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FIGURE 11.45
Central differences with three segments.
Three Segments
3
,
Fig. 11.45)
For three segments,
h
(
M
=
=
length
/
M
=
L
/
3. Then, with (1) discretized using (2) at nodes 1
and 2,
2
h
2
(w
1
h
4
(w
−
1
)
+
α
−
4
w
+
6
w
−
4
w
+
w
−
2
w
+
w
)
=
0
0
1
2
3
0
1
2
(6)
2
h
2
(w
1
−
1
h
4
(w
0
−
w
3
+
w
4
)
+
α
4
w
1
+
6
w
2
−
4
2
w
2
+
w
3
)
=
0
These reduce to
2
h
2
2
h
2
(
7
w
−
4
w
)
+
α
(
−
2
w
+
w
)
=
0
,
(
−
4
w
+
5
w
)
+
α
(w
−
2
w
)
=
0
(7)
1
2
1
2
1
2
1
2
2
h
2
,
With
λ
=
α
7
w
1
w
2
0
0
−
2
λ
−
4
+
λ
=
(8)
−
4
+
λ
5
−
2
λ
K U 0
The determinant of these equations provides the characteristic equation from which the
critical load can be calculated. Thus,
=
7
−
2
λ
−
4
+
λ
2
0or19
−
16
λ
+
3
λ
=
0
(9)
−
4
+
λ
5
−
2
λ
±
√
28
Then
λ
1
,
2
=
16
/
6
/
36 or
λ
1
=
3
.
549
,
λ
2
=
1
.
785. The lower value of
λ
leads to the
critical load.
9
EI
063
EI
L
2
P
cr
=
1
.
785
·
L
2
=
16
.
(10)
Collatz (1960) suggests that an improved eigenvalue can be obtained from two (
n
=
2
and
n
=
3) finite difference calculations using
h
1
h
2
−
P
cr
=
P
1
+
h
1
(
P
1
−
P
2
)
(11)
For the beam of this example,
12
063
EI
9
5
4
31
EI
L
2
P
cr
=
+
.
L
2
=
19
.
(12)
This is 4.4% in error relative to the exact solution of
P
cr
=
20
.
19
EI
/
L
2
.
Four Segments
(
M
=
4, Fig. 11.46)
For four segments,
h
=
L
/
4, and at nodes 1, 2, and 3,
2
h
2
(w
1
h
4
(w
−
1
)
+
α
−
4
w
+
6
w
−
4
w
+
w
−
2
w
+
w
)
=
0
0
1
2
3
o
1
2
2
h
2
(w
1
h
4
(w
)
+
α
−
4
w
+
6
w
−
4
w
+
w
−
2
w
+
w
)
=
0
(13)
0
1
2
3
4
1
2
3
2
h
2
(w
1
h
4
(w
)
+
α
−
4
w
+
6
w
−
4
w
+
w
−
2
w
+
w
)
=
0
1
2
3
4
5
2
3
4
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