6.3 Answers for Problems from Chap. 3
141
node 7:
E I Y
X 3 (u 9 − 4u 8 + 6u 7 − 4u 6 + u 5 ) = −q 0 X ,
(6.162)
node 8:
E I Y
X 3 (u 10 − 4u 9 + 6u 8 − 4u 7 + u 6 ) = −q 0 X ,
(6.163)
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
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
= −
F 0 X
3
E I Y
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
0
0
0
1
0
0
0
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
.
(6.164)
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
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
= −
F 0 L
3
E I Y
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
1
1024
3
1024
5
1024
3
512
5
1024
3
1024
1
1024
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
,
(6.165)
and the relative error in the middle of the beam is obtained as [1]:
relative error =
3
512
−
1
192
1
192
× 100 = 12.5% .
(6.166)
3.22 Finite difference approximation of a cantilevered beam based on five
domain nodes with imposed tip displacement
The finite difference discretization of the cantilevered beam is shown in Fig. 6.10 for
five domain nodes.
Evaluation of the finite difference approximation of the fourth-order differential
equation according to Eq. (3.9) at the inner nodes i = 2, . . . , 4 gives:
node 2:
E I Y
X 3 (u 4 − 4u 3 + 6u 2 − 4u 1 + u 0 ) = 0 ,
(6.167)
node 3:
E I Y
X 3 (u 5 − 4u 4 + 6u 3 − 4u 2 + u 1 ) = 0 ,
(6.168)