Solid Slab Calculator

How to Use This Calculator

This Solid Slab Calculator helps you estimate the required longitudinal reinforcement for a simply supported solid concrete slab (laje maciça) under uniform loading. You need to provide the slab dimensions, concrete strength, and applied loads. The calculator evaluates the maximum bending moment using the smaller span and then computes the steel area based on equilibrium of forces in the section.

Start by measuring or defining the clear spans of the slab: Lx is the shorter span, Ly the longer span. The slab thickness must be at least 7 cm for roof slabs and 10 cm for floor slabs according to most standards. Enter the concrete characteristic compressive strength (fck) in MPa — typical values are 20, 25, or 30 MPa. The additional load includes the self‑weight of covering, partitions, and live loads (e.g., 2 kN/m² for residential occupancy plus 1 kN/m² for finishing).

After filling all fields, click the calculate button. The result shows the required area of steel per meter width (mm²/m) and a suggested bar diameter, assuming a 20 mm cover and fyk = 500 MPa (CA‑50 steel). The calculator automatically applies minimum reinforcement (0.15% of concrete section) when the calculated area is lower.

Formula and Methodology

The design is based on the simplified bending theory for a one‑meter‑wide strip. The ultimate moment is: M = (q_total × Lx²) / 8, where q_total is the factored total load per meter (self‑weight + additional load + live load). The self‑weight is calculated from the slab volume × 25 kN/m³ (concrete density).

Then the required steel area As is obtained by solving the equilibrium equation: 0.85·fc·b·d·(d − 0.5·x) = M, with x being the neutral axis depth, and As = (0.85·fc·b·x) / fyd. The effective depth d is taken as total thickness minus 20 mm cover minus half the bar diameter (approximated as 20 mm). The concrete design strength fc = 0.85·fck/1.4, and steel design strength fyd = 500/1.15 MPa.

If the calculated As is less than the minimum area (0.15% of b·d for slabs in flexure), the minimum value is used. The suggested bar diameter is chosen to match the required area using a simple bar spacing criterion.

Practical Examples

Example 1: Lx = 4.0 m, Ly = 5.0 m, thickness = 12 cm, fck = 25 MPa, additional load = 3 kN/m².

• Self‑weight = 0.12 m × 25 kN/m³ = 3 kN/m².
• Total load = 3 + 3 + 2 (live load) = 8 kN/m² (assuming q=8 for simplicity).
• M = 8 × 4² / 8 = 16 kNm/m.
• d = 120 − 20 = 100 mm = 0.10 m.
• fc = 0.85 × 25 / 1.4 = 15.18 MPa.
• Solving gives As ≈ 350 mm²/m → φ10 mm every 22 cm (or φ8 mm every 14 cm).

Example 2: Lx = 5.5 m, Ly = 6.0 m, thickness = 15 cm, fck = 20 MPa, additional load = 2 kN/m².

• Self‑weight = 0.15 × 25 = 3.75 kN/m².
• Total load = 3.75 + 2 + 1.5 = 7.25 kN/m².
• M = 7.25 × 5.5² / 8 = 27.4 kNm/m.
• d = 150 − 20 = 130 mm.
• fc = 0.85 × 20 / 1.4 = 12.14 MPa.
• As ≈ 490 mm²/m → φ12.5 mm every 25 cm.

Tips and Best Practices

• Always verify that the slab thickness satisfies deflection limits: typically Lx/35 for simply supported slabs.
• Consider crack control: maximum bar spacing ≤ 20 cm for exposed slabs.
• Remember that this calculator does not check shear or punching shear; ensure slab thickness is adequate for shear forces.
• For two‑way slabs (Ly/Lx ≤ 2), distribute reinforcement in both directions; this calculator gives the reinforcement for the critical direction.
• Consult a licensed structural engineer for final design and detailing.