Two-Way Slab Design per ACI 318-19: Step-by-Step Worked Example
Complete worked example of a 6m × 6m interior flat plate panel per ACI 318-19 Direct Design Method — from minimum thickness through reinforcement selection.
This article walks through a complete two-way flat plate design for a square interior panel using the ACI 318-19 Direct Design Method (DDM), as defined in §8.10. Every number is calculated from scratch.
Problem Statement
| Parameter | Value |
|---|---|
| Panel size (ln × ln) | 6.0 m × 6.0 m clear span |
| Concrete strength f’c | 25 MPa (3,625 psi) |
| Steel yield strength fy | 420 MPa (60,900 psi) |
| Slab system | Flat plate (no drop panels, no column capitals) |
| Panel type | Interior (all four edges continuous) |
| Loads | Dead: 7.0 kPa (superimposed) + self-weight; Live: 3.0 kPa |
Step 1 — Minimum Thickness (Table 8.3.1.1)
For an interior flat plate panel, the minimum thickness from ACI 318-19 Table 8.3.1.1 is:
h_min = ln × (0.8 + fy/200,000) / 33
Converting fy to psi: 420 MPa × 145.038 = 60,916 psi
ln in inches: 6.0 m / 0.0254 = 236.2 in (using clear span)
h_min = 236.2 × (0.8 + 60,916/200,000) / 33
= 236.2 × (0.8 + 0.305) / 33
= 236.2 × 1.105 / 33
= 261.0 / 33
= 7.91 in → round up to 8 in (203 mm)
Minimum slab thickness = 203 mm. Use h = 210 mm (round to nearest 10 mm construction increment).
Step 2 — Self-Weight and Factored Load
Self-weight: 0.210 m × 24 kN/m³ = 5.04 kN/m²
Superimposed dead: 7.0 kN/m²
Total dead load wD = 5.04 + 7.0 = 12.04 kN/m²
Live load wL = 3.0 kN/m²
Factored load per ACI 318-19 §5.3.1 (Comb. 1.2D + 1.6L):
wu = 1.2 × 12.04 + 1.6 × 3.0
= 14.45 + 4.80
= 19.25 kN/m²
Step 3 — Total Static Moment M₀ (§8.10.3.2)
The total static moment for one panel strip:
M₀ = wu × l₂ × ln² / 8
Where:
- l₂ = span in the perpendicular direction = 6.0 m (centre-to-centre for interior panels = clear span + column width ≈ 6.0 m assuming 300 mm square columns)
- ln = clear span in design direction = 6.0 m
M₀ = 19.25 × 6.0 × 6.0² / 8
= 19.25 × 6.0 × 36.0 / 8
= 4,158 / 8
= 519.8 kN·m
Total static moment M₀ = 519.8 kN·m per panel strip.
Step 4 — Distribute M₀ to Column and Middle Strips (§8.10.4–8.10.5)
For an interior span of a flat plate (no beams, αf = 0):
| Strip | Negative moment | Positive moment |
|---|---|---|
| Column strip | 0.75 × 0.65 M₀ | 0.60 × 0.35 M₀ |
| Middle strip | 0.25 × 0.65 M₀ | 0.40 × 0.35 M₀ |
The distribution at interior supports (§8.10.4.2): 65% negative, 35% positive.
Negative moments:
- Column strip: 0.75 × (0.65 × 519.8) = 0.75 × 337.9 = 253.4 kN·m
- Middle strip: 0.25 × 337.9 = 84.5 kN·m
Positive moments:
- Column strip: 0.60 × (0.35 × 519.8) = 0.60 × 181.9 = 109.2 kN·m
- Middle strip: 0.40 × 181.9 = 72.7 kN·m
Step 5 — Effective Depth and Reinforcement
Assume #13 bars (13 mm dia), clear cover = 20 mm:
d = h - cover - bar_dia/2
= 210 - 20 - 13/2
= 210 - 20 - 6.5
= 183.5 mm ≈ use d = 183 mm
Column strip width = l₂/2 = 6.0/2 = 3.0 m each side, total column strip width = 3.0 m.
Middle strip width = 6.0 - 3.0 = 3.0 m.
Column strip negative moment — As calculation:
Using the approximate method (low ρ, a ≈ As·fy / 0.85·f’c·b):
Mu_neg,col = 253.4 kN·m per 3.0 m strip
ρ_required = [0.85 f'c/fy] × [1 - √(1 - 2Mu/(φ·0.85·f'c·b·d²))]
b = 3,000 mm, d = 183 mm, f'c = 25 MPa, fy = 420 MPa, φ = 0.90
Rn = Mu / (φ·b·d²)
= (253.4 × 10⁶) / (0.90 × 3,000 × 183²)
= 253.4 × 10⁶ / (0.90 × 3,000 × 33,489)
= 253.4 × 10⁶ / 90,420,300
= 2.80 MPa
ρ = (0.85 × 25/420) × [1 - √(1 - 2×2.80/(0.85×25))]
= 0.05060 × [1 - √(1 - 0.2635)]
= 0.05060 × [1 - √0.7365]
= 0.05060 × [1 - 0.8582]
= 0.05060 × 0.1418
= 0.00717
As = ρ × b × d = 0.00717 × 3,000 × 183 = 3,939 mm²
Minimum As (§8.6.1.1): ρ_min = 0.0018 (for fy = 420 MPa, Grade 60), As_min = 0.0018 × 3,000 × 210 = 1,134 mm² — flexural demand governs.
Bar selection: Use #13 @ 150 mm → As = (3,000/150) × 132 = 2,640 mm²/strip.
Recalculate with #13: As per m = 132 mm²/bar. Bars needed = 3,939/132 = 29.8 → 30 bars over 3.0 m = 1 bar per 100 mm.
→ #13 @ 100 mm o.c. for column strip negative steel (As = 3,960 mm²/strip).
Summary Table
| Location | Mu (kN·m) | As_req (mm²/strip) | Bar selection |
|---|---|---|---|
| Col. strip negative | 253.4 | 3,939 | #13 @ 100 mm |
| Middle strip negative | 84.5 | 1,177 | #13 @ 250 mm |
| Col. strip positive | 109.2 | 1,476 | #13 @ 200 mm |
| Middle strip positive | 72.7 | 972 | #13 @ 250 mm |
All bars are Grade 420 (fy = 420 MPa), slab h = 210 mm.
What Civora Does Automatically
This calculation took about 45 minutes by hand. Civora runs it in under a second, checks punching shear at every column (the check most commonly missed), generates the reinforcement layout drawing, bar bending schedule, and a full calculation report — all in one click.