ASCE7-10 Section 27.4.7 Minimum Wind Loads on Open Buildings

[JohnElder]

I have a question on how to apply the minimum wind load of 16psf per section 27.4.7 of ASCE7-10 on open buildings. (This is for the design of Solar carport type structure).

My design values are as follows:
Wind Speed V=100 MPH
Exposure = B
Risk Category = I
Enclosure Classification = Open Building
Average Roof Height = 12ft
Roof Angle = 5 Degrees
Kd = 0.85
Kz = 0.57 (ASCE7-10 Table27.3-1)
Kzt=1.0
G = 0.85
Monoslope Roof

Therefore, qh = 0.00256KzKztKdV^2 = 12.4 PSF

This value is less than that of section 27.4.7. Section 27.4.7 states that "The design wind force for open buildings shall be not less than 16 psf multiplied by the area Af". This section does not say to use a minimum "Velocity Pressure, qh" of 16 psf. Therefore, when designing an open building for the MWFRS wind loads per section 27.4.3 and Figure 27.4-4, I am assuming that I use the qh = 12.4 PSF and not 16PSF.

That is where I am confused, how do I apply the minimum 16 PSF wind load to the structure? Do I apply a uniform load of 16 PSF in addition to the pressures of section 24.7.3 or am I suppose to use qh=16 PSF in the loading of section 27.4.3?

 

[Yao1989]

looks to me you use qh = 12.4psf, but your final p (p=qh*G*CN) should not be less than 16psf. I have only used the ASCE open structure countable times though (with one hand) so you may want to get more opinions.

 

[MacGruber22]

Wind pressure is = qh (qz) x (pressure coefficient). It seems you have not calculated the pressure coefficient which will increase the design pressure based on the geometry of your open structure, accounting for the various roof zones. Once you have calculated that, compare to the minimum pressures. For an open structure, it is likely the actual design pressures will be far greater than the minimum required. Don't forget that structural elements with effective wind areas less than 700 ft^2 have to be considered for components and cladding pressures.

"It is imperative Cunth doesn't get his hands on those codes."

 

[mike20793]

You found the pressure at the height, h, but you still have to multiply it by the gust factor and pressure coefficient (this equation is found in section 27.4.3) to obtain the net design wind pressure, p. The value for p cannot be less than 16 psf.

 

[JohnElder]

I guess I wasn't clear. For open buildings, there are two pressures, not just one. When I apply G=0.85 and the Cnw and Cnl factors from figure 27.4-4, I get the following pressures for a single load case:

Load Case A: Wind Direction=0 Degree: Cnw = 1.2, Cnl=0.3; Therefore Pnw=12.7psf and Pnl=3.2psf

The carport structures are "T" shapes (See attachment). The unbalanced loading of figure 27.4-4 will induce a moment on the structure unlike a uniform 16PSF structure. I guess I would need to check both the unbalanced loading above and a uniform 16psf? I just want to be sure that I don't use the minimum 16PSF pressure when calculating p with the Cnw and Cnl coefficients.

 

[mike20793]

The 16 psf minimum is a horizontal load applied to the vertical projection of the roof. The load case is separate from the pressures acting normal to the roof surface.