Perpendicular Roof Wind Pressures

[CTSeng]

Both the MWFRS and C&C tables in ASCE 7 have notes stating 'plus or minus signs signify pressures acting toward and away from the surfaces, respectively' or 'Roof – Net pressure (sum of external and internal pressures) applied normal to all roof surfaces.'

If I consider the pressure normal to the surface and have a steep slope roof can I then trig out the horizontal and vertical elements to design the lateral and uplift items? I always had the understanding that a pressure acts in all directions like water pressure and applied the ASCE 7-98 pressures as is on both the horizontal and vertical projections.

Thanks

 

[ishvaaag]

I think you can as long the combination of your vertical and lateral forces amount to that specified.

 

[Lutfi]

You should consider the X and Y components and its effect on the structure.

 

[egc]

If I understand your question right, the pressure values that ASCE7-XX use are given based on the stated conditions (i.e., they act normal to the roof surface). So you can trig out (cos(alpha)/sin(aplha) where alpha is the roof angle) the X (lateral) and Y (uplift) components. I don't believe that the "water" analogy is appropriate because in this case the fluid (air) is moving. Moving water (waves?) generate a lot more dynamic loading then static head pressure values.

I just hang here (occasionally)

 

[daicsa]

I agree with Lutfi. You must consider the X and Y components of the pressure on the roof .

 

[dpa]

Even though pressure acts in all directions the structures which are subjected to pressure react perpendicular and parallel with their own surfaces. Do trig it out. Nice terminology trig it out. I'll be using that in the future.

 

[chichuck]

CTseng,

I think what you said you've done in the past is correct.

Consider as an example a gabled frame: 50'wide, 20' eave height, 4:12 slope on roof.

On each roof beam, hor projection is 25', vertical projection is 8.33' Length of each roof beam is 25.31'

Let's assume that the ASCI-7 wind pressure on windward walls is +20psf and pressure perpendicular to the roof surface is +15psf.

To get the total vertical component of the force on (half) the roof, you take the vertical component of that 15psf on a length of 25.31'.

15psf * (25/25.31) * 25.31' = 375 lb (* the bay spacing)

to get the horizontal component of the force on (half) the roof,

15psf * (8.33/25.31) * 25.31' = 60 lb (* the bay spacing). On a gable roof, the horizontal component of the force on the other roof beam is not in the opposite direction, because there is usually suction on that part of the roof, and it is of a different magnitude from this force. So you'd have to figure that one out separately.

This would get you to the vertical forces and horizontal forces for the frame, as long as you have correctly included the internal suctions or pressures.

If these horizontal and vertical loads are correctly apportioned to the joints of the frame, you could, I suppose do a moment distribution to get member moments/shears/reactions, etc. But to do that you would also need to get FEM's and that would require normal loads to get to the moments. I bet most of us would be using a computer program to analyze that, and those also use as input the normal loads. You could use these to figure support reactions to use for foundation design.

But note this would not be the right values to use for roof girts, roof decks, wall girts or wall panels. For those you would have to use the components and cladding pressure coefficients. And, for those, you would need to use pressures normal to the surfaces, since that is the direction of the loads and the bending for these members.

I'm not sure what "lateral and uplift items" you could design based on these numbers. Can you elaborate?

Oh, and one last bit of advice: be very careful to get all the load combinations that are in ASCE 7-XX so that you get the most critical one.


Regards,

chichuck

 

[CTSeng]

chichuck,
So what you are saying is I can separate the X and Y components but that they should be applied to the actual area or sloping area of the roof. So for your 50’ wide 4:12 roof w/ 15psf normal pressure the horizontal would be 4.74psf and the vertical 14.23psf. Then to get the total vertical component of the force on (half) the roof:
14.23psf * 26.35’ = 375plf * (bay spacing) [note: the length should be 26.35’ not 25.31’]

and the total horizontal component of the force on (half) the roof:
4.74psf * 26.35’ = 125plf * (bay spacing) [note: the length should be 26.35’ not 25.31’]

This is basically saying I should use the 15psf similar to a water pressure acting in all directions so the total vertical component of the force on (half) the roof:
15psf * 25’ = 375plf * (bay spacing) SAME AS ABOVE

and the total horizontal component of the force on (half) the roof:
15psf * 8.33’ = 125plf * (bay spacing) SAME AS ABOVE

I guess what I wanted to know when starting this thread is if I could apply the X and Y components to the X and Y projections. So the total vertical component of the force on (half) the roof:
14.23psf * 25’ = 355.75plf * (bay spacing)

and the total horizontal component of the force on (half) the roof:
4.74psf * 8.33’ = 39.5plf * (bay spacing)

As you can see the horizontal would have the potential of being greatly reduced which would help for the Main Wind System. This is the primary reason for raising this question although I would also make use of the reduction in uplift to size anchorage using Component and Cladding pressures.

I actually tested my theory with a spreadsheet I have developed and have learned that applying the X and Y components to the X and Y projections is appropriate. As I increase the roof pitch to a virtually vertical wall the MWFRS roof pressures equate to the wall pressures. It’s not quite as simple with C&C pressures given the multiple charts for differing roof slopes but I feel it still is the correct way to determine the wind loads.

 

[unclesyd]

There is a nice lttle program at:

http://www.archonengc.com

This program will get you in the ballprak.
It isn't very expensive about $40 USD and you can download it.

 

[unclesyd]

Correct URL

http://www.archoneng.com

 

[chichuck]

CTSeng,

yep, that's what I am saying. Its just statics. The 15psf load is specified as normal to that (sloping) roof surface. (See ASCE 7-98 Figure 6-3) To figure out the H & V components, you can take components of the force, as I detailed, or you can take the 15psf on the H & V projections, they are mathematically the same. You can figure it out either way. But what you are trying to do is statically incorrect.

Don't know about your spreadsheet, and how it verifies that it is appropriate to apply H & V components to H & V projections. You didn't elaborate. I can see that as you change the geometry, that is to increase roof slope angle, the total length of roof beam increases, the hor. component of the 15psf increases, and the vertical component of the 15psf decreases. When you approach angle = 90 degrees, you approach 15psf hor component and 0psf vertical component. Is that what you mean when you say MWFRS roof pressures equate to wall pressures?

Please note that (in ASCE 7-XX) the 15psf on the roof will change as the roof slope changes. Do that before you figure components in your spreadsheet. Also note that roof pressures are based on a different Cp values than are the wall pressures. (See ASCE 7-98 Figure 6-3). At a roof slope of 90 degrees (not possible, I know.) the Cp for the roof pressure is .01 * 90 = 0.90. For the windward wall Cp is 0.80. So I don't think the roof pressure would equal the wall pressure for that hypothetical case.

Regards,

chichuck