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

Ribbed slab solution procedure10 steps — the same route, four changes

The solution procedure is the recipe. 10 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.

The flagship exercise: a ribbed slab on 3 beams · 6.3 + 6.3 m · h = 30 · t_f = 5 · b_w = 15 · b_f = 65 · B30

Static schemes

Goal a scheme for the slab — and in a ribbed slab also for the hidden beams

What is special in a ribbed slab
Rib	a small beam with the rib width — the reinforcement sits in it
Block	filling between the ribs — does not carry, only weighs
Hidden beam	a beam at the slab depth — its scheme is needed for h₀

Order of work

Look at the plan: the slab rests on beams. The spanning direction — perpendicular to the beams.
Draw a static scheme for every slab strip: every beam that crosses it = a support. A free end = a cantilever.
Measure every span from beam centre to beam centre, and write it on the scheme in metres.
Count: how many supports, is there a cantilever, and identify the scheme type in the coefficient table.
Also draw a scheme for every hidden beam the ribs rest on — its spans go into the estimated thickness.

In the flagship exercise

Scheme 1: 3 supports · 6.3 + 6.3 m
The hidden beams: in the flagship exercise the thickness is given — their scheme is not needed for h₀.

⚠ 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.

Equivalent spans L₀

Goal L₀ for every span — of the slab and of the hidden beams

L₀ = coefficient · L

L_0(max) = the largest in the slab · L_0,beam(max) = the largest in the hidden beams

L: The span from beam centre to beam centre (cm)
coefficient: by the position of the span in the scheme — the same table as in the solid slab

Equivalent span coefficients — by scheme
Scheme	Coefficients (from left to right)
2 supports	1
3 supports	0.8 · 0.8
4 supports	0.8 · 0.6 · 0.8
Cantilever + span	2.2 · 0.9
Cantilever + 2 spans	2.2 · 0.7 · 0.8
Cantilever + span + cantilever	2.2 · 0.8 · 2.2
Continuity fixity = an end that continues into another span. More schemes — in the BST tables appendix.

Order of work

For every span in the slab schemes: L₀ = coefficient × L (cm). Mark the largest.
For every span in the hidden-beam schemes: the same calculation. Mark the largest separately.
Both values go into the assumed thickness — each with its own divisor.

In the flagship exercise

Scheme 1 · span 1L₀ = c · L = 0.8 · 630 = 504 cm
Scheme 1 · span 2L₀ = c · L = 0.8 · 630 = 504 cm
The largest equivalent spanL_0(max) = 504 cm

⚠ Common mistake Using the L₀ of the hidden beam in the thickness check. The check (step 5) — with L₀(max) of the slab only.

03
Assumed thickness h₀ and the section

Goal choose an initial thickness and define the rib section

h₀ = max(L_0(max)18 , L_0,beam(max)14) ≥ 20
b_f = B + b_w
t_f
Thickness of the top flange (cm) — given, usually 5
b_w
Rib width (cm) — given, usually 15
B
Block width (cm)
b_f
The width one rib carries — from mid-block to mid-block

Order of work
1. Thickness given? Skip h₀ — but define the section.
2. h₀ = the larger of L₀(max)/18 of the slab and L₀,beam(max)/14 of the hidden beam. Round up to a whole centimetre, minimum 20.
3. Section not given → take t_f = 5, b_w = 15, b_f = 65 (block 50).
4. Draw the section: flange, rib, block — and mark h, t_f, b_w, b_f.
In the flagship exercise
- Slab thickness — givenh = 30 cm
- The section is given: t_f = 5, b_w = 15, b_f = 65 cm (block 50).
⚠ Common mistake Rounding h₀ to the nearest 5 cm. With ribs, round to a single centimetre — 26.55 → 27, not 30.
04
Equivalent thickness, self-weight and service load

Goal convert the ribbed section into an equivalent concrete thickness and sum the loads

t_eq = t_f + b_w · (h − t_f)b_f + 1
S.w = 2.5 · t_eq + 0.5 · (h − t_eq)100
f_ser = S.w + Δg + q
t_eq
Thickness of the concrete alone, spread over the full width (cm); +1 for safety
2.5 · 0.5
Unit weight: concrete · block (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. Equivalent t: the flange + the proportional part of the rib + 1 cm.
2. Self-weight: the concrete (equivalent t × 2.5) plus the block (the rest × 0.5), divided by 100.
3. f_ser = S.w + Δg + q — as in the solid slab, in t/m².
4. Section not given: t_f = 5, b_w = 15, b_f = 65 → equivalent t = 5 + 15·(h − 5)/65 + 1.
In the flagship exercise
- Equivalent thicknesst_eq = t_f + b_w · (h − t_f)b_f + 1 = 5 + 15 · (30 − 5)65 + 1 = 11.77 cm
- Self-weightS.w = 2.5 · t_eq + 0.5 · (h − t_eq)100 = 2.5 · 11.77 + 0.5 · (30 − 11.77)100 = 0.385 t/m²
- Service loadf_ser = S.w + Δg + q = 0.385 + 0.2 + 0.3 = 0.885 t/m²
⚠ Common mistake Calculating the self-weight as in the solid slab: 2.5 · h / 100.
Why it is a mistake
Most of the section is block (0.5 t/m³), not concrete. A 30 cm ribbed slab weighs 0.385 t/m², not 0.75 — almost a factor of 2, which enters everything that follows.
How to avoid it
First the equivalent t, then two parts in the weight: concrete and block.
In the flagship exercise
Equivalent t = 5 + 15·25/65 + 1 = 11.77 · S.w = (2.5·11.77 + 0.5·18.23)/100 = 0.385 t/m² · f_ser = 0.885.
05
Thickness check — K₁₁ from the table

Goal make sure the thickness meets deflection — with the K₁₁ of a ribbed slab

K₁₂ = 24.4∛(f_ser)
ht_f , b_fb_w → K₁₁ from the table
h_req = L₀(max)K₁₁ · K₁₂ · K₁₃ ≤ h
K₁₁
Ribbed slab coefficient — by the two section ratios, from the K₁₁ table
K₁₂
Load coefficient — by the service load
K₁₃
Aggregate and concrete — from the table

K₁₃ — aggregate and concrete coefficient
B20 B30 B40 B50
Limestone aggregate 0.97 1 1.02 1.05
Dolomite aggregate 1.01 1.04 1.07 1.09

Order of work
1. K₁₂ = 24.4 / ∛f_ser.
2. Calculate h/t_f (for example 30/5 = 6) and b_f/b_w (65/15 = 4.33). Look it up in the K₁₁ table for ribs — row by h/t_f, column by b_f/b_w. A value between columns → round to the nearest value (BST: 4.33 → 4.3).
3. Required h = L₀(max) / (K₁₁·K₁₂·K₁₃), round up. Passes: ≤ h.
4. Fails: given thickness → 'does not meet the deflection requirements'. h₀ → +1 cm and back to step 4 (equivalent t and S.w change).
In the flagship exercise
- Load coefficientK₁₂ = 24.4∛(f_ser) = 24.4∛(0.885) = 25.41
- Section ratios for the K₁₁ tableht_f = 305 = 6
- b_fb_w = 6515 = 4.33
- Ribbed slab coefficient — from the tableK₁₁ = 0.747
- Aggregate and concrete coefficientK₁₃ = 1
- Required thicknessh_req = L₀(max)K₁₁ · K₁₂ · K₁₃ = 5040.747 · 25.41 · 1 = 26.55 cm
- Thickness check (rounding up)h_req = 26.55 → 27 ≤ h = 30 cm ✓
⚠ Common mistake Taking K₁₁ = 1 as in the solid slab.
Why it is a mistake
With ribs K₁₁ is less than 1 (about 0.75) — the slab is more flexible, and the required h is larger. With K₁₁ = 1 the check 'passes' a slab that fails.
How to avoid it
Ribs = the K₁₁ table. Two ratios, one value. The full table is in the BST tables appendix.
In the flagship exercise
h/t_f = 6, b_f/b_w = 4.33 → K₁₁ = 0.747 · required h = 504 / (0.747 · 25.41 · 1) = 26.55 → 27 ≤ 30 ✓
06
Design loads — and per rib

