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LecturesModule 26

RC — One-Way Slabs

Flexural design, temperature/shrinkage steel, and minimum thickness per ACI Chapters 7 and 24.

AISC Reference Box
  • ACI 318-19 Ch. 7 — One-Way Slabs
  • ACI Table 7.3.1.1 — Minimum Slab Thickness (deflection)
  • ACI §7.6.1 — Minimum Flexural Reinforcement
  • ACI §24.4 — Shrinkage and Temperature Reinforcement
  • ACI §7.7 — Reinforcement Detailing — Spacing

Lecture Notes

A one-way slab spans primarily in one direction — it is supported on two opposite edges, or has a long-to-short panel ratio L/S > 2 when supported on four edges. Design proceeds per ACI 318-19 Chapter 7 (one-way slabs) using a 12-inch-wide design strip; loads and steel are reported per foot of width.

Minimum thickness (ACI Table 7.3.1.1) to avoid explicit deflection calculations: simply supported h ≥ L/20, one end continuous L/24, both ends continuous L/28, cantilever L/10. For fy ≠ 60 ksi, multiply by 0.4 + fy/100,000.

Flexural reinforcement is designed exactly like a beam with b = 12 in. (per foot of width). The minimum steel for slabs is governed by temperature and shrinkage requirements rather than the flexural minimum (ACI §7.6.1.1): As,min = ρmin·b·h, with ρmin = 0.0020 for Grade 40/50, 0.0018 for Grade 60, and 0.0018·60,000/fy for higher grades (not less than 0.0014).

Maximum bar spacing for principal flexural steel (ACI §7.7.2.3): s ≤ min(3·h, 18 in.). For shrinkage and temperature steel (ACI §24.4.3.3): s ≤ min(5·h, 18 in.). Cover for slabs not exposed to weather is 0.75 in.

Shear in one-way slabs (ACI §7.5.3): φVc = φ·2·λ·√f'c·b·d (psi). One-way slabs are designed so that Vu ≤ φVc — stirrups are essentially never used in normal building slabs. If shear governs, increase the slab thickness.

Temperature and shrinkage reinforcement runs PERPENDICULAR to the main flexural steel to control cracking due to volumetric changes. It is uniformly distributed across the slab and must satisfy the same ρmin (0.0018 for Grade 60).

Project case study — Cardinal Square RC — 4-story cast-in-place concrete 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 @ 30 ft × 30 ft. Stories: 4 @ 13 ft (52 ft roof). Floor: 7 in one-way slab on RC beams (b = 14 in, h = 24 in) framing to RC columns (16 in × 16 in tied) on grid. Roof: same beam grid + 6 in slab. Materials: Normal-weight concrete f'c = 4,000 psi (λ = 1.0). Reinforcement Grade 60 (fy = 60,000 psi); ties/stirrups Grade 60 #3 / #4. Clear cover: 1.5 in interior beams/slabs, 2 in columns, 3 in footings (ACI §20.5.1).

Chapter 26 — One-way slab between beams
Typical 10 ft clear-span one-way slab, h = 7 in, d ≈ 5.75 in (#4 bars, ¾ in cover)
Demand carried forward
Per 12-in strip: wD = (7/12)·0.15 + 0.025 (SDL) = 0.113 klf wL = 0.050 klf. wu = 1.2·0.113 + 1.6·0.050 = 0.216 klf Mu (interior span, ACI coefficient 1/14) = wu·ℓn²/14 = 0.216·10²/14 ≈ 1.54 k-ft per ft.
This chapter contributes
Sizes main steel: required As ≈ Mu·12/(φ·fy·(d − a/2)) ≈ 0.06 in²/ft → #4 @ 16 in (As = 0.15 in²/ft) governed by As,min = 0.0018·b·h = 0.0018·12·7 = 0.151 in²/ft. Provides temperature/shrinkage steel #4 @ 18 in (As ≥ 0.0018·Ag) in the transverse direction (ACI §24.4).
Feeds into next chapter
Slab self-weight from this chapter is the dead load wD used in Chapter 22 (beam flexure) — closing the load path.
12-in design strip (per foot)h ≥ L/(20, 24, 28, 10) — Gr 60T&S steel ρmin = 0.0018·b·h, s ≤ min(5h, 18 in.)
Design per 12-in strip; h ≥ L/(20, 24, 28, 10); T&S steel ρmin = 0.0018·b·h perpendicular.

