Using DL to resist overturning on wood shear wall?
Using DL to resist overturning on wood shear wall?
(OP)
When you are calculating the overturning hold-down forces on a wood shear wall segment, do you treat the entire wall segment as a rigid body and reduce the hold-down forces based on the dead load (self-weight or from floor above) across the entire wall segment?
I always do this for a masonry or concrete shear wall since the wall is truly one integral rigid body. However, a wood wall is an assembly of vertical studs, blocking, and sheathing. I have questions on whether or not dead load on a wall stud in the middle of a long wall segment is truly capable of reducing the overturing uplift forces of the chord studs at the end of the wall segment. However, I haven't really seen a discussion of this issue. In Breyer's wood book, he indicates that you should reduce the forces due to the dead load, but without any discussion of my questions above.
Thoughts? TIA.
I always do this for a masonry or concrete shear wall since the wall is truly one integral rigid body. However, a wood wall is an assembly of vertical studs, blocking, and sheathing. I have questions on whether or not dead load on a wall stud in the middle of a long wall segment is truly capable of reducing the overturing uplift forces of the chord studs at the end of the wall segment. However, I haven't really seen a discussion of this issue. In Breyer's wood book, he indicates that you should reduce the forces due to the dead load, but without any discussion of my questions above.
Thoughts? TIA.






RE: Using DL to resist overturning on wood shear wall?
I use .6 x wall weight and use create a couple that will help reduce the overturning. The couple on arm of the couple is about the center line of the shear wall segment and the other is centerd on the hold down anchor.
RE: Using DL to resist overturning on wood shear wall?
If you apply that same principle o your wood shearwalls and consider the top plate to span between the two chords (an extreme example, I know, but just to get started to make the point), you'll get a reaction at the tension chord and the compression chord. The resulting effect is the same since you're simply taking moments about a corner of the shearwall. Now if you take that same principle and apply it to a tip plate spanning between studs at 16" o.c. I think you'll find the same thing.
If the studs line up then I think the answer is pretty clear.
I also want to point out that I don't think this would be the same if the wall had multiple hold-downs along it's length.
RE: Using DL to resist overturning on wood shear wall?
Mike McCann
MMC Engineering
Motto: KISS
Motivation: Don't ask
RE: Using DL to resist overturning on wood shear wall?
RE: Using DL to resist overturning on wood shear wall?
DaveAtkins
RE: Using DL to resist overturning on wood shear wall?
RE: Using DL to resist overturning on wood shear wall?
It is a common practice to use deadload to resist overturning for wood walls, but unless it is a concentrated load just about right at the end of the wall, it is almost always in error. I had a professor swear that he would come get us if he ever caught us doing that.
Let me know if you want more explanation.
RE: Using DL to resist overturning on wood shear wall?
RE: Using DL to resist overturning on wood shear wall?
I would love to debate your professor about this. If you look at each 4' width of shear wall as an individual section (because plywood or OSB positioned vertically is 4' wide), and look at an individual section as it overturns, the "back" side of the section tries to pick up the 4' section of wall "behind" it. But it can't, because the "front" of that section of wall is pushing down, due to its own overturning.
And so, the tension and compression forces only occur at each end of the entire shear wall.
DaveAtkins
RE: Using DL to resist overturning on wood shear wall?
Yes, that is exactly what is happening. So, take two things from that.
1) The individual panels are rotating. One panel is providing the compression to resist the tension for the next. So, each is rotating about its center.
2) If the first panel does not have the tension resisted at its end, it cannot resist its full amount. In other words, you have two options if you decide to use deadload in the middle of the wall. 1) you can pretend that the shearwall is shorter in length 2) you can say the wall before that deadload is less affective.
Think of it this way, pretend there is no deadload. Place a holddown in the middle of the length of the wall. It isn't going to matter if the holddown is twice the capacity. The shearwall isn't as strong as if the holddown were at the end.
RE: Using DL to resist overturning on wood shear wall?
