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 exact solution would be obtained at p equals 1.

r Subscript i Baseline equals upper N Subscript i Baseline left-parenthesis xi overbar right-parenthesis where xi overbar equals upper Q Superscript negative 1 Baseline left-parenthesis x overbar right-parenthesis equals negative 1 slash 2 period

      Therefore the coefficients of the shape functions can be written as a Subscript i Baseline equals r Subscript i plus 2 Baseline slash 2 (i equals 1 comma 2 comma ellipsis comma p minus 1) where the variables are renumbered through shifting the indices to account for the boundary conditions: a 1 equals a 2 equals 0. Hence

u Subscript upper F upper E Baseline equals one half sigma-summation Underscript i equals 1 Overscript p minus 1 Endscripts upper N Subscript i plus 2 Baseline left-parenthesis xi overbar right-parenthesis upper N Subscript i plus 2 Baseline left-parenthesis xi right-parenthesis

      and the QoI is:

u prime Subscript upper F upper E Baseline left-parenthesis 0 right-parenthesis equals sigma-summation Underscript i equals 1 Overscript p minus 1 Endscripts upper N Subscript i plus 2 Baseline left-parenthesis xi overbar right-parenthesis StartFraction d upper N Subscript i plus 2 Baseline Over d xi EndFraction vertical-bar Subscript xi equals negative 1 Baseline period

      From the definition of Ni in eq. (1.53) we have

period vertical-bar times times dN plus plus i 2 times times d xi Subscript xi equals negative 1 Baseline equals StartRoot StartFraction 2 i plus 1 Over 2 EndFraction EndRoot upper P Subscript i Baseline left-parenthesis negative 1 right-parenthesis equals StartRoot StartFraction 2 i plus 1 Over 2 EndFraction EndRoot left-parenthesis negative 1 right-parenthesis Superscript i

      and the QoI can be written as

u prime Subscript upper F upper E Baseline left-parenthesis 0 right-parenthesis equals sigma-summation Underscript i equals 1 Overscript p minus 1 Endscripts upper N Subscript i plus 2 Baseline left-parenthesis xi overbar right-parenthesis StartRoot StartFraction 2 i plus 1 Over 2 EndFraction EndRoot left-parenthesis negative 1 right-parenthesis Superscript i Baseline equals one half sigma-summation Underscript i equals 1 Overscript p minus 1 Endscripts left-parenthesis negative 1 right-parenthesis Superscript i Baseline left-parenthesis upper P Subscript i plus 1 Baseline left-parenthesis xi overbar right-parenthesis minus upper P Subscript i minus 1 Baseline left-parenthesis xi overbar right-parenthesis right-parenthesis

      The indirect method is based on eq. (1.18) which, applied to this example, takes the form

integral Subscript 0 Superscript 1 Baseline u prime v Superscript prime Baseline d x equals integral Subscript 0 Superscript 1 Baseline delta left-parenthesis x overbar right-parenthesis v d x plus left-parenthesis u prime v right-parenthesis Subscript x equals 1 Baseline minus left-parenthesis u prime v right-parenthesis Subscript x equals 0 Baseline period

      Selecting v equals 1 minus x and rearranging the terms we get

u prime left-parenthesis 0 right-parenthesis equals v left-parenthesis x overbar right-parenthesis plus integral Subscript 0 Superscript 1 Baseline u prime d x equals v left-parenthesis x overbar right-parenthesis equals 0.75 Graph depicts example 1.9. Values of uFE′(0) computed by the direct method.
computed by the direct method.

      which is the exact solution. The choice v equals 1 minus x was exceptionally fortuitous because it happens to be the Green's function (also known as the influence function) for u prime left-parenthesis 0 right-parenthesis. Therefore the extracted value is independent of the solution u element-of upper E Superscript 0 Baseline left-parenthesis upper I right-parenthesis.

      Let us choose v equals 1 minus x squared for the extraction function. In this case

u prime left-parenthesis 0 right-parenthesis equals v left-parenthesis x overbar right-parenthesis minus integral Subscript 0 Superscript 1 Baseline u prime v Superscript prime Baseline d x equals StartFraction 15 Over 16 EndFraction plus 2 integral Subscript 0 Superscript 1 Baseline u prime x d x period

      Substituting u prime Subscript upper F upper E for u prime:

StartLayout 1st Row 1st Column integral Subscript 0 Superscript 1 Baseline u prime Subscript upper F upper E Baseline x d x equals 2nd Column sigma-summation Underscript i equals 1 Overscript p minus 1 Endscripts StartFraction upper N Subscript i plus 2 Baseline left-parenthesis xi overbar right-parenthesis Over 2 EndFraction StartRoot StartFraction 2 i plus 1 Over 2 EndFraction EndRoot integral Subscript negative 1 Superscript 1 Baseline upper P Subscript i Baseline left-parenthesis xi right-parenthesis StartFraction 1 plus xi Over 2 EndFraction d xi 2nd Row 1st Column equals 2nd Column one fourth sigma-summation Underscript i equals <a href=СКАЧАТЬ