Finite Element Analysis. Barna Szabó

Finite Element Analysis

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"/>

and by indirect computation:

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

Example 1.9 The following example illustrates that the indirect method can be used for obtaining the QoI efficiently and accurately even when the discretization was very poorly chosen. We will consider the problem

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 . Then the exact solution would be obtained at p equals 1 .

If we use the Legendre shape functions then the coefficient matrix displayed in eq. (1.66) will be perfectly diagonal. The first two rows and columns will be zero on account of the boundary conditions and the diagonal term will be 2. Referring to eq. (1.75) the right hand side vector will be

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

where we made use of eq. (1.55). The relationships between the polynomial degree ranging from 2 to 100 and the corresponding values of the QoI computed by the direct method are displayed in Fig. 1.7. It is seen that convergence to the exact value u prime Subscript upper E upper X Baseline left-parenthesis 0 right-parenthesis equals 0.75 is very slow.

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.

Figure 1.7 Example 1.9. Values of Скачать книгу