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$            RIGID FORMAT No. 11, Frequency Response - Modal Analysis
$       Frequency Response and Random Analysis of a Ten-Cell Beam (11-1-1)
$ 
$ A.  Description
$ 
$ This problem demonstrates the frequency response solution of a structure using
$ uncoupled modal formulation. With modal formulation, the structural degrees of
$ freedom used in the solution are the uncoupled modal displacements. The
$ solution equations are simple and efficient. The saving in time, however, is
$ offset by the operations necessary to extract the modes, transform the loads
$ to modal coordinates, and transform the modal displacements to structural
$ displacements.
$ 
$ This problem also illustrates the various methods of applying frequency
$ response loads. Loads may be input as complex numbers, with phase lag angles
$ and/or time lag factors. The loads may be added together for each subcase.
$ 
$ The structure to be solved consists of a beam with simple supports on the end.
$ The parameters selected produce natural frequencies of 50, 200, 450, and 800
$ cps. The applied loads for the three subcases are applied to the center with
$ variations in phase angles, time lags, and input formats. The first two
$ subcases use three loaded points which, in essence, simulate a load on the
$ center.
$ 
$ Included in the structural representation is a "general element" representing
$ the first two cells of the ten-cell bean. The flexibility matrix, [Z], of the
$ element represents the displacements of grid points 2 and 3 when point 1 is
$ fixed. The rigid body matrix, [S], represents the rigid body motions of points
$ 2 and 3 when point 1 is displaced in the x, z, or theta  directions.
$                                                        y
$ 
$ The random analysis data consists of a flat power spectral density function
$ ("white noise") for the three loading subcases. The first subcase spectral
$ density is connected to the third subcase spectral density, simulating two
$ interdependent probability functions. The XY-plotter is used to plot the
$ displacement and acceleration power spectral density function of grid 6
$ (center of the beam). The displacement autocorrelation function is also
$ plotted for the same point. All values are tabulated in the printout.
$ 
$ A static analysis restart of the frequency response problem is demonstrated.
$ Gravity and element enforced deformation loads are used with a change in the
$ single-point constraints.
$ 
$ B. Input
$ 
$ 1. Parameters:
$ 
$    l  =  20            - length
$ 
$    I  =  .083          - bending inertia
$     1
$ 
$    A  =  21.18922      - cross sectional area
$ 
$                   6
$    E  =  10.4 x 10    - modulus of elasticity
$ 
$                    -3
$    p  =  .2523 x 10    - mass density
$ 
$    M  =  pAl           - total mass
$ 
$ 2. Constraints:
$ 
$    u   =  theta  = theta  = 0    - all points
$     y          x        z
$ 
$    u   = u   = u    = 0          - frequency response
$     x1    z1    z11
$ 
$    u   = u   = u    = u    = 0   - static analysis
$     x1    z1    x11    z11
$ 
$ 3. Modal Data:
$ 
$    Interval: 40 < f < 1000 cps
$ 
$    Normalization: Modal Mass  =  1.0
$ 
$    Number of modes used in formulation: 4
$ 
$                                   -4
$    Modal Damping ratio: g = 4 x 10   f
$ 
$ 4. Loads, Frequency Response:
$ 
$    The loading functions for subcase 1 are:
$ 
$      P     =  50
$       z,5
$ 
$      M     =  -100
$       y,5
$ 
$      P     =  50 + 100 (cos 60 degrees + i sin 60 degrees)
$       z,6          ---------------------------------------
$                                    SET 6
$ 
$      P     =  50
$       z,7
$ 
$      M     =  100
$       y,7
$ 
$    The loading for subcase 2 is:
$ 
$      P     =  50
$       z,5
$ 
$      M     =  -100
$       y,5
$ 
$      P     =  50 + 100 (cos2f degrees - i sin2f degrees)
$       z,6          -------------------------------------
$                           SET 7, tau = .005555
$ 
$      P     =  50
$       z,7
$ 
$      M     =  100
$       y,7
$ 
$    The load for subcase 3 is:
$ 
$      P     =  2[75 + 50i(cos30 degrees - i sin30 degrees)] = 200 + 86.6i
$       z,6
$ 
$    Note: At f = 30 cps the three subcases are nearly identical.
