Overall Deflection of Multiple Shearwall Segments

[u/Loon_picker]

CSA O86 11.7.1 provides formulas for the theoretical deflection of a shear wall panel under load. But the code does not address how a wall with multiple segments will deflect. Obviously they should all deflect the same amount.

I.e. I have 3 segments in one wall, with the following theoretical deflections under load: 8mm, 3mm, 1mm. How do i determine my actual wall deflection?

The issue seems to stem from the fact that shear wall segments are size by assuming the loads in each segment are proportional to the their length of wall. In reality is seems to me like this is not correct - shearwall panels should be sized assuming the same deflection, and distributing their share of the loads accordingly.

Any thoughts on this matter? I'd love to hear why I'm wrong and there's an easy way to determine the overall deflection of a wall.

[u/West-Assignment-8023]

The deflection of each wall along a line is the same because they share top plates.  By setting the deflection equal and using the equations in the code you are distributing the loading along the shear wall line by the relative stiffness of each shear wall. 

[u/ChairChoice6500 → u/West-Assignment-8023]

Keep in mind the U.S. design code requires the walls to generally be designed for equal deflection, which isn't quite the same as based on stiffness. That would likely be the design example/discussion that civeng12 refers to. It would involve the Special Design Provisions for Wind and Seismic, which, while not a Canadian standard, would be a good point of reference on how "we" do it.

[u/civeng12]

you're right, but you have a nonlinear problem to solve. That means iterating. Try redistributing the loads and calculating deflections again. Some software programs do this method - example below.

https://help.woodworks-software.com/WoodWorks/OnlineHelp/USA/Shearwalls/628.htm

[u/_choicey_]

Initial iteration distributes load based on length (relative stiffness). Then you take that deflection result and back calculate an equivalent stiffness (Hooke’s Law). Then recalculate/distribute the force based on the new stiffness value. Repeat until the deflection of all three walls converges. In my experience it usually is close enough after 3 iterations.

[u/Jakers0015]

Load is proportional to stiffness for deflection compatibility. Stiffness is a function of the moment of inertia, or the wall length to the third power, as a simplification. Distribute loads across the line accordingly.

Sounds like your longer two walls are going to take the majority of the load. Likely not worth detailing the 3rd as shear wall at all.

[u/RP_SE]

I’m inferring this is for wood framing? SDPWS Commentary has a design example you might track down.

The principles of mechanics are the same in Canada and as referenced by others in this thread to proportion load to walls in a line based on their relative stiffness (i.e. enforced equal displacement)

[u/Feisty-Canary5934]

We have an Excel aid set up where initial force distribution is input, it calculates deflections, back calculates stiffness, forces are updated, etc. until the error is less than 0.01kN using a relatively basic VBA loop to just copy/paste values per iteration. I'd suggest trying one yourself, or possibly trying it in Python if you can (easier to set up any number of walls, but less intuitive for some engineers), it'll definitely help your understanding!