The Finite Difference Method - Mechanics of Structures: Variational and Computational Methods - page 506

Some important observations can be made. Matrix K is banded, symmetric, positive defi-
nite, and tridiagonal (non-zero elements occur only on the main diagonal and on one subdiag-
onal above and below). The solution can be efficiently computed using a solution algorithm
specially designed to solve such a system of equations. Recall (Table 8.1) that the error of
this finite difference approximation is 0 h 2

This implies that a reduction in the grid size
will lead to a reduction in the error incurred and an improvement in the solution accuracy.

( ).

Accuracy Comparison for Several Values of M

ξ Exact M = 2 M = 4 M = 8 M = 16

(% error) (% error) (% error) (% error)

0 . 25 0 . 013149 0 . 013917 0 . 013333 0 . 01320

( 5 . 8% ) ( 1 . 5% ) ( 0 . 37% )

(24)

0 . 50 0 . 017601 0 . 020833 0 . 018452 0 . 017817 0 . 01765

( 18 . 4% ) ( 4 . 8% ) ( 1 . 2% ) ( 0 . 31% )

0 . 75 0 . 012026 0 . 012567 0 . 012165 0 . 012062

( 4 . 6% ) ( 1 . 2% ) ( 0 . 29% )

It is apparent that the greater the number of subdivisions, the smaller the error. An accuracy
to within about 0.1% would be achieved with M Although the error of the finite

difference procedure decreases with increasing M, other errors, e.g., roundoff error, can be
expected to increase.

= 32 .

Higher Order Schemes
If additional grid points are included in the difference equations, improved results can be
expected. That is, for the same number of beam subdivisions, more accurate results will
be obtained. It can also be reasoned that for the same degree of accuracy, a higher order
technique requires fewer subdivisions of the structure than the simple central difference
technique.
From Table 8.1, we choose a central difference expression

1
12 h 2 [

u i = − u i − 2 + 16 u i − 1 − 30 u i + 16 u i + 1 − u i + 2 ] (25)

which involves five grid points. As an example, use four subdivisions ( M = 4 ) as shown

in Fig. 8.5d. Then

ξ i = i / M = i / 4 .

From (7) and (25),

1

u 1 =

i = 1:

2 [

− u + 16 u 0 − 30 u 1 + 16 u 2 − u 3 ]

− 1

12 ( 0 . 25 )

=− ( 0 . 25 ) 0 . 75 / 1 . 25 =− 0 . 15

1

u 2 =

i = 2:

2 [

− u 0 + 16 u 1 − 30 u 2 + 16 u 3 − u 4 ]

12 ( 0 . 25 )

(26)

=− ( 0 . 5 ) 0 . 5 / 1 . 5 =− 0 . 166667

1

u 3 = i = 3:

2 [

− u 1 + 16 u 2 − 30 u 3 + 16 u 4 − u 5 ]

12 ( 0 . 25 )

=− ( 0 . 75 ) 0 . 25 / 1 . 75 =− 0 . 107143

We have three equations for the seven unknowns u − 1 ,u 0 ,u 1 ,u 2 ,u 3 ,u 4 , and u 5 . As shown

−

in Fig. 8.5d, fictitious nodes 1 and 5 beyond the ends of the beam have been included.