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Composite Beams

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Why Structural Steel?

Steel beams acting compositely with concrete slabs, shear studs (Chapter I).

⏱ 110 min★ Advanced✎ 3 objectives
§01Section 01

Engineering story

Engineering story
Chapter 11 · Composite Beams

Steel beams acting compositely with concrete slabs, shear studs (Chapter I).

A real project narrative for this chapter will be authored as this chapter migrates to the v3.0 structured schema.

§02Section 02

Learning objectives

After this chapter you will be able to
  • Compute PNA in a composite beam
  • Size shear studs per AISC I8
  • Check pre-composite construction
§03Section 03

Engineering motivation

§04Section 04

Failure mechanisms

Failure mechanisms & lessons learned

Photographs and lessons-learned case studies for this topic will be added during chapter migration.

AISC Reference Box
  • AISC 360-22 — Specification chapter governing this topic
  • AISC Manual 16th Ed. — Design tables and worked examples

Why This Chapter Matters

Composite action between a steel beam and a concrete slab can double the flexural capacity of the bare beam. Shear studs, plastic neutral axis location, and effective slab width all change the design.

Learning Objectives

  • Determine effective slab width beff per AISC I3.1a.
  • Locate the plastic neutral axis (PNA) — in slab, top flange, or web.
  • Compute Mn for fully composite and partially composite sections.
  • Design shear-stud connectors per AISC I8: Qn, spacing, edge distance, number.
  • Check construction (non-composite) strength.

Where This Chapter Is Used

Every floor beam in a modern office building with a concrete-on-metal-deck slab — used in the Chapter 20 capstone floor system.

ANSI / AISC 360-22Specification for Structural Steel Buildings16.1-126 to 16.1-159
Chapter
I
AISC 360-22

Chapter I. Design of Composite Members

Use this reference to flip directly to the correct page of the AISC 360-22 Specification while solving problems in this course chapter.

§Section titlePage
I1General Provisions16.1-126
I2Axial Force (Composite Columns — Encased/Filled)16.1-128
I3Flexure — Steel Beam + Concrete Slab (Effective Width, PNA, Mn)16.1-138
I4Combined Flexure and Axial Force16.1-149
I6Load Transfer16.1-152
I8Steel Anchors in Composite Beams (Shear Studs Qn)16.1-154

Companion reference: AISC Manual Part 3 — Composite Beam Design Tables

Lecture Notes

Chapter 11 — Composite Beams (AISC 360-22 Chapter I)

Chapter focus. A steel beam supporting a concrete slab is often made composite: shear studs weld the two together so the slab takes compression and the steel takes tension. That typically doubles the effective moment capacity for the same steel weight. AISC Chapter I covers the plastic-neutral-axis analysis, stud strength, and deflection with the composite Ieff.

1. Composite Action

Shear studs welded to the top flange transfer horizontal shear from a concrete slab to the steel beam so the two act as one composite section — much stiffer and stronger than the bare beam.

Composite beam — plastic stress distribution (full composite) Concrete slab (0.85 f'c compression block) W-shape (Fy tension) PNA
Full composite action: concrete resists compression, steel resists tension.

2. Effective Slab Width (§I3.1a)

beff = min( L/8, s/2 interior, sedge + L/8 )

3. Horizontal Shear V'

V' = min( 0.85·f'c·Ac, Fy·As, ΣQn )
  • Full composite: ΣQn ≥ min(concrete, steel).
  • Partial composite: 0.25·min ≤ ΣQn < min; interpolate φMn.

4. Stud Strength (§I8.2a)

Eq. I8-1Qn = 0.5·Asa·√(f'c·Ec) ≤ Rg·Rp·Asa·Fu

Rg, Rp depend on deck orientation and stud position (Table I8.1). Typical 3/4 in stud on 3 in VLI20 deck ⊥ beam: Qn ≈ 17.2 k.

5. Plastic Neutral Axis

Compare C = 0.85·f'c·beff·a to T = Fy·As:

  • C ≥ T → PNA in slab; a = T/(0.85·f'c·beff).
  • C < T → PNA in steel section; iterate flange/web.
Mn = C·(d/2 + tslab − a/2) (PNA in slab)

Additional Design Aids & Stratified Equations

Composite floor — stud layout on top flange Concrete slab (beff) Steel beam top flange · 3/4″ studs @ 6″ o.c. Vmax → studs from support to Mmax
Uniform stud spacing on the top flange between support and maximum-moment point.
Number of studs each side of M_maxN = V' / Q_n
Concrete modulus (NWC)E_c = w_c^{1.5} · 33 · √f'_c (psi) → 4-ksi NWC gives ≈3,644 ksi

⚠ Common mistakes

  • Using slab thickness above ribs instead of full slab depth for Ac.
  • Forgetting RgRp for deck ⊥ beam.
  • Assuming full composite when ΣQn < controlling min.
  • Ignoring construction-stage bare-beam stresses.

📖 Using the AISC Manual — Composite Beams

Companion reference: Manual Part 3 — Table 3-19 (composite W-shapes with concrete slab) & 3-20 (lower-bound φMn) & 3-21 (shear-stud strength Qn). In practice, engineers rarely compute every quantity from first principles — the AISC Steel Construction Manual (16th Ed.) tabulates φRn (or Rn/Ω) for every rolled shape so you can pick a member in seconds. Formulas remain essential for understanding, verifying, and for anything the tables do not cover.

TableGives youHow to read
3-19φMn for each W + slab combo at 7 degrees of composite action (25%→100%)Enter with beam size and Y2 = distance from top of steel to concrete compression block centroid. Read φMn at the level of composite action closest to your required moment.
3-20Lower-bound φMn (partial composite)Same axis; use when connector count is limited by deck geometry.
3-21Nominal stud strength Qn vs. f'c and stud sizeRead Qn → number of studs per half-span = ΣQn/Qn.

Typical workflow: pick required φMn → enter Table 3-19 at Y2≈5.5″ (4.5″ slab − 1″ compression) → find lightest W meeting φMn at 100% composite → from Table 3-21 count studs. Done.

Table vs. formula — which to use? Use tables to select a shape quickly. Use formulas to verify odd geometry, non-standard grades (Fy≠50 ksi), unusual K-factors or Lb, and to answer exam problems that hand you a section not in the current Manual.
Project case study — Cardinal Square — 4-story braced-frame office

Every chapter's worked example is one step in the design of the same building: Plan: 4 bays N–S × 3 bays E–W, each 30 ft × 30 ft. Stories: 4 @ 13 ft (52 ft roof). Composite floor: 4.5 in NW concrete on 3 VLI20 deck. Roof: 1.5 in B-deck + insulation + membrane. Materials: Wide-flange members A992 (Fy = 50 ksi, Fu = 65 ksi). Plates A572 Gr. 50. HSS bracing A500 Gr. C. Bolts A325-N 7/8 in dia. Welds E70XX. Concrete f'c = 4 ksi. Anchor rods F1554 Gr. 36.

Chapter 11 — Composite filler beam
Filler beam from Chapter 6 made composite with the 4.5-in slab
Demand carried forward
Same Mu = 191 k-ft, but now with composite action.
This chapter contributes
Sizes shear studs (3/4 in dia, 4.5 in long) to develop full composite action. Re-uses the AISC Manual Table 3-19 lookup with ΣQn.
Feeds into next chapter
Stud layout becomes a quantity on the structural drawings.
Concrete slab (t = 4.5")Composite beam w/ shear studs
Composite beam: concrete slab + W-shape connected by headed shear studs (AISC Ch. I).

Formula Sheet

NameEquationAISC Ref
Design strengthφ Rn ≥ RuAISC 360-22 B3.1

Worked Example

Worked Example 11.1 — Composite W18×35 Filler Beam

Given

  • W18×35 filler beam, L = 30 ft, spacing s = 10 ft, A992 (Fy = 50 ksi).
  • As = 10.3 in², d = 17.7 in.
  • 4.5 in NW concrete on 3 in deck ⊥ beam; f'c = 4 ksi; tc = 4.5 in above deck ribs.
  • Mu = 191 k-ft (from Ch 6).
Composite W18×35 — slab + deck + studs, PNA in slab shear studs Q_n b_eff = min(L/8, s) t_c = 4.5″ deck 3″ d = 17.7″ C = 0.85 f'_c · b_eff · a T = A_s F_y = 515 k a = min(A_s F_y, 0.85 f'_c b_eff t_c) / (0.85 f'_c b_eff) ; M_n = T·(d/2 + t_c + t_deck − a/2)
Figure 11.1a — Composite section: slab, deck ribs, shear studs, stress block

Step 1 — Effective Width

§I3.1abeff = min(L/8, s/2·2) = min(L/8, s)
beff = min(L/8, s) = min(45 in, 120 in) = 45 in

Step 2 — Horizontal Shear V'

§I3.2d(1)V' = min(As·Fy, 0.85 f'c·beff·tc)
Concrete: 0.85·4·(45·4.5) = 688 k
Steel: 50·10.3 = 515 k
V' = min = 515 k (steel governs)

Step 3 — Stud Count

FormulaNhalf-span = V' / Qn ;  Ntotal = 2·Nhalf-span
Qn = 17.2 k/stud → half-span N = 515/17.2 = 30 → total 60 studs = one per 6 in.

Step 4 — Depth of Stress Block

Formulaa = V' / (0.85 f'c · beff); PNA in slab if a ≤ tc
a = 515 / (0.85·4·45) = 3.37 in < 4.5 → PNA in slab ✓

Step 5 — φMn

FormulaMn = V'·(d/2 + tc + tdeck − a/2);  φMn = 0.90 Mn
Lever arm y = d/2 + tc + tdeck − a/2 = 8.85 + 4.5 + 3 − 1.685 = 14.67 in
Mn = 515·14.67 / 12 = 629 k-ft
φMn = 0.90·629 = 566 k-ft ≥ 191 ✓
Composite W18×35 utilization = 0.34 (vs 0.75 bare). Live-load deflection (Ch 18) usually governs at this ratio.

Additional Worked Examples

Textbook — Aghayere & Vigil (2009)

Chapter 7 covers composite beams (Chapter I of AISC 360-22). The steel beam acts compositely with the concrete slab through shear studs welded through the deck. Full composite action uses ΣQn ≥ min(0.85 fc'·Ac, Fy·As).

Example 7-1 — Effective slab width

Example 7-1 — Effective slab width

Setup. Interior W18×35 beam, span L = 30 ft, beam spacing 10 ft, slab cover 6 in.

AISC Reference: AISC §I3.1.1a

Numerical practice

be?

  1. A. 60 in
  2. B. 75 in (Answer)
  3. C. 90 in
  4. D. 120 in

Step-by-step solution

be = min(L/8·2, s) = min(2·(30·12)/8, 10·12) = min(90, 120) = 90 in (governs by L/4 effective limit).

Example 7-2 — Shear stud strength

Example 7-2 — Shear stud strength

Setup. 3/4" diameter shank (Asc = 0.442 in²), fc' = 4 ksi, NWC (wc = 145 pcf), single stud per rib transverse to deck (Rg = 1.0, Rp = 0.75).

AISC Reference: AISC §I8.2

Numerical practice

Qn per stud?

  1. A. 12 k
  2. B. 17 k (Answer)
  3. C. 21 k
  4. D. 26 k

Step-by-step solution

Ec = 145^1.5·√4 = 3492·2 = 3492 ksi; 0.5·0.442·√(4·3492) = 26.1 k; cap: 1.0·0.75·0.442·65 = 21.5 k; lower → 21.5 k. (Some tables limit further to 17 k for typical deck.)

Example 7-3 — Full composite stud count

Example 7-3 — Full composite stud count

Setup. W18×35, As = 10.3 in², Fy = 50 ksi; slab strength 0.85·fc'·be·t = 0.85·4·90·4 = 1224 k.

AISC Reference: AISC §I3.2d

Numerical practice

Number of 3/4" studs (Qn = 17.5 k) required on each side of max moment for full composite?

  1. A. 12
  2. B. 16
  3. C. 20
  4. D. 30 (Answer)

Step-by-step solution

ΣQn = min(1224, 50·10.3 = 515) = 515 k. n = 515/17.5 = 29.4 → 30 studs each side of Mmax.

Example 7-6 — Composite Mn

Example 7-6 — Composite Mn

Setup. Same W18×35 fully composite with 4-in NWC slab on 3-in metal deck; ΣQn = 515 k.

AISC Reference: AISC Manual Table 3-19

Numerical practice

Read φMn from Manual Table 3-19 (closest)?

  1. A. 340 k·ft
  2. B. 420 k·ft
  3. C. 510 k·ft (Answer)
  4. D. 590 k·ft

Step-by-step solution

Table 3-19 for W18×35 with ΣQn = 515 k and Y2 ≈ 5 in gives φMn ≈ 510 k·ft (≈ 70% gain over the bare-steel φMp ≈ 290 k·ft).

FE-Style Worked Examples(6)

Each example mirrors the NCEES FE Civil Reference Handbook style: brief givens, a labeled figure, AISC section reference, step-by-step numeric solution, and a single boxed answer.

Given
Interior beam, span L=30 ft, beam spacing s=10 ft.
AISC Reference
AISC §I3.1a
Step-by-step solution
  1. be
    min(L/8 each side, s/2 each side) summed = min(30·12/8, 10·12/2) = min(45 in, 60 in)·2 = 90 in
Answer be = 90 in (7.5 ft).
Effective slab width
Problem statement image
Concrete slab (t = 4.5")Composite beam w/ shear studs
DIMDimensions from the problem statement
L = 30fts = 10ft
Composite beam cross-section
  • Concrete slab t_c above a wide-flange girder
  • Effective width b_eff = min(L/4, s, …) per AISC §I3.1a
  • Shear studs distribute horizontal shear along the flange
  • PNA location determines the plastic moment M_p

Graded Chapter Quiz(13 FE-style questions · AISC Manual required)

These questions reference AISC Steel Construction Manual (16th ed.) — sections, equations, and tables are cited explicitly. Use a calculator. Each question offers a clue you may reveal before answering. Submissions are recorded to your account once signed in.

C11-01AISC 360-22 §I3.1a
1. Interior composite beam W18×35, L = 30 ft span, beams @ 10 ft o.c., 4.5-in NW slab. Compute the effective flange width beff per §I3.1a.
Composite beam section 4.5 in slab, f′c shear studs A992 W-shape
C11-02AISC 360-22 §I3.2d
2. Same beam. Compute the maximum tensile force in the steel Tmax = As·Fy.
Composite beam section 4.5 in slab, f′c shear studs A992 W-shape
C11-03AISC 360-22 §I3.2d
3. Same beam. Maximum concrete compressive force Cmax = 0.85·f'c·beff·tslab (t = 4.5 in above deck).
Composite beam section 4.5 in slab, f′c shear studs A992 W-shape
C11-04AISC 360-22 §I3.2d
4. Since Tmax (515) < Cmax (1,530), the beam is 'fully composite' with T = 515 k. Compute the concrete stress-block depth a.
Composite beam section 4.5 in slab, f′c shear studs A992 W-shape
C11-05AISC 360-22 §I3.2d
5. Compute the nominal moment Mn about the PNA and then φbMn (fully composite, W18×35 d = 17.7, tslab = 5, a = 1.68).
Composite beam section 4.5 in slab, f′c shear studs A992 W-shape
C11-06AISC Manual Table 3-21
6. AISC Manual Table 3-21 gives Qn for a ¾-in Ø headed stud in 4-ksi NW concrete ≈:
C11-07AISC 360-22 §I8
7. Number of studs required BETWEEN Mmax and each support to develop full composite action (T = 515 k, Qn = 21.5).
Composite beam section 4.5 in slab, f′c shear studs A992 W-shape
C11-08AISC 360-22 §I8.2c
8. Deck ribs run PARALLEL to the beam (typical girder). Per §I8.2c the reduction factor Rp for stud strength is:
C11-09AISC 360-22 §I3.2d(2)
9. Partial composite action per §I3.2d(2) is allowed as long as:
C11-10AISC 360-22 Comm. I3
10. During construction (before concrete cures), the beam must support its own weight + wet concrete acting on the STEEL SECTION ONLY. For deflection check use:
C11-11AISC 360-22 §I8.2c
11. Formed deck PERPENDICULAR to beam, 3-in deck, 1 stud per rib: Rp per §I8.2c:
C11-12AISC Manual Part 3
12. Recommended camber (AISC Manual Part 3):
C11-13AISC 360-22 §I4
13. In a composite beam, web shear is designed using:
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§11Section 11

Chapter summary

Formula sheet
  • Design strength
    φ Rn ≥ Ru
    AISC 360-22 B3.1
Engineering checklist
  • Module 11: Composite Beams
  • Key limit states and AISC references are listed in the reference box.
  • Use φRn ≥ Ru for every check.
  • Verify section properties with the official AISC Manual.
Professional tips
  • • Mixing ASD and LRFD load combinations in the same problem.
  • • Using nominal strength Rn instead of design strength φRn.
  • • Forgetting to check every limit state listed in the AISC chapter.
§13Section 13

FE exam preparation

FE exam preparation
Concept review
Concept summary coming soon.
Calculator tips

Calculator tips coming soon.

Common exam traps

Traps coming soon.

Time management

Aim for ~3 minutes per FE problem; skip and return to any item that takes longer than 5 minutes.