[WOOD] In plane force distribution based on 4.3.5.5.1 exception
[WOOD] In plane force distribution based on 4.3.5.5.1 exception
(OP)
When using the exception to deformation compatibility for in plane shear distribution, do you apply the capacity reduction for high aspect ratio shear walls (h/b>2) after or before force distribution?
Assume all shear walls in a line are of the same construction.
If you apply the capacity reduction to high aspect ratio shear walls after force distribution, you distribute based on wall length since all walls have the same unit shear capacity. This equates to the traditional method noted in Breyer. Then you check actual shear vs allowable shear. Actual shear is the same in all walls. Allowable shear is reduced for the the high aspect ratio shear walls (h/b>2). That is how it is shown in this APA presentation.
However, I believe you need to apply the capacity reduction to high aspect ratio shear walls (h/b>2) before force distribution. See excerpt from 2021 SPDWS example C4.3.5.5.1-2 below.
I'm thinking something like this:
ASD design
V wind = 2000 #
15/32" sheath, 8d common, 6" edge nailing
v allow = 730/2 = 365 plf
Wall 1
h = 8 ft
b = 8 ft
h/b = 1.00 < 2
No capacity reduction
Design capacity = 8*365 = 2920 #
Wall 2
h = 8 ft
b = 3 ft
h/b = 2.67 > 2
Capacity reduction = 2*3/8 = 0.75
Design capacity = 3*365*0.75 = 821 #
Distribute forces based on design capacity
V1 = 2000*2920/(2920+821) = 1561 #
V2 = 2000*821/(2920+821) = 439 #
v1 = 1561/8 = 195 plf < 365 plf ok
v2 = 439/3 = 146 plf < 0.75*365 = 274 plf ok
Assume all shear walls in a line are of the same construction.
If you apply the capacity reduction to high aspect ratio shear walls after force distribution, you distribute based on wall length since all walls have the same unit shear capacity. This equates to the traditional method noted in Breyer. Then you check actual shear vs allowable shear. Actual shear is the same in all walls. Allowable shear is reduced for the the high aspect ratio shear walls (h/b>2). That is how it is shown in this APA presentation.
However, I believe you need to apply the capacity reduction to high aspect ratio shear walls (h/b>2) before force distribution. See excerpt from 2021 SPDWS example C4.3.5.5.1-2 below.
I'm thinking something like this:
ASD design
V wind = 2000 #
15/32" sheath, 8d common, 6" edge nailing
v allow = 730/2 = 365 plf
Wall 1
h = 8 ft
b = 8 ft
h/b = 1.00 < 2
No capacity reduction
Design capacity = 8*365 = 2920 #
Wall 2
h = 8 ft
b = 3 ft
h/b = 2.67 > 2
Capacity reduction = 2*3/8 = 0.75
Design capacity = 3*365*0.75 = 821 #
Distribute forces based on design capacity
V1 = 2000*2920/(2920+821) = 1561 #
V2 = 2000*821/(2920+821) = 439 #
v1 = 1561/8 = 195 plf < 365 plf ok
v2 = 439/3 = 146 plf < 0.75*365 = 274 plf ok
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RE: [WOOD] In plane force distribution based on 4.3.5.5.1 exception
RE: [WOOD] In plane force distribution based on 4.3.5.5.1 exception
I'd state it differently. I don't think a "rigorous analysis" does a very good job of accounting for how much more non-linearity you get from a skinny shear wall. Elongation of the hold downs and such. Now, I'm sure the CURREE folks who did all the testing can come up with a rigorous non-linear model that is very complicated that gets close to the actual behavior shown in the tests. But, many people would use what they THINK is a truly rigorous model and still not penalize these skinny walls enough.
The reality is that the code writers are penalizing these types of shear walls based on observed behavior in past earthquakes that agrees with testing. That these narrow walls don't take as much load as most engineers think they would. Ergo, they came up with this method to penalize them based on their aspect ratio.
RE: [WOOD] In plane force distribution based on 4.3.5.5.1 exception
RE: [WOOD] In plane force distribution based on 4.3.5.5.1 exception
Meaning there are many different forms of a "rigorous" analysis engineer might use that still wouldn't properly capture the softening effect of the high aspect ratio walls.
RE: [WOOD] In plane force distribution based on 4.3.5.5.1 exception
(just venting now)
None of these references show how the distribution method would apply for a given force. They only look at capacity. Which I understand can be used to infer, but the point of a worked example is to demonstrate practical use. Come on.
2021 SDPWS example C4.3.5.5.1-2 (p47) (only looks at capacity)
APA 2021 SDPWS (incorrect distribution)
AWC 2015 SPDWS (only looks at capacity, also poor handling of topic)
Woodworks 2015 SDPWS (p68-73) (only looks at capacity)