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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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