Formula Sheet

NameEquationAISC Ref
Min. thickness (one end continuous, Gr 60)h ≥ L / 24ACI Table 7.3.1.1
Design stripb = 12 in. (per foot of width)ACI Ch. 7
Whitney block (per foot)a = (As·fy)/(0.85·f'c·b); Mn = As·fy·(d − a/2)ACI §22.2.2.4
Shrinkage / temperature ρ (Gr 60)ρmin = 0.0018 → As,min = 0.0018·b·hACI §24.4.3.2
Max spacing — flexures ≤ min(3h, 18 in.)ACI §7.7.2.3
Max spacing — T&Ss ≤ min(5h, 18 in.)ACI §24.4.3.3
Slab shearφVc = 0.75·2·λ·√f'c·b·d (psi)ACI §22.5.5.1

Worked Example

One-way slab — thickness and reinforcement for a 14 ft span

Given
Interior one-way slab, both ends continuous, L = 14 ft = 168 in. Superimposed dead load DL = 25 psf, live load LL = 50 psf. Materials f'c = 4000 psi, fy = 60,000 psi. Clear cover 0.75 in.
Load combination
wu = 1.2·DL + 1.6·LL (slab self-weight added once h is chosen).
Required strength
Select h, design positive-moment steel per foot, and temperature steel.
Limit states
  • Min. thickness for deflection
  • Flexural strength φMn
  • Minimum flexural / T&S steel
  • Max bar spacing
AISC reference
ACI 318-19 Ch. 7
Solution steps
  1. 1. Given
    L = 14 ft = 168 in. (both ends continuous). DL = 25 psf (superimposed). LL = 50 psf. f'c = 4000 psi, fy = 60,000 psi. Cover = 0.75 in.
  2. 2. Find
    Slab thickness h, main steel As per foot, T&S steel.
  3. 3. ACI reference
    ACI Table 7.3.1.1 (h-min) §22.2.2.4 (flexure) §24.4.3 (T&S).
  4. 4. Assumptions
    Tension-controlled (verify εt). Concrete unit weight 150 pcf. Design per 12 in. wide strip.
  5. 5. Minimum thickness
    h ≥ L/28 (both ends continuous). h ≥ 168/28. h ≥ 6.0 in. Use h = 6.0 in.
  6. 6. Self-weight & wu
    Slab self-weight = (6/12)·150 = 75 psf. DL,total = 75 + 25 = 100 psf. wu = 1.2·100 + 1.6·50. wu = 120 + 80. wu = 200 psf = 0.200 ksf.
  7. 7. Factored moment per ft
    Continuous interior, approx. Mu = wu·L²/14 (ACI coeff., interior span +M). Mu = 0.200·(14)²/14. Mu = 0.200·14. Mu = 2.80 k·ft/ft = 33.6 k·in/ft.
  8. 8. Effective depth d
    Use #4 main bars (db = 0.50 in.). d = h − cover − db/2. d = 6.0 − 0.75 − 0.25. d = 5.00 in.
  9. 9. Required As (trial)
    Estimate jd ≈ 0.95·d = 4.75 in. As ≈ Mu / (φ·fy·jd). As ≈ 33.6 / (0.90·60·4.75). As ≈ 33.6 / 256.5. As ≈ 0.131 in²/ft.
  10. 10. Verify with exact equation
    a = (As·fy)/(0.85·f'c·b). a = (0.131·60,000)/(0.85·4000·12). a = 7860 / 40,800. a = 0.193 in. Mn = As·fy·(d − a/2). Mn = 0.131·60·(5.00 − 0.096). Mn = 0.131·60·4.904. Mn = 38.5 k·in/ft. φMn = 0.90·38.5 = 34.7 k·in/ft ≥ 33.6 ✓
  11. 11. Minimum flexural steel
    As,min = 0.0018·b·h = 0.0018·12·6.0 = 0.130 in²/ft. 0.131 ≈ 0.130 ✓ (T&S governs).
  12. 12. Choose bars
    Use #4 @ 18 in. o.c. → As = 0.20·(12/18) = 0.133 in²/ft. max s = min(3·6.0, 18) = 18 in. ✓
  13. 13. Temperature & shrinkage steel
    As,TS = 0.0018·12·6.0 = 0.130 in²/ft. Use #4 @ 18 in. o.c. each face/top → As provided 0.133 in²/ft ✓ max s = min(5·6.0, 18) = 18 in. ✓
Final design decision
Use h = 6.0 in. one-way slab. Main flexural steel: #4 @ 18 in. o.c. bottom (positive M). Temperature & shrinkage steel: #4 @ 18 in. o.c. perpendicular to span. φMn = 34.7 k·in/ft ≥ Mu = 33.6 k·in/ft.
Common mistakes in this example
  • Forgetting to add slab self-weight to DL before factoring.
  • Designing flexural steel on a per-bar basis instead of in²/ft.
  • Using 3h max spacing for T&S steel (T&S uses 5h, not 3h).
  • Mixing units in the Whitney block (lb vs kip).

