Finite Element Analysis. Barna Szabó

Finite Element Analysis

u Subscript upper E upper X Baseline phi Subscript i Superscript double-prime Baseline d x period"/>

Since phi double-prime Subscript i Baseline equals 0 for i equals 1 and i equals 2 , we have:

StartLayout 1st Row 1st Column r 1 Superscript left-parenthesis k right-parenthesis Baseline equals 2nd Column minus StartFraction 1 Over script l Subscript k Baseline EndFraction left-parenthesis kappa u Subscript upper E upper X Baseline right-parenthesis Subscript x equals x Sub Subscript k plus 1 plus StartFraction 1 Over script l Subscript k Baseline EndFraction left-parenthesis kappa u Subscript upper E upper X Baseline right-parenthesis Subscript x equals x Sub Subscript k plus StartFraction script l Subscript k Baseline Over 2 EndFraction integral Subscript negative 1 Superscript 1 Baseline left-parenthesis c u Subscript upper E upper X Baseline right-parenthesis Subscript x equals upper Q Sub Subscript k Subscript left-parenthesis xi right-parenthesis Baseline upper N 1 d xi 2nd Row 1st Column r 2 Superscript left-parenthesis k right-parenthesis Baseline equals 2nd Column StartFraction 1 Over script l Subscript k Baseline EndFraction left-parenthesis kappa u Subscript upper E upper X Baseline right-parenthesis Subscript x equals x Sub Subscript k plus 1 minus StartFraction 1 Over script l Subscript k Baseline EndFraction left-parenthesis kappa u Subscript upper E upper X Baseline right-parenthesis Subscript x equals x Sub Subscript k plus StartFraction script l Subscript k Baseline Over 2 EndFraction integral Subscript negative 1 Superscript 1 Baseline left-parenthesis c u Subscript upper E upper X Baseline right-parenthesis Subscript x equals upper Q Sub Subscript k Subscript left-parenthesis xi right-parenthesis Baseline upper N 2 d xi EndLayout

and for i greater-than-or-equal-to 3 we have:

where upper P Subscript i minus 2 Baseline left-parenthesis xi right-parenthesis is the Legendre polynomial of degree and eq. (D.10) was used.

Since the exact solution is known, the exact value of the potential energy can be determined for any set of values of α, κ, c and script l . When κ and c are both constants then

(1.107)

The exact values of the potential energy for the data kappa equals 1 , c equals 50 and script l equals 1 and various values of α are shown in Table 1.2.

Table 1.2 Exact values of the potential energy for

and

α		α
0.600	−2.3728354978	1.000	−1.0000000000
0.700	−1.7571858289	1.500	−0.5104166667
0.800	−1.4176885916	2.000	−0.3047619048
0.900	−1.1799028822	3.000	−0.1420634921

When α is a fractional number then derivatives higher than α will not be finite in x equals 0 . In the range 0.5 less-than alpha less-than 1 the first derivative in the point is infinity. This range of α has considerable practical importance because the exact solutions of two‐ and three‐dimensional problems often have analogous terms.

When α is an integer then all derivatives of Скачать книгу