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A complete solution for a simple beam with no loading would be a third order polyno-
mial with a term including
3
Hence, it should not be surprising when the trial solution
proposed here using (2) does not lead to the exact solution. However, note that the polyno-
mial of (2) is indeed a solution of the homogeneous beam relationship
ξ
.
i
v
.
In Eq. (7.84) [or Eq. (7.81)] retain only those terms corresponding to the actual boundary
conditions f
or
the
p
ro
ble
m
. A
lso, since there are no applied forces or displacements on the
boundaries,
w
=
0
w
=
θ
=
=
=
.
M
V
0
Equation (7.81) then reduces to
δθ(
)
(
)
−
δ
(
)w(
)
+
δ
(
)w(
)
+
δ
(
)θ(
)
=
L
M
L
V
L
L
V
0
0
M
0
0
0
(4)
or, in the notation of Eq. (7.84),
N
u
(
)w
(
N
T
u
N
T
u
N
T
u
)w
(
L
L
)
+
(
L
)w(
L
)
−
(
0
)w(
0
)
+
(
0
0
)
=
0
(5)
To use (5), we will need
p
0
L
4
120
EI
(
2
]
2
]
4
5
N
u
=
[
ξξ
w(
x
)
=
[
ξξ
w
+
5
ξ
−
ξ
)
p
0
L
3
24
EI
(
1
L
[1
1
L
[1
N
u
=
w
(
3
4
2
ξ
]
x
)
=
2
ξ
]
w
+
4
ξ
−
ξ
)
(6)
p
0
L
2
6
EI
1
L
2
[0
1
L
2
[0
N
u
=
w
(
2
3
2]
x
)
=
2]
w
+
(
3
ξ
−
ξ
)
1
L
3
[0
N
u
=
0]
Substitute these expressions in (5).
1
2
[0
0
2
[1
p
0
L
4
3
EI
2]
w
+
+
0]
w
=
0
or
02
04
00
20
1
p
0
L
4
EI
/
3
+
w
=−
(7)
2
/
3
Finally, we obtain a system of equations for
w
,
02
24
w
−
p
0
L
4
EI
1
/
3
1
=
(8)
w
−
2
/
3
2
which has the solution
w
1
w
2
0
−
p
0
L
4
EI
=
(9)
1
/
6
The approximate deflection then becomes
p
0
L
4
120
EI
(
−
2
4
5
w(
x
)
=
20
ξ
+
5
ξ
−
ξ
)
(10)
It is clear that this solution does not satisfy all boundary conditions. For example,
w(
L
)
=
0
.
Therefore, we will try to improve the trial solution.
For a second trial solution, use
w
1
2
3
2
3
]
w
=
N
u
w
=
ξ w
1
+
ξ
w
2
+
ξ
w
3
=
[
ξξ
ξ
w
(11)
2
w
3
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