Page 49 - Using ANSYS for Finite Element Analysis A Tutorial for Engineers
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36  •   Using ansys for finite element analysis

                 f  1 ()  =− 21 10 * d
                            3
                         *
                  x 1          2 x
                 f  1 ()  =− 375  kN
                        .
                  x 1
                 f  1 ()  =− f  1 ()  =  375  kN
                             .
                 2 x    x 1
                The second element between nodal 2, 3:
                 f  x       21  − 21  d  x 
                     
                   2 ()
                
                  2 2 ()  =  10 *     d  2 =  
                          3
                  f   3 x     − 21  21   3 x  0 
                 f  2 ()  =  21 10 * d
                           3
                        *
                 2 x          2 x
                 f  2 ()  =  375  kN
                       .
                 2 x
                 f  2 ()  =− f  2 ()  =− 375  kN
                               .
                 3 x    2 x
                The third element between nodal 2, 4:
                     
                 f  x       21  − 21  d  x 
                   2 ()
                
                  2 2 ()  =  10 *  −    d  2 =  
                          3
                  f   4 x      21  21   4 x  0 
                 f  2 ()  =  21 10 * d
                           3
                        *
                 2 x          2 x
                 f  2 ()  =  375  kN
                       .
                 2 x
                 f  2 ()  =− f  2 ()  =− 375  kN
                               .
                 4 x    2 x
                The fourth element between nodal 2, 5:

                 f  x       21  − 21  d  x 
                     
                   2 ()
                
                          3
                  2 2 ()  =  10 *     d  2 =  
                  f   5 x     − 21  21   5 x  0 
                 f  2 ()  =  21 10 * d
                           3
                        *
                 2 x          2 x
                 f  2 ()  =  375  kN
                       .
                 2 x
                 f  2 ()  =− f  2 ()  =− 375  kN
                               .
                 5 x    2 x
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