IBC 2009 Wood Shear Wall Holddown Design - Dead Load Restoring Moment?

[MJB315]

All,

Assume you are designing an individual, non-perforated, wood shear wall. How do you calculate the holddown force for the chord members?

It sounds like a simple question (and I'm sure there is a simple answer), but I am confused.

I get into the sources of my confusion below (for this one issue alone; time prevents a discussion about my general confusion), but I am torn between the following three approaches:

1. Design for the overturning moment alone, neglecting any dead-load restoring moments (probably too conservative);
2. Design for the overturning moments, subtracting off the dead-loads supported by the chord member ALONE (probably correct);
3. Design for the overturning moments, subtracting off the dead-loads supported by the entire shear wall as if it was a rigid body (probably unconservative, but what I have always seen done).

Thoughts?



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My confusion stems from partially from the code, partially from articles, and partially from experience.

First, in the ANSI/AF&PA Special Design Provisions for Wind and Seismic (SDPWS) 2008, which is referenced by IBC2009, the tension force in the chord member is calculated by:

4.3.6.1.1 Tension and Compression Chords: Tension force, T, and a compression force, C, resulting from shear wall overturning forces at each story level shall be calculated in accordance with the following: T = C = v*h (where v is the unit shear at the top of the wall).

Ok. Even though it seems a little fast and loose, it's clear enough that tension in the chord is due to the overturning forces.

SDPWS 2008 (4.3.6.4.2) sas something along the lines of:

"Where the dead load stabilizing moment is not sufficient to prevent uplift due to overturning moments on the wall, an anchoring device shall be provided at each end of the wall."

Ok, seems reasonable. But I don't see anywhere that defines the dead load stabilizing moment.

Structural Magazine and Simpson Strongtie (who would know better?) get into the issue a bit in an August 2011 article (see link below), but they say on page 2:

"Designers should be cautioned that these equations do not include several factors that impact the design of the framing members and connections. ... In addition, dead load above the shear wall end posts can reduce the tension force and increase the compression force..."

Ok, this seems to say that only the dead load in the posts subtract away from the anchorage forces.

But, traditional textbooks (and the way I have been taught by others) consider the entire dead-load in the wall when figuring restoring moments, as shown in the second link below.

Thoughts?

http://www.strongtie.com/ftp/articles/AnchorageOfWoodShearWalls.pdf

http://books.google.com/books?id=ZX...ear wall dead load stabilizing moment&f=false

"We shape our buildings, thereafter they shape us." -WSC

 

[JAE]

For me: 0.6D +/- W

Per ASCE 7.

Use statics/load path to determine what applicable dead loads are applied to the wall and in what fashion/location.

 

[msquared48]

What JAE said, but using any dead load the wall sees, including it's own dead load. It can and does make a whale of a difference for load bearing walls, not so much, if any, for non-bearing walls.

Mike McCann
MMC Engineering

http://mmcengineering.tripod.com

 

[RFreund]

This is a good question and one that seems to come up semi frequently. I mean even though your only asking for a solid wall i think it is equally applicable to a perforated shear wall. I mean I think realistically the tension chordl will see the tension force before lifting up one or two panels (4-8 ft) but for the wall to actually overturn it would need to lift up the dead load of wall. I agree with JAE about using 0.6D +- W but how much DL is trib to the Tension chord.

What if you make a section cut through the wall say 10' from the tension chord. The you analyse it as a beam. So you have an upward force (the tension force)at the end. Then you have a uniformly distributed dead load on top of the wall/beam. You then analyse it as a cantilever beam 10' long with the point load at the end and the distributed load. Then you adjust the length until the deflection reaches the slip listed for the hold down. The problem with this is that I'm not sure how you would calc the deflection or even what you would use for your I given the orientation of the sheathing. Ehh... yeah not sure this is the way to go, but it's a thought!

