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Substituting the above values into Equation 2.8 to determine the design wind pressure at 100 ft above grade gives 92=9.K.K.KK = 25.60(0.99)(1)(0.85)(0.91) 19.6 lb/ft² Note: To compute wind pressures at other elevations on the windward side, the only factor that changes in the above equation is K₂, tabulated in Table 2.4. For example, at an elevation of 50 ft, K₂= 0.81 and q₂ = 16.04 lb/ft². Determine the design wind pressure on the windward face

Substituting the above values into Equation 2.8 to determine the design wind pressure at 100 ft above grade gives 92=9.K.K.KK = 25.60(0.99)(1)(0.85)(0.91) 19.6 lb/ft² Note: To compute wind pressures at other elevations on the windward side, the only factor that changes in the above equation is K₂, tabulated in Table 2.4. For example, at an elevation of 50 ft, K₂= 0.81 and q₂ = 16.04 lb/ft². Determine the design wind pressure on the windward face

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Pls explain where the 11.3 lb/ft^2 came from
STEP 3 STEP 4 STEP 5 Substituting the above values into Equation 2.8 to determine the design wind pressure at 100 ft above grade gives 92=9.KKKK = 25.60(0.99)(1)(0.85)(0.91) = 19.6 lb/ft² Note: To compute wind pressures at other elevations on the windward side, the only factor that changes in the above equation is K₂, tabulated in Table 2.4. For example, at an elevation of 50 ft, K₂ = 0.81 and q₂ = 16.04 lb/ft². windward face Determine the design wind pressure on the windward face AB, using Equation 2.9. Gust factor G= 0.85, read Cp = 0.8 (from Table 2.7). Substi- tuting into Equation 2.9 produces p=qGC, 19.6(0.85)(0.8) = 13.3 lb/ft² Determine the wind pressure on the leeward side: C₂= -0.5 (Table 2.7) and G=0.85 p=q₂GC, 19.6(0.85)(-0.5)= -8.33 lb/ft² Compute the wind pressure on the two sides perpendicular to the wind: O C₂ = -0.7 G=0.85 p=q₂GCp = 19.6(0.85)(-0.7) = -11.66 lb/ft² The distribution of wind pressures is shown in Figure 2.17b. wind= 130 mph G. B = 60' A F = (a) leeward face D E L = 60' 100' 13.3 lb/ft² 11.66 lb/ft² B с 8.33 lb/ft² 11.66 lb/ft2 D 11.3 lb/ft² (b) Figure 2.17: Variation of wind sides of buildings.
Transcribed Image Text:STEP 3 STEP 4 STEP 5 Substituting the above values into Equation 2.8 to determine the design wind pressure at 100 ft above grade gives 92=9.KKKK = 25.60(0.99)(1)(0.85)(0.91) = 19.6 lb/ft² Note: To compute wind pressures at other elevations on the windward side, the only factor that changes in the above equation is K₂, tabulated in Table 2.4. For example, at an elevation of 50 ft, K₂ = 0.81 and q₂ = 16.04 lb/ft². windward face Determine the design wind pressure on the windward face AB, using Equation 2.9. Gust factor G= 0.85, read Cp = 0.8 (from Table 2.7). Substi- tuting into Equation 2.9 produces p=qGC, 19.6(0.85)(0.8) = 13.3 lb/ft² Determine the wind pressure on the leeward side: C₂= -0.5 (Table 2.7) and G=0.85 p=q₂GC, 19.6(0.85)(-0.5)= -8.33 lb/ft² Compute the wind pressure on the two sides perpendicular to the wind: O C₂ = -0.7 G=0.85 p=q₂GCp = 19.6(0.85)(-0.7) = -11.66 lb/ft² The distribution of wind pressures is shown in Figure 2.17b. wind= 130 mph G. B = 60' A F = (a) leeward face D E L = 60' 100' 13.3 lb/ft² 11.66 lb/ft² B с 8.33 lb/ft² 11.66 lb/ft2 D 11.3 lb/ft² (b) Figure 2.17: Variation of wind sides of buildings.
3 → Ixvii (67 of 790) Solution STEP 1 ▸ Determine the wind pressure distribution on the four sides of an eight-story hotel located on relatively flat ground approximately 2500 ft above sea level; the Risk Category II basic wind speed is 100 mph. Consider the case of a strong wind acting directly on face AB of the building in Figure 2.17a. Assume the building is classified as stiff because its natural period is less than 1 s; therefore, the gust factor G equals 0.85. The building is located in an urban area, so exposure Bapplies. Since the building is located on level ground, K₂=1. TEP 2 с Compute the static wind pressure using Equation 2.6a: qs = 0.00256V² = 0.00256(100)² = 25.60 lb/ft² Compute the magnitude of wind pressure on the windward side at the top of the building, 100 ft above grade, using Equation 2.7a. K₂ = 0.99 (Figure 2.15a or Table 2.4) K₂ = 1 (level ground) Ka=0.85 (Table 2.5) K = 0.91 (Table 2.6 or Figure 2.15b) A ubstituting the above values into Equation 2.8 to determine ne design wind pressure at 100 ft above grade gives 9₂=9sK₂K₂KK₁ FINAL PAGES 2.8 Wind Loads 30/07/19 ap 53
Transcribed Image Text:3 → Ixvii (67 of 790) Solution STEP 1 ▸ Determine the wind pressure distribution on the four sides of an eight-story hotel located on relatively flat ground approximately 2500 ft above sea level; the Risk Category II basic wind speed is 100 mph. Consider the case of a strong wind acting directly on face AB of the building in Figure 2.17a. Assume the building is classified as stiff because its natural period is less than 1 s; therefore, the gust factor G equals 0.85. The building is located in an urban area, so exposure Bapplies. Since the building is located on level ground, K₂=1. TEP 2 с Compute the static wind pressure using Equation 2.6a: qs = 0.00256V² = 0.00256(100)² = 25.60 lb/ft² Compute the magnitude of wind pressure on the windward side at the top of the building, 100 ft above grade, using Equation 2.7a. K₂ = 0.99 (Figure 2.15a or Table 2.4) K₂ = 1 (level ground) Ka=0.85 (Table 2.5) K = 0.91 (Table 2.6 or Figure 2.15b) A ubstituting the above values into Equation 2.8 to determine ne design wind pressure at 100 ft above grade gives 9₂=9sK₂K₂KK₁ FINAL PAGES 2.8 Wind Loads 30/07/19 ap 53
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