# ECIV 720 A Advanced Structural Mechanics and Analysis(1)

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```					          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 Dr
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