EIT

http://www.HowToEngineer.com

 

[pittguy12]

I agree, 0.6DL - WL (or E) controls the design of the tension chord. Since both ends effectively act as tension cords if the load reverses, this controls design of the uplift anchor on both ends. I would probably use self weight of the wall only and see where I was. If the difference between using no DL, self weight, or DL from above ultimately comes down to a small difference in anchor bolt size anyway, I would specificy the larger bolt and call it a day.

As far as subtracting off only the portion of the DL tributary to the chord, I believe this is too conservative. Your shear wall is fastened together for the specific purpose of acting as a single unit. Why not take advantage of this. As long as you meet the criteria for W/L ratios and have sized your fasteners properly, I think you can assume all DL (how much ever you decide to take) on the wall resists uplift.

Coincidently, the compression chord would also need checked with this equation but I imagine another load case, DL + 0.75LL + 0.75SL + 0.75SL for instance, would control its design. Additionally, if your compression chord is acting at the corner of two shear walls, it may be subject to a combined loading per the wind or seismc load cases or seismic (depending on building type and SDC) if you are using ASCE 7.

Long winded responses...sorry...ultimately I guess I would check the worst case first and see where I was. If I end up with needing a 5/8" dia bolt instead of a 1/2" dia bolt then I am not going to get too worked up about that. At least I have not been involved in a project where something like this would make a difference ultimately.

 

[MJB315]

Thanks all- I appreciate your responses. I don't want to fire up an unnecessary debate/conversation about something trivial. But I also want to try to flush this out as best as I can while its cued up in my mind.

I agree that many times the difference in choosing a holddown is minor, and its appropriate (and probably smart) to choose one that's stronger and only marginally more expensive. JAE- right on: using a rational analysis to determine the load paths and using 0.6D+W controls.

But I guess my question is shifting towards: How does a shear wall fail? Does anyone have access to (or have seen) research on it? It seems that we would need to know the mechanism in order to rationally analyze the forces working on the wall.

Assuming the tension in the chord controls, I would assume that the wall fails in one of two ways:

1. The sheathing causes the entire wall to act like a rigid body, the entire wall "rotates about" the compression chord, and the holddown locally fails. (This would assume that the wall assembly is capable of acting like a beam, and all that goes into it.)

2. The sheathing is not rigid enough to have the wall act like a rigid body. But the sheathing is engaged to transfer the horizontal shear forces, deforming in the shape of a rhombus/parallelogram, and thereby causing tension/compression forces in the chords. This tension force is only resisted by the dead load in the immediate vicinity of the post, and if the net uplift it too much, the holddown locally fails.

Any ideas?

"We shape our buildings, thereafter they shape us." -WSC

 

[pittguy12]

Excellent question...

I do not know the definitive answer but I do remember reading some rationale behind the increased anchorage requirements of the IBC. Apparently research done after the northridge earthquake in the 90's pointed to several failures of the sill plate being improperly anchored or failure of the sill plate itself where anchored. Whether this was a shear or tension failure I am not sure. However the conclusions lead to the requirements in the 2006 IBC and beyond for specific plate washer requirements at sill anchorage locations.

 

[Cadair]

The entire wall does not act like a rigid body. Each panel is more or less independently rotating. Only the area around the chord should be used to resist uplift.

 

[MiketheEngineer]

Some codes only allow for 2/3 of the dead load as a "help". Some don't allow the "overturning" bolts to act as shear bolts also.

Check carefully.

 

[MJB315]

Cadair,

Do you know of anything in the code / research that supports that? I believe it, but it'd be good to see it.

"We shape our buildings, thereafter they shape us." -WSC

 

[msquared48]

Although not really precise, the .6 factor is there to try to get a safety factor of 1.5 against overturning.

You could do the same thing taking 90% of the dead load, and making sure the righting moment exceeded the overturning moment by 50%. Same effect in the end. You would also have to increase your uplift by 50% to accommodate the holddown FS of 1.5 too.

