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where
m
22
k
21
k
22
2
m
22
T
2
=
+
=
+
m
m
11
m
11
k
21
k
12
k
22
Equation (10.23) contains only one variable
u
1
, and, hence, the original two-DOF system is
reduced to a single-DOF system.
In this procedure, Eq. (10.18) from the static case is used to reduce the DOF of the dynamic
equations. This means that it is supposed that DOF 2 is related to DOF 1 through the static
stiffness properties of the structure. The inertia property of DOF 2 is ignored, which implies
that the structural inertia of DOF 2 does not affect the deformation shape of the system. Note
that this is not the same as simply setting
m
2
=
T
2
k
22
k
=
k
11
+
Tk
21
+
Tk
12
+
=
k
11
−
0. An application of the static relation of
Eq. (10.18) to the dynamic case was introduced by Guyan (1965), and, hence, this kinematic
condensation is often referred to as
Guyan reduction.
Reduction of the number of DOF in a system can be considered in terms of dependent
and independent DOF. In the above case,
u
1
is the independent (
active
or
master
) coordinate,
while
u
2
is the dependent (or
slave
) coordinate. In practice, it is common to let the translation
DOF be the master coordinates and the rotary DOF be the slave coordinates.
Consider now a multiple DOF system. Define
V
a
to be the active or master set of dynamic
DOF and
V
0
to be the omitted set of DOF of the system, i.e., the DOF to be condensed.
Rewrite the governing equations
M V
+
KV
=
0
(10.24)
with
V
a
V
0
V
=
This form is achieved by interchanging some rows and columns of the original
M
and
K
in
Eq. (10.16). Partition the resulting
M
and
K
as
M
aa
and
K
aa
M
a
0
K
a
0
M
=
K
=
(10.25)
M
0
a
M
00
K
0
a
K
00
Then Eq. (10.24) appears as
M
aa
V
a
V
0
K
aa
V
a
V
0
0
0
M
a
0
K
a
0
+
=
(10.26)
M
0
a
M
00
K
0
a
K
00
Note for a lumped mass model that
M
a
0
For Guyan reduction, we will neglect
the dynamic effect due to the mass associated with
V
0
=
M
0
a
=
0
.
.
Then the second equation of Eq.
(10.26) provides
K
0
a
V
a
+
K
00
V
0
=
0
This relation can be solved for
V
0
in terms of
V
a
to obtain
K
−
1
=−
V
0
00
K
0
a
V
a
(10.27)
Rewrite this as
I
−
V
a
V
0
=
TV
a
,
T
=
(10.28)
K
−
1
00
K
0
a
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