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Direct Numerical Integration Methods ———  333


                                       Table 6.4  Algorithm based on Wilson Theta method
                        (a)  Initial Computations:
                            1.  Form stiffness [K], mass [M] and damping [C] matrices


                                         X
                                                     { }  .
                                            , X
                            2.  Initialize { } { }  and X    0
                                               0
                                          0
                            3.  Select time step  t∆ and calculate integration constants,
                                θ =1.4(say):
                                            6        3               θ∆t      a       − a
                                     a =       ;  a =   ;  a =  2 ;  a =  ;  a =  0  ;  a =  2  ;
                                                              a
                                      0
                                         ( ∆ ) tθ  2  1  θ∆t  2  1  3  2   4  θ    5   θ
                                            3      ∆       ∆t  t 2
                                     a =−    ; a =   ; a =    .
                                         1
                                      6
                                                7
                                                       8
                                            θ      2       6
                            4.   Form effective stiffness matrix:
                                         [] a M+  0 [ ] a C+  1 [ ]
                                      K =
                                            K
                                     
                                           
                                                
                                                      L D
                            5.   Triangularize  K   :  K = [][ ][ ] L  T
                                           
                                                
                        (b)  For each time step:
                            1.  Calculate effective force vector at time t + t∆ :

                                            =
                                                                               +
                                                           −
                                                              F
                                                                    M
                                                                            X
                                      F
                                                                 )
                                                     { t+
                                     { t+ θ∆t } {} θF +  ( F  ∆t }{} + [ ]  (a 0 { } a 2 { } { } )X    t  +  2 X    t
                                                                             t
                                                               t
                                                t
                                                                 + [] (C  a X 1  +  2 X    t  +  3 { } )X       t
                                                           { } { } a
                                                          1
                            2.   Solve for displacements at time  t +θ∆ t:
                                        {   X t+θ ∆   } { t+θ ∆t  t }
                                                 =
                                                    F
                                      K

                                            , X
                                          X
                            3.  Calculate {} {}  and X
                                                     {} at time  t + t∆ :


                                                                 +
                                                         +
                                                   −
                                                     X
                                { X    t+ ∆t } =  a 4  { ( X t+ θ∆t } { }) a 5 { } a 6 { }
                                                              X
                                                                      X
                                                               t
                                                       t
                                                                       t



                                             +
                                                        +
                                       =
                                                          X
                                          X
                                { X   t+ ∆t } { } a 7  { ( X t+ ∆t } { } )
                                           t
                                                            t



                                                                +
                                       =
                                             +
                                          X
                                                 { } a
                                {X t+ ∆t } { } ∆t X    t  +  8  { ( X    t+ ∆t } { } )
                                                                 2 X
                                                                     t
                                           t
                   6.5.3  Newmark Beta Method
                   The Newmark Beta integration method is also based on the assumption that the acceleration varies linearly
                   between two instants of time. Two parameters  α  and  β are used in this method, which can be changed to
                   suit the requirements of a particular problem. The expressions for velocity and displacements are given by



                                  X    t+ ∆  =  X +    (1− α )X + α X       t+    ∆t            ...(6.69)
                                                           t
                                                    t
                                          t
                                                          ∆t



                                  X   =  X +  X    ∆    1  −β   X + β X    t+    ∆t +    t 2  ...(6.70)
                                   t+
                                              t
                                          t
                                    ∆
                                                     2    t   ∆t  t 
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