$-------------------------------------------------------------------------------
$                       RIGID FORMAT No. 1, Static Analysis
$       Simply-Supported Rectangular Plate with a Thermal Gradient (1-11-1)
$   Simply-Supported Rectangular Plate with a Thermal Gradient (INPUT, 1-11-2)
$ 
$ A. Description
$ 
$ This problem illustrates the solution of a general thermal load on a plate 
$ with the use of an equivalent linear thermal gradient. The thermal field is a 
$ function of three dimensions, demonstrated by the TEMPP1 card. The plate is 
$ modeled with the general quadrilateral, QUAD1, elements. Two planes of 
$ symmetry are used. This problem is repeated via the INPUT module to generate 
$ the QUAD1 elements. 
$ 
$ B. Input
$ 
$                    5            2
$      E  =  3.0 x 10  pounds/inch      (Young's modulus)
$ 
$      v  =  0.3                        (Poisson's ratio)
$                           2     4
$      p   =  1.0 pound-sec. /inch      (Mass density)
$ 
$  alpha   =  0.01 inch/deg. F/inch     (Thermal expansion coefficient)
$ 
$      T   =  0.0 deg. F                (Reference temperature)
$       R
$ 
$      T   =  2.5 deg. F                (Temperature difference)
$       o
$ 
$      a   =  10.0 inch                 (Width)
$ 
$      b   =  20.0 inch                 (Length)
$ 
$      t   =  0.5 inch                  (Thickness)
$ 
$ The thermal field is
$ 
$                   pix       piy  2z 3
$      T  =  T (cos ---) (cos ---)(--)  
$             o      a        b     t
$ 
$                      pix        piy   3
$ and     =  160.0(cos ---)  (cos ---) z  deg. F
$                      10         20
$ 
$ C. Theory
$ 
$ The plate was solved using a minimum energy solution. The net moments, 
$ {M }, in the plate are equal to the sum of the elastic moments, {M }, and the 
$   N                                                               e
$ thermal moments, {M }.
$                    t
$ 
$      {M }  =  {M } + {M }                                                  (1)
$        N        t      e
$ 
$ where the thermal moment is
$ 
$                                                     +
$                              +   +                  |
$                              | 1 |     pix      piy |
$      {M }  =  alphaT' D(1+v) | 1 | cos ---  cos --- |
$        t            o        | 0 |      a        b  |
$                              +   +                  |
$                3                                    |                      (2)
$              Et                                     |
$ and  D =  ---------                                 |
$                 2                                   |
$           12(1-v )                                  |
$                                                     +
$ 
$ and T' = 6T /5t is the effective thermal gradient.
$      o     o
$ 
$ The elastic moment is defined by the curvatures, x, with the equation:
$ 
$                 +               +
$                 | x   +  vx     |
$                 |  x       y    |
$                 |               |
$      {M }  =  D | x   +  vx     |                                          (3)
$        e        |  y       x    |
$                 |               |
$                 |    (1-v)      |
$                 |   ------- x   |
$                 |      2     xy |
$                 +               +
$ 
$ Assuming a normal displacement function, W, of
$ 
$ 
$                                         npix      mpiy
$      W  = sum from n sum from m W   cos ----  cos ----                     (4)
$                                  nm      a         b
$ 
$ 
$ then                                                      
$                                                                         +
$            2                                                            |
$           a W                              2     n  2     npix     mpiy |
$    x   =  ---  = - sum from n sum from m pi W  ( - )  cos ---- cos ---- |
$     x       2                                nm  a         a        b   |
$           ax                                                            |
$                                                                         |
$            2                                                            |
$           a W                              2     m  2     npix     mpiy |
$    x   =  ---  = - sum from n sum from m pi W  ( - )  cos ---- cos ---- |  (5)
$     y       2                                nm  a         a        b   |
$           ay                                                            |
$                                                                         |
$            2                                                            |
$           a W                              2    nm      npix     mpiy   |
$    x   =2 ---- = - sum from n sum from m pi W  (--) sin ---- sin ----   |
$     xy    axay                               nm ab       a        b     |
$                                                                         +
$ The work done by the thermal load is:
$ 
$                             T          1                    T       
$    U  =  integral from A {X} {M } dA + - integral from A {X} {M } dA       (6)
$                                t       2                       e    
$ 
$ where A is the surface area. Performing the substitution and integrating 
$ results in the energy expression: 
$  
$                            2   2  2
$            alphaT' D(1+v)pi  (a +b )
$                  o                      D  
$      U = - ---------------------  W   + -- sum from n=1 to infinity
$                   4ab              11   2
$ 
$                                                2                           (7)
$                                 4    ( 2     2)
$                                piab  (n     m )   2
$      sum from n=1 to infinity  ----  (--  + --)  W
$                                 4    ( 2     2)   nm
$                                      (a     b )
$ 
$ The static solution exists at a minimum energy:
$ 
$      aU
$      -----  =  0                                                           (8)
$      aW
$        nm
$ 
$ This results in all but W   equal to zero. The displacement function is 
$                          11
$ therefore: 
$ 
$                               2 2  
$                 alphaT' (1+v)a b 
$                       o             pix      piy
$      W(x,y)  =  --------------  cos ---  cos ---                           (9)
$                   2  2    2          a        b
$                  ã (a  + b )
$ 
$ Solving for moments by differentiating W and using equation (3) results in the 
$ equations for element moments: 
$                        +           +
$                        |     2   2 |
$                        |    b +va  |     pix     piy
$      M  = alphaT'D(1+v)|1 - -------| cos --- cos ---                      (10)
$       x         o      |     2  2  |      a       b
$                        |    a +b   |
$                        +           +
$                        +           +              
$                        |     2   2 |              
$                        |    a +vb  |     pix     piy
$      M  = alphaT'D(1+v)|1 - -------| cos --- cos ---                      (11)
$       y         o      |     2  2  |      a       b 
$                        |    a +b   |              
$                        +           +              
$ 
$                        2
$            alphaT'D(1-v )ab
$      M           o               pix     piy
$       xy = ----------------- sin --- cos ---                              (12)
$              2      2             a       b 
$             a   +  b
$ 
$ D. Results
$ 
$ The maximum errors for displacements, constraint forces, element forces, and 
$ element stresses are listed in Table 1. 
$ 
$               Table 1. Comparison of Analytical and NASTRAN Results
$              --------------------------------------------------------
$               CATEGORY       MAXIMUM         MAXIMUM                  
$                              ANALYTICAL      NASTRAN         PERCENT 
$                              VALUE           DIFFERENCE      ERROR    
$              --------------------------------------------------------
$              Displacement              -1               -3            
$                             6.2898 x 10     11.5464 x 10     -0.25   
$              -------------------------------------------------------- 
$              Constraint                                              
$              Force              150.0           -.9594       -0.65    
$              --------------------------------------------------------
$              Element Mom.,             2                              
$              M              1.4770 x 10        -1.1767       -0.80   
$               x                                                       
$              --------------------------------------------------------
$              Element                     3                            
$              Stress         7.764618 x 10     -90.33275      -1.16   
$              --------------------------------------------------------
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