FE-Style Worked Examples (5)

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
Clear span L = 14 ft; simply supported; fy = 60 ksi; NWC.
AISC Reference
ACI 318-19 Table 7.3.1.1
Step-by-step solution
  1. 1. Problem Statement
    Pick the minimum slab thickness to skip deflection computation.
  2. 2. Variables & Values
    L = 14 ft = 168 in. simply supported Grade 60.
  3. 3. Approach
    Use h ≥ L/20 from Table 7.3.1.1 (Gr 60 baseline, no modifier).
  4. 4. Formula Setup
    h_min = L/20
  5. 5. Substitution
    h_min = 168/20 = 8.4 in.
  6. 6. Solution Steps
    Use h = 8.5 in (round to 1/2 in.).
Answer h ≥ 8.5 in.
ΔLive-load deflection check

Practice Problems

  1. [E] State min thickness h ≥ L/(20, 24, 28, 10) for Gr 60.
  2. [E] State design strip b = 12 in.
  3. [E] T&S ρmin = 0.0018 for Gr 60.
  4. [E] Max flexural bar spacing min(3h, 18 in.).
  5. [E] Max T&S bar spacing min(5h, 18 in.).
  6. [M] Design one-way slab simply supported L = 12 ft, DL_SI = 20 psf, LL = 100 psf, f'c = 4, fy = 60.
  7. [M] Min thickness for cantilever one-way slab L = 6 ft, fy = 60.
  8. [M] T&S steel for 8 in. slab, 12 ft strip, fy = 60.
  9. [M] φMn per foot for a 6 in. slab with #4 @ 12 in. o.c., d = 5, f'c = 4.
  10. [M] φVc per foot for a 6 in. slab carrying Vu = 1.8 k/ft, f'c = 4.
  11. [H] Continuous interior one-way slab L = 14 ft, DL_SI = 25, LL = 50, f'c = 4, fy = 60 — full design w/ T&S.
  12. [H] One-way slab with 200 psf live (parking). Required h, As, T&S.
  13. [H] Long-term + creep deflection per §24.2 using λΔ = 2.0.
  14. [H] Stair flight as a one-way slab: sloped 12 ft, LL = 100 psf. Design h, As.
  15. [H] One-way slab with a 4 ft opening: strip reduction; check shear/bending around opening.
Structured Clues
  • Design per 12-in. wide strip → report in²/ft.
  • T&S steel runs perpendicular to flexural steel with ρ = 0.0018 (Gr 60).
  • Max spacing: 3h (flexure) vs 5h (T&S) — they're different.
Code References
  • ACI 318-19 Ch. 7
  • Table 7.3.1.1
  • §24.4

Quiz

1. Minimum thickness of an interior continuous one-way slab (Gr 60) is:
2. Temperature and shrinkage steel ratio for Grade 60 is:
3. Max spacing of principal flexural steel in a slab is:
4. A one-way slab is recognized when the panel aspect ratio L/S satisfies:
5. Slab shear strength φVc per foot is (psi):
6. Minimum thickness h (in.) for an interior continuous slab, L = 16 ft, Grade 60:
7. Slab self-weight (psf) for h = 7 in., NWC at 150 pcf:
8. Factored load wu (psf) for DL = 100 (incl. SW), LL = 80:
9. T&S steel area (in²/ft) for h = 7 in. slab, Grade 60:
10. Required spacing of #4 (Ab = 0.20 in²) T&S bars per foot for the above slab:
11. Maximum spacing of principal flexural steel for h = 7 in.:
12. Maximum spacing of T&S steel for h = 7 in.:
13. A panel is considered one-way when the long/short ratio L/S is:
14. Effective depth d (in.) for h = 7, cover = 0.75, #4 main bars:
15. Cantilever one-way slab minimum thickness for L = 5 ft, Grade 60:

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 rc — one-way slabs. 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.