The Finite Element Method AP ractical Course by pYJ3t1

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									The Finite Element Method
              A Practical Course



CHAPTER 5:

FEM FOR 2D SOLIDS
                      CONTENTS
   INTRODUCTION
   LINEAR TRIANGULAR ELEMENTS
    –   Field variable interpolation
    –   Shape functions construction
    –   Using area coordinates
    –   Strain matrix
    –   Element matrices
   LINEAR RECTANGULAR ELEMENTS
    –   Shape functions construction
    –   Strain matrix
    –   Element matrices
    –   Gauss integration
    –   Evaluation of me
                      CONTENTS
   LINEAR QUADRILATERAL ELEMENTS
    – Coordinate mapping
    – Strain matrix
    – Element matrices
    – Remarks
   HIGHER ORDER ELEMENTS
   COMMENTS (GAUSS INTEGRATION)
   CASE STUDY
           INTRODUCTION
 2D solid elements are applicable for the analysis
  of plane strain and plane stress problems.
 A 2D solid element can have a triangular,
  rectangular or quadrilateral shape with straight or
  curved edges.
 2D solid element can deform only in the plane of
  the 2D solid.
 At any point, there are two components in x and y
  directions for the displacement as well as forces.
             INTRODUCTION
   For plane strain problems, the thickness of the
    element is unit, but for plane stress problems, the
    actual thickness must be used.
   In this course, it is assumed that the element has a
    uniform thickness h.
   Formulating 2-D elements with given variation of
    thickness is also straightforward, as the procedure
    is the same as that for a uniform element.
2D solids – plane stress and plane strain




 Plane stress                Plane strain
LINEAR TRIANGULAR ELEMENTS
 Less accurate than quadrilateral elements
 Used by most mesh generators for complex
  geometry
 Linear triangular element

    Nodes   Triangular elements
                                  y, v                3 (x3, y3)
                                                        (u3, v3)
                                                                   fsy
                                                                         fsx
                                                       A
                                                                          2 (x2, y2)
                                                                            (u2, v2)
                                         1 (x1, y1)
                                           (u1, v1)                            x, u
Field variable interpolation
U h ( x, y )  N( x, y )de
              u1         
              v           displaceme nts at node 1
               1         
              u 
                         
where    de   2          displaceme nts at node 2
              v 2        
              u 3        
                          displaceme nts at node 3
               v3 
                         
                                                   y, v                3 (x3, y3)
   N1    0     N2     0       N3    0                                  (u3, v3)
N
  0      N1    0      N2      0     N3 
                                        
                                                                                    fsy
                                                                                          fsx
      Node 1      Node 2          Node 3                                A
                                                                                           2 (x2, y2)
      (Shape functions)                                                                      (u2, v2)
                                                          1 (x1, y1)
                                                            (u1, v1)                            x, u
Shape functions construction

Assume,
N1  a1  b1 x  c1 y   Ni  ai  bi x  ci y       i= 1, 2, 3
N2  a2  b2 x  c2 y                          ai 
                                                
                        or    Ni  1 x     y bi   pT 
N3  a3  b3 x  c3 y                          c 
                                       pT       i
                                                
Shape functions construction

Delta function property:
                                                  N1 ( x1 , y1 )  1
                   1   for i  j
Ni ( x j , y j )                                N1 ( x2 , y2 )  0
                   0    for i  j
                                                  N1 ( x3 , y3 )  0

Therefore, N1 ( x1 , y1 )  a1  b1 x1  c1 y1  1
           N1 ( x2 , y2 )  a1  b1 x2  c1 y2  0
           N1 ( x3 , y3 )  a1  b1 x3  c1 y3  0

                     x2 y3  x3 y2        y2  y3        x3  x2
Solving,        a1                , b1          , c1 
                         2 Ae              2 Ae           2 Ae
    Shape functions construction
                    1 x1      y1
           1    1                 1
    Ae      P  1 x2         y2  [( x2 y3  x3 y2 )  ( y2  y3 ) x1  ( x3  x2 ) y1 ]
           2    2                 2
                  1 x3        y3

