Information Technology Reference
In-Depth Information
FIGURE 5.28
Frame of Example 5.7.
Element 1:
Use
k
1
and
k
1
as given by Eqs. (1) and (3) of Example 5.5. For the first loading case, take
p
i
0
from Chapter 4, Table 4.2.
0
0
2
√
3
2
−
−
1
/
6
0
0
.
p
0
p
i
0
=−
=−
(1)
0
2
2
√
3
−
1
−
/
−
−
6
2
.
0
Using Eq.(2) of Example 5.5, the loading is transformed to global coordinates as
0
−
√
3
2
−
3
.
.
0
T
1
T
p
i
0
p
10
=−
=
(2)
−
3
.
0
√
3
−
−
2
.
0
Element 2:
Since there is no distributed loading on element 2,
p
20
=
.
0
The local and global coordinate
=
k
2
systems coincide so that
k
2
.
=
.
=
·
m
2
, and
EA
=∞
Use
3
0m,
EI
4712 kN
, giving
∞
0
0
−∞
0
0
0
2094
.
2
−
3141
.
30
−
2094
.
2
−
3141
.
3
0
−
3143
.
3
6282
.
7
0
3141
.
3
3141
.
3
k
2
=
(3)
−∞
0
0
∞
0
0
−
.
.
.
.
0
2094
2
3143
3
0
2094
2
3141
3
0
−
3141
.
3
3141
.
3
0
3141
.
3
6282
.
7
Search WWH ::
Custom Search