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# RE: seismic design manual - volume 1 - seaoc

• To: seaint(--nospam--at)seaint.org
• Subject: RE: seismic design manual - volume 1 - seaoc
• From: "Chris Roper" <christophermroper(--nospam--at)hotmail.com>
• Date: Fri, 12 Sep 2003 16:20:05 +0000

I would be able to see your reasoning about the distribution if the definition of hx did not have the word "attachment" in it, but it does. To me then, you have a rectangular distribution defined by the attachment at the top of panel, and a rectangular distribution defined by the attachement at the bottom of panel. The resulting rectangular distribution of seismic force used for design of the panel between those two points is the average of those two distributions. The coefficient doesn't increase as you go up. The wording isn't such that "hx is the elevation of any given point with respect to hx." hx is a fixed point and so is hr.
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Hopefully we're not trying to argue the same point, but in different ways.

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I agree with you 1000% about your concerns of infinitely laborious analysis becoming standard practice... for what purpose... to precisely analyze structures for lateral loads that are frankly, our best guess. It doesn't make sense to me either. We have to keep in mind though that may of the people posting questions to this list are at a different level in their career and knowledge. Sometimes you have to analyze things to death a few times to realize the kinds of things that do and don't control a design. It's a constant learning process. I would just hate to see people be affraid to post questions to this list because they're affraid of being made to look stupid (not that that's what your intent was.)
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```From: "Bill Allen" <T.W.Allen(--nospam--at)cox.net>
To: <seaint(--nospam--at)seaint.org>
Subject: RE: seismic design manual - volume 1 - seaoc
Date: Fri, 12 Sep 2003 08:31:13 -0700

Chris-

When designing the wall element, h_sub_x divided by h_sub_r is relevant.

See SDM Vol 1, p. 112.

Seismic coefficient at Roof = 0.533 (hx/hr =36/36)

Seismic coefficient at 2nd floor = 0.311 (hx/hr =16/36)

Seismic coefficient at first floor = 0.28 (hx/hr =0/36, but the cutoff
of 0.7 x 0.4 controls)

"Using the average..." to me considers the triangular distribution of
the seismic forces. [1+3 x h(x)/hr] is a linear expression. These are
not two rectangular distributions, but one rectangular distribution
between grade and the 2nd floor based on an increasing coefficient where
the average is taken and one rectangular distribution between the 2nd
floor and roof based on an increasing coefficient  where the
average(0.533 + 0.311)/2 = 0.422 is taken. IMO, this is another
simplificaton (making the reactions even more *theoretically*
erroneous), but O.K. in my book since it makes the problem simpler.

You do make a good point, though. IMO, this section of the 1997 UBC (I
call the Bob Bachmann Formula) did a huge disservice to the engineering
community by taking something relatively simple and making it overly
complicated and difficult to interpret. Before the SDM, I could get two
or three interpretations on the proper coefficient when designing a mass
wall. One (IMO legitimate) argument would be to design for a trapezoidal
force, but the SDM simplified it (fortunately, there were some rational
engineers with good common sense writing the example problems), but that
in itself was interpretive. As we all know, when you publish a code
subject to interpretation, there will always be disagreements and cause
for concern in a court of law or just trying to get plan check approval.

Yes, he can use whatever method he wants since he does not work for me.
I don't really care. He can turn it into a F.E.M. thesis for all I care.
But, when someone promotes a method of practice in public, I want to
offer my objection, just so that something rediculous (not saying that
this particular issue is) doesn't become "standard practice" like that
Rigid Wood  Diaphragm B.S. has/had become.

