IBC Seismic Effect E
IBC Seismic Effect E
(OP)
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.
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.






RE: IBC Seismic Effect E
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.
RE: IBC Seismic Effect E
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!
RE: IBC Seismic Effect E
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!
So just keep remembering that the "equation" for E is really like an inclusion list of what "effects" need to be included in the code required combinations whenever you see E in the combo.
RE: IBC Seismic Effect E
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 remember though)
Any help would be greatly appreciated.
RE: IBC Seismic Effect E
RE: IBC Seismic Effect E
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 member sizes that come from the two different combos is big...do you agree...does that seem odd...it does to me.
Thanks again.
RE: IBC Seismic Effect E
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.
RE: IBC Seismic Effect E
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 members which are also most affected by the lateral combo anyway.
But I agree....the code is really warped in that there are WAY too many load combinations and these cause us to spend a lot more time calculating than I think is really necessary.
RE: IBC Seismic Effect E