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RE: Commercial Bldg. - Second Floor Loads

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While I don't agree with his approach, Tim is correct.  Usually the smaller
the tributary area, the larger the load.  Odds are that a small are may see
the full design load but a large area will not.  Hence, live loads can be
reduced for larger tributary areas.  IIRC, 900 ft^2 is where the reduction
begins (check chapter 16).  One thing you haven't mentioned yet is the point
load.  Standard office loading requires a 2,000 pound point to be spread
over an area 2 foot square (not 2 ft^2) and may be placed at any location.
Review the loading table in chapter 16 carefully.  ASCE 7 has a decent
commentary on where some of the loads came from.

Lastly, 14 PSF seems a little low for DL in a TV studio.  Have you
considered mechanical equipment such as lighting, cable trays, etc?  You
might also ask if you mezzanine will be supporting cameras.  I did a TV
studio recently and never got a straight answer from the owner, but I would
bet there is special deflection criteria for cameras.

Hope this helps a little,

Jake Watson, P.E.
Salt Lake City, UT

-----Original Message-----
From: DRPFLY(--nospam--at)aol.com [mailto:DRPFLY(--nospam--at)aol.com]
Sent: Wednesday, September 25, 2002 5:18 PM
To: seaint(--nospam--at)seaint.org
Subject: Re: Commercial Bldg. - Second Floor Loads


Thanks for all your helpful and contradictory advice.

The commercial building with the new mezzanine is currently owned by  a TV
station and the architect suggested the 100 PSF to cover all future uses
including, I guess, moving the partitions.  There are a couple of stairways
(exit facilities) plus a walkway around part of the mezzanine.  I think
using
a blanket LL = 100 PSF is okay for the all the intended uses.

According to the UBC 1606, partition loads are considered Dead Loads and if
they are movable I am obligated to include this in the regular floor dead
loads.  This assumes that partitions are stacked side-by-side the whole
length and width of the tributary floor area.  Seems a bit heavy-handed to
me, but I want to do the right thing.

On the other hand, it seems to me a linear truss can carry LL=100 PSF and DL
= 14 PSF but not partition load at the same time.

So this is what I'm going to do:

Truss design:  DL = 14 PSF
                      LL =  100 PSF

Floor beams:  DL = 14 + 20 PSF
                      LL = 100 PSF
I think that covers all the bases, don't you?  (Oh no, a sports cliche!)
Come on Angels!

Regards,
Denise Poeltler
The Flying Buttress
Palm Desert, CA

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