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Re: Shear Wall Design/Billy Bob
[Subject Prev][Subject Next][Thread Prev][Thread Next]- To: "'seaint(--nospam--at)seaint.org'" <seaint(--nospam--at)seaint.org>
- Subject: Re: Shear Wall Design/Billy Bob
- From: "Swingle, Mark" <Mark.Swingle(--nospam--at)dgs.ca.gov>
- Date: Thu, 18 Nov 1999 11:59:56 -0800
- Cc: "'mswingle(--nospam--at)earthlink.net'" <mswingle(--nospam--at)earthlink.net>
Andrew, Yes and yes. You may use "the force obtained from the vertical distribution of my base shear" AND you may divide it by 1.4, according to my interpretation of the 1997 UBC. Now I will give you the long answer, assuming wood structural panel shear walls: Calculate the base shear V per 1630.2.1, using all appropriate factors (Na, Ca, I, and R=5.5). Eh=V per 1630.1.1. Hopefully rho=1. If not, add walls to make it so. Since E=rho(E) + Ev, this reduces to E=V since Ev=0 for ASD per same code section. Remember, E=V for this case (anytime you use ASD and rho=1). You can choose to divide by 1.4 now or later, knowing that you will later use the load combinations of 1612.3.2, in which E is always divided by 1.4. I typically do it at this point, since the forces will then be identical to 1994 UBC (unless Na>1) for most short-period buildings. If you do it now, don't forget later on that you already did, and also don't forget to "put the 1.4 back in" before using Em or section 1612.4, since Em is never divided by 1.4 in any load combination. [Em may come up if you have discontinuous shear walls per 1630.8.2, but does not apply to collectors in bldgs with wood shear walls per exception to 1633.2.6.] Then calculate Fx for each story per 1630.5. Ft will always be zero (per 1630.5) for a wood building since T<0.7 sec for all buildings under 54 ft tall per Method A of 1630.2.2. Then calclulate Vx for each story as the sum of Fx for that story and all above it per 1630.6. Distribute Vx at each level to the shear walls any way you wish (I recommend tributary area). The force to each wall is now E/1.4, unless you haven't yet divided by 1.4. Now calculate unit shears and use the table values from Table 23-II-I-1 directly. Note, however, that you must use the load combinations from 1612.3.2 (NOT 1612.3.1) in order to use the table values directly, since the table values have the one-third increase in allowable stress already included (see footnote 1 of the table). The load combinations of 1612.3.1 do not allow a one-third increase in allowable stress. Equation 33-1 in Section 1633.2.9.2 is applicable to diaphragm design only (including connection of the diaphragm to the shear walls, either directly, or with the assistance of collectors). These forces are not used for shear wall design. Mark Swingle, SE Oakland, CA These are my own opinions and not necessarily those of my employer. ---------------------------------------- On 17 Nov 1999 Andrew Arnold wrote: (cut)....Now, back to the purpose of this list server. I had already taken into consideration most of what advice was offered, now the question still remains about the allowable stress design portion in the code (Sec. 1612.3). Does anybody know if I may take my computed lateral force at the diaphragm level in question (the force obtained from the vertical distribution of my base shear) and divide it by 1.4 as the load combination in 1612.3.1 implies (Eq. 12-9), in order to specify the nailing schedules for my wood diaphragm? Andrew Arnold, EIT Arnold Engineering
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