Hidden-beam width solution procedure6 steps — the depth is given, the width from the deflection
The solution procedure is the recipe. 6 fixed steps — whatever exercise is in front of you.
Scheme → L₀ → Loads → Governing segment → K₁₂ → b
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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01
Static scheme of the beam
Goal a support at every column, a cantilever at a free end
Hidden beam Depth = the slab thickness — given, not checked Width the unknown — comes out of the deflection Load from the slab only — the self-weight is already inside the slab Order of work
- Mark the beam on the plan; a support at every column, the spans from centre to centre in metres.
- Next to every segment: which slab scheme rests on it from each side.
In the flagship exercise
- The beam: cantilever–4 · 4–5 · 5–6 — a support at every column
- The depth = the slab thickness: h = 33 cm.
⚠ Common mistake Adding self-weight to a hidden beam. It is part of the slab — its weight is already in the S.w of the slab.
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02
Equivalent spans L₀
Goal L₀ for every segment
L0 = coefficient · L- L
- The span from column centre to column centre (cm)
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 segment: coefficient × span, in centimetres. Cantilever × 2.2.
- Write it in the table, a row for every segment.
In the flagship exercise
- Equivalent span — cantilever–4L0,cantilever–4 = c · L = 2.2 · 140 = 308 cm
- Equivalent span — 4–5L0,4–5 = c · L = 0.7 · 590 = 413 cm
- Equivalent span — 5–6L0,5–6 = c · L = 0.8 · 460 = 368 cm
⚠ Common mistake Forgetting that the cantilever gets 2.2 — it is almost always the governing segment.
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03
Service load for every segment
Goal what arrives from the slab — segment by segment
fser,beam = fser,slab · (coefficient · Lleft + coefficient · Lright)- fser,slab
- The service load of the slab (t/m²)
- coefficient
- The transfer coefficient of the span next to the beam
- L
- The real span of the slab (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
- For every segment: on each side coefficient × span (m), add up, multiply by the fser of the slab.
- The result in t/m. A segment facing an opening: 0 — but in a hidden beam that is rare.
In the flagship exercise
- Segment cantilever–4qcantilever–4 = fser · Σ(c · L) = 0.847 · (0.6 · 6.3 + 0.6 · 6.3) = 6.403 t/m
- Segment 4–5q4–5 = fser · Σ(c · L) = 0.847 · (0.6 · 6.3 + 0.6 · 6.3) = 6.403 t/m
- Segment 5–6q5–6 = fser · Σ(c · L) = 0.847 · (1 · 1.8 + 0.5 · 6.3) = 4.193 t/m
⚠ Common mistake Using fd,max. For the beam width — fser (deflection = service state).
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04
The governing segment
Goal which segment dictates the width
L0 · ∛(fser) → the largest governs- L0
- The equivalent span of the segment (cm)
- fser
- The service load of that segment (t/m)
Order of work
- For every segment: L₀ × ∛fser. The largest governs.
- Continue with the L₀ and the fser of the governing segment only.
In the flagship exercise
- Segment cantilever–4L0 · ∛(fser) = 308 · ∛(6.403) = 571.93
- Segment 4–5L0 · ∛(fser) = 413 · ∛(6.403) = 766.91
- Segment 5–6L0 · ∛(fser) = 368 · ∛(4.193) = 593.41
⚠ Common mistake Choosing the segment with the largest load, or the longest one — without the product.
- Why it is a mistake
- Both factors play: a short cantilever with 2.2 against a long span with 0.7. Only the product decides.
- How to avoid it
- A row for every segment, a product for every row, mark the largest.
- In the flagship exercise
- cantilever–4: 308 · ∛6.4 = 571.9 · 4–5: 413 · ∛6.4 = 766.9 · 5–6: 368 · ∛4.19 = 593.4 → 4–5 governs.
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05
K₁₂ from the given depth
Goal this time K₁₂ comes out of the deflection formula — not out of the load
K11 = 1K12 = L0K11 · h · K13- h
- The beam depth = the slab thickness (cm)
- L0
- The equivalent span of the governing segment (cm)
- 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
- The depth is known, so flip the deflection formula: K₁₂ = L₀ / (K₁₁ · h · K₁₃).
- K₁₁ = 1 (rectangular section). K₁₃ from the table — usually 1.
In the flagship exercise
- Rectangular beamK11 = 1
- Aggregate and concrete coefficientK13 = 1
- Beam depth = slab thicknessh = 33 cm
- Load coefficientK12 = L0K11 · h · K13 = 4131 · 33 · 1 = 12.52
⚠ Common mistake Calculating K₁₂ = 5.25 · ∛(b / fser) — but b is the unknown. In a hidden beam K₁₂ comes out of the deflection.
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06
Beam width
Goal from K₁₂ and the load — the required width, rounded to 5
breq = fser · (K125.25)³- fser
- The service load of the governing segment (t/m)
- breq
- The width of the hidden beam (cm)
Order of work
- b = fser · (K₁₂ / 5.25)³ — the K₁₂ formula of a beam, flipped.
- Round up in steps of 5 cm.
- Final answer: h / b — the depth as the slab thickness, the width calculated.
In the flagship exercise
- Required widthbreq = fser · (K125.25)³ = 6.403 · (12.525.25)³ = 86.84 cm
- Beam width (steps of 5 cm)breq = 86.84 → b = 90 cm ✓
⚠ Common mistake Rounding down, or to a single centimetre.
- Why it is a mistake
- Missing width = deflection above the allowed. And a beam is cast in round dimensions — 86.84 does not exist on site.
- How to avoid it
- Always up, always to 5: 86.84 → 90.
- In the flagship exercise
- K₁₂ = 413 / (1 · 33 · 1) = 12.52 · b = 6.403 · (12.52/5.25)³ = 86.84 → b = 90 cm.
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