Isoparametric Elements 4.pdf

Institute of Structural Engineering Page 1
Method of Finite Elements I
Chapter 4
Isoparametric Elements

30-Apr-10
Today’s Lecture Contents
• The Shape Function
• Isoparametric elements
• 1D Demonstration: Bar elements
• 2D Demonstration: Triangular elements

30-Apr-10
Shape functions comprise interpolation functions which relate the
variables in the finite element with their values in the element nodes. The
latter are obtained through solving the problem using finite element
procedures.
In general we may write:
where:
is the function under investigation (for example:
displacement field)
is the shape functions matrix
is the vector of unknowns in the nodes
(for example: displacements)
The Shape Function
 
( ) i i
i
u x H x u u
 
 H
( )
u x
( )
x
H
 
1 2
T
N
u u u u


30-Apr-10
Regardless of the dimension of the element used, we have to bear in
mind that Shape Functions need to satisfy the following constraints:
• in node i has a value of 1 and in all other nodes assumes a
value of 0.
• Furthermore we have to satisfy the continuity between the
adjoining elements.
Example: rectangular element with 6 nodes
The Shape Function
 
i
H x
H4 H5

30-Apr-10
Usually, polynomial functions are used as interpolation functions,
for example:
The Shape Function
 
0
N
i
n i
i
P x a x

 
where n is the order of the polynomial; is equal to the
number of unknowns in the nodes (degrees of freedom).
In MFE we use three different polynomials:
 Lagrange
 Serendipity and
 Hermitian polynomials.

30-Apr-10
Lagrange Polynomials
The Shape Function
A function φ(x) can be approximated by a polynomial of the order m
and the values of φ(x) in those m+1 points:
     
1
1
m
i
i
i
x x L x
  


  
where:
 
    
    
1
1 2 1
1 1 2 1
m
j m
i
j i j i i i m
i j
x x x x x x x x
L x
x x x x x x x x


 

   
 
   


30-Apr-10
The Shape Function
A function φ(x) can be approximated by a polynomial of the order m
and the values of φ(x) in those m+1 points:
f x
( )»f x
( )= Li
x
( )fi
i=1
m+1
å

30-Apr-10
Lagrange
Polynomials
Example

30-Apr-10
Serendipity Polynomials
These functions are similar to Lagrangian polynomials, but for their
incompleteness. Due to this fact we do not need to introduce
additional inner nodes, as for Lagrangian polynomials of higher order.
Serendipity Polynomials

30-Apr-10
Hermitian Polynomials
Lagrangian polynomials and serendipity functions provide a C0
continuity. If we additionally need continuity of the first derivatives
between the finite elements we use Hermitian polynomials.
A Hermitian polynomial of the order n, Hn(x), is a 2n+1 order
polynomial. For example a Hermitian polynomial of the first order is
actually a third order polynomial.
Let us consider a bar element with nodes on its ends. Unknowns are
values of the function φ in the nodes 1 and 2, φ1 and φ2, and first
derivatives of φ in respect to x , φ1,x and φ2,x
Remember: The 1st derivative of
displacement, corresponds to rotation

30-Apr-10
Hermitian Polynomials

30-Apr-10
In general, we would like to be able to represent any element in a
standardized manner – introducing a transformation between the
natural coordinates and the real coordinates (global or local)
Different schemes exist for establishing such transformations:
1 sub parametric representations (less nodes for
geometric than for displacement representation)
2 isoparametric representations (same nodes for both
geometry and displacement representation)
3 super parametric representations (more nodes for
geometric than for displacement representation)

30-Apr-10
Displacement fields as well as the geometrical representation of the
finite elements are approximated using the same approximating
functions – shape functions
x
y
r
s
1, -1
1, 1
-1, 1
-1, -1
1 1
ˆ ˆ
,
u v
2 2
ˆ ˆ
,
u v
3 3
ˆ ˆ
,
u v
4 4
ˆ ˆ
,
u v
1
4
2
3
This transformation allows us to refer to similar elements (eg. truss, beams, 2D
elements, in a standard manner, using the natural coordinates (r,s), without
every time referring to the specific global coordinate system used (x,y).

