Transcript FE Modification of the Rayleigh-Ritz Method

Variational
and Weighted
Residual Methods
FE Modification of the
Rayleigh-Ritz Method
In the Rayleigh-Ritz method
A single trial function is applied throughout the entire region
Trial functions of increasing complexity are required to model all
but the simplest problems
The FE approach
uses comparatively simple trial functions that are applied piecewise to parts of the region
These subsections of the region are then the finite elements
2
FE Modification of the
Rayleigh-Ritz Method
Consider the problem of 1-D heat flow, the functional to be
extremised is
2


  d 

      k 
  Q  x   dx  k  r  r

W 
  dx 

where the integral over W corresponds to the length of the region
and Neumann boundary conditions are specified at one end, G,of
the region
3
FE Modification of the
Rayleigh-Ritz Method
The length over which the solution is required, is divided up into
finite elements
In each element the value of  is found at certain points called
nodes
Two nodes will mark the extremities of the element
Other nodes may occur inside the element
4
FE Modification of the
Rayleigh-Ritz Method
Let the unknown temperatures at the nodes of the element e be
     i
e
 
e
 i 1 ....  i  n 
T
 i 


 i 1


 . 


.


 . 


  i  n 
where n+1 is the number of nodes in each element.
5
FE Modification of the
Rayleigh-Ritz Method
The temperature at any other position in the element is represented
in terms of the nodal values {}e and shape functions associated with
each node
 
N
b
 b   N  
e
b
where Nb is the shape function associated with the node b and
b=i ... i+n and [N] is the corresponding row matrix.
6
FE Modification of the
Rayleigh-Ritz Method
Let us write the trial function  over the entire region
 
N
g

 in the form


where the summation is over all the nodes in
W.
7
FE Modification of the
Rayleigh-Ritz Method
The global shape functions N g have been used to take into
account the contribution from  to  over the entire region W
The global shape functions over much of W will be zero
For interior nodes of an element N g will be non-zero only within
that element
End nodes of an element will have non-zero values over the two
elements sharing the node.
8
FE Modification of the
Rayleigh-Ritz Method
For example :
g
N i n
g
is non-zero only in elements e and e+1.
g
g
N i 1 , N i  2 , . . . . N i  n 1
will be non-zero only in
element e.
9
FE Modification of the
Rayleigh-Ritz Method
Neglecting for the moment, consideration of the first and last
elements of the region
Write the Rayleigh-Ritz statement in which the nodal values are the
adjustable parameters.
Consider the nodes i...i+n belonging to element e
10
FE Modification of the
Rayleigh-Ritz Method
   
 i


    0
 i  elem ent e 1 elem ent e 
   

   
 i  n
 
 


    0: b  i  1 . . . . . i  n  1
  elem en t e 


 i  n

  
 elem ent e

 0
elem ent e  1 
where for example element e stands for
  d 

  Q  x  dx
 k 
 dx 

2

over the element e
11
FE Modification of the
Rayleigh-Ritz Method
Since

 i

 i

is an expression involving {e-1
elem ent e  1

involves {e
elem ent e
and there is no relationship between {e-1 and {e ,both
expressions must be equal to zero
12
FE Modification of the
Rayleigh-Ritz Method
Let us
focus on the terms containing an integral over the element e
Drop the superscript g on the shape functions
Suppose that the element extends from x=xe to x=xe+h
No loss in generality is incurred if we
Shift the origin to x=xe
Take the element to extend rather from 0 to h
13
FE Modification of the
Rayleigh-Ritz Method
The function can be written as,
   
 

  k  d
e 
e 

 Q  x   N     dx 
 N   
  

   0  2  dx
 
k
2
 = i ...i+n
where
Note that

x


 N   

x
e
 i 
  dN 
e
 dN i dN i  1


,
.....  .  
 
 dx



dx

  dx 
 i  n 
14
FE Modification of the
Rayleigh-Ritz Method
Also, noting
   
    

