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Re: More problems with the 1997 UBC

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I suggest a couple of points of relief, until some of the others on the cc
list weigh in:

1.  I believe that the provisions of Footnote 14 of Table 16-O are intended
to address the dynamic impact effects of many isolator types with the
"stops"  that provide limits for the lateral motions.  This amplification
(due to unrestrained motion prior to impact with the "stops") is in addition
to the in-structure amplification of the accelerations input to the base of
the supporting structure, and can actually be considerably higher than is
addressed in the footnote.  There may be isolator types applicable to your
case with integral or independent "snubbers" that provide lateral restraint
with minimal force amplification due to impact.

2. Not all drilled-in-concrete anchors are necessarily shallow or of the
expansion (friction) type.  Have you considered the undercut types?

Where are the SQUG rules when you need them??

Mark D. Anderson

Subject: More problems with the 1997 UBC


> For those of you who are interested...
>
> Here's a good example of how to use the 1997 UBC for equipment design.  If
> you choose to accept the challenge, I urge you to run an independent
> calculation by yourself to see what answer you get...  Then compare it to
my
> solution, which has been independently checked by two other structural
> engineers.
>
>
> PROBLEM:
>
> I have a client who wishes to hang a 800 pound piece of equipment from the
> ceiling somewhere in an existing 30 story building.  For what lateral
force
> should I design it, using the 1997 UBC and Allowable Stress Design (ASD)?
>
> Given:
>
> Zone 4, Soil Type D, no near field effects, life-safety equipment
>
> Ca = 0.44
> Ip = 1.5
>
>
>
>
>
>
>
> SOLUTION:
>
> Using Equation 32-1:
>
> Fp = (4 * 0.44 * 1.5 * 0.8 kips)/1.4  = 1.51 kips (1.88 g)
>
> That seems like an awfully large force for a piece of equipment hanging
from
> a ceiling.  This equipment is suspended less than two feet below the
> concrete floor slab...  It seems really unlikely that the equipment needs
to
> be elastically designed for accelerations this large.  Note that if we had
> been located near a Type A Fault, the design accelerations would have been
> another 50% larger!
>
>
> Maybe I can use the more complicated, but more "rational" formula,
Equation
> 32-2, which should give me a lower design force, correct?
>
> ap = 1.0 and Rp = 3.0 (equipment hung from ceiling)
>
> And since my client would like to place this equipment on each floor and
> would like one single design, rather than a separate design for each
floor,
> hx/hr = 1.0
>
> Fp = (1.0 * 0.44 * 1.5 * (1 + 3*1.0)/3.0 * 0.8 kips) / 1.4 = 0.50 kips
(0.63
> g)
>
> Which isn't too bad...
>
> But wait, since this equipment has vibration isolators, ap = 2.5 and Rp =
> 1.5, per footnote 14, which I almost missed!  (There are so many footnotes
> that they take up almost a page themselves, even when formatted in a
really
> small font!)
>
> Fp = (2.5 * 0.44 * 1.5 * (1 + 3*1.0)/1.5 * 0.8 kips) / 1.4 = 2.5 kips
(3.13
> g)
>
> Which is much higher than the simple equation 32-1...
>
> There's also another provision in footnote #14!
>
> Since I have to hang this from the ceiling, I have to use either expansion
> anchors or shallow anchors, so the forces for the connections are doubled
> (ap = 5.0).
>
> Fp = 5.0 kips (6.25 g).
>
> Even if I am only at midheight in the building, the "rational" formula
> design force is far grater than that produced by the simple formula.
>
> It seems that the dynamic and supposedly rational formula 32-2 can break
> down fairly quickly and that I am reduced to designing this equipment for
> 1.88 g.
>
> BUT WAIT!  This footnote #14 to Table O, which is ONLY referenced by
> Equation 32-2, back-references Equation 32-1 and requires me to further
> double my forces for the anchors, even if I use only Equation 32-1.  This
is
> an incredible error in the code, since I never had to use Equation 32-2
(and
> consequently Table O) in the first place!  Unless I tried to use the
> "alternate" procedure, I would never even know that the code requires me
to
> double the forces yet again...
>
>
> THEREFORE, the design force for the anchors for this equipment must be
> designed for
>
> Fp = (4 * 0.44 * 1.5 * 0.8 kips) * 2.0 /1.4 = 3.02 kips (3.77 g)
>
> Since connections typically have a factor of safety of at least 3, this
> equipment should be good for at least 11.31 g.  What a relief!  I was
really
> concerned that this equipment, hung less than two feet from the underside
of
> a slab in a 30 story building was going to experience accelerations of
that
> magnitude!
>
> If the building had been located within 2 km of a Type A fault, I would
have
> to design for 5.66 g and the equipment would be able to withstand 16.97 g.
>
>
> For comparison, the forces in the 1994 UBC would only be
>
> Z = 0.4
> Cp = 0.75 * 2 (Non-rigid equipment)
> Ip = 1.5
>
> Fp = 0.4 * 0.75 * 2 * 1.5 * 0.8 kips = 0.72 kips (0.90 g).
>
> Which means that the equipment should be able to withstand 2.7 g.
>
>
> For this example, the 1997 UBC forces are more than four (or six,
depending
> on location) times larger than the design forces of the 1994 UBC!
>
>
> Based on this example and a number of other similar examples, here are my
> conclusions regarding the 1997 UBC non-structural provisions:
>
> The provisions are much more complicated.
> The provisions are much more difficult to understand.
> The provisions can produce nonsensical results.
> The provisions contain outright errors.
>
> It is my opinion that SEAOC should work to restore the 1994 UBC
provisions.
>
> -Gary
>
> Gary R. Searer, P.E., S.E.  -  gsearer(--nospam--at)wje.com
> Wiss, Janney, Elstner Associates, Inc.
>
>