140
6 Answers to Supplementary Problems
Thus, we can apply the finite difference approximation of the fourth-order differential equation according to Eq. (3.9). Writing this statement 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 ,
(6.151)
node 3:
E I Y
X 3 (u 5 − 4u 4 + 6u 3 − 4u 2 + u 1 ) = −q 0 X = −F 0 ,
(6.152)
node 4:
E I Y
X 3 (u 6 − 4u 5 + 6u 4 − 4u 3 + u 2 ) = 0 ,
(6.153)
or under consideration of the boundary conditions, i.e. u 1 = u 5 = 0, u 0 = u 2 and
u 6 = u 4 , in matrix notation:
⎡
⎣
7 −4 1
−4 6 −4
1 −4 7
⎤
⎦
⎡
⎣
u 2
u 3
u 4
⎤
⎦ = −
F 0 X
3
E I Y
⎡
⎣
0
1
0
⎤
⎦ .
(6.154)
The solution of this linear system of equations gives the unknown nodal values as:
⎡
⎣
u 2
u 3
u 4
⎤
⎦ = −
F 0 L
3
E I Y
⎡
⎢
⎣
1
256
1
128
1
256
⎤
⎥
⎦ ,
(6.155)
and the relative error in the middle of the beam is obtained as [1]:
relative error =
1
128
−
1
192
1
192
× 100 = 50.0% .
(6.156)
Writing the finite difference statement at the inner nodes i = 2, . . . , 8 gives:
node 2:
E I Y
X 3 (u 4 − 4u 3 + 6u 2 − 4u 1 + u 0 ) = −q 0 X ,
(6.157)
node 3:
E I Y
X 3 (u 5 − 4u 4 + 6u 3 − 4u 2 + u 1 ) = −q 0 X ,
(6.158)
node 4:
E I Y
X 3 (u 6 − 4u 5 + 6u 4 − 4u 3 + u 2 ) = −q 0 X ,
(6.159)
node 5:
E I Y
X 3 (u 7 − 4u 6 + 6u 5 − 4u 4 + u 3 ) = −q 0 X ,
(6.160)
node 6:
E I Y
X 3 (u 8 − 4u 7 + 6u 6 − 4u 5 + u 4 ) = −q 0 X ,
(6.161)