Mike McCann
MMC Engineering

http://mmcengineering.tripod.com

 

[Cadair]

It is what I was taught at school from Dr. Dan Dolan of Washington State University. However, if you call the American Wood Council, who make the SDPWS, they say they are working on a publication along those lines.

 

[dcarr82775]

0.6D is the same as using 90% of dead load to resist overturning while shooting for a 1.5 factor of safety.

As for academia suddenly deciding the way shear walls have been designed for decades is crap....God I hope they don't go down that road.

 

[hawkaz]

I think that report is being funded by the Masonry Institute

 

[Cadair]

The "academia suddenly changing things" brings up a good point. If you feel it is still on topic, I pose the following hypothetical.

Say that a change would better reflect reality. What does that mean? Should it change because after all, theoretically we are under designing uplift restraint? Then again, its been done that way for years, and I for one, don't know that it has caused problems. Should we wait and see if there are problems?

If academia had wanted to change baseplates or anchor bolt washers prior to the Northridge earthquake, would we have resisted that?

 

[MJB315]

Dcarr, I understand your sentiment- but academia and practice are two sides of the coin.

Practice utilizes the current body of knowledge; Academia adds to it.

Someone has to check out our details / systems / procedures to see if they actually perform the way we think...because frankly, practice doesn't really have the time.

I agree that there's rarely need to reinvent the wheel, provided the wheels in service are actually round.

"We shape our buildings, thereafter they shape us." -WSC

 

[RFreund]

it will be interesting to see if the report considers the shear force as a point load or a uniformly distributed load along the wall. if you had a test mechanism that applied a point road I would imagine you would see each panel acting individually. Where is if you had a uniform load you may see more of a rigid body . then again maybe not

EIT

http://www.HowToEngineer.com

 

[abusementpark]

It sounds like a simple question (and I'm sure there is a simple answer), but I am confused.

Unfortunately, there isn't really a simple answer and your confusion is warranted. I posed this same question on this board over a year ago and did not find a consensus. See link below.

http://www.eng-tips.com/viewthread.cfm?qid=290482

I ultimately contacted one of the authors of a wood textbook. They indicated to me that, yes, the traditional practice has been to consider the dead load of the entire system (i.e.- treat it as a rigid body) for overturning/uplift calculations. However, it is also known from experimental testing that the wood elements in a shearwall clearly behaves as deformable materials (at least at ultimate capacity), and it is not necessarily true that the simple rigid body analysis adequately models the ultimate structural behavior. So there is the difficult question of how do you accurately model the effect of distributed dead loads.

Another factor to consider: what about the anchor bolts distributed along the length of the shear wall that are only designed to resist the shear? In reality, how much to they contribute to overturning resistance and how do they affect the amount of DL reduction that makes it to the tension chord of the wall?

 

[RFreund]

Sorry for the grammatical errors in my last post, it was sent from my phone. This is one of my favorite questions that I wish I knew more about (offset surcharge pressures on retaining walls are another) and I do remember reading abusementpark's post last year. It would be interesting to see a more in-depth publication on this topic. However in regards to academia modeling and changing the design method that has seem to work for so many years - I wonder if the testing is really modeling what is really happening. I mean how are they distributing the forces to the shear wall, when testing? What are the effects of the anchor bolts, etc...

EIT

http://www.HowToEngineer.com

 

[Cadair]

RFreund,

The best I can offer is to say how the load was being applied to walls when I was testing them in Grad school. There were two common methods for attachment. The first it to attach a metal section (usually a C-section) to the top plate of an 8' wall and use screws down its length to distribute the load. One end of the channel was attached to a hydraulic actuator. This method became less popular because it was feared that the continuous channel significantly added to the stiffness of the top plate and add dead load.. So the other method was to basically put a holddown like device at one end of the wall and attach the actuator to that. Tests have been run with both methods. There isn't too much significant difference between the two when a wall is fully restrained with holddowns.