Page 344 - MATLAB an introduction with applications
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Direct Numerical Integration Methods ———  329



                                      X =  X  −  ∆tX    +  ∆  2  X     –  ∆t  t 3  X .              ...(6.44)
                                                  t+
                                          t+
                                                   ∆t
                                           ∆ t
                                                    t
                                                        2  t+ ∆  t  6  t+ ∆  t
                                                             t ∆
                                                                         t ∆

                                                  t ∆
                                  X    =  X  −  (2 ) X    +  (2 ) 2  X       –  (2 ) 3  X           ...(6.45)
                                     ∆
                                                     t+
                                          t+
                                   t−
                                            t
                                           ∆t
                                                      ∆
                                                       t
                                                            2    t+ ∆  t  6   t+ ∆  t
                                                                         t ∆
                                                            t ∆

                                                 t ∆
                                 X    =  X   −  (3 ) X    +  (3 ) 2    X    –  (3 ) 3     X       .    ...(6.46)
                                          t+
                                                     t+
                                           ∆ t−∆
                                    2 t
                                                       t
                                                      ∆t
                                                            2    t+ ∆  t  6    t+ ∆  t

                   Solving Eqs. (6.44) to (6.46) for  X t+ ∆t  and X   t+∆ t  , we get

                                  X t+ ∆  =  1  (2X t+ ∆t  t  − 5X +  4X t−∆ t  –X t –2 t  )        ...(6.47)
                                                                  ∆
                                                      t
                                        ∆t 2
                                         1
                                  X    t+ ∆  =  6 t ∆  (11X t+ ∆t  − 18X +  9X t− ∆ t  t  −  2X t− 2 t ∆  )  ...(6.48)
                                                       t
                   The difference formulas in the Houbolt algorithm are, therefore, given by
                                                                    −
                                                    −
                                                                            }
                                                       { } 4 X
                                { X    t+ ∆  } =  1 2  2   {X t+ ∆t  t } 5 X t  +  {  t− ∆  t } {X t− 2 t ∆   ...(6.49)
                                        ∆t
                                         1
                                                                        {
                                                                              }
                                                                      −
                                                     −
                                                        { } 9 X
                                { X   t+ ∆  } =  6 t ∆    11 {X t+ ∆t  t } 18 X t  +  {  t− ∆  t } 2 X t− 2 t ∆   ...(6.50)
                                                          {

                   By substituting the expressions for { X t+ ∆  }  and  X   t+ ∆t  t } from (6.49) and (6.50), respectively,  into (6.13), we
                   get
                            
                             M    {   X t+ ∆t } { F t+ ∆t }                                       ...(6.51)
                                       =
                            
                                                       F
                         
                   where  M  is the effective mass matrix and { t+ ∆t } is the effective force vector.
                                          2       11
                                   
                                    M  =  ∆t 2 [ ] + 6 t ∆  [] [ ]                                 ...(6.52)
                                                     C +
                                                          K
                                             M
                                      
                                   
                                                               
                                                          3
                                { F t+∆ =  F  ∆t } +     ∆t 5 2 [ ] +  ∆t [] { }
                                                                 X
                                                     M
                                                             C
                                                               
                                     t
                                      } { t+
                                                               
                                                                  t
                                          4       3              1       1    
                                                                      M
                                                                               C
                                                      C
                                             M
                                                        
                                                                                 
                                                   –    ∆  2  [ ] +  2 t ∆  [] {X t− ∆t } +    ∆t  t 2 [ ] +  3 t ∆  [] {X t− 2 t ∆  }  ...(6.53)
                                                                                 
                                                        
                   Note that the equilibrium equation at time  t + ∆ ,  Eq. (6.51) is used in finding the solution for {X t+ ∆t }. For
                                                         t
                   this reason, this method is called an implicit integration method. It can be seen that the velocities and
                   accelerations at time  t +  ∆t are obtained by substituting for {X t+ ∆t } in (6.50) and (6.49) respectively. Also
                   that a knowledge of  XX t− ∆t   and X t –2 t ∆  is needed to find solution for  {X t+ ∆t }. Since there is no direct
                                       ,
                                      t
                                                 {
                   method available to find {X t –t ∆  } and  X t –2 t ∆  } , initially we use the central difference method to find solution
                   at time  ∆t and 2∆t. This makes the method non-self starting. The method also requires large computer
                   storage to store displacements for the previous time steps. The step by step procedure to be used in the
                   Houbolt method is summarized in Table 6.3. A basic difference between the Houbolt method in Table 6.3
                   and the central difference method in Table 6.2 is the appearance of the stiffness matrix K as a factor to the
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