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Wind pressure on 36 m building, core city

Short answer

Design wind speed Vz = Vb×k1×k2×k3 = 47×1×0.966×1 = 45.4 m/s. Design pressure pz = 0.6Vz² = 1.237 kN/m². Net coefficient (Cpe - Cpi) = -1.30 gives roof suction pressure 1.608 kN/m². Force on one truss slope (area 91.4 m²) is 1.30×91.4×1.237 = 147.0 kN.

The building is to be constructed in core city area for a 50 years life. The size of the building is over 30 m. The height of the building is 36 m and is classified as 1st category building. Determine the wind pressure at the site and force on the truss.

Assumptions (data not given in the question). VbV_b is not stated; take 47 m/s (Kathmandu region, IS 875 Part 3 Fig. 1). The building is in a core city, so terrain Category 4 (large city centre with obstructions). Height 36 m, so the greatest dimension is 36 m and the structure is Class B (20 m to 50 m). A design life of 50 years gives k1=1.0k_1 = 1.0. The roof truss is taken on a plan width w=30w = 30 m with a low roof pitch of α=10∘\alpha = 10^\circ, truss spacing 6 m and normal permeability. k3=1k_3 = 1 (flat site).

Step 1: Design wind speed (IS 875 Part 3:1987, cl. 5.3)

Vz=Vb k1k2k3V_z = V_b\,k_1k_2k_3
  • Basic wind speed Vb=47V_b = 47 m/s.
  • Risk coefficient k1=1.00k_1 = 1.00 (design life 50 years, Table 1).
  • Terrain, height and size factor k2k_2: Category 4, Class B, at height 36 m, interpolating between 30 m (0.93) and 50 m (1.05): k2=0.966k_2 = 0.966 (Table 2).
  • Topography factor k3=1.000k_3 = 1.000 (flat site, slope less than 3°).
Vz=47×1.00×0.966×1.000=45.40 m/spz=0.6Vz2=0.6×45.402=1237 N/m2=1.237 kN/m2\begin{aligned} V_z &= 47\times1.00\times0.966\times1.000 = 45.40\ \text{m/s} \\ p_z &= 0.6V_z^2 = 0.6\times45.40^2 = 1237\ \text{N/m}^2 = 1.237\ \text{kN/m}^2 \end{aligned}

Step 2: Pressure coefficients (Table 5, pitched roof of rectangular clad building)

Height ratio h/w=1.20h/w = 1.20 (band 1/2<h/w≤3/21/2 < h/w \le 3/2), roof angle α=10.0∘\alpha = 10.0^\circ. The values for α=10∘\alpha = 10^\circ are read directly from the table.

Internal pressure coefficient Cpi=±0.2C_{pi} = \pm0.2 (normal permeability), cl. 6.2.3.

SurfaceCpeC_{pe}Cpe−CpiC_{pe} - C_{pi} (Cpi=+0.2C_{pi}=+0.2)Cpe−CpiC_{pe} - C_{pi} (Cpi=−0.2C_{pi}=-0.2)pdp_d (N/m²), Cpi=+C_{pi}=+pdp_d (N/m²), Cpi=−C_{pi}=-
Windward slope EF (wind 0°)-1.10-1.30-0.90-1608-1113
Leeward slope GH (wind 0°)-0.60-0.80-0.40-989-495
Slope EG (wind 90°)-0.80-1.00-0.60-1237-742
Slope FH (wind 90°)-0.60-0.80-0.40-989-495

Design pressure normal to the roof: pd=(Cpe−Cpi) pzp_d = (C_{pe} - C_{pi})\,p_z (positive = towards the surface, negative = suction). The most severe case is the windward slope EF (wind 0°), with Cpe−Cpi=−1.30C_{pe} - C_{pi} = -1.30, so pd=−1608p_d = -1608 N/m² =−1.608= -1.608 kN/m².

Step 3: Wind force on the roof / truss

F=(Cpe−Cpi) A pzF = (C_{pe} - C_{pi})\,A\,p_z, acting normal to the roof surface. The most severe net coefficient, −1.30-1.30, is used for the maximum force.

  • One slope of a truss (span 30 m, slope length 15/cos⁡10∘=15.2315/\cos10^\circ = 15.23 m, spacing 6 m): A=91.4A = 91.4 m², F=1.30×91.4×1.237=147.0F = 1.30\times91.4\times1.237 = 147.0 kN (suction, i.e. acting away from the roof).

Answer

  • Design wind speed Vz=45.4V_z = 45.4 m/s; design wind pressure pz=1237p_z = 1237 N/m² (1.24 kN/m²).
  • Net design pressure on the roof: up to 16081608 N/m² (1.61 kN/m²), suction.
  • Force on one slope of a truss (span 30 m, slope length 15.23 m, spacing 6 m): 147.0 kN.

✓ Answer checked by Claude Opus 5.5 (high effort)

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