SHELL MESHING DOUBTS

[analyst89]

Why do we use quad and trias in shell meshing for automotive analysis? why not other 2-D elements?

 

[rb1957]

'cause they account for 99.99999% of 2D elements.

there is a tiny population of non-quad elements, used for transition. but you can mesh any surface with tris (remember tri3s "suck", tri6s are much better) and most surfaces with quads (the fourth node on the quad can create warped elements, but if this isn't excessive it isn't a problem).

 

[analyst89]

thanks, but why there is no use of penta 2-D elenments in shell mesh?

 

[jhardy1]

Yeah, if you had access to pentagonal and hexagonal 2D elements, it would be SOOOOO much easier to model and analyse soccer balls and buckyballs!

Why stop at pentagons? Why not heptagons, octagons, dodecagons, tetradecagons, heptadecagons, ...

In the absence of pentagonal and hexagonal elements etc, I guess you will just have to mesh with TRIs and QUADs. You can make a pentagon out of 5 triangles (or if you don't want to generate an extra internal node, try three triangles, or one quad and a triangle), while there are several ways to make a hexagonal "element" (such as 6 triangles, one quad and two triangles, two quads and two triangles, ...)

http://julianh72.blogspot.com

 

[rb1957]

'cause it's easy to relate stress and strains inside a quadrilateral.

why a pentagon ? it's harder to have 5 points on a plane than 4.

of course you could code for yourself a generally spherical 5 noded pentagon element

 

[bgatto]

The open-source FEA software called "Z88" uses kinda 20 types of elements (2D and 3D). You can download the manual and read more about these elements, maybe you will find the answer of your question there.

This is the direct link to the manual:

http://www.z88.uni-bayreuth.de/z88v13/download/z88mane.pdf

 

[ThisIsMe86]

Hi Analyst89,

I am stretching my logical reasoning capability here. So I might be wrong.

The reason i think is because of the node count. If you refer back to the theory of FEA, you end up solving matrices for an element of 2 nodes (that is the starting point FEA theory that everyone studied I guess). While doing so, we combine matrices for both the nodes and endup with a bigger matrix. This should mean the higher the number of nodes, the bigger is our stiffness matrix and hence more complications and solution time.

Keeping that in mind, the least number of nodes that we can use in 2D is 3. But, tria's are not recommended for analysis for well known reasons. So the next available option is quad (4 nodes).

to all: Please feel free to correct me if I am wrong. I wrote this by combining everything I have learnt (which is just a drop in ocean)in FEA.


- This is me

 

[rb1957]

this_is_me ...
i don't think so. a model of 4 5-noded 2d shells would have the same number of freedoms as 5 4-noded shells ... how they relate to one another would (of course) be different.

 

[mcherry4]

Correct me if I'm wrong, but it seems to me that higher numbers of nodes leads to less sparse stiffness matrices. If you consider the amount of shape functions that contribute to the nodal displacement at a specific node, it can get pretty substantial. Then again, if you're trying to represent a very complex surface, the savings in the reduced number of elements might be a good trade off.