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We know that the displacements at
Now, we
have three equations for five unknowns. Two additional conditions will be required. Recall
that the static boundary conditions are
M
ξ
=
0 and 1 must be zero. Thus,
u
0
=
u
4
=
0
.
0or
u
0
=
u
4
=
(ξ
=
0
)
=
M
(ξ
=
1
)
=
0
.
The
central difference expressions are
u
0
=
(
h
2
1
/
)(
u
−
2
u
0
+
u
1
)
−
1
(27)
u
4
=
(
h
2
1
/
)(
u
3
−
2
u
4
+
u
5
)
Since
u
0
=
u
4
=
u
0
=
u
4
=
0
,
(27) gives
u
=−
u
1
and
u
5
=−
u
3
(28)
−
1
Now (26) provides three equations for the three unknowns
u
1
,u
2
,
and
u
3
.
The somewhat cumbersome treatment of the fictitious node points can be avoided by
using forward and backward difference expressions near the boundary. Thus, if we use
(Table 8.1)
u
1
=
(
12
h
2
1
/
)(
11
u
0
−
20
u
1
+
6
u
2
+
4
u
3
−
u
4
)
(29a)
for
i
=
1 of (26), and
u
3
=
(
12
h
2
1
/
)(
−
u
0
+
4
u
1
+
6
u
2
−
20
u
3
+
11
u
4
)
(29b)
for
i
=
3 of (26), then no nodes occur beyond the ends of the beam. In this case, with
u
0
0
,
we again have three equations for three unknowns.
From (28) and (26), we obtain the stiffness-like equations
=
u
4
=
−
29
16
−
1
u
1
u
2
u
3
0
.
1125
=−
16
−
30
16
0
.
1250
−
1 6
−
29
0
.
08035
(30)
K
U
=
P
which have the solutions
u
1
=
0
.
013183
,
2
=
0
.
017617
,
3
=
0
.
012033
(31)
These correspond to errors of 0.27%, 0.095%, and 0.086%, respectively. This is a significant
improvement over the use of the simple central difference formula for
M
.
The better results of the improved difference expression of a higher order scheme are to
be expected, since the error is 0
=
4
h
4
h
2
).
For an arbitrary
M,
the higher order difference technique, using the central difference
expression (25), provides the general expression
(
)
,
while that for the simple central difference is 0
(
M
3
u
i
−
2
+
16
u
i
−
1
−
30
u
i
+
16
u
i
+
1
−
u
i
+
2
=−
12
i
(
M
−
i
)/
(
M
+
i
)
(32)
This will lead to a banded matrix of bandwidth 5.
Multiple Position Difference Method
Choose from Table 8.2 the multiple difference expression for the second derivative
12
h
2
(
u
i
−
1
+
10
u
i
+
u
i
+
1
=
u
i
−
1
−
2
u
i
+
u
i
+
1
)
(33)
Choose four subdivisions (Fig. 8.5b) so that
M
=
4
,h
=
1
/
4
,
and
ξ
=
i
/
4
.
From (33),
i
u
0
+
10
u
1
+
u
2
=
(
i
=
1:
12
)(
16
)(
u
0
−
2
u
1
+
u
2
)
u
1
+
10
u
2
+
u
3
=
(
i
=
2:
12
)(
16
)(
u
1
−
2
u
2
+
u
3
)
(34)
u
2
+
10
u
3
+
u
4
=
(
i
=
3:
12
)(
16
)(
u
2
−
2
u
3
+
u
4
)
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