  2

  x    
    x 
2
e dN 
   
 dN 

   2
  
 x  
dx
 dx 
Since

Hence
    dN 

 
    x 
dx
e
 dN 
 
  
 x  dx 
15
FE Modification of the
Rayleigh-Ritz Method
So, differentiating under the integral sign, we have
   
 
h


0
e  dN  
  dN 




 Q  x  N  dx  0

k 
 dx 
  dx 

Hence
h
e
 dN    dN 

k 
  dx 


 dx   dx 
0
h
 Q  x N

dx
0
16
FE Modification of the
Rayleigh-Ritz Method
This equation is one in the set of n+1 simultaneous equations
obtained by letting  run through the values i...i+n :
 k i ,i






 
k i ,i  1
.
.
k i  1 ,i  1
.
.

.
.


.



k i ,i  n    i   Fi e 


 
k i  1 ,i  n 


 

 
. 



 

 
. 


 
k i  n ,i  n    i  n   Fi e n 
17
FE Modification of the
Rayleigh-Ritz Method
where
h
F 
e
 Q  x N

dx
0
 N   N 
 k 
 
 dx
 x   x  b
0

t
and
kb
In the end elements, where Neumann boundary conditions may
have to be considered, there is an additional term

 
k
r
 r   k  r N  ,r
where Nr is the value of N on the boundary G
18
FE Modification of the
Rayleigh-Ritz Method
If there are two 2-noded elements, labelled m and n, with nodes i,
i+1 and i+2, assembly of the element matrices is as before. Then
for the first element m
 k im, i
 m
 k i  1, i
m
k i , i  1    i   Fi m 

 m
m
k i  1, i  1    i  1   Fi  1 
and similarly for element n
 k in 1, i 1
 n
 k i  2 ,i 1
n
k i 1, i  2    i  1   Fi n 1 

 n 
n
k i  2 , i  2   i  2   Fi  2 
19
FE Modification of the
Rayleigh-Ritz Method
By combining these two matrix equations
 k im, i
 m
 k i  1, i

 0
m
k i ,i 1
k
m
i  1, i  1
n
m
   

F
i
i

  m
n
n 
k i  1,i  2    i  1    Fi  1  Fi  1 
    F n

n
k i  2 , i  2   i  2  
i2

0
 k i  1, i  1
n
k i  2 ,i 1

The global assembly matrix is built up in this way
The boundary conditions on the extreme elements are inserted
The set of equations is solved for the unknown values of 
20
Example 3
Find an approximate solution to Example 1 for the rod of constant
cross section using three linear elements of equal length.
21
FE Modification of the
Rayleigh-Ritz Method
All elements will have the same stiffness matrix
The coordinate origin is to be at node 1
The shape function in element 1 for node 1 is
N1  1 
x
h
For node 2
N2 
x
h
22
FE Modification of the
Rayleigh-Ritz Method
From the trial function
h
k / b
 dN    dN b 

 k 
 dx
 dx   dx 
0

k
for
 b
for
 b
h

k
h
and
h
F 
e
 Q  x N
0

dx  Q
h
for   1, 2
2
23
FE Modification of the
Rayleigh-Ritz Method
For element 1 we have
k 1

h 1
Qh
 k  1 N 1,r
 1   1  
2
   
Qh
 1  2  

2





where N1,r is the value of N1 (the value is 1) at node 1
For element 2
k 1

h 1

 1  2  
   
 1   3  

h
2
h

2
24
FE Modification of the
Rayleigh-Ritz Method
For element 3
k 1

h 1
h


Q
 1   3  

2

    h
 1  4   Q  k  
2
 2

where N4,r is the value of N4 at node 4 (N is equal to 1)
25
FE Modification of the
Rayleigh-Ritz Method
Assembling the matrices, we have
1

k 1
h0

0
1
0
2
1
1
2
0
1
 h



k

1 
0   1   2

  
0  2  
Qh


 

