Multistory wood shear walls, overturning and uplift

[work2play]

Hello, I am new to this site and really found it to be informative so far. I have a question regarding multistory wood shear walls and uplilft.

Two methods for determining overturning and uplift are described below for a two story building. Dead loads are neglected for the description.

Method 1) Loads at roof and floor levels are multiplied by the respective height above ground (x=floor to floor height) and divided by the wall length to determine overturning.
Uplift = [Roof*(2x) + Floor*(x)]/L

Method 2) Story shears are used at each level to determine uplift at each level. Second floor uplift = (Roof*x)/L and uplift at first floor = [(Floor+Roof)*x]/L. If uplift occurs at the second floor this force is added to the first floor reaction to determine what hold down, if any, is required at the foundation.

Design of Wood Structures by Breyer says to examine the stability of the individual shear walls. If the individual parts are stable, the entire building is stable. It goes on to say that overturning of the entire building (such as steel rigid frames) is not done for a wood shearwall building.

Would you interpret this to mean you would use method 2 described above for wood framed buildings or still use method 1? What approach do other people use to determine hold downs and uplift forces at wood shear walls? I have found one example on the internet by Philip Line of a two story shear wall. There is not a large amount of information out there for multistory wind shear walls.

Please let me know what you think and if you would like any clarifications. Sorry for being longwinded.

Thank you.

 

[Taro]

I believe your two methods produce identical results. Just rearrange the equations using the distributive property of algebra to prove this.

 

[whyun]

I am in agreement with Taro. Both methods produce identical results in terms of shear per lineal foot and the uplift force at the lowest story considered. Be sure to add appropriate tributary deadloads at both floors for method 1.

 

[work2play]

That is correct if we have uplift at the second floor for method 2. I should have clarified it better, if we assume dead load is used and there is no uplift at the second floor, then you will get an increased uplift (about 40%??or more) using method 1. How would this be resolved?

 

[RARWOOD]

My interpertation of Breyer's statement on overturning would be that in a wood building you have a flexiable diaphragm, so that an indvidual shear wall only receives shear load from the diaphragms that frame into it.

In a building with a flat plate concrete slab I would consider the diaphragm to be rigid. In my design experience for this type of building I would distribute load into the indvidual shear walls based on their stiffness. The assumption I followed was with a series of short walls and long wal, as the shorter walls are loaded they begin to deflect as they do this the rigid diaphragm starts transferring load to the longer walls.

In a wood framed building you could have shear walls on the two ends and one in the middle. One end wall and the middle wall could be the full width of the building while the remaining end wall could be half the building width. With a flexiable diaphragm, the diaphragm act as single span beams between the shear walls. As a result the two end walls would carry the same shear load. The short wall could be heavily loaded while the other two walls were lightly loaded. In this case as long as all three walls have sufficient resisting load to prevent overturning the building will be stable.

 

[whyun]

If there is no uplift at the second floor, apply the downward force from the upper level to the free-body diagram for the lower level.

 

[work2play]

Would you calculate the uplift force at the first floor of a multistory building by using the first floor wall height with the total shear at that point with a flexible diaphragm? Each level will be considered seperate in terms of determining uplift at each floor this way and you will sum the uplift forces as you go down, ignoring compression when it occurs?

 

[whyun]

Yes. shear applied at the lower level should be the cumulative shear from all floors above. Technically you should include both the compression and tension/compression reactions at each end of the upper shear wall and apply them as imposed loads on the lower wall in question.

For simplicity, the compression loads on the compression end can be ignored because moment arm is zero.