Area of triangle           Moment matrix

    Substitute a1, b1 and c1 back into N1 = a1 + b1x + c1y:
            1
    N1         [( y2  y3 )( x  x2 )  ( x3  x2 )( y  y2 )]
           2 Ae
Shape functions construction
Similarly,
                               1
N 2 ( x1 , y1 )  0   N2          [( x3 y1  x1 y3 )  ( y3  y1 ) x  ( x1  x3 ) y ]
                              2 Ae
N 2 ( x2 , y2 )  1
                             1
N 2 ( x3 , y3 )  0             [( y3  y1 )( x  x3 )  ( x1  x3 )( y  y3 )]
                            2 Ae
                            1
N3 ( x1 , y1 )  0    N3       [( x1 y2  x1 y1 )  ( y1  y2 ) x  ( x2  x1 ) y ]
                           2 Ae
N3 ( x2 , y2 )  0
                               1
N3 ( x3 , y3 )  1                [( y1  y2 )( x  x1 )  ( x2  x1 )( y  y1 )]
                              2 Ae
Shape functions construction

Ni  ai  bi x  ci y
               1
where    ai       ( x j yk  xk y j )    i= 1, 2, 3
              2 Ae
               1                          J, k determined from cyclic
         bi       ( y j  yk )           permutation
              2 Ae
                                                       i = 1, 2
               1                                     i
         ci       ( xk  x j )
              2 Ae
                                               k          j
                                    k = 3, 1                  j = 2, 3
           Using area coordinates
              Alternative method of constructing shape
               functions

                                       1 x       y
                                      1              1
                2-3-P:        A1      1 x2     y2  [( x2 y3  x3 y2 )  ( y2  y3 ) x  ( x3  x2 ) y ]
y                                     2              2
                                        1 x3     y3
                      k, 3
                                                          A1
                                                     L1 
                                                          Ae
                                                                                             A2
                          A1           Similarly,  3-1-P                 A2            L2 
                  P                                                                          Ae
                                                                                             A3
    i, 1                                               1-2-P             A3            L3 
                                      j, 2                                                   Ae
                                             x
Using area coordinates
Partitions of unity:    L1  L2  L3  1

                     A1 A2 A3 A1  A2  A3
       L1  L2  L3                    1
                     Ae Ae Ae      Ae

Delta function property: e.g. L1 = 0 at if P at nodes 2 or 3

Therefore,   N1  L1 , N 2  L2 , N3  L3

                  U h ( x, y )  N( x, y )de
Strain matrix
       u                                           
 xx                                             0
        x                                 x        
       v                                         
 yy                  LU     where L   0
        y
                                                  y 
                                                      
       u v                              
 xy                                             
        y  x                                       
                                           y
                                                  x 
                                                      
  LU  LNd e  Bd e
                  
                0
          x                    b1 0 b2 0 b3 0 
                
B  LN   0        N       B   0 c1 0 c2 0 c3 
                                                        
                y 
         
                                  c1 b1 c2 b2 c3 b3 
                                                        
                 
                   
          y
                x 
                                 (constant strain element)
Element matrices
                        h
k e   B cBdV   (  dz )BT cBdA   hBT cBdA
          T
                        0
    Ve            Ae                    Ae

                                       Constant matrix
 k e  hAe BT cB


                        h
me   N NdV 
              T
                          dz NT NdA   hNT NdA
                        0
     Ve            Ae                   Ae
Element matrices
For elements with uniform density and thickness,
              N1 N1     0       N1 N 2    0       N1 N 3     0 
              0        N1 N1     0       N1 N 2    0       N1 N 3 
                                                                  
              N 2 N1     0      N2 N2      0      N 2 N3     0 
me  h                                                           dA
        Ae    0        N 2 N1     0      N2 N2      0      N 2 N3 
              N 3 N1     0      N3 N 2     0      N3 N3      0 
                                                                  
              0
                       N3 N1      0      N3 N 2     0      N3 N3 

Eisenberg and Malvern (1973):

                                           m! n! p!
                    L1 Ln L3 dA 
                      m     p
                                                        2A
                                       (m  n  p  2)!
                         2
                 A
                                    y, v                3 (x3, y3)
Element matrices                                          (u3, v3)
                                                                     fsy
                                                                           fsx
         2 0 1 0 1 0                                   A
           2 0 1 0 1
                                                                          2 (x2, y2)
     hA      2 0 1 0                                                       (u2, v2)
me                                      1 (x1, y1)
     12         2 0 1                      (u1, v1)                            x, u
          sy.     2 0
                     
         
                    2
                      

                f sx 
 fe   NT
                 dl                     0
       l  2 3
                f sy                    0
                                           
                                  1        fx 
                                           
 Uniform distributed load:   f e  l 2 3  
                                  2        fy
                                           fx 
                                           
