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The BST binders · Slabs binder · Solution procedure — the order of steps · updated 27.9.2026

Solid slab solution procedure9 steps — from the scheme to the bars

The solution procedure is the recipe. 9 fixed steps — whatever exercise is in front of you.

Schemes → L₀ → Thickness → Loads → M → Reinforcement → Transfer

In every step: goal, formula, order of work, common mistake — and the flagship exercise solved next to it, in the BST engine.

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The flagship exercise: a continuous slab on 3 beams · 4 + 4 m · q = 0.3 · Δg = 0.2 t/m² · B30
  1. 01

    Static schemes

    Goal turn the plan into schemes — supports, spans and cantilevers

    What you see in the plan
    BeamA support of the slab
    An edge without a beamCantilever (equivalent span × 2.2)
    Direction of tensionPerpendicular to the beams — the spans are measured along it

    Order of work

    1. Look at the plan: the slab rests on beams. The spanning direction — perpendicular to the beams.
    2. Draw a static scheme for every slab strip: every beam that crosses it = a support. A free end = a cantilever.
    3. Measure every span from beam centre to beam centre, and write it on the scheme in metres.
    4. Count: how many supports, is there a cantilever, and identify the scheme type in the coefficient table.

    In the flagship exercise

    • Scheme 1: 3 supports · 4 + 4 m

    ⚠ Common mistake Measuring the span from the face of the beam to the face of the next beam (the clear span), or forgetting a beam that crosses the strip.

    Why it is a mistake
    The equivalent span, the required thickness and the moments — all depend on L. A span that is too short reduces everything, and the slab comes out thinner than required.
    How to avoid it
    Always from beam centre to beam centre. Before moving on, count: every beam the strip crosses is a support, and a free end is a cantilever — with no support.
  2. 02

    Equivalent spans L₀

    Goal calculate the equivalent span for every span, and find the largest

    L0 = coefficient · L
    L0(max) = the largest of all the schemes
    L
    The span from beam centre to beam centre (cm)
    coefficient
    By the position of the span in the scheme — from the table
    Equivalent span coefficients — by scheme
    SchemeCoefficients (from left to right)
    2 supports1
    3 supports0.8 · 0.8
    4 supports0.8 · 0.6 · 0.8
    Cantilever + span2.2 · 0.9
    Cantilever + 2 spans2.2 · 0.7 · 0.8
    Cantilever + span + cantilever2.2 · 0.8 · 2.2
    Continuity fixity = an end that continues into another span. More schemes — in the BST tables appendix.

    Order of work

    1. For every span: find its coefficient in the table by the scheme and its position in it (edge / interior / cantilever).
    2. Multiply: L₀ = coefficient × L, in centimetres.
    3. Mark L₀(max) — the largest of all the spans in all the schemes. It sets the thickness of the whole slab.
    4. A scheme with a cantilever: the cantilever gets 2.2 — it usually governs, even when it is short.

    In the flagship exercise

    • Scheme 1 · span 1L0 = c · L = 0.8 · 400 = 320 cm
    • Scheme 1 · span 2L0 = c · L = 0.8 · 400 = 320 cm
    • The largest equivalent spanL0(max) = 320 cm

    ⚠ Common mistake Taking the longest span instead of the largest equivalent span.

    Why it is a mistake
    A 1.5 m cantilever gives L₀ = 330 cm — more than a 4 m end span (320 cm). The thickness is set by the equivalent span, not by the length.
    How to avoid it
    Calculate L₀ for every span without exception, and only then choose the largest.
    In the flagship exercise
    Two 4 m spans in a 3-support scheme: 0.8 · 400 = 320 cm for each — L₀(max) = 320 cm.
  3. 03

    Estimated thickness h₀

    Goal choose an initial thickness — only when the thickness is not given

    h0 = L0(max)25 ≥ 15
    L0(max)
    The largest equivalent span (cm)
    15
    Minimum thickness for a solid slab (cm)

    Order of work

    1. Thickness given in the question? Skip — go to the check with the given h.
    2. Not given: h₀ = L₀(max) / 25, and round up to a whole centimetre.
    3. Came out less than 15 → take 15.
    4. h₀ is an educated guess. The check in step 5 will confirm it or increase it.

