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Wind pressure on high-rise building, 50 m

Short answer

Given Vb=47 m/s, k1=1, k2=1.05 (Category 4, Class B, 50 m), k3=1: Vz=49.35 m/s, pz=1.461 kN/m². For h/w=1.67, net coefficient -1.0 gives suction pressure 1.461 kN/m². Force on one slope (area 15.23×5=76.2 m²): 1.0×76.2×1.461=111.3 kN; whole truss 222.5 kN.

The high rise building at Kathmandu is to be constructed for a 50 years life; the size of the building is over 30 m. The height of the building is 50 m. Determine the wind pressure at the site and force on the truss. Where basic wind speed of Kathmandu is 47 m/sec. (Building size 40×30 m², at Sundhara, Kathmandu.)

Assumptions. Vb=47V_b = 47 m/s (given). Kathmandu high-rise: terrain Category 4 (city centre). Height 50 m, so the greatest dimension is 50 m, which does not exceed 50 m: Class B. Life 50 years: k1=1.0k_1 = 1.0; flat site: k3=1k_3 = 1. Plan 40 m × 30 m, so w=30w = 30 m and h/w=50/30=1.67h/w = 50/30 = 1.67. The roof truss is taken to span the 30 m width with a low slope α=10∘\alpha = 10^\circ, truss spacing 5 m and normal permeability.

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 50 m: k2=1.05k_2 = 1.05 (Table 2, read directly).
  • Topography factor k3=1.000k_3 = 1.000 (flat site).
Vz=47×1.00×1.050×1.000=49.35 m/spz=0.6Vz2=0.6×49.352=1461 N/m2=1.461 kN/m2\begin{aligned} V_z &= 47\times1.00\times1.050\times1.000 = 49.35\ \text{m/s} \\ p_z &= 0.6V_z^2 = 0.6\times49.35^2 = 1461\ \text{N/m}^2 = 1.461\ \text{kN/m}^2 \end{aligned}

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

Height ratio h/w=1.67h/w = 1.67 (band 3/2<h/w<63/2 < h/w < 6), roof angle α=10.0∘\alpha = 10.0^\circ.

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°)-0.70-0.90-0.50-1315-731
Leeward slope GH (wind 0°)-0.60-0.80-0.40-1169-585
Slope EG (wind 90°)-0.80-1.00-0.60-1461-877
Slope FH (wind 90°)-0.80-1.00-0.60-1461-877

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 slope EG (wind 90°), with Cpe−Cpi=−1.00C_{pe} - C_{pi} = -1.00, so pd=−1461p_d = -1461 N/m² =−1.461= -1.461 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, −1.00-1.00, is used for the maximum force.

  • One slope of one truss (slope length 15/cos⁡10∘=15.2315/\cos10^\circ = 15.23 m, spacing 5 m): A=76.2A = 76.2 m², F=1.00×76.2×1.461=111.3F = 1.00\times76.2\times1.461 = 111.3 kN (suction, i.e. acting away from the roof).

  • Whole truss (both slopes): A=152.3A = 152.3 m², F=1.00×152.3×1.461=222.5F = 1.00\times152.3\times1.461 = 222.5 kN (suction, i.e. acting away from the roof).

Answer

  • Design wind speed Vz=49.4V_z = 49.4 m/s; design wind pressure pz=1461p_z = 1461 N/m² (1.46 kN/m²).
  • Net design pressure on the roof: up to 14611461 N/m² (1.46 kN/m²), suction.
  • Force on one slope of one truss (slope length 15.23 m, spacing 5 m): 111.3 kN.
  • Force on the whole truss (both slopes): 222.5 kN.

✓ Answer checked by Claude Opus 5.5 (high effort)

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