Goal design load per square metre, then for one rib

g = Δg + S.w
f_d,min = 1.2 · g (scheme with a cantilever: 1.0 · g)
f_d,max = 1.4 · g + 1.6 · q
f_{d,max per rib} = f_d,max · b_f
b_f
Width the rib carries — in metres! (65 cm = 0.65 m)
f_{d,max per rib}
Distributed load on the rib (t/m) — with it we move on to the statics formulas

Order of work
1. g = Δg + S.w by the final thickness. f_d,max = 1.4·g + 1.6·q (t/m²).
2. Multiply by b_f in metres: the load one rib carries, in t/m.
3. From here the rib is a beam: moments and reinforcement — for one rib, not per metre.
4. f_d,min (1.2·g or 1.0·g with a cantilever) — only for unfavourable load cases, also × b_f.
In the flagship exercise
- Dead loadg = Δg + S.w = 0.2 + 0.385 = 0.585 t/m²
- Minimum design load — regular schemef_d,min = 1.2 · g = 1.2 · 0.585 = 0.702 t/m²
- Maximum design loadf_d,max = 1.4 · g + 1.6 · q = 1.4 · 0.585 + 1.6 · 0.3 = 1.299 t/m²
- Design load on a rib (width b_f in metres)f_d,max per rib = f_d,max · b_f = 1.299 · 0.65 = 0.844 t/m
⚠ Common mistake Multiplying by b_f in centimetres (65) instead of metres (0.65) — the load per rib comes out 100 times too large.

K₁₃ — aggregate and concrete coefficient
	B20	B30	B40	B50
Limestone aggregate	0.97	1	1.02	1.05
Dolomite aggregate	1.01	1.04	1.07	1.09

Design moments per rib

Goal moments in the span and over the supports — for one rib

M_d = coefficient · F · L²

F = f_{d,max per rib}

L: The real span in metres
F: Design load per rib (t/m)
coefficient: From the scheme table — negative over a support, positive in the span

Moments in a continuous slab — coefficient × F × L²
Scheme	Over the support	In the span
2 supports	—	F·L² / 8
3 supports, equal spans	−F·L² / 8	0.07·F·L²
4 supports, equal spans	−0.1·F·L²	0.08 · 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

The same table as for the solid slab — only F is the load per rib.
Over every internal support: negative. In every span: positive. Cantilever: −F·L²/2.
The result is in t·m for one rib. The largest of each sign goes to the reinforcement.

In the flagship exercise

Negative moment over the middle support BM_d = −0.125 · F · L² = −0.125 · 0.844 · 6.3² = -4.187 t·m
Maximum moment in span 1M_d = 0.07 · F · L² = 0.07 · 0.844 · 6.3² = 2.345 t·m
Maximum moment in span 2M_d = 0.07 · F · L² = 0.07 · 0.844 · 6.3² = 2.345 t·m

⚠ Common mistake Substituting L₀ instead of L. The moment — with the real span in metres.

08
Positive reinforcement — in the span (T-section)

Goal bottom reinforcement per rib: the compression is in the wide flange

d = h − 3
A_s,min = ρ · b_w · d
ω = 1 − √(1 − 2 · M_d · 10⁴b_f · d² · f_cd)
x = ω · d ≤ t_f
A_s = M_d · 10⁵0.95 · d · f_sd ≥ A_s,min
ω < 0.1 → ω = 0.1 · x > t_f — the compression goes down into the rib: calculate as a rectangular section of width b_w (as in step 9)
M_d
Design moment of the segment (t·m)
d
Effective depth — to the centre of the reinforcement (cm)
f_cd
Concrete strength: 130 kg/cm² for B30
f_sd
Steel strength: 4350 kg/cm² (P500)
b_f
Compression width in the span — the whole flange (cm)
x
Depth of the compression zone (cm)

Minimum reinforcement coefficient
B30 B40 B50
ρ 0.0013 0.00157 0.0018
Bar selection for a rib — As (cm²)
Φ10 Φ12 Φ14 Φ16
2 1.57 2.26 3.08 4.02
3 2.36 3.39 4.62 6.03
4 3.14 4.52 6.16 8.04
Minimum 2 bars in a rib. Bar area: Φ8 0.50 · Φ10 0.79 · Φ12 1.13 · Φ14 1.54 · Φ16 2.01

