Page 35 - Using ANSYS for Finite Element Analysis A Tutorial for Engineers
P. 35

22  •   Using ansys for finite element analysis
                1.2.3   solution of set of linear algeBraic
                       equations

                1.2.3.1  cramer’s rule

                Consider a set of linear algebraic equations written as follows:
                In matrix notation: [a]{x} = {c}

                Or, in index notation: ∑ n j=1  ax = c i
                                           j
                                        ij
                         
                Let matrix  d  i ()  be matrix [a] with column i replaced by [c]. Then:
                            
                                              d  i ()
                                          x =  a
                                           i


                Example:
                Consider a set of three linear algebraic equations given as follows:

                                       −x  + x  − x
                                            3
                                                 2
                                         1    2    3
                                       2x − 4x + 2x
                                         1    2    3

                                          4x + x
                                            2   3
                    In matrix form:


                                                      2
                                             2
                                    −  1  3  −  x 1    
                                               x     
                                                      1
                                     2  −4  2     2  =  
                                               
                                                      
                                                  
                                     0  4  1   x     
                                                      3
                                                3
                    Solution using Cramer’s rule:
                     −1  3  −2
                                                − ) +− ( )(
                                                           0
                                                                 − −16 10
                                    ( 4 8
                                  1
                                            3
                 a = 2  −4   2  = − ( ) −− ) −()(20   28  − ) = 126    =
                     0   4   1
                                d  1 ()    2   3  − 2  −
                            x =     =   1  1 − 4   2 =  41  =  41 .
                             1   a     − (  10)        − 10
                                           3   4   1
   30   31   32   33   34   35   36   37   38   39   40