## 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

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?

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?

TTFN (ta ta for now)

I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg

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

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?

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?

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?

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?

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?

I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg

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

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

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

"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?

I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg