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More Windy Componets & Cladding

More Windy Componets & Cladding

More Windy Componets & Cladding

(OP)
Wind loading on a large garage door (horizontal slider).

Given the following parameters:

V = 90 mph                
I = 1.0
Kd = .85
Kz = .98
Kzt = 1.44

ASCE 7-02 Section 6.5.10:

qz = (.00256)*Kz*Kzt*Kd*V^2*I
   = (.00256)(.98)(1.44)(.85)(90^2)(1.0)
   = 24.87 psf

Now for the C & C design wind pressure for a low rise building (< 60') see

ASCE 7-02 Section 6.5.12.4:

p = qh[GCp-GCpi]   where

GCpi  =   .55
GCp   =  0.90
      = -1.00

then p = 24.87*[0.90 -.55] = 8.71 psf

or
        p = 24.87*[-1.0 – (-0.55)] = -11.19 psf

This just doesn't see right to me for the wind load on a large garage door.  The magnitude of the difference between positive and negative pressure seems about right, but I'm thinking the values should be about double what I get; what do you fellow designers think?  Where am I going wrong?
 

RE: More Windy Componets & Cladding

You have the internal pressure coefficients reversed.  When there is external pressure on the door, assume there is internal suction (0.9 + 0.55).  Likewise, with external suction, internal pressure (-1 - 0.55).

DaveAtkins

RE: More Windy Componets & Cladding

(OP)
OK, so switching those around results in:

p = 24.87*[0.90 -(-.55] = 36.06 psf

or

p = 24.87*[-1.0 – (0.55)] = -38.54 psf

I fill better, although that's a pretty hefty wind load. It is about 18 psf more than in Table 1609.6.2.1(2) for Walls in Zone 4.  Any thoughts on that?

Thanks Dave!

RE: More Windy Componets & Cladding

Those are large C&C wind loads but it appears the cause of their magnitude is because you have a topo factor (Kzt = 1.44) that has driven them up a bit.

 

RE: More Windy Componets & Cladding

On large flexible doors, you will end up with catenary forces on the ends, meaning you will have to design the end connections for shear, but also axial forces.

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