ASCE 7-05 Effective Wind Area 1

[Heck83]

Hi, I am new to ASCE 7-05, Eng Tips, and my internship as a structural engineer. I am trying to understand the effective wind area. I am given to understand that it is the area of, say, a wall in which a window is located. So, it is the area of the wall and not the window?

Thanks for helping a beginning structural engineer!

 

[Stenbrook]

The effective are is dependent on what you are designing for. It is built under the assumption that the larger the area is, the less likely that the maximum wind pressure would be applied to the entire surface. So if you are designing a beam, the effective area is the tributary area for that beam (the span multiplied by the trib width). There is a provision in the ASCE 7-05 that states " the effective wind area.. is the span length multiplied by the effective width that need not be less than one-third the span length. For Cladding fasteners, the effective wind area shall not be greater than the area that is tributary to an individual fastener." You can find that on the second page of the wind chapter under the definition for Effective Wind Area, A.

For the design of the wall, if you are designing wall studs, I usually use the height of the wall squared divided by 3 as my area. If I am designing a concrete tilt wall panel than I will design using the area of the panel.

I hope that helps!

 

[Heck83]

Thank you so much! I have been wrong! So, if you have a window in a wall, the design for its fasteners, would be the area of the window? And if I am designing a wall with two shearwalls on each end, then I would use the area of the total wall? Stenbrook, where does the wall squared come from?

One more question... Why are there no leeward pressures for walls given in the simplified version
(table 6-2 (method I) of ASCE 7-05?

Thanks!!!

 

[Stenbrook]

So when designing with wind there are two different types of wind pressures. There is Main Wind Force Resisting System (MWFRS) and Components and Cladding (C&C). When you design the lateral resisting system of the structure (which would include shear walls, braces, moment frames,diaphragms, etc.) you would use the MWFRS pressures which are usually a little bit lower than C&C pressures as they are assumed to be acting on the entire structure. To the best of my knowledge you don't use the effective area to determine the MWFRS pressures. As for the window and its fasteners, this would be considered cladding and thus you would use C&C pressures. To determine the pressure you would need to know the effective area. This effective area for the fastener would be the tributary area for the fastener, not the area of the window. This is usually less than 10ft^2 from my experience but you can run the calculations.

I don't use the simplified method much if at all in figure 6-2, but my assumption is that they are assuming that both the windward and leeward wall are the same height and have combined the pressures together for the one side. So if your walls are different heights, you will most likely have to use a different method to calculate the pressures.I usually use Method 2 in Figure 6-6

 

[Heck83]

What I have been doing is using the simplified version for C&C and the analytical method for MWFRS. As far as C&C my concern is what psf rating to spec the windows, so in that case the effective area would be the area of the window?

Also, I'm curious to know where you got the wall^2/3 for the MWFRS tributary area.

Thanks again, I am getting a lot out of this dialogue.

 

[Stenbrook]

If you are rating the window then yes I would use the area of the window as the effective area. The wall height is considered the span length for a stud. The ASCE says that the tributary width need not be less than one third the span as I mentioned in my earlier post. So if we say the span or the height of the stud is h, your effective area is one of two options. (h)*trib width or (h)*(h/3)=(h^2)/3. For most studs the spacing is close to 12-24 in o.c. If the height is let's say 10'-0", then the h/3=3'-4" which is greater than the spacing and thus control in the effective area equation. So you would use (h^2)/3

 

[Heck83]

Perfect! Thanks!

 

[Heck83]

So if you have a wood-framed roof that is sloping downward, then the effective wind area would be continuously decreasing with the length of the rafters? Is this true?

 

[Angre]

Hi Heck83,

For smaller areas, we take the so-called "lesser time average".Let me explain.

The wind velocity to be taken is a function of height( well understood by all) and time t(not so well understood.
Let me illustrate with an example.
Given :- Wind velocity V[sub]basic[/sub] of 35 m/s at 10m above ground averaged over t[sub]basic[/sub] of 1 hour.

Design the following :-
1) Design a structure at height of 50m and time average of 3 sec.For structures with small areas averaging time is less.
2)Design a structure at height of 45m and time average of 3 minutes.For structures with large areas averaging time is greater.

So WindVel(V[sub]basic[/sub],t[sub]basic[/sub],10m,50m,3sec)= say 47 m/s for case 1.(The expression to calculate the functionis given in API RP-2A.
Similarly
WindVel(V[sub]basic[/sub],t[sub]basic[/sub],10m,45m,3 minutes)= say 44 m/s.

Rgds

 

[Stenbrook]

Angre,

I don't disagree with the difference in wind pressures based on time, however in the US all wind calculations are based on a 3-second gust regardless of the size of the structure. The effective area does modify the pressure but I don't know that it is specifically because of time. I don't know of any provision in the code that says if the structure is larger you use a corresponding averaging time.

As for your question Heck83, if you are designing the rafters then the effective area for each one would be different based on the span and the tributary width as we have discussed earlier. You also have to pay attention to what zone the rafter falls in (zone 1, zone 2, and zone 3) as the pressures vary greatly between the zones. You can see how these zones are defined in figure 6-3 of the ASCE 7-05

 

[Heck83]

Angre, thanks for your contribution to the discussion.

Stenbrook, do you think it is not acceptable to consider a plane of a roof to be a system, and therefore use the total area of the plane? Of course, this is paying proper attention to which zone(s) lie in the plane.

 

[Stenbrook]

It is acceptable to use the plane of a roof as a system when you are designing the diaphragm for the roof which is part of your MWFRS pressures(this is assuming a large effective area already). But, when you are designing an individual member such as a beam, you can only consider the effective area (the span multiplied by its tributary width) for that beam. You must also use components and cladding wind pressures for the beam.

 

[Heck83]

That seems pretty conclusive. How about a case where you have an effective area less than 10 ft^2? Can that be used as a minimum, would one have to extrapolate? I'm assuming that, for the the simplified version, that you can interpolate for intermediate values. However, I do not see a provision for that. I have also been doing the same for intermediate wind velocities, though I guess a more accurate method would be to perform parabolic interpolation since the pressure is proportionate to velocity squared. Again, I do not see provisions for this.

 

[Stenbrook]

So there is literally no difference between the simplified method 1 and method 2 when it comes to what the pressures come out to. The simplified method just puts common points in a chart. If you have a value that isn't listed in the charts, than you have two options. The first is to interpolate between the values in the chart to what you need. The second is to use method 2. Any effective area less than 10 ft^2, you would use 10ft^2 for. You can interpolate up until you reach 500 ft^2. If you look at figure 6-11A it shows how your Cp value changes linearly from 10 to 500 ft^2 and is constant below 10 and above 500. That is how the simplified charts are behaving as well.

 

[Heck83]

Yes, I see that now. I also notice that there is only a windward pressure on the MWFRS in the simplified method. Do they combine the windward and leeward pressures into one windward pressure?