Practical guide to one-way and two-way solid slab design according to ACI 318-19M. Covers thickness requirements, load combinations, flexural and shear design (including punching) with clear formulas and an example.
1. Slab System Classification
One-way slab: Main flexure spans in one direction only (ratio of longer to shorter span ≥ 2.0). Two-way slab: Significant bending in both directions (span ratio < 2.0). Refer to ACI 318-19M §13 and §8.
2. Minimum Thickness Requirements
Serviceability deflection control can be satisfied by meeting the minimum thickness in ACI 318-19M §24.2.2 (Table 24.2.2). Typical limits:
| Member/Support Condition | One-Way Slab | Two-Way Slab (without drop panels) | Reference |
|---|---|---|---|
| Simply supported | L/20 | h ≥ span/36 | ACI Table 24.2.2 & §8.3 |
| One end continuous | L/24 | h ≥ span/40 | ACI Table 24.2.2 |
| Both ends continuous | L/28 | h ≥ span/45 | ACI Table 24.2.2 |
| Cantilever | L/10 | — | ACI Table 24.2.2 |
3. Load Combinations
Use strength load combinations per ACI 318-19M §5.3.1. For gravity loading:
wu = 1.2 wD + 1.6 wL
4. Flexural Design
4.1 One-Way Slab Strip
Design a 1 m wide strip as a rectangular section. Determine factored moment envelope and compute required steel area using strength reduction factor φ = 0.9 for flexure.
Mn = As · fy · (d - a/2), a = As·fy / (0.85 f'c b)
4.2 Two-Way Slab Moments
Use direct design method (DDM) or equivalent frame method (EFM) per ACI 318-19M §8.10–§8.12. Total static moment:
MT = wu · Lx · Ly · (Lx^2 + Ly^2) / 8
Distribute MT to negative/positive moments and to column/ middle strips using coefficients per code.
5. Shear Design
5.1 One-Way Shear
φVn ≥ Vu, Vc = 0.17 λ √f'c · b · d
5.2 Punching Shear (Two-Way)
φVn ≥ Vu, Vc = 0.33 λ √f'c · bo · d
Where bo is the critical perimeter at 2d from the column face. Add shear reinforcement if required (stud rails or stirrups) per §22.6.
6. Minimum and Maximum Reinforcement
Provide minimum steel per §8.6.1 and §24.4:
- Tension steel minimum: ρmin = max(3√f'c/fy, 200/fy)
- Spacing limits: s ≤ min(3h, 450 mm) for main bars; distribution steel per §8.7
- Bar cover: ≥ 20–40 mm depending on exposure (§20.6)
7. Example: One-Way Slab Design
7.1 Problem Statement
Interior one-way slab strip spanning L = 4.5 m (both ends continuous). Material strengths: f'c = 28 MPa, fy = 420 MPa. Service loads over tributary area: wD = 3.5 kN/m² (superimposed dead, excluding self-weight), wL = 2.5 kN/m². Design per meter strip (b = 1000 mm).
7.2 Thickness, Self-Weight, and Factored Loads
Minimum thickness (Table 24.2.2): h ≥ L/28 = 4500/28 = 161 mm → select h = 170 mm.
Effective depth (assume 10 mm bars, 20 mm cover): d ≈ h − 30 ≈ 140 mm.
Self-weight: ρ ≈ 24 kN/m³ → wself = 0.17 × 24 = 4.08 kN/m².
Total dead load: w̄D = 3.5 + 4.08 = 7.58 kN/m² → per meter strip wD = 7.58 kN/m.
Live load per meter strip: wL = 2.5 kN/m.
Factored load (ACI §5.3.1): wu = 1.2wD + 1.6wL = 1.2(7.58) + 1.6(2.5) = 9.10 + 4.00 = 13.10 kN/m.
Effective depth (assume 10 mm bars, 20 mm cover): d ≈ h − 30 ≈ 140 mm.
Self-weight: ρ ≈ 24 kN/m³ → wself = 0.17 × 24 = 4.08 kN/m².
Total dead load: w̄D = 3.5 + 4.08 = 7.58 kN/m² → per meter strip wD = 7.58 kN/m.
Live load per meter strip: wL = 2.5 kN/m.
