I don't usually give LPS for questions, but that's a doozy. I see a lot of people have written nearly identical papers plotting max stress or eigenvalue(1) as a function of mesh density, I have yet to see a good explanation.

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Greg Locock


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Is it as simple as springs in parallel?

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Greg Locock


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It is explained in various FEM theory texts. And not all element formulations converge from below. In general the element formulations are overly stiff due to approximatations of the displacement field within the element.

Indeed, this is explained in some books about FEM. For example K.J. Bathe ("Finite Element Procedures") proves that the strain energy in FEM solution is always smaller or equal to the strain energy in exact solution.

"not all element formulations converge from below"
"the strain energy in FEM solution is always smaller or equal to"

Only one of those statements can be correct.

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Greg Locock


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Maybe K.J. Bathe was referring only to "standard" FEM formulation without considering some special forms that are exceptions to this rule. Anyway, in the book it is clearly stated that FEM converges from below.

I have always been taught that the solution converges from below as well. It's just not clear why that's the case. I'll check out the Bathe book and see if there is a good explanation.

“The most successful people in life are the ones who ask questions. They’re always learning. They’re always growing. They’re always pushing.” Robert Kiyosaki

Apologies in advance to all FE purists.  My overly simplistic explanation of this is that, in effect, through the interpolation functions etcetera, we are mathematically constraining the element to take up a particular type of shape rather than allowing it to take up the shape it would prefer.  This artificial constraint manifests itself as a stiffening effect on the element.  So our model will be stiffer than it would ideally be, and thus our calculated displacements will be smaller than they ought to be.

How does this affect p-element solvers?

Denial - nicely put. In my mind I had a right angled 2d element, under some loads at each vertex. Now split the hypotenuse and create two new triangles. Under the same loads, to get the same deflection at the vertices of the original triangle, we'd have to supply more forces at the new node if we want the hypotenuse to stay straight.

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Greg Locock


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Sorry for the math-dump but this is perhaps the most concise way to state the answer.

Adapted from Hughes "The Finite Element Method", Chapter 4.

Consider the exact problem written in weak form:

Find u ∈ S such that for all w ∈ V :
a(w, u) = (w, f) + (w, g)_Γ


The approximate finite element problem is:

Find u^h ∈ S^h such that for all w^h ∈ V^h :
a(w^h, u^h) = (w^h, f) + (w^h, g)_Γ

And assume the following:

• S^h ⊂ S
• V^h ⊂ V
• a(·,·), (·,·), and (·,·)_Γ are symmetric and bilinear
• a(·,·) and ||·||_m define equivalent norms on V

Theorem: Let e = u^h-u denote the error in the finite element approximation. Then,

1. a(w^h, e) = 0, for all w^h in V^h
2. a(e, e) ≤ a(U^h-u, U^h-u), for all U^h ∈ S^h

Part (a) means that the error is orthogonal to the subspace V^h ⊂ V, in other words "u^h is the projection of u onto S^h with respect to a(·,·) "

Part (b) is the best approximation property: "there is no member of S^h that is a better approximation to u (with respect to the energy norm a(·,·) ) than u^h "

From here you can prove the corollary:
a(u, u) = a(u^h, u^h) + a(e, e)

And rearranging to
a(e, e) = a(u, u) - a(u^h, u^h)
reveals that the energy of the error equals (minus) the error of the energy

It is also a direct consequence of the corollary that
a(u^h,u^h)≤ a(u, u)

That is that the approximate solution underestimates the strain energy