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with the boundary conditions
u
(
0
)
=
u
(
1
)
=
0
(8)
In establishing (7), use is made of
d
/
dx
=
(
d
/
d
ξ)(
d
ξ/
dx
)
=
(
1
/
L
)(
d
/
d
ξ)
(9)
The exact solution to (7) and (8), which can be found by simple integration, gives
=
.
=
.
=
.
u
ξ
=
1
/
4
0
013149
,
0
017601
,
0
012026
(10)
ξ
=
1
/
2
ξ
=
3
/
4
which will be useful for comparison.
Simple Central Difference Technique
Use the central difference form of Eq. (8.2)
1
h
2
u
i
=
(
u
i
−
1
−
2
u
i
+
u
i
+
1
)
(11)
Divide the span of the beam into
M
segments of equal length. With
ξ
=
x
/
L,
the length of
the beam is 1.
Case a (Fig. 8.5a, M
2)
Begin with just two segments
=
(
M
)
. Then
h
=
length
/
M
=
1
/
2
=
0
.
5
.
For the configuration
of Fig. 8.5a.
1
u
1
=
5
2
(
u
0
−
2
u
1
+
u
2
)
=
4
(
0
−
2
u
1
−
0
)
=−
8
u
1
(12)
0
.
5 and
u
1
=−
ξ
=
.
From (7), with
0
8
u
1
,
−
8
u
1
=−
0
.
5
(
1
−
0
.
5
)/(
1
+
0
.
5
)
(13)
or
020833 (14)
It follows from the exact solution (10) that the error incurred with this simple configuration
is
u
1
=
0
.
0
.
020833
−
0
.
017601
×
100
=
18
.
4%
(15)
0
.
017601
Case b (Fig. 8.5b, M
4)
As a second case, divide the beam into four segments. Then
M
=
=
4
,h
=
0
.
25
.
From (7)
and (11),
u
1
=
(
25
2
i
=
1:
1
/
0
.
)(
u
0
(
=
0
)
−
2
u
1
+
u
2
)
=−
0
.
25
(
1
−
0
.
25
)/(
1
+
0
.
25
)
=−
0
.
150
u
2
=
(
25
2
=
/
.
)(
−
+
)
i
2:
1
0
u
1
2
u
2
u
3
=−
.
(
−
.
)/(
+
.
)
=−
.
0
5
1
0
5
1
0
5
0
166667
u
3
=
(
25
2
i
=
3:
1
/
0
.
)(
u
2
−
2
u
3
+
u
4
(
=
0
))
=−
0
.
75
(
1
−
0
.
75
)/(
1
+
0
.
75
)
=−
0
.
107143
(16)
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