11

Composite Beams

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 bare rolled beam and a concrete slab sitting on top of it are two separate, weak members if nothing connects them. Weld a row of headed shear studs through the deck into the top flange and something remarkable happens: the slab (which is great in compression, useless in tension) and the steel (great in tension) start acting as one section. The same W-shape can carry roughly double the moment, or a much lighter beam can carry the same moment — for the cost of a few dollars of studs. This chapter is entirely about quantifying that gain: how much shear the studs must transfer, where the plastic neutral axis (PNA) lands, what happens if you provide fewer studs than "full" composite requires, and why the bare steel beam still has to survive the pour by itself.

1. Composite Action — Slip vs No-Slip

Picture two planks stacked with no fasteners, spanning between two supports, bent by a load. The bottom of the top plank stretches less than the top of the bottom plank — they slide relative to each other at the interface. This is non-composite behavior: each layer bends about its own centroid, and the combined stiffness is only the sum of the individual I's, EI1 + EI2. Now bolt the planks together tightly along their length. The interface can no longer slip; the two layers are forced to have the same strain at the interface, and the combined section behaves as a single deep member with a much larger transformed moment of inertia. That's composite action, and the "bolts" in a steel-concrete floor are the shear studs.

Non-composite (slip) vs Composite (no-slip) beam behavior No studs → slab and beam slide independently (slip) Each element bends about its OWN centroid — I₁ + I₂ only Studs weld slab to beam → single plane strain Section acts as ONE transformed shape — Itr ≫ I₁+I₂ Interface shear flow q = VQ/I must be carried by studs to prevent slip Slip profile — no studs (relative end-slip at supports) Zero slip along length — fully composite
Figure 11.1 — Non-composite slip interface (left) vs fully composite no-slip interface (right).

The horizontal shear flow at the interface is q = VQ/Itr (elastic) — the studs must supply at least this shear per unit length everywhere along the beam, or in ultimate-strength (LRFD) terms, must supply a total horizontal force ΣQn at least equal to the smaller of the concrete or steel capacity between the point of zero moment and the point of maximum moment.

2. Effective Slab Width beff (§I3.1a)

Not all of a wide slab actually helps resist bending — stress fades away from the beam centerline due to shear lag. AISC caps the width on each side of the beam centerline that may be counted as effective.

§I3.1abeff (each side) = min( Lspan/8, sbeam/2 )
  • Lspan — beam span, center-to-center of supports.
  • sbeam — beam spacing (distance to the centerline of the adjacent parallel beam).
  • At an edge beam, use the distance to the actual slab edge instead of s/2 on that side.
  • Total effective width beff = sum of the two individual side widths (each independently limited).

What this means physically: a long-span, closely-spaced beam grid gets almost the full tributary width; a short-span, widely-spaced grid is capped by L/8 on each side and the extra slab width beyond that contributes nothing to bending strength (though it still carries gravity load to the beam).

3. Plastic Neutral Axis — Three Cases

Composite flexural strength (§I3.2a) is found the same way as a plastic-moment calculation for a bare shape — split the section into a compression zone and tension zone at full yield, and find where the two force resultants balance. The complication is that the compression zone is now concrete (weak in tension, strong-ish in compression, taken at a rectangular stress block of 0.85f'c) sitting on top of a steel shape. Work down from the top of the slab, comparing the maximum available concrete compression force to the maximum available steel tension force:

Plastic neutral axis (PNA) location — three cases Case (a): PNA in slab C(concrete,partial) = T(steel,full) Case (b): PNA in top flange C(slab,full) between T(webFy)&T(flange) Case (c): PNA in web C(slab,full)+C(flange) = T(rest of web) Rule: compare C_max(concrete) to T_max(steel) working DOWN from the top of slab; each region checked in turn. AISC Manual Tables 3-19/3-20 tabulate φM_n directly for these three PNA locations (Y1, Y2 rows).
Figure 11.2 — The PNA can land in the slab, in the top flange, or in the web, depending on relative concrete vs steel capacity.
  1. PNA in the slab (most common for typical floor beams) — the full steel section yields in tension, T = AsFy, and only part of the slab depth a is needed in compression to balance it: C = 0.85f'cbeffa = T.
  2. PNA in the top flange — even using the ENTIRE slab depth in compression, Cslab < Tfull steel. The PNA moves down into the flange; part of the flange goes into compression, adding to C, until equilibrium.
  3. PNA in the web — heavy/deep steel section relative to a thin slab; the entire top flange plus part of the web must go into compression to balance the tension force in the rest of the web plus bottom flange.
§I3.2aMn = ΣFi·(distance between the compression and tension force resultants)

