1.6 A beautiful idea that does not work

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• An FE solution cannot be improved on the same mesh.

Once it is understood that the error of an FE solution can be traced back to

deviations in the load, could the situation not be improved by applying the

residual forces p − ph, solving this load case again with finite elements, and

repeating this loop as long as the error is greater than a preset error margin

ε?

This idea does not work, because the residual forces

16 1 What are finite elements?

Fig. 1.11. The FE solution

of this load case is zero

r = p − ph (1.51)

leave no traces on the mesh, i.e., all the equivalent nodal forces fr

i vanish,

fr

i =

_ l

0

pϕi dx −

_3

j=1

fj · ϕi(xj) = fi − fi = 0 for all ϕi , (1.52)

so that the rope will not deflect, because zero nodal forces mean zero deflection:

Ku = 0 ⇒ u = 0 . (1.53)

This riddle is easily solved by recalling that the exact curve w is projected

onto the trial space Vh. But because the error w − wh is orthogonal (in the

energy sense) to the space Vh,

_ l

0

H(w

_ − w

_

h) ϕi

_

dx = 0 for all ϕi , (1.54)

it casts no shadow, i.e., e = 0.

It follows that there are load cases which cannot be solved on an FE mesh

(see Fig. 1.11) namely all load cases where the load p is so arranged that it

contributes no work. This is the case if all equivalent nodal forces fi are zero:

fi = δWe(p, ϕi) = 0, i= 1, 2, . . . n . (1.55)

Loads that happen to be parallel to the line of sight have a “null shadow”.

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