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ATC FEMA 273 Technical Seminar Series

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ATC FEMA 273 Technical Seminar Series

Applied Technology Council ( has announced a technical
series on FEMA 273, "Guidelines for the Sesimic Rehabilitation of Buildings"
for Salt Lake City, Utah on November 19-20,1998 and Portland, Oregon on
December 8-9, 1998.  
For additional information, contact ATC web site ( or e-
mail: atc@atcouncil org.  The seminar registration is very reasonable ($100 or
less if you are a member of the local Structural Engineers Association) and
includes seminar notes, FEMA 273/274 documents and a working dinner (cold)
each day.

At the ATC FEMA 273 Seminar held in San Diego, California on November
5-6,1998, a Wood Frame Example was presented.  This Wood Frame Example was a
three story (36 feet high) by 75 foot wide by 120 foot long wood frame
building braced by three story plywood shear walls.

This building had a calculated weight = 864 kips and a building period = 0.88
seconds.  The "Pseudo lateral load" (Base Shear), V = C1*C2*C3*Sa*W =
1.25 x 0.740 x 864 = 799 kips or 0.925 x 864 = 799 kips.  (Note: 92.5%

The "Pseudo lateral loads" on a single three story plywood shear wall 24 feet
long are as follows:  At the Roof = 50.1 kips, 3rd Floor = 57.3 kips, and 2nd
Floor  = 25.5 kips.

The total base shear on this single three story 24 foot long plywood shear
wall = 132.9 kips.  The 3rd story plywood shear = 2,400 lb. per ft., 2nd story
plywood shear = 4,470 lb. per ft,. and the 1st story plywood shear = 5, 450
lb. per ft.

The ATC Instructor (based on a copy of the Wood Frame Building Example handed
out at the Seminar) elected to only evaluate the 2nd story portion of the
plywood shear wall.  At the 2nd story, the plywood shear  = 4,470 lb. per ft.
and the tension (tie-down) uplift force at 2nd floor at the edge of the
plywood shear  wall = 18.9 kips.

Following the same procedures used by the ATC Seminar Instructor for the Wood
Frame Building, the 1st story plywood shear = 5, 450 lb. per ft. and the
tension (tie-down) uplift force at the edge of the plywood shear wall = 36.5

Following the "m" value reduction factors in FEMA 273, produced a design with
a 15/32" thick, Structural 1 Plywood with 10d @ 2 inches on center (a nail
size and close spacing I would never use) in the 1st story and 10d @ 3 inches
on center in the 2nd and 3rd stories.

The tie-down from the plywood shear wall to the reinforced concrete foundation
at the first floor was a Simpson HD-8A, using 3 1/2 inch wood members, which
has a FEMA 273 yield value = (7,460 x 2.8) / 1.33 = 15, 700 lbs.  This
resulted in a Demand-Capacity Ratio of 36.5 kips / 15.7 kips = 2.3, which is
less than an "m" value in FEMA 273, Table 8-1 = 2.8.

If the maximum design tension tie-down force that the reinforced concrete
foundation must be designed for is limited by the tension yield value of the
Simpson HD-8A = 15, 700 lbs., then 104 cubic feet or 3.9 cubic yards of
concrete are required to "tie-down" the plywood shear wall at the foundation
level at each end of the shear wall. 

Do the plywood shear values and the tension (tie-down) forces that must be
resisted by the reinforced concrete foundation look reasonable?  Or does this
Wood Frame Example illustrate again (in my judgment) one of the fundamental
problems with FEMA 273, Linear Static Procedues, how are the overturning
forces accounted for using a "rational engineering explanation",  when the
"Pseudo lateral loads" are so large?

I have sent a copy of my calculations to the ATC Seminar Instructor for the
Wood Frame Building Example and hope he will address in his forthcoming
lectures and revised seminar notes the problems of the large plwood shearing
stresses and large tie-down forces in the 1st story and how these tie-down
forces can be resisted.

Frank E. McClure      FEMCCLURE(--nospam--at)  November 16, 1998