$ 
$ 5. Random Analysis Data
$ 
$    The nonzero factors for the three subcases are:
$ 
$      S   =  50              |
$       11                    |
$                             |
$      S   =  S   =  50       |
$       13     31             |
$                             |   0 < f < 100
$      S   =  100             |
$       22                    |
$                             |
$      S   =  50              |
$       33                    |
$ 
$      S   =  0 ,  f > 100
$       ij
$ 
$    The time lags selected for the autocorrelation function calculations are:
$ 
$      T =  0.0, 0.001, 0.002,..., 0.1
$ 
$ 6. Static Loads for Restart
$ 
$    The problem is run first as a frequency response analysis. It is restarted
$    as a static analysis with the following loads:
$ 
$      Gravity vector:       g   =  32.2
$                             z
$ 
$      Element Deformation:  delta   = 0.089045 (expansion)
$                                 10
$ 
$ C. Theory
$ 
$    1. The theoretical eigenvalue data according to Reference 8 is
$ 
$             2  2
$            n pi
$     f   =  ------- sqrt(EI/A)  =  50, 200, 450, 800 ... (natural freqs.)   (1)
$      n           2
$            (2pi)l
$ 
$     m   =  1.0    (modal mass)                                             (2)
$      n
$ 
$                                             2 n pi x    -1/2
$     phi (x)  =  [integral from o to l pA sin  ------ dx]
$        n                                         l
$ 
$                      n pi x                  n pi x
$                  sin(------) = sqrt(2/M) sin(------)  (mode shape)         (3)
$                        l                       l
$ 
$    2. The theoretical frequency response at the center point is essentially
$       the response of the first mode which is
$ 
$                 sum from j phi   P (w) phi
$                               1,6 j       1,j
$       u (w)  =  -----------------------------    (j = deg. of freedom no.) (4)
$        6               2    2
$                    m (w  - w  + igww )
$                     1  i            1
$ 
$       At the first natural frequency of 50 cps, the response will be nearly
$       equal to the response of the first mode. The responses at the center
$       point for the three subcases are
$ 
$        1    3   94.764 + 41.033i
$       u  = u  = ----------------     (Subcases 1 and 3)                    (5)
$        6    6           2
$                  (50 - f ) + if
$ 
$        2   23.691(3 + 2cos2f - 2i sin2f)
$       u  = -----------------------------    (Subcase 2)                    (6)
$        6              2
$                (50 - f ) + if
$ 
$    3. The random analysis is explained in Reference 15. The power spectral
$       response coefficients for the three subcases are given by the matrix:
$ 
$                     +                  +
$                     |  0.5    0    0.5 |
$                     |                  |
$       [S ]  =  100  |   0    1.0    0  |                                   (7)
$         i           |                  |
$                     |  0.5    0    0.5 |
$                     +                  +
$ 
$       If {H } is the vector of the responses of a point, i, to the three
$            j
$       loading cases, the power spectral response, S  is
$                                            j
$ 
$               _  T                _
$       S   =  {H } [S]{H }        (H  is the complex conjugate)             (8)
$        j       j       j           j
$ 
$       or
$ 
$                          2     _      _            2         2
$       S   =  100[0.5|H  | +0.5(H  H  +H  H  )+|H  | +0.5|H  | ]            (9)
$        j              1j        1j 3j  3j 1j    2j        3j
$ 
$       Since H   = H  , then:
$              1j    3j
$ 
$                      2           2
$       S   =  200|H  |  + 100|H  |                                         (10)
$        j          1j          2j
$ 
$       The mean square response is obtained by integrating the power spectral
$       density over the frequency. In this particular case the frequency
$       increments are uniform and the mean square response is simply
$ 
$       E   =  sum from i pi [S (f   ) - S (f )] gradient f                 (11)
$        i                     j  i+1     j  i
$ 
$       The analytic solution for the displacement spectral density response of
$       the center point due to the first mode is:
$ 
$                             4                3
$                 200(1.066x10 ) + 100(.5613x10 )(13 + 12cos2f)
$       S (f)  =  ---------------------------------------------  =
$        j                      2    2 2    2
$                           [(50  - f )  + f ]
$ 
$                         6           6
$                 2.862x10  + .6735x10  cos2f
$                 ---------------------------                               (12)
$                       4        2     4
$                     (f  - 4999f  + 50 )
$ 
$       The mean deviation, sigma , is
$                                j
$ 
$                            E
$                             i
$       sigma   =  sqrt(--------------)                                     (13)
$            j          2 pi (f  - f )
$                              n    o
$ 
$       where f  and f  are the upper and lower frequency limits.
$              n      o
$ 
$    4. The results of the static analysis restart are
$ 
$       a. The gravity load produces normal displacements (in the z direction)
$          and element moments as follows:
$ 
$                    pAgx   3      2    3
$          u (x)  =  ---- (l  - 2lx  + x )                                  (14)
$           z        24EI
$ 
$                    pAg   2
$          M (x)  =  --- (x  - lx)                                          (15)
$           1         2
$ 
$       b. The element deformation produces the following axial forces and
$          displacements:
$ 
$                    delta 10
$          F   =  AE --------                                               (16)
$           x           l
$ 
$                   F
$                    x
$          u   =  - -- x     (x < 18)                                       (17)
$           x       AE
$ 
$ D. Results
$ 
$ The responses at the center point for Subcases 1 and 3 are:
$ 
$       ----------------------------------------------
$ 
$        f    u  (one mode)         u  (NASTRAN)
$              6                     6
$       ----------------------------------------------
$        0    .0413 @ 23.42 deg.    .0429 @ 22.9 deg.
$ 
$       30    .0646 @ 22.34 deg.    .0668 @ 21.8 deg.
$ 
$       50    2.066 @ 293.42 deg.   2.074 @ 281.5 deg.
$       ----------------------------------------------
$ 
$ The response at the center point for Subcase 2 is:
$ 
$       ----------------------------------------------
$ 
$        f    u  (one mode)         u  (NASTRAN)
$              6                     6
$       ----------------------------------------------
$        0    .047 @ 0 deg.         .049 @ 0 deg.
$ 
$       30    .0646 @ -22.34 deg.   .0668 @ -23.97 deg.
$ 
$       50    1.565 @ 233.4 deg.    1.577 @ 223.0 deg.
$       ----------------------------------------------
$ 
$ In numerical terms, the displacements of the center point (x = l/2) are:
$ 
$                  Theoretical         NASTRAN
$ 
$                          -2                    -2
$        u    =  4.452 x 10            4.435 x 10
$         x6
$                          -4                    -4
$        u    =  4.155 x 10            4.121 x 10
$         z6
$ 
$ The element forces at the center of the beam are:
$ 
$                  Theoretical         NASTRAN
$ 
$                           6                     6
$        F    =  -.9811 x 10           -.9848 x 10
$         x5
$ 
$        M    =  -8.607                -8.607
$         6
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