# RE: Diaphragm Design Per 1633.2.9 Using ASD

• To: <seaint(--nospam--at)seaint.org>
• Subject: RE: Diaphragm Design Per 1633.2.9 Using ASD
• Date: Wed, 19 Jan 2000 15:53:47 -0800
```Ed,
With that said, the reduction E/1.4 for ASD goes back to section 1612.3.1
"Where allowable stress design (working stress design) is used, structures
and all portions thereof shall resist the most critical effects resulting
from the following combinations of loads:
D						(12-7)
D + L + (Lr or S)				(12-8)
D+(W or E/1.4)				(12-9)
0.9D+/-E/1.4				(12-10)
D+0.75[L+(Lr or S)+(W or E/1.4)]	(12-11)
No increase in allowable stresses shall be used with these load
combinations except as specifically permitted by Section 1809.2."

-----Original Message-----
From: Edward Arrington [mailto:Earringt(--nospam--at)ENG.CI.LA.CA.US]
Sent: Wednesday, January 19, 2000 2:45 PM
To: seaint(--nospam--at)seaint.org
Subject: Diaphragm Design Per 1633.2.9 Using ASD

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".
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?

Regards,
ed arrington

```