 2: Composite Section Classification

For the plastic stress distribution to apply (AISC I3.2a), the web of the steel section must be compact per AISC Table B4.1b:

h/tw = 17.7 / 0.300 = 59.0 (approximate web height)
lambda_pw = 3.76 × sqrt(E/Fy) = 3.76 × sqrt(29,000/50) = 90.6

h/tw = 59.0 < 90.6: Web is compact. Plastic stress distribution on the composite section is permitted.

Additionally, the ratio of concrete area to steel area per AISC I1.2 must be within the tested range. For beams with metal deck ribs perpendicular to the beam, the concrete below the top of the deck rib is neglected per AISC I3.2c.

Step 3: Factored Loads — Strength Limit State (LRFD)

Superimposed loads (post-construction, composite section resists):

w_superimposed = (25 + 50) psf × 10 ft trib = 750 plf = 0.750 klf

w_u = 1.2 × 25 × 10 + 1.6 × 50 × 10 = 1.2 × 250 + 1.6 × 500 = 300 + 800 = 1,100 plf = 1.100 klf

Maximum factored moment (simply supported, uniform load):

M_u = w_u × L^2 / 8 = 1.100 × 30^2 / 8 = 1.100 × 900 / 8 = 123.75 kip-ft

Step 4: Plastic Moment Capacity of Composite Section (AISC I3.2)

The plastic stress distribution determines the nominal flexural strength Mn. The analysis involves finding the plastic neutral axis (PNA) location and computing the internal force couple.

Step 4a: Forces in the Composite Section

Force in the concrete slab at full plasticity (AISC I3.2d, Compression):

Cc = 0.85 × f'c × (be × t_slab)
   = 0.85 × 4 ksi × (90 in. × 4.5 in.)
   = 0.85 × 4 × 405
   = 1,377 kips

Force in the steel section at full yield:

Ts = Fy × As = 50 ksi × 10.3 in^2 = 515 kips

Since Cc (1,377 kips) is greater than Ts (515 kips), the plastic neutral axis (PNA) lies within the concrete slab. The concrete compression block depth is:

a = Ts / (0.85 × f'c × be)
  = 515 / (0.85 × 4 × 90)
  = 515 / 306
  = 1.683 in.

The PNA is 1.68 in. below the top of the slab. This is within the 4.5 in. slab thickness, confirming PNA in slab.

Step 4b: Lever Arm and Nominal Moment

For PNA in the slab, the internal force couple is between the full tensile yield force in the steel beam (Ts) acting at mid-depth of the steel section, and the concrete compression force (Cc = Ts) acting at mid-depth of the compression block.

Lever arm from the steel centroid to the centroid of the concrete compression block:

d1 = d/2 + t_slab - a/2
   = 17.7/2 + 4.5 - 1.683/2
   = 8.85 + 4.5 - 0.842
   = 12.51 in.

Nominal plastic moment of the composite section:

Mn = Ts × d1 = 515 kips × 12.51 in. = 6,443 kip-in. = 536.9 kip-ft

Step 4c: Design Flexural Strength (phi = 0.90 per AISC I3.2):

phi_b × Mn = 0.90 × 536.9 = 483.2 kip-ft

Check: Mu = 123.75 kip-ft < phi_b*Mn = 483.2 kip-ft. Composite flexure OK. Utilization = 123.75/483.2 = 0.256.

Step 5: Shear Stud Design (AISC I8.2)

Shear connectors transfer horizontal shear between the concrete slab and the steel beam. For full composite action, the number of studs between the point of maximum moment and zero moment must resist the entire horizontal shear force.

Step 5a: Horizontal Shear Force

The horizontal shear force to be transferred per AISC I8.2c is the smaller of:

V'_concrete = 0.85 × f'c × Ac = 0.85 × 4 × 405 = 1,377 kips
V'_steel = Fy × As = 50 × 10.3 = 515 kips

V' = min(1,377, 515) = 515 kips

Step 5b: Shear Strength of One Stud (AISC I8.2a)

Use 3/4 in. diameter × 3-1/2 in. long headed stud anchors. The stud is welded through the metal deck (ribs perpendicular to the beam).

