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Lecture Notes: Introduction to Finite Element Method

Chapter 4. FE Modeling and Solution Techniques

III. Equation Solving

Direct Methods (Gauss Elimination):

Solution time proportional to NB2 (N is the dimension of the matrix, B the bandwidth)

Suitable for small to medium problems, or slender structures (small bandwidth)

Easy to handle multiple load cases

Iterative Methods:

Solution time is unknown beforehand

Reduced storage requirement

Suitable for large problems, or bulky structures (large bandwidth, converge faster)

Need solving again for different load cases

© 1997-2002 Yijun Liu, University of Cincinnati

109

Lecture Notes: Introduction to Finite Element Method

Chapter 4. FE Modeling and Solution Techniques

Gauss Elimination - Example:

8 2

 

0 x1

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2 4

3

x2

 

=

1

 

 

0

3

 

3

x

3

 

 

 

3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Forward Elimination:

 

 

 

 

 

 

 

 

 

(1) 8

2

 

 

0

 

2

 

Form

(2)

2

4

 

 

3

 

1

;

 

(3)

 

0

3

 

 

 

3

 

3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(1) + 4 x (2) (2):

 

 

 

 

 

 

 

 

 

 

(1)

8

2

 

 

 

0

 

 

2

 

 

 

(2)

0

14

 

12

 

2

;

 

 

(3)

 

 

3

 

 

3

 

 

 

 

 

 

0

 

 

 

 

3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(2) + 143 (3) (3):

(1)

8

2

0

2

 

(2)

0

14

12

2

;

(3)

 

0

2

 

 

0

12

 

 

 

 

 

 

 

Back Substitution:

x3 =12 / 2 = 6

x2 = (2 +12x3 ) /14 = 5 x1 = (2 + 2x2 ) / 8 =1.5

or Ax = b.

1.5 or x = 5 .

6

© 1997-2002 Yijun Liu, University of Cincinnati

110

Lecture Notes: Introduction to Finite Element Method

Chapter 4. FE Modeling and Solution Techniques

Iterative Method - Example:

The Gauss-Seidel Method

Ax = b (A is symmetric)

N

 

 

or aij x j

= bi ,

i =1,2,..., N.

j =1

 

 

Start with an estimate x( 0 ) and then iterate using the following:

(k +1)

 

1

 

i 1

(k +1)

 

N

(k )

 

 

aij x j

 

aij x j

 

xi

=

 

bi

 

 

,

aii

 

 

 

 

j =1

 

 

j =i +1

 

 

for i =1, 2,..., N.

In vector form,

x(k +1) = AD 1 [b AL x(k +1) ALT x(k ) ],

where

AD = aii is the diagonal matrix of A,

A L is the lower triangular matrix of A,

such that A = AD + AL + ALT .

Iterations continue until solution x converges, i.e.

 

 

 

x(k +1) x(k )

 

 

 

 

ε ,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

x(k )

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

where ε is the tolerance for convergence control.

© 1997-2002 Yijun Liu, University of Cincinnati

111

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