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expressed as
w
0
+
w
0
+
w
L
+
w
L
=
w
T
N
u
N
u
=
w(
x
)
=
H
1
w
+
H
2
H
3
H
4
w
+
H
5
H
6
w
(10)
0
L
w
=
w
|
w
=
w
|
with
L
,
etc. Chapter 4, Section 4.4.2 may be helpful in understanding
the use of Hermitian polynomials. Expressions for
H
j
i
0
,
0
x
=
L
x
=
can be found in
standard references or derived using the procedures of Chapter 4. Because of the boundary
conditions of the beam (Fig. 7.1),
as functions of
ξ
w
=
w
0
=
w
=
w
L
=
.
0
Thus,
0
L
H
3
w
0
+
H
5
w
L
=
w
T
N
u
w
=
N
u
w
=
(11)
with
1
2
ξ
5
3
2
ξ
3
2
ξ
1
2
ξ
2
3
4
3
4
5
N
u
=
−
+
−
(
−
4
ξ
+
7
ξ
−
3
ξ
)
w
0
w
L
w
=
Insertion of (11) into (2) leads to the system of equations for
w
,
L
3
34
1
p
0
L
EI
/
6
w
=
(12)
4
192
−
8
/
6
which has the solution
p
0
L
4
120
EI
[8
1]
T
w
=
−
(13)
Substitution of this expression in (11) gives
p
0
L
4
120
EI
(
2
3
4
5
w
=
4
ξ
−
8
ξ
+
5
ξ
−
ξ
)
(14)
The results can be compared with the exact solution, which has been derived several
times in this work.
A. Exact solution:
p
0
L
4
120
EI
(
2
3
4
5
w
=
4
ξ
−
8
ξ
+
5
ξ
−
ξ
)
(15)
B. Trial deflection with
m
=
1:
p
0
L
4
108
EI
(
2
3
4
w
=
3
ξ
−
5
ξ
+
2
ξ
)
(16)
C. Trial deflection with
m
=
2:
p
0
L
4
120
EI
(
2
3
4
5
w
=
4
ξ
−
8
ξ
+
5
ξ
−
ξ
)
(17)
D. Trial deflection based on fifth degree Hermitian polynomials:
p
0
L
4
120
EI
(
2
3
4
5
w
=
4
ξ
−
8
ξ
+
5
ξ
−
ξ
)
(18)
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