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# Re: '97 UBC Lateral Design - Envelope Solutions???? - Part 2

• To: "'seaint(--nospam--at)seaint.org'" <seaint(--nospam--at)seaint.org>
• Subject: Re: '97 UBC Lateral Design - Envelope Solutions???? - Part 2
• From: "Swingle, Mark" <Mark.Swingle(--nospam--at)dgs.ca.gov>
• Date: Wed, 29 Sep 1999 09:36:17 -0700

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PART 2 OF 2
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Re: '97 UBC Lateral Design - Envelope Solutions????

Mark Swingle (MS) continues his response to Dennis Wish's (DW) post:

(MS) Let's take the first wall mentioned above: "....in one wall, the
flexible
analysis yielded 21,371 pounds of shear into this lline of resistance. The
same grid line, under torson analysis 1,523 pounds of shear."

Let's see what the tributary area is that this wall is resisting:

OK, I will make the following assumptions:
1.  3 story building
2.  No irregularities (continuous 3-story wall)
3.  DL of roof and 2 floors = 15 psf each
4.  DL of int and ext walls tributary to each level (3 levels
above the base) is 10 psf as a horiz distr load.
5.  Zone 4
6.  Na = 1.5 (near-fault)
7.  R = 5.5
8.  rho = 1

Therefore

Total trib weight = (45 + 30) = 75 psf
E = 0.30W or E/1.4 = 0.21W   (by ASD method)

So, the tributary force to any wall on the first floor is:

F = 0.21(75 psf) = 16 psf = 0.016 ksf

In your tributary analysis, therefore, THIS wall is resisting 21k/0.016ksf =
1,333 SF of tributary area.  Now, assuming that this area is a rectangle
with an aspect ratio of 2:1, this would yield a rectangle that is 15 ft x 30
ft.  Now, assuming that the adjacent walls on each side are equidistant from
this wall, that means that the diaphragm span on EACH side is 15 feet.

****(Note that if I had assumed a one-story building, or Zone 4, or a
lighter building, or not near-fault, or any combination, this tributary area
would be much LARGER.  My assumptions are very conservative with regard to
the points I will make below.)****

Let's assume the wall is sheathed with 1/2" Struct I, nailed w/ 10d @3" oc.

With an allowable capacity of 665 plf this requires a wall with a length
(per
your flexible diap (tributary) analysis) of 21.3k/0.665klf = 32 feet.  So
this
wall is the entire length of the rectangle.  (Remember that your design base
shear is probably lower and/or this may not be a 3-story building, so your
area may be much larger.  And also remember that if I assumed lighter
nailing, the wall would be LONGER.)

No matter how you slice it, this is a STIFF wall.

But here's the point (at last, you say!!!!).  You say that for the rigid
case, the force is only 1.5 kips.  For the purposes of this example, this is
essentially equal to zero.  That means that the adjacent walls (or some
other walls somewhere in the builing) ARE SO MUCH STIFFER THAN THIS WALL
that they take practically all of it's tributary load!!!!  How can there
possibly be other walls that are stiffer than a 32 ft wall (or 3 ten-ft
walls, whatever)????

This is just not possible.

In addition, this example means that for the rigid case, your wood diaphragm
is so stiff that it can span 30 feet (skipping "over" the wall that has no
force in it now) with almost no deflection (otherwise it would put load on
the wall).

This is just not possible.

------------------------------
Dennis continues:
(DW) "If we follow your strict interpretation of the the code method so that
if the diaphragm is "not-flexible" we distribute shear by torsional analysis
-
how many of us would feel justified in ignoring the flexible resulsts when
the discrepancy between the two is 18 times larger than what we chose to
resist. Remember, it is not a clear cut case of one analysis yielding more
restrictive answers - this depends on the wall stiffness and the location of
the applied loads. In my example there is sufficient different on some walls
to warrant compliance with the flexible analysis while other walls require
compliance with rigid."

Mark responds:
(MS)  I never said that if the diaphragm fails the flexible test that one
needs to analyze it by rigid diaphragm analysis.  Reread what I said.  I
never said to ignore the flexible case.  I said that in the worst case, one
may want to adjust the design shears TOWARD the infinitely stiff case
BECAUSE THE CODE REQUIRES IT.

Even if your example above is correct, that the other walls ARE that much
stiffer, my point is that no wood diaphragm is stiff enough to redistribute
the shears ALL THE WAY to the 100% rigid analysis results.  I think my
example above demonstrates this, since none of us can imagine a wood
diaphragm with a span of 30 feet (OR MORE) not deflecting very much.

EVEN IF YOUR WALL STIFFNESSES ARE SO RADICALLY VARYING, AND EVEN IF THE
DIAPHRAGM FAILS THE 2:1 TEST, IT WILL STILL BE FLEXIBLE ENOUGH TO DEFLECT
AND THEREFORE TO LOAD THE WALL THAT YOU SAY HAS NO LOAD.

Please take this in the spirit in which it was intended.  I am simply
looking for rationality.  This is not a personal attack.

I will repeat what I said at the beginning:  Now that you have volunteered
to be on the panel this weekend, you have to be prepared to stand by your
statements more than ever before.  You are now in a similar position to that
of the committee members of the past who have been criticized by you (and
others) for not being rational.

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END OF PART 2 OF 2
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Mark Swingle
Oakland, CA

Disclaimer:  These are my own opinions and are subject to change.

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