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# Hex element V/S Tetrahedron elments

Mon, 2009-03-02 02:54 - g_rakesh2

what is the advantage of brick elements (hex element) over tetrahedron Element. Most of Finite element softwares use automatic meshing to create tetrahedron mesh. which is the better brick or tet?

thanks & regards,

Rakesh

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## Comments

## As far as I know there are

As far as I know there are quite robust meshing routines for volumes using tetrahedron elements. Although there are meshing routines for volumes using brick elements, they are not as robust and may not always work for complex volumes. It is for this reason that tetrahedrons are useful.

However, the problem is that traditionally tetrahedrons used as finite elements are not as well behaved as the brick element. Hence, you have to be careful with tetrahedron elements.

Below is a quote from the abstract of Michael Puso's paper "A stabilized nodally integrated tetrahedral", Int. J. Numer. Meth. Engng 2006; 67:841–867, that says it all very well.

"It is well known that linear tetrahedral elements perform poorly in problems with plasticity, nearly incompressible materials, and acute bending. For a variety of reasons, low-order tetrahedral elements are preferable to quadratic tetrahedral elements; particularly for nonlinear problems. But the severe locking problems of tetrahedrals have forced analysts to employ hexahedral formulations for most nonlinear problems. On the other hand, automatic mesh generation is often not feasible for building many 3D hexahedral meshes."

Puso has another more recent paper that also shows how to overcome these problems of tetrahedrals using nodal integration and stabilization techniques. (see, Int. J. Numer. Meth. Engng 2008; 74:416–446)

I hope that helps answer your question.

regards,

Louie

## re:Louie

g_rakesh2

B.E. Mechanical

M-Tech CAD/CAM & AUTOMATION

## thanks

Replying very late sorry for the same but its not clear that which is better? tet or hex elements?

One moe thing I want to ask you between tri elements and quad elements which is better and why?

g_rakesh2

B.E. Mechanical

M-Tech CAD/CAM & AUTOMATION

## Linear Tet & linear triangles are constant strain elements

Hi Rakesh,

Any FE book should tell you that the basic linear triangle and tetrahedron elements are constant strain elements (Displacement interpolation is linear, and hence the strains/stresses are constant in any element). Obviously, you will need an extremely fine mesh near the locations where stress/strain gradients are present, with these elements. So the performance of bilinear rectangles or trilinear brick elements are expected to be much better compared to these.

However, if you go for the more advanced elements (higher order, stabilized, reduced integration, hybrid etc.), the comparison may not be so straight forward.

Jayadeep

## Tetrahedral elements and Brick elements

Hello Rakesh,

The constant Strain Tetrahedron has 12 degrees of freedom (dof) but the brick element has 24 dof. The other thing is that in general you need to use Gaussian integration points for the brick element to calculate the terms of your matrices and vectors which takes more cpu time. Approximately, it takes the same time for the curved brick elements. Before 2013, I have not found a valid method to calculate the shape functions of a tetrahedron. Some existing formula give negative volumes !

The new method that I have developed in 2013 computes the shape functions of the constant strain tetrahedron efficiently with direct matrix solving and after that the corresponding matrices (stiffness, .......) with direct matrix products : see my recent publication :

" Dynamic Detection of Reinforced Concrete Bridge Damage by Finite Element Model Updating ", Mohammed Lamine Moussaoui, Mohamed Chabaat and Abderrahmane Kibboua, Journal of Mechanics Engineering and Automation, Vol 04, Number 1, PP 40-45, January 2014, David Publishing Company, USA

http://www.davidpublishing.com

It is also important to notice that the choice of the tetrahedron corresponds to that the tetrahedral elements fit very well any arbitrary shaped volume, eventually with curvatures, to be analyzed. This is the economic one.

Best regards