## Why is Finite Element Analysis an Upper Bound Method

## Why is Finite Element Analysis an Upper Bound Method

(OP)

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

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.

“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

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.

“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

## RE: Why is Finite Element Analysis an Upper Bound Method

Cheers

Greg Locock

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## RE: Why is Finite Element Analysis an Upper Bound Method

Cheers

Greg Locock

New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?

## RE: Why is Finite Element Analysis an Upper Bound Method

## RE: Why is Finite Element Analysis an Upper Bound Method

## RE: Why is Finite Element Analysis an Upper Bound Method

"the strain energy in FEM solution is always smaller or equal to"

Only one of those statements can be correct.

Cheers

Greg Locock

New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?

## RE: Why is Finite Element Analysis an Upper Bound Method

## RE: Why is Finite Element Analysis an Upper Bound Method

“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

## RE: Why is Finite Element Analysis an Upper Bound Method

## RE: Why is Finite Element Analysis an Upper Bound Method

## RE: Why is Finite Element Analysis an Upper Bound Method

Greg Locock

## RE: Why is Finite Element Analysis an Upper Bound Method

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

exact problemwritten in weak form:Find

∈usuch that for allS∈w:Va(

,w) = (u,w) + (f,w)g_{Γ}The approximate finite element problem is:

Find

∈u^{h}such that for allS^{h}∈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_{Γ}are symmetric and bilinear_{m}define equivalent norms onVTheorem:Letdenote thee = u^{h}-uerrorin the finite element approximation. Then,- a(
- a(

Part (a) means that the error is orthogonal to the subspace,w^{h}) = 0, for alleinw^{h}V^{h},e) ≤ a(e-U^{h},u-U^{h}), for allu∈U^{h}S^{h}⊂V^{h}, in other words "Vis the projection ofu^{h}ontouwith respect to a(·,·) "S^{h}Part (b) is the best approximation property: "there is no member of

that is a better approximation toS^{h}(with respect to the energy norm a(·,·) ) thanu"u^{h}From here you can prove the corollary:

a(

,u) = a(u,u^{h}) + a(u^{h},e)eAnd rearranging to

a(

,e) = a(e,u) - a(u,u^{h})u^{h}reveals that

the energy of the error equals (minus) the error of the energyIt is also a direct consequence of the corollary that

a(

,u^{h})≤ a(u^{h})u,uThat is that the approximate solution underestimates the strain energy