$-------------------------------------------------------------------------------
$                   RIGID FORMAT No. 2, Inertia Relief Analysis
$        Inertia Relief Analysis of a Circular Ring Under Concentrated and
$                            Centrifugal Loads (2-1-1)
$ 
$ A. Description
$ 
$ This problem illustrates the use of inertia relief analysis to solve a free-
$ body problem. In inertia relief the structure is under constant acceleration
$ due to the applied loads; the reactions to the applied load are due to the
$ masses of the structure. Ficticious, nonredundant support points must be
$ provided to define a reference system attached to the body. The dlsplacements
$ of the body are measured relative to the supported coordinates.
$ 
$ The structure consists of a spinning ring with constant radial load applied to
$ one point. The rotational velocity creates centrifugal loads and the point
$ load causes inertia reactions. The actual dynamic motion of the whole
$ structure is a cyclic motion of the center point coinciding with the rotation
$ of the ring. The displacements measured by the inertia relief analysis,
$ however, will be the static motion relative to the support point
$ dlsplacements.
$ 
$ The displacements are defined in a cylindrical coordinate system (u1 = u ,
$                                                                    1    r
$ u  = u ,   u3 = u ). The elements used are BAR elements wIth a large cross-
$  2    phi        z
$ sectional area to minimize axial deformations. The BARs were offset a uniform
$ radial distance from the grid points to demonstrate the offset option of the
$ BAR element.
$ 
$ B. Input
$ 
$ 1. Parameters:
$ 
$    R = 10.0 (radius at end of BAR elements)
$ 
$    R  = 11.0 (Radius at grid points)
$     1
$ 
$    I = 10.0 (Bending inertia)
$ 
$    p = 0.5 (Mass density)
$ 
$    E = 1000. (Modulus of elasticity)
$ 
$    A = 1000. (Cross-sectional area)
$ 
$ 2. Loads:
$ 
$    P     = 100
$     r,13
$ 
$    f = 1.59 cps (Rotational velocity, w = 1.0 radians per second)
$ 
$ 3. Supports:
$ 
$    a) The u    direction is supported to restrict vertical translation.
$            r,1
$ 
$    b) The u      and u     directions are supported to restrict rotation and
$            phi,1      phi,13
$       horizontal translation.
$ 
$ 4. Grid Point Weight Generator Input:
$ 
$    Weight and moment of inertia are defined relative to point 19.
$ 
$ C. Theory
$ 
$ 1. The Element Forces and Moments may be solved by the following analysis, as
$    explained in Reference 7, Chapter 12.
$ 
$    a) Using symmetry the structure may be defined by the free-body diagram.
$ 
$       The static equilibrium equations at any angle are
$ 
$         A = A  cos phi + mu phi sin phi (Axial Force)                      (1)
$              o
$ 
$         V = A  sin phi + mu phi cos phi (Shear)                            (2)
$              o
$ 
$         M = M  + r[mu(1 - cos phi - phi sin phi) +
$              o
$ 
$         A (1 - cos phi)]  (Bending Moment)                                 (3)
$          o
$ 
$    b) Using energy and Castigliano's Theorem:
$ 
$               R                        2
$         U  = ---  integral o to pi of M  d phi                             (4)
$              2EI
$ 
$         deltaU
$         ------ = 0                                                         (5)
$         deltaM
$               o
$ 
$         deltaU
$         ------ = 0                                                         (6)
$         deltaA
$               o
$ 
$       These are the deflections at the bottom which are fixed. The resulting
$       two equations are used in step c.
$ 
$    c) Solving the equations in (b) gives the redundant forces:
$ 
$                 1          F
$         A  = - --- mu = - ---                                              (7)
$          o      2         4pi
$ 
$              Rmu   FR
$         M  = --- = ---                                                     (8)
$          o    2     4
$ 
$    d) Adding a dummy load and solving the problem with the above boundary
$       conditions gives the displacement due to the point load:
$ 
$                    3      2
$                  FR     pi
$         delta  = ---  ( -- - 1 )                                           (9)
$              f   piEI   8
$ 
$    e) The axial stress and radial displacement due to the centrifugal load is
$ 
$ 
$                    2  2           2
$         delta  = pR  w  = 5.0 x 10                                        (10)
$              w
$ 
$                     3 2
$                  2pR w
$         delta  = ------- = 1.0                                            (11)
$              w      E
$ 
$ D. Results
$ 
$ 1. The total result of summing the two loads is
$ 
$               -----------------------------------------------------
$                                                 THEORY    NASTRAN
$               -----------------------------------------------------
$                delta = Displacement u              1.75      1.734
$                                      r,13
$                M  = Moment BAR #1, end A         -79.5     -80.48
$                 o
$                M  = Moment BAR #12, end B       -238.5    -236.0
$                 1
$               -----------------------------------------------------
$ 
$ 2. The structural mass characteristics as calculated by the grid point weight
$    generator are
$ 
$               -----------------------------------------------------
$                         THEORETICAL                    NASTRAN
$               -----------------------------------------------------
$                X   = 11.0 from point 19                  11.0
$                 CG
$                              4               4                    4
$                Mass = pi x 10  = 3.14159 x 10          3.1326 x 10
$ 
$                            pi     6              6               6
$                I   = I   = -- x 10  = 1.5708 x 10     1.5663 x 10
$                 xx    yy   2
$ 
$                             6               6                     6
$                I   = pi x 10  = 3.14159 x 10           3.1326 x 10
$                 zz
$               -----------------------------------------------------
$ 
$                (Inertias are about center of gravity)
$ 
$    NASTRAN gives slightly different answers due to the polygonal shape of the
$    finite element model.
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