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ECIV 720 A Advanced Structural Mechanics and Analysis(1)

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ECIV 720 A Advanced Structural Mechanics and Analysis(1) Powered By Docstoc
					          ECIV 720 A
  Advanced Structural Mechanics
          and Analysis

Lecture 15:
Quadrilateral Isoparametric Elements
   (cont’d)
   Force Vectors
   Modeling Issues
Higher Order Elements
    Integration of Stiffness Matrix

            1 1
      k e  t   B DB det Jdd
                    T

            1 1

 T
B (8x3)
D (3x3)                 ke (8x8)
B (3x8)
        Integration of Stiffness Matrix

  Each term kij in ke is expressed as

              3 T 3          
       1 1
kij  t     Bim  Dml Blj  det J  , dd 
        1 1 m 1 l 1      
      1 1
    t   g  , dd
     1 1
  Linear Shape Functions is each Direction
    Gaussian Quadrature is accurate if
    We use 2 Points in each direction
                Integration of Stiffness Matrix
                               
                                                            w1  1
  1   1 3
                                                            w2  1
                                                   
  1   1 3
                                                 1 1
                                                t   g  , dd 
                  1   1 3   2  1 3
                                                 1 1

w1w1 g 1 ,1   w2 w1 g 2 ,1   w1w2 g 1 ,2   w2 w2 g 2 ,2 

               g 1 ,1   g 2 ,1   g 1 ,2   g 2 ,2 
      Choices in Numerical Integration



• Numerical Integration cannot produce exact
  results
• Accuracy of Integration is increased by using
  more integration points.
• Accuracy of computed FE solution DOES
  NOT necessarily increase by using more
  integration points.
                               FULL Integration



• A quadrature rule of sufficient accuracy to
  exactly integrate all stiffness coefficients kij

• e.g. 2-point Gauss rule
                                         exact for polynomials
 1   1 3                               up to 2nd order
                                      
 1   1 3



              1   1 3   2  1 3
   Reduced Integration, Underintegration
Use of an integration rule of less than full order

Advantages
• Reduced Computation Times
• May improve accuracy of FE results
• Stabilization

Disadvantages
  Spurious Modes
  (No resistance to nodal loads that tend to activate the
  mode)
               Spurious Modes
   Consider the 4-node plane stress element
                  1

                                     t=1
       1                             E=1
                                     v=0.3



8 degrees of freedom            8 modes

           Solve Eigenproblem
            Spurious Modes




     1  0
Rigid Body Mode              2  0
                     Rigid Body Mode
              Spurious Modes




     3  0
Rigid Body Mode
                Spurious Modes




4  0.495                   5  0.495
Flexural Mode                    Flexural Mode
  Spurious Modes




6  0.769
Shear Mode
            Spurious Modes




7  0.769              8  1.43
                    Uniform Extension Mode
Stretching Mode
                          (breathing)
             Element Body Forces
   Total Potential          Galerkin
      1 T
    ε DεtdA         0              σ εφ dV
                                               T

    e 2
        Ae                           Ve
                             e

      u ftdAT
                                φ fdV           T
          Ae                              Ve
      e                          e

      u TtdlT
                                φ TdA           T
          le                              Ae
      e                          e

     u PiT
           i                   φ Pi      T
                                           i
      i                          i
                        Body Forces
                   Integral of the form


    u f det Add                        φ f det Add
       T                                       T
Ae                                    Ae

                                                             q1 
    u   N1     0    N2   0    N3       0        N4   0  
                                                         
    v   0      N1   0    N2   0        N3       0    N4  
                                                            q8 

                                                            1 
     x   N1   0    N2    0   N3        0       N4   0  
                                                         
     y   0    N1    0   N2    0       N3        0   N4  
                                                            8 
                   Body Forces
               In both approaches

                       f x N1 det Add 
                       Ae                 
                       f y  N1 det Add 
                       Ae                  
    WP  q1     q8  f x N 2 det Add 
                      
                            Ae             
                                          
                      f                    
                       Aey  N 4 det Add 
                                           

Linear Shape Functions

Use same quadrature as stiffness maitrx
               Element Traction
   Total Potential          Galerkin
      1 T
    ε DεtdA         0              σ εφ dV
                                               T

    e 2
        Ae                           Ve
                             e

      u ftdAT
                                φ fdV           T
          Ae                              Ve
      e                          e

      u TtdlT
                                φ TdA           T
          le                              Ae
      e                          e

     u PiT
           i                   φ Pi      T
                                           i
      i                          i
                       Element Traction
    Similarly to triangles, traction is applied
    along sides of element               N 01
               3       v
                                           N 2  1   
                                                1
                           Ty
                                                2
                               Tx u
                                           N 3  1   
                                                1
4
                                               2
                                           N4  0
                                2
    1                               WP T23   u TtdlT
                                                 le
                            Traction
                                                          q1 
        u  0 0 N 2           0        N3   0     0 0  
                                                     
         v  0 0 0           N2        0    N3    0 0  
                                                         q8 
  For constant traction along side 2-3
                                 0
                                 0
                                                 Traction
                                 Tx 
                       t e l2 3                 components
WP T23  q1    q8            Ty 
                           2                       along 2-3
                                 0
                                  
