Information Technology Reference
In-Depth Information
where, with boundary conditions taken into account,
m
1
m
11
0
0
M
=
=
0
m
2
0
m
22
k
1
k
11
+
k
2
−
k
2
k
12
K
=
=
−
k
2
k
2
k
21
k
22
and
u
1
u
2
V
=
Assume that
m
22
is very small compared to
m
11
, and, hence, the DOF associated with
m
22
is to be condensed out. Before treating kinematic condensation, consider briefly static
condensation. The displacement equations for the static response of a general two-DOF
system can be written as
k
11
u
1
u
2
P
1
P
2
k
12
=
(10.17)
k
21
k
22
The second equation of Eq. (10.17) can be used to express
u
2
in terms of
u
1
P
2
k
22
+
u
2
=
Tu
1
(10.18)
where
T
Use this relationship in the first equation in Eq. (10.17) to eliminate
u
2
(back substitution) to obtain
=−
k
21
/
k
22
.
k
12
k
22
P
2
(
k
11
+
k
12
T
)
u
1
=
P
1
−
(10.19)
This is referred to as a statically condensed relation.
Now, return to the dynamic case where the equations of motion are given by Eq. (10.16).
If the relationship between
u
1
and
u
2
during the dynamic response is assumed to be the
same as for the static case, then the variable
u
2
can be eliminated from Eq. (10.16). That
is, the static relationship of Eq. (10.18) will be assumed to apply for dynamic motion. Set
P
2
=
0
,
so that Eq. (10.18) provides
u
2
=
Tu
1
and
u
2
=
T u
1
(10.20)
Substitute Eq. (10.20) into Eq. (10.16) giving
m
11
u
1
T u
1
k
11
u
1
Tu
1
0
0
0
k
12
+
=
(10.21)
0
m
22
k
21
k
22
Premultiply Eq. (10.21) by [1
T
] to obtain
T
]
m
11
u
1
T u
1
T
]
k
11
u
1
Tu
2
T
]
0
0
0
k
12
[1
+
[ 1
=
[ 1
(10.22)
0
m
22
k
21
k
22
or
+
=
m u
1
ku
1
0
(10.23)
Search WWH ::
Custom Search