3-3 - 168 Finite Element Analysis and Design 3 Using the...

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168 Finite Element Analysis and Design 3. Using the Galerkin method, solve the following differential equation with the approximate solution in the form of 2 1 2 ( ) u x c x c x . Compare the approximate solution with the exact one by plotting them on a graph. Also compare the derivatives / du dx and / du dx . 2 2 2 0, 0 1 (0) 0 Boundary conditions (1) 1 d u x x dx u du dx Solution: (a) Exact solution: By integrating the governing equation twice and applying two boundary conditions, the exact solution becomes 4 1 4 ( ) 12 3 u x x x   (b) Galerkin method: For the given form of approximation, the trial functions are 2 1 2 ( ) , ( ) x x x x From Eqs. (3.13) and (3.14), the coefficient matrix and the vector on the RHS becomes 1 2 11 1 0 1 12 21 1 2 0 1 2 4 22 2 3 0 ( ) 1 1 ( ) K dx K K dx K dx     1 2 1 1 1 1 0 1 2 2 2 2 2 0 5 ( ) (1) (1) (0) (0) 4 6 ( ) (1) (1) (0) (0) 5 F x dx u u F x dx u u Thus, the matrix becomes: 7 5 1 1 4 5 4 6 3 2 2 3 5 20 1 1 1 c c c c  
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