Factored load (ACI §5.3.1): wu = 1.2wD + 1.6wL = 1.2(7.58) + 1.6(2.5) = 9.10 + 4.00 = 13.10 kN/m.
7.3 Design Moments (Continuous One-Way Slab)
Using common elastic coefficients for interior spans: M- ≈ wuL²/12 at supports, M+ ≈ wuL²/16 at midspan.
L² = (4.5)² = 20.25 m².
M- = wuL²/12 = 13.10 × 20.25 / 12 = 22.12 kN·m per meter.
M+ = wuL²/16 = 13.10 × 20.25 / 16 = 16.58 kN·m per meter.
M- = wuL²/12 = 13.10 × 20.25 / 12 = 22.12 kN·m per meter.
M+ = wuL²/16 = 13.10 × 20.25 / 16 = 16.58 kN·m per meter.
7.4 Flexural Reinforcement Design
Use φ = 0.9 for flexure. Approximate lever arm z ≈ 0.9d = 0.9 × 140 = 126 mm.
As ≈ Mu × 10⁶ / (φ × fy × z).
As- = 22.12 × 10⁶ / (0.9 × 420 × 126) = 22.12e6 / 47,628 ≈ 464 mm²/m.
As+ = 16.58 × 10⁶ / (0.9 × 420 × 126) = 16.58e6 / 47,628 ≈ 348 mm²/m.
As ≈ Mu × 10⁶ / (φ × fy × z).
As- = 22.12 × 10⁶ / (0.9 × 420 × 126) = 22.12e6 / 47,628 ≈ 464 mm²/m.
As+ = 16.58 × 10⁶ / (0.9 × 420 × 126) = 16.58e6 / 47,628 ≈ 348 mm²/m.
7.5 Minimum Steel and Bar Selection
Temperature/shrinkage steel for slabs (ACI §24.4.3): As,min ≥ 0.0018 b h = 0.0018 × 1000 × 170 = 306 mm²/m.
Choose practical bar layouts per meter strip:
• Negative region (supports): 12 mm @ 225 mm → A = (π×12²/4) × (1000/225) = 113 × 4.44 = 502 mm²/m ≥ 464 mm²/m ✓
• Positive region (midspan): 10 mm @ 250 mm → A = (π×10²/4) × (1000/250) = 78.5 × 4 = 314 mm²/m → slightly below 348 mm²/m → use 10 mm @ 225 mm → 78.5 × 4.44 = 349 mm²/m ✓
Both exceed As,min = 306 mm²/m.
• Negative region (supports): 12 mm @ 225 mm → A = (π×12²/4) × (1000/225) = 113 × 4.44 = 502 mm²/m ≥ 464 mm²/m ✓
• Positive region (midspan): 10 mm @ 250 mm → A = (π×10²/4) × (1000/250) = 78.5 × 4 = 314 mm²/m → slightly below 348 mm²/m → use 10 mm @ 225 mm → 78.5 × 4.44 = 349 mm²/m ✓
Both exceed As,min = 306 mm²/m.
7.6 Shear and Deflection Checks
One-way shear at d from face (conservative at midspan for strip design): Vu ≈ wu L / 2 = 13.10 × 4.5 / 2 = 29.5 kN per meter.
Concrete shear capacity (ACI §22.5): Vc = 0.17 λ √f'c b d = 0.17 × 1.0 × √28 × 1000 × 140 = 0.17 × 5.292 × 140,000 ≈ 126 kN.
φVc (φ = 0.75) = 94.5 kN > 29.5 kN ✓ No stirrups required in slab.
Deflection: h = 170 mm ≥ L/28 = 161 mm ✓ (Table 24.2.2). Serviceability satisfied.
Concrete shear capacity (ACI §22.5): Vc = 0.17 λ √f'c b d = 0.17 × 1.0 × √28 × 1000 × 140 = 0.17 × 5.292 × 140,000 ≈ 126 kN.
φVc (φ = 0.75) = 94.5 kN > 29.5 kN ✓ No stirrups required in slab.
Deflection: h = 170 mm ≥ L/28 = 161 mm ✓ (Table 24.2.2). Serviceability satisfied.
