Beam depth solution procedure7 steps — segment by segment up to the check
The solution procedure is the recipe. 7 fixed steps — whatever exercise is in front of you.
Scheme → L₀ → h₀ → Loads → Governing segment → K₁₂ → Check
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 beam = a scheme of its own: every column is a support
What counts Column A support — measure from column centre to column centre An end without a column Cantilever Segment The part between two columns — every segment has its own load Order of work
- Mark the beam on the plan, and colour every segment differently.
- Draw a scheme: a support at every column, the spans in metres from centre to centre.
- Next to every segment write: which slab rests on it from the left and from the right — or 'opening' (no slab).
- A beam has uniformly distributed loads only — every segment with its own value.
In the flagship exercise
- The beam: 3–2 · 2–1 — a support at every column
- Segment 3–2 faces an opening (no slab) · segment 2–1 carries a slab on one side.
⚠ Common mistake Drawing a support at every crossing beam, as in a slab. In a beam — only columns are supports.
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02
Equivalent spans L₀
Goal L₀ for every segment of the beam
L0 = coefficient · L- L
- The span from column centre to column centre (cm)
- coefficient
- By the position of the segment in the scheme — from the table
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: the coefficient from the table × the span, in centimetres.
- Write it in a table — a row per segment. The table fills up in the next steps.
In the flagship exercise
- Equivalent span — 3–2L0,3–2 = c · L = 0.8 · 340 = 272 cm
- Equivalent span — 2–1L0,2–1 = c · L = 0.8 · 460 = 368 cm
⚠ Common mistake One coefficient for the whole beam. Every segment gets the coefficient of its position: end 0.8, inner 0.6, cantilever 2.2.
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03
Estimated depth h₀
Goal an initial depth — when the depth is not given
h0 = L0(max)10 ≥ 30- L0(max)
- The largest equivalent span in the beam (cm)
- 30
- Minimum beam depth (cm)
Order of work
- Depth given? Skip — check with the given h.
- Not given: h₀ = L₀(max) / 10, round up in steps of 5 cm, minimum 30.
- The beam width b — given (usually 20–30 cm).
In the flagship exercise
- h₀ = 368 / 10 = 36.8 → 40 cm. The exercise gives h = 50, which was checked and grew by iteration to 55:
- Estimated h — iteration 2h = 50 + 5 = 55 cm
⚠ Common mistake Rounding to a single centimetre. In a beam round to 5 cm: 36.8 → 40.
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04
Service load for every segment
Goal self-weight + what comes from the slab — segment by segment
S.w = 2.5 · h100 · b100fser,beam = ffrom slab + S.wffrom slab = fser,slab · (coefficient · Lleft + coefficient · Lright)- S.w
- Self-weight of the beam (t/m)
- b, h
- Width and depth of the beam (cm)
- fser,slab
- The service load of the slab (t/m²)
- coefficient
- The transfer coefficient of the span next to the beam
Order of work
- S.w — by the estimated (or given) h and b.
- For every segment: transfer from the slab — from each side coefficient × span (m) × fser of the slab. A segment facing an opening gets 0.
- fser of the segment = from the slab + S.w. Write it in the table, in t/m.
- A beam-only exercise (no slab): fser = S.w + Δg + q, which are given in t/m.
In the flagship exercise
- Segment 3–2 — an opening, no slab scheme rests on itq3–2 = 0 t/m
- Segment 2–1q2–1 = fser · Σ(c · L) = 1 · (0.5 · 4.6) = 2.3 t/m
- Self-weight of the beamS.w = 2.5 · h100 · b100 = 2.5 · 55100 · 30100 = 0.413 t/m
- Service load — 3–2fser,3–2 = q + S.w = 0 + 0.413 = 0.413 t/m
- Service load — 2–1fser,2–1 = q + S.w = 2.3 + 0.413 = 2.713 t/m
⚠ Common mistake Giving all the segments the same load.
- Why it is a mistake
- A segment facing an opening carries only its own weight. A segment carrying a slab on both sides — several times more. The governing segment is set exactly by this difference.
- How to avoid it
- A table: a row per segment, a column for L₀, for fser and for the product. Fill it row by row.
- In the flagship exercise
- h = 55: S.w = 2.5·0.55·0.3 = 0.413 t/m · segment 3–2 (opening): 0.413 · segment 2–1: 2.3 + 0.413 = 2.713 t/m.
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05
The governing segment
Goal which segment dictates the depth
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 × the cube root of fser.
- The largest product = the governing segment. Continue only with it: its L₀ and its fser.
- Not necessarily the long span — a short span with a heavy load can govern.
In the flagship exercise
- Segment 3–2L0 · ∛(fser) = 272 · ∛(0.413) = 202.56
- Segment 2–1L0 · ∛(fser) = 368 · ∛(2.713) = 513.25
⚠ Common mistake Choosing the longest segment without calculating.
- Why it is a mistake
- The depth also depends on the load: ∛ of 8 times the load = 2 times. A short, loaded segment can need more depth than a long, empty one.
- How to avoid it
- Calculate the product for every segment, always. It is one row in the table.
- In the flagship exercise
- 3–2: 272 · ∛0.413 = 202.6 · 2–1: 368 · ∛2.713 = 513.3 → segment 2–1 governs.
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06
K₁₂, K₁₁, K₁₃
Goal the coefficients — with the beam width and the load of the governing segment
K12 = 5.25 · ∛(bfser)K11 = 1- b
- The beam width (cm)
- fser
- The service load of the governing segment (t/m)
- K11
- A beam with a rectangular section = 1
- 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
- K₁₂ = 5.25 · ∛(b / fser) — note: b divided by fser, of the governing segment.
- K₁₁ = 1 (rectangular section). K₁₃ from the table.
In the flagship exercise
- Load coefficientK12 = 5.25 · ∛(bfser) = 5.25 · ∛(302.713) = 11.7
- Rectangular beamK11 = 1
- Aggregate and concrete coefficientK13 = 1
⚠ Common mistake Flipping the fraction: fser / b. In a beam — b on top, fser at the bottom.
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07
Depth check
Goal required h ≤ h — or an iteration of 5 cm
hreq = L0(gov)K11 · K12 · K13 ≤ h- L0
- The equivalent span of the governing segment (cm)
- h
- The given or estimated depth (cm)
Order of work
- Required h = L₀ (governing) / (K₁₁·K₁₂·K₁₃), round up.
- Passes: required h ≤ h. A given depth that fails → 'the beam does not meet the deflection requirements'.
- h₀ that fails → increase by 5 cm and go back to step 4: S.w changes, fser changes, maybe the governing segment too.
- Write the final answer: b / h in centimetres.
In the flagship exercise
- Required depthhreq = L0K11 · K12 · K13 = 3681 · 11.7 · 1 = 31.45 cm
- Depth check (rounding up)hreq = 31.45 → 32 ≤ h = 55 cm ✓
⚠ Common mistake h₀ failed, increase by 5 — and check only the required-h line.
- Why it is a mistake
- New h = new self-weight = new fser in every segment = new K₁₂. Even the governing segment can change.
- How to avoid it
- Iteration = go back to step 4 and fill the table again. BST marks every iteration with a number.
- In the flagship exercise
- h = 55: K₁₂ = 5.25 · ∛(30/2.713) = 11.7 · required h = 368 / 11.7 = 31.45 → 32 ≤ 55 ✓ — and you can go lower: a try with 35 also passes.
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