Shear wall dead weight & overturning 1

[soiset]

When designing shear walls, I have typically used only the dead weight of the wall itself (studs & sheathing) to calc the required holddowns. Of course, the wall could never lift independently of the rest of the bldg. What is a more realistic approach to obtaining the actual deadweight that resists overturning? For example, if an interior shear wall intersects with an exterior wall, how should one consider the dead weight of the exterior wall?

Chris

 

[JAE]

We have traditionally used a width of adjoining wall equal to the height of the shearwall.

For example, if you have a shearwall on the end of a 60 foot long building, say its 20 feet long and 10 feet high, then we would include a 10 foot length of the intersecting wall (10 ft. out of the 60 ft) in our dead weight calculations. We also include floor or roof dead load in this as well.

Just make sure that the CONNECTION of this intersecting wall is adequate for its 10 ft. length of weight.

 

[soiset]

JAE,
That is an interesting approach, and I can see an intuitive justification for the height=length approach. Do you know of an experimental or rational justification for your approach? How do you handle an interior shearwall that intersects an exterior wall? Do you take lengths of the exterior wall in both directions equal to the height of the shear wall? For example, would a 10' high shear wall be held down by the weight of 20' feet of exterior wall (10' in both directions)? Are you also saying that you would count the dead weight of the floor bearing on that length of exterior wall? Also, what nature of attachment to that exterior wall would be required?
Thank you for your response.

 

[JAE]

I can't point to any specific research on this subject. Just common practice that I've been taught over the years and used on a number of projects....but used discriminatively.

Yes, for an interior shearwall intersecting an exterior wall, we'd used the weight of the exterior wall in both directions.

The floor/roof weights would also be used, but only if they are bearing ON the exterior wall, not hanging from them below.

The connection, then, between the exterior "weight" and the shearwall, would have to be able to support the dead weight you are counting on. So if you have an exterior wall that is 10 feet long, and 10 feet high (10 x 10 x 9 psf weight = 900 lbs.) The 900 lbs would have to be transferred to the shearwall across the 10 foot height = 90 plf. This is usually accomplished through the blocking at the intersection and the sheathing in both walls. Thus, we make sure that the exterior wall is properly sheathed with gyp-board or plywood that can handle 90 plf.

 

[soiset]

I would guess that you only measure usable exterior wall length (for deadload) to the first opening (door or window). So if there is an exterior door 2' from the intersection of the interior shear wall and the exterior wall, you would only benefit from the dead weight of 2' of exterior wall and the roof or floor bearing on that length of wall, correct?

 

[JAE]

For the weight of the wall itself...correct..the 2 ft. would apply. However, the lintel over the opening contributes half the opening span into that 2 ft. segment so roof or floor load above would also follow that load path and go into the segment.

None of this, obviously, is of an exact nature. We just try to err on the conservative side while taking advantage of obvious weight that the shearwall would have to overcome to overturn. Using this method definitely decreases the number and size of holddowns.

 

[whyun]

This may be obvious but if the end conditions are different, overturning calculations should be done twice. Load in one direction and opposite direction. Holddown sizes will be different at each ends.

Also, when including the floor dead load and roof dead load as a part of the resisting moment, one must look at how conservative the loading was assumed in the beginning. For framing design, the design loads are often overestimated. But in overturning design, using this conservative dead load will result in an unconservative design for holdowns. Different codes have different factors to be applied to reduce the dead load resistance to overturning. UBC97 applies 0.9, previous versions has 0.85 for resistance to seismic overturning. JAE, in this case, would you take 0.9 of the accurate DL or 0.9 of the original conservatively estimated dead load?

 

[soiset]

For overturning of lateral resisting elements, UBC97 requires a 0.67 factor for the dead load. For multi-story buildings, I multiply the uplift at the shearwall end by 1.5, then compare it to the dead load resisting. If the dead load is higher, no hold down is required, and I carry the difference (dead load minus 1.5 x uplift) down to the next floor. When a hold down is required, I use the actual uplift load minus the actual dead load for the required hold down size (because the hd capacities listed are for working loads) and again carry the difference down to the next floor (actual uplift minus actual dead load). I just make sure I have the actual dead loads carefully calculated by material weights.

 

[tony3]

don't forget the wind uplift on the roof...

 

[JAE]

I agree with the last three posts....good points.

 

[boo1]

Do you calculate the mass of the foundation to resist the overturn moment?

 

[SacreBleu]

We consider exact dead load, including weight of wall an any beams/headers or girder truss etc. With an intersecting wall, we add only 4' of its tributary dead load, including its roof or floor trib dead load. (tee-condition) or 2' (corner condition).
The dead load is reduced according to the applicable code (IBC went overly conservative with its 0.6 x dead load for seismic).

 

[whyun]

Re boo's question:
weight of the foundation shall be considered for "overall" stability of the system but for anchorage/holdown design, it can not be included.

Re SacBleu's comment:
I agree that 0.6 dead load is very conservative. In the UBC, it used to be 0.85 for a long time, then it was changed to 0.9 in 1997. We'll have to wait and see what this factor will be in California for the next code cycle.

 

[SacreBleu]

Additionally, where there is a very small dead load, the Factor of Safety is actually smaller when using the classic concept of keeping the applied overturn moment at service load, and reducing the dead load by a factor.
If we multiply the applied lateral load by a factor to increase the overturn moment, and use a smaller reduction of dead load, we get a more uniform Factor of Safety against overturn, but the "bookkeeping" becomes a mess unless we use canned software or Excel.

 

[14159]

You can include any dead weight that is vertically supported by your shear wall, not just the wall weight. For example, if you have joists framing into the wall, then the tributary floor dead load can be included. Your exterior wall probably can't be included for shear wall hold-down design because it's probably only laterally connected to your shear wall. If the higher cladding is somehow vertically supported by the shear wall, then its weight can be included also. You gotta know how it goes together well enough to know what is actually holding the shear wall down and what is not.

DBD

 

[SacreBleu]

DBDavis:
We have a special detail that connects an interior shear wall to the exterior wall channel (Tee-condition). The number of nails transfers the tributary dead load of the exterior wall.