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 :
Porosity : 93%
Pores/cm : 4
Purity : 98.5%
Grade : Aluminum 6101
For Copper:
Thickness : 6.35mm
Bulk density :
Porosity : 91%
Pores/cm : 4
Purity : 99.9%

Comments 0

5 Answers

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.

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

Comments 1

do you know how to create in abaqus porous aluminum foam model

Comments 0

i want to draw 2 spheres in porous medium, can anybody help me?

Comments 0