In practice, AISC Manual Tables 3-19 (Fy = 50 ksi) tabulate φbMn directly versus PNA location (rows labeled by distance Y1 below the top of steel) and percent composite action — this chapter shows the underlying hand calculation those tables automate.

4. Full vs Partial Composite Action

"Full" composite means enough studs are provided that the horizontal shear transferred, ΣQn, equals or exceeds the smaller of the two possible force limits — concrete crushing or steel yielding:

§I3.2d(1)(2)V' = ΣQn ≤ min( 0.85·f'c·Ac, Fy·As )
  • Ac = effective concrete area = beff·tc (concrete above the deck flutes only, ignore concrete inside/below ribs).
  • As = area of the steel shape.
  • Full composite: enough studs that ΣQn ≥ min(concrete, steel) — the PNA analysis of §3 above applies directly.
  • Partial composite: fewer studs are provided (often to save cost when deflection, not strength, controls). Use ΣQn actually installed as the horizontal force V' in the PNA analysis — the effective compression block a becomes a = ΣQn/(0.85f'cbeff), smaller than the full-composite case, and Mn drops smoothly with % composite.
  • AISC requires at least 25% composite action (ΣQn ≥ 0.25·min) for the interpolation approach in the Manual tables to be valid.

What this means physically: partial composite design trades some flexural strength for fewer, cheaper studs — a common owner-driven cost decision when the beam has strength to spare after other checks.

5. Shear Stud Strength Qn (§I8.2a)

Each headed stud has a nominal shear (horizontal shear-transfer) capacity that is the lesser of the stud shearing off and the surrounding concrete crushing/splitting:

Eq. I8-1Qn = 0.5·Asa·√(f'c·Ec) ≤ Rg·Rp·Asa·Fu
  • Asa — cross-sectional area of the stud shank (for a ¾″ stud, Asa = 0.442 in²).
  • Ec — concrete modulus, Ec = wc1.5·33·√f'c (psi, wc in pcf); for 145 pcf NWC at f'c = 4 ksi, Ec ≈ 3,644 ksi.
  • Fu — stud material tensile strength, typically 65 ksi.
  • Rg — deck-geometry reduction factor (Table I8.2a): 1.0 for solid slab or one stud per rib with deck ∥ beam; 0.85 for two studs per rib; 0.7 for three-plus studs per rib.
  • Rp — stud position factor: 0.75 for studs welded through deck (deck ⊥ or ∥, strong or weak position); 1.0 in a few specific favorable deck configurations noted in Table I8.2a — most practical designs use 0.75 (⊥ deck, "not through-deck welded in a haunch," standard case) or 0.6 (weak position, ⊥, unfavorable).

For a common case — ¾ in. dia. stud, deck ⊥ to the beam, standard weak-position welding, 3 in. deck, f'c = 4 ksi NWC, one stud per rib — Qn works out to roughly 17–21 kips depending on Rp; the AISC Manual Table 3-21 tabulates common combinations directly so hand computation of Eq. I8-1 is rarely needed in practice, but every input above should be checked against the actual deck/stud submittal.

6. Deck Orientation Effects

Steel deck ribs run either parallel or perpendicular to the supporting beam, and this changes both Ac and Qn:

  • Deck ⊥ to beam (most common for filler beams) — ribs interrupt the slab, so only concrete above the deck flutes (tc) counts for Ac; studs sit inside individual ribs, subject to Rg, Rp reductions above.
  • Deck ∥ to beam (typical for girders) — the rib runs along the beam, so the concrete inside the rib directly over the beam can be included in Ac if the rib width wr ≥ 2 in.; Rg = Rp = 1.0 unless the average rib width < 2 in.
§I3.2chr ≤ 3 in.; wr ≥ 2 in. (typical composite deck limits for Chapter I to apply directly)

7. Horizontal Shear Force Transfer & Stud Layout

Studs need not be uniformly spaced by calculation — AISC only requires the CUMULATIVE ΣQn between any point and the nearest point of zero moment to be at least equal to the moment demand at that point (§I8.3), but uniform spacing between the maximum-moment section and each end is standard practice and satisfies this automatically for typical uniform-load beams.

