Why is Finite Element Analysis an Upper Bound Method 1

MegaStructures · Apr 23, 2021

Here's a pretty easy one I can't wrap my head around:

For finite element analysis of a mechanical stress problem why does deflection and stress always increase towards an exact solution?

I understand that the behavior of elements in between nodes are approximated by linear or quadratic functions and cannot exactly represent an irregular shape, so finite elements must be made smaller until the deflection profile required between the nodes can adequately represented by these shape functions; however, I don't understand why this increasingly more accurate shape approximation always produces higher stress/deflections.

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GregLocock · Apr 23, 2021

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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GregLocock · Apr 23, 2021

Is it as simple as springs in parallel?

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SWComposites · Apr 23, 2021

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.

FEA way · Apr 24, 2021

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.

GregLocock · Apr 24, 2021

"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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FEA way · Apr 24, 2021

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.

MegaStructures · Apr 24, 2021

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.

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Denial · Apr 28, 2021

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.

3DDave · Apr 28, 2021

How does this affect p-element solvers?

GregLocock · Apr 29, 2021

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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Coreform_Greg · Apr 29, 2021

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:
[ul]
[li]S^h ⊂ S[/li]
[li]V^h ⊂ V[/li]
[li]a(·,·), (·,·), and (·,·)_Γ are symmetric and bilinear[/li]
[li]a(·,·) and ||·||_m define equivalent norms on V[/li]
[/ul]

Theorem: Let e = u^h-u denote the error in the finite element approximation. Then,
[ol a]
[li]a(w^h, e) = 0, for all w^h in V^h[/li]
[li]a(e, e) ≤ a(U^h-u, U^h-u), for all U^h ∈ S^h[/li]
[/ol]

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

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Why is Finite Element Analysis an Upper Bound Method 1

MegaStructures

Structural

GregLocock

Automotive

GregLocock

Automotive

SWComposites

Aerospace

FEA way

Mechanical

GregLocock

Automotive

FEA way

Mechanical

MegaStructures

Structural

Denial

Structural

3DDave

Aerospace

GregLocock

Automotive

Coreform_Greg

Mechanical

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