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 will be concerned with the evaluation of the integral on the kth element:

integral Subscript x Subscript k Baseline Superscript x Subscript k plus 1 Baseline Baseline kappa left-parenthesis x right-parenthesis u prime Subscript n Baseline v Subscript n Superscript prime Baseline d x equals integral Subscript x Subscript k Baseline Superscript x Subscript k plus 1 Baseline Baseline kappa 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 StartFraction d upper N Subscript j Baseline Over d x EndFraction right-parenthesis left-parenthesis sigma-summation Underscript i equals 1 Overscript p Subscript k Baseline plus 1 Endscripts b Subscript i Baseline StartFraction d upper N Subscript i Baseline Over d x EndFraction right-parenthesis d x period

      (1.63)d x equals StartFraction x Subscript k plus 1 Baseline minus x Subscript k Baseline Over 2 EndFraction d xi identical-to StartFraction script l Subscript k Baseline Over 2 EndFraction d xi

      where script l Subscript k Baseline equals Overscript def Endscripts x Subscript k plus 1 Baseline minus x Subscript k is the length of the kth element. Also,

StartFraction d Over d x EndFraction equals StartFraction d Over d xi EndFraction StartFraction d xi Over d x EndFraction equals StartFraction 2 Over x Subscript k plus 1 Baseline minus x Subscript k Baseline EndFraction StartFraction d Over d xi EndFraction identical-to StartFraction 2 Over script l Subscript k Baseline EndFraction StartFraction d Over d xi EndFraction dot

      Therefore

integral Subscript x Subscript k Baseline Superscript x Subscript k plus 1 Baseline Baseline kappa left-parenthesis x right-parenthesis u prime Subscript n Baseline v Subscript n Superscript prime Baseline d x equals StartFraction 2 Over script l Subscript k Baseline EndFraction integral Subscript negative 1 Superscript plus 1 Baseline kappa 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 StartFraction d upper N Subscript j Baseline Over d xi EndFraction right-parenthesis left-parenthesis sigma-summation Underscript i equals 1 Overscript p Subscript k Baseline plus 1 Endscripts b Subscript i Baseline StartFraction d upper N Subscript i Baseline Over d xi EndFraction right-parenthesis d xi period

      We define

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

      and write

      (1.65)integral Subscript x Subscript k Baseline Superscript x Subscript k plus 1 Baseline Baseline kappa left-parenthesis x right-parenthesis u prime Subscript n Baseline v Subscript n Superscript prime 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 k Subscript i j Superscript left-parenthesis k right-parenthesis 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:

      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 СКАЧАТЬ