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AWC - 2008 Wind and Seismic 4.2.7.3

AWC - 2008 Wind and Seismic 4.2.7.3

(OP)
This section discuses diaphragms diagonally sheathed with double-layer of lumber.

It states that "Each chord shall be considered as a beam with uniform load per foot equal to 50% of the unit shear due to the diaphragm action. The load Shall be assumed as acting normal to hte chord in the plane of the diaphragm in either direction."

I'm not sure I follow what they mean here...

Maybe someone can shed some light on this?

Thanks!

EIT
www.HowToEngineer.com

RE: AWC - 2008 Wind and Seismic 4.2.7.3

Per the 2006 IBC

"2306.3.5 Double diagonally sheathed lumber diaphragms.

Double diagonally sheathed lumber diaphragms shall be constructed of two layers of diagonal sheathing boards at 90 degrees (1.57 rad) to each other on the same face of the supporting members. Each chord shall be considered as a beam with uniform load per foot equal to 50 percent of the unit shear due to diaphragm action. The load shall be assumed as acting normal to the chord in the plan of the diaphragm in either direction. The span of the chord or portion thereof shall be the distance between framing members of the diaphragm, such as the joists, studs and blocking that serve to transfer the assumed load to the sheathing. The shear capacity of double diagonally sheathed diaphragms of Southern pine or Douglas fir-larch shall not exceed 600 plf (8756 kN/m) of width. The shear capacity shall be adjusted by reduction factors of 0.82 for framing members of species with a specific gravity equal to or greater than 0.42 but less than 0.49 and 0.65 for species with a specific gravity of less than 0.42, as contained in the AF&PA NDS. Nailing of diagonally sheathed lumber diaphragms shall be in accordance with Table 2306.3.3."

Garth Dreger PE - AZ Phoenix area
As EOR's we should take the responsibility to design our structures to support the components we allow in our design per that industry standards.

RE: AWC - 2008 Wind and Seismic 4.2.7.3

(OP)
Garth, thanks for the response.

I'm still not sure I fully understand. What do they mean by each chord shall be considered as a beam with uniform load per foot equal to 50% of the unit shear due to the diaphragm action.

So say for example you have a flat roof that is 75' long x 40' deep with 2x flat rafters @ 24" oc, bearing on 2x6 wood stud walls. You're diaphragm load is 250plf and you consider the double 2x6 top plate to act as a chord. So for wind perpendicular to the 100' dimension... the diaphragm shear is 250 plf * 75' / (2 * 40) = 235plf. Chord force is 250plf * 75^2 / 8 *(1/40) = 4,400 lb

So the double top plate is now is evaluated with 235plf uniform load applied in the plane of the diaphragm and the tension/compression force. The chord spans between the 2x rafters?

Why is this?

EIT
www.HowToEngineer.com

RE: AWC - 2008 Wind and Seismic 4.2.7.3

First off, you should design per engineering principles to meet the building code not from the building code.

Per engineering principles how do you get the lateral force into the diaphragm. Some of the force is applied directly to the diaphragm. Some is transferred into the diaphragm. Consider the top plate of the walls. The normal wind load is applied to the wall sheathing which transfers it to the top plate. The top plate then transfer it to the joists/trusses supporting the diaphragm or to blocking diagonal bracing etc. then to the diaphragm. The top plate (chord) must act as a beam (with tension form the diaphragm action) to transfer the force from the studs to the joists. IMHO.

The building code is and always have been poorly stated in this.

But I could be wrong. After all, the Building Official is the authority on the building code.

Garth Dreger PE - AZ Phoenix area
As EOR's we should take the responsibility to design our structures to support the components we allow in our design per that industry standards.

RE: AWC - 2008 Wind and Seismic 4.2.7.3

(OP)
Thanks again for the response.

I agree with your statement of load path, however that doesn't seem to be the code is saying.

Quote (woodman)

The normal wind load is applied to the wall sheathing which transfers it to the top plate
Yes, to me this is the "applied diaphragm load" in my example above. But the code says the unit shear. Unless do they mean the unit shear of the diaphragm at that location? meaning if you were to draw a shear diaphragm you take this value and apply it to the chord (top plate).




EIT
www.HowToEngineer.com

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