Using DL to resist overturning on wood shear wall? 2

[abusementpark]

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.

 

[OHIOMatt]

The idea of making it a shear wall is that it will act as a single unit. If it did not, you would not be able to get the force to the hold down.

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.

 

[Lion06]

I haven't drawn out a FBD to verify, but I don't think it will matter (at least not by much depending on whether the studs line up). I say that for one reason and here i'll use an example. If you have a simply supported beam with multiple point loads and varying line loads you can find the reactions by making all of those load conditions into a single point load acting at a specific location.

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.

 

[msquared48]

If the desd load is substantial and will reduce the holddown, I use it. Otherwise not.

Mike McCann
MMC Engineering
Motto: KISS
Motivation: Don't ask

 

[larsacious]

The sheathing that you use to resist the shear is stiff and will behave as a rigid body just like the floor diaphragm.

 

[DaveAtkins]

Yes, I remember seeing an example on perforated shear wall design that assumes the entire wall is one unit, with the dead load counteracting the overturning.

DaveAtkins

 

[a2mfk]

Usually for me in wind country roof dead load is obviously negated by the wind pressure but I use a conservative roof dead load and floor load x 0.6D to reduce the net overturning. However, with most wood frame structures this amount does not add up to much as Mike said...

 

[Cadair]

I would not. At least I wouldn't in the example you gave of a long shearwall. To think of a shearwall as a simply supported beam is incorrect. It is more like a beam on an elastic foundation. A load in the middle just isn't going to affect the ends unless huge displacements happen. Also, wood shearwalls do not behave as a rigid body. The individual panels do, but not the wall as a whole. The panels individually rotate, so that deadload will affect the nearest panel (or more "down load") but that is about it, especially with openings.

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.

 

[MiketheEngineer]

I think IBC allows only .6 of the Dead Load - I guess based on the assumption that we over estimate the weights of materials..

 

[DaveAtkins]

Cadair,

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

 

[Cadair]

DaveAtkins,

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.

 

[DaveAtkins]

From the original post, I assumed the dead load was uniform along the entire length of the wall. If you check statics of each free body diagram, they add up in such a way that the entire wall acts as a unit.

DaveAtkins

 

[Lion06]

For the record, I was not suggesting that a shearwall could be looked at as a simply supported, I was merely using an analogy for the top plate only of the shearwall for the sole purposes of where you might apply a resisting DL from above.

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).

 

[Cadair]

I agree with what DaveAtkins and PAStructuralPE are saying, it is just to what extent.

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?

 

[ron9876]

If interior studs are connected to the floor and to the top plate it looks to me like the studs in compression would be much stiffer than the wall (acting as a unit) would be in bending. So it seems to me that the majority of the vertical load would be attracted by the interior studs.

 

[DaveAtkins]

In order to pick up the back end of a shear wall which has a 300# man standing on the middle of the wall, you would need to lift the man up. So, yes, a concentrated load on the middle of a shear wall resists overturning. The wall, being 8' deep, is stiff enough to not fail in bending as it picks up and rotates about the front end of the wall.

DaveAtkins

 

[Cadair]

DaveAtkins is correct that you would need to pick the man up, and the wall probably is stiff enough in bending. But once it is being used in bending, it is no longer being used for shear. Also, before you actually get to the point of picking the guy up, you are going to engage the nails along the bottom of the panel to start resisting uplift.

I'm attaching some statics examples to show my point. Sorry for their messiness. I did them quick.

 

[DaveAtkins]

I disagree with page 2 of your calcs. When DL is used to resist OTM, T is no longer equal to C. The DL reduces T and increases C. This satisfies statics.

DaveAtkins

 

[Cadair]

I apologize. I was going too fast. Try this for page 2. Same result though.

 

[DaveAtkins]

Moments are not in equilibrium because you forgot to put the external "D/2" force on each free body diagram.

DaveAtkins