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 Blank 2nd Column Blank 3rd Column 2 slash 5 4th Column 0 5th Column negative 1 slash 5 StartRoot 21 EndRoot 6th Column 0 6th Row 1st Column Blank 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 6th Column Blank 7th Row 1st Column Blank 2nd Column Blank 3rd Column Blank 4th Column 2 slash 21 5th Column 0 6th Column negative 1 slash 7 StartRoot 45 EndRoot 8th Row 1st Column Blank 2nd Column Blank 3rd Column left-parenthesis s y m period right-parenthesis 4th Column Blank 5th Column Blank 6th Column Blank 9th Row 1st Column Blank 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column 2 slash 45 6th Column 0 10th Row 1st Column Blank 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 6th Column Blank 11th Row 1st Column Blank 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 6th Column 2 slash 77 EndMatrix"/>

      (1.71)StartLayout 1st Row 1st Column m Subscript i i Superscript left-parenthesis k right-parenthesis Baseline equals 2nd Column StartFraction c Subscript k Baseline script l Subscript k Baseline Over 2 EndFraction StartFraction 1 Over 2 left-parenthesis 2 i minus 3 right-parenthesis EndFraction integral Subscript negative 1 Superscript plus 1 Baseline left-parenthesis upper P Subscript i minus 1 Baseline left-parenthesis xi right-parenthesis minus upper P Subscript i minus 3 Baseline left-parenthesis xi right-parenthesis right-parenthesis squared d xi 2nd Row 1st Column equals 2nd Column StartFraction c Subscript k Baseline script l Subscript k Baseline Over 2 EndFraction StartFraction 2 Over left-parenthesis 2 i minus 1 right-parenthesis left-parenthesis 2 i minus 5 right-parenthesis EndFraction comma i greater-than-or-equal-to 3 EndLayout

      and all off‐diagonal terms are zero for i greater-than-or-equal-to 3, with the exceptions:

      (1.72)m Subscript i comma i plus 2 Superscript left-parenthesis k right-parenthesis Baseline equals m Subscript i plus 2 comma i Superscript left-parenthesis k right-parenthesis Baseline equals minus StartFraction c Subscript k Baseline script l Subscript k Baseline Over 2 EndFraction StartFraction 1 Over left-parenthesis 2 i minus 1 right-parenthesis StartRoot left-parenthesis 2 i minus 3 right-parenthesis left-parenthesis 2 i plus 1 right-parenthesis EndRoot EndFraction comma i greater-than-or-equal-to 3 period

m Subscript i j Superscript left-parenthesis k right-parenthesis Baseline equals StartFraction c Subscript k Baseline script l Subscript k Baseline Over 2 EndFraction integral Subscript negative 1 Superscript 1 Baseline upper N Subscript i Baseline upper N Subscript j Baseline d xi almost-equals StartFraction c Subscript k Baseline script l Subscript k Baseline Over 2 EndFraction w Subscript i Baseline delta Subscript i j

      where wi is the weight of the ith Lobatto point. There is an integration error associated with this term because the integrand is a polynomial of degree 2 p. To evaluate this integral exactly n greater-than-or-equal-to left-parenthesis 2 p plus 3 right-parenthesis slash 2 Lobatto points would be required (see Appendix E), whereas only p plus 1 Lobatto points are used. Throughout this book we will be concerned with errors of approximation that can be controlled by the design of mesh and the assignment of polynomial degrees. We will assume that the errors of integration and errors in mapping are negligibly small in comparison with the errors of discretization.

      Exercise 1.9 Assume that c left-parenthesis x right-parenthesis equals c Subscript k is constant on Ik. Using the Lagrange shape functions of degree p equals 3, with the nodes located in the Lobatto points, compute m 33 Superscript left-parenthesis k right-parenthesis numerically using 4 Lobatto points. Determine the relative error of the numerically integrated term. Refer to Remark 1.6 and Appendix E.

      Exercise 1.10 Assume that c left-parenthesis x right-parenthesis equals c Subscript k is constant on Ik. Using the Lagrange shape functions of degree p equals 2, compute m 11 Superscript left-parenthesis k right-parenthesis and m 13 Superscript left-parenthesis k right-parenthesis in terms of ck and ℓk.

      1.3.4 Computation of the right hand side vector

      Computation of the right hand side vector involves evaluation of the functional upper F left-parenthesis v right-parenthesis, usually by numerical means. In particular, we write:

      The element‐level integral is computed from the definition of vn on Ik:

      (1.74)integral Subscript x Subscript k Baseline Superscript x Subscript k plus 1 Baseline Baseline f left-parenthesis x right-parenthesis v Subscript n Baseline d x equals StartFraction script l Subscript k Baseline Over 2 EndFraction integral Subscript negative 1 Superscript plus 1 Baseline f left-parenthesis upper Q Subscript k Baseline left-parenthesis xi right-parenthesis right-parenthesis left-parenthesis sigma-summation Underscript i equals 1 Overscript p Subscript k plus 1 Baseline Endscripts b Subscript i Superscript 
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