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

By Petr Krysl

Version of a taut cord -- the strategy of Galerkin -- Statics and dynamics examples for the cord version -- Boundary stipulations for the version of a taut twine -- version of warmth conduction -- Galerkin process for the version of warmth conduction -- Steady-state warmth conduction strategies -- brief warmth conduction suggestions -- increasing the library of point varieties -- Discretization mistakes, mistakes keep watch over, and convergence -- version of elastodynamics -- Galerkin formula for elastodynamics -- Finite components for real three-D difficulties -- interpreting the stresses -- aircraft pressure, aircraft tension, and axisymmetric types -- Consistency + balance = convergence

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0 However, this time FL = 0 and there will be a contribution of the first term in the final equations. Also note that the two test functions Nj , j = 1, 2 must not be simultaneously zero at x = L. If that was the case, all effects of the applied force FL would be erased from the formulation, and we couldn’t possibly get a meaningful result. Substituting the test and trial function we have L FL N1 (L) − 0 FL N2 (L) − 0 L ∂N1 ∂N1 P a1 dx − ∂x ∂x ∂N2 ∂N1 P a1 dx − ∂x ∂x The Matlab symbolic algebra reads L 0 0 L ∂N1 ∂N2 P a2 dx = 0 , ∂x ∂x ∂N2 ∂N2 P a2 dx = 0 , ∂x ∂x 26 Thermal and Stress Analysis with the FEM >> syms L P q x FL real q=0; N1=sin(pi*x/L); N2=sin(3/2*pi*x/L); K=int(diff( [N1;N2])*P*diff([N1,N2]),0,L) F=q*int([N1;N2],0,L)+ [FL*subs(N1,x,L);FL*subs(N2,x,L)] a=K\F K = [ (P*pi^2)/(2*L), (9*pi*P)/(5*L)] [ (9*pi*P)/(5*L), (9*P*pi^2)/(8*L)] F = 0 -FL a = (80*FL*L)/(P*pi*(25*pi^2 - 144)) -(200*FL*L)/(9*P*(25*pi^2 - 144)) This yields the approximate solution as shown in the figure below.

For one thing, w3 on the left-hand side is known, while R0 on the right-hand side is unknown. Because of the way in which we numbered the unknowns, we can solve this system by partitioning. We will split the matrices and vectors by cutting between the second and third column on the left and between the second and third row on both sides as     2 , −1 −1 2P  qL/2  w1 + 2P   w3 = w2 0 qL/4 + FL L −1 , 1 L and 2P L −1 , 0 2P w1 + w2 L 1 w3 = qL/4 + R0 Visually comparing these two equations with the unpartitioned original matrix equation we can see that nothing changed.

The resulting system of equations is written as [K][w] = [L] where the dimension of the stiffness matrix is N × N , N being the total number of nodes. Let us call Nd the number of prescribed displacements, and Nf the number of unknown degrees of freedom. Evidently we have N = Nd + Nf . We shall use the convention of numbering first the unknown degrees of freedom, and only then the prescribed degrees of freedom. e. with prescribed degrees of freedom); and [Kdd ] is the stiffness matrix that links the prescribed displacements [wd ] to the forces acting on the nodes with supports.

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