Finite Element Analysis. Barna Szabó

Finite Element Analysis

multiplying eq. (1.133) by a test function v element-of upper E Superscript 0 Baseline left-parenthesis upper I right-parenthesis

and integrating by parts:

(1.134) integral Subscript 0 Superscript script l Baseline left-parenthesis kappa u prime v Superscript prime Baseline plus c u v right-parenthesis d x equals minus integral Subscript 0 Superscript script l Baseline mu StartFraction partial-differential squared u Over partial-differential t squared EndFraction v d x period

We now introduce u equals upper U left-parenthesis x right-parenthesis upper T left-parenthesis t right-parenthesis where upper U element-of upper E Superscript 0 Baseline left-parenthesis upper I right-parenthesis , upper T element-of upper C squared left-parenthesis 0 comma infinity right-parenthesis . This is known as separation of variables. Therefore we get

(1.135) upper T integral Subscript 0 Superscript script l Baseline left-parenthesis kappa upper U prime v prime plus c upper U v right-parenthesis d x equals minus StartFraction partial-differential squared upper T Over partial-differential t squared EndFraction integral Subscript 0 Superscript script l Baseline mu upper U v d x

which can be written as

(1.136)

Since the functions on the left are independent of t, the function T depends only on t, both expressions must equal some positive constant denoted by omega squared . That constant has to be positive because the expression on the left holds for all v element-of upper E Superscript 0 Baseline left-parenthesis upper I right-parenthesis and if we select v equals upper U then the expression on the left is positive.

The function upper T left-parenthesis t right-parenthesis satisfies the ordinary differential equation

(1.137) StartFraction partial-differential squared upper T Over partial-differential t squared EndFraction plus omega squared upper T equals 0

the solution of which is

(1.138) upper T equals a cosine left-parenthesis omega t right-parenthesis plus b sine left-parenthesis omega t right-parenthesis

where ω is the angular velocity (rad/s). Alternatively ω is written as omega equals 2 pi f where f is the frequency (Hz).

To find ω and U we have to solve the problem

(1.139) integral Subscript 0 Superscript script l Baseline left-parenthesis kappa upper U prime v Superscript prime Baseline plus c upper U v right-parenthesis d x minus omega squared integral Subscript 0 Superscript script l Baseline mu upper U v d x equals 0 for all v element-of upper E Superscript 0 Baseline left-parenthesis upper I right-parenthesis

which will be abbreviated as

(1.140) upper B left-parenthesis upper U comma v right-parenthesis minus omega squared upper D left-parenthesis upper U comma v right-parenthesis equals 0 for all v element-of upper E Superscript 0 Baseline left-parenthesis upper I right-parenthesis period

There are infinitely many solutions called eigenpairs left-parenthesis omega Subscript i Baseline comma upper U Subscript i Baseline right-parenthesis , i equals 1 comma 2 comma ellipsis comma infinity . The set of eigenvalues is called the spectrum. If Ui is an eigenfunction and α is a real number then alpha upper U Subscript i is also an eigenfunction. In the following we assume that the eigenfunctions have been normalized so that

upper D left-parenthesis upper U Subscript i Baseline comma upper U Subscript i Baseline right-parenthesis identical-to integral Subscript 0 Superscript script l Baseline mu upper U Subscript i Superscript 2 Baseline d x equals 1 period

If the eigenvalues are distinct then the corresponding eigenfunctions are orthogonal: Let left-parenthesis omega Subscript i Baseline comma upper U Subscript i Baseline right-parenthesis and be eigenpairs, i not-equals j . Then from eq. (1.140) we have

StartLayout 1st Row 1st Column upper B left-parenthesis upper U Subscript i Baseline comma upper U Subscript j Baseline right-parenthesis minus omega Subscript i Superscript 2 Baseline upper D left-parenthesis upper U Subscript i Baseline comma upper U Subscript j Baseline right-parenthesis 2nd Column equals 0 2nd Row 1st Column upper B left-parenthesis upper U Subscript j Baseline comma upper U Subscript i Baseline right-parenthesis minus omega Subscript j Superscript 2 Baseline upper D left-parenthesis upper U Subscript j Baseline comma upper U Subscript i Baseline right-parenthesis 2nd Column equals 0 period EndLayout

Subtracting the second equation from the first we see that if omega Subscript i Baseline not-equals omega Subscript j then Ui and Uj are orthogonal functions:

(1.141) upper D left-parenthesis upper U Subscript i Baseline comma upper U Subscript j Baseline right-parenthesis identical-to integral Subscript 0 Superscript script l Baseline mu upper U Subscript i Baseline upper U Subscript j Baseline d x equals 0

and hence upper B left-parenthesis upper U Subscript i Baseline comma upper U Subscript j Baseline right-parenthesis equals 0 .

Importantly, it can be shown that any function Скачать книгу