Finite Element Analysis. Barna Szabó

Finite Element Analysis

Baseline left-parenthesis 0 right-parenthesis"/> computed by the direct method.

which is the exact solution. The choice v equals 1 minus x was exceptionally fortuitous because it happens to be the Green's function (also known as the influence function) for u prime left-parenthesis 0 right-parenthesis . Therefore the extracted value is independent of the solution u element-of upper E Superscript 0 Baseline left-parenthesis upper I right-parenthesis .

Let us choose v equals 1 minus x squared for the extraction function. In this case

u prime left-parenthesis 0 right-parenthesis equals v left-parenthesis x overbar right-parenthesis minus integral Subscript 0 Superscript 1 Baseline u prime v Superscript prime Baseline d x equals StartFraction 15 Over 16 EndFraction plus 2 integral Subscript 0 Superscript 1 Baseline u prime x d x period

Substituting u prime Subscript upper F upper E for u prime :

StartLayout 1st Row 1st Column integral Subscript 0 Superscript 1 Baseline u prime Subscript upper F upper E Baseline x d x equals 2nd Column sigma-summation Underscript i equals 1 Overscript p minus 1 Endscripts StartFraction upper N Subscript i plus 2 Baseline left-parenthesis xi overbar right-parenthesis Over 2 EndFraction StartRoot StartFraction 2 i plus 1 Over 2 EndFraction EndRoot integral Subscript negative 1 Superscript 1 Baseline upper P Subscript i Baseline left-parenthesis xi right-parenthesis StartFraction 1 plus xi Over 2 EndFraction d xi 2nd Row 1st Column equals 2nd Column one fourth sigma-summation Underscript i equals 1 Overscript p minus 1 Endscripts upper N Subscript i plus 2 Baseline left-parenthesis xi overbar right-parenthesis StartRoot StartFraction 2 i plus 1 Over 2 EndFraction EndRoot integral Subscript negative 1 Superscript 1 Baseline upper P Subscript i Baseline left-parenthesis xi right-parenthesis left-parenthesis upper P 0 left-parenthesis xi right-parenthesis plus upper P 1 left-parenthesis xi right-parenthesis right-parenthesis d xi equals negative three thirty-seconds dot EndLayout

Taking the orthogonality of the Legendre polynomials (see eq. (D.13)) into account, the sum has to be evaluated only for p equals 2 . The extracted value of u prime Subscript upper F upper E Baseline left-parenthesis 0 right-parenthesis for p greater-than-or-equal-to 2 is (31.25% error).

An explanation of why the extraction method is much more efficient than direct computation is given in Section 1.5.4.

Exercise 1.16 Find u prime Subscript upper F upper E Baseline left-parenthesis 0 right-parenthesis for the problem in Example 1.7 by the direct and indirect methods. Compute the relative errors.

Exercise 1.17 For the problem in Example 1.9 let v equals 1 minus x cubed be the extraction function. Calculate the extracted value of u prime Subscript upper F upper E Baseline left-parenthesis 0 right-parenthesis for p greater-than-or-equal-to 3 .

Nodal forces

The vector of nodal forces associated with element k, denoted by left-brace f Superscript left-parenthesis k right-parenthesis Baseline right-brace , is defined as follows:

(1.88)

where left-bracket upper K Superscript left-parenthesis k right-parenthesis Baseline right-bracket is the stiffness matrix, left-brace a Superscript left-parenthesis k right-parenthesis Baseline right-brace is the solution vector and left-brace r overbar Superscript left-parenthesis k right-parenthesis Baseline right-brace is the load vector corresponding to traction forces, concentrated forces and thermal loads acting on element k.

The sign convention for nodal forces is different from the sign convention for the bar force: Whereas the bar force is positive when tensile, a nodal force is positive when acting in the direction of the positive coordinate axis.

Exercise 1.18 Assume that hierarchic basis functions based on Legendre polynomials are used. Show that when κ is constant and c equals 0 on Ik then

f 1 Superscript left-parenthesis k right-parenthesis Baseline plus f 2 Superscript left-parenthesis k right-parenthesis Baseline equals r 1 Superscript left-parenthesis k right-parenthesis Baseline plus r 2 Superscript left-parenthesis k right-parenthesis

Geometric representation of exercise 1.8 and provide citation. Notation.

Figure 1.8 Exercise 1.18. Notation.

independently of the polynomial degree pk. For sign convention refer to Fig. 1.8. Consider both thermal and traction loads. This exercise demonstrates that nodal forces are in equilibrium independently of the finite element solution. Therefore equilibrium of nodal forces is not an indicator of the quality of finite element solutions.

1.5 Estimation of error in energy norm

We have seen that the finite element solution minimizes the error in energy norm in the sense of eq. (1.48). It is natural therefore to use the energy norm as a measure of the error of approximation. There are two types of error estimators: (a) A priori estimators that establish the asymptotic

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