Line load under shear wall?

[Bubik]

Hi everyone

I have been tasked to provide the value of a line load under a timber wall subjected to wind load. The wall will sit on a beam.
I know how to calculate pressure under shear wall/retaining wall, but I have no clue how to approach the problem. All they need is the line load from wind under the wall.

Thank you!!

 

[jayrod12]

There probably isn't one. There will be a horizontal shear from wind, concentrated vertical load at the ends of the shear wall from wind, and vertical line loads from any gravity load supported by the wall.

Potentially there is a vertical line load from wind, but it would be uplift.

 

[Ganesh Persaud]

I’ll have to agree with jayrod12, The wind load applied to the area of wall will have an inplane horizontal shear.

However, you’re looking for the wind UDL load that would be acting on the beam, correct?

Treating the wall as an element with no openings, if you take the wind pressure multiplied by the Height of wall, you would get the wind inplane (horizontal) load component acting on the beam in units of force/length.

 

[jayrod12]

To be clear, the only UDL from wind that would be acting on the beam would be either a lateral load applying axial load to the beam, or an uplift load. I do not see a scenario where there is a udl on the beam in the downward direction from wind.

 

[JoshPlumSE]

Bubik -

So, you've got a wood shear wall above and it's being supported by a beam. You want to convert the load from the wall into a loading on the beam.

There are a couple of ways to do this:
1) The typical way would be the following:
a) Add a distributed shear at the base of the wall into the beam. That's the easy part.
b) Convert the moment at the base of the wall into a tension / compression force couple at the hold down / compression chord locations for the wall.

2) Another way to do this would be to put this in as a triagular distributed load at the beam. Where the moment from the triangular loading is equal to the moment caused by the shear force times the height of the wall. This is simple statics (Sum of the moments = 0). Though I should point out that item 1) is typically how wood design of shear walls is performed.

 

[Bubik]

Thank you all, it's great!!

 

[dold]

The quickest way to find bearing pressure is trusty old (P/A ± M/S). It's just superimposing bearing from a uniform distributed load + the triangular load(s) that JoshPlum shows. But it sounds like you've got that part under control.

Although I would have to disagree with the notion that wind loads can't cause an inward pressure and load a wall in the direction of gravity. Or maybe i'm misinterpreting something.

 

[jayrod12]

Maybe it's just my experience, but wind loads on standard roof shapes/profiles only apply suction. I didn't say in my responses that it wasn't possible, but I do not see a situation utilizing wood shear walls, that would result in a downward uniform load on a shear wall from only the wind load. I see either uplift, or a racking load resulting in horizontal shear and a T/C couple at the ends of the walls.