                                           fy
                                           
    LINEAR RECTANGULAR
         ELEMENTS
 Non-constant strain matrix
 More accurate representation of stress and strain
 Regular shape makes formulation easy
Shape functions construction

  Consider a rectangular element


       u1                                 y, v
                  displacements at node 1
      v                                                             
       1                                               4 (x4, y4)             3 (x3, y3)
      u2     
                displacements at node 2                 (u4, v4)               (u3, v3)
      v      
 de   2                                                                      fsy
      u3                                         2b
       v3       displacements at node 3                                               fsx
             
                                                                      2a
      u4     
      v         displacements at node 4               1 (x1, y1)             2 (x2, y2)
       4     
                                                           (u1, v1)               (u2, v2)

                                                                                        x, u
Shape functions construction
                                                                  4 (1, +1)                 3 (1, +1)
  y, v                                                              (u4, v4)                   (u3, v3)
                            
              4 (x4, y4)             3 (x3, y3)
                (u4, v4)               (u3, v3)                                    
                                     fsy
         2b                                   fsx                  2
                                                                                       
                           2a
              1 (x1, y1)             2 (x2, y2)                                2
                (u1, v1)               (u2, v2)                   1 (1, 1)                 2 (1, 1)
                                                                    (u1, v1)                   (u2, v2)
                                             x, u

      x  ( x2  x1 ) / 2      y  ( y2  y1 ) / 2
                        , 
              a                        b

                                                       N1   0          N2         0 N3         0         N4   0
U ( x, y )  N( x, y )de where
  h
                                                    N
                                                      0     N1         0          N2 0         N3        0    N4 
                                                                                                                  
                                                         Node 1           Node 2            Node 3         Node 4
Shape functions construction
N 1  1 (1   )(1   )
      4

N 2  1 (1   )(1   )
      4

N 3  1 (1   )(1   )
                                          N j  1 (1   j  )(1   j )
                                                4
      4

N 4  1 (1   )(1   )                                                        4 (1, +1)                3 (1, +1)
      4                                                                           (u4, v4)                  (u3, v3)

N 3 at node 1  1 (1   )(1   )  1  0
                4
                                   1                                                          
N 3 at node 2  1 (1   )(1   )  1  0
                4                                   Delta function
                                   1                                          2
N 3 at node 3  1 (1   )(1   )  1  1         property                                         
                4
                                   1

N 3 at node 4  1 (1   )(1   )  1  0
                4
                                                                                             2
                                   1                                          1 (1, 1)                2 (1, 1)
 4                                                                                (u1, v1)                  (u2, v2)

        Ni  N1  N 2  N 3  N 4
i 1                                                                                 Partition of
 [(1   )(1   )  (1   )(1   )  (1   )(1   )  (1   )(1   )]
     1
     4                                                                               unity
 1 [2(1   )  2(1   )]  1
  4
Strain matrix

           1
              a     0     1
                           a      0      1
                                          a    0      1
                                                      a      0 
        1          1            1         1           1 
B  LN   0        b     0      b     0      b     0       b 
        4 1
                  1
                    a
                           1
                           b
                                  1    1   1   1
                                                             a 
                                                              1
          b                       a      b     a     b           


Note: No longer a constant matrix!
Element matrices

  x a,   y b        dxdy = ab dd

Therefore,

                       1 1
k e   hB cBdA  
         T
                              abhB T cBdd
                       1 1
     A

                       h                           1       1
me   N NdV    dz N NdA   hN NdA  
         T                       T             T
                                                                abh NT Nd d
                       0                           1   1
     V             A                    A
Element matrices

               f sx 
fe   N T
                dl
         2 3
               f sy 
      l



For uniformly distributed load,
       0                        y, v
                                                            
       0                                    4 (x4, y4)             3 (x3, y3)
                                              (u4, v4)               (u3, v3)
        fx                                                         fsy
                                       2b                                   fsx
        fy
f e  b                                                  2a
        fx                                  1 (x1, y1)             2 (x2, y2)
        fy                                    (u1, v1)               (u2, v2)
        
       0                                                                   x, u
       0 
        
Gauss integration

   For evaluation of integrals in ke and me (in practice)