Order of work
1. d = h − 3. A_s,min = ρ · b_w · d — per rib (b_w, not 100).
2. ω with b_f: the whole flange is in compression. Came out less than 0.1 → 0.1.
3. Check x = ω · d ≤ t_f: does the compression zone stay in the flange? Yes → continue.
4. A_s = M_d · 10⁵ / (0.95 · d · f_sd), at least A_s,min. Choose bars for the rib — minimum 2.
5. Distribution rib (perpendicular to the ribs): every 2 m if q > 0.3 t/m², otherwise every 2.5 m; 2 bars at the top and 2 at the bottom, total = the largest positive A_s in the slab.
In the flagship exercise
Bottom reinforcement · 2Φ12
- Positive moment · bottom reinforcementM_d = 2.345 t·m
- Effective depthd = h − 3 = 30 − 3 = 27 cm
- Minimum reinforcementA_s,min = ρ · b_w · d = 0.0013 · 15 · 27 = 0.527 cm²
- Compression width — b_f (positive moment)ω = 1 − √(1 − 2 · M_d · 10⁴b_f · d² · f_cd) = 1 − √(1 − 2 · 2.345 · 10⁴65 · 27² · 13) = 0.0388
- Depth of the compression zonex = ω · d = 0.1 · 27 = 2.7 cm
- The compression zone is within the flangex = 2.7 ≤ t_f = 5 cm ✓
- Required reinforcementA_s,req = M_d · 10⁵0.95 · d · f_sd = 2.345 · 10⁵0.95 · 27 · 4350 = 2.102 cm²
- Governing reinforcementA_s = max(A_s,min , A_s,req) = 2.102 cm²
- Bar selection for a rib2Φ12 → A_s = 2.26 ≥ 2.1 cm² ✓
⚠ Common mistake Calculating ω in the span with b_w (15) instead of b_f (65).
Why it is a mistake
In the span the compression is at the top, in the wide flange. With b_w ω comes out several times larger, and sometimes 'there is no ω' — and the student enlarges the section for no reason.
How to avoid it
Span = positive = compression at the top = b_f. Support = negative = compression at the bottom, in the rib = b_w.
In the flagship exercise
ω = 0.0388 → 0.1 · x = 0.1 · 27 = 2.7 ≤ 5 ✓ · A_s = 2.345·10⁵ / (0.95 · 27 · 4350) = 2.10 cm² → 2Φ12 (2.26).
09
Negative reinforcement — over the support (rectangular section)

Goal top reinforcement per rib: the compression is in the narrow rib

A_s,min = ρ · b_w · d
ω = 1 − √(1 − 2 · M_d · 10⁴b_w · d² · f_cd)
A_s = M_d · 10⁵d · (1 − 0.5ω) · f_sd ≥ A_s,min
0.1 ≤ ω ≤ 0.4 · ω > 0.4 → the rib is not enough for the compression: flip a block next to the support (the rib widens) and continue with ω = 0.4
M_d
Design moment of the segment (t·m)
d
Effective depth — to the centre of the reinforcement (cm)
f_cd
Concrete strength: 130 kg/cm² for B30
f_sd
Steel strength: 4350 kg/cm² (P500)
b_w
Compression width over the support — the rib only (cm)

Bar selection for a rib — As (cm²)
Φ10 Φ12 Φ14 Φ16
2 1.57 2.26 3.08 4.02
3 2.36 3.39 4.62 6.03
4 3.14 4.52 6.16 8.04
Minimum 2 bars in a rib. Bar area: Φ8 0.50 · Φ10 0.79 · Φ12 1.13 · Φ14 1.54 · Φ16 2.01

