Wind pressure on roof truss, 20 m span, 30 deg

Assumptions. Span 20 m, roof slope $30^\circ$ (rise $= 10\tan30^\circ = 5.77$ m). The height "of the truss roof above the support" is taken as the eaves height $h = 10$ m, so $h/w = 10/20 = 0.5$ and the mean roof height $= 10 + 5.77/2 = 12.9$ m. Kathmandu: $V_b = 47$ m/s; town terrain with closely spaced obstructions: Category 3; greatest dimension 20 m, so Class B (20 m to 50 m) is taken. Life 50 years ($k_1 = 1$), flat site ($k_3 = 1$), normal permeability.

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

$$V_z = V_b\,k_1k_2k_3$$

• Basic wind speed $V_b = 47$ m/s.
• Risk coefficient $k_1 = 1.00$ (50-year life; Table 1).
• Terrain, height and size factor $k_2$: Category 3, Class B, at height 12.9 m: interpolating between 10 m (0.88) and 15 m (0.94): $k_2 = 0.915$ (Table 2).
• Topography factor $k_3 = 1.000$ (flat site).

$$\begin{aligned} V_z &= 47\times1.00\times0.915\times1.000 = 42.99\ \text{m/s} \\ p_z &= 0.6V_z^2 = 0.6\times42.99^2 = 1109\ \text{N/m}^2 = 1.109\ \text{kN/m}^2 \end{aligned}$$

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

Height ratio $h/w = 0.50$, roof angle $\alpha = 30.0^\circ$. The values are read directly from the $\alpha = 30^\circ$ row, so no interpolation is needed.

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

Design pressure normal to the roof: $p_d = (C_{pe} - C_{pi})\,p_z$ (positive = towards the surface, negative = suction). The most severe case is slope EG (wind 90°), with $C_{pe} - C_{pi} = -0.90$, so $p_d = -998$ N/m² $= -0.998$ kN/m².

Answer

• Design wind speed $V_z = 43.0$ m/s; design wind pressure $p_z = 1109$ N/m² (1.11 kN/m²).
• Net design pressure on the roof: up to $998$ N/m² (1.00 kN/m²), suction. The largest downward (positive) pressure is $+222$ N/m², on the windward slope with $C_{pi} = -0.2$.