Isoparametric Elements 4.pdf

FEM

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

2. Institute of Structural Engineering Page 2
Method of Finite Elements I
30-Apr-10
Today’s Lecture Contents
•The Shape Function
•Isoparametricelements
•1D Demonstration: Bar elements
•2D Demonstration: Triangular elements

3. Institute of Structural Engineering Page 3
Method of Finite Elements I
30-Apr-10
Shape functions comprise interpolation functions whichrelate 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)
TheShape Function()
ii
i
uxHxuu H ()ux ()xH  
12
T
N
uuuu

4. Institute of Structural Engineering Page 4
Method of Finite Elements I
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 ihas 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
TheShape Function
i
Hx
H
4 H
5

5. Institute of Structural Engineering Page 5
Method of Finite Elements I
30-Apr-10
Usually, polynomial functions are used as interpolation functions,
for example:
TheShape Function
0
N
i
ni
i
Pxax


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
Hermitianpolynomials.

6. Institute of Structural Engineering Page 6
Method of Finite Elements I
30-Apr-10
Lagrange Polynomials
TheShape 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
xxLx



where:
   
   
1
1 2 1
1 1 2 1
m
j m
i
j ij i i im
ij
xx xxxxxx
Lx
xxxxxxxx


 

 




7. Institute of Structural Engineering Page 7
Method of Finite Elements I
30-Apr-10
Lagrange Polynomials
TheShape 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:
fx()»fx()=L
i
x()f
i
i=1
m+1
å

8. Institute of Structural Engineering Page 8
Method of Finite Elements I
30-Apr-10
Lagrange
Polynomials
Example

9. Institute of Structural Engineering Page 9
Method of Finite Elements I
30-Apr-10
Serendipity Polynomials
ThesefunctionsaresimilartoLagrangianpolynomials,butfortheir
incompleteness.Duetothisfactwedonotneedtointroduce
additionalinnernodes,asforLagrangianpolynomialsofhigherorder.
Serendipity Polynomials
Lagrange Polynomials

10. Institute of Structural Engineering Page 10
Method of Finite Elements I
30-Apr-10
HermitianPolynomials
LagrangianpolynomialsandserendipityfunctionsprovideaC
0
continuity.Ifweadditionallyneedcontinuityofthefirstderivatives
betweenthefiniteelementsweuseHermitianpolynomials.
AHermitianpolynomialoftheordern,H
n
(x),isa2n+1order
polynomial.ForexampleaHermitianpolynomialofthefirstorderis
actuallyathirdorderpolynomial.
Letusconsiderabarelementwithnodesonitsends.Unknownsare
valuesofthefunctionφinthenodes1and2,φ
1andφ
2,andfirst
derivativesofφinrespecttox,φ
1,xandφ
2,x
Remember: The 1
st
derivative of
displacement, corresponds to rotation

11. Institute of Structural Engineering Page 11
Method of Finite Elements I
30-Apr-10
HermitianPolynomials

12. Institute of Structural Engineering Page 12
Method of Finite Elements I
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)
Isoparametric Elements

13. Institute of Structural Engineering Page 13
Method of Finite Elements I
30-Apr-10
Displacement fields as well as the geometrical representation of the
finite elements are approximated using the same approximating
functions –shape functionsx
y
r
s
1, -1
1, 1-1, 1
-1, -1 11
ˆˆ,uv 22
ˆˆ,uv 33
ˆˆ,uv 44
ˆˆ,uv 1 4 2 3
Isoparametric Elements
This transformation allows us to refer to similar elements (eg. truss, beams, 2D
elements, in a standardmanner, using the natural coordinates (r,s), without
every time referring to the specific global coordinate system used (x,y).

14. Institute of Structural Engineering Page 14
Method of Finite Elements I
30-Apr-10
Isoparametric elements can be one-, two-or three-dimensional:
The principle is to assure that the value of the shape function h
i
is equal to one in node iand equals zero in other nodes. 1 1 1
; ;
n n n
ii ii ii
i i i
xhxyhyzhz
  
   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
  
    
Isoparametric Elements

15. Institute of Structural Engineering Page 15
Method of Finite Elements I
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!
Isoparametric Elements

16. Institute of Structural Engineering Page 16
Method of Finite Elements I
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


 
       
   
    
   
        

       
   
       
          
Isoparametric Elements

17. Institute of Structural Engineering Page 17
Method of Finite Elements I
30-Apr-10
We recall the Jacobianoperator J:1


  

JJ
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


 
       
   
    
   
        

       
   
       
          
Isoparametric Elements

18. Institute of Structural Engineering Page 18
Method of Finite Elements I
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
Isoparametric Elements

19. Institute of Structural Engineering Page 19
Method of Finite Elements I
30-Apr-10
Bar Element

20. Institute of Structural Engineering Page 20
Method of Finite Elements I
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
ˆu 2
ˆu y x 1
x 2
x 0r 1r 1r 12
2
1
11
ˆˆ(1 ) (1 )
22
ˆ
ii
i
u r u r u
hu

   
 12
2
1
11
(1)(1)
22
ii
i
xrxrx
hx




21. Institute of Structural Engineering Page 21
Method of Finite Elements I
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-coordinatedu du dr
dx dr dx
 1 2 21
1 2 21
21 21
21
11 1
ˆˆˆˆ((1)(1))()
22 2
11 1
((1)(1))()
22 22
ˆˆˆˆ()()
()
dud
ruruuu
drdr
dxd L
rxrxxx
drdr
uuuudu
dxxxL






 1
ˆu 2
ˆu y x 1
x 2
x 0r 1r 1r

22. Institute of Structural Engineering Page 22
Method of Finite Elements I
30-Apr-10
Bar Element
The strain-displacement matrix then becomes:
and the stiffness matrix is calculated as: 
1
11
L
B  
1
2
1
1
2 1
1
11,
1 2
11 11