 1  3  
Qh

  

 1   4 
Qh
 k 2 

 2

26
FE Modification of the
Rayleigh-Ritz Method
Given 1  4  0 we can solve for 2 and 3 from rows 2 and
3
2
 0  2 2   3

k
  2  2 3   0 
from which
since
2  3  0 
h
h 2
k
h
h 2
 0 
k
4QL
3
9k
2L
3
27
Comparison of FE
and Exact Solution
28
Comparison of FE
and Exact Solution
It can be seen that,
This is the same as the exact solution for the nodal values
The finite-element approximation deviates from the exact solution
between the nodes
As the number of elements is increased, the deviation from the
exact results at the non-nodal positions decreases
29
Natural Coordinates and
Quadratic Shape Functions
For convenience, a dimensionless coordinate x is used rather
than x so that over the length of a 1-D element the value of x runs
from +1 to -1
In the previous example if x is measured from node 1, then in
element 1,
2x

x  
 1
 h

The shape functions become
N1 
1
N2 
1
2
2
1  x 
1  x 
30
Natural Coordinates and
Quadratic Shape Functions
Since
dx
dx

2
h
The trial function becomes
kb
 N    N b 



 dx

h  1  x   x 
2k
1
1
and
F 
e
 Q x N

dx
1
31
Natural Coordinates and
Quadratic Shape Functions
Higher-order shape functions allow the variable to alter in a more
complicated fashion within an element (fewer quadratic than linear
elements are required but with a higher amount of computation per
element)
There are two methods used to obtain good precision in FE
packages
p-approach: better precision by using shape functions of
increasing complexity
h-approach: better precision is obtained by mesh refinement
32
Natural Coordinates and
Quadratic Shape Functions
The shape functions for a Quadratic 1-D element, which has three
nodes, are
N1 
1
2
x  x  1
N2  1  x
N3 
1
2
2
x  x  1
33
Stiffness Matrix for 1-D
Quadratic Element (HC)
From the trial function, the components of the 3x3 element stiffness
matrix satisfy the condition
kb
 N    N b 



 dx

h  1  x   x 
2k
1
;
 , b  1, 2 , 3
Now
N1
x
x
1
N 2
2
x
 2x
N 3
x
x
1
2
34
Stiffness Matrix for 1-D
Quadratic Element (HC)
Hence
k 12 
and so on, so that
1
2
1
7



k 11 
x
dx 
h 1 
2
6
2k
1
1
4


x


2
x
d
x



h 1 
2
3
2k
 7
 6
2k  4
k 

h
3
 1

 6

4
3
8
3
4

3
1 
6 
4
 
3
7 

6 
Note that the matrix is symmetrical
35
Stiffness Matrix for 1-D
Quadratic Element (HC)
To evaluate
F 
e
h
1
Q x N

2

dx
1
With Q a constant
F2 
hQ
e
F3 
e
1
h
1

x
x

1
d
x

Q




2 1  2
6

F1 
hQ
e
2
1
 1  x d x
2
1
 2Q
h
3
1
h
1

x
1

x
d
x

Q




2 1  2
6

hQ
36
Stiffness Matrix for 1-D
Quadratic Element (HC)
d
If  1 and  2 are the values of
equations to be solved
 7
 6
2k  4

h
3
 1

 6

4
3
8
3
4

3
dx
at x = ± L, then the
1 
Qh

 k 1

6   1   6
4    2Qh 
   2   

3   
3


7  3 
Qh


 k 3 
6 
 6

37
Stiffness Matrix for 1-D
Quadratic Element (HC)
With 1  3 0
expanding row 2 and solving for 2 gives
2  0 
QL
2
2k
Since h=2L which is the exact solution,  1 can be found by
substituting the value of 2 and expanding row 1
Again, the exact value is obtained
1 
QL
k
The estimate of  1 approaches the exact value as the number of
elements is increased
38