Wind load ASCE

[mafiucivil]

I would be grateful if you can help me to solve some doubts, about the ASCE-7-05. I'm trying to obtain the loads for a low-rise building with the most precise method.

1. In the analytical method, there is in Figure 6-10 a way to obtain the External Pressure Coefficient (Cp) for low-rise buildings, but in figure 6-6 this Coefficient can be obtained for all heights.
I noted that the results aren’t similar because the Coefficients for low-rise buildings (analytical method) are more significant in the corners
Which would be the most precise method of both?

2. In the ASCE a minimal load is specified in section 6.1.4.1 for the Main Wind-Force Resisting System (MWFRS), this load shall not be less than 0.48kN/m2 multiplied by the area projected onto a vertical plane normal to the wind direction. If I’m working with the simplified method and I have a load less than this minimum, the correct procedure is to multiply the value that I have with a factor to obtain the (Area projected) x (0.48kN/m2), this is for windward, leeward and roof loads or only the horizontal loads?

 

[OldPaperMaker]

mafiucivil,
For the answers to both of your questions I suggest that you check out the Commentary to ASCE 7-05.
The answer to your first question starts on page 297. The answer to your second question is found in Figure C6-1 on page 307.

I hope that this helps.

 

[mafiucivil]

OldPaperMaker, thank you for your answer. I read the commentary and understand that for low rise building it's more precise the method for h<60ft (figure 6-10).

But my second question I still don't understand what I have to do. In the figure C6-1 it's clear that you need to apply 10 psf as a minimum load on the projected area onto a vertical plane, but I don't know the correct procedure to obtain the correct force if I have less than 10 psf. For example I have

kh=0.7
qh=21.44kgf/m2
h=4.83m
B=15m
L=30m

gable roof with
eave=4m
max_height=5.65m
angle=12.4
a=1.5m

(I'm using a very low velocity of 24.1 m/s or 53.7mph, because the new code in my country allow that velocity and must be calculeted with the asce7)

(Load_kgf/m2)x(Area projected _m^2)=
1: 6.09 x 108 = 657.7
2: -18.65sin(12.4) x 44.55 = -178.4
3: 12.95*sin(12.4) x 44.55 = 123.9 (positive in one dir)
4: 11.56 x 108 = 1248
1E: 11.23 x 12 = 134.8
2E: -26.8*sin(12.4) x 4.95 = -28.5
3E: 16.9*sin(12.4) x 4.95 = 17.9 (positive in one dir)
4E: 15.3 x 12 = 183.6

The total force is 2159 kgf, and if I multiply the 10psf x Area = 8296 kgf.

The question is, I need to multiply all the pressures by (Ftot/Fmin) ?


*my english isn't very good, i'm still learning it. I hope that you understand.

 

[mafiucivil]

In the last sentece I meant Fmin/Ftot

 

[a2mfk]

If you have less than the min of 10psf, you just use 10psf. Unless of course a negative pressure will provide a worse effect on the structure, that of course is for you to decide.

 

[mafiucivil]

thank you both for your answers, ok so I can design the MWFRS with the 10psf all over the height, but what happen with the roof? Should I use the pressures 2 and 3 or a bigger?

 

[OldPaperMaker]

I don't know how to help you out on your question. In the US the minimum wind velocity is 85 MPH (miles per hour). In my work all of the ASCE methods, except the 10 psf minimum noted in section 6.1.4.1, result in a uplift on the windward & lee side of the roof that is greater than 10 psf. Therefore, I have never had to deal with this situation.
It is probably safe to apply the 10 psf to the horizontal projection of the roof.