Mathematical Physics - Volume II - Numerical Methods

Chapter 2. Finite element method

44

where indices 1 and 2 are related to the nodes of a finite element, and P ( S e 2 ) are the fluxes of stress P = ku ′ in these nodes. When applied to other types of finite elements, the indices in these equations need to be changed in accordance with the way in which the nodes are designated in that case. For example, if the finite element is introduced between nodes 6 and 7, u e 1 becomes u 6 , u e 2 becomes u 7 , P ( S e 1 ) is the value of ku ′ when approaching node 6 from the right and P ( S e 2 ) is its value when approaching node 7 from the left. Let us now consider the four finite element mesh for a bar with two nodes, Fig. ?? , i.e. for a total of 5 nodes. The global stiffness matrix will be a 5 × 5 matrix, and the equations needed to form it are, in addition to the ones mentioned in (2.88): second finite element k 2 11 u 2 + k 2 12 u 3 = f 2 1 + P ( z + 1 ) k 2 21 u 2 + k 2 22 u 3 = f 2 2 − P ( z − 2 ) (2.89) third finite element k 3 11 u 3 + k 3 12 u 4 = f 3 1 + P ( z + 2 ) k 3 21 u 3 + k 3 22 u 4 = f 3 2 − P ( z − 3 ) (2.90) fourth finite element k 4 11 u 4 + k 4 12 u 5 = f 4 1 + P ( z + 3 ) k 4 21 u 4 + k 4 22 u 5 = f 4 2 − P ( z − 4 ) , (2.91) which results in the following matrix equation: 1 ) and P ( S e

   k 1 11 k 1 21 0 0

  

   u 1 u 2 u 3 u 4 u 5

   =

1 12

k

0

0 0

0 0 0

1 22 + k

2 11

2 12

k

k

k 2

2 22 + k

3 11

3 12

k

k

21

k 3

3 22 + k

4 11

4 12 4 22

0 0

k

k k

21

k 4

0

0

21

(2.92)

   f 1 f 2 f 2

  

f 1 1 + P ( 0 )

2 1 + [ | P ( z 1 ) | ] 3 1 + [ | P ( z 2 ) | ] 1 4 + [ | P ( z 3 ) | ]

2 + f 2 + f 3 + f

=

f 2 4 − P ( z 4 = l )

What remains now is to incorporate displacement boundary conditions, since stress boundary conditions are already a part of the finite element equations. If, for example, we define that u ( 0 ) = u 0 and u ( l ) = u l then equation (2.92) has two less unknown displacements, and can be written as:   k 1 21 k 1 22 + k 2 11 k 2 12 0 0 0 k 2 21 k 2 22 + k 3 11 k 3 12 0 0 0 k 3 21 k 3 22 + k 4 11 k 4 12     u 2 u 3 u 4   =

=   f 2

 

f 2

1

1 + f

2

ˆ f

3 1 +

(2.93)

2 + f

f 2

1

3 + f

4

with two auxiliary equations:

k 1

1 12 u 2 = f 4 22 u 5 = f

1 1 + P ( 0 ) ⇒ P ( 0 ) = k 1 2 4 − P ( l ) ⇒ P ( l ) = − k 4

1 12 u 2 − f

1

11 u 1 + k

11 u 1 + k

1

(2.94)

k 4

4 22 u 5 + f

2 4 ,

21 u 4 + k

21 u 4 − k

with unknown values of P ( 0 ) and P ( l ) .