Finite Element Analysis. Barna Szabó

Finite Element Analysis - Barna Szabó


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(1.60), the standard coordinate xi 0 element-of upper I Subscript st corresponding to x0 is determined:

      and u Subscript upper F upper E Baseline left-parenthesis x 0 right-parenthesis is computed from

      (1.84)u Subscript upper F upper E Baseline left-parenthesis x 0 right-parenthesis equals sigma-summation Underscript j equals 1 Overscript p Subscript k Baseline plus 1 Endscripts a Subscript j Superscript left-parenthesis k right-parenthesis Baseline upper N Subscript j Baseline left-parenthesis xi 0 right-parenthesis period

      Direct computation of bold-italic u bold-italic prime Subscript bold-italic upper F upper E Baseline left-parenthesis bold-italic x bold 0 right-parenthesis

      (1.85)left-parenthesis StartFraction d u Subscript upper F upper E Baseline Over d x EndFraction right-parenthesis Subscript x equals x 0 Baseline equals StartFraction 2 Over script l Subscript k Baseline EndFraction left-parenthesis StartFraction d u Subscript upper F upper E Baseline Over d xi EndFraction right-parenthesis Subscript xi equals xi 0 Baseline equals StartFraction 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 <hr><noindex><a href=Скачать книгу