Finite Element Analysis - Barna Szabó

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left-parenthesis negative 2 right-parenthesis EndFraction exp left-parenthesis 2 x right-parenthesis period"/>

      Using five elements of equal length on the interval upper I equals left-parenthesis 0 comma 1 right-parenthesis and p equals 1 assigned to each element, find the finite element solution for this problem.

left-bracket upper C Superscript left-parenthesis k right-parenthesis Baseline right-bracket equals Start 2 By 2 Matrix 1st Row 1st Column 79 slash 15 2nd Column negative 73 slash 15 2nd Row 1st Column negative 73 slash 15 2nd Column 79 slash 15 EndMatrix comma k equals 1 comma 2 comma ellipsis 5

      where we used kappa Subscript k Baseline equals 1, c Subscript k Baseline equals 4, script l Subscript k Baseline equals 1 slash 5. The assembled unconstrained coefficient matrix is:

left-bracket upper C right-bracket equals Start 6 By 6 Matrix 1st Row 1st Column 79 slash 15 2nd Column negative 73 slash 15 3rd Column 0 4th Column 0 5th Column 0 6th Column 0 2nd Row 1st Column negative 73 slash 15 2nd Column 158 slash 15 3rd Column negative 73 slash 15 4th Column 0 5th Column 0 6th Column 0 3rd Row 1st Column 0 2nd Column negative 73 slash 15 3rd Column 158 slash 15 4th Column negative 73 slash 15 5th Column 0 6th Column 0 4th Row 1st Column 0 2nd Column 0 3rd Column negative 73 slash 15 4th Column 158 slash 15 5th Column negative 73 slash 15 6th Column 0 5th Row 1st Column 0 2nd Column 0 3rd Column 0 4th Column negative 73 slash 15 5th Column 158 slash 15 6th Column negative 73 slash 15 6th Row 1st Column 0 2nd Column 0 3rd Column 0 4th Column 0 5th Column negative 73 slash 15 6th Column 79 slash 15 EndMatrix dot

      Upon enforcement of the Dirichlet conditions the system of equations is

left-bracket upper C right-bracket equals Start 4 By 4 Matrix 1st Row 1st Column 158 slash 15 2nd Column negative 73 slash 15 3rd Column 0 4th Column 0 2nd Row 1st Column negative 73 slash 15 2nd Column 158 slash 15 3rd Column negative 73 slash 15 4th Column 0 3rd Row 1st Column 0 2nd Column negative 73 slash 15 3rd Column 158 slash 15 4th Column negative 73 slash 15 4th Row 1st Column 0 2nd Column 0 3rd Column negative 73 slash 15 4th Column 158 slash 15 EndMatrix Start 4 By 1 Matrix 1st Row a 2 2nd Row a 3 3rd Row a 4 4th Row a 5 EndMatrix equals Start 4 By 1 Matrix 1st Row 73 slash 13 2nd Row 0 3rd Row 0 4th Row 146 slash 15 EndMatrix left-bracket upper C right-bracket equals Start 6 By 6 Matrix 1st Row 1st Column 1 2nd Column 0 3rd Column 0 4th Column 0 5th Column 0 6th Column 0 2nd Row 1st Column 0 2nd Column 158 slash 15 3rd Column negative 73 slash 15 4th Column 0 5th Column 0 6th Column 0 3rd Row 1st Column 0 2nd Column negative 73 slash 15 3rd Column 158 slash 15 4th Column negative 73 slash 15 5th Column 0 6th Column 0 4th Row 1st Column 0 2nd Column 0 3rd Column negative 73 slash 15 4th Column 158 slash 15 5th Column negative 73 slash 15 6th Column 0 5th Row 1st Column 0 2nd Column 0 3rd Column 0 4th Column negative 73 slash 15 5th Column 158 slash 15 6th Column 0 6th Row 1st Column 0 2nd Column 0 3rd Column 0 4th Column 0 5th Column 0 6th Column 1 EndMatrix Start 6 By 1 Matrix 1st Row a 1 2nd Row a 2 3rd Row a 3 4th Row a 4 5th Row a 5 6th Row a 6 EndMatrix equals Start 6 By 1 Matrix 1st Row 1 2nd Row 73 slash 15 3rd Row 0 4th Row 0 5th Row 146 slash 15 6th Row 2 EndMatrix

      where the first and sixth equations are placeholders for the boundary conditions a 1 equals 1, a 6 equals 2. The solution is:

StartSet a EndSet equals left-brace 1.0000 0.8784 0.9012 1.0722 1.4194 2.0000 right-brace Superscript upper T Baseline period

      Exercise 1.14 Solve the problem in Example 1.7 with the boundary conditions u left-parenthesis 0 right-parenthesis equals 1, u prime left-parenthesis 1 right-parenthesis equals 3.6.

      Exercise 1.15 Solve the problem in Example 1.7 with the boundary conditions u prime left-parenthesis 0 right-parenthesis equals negative 1, u left-parenthesis 1 right-parenthesis equals 2.

      Following assembly of the coefficient matrix and enforcement of the essential boundary conditions (when applicable) the resulting system of simultaneous equations is solved by one of several methods designed to exploit the symmetry and sparsity of the coefficient matrix. The solvers are classified into two broad categories; direct and iterative solvers. Optimal choice of a solver in a particular application is based on consideration of the size of the problem and the available computational resources.

      At the end of the solution process the finite element solution is available in the form

      (1.81)u Subscript upper F upper E Baseline equals sigma-summation Underscript j equals 1 Overscript upper N Subscript u Baseline Endscripts a Subscript j Baseline phi Subscript j Baseline left-parenthesis x right-parenthesis

      where the indices reference the global numbering and Nu is the number of degrees of freedom plus the number of Dirichlet conditions.

      The basis functions are decomposed into their constituent shape functions and the element‐level solution records are created in the local numbering convention. Therefore the finite element solution on the kth element is available in the following form:

      (1.82)u Subscript upper F upper E Superscript left-parenthesis k right-parenthesis Baseline 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 right-parenthesis period

      1.4.1 Computation of the quantities of interest

      The computation of typical engineering quantities of interest (QoI) by direct and indirect methods is outlined in this section.

      Computation of uFE(x0)