                             1                  m
    In 1 direction: I           f ( )d   w j f ( j )
                         1
                                                 j 1


    m gauss points gives exact solution of
    polynomial integrand of n = 2m - 1

                                                          nx     ny
                             1       1
    In 2 directions: I                  f ( , )d d            wi w j f (i , j )
                         1       1
                                                          i 1   j 1
Gauss integration
 m        Gauss Point j            Gauss Weight wj   Accuracy order
                                                            n
 1   0                          2                           1

 2   -1/3, 1/3                1, 1                        3

 3   -0.6, 0, 0.6             5/9, 8/9, 5/9               5

 4   -0.861136, -0.339981,      0.347855, 0.652145,         7
     0.339981, 0.861136         0.652145, 0.347855
 5   -0.906180, -0.538469, 0,   0.236927, 0.478629,         9
     0.538469, 0.906180         0.568889, 0.478629,
                                0.236927
 6   -0.932470, -0.661209,      0.171324, 0.360762,        11
     -0.238619, 0.238619,       0.467914, 0.467914,
     0.661209, 0.932470         0.360762, 0.171324
Evaluation of me

             4 0 2 0 1 0 2 0
               4 0 2 0 1 0 2
                             
                  4 0 2 0 1 0
                             
        hab        4 0 2 0 1
   me 
         9            4 0 2 0
                             
                        4 0 2
              sy.         4 0
                             
             
                            4
                              
Evaluation of me
               1 1
mij  hab               N i N j dd
              1 1

       hab       1                                1
   
      16 1           (1   i )(1   j )d 
                                                1
                                                         (1  i )(1   j )d

     hab
         (1  1  i j )(1  1 i j )
               3              3
       4
                  hab                                                 hab
E.g. m33                    (1   1  1)(1   1  1)  4 
                                  1
                                  3
                                                    1
                                                    3
                       4                                                 9

  Note: In practice, gauss integration is often used
    LINEAR QUADRILATERAL
          ELEMENTS
 Rectangular elements have limited application
 Quadrilateral elements with unparallel edges are
  more useful
 Irregular shape requires coordinate mapping
  before using Gauss integration
Coordinate mapping
  y     4 (x4, y4)
                     3 (x3, y3)
                                                  
                                          1,
                                       4 ( +1)        3 (1, +1)
                                                                 

                          2 (x2, y2)
        1 (x1, y1)
                             x         1 ( 
                                          1, 1)       2 (1, 
                                                            1)


      Physical coordinates              Natural coordinates

  U h ( , )  N( , )d e (Interpolation of displacements)

  X( , )  N( , )xe (Interpolation of coordinates)
Coordinate mapping
                                 x1           
                                y                coordinate at node 1
                                                
X( , )  N( , )xe            1
                                 x2           
                                                 coordinate at node 2
              x                y2           
where       X  ,        xe   
               y               x3           
                                 y3              coordinate at node 3
                                              
                                 x4           
N1  1 (1   )(1   )         y                coordinate at node 4
     4                           4            
N 2  1 (1   )(1   )
      4                                   4
N 3  1 (1   )(1   )
      4                           x   N i ( ,  ) xi
                                         i 1
N 4  1 (1   )(1   )
      4
                                          4
                                  y   N i ( , ) y i
                                         i 1
Coordinate mapping
                                           4
Substitute 1 into x   N i ( , ) xi
                                          i 1

x  1 (1   ) x2  1 (1   ) x3
    2               2                                    x  1 ( x 2  x3 )  1  ( x3  x 2 )
                                                             2                2
                                               or
y  (1   ) y2  (1   ) y3
     1
     2
                     1
                     2                                   y  1 ( y 2  y3 )  1 ( y3  y 2 )
                                                             2                2


                                    ( y3  y2 )
Eliminating ,               y                 {x  1 ( x2  x3 )}  1 ( y2  y3 )
                                    ( x3  x2 )      2                2



                         y   4 (x4, y4)
                                                 3 (x3, y3)
                                                                              
                                                                     4 (1, +1)     3 (1, +1)
                                                                                               

                                                      2 (x2, y2)
                             1 (x1, y1)
                                                        x             1 (1, 1)   2 (1, 1)
Strain matrix
N i N i    x N i    y               N i     N i 
                                              x 
    x        y    
                                  or           J      
N i N i    x N i     y              N i     N i 
                                              y 
    x      y                                   

              x     y 
                    
where      J                 (Jacobian matrix)
              x     y 
                    
                        
                                     N1   N 2    N3   N 4   x1   y1 
                                                       x2    y2 
Since X( , )  N( , )xe ,     J                                    
                                     N1   N 2    N3   N 4   x3   y3 
                                                                       
