130
6 Answers to Supplementary Problems
E I Y
X 2
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−2 1 0 0 0 0 0 0 0
1 −2 1 0 0 0 0 0 0
0 1 −2 1 0 0 0 0 0
0 0 1 −2 1 0 0 0 0
0 0 0 1 −2 1 0 0 0
0 0 0 0 1 −2 1 0 0
0 0 0 0 0 1 −2 1 0
0 0 0 0 0 0 1 −2 1
0 0 0 0 0 0 0 1 −2
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u 2
u 3
u 4
u 5
u 6
u 7
u 8
u 9
u 10
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=
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M 0
M 0
M 0
M 0
M 0
M 0
M 0
M 0
M 0
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,
(6.92)
from which its solution is obtained as:
u 2 = −
9M 0 L
2
200E I Y
, u 3 = −
2M 0 L
2
25E I Y
, u 4 = −
21M 0 L
2
200E I Y
,
(6.93)
u 5 = −
3M 0 L
2
25E I Y
, u 6 = −
M 0 L
2
8E I Y
.
(6.94)
The system of linear equation based on the 4th order PDE is given by:
E I Y
X 3
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5 −4 1 0 0 0 0 0 0
−4 6 −4 1 0 0 0 0 0
1 −4 6 −4 1 0 0 0 0
0 1 −4 6 −4 1 0 0 0
0 0 1 −4 6 −4 1 0 0
0 0 0 1 −4 6 −4 1 0
0 0 0 0 1 −4 6 −4 1
0 0 0 0 0 1 −4 6 −4
0 0 0 0 0 0 1 −4 5
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u 2
u 3
u 4
u 5
u 6
u 7
u 8
u 9
u 10
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=
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−M 0 //X
0
0
0
0
0
0
0
−M 0 //X
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, (6.95)
which gives the same results as in the case of the 2nd order PDE. Furthermore, the
finite difference and analytical solutions are again identical, cf. [1].
3.14 Finite difference approximation of a simply supported beam—displacement,
bending moment and shear force distribution
The single force F 0 is understood as the integral value of a distributed load q 0 , which
is acting over a length of X .
The FD solution for the displacement can be taken from Example 3.1a as:
u 1 = u 5 = 0 , u 2 = u 4 =
F 0 L
3
64E I Y
, u 3 =
3F 0 L
3
128E I Y
.
(6.96)
FD moment distribution: