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LecturesModule 11

Composite Beams

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

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

Lecture Notes

This module introduces composite beams. Lecture content here covers the governing physics, LRFD philosophy, and how the relevant AISC 360-22 chapter organizes the limit states.

Instructors can replace this text in Admin Mode. Each section is structured around: (1) behavior, (2) failure modes, (3) AISC limit-state equations, (4) design workflow, (5) detailing requirements.

A short comparison to ASD is included only where the resistance factor / safety factor relationship clarifies the LRFD design check.

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

Composite Beams

Given
Replace with project-specific given data (loads, geometry, material).
Load combination
Controlling LRFD load combination from ASCE 7.
Required strength
Compute required strength Ru from the controlling combination.
Limit states
  • Limit state 1
  • Limit state 2
AISC reference
AISC 360-22 — applicable chapter
Solution steps
  1. 1. Required strength
    Compute Ru.
  2. 2. Trial section
    Pick a trial from AISC shape tables Instructor should verify with official AISC Manual.
  3. 3. Check each limit state
    Apply φ Rn ≥ Ru for every governing limit state.
  4. 4. Iterate
    Resize until the most economical section satisfies all checks.
Final design decision
Select the lightest section that satisfies all LRFD limit states.
Common mistakes in this example
  • Skipping a limit state
  • Using the wrong φ factor
  • Forgetting serviceability checks

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).
Concrete slab (t = 4.5")Composite beam w/ shear studs

Textbook — Aghayere & Vigil (2009)(4 worked examples with figures + numerical answers)

Worked examples scanned directly from the CEGR 436 course textbook. Each card shows the original page (figure + full step-by-step solution) and adds an FE-style numerical multiple-choice prompt with answer key.

Chapter summary

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).

  • Effective slab width be ≤ min(L/8, sclear/2, edge dist) on each side (AISC §I3.1).
  • Stud strength: Qn = 0.5·Asc·√(fc'·Ec) ≤ Rg·Rp·Asc·Fu (AISC §I8).
  • Full composite: ΣQn = min(0.85·fc'·be·t, As·Fy).
  • Partial composite OK if ΣQn ≥ 25% of full; reduces stud count.
  • Construction stage: check non-composite beam for wet-concrete + construction live load.
  • Deflection: use lower-bound or effective Ieff per Manual Part 3 commentary.
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?
Textbook page 299 — Effective slab width
Aghayere & Vigil (2009), p. 299 — full worked solution & sketch.

Practice Problems

  1. [E] State AISC Chapter I governs composite construction.
  2. [E] Define effective slab width be per §I3.1a.
  3. [E] State φb = 0.90 for composite flexure.
  4. [E] State shear-stud diameter range (1/2 to 7/8 in.).
  5. [E] Define full vs partial composite action.
  6. [M] Effective slab width for W21x50 at 9 ft o.c. on 30 ft span.
  7. [M] Locate PNA for fully composite W21x50, 4.5 in. slab, f'c = 4 ksi.
  8. [M] Number of 3/4 in. studs (Qn = 21.5 k) for full composite action of W18x35 (As = 10.3 in²).
  9. [M] φMn for fully composite W24x55, 4 in. slab, f'c = 4 ksi, 30 ft, 10 ft o.c.
  10. [M] Check construction-stage deflection of W18x35 alone before slab cures.
  11. [H] Design partially composite W21x50 with ΣQn = 0.5·As·fy; quantify weight savings.
  12. [H] Concrete-encased composite column W12x72 in 24 x 24 in. concrete, f'c = 5 ksi. Compute Po per §I2.1b.
  13. [H] Long-term + creep deflection: composite vs non-composite, 30 ft, wL = 1.0 k/ft.
  14. [H] Stud distribution between zero- and max-moment points of a continuous composite beam.
  15. [H] Composite beam with web opening 12 x 24 in. at quarter points — check per AISC DG-2.
Structured Clues
  • Effective slab width be = min(L/4, c-c spacing, edge distance).
  • Locate PNA in slab if ΣAs·fy ≤ 0.85·f'c·be·tslab; otherwise inside the steel.
  • Full composite requires ΣQn ≥ min(As·fy, 0.85·f'c·be·tslab).
Code References
  • AISC 360-22 Ch. I
  • AISC Manual Part 3, Tables 3-19 to 3-21

Quiz

1. Which AISC 360-22 chapter primarily governs composite beams?
2. In LRFD, the basic design inequality is:

Common Student Mistakes

  • 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.

"Professor Explains" Script

Today we're talking about composite beams. Think of this topic as one step in the LRFD workflow: identify the demand, identify the limit states from the relevant AISC chapter, then check that φ·Rn is at least equal to Ru. We'll walk through the failure modes, the equations, and a worked example. Pay close attention to where the resistance factor changes — that's where students lose points on exams.