    In the flagship exercise

    • Initial thicknessh0 = max(L0(max)25 , 15) = max(32025 , 15) = 15 cm

    ⚠ Common mistake Rounding down. An estimated thickness is always rounded up — a thickness that is too small fails the check.

  4. 04

    Service load

    Goal add up all the loads without safety factors

    S.w = 2.5 · h100
    fser = S.w + Δg + q
    h
    Slab thickness (cm)
    2.5
    Unit weight of concrete (t/m³)
    S.w
    Self-weight (t/m²)
    Δg
    Additional dead load — flooring, plaster (t/m²)
    q
    Live load (t/m²)

    Order of work

    1. Self-weight: the thickness in centimetres times 2.5, divided by 100 — gives t/m².
    2. Add the additional dead load Δg and the live load q, both from the question.
    3. fser is used for the thickness check (deflection) only — without safety factors.

    In the flagship exercise

    • 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

    ⚠ Common mistake Multiplying fser by 1.4 and 1.6 already here.

    Why it is a mistake
    The safety factors belong to the design (ultimate) loads. The deflection check is done at the service state — real loads.
    How to avoid it
    Two different loads, two names: fser for the thickness, fd for the reinforcement. Write the name next to each one.
    In the flagship exercise
    S.w = 2.5 · 15 / 100 = 0.375 · fser = 0.375 + 0.2 + 0.3 = 0.875 t/m² — without factors.
  5. 05

    Thickness check — K₁₁, K₁₂, K₁₃

    Goal make sure the thickness meets the deflection requirement of SI 466

    K12 = 24.4∛(fser)
    K11 = 1
    hreq = L0(max)K11 · K12 · K13 ≤ h
    K12
    Load coefficient — by the service load
    K11
    Slab-type coefficient: one-way solid slab = 1
    K13
    Aggregate and concrete coefficient — from the table
    K₁₃ — aggregate and concrete coefficient
    B20B30B40B50
    Limestone aggregate0.9711.021.05
    Dolomite aggregate1.011.041.071.09

    Order of work

    1. K₁₂ = 24.4 divided by the cube root of fser.
    2. K₁₁ = 1 (solid slab). K₁₃ — by the aggregate type and the concrete in the table.
    3. h required = L₀(max) divided by the product K₁₁·K₁₂·K₁₃, and round up.
    4. Passes: h required ≤ h. Fails with a given thickness → write 'the slab does not meet the deflection requirements'. Fails with h₀ → increase by 1 cm and go back from step 4 (S.w changes!).
    5. h larger than h required by more than 5 cm? Worth reducing and recalculating — saves concrete.

    In the flagship exercise

    • 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 ✓

    ⚠ Common mistake Increasing h after a failed check, but continuing with the old fser.

    Why it is a mistake
    New h = new self-weight = new fser = new K₁₂. The check with the old numbers is worth nothing.
    How to avoid it
    Every iteration starts from the self-weight. BST marks 'iteration 2' and recalculates everything.
    In the flagship exercise
    K₁₂ = 24.4 / ∛0.875 = 25.51 · h required = 320 / (1 · 25.51 · 1) = 12.54 → 13 ≤ 15 ✓ — passes in the first iteration.
  6. 06

    Design loads

    Goal move from the service state to the ultimate state — with the safety factors

    g = Δg + S.w
    fd,min = 1.2 · g (scheme with a cantilever: 1.0 · g)
    fd,max = 1.4 · g + 1.6 · q
    g
    The whole dead load: self-weight + Δg (t/m²)
    1.4 · 1.6
    The safety factors: dead · live
    fd,min
    For the 'light' spans in critical loading patterns (checkerboard)

    Order of work

    1. g = Δg + S.w — by the final thickness after the check.
    2. fd,max = 1.4·g + 1.6·q — the load for the reinforcement. All the moments are calculated with it.
    3. fd,min = 1.2·g (regular scheme) or 1.0·g (scheme with a cantilever) — only when a critical loading pattern is asked for: fd,max on the span being checked and on both sides of the support being checked, fd,min on the rest.
    4. The units: t/m² — load on a square metre of slab.

