Information Technology Reference
In-Depth Information
Substitution of these expressions into Eq. (11.76), with
N
0
=−
P
, leads to
v
T
[
EI
G
T
k
EI
G
P
G
T
k
P
G
]
v
−
δ
W
=
δ
+
=
0
(2)
with
2
3
4
2
3
3
2
12
5
2
ξ
6
ξ
12
ξ
1
L
2
1
1
1
L
N
u
N
u
Ld
3
4
5
k
P
=
ξ
=
2
ξ
6
ξ
12
ξ
Ld
ξ
=
1
2
6
5
(3)
2
0
0
4
5
6
2
ξ
6
ξ
12
ξ
2
5
12
7
1
2
00 8
00 6
0024
0024
ξ
1
L
4
1
1
1
L
3
N
u
N
i
u
3
k
EI
=
Ld
ξ
=
0024
ξ
Ld
ξ
=
(4)
0
0
4
0024
ξ
/
5
0857
L
3
This gives 1
.
8
EIL
−
P
0
.
=
0or
1
.
8
EI
L
2
=
21
EI
L
2
P
cr
=
(5)
0
.
0857
Consider a two-term assumed displacement that satisfies all the boundary conditions.
2
3
4
3
4
5
w
=
(
3
ξ
/
2
−
5
ξ
/
2
+
ξ
)w
1
+
(
4
ξ
/
3
−
7
ξ
/
3
+
ξ
)w
2
(6)
For this assumed displacement,
3
/
2
0
w
1
w
G
−
/
/
5
24
3
v
=
=
(7)
1
−
7
/
3
2
0
1
=
ξ
2
ξ
3
ξ
4
ξ
5
]
N
u
[
N
u
=
[
2
ξ
3
ξ
2
4
ξ
3
5
ξ
4
]
/
L
N
u
=
2
3
]
L
2
[26
ξ
12
ξ
20
ξ
/
(8)
N
u
=
2
]
L
3
[0624
ξ
60
ξ
/
N
i
u
=
L
4
[0024120
ξ
]
/
2
PL
2
With these expressions and
λ
=
ε
=
/
EI
, Eq. (11.76) leads to
1
0
0
0
.
80
0
.
80
.
0857
0
.
0429
−
λ
=
(9)
0
.
80
0
.
6095
0
.
0429
0
.
0254
λ
2
−
λ
+
.
=
.
or
8714
1359
2
0
Then
35
EI
L
2
P
cr
=
20
.
(10)
L
2
as compared to the exact solution of
P
cr
=
20
.
19
EI
/
.
Note that Galerkin's method appears
to provide an upper bound to the exact solution.
Search WWH ::
Custom Search