$-------------------------------------------------------------------------------
$           RIGID FORMAT No. 9, Transient Analysis - Direct Formulation
$               Transient Analysis with Direct Matrix Input (9-1-1)
$ 
$ A. Description
$ 
$ This problem demonstrates the capability of NASTRAN to perform transient
$ analysis on a system having nonsymmetric stiffness, damping, and mass
$ matrices. The problem also illustrates the use of time step changes, selection
$ of printout intervals, application of loads, initial conditions, and a simple
$ curve plot package.
$ 
$ The matrices and loads used are actually the product of a transformation
$ matrix and diagonal matrices. The resulting answers are easily calculated
$ while the input matrices are of general form. The matrix equation solved is
$ 
$        ..         .
$    [M]{u}  +  [B]{u}  +  [K]{u}  =  {P(t)}                                 (1)
$ 
$ The problem is actually four disjoint single degree of freedom problems which
$ have been transformed to a general matrix problem.
$ 
$ The resulting diagonal matrices are pre-multiplied by the matrix:
$ 
$            +                    +
$            |  2   -1     0    0 |
$            |                    |
$            | -1    2    -1    0 |
$    [X]  =  |                    |                                          (2)
$            |  0   -1     2   -1 |
$            |                    |
$            |  0    0    -1    2 |
$            +                    +
$ 
$ The answers for the disjoint problem above will be the same as for the general
$ matrix problem since the general case:
$ 
$             ..        .
$    [X]([M ]{u} + [B ]{u} + [K ]{u}  =  [X]{P}                              (3)
$          o         o         o
$ 
$ has the same results as the disjoint case:
$ 
$         ..        .
$    [M ]{u} + [B ]{u} + [K ]{u}  =  {P}                                     (4)
$      o         o         o
$ 
$ B. Input
$ 
$ 1. The actual matrix input is:
$ 
$            +                    +
$            | 20   -1.5   0    0 |
$            |                    |
$            |-10    3.0  -4    0 |
$    [M]  =  |                    |
$            |  0   -1.5   8    0 |
$            |                    |
$            |  0    0.0  -4    0 |
$            +                    +
$ 
$            +                    +
$            |  0   -15    0    0 |
$            |                    |
$            |  0    30   -24   0 |
$    [B]  =  |                    |
$            |  0   -15    28  -2 |
$            |                    |
$            |  0    0    -24   4 |
$            +                    +
$ 
$            +                      +
$            | 2000   0     0    0  |
$            |                      |
$            | -1000  0    -100  0  |
$    [K]  =  |                      |
$            |   0    0     200 -20 |
$            |                      |
$            |   0    0    -100  40 |
$            +                      +
$ 
$ 2. The initial conditions are:
$                        .
$    u    =  0           u    =  10.0
$     10                  10
$                        .
$    u    =  0           u    =  0.5
$     11                  11
$                        .
$    u    =  0           u    =  0
$     12                  12
$                        .
$    u    =  -10.0       u    =  0
$     13                  13
$ 
$ 3. At t = 1.0 a step load is applied to each point. The load on the uncoupled
$    problems is:
$ 
$             | 0   |
$             | 1.5 |
$      P   =  | 4.0 |
$       o     | 20  |
$ 
$    The transformed load is:
$ 
$                          | -1.5  |
$                          | -1.0  |
$      {P}  =  [X]{P }  =  | -13.5 |
$                   o      | 36.0  |
$ 
$ C. Theory
$ 
$ The results are responses of single degree of freedom systems. Equations are
$ given in Reference 12, Chapter 9.
$ 
$   0 < t < 1.0 , delta t  =  .005
$ 
$                                .
$     u   = sin 10t              u   = 10 cos 10t
$      10                         10
$ 
$                     -10t       .         -10t
$     u   = 0.05(1 - e    )      u   = 0.5e
$      11                         11
$                                .
$     u   = 0                    u   = 0
$      12                         12
$ 
$               -10t             .         -10t
$     u   = -10e                 u   = 100e
$      13                         13
$ 
$   t > 1.0 , delta t  =  .015
$ 
$     u   = sin 10t
$      10
$                     -10t                     -10(t-1)
$     u   = 0.05(1 - e    ) + 0.1(t - 1.1 + .1e        )
$      11
$                      -3t
$     u   = 0.04 {1 - e   [cos4(t-1) + 3/4 sin4(t-1)]}
$      12
$ 
$               -10t        -10(t-1)
$     u   = -10e     + 1 - e
$      13
$ 
$ D. Results
$ 
$ The deviations of the NASTRAN results and the theoretical response are due to
$ the selection of time steps. For instance point 11 has a time constant equal
$ to two time steps. The initial error in velocity due to the first step causes
$ the displacement error to accumulate. Using a smaller time step has resulted
$ in much better results.
$ 
$ APPLICABLE REFERENCES
$ 
$ 12. H. Yeh and J. I. Abram, MECHANICS OF SOLIDS AND FLUIDS. Vo1. I, Particle
$     and Rigid Body Mechanics. NcGraw-Hill, 1960.
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