Wind load ASCE
Wind load ASCE
(OP)
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?
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?






RE: Wind load ASCE
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.
RE: Wind load ASCE
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.
RE: Wind load ASCE
RE: Wind load ASCE
RE: Wind load ASCE
RE: Wind load ASCE
It is probably safe to apply the 10 psf to the horizontal projection of the roof.