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RE: wind loads

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Take my opinions with a  grain of salt, as it seems just with seismic there are many ways of interpreting wind load provisions in codes, often in the same office :) This is just what I would do....

1.	Section 1609.1.2 of the 2000 IBC states the wind loads used in
the design of the main wind-force-resisting system shall not be less then 10 psf multiplied by the area of the building or structure projected on a vertical plan normal to the wind direction (and similarly 10 psf is the minimum wind load for components and cladding and also for open structures).  However, footnote c in table 1609.6.2.1(1) states if the pressure is less than 0, use 0 for the horizontal roof loads for the end zone and the interior zone for MWFRS.  It seems to me there is a contradiction between them.  What do you use and why. 

-From Table 1609.6.2.1(1), the second column, under HORIZONTAL LOADS. These pressures are the sum of the windward and leeward forces in the HORIZONTAL direction. Rather then apply a negative value to the vertical roof projection, which would subtract from your positive wall loads, you use "0" which is the equivalent of neglecting it. This results in a larger roof diaphragm and base shear then if you subtracted the roof horizontal pressure. 

2.	When the forces are developed do you apply them uniformly across
the building width and then the average plf is applied to the shear walls [if you calculate the end zone width and the interior zone width multiplied by the appropriate height (gives area) times value in the table (psf) gives the resultant force in #'s then sum the forces from both zones and divide by the building width to get average plf]  OR do you calculate the plf in the end zone and in the interior zone and then apply it to the tributary area and then distribute the force into the shear walls.

-calculate the plf in the end zone and in the interior zone and then apply it to the tributary area and then distribute the force into the shear walls.  (Method differs depending on flexible vs rigid diaph.)

3.	If it worked out that the calculated edge strip (a) is 3' and
therefore the width of the end zone is 2a =3D 6', and the building is = 50' wide would there be a 6' end zone on each end and the width of the interior zone be 38'?  See figure 1609.6(1) or would there be a 3' end zone on each end and the interior zone would be 54'? 

2a= EZ, so for your case the interior zone would be 50ft- 2x(2a), or 38ft, with 6ft EZ on each end...

Andrew Kester, EI 
Project Engineer

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