Practical ruleN (studs, each side of Mmax) = V' / Qn; total studs = 2N (simple span, single Mmax)
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
Figure 11.3 — Stud layout along the top flange; the required total is split evenly on each side of the maximum-moment section for a symmetric simple span.

8. Construction Stage (Unshored) Checks

Unless temporary shoring is specified, the steel beam alone — with zero composite action — must carry the wet concrete weight, the metal deck, and a construction live load (commonly 20 psf per ASCE 7 / SDI) before the concrete has cured enough to act compositely.

Unshored construction stage — bare steel beam alone carries wet concrete w_D,wet concrete + w_construction live (20 psf) — bare steel section only Checks: φ_bM_p (compact bare beam) and Δ_construction ≤ L/360 (or per spec) Studs add no stiffness here — concrete has not cured to composite strength yet.
Figure 11.4 — Unshored construction stage: the bare steel section alone resists wet concrete + construction live load.
Ch. FφbMp = φb·Fy·Zx (bare compact beam, adequate lateral bracing during pour) ≥ Mu,constr

Lateral bracing of the compression (top) flange during the pour is provided only by the deck attachment and temporary shoring/edge form ties — this must be verified separately (often more conservative Lb than the final composite condition, where the hardened slab braces the top flange continuously).

9. Shear Strength — Steel Web Alone

Composite action increases flexural (bending) capacity dramatically, but it does essentially nothing for shear — the concrete slab is not relied upon to resist beam shear. AISC takes the ENTIRE beam shear on the bare steel web, exactly as in Chapter G, regardless of composite percentage:

Ch. GφvVn = φv·0.6·Fy·Aw·Cv1 (steel web only, no concrete contribution)

10. Deflection with the Lower-Bound Moment of Inertia ILB

For live-load (post-composite) deflection, using the fully-transformed elastic Itr over-predicts stiffness for a partial-composite or a not-yet-fully-cracked section. AISC Commentary provides the lower-bound moment of inertia ILB — a closed-form transformed-section value based on the actual PNA location for the installed ΣQn, giving a conservative (slightly larger) deflection than the fully elastic transformed section:

Commentary I3.2ILB = Ix + As(YENA − d/2)² + (ΣQn/Fy)(2d + Y2 − YENA)
  • Ix, As, d — bare steel shape properties.
  • Y2 — distance from top of steel to the top of the slab (tc + tdeck, or slab top to steel top).
  • YENA — distance from bottom of the steel section (or top of steel, per Manual convention) to the elastic neutral axis of the transformed section, computed from ΣQn/Fy acting as an "equivalent concrete area at the top of steel."
  • AISC Manual Table 3-20 tabulates ILB directly for standard W-shapes vs percent composite — matching this formula.

What this means physically: ILB is always between Ix (bare steel, lower bound) and Itr (fully composite elastic, upper bound) — it recognizes that concrete cracking and imperfect stud slip make the beam a little less stiff than idealized elastic transformed-section theory predicts.

QuantityGovernsSection
beffmin(L/8, s/2) per side§I3.1a
Horizontal shear V'min(0.85f'cAc, FyAs, ΣQn)§I3.2d
Stud strength Qnmin(shear fracture, concrete crushing)§I8.2a, Eq. I8-1
φb (flexure, composite)0.90§I3.2a
φv (shear, steel web)0.90 (1.00 for some rolled W ≤ h/tw ≤ 2.24√(E/Fy))Ch. G
Minimum composite %25% of min(concrete, steel)§I3.2d User Note

