huangyhg

more windy componets & cladding
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?

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
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!
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.

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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