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ASCE FIG. 27.4-8 Design wind load cases

ASCE FIG. 27.4-8 Design wind load cases

ASCE FIG. 27.4-8 Design wind load cases

(OP)
Good afternoon all, I need help understanding how the roof wind loading works with these loading cases. My superior said he doesn't normally use this method. If some of you are familiar with using MWFRS part 1, please help. Attached is the particular page associated with this figure.

RE: ASCE FIG. 27.4-8 Design wind load cases

What program are you applying these loads in? When I use RISA, I just find the total wind load (not the pressure) and apply it at the geometric center in both directions. Then you take 75% of that load (in each direction) and apply it with an eccentricity equal to 15% of the direction perpendicular to the applied wind load.

You can capture the eccentricity using a distributed load by having a trapezoidal load equal to 75% of the uniform load plus the resulting moment caused by said eccentricity.

RE: ASCE FIG. 27.4-8 Design wind load cases

(OP)
I'm doing it by hand, am I not reading it correctly? It just says plx and ply, nothing about how it works with the values from the roof. So is it entirely a wall load? If so, why would you neglect the roof loads?

RE: ASCE FIG. 27.4-8 Design wind load cases

No, the same principles work at the roof. It's easier to apply as concentrated loads in my opinion. You have q, so you can multiply q times the projected roof area to get the roof portion of the wind load. To get the rest of the load at the roof, you multiply q times the building width perpendicular to the direction of applied load times half the floor height supporting the roof.

RE: ASCE FIG. 27.4-8 Design wind load cases

If you are checking diaphragm connections, or metal deck-to-steel attachments, you can use this as shown. Otherwise, you can simplify it as mike indicates.

RE: ASCE FIG. 27.4-8 Design wind load cases

Look at ASCE 7-10 appendix D for buildings exempted from torsional load cases.

One time I hand calculated all of the wind load cases for a semi-rigid diaphragm on a fairly irregular building because RAM Structural System did not automatically distribute the loads properly. After all of the hand calcs, one frame ended up with a less than 1% increase in force than case 1.

RE: ASCE FIG. 27.4-8 Design wind load cases

(OP)
Thanks for the clear up all. WannabeSE thanks for the appendix reference it kind of maps it out for me.

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