                                                        x4
                                                                      y4 
                                                                           
Strain matrix
              N i       N i 
              x           (Relationship between differentials of shape
Therefore,         J 
                       1
                                  functions w.r.t. physical coordinates and
              N i       N i 
              y           differentials w.r.t. natural coordinates)
                              

                       Replace differentials of Ni w.r.t. x and y
                       with differentials of Ni w.r.t.  and 


          x  0 
B  LN   0
               y  N
                    
          y  x 
                   
Element matrices
Murnaghan (1951) : dA=det |J | dd

          1       1
ke                   hBTcBdet J d d
         1      1



                               h
m e   N NdV    dxN T NdA   hN T NdA
               T
                               0
     V                     A               A
         1 1
              hN T Ndet J dd
      1 1
Remarks

   Shape functions used for interpolating the coordinates are
    the same as the shape functions used for the interpolation
    of the displacement field. Therefore, the element is called
    isoparametric element.
   Note that the shape functions for coordinate interpolation
    and displacement interpolation do not have to be the same.
   Using the different shape functions for coordinate
    interpolation and displacement interpolation, respectively,
    will lead to the development of so-called subparametric or
    superparametric elements.
          HIGHER ORDER ELEMENTS
             Higher order triangular elements
                             (0,0,p)

                                                       nd = (p+1)(p+2)/2
              (1,0,p1)                (0,1,p1)
                                                       Node i,               I J K  p
          (2,0,p2)
                                                       Argyris, 1968 :
                                       L1
                  L2
                          i (I,J,K)
                                                                Ni  lII ( L1 )lJJ ( L2 )lK ( L3 )
                                                                                          K


                               L3                  (0,p1,1)               ( L  L 0 )( L  L 1 )     ( L  L (  1) )
                                                            l ( L ) 
                                                                          ( L   L 0 )( L   L 1 )   ( L   L (  1) )
(p,0,0)   (p1,1,0)                                   (0,p,0)
HIGHER ORDER ELEMENTS
    Higher order triangular elements (Cont’d)
    y, v           3           N1  (2 L1  1) L1
                               N 4  4 L1 L2
                           5
               6

                                2                                                                  1
                                                    y, v               3                    N1      (3L1  1)(3L1  2) L1
                       4                                                                           2
           1                                                                    7                  9
                                         x, u                      8                         N4     L1 L2 (3L1  1)
                                                                                        6          2
                                                               9                                N10  27 L1 L2 L3
                                                                           10
                                                                                            2
       Quadratic element                                   1           4
                                                                                    5

                                                                                                       x, u



                                                           Cubic element
        HIGHER ORDER ELEMENTS
           Higher order rectangular elements
                                        Lagrange type:
                                       (Zienkiewicz et al., 2000)
(0,m)                           (n,m)

                                               Ni  N I1D N J D  lIn ( )lJm ( )
                                                            1

                         i(I,J)
                                                               (  0 )(  1 ) (   k 1 )(   k 1 ) (   n )
                                             lkn ( ) 
                                                            ( k  0 )( k  1 ) ( k   k 1 )( k   k 1 ) ( k   n )
                  0




(0,0)                             (n,0)
HIGHER ORDER ELEMENTS
   Higher order rectangular elements(Cont’d)
        4    7       3
       J=2
                          

                               
                                          6
                                              (9 node quadratic element)
       J=1     8     9


                                                                                       1
       J=0                                              N 5  N1 D ( ) N 0 D ( )   (1   )(1   )(1   )
                                                                1         1

              1           5          2                                                 2
             I=0         I=1        I=2
                                                                                     1
                                                        N 6  N 2 D ( ) N1 D ( )   (1   )(1   )(1   )
                                                                1         1

                              1                                                      2
N1  N 0 D ( ) N 0 D ( )   (1   ) (1   )
       1          1
                                                                                     1
                              4                         N 7  N1 D ( ) N 2 D ( )  (1   )(1   )(1   )
                                                                1         1

                                1                                                    2
N 2  N 2 D ( ) N 0 D ( )    (1   ) (1   )
        1          1
                                                                                       1
                                4                       N8  N 0 D ( ) N11D ( )    (1   )(1   )
                                                                1

