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Is this the right way to calculate volume fraction?

Is this the right way to calculate volume fraction?

Is this the right way to calculate volume fraction?

(OP)
I exploit the symmetry of fig.1 to simplify it to fig.2. This figure has a volume fraction of 0.2, and I need to find a relation between t and L. Have I done it right?


Volume of cube in the middle: (L-2*t)^3

And 4 rectangular plates around the cube in the middle: t*(L-2*t)^2

Volume of fig 2 is then: L^3 – ((L-2*t)^3 + t*(L-2*t)^2)

So volume fraction is that volume divided by the unit cell cube volume, giving:

(L^3 – ((L-2*t)^3 + t*(L-2*t)^2))/L^3 = 0.2

Which gives the relation:

t = 0.0428787 L

RE: Is this the right way to calculate volume fraction?

Figure 2 is not a unit cell, in that you cannot stack those next to each other to produce figure 1.

The corners of the unit cell are in the center of where the rods intersect.

Also, your last equation cannot be close to correct because the relationship between t and L is not a constant. L and t can be any two numbers, as long as L>2t.

RE: Is this the right way to calculate volume fraction?

He's given a fixed volume fraction, though. Nevertheless, the final equation is wrong for another reason, namely, the numerator should only contain the volume associated with the beams. Something like below. There are 12 beams and 8 corner cubes



TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg
FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers Entire Forum list http://www.eng-tips.com/forumlist.cfm

RE: Is this the right way to calculate volume fraction?

"(L.2t)" should be (L-2t)

L is CL to CL and t is the width of an arm, and the arms are in the edges of the cube, then the empty central cube is (L-t)^3 ... t = 2*(t/2)
and there are 6 side (face) voids (L-t)^2*(t/2) ...
(L-t)^3+6*((L-t)^2*(t/2)) = 0.8*L^3
(L-t)^2*((L-t)+3t) = 0.8*L^3 or
(1-(t/L))^2*(1+2*(t/L)) = 0.8
t/L = 0.287141
(if an arm is 2t wide, then t = 0.14357

or the filled space is 20% ...
4*L*(t/2)^2 + 8*(L-t)*(t/2)^2 = 0.2*L^3
(t/L)^2 *(1+2*(1-(t/L))) = 0.2

if the arms are central to the cube (but then there's on "central void") ...
3*L*t^2 - 2*t^3 = 0.2*L^3
3(t/L)^2 -2(t/L)^3 = 0.2
huh!, exactly the same ...


another day in paradise, or is paradise one day closer ?

RE: Is this the right way to calculate volume fraction?

so you're defining the arms as 2t wide ?
OP's sketch defines as "t" wide.

another day in paradise, or is paradise one day closer ?

RE: Is this the right way to calculate volume fraction?

ok, but you can see the confusion.

for my money the interesting thing was modelling as a central stiffener gave the same result and a simple neat expression ...
3(t/L)^2 -2(t/L)^3 = 0.2

another day in paradise, or is paradise one day closer ?

RE: Is this the right way to calculate volume fraction?

Actually, the OP's image shows his "unit cell," wherein I presumed that the "t" was one-half of the thickness of the beams in the full-up structure, so I was simply following his lead.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg
FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers Entire Forum list http://www.eng-tips.com/forumlist.cfm

RE: Is this the right way to calculate volume fraction?

yeah, but he shows the width of a leg being "t".

another day in paradise, or is paradise one day closer ?

RE: Is this the right way to calculate volume fraction?

to the OP ...

"Volume of cube in the middle: (L-2*t)^3" ... no, if L is CL to CL then it's L-2(*(t/2) = (L-t)^3

"And 4" ... no, there are 6 (one at each face of the cube) ...
"rectangular plates around the cube in the middle: t*(L-2*t)^2" ... no, t/2*(L-t)^2

another day in paradise, or is paradise one day closer ?

RE: Is this the right way to calculate volume fraction?

The unit cell would have the legs "t" being half the width of the full structure's, I think, since one would slice the structure right down middle of the width of each full sized leg

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg
FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers Entire Forum list http://www.eng-tips.com/forumlist.cfm

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