Re: Seismic Design Example

Sharon

 

Vertical component of earthquake.  .2Sds...... .2x1.19x3= .714D

 

 

In a message dated 2/15/2008 3:37:51 P.M. Pacific Standard Time, ECVAl3@aol.com writes:

For those of you that have and understand the "2006 IBC Structural/Seismic Design Manual Vol.2" I have a question.

On page 32 to determine the Anchorage at A on a perforated shear wall:

Going from the formula  (0.9-0.2*SDS)*D + Wo*QE

Wo = 3.0,   SDS = 1.19

they suddenly go to the formula  (0.9-0.2*SDS)*D + Wo*E

Putting the numerical values in they get  (0.9-0.2*1.19)*D + 0.66*D

 

My question is, how did they get 0.66*D from Wo*QE or Wo*E  ?

Can anyone shed some light on this for me?

 

Thank you,

S.Macie

 

Joe Venuti
Johnson & Nielsen Associates
Palm Springs, CA

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