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# Re: Masonry fireplaces and chimneys, Fp calcs

• To: <seaint(--nospam--at)seaint.org>
• Subject: Re: Masonry fireplaces and chimneys, Fp calcs
• From: Samir Ghosn <sghosn(--nospam--at)harris-assoc.com>
• Date: Tue, 25 Feb 2003 09:26:27 -0800

```Thor
You have not gone crazy at all.  The codes have a tendency to do that some
times especially when you find out that the prescriptive package, section
3102.4.3 would give you a method of providing seismic anchorage without
having to calculate the forces per the equation you cited.  I am not
surprised that most engineers would elect to use the prescriptive package
as it meets the code intent. If you are doing work in LA City or county you
might find that masonry chimneys are prohibited just because they don't
figure and yes if the contractor said I have done it like this for the last
20 years, he is probably not lying on this one.  One thing remains true is
that masonry chimney will invariably pull away and land on the roof any
way,or the ground.  I have witnessed boht conditions after Norhridge Quake.
Good luck,
Samir Y. Ghosn, P.E.
. ..At 09:22 PM 2/24/2003 -0800, you wrote:
>A couple of new Code editions have come along since I moved from the worst
>seismic zone to the second-worst, whereupon wind forces came to govern most
>of my lateral designs.
>
>In the 1991 UBC, the equation for figuring F-sub-p for seismic force on a
>portion of a structure was :  Fp=Z*I*Cp*Wp
>
>In Zone 4 for things besides cantilevered  parapets, this typically gave a
>value of Fp=0.3*Wp.  (0.225*Wp in Zone 3.....)
>
>The 1997 UBC and the beloved 2001 Calif. Bldg. Code give one formula for
>Fp=4*Ca*Ip*Wp, which gives a value in Zone 3 of Fp=4*0.36*I*Wp/1.4  for ASD,
>or Fp=1.03*Wp for typical soil and importance factors.
>
>This more than quadruples the design lateral force, from what I can tell
>after reviewing the SEAOC Blue Book and triple-checking myself....  What I
>am hoping is that someone on the List  will really embarrass me by saying,
>"you forgot to multiply by the 'Omicron-Theta-prime-sub-nought factor' that
>will bring your forces back down so Fp=0.38*Wp."
>
>I just designed a replacement chimney for a house--the original suffered
>from a car crash--and the footing needed to be 9 x 11 feet to resist
>overturning.  The existing house has skip-sheathing on the roof, so I could
>not anchor at the roof diaphragm level.  The contractor does not want to
>undermine half the house to pour this enormous footing.  ("Good-old-days"
>footing size would have been about 4 x 5 feet, 1/5 the concrete and
>excavation)
>
>I envision the Fp from masonry chimneys  exceeding the base shear for the
>entire house in some cases.  For this particular chimney, if  I actually had
>a roof diaphragm, I would need to strap the chimney to the roof framing for
>
>Please tell me whether I have finally gone crazy on my own, or we have all
>gone crazy together.
>
>Thor Matteson
>
>http://www.shearwalls.com
>
>
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