Finite Element Analysis. Barna Szabó
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Название: Finite Element Analysis

Автор: Barna Szabó

Издательство: John Wiley & Sons Limited

Жанр: Физика

Серия:

isbn: 9781119426462

isbn:

СКАЧАТЬ target="_blank" rel="nofollow" href="#fb3_img_img_2f14ff38-e6d2-50a0-a1d3-e7b3fb56f040.png" alt="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:

      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)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 sigma-summation Underscript i equals 1 Overscript p Subscript k Baseline plus 1 Endscripts sigma-summation Underscript j equals 1 Overscript p Subscript k Baseline plus 1 Endscripts m Subscript i j Superscript left-parenthesis k right-parenthesis Baseline a Subscript j Baseline b Subscript i Baseline equals StartSet b EndSet Superscript upper T Baseline left-bracket upper M Superscript left-parenthesis k right-parenthesis Baseline right-bracket StartSet a EndSet

      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

left-bracket upper M Superscript left-parenthesis k right-parenthesis Baseline right-bracket equals Start 4 By 4 Matrix 1st Row 1st Column m 11 Superscript left-parenthesis k right-parenthesis Baseline 2nd Column m 12 Superscript left-parenthesis k right-parenthesis Baseline 3rd Column midline-horizontal-ellipsis 4th Column m Subscript 1 comma p Sub Subscript k Subscript plus 1 Baseline 2nd Row 1st Column m 21 Superscript left-parenthesis k right-parenthesis Baseline 2nd Column m 22 Superscript left-parenthesis k right-parenthesis Baseline 3rd Column midline-horizontal-ellipsis 4th Column m Subscript 2 comma p Sub Subscript k Subscript plus 1 Baseline 3rd Row 1st Column vertical-ellipsis 2nd Column Blank 3rd Column down-right-diagonal-ellipsis 4th Column vertical-ellipsis 4th Row 1st Column m Subscript p Sub Subscript k Subscript plus 1 comma 1 Superscript left-parenthesis k right-parenthesis Baseline 2nd Column m Subscript p Sub Subscript k Subscript plus 1 comma 2 Superscript left-parenthesis k right-parenthesis Baseline 3rd Column midline-horizontal-ellipsis 4th Column m Subscript p Sub Subscript k Subscript plus 1 comma p Sub Subscript k Subscript plus 1 EndMatrix dot

      Example 1.4 When c left-parenthesis x right-parenthesis equals c Subscript k is constant on Ik and the Legendre shape functions are used then the element‐level Gram matrix is strongly diagonal. For example, for p Subscript k Baseline equals 5 the Gram matrix is: