Some times tetrahedral mesh is better and some times hexahedral mesh elements are approximate the material in better way. I want to know the conditions. Advantages and disadvantages of these type of solid 3D elements.
Tetrahedral elements can fit better complex geometry. However, when you integrate the shape functions with points of Gauss it is less accurate than hexahedral elements. In addition, one of the factors that determines the quality of your mesh is the distortion of your elements. The reason for this lays on the mapping from real to natural space of integration. To sum up, if your geometry is simple, the best option is to mesh it with hexahedral elements. If it is not possible (curved geometries, accute angles or similar) then go with tetrahedal but controlling the distortion of the elements.
Tetrahedral elements can fit better complex geometry. However, when you integrate the shape functions with points of Gauss it is less accurate than hexahedral elements. In addition, one of the factors that determines the quality of your mesh is the distortion of your elements. The reason for this lays on the mapping from real to natural space of integration. To sum up, if your geometry is simple, the best option is to mesh it with hexahedral elements. If it is not possible (curved geometries, accute angles or similar) then go with tetrahedal but controlling the distortion of the elements.
Hexahedra meshes are economic with the number of elements because the same degrees of freedom (or for 8 nodes) for one Hexaedron corresponds to six Tetrahedra. It is obvious that increasing the number of elements will not increase the size of the global finite element matrices but the computations for one hexahedron are generated also for six tetrahedra. This step has to be compared in cpu time in order to state if it is interesting to use hexahedra than constant strain tetrahedra knowing that curved or linear hexahedra use Gauss intergration points to generate the element characteristics (stiffness, mass, ..) and tetrahedra use exact formula without any integration to get the same characteristics. Researchers have always used tetrahedra elements because they fit very well arbitrary shaped geometries with their simple computations.
For the same degree of polynomial the finite element space generated by hexahedral elements is more rich than the space generated by tetrahedral elements. However the tetrahedral elements are best to model complex geometry domain with little distortion of mesh. Moreover, the computational cost for assembling the global stiffness matrix for tetrahedral elements is lower because there is not necessary numerical integration.
1) You can modify a Hex mesh much easier by moving a line of nodes. So if you want to make a wall thinner or something shorter, it is easier.
2) You need less Hex8s to get the same accuracy as with Tet10s. And you should never model with Tet4s unless you need to attach to quad4s and then you should use a lot of tet4s otherwise your mesh will be too stiff.
Try all of these elements with a simple cantilever beam problem for which there is a closed form solution. See how many or how few of each element type you need to get the closed form solution.
How come no one is talking about the incompressibility issue with tetrahedral elements!? Yes, they fit the geometry and all really well, but what do you do when the poissons ratio of the material (lets say, like Rubber) approaches 0.5?
Volumetric locking takes place, and tet elements start acting too stiff and underestimate the stresses/strains very much!
Am I the only one who thinks this is the MAIN problem with tet meshes? Please correct me if I am wrong.
I have a question about meshing an element in abaqus, I am wondering if it is possible that we use hex for some part of the model and tet for another parts of it?