                              1                                                        2
N 3  N 2 D ( ) N 2 D ( )   (1   )(1   )
        1          1
                                                        N 9  N1 D ( ) N1 D ( )  (1   2 )(1   2 )
                                                                1         1
                              4
                                1
N 4  N 0 D ( ) N 2 D ( )    (1   )(1   )
        1          1

                                4
            HIGHER ORDER ELEMENTS
                Higher order rectangular elements(Cont’d)
                 Serendipity type:
                =1        
        4              7       3

                                     N j  1 (1   j )(1   j )( j   j  1)
                                                    4                                           j  1, 2, 3, 4
                                   6          N j  1 (1   2 )(1   j )
                                                    2                                           j  5, 7
    8                  0               
                                              N j  1 (1   j )(1   2 )
                                                    2                                           j  6,8

        1                  5   2
                =1

        (eight node quadratic element)
HIGHER ORDER ELEMENTS
    Higher order rectangular elements(Cont’d)

                                  N j  32 (1   j )(1   j )(9 2  9 2  10)
                                         1

4        10       9   3
                                                  for corner nodes j  1, 2, 3, 4

                          8
                                   N j  32 (1   j )(1   2 )(1  9 j )
                                         9
    11
                              
                                                  for side nodes j  7, 8, 11, 12 where  j  1 and  j   1
                                                                                                             3
    12                    7
                                   N j  32 (1   j )(1   2 )(1  9 j )
                                         9



                  6   2
                                                  for side nodes j  5, 6, 9, 10 where  j   1 and  j  1
                                                                                               3
1        5



(twelve node cubic element)
ELEMENT WITH CURVED
      EDGES
     3                               3
                 5
                                                 5
                                 6
     6
                                                         2
         4               2                   4
                             1
     1




                                 4       7           3
                     3
     4   7


     8                   6                               6
                             8

 1                   2
             5
                             1           5               2
COMMENTS (GAUSS INTEGRATION)

   When the Gauss integration scheme is used, one has to
    decide how many Gauss points should be used.
   Theoretically, for a one-dimensional integral, using m
    points can give the exact solution for the integral of a
    polynomial integrand of up to an order of (2m1).
   As a general rule of thumb, more points should be used for
    higher order of elements.
         COMMENTS (GAUSS
           INTEGRATION)
   Using smaller number of Gauss points tends to counteract
    the over-stiff behaviour associated with the displacement-
    based method.
   Displacement in an element is assumed using shape
    functions. This implies that the deformation of the element
    is somehow prescribed in a fashion of the shape function.
    This prescription gives a constraint to the element. The so-
    constrained element behaves stiffer than it should be. It is
    often observed that higher order elements are usually softer
    than lower order ones. This is because using higher order
    elements gives less constraint to the elements.
         COMMENTS (GAUSS
           INTEGRATION)
   Two gauss points for linear elements, and two or three
    points for quadratic elements in each direction should be
    sufficient for many cases.
   Most of the explicit FEM codes based on explicit
    formulation tend to use one-point integration to achieve the
    best performance in saving CPU time.
              CASE STUDY
   Side drive micro-motor
        CASE STUDY
10N/m
                        Elastic Properties of Polysilicon
        10N/m
                  Young’s Modulus, E             169GPa

                   Poisson’s ratio,              0.262

          10N/m       Density,                2300kgm-3
  CASE STUDY




Analysis no. 1: Von Mises stress
  distribution using 24 bilinear
quadrilateral elements (41 nodes)
    CASE STUDY




 Analysis no. 2: Von Mises stress
  distribution using 96 bilinear
quadrilateral elements (129 nodes)
   CASE STUDY




 Analysis no. 3: Von Mises stress
  distribution using 144 bilinear
quadrilateral elements (185 nodes)
  CASE STUDY




Analysis no. 4: Von Mises stress
distribution using 24 eight-nodal,
 quadratic elements (105 nodes)
  CASE STUDY




 Analysis no. 5: Von Mises stress
distribution using 192 three-nodal,
 triangular elements (129 nodes)
             CASE STUDY
                              Total number    Maximum
Analysis   Number / type of
                               of nodes in    Von Mises
  no.         elements
                                  model      Stress (GPa)

              24 bilinear,
   1                              41           0.0139
             quadrilateral

              96 bilinear,
   2                              129          0.0180
             quadrilateral

             144 bilinear,
   3                              185          0.0197
             quadrilateral

            24 quadratic,
   4                              105          0.0191
            quadrilateral

   5          192 linear,         129          0.0167
              triangular

								
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