ibc seismic effect e

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ibc seismic effect e
having calculated the seismic base shear v and distributed it among the floor levels, i would like to use a computer model of my structure to solve for the member loads resulting from the ibc2000 asd load combinations (section 1605.3.1)  section 1617.1 gives equations for e depending on whether the dl are additive or couteractive, i suppose they could be either depending on direction of eq force.
does anyone have suggestions on solving earthquake combinations using computer models? the only way i see is to solve the model with eq shears only, then by hand taking the member vertical loads, which would be qe in sec 1617.1, and solving for e.  then combining e as indicated in 1605.3.1 with other loads. this is cumbersome, but how else can you calculate e?
any help would be appreciated.
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qe is not the vertical load effect but the horizontal load effect.  the vertical load effect from seismic actions is the second part of equations 16-28 and 16-29 (0.2 x sds x d)
the effect - e is simply the combination of horizontal and vertical loads and equations 16-28 and 16-29 are the definitiion of e.  
so in your computer model, you define seperately the dead load, live load, seismic load, etc. as individual load cases.
your program should have a means to combine those individual loads into combinations per 1605.3.1.
suppose you have the following cases:
dead load - lc1  (all vertical loads)
live load - lc2  (all vertical loads)
seismic load - lc3  (this is qe - all horizontal loads)
the e is defined as:
(assume rho = 1.0 for simplicity)
e = lc3 + (0.2 x sds x lc1)
e is not the algebraic summation of lc3 and lc1 - rather, it implies that the two effects from horiz. and vert. loads lc3 and lc1 are included together.
then your combinations from 1605.3.1 would be
lc1
lc1 + lc2
lc1 + 0.7 x e
    and this equals lc1 + 0.7(lc3) +/- 0.7(0.2 x sds x lc1)
(0.6 x lc1) + 0.7e
    and this equals 0.6(lc1) + 0.7(lc3) +/- 0.7(0.2 x sds x lc1)
all of this can occur within your program - there should be no reason to do anything by hand.
thank you very much! your answer is very clear, but if you could confirm, qe is the same as the fx's (eq 16-41)  i think this was the part i was miss understanding.
in fact now that i type it, qe really is not exactly fx's it is the effect of the fx's but as long as it is treated in the model load combinations and factors separate from the vertical loads and separate from the vertical load portion of e the math works out (as you described in your response).  confusing to write but i think i understand.
thanks!
yes, you can think of qe as the same as the combined effects of all the fx's.  another way to look at it is that qe could be the shear in a beam somewhere, that is the result of the fx's.  or, it could be the moment in a column, etc. due to the fx's....
when they first came out with the e equation i was stumped at first to see an algebraic sum of a seismic load (that i normally think of as horizontal) added to a vertical dead load effect from seismic (a percent of dl)....how do you add vertical loads and horizontal loads?  a major violation of physics!
resurect an old question.
we have been using the e=rho*qe(+-)0.2sds*d as described in the code.  about a month ago i thought i read some where that under certain conditions the vertical component could be taken as zero.  for the life of me i cannot recall where i read it, or even if i read it.  can anyone help? i want to say that whatever the conditions were they would typically apply to my work, perhaps it was when elfp was used (can't re  
you can take the vertical component of seismic (.2sds x d) as zero for asd design (vs. lrfd).  it shows up in the definition of e in the code.
thanks jae for all of your help reference this issue.  i am not able to find the code reference you mention. can you tell me the code section that indicates that for asd the vertical load can be zero.
let me ask you your opinion on a separate issue.  there are 2 groups of load combinations in the ibc for asd.  the first has a load combo d+l+.7e+lr, this combo does not allow the 1/3 increase for stresses, and for where i'm at s is pretty small and lr ends up being 20 psf.  a full live load on the roof during a seismic event seems unlikely.  
the alternate group of load combinations has a similar load combo d+l+s+e/1.4, this combo does allow the 1/3 increase. also this combo does not include the roof live load (only a relatively small snow load for my area) why ever use the first set of load combos?  sometimes, i suppose the snow load may be heavier than 20 psf lr but to not get to use the 1/3 increase is a big hit.  what i'm getting at is that the resulting   
zzzzzzzzz
i guess that the reference to ignoring this vertical seismic effect only showed up in the 1997 ubc, not the ibc 2000.
sorry about that.
on your load combination question - there has been some previous eng-tips discussions on this topic - i believe that the icc has included two different asd combos to allow a transition time to eventually create one set (someone out there know if the 2003 ibc has one or two asd combo sets?)
yes, there is a difference - is it that big?
for the gravity only case - both have d+l+lr  and d+l+s without any 1/3 increase, which would give you the same beam size.
for the lateral case, there is a difference in the e case:
first group:
d+l+lr+0.7e  with no 1/3 increase
d+l+s+0.7e  with no 1/3 increase (s is small)
and
second group:
d+l+s+e/1.4  with a 1/3 increase (s is small)
when comparing the first group, with the s load, there obviously is a 1.33 difference in the result, but your earlier gravity load combo, you have erased that difference.   also, the lr value can be significantly reduced for your columns and girders that take larger tributary areas so your lr and s may be more equal for these large area   
the "2000 ibc structural application guide" page 28 does a nice job of explaining "e". also, the "2000 ibc handbook-structural provisions" has good examples showing the application of load combinations.