A pragmatic introduction to the finite element method for by Petr Krysl

By Petr Krysl

Version of a taut cord -- the strategy of Galerkin -- Statics and dynamics examples for the twine version -- Boundary stipulations for the version of a taut cord -- version of warmth conduction -- Galerkin process for the version of warmth conduction -- Steady-state warmth conduction options -- brief warmth conduction suggestions -- increasing the library of point varieties -- Discretization errors, errors regulate, and convergence -- version of elastodynamics -- Galerkin formula for elastodynamics -- Finite parts for precise three-D difficulties -- studying the stresses -- airplane pressure, aircraft rigidity, and axisymmetric types -- Consistency + balance = convergence

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Clearly the largest error is just a little bit over 1%. 6 ξ Exercise 9. 8 1 24 Thermal and Stress Analysis with the FEM Solve for the approximate deflection of a prestressed table with uniform load, simply supported at x = 0 and force-free boundary condition at x = L, using the Galerkin method. Take as the trial function basis the to functions N1 (x) = x and N2 (x) = x2 . The test functions are Nj , j = 1, 2. Solution: We are solving the boundary value problem: P w′′ + q = 0 w(0) = 0 w′ (L) = 0 The trial function is w(x) = a1 N1 (x) + a2 N2 (x) Note that the trial function satisfies the essential boundary condition w(0) = 0 because N1 (0) = N2 (0) = 0 The coefficients a1 , a2 are the unknowns.

9. Fig. 9. The relationship of Galerkin and finite element methods (FEM). The intersection of the two sets are the Galerkin Finite Element Methods (GFEM). 12 Element-by-element computations An especially powerful organizing principle in the finite element method that makes it easy to automate is that the computation of the necessary matrices and vectors can be carried out elementby-element. First an example of such a calculation. 10 and identify opportunities for element-by-element computations.

14) simplifies for statics (zero accelerations) and no natural boundary conditions to L − 0 ∂ηj (x) ∂w(x) P dx + ∂x ∂x L ηj (x)q dx = 0 for j = 1, 2 . 0 Substituting the test and trial function we have L − 0 − 0 L ∂N1 ∂N1 P a1 dx + − ∂x ∂x ∂N2 ∂N1 P a1 dx + − ∂x ∂x L 0 L 0 ∂N1 ∂N2 P a2 dx + ∂x ∂x ∂N2 ∂N2 P a2 dx + ∂x ∂x L N1 (x)q dx = 0 , 0 L N2 (x)q dx = 0 , 0 and simplifying further L − 0 L − 0 ∂N1 ∂N1 P dx a1 + − ∂x ∂x ∂N2 ∂N1 P dx a1 + − ∂x ∂x L 0 L 0 ∂N1 ∂N2 P dx a2 + ∂x ∂x ∂N2 ∂N2 P dx a2 + ∂x ∂x L N1 (x)q dx = 0 , 0 L N2 (x)q dx = 0 , 0 Here the terms L Ki,j = 0 ∂Ni ∂Nj P dx ∂x ∂x are the elements of the stiffness matrix, and L Fi = Ni (x)q dx 0 are the elements of the load vector.

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