Stud cross-sectional area:

Asc = pi × d_stud^2 / 4 = pi × 0.75^2 / 4 = 0.442 in^2

Concrete modulus of elasticity (ACI 318-19 Eq. 19.2.2.1.b for normal-weight concrete):

Ec = wc^1.5 × 33 × sqrt(f'c) = 145^1.5 × 33 × sqrt(4,000) = 1,746 × 33 × 63.25 = 3,644 ksi

Nominal strength of one stud (AISC Equation I8-1):

Qn1 = 0.5 × Asc × sqrt(f'c × Ec)
    = 0.5 × 0.442 × sqrt(4 × 3,644)
    = 0.5 × 0.442 × sqrt(14,576)
    = 0.5 × 0.442 × 120.7
    = 26.7 kips

Upper limit (AISC Equation I8-3, with Rg = 1.0 and Rp = 1.0 for no deck profile reduction for the base case):

Qn2 = Rg × Rp × Asc × Fu_stud
    = 1.0 × 1.0 × 0.442 × 65
    = 28.7 kips

For ribs perpendicular: Per AISC I8.2a, when the deck ribs are perpendicular to the steel beam, a reduction factor applies:

Rg = 1.0 (one stud per rib)
Rp = 0.6 × (wr/hr) × [(Hs/hr) - 1.0] <= 1.0

For 20-gage deck: wr = 6 in. (average rib width), hr = 2 in. (rib height), Hs = 3.5 in. (stud length after welding, typically 3 in. above deck):

Rp = 0.6 × (6/2) × [(3.5/2) - 1.0] = 0.6 × 3 × [1.75 - 1.0] = 0.6 × 3 × 0.75 = 1.35

Rp is capped at 1.0. So Rp = 1.0 for this configuration.

Qn = min(26.7, 28.7) = 26.7 kips per stud

Step 5c: Number of Studs Required

For a simply supported beam with uniform load, the region between the maximum moment (midspan) and zero moment (support) is half the span. The required number of studs for that half-span is:

N_req_half = V' / (Qn) = 515 / 26.7 = 19.3 studs

For the full beam, total studs required:

N_total = 2 × 20 = 40 studs (round up to even number)

With 40 studs distributed over the full span (30 ft), spacing = 360 in. / (40/2 - 1) roughly 18 in. spacing, which is practical. Minimum spacing per AISC I8.2d is 6 times the stud diameter = 6 × 0.75 = 4.5 in.

Check: Use 40 × 3/4 in. diameter × 3-1/2 in. headed studs in pairs at approximately 18 in. spacing. Shear stud design OK.

Step 6: Deflection Check — Construction Stage (Steel Beam Alone)

During construction, the unshored steel beam alone carries the wet concrete and deck weight.

Construction load:

w_construction = 57.4 psf × 10 ft = 574 plf = 0.574 klf (unfactored construction load)

Deflection of steel beam alone under wet concrete:

delta_c = 5 × w × L^4 / (384 × E × Is)
        = 5 × (574/12) × (360)^4 / (384 × 29,000 × 510)
        = 5 × 47.83 × 1.6796e10 / (384 × 29,000 × 510)
        = 4.016e12 / 5.679e9
        = 0.707 in.

A common camber of 75% of the dead load deflection = 0.75 × 0.707 = 0.53 in. (specify 1/2 in. camber for practical fabrication). Camber eliminates the visual sag and provides a level floor after concrete placement.

Step 7: Deflection Check — Composite Stage (Superimposed Loads)

Post-construction, the composite section resists superimposed dead and live loads. The composite moment of inertia is significantly larger than the steel section alone.

Effective moment of inertia of composite section (AISC Commentary I3.1):

The transformed section method is used. The modular ratio n = Es/Ec = 29,000/3,644 = 7.96. The effective concrete width transformed to steel is be/n = 90/7.96 = 11.3 in.

For preliminary design, the lower-bound moment of inertia for a partially composite beam can be estimated. For full composite action with PNA in the slab, the effective moment of inertia is approximately:

I_eff ≈ Is + (0.5 × Itr)

where Itr is the moment of inertia of the transformed composite section including the cracked concrete contribution. For a W18x35 with 4.5 in. slab and 90 in. effective width:

I_lower_bound = Is + (As × d^2) / (4 × n) ≈ 510 + (10.3 × 17.7^2) / (4 × 7.96)
              ≈ 510 + (10.3 × 313.3) / 31.84
              ≈ 510 + 3,227 / 31.84
              ≈ 510 + 101
              ≈ 611 in^4

For a more precise estimate, the transformed cracked section moment of inertia is approximately:

I_tr ≈ 1,250 in^4 (conservative for hand calculation)

Effective moment of inertia for deflection (AISC Commentary Eq. C-I3-1):

I_eff = Is + (I_tr - Is) × (sum Qn / Cf)^0.5

For full composite action: sum Qn / Cf = 1.0, so I_eff = Is + (I_tr - Is) = I_tr.

Using I_tr ≈ 1,250 in^4:

Live load deflection:

delta_LL = 5 × (50 × 10 / 12) × 360^4 / (384 × 29,000 × 1,250)
         = 5 × 41.67 × 1.6796e10 / (384 × 29,000 × 1,250)
         = 3.499e12 / 1.392e10
         = 0.251 in.

Allowable live load deflection for floor beams: L/360 = 360/360 = 1.0 in.

delta_LL = 0.251 in. < 1.0 in. Live load deflection OK. Utilization = 0.251.

