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Statics Help

Statics Help

Statics Help

I have a very simple question that I believe anyone can figure out, except me. Thanks for help.

I was studying basic earthquake design by chapter 11 of Concrete Design Handbook by the CPCA. In the symmetric building, the author talks about eccentric loads. The torsional eccentricity moment is 3002.7 kNm
The building is an 7 - 6m bay symmetric frame (8 rows of columns), totalling 42 m. How do I find the shear force in the column located in the first frame. The author says it is moment/72=42kN
Anyone knows where this 72 comes from?
Thanks a lot.

RE: Statics Help

You need a 3-D analysis for such a frame structure and the shear forces at any level can be read from the computer output.
does that help?

RE: Statics Help

Thanks, but I just want to go through the hand calc's and understand the statics behind that number. Anybody owns the COncrete Design Handbook by the Canadian Portland Cement Association?

RE: Statics Help

Inthe assumption of rigid behaviour and for equal stiffness of the frames...

you port the eccentric shear to cog=center of torsion for this case and first allocate its equal share to every frame.

Then a number of pairs of linearly growing from the center shears at the frames need be in equilibrium with the moment, which will give you the worse shear forces from moment coming from eccentricity in the extreme frames.

Then you add the equitative shear with that of moment.

But of course this is no substitute for 3D analysis in that this can capture more realistic allocation of the shears and hopefully more reliably solve more complicated cases.

RE: Statics Help


I think the 72 value used in the formula is the section modulus of the 8 columns:

2 @ 3^2    =     18 col-m^2
2 @ 9^2    =    162
2 @ 15^2   =    450
2 @ 21^2   =    882
   I       =   1,512 col-m^2

   S       =   1,512/21  =  72 col-m

Hope this will help


RE: Statics Help


I was about to do an identical calculation to dlew; I am confident that is what the author was intending.

BUT.. The calulation is correct in finding the component of shear force due to the eccentric moment, but it is not the TOTAL shear in the column.  To get the total shear in the column you have to add the 42 kN to the average shear per column (horizontal load /8).

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