How to model A porous media in solidworks?
I want to model a simple porous metal with porosity 93% and 91%.
Can anybody help me?
(aluminium foam 93%porosity)
(copper foam 91%porosity)
*Size 150mm x150mm and thickness 6.35mm
*Solidwork software/ Ansys Fluent
For Aluminium:
Thickness : 6.35mm
Bulk density : 0.2g.cm-3
Porosity : 93%
Pores/cm : 4
Purity : 98.5%
Grade : Aluminum 6101
For Copper:
Thickness : 6.35mm
Bulk density : 0.8g.cm-3
Porosity : 91%
Pores/cm : 4
Purity : 99.9%
5 Answers
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1.- make a rectangle of 150 x 150 size in the horizontal plane.
2.- extrude it at 6,35 mm of height.
3.- draw a circle in the box superior corner, of a diameter minor than 2 (you can calculate an initial value of this diameter of the hole, as I will explain to you at the end of this comment).
4.- draw a line in the middle of the circle
5.- cut or trim the middle of the circle
6.- make a cut by revolution (revolve) the semicircle around the middle line, obtaining a spherical hole
7.- array a matrix in the x and y direction each 2 mm by 2 of separation, along all the block
8.- array the matrix in the z axis to obtain the foam
9.- go to calculate, properties and select the solid, in edit, input the density properties and then click in recalc.
10.- take a look of the values
11.- resize the diameter of the spherical hole to go a better values in the properties until you get the right numbers you looking for.
12..- safe the objet as a .sat or .igs objet and then you can open your objet in ansys to make it an analysis.
HOW TO CALCULATE THE INITIAL VALUE OF THE CIRCLE DIAMETER:
You can calculate the radius of the spherical hole if you 1rst make a 10x10x6.35 box and count how much complete spherical holes have the section box. If each hole is separated each 2mm (because the premise of 4 hole each centimeter), then the 10x10x6.35 piece have 86 holes.
The volume of the 10x10x6,35 is equal to 635mm^3.
As we know, the 93% of the volume is air. Then, the 7% of 635 is metal. Solving that, you get: 635x0.07=44.45mm^3 of metal and 635x0.93=590.55mm^3 of air.
By the division of that air volume with the number of holes by that representative piece, you get: 590.55/86=6.8669mm^3 by each spherical hole.
By the other hand, you know that the volume of a sphere is V=4/3*phi*r^3
Then: 6.8669=4/3*phi*r^3
so: r=0.5464mm.
Use so the diameter of 1.0929mm by each hole so you get a nice initial aproximation to the 93% of the porosity
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