Design Workflow

  1. Determine beff from span and beam spacing (§I3.1a).
  2. Compute Ac = beff·tc and both limiting horizontal shear forces 0.85f'cAc and FyAs; take the smaller as the full-composite target V'.
  3. Select a trial stud size/spacing, compute Qn per Eq. I8-1 with the correct Rg, Rp for the deck orientation, and determine ΣQn installed.
  4. Locate the PNA using the actual ΣQn (full or partial) and compute φbMn; compare to Mu.
  5. Check bare-steel shear φvVn against Vu (composite action does not help shear).
  6. Check the UNSHORED construction stage: bare beam φbMp and Δconstr against wet-concrete + construction live load, including top-flange bracing during the pour.
  7. Check post-composite live-load deflection using ILB against the applicable limit (Ch. 18) and specify camber if needed.

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
Exam / FE tip. Always find beff and Ac FIRST, then compare 0.85f'cAc to FyAs before touching the stud count — whichever is smaller is your target horizontal force for 100% composite. If the problem gives you a specific number of studs already installed, use ΣQn actually provided (not the theoretical full-composite value) as V' in every subsequent step.

⚠ Common mistakes

  • Using the full slab thickness (including concrete inside the deck ribs) instead of just tc above the ribs for Ac.
  • Forgetting Rg, Rp reductions for deck ⊥ to the beam — Qn can drop 25–40% from the bare-plate value.
  • Assuming full composite action when the actual stud count gives ΣQn below min(concrete, steel) — must redo the PNA analysis with the ACTUAL ΣQn.
  • Ignoring the unshored construction-stage check — the bare beam alone, without any composite benefit, must survive the wet-concrete pour.
  • Using Ix (bare steel) or the fully-transformed Itr for live-load deflection instead of the code-intended ILB — both give unconservative or overly optimistic answers relative to Manual practice.
  • Assuming concrete helps resist beam shear — Chapter I never increases Vn; shear is 100% steel web.

📖 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

Chapter 11 — Worked Examples (Composite Beams)

Worked Example 11.1 — Composite φMn, PNA in the Slab

Adapted in style from AISC Design Examples V16.0, Part I (composite beam methodology).

  • W18×35, A992 (Fy=50 ksi): As=10.3 in², d=17.7 in.
  • L = 30 ft simple span, beam spacing s = 10 ft.
  • 3-in composite deck ⊥ beam, tc = 4.5 in normal-weight concrete above deck, f'c = 4 ksi.
  • Mu = 191 kip-ft (governing factored moment, from tributary loads).
Ex.11.1 — Composite W18×35, PNA in slab L=30 ft, s=10 ft, f'c=4 ksi, tc=4.5 in
Figure 11.1a — Composite section geometry.
Step 1 — Effective width (§I3.1a).
§I3.1abeff = min(L/8 per side ×2, s) = min(2×45, 120) = min(90,120) = 90 in
beff = 90 in.
Step 2 — Horizontal shear force limits (§I3.2d).
Eq. §I3.2dV' = min(0.85 f'c Ac, Fy As)
Concrete: 0.85(4)(90×4.5) = 1,377 k. Steel: 50(10.3) = 515 k. → V' = 515 k (steel governs; full composite).
Step 3 — Depth of stress block a.
Formulaa = V'/(0.85 f'c beff)
a = 515/(0.85×4×90) = 1.68 in < tc = 4.5 in → PNA in slab ✓
Step 4 — Nominal & design moment.
§I3.2aMn = V'(d/2 + tdeck + tc − a/2); φb = 0.90
Lever arm = 8.85 + 3 + 4.5 − 0.84 = 15.51 in. Mn = 515×15.51/12 = 665 k-ft. φMn = 0.90×665 = 599 k-ft.
Limit stateCapacityDemandDCR
Composite flexure φMn599 k-ft191 k-ft0.32
Governing limit state: composite flexure with PNA in the slab. φMn = 599 k-ft ≥ Mu = 191 k-ft ✓ — W18×35 is adequate (and lightly loaded at DCR = 0.32, typical when deflection/vibration size the beam).

Engineering interpretation: because the steel governs V' (515 k < 1,377 k concrete capacity), full composite action here only "uses" a small compression depth a = 1.68 in — well within the slab, confirming the simplest PNA case applies.

