# RE: Diaphragm Design Per 1633.2.9 Using ASD

• To: seaint <seaint(--nospam--at)seaint.org>
• Subject: RE: Diaphragm Design Per 1633.2.9 Using ASD
• Date: Sun, 21 Jan 1996 07:36:21 -0800
```Ed,
You answered your own question.  Fpx does equal Eh.  So any design
force derived from Fpx is to be multiplied by rho to get E, as you
stated.  Then use the load combos in 1612.  If you want ASD, every
ASD combo with E is of the form E/1.4.   If you want to use the 1/3
increase, then simply use the combos in 1612.3.2 (instead of
1612.3.1), since it IS allowed in there.

Alternately, if your force is subject to the Em provisions (such as a
collector force derived from Fpx), then the force is Eh, which must
be multiplied by omega to get Em (instead of by rho to get E).  Then
use the load combos of 1612.4, which CANNOT be divided by 1.4.  At
this point the demand forces are the same for LRFD and for ASD.  If
you want to use LRFD, this is your factored force to be compared to
nominal capacity.  If you want to use ASD, you get to use an
allowable stress increase of 1.7 INSTEAD of 1.33, but you can use an
additional 1.33 FOR WOOD ONLY since it is C(D), not a "1/3 increase".

Ben,
All earthquake loads must be put in the form of E so that they can be
used in the load combinations of 1612.  The definition of E includes
rho times Eh, so unless the code specifically deems rho to equal 1.0
for diaphragms, then a rho of greater than 1.0 must be used.  Mike
Valley's point merely confirms the point that such a specific
sentence was deliberately left out of the 97 UBC.

In Section 1630.1.1, Eh is defined as "the earthquake load due to the
base shear, V, as set forth in Section 1630.2 or .... Fp as .... in
Section 1632."  These references are simply directing the user to the
DEFINITIONS of V and Fp.  That is just the starting point, since ALL
earthquake forces in Chapter 16 are a function of one of these two.

Ft, Fx, and Fpx are all a function of V, which is defined in Section
1630.2.  Only Fp is not a function of V; it has its own equation,
which is in Section 1632.

Simply because 1630.2 is mentioned and not 1633.2.9 specifically does
not mean that diaphragms are not subject to rho.  After all, Fx is
not defined in 1630.2, but rather in 1630.5.

So, in summary,

ALL earthquake loads are a function of either V or Fp.  ALL of these
loads are equivalent to Eh, and all are thus multiplied by rho to get
E, or multiplied by omega to get Em.

This brings me back to the original point in my first post on "Rho
for diaphragms", which is that the Seismic Design Manual is
incorrect, IMHO, to state that rho does not apply to diaphragms.

The authors should be interpreting the code rather than offering
their personal opinions about what it SHOULD have said.  The 97 UBC
is now a legal document.

Mark Swingle, SE

------------------------------

On 19 Jan 00 Edward Arrington wrote:

<<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

----------------------------------------------

On 14 Jan 00 Ben Yousefi wrote:
Subject: RE: Rho for diaphragms

<<Mark

<<There are no sections in the code that specifically exempt the
diaphragm from application of Rho. However, if you look at 1630.1.1
under definition of Eh (which Rho gets applied to) it refers to
earthquake loads as set forth in 1630.2 and 1632. The provisions for
diaphragm design are in 1633.2.9.

<<Therefore, I would conclude that Rho need not be applied to the
diaphragm.

<<Ben Yousefi, SE
<<San Jose, CA

-------------------------------------

and Michael Valley wrote:

<<Because the INDIVIDUALS who work on the Codes, Blue Book, and
Seismic Design Manual are not the same (even though they may all
represent SEAOC), there are often subtle and sometimes glaring
differences in these documents.

<<During balloting of the 2000 IBC, a proposal was made (S89-99) to
"fix" part of the redundancy section by adding the sentence "For
design of diaphragms and collectors, a rho of 1.0 may be used."  The
proposal was disapproved by the committee (including influential
members of SEAOC); the committee's response was "The reliability
factor should apply to the design of diaphragms and collectors."
Apparently there is not consensus on this matter (even within
SEAOC).

<<Although I believe that the idea of requiring redundancy is sound,
I must agree with Charles Greenlaw that the present codification of
redundancy is "a regular Rosemary's Baby."

```