DaveAtkins
RE: Using DL to resist overturning on wood shear wall?
I don't know that I believe that a DL on top of a wood shearwall contributes nothing to resisting overturning. I understand it's not a truly rigid body, but for the magnitude of loads under consideration it's a reasonable approximation. Joints are typically staggered, so it's not like there is a continuous vertical joint at any given stud.
To say that the load can't distribute to anything beyond the sheathing to which the individual stud is fastened discounts the diaphragm behavior completely for vertical loads (which is sometimes more rigid than for lateral loads depending on the aspect ratio). If you have a floor diaphragm and you grab the end of a single joist and yank on it to put it in tension that force is resisted by the diaphragm, not by the connection of the joist at the opposite end of whatever it happens to be connected to.
I guess part of the question is what does the OP consider a "long" wood shearwall. Honestly, if your shearwall is so long and short that this is a concern, you probably don't have very high uplift forces anyway.
I'm picturing myself standing on a platform next to the top of a wood shearwall, say 16' long and 8'high (just to make sure we have joints in the wall). If I can push that wall over (in-plane) with a force of (just say) 200 pounds with just the wall standing free (but stabilized out-of-plane) and then a 300 pound guy stands on top of the wall, 8' from each end (in the center), I just don't believe that I'll be able to push that wall over nearly as easily. In fact, I think it will be significantly more difficult (without diving into the numbers).
RE: Using DL to resist overturning on wood shear wall?
I'll try to explain it this way. We agreed that it was the ends that had the compression and the tension. So, my question is how is a force in the middle going to help resist the tension? What is the loadpath? It cannot be the sheathing because you are already using that to resist shear. (There might be some residual capacity, but how much?) The basically leaves the double top plate in weak axis bending. It isn't going to send it over because there is much stiffer studs right beneath it. (Which is why the diaphragm analogy doesn't quite work. Diaphragms don't have one whole side supported, or it wouldn't act like a beam.)
A uniform load would help resist uplift, but only the load near the uplift.
Part of the problem might be that standards for engineered shearwalls only recognize fully restrained walls. In PAStructuralPE's example, he went from unrestrained to partially restrained. Neither of those are really allowed. Deadload helps an unrestrained wall, but that wasn't an option to begin with. In other words, if you already had a full holddown, would the guy standing on it make a difference?
RE: Using DL to resist overturning on wood shear wall?
RE: Using DL to resist overturning on wood shear wall?
DaveAtkins
RE: Using DL to resist overturning on wood shear wall?
I'm attaching some statics examples to show my point. Sorry for their messiness. I did them quick.
RE: Using DL to resist overturning on wood shear wall?
DaveAtkins
RE: Using DL to resist overturning on wood shear wall?
RE: Using DL to resist overturning on wood shear wall?
DaveAtkins
RE: Using DL to resist overturning on wood shear wall?
RE: Using DL to resist overturning on wood shear wall?
Perhaps I am missing something. If so, could you show me with a diagram?
RE: Using DL to resist overturning on wood shear wall?
RE: Using DL to resist overturning on wood shear wall?
DaveAtkins
RE: Using DL to resist overturning on wood shear wall?
If you like, you can see the attached, but it is just statics and algebra from your own sketch. If you follow the previous paragraph, there is no reason to.
RE: Using DL to resist overturning on wood shear wall?
Now everything is in equilibrium.
DaveAtkins
RE: Using DL to resist overturning on wood shear wall?
RE: Using DL to resist overturning on wood shear wall?
1) It isn't in equilibrium. Take the moments about the center of the panel.
2) How do you expect to transfer those forces? All you have done is confuse the issue by showing them like that. The issue is the forces on the external forces on the panel. The panel only has dowel fasteners around its edge. Dowel fasteners do not transfer moments. They transfer a force-couple which causes a moment. Similarly when a concentrated load is placed on a row of fasteners, it is distributed along the row. (Lets just assume uniformly for ease.) This uniform lateral load is what is of interest. It is what shearwalls are designed around, the panel/nail interaction capacity.
