Diaphragm (Distributed + Point Load)

[EngStuff]

My question is regarding point load transfer with a distributed load and a notch. I also drew an image. see below

A quick description

Working on a custom home. Stuck with this layout so cannot add or remove shear walls/beams columns.
1. Blue walls are first floor shear walls.
2. Green wall is 2nd floor shear wall (non load bearing).
3. 200 plf wind load with a 5k center load from the 2nd floor shear wall.
4. 2nd floor shear wall is supported by a beam/column(not a moment frame).



Beam/column will not resist shear and will not be a moment frame due to the fact if it is, i would have to take a strut all the way to the right side. So it will only see uplift and downward force.

Note: I am following part of terry malones example.

My thoughts + question

1. Can we stop the strut for the 2nd floor shear wall @ collector "c1" on d2. Also, if we were to do that, do all the 5k load go into the diaphragm d2 and none on diaphragm d1. If we can, I don't see how I can get 200 plf distributed between d1 and d2 and 5000 only on d1.

Part of me thinks that 5k load will be distributed throughout d1 and d2. but my other thought is how can it unless i place a strut down the center the full length which I am not going to do that?

2. Do I take the strut s1 into the transfer diaphragm and all the 5k load transfers within the TD1?

Pretty much stuck on how to do this. I think I got the layout done correctly, but don't have the right idea on how to transfer the 5k force.

Thanks

 

[SJBombero]

No matter what, because of the notch, D1 is necessary to transfer the second floor shear into SW1. I see two load paths.
1. An adequately developed strap extending the strut into D1.
2. If D2 has adequate capacity to develop 5000 lb of the length of teh 2nd floor shear wall attachment, 1/2 can go into each boundary to be collected into SW1 and SW2. But because of the notch, that 2500 lb + whatever is collected from the 1st floor diaphragm needs to be developed into the diaphragm in order to get to SW1.

1. Can we stop the strut for the 2nd floor shear wall @ collector "c1" on d2. Also, if we were to do that, do all the 5k load go into the diaphragm d2 and none on diaphragm d1. If we can, I don't see how I can get 200 plf distributed between d1 and d2 and 5000 only on d1.

This is load path 2. Diaphragm D2 would then have 2500 lb at the boundary + whatever it collected from the 200 plf lateral.

2. Do I take the strut s1 into the transfer diaphragm and all the 5k load transfers within the TD1?

This is load path 2 except you want to shorten the drag connection over the depth of TD1. I don't have Terry Malone's book here at the moment to review your layout. I believe the transfer diaphragm should still meet aspect ratio limits because you are presuming that this specific section of the diaphragm is stiff enough to act like a beam and transfer shear forces rigidly. If the aspect ration is slender like you are showing, it wouldn't work. The transfer diaphragm also has to be deep enough to develop a drag strut. If you wanted to develop S1 into TD1 you would end up with 5000 lb / 10 ft = 500 plf diaphragm shear (assuming TD1 is 10 ft deep). It is better just to extend S1 as deep into D1 as you need to so that the total diaphragm shear is within the shear capacity of the diaphragm.

 

[EngStuff]

My drawing wasn't good. My TD was a 2:1 Ratio so I wouldn't have to block it. See the image below, I show the full dimensions and calculations to get my shear forces. I want to know if I got the correct idea on transferring them using load path "2" without extending S1 into the TD and only having S2 into TD. Also I didn't continue my calculations on the T=C forces on C1 and C2 since those are straight forward.

picture looks unclear on my preview so I cut it down to 3 pics as well.

Excuse me for the messy writing, it's really late night and it's on a 60-70 degree monitor(stand hinge broken)

 

[SJBombero]

The number look like what I would expect. You can check by adding up all the inputs from the diaphragm and collector to SW1 and SW2 and make sure it adds up to 13200.

 

[EngStuff]

My confusion is at the beginning.

Pretty much I have it where D2 sees it's share of the distributed load and the 5k point load. D1 only sees it's share of the distributed load. I am having a hard time wrapping my head on this. why doesn't D1 see it's share of the point load? like if we are doing the load distribution within the diaphragm.

On my second line of calculation I have (A-B)(2-4) = 151 plf, per the 200 distributed load. shouldn't it also include the share of the 5k load? why or why not?

 

[SJBombero]

D1 does not experience any of the diaphragm shear from the point load because the load is already in the collector (double top plate if this is a wood house) by that point.

I don't understand what you are doing with the (A-B)(2-4). The diaphragm shear from the 200 plf lateral load should be 0.5 x 200 plf x 41' / 80' = 52.25 plf. If you add up your shears (A-B)(1-2) and (A-B)(2-4) you get a total lateral reaction 45 plf x 18' + 151 plf x 42' = 20k total lateral load at the first floor not yet including the point load. I don't think that is correct.

 

[ProgrammingPE]

I don't understand what you are doing with the (A-B)(2-4). The diaphragm shear from the 200 plf lateral load should be 0.5 x 200 plf x 41' / 80' = 52.25 plf. If you add up your shears (A-B)(1-2) and (A-B)(2-4) you get a total lateral reaction 45 plf x 18' + 151 plf x 42' = 20k total lateral load at the first floor not yet including the point load. I don't think that is correct.

In EngStuff's calc, the 45 plf and 151 plf is how much of the 200plf linear load is applied to diaphragm (A-B)(1-2) and diaphragm (A-B)(2-4) based on their depths. It's the linear load, not the shear at the ends.

I noticed two issues in the calc. First, in the calculation va = -3178.4/42 = -73.8 plf, where did the 3178.4 come from? Shouldn't this be using the Ra value of -3101 calculated in the line before?

Second, the 77 plf and 73.8 plf were determined for diaphragm D1 using its depth of 42 feet, but then those values get combined with the transfer diaphragm end reactions which are based on its depth of 20 feet. These are shears for two different locations. You do need to account for the 200 plf linear load at the end of TD1, but to do this, you would either use the combined depth of 62 feet or equivalently you would have to figure out how much of the 151 plf linear load that you used for D1 should be applied to TD1 instead.


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[EngStuff]

I noticed two issues in the calc. First, in the calculation va = -3178.4/42 = -73.8 plf, where did the 3178.4 come from? Shouldn't this be using the Ra value of -3101 calculated in the line before?

Second, the 77 plf and 73.8 plf were determined for diaphragm D1 using its depth of 42 feet, but then those values get combined with the transfer diaphragm end reactions which are based on its depth of 20 feet.

You're right It should be 3101 lbs and it should be the 62 feet. They drop down to 50.1 and 52.3 respectively.

Out of curiosity, what if we have a opening inside the TD1? Are we allowed to do that? or do we just decrease the size of the TD1 which causes an increase of loads.

Thanks!