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.

 

[ron9876]

What I was trying to say above is that for the wall to transfer vertical loads to the end it must act like a beam. That means that strain is required to get stress. If the studs are attached to the floor the available movement is limited to the deflection/movement of nails/studs/plywood. Don't see how that could work in the real world.

 

[Cadair]

I'm not seeing it. I'm looking at the individual panels, which cannot resist concentrated loads, only shear through their nails around the edges. In other words, the shear is the concentrated loads. They are just distributed through the length or height of that panel portion. For example, if you sum the two shears where the two panels meet, you get a net downward force of D, the external force. As you can see in the calc, the right panel has to resist more than its tabulated shear value.

Perhaps I am missing something. If so, could you show me with a diagram?

 

[Cadair]

Rereading my last post, I realize I didn't say it clearly. My diagram is showing only external forces. It is just external forces for the panels.

 

[DaveAtkins]

Everything is in equilibrium--see attachment.

DaveAtkins

 

[Cadair]

I disagree; your wall is no different than mine. Take the moment about any corner of your individual panel, and it doesn't work. Also, the if the wall is resisting its horizontal capacity, then you've overloaded it on the vertical shears.

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.

 

[DaveAtkins]

I looked at the free body diagram again--I forgot to show some of the forces and the moment from the adjacent panel.

Now everything is in equilibrium.

DaveAtkins

 

[a2mfk]

This morphed into a good theoretical shear wall discussion, but the original question about dead loads brings me back to square one. Most wood framed buildings use light weight wood framed floors with smaller spans (that steel framed buildings) and interior bearing walls. So dead load resistance will be fairly minimal in most cases, which is probably why some engineers outright neglect it anyway.

 

[Cadair]

DaveAtkins, your last attachment was really confusing to me.

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.

 

[Den32]

0.6 DL can be taken into account to resist wall uplift, within reason (use judgement).

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'

 

[DaveAtkins]

When you draw a free body diagram, you must include all forces and moments acting on the body at the cut. That is what I did--and it IS in equilibrium. If you don't include the effect the holddown force has at the cut (which is a moment), then you don't have a true free body diagram.

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

 

[Cadair]

It isn't in equilibrium. See pages 1-3 of attached. The problem is that you are hiding it in the moment.

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.

 

[DaveAtkins]

You made a mistake summing moments (see attached). The free body is in equilibrium.

I have also attached Section 2305.3.7 from the IBC which specifically allows the use of 0.6DL to reduce overturning.

DaveAtkins

 

[Cadair]

I apologize. I did get my sign wrong. However, you haven't explained how that moment is transferred, so when it is actually shown as a force, it proves to not be in equilibrium. (As shown on page 3 of the previous attachment.)

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.

 

[DaveAtkins]

I resolved the moment into a force couple, and the free body is still in equilibrium (see attached).

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

 

[Cadair]

Apparently I cannot sum moments anymore. (I think it is because I'm used to seeing it in terms of shear, not force. I apologize again.)

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?

 

[abusementpark]

Unfortunately, it sounds like there are a lot of mixed opinions on this topic. I don't guess APA or any other technical organization has commented on this issue.

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.

 

[Cadair]

The American Wood Council, the people that make the NDS and Special Design Provisions for Wind and Seismic, are in the process of writing a technical report on how deadload should be dealt with. It also addresses openings without strapping. Basically it is a mechanics model to handle these things.

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.

 

[slomobile]

Most everyone here has recommended the use of 0.6D for overturning resistance in wood ASD design; this is based on ASD load combos in both ASCE 7 & the IBC I presume. No one has mentioned the requirement that the vertical component of seismic must be applied also, 0.2Sds (ASCE 7-05, 12.4.2.2). So if your force comes from seismic lateral loads, your dead load resistance will effectively be reduced by that amount. 0.2Sds is a strength level force, so for arguments sake assume an Sds=1.0, therefore vertical component of Dead Load = 0.2x1.0x0.7=0.14, so your effective dead load is actually, 0.6-0.14 = 0.46

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?

 

[DaveAtkins]

It will be interesting to see what they propose.

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

 

[Cadair]

It will be interesting to see what AWC ends up with. I wish I even knew when it is supposed to be done.

@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.