3) Distributing the concentrated loads you show as uniform loads, and assuming that the moment is transferred through the top and bottom plate (the only 2 common members that the two panels share, which means the only why a force-couple can be created,) the deadload adds to the shear seen in the panel. This means that if you used deadload from the middle of the wall, you would have to reduce the tabulated capacity of the wall. You have not addressed this. (This gets back to "If you use the panels to pick up the load as a beam, that takes away from their ability to resist shear.")
I appreciate the time you've spent on this. I do not wish to be giving people bad advice, but I don't want engineers under designing shearwalls either.
RE: Using DL to resist overturning on wood shear wall?
Remember, even on a long wall, the Dead Load (DL) is typically a distributed load (not a concentrated load). I have used a portion (at least equal to the wall height) of the distributed dead load on long walls (i.e. if wall is 40' long by 10' tall, use the 10' of wall closest to your holddown location for dead load) since wall sheathing will span/arch this dead load to the holddown stud. Draw a Diagram with all forces as Dave Atkins has down.
So, I basically agree with DaveAtkins. Except, I typically don't use the Full Dead Load when calc'ing compression stud force (since dead load typically goes directly down in nearest field studs, then down to foundation). Think of a concrete beam stress diagram (concentrated tension and distributed compression) - the rebar is the 'holddown'
RE: Using DL to resist overturning on wood shear wall?
Another way to look at this is--when a horizontal shear pushes on a sheathing panel, it wants to overturn. The sheathing shears along the top plate, both edge studs, and the sill plate. But as the sheathing tries to pick up the stud at the back side of the panel, it cannot--because there is a dead load pushing down on the stud. The dead load does not enter the sheathing, so it does not add to the shear in the sheathing. It stays in the stud, and reduces overturning on the panel.
DaveAtkins
RE: Using DL to resist overturning on wood shear wall?
I'm not saying that you don't show all the forces, I'm saying that you can sum them. Just like you can sum them to find the resultant force. Also, showing moments is just a shorthand method that can sometimes lead to trouble (such as this case.) It would be like saying, "No my beam-column isn't over loaded in compression, it only has this much axial and that much moment. Two completely different things." Moments are just shorthand for some form of force couple. To know what is really going on, show the force couple.
As far as the summing of the forces, if you aren't saying that the overturning is being resisted by the panel, then it isn't needed.
Now it is just an issue of loadpath. If no holddown or an inadequate holddown to resist the entire OT force is designed, then there are only 3 options to resist the rest of the force. The bottom plate, the top plate and the panel are the only member that touch the stud, so the force has to go through them. It will go to the stiffest. See page 4 of the attached for that breakdown. Hopefully you will see that the panel is stiffer than the top plate in weak-axis bending, so it has to be the panel. Since your method requires it not to go through the panel, it doesn't work. I've shown that by going through the panel, it lessens the ability of the panel to resist the shear force.
RE: Using DL to resist overturning on wood shear wall?
I have also attached Section 2305.3.7 from the IBC which specifically allows the use of 0.6DL to reduce overturning.
DaveAtkins
RE: Using DL to resist overturning on wood shear wall?
As far as Section 2305.3.7, it has no bearing over my contention. All Section 2305.3.7 says is that if deadload can resist uplift, you can use it. I'm saying that when the deadload is away from the area being uplifted, it cannot resist it so according to Section 2305.3.7, you cannot use it.
I'll give in if you can tell me two things and have them make sense. My contention rests on the resulting consequences of these questions.
1) How is that moment being transferred? The actual forces going through the connections or inside of members. Shear, tension and compression forces only please. No moments.
2) Can you honestly say that the top plate in weak-axis bending is going to be stiffer than the panel with its fasteners? (See page 4 of previous attachment.)
As always, your time is appreciated.
RE: Using DL to resist overturning on wood shear wall?
