Shear Wall Aspect Ratio with Perforated design and other conditions 1

[Tahoebrian]

Hi all,

I am going through and rethinking my shear wall approach and have come across a couple questions that I cannot find a definite answer on. I am hoping some of you can share your thoughts.

First question.. the 2008 SDPWS states that the aspect ratio (paraphrasing) is for high aspect ratio wall segments relative to other wall segments with lower ratio IN THE SAME LINE. Now suppose you have a wall line that does not have any lower ratio wall segments.. say for example just 2 segments at 3.5:1 (typical for the front of a residential garage). It seems like the aspect ratio would not be applicable since there are only two wall segments of the same ratio. Thoughts?

Second Question.. When calculating the wall height aspect ratio for a perforated shear wall, what would be a suitable value for the b term (shearwall length). It seems reasonable that the b term would be the sum of wall segments in a perforated assembly but I can find nothing mentioning this.

Thanks a bunch.

 

[phamENG]

For starters, I don't have a copy of the 2008, just the 2015. So my references will be from there. You can access it free here.

I'm not sure I understand your first question. Are you referring to "Shear Wall Aspect Ratios and Capacity Adjustments" in 4.3.4.3? If so, it absolutely applies in the situation you've mentioned. You have to multiply all shear walls with an aspect ratio greater than 2:1 but 2x the inverse of your largest aspect ratio in that line. In your stated example, the largest aspect ratio is still used even though they are all the same.

For your second question, each indiviidual segment on the line will have its own aspect ratio, and b will be different for each segment. (Figure 4C on page 29).

For instance, if you have a wall that's 12 feet long with a 3' window located 4' from one end and a wall height of 9', you'll have two segments with aspect ratios of 9/4=2.25:1 and 9/5=1.8:1. The first one will get an aspect ratio adjustment, so when you sum the total length you'll end up with 5+4*(2*4/9)=5+3.56=8.56ft for SUM(L).

Hope that helps. If I missed the mark, let me know.