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Re: Displaced Center of Mass Question

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Seaintonln(--nospam--at)aol.com wrote:
> 
> The design examples from the Feb 1998 SEAOC Wood Seminar for the 97 UBC (and
> the draft of the ICBO Design Manual Volume II) note "by displacing the center
> of mass by 5% can result in the C.M. being on either side of the C.R. and can
> produce added torsional shears in all walls."
> 
> However, the Reinforced Masonry Engineering Handbook by James Amrhein (5th
> edition 1994 UBC compliant) simply adds the 5% of the diaphragm depth
> perpendicular to the direction of loading and adds it to the difference
> between the C.M. and C.R.. - leaving the displaced C.M. in only one location.
> 
> Which is considered the standard of practice in rigid diaprhagm analysis?  If
> the more involved method is applied in wood construction, has anyone been
> able to calculate a significant difference in added shear from torsion?  If
> so, how many of these buildings were residential (single and mulitple
> residential) etc.
> 
> Dennis Wish PE

I haven't applied it to a residence yet they don't appear to be
enforcing that provision here in Utah.  However, the 5% is a relatively
simple issue to address.  If the difference between your center of mass
(CM) and your center of rigidity (CR) is greater than your 5%, then
simply add to the eccentricity.  If the 5% eccentricity is less than the
difference then you must move it both directions.  Here why: if you use
the first case, by decreasing the eccentricity you are simply decreasing
the applied torsion.  During the second case, you actually will reverse
the direction of the torsional forces.  This means that the walls with
an innitail negative force, will now have a positive force.  See below.

Sample calc:

say 100'x100' bldg.   	CM: X=50', Y=50' (If you find this building let
me know :)
			CR: X=50', Y=48'

So 5% = (100)*(0.05) = 5'
Torsion ecc e = 50'-48' = 2'

Case 1:

Locate new CMy = 48'+5' = 53' (Above CR will create a positive moment)

Case 2:

Locate new CMy = 48'-5' = 43' (Below CR will create a negative moment)

A posative moment will add torsional forces to direct shear forces to
wall "A" and a negative moment will subtract torsional forces (neglect)
from direct shear.  I hope this helps clarify the idea a little bit.  By
the way, there are errors in the example you are looking at.  I can't
rember what they are off the top of my head, but be careful.

Enjoy, 	Jake Watson, E.I.T.
	Salt Lake City, UT