Common mistakes: using Ac based on total slab depth instead of tc above the deck; forgetting to check a ≤ tc before using the "PNA in slab" formula.

Worked Example 11.2 — Stud Count & Partial Composite Action

Continuation of the W18×35 beam of Example 11.1, now checked for a reduced stud count.

  • Same beam/slab as Ex.11.1. Full-composite target V' = 515 k (from Step 2 above).
  • ¾-in dia. studs, Fu = 65 ksi, Asa = 0.442 in², deck ⊥ beam, one stud per rib: Rg = 1.0, Rp = 0.75.
  • Ec = wc1.5·33·√f'c = 1451.5·33·√4,000 psi = 3,644 ksi.
  • Contractor proposes 28 studs total (14 each side of midspan).
Ex.11.2 — Partial-composite stud count 28 studs total (14 each side of midspan) — 65% composite
Figure 11.2a — Reduced stud layout, 14 studs each half-span.
Step 1 — Stud strength Qn (Eq. I8-1).
Eq. I8-1Qn = min[0.5Asa√(f'cEc), RgRpAsaFu]
0.5(0.442)√(4×3,644) = 0.221×120.7 = 26.7 k. RgRpAsaFu = 1.0×0.75×0.442×65 = 21.5 k. → Qn = 21.5 k/stud (fracture governs).
Step 2 — Installed ΣQn and % composite. ΣQn = 14 studs (half span) × 21.5 = 301 k. % composite = 301/515 = 58% — above the 25% minimum, so partial-composite interpolation is valid.
Step 3 — Recompute PNA with the ACTUAL ΣQn = 301 k (governs over 515 k).
§I3.2d partiala = ΣQn/(0.85f'cbeff)
a = 301/(0.85×4×90) = 0.984 in < tc → PNA still in slab.
Step 4 — φMn with partial composite action. Lever arm = 8.85+3+4.5−0.49 = 15.86 in. Mn = 301×15.86/12 = 398 k-ft. φMn = 0.90×398 = 358 k-ft.
Composite levelΣQnφMn
100% (60 studs, Ex.11.1)515 k599 k-ft
58% (28 studs, this example)301 k358 k-ft
With 28 studs (58% composite), φMn = 358 k-ft ≥ Mu = 191 k-ft ✓. Fewer studs than full composite still satisfies strength; only stiffness (Example 11.4) needs re-checking with the reduced ILB.

Engineering interpretation: partial composite design saved 32 studs (60→28) while still meeting the strength demand with margin, illustrating why designers rarely specify 100% composite unless deflection or vibration requires the extra stiffness.

Common mistakes: using the full-composite V' = 515 k instead of the actual installed ΣQn = 301 k when recomputing a and Mn; forgetting the 25% minimum composite action threshold.

Worked Example 11.3 — Unshored Construction-Stage Check

Bare-steel check of the same W18×35 before the slab cures, per AISC Chapter F.

  • W18×35: Zx = 66.5 in³, Fy = 50 ksi, compact, fully braced by deck attachment at ≤ 6 ft during pour (Lb ≤ Lp).
  • Wet concrete: 4.5 in slab (145 pcf) + 3 in deck (self-weight 3 psf) over 10-ft tributary width.
  • Construction live load = 20 psf (SDI/ASCE 7 minimum during placement).
  • L = 30 ft simple span.
Ex.11.3 — Unshored construction-stage bare beam w_D(wet conc)=0.469 klf + 20 psf constr. LL → w_u = 0.75 klf
Figure 11.3a — Unshored bare beam under wet concrete + construction live load.
Step 1 — Construction-stage loads. Wet concrete: (4.5/12)(145) = 54.4 psf; + deck 3 psf = 57.4 psf → ×10 ft = 0.574 klf. Construction LL: 20 psf ×10 ft = 0.20 klf. wu = 1.2(0.574) + 1.6(0.20) = 0.689 + 0.320 = 1.01 klf.
Step 2 — Required moment.
Simple beamMu = wuL²/8
Mu = 1.01(30)²/8 = 113.6 k-ft.
Step 3 — Bare-steel flexural capacity (Ch. F, compact, braced).
Eq. F2-1φbMp = φbFyZx
Mp = 50(66.5)/12 = 277 k-ft. φMp = 0.90(277) = 249 k-ft ≥ 113.6 k-ft ✓.
Step 4 — Construction-stage deflection (informational). Δ = 5(0.574)(30×12)⁴/(384×29,000×510)/12³... using Ix=510 in⁴: Δ ≈ 0.62 in ≈ L/580, typically within the L/360 (or contractor-specified) limit for wet-concrete ponding effects — verify ponding per SDI if Δ > ¾ in.
Limit stateCapacityDemand
Bare-beam flexure φMp249 k-ft113.6 k-ft
Construction deflectionL/360 = 1.0 in0.62 in
Bare W18×35 is adequate for the unshored pour: φMp = 249 k-ft ≥ 113.6 k-ft ✓, and construction deflection is within typical limits. No temporary shoring required for this span/loading combination.

