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RE: Diaphragm Design Per 1633.2.9 Using ASD
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- Subject: RE: Diaphragm Design Per 1633.2.9 Using ASD
- From: Mark Swingle <mswingle(--nospam--at)earthlink.net>
- 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". 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? <<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."
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