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Staggered Stud Wall LTB

XR250

Structural
Jan 30, 2013
6,127
Got a customer wanting to use a staggered stud wall for thermal break purposes. It will use 2x8 plates and 2x6 @ 24" O.C. at the exterior and 2x4 @ 24" O.C. on the interior. The wall is about 11 ft. tall. has a 15 psf wind and about 500 plf axial. Have we decided that fastening one "flange" of the stud to the exterior sheathing is sufficient to brace for LTB?
I'd like to add some mid-height blocking just because or a 2x4 brace between the 2x4 and 2x6 but I am sure the customer will pushback.
 
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Sheathing on one face is sufficient for axial bracing

For LTB from the wind pressure bending you'd need sheathing on the compression face of the stud or alternately blocking developed into the diaphragm.
 
Sheathing on one face is sufficient for axial bracing

For LTB from the wind pressure bending you'd need sheathing on the compression face of the stud or alternately blocking developed into the diaphragm.
Well, found this in the NDS...
“When stud walls in light frame construction are adequately sheathed on at least one side, the depth, rather than breadth of the stud, shall be permitted to be taken as the least dimension in calculating the l_e/d ratio. The sheathing shall be shown by experience to provide lateral support and shall be adequately fastened.”
 
What I have done in the past is providing blocking but on the vertical, so that the air gap remains between the blocking and the stud of the other face of wall. Blocking gets fastened into the sheathing. So you're effectively bracing the first 1 1/2" of stud instead of just relying on sheathing. Here's an image:

staggered wall.png

Mine was a 2x8 stud cavity with 2x6 studs, but should work for you. My studs were also not laterally loaded for wind (these were only used at load bearing partitions on the interior of the building for sound attenuation, but I think this would be sufficient as long as you sized each stud to take full 24" of wind, instead of 12" since the outside face doesn't touch the inside stud.

This shouldn't affect their overall goal of thermal break, so they might not push back as hard as you'd initially thought.
 
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Our code allows tension face restraint to count (compression side would obviously be even better)
I've always interpreted 'continuously restrained' to allow sheathing with discrete fasteners at 150-300mm crs max in the context of residential framing
1749594132423.png
 
Well, found this in the NDS...
“When stud walls in light frame construction are adequately sheathed on at least one side, the depth, rather than breadth of the stud, shall be permitted to be taken as the least dimension in calculating the l_e/d ratio. The sheathing shall be shown by experience to provide lateral support and shall be adequately fastened.”
This and @EngDM post address axial compression buckling, Cp, for the wind suction case placing the wall interior stud face in compression neither option would be considered bracing for the computation of the CL factor.
 
Our code allows tension face restraint to count (compression side would obviously be even better)
I've always interpreted 'continuously restrained' to allow sheathing with discrete fasteners at 150-300mm crs max in the context of residential framing
View attachment 10646
In the formula (that I am not familiar with), is higher S1 good or bad? S1 would increase for deeper studs if d=depth and b = thickness is held constant.

In the formula that XR250 shows (from Appendix A of NDS), le/d would get smaller with deeper studs if d = depth is substituted for le/b, then I assume the allowed stress changes in the Rb portion of allowed bending as shown below. Rb would go down, thus increasing FbE.

1749600032931.png
The Appendix A provision appears to be saying if your sheathing is not easy to distort, it will be more difficult to buckle the compression face of the stud. I am still trying to picture why a deeper stud is less likely to buckle than a shallow one under this appendix option and therefore allowed to have more bending stress.
 
Ron, higher S1 = bad (number = how slender it is, higher = more slender)
I suppose I left out the useful context of the normal equation - basically, reverting to 3*d/b is hugely favourable over how the same beam would calculate out if it didn't have tension edge restraint

1749603727218.png
 
This and @EngDM post address axial compression buckling, Cp, for the wind suction case placing the wall interior stud face in compression neither option would be considered bracing for the computation of the CL factor.
I'm confused, Celt. Looks like CL is based on Rb which is based on le b/d.
Also..
1749641498350.png
 
Did not even see 4.4.1.2 but looks like a 2x6 works if the ends are nailed
 
Is there drywall on the inside? These staggered studs are the norm in hotels and similar occupancies.
 
I'm confused, Celt. Looks like CL is based on Rb which is based on le b/d.
Also..
View attachment 10667
For Le my interpretation has been:
Axial
Le = Ke L from section 3.7.1.2 and for a wall stud have taken Ke as 1.0, then using the A.11.3 to justify one side sheathed as bracing the narrow face so only need Le/d

Bending
When the compression is not braced and not taking advantage of 4.4.1
Le for use in the Rb formula should come from Table 3.3.3, for uniform wind pressure usually ends up as the 1.63Lu +3d formula, where Lu is the stud height.
 

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