Slab reinforcement Buildingd, ω, As — and a bar every how many cm

What does each part do? (30 seconds)

• 10⁵ — unit conversion: t·m → kg·cm.
• ω — how much of the section is in compression. (1 − 0.5ω)·d = the lever arm between the compression in the concrete and the tension in the steel.
• f_cd — design strength of the concrete (B30: 13). f_sd — design strength of the steel (P500: 4350 kg/cm²). Some booklets write 435 and 10⁴ instead of 4350 and 10⁵ — same result.
• Other concrete: B40 — f_cd = 17.4, ρ = 0.00157 · B50 — f_cd = 21.7, ρ = 0.0018.
• A_s,min — even when the moment is small, a slab never goes out without a minimum mesh.

The slab from the path: two moments, two meshes

• Top
  M_d = 2.57 → ω = 0.148 → A_s = 5.318 cm²/m → Φ12@21 5.38 ≥ 5.32 ✓
• Bottom
  M_d = 1.439 → ω = 0.080 → 0.1 → A_s = 2.902 cm²/m → Φ8@17 2.96 ≥ 2.90 ✓
• Minimum
  0.0013 · 100 · 12 = 1.56 cm²/m — smaller than both, does not govern

That's it. Φ12 every 21 cm at the top over the support, Φ8 every 17 cm at the bottom in the spans.

The full calculation lines

Top reinforcement

• Negative moment · top reinforcementM_d = 2.57 t·m
• Effective depthd = h − 3 = 15 − 3 = 12 cm
• Minimum reinforcementA_s,min = ρ · 100 · d = 0.0013 · 100 · 12 = 1.56 cm²/m
• ω = 1 − √(1 − 2 · M_d · 10⁴100 · d² · f_cd) = 1 − √(1 − 2 · 2.57 · 10⁴100 · 12² · 13) = 0.1483
• Required reinforcementA_s,req = M_d · 10⁵d · (1 − 0.5ω) · f_sd = 2.57 · 10⁵12 · (1 − 0.5 · 0.1483) · 4350 = 5.318 cm²/m
• Governing reinforcementA_s = max(A_s,min , A_s,req) = 5.318 cm²/m
• Bar selectionΦ12@21 → A_s = 5.38 ≥ 5.32 cm²/m ✓

Bottom reinforcement

• Positive moment · bottom reinforcementM_d = 1.439 t·m
• Effective depthd = h − 3 = 15 − 3 = 12 cm
• Minimum reinforcementA_s,min = ρ · 100 · d = 0.0013 · 100 · 12 = 1.56 cm²/m
• ω = 1 − √(1 − 2 · M_d · 10⁴100 · d² · f_cd) = 1 − √(1 − 2 · 1.439 · 10⁴100 · 12² · 13) = 0.0801
• Required reinforcementA_s,req = M_d · 10⁵d · (1 − 0.5ω) · f_sd = 1.439 · 10⁵12 · (1 − 0.5 · 0.1) · 4350 = 2.902 cm²/m
• Governing reinforcementA_s = max(A_s,min , A_s,req) = 2.902 cm²/m
• Bar selectionΦ8@17 → A_s = 2.96 ≥ 2.9 cm²/m ✓

How to read the bar table

Area of one bar (cm²): Φ8 — 0.503 · Φ10 — 0.785 · Φ12 — 1.131 · Φ14 — 1.539 · Φ16 — 2.011. A bar every s cm gives per 1 m: A_s = a · 100 / s. Here is part of the table (cm²/m):

Reinforcement area per 1 m by diameter and spacing

Diameter	@10	@15	@20	@25
Φ8	5.03	3.35	2.51	2.01
Φ10	7.85	5.24	3.93	3.14
Φ12	11.31	7.54	5.65	4.52
Φ14	15.39	10.26	7.69	6.15

The way: choose a diameter → s = a · 100 / A_s → round down to a whole cm, and no more than 25. Top: 113.1 / 5.318 = 21.3 → Φ12@21 ✓ · Bottom: 50.3 / 2.902 = 17.3 → Φ8@17 ✓

Distribution reinforcement: in the direction perpendicular to the main reinforcement: A_s = max(0.2 · A_s,main , 0.5 · A_s,min), spacing up to min(30 , 3d) cm. It spreads the load and holds the main bars in place.

Which diameter? Any diameter that gives a reasonable spacing (not too dense) works. BST picks the option with the smallest excess above the required.

5 traps

• d = h. Always h − 3.
• ω = 0.08 and carrying on. Less than 0.1 → substitute 0.1.
• 10⁴ and 10⁵ get swapped. 10⁴ in the ω formula, 10⁵ in the A_s formula.
• Spacing of 30 cm. The maximum is 25.
• Forgetting A_s,min. When the moment is small — the minimum governs.