aaput
Materials
- Aug 17, 2011
- 3
Hello,
I kindly ask for a tip. On the attached image we can see points A, B, C, D being vertices of a regular tetrahedron. In fact these points denote the center of mass of respective rigid body. Point E is a small rigid body (relative to the others) playing role of the center of mass of whole tetrahedron.
So, the 5 points are rigid inclusions in the elastic matrix. My goal is to force the points to deform in particular manner. Namely, the tetrahedron has to remain regular. So the transformation has to be isometric one.
My first idea was to formulate MPC constraints which I did with a little help from Mathematica. Because Abaqus in this case uses elimination method for constraints I was not able to control the distances (BC=AB, CD=AB, DA=AB). This if why I introduced the point E.
I thought that good idea was to extract change in length and rotation from deformation of pair AE and then apply the transformations to the remaining pairs. Of course E would still be center of the new tetrahedron.
But this approach seems to fail in the case of small displacements probably due to numerical errors. Moreover it was hard to obtain compact form of the combination of a few affine transformations (translations and rotation) and the simplifications of the formulas took ages as well.
Then I realized that CYLINDRICAL connector section is a good candidate. This constraint applied to pairs AE,...,DE keeps the angles but of course not the lengths.
Finally, is it possible to extract the length of deformed "wire" AE and tell the others to be the same length simultaneously being constrained by mentioned cylindrical connector? There is still one dependent DOF in each BE,CE,DE.
Any other ideas of isometric transformation appreciated
thanks in advance
Artur
PS. in two-dimensional world (with rigid inclusions placed in the vertices of a triangle) approach "equal lengths" works fine, because we deal with 2 dimensions and 2 equations.
I kindly ask for a tip. On the attached image we can see points A, B, C, D being vertices of a regular tetrahedron. In fact these points denote the center of mass of respective rigid body. Point E is a small rigid body (relative to the others) playing role of the center of mass of whole tetrahedron.
So, the 5 points are rigid inclusions in the elastic matrix. My goal is to force the points to deform in particular manner. Namely, the tetrahedron has to remain regular. So the transformation has to be isometric one.
My first idea was to formulate MPC constraints which I did with a little help from Mathematica. Because Abaqus in this case uses elimination method for constraints I was not able to control the distances (BC=AB, CD=AB, DA=AB). This if why I introduced the point E.
I thought that good idea was to extract change in length and rotation from deformation of pair AE and then apply the transformations to the remaining pairs. Of course E would still be center of the new tetrahedron.
But this approach seems to fail in the case of small displacements probably due to numerical errors. Moreover it was hard to obtain compact form of the combination of a few affine transformations (translations and rotation) and the simplifications of the formulas took ages as well.
Then I realized that CYLINDRICAL connector section is a good candidate. This constraint applied to pairs AE,...,DE keeps the angles but of course not the lengths.
Finally, is it possible to extract the length of deformed "wire" AE and tell the others to be the same length simultaneously being constrained by mentioned cylindrical connector? There is still one dependent DOF in each BE,CE,DE.
Any other ideas of isometric transformation appreciated
thanks in advance
Artur
PS. in two-dimensional world (with rigid inclusions placed in the vertices of a triangle) approach "equal lengths" works fine, because we deal with 2 dimensions and 2 equations.