And I still contend that the sheathing, when it tries to overturn, cannot pick the stud up that has dead load on it. This, in turn, reduces overturning on the entire shear wall.
DaveAtkins
RE: Using DL to resist overturning on wood shear wall?
I guess I'm not understanding what you are saying.
1) When you resolved the moment into the force couple, see how it adds DL/4H to the shear that segment has to resist at the top? I've been under the impression that you say that it will not increase the shear load at the top. You show it doing so. (With a holddown, that segment would only resist V/2 or half the shearforce. Now it has to resist V/2 + DL/4H). Sum your V and V/2 -DL/4H and this shows up.
2) How does the stud that is being overturned have deadload on it? The deadload is L/2 away. There has to be a loadpath. What is it?
RE: Using DL to resist overturning on wood shear wall?
I've usually erred on the side of safety and assumed no dead load reduction on a hold-down due to rigid body action of a wood shear wall segment. However, there must be some resistance, particularly if the wall has a large amount of floor or roof load on it as well. For example, consider a 3 or 4 story wood shear wall that supports significant dead load from each floor. At some point it becomes a little absurd to design for no dead load reduction. However, I am still not convinced that treating the entire walls segment as a complete rigid body is a safe assumption. Dealing with large concentrated loads at various locations along a wall can be even more murky.
Somebody needs to get in a lab and do some testing.
RE: Using DL to resist overturning on wood shear wall?
It isn't going to treat the entire wood shearwalls as a rigid body. That isn't how they work. Each panel is a rigid body, so only the deadload that would be useful for the panel having to resist uplift can be used. It is probably going to look somewhat similar the last example in my first attachment.
RE: Using DL to resist overturning on wood shear wall?
As long as you don't meet the exceptions of course.
On a side note, if caair is to be taken literally that dead load near the center of long wall cannot help the overturning of a timber shearwall, then does that mean also that the process of including holdowns on the ends of the walls at the vertical chords is pointless because the wall will rip itself apart in the middle? it seems this is logical from the arguments presented, because if the wall cannot act as a unit, then how would a long wall ever engage the holddowns at each end? Is there a need to add holddowns at some regular interval on a long wall?
RE: Using DL to resist overturning on wood shear wall?
But I still won't require holddowns at the ends of shear walls that have sufficient dead load to prevent overturning. If I did, virtually every building would require holddowns at all shear walls. And it makes no sense that a long shear wall with a lot of roof and/or floor dead load will overturn.
DaveAtkins
RE: Using DL to resist overturning on wood shear wall?
@slomobile
No, only including deadload at the ends does not have an effect on the middle like that. Each panel is looked at individually. One panel provides the tension restraint to the next by compression on the shared stud. As long as everything is consistent, it works out. (Openings, sudden uplifts from outside source, etc. are what I mean by inconsistencies.)
@DaveAtkins
I'm not sure if every building would or wouldn't need a holddown. The method doesn't so much require holddowns, but reduces capacity if they aren't there. It also increases capacity for sheathing above and below openings that before didn't count for anything. It definitely would be a change though, you are right.
Big changes like this happen. Before the Northridge Earthquake, a lot of things were different, especially in steel. Maybe this time they figured things out before an event.
One way or another, it is obvious that if AWC expects this method to be widely accepted, they face quite a bit of educational hurdles.
RE: Using DL to resist overturning on wood shear wall?
Unless I am mistaken, I don't remember many properly constructed plywood shear wall buildings having structural failure issues at Northridge or Loma Prieta. Many multiple story wood buildings that had issues were sitting on top of concrete parking levels. These parking levels pancaked, but the wood above maintained its integrity for the most part (although at a lower level than it was constructed at). I heard of emergency workers driving past buildings multiple times before they realized that what was originally 3 stories of wood on a parking level was now 3 stories of wood on rubble.
If there is a valid reason to change the codes, I'm all for it, but let's not change codes just for the sake of change (or to sell books).