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Re: UBC'97, Orthogonal Effects
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- Subject: Re: UBC'97, Orthogonal Effects
- From: Jake Watson <jwatson(--nospam--at)inconnect.com>
- Date: Tue, 11 Apr 2000 21:20:16 -0600
Let me preface by emphasizing I am an E.I.T., but I will take a stab at it. I have been doing similar calcs recently. Question 1: Try this as your building (If only it were shaped like this....) _________________ North End | | | | | | | | | | <---- Vu | | | | | | |_______________| South End Because the force is due East, the transverse END (not direction "...at one end of the structure transverse to an axis....") would be the North and South ends of the building. This means that if applying the force through the center of mass displaces the North end 1" and the South end 1.2", you must amplify the force. A direct answer; no, you do not have to calculate deflection of Y to determine torsion for X. Question 2: Under dynamic provisions, you must consider orthogonal effects. You haven't mentioned if this is dynamic or static procedure. I'm not so sure about static, but under dynamic you have two choices. You can use square root of the sum of the squares (SRSS) or you may take 100% of your force in the X direction and 30% of the force in the Y direction. I don't have a UBC in front of me or I would quote you section numbers. This is all un ch 16 somewhere. If you are doing dynamic, I recommend you go to CSI's web site and read some of their papers. (http://www.csiberkeley.com/Tech_Info/edwilson.htm) Secondly, you have to ask yourself a question, "Can I have a structure twist only along the X axis?" Torsion by definition is rotation, not translation. My opinion here, no code to back it up - you should consider it torsionally irregular in both directions. Question 3: Sorry, not a clue. But when in doubt, be conservative. Easier said than done...... :) One last thought - if you are using concrete shearwalls in a C shape be aware that ACI318-99 treats perpendicular joined walls differently. If I remember right (please check this for yourself) ACI318-99 will let you use 15% of the clear height for the maximum flange width to draw forces. For example, on my box above, if the North wall is 50 feet long and 10 feet high I can only use 1.5 feet for my flange width. ACI318-95 gives you different percentages for compression and tension (makes dynamic analysis next to impossible). Best of luck to ya. Jake Watson, E.I.T. Salt Lake City, UT
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- UBC'97, Orthogonal Effects
- From: Alexander Sasha Itsekson
- UBC'97, Orthogonal Effects
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