Centroidal axis of Shear Perimeter 2

[XinLok]

I am reading one example about get the location of centrical axis of shear perimeter.

In this example, they are telling the distance from line A-B to the centroid of the shear perimeter is as below:

I am not understanding how this been calculated, can anyone support please???

How to get the location of Centroid axis of shear perimeter please

 

[Celt83]

Reference Appendix C of ACI 421.2R-20 or Chapter 5 of Structural Design Guide to the ACI Building Code, 4th Edition by Hoffman, Gustafson, and Gouwens

 

[Luceid]

It's basically a normal centroid of area calculation - just with the areas going in and out of the page.

In this case - working backwards it seems like the effective depth of the slab "d" is 32.5 inches

Area of Side BC = 32.25*32.5
Area of Side AB = 56.5*32.5
Area of Side AD = 32.25*32.5

Measuring from line AB as x=0, to the left, the centroid of each area is:

BC: -32.25/2
AB: 0
AD: -32.25/2

So your sum of A*y divided by sum of A is

[ (32.25*32.5)*(-32.25/2) + (56.5*32.5)*0 + (32.25*32.5)*(-32.25/2) ] / [ (32.25*32.5) + (56.5*32.5) + (32.25*32.5) ]

Which simplifies down to

[ 2*(32.25*32.5) * (-32.25/2) ] / [ 32.5 * (32.25+56.5+32.25) ] which hopefully equals -8.6", measured from the A-B line

 

[XinLok]

Well explained Luceid.
Thank you very much.