Recent content by damascus

This question is vague, but this is how I would interpret it if force to solve it: 8' X 200PLF = 1600# (since why else would they give you length of wall and linear #/ft) Then its just a force triangle, 8'/1600# = 17'/X X = 3400# Which is pretty much what you already did. But this is a bad...

If it helps I think the answer on the right is the better configuration.

J = Summation of all walls being considered  ((distance of wall to the center of rigidity)^2 X rigidity of wall)  J = Summation: r^2 x R Your example all the distances are 5'  and all four walls have same rigidity = 1. J = 4(2.5^2 x 1) or 4(7.5^2 x 1) They are both symmetrical with 4 walls...

I have a educated guess, and I have a hunch it has to do with torsion. I am guessing if you said all the walls have same rigidity, center of mass, and center of gravity then you would get a different J (torsional rigidity). The higher the J is the lower the shear due to torsion is for each...

When I passed the vertical, it was released on same day as PE. But it probably varies with each cycle. 

Thanks for the response, follow up question: If you have a 4 bay frame with 2 brace frames (both 20') and considered to have the same rigidity are the lateral loads split evenly between them.  Therefore each would take 5k of the total 10k of the last problem. And if the frames are different...

Quick question on horizontal forces in collector. Given: 3 bay frame, middle bay is steel brace frame, each bay is 20' long, lateral force is 10k at this frame (wind or seismic) Vs= 10k/60' = 0.166 k/ft (force through diaphragm) Vr=10k/20' = 0.5 k/ft (brace frame resisting) Difference =...