7.7 One-Way Slab Illustrations
Elevation: Span, Loads, and Support Condition
Plan: One-Way Strip and Bar Zones
Section: Thickness, Effective Depth, and Bar Positions
One-Way Slab Bending Moment Diagram (Continuous)
8. Example: Two-Way Slab (Interior Panel)
Interior panel 6.0 m (x) × 5.0 m (y), same service loads; use h = 170 mm (d ≈ 140 mm). Design by Direct Design Method (DDM) overview with practical per-meter design moments.
8.1 Factored Load and Static Moment
Use wu from Section 7 (13.10 kN/m acting in both directions).
Total static moment per ACI DDM: MT = wu Lx Ly (Lx² + Ly²)/8 = 13.10×6×5×(36+25)/8 = 13.10×30×61/8 = 299.6 kN·m (≈ 300 kN·m).
Total static moment per ACI DDM: MT = wu Lx Ly (Lx² + Ly²)/8 = 13.10×6×5×(36+25)/8 = 13.10×30×61/8 = 299.6 kN·m (≈ 300 kN·m).
8.2 Distribution of Moments
Typical DDM distributions (interior panel): 65% negative at supports, 35% positive at midspans. Proportion to each direction by Ly²:(Lx²+Ly²) and Lx²:(Lx²+Ly²) respectively; then split into column/middle strips (column strip ≈ 0.25 of panel width, receives 75% of negative and 60% of positive moments). Values below illustrate the process; final detailing follows local practice and ACI tables.
Direction x share: αx = Ly²/(Lx²+Ly²) = 25/61 = 0.4098 → MT,x = 300×0.4098 = 123 kN·m.
Direction y share: αy = Lx²/(Lx²+Ly²) = 36/61 = 0.5902 → MT,y = 300×0.5902 = 177 kN·m.
Negative/positive split:
x: M- = 0.65×123 = 79.9 kN·m; M+ = 0.35×123 = 43.1 kN·m.
y: M- = 0.65×177 = 115.1 kN·m; M+ = 0.35×177 = 61.9 kN·m.
Column strip shares (for each direction total):
x: Mcol,- = 0.75×79.9 = 59.9; Mcol,+ = 0.60×43.1 = 25.9 kN·m.
y: Mcol,- = 0.75×115.1 = 86.3; Mcol,+ = 0.60×61.9 = 37.1 kN·m.
Each direction has two column strips per panel; distribute accordingly for detailing. Middle-strip moments are the remainders.
Direction y share: αy = Lx²/(Lx²+Ly²) = 36/61 = 0.5902 → MT,y = 300×0.5902 = 177 kN·m.
Negative/positive split:
x: M- = 0.65×123 = 79.9 kN·m; M+ = 0.35×123 = 43.1 kN·m.
y: M- = 0.65×177 = 115.1 kN·m; M+ = 0.35×177 = 61.9 kN·m.
Column strip shares (for each direction total):
x: Mcol,- = 0.75×79.9 = 59.9; Mcol,+ = 0.60×43.1 = 25.9 kN·m.
y: Mcol,- = 0.75×115.1 = 86.3; Mcol,+ = 0.60×61.9 = 37.1 kN·m.
Each direction has two column strips per panel; distribute accordingly for detailing. Middle-strip moments are the remainders.
8.3 Simplified Per-Meter Steel Design (Check)
For quick sizing, many offices also check per-meter design moments using elastic coefficients: M- ≈ wuL²/12 and M+ ≈ wuL²/16 for each direction. This produces reasonable reinforcement demands that are then adjusted to meet DDM strip allocations.
For Lx = 6 m: M- = 13.10×36/12 = 39.3 kN·m/m; M+ = 13.10×36/16 = 29.5 kN·m/m.
For Ly = 5 m: M- = 13.10×25/12 = 27.3 kN·m/m; M+ = 13.10×25/16 = 20.5 kN·m/m.
With d = 140 mm and z ≈ 126 mm, φ = 0.9:
As (mm²/m) = Mu×10⁶ / (0.9×420×126).
• x-direction: As- = 39.3e6/47,628 = 825; As+ = 29.5e6/47,628 = 619 mm²/m.
• y-direction: As- = 27.3e6/47,628 = 573; As+ = 20.5e6/47,628 = 430 mm²/m.
