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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: "Bill Marczewski" <bmarczewski(--nospam--at)pndast.com>
• Date: Tue, 25 Feb 2003 10:10:16 -0800

Title: RE: Masonry fireplaces and chimneys, Fp calcs

Thor,

I wish I knew of the 'Omicron-Theta-prime-sub-nought factor' that you wish you were missing.  I'm not sure what the California Building Code says regarding Fp, however I do believe you are correct with Fp(min) = 0.7*Ca*Ip*Wp, and the upper bound being Fp(max) = 4.0*Ca*Ip*Wp.  Did you evaluate '97 UBC equation 32-2 to see what range it falls into?  I thought I was crazy and that my client would definately find another engineer next time around when I specified restraint using these equations.  I think in summary we went crazy together.  Sorry about that.

Bill S. Marczewski, P.E.
Peratrovich, Nottingham & Drage, Inc.
503-325-1250 Main; 503-325-9789 Fax
bmarczewski(--nospam--at)pndast.com
www.pnd-anc.com

-----Original Message-----
From: Thor Matteson, SE [mailto:matteson(--nospam--at)yosemite.net]
Sent: Monday, February 24, 2003 9:22 PM
To: SEAINT
Subject: Masonry fireplaces and chimneys, Fp calcs

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

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