30-Apr-10
Isoparametric elements can be one-, two- or three-dimensional:
The principle is to assure that the value of the shape function hi
is equal to one in node i and equals zero in other nodes.
1 1 1
; ;
n n n
i i i i i i
i i i
x h x y h y z h z
  
  
  
1 1 1
ˆ ˆ ˆ
; ;
n n n
i i i i i i
i i i
u hu v hv w h w
  
  
  

30-Apr-10
In order to establish the stiffness matrixes we must differentiate
the displacements with respect to the coordinates (x, y, z).
Since the shape functions are usually defined in natural
coordinates we must introduce the necessary coordinate
transformation between natural and global or local coordinate
system. For example:
x y z
r x r y r z r
x y z
s x s y s z s
x y z
t x t y t z t
   
   
   
      
  
      
      
  
      
      
  
      
Chain rule of differentiation!

30-Apr-10
Written in matrix notation:
x y z
r x r y r z r
x y z
s x s y s y s
x y z
t x t y t z t
   
   
   
      
  
      
      
   
      
      
  
      
x y z
r r r r x
x y z
s s s s y
x y z
t t t t z
 
 
 
 
    
   
 
   
    
 
   
    
 
   
  
   
    
 
   
    
 
   
 
   
   
    
 

30-Apr-10
We recall the Jacobian operator J:
1

 
  
 
J J
r x x r
 
 
x y z
r r r r x
x y z
s s s s y
x y z
t t t t z
 
 
 
 
    
   
 
   
    
 
   
    
 
   
  
   
    
 
   
    
 
   
 
   
   
    
 

30-Apr-10
Let us now consider the derivation of the stiffness matrix K.
Firstly, we write for the strain matrix (using matrix B):
and than we can write up the integrals for calculating the stiffness
matrix:
ˆ

ε Bu
det
T
V
T
V
dV
dr ds dt




K B CB
B CB J

30-Apr-10
Bar Element

30-Apr-10
Bar Element
The relation between the x-coordinate
and the r-coordinate is given as:
The relation between the displacement
u and the nodal displacements are
given in the same way:
1
û 2
û
y
x
1
x
2
x
0
r 
1
r   1
r 
1 2
2
1
1 1
ˆ ˆ
(1 ) (1 )
2 2
ˆ
i i
i
u r u r u
hu

   
 
1 2
2
1
1 1
(1 ) (1 )
2 2
i i
i
x r x r x
h x

   
 

30-Apr-10
Bar Element
We need to be able to establish the strains – meaning we need to
be able to take the derivatives of the displacement field in regard
to the x-coordinate
du du dr
dx dr dx
  
1 2 2 1
1 2 2 1
2 1 2 1
2 1
1 1 1
ˆ ˆ ˆ ˆ
( (1 ) (1 ) ) ( )
2 2 2
1 1 1
( (1 ) (1 ) ) ( )
2 2 2 2
ˆ ˆ ˆ ˆ
( ) ( )
( )
du d
r u r u u u
dr dr
dx d L
r x r x x x
dr dr
u u u u
du
dx x x L

     
      

 
  

1
û 2
û
y
x
1
x
2
x
0
r 
1
r   1
r 

30-Apr-10
Bar Element
The strain-displacement matrix then becomes:
and the stiffness matrix is calculated as:
 
1
1 1
L
 
B
 
1
2
1
1
2 1
1
1 1 ,
1 2
1 1 1 1
1 1 1 1
2
AE dx L
dr
L dr
AE L AE
r
L L



 
   
 
 

 
   
 
   
 
   

