OpenSpecEngineering

Finds the wind velocity pressure at a height on a structure, and the design pressure on a surface with a known pressure coefficient, by ASCE/SEI 7-22 Chapter 26 [1]. Structural engineers and cladding designers use it to load walls, roofs and frames.

How it is calculated

The velocity pressure (Eq. 26.10-1.SI) and the design pressure are

q_z = 0.613·K_z·K_zt·K_e·V²
p = K_d·q_z·GC_p − K_d·q_z·(±GC_pi)

where q_z is in N/m² and V is the basic wind speed in m/s for the building’s risk category, from the maps (Figs. 26.5-1A to 26.5-1D) or the ASCE Hazard Tool. K_z rises with height and with open terrain. The page takes it from Table 26.10-1, interpolated between heights, or from the table’s note 1 formula with the exposure constants α and z_g of Table 26.11-1; the two differ by up to about 0.015, mostly in exposure D. K_zt is the topographic factor (1.0 on level ground) and K_e the ground elevation factor of Table 26.9-1.

K_d is the directionality factor of Table 26.6-1. In 7-22 it is no longer part of q_z; Chapters 27 to 30 apply it in the loads, so the page multiplies both pressure terms by it. GC_p is the product of gust effect factor and external pressure coefficient for the surface, from Chapter 27, 28, 29 or 30. GC_pi is the internal pressure coefficient of Table 26.13-1, taken here at the same q_z. Positive p pushes on the surface; negative p pulls away from it.

Limits and common mistakes

• Enter GC_p with its sign: windward walls are positive, leeward walls and most roofs negative.
• Leeward walls, side walls, roofs and the internal pressure all use q_h at the mean roof height. Set the height to h for those.
• The table stops at 500 ft (152.4 m); use the formula above that.
• For the Chapter 28 envelope procedure, exposure B takes K_z = 0.70 below 30 ft (9.1 m). The page does not apply that floor.
• The result is a strength-level pressure. Multiply by 0.6 for allowable stress design.

Worked example

An enclosed three-story office in open country (exposure C) has a 12 m mean roof height and a 51 m/s design wind speed. Find the main-frame pressure on the windward wall at the top of the building, where G = 0.85 and C_p = 0.8 give GC_p = 0.68.

1. Enter Basic Wind Speed 51, choose Exposure Category C and enter Height Above Ground 12 m.
2. Keep Exposure Coefficient K_z From Table 26.10-1, K_zt 1.0, K_e = 1.0 and Structure Type Buildings — main wind force resisting system (K_d 0.85).
3. Enter External Pressure Coefficient × Gust Effect 0.68 and keep Enclosure Classification Enclosed (GC_pi 0.18).
4. Press Solve. The Solution shows: Velocity pressure exposure coefficient Kz: 1.03622
   Velocity pressure qz = 0.613·Kz·Kzt·Ke·V²: 1,652.16 N/m²
   Directionality-adjusted velocity pressure Kd·qz: 1,404.34 N/m²
   Design pressure p (with −GCpi): 1,207.73 N/m²
   Design pressure p (with +GCpi): 702.169 N/m²

Check by hand. 12 m is 39.37 ft, so K_z = 0.98 + 0.937 × (1.04 − 0.98) = 1.0362. Then q_z = 0.613 × 1.0362 × 51² = 1,652 N/m², and 0.85 × 1,652 = 1,404 N/m². The wall takes 1,404 × (0.68 + 0.18) = 1,208 N/m² when the building’s inside is under suction, about 1.2 kPa or 25 psf.