Regards,

T. William (Bill) Allen, S.E. (CA #2607)
V/F (949) 248-8588
San Juan Capistrano, CA
http://members.cox.net/ballense/

:-----Original Message-----
:From: Chris Roper [mailto:christophermroper(--nospam--at)hotmail.com]
:Sent: Friday, September 12, 2003 7:47 AM
:To: seaint(--nospam--at)seaint.org
:Subject: RE: seismic design manual - volume 1 - seaoc
:
:Bill,
:
:Since when have the forces from UBC equation 32-2 been distributed
:vertically in a triangular fashion based on hx/hr?  That would be the
case
:if hx were variable up the height of the panel.  As last I checked,
it's
:the
:fixed elevation at point of attachment and it doesn't vary, just like
hr is
:a fixed point that doesn't vary.  What you're really doing is taking
the
:average of two rectangular distributions.
:
:Also, if a guy wants to sharpen his pencil, why do we have to badger
him
:for
:it.  He's not working for you, so what care do you have?
:
:
:>From: "Bill Allen" <T.W.Allen(--nospam--at)cox.net>
:>To: <seaint(--nospam--at)seaint.org>
:>Subject: RE: seismic design manual - volume 1 - seaoc
:>Date: Fri, 12 Sep 2003 07:15:03 -0700
:>
:>I'll respond to my part.
:>
:>Your post did not mention that the wall was concrete nor did I take
the
:>time to thumb through the SDM (you didn't mention which example it
was,
:>which would have made it easier), but I have now. I believe you're
:>referring to Example 36 on page 113.
:>
:>I STILL believe you are taking a simple problem and making it
:>complicated. No sarcasm intended.
:>
:>For a simple span beam, M = WL^2/8.
:>
:>For a continuous beam, the negative moment (at the 2nd floor support),
M
:>= WL^2/8 (if the load is the same on both spans).
:>
:>If the wall is reinforced at the middle of the section, considering
the
:>spans as simple spans, the analysis is conservative and you will have
:>saved some (IMO useless) analysis time for more important things.
:>
:>I find it interesting that you are not concerned about the reactions
:>considering continuous spans, because the load at the 2nd floor is 25%
:>higher when considering continuity.
:>
:>The most important point I want to make (and it is a serious one, not
a
:>sarcastic one) is that it might be a mistake to get one's nose too
close
:>to the paper and try to make the problem more complicated than it
should
:>be. Just look at the loads you are applying to the wall. How accurate
do
:>you believe they are? If the forces (themselves a good guess at best,
an
:>arbitrary factor x DL at worst) are distributed verically in a
:>triangular fashion (based on the function of h_sub_x over h_sub_r) and
:>we are only taking the average between the points of support (again,
for
:>simplicity), how accurate can the analysis be? I sincerely believe you
:>are trying to make a silk purse out of a sow's ear.
:>
:>Yes, I agree that one should think about continuity in the design, but
:>when the analysis is based on simple spans which produces conservative
:>results, I don't see taking the analysis any further.
:>
:>Regards,
:>
:>T. William (Bill) Allen, S.E. (CA #2607)
:>V/F (949) 248-8588
:>San Juan Capistrano, CA
:>http://members.cox.net/ballense/
:>
:>:-----Original Message-----
:>:Sent: Thursday, September 11, 2003 9:52 PM
:>:To: seaint(--nospam--at)seaint.org
:>:Subject: Re: seismic design manual - volume 1 - seaoc
:>:
:>:here's my thoughts to the 3 (mostly) serious responses...
:>:
:>:************* Jake Watson, P.E. **************
:>:
:>:What is more conservative?  If you place the reinforcing in the
center
:>or
:>:the wall, you will end up with more steel by assuming two simple
spans.
:>:Secondly, you would need to do a cracked analysis and take into
account
:>the
:>:stiffness of each floor if you want a true continuous two span wall.
:>The
:>:double simple span is not necessarily correct, just more conservative
:>and
:>:significantly easier to calc.
:>:
:>:************* Jake Watson, P.E. **************
:>:
:>:**da**  jake, i agree - thank you **da**
:>:
:>:
:>:
:>:
:>:However, most engineers, when faced with a
:>:tributary-area problem like this, would figure the
:>:
:>:**da** that, in itself, doesn't make it right, does it?  **da**
:>:
:>:In other words, the far simpler
:>:tributary-area method is more accurate than a
:>:continuous-beam analysis.
:>:
:>:**da** is it? what about the negative moment at the floor level?
this
:>is a
:>:wall design, not a slab on grade, elevated slab, roof reaction
design.
:>:**da**
:>:
:>:This, laziness, and
:>:historical acceptability are the reasons the far less
:>:complex simple-span distribution continues to be used.
:>:
:>:**da** not by me... **da**
:>:
:>:But yes, you're right, in this case a continuous-beam
:>:analysis would probably more accurately represent the
:>:reactions.
:>:
:>:**da**  like i said above, i don't care about the reactions at this
:>point,
:>:i'm just trying to find the real flexural stresses in the wall...
:>mike,
:>:thank you for your responses.... **da**
:>:
:>:
:>:
:>:**T. William (Bill) Allen, S.E. ***
:>:
:>:> The load is applied to the studs which act as simple "beams",
:>spanning
:>:> from sill plate to top plate. Trib load to the foundation, 2nd
floor
:>and
:>:> roof, respectively.
:>:
:>:**da** i would agree wholeheartedly if the second floor interrupted
the
:>:wall
:>:and there were 2 physical wall elements, one above the floor and one
:>below.
:>:but we are talking about a monolithic wall with full height
continuity
:>:**da**
:>:>
:>:> We've got plenty of things to worry about rather than try to make
the
:>:> vertical distribution of loads act in accordance with a continuous
:>beam
:>:> distribution.
:>:
:>:**da**  you aren't serious are you? **da**
:>:
:>:> Let's find something else to make complicated, shall we?
:>:
:>:**da** i think i'm supposed to take this as sarcasm, right?  you can
:>call
:>:me
:>:silly but i have a real soft spot for mechanics and free body
diagrams.
:>:continuous beam analysis isn't THAT hard, is it?  **da**
:>:
:>:
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