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Diaphragm Design Per 1633.2.9 Using ASD

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Well, while many have been discussing the above topic, particularly as it
relates to the use of rho =1.0, I was wondering if I can get folks input on
calculating the design load for a diaphragm, based on ASD, while trying to
adhere to the Code as much as possible.  Here are my conclusions/confusion:

The Code Path:
My diaphragm design force of 1633.2.9 is Fpx.
Fpx is based on Fi.  
Fi is based on V per Equ.30-15. 
V, per 1630.2.1, is a Strength based force.  
Therefore, since Fpx is based on V, then Fpx is also a strength based force.  
So what does the code provide in order to convert from USD to ASD?  
Well it seems to me that the only avenue available are the load combinations as
indicated per 1629.1, "Basis of Design".
But, these load combo's use "E", instead of Fpx!!!
I might assume that Fpx of 1633.2.9 is equal to Eh except that Eh is defined
as, ". . . or the design lateral force, Fp, as set forth in Section 1632. 
Since the diaphragm design force is in 1633, Fpx does not equal Eh.  

Conclusion: ???????

I know that ultimately I need to reduce Fpx by 1.4 to design for ASD, but which
section in the code will allow me to do this directly?

Assuming I would conclude that Equ. 30-1 does apply and so Fpx does equal Eh,
and E=(1.0 to 1.5)*Eh=Fpx, do I then set LL and SL to zero in the load
combinations of 1612.3?  What about "D", do I just set it to zero to because it
does not apply? Does the code specifically allows me to do this?

ed arrington