Step 8: Summary of Design Checks

All checks pass. The W18x35 composite beam with 4.5 in. slab and 40 shear studs provides adequate strength and stiffness for the 30 ft span at 10 ft spacing. Composite action increases the moment capacity by over 300% compared to the non-composite W18x35 (Mn_non-composite = Mp = 277 kip-ft).

AISC 360-22 Chapter I — Composite Member Design Overview

AISC 360-22 Chapter I covers the design of composite steel-concrete members. The chapter is organized as:

• I1 — General Provisions: Scope, definitions, and references. Composite members must have interaction between the steel and concrete elements designed to transfer forces through direct bearing, shear connectors, or bond.
• I2 — Axial Members: Design of composite columns including concrete-filled HSS and steel-encased concrete sections.
• I3 — Flexural Members: Design of composite beams with the concrete slab in compression. Plastic stress distribution (I3.2a) or elastic strain compatibility may be used.
• I4 — Combined Flexure and Axial Force: Beam-column interaction for composite members.
• I5 — Load Transfer: Force transfer between steel and concrete elements at connections.
• I6 — Detailing Requirements: Minimum reinforcement, cover, and shear connector placement.
• I8 — Shear Connectors: Strength, placement, and detailing of headed stud anchors and channel connectors.

Composite Beam Optimization Principles

• Partial vs. Full Composite Action: Full composite action requires enough shear studs to develop the full plastic capacity of the section. Partial composite action (fewer studs) may be permitted per I3.2d when V' is not fully developed, but the moment capacity is interpolated between the steel-only strength and the full composite strength. Most designs target 50-100% composite action.
• Shored vs. Unshored Construction: Shored construction eliminates construction-stage steel beam stresses but requires temporary support systems. Unshored construction is more economical for typical multi-story buildings; the steel beam must be checked for construction loads alone.
• Camber for Long Spans: For spans over 25 ft, cambering the steel beam (typically 75-80% of dead load deflection) provides a level floor after concrete placement. Specify camber on the fabrication drawings.
• Deck Orientation: Metal deck ribs perpendicular to the beam reduce the concrete area available for compression (concrete below the top of the rib is neglected per I3.2c). Deck ribs parallel to the beam are more efficient for composite action because the full slab thickness contributes.

Frequently Asked Questions

How is the effective width of a composite beam slab determined?

Per AISC 360-22 Section I3.1a, the effective width of the concrete slab on each side of the beam centerline is the minimum of: one-eighth of the beam span, one-half the distance to the adjacent beam web, and the distance to the slab edge. The total effective width is twice this value plus the beam flange width (or the full slab width for edge beams).

How many shear studs are required for full composite action?

Per AISC 360-22 Section I8.2c, the number of shear connectors required between the point of maximum moment and zero moment shall equal the horizontal shear force V' divided by the nominal shear strength of one connector Qn. V' is the smaller of 0.85f'cAc and FyAs. The connector strength Qn per AISC I8.2a is Qn = 0.5 × Asc × sqrt(f'c × Ec) ≤ Rg × Rp × Asc × Fu for headed stud anchors.

What is the difference between shored and unshored composite construction?

In shored construction, the steel beam is temporarily supported during concrete placement and curing. The steel beam carries no dead load from the wet concrete; once shores are removed, the composite section resists all loads. In unshored construction, the steel beam alone carries the wet concrete dead load, and only superimposed dead and live loads are resisted by the composite section. Unshored construction is more common in practice but requires checking the steel beam alone for construction loads.

How do metal deck ribs affect shear stud capacity?

When deck ribs are perpendicular to the steel beam, a reduction factor Rp is applied to the stud strength per AISC I8.2a. Rp = 0.6 × (wr/hr) × [(Hs/hr) − 1.0] ≤ 1.0, where wr is the average rib width, hr is the nominal rib height, and Hs is the stud length after welding. For ribs parallel to the beam, a reduction factor based on the rib width-to-depth ratio applies.

Does composite action eliminate the need for camber?

No. Camber is required to compensate for the dead load deflection of the steel beam alone during construction (unshored) and for the long-term creep deflection of the composite section. Typically, 75-80% of the pre-composite dead load deflection is compensated by camber. The remaining deflection and live load deflection are accommodated in the composite stage.

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Related References

• AISC Beam Design Example — W24x55
• Beam Formulas Reference
• Steel Section Properties Reference
• Shear Lag Factor Reference
• AISC Steel Construction Tables
• How to Verify Calculations
• Steel Floor Systems Reference
• Steel Sustainability Reference

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This page is for educational and reference use only. It does not constitute professional engineering advice. All design values must be verified against the applicable standard and project specification before use. The site operator disclaims liability for any loss arising from the use of this information.