138
6 Answers to Supplementary Problems
node 5:
E I Y
X 3 (u 7 − 4u 6 + 6u 5 − 4u 4 + u 3 ) = −q 0 X ,
(6.144)
node 6:
E I Y
X 3 (u 8 − 4u 7 + 6u 6 − 4u 5 + u 4 ) = −q 0 X ,
(6.145)
node 7:
E I Y
X 3 (u 9 − 4u 8 + 6u 7 − 4u 6 + u 5 ) = −q 0 X ,
(6.146)
node 8:
E I Y
X 3 (u 10 − 4u 9 + 6u 8 − 4u 7 + u 6 ) = −q 0 X
(6.147)
or under consideration of the boundary conditions, i.e. u 1 = u 9 = 0, u 0 = u 2 and
u 10 = u 8 , in matrix notation:
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
7 −4 1 0 0 0 0
−4 6 −4 1 0 0 0
1 −4 6 −4 1 0 0
0 1 −4 6 −4 1 0
0 0 1 −4 6 −4 1
0 0 0 1 −4 6 −4
0 0 0 0 1 −4 7
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
u 2
u 3
u 4
u 5
u 6
u 7
u 8
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
= −
q 0 X
4
E I Y
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
1
1
1
1
1
1
1
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
.
(6.148)
The solution of this linear system of equations gives the unknown nodal values as:
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
u 2
u 3
u 4
u 5
u 6
u 7
u 8
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
= −
q 0 L
4
E I Y
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
21
32768
7
4096
85
32768
3
1024
85
32768
7
4096
21
32768
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
,
(6.149)
and the relative error in the middle of the beam is obtained as [1]:
relative error =
3
1024
−
1
384
1
384
× 100 = 12.5% .
(6.150)
3.21 Finite difference approximation of a fixed-ended beam with a single load
The finite difference discretization of the fixed-ended beam is shown in Fig. 6.8 for
five and nine domain nodes.
The solution approach is based on the idea that the single force F 0 can be modeled
as a distributed load q 0 , which acts over a length of X , see Fig. 6.9.