Tham khảo tài liệu 'finite element analysis - thermomechanics of solids part 11', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 11 Solution Methods for Linear Problems NUMERICAL METHODS IN FEA Solving the Finite-Element Equations Static Problems Consider the numerical solution of the linear system Ky f in which K is the positive-definite and symmetric stiffness matrix. In many problems it has a large dimension but is also banded. The matrix can be triangularized K LLT in which L is a lower triangular nonsingular matrix zeroes in all entries above the diagonal . We can introduce z Lty and obtain z by solving Lz f. Next Y can be computed by solving LTY z. Now Lz f can be conveniently solved by forward substitution. In particular Lz f can be expanded as starting from the upper-left entry using simple arithmetic z1 f1 T11 z2 f2 - T21z1 T22 Z3 f T3izi T32Z2 T33 . Next the equation LY z can be solved using backward substitution. The equation is expanded as T11 0 0 0 T12 . . . T1n ĩi f f i T22 . . . . 2 . 0 . . . . Y3 . . 1 1 1 . . t n 2 n 2 t n 2 n 1 t n 2 n . fn 2 . . 0 I c n 1 n 1 n 1 n . fn 1 0 Tnn _ Y n J fn J 153 2003 by CRC CRC Press LLC 154 Finite Element Analysis Thermomechanics of Solids Starting from the lower-right entry solution can be achieved using simple arithmetic as Yn fjlm Y n-1 fn-1 ln-1 1Yn yin-1 n-V Y n-2 fn-2 ln-2 nYn ln-2 n-1Y n n 2 In both procedures only one unknown is encountered in each step row . Matrix Triangularization and Solution OF Linear Systems We next consider how to triangularize K. Suppose that the upper-left j - 1 X j -1 block Kj-1 has been triangularized K j-1 L j-1Lr-1. In determining whether the j X j block Kj can be triangularized we consider K j kT k j L j-1 0 rLT L j-1 À j k j _ 3T ljj _ 0T j K in which kj is a j - 1 X 1 array of the first j - 1 entries of the j h column of Kj. Simple manipulation suffices to furnish kj and ljj. k Lj-A ljj V kjj- j Note that Ầj can be conveniently computed using forward substitution. Also note that l Ik. - kTK-1k . The fact that K 0 implies that l is real. Obviously j A jj. j .