Engineering interpretation: the construction stage often governs beam SELECTION for longer unshored spans, even though it uses a much smaller load than the final composite condition — because zero composite stiffness/strength is available yet.

Common mistakes: forgetting the construction live load entirely; assuming the deck bracing is adequate without checking the actual unbraced length against Lp for the bare shape; skipping the ponding deflection check on long unshored spans.

Worked Example 11.4 — Composite Deflection Using ILB

Live-load deflection check for the fully-composite W18×35 of Example 11.1 (60 studs, 100% composite).

  • W18×35: Ix = 510 in⁴, As = 10.3 in², d = 17.7 in.
  • 100% composite (ΣQn=515 k = FyAs); Y2 = tc+tdeck = 4.5+3 = 7.5 in.
  • From AISC Manual Table 3-20 methodology (or direct computation): ILB ≈ 1,250 in⁴ for this beam at 100% composite.
  • Service live load wL = 0.5 klf, L = 30 ft. Limit: ΔL ≤ L/360.
Ex.11.4 — Composite deflection with I_LB I_LB = 1,250 in⁴, w_L = 0.5 klf, L=30 ft
Figure 11.4a — Composite beam deflected shape using I_LB.
Step 1 — Lower-bound moment of inertia (Commentary I3.2).
Commentary Eq. C-I3-4ILB = Ix + As(YENA−d/2)² + (ΣQn/Fy)(2d+Y2−YENA)
Using YENA (from steel-top datum, computed from force balance of As and ΣQn/Fy) ≈ 5.75 in and ΣQn/Fy = 515/50 = 10.3 in²: ILB = 510 + 10.3(5.75−8.85)² + 10.3(2×17.7+7.5−5.75) ≈ 510+99+374 ≈ 1,250 in⁴ (matches Manual Table 3-20 for this section at full composite).
Step 2 — Live-load deflection.
Simple beam, uniform loadΔL = 5wLL⁴/(384EILB)
L = 360 in, wL = 0.5/12 = 0.0417 k/in. ΔL = 5(0.0417)(360)⁴/(384×29,000×1,250) = 5(0.0417)(1.68×10¹⁰)/(1.393×10¹⁰) = 0.628 in.
Step 3 — Compare to limit. L/360 = 360/360 = 1.00 in. ΔL = 0.628 in < 1.00 in ✓.
Property usedValueEffect on Δ
Ix (bare steel)510 in⁴Δ would be 2.45× too large (unconservative for stiffness credit)
ILB (code-intended)1,250 in⁴ΔL = 0.628 in — governs design
Governing check: live-load deflection using ILB. ΔL = 0.628 in ≤ L/360 = 1.00 in ✓. Composite action roughly 2.5× the bare-steel stiffness, consistent with typical composite floor beam behavior.

Engineering interpretation: ILB is deliberately conservative relative to the fully-transformed elastic Itr because it accounts for concrete cracking and stud flexibility — always use ILB (or the Manual table value), never Ix or Itr, for composite beam deflection.

Common mistakes: using Ix (bare steel) for post-composite deflection, dramatically over-predicting Δ; using the fully-transformed elastic Itr instead of ILB, under-predicting Δ; mixing units of L (ft vs in) inside the Δ formula.

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.