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Of course, the displacement boundary conditions are
u
0
=
u
4
=
0
.
The derivatives on the
left-hand side are replaced by the values obtained from (7) with
M
=
4
,i
=
0
,
1
,
2
,
3
,
and
4. This gives
u
0
=
u
1
=−
u
2
=−
u
3
=−
u
4
=
0
,
0
.
15
,
0
.
16667
,
0
.
107143
,
0
Substitution of these values in (34) yields
10
−
3
−
120
1
u
1
u
2
u
3
8
.
68
=−
21
01
−
10
.
02
(35)
−
2
6
.
45
K
U
=
P
with the solution
u
1
=
0
.
013132
,u
2
=
0
.
0175843
,
and
u
3
=
0
.
0120155
.
The associated errors
of
082%
,
respectively, indicate that only a slight improvement
occurs in the results relative to the higher order scheme.
The general expression for this beam, using the multiple position difference form of (33),
would be
−
0
.
13%
,
−
0
.
095%
,
and
−
0
.
ξ
i
(
1
−
ξ
i
−
1
)
10
ξ
i
(
1
+
ξ
i
+
ξ
i
+
1
(
−
ξ
i
)
1
−
ξ
i
+
1
)
h
2
u
i
−
1
−
2
u
i
+
u
i
+
1
=−
(
/
12
)
+
ξ
i
−
1
+
(36)
1
1
1
+
ξ
i
+
1
Accuracy Comparison
Results for the three techniques for
M
=
4 and 8 are summarized below
Simple Central
Higher Order
Multiple Position
Differences
Schemes
Differences
ξ
Exact
M
=
4
M
=
8
M
=
4
M
=
8
M
=
4
M
=
8
(% error)
(% error)
(% error)
(% error)
(% error)
(% error)
0
.
25
0
.
013149
0
.
013917
0
.
013333
0
.
013183
0
.
013152
0
.
013132
0
.
013148
(37)
(
5
.
8%
)
(
1
.
5%
)
(
0
.
27%
)
(
0
.
018%
)
(
−
0
.
13%
)
(
−
0
.
0076%
)
0
.
50
0
.
017601
0
.
018462
0
.
017817
0
.
017617
0
.
017602
0
.
0175843
0
.
017600
(
4
.
8%
)
(
1
.
2%
)
(
0
.
095%
) (
0
.
0047%
) (
−
0
.
095%
)
(
−
0
.
0066%
)
0
.
75
0
.
012026
0
.
012567
0
.
012165
0
.
012033
0
.
012027
0
.
0120155
0
.
0120255
(
4
.
6%
)
(
1
.
2%
)
(
0
.
086%
) (
0
.
0055%
) (
−
0
.
082%
)
(
−
0
.
0055%
)
Several conclusions can be drawn from these results. The simple central difference form-
ula provides a result within 5% with relative efficiency (bandwidth
3). This accuracy may
be adequate for many purposes. The error decreases somewhat if the more complicated
methods are employed. For
M
=
=
4
,
the errors for the higher order and multiple position
methods are about 1
/
50 the error of the simple central difference. For
M
=
8
,
the change is
about 1
/
200
.
EXAMPLE 8.2 Beam with Linearly Varying Loading
Apply finite differences to obtain the response of the beam of Fig. 8.6. This beam has been
used as an example in several earlier chapters.
Divide the beam into five segments, so that
h
=
L
/
5
.
The local governing equation is
i
v
EI
w
(
x
)
=
p
z
(
x
)
=
p
0
(
1
−
x
/
L
)
(1)
with boundary conditions
w(
0
)
=
0
,
θ(
0
)
=
0
,
w(
L
)
=
0
,
and
M
(
L
)
=
0
.
At particular points,
(1) becomes
v
i
EI
w
=
p
i
(2)
i
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