K J J
K K

30-Apr-10
The natural coordinates for the 3-node bar element:
Bar Elements
Node 2
Node 1 Node 3
0.3L 0.7L
1
0
r
x
 

0
0.3
r
x L


1
r
x L


1
r   0
r  1
r 
1
1
r   0
r  1
r 
1
2
2 1
h r
 
2
1
1 1
(1 ) (1 )
2 2
h r r
   
1
r   0
r  1
r 
1
2
3
1 1
(1 ) (1 )
2 2
h r r
   

30-Apr-10
The natural coordinates for the generalized bar element:
Bar Elements
Node 1
1
r   0
r  1
r 
1
r  
1
3
r   1
r 
1
3
r 
Three nodes
Four nodes
Node 3 Node 2
Node 4

30-Apr-10
The shape functions for the
generalized bar element:
Bar Elements
1
2
2
3
3 2
4
1
(1 )
2
1
(1 )
2
(1 )
1
( 27 9 27 9)
16
h r
h r
h r
h r r r
 
 
 
    
2
2
1
(1 )
2
1
(1 )
2
r
r
 
 
3 2
3 2
3 2
1
( 9 9 1)
16
1
(9 9 1)
16
1
(27 7 27 7)
16
r r r
r r r
r r r
    
   
   
Include only if
node 3 is present
Include only if nodes 3 and 4
are present
Shape functions

30-Apr-10
Let us now move in the 2D. The natural coordinates for the
standard triangular element (r and s) may be represented as:
and the shape functions may be given as:
r
s
1 2
3
(0,0) (1,0)
(0,1)
1
2
3
1
h r s
h r
h s
  


Triangular Elements

30-Apr-10
Let us establish the stiffness matrix for a triangular constant strain
element:
4
3
x r s
y s
 

r
s
1 2
3
(0,0) (1,0)
(0,1)
1 2
3
(0,0) (4,0)
(1,3)
x
y
3 3
1 1
x ,
i i i i
i i
h x y h y
 
 
 
1
2
3
1
h r s
h r
h s
  


Triangular Elements

30-Apr-10
x y z
r r r r x
x y z
s s s s y
x y z
t t t t z
 
 
 
 
    
   
 
   
    
 
   
    
 
   
  
   
    
 
   
    
 
   
 
   
   
    
 
In this case (2D) we obtain for the Jacobi matrix :
4
3
x r s
y s
 

1
4 0
1 3
3 0
1
1 4
det
3 0
1
1 4
12

 
  
 

 
  

 
 
  

 
J
J
J
Triangular Elements
1
x r

 

 
J

30-Apr-10
Further on, we obtain for the interpolation matrix H and for
displacement field (u, v) in element:
(1 ) 0 0 0
0 (1 ) 0 0
r s r s
r s r s
 
 
  
 
 
H
1 1 2 2 3 3
1 1 2 2 3 3
ˆ ˆ ˆ ˆ ˆ ˆ
(1 ) 0 0 0
ˆ ˆ ˆ ˆ ˆ ˆ
0 (1 ) 0 0
u r s u v ru v su v
v u r s v u rv u sv
       
       
Triangular Elements

30-Apr-10
For plane stress problems one has
, ,
xx yy xy
u v u v
x y y x
  
   
   
   
r s r s
x x r x s x x x r
r s r s
y y r y s y y y s
       
     
 
        
       
 
   
   

        
     

 
     
       
 
 
    
 
   
  
   
   
   
 
  
   
 
   
 
  
   
   
   
   
 
  
   
   
x y
x x
r r r
x y
y y
s s s
J
Triangular Elements

30-Apr-10
1 1 2 2 3 3
1 1 2 2 3 3
ˆ ˆ ˆ ˆ ˆ ˆ
(1 ) 0 0 0
ˆ ˆ ˆ ˆ ˆ ˆ
0 (1 ) 0 0
u r s u v ru v su v
v u r s v u rv u sv
       
       
, ,
xx yy xy
u v u v
x y y x
  
   
   
   
1 2 1 3
1 3 1 2
ˆ ˆ ˆ ˆ
,
ˆ ˆ ˆ ˆ
,
u u
u u u u
r s
v v
v v v v
s r
 
     
 
 
     
 
Having in mind that
one obtains
Triangular Elements

30-Apr-10
and so
with
1
1
1 0 1 0 0 0
ˆ
1 0 0 0 1 0
0 1 0 1 0 0
ˆ
0 1 0 0 0 1
u
x
u
y
v
x
v
y



 
  
 

    
 
   
 

 

 
  
 

    
 
   
 

 
J u
J u
1 3 0
1
1 4
12
  
  

 
J
Triangular Elements
u/r
u/s
v/r
v/s

30-Apr-10
one gets
3 0 1 0 1 0 0 0
1
ˆ
1 4 1 0 0 0 1 0
12
3 0 0 1 0 1 0 0
1
1 4 0 1 0 0 0
12
u
x
u
y
v
x
v
y

 
  
   

      
  
     
 

 

 
  
 

    
  
   
 

 
u
ˆ
1
 
 
 
u
Triangular Elements

30-Apr-10
and
3 0 3 0 0 0
1
ˆ
3 0 1 0 4 0
12
0 3 0 3 0 0
1
ˆ
0 3 0 1 0 4
12
u
x
u
y
v
x
v
y

 
  
 

    
  
   
 

 

 
  
 

    
  
   
 

 
u
u
3 0 3 0 0 0
1
0 3 0 1 0 4
12
3 3 1 3 4 0

 
 
  
 
 
  
 
B
Triangular Elements
strain vector does not
depend on r or s so it is
constant strain triangle
u/x
v/y
u/y + v/x

30-Apr-10
We now calculate the stiffness matrix as:
det
T T
V V
dV dr ds dt
 
 
K B CB B CB J
2
3 0 3
0 3 3
1 0 3 0 3 0 0 0
3 0 1
1 0 0 3 0
0 1 3
144(1 )
1
0 0
0 0 4
2
0 4 0




 
 
   
 
    
 
  

 
   

    

   
 
 
 

S
E t
K 1 0 4 det
3 3 1 3 4 0
 
 

 
 
  
 
drds
J
Triangular Elements

30-Apr-10
which yields
2
3 3 3 0 4
3 3 3 1 0 4
12(1 )
3(1 ) 3(1
2
  
 

 
  

   

   

T
S
E t
K B det
) (1 ) 3(1 ) 4(1 )
0
2 2 2 2
  
 
 
 
 
 
   
 
 
drds
J
Triangular Elements

30-Apr-10
and further
2
9(1 ) 9(1 ) 3(1 ) 9(1 ) 12(1 )
9 9 9 3 12
2 2 2 2 2
9(1 ) 3(1 ) 9(1 ) 12(1 )
9 9 3
2 2 2 2
12(1 )
    
  
   


     
     
    
   


 
S
E t
K
12
(1 ) 3(1 ) 4(1 )
9 3 12
2 2 2
Symmetrical
  
 

   
  
9(1 ) 12(1 )
1 4
2 2
16(1 )
0
2
 

 
 

16
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
drds
Triangular Elements

30-Apr-10
finally delivering the stiffness matrix for plane stress problem:
2
9(1 ) 9(1 ) 3(1 ) 9(1 ) 12(1 )
9 9 9 3 12
2 2 2 2 2
9(1 ) 3(1 ) 9(1 ) 12(1 )
9 9 3
2 2 2 2
24(1 )
    
  
   


     
     
    
   



E t
K
12
(1 ) 3(1 ) 4(1 )
9 3 12
2 2 2
Symmetrical
  
 

   
  
9(1 ) 12(1 )
1 4
2 2
16(1 )
0
2
 

 
 

16
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Triangular Elements

Isoparametric Elements 4.pdf

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