    In the flagship exercise

    • Dead loadg = Δg + S.w = 0.2 + 0.375 = 0.575 t/m²
    • Minimum design load — regular schemefd,min = 1.2 · g = 1.2 · 0.575 = 0.69 t/m²
    • Maximum design loadfd,max = 1.4 · g + 1.6 · q = 1.4 · 0.575 + 1.6 · 0.3 = 1.285 t/m²

    ⚠ Common mistake Forgetting the self-weight inside g. The dead load is Δg + S.w, not Δg alone.

  7. 07

    Design moments

    Goal find the moments in the span and over the supports without drawing diagrams

    Md = coefficient · F · L²
    F = fd,max
    L
    The real span in metres — not the equivalent one!
    F
    The maximum design load (t/m²) — for a strip 1 m wide
    coefficient
    From the scheme table: negative over a support, positive in the span
    Moments in a continuous slab — coefficient × F × L²
    SchemeOver the supportIn the span
    2 supports—F·L² / 8
    3 supports, equal spans−F·L² / 80.07·F·L²
    4 supports, equal spans−0.1·F·L²0.08 · 0.025 · 0.08
    Cantilever−F·L² / 2—
    Unequal spans and more schemes — in the tables appendix. F = the maximum design load (t/m² in a solid slab, t/m per rib in a ribbed slab).

    Order of work

    1. Identify the scheme in the table (number of supports, equal spans, cantilever).
    2. Over every interior support: negative moment = coefficient · F · L². In every span: positive moment = coefficient · F · L².
    3. Cantilever: M = −F · L² / 2 over its support — always, without a table.
    4. For every moment write: where it is, sign, value in t·m per metre of width. The largest of each sign goes to the reinforcement.

    In the flagship exercise

    • Negative moment over the middle support BMd = −0.125 · F · L² = −0.125 · 1.285 · 4² = -2.57 t·m
    • Maximum moment in span 1Md = 0.07 · F · L² = 0.07 · 1.285 · 4² = 1.439 t·m
    • Maximum moment in span 2Md = 0.07 · F · L² = 0.07 · 1.285 · 4² = 1.439 t·m

    ⚠ Common mistake Substituting the equivalent span L₀ in the formula instead of the real span.

    Why it is a mistake
    L₀ is meant for the deflection check only. The moment depends on the real span — L₀ is smaller and reduces the moment by (coefficient)².
    How to avoid it
    Two columns in the notebook: L₀ in centimetres for the thickness, L in metres for the moments.
    In the flagship exercise
    3 supports, equal spans: MB = −0.125 · 1.285 · 4² = −2.57 t·m · M span = 0.07 · 1.285 · 4² = 1.439 t·m.
  8. 08

