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Fw: Application of 3Rw/8 - How to apply

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Dennis:

I feel it is important to remember that the 3(Rw/8) factor is meant to give
an estimate of the MAXIMUM EXPECTED DISPLACEMENT of the structure during an
actual earthquake.  In your case, using an Rw of 6, SEAOC is saying that if
you had a displacement of 1" under the reduced design base shear, then you
should expect at least 2.25" of displacement of the frame during the actual
earthquake.

If your columns/beams are capable of going to delta-max without yielding,
then an elastic analysis will tell you what the axial/shear/moment
situation will be in your structure at that displacement.  If the girders
cannot achieve delta-max without yielding, then an elastic analysis will
overestimate the moments/shears that the girder can input into the system.

I think that the important point to keep in mind is that if you have an
yielding mechanism (nominal strength<strength required at delta-max), then
you will need to check if that mechanism is desirable or not.  I have most
of my experience in R/C, so when I am checking R/C columns, I would list
diagonal tension as undesirable and flexural+axial as possibly OK.  If a
column cannot take the deformation at delta-max without some yielding in
flexure, I can check that the plastic rotation capacity is sufficient to
permit yielding.  BUT, if it cannot take the shear associated with
delta-max (corrected for flexural yielding of column if that is the case),
then that is definitely NOT OK and something needs to change in the column.

My point is that it doesn't make sense to me to design one part of the
system (i.e. the joint) for 3(Rw/8) if:

1) the girder cannot input the moments/shears into the joint that an
ELASTIC analysis says it will because it will be yielding prior to
delta-max.  The results of an elastic analysis should be corrected to get a
better idea of the actual forces manifested in the joint.  Or, you could
just overdesign the joint (no harm in that IMO).

or

2) if the girder or column will yield in an undesirable manner (either
unstable or not desirable from a hysteretic point of view) prior to
achieving delta-max.  This would be because the system would already have
problems and the joint strength would become a non-issue.

I realize that you asked for a quick answer, and this isn't one, but this
particular aspect of the code is very interesting to me as it mimics
displacement based design methods, which I feel are much more rationale
than the current force based approach.

One last point: I gather that you are designing this frame for a retrofit
measure.  Have you checked the rest of the structure for the same level of
magnified DISPLACEMENT-->3(Rw/8)?  The rest of your structure will be
displacing as well, and if you are retrofitting it, then it is obviously
deficient in some manner.

Hope this helps (or at least doesn't hurt!)

T. Eric Gillham PE
----------
> From: Dennis S. Wish PE <wish(--nospam--at)cwia.com>
> To: seaint(--nospam--at)seaint.org
> Subject: RE: Application of 3Rw/8 - How to apply
> Date: Thursday, September 24, 1998 6:26 AM
> 
> Dick,
> I'm still not clear on this one. You have stated pretty much what is
written
> in 2211.6 and the wording of this is what is confusing to me. I need a
> practicle example. I would assume that you would first determine the
> appropriate member size based upon applying the original load to the
frame.
> The 3Rw/8 term would be applied to the design of the beam to column
> connection and therefore would be applied to the moment and shear at that
> node. The initial load is not increased only the stress in the joint.
> Is this a correct assumption?
> 
> Dennis
> 
> -----Original Message-----
> From: Horning, Dick/CVO [mailto:dhorning(--nospam--at)CH2M.com]
> Sent: Wednesday, September 23, 1998 1:02 PM
> To: 'seaint(--nospam--at)seaint.org'
> Subject: RE: Application of 3Rw/8 - How to apply
> 
> 
> Rw = 6, not 8, for steel OMRF.  Framing elements not part of the lateral
> force resisting system must be able to support gravity loads under a
> displacement of 3Rw/8 times the calculated displacement of your moment
> frame.  You don't need to multiply your moment frame forces by it.
> 
> > ----------
> > From: 	Dennis S. Wish PE[SMTP:wish(--nospam--at)cwia.com]
> > Sent: 	Wednesday, September 23, 1998 12:38 PM
> > To: 	SEA International List
> > Subject: 	Application of 3Rw/8 - How to apply
> >
> > I design very little steel so I apologize up front for my lack of
> > understanding regardless of the number of posts which have been sent on
> > this issue. I am in need of a fast answer and would appreciate any
> > responses.
> >  
> > I have designed a simple (Ordinary) moment frame for a seismic
retrofit.
> > The frame is one bay - 25 feet between columns and 12 feet in height. 
The
> > lateral load applied from the results of the UCBC design is 10.57 Kips.
> > The frame is pinned at the base and therefore develops moment in the
> > connection of beam to column.
> >  
> > Do I apply 3Rw/8 to the moment at the connection or is it applied to
both
> > the moment and the shear at the end of the beam?
> >  
> > Also, is it correct to assume Rw is equal to 8 for use with an Ordinary
> > Moment Frame?
> >  
> > I appreciate any fast responses.
> >  
> > thanks
> > Dennis Wish PE
> >  
> >
> 
> 
> 
> 
>