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IBC 2006/ASCE 7-05 Load combo question

IBC 2006/ASCE 7-05 Load combo question

IBC 2006/ASCE 7-05 Load combo question

(OP)
Hi, question about code interpretation.
In ASCE 7-05 Section 12.4.2.3 discusses seismic load combinations where Sds and rho are included.  In reviewing IBC 2006, I do not see anything that says we need to use the combos in section 12.4.2.3 for seismic design.  I only see discussion about load combos with the overstrength factors.  How do you interpret IBC06?  Should these load combos from ASCE 7-05 be used in lieu of the standard 1.2D+L+1.0EQ and 0.9D+1.0EQ?

Any help is greatly appreciated.
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RE: IBC 2006/ASCE 7-05 Load combo question

They're the same…..

IBC definition of  E:  "Combined affect of horizontal and vertical earthquake-induced forces as defined in Section 12.4.2 of ASCE 7" (Section 1602)

ASCE says E = Eh + Ev

Eh = rho*QE and Ev = .2*SDS*D (ASCE 12.4.2)

So:
    1.2*D + 1.0*E + L
= 1.2*D + 1.0*(rho*QE + 0.2*SDS*D) + L
= 1.2*D + 1.0*rho*QE + 1.0*0.2*SDS*D + L
= 1.2*D + 1.0*0.2*SDS*D + 1.0*rho*QE + L
= (1.2 + 0.2*SDS)*D + 1.0*rho*QE + L  ----> Which is ASCE 12.4.2.3

I think the other equations fall out the same way.

If you are in a lower seismic zone, it really doesn't matter too much……

For example, say Seismic Design Category = C and SDS = .20

Then:

rho =1.0 (ASCE 12.3.4.1) and Eh = QE = normal horizontal effects from V
Ev =.2*SDS*D = .04*D.

So the horizontal E just equals Eh and the vertical E is just adding 4% to the Dead Load…..

Now if your really in a low seimsic zone and SDS <= .125, Ev can = 0 and your really back to 1.2*D + 1.0*E + L where E just equals Eh

RE: IBC 2006/ASCE 7-05 Load combo question

Note, for simplification, I ignored the 0.2S which is to be included in the same way per IBC 1605.2.1, ASCE 2.3.2, and ASCE 12.4.2.3.

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