3.4 Solved Problems
47
node 2:
E I Y
X 3 (u 4 − 4u 3 + 6u 2 − 4u 1 + u 0 ) = 0 ,
(3.28)
node 3:
E I Y
X 3 (u 5 − 4u 4 + 6u 3 − 4u 2 + u 1 ) = −F 0 (= q 0 X ) ,
(3.29)
node 4:
E I Y
X 3 (u 6 − 4u 5 + 6u 4 − 4u 3 + u 2 ) = 0 .
(3.30)
The vertical displacement is zero at both ends and it can be immediately concluded
that u 1 = u 5 = 0. The fictitious nodes i = 0 and i = 6 outside the domain can be
eliminated based on the boundary condition that the moment must be equal to zero
at the supports, i.e. M Y (X = 0) = M Y (X = L) = 0. Application of the bending
differential equation in the form with the bending moment according to Table 3.1,
i.e. E I Y
d
2 u
dX 2 = −M Y , and the centered finite difference approximation of the second
order derivative according to Table 1.1, the following two conditions can be derived:
E I Y
d
2 u
dX 2
1
= E I Y
u 2 − 2u 1 + u 0
X 2
= 0 ,
(3.31)
E I Y
d
2 u
dX 2
5
= E I Y
u 6 − 2u 5 + u 4
X 2
= 0 ,
(3.32)
from which the two conditions u 0 = −u 2 and u 6 = −u 4 can be obtained. Introducing
these relationships for the fictitious nodes and the condition of zero displacement at
the supports into the system of equations according to (3.28)–(3.30) gives:
node 2:
E I Y
X 3 (5u 2 − 4u 3 + u 4 ) = 0 ,
(3.33)
node 3:
E I Y
X 3 (−4u 2 + 6u 3 − 4u 4 ) = −F 0 ,
(3.34)
node 4:
E I Y
X 3 (u 2 − 4u 3 + 5u 4 ) = 0 ,
(3.35)
or in matrix notation:
E I Y
X 3
⎡
⎣
5 −4 1
−4 6 −4
1 −4 5
⎤
⎦
⎡
⎣
u 2
u 3
u 4
⎤
⎦ =
⎡
⎣
0
−F 0
0
⎤
⎦ .
(3.36)
The solution of this linear system of equations gives the unknown nodal values as:
u 2 = −
F 0 X
3
E I Y
, u 3 = −
3F 0 X
3
4E I Y
, u 4 = −
F 0 X
3
E I Y
,
(3.37)
or with X =
L
4
as:
47
node 2:
E I Y
X 3 (u 4 − 4u 3 + 6u 2 − 4u 1 + u 0 ) = 0 ,
(3.28)
node 3:
E I Y
X 3 (u 5 − 4u 4 + 6u 3 − 4u 2 + u 1 ) = −F 0 (= q 0 X ) ,
(3.29)
node 4:
E I Y
X 3 (u 6 − 4u 5 + 6u 4 − 4u 3 + u 2 ) = 0 .
(3.30)
The vertical displacement is zero at both ends and it can be immediately concluded
that u 1 = u 5 = 0. The fictitious nodes i = 0 and i = 6 outside the domain can be
eliminated based on the boundary condition that the moment must be equal to zero
at the supports, i.e. M Y (X = 0) = M Y (X = L) = 0. Application of the bending
differential equation in the form with the bending moment according to Table 3.1,
i.e. E I Y
d
2 u
dX 2 = −M Y , and the centered finite difference approximation of the second
order derivative according to Table 1.1, the following two conditions can be derived:
E I Y
d
2 u
dX 2
1
= E I Y
u 2 − 2u 1 + u 0
X 2
= 0 ,
(3.31)
E I Y
d
2 u
dX 2
5
= E I Y
u 6 − 2u 5 + u 4
X 2
= 0 ,
(3.32)
from which the two conditions u 0 = −u 2 and u 6 = −u 4 can be obtained. Introducing
these relationships for the fictitious nodes and the condition of zero displacement at
the supports into the system of equations according to (3.28)–(3.30) gives:
node 2:
E I Y
X 3 (5u 2 − 4u 3 + u 4 ) = 0 ,
(3.33)
node 3:
E I Y
X 3 (−4u 2 + 6u 3 − 4u 4 ) = −F 0 ,
(3.34)
node 4:
E I Y
X 3 (u 2 − 4u 3 + 5u 4 ) = 0 ,
(3.35)
or in matrix notation:
E I Y
X 3
⎡
⎣
5 −4 1
−4 6 −4
1 −4 5
⎤
⎦
⎡
⎣
u 2
u 3
u 4
⎤
⎦ =
⎡
⎣
0
−F 0
0
⎤
⎦ .
(3.36)
The solution of this linear system of equations gives the unknown nodal values as:
u 2 = −
F 0 X
3
E I Y
, u 3 = −
3F 0 X
3
4E I Y
, u 4 = −
F 0 X
3
E I Y
,
(3.37)
or with X =
L
4
as: