# Why do the shape of finite elements matter?

I have used FEA for a couple of years now, but using it and using it correctly are two different things, safety factor is not the solution to everything. I have the feeling I won't be using it right unless I have a clear answer to that question:

I am aware elements must be close to their ideal shape (based on the Jacobian only?) in order to get accurate results.. But why? Since I understand it comes from a coordinate transform, unless two vectors of the element become colinear shouldn't the results be accurate no matter its shape?

A step-by-step answer based on an illustrated example (arbitrary stress distribution) would be ideal, especially given that it is a relatively common question (but never well answered from what I have seen).

• P.S: This is a "repost" of the Physics stackexchange, I had initially posted it in the wrong one. – Mister Mystère May 4 '14 at 14:27

You're right, the Jacobian would be invertible unless you have colinear/coplanar elements, i.e., degenerate elements. But this is the case when you have exact arithmetic. In real FEM programs you use floating point arithmetic and then, you can have Jacobians that are nearly zero. Probably, you can still invert this $2\times2$ or $3\times3$ matrix, but the error will accumulate in your model as a whole. Finally... getting an ill-conditioned matrix.