Bar options per meter (meet As,min = 306 mm²/m):
• Top over columns (x): 12 mm @ 150 → 113×6.67 = 754 mm²/m (increase local top steel within column strip to meet DDM).
• Midspan (x): 12 mm @ 200 → 113×5 = 565 mm²/m ≥ 619? adjust to 12 @ 180 → 113×5.56 = 628 mm²/m ✓
• Top over columns (y): 12 mm @ 200 → 565 mm²/m ≥ 573? use 12 @ 190 → 113×5.26 = 595 mm²/m ✓
• Midspan (y): 10 mm @ 225 → 78.5×4.44 = 349 mm²/m ≥ 430? use 10 @ 180 → 78.5×5.56 = 436 mm²/m ✓
Final bar spacing and lengths should follow strip allocations from 8.2 (higher top steel in column strips, curtailed into middle strips) and development length requirements.
For Ly = 5 m: M- = 13.10×25/12 = 27.3 kN·m/m; M+ = 13.10×25/16 = 20.5 kN·m/m.
With d = 140 mm and z ≈ 126 mm, φ = 0.9:
As (mm²/m) = Mu×10⁶ / (0.9×420×126).
• x-direction: As- = 39.3e6/47,628 = 825; As+ = 29.5e6/47,628 = 619 mm²/m.
• y-direction: As- = 27.3e6/47,628 = 573; As+ = 20.5e6/47,628 = 430 mm²/m.
Bar options per meter (meet As,min = 306 mm²/m):
• Top over columns (x): 12 mm @ 150 → 113×6.67 = 754 mm²/m (increase local top steel within column strip to meet DDM).
• Midspan (x): 12 mm @ 200 → 113×5 = 565 mm²/m ≥ 619? adjust to 12 @ 180 → 113×5.56 = 628 mm²/m ✓
• Top over columns (y): 12 mm @ 200 → 565 mm²/m ≥ 573? use 12 @ 190 → 113×5.26 = 595 mm²/m ✓
• Midspan (y): 10 mm @ 225 → 78.5×4.44 = 349 mm²/m ≥ 430? use 10 @ 180 → 78.5×5.56 = 436 mm²/m ✓
Final bar spacing and lengths should follow strip allocations from 8.2 (higher top steel in column strips, curtailed into middle strips) and development length requirements.
8.4 Punching Shear Check (Quick)
At an interior column, take tributary reaction Ru ≈ wu × (Lx×Ly)/4 = 13.10×30/4 = 98.3 kN (per column).
Critical perimeter at 2d: bo ≈ 4(ccol + 2d). For a 400×400 mm column and d = 140 mm: bo = 4(400 + 280) = 2,720 mm.
Two-way concrete shear (ACI §22.6): Vc = 0.33 λ √f'c bo d = 0.33×5.292×2,720×140 ≈ 658 kN; φ = 0.75 → φVc ≈ 494 kN ≫ Ru ✓
Critical perimeter at 2d: bo ≈ 4(ccol + 2d). For a 400×400 mm column and d = 140 mm: bo = 4(400 + 280) = 2,720 mm.
Two-way concrete shear (ACI §22.6): Vc = 0.33 λ √f'c bo d = 0.33×5.292×2,720×140 ≈ 658 kN; φ = 0.75 → φVc ≈ 494 kN ≫ Ru ✓
8.5 Illustrative Diagrams (Geometry, Strips, Reinforcement)
Slab Geometry and Support Layout (Plan)
Direct Design Method Strip Layout (Plan)
Indicative Reinforcement Layout (Plan)
Punching Shear Critical Perimeter (Interior Column)
Two-Way Slab Directional Moment Diagrams (Per Meter)
9. Detailing Summary
- Provide continuous top bars over supports and negative moment regions
- Bottom bars continuous over midspan; development length per §25.4
- Two-way slabs: adhere to strip widths (column/middle) and distribution
- Provide temperature/shrinkage reinforcement per §24.4.3
10. References
| Reference | Description |
|---|---|
| ACI 318-19M | Building Code Requirements for Structural Concrete |
| ACI 318R-19 | Commentary on Building Code Requirements for Structural Concrete |
| ACI SP-17 | ACI Design Handbook |