Order of work
1. ω with b_w: over the support the compression is at the bottom, in the rib.
2. ω between 0.1 and 0.4 → A_s = M_d · 10⁵ / (d · (1 − 0.5ω) · f_sd).
3. ω > 0.4 → 'flipping a block': next to the support remove a block and cast solid concrete, and continue the calculation with ω = 0.4.
4. Choose bars for the rib (≥ 2), and draw: top reinforcement over the support, bottom in the span.
In the flagship exercise
Top reinforcement · 4Φ12
- Negative moment · top reinforcementM_d = 4.187 t·m
- Effective depthd = h − 3 = 30 − 3 = 27 cm
- Minimum reinforcementA_s,min = ρ · b_w · d = 0.0013 · 15 · 27 = 0.527 cm²
- Compression width — b_w (negative moment)ω = 1 − √(1 − 2 · M_d · 10⁴b_w · d² · f_cd) = 1 − √(1 − 2 · 4.187 · 10⁴15 · 27² · 13) = 0.359
- Required reinforcementA_s,req = M_d · 10⁵d · (1 − 0.5ω) · f_sd = 4.187 · 10⁵27 · (1 − 0.5 · 0.359) · 4350 = 4.345 cm²
- Governing reinforcementA_s = max(A_s,min , A_s,req) = 4.345 cm²
- Bar selection for a rib4Φ12 → A_s = 4.52 ≥ 4.35 cm² ✓
⚠ Common mistake ω came out 0.55 over the support — and the student adds compression reinforcement as in a solid slab.
Why it is a mistake
With ribs the solution is to change the section: flipping a block. Compression reinforcement in a 15 cm rib is not practical, and it is not what the standard requires here.
How to avoid it
Ribs: ω > 0.4 → flipping a block + ω = 0.4. Solid: ω > 0.4 → compression reinforcement. Two worlds, two solutions.
In the flagship exercise
ω = 0.359 ≤ 0.4 ✓ · A_s = 4.187·10⁵ / (27 · 0.82 · 4350) = 4.35 cm² → 4Φ12 (4.52).

Minimum reinforcement coefficient
	B30	B40	B50
ρ	0.0013	0.00157	0.0018

Bar selection for a rib — As (cm²)
	Φ10	Φ12	Φ14	Φ16
2	1.57	2.26	3.08	4.02
3	2.36	3.39	4.62	6.03
4	3.14	4.52	6.16	8.04
Minimum 2 bars in a rib. Bar area: Φ8 0.50 · Φ10 0.79 · Φ12 1.13 · Φ14 1.54 · Φ16 2.01

Bar selection for a rib — As (cm²)
	Φ10	Φ12	Φ14	Φ16
2	1.57	2.26	3.08	4.02
3	2.36	3.39	4.62	6.03
4	3.14	4.52	6.16	8.04
Minimum 2 bars in a rib. Bar area: Φ8 0.50 · Φ10 0.79 · Φ12 1.13 · Φ14 1.54 · Φ16 2.01

Load transfer to the hidden beam

Goal how much load each hidden beam receives from the slab — when required

f_{to beam} = f · (left coeff. · L_left + right coeff. · L_right)

f: The service load for the beam width · the maximum design load for its reinforcement (t/m²)
L: The real span next to the beam (m)

Transfer coefficients — how much of the slab reaches each beam
Scheme	To the end beam	To an inner beam
2 supports	0.5	—
3 supports	0.4	0.6 + 0.6
4 supports	0.4	0.5 + 0.5 (inner)
Cantilever	The whole cantilever — 1	—
List of all the schemes: /learn/slabs/tables/ · the coefficient refers to the span next to the beam.

Order of work

The same calculation as in a solid slab: on each side coefficient × span, add up, multiply by the slab load (t/m²).
The result in t/m on the hidden beam — continue to the beam-width solution procedure.
In a ribbed slab the beam is hidden: its depth = the slab thickness, so the width is calculated.

In the flagship exercise

In the flagship exercise there is no hidden beam to calculate — the load transfer is demonstrated in the beam-width solution procedure.

⚠ Common mistake Transferring the load per rib (t/m) to the beam instead of the slab load (t/m²). The transfer — always from the slab's t/m².

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Slabs binder · BST · beamsolvertool.com/en/learn/sdp/slab-ribbed/

Solve a ribbed slab in BST BST guides you by the same procedure — one rib, step by step

All pages as a print-ready PDF — in the Binders to download of BST PRO

Ribbed slab solution procedure10 steps — the same route, four changes

Static schemes

Equivalent spans L₀

Assumed thickness h₀ and the section

Equivalent thickness, self-weight and service load

Thickness check — K₁₁ from the table

Design loads — and per rib

Design moments per rib

Positive reinforcement — in the span (T-section)

Negative reinforcement — over the support (rectangular section)

Load transfer to the hidden beam