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FIGURE 5.24
Linearly varying load.
where
10
−
p
a
p
b
N
p
=
[1
ξ
]
G
p
=
p
p
=
(3)
11
We wish to calculate [Eq. (4.58)]
G
T
b
a
G
T
1
0
p
i
0
N
u
p
z
dx
N
u
N
p
=
=
G
p
p
p
d
ξ
(4)
where
G
and
N
u
are taken from Chapter 4, Eq. (4.47a). We find
.
1
1
/
2
1
ξ
ξ
.
1
1
1
/
2
1
/
3
N
u
N
p
d
ξ
=
[1
ξ
]
d
ξ
=
(5)
.
2
1
/
3
1
/
4
0
0
3
ξ
.
1
/
4
1
/
5
Carry out the multiplications indicated in (4) and find
7
/
20
3
/
20
p
a
p
b
−
1
/
20
−
1
/
30
p
i
0
=
(6)
3
/
20
7
/
20
1
/
30
1
/
20
For the two-element beam of Fig. 5.5:
Element 1, with
p
a
=
p
0
,p
b
=
p
0
/
2
Element 2, with
p
a
=
p
0
/
2
,p
b
=
0
=
=
51
−
21
−
p
a
p
b
p
b
p
c
p
0
120
p
0
120
8
39
7
3
9
2
p
1
0
p
2
0
(
)
=
(
)
=
(7)
The global matrix will appear as
P
0
P
∗
−
=
KV
(8)
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