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Lecture 24 – Wind Loads

In many parts of the country, high wind is the most severe load that affects abuilding. In fact, annually, wind damage to buildings and other structuresexceeds all other natural disasters combined. Two of the worst natural disastersever to strike the USA, in terms of damage, was Hurricane Andrew which struck

southern Florida in 1992, causing $30 BILLION, and Hurricane Katrina strikingthe Gulf Coast in 2006 causing $100 BILLION (most of this damage due toflooding). Fortunately there were relatively few deaths. A hurricane strikingBangladesh in 1991 killed at estimated 139,000 people. Below are severalpictures of hurricane damage:

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Wind Loads per IBC 

Section 1609 of the IBC dictates that wind loading shall be determined as per Chapter 6 of ASCE 7. There are 2 methods that may be used:

1) Chapter 6.4 – Method 1 – “Simplifies Procedure” (See Lecture 25)2) Chapter 6.5 – Method 2 – Analytical Procedure”

 ASCE 7 – Chapter 6.5 – Method 2 “ Analyt ical Procedure” 

The velocity pressure, qz, evaluated at height “z” shall be calculated from thefollowing formula:

qz = 0.00256KzKztKdV2Iw 

where:qz = velocity pressure at height “z” in lbs. per square footKz = velocity pressure exposure coefficient evaluated at height “z” and

Exposure B, C or D (from IBC 1609.4) (see ASCE 7 Table 6-3 below)Kzt = topographic factor, wind speed increases if traveling up steep hillsKd = wind directionality factor, ranging from 0.85 (buildings) → 0.95 (chimneys) V = basic wind speed in MPH for 3-second gust at 33 ft. above ground in

Category C as shown in IBC Figure 1609Iw = Importance factor for wind design per building category as defined in IBC

Table 1604.5:

Category: Non-Hurricane ProneRegions with V = 85 –100 mph and Alaska

Hurricane ProneRegions wi th V > 100 mph

I

0.87 0.77II 1.00 1.00III 1.15 1.15IV 1.15 1.15

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 And the design wind pressure, p, to be applied to the Main Wind Force ResistingSystem (MWFRS), of a building, is given by the following formula:

p = qzGCp – qz(GCpi)

where:p = design wind pressure for main wind force-resisting system (MWFRS), psf qz = velocity pressure at height “z” (see above)G = 0.85Cp = external pressure coefficient (see ASCE 7 Figure 6-6 below)

GCpi = 0.0 for open buildings= +0.55 or -0.55 for partially enclosed buildings= +0.18 or -0.18 for enclosed buildings

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ExampleGIVEN: The gable-end fire-rescue building as shown below. The Category IV building is located along the coastline of Miami, Florida.

REQUIRED: Determine the wind pressure on the MWFRS of the following:a) Windward wallb) Leeward wallc) Windward roof 

d) Leeward roof e) Gable end walls

1) Determine the mean roof height above ground:

Mean roof ht, “h” = 14’ + ½(12’)= 20’

14’

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2) Determine the Exposure Category f rom IBC Section 1609.4:

Use Exposure Category “C” since building is located on Miamicoast.

3) Determine the veloci ty pressure, qz :

qz = 0.00256KzKztKdV2Iw 

where:Kz = 0.90 from ASCE 7 Table 6-3 (ht. = 20’, Exposure C)Kzt = 1.0 (assuming flat terrain)Kd = 0.85 (Building structure)V = 150 MPH (from IBC Figure 1609)Iw = 1.15 since it is Cat. IV (from IBC Table 1604.5)

qz = 0.00256(0.90)(1.0)(0.85)(150 MPH)2(1.15)= 50.7 psf 

4) Determine the wind pressure on the Windward wall:

p = qzGCp – qz(GCpi)

where:qz = 50.7 psf G = 0.85Cp = 0.8 (from ASCE 7 Figure 6-6)(GCpi) = +0.18 since it is an enclosed building

p = 50.7 psf(0.85)(0.8) – 50.7(+0.18) p = 43.6 psf or 25.4 psf 

5) Determine the wind pressure on the Leeward wall:

Use same values as above, except Cp Determine L/B = 40’/100’

= 0.4

Use Cp = -0.5 for L/B of 0 → 1 (from ASCE 7 Figure 6-6)

p = qzGCp – qz(GCpi)

p = 50.7 psf(0.85)(-0.5) – 50.7(+0.18) p = -30.7 psf or -12.4 psf 

Note: The negative sign indicates a negative “ suction” force.

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6) Determine the wind pressure on the Windward roof:

Use same values as above, except Cp 

First, determine the h/L ratio = 20’/40’= 0.5

Next, determine the roof angle θ as measured from the horizontal:

Determine Cp from ASCE 7 Figure 6-6 for “Wind Normal to Ridge for θ ≥100”:

Cp = -0.2 and 0.3 (interpolated for h/L = 0.5 and θ = 310)

Determine the wind pressure p as follows:

p = qzGCp – qz(GCpi)

p = 50.7 psf(0.85)(-0.2) – 50.7 psf(+0.18) p = -17.7 psf or +0.5 psf 

or 

p = 50.7 psf(0.85)(0.3) – 50.7 psf(+0.18) p = 22.1 psf or 3.8 psf 

20’-0”

12’-0”θ 

θ = tan-1(12/20)= 310 

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7) Determine the wind pressure on the Leeward roof:

Use same values as above, except Cp 

Determine Cp from ASCE 7 Figure 6-6 for “Wind Normal to Ridge for θ ≥100”:

Cp = -0.6 for θ> 200 

Determine the wind pressure p as follows:

p = qzGCp – qz(GCpi)

p = 50.7 psf(0.85)(-0.6) – 50.7 psf(+0.18) p = -35.0 psf or -16.7 psf 

8) Determine the wind pressure on the Gable end walls:

Use same values as above, except Cp 

Determine Cp from ASCE 7 Figure 6-6 for “Side Walls”

Cp = -0.7 for all values of L/B

Determine the wind pressure p as follows:

p = qzGCp – qz(GCpi)

p = 50.7 psf(0.85)(-0.7) – 50.7 psf(+0.18) p = -39.3 psf or -21.0 psf 

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9) Draw “ Summary Sketches” showing WORST-CASE LOADS: