Finite Element Analysis - Barna Szabó

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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 the 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.