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Re: IBC Sec. 1617.1
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- Subject: Re: IBC Sec. 1617.1
- From: Scott Maxwell <smaxwell(--nospam--at)engin.umich.edu>
- Date: Fri, 12 Sep 2003 14:22:47 -0400 (EDT)
Jason: To me, the definition of E just after Equation 16-28 should help provide some clarification. It states: "E = The combined effect of horizontal and VERTICAL earthquake-induced forces." (emphasis is mine) Add to that, that QsubE is defined as: "QsubE = The effect of horizontal seismic forces." To me, that makes the 0.2SsubDS*D term a vertical component term. Further more, this is consistant with how the 1997 UBC deals with the similar situation (and the 1997 UBC is to some degree at least a minimal basis for what is in the 2000 IBC). The similar equation in the UBC (eq 30-1) is E = rho*Eh + Ev. And Ev is clearly defined as the vertical component of the earthquake ground motion ("Ev = the load effect resulting from the vertical component of the earthquake ground motion and is equal to an addition of 0.5*Ca*I*D to the dead load effect, D, for Strength Design, and may be taken as zero for Allowable Stress Design.") And last, since the 2000 IBC Seismic provisions are basically based off of the NEHRP Provisions, you could get a hold of the 1997 NEHRP provision and see what the commentary says on this section. HTH, Scott Ypsilanti, MI On Fri, 12 Sep 2003, Jason Kilgore wrote: > I normally work with buildings with a seismic design category of A, B, or C, > so I rarely use Equations 16-28 through 16-31. (rho = 1, I don't use > section 1617.1.2, and 0.2*Sds = about 0.02). > > Now I've got a problem with an Sds of 68%, so the second term in those > equations is much higher. > > It's obvious that the first term acts horizontaly (Qe). I've always worked > on the assumption that the second term acts vertically. For example, take a > drag strut with vertical and horiz. loads. Assume rho = 1.5. > > E = rho*Qe +/- 0.2*Sds*D > E = 1.5*Qe +/- 0.136*D > > You would then plug this value of E into, for example, Eq. 16-5: > F = 1.2*D + 1*E + 0.5*L + 0.2*S > > F = 1.2*D + 1.5*Qe + 0.136*D + 0.5*L + 0.2*S > > F = 1.336*D + 1.5*Qe + 0.5*L + 0.2*S > > In other words, the "D" value from equations 16-28 through 16-31 are added > to the "D" in whichever final load equation you use. > > Recently it was pointed out to me that the D in those equations is actualy a > horizontal force, and should be directly added to Qe to get a purely > horizontal "E" force. In other words, part of the horizontal load in a > member is from the seismic base shear, and part is caused directly by the > dead weight (and supported dead weight) of the member itself. > > I think this person is wrong, but would like other expert opinions. > Thanks, > > ---- > Jason Kilgore > Leigh & O'Kane, L.L.C. > ******* ****** ******* ******** ******* ******* ******* *** * Read list FAQ at: http://www.seaint.org/list_FAQ.asp * * This email was sent to you via Structural Engineers * Association of Southern California (SEAOSC) server. To * subscribe (no fee) or UnSubscribe, please go to: * * http://www.seaint.org/sealist1.asp * * Questions to seaint-ad(--nospam--at)seaint.org. Remember, any email you * send to the list is public domain and may be re-posted * without your permission. Make sure you visit our web * site at: http://www.seaint.org ******* ****** ****** ****** ******* ****** ****** ********
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