Probleemomschrijving
Bedrijfshal in Rotterdam, NL (Windzone 2, v_b,0 = 25 m/s). Grondvlak: 30m × 50m, nokhoogte h = 12m, plat dak. Terrein: open platteland ( terreincategorie II).
Berekening Stappen — 5 stappen
§4.2
Basiswindsnelheid v_b
Afleiding
v_b,0 = 25 m/s (NL Zone 2, NEN-NA EN 1991-1-4 NA §4.2)
c_dir = 1.0 (no directional factor in NL NA)
c_season = 1.0 (no seasonal reduction in NL NA)
v_b = c_dir × c_season × v_b,0 = 25 m/s
Formule
v_b = c_{dir} c_{season} v_{b,0}
Resultaat
v_b = 25 m/s
§4.3
Ruheidsfactor c_r(z)
Afleiding
Terrain category II: z_0 = 0.05m, z_min = 5m
z_e = h = 12m > z_min = 5m
c_r(z) = k_r × ln(z/z_0) = 0.19 × ln(12/0.05)
= 0.19 × ln(240)
= 0.19 × 5.48 = 1.042
Air density ρ = 1.25 kg/m³ (standard)
q_p(z_e) = (1+0.2×A) × 0.5×ρ×v_m² = use this
v_m(z) = c_r(z) × v_b = 1.042 × 25 = 26.05 m/s
Formule
v_m(z) = c_r(z) c_0(z) v_b
Resultaat
v_m = 26.05 m/s
§4.4
Piekwindsnelheidsdruk q_p(z)
Afleiding
I_v(z) = σ_v/v_m = k_1/(c_0×ln(z/z_0)) = 1.0/(1×5.48) = 0.183
k_1 = 1.0 (turbulence factor)
q_p(z) = [1 + 2×k_1×I_v(z)] × 0.5×ρ×v_m²
= [1 + 2×0.183] × 0.5×1.25×26.05²
= 1.366 × 0.625 × 678.6
= 579 N/m²
= 0.579 kN/m²
Formule
q_p(z) = [1 + 7 I_v(z)] \cdot \frac{1}{2} \rho v_m^2
Resultaat
q_p(z_e) = 0.579 kN/m²
§5.3
Externe druk op wanden c_pe
Afleiding
d/h = 50/12 = 4.17, b/h = 30/12 = 2.5
Reference height h = 12m
Zone D (windward): c_pe,10 = +0.7
Zone A (side): c_pe,10 = -1.1 (h/b < 4 → interpolate)
Zone E (leeward): c_pe,10 = -0.4
Zone F/G (corners): c_pe,10 = -1.1 (local on edges)
c_pe,1 = c_pe,10 + Δc_pe,D (if d/h > 4 use full reference)
For cladding design, use cpe,1 for area < 1m²
Formule
w_E = q_p(z_e) \cdot c_{pe}
Resultaat
Zone D: 0.58×0.7 = 0.41 kN/m², Zone E: -0.23 kN/m²
§5.3.2
Ontwerpwinddruk op gevel
Afleiding
With ±0.2 internal pressure:
c_pi = +0.2 (worst case for outward load)
w_total = q_p × (c_pe - c_pi) = 0.579×(0.7 - 0.2) = 0.290 kN/m²
c_pi = -0.2 (worst case for inward load)
w_total = 0.579×(-1.1 - (-0.2)) = -0.579×0.9 = -0.521 kN/m²
c_f = 1.2 (friction coefficient, if applicable)
Cladding panel design: w_design = 0.52 kN/m² (inward)
= 0.29 kN/m² (outward) — governs for fixings
Use spreadsheet and structural glazing program for area-dependent cpe
Friction loads: F_fr = q_p(z) × c_f × A_rear for area > 10m²
Formule
w = q_p(z_e) (c_{pe} - c_{pi})
Resultaat
Cladding design: 0.52 kN/m² inward, 0.29 kN/m² outward