    Reinforcement — top and bottom

    Goal calculate the reinforcement area per metre of slab and choose bars

    d = h − 3
    As,min = ρ · 100 · d
    ω = 1 − √(1 − 2 · Md · 10⁴100 · d² · fcd)
    As = Md · 10⁵d · (1 − 0.5ω) · fsd ≥ As,min
    ω < 0.1 → take ω = 0.1 · ω > 0.4 → the section is not enough for compression: add compression reinforcement (BST calculates: Mcd,max = 0.32·100·d²·fcd·10⁻⁴, ΔM = Md − Mcd,max, A's = ΔM·10⁵ / (fsd · (h − 6)), As = Mcd,max·10⁵ / (fsd · d · 0.8) + A's)
    Md
    Design moment of the segment (t·m)
    d
    Effective depth — to the centre of the reinforcement (cm)
    fcd
    Concrete strength: 130 kg/cm² for B30
    fsd
    Steel strength: 4350 kg/cm² (P500)
    ρ
    Minimum reinforcement coefficient — from the table
    100
    Strip width (cm) — calculated per metre of slab
    Minimum reinforcement coefficient
    B30B40B50
    ρ0.00130.001570.0018
    Design strength of concrete
    B30B40B50
    fcd1317.421.7
    MPa · in BST: 130 / 174 / 217 kg/cm²
    Bar selection per metre of slab — As (cm²/m)
    @ (cm)Φ8Φ10Φ12Φ14
    105.037.8511.3115.39
    153.355.247.5410.26
    202.513.935.657.69
    252.013.144.526.15
    As = area of one bar × 100 / the spacing. The full table (7–30 cm) — in the tables appendix.

    Order of work

    1. d = h − 3 (3 cm cover to the centre of the bars).
    2. As,min = ρ · 100 · d — the floor you never go below.
    3. ω for every moment: substitute Md, d, fcd. Came out less than 0.1 → 0.1.
    4. As = Md · 10⁵ / (d · (1 − 0.5ω) · fsd). Take the larger of As and As,min.
    5. Choose a bar and spacing from the table: As provided ≥ As, spacing up to 25 cm. Negative moment → top reinforcement, positive → bottom.
    6. Distribution reinforcement (perpendicular): As = max(0.2 · As main, 0.5 · As,min), spacing up to min(30, 3d).

    In the flagship exercise

    Top reinforcement · Φ12@21

    • 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 · Φ8@17

    • 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 ✓

    ⚠ Common mistake Forgetting the 10⁵ (or 10⁴) in the formula for ω and As, or mixing them up.

    Why it is a mistake
    Md in t·m, d in cm, fcd in kg/cm² — without the unit factor the result is 10,000 times too small. You get reinforcement of 0.0005 cm² and nobody notices.
    How to avoid it
    In BST: ω with 10⁴ and fcd = 13; As with 10⁵ and fsd = 4350. Quick check: As in an ordinary slab comes out between 2 and 15 cm²/m.
    In the flagship exercise
    Over B: d = 12, ω = 0.148, As = 2.57·10⁵ / (12 · 0.926 · 4350) = 5.32 cm²/m → Φ12@21 (5.38). In the span: 2.90 → Φ8@17 (2.96).
  9. 09

    Load transfer to the beams

    Goal calculate how much load each beam gets from the slab — when required

    fto beam = f · (left coeff. · Lleft + right coeff. · Lright)
    f
    Service load for the beam dimensions · maximum design load for the beam reinforcement (t/m²)
    L
    The real span next to the beam (m)
    coefficient
    The load-transfer coefficient of that span — from the table
    Transfer coefficients — how much of the slab reaches each beam
    SchemeTo the end beamTo an inner beam
    2 supports0.5—
    3 supports0.40.6 + 0.6
    4 supports0.40.5 + 0.5 (inner)
    CantileverThe whole cantilever — 1—
    List of all the schemes: /learn/slabs/tables/ · the coefficient refers to the span next to the beam.

    Order of work

    1. Mark the beam on the plan. On each side of it — which slab scheme rests on it, and which span.
    2. From each side: the load-transfer coefficient × the span (m). A cantilever gives its whole length × 1.
    3. Add the two sides and multiply by the slab load: fser when calculating the beam depth or width, fd,max when calculating the beam reinforcement.
    4. The result in t/m — a distributed load along the beam, from which you continue to the beam-depth solution procedure.

    In the flagship exercise

    • In the flagship exercise there is no beam to calculate — the load transfer is shown in the beam-depth solution procedure: segment 2–1 gets fser · (0.5 · 4.6).

    ⚠ Common mistake Transferring 0.5 of the span to the end beam. In a 3-support scheme the end beam gets 0.4 — the middle support takes 0.6 from each side.

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