                                 
                                 0
                                  
                                     Stresses
                                                              More Accurate at
1   1 3
                                                              Integration points
                                                                
1   1 3
                                                               Stresses are
                                                               calculated at
             1   1 3       2  1 3                         any ,
                                    1        0       1         0      1    0  1    0 

σ  DBq
                                                                                                             0 
                                 1  1        0       1        0      1    0       1   
                          G                                                                                      
                                 4 0           1        0       1    0 1    0                1   
                                   
                                    0          1        0       1       0    1    0         1    
                                                                                                                    

                                              J 22  J12                    0          0 
                                         1 
                                 A         0      0                     J 21      J11 
                                       det J
                                              J 21 J11
                                                                           J 22       J12 
                                                                                            
         Modeling Issues: Nodal Forces
    In view of…

      1 T
    ε DεtdA
    e 2
        Ae


        u ftdAT            A node should be
            Ae                placed at the location
        e

        u TtdlT            of nodal forces
            le
        e

       u PiT
             i
        i
Or virtual potential energy
Modeling Issues: Element Shape

     Square : Optimum Shape
     Not always possible to use

       Rectangles:         Larger ratios
       Rule of Thumb       may be used
       Ratio of sides <2   with caution


        Angular Distortion
        Internal Angle < 180o
    Modeling Issues: Degenerate Quadrilaterals
Coincident Corner Nodes
           4                                   4


           x             3
                                               x
                   x

       x                                   x       x
               x                                   x
1                                      1

                   2                                   2   3


                                       Integration Bias

                       Less accurate
    Modeling Issues: Degenerate Quadrilaterals
Three nodes collinear
          4
                                Integration Bias
          x           3                                    3
                  x
                                           4           x
                                           x
      x                               x
              x                                x
1                                 1

                  2                                2




                      Less accurate
  Modeling Issues: Degenerate Quadrilaterals


        2 nodes




Use only as necessary to improve representation of
geometry
      Do not use in place of triangular elements
              A NoNo Situation
y
                          3   (7,9)

                                             

             (6,4)                            
                      4

                                          Parent

      1                               2
                                          J singular
    (3,2)                        (9,2)

                                                   x
            All interior angles < 180
Another NoNo Situation


                            




                            

                         x, y
                         not uniquely
                         defined
              FEM at a glance

It should be clear by now that the cornerstone
in FEM procedures is the interpolation of the
displacement field from discrete values

                     m
      ux, y, z    N i x, y, z u i
                    n 1


Where m is the number of nodes that define
the interpolation and the finite element and N
is a set of Shape Functions
             FEM at a glance

                        m=2
1=-1          2=1




1               3              2
                                      m=3
 1=-1                        2=1
                           FEM at a glance
                      q6
                  3
                           q5
                                      m=3
    q2
              v
                      u     q4
         q1
    1                            q3
                           2

                                           1


                                                        4



                                 m=4                    
                                        2




                                                   3
                FEM at a glance
In order to derive the shape functions it was
assumed that the displacement field is a
polynomial of any degree, for all cases considered


1-D    ux   a0  a1 x  a2 x   an x
                                2          n




2-D   ux, y   a0  a1 x  a2 y  a3 xy  ...
Coefficients ai represent generalized coordinates
                  FEM at a glance
For the assumed displacement field to be admissible
we must enforce as many boundary conditions as the
number of polynomial coefficients
   1       e.g.      3                 2

   1=-1                            2=1

  ux  x1   a0  a1 x1  a x  u1
                               2
                             2 1

  ux  x2   a0  a1 x2  a x  u2
                               2
                             2 2

  ux  x3   a0  a1 x3  a x  u3
                               2
                             2 3
                FEM at a glance
This yields a system of as many equations as
the number of generalized displacements
        Ceofficient       a0  u1 
                               
          Matrix            
    
          ( x, y , z )    an  un 
                             
         that can be solved for ai

            a0            u1 
                      1     
                C    
            a            u 
             n            n
              FEM at a glance

Substituting ai in the assumed displacement field

   ux   a0  a1 x  a2 x   an x
                            2          n



and rearranging terms…
                   m
          ux    N i x u i
                  n 1
                     FEM at a glance

  1                 3                 2
                                            u()=a0+a1  +a2  2
   1=-1                            2=1


