Shear Wall Problem 2

[JKW05]

A 6'x6' entry addition to an existing single story wood framed structure has been constructed. Now I have been asked to show that this is adequate for wind loads.

Here is the plan:

____________________
| |
| |
|_ _|

There are no openings on the sides; a 3' wide door occurs in the front wall (opposite the original building). Walls are 8' high, 2x4@16" o.c., fully sheathed with structural panels. Floor is slab on grade. Code is 2006 IBC. Seismic Design Category A.

IBC 2308.9.3 requires braced wall panels at least 48" wide at each end; I only have 18" each side of the door.

IBC 2308.9.3.1 allows 2'-8" panels; I only have 18"

IBC 2308.9.3.2 allows 16" panels, but the header must be at least 6' between panels; I would only have 3'-4" between the 16" panels.

So I give up on "Conventional Light Frame Construction"

IBC 2305.3.4 limits the aspect ratio of a shear wall pier to 3.5:1; Each side of the door is 8 ft./1.5 ft. = 5.3:1

Trying a perforated shear wall doesn't work: 2305.3.5.1 defines the height of the perforated shear wall segment to be the same as the shear wall pier.

Assuming I could make the roof work as a diaphragm (the entry has a cathedral ceiling) to get the load back into the original building doesn't pass either….IBC 2305.2.5 limits the length of a cantilevered diaphragm to 2/3 of the diaphragm width, so the maximum cantilever permitted is 4 feet; I have 6.

This is a simple wood box, with a hole in one side. There are countless "tool sheds" built to a lesser construction that have been standing for ions. It can't be this hard…..help!!

 

[akastud]

JKW 05,

Two thoughts:

1. Try shear transfer around openings. Ratio then becomes height of door / 1.5'.

2. Assume the diaphragm is rigid, it will definitely meet the definition of a rigid diaphragm.

akastud

 

[msquared48]

The roof diaphragm is small enough by code to cantilever back to the main structure with the proper reinforcing, without the shear walls.

In lieu of this, you could try one or two Simpson Strong Wall panels, but they are pricey.

Mike McCann
McCann Engineering

 

[JKW05]

Thanks for your quick responses guys!

The cantilevered rigid diaphragm approach was the first thing I thought of, but then I saw the (2/3)*W maximum in Section 2305.2.5 which I take as meaning I would be limited to 4 feet, not the 6 feet that I have.

aka...even just with the door height the ratio is ~4.4 (6.66'/1.5'), which exceeds the 3-1/2

I'm sure this structure works, but it seems like the most basic structures are often harder to "engineer" than the larger ones.

 

[DaveAtkins]

This is not a cantilevered diaphragm--it is an open front structure. L/W can be 1.5, and you have L/W = 1. This small diaphragm can rack the loads back to the main building, and the torsion can be resisted by the side walls.

DaveAtkins

 

[akastud]

JKW 05,

What Dave Atkins is saying and what I was implying by rigid diaphragm is not a cantilevered diaphragm, but a rigid diaghragm that with a center of rigidity and different center of mass. When you calculate the center of rigidity you will find that it ends up on the wall at the existing structure. All of the shear will go directly into the wall and there will be a net moment on the diaphragm that can be resisted by the two sidewalls.

akastud

 

[msquared48]

OK Dave. I see that. I was not considering using the two walls normal to the house to resist the torsion, which is possible here.

Mike McCann
McCann Engineering

 

[JKW05]

Now I see it....thanks for all your help!

JW