Finite Element Analysis. Barna Szabó

Finite Element Analysis

Baseline a Subscript j Baseline b Subscript i Baseline identical-to StartSet b EndSet Superscript upper T Baseline left-bracket upper K Superscript left-parenthesis k right-parenthesis Baseline right-bracket StartSet a EndSet period"/>

The terms of the stiffness matrix k Subscript i j Superscript left-parenthesis k right-parenthesis depend on the the mapping, the definition of the shape functions and the function kappa left-parenthesis x right-parenthesis . The matrix left-bracket upper K Superscript left-parenthesis k right-parenthesis Baseline right-bracket is called the element stiffness matrix. Observe that k Subscript i j Superscript left-parenthesis k right-parenthesis Baseline equals k Subscript j i Superscript left-parenthesis k right-parenthesis Baseline comma that is, left-bracket upper K Superscript left-parenthesis k right-parenthesis Baseline right-bracket is symmetric. This follows directly from the symmetry of upper B left-parenthesis u comma v right-parenthesis and the fact that the same basis functions are used for un and vn.

In the finite element method the integrals are evaluated by numerical methods. Numerical integration is discussed in Appendix E. In the important special case when kappa left-parenthesis x right-parenthesis equals kappa Subscript k is constant on Ik, it is possible to compute left-bracket upper K Superscript left-parenthesis k right-parenthesis Baseline right-bracket once and for all. This is illustrated by the following example.

Example 1.3 When kappa left-parenthesis x right-parenthesis equals kappa Subscript k is constant on Ik and the Legendre shape functions are used then, with the exception of the first two rows and columns, the element stiffness matrix is perfectly diagonal:

(1.66)

Exercise 1.8 Assume that kappa left-parenthesis x right-parenthesis equals kappa Subscript k is constant on Ik. Using the Lagrange shape functions displayed in Fig. 1.3 for p equals 2 , compute k 11 Superscript left-parenthesis k right-parenthesis and k 13 Superscript left-parenthesis k right-parenthesis in terms of κk and ℓk.

Computation of the Gram matrix

The second term of the bilinear form is also computed as a sum of integrals over the elements:

(1.67)

We will be concerned with evaluation of the integral

StartLayout 1st Row 1st Column integral Subscript x Subscript k Baseline Superscript x Subscript k plus 1 Baseline Baseline c left-parenthesis x right-parenthesis u Subscript n Baseline v Subscript n Baseline d x equals 2nd Column integral Subscript x Subscript k Baseline Superscript x Subscript k plus 1 Baseline c left-parenthesis x right-parenthesis left-parenthesis sigma-summation Underscript j equals 1 Overscript p Subscript k Baseline plus 1 Endscripts a Subscript j Baseline upper N Subscript j Baseline right-parenthesis left-parenthesis sigma-summation Underscript i equals 1 Overscript p Subscript k Baseline plus 1 Endscripts b Subscript i Baseline upper N Subscript i Baseline right-parenthesis d x 2nd Row 1st Column equals 2nd Column StartFraction script l Subscript k Baseline Over 2 EndFraction integral Subscript negative 1 Superscript plus 1 Baseline c left-parenthesis upper Q Subscript k Baseline left-parenthesis xi right-parenthesis right-parenthesis left-parenthesis sigma-summation Underscript j equals 1 Overscript p Subscript k Baseline plus 1 Endscripts a Subscript j Baseline upper N Subscript j Baseline right-parenthesis left-parenthesis sigma-summation Underscript i equals 1 Overscript p Subscript k Baseline plus 1 Endscripts b Subscript i Baseline upper N Subscript i Baseline right-parenthesis d xi period EndLayout

Defining:

(1.68) m Subscript i j Superscript left-parenthesis k right-parenthesis Baseline equals StartFraction script l Subscript k Baseline Over 2 EndFraction integral Subscript negative 1 Superscript 1 Baseline c left-parenthesis upper Q Subscript k Baseline left-parenthesis xi right-parenthesis right-parenthesis upper N Subscript i Baseline upper N Subscript j Baseline d xi

the following expression is obtained:

(1.69)

where StartSet a EndSet equals left-brace a 1 a 2 ellipsis a Subscript p Sub Subscript k Subscript plus 1 Baseline right-brace Superscript upper T , StartSet b EndSet Superscript upper T Baseline equals left-brace b 1 b 2 ellipsis b Subscript p Sub Subscript k Subscript plus 1 Baseline right-brace and

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