Variational and Weighted Residual Methods

Variational and Weighted

Residual Methods

2

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 piece-

wise to parts of the region

These subsections of the region are then the finite elements

3


Consider the problem of 1-D heat flow, the functional to be

extremised is

where the integral over corresponds to the length of the region

and Neumann boundary conditions are specified at one end,

,of the region

2

r r

dk Q x dx kdx

4


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

5


Let the unknown temperatures at the nodes of the element e be

where n+1 is the number of nodes in each element.

e

i i i n

T 1....

1

.

.

.

i

i

e

i n

6


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

where N is the shape function associated with the node and

=i ... i+n and [N] is the corresponding row matrix.

N Ne

7


Let us write the trial function over the entire region in the form

where the summation is over all the nodes in .

N g

8


The global shape functions have been used to take into

account the contribution from to over the entire region

The global shape functions over much of will be zero

For interior nodes of an element 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.

gN

gN

9


For example :

is non-zero only in elements e and e+1.

will be non-zero only in

element e.

Ni ng

N N Nig

ig

i ng

1 2 1, , ....

10


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

11


where for example element e stands for

over the element e

i n i n element e element e

1

0

element e

i i n0 1 1: .....

i i element e element e

1

0

kddx

Q x dx

2

12


Since

i element e

1

i element e

is an expression involving {e-1

involves {e

and there is no relationship between {e-1 and {e ,both

expressions must be equal to zero

13


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

14


The function can be written as,

k d

dxN Q x N dx

e ek

2

2

0

= i ...i+n where

Note that

x x

NdN

dx

dN

dx

dN

dxe i i

i

i n

e

, ..... .1

15


Also, noting

Since

Hence

x x x

dN

dx

dN

dxe

2

2 2

x

dNdx

e

x

dN

dx

16


So, differentiating under the integral sign, we have

kdN

dx

dN

dxdx Q x N dxe

hh

00

Hence

kdN

dx

dN

dxQ x N dx

eh

0

0

17


This equation is one in the set of n+1 simultaneous equations

obtained by letting run through the values i...i+n :

k k k

k k

k

F

F

i i i i i i n

i i i i n

i n i n

i

i n

ie

i ne

, , ,

, ,

,

. .

. .

. . .

. .

1

1 1 1

18


where

and

In the end elements, where Neumann boundary conditions may have to be

considered, there is an additional term

where Nr is the value of N on the boundary

F Q x N dxe

h

0

k kN

x

N

xdx

t

0

k k Nr r r r ,

19


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

and similarly for element n

11, 1 1, 2 1

22, 1 2, 2 2

n n nii i i i i

n n nii i i i i

k k F

k k F

, , 1

11, 1, 1 1

m m mii i i i i

m m mii i i i i

k k F

k k F

20


By combining these two matrix equations

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

, , 1

1, 1, 1 1, 1 1, 2 1 1 1

2 22, 1 2, 2

0

0

m m mi i i i i im m n n m ni i i i i i i i i i i

nn ni ii i i i

k k F

k k k k F F

Fk k

21

Example 3

Find an approximate solution to Example 1 for the rod of constant

cross section using three linear elements of equal length.

22


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

For node 2

Nxh1 1

Nxh2

23


From the trial function

and

k kdN

dx

dN

dxdx

k

hk

h

h

/

0

for

for

F Q x N dx Qhfore

h

0 2

1 2,

24


For element 1 we have

k

h

Qhk N

Qhr

1 1

1 12

2

1

2

1 1

,

where N1,r is the value of N1 (the value is 1) at node 1

For element 2

k

h

h

h

1 1

1 12

2

2

3

25


For element 3

where N4,r is the value of N4 at node 4 (N is equal to 1)

k

h

Qh

Qhk

1 1

1 12

2

3

42

26


Assembling the matrices, we have

11

2

3

42

1 1 0 0 21 2 1 0

0 1 2 1

0 0 1 1

2

hk

Qhk

Qhh

Qhk

27


Given we can solve for and from rows 2

and 3

from which

since

0 2 3

2

2 3 0

2

2

2

h

k

h

k

2 3 0

2

0

34

9

h

k

QL

k

hL

2

3

28

Comparison of FE and Exact Solution

29

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

30

Natural Coordinates and Quadratic Shape Functions

For convenience, a dimensionless coordinate is used rather

than x so that over the length of a 1-D element the value of runs

from +1 to -1

In the previous example if x is measured from node 1, then in

element 1,

The shape functions become

21

x

h

N

N

1

2

1

21

1

21

31


Since

The trial function becomes

and

d

dx h

2

kk

h

N Nd

2

1

1

F Q N de

1

1

32


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

33


The shape functions for a Quadratic 1-D element, which has three

nodes, are

N

N

N

1

22

3

1

21

1

1

21

34

Stiffness Matrix for 1-D Quadratic Element (HC)

From the trial function, the components of the 3x3 element stiffness

matrix satisfy the condition

Now

kk

h

N Nd

2

1 2 31

1

; , , ,

N N N1 2 31

22

1

2

35


Hence

and so on, so that

Note that the matrix is symmetrical

kk

hd

kk

hd

11

2

1

1

12

1

1

2 1

2

7

6

2 1

22

4

3

kk

h

2

7

6

4

3

1

64

3

8

3

4

31

6

4

3

7

6

36


To evaluate

With Q a constant

FhQ N de

2 1

1

FhQ

d Qh

FhQ

d Qh

FhQ

d Qh

e

e

e

1

1

1

22

1

1

3

1

1

2

1

21

6

21 2

3

2

1

21

6

37


If and are the values of at x = ± L, then the equations to be solved

dx

d

2

7

6

4

3

1

64

3

8

3

4

31

6

4

3

7

6

62

3

6

1

2

3

1

3

k

h

Qhk

Qh

Qhk

1 2

38


With expanding row 2 and solving for gives

Since h=2L which is the exact solution, can be found by

substituting the value of and expanding row 1

Again, the exact value is obtained

The estimate of approaches the exact value as the number of elements is increased

1

1

2 0

2

2

QL

k

1 QL

k

Date post:	18-Jan-2016
Category:	Documents
Upload:	gino
View:	158 times
Download:	10 times

Variational and Weighted Residual Methods

Documents