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FIGURE 11.46
Central differences with four segments.
so that
(
7
w
−
4
w
+
w
)
+
λ(
−
2
w
+
w
)
=
0
1
2
3
1
2
(
−
4
w
+
6
w
−
4
w
)
+
λ(w
−
2
w
+
w
)
=
0
(14)
1
2
3
1
2
3
(w
−
4
w
+
5
w
)
+
λ(w
−
2
w
)
=
0
1
2
3
2
3
and
(
7
−
2
λ)
(
−
4
+
λ)
(
1
+
0
)
w
0
0
0
1
=
(
−
4
+
λ)
(
6
−
2
λ)
(
−
4
+
λ)
w
(15)
2
(
1
+
0
)
(
−
4
+
λ)
(
5
−
2
λ)
w
3
K
U
=
0
2
3
Then 44
−
68
λ
+
30
λ
−
4
λ
=
0, which gives
λ
=
1
.
111. The critical load would then be
1
111
16
EI
L
2
78
EI
L
2
P
cr
=
1
.
=
17
.
(16)
With Collatz's improvement,
EI
L
2
=
99
EI
L
2
P
cr
=
(
16
.
06
+
3
.
93
)
19
.
(17)
This differs from the exact solution by 1.0%.
Galerkin's Method
Any of the weighted-residual procedures of Chapter 7 can also be employed to compute
buckling loads. To illustrate this, consider Galerkin's method which begins with the global
form of the governing differential equation (Eq. (11.31)(
EI
w
)
−
N
0
w
=
0
)
for a beam
element in the form
b
a
δw(
b
a
δw
w
)
dx
N
0
w
dx
EI
−
=
0
(11.75)
and utilizes a trial function that satisfies all boundary conditions. With
w
=
N
u
Gv
,
v
T
G
T
N
u
EI
N
i
u
dx
G
G
T
N
u
N
0
N
u
dx
G
v
δ
−
=
0
(11.76)
EXAMPLE 11.17 A Fixed-Hinged Column
For the fixed-hinged column of Fig. 11.28, choose the assumed displacement
4
w
=
w
(ξ
−
1
3
2
2
.
5
ξ
+
1
.
5
ξ
)
,
ξ
=
x
/
L
and note that this satisfies the force boundary condition
M
b
=
0,
as well as all displacement boundary conditions (
w
=
θ
=
w
=
0). For this trial solution,
a
a
b
v
=
w
1
2
3
4
]
N
u
=
[
ξ
ξ
ξ
N
u
=
L
2
2
]
[26
ξ
12
ξ
/
(1)
N
i
u
=
L
4
[0024]
/
51
]
T
=
[
1
.
5
−
2
.
G
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