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 2 Over script l Subscript k Baseline EndFraction sigma-summation Underscript j equals 1 Overscript p Subscript k Baseline plus 1 Endscripts a Subscript j Superscript left-parenthesis k right-parenthesis Baseline left-parenthesis StartFraction d upper N Subscript j Baseline Over d xi EndFraction right-parenthesis Subscript xi equals xi 0"/>

      where script l Subscript k Baseline equals Overscript def Endscripts x Subscript k plus 1 Baseline minus x Subscript k. The computation of the higher derivatives is analogous.

      Remark 1.8 When plotting quantities of interest such as the functions u Subscript upper F upper E Baseline left-parenthesis x right-parenthesis and u prime Subscript upper F upper E Baseline left-parenthesis x right-parenthesis, the data for the plotting routine are generated by subdividing the standard element into n intervals of equal length, n being the desired resolution. The QoIs corresponding to the grid‐points are evaluated. This process does not involve inverse mapping. In node points information is provided from the two elements that share that node. If the computed QoI is discontinuous then the discontinuity will be visible at the nodes unless the plotting algorithm automatically averages the QoIs.

      Indirect computation of bold-italic u bold prime Subscript bold-italic upper F upper E Baseline left-parenthesis bold-italic x bold 0 right-parenthesis in node points

      The first derivative in node points can be determined indirectly from the generalized formulation. For example, to compute the first derivative at node xk from the finite element solution, we select v equals upper N 1 left-parenthesis upper Q Subscript k Superscript negative 1 Baseline left-parenthesis x right-parenthesis right-parenthesis and use

      (1.86)integral Subscript x Subscript k Baseline Superscript x Subscript k plus 1 Baseline Baseline left-parenthesis kappa u prime Subscript upper F upper E Baseline v Superscript prime Baseline plus c u Subscript upper F upper E Baseline v right-parenthesis d x equals integral Subscript x Subscript k Baseline Superscript x Subscript k plus 1 Baseline Baseline f v d x plus left-bracket kappa u prime Subscript upper F upper E Baseline v right-bracket Subscript x equals x Sub Subscript k plus 1 Subscript Baseline minus left-bracket kappa u prime Subscript upper F upper E Baseline v right-bracket Subscript x equals x Sub Subscript k Subscript Baseline period

      Test functions used in post‐solution operations for the computation of a functional are called extraction functions. Here v equals upper N 1 left-parenthesis upper Q Subscript k Superscript negative 1 Baseline left-parenthesis x right-parenthesis right-parenthesis is an extraction function for the functional minus left-bracket kappa u prime Subscript upper F upper E right-bracket Subscript x equals x Sub Subscript k. This is because v left-parenthesis x Subscript k Baseline right-parenthesis equals 1 and v left-parenthesis x Subscript k plus 1 Baseline right-parenthesis equals 0 and hence

      (1.87)StartLayout 1st Row 1st Column minus left-bracket kappa u prime Subscript upper F upper E right-bracket Subscript x equals x Sub Subscript k Baseline equals 2nd Column integral Subscript x Subscript k Baseline Superscript x Subscript k plus 1 Baseline left-parenthesis kappa u prime Subscript upper F upper E Baseline v prime plus c u Subscript upper F upper E Baseline v right-parenthesis d x minus integral Subscript x Subscript k Baseline Superscript x Subscript k plus 1 Baseline f v d x 2nd Row 1st Column equals 2nd Column sigma-summation Underscript j equals 1 Overscript p Subscript k Baseline plus 1 Endscripts c Subscript 1 j Superscript left-parenthesis k right-parenthesis Baseline a Subscript j Superscript left-parenthesis k right-parenthesis minus r 1 Superscript left-parenthesis k right-parenthesis EndLayout

      where, by definition; c Subscript i j Superscript left-parenthesis k right-parenthesis Baseline equals k Subscript i j Superscript left-parenthesis k right-parenthesis Baseline plus m Subscript i j Superscript left-parenthesis k right-parenthesis.

      Example 1.8 Let us find u prime Subscript upper F upper E Baseline left-parenthesis 1 right-parenthesis for the problem in Example 1.7 by the direct and indirect methods. In this case the exact solution is known from which we have u prime Subscript upper E upper X Baseline left-parenthesis 1 right-parenthesis equals 3.5978. By direct computation:

u prime Subscript upper F upper E Baseline left-parenthesis 1 right-parenthesis equals StartFraction 2 Over script l 5 EndFraction left-parenthesis StartFraction d u Subscript upper F upper E Baseline Over d xi EndFraction right-parenthesis Subscript xi equals 1 Baseline equals 5 left-parenthesis a 6 minus a 5 right-parenthesis equals 2.9028 left-parenthesis 19.32 percent-sign error right-parenthesis u prime Subscript upper F upper E Baseline left-parenthesis 1 right-parenthesis equals minus StartFraction 73 Over 15 EndFraction a 5 plus StartFraction 79 Over 15 EndFraction a 6 equals 3.6254 left-parenthesis 0.77 percent-sign error right-parenthesis period integral Subscript 0 Superscript script l Baseline u prime v Superscript prime Baseline d x equals integral Subscript 0 Superscript script l Baseline delta left-parenthesis x minus x overbar right-parenthesis v d x equals v left-parenthesis x overbar right-parenthesis comma u left-parenthesis 0 right-parenthesis equals u left-parenthesis script l right-parenthesis equals 0

      where δ is the delta function, see Definition A.5 in the appendix. Let us be interested in finding the approximate value of u prime left-parenthesis 0 right-parenthesis. The data are script l equals 1 and x overbar equals 1 slash 4. We will use one finite element and p equals 2 comma 3 comma ellipsis This is a poorly chosen discretization because the derivatives of u are discontinuous in the point x equals x overbar, whereas all derivatives of the shape functions are continuous. The proper discretization would have been to use two or more finite elements with a node point in x equals x overbar. Then СКАЧАТЬ