1 1 1 12
AE dxL
dr
L dr
AEL AE
r
LL



  



 

 
 

K JJ
KK

23. Institute of Structural Engineering Page 23
Method of Finite Elements I
30-Apr-10
The natural coordinates for the 3-node bar element:
Bar ElementsNode2Node1 Node30.3L 0.7L 1
0
r
x

 0
0.3
r
xL

 1r
xL

 1r 0r 1r 1 1r 0r 1r 1 2
2
1hr 2
1
11
(1 ) (1 )
22
h r r    1r 0r 1r 1 2
3
11
(1 ) (1 )
22
h r r   

24. Institute of Structural Engineering Page 24
Method of Finite Elements I
30-Apr-10
The natural coordinates for the generalized bar element:
Bar Elements
Node 11r 0r 1r 1r 1
3
r 1r 1
3
r
Threenodes
Fournodes
Node 3 Node 2Node 4

25. Institute of Structural Engineering Page 25
Method of Finite Elements I
30-Apr-10
The shape functions for the
generalized bar element:
Bar Elements1
2
2
3
32
4
1
(1 )
2
1
(1 )
2
(1 )
1
( 27 9 27 9)
16
hr
hr
hr
h r r r



     2
2
1
(1 )
2
1
(1 )
2
r
r

 32
32
32
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
    
   
   
Includeonlyif
node3 ispresent
Includeonlyifnodes3 and4
arepresent
Shape functions

26. Institute of Structural Engineering Page 26
Method of Finite Elements I
30-Apr-10
Let us now move in the 2D. The naturalcoordinates for the
standardtriangular 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
1h r s
hr
hs
  


Triangular Elements

27. Institute of Structural Engineering Page 27
Method of Finite Elements I
30-Apr-10
Let us establish the stiffness matrix for a triangular constant strain
element:4
3
x r s
ys

 r s 1 2 3 (0,0) (1,0) (0,1) 1 2 3 (0,0) (4,0) (1,3) x y 33
11
x,
ii ii
ii
hxyhy

 1
2
3
1h r s
hr
hs
  


Triangular Elements

28. Institute of Structural Engineering Page 28
Method of Finite Elements I
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
ys

 1
40
13
301
14det
301
1412














J
J
J
Triangular Elements1
xr



J

29. Institute of Structural Engineering Page 29
Method of Finite Elements I
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 ˆ0 (1 ) 0 0
u r s u v ru v su v
v u r s v u rv u sv
       
       
Triangular Elements

30. Institute of Structural Engineering Page 30
Method of Finite Elements I
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
            

     
        
    
            
                
        
      
  
      
        
              
xy
xxr r r
xy
yys s s
J
Triangular Elements

31. Institute of Structural Engineering Page 31
Method of Finite Elements I
30-Apr-101 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 ˆ0 (1 ) 0 0
u r s u v ru v su v
v u r s v u rv u sv
       
        , ,
xx yy xy
uvuv
xyyx



 1 2 1 3
1 3 1 2
ˆ ˆ ˆ ˆ,
ˆ ˆ ˆ ˆ,
uu
u u u u
rs
vv
v v v v
sr

     


     

Having in mind that
one obtains
Triangular Elements

32. Institute of Structural Engineering Page 32
Method of Finite Elements I
30-Apr-10
and so
with1
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







  






  

Ju
Ju 1
301
1412
 



J
Triangular Elements
u/r
u/s
v/r
v/s

33. Institute of Structural Engineering Page 33
Method of Finite Elements I
30-Apr-10
one gets3 0 1 0 1 0 0 01
ˆ
1 4 1 0 0 0 1 012
3 0 0 1 0 1 0 01
1 4 0 1 0 0 0 12
u
x
u
y
v
x
v
y


   

   
     






  

u
ˆ
1



u
Triangular Elements

34. Institute of Structural Engineering Page 34
Method of Finite Elements I
30-Apr-10
and3 0 3 0 0 01
ˆ
3 0 1 0 4 012
0 3 0 3 0 01
ˆ
0 3 0 1 0 412
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 ror sso it is
constant strain triangle
u/x
v/y
u/y+ v/x

35. Institute of Structural Engineering Page 35
Method of Finite Elements I
30-Apr-10
We now calculate the stiffness matrix as:det
TT
VV
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 3144(1 )
1
00 0 0 4
2
0 4 0





 

  
 
  

  
 



S
Et
K 1 0 4 det
3 3 1 3 4 0




  

drdsJ
Triangular Elements

36. Institute of Structural Engineering Page 36
Method of Finite Elements I
30-Apr-10
which yields2
3 3 3 0 4
3 3 3 1 0 4
12(1 )
3(1 ) 3(1

2
  



  

   

   

T
S
Et
KB det
) (1 ) 3(1 ) 4(1 )
0
2 2 2 2
  




    


drdsJ
Triangular Elements

37. Institute of Structural Engineering Page 37
Method of Finite Elements I
30-Apr-10
and further2
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
Et
K
12
(1 ) 3(1 ) 4(1 )
9 3 12
2 2 2
Symmetrical
  


   
  
9(1 ) 12(1 )
1 4
22
16(1 )
0
2






16















drds
Triangular Elements

38. Institute of Structural Engineering Page 38
Method of Finite Elements I
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 )
    
  
   


     
     
    
   



Et
K
12
(1 ) 3(1 ) 4(1 )
9 3 12
2 2 2
Symmetrical
  


   
  
9(1 ) 12(1 )
1 4
22
16(1 )
0
2






16















Triangular Elements