                                       u    N1  u1 
u(-1)=a0 -a1 +a2 =u1

u(1)=a0 +a1 +a2 =u2           …                N 2  u2 
                                               N 3  u3
u(0)=a0 =u3

N1      1                         N 2     1   
           1                                          1
           2                                          2
                   N3    1   1   
         Let’s go through the exercise


    1                                    2

    x1                                x2

Assume an incomplete form of quadratic variation



             u x   a0  a2 x   2
  Incomplete form of quadratic variation
   1                               2

    x1                            x2
          We must satisfy

ux1   a0  a2 x1  u1
                 2



ux2   a0  a2 x2  u2
                  2



                1 x  a0  u1 
                      2

                         
                      1
                     2
                1 x   a1  u2 
                      2
Incomplete form of quadratic variation

     1 x  a0  u1 
             2

              
             1
            2
     1 x   a1  u2 
             2


 And thus,


a0      1       x  x  u1 
                   2       2

  2 2                   
                   2       1
                 
 a1  x2  x1    1 1  u2 
 Incomplete form of quadratic variation


     x u x u
       2       2                 u1  u2
a0    2 1     1 2          a1  2 2
       x x
         2
         2     1
                2                x2  x1
  And substituting in

       u x   a0  a2 x 
                        2


        2
         x u  x u  u1  u2 2
                2
u x  2 1
                   2
                1 2
                             x
          2
          2x x 1
                 2
                    x2  x12
 Incomplete form of quadratic variation


         x u  x u  u1  u2 2
          2         2
u x    2 1
                   2
                    1 2
                             x
           x x
            2
            2       x2  x1
                    1
                     2     2




 Which can be cast in matrix form as

         x  x
          2     2
                      x x 2   2
                                     u1 
u x   2                1
                                     u 
          x  x1      x  x1
          2     2         2 2
          2               2          2 
          Isoparametric Formulation

The shape functions derived for the interpolation
of the displacement field are used to interpolate
geometry

     1                                         2

     x1                                    x2

       x2  x 2
         2
                      x x2    2
                                      x1 
    x 2                 1
                                    x 
       x2  x1        x  x1
               2         2  2
                         2           2 
            Intrinsic Coordinate Systems

   Intrinsic coordinate systems are introduced
   to eliminate dependency of Shape functions
   from geometry
            
                               N i  1  i 1  i 
 4 (-1,1)       3 (1,1)             1
                                    4
                                      The price?
                           
                               Jacobian of transformation

1 (-1,-1)       2 (1,-1)
                           Great Advantage for the money!
     Field Variables in Discrete Form

Geometry
        x  Nxn
                             (cartesian ) 
                        J                 
Displacement                 (intrinsic ) 
       u  Nun



Strain Tensor        Stress Tensor
e = B un              s = DB un
                     FEM at a glance

Element Strain Energy
     1 T        1 T
U e   ε σDdV  u e k eu e
     2 Ve       2
Work Potential of Body Force

W   f     u fdV  u
                 T         T
                           e    
                                      T
                                  N fdV
           Ve                   Ve

Work Potential of Surface Traction

W   f     u TdS  u
                 T          T
                            e   
                                      T
                                  N TdS   etc
            Se                   Se
            Higher Order Elements

 Quadrilateral Elements

     Recall the 4-node

                          Complete Polynomial
                     u ,   a1  a2  a3  a4

                   4 generalized displacements ai


4 Boundary Conditions for admissible displacements
              Higher Order Elements

Quadrilateral Elements
Assume Complete Quadratic Polynomial
u  ,   a1 
          a2  a3  a4 
          a5  a6  a7   a8   a9
              2     2      2   2   2         2



                    9 generalized displacements ai
                   9 BC for admissible displacements
9-node quadrilateral

    9-nodes x 2dof/node = 18 dof

   BT18x3    D3x3      B3x18

         ke 18x18
            9-node element Shape Functions

    Following the standard procedure the shape
    functions are derived as
4                         3
                                      Corner Nodes

                                      N i      i    i 
        7                                  1
                                                                          i  1,2,3,4
        9             6                    4
    8
                                     Mid-Side Nodes
        5
                                           2
                                               
                                      N i  1  i   i 
                                           1            2         2
                                                                      
1
    Middle Node
                          2
                                             i 1  i   i 1  i 
                                               i  5,6,7,8
        
Ni  1   1  
             2
                            2
                                     i9
        9-node element – Shape Functions

   Can also be derived from the 3-node axial element


          1                3                     2

          1=-1                                2=1

N1      1                            N 2     1   
           1                                             1
           2                                             2
                  N3    1   1   
    Construction of Lagrange Shape Functions
                          N1      1   
                                     1
N1      1   
           1                         2
           2
                      1 (-1,-1)




                                                      
          
                   (1,)
                                   (1,1)

