BST — Beam Solver Tool ← Slabs binder

The BST binders · Slabs binder · The route · updated 27.9.2026

Solid slabFrom the data to the bars — in one exercise

Solid slab = a full concrete plate that rests on beams and works in one direction. Everything is calculated for a 1 m width of slab.

Equivalent span → Thickness → Design loads → Moments → Reinforcement → Transfer to the beams

6 steps, always in the same order. Each step has its own page in the binder — here you see them all working together.

  1. 01

    What is given in the exercise — and what each letter means

    • q — live load (people, furniture): 0.3 t/m²
    • Δg — additional dead load (flooring, plaster, partitions): 0.2 t/m²
    • Concrete and aggregate — B30, limestone aggregate. They set K13 and the reinforcement data.
    • The scheme — how the slab sits on the beams: here 3 supports (A, B, C) and two spans of 4 m.
    • Self-weight S.w — not given. It depends on the thickness, so it is calculated along the way.
    q = 0.3 · Δg = 0.2 t/m²ABC4 m4 m
    Slab scheme — like a continuous beam 1 m wide · scroll the drawing sideways
  2. 02

    6 steps — the BST numbers

    • Equivalent span
      L0 = c · L = 0.8 · 400 = 320 cm in every span (c — a coefficient by the type of scheme) → L0(max) = 320 How? →
    • Thickness
      Initial thickness h0 = max(⌈320/25⌉ , 15) = 15 cm → S.w = 0.375 → hreq = 12.54 → 13 ≤ 15 How? → Passes ✓
    • Design loads
      g = Δg + S.w = 0.2 + 0.375 = 0.575 → Fd,max = 1.4·0.575 + 1.6·0.3 = 1.285 t/m² How? →
    • Moments
      With F = Fd,max, L = 4 m: over B −0.125·F·L² = −2.57 · in the span 0.07·F·L² = 1.439 t·m How? →
    • Reinforcement
      d = h − 3 = 12 · top over B: Φ12@21 · bottom in the spans: Φ8@17 (there ω came out 0.08 — below the minimum, so 0.1 is substituted) How? →
    • To the beams
      With the service load fser = q + Δg + S.w = 0.875: beam B receives 0.875 · (0.6·4 + 0.6·4) = 4.2 t/m, beams A and C — 0.875 · 0.4·4 = 1.4 t/m How? →
    h = 15 Φ8@17 in the spans Φ12@21 over B
    The section: top reinforcement over the middle support, bottom in the spans

    That's it. A complete slab: 15 cm thick, two reinforcement meshes — and a load ready for every beam.

    All the calculation lines (as in the BST PDF)

    Thickness

    • Initial thicknessh0 = max(L0(max)25 , 15) = max(32025 , 15) = 15 cm
    • Self-weightS.w = 2.5 · h100 = 2.5 · 15100 = 0.375 t/m²
    • Service loadfser = S.w + Δg + q = 0.375 + 0.2 + 0.3 = 0.875 t/m²
    • Load coefficientK12 = 24.4∛(fser) = 24.4∛(0.875) = 25.51
    • Slab type coefficient — solid, one-way spanningK11 = 1
    • Aggregate and concrete coefficientK13 = 1
    • Required thicknesshreq = L0(max)K11 · K12 · K13 = 3201 · 25.51 · 1 = 12.54 cm
    • Thickness check (rounding up)hreq = 12.54 → 13 ≤ h = 15 cm ✓

    Design loads

    • Dead loadg = Δg + S.w = 0.2 + 0.375 = 0.575 t/m²
    • Maximum design loadfd,max = 1.4 · g + 1.6 · q = 1.4 · 0.575 + 1.6 · 0.3 = 1.285 t/m²

    Top reinforcement

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

    Bottom reinforcement

    • Positive moment · bottom reinforcementMd = 1.439 t·m
    • Effective depthd = h − 3 = 15 − 3 = 12 cm
    • Minimum reinforcementAs,min = ρ · 100 · d = 0.0013 · 100 · 12 = 1.56 cm²/m
    • ω = 1 − √(1 − 2 · Md · 10⁴100 · d² · fcd) = 1 − √(1 − 2 · 1.439 · 10⁴100 · 12² · 13) = 0.0801
    • Required reinforcementAs,req = Md · 10⁵d · (1 − 0.5ω) · fsd = 1.439 · 10⁵12 · (1 − 0.5 · 0.1) · 4350 = 2.902 cm²/m
    • Governing reinforcementAs = max(As,min , As,req) = 2.902 cm²/m
    • Bar selectionΦ8@17 → As = 2.96 ≥ 2.9 cm²/m ✓
  3. 03

    5 traps everyone falls into

    • fser instead of Fd,max in the moments. fser — for the thickness and the load transfer. Fd,max — for the moments and the reinforcement.
    • 1.4 and 1.6 get swapped. 1.4 on the dead load (g), 1.6 on the live load (q).
    • d = h. The effective depth is d = h − 3 — to the centre of the reinforcement.
    • Forgetting the top reinforcement. Negative moment over a support → the reinforcement is at the top.
    • Rounding hreq down. Always up: 12.54 → 13.

    Bonus: Beam? Slab? Same statics. The 0.125·F·L² of a single span is the qL²/8 from the beams binder.

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Slabs binder · BST · beamsolvertool.com/en/learn/slabs/solid-slab/

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