Mathematical Physics - Volume II - Numerical Methods

Chapter 3. Comparison of finite element method and finite difference method

52

Now, we can apply the finite difference method in order to solve equation (204), by introducing nodes as seen in Figure 3.1, including hypothetical nodes ( − 1 ) and ( 4 ) . For nodes ( 1 ) and ( 2 ) , we obtain that:

L 3

L 3

L 3

L 3

L 3

-1

0

1

2

3

4

w

w

w

w

w

w

-1

0

1

2

3

4

Figure 3.1: Finite difference mesh.

W − 1 − 4 W 0 + 6 W 1 − 4 W 2 + W 3 = β 4 W 0 − 4 W 1 + 6 W 2 − 4 W 3 − W 4 = β 4

1 W 1 , 1 W 2 ,

(3.24)

where β 1 = L 3 β . Boundary conditions are W 0 = W 3 = 0 and dW dx = 0 in nodes ( 0 ) and ( 3 ) , i.e. W − 1 = W 1 and W 2 = W 4 . By replacing boundary conditions in (3.24) we obtain a system of homogeneous linear algebraic equations with two unknowns, W 1 and W 2 : 7 W 1 − 4 W 2 = β 4 1 W 1 , − 4 W 2 + 7 W 2 = β 4 1 W 2 , (3.25) and by solving of the above, we obtain the following values:

L 2 r L 2 r

W 1 W 2 W 1 W 2

= =

1 1 1 − 1

EI m EI m

15 , 59

ω 1 =

(3.26)

29 , 85

ω 2 =

Solution accuracy can be improved by introducing additional nodes.

3.1.2 Finite element method solution

The beam is divided into an adequate number of subdomains (finite elements), in this case two, for which the following (cubic) interpolation is introduced:

e ) 1 ( 2 ξ

3 − 3 ξ 2 + 1 )+ W ( e ) 3 − 2 ξ 2 + ξ )+ lW ( e )

2 − 2 ξ 3 )+

Ω ( x ) = W (

3 ( 3 ξ

(3.27)

e ) 2 ( ξ

3 − ξ 2 ) ,

+ lW (

4 ( ξ

where ξ = x / l , x is the longitudinal coordinate, l is the element length, Figure 3.2.