RC Slab Designer (ACI 318-19)

How the design works

The slab is designed as a 1 m wide strip. In coefficient mode the factored load wu = max(1.2D + 1.6L, 1.4D) is turned into moments with the ACI simplified coefficients (6.5.2), for example wu·ln²/14 at midspan of an end span with an integral support and wu·ln²/10 at the first interior support of three or more spans. For two-way slabs, or any slab analysed in software, choose "My analysis" and enter the moments per metre width directly.

For each location the tool finds the required steel, applies the 0.0018bh minimum, and chooses the bar spacing limited by 3h or 450 mm (7.7.2.3) and by crack control (24.3.2). It then confirms φMn ≥ Mu and that the section is tension-controlled. Distribution steel is 0.0018bh at no more than 5h or 450 mm.

ACI 318-19 points

• Minimum flexural steel is 0.0018·b·h for all steel grades (7.6.1.1); the old 0.0014 allowance for high-strength steel is gone.
• Slabs must be tension-controlled (7.3.3.1).
• One-way shear without stirrups uses Table 22.5.5.1(c): Vc = 0.66λs·λ·ρw^1/3·√f′c·b·d. Lightly reinforced slabs get noticeably less shear capacity than under ACI 318-14.

Worked example

End span, 4.0 m clear, three or more spans, h = 175 mm, SDL 1.5 and LL 3.0 kN/m², f′c 28 MPa, fy 420 MPa, Ø12 bars, 20 mm cover (d = 149 mm). D = 4.20 + 1.5 = 5.70 kN/m², wu = 1.2 × 5.70 + 1.6 × 3.0 = 11.64 kN/m². Mu+ = 11.64 × 4² / 14 = 13.30 kN·m/m and Mu− at the first interior support = 11.64 × 4² / 10 = 18.62 kN·m/m. At midspan the minimum steel (0.0018 × 1000 × 175 = 315 mm²/m) governs over the 240 mm²/m required; over the support 337 mm²/m is needed. Both give Ø12 @ 300 mm, limited by crack control (fs = 2/3fy → 300 mm), with φMn = 20.8 kN·m/m. Distribution steel is Ø10 @ 240. Shear: Vu = 25.0 kN/m at d ≤ φVc = 53.2 kN/m. Minimum thickness ln/24 = 167 mm ≤ 175 mm.

Cross-check: for a 250 mm slab with Ø12 @ 200 (d = 210 mm, f′c 40 MPa) the tool gives a = 6.99 mm and φMn = 44.14 kN·m/m, the same as an independent verified spreadsheet.

FAQ