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A Second Iteration
2
,
To observe the effect of further iterations, return to (4) and determine
w
x
x
0
(
C
60
2
4
x
6
15
Lx
5
15
L
2
x
4
2
L
3
x
3
6
L
4
x
2
w
=
−
+
+
−
)
dx dx
0
2
L
3
L
x
0
(
x
3
4
x
6
15
Lx
5
15
L
2
x
4
2
L
3
x
3
6
L
4
x
2
+
−
+
+
−
)
dx dx
0
2
L
2
L
0
x
0
(
dx dx
3
x
2
4
x
6
15
Lx
5
15
L
2
x
4
2
L
3
x
3
6
L
4
x
2
−
−
+
+
−
)
C
8400
(
10
x
8
50
Lx
7
70
L
2
x
6
14
L
3
x
5
70
L
4
x
4
13
L
5
x
3
39
L
6
x
2
=
−
+
+
−
−
+
)
(16)
From Eq. (11.74),
4
x
4
15
Lx
3
15
L
2
x
2
2
L
3
x
6
L
4
140
(
−
+
+
−
)
EI
L
2
P
cr
=−
(17)
(
10
x
6
−
50
Lx
5
+
70
L
2
x
4
+
14
L
3
x
3
−
70
L
4
x
2
−
13
L
5
x
+
39
L
6
)
This can be evaluated, giving
x
0
L
/
4
L
/
2
3
L
/
4
L
(18)
P
cr
L
2
EI
21.54
21.03
20.40
20.08
20.00
% Error
6.67
4.17
1.06
−
0
.
55
−
0
.
94
The mean value would be
P
cr
L
2
EI
1
5
(
=
21
.
54
+
21
.
03
+
20
.
40
+
20
.
08
+
20
.
00
)
=
20
.
61
(
2
.
09% Error
)
(19)
From (11)
38
EI
L
2
P
cr
=
20
.
(
0
.
92% Error
)
(20)
and from the Rayleigh quotient of (14),
196
EI
L
2
P
cr
=
20
.
(
0
.
007% Error
)
(21)
Finite Differences
As is to be expected, the finite difference expressions of Chapter 8 can be employed to
compute the critical loading. Simply use the finite difference relations to discretize the gov-
erning equations [Eq. (11.31)]. Apply the boundary conditions and then find the critical load
from a characteristic equation formed of the determinant of the finite difference equations.
EXAMPLE 11.16 A Fixed-Hinged Column
Use finite difference equations to compute the critical load for the fixed-hinged column of
Fig. 11.28.
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