Determining the Centroid of a 3D Section

[CivilSigma]

Homework Statement

[/B]
I am having a problem understanding a calculation performed as part of a bigger solution in the design of Slabs.

That is, how to determine the centroid of the critical shear section, which consist of 3 planes intersecting to form a 3D model (please see attached picture). For this particular case, there is no fourth shear plane that encloses with the other planes. So, the 3D figure is "open" on one side.

So excluding details of design, I am having trouble understanding how the authors compute the shear centroid which is just the location of the centroid on the top.

Homework Equations

Please see attached picture.
https://imgur.com/a/dxwWD

The author defines 'e' as the location of the centroid of critical shear, and the value of e is:
$$e=\frac{805^2}{2 \cdot 805+610}$$

Can one please explain the above formula? I presume the denominator is the perimeter of the shape, but I cannot figure out what the numerator represents.


Thank you.

 

[Dr.D]

Define the two long parallel legs as length L, and the short connecting legas W. Then the total length is
Ltotal = 2*L + W
Now, calculate moments about the short edge:
Mom = 2 * int(x dx, x = 0 .. L) = 2 * (x^2/2) between 0 and L = 2 * (L^2/2) = L^2
By definition, the centroid is such that
Mom = e * Ltotal or
e = Mom/Ltotal =L^2/(2*L+W)
Then plug in the numbers.

 

[CivilSigma]

Te moment calculations make sense, thank you.

But why is the definition Moment = e * Perimeter ? This seems odd to me, can you please explain it further?

Can you please recommend some articles to read , I'm not sure what to Google as searching for centroid calculation gives me the concepts in 2D which I am familiar with.

 

[Dr.D]

Re-think this as a center of mass calculation. Assign a value mu = mass/length that is the same for all three sides.
The first step is the calculation of the total mass
Mtotal = mu*(2*L+W)
Now calculate the moment of the mass distribution,
Mom = 2 * int(mu*x dx, x = 0 .. L) = mu*2 * (x^2/2) between 0 and L = mu*2 * (L^2/2) = mu*L^2
Finally, make the CM location calculation by
e = Mom/Mtotal and note that mu is in both numerator and denominator and thus drops out.

 

[CivilSigma]

I am trying to follow this, but I have some more questions, if I may:

1. Where does the 2 come from in the integral ? Were the integral bounds -L to L ? If that's the case, can you please clarify the axis location?

2. By assigning mu = mass / length, are we treating the 3D plane as a line surface in 2D. And why is mu the same for all three edges? Is it because the material is uniform, therefore the "density" mu (but mu has units [Mass] / [ Length] not [Length ^3] ) is therefore equal per edge?

3. If we repeat the same calculation but now consider a fourth plane near the free end, then e = 805/2 =402.5. But when I try this using the above method,
e = L^2 / (2L+2W) = 805^2 / (2*805+2*610) = 229 ?

 

[Dr.D]

I think maybe it is time for me to let someone else try to explain this to you.

 

[CivilSigma]

I think maybe it is time for me to let someone else try to explain this to you.

Am I really that off on all 3 points?

 

[CivilSigma]

Please Dr. D, if you can just provide me with a reference to read or to research, I would appreciate it.

I need to figure this out to complete the problem I am working on

 

[Dr.D]

I really do not think you need a reference. I think that what is needed is for YOU to THINK about this simple problem. Start with the definition of a centroid and work from there, forgetting all reference to formulae, examples, etc. How would YOU solve this problem? If you don't know the definition of a centroid, look it up (the 'net is your friend).

 

[CivilSigma]

I really do not think you need a reference. I think that what is needed is for YOU to THINK about this simple problem. Start with the definition of a centroid and work from there, forgetting all reference to formulae, examples, etc. How would YOU solve this problem? If you don't know the definition of a centroid, look it up (the 'net is your friend).

Ha! I get it now !
I really needed to think about it :)

What confused me was that I was trying to find the centroid of the plane in 2D that lays on top.

But now I understand that the "shape" we have is simply of 3 sides folded. Then the centroid of that section is:

$$ \bar{y} = \frac{\int ydA}{\int dA}$$

$$\bar{y} = \frac{ 2* \cdot \int_0^L y \cdot depth \cdot dy + 0} {2L \cdot depth + W \cdot depth}$$

2 times the integral since we have two identical side faces
and the third face's moment is 0 because I am taking moments about that axis
Depth factor cancels out to give us the location of the centroid from the short axis as:

$$\bar{y} = \frac{L^2}{2L+W}$$


Thank you again Dr.D !