Day 0Air 9 CConcrete -GBP 0
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House Build Sequence Lab

Design the house, then sequence the trades. Concrete must gain strength before it is loaded; joists and roof trusses are checked against dead, imposed and snow loads.

Scenarios

Design

Build sequence

Live checks (Eurocode-style)

under 70%70-100%over 100%

Concrete curing

measured fcm(t)forecast at the mean temperaturedashed: trade gates

Schedule and cost

Experiments: independent numerical checks

Press a button. Each experiment recomputes a result by a different route and prints the error.

Site log

The science

Concrete strength. Strength grows with equivalent age. The temperature-adjusted age is the sum of the elapsed time weighted by an Arrhenius rate that equals 1 at 20 C (EN 1992-1-1 Annex B style):

t_eq = Σ exp(13.65 - 4000/(273+T)) · Δt   ;   f_cm(t) = exp{ s [1 - √(28/t_eq)] } · f_cm28

f_cm28 = f_ck + 8 MPa; s = 0.38 (slow), 0.25 (normal), 0.20 (rapid cement). The Nurse-Saul index M = Σ(T + 10)Δt is shown for comparison. Simplifications: hydration is switched off at or below 0 C; if the concrete freezes before it has 5 MPa, 25% of its strength is lost for good; hydration heat, size effect and curing method are ignored.

Trade gates. Formwork/light floor loads need 5 MPa. For walls, roof and cladding this lab uses its own rule: required strength = 6 x (1.35 x line load x 0.6 m / 38x140 mm stud contact area), at least 5 MPa. The handover needs 70% of f_cm28. These are teaching rules, not code clauses. Loading earlier is possible with the override, and costs a crack repair (4 days, GBP 3500).

Floor joist (simply supported, uniform load, EN 1995-1-1):

M = wL²/8   σ = M/W, W = bh²/6   f_m,d = k_mod k_h f_m,k / γ_M   (γ_M = 1.3)
w = 1.35 g + 1.5 q   k_mod = 0.6 (permanent), 0.8 (medium-term imposed)
δ = 5wL⁴/(384 E I) · (1+k_def) ≤ L/250   (k_def = 0.6, ψ_2 = 0.3)
τ = 1.5 V /(0.67 b h) vs k_mod f_v,k/γ_M

Lateral-torsional buckling is assumed prevented by the deck and blocking; vibration, notches and bearing length are not checked.

Snow. s = μ1 Ce Ct sk with Ce = Ct = 1; μ1 = 0.8 up to 30 degrees, then 0.8(60 - α)/30, zero at 60 degrees (EN 1991-1-3). Two arrangements are analysed: uniform and unbalanced (0.5 μ1 on one slope). Wind, drift and sliding snow are not modelled.

Roof truss. A pin-jointed Howe truss (13 members). The solver assembles the 16 joint equations and solves them by Gaussian elimination for the member forces and the reactions. Combinations: 1.35 G (k_mod 0.6) and 1.35 G + 1.5 S (k_mod 0.9, short-term). Tension: N/A ≤ k_mod k_h f_t,0,k/γ_M. Compression uses the EN 1995 column factor k_c from the relative slenderness λrel = (l/i)/π · √(f_c,0,k/E0.05); the chords add the bending between joints M = wl²/8 (continuity ignored, which is conservative). Not modelled: nail-plate joints and slip, truss deflection and creep, out-of-plane bracing design (assumed adequate), wind uplift, the ceiling and water-tank loads beyond a fixed 0.35 kN/m².

Footings. Allowable soil bearing under characteristic loads of roof, floor, wall and footing weight. Settlement, frost depth and eccentricity are not checked.

Money and time. Prices (GBP) and crew days are illustrative round numbers, not a quotation. This is a teaching model, not a design tool and not a substitute for a structural engineer.

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