                                 1           1           
  N1  ,   N1  N1      1      1   
                                 2           2           

             N i      i    i  i  1,2,3,4
                    1
                    4
N1,2,3,4 Graphical Representation
N5,6,7,8 Graphical Representation
N9 Graphical Representation
      Polynomials & the Pascal Triangle
 ux, y   a0  a1 x  a2 y  a3 xy an x                      n

               Pascal Triangle                            Degree
                     1                                      0
                      x        y                            1
               x2         xy         y2                     2
          x3       x 2y        xy2        y3                3
     x4     x 3y      x2y2         xy3          y4          4
x5   x 4y      x3y2        x2y3           xy4        y5     5

          …….
               Polynomials & the Pascal Triangle
          To construct a complete polynomial
                                               ux, y   a0  a1 x  a2 y  a3 xy
                          1               Q1                  4-node Quad
                      x        y                  Q2

                x2        xy         y2                    u  x, y  
          x3       x 2y        xy2        y3               a0 1 

     x4                                                    a1 x  a2 y 
            x 3y      x2y2         xy3          y4
                                                           a3 x 2  a4 xy  a5 y 2 
x5   x 4y       x3y2       x2y3           xy4        y5 a6 x 3  a7 x 2 y  a8 xy2  a9 y 3
                                                           9-node Quad
          …….
                                                                   etc
                          Incomplete Polynomials

                           1
                      x        y
                                                          ux, y   a0  a1 x  a2 y
               x2         xy         y2
                                                          3-node triangular
          x3       x 2y        xy2        y3
     x4     x 3y      x2y2         xy3          y4
x5   x 4y      x3y2         x2y3          xy4        y5

          …….
                          Incomplete Polynomials

                           1
                      x        y                         u  x, y  
               x2         xy         y2                  a0 1 
          x3       x 2y        xy2        y3             a1 x  a2 y 
     x4     x 3y      x2y2         xy3          y4       a3 x 2  a4 xy  a5 y 2 
x5   x 4y      x3y2         x2y3          xy4        y5 a6 x 2 y  a7 xy2

          …….
                 8-node quadrilateral
Assume interpolation
                             4                   3
                                         
u  x, y                           7
a0 1                                        6
                                 8
a1 x  a2 y                                         
a3 x 2  a4 xy  a5 y 2             5

a6 x y  a7 xy
     2           2
                             1                   2


8 coefficients to determine for admissible displ.
8-node quadrilateral

    8-nodes x 2dof/node = 16 dof

   BT16x3    D3x3      B3x16

         ke 16x16
            8-node element Shape Functions

    Following the standard procedure the shape
    functions are derived as
4                   3
                                      Corner Nodes
        7
                            N i  1  i 1  i i  i  1
                                 1
                6
    8                            4
                        
                                 i  1,2,3,4
        5
1                   2             Mid-Side Nodes


                     2
                                            
                N i  1  i  i  1  i   i 
                     1                           2        2
                                                                 
                       i  5,6,7,8
N1,2,3,4 Graphical Representation
N5,6,7,8 Graphical Representation
                          Incomplete Polynomials

                           1
                      x        y
               x2         xy         y2                   u  x, y  
          x3       x 2y        xy2        y3              a0 1 
     x4     x 3y      x2y2         xy3          y4        a1 x  a2 y 
                                                          a3 x  a4 xy  a5 y
                                                              2                 2
x5   x 4y      x3y2         x2y3          xy4        y5

          …….
                  6-node Triangular
Assume interpolation
                                      3

 u  x, y  
                                6         5
 a0 1 
 a1 x  a2 y 
 a3 x 2  a4 xy  a5 y 2
                            1       4         2


6 coefficients to determine for admissible displ.
                    6-node triangular


        3              6-nodes x 2dof/node = 12 dof

                     BT12x3     D3x3      B3x12
    6       5

                            ke 12x12
1       4       2
            6-node element Shape Functions

    Following the standard procedure the shape
    functions are derived as
        3
                              Corner Nodes

                         Ni  Li 2Li  1   i  1,2,3
    6          5
                          Mid-Side Nodes
                            N 4  4L1L2
1       4          2        N5  4 L2 L3
Li:Area coordinates         N 6  4 L3 L1
          Other Higher Order Elements

                          1                     12-node quad
                      x        y
                                                          4      3
               x2         xy         y2
          x3       x 2y        xy2        y3
     x4     x 3y      x2y2         xy3          y4                    
x5   x 4y      x3y2        x2y3           xy4        y5
                                                          1       2
          …….
          Other Higher Order Elements

                          1                     16-node quad
                      x        y
                                                          4      3
               x2         xy         y2
          x3       x 2y        xy2        y3
     x4     x 3y      x2y2         xy3          y4                    
x5   x 4y      x3y2        x2y3           xy4        y5
                                                          1       2
                      x3y2
          …….

				
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Jun Wang Jun Wang Dr
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