Information Technology Reference
In-Depth Information
so that the complete trial function is
p
0
L
4
120
EI
(
2
3
4
5
w(
x
)
=
ξ w
+
ξ
w
+
ξ
w
+
ξ
−
ξ
)
5
(12)
1
2
3
We will need the derivatives
p
0
L
3
24
EI
(
1
L
[1
1
L
[1
N
u
=
2
]
w
(
2
]
3
4
ξ
ξ
)
=
ξ
ξ
+
ξ
−
ξ
)
2
3
x
2
3
w
4
(13)
1
L
2
[026
1
L
2
[026
p
0
L
2
6
EI
N
u
=
w
(
2
3
ξ
]
x
)
=
ξ
]
w
+
(
3
ξ
−
ξ
)
Introduction of these expressions into Eq. (7.84) in the form
N
u
)w
(
N
T
u
N
T
u
N
T
u
)w
(
(
1
1
)
+
(
1
)w(
1
)
−
(
0
)w(
0
)
+
(
0
0
)
=
0
(14)
leads to
[026]
[111]
1
2
3
0
0
6
0
2
0
p
0
L
4
3
EI
p
0
L
4
30
EI
[100]
w
+
+
w
+
+
w
=
0
or
−
1
/
3
02 6
0412
0618
000
000
666
000
200
000
!
"
p
0
L
4
EI
+
+
w
=
−
2
/
3
−
1
/
5
−
1
This reduces to the expression
=
02 6
24 2
61224
w
−
1
/
3
1
p
0
L
4
EI
w
−
2
/
3
(15)
2
w
−
6
/
5
3
which gives the free parameters
w
0
1
p
0
L
4
EI
=
w
1
/
30
(16)
2
w
3
−
1
/
15
This solution corresponds to the polynomial
p
0
L
4
120
EI
(
2
3
4
5
w(
x
)
=
4
ξ
−
8
ξ
+
5
ξ
−
ξ
)
(17)
which, as we learned earlier, is the exact solution for the displacement.
To compare the results, we label the solutions as
A. Exact solution given by (17). Also the same as case C
B. First Trial Deflection:
w(
p
0
L
4
2
4
5
x
)
=
/(
120
EI
)(
−
20
ξ
+
5
ξ
−
ξ
)
w(
)
=
p
0
L
4
/(
)(
ξ
2
−
ξ
3
+
ξ
4
−
ξ
5
)
C. Second Trial Deflection:
x
120
EI
4
8
5
Search WWH ::
Custom Search