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0
,
and ignore the rows
corresponding to the unknown reactions. This leads to the generalized linear eigenvalue
problem
Apply the displacement boundary conditions to
(
K
−
λ
M
)
φ
=
−
λ
=
2
2
4
6
2
0
0000
0200
0000
0000
θ
a
w
6
24
0
−
6
EI
ρ
2
b
0
(3)
2
2
2
3
2
08
2
θ
b
2
2
θ
0
−
6
2
4
c
2
. The frequencies can be found from this relationship using a standard eigen-
value solution procedure.
We find
where
λ
=
ω
=
ω
1
ω
=
179
.
408 rad
/
sec
or
f
1
π
=
28
.
55 Hz
(4)
1
2
As expected, the fundamental frequency derived using the lumped mass model is lower
than the exact value of
ω
=
.
74 rad/sec. Note that only one frequency is obtained. This
is expected, since for this simple model only one mass is able to move, i.e., this is a single
DOF system.
180
1
EXAMPLE 10.7 Natural Frequencies of a Frame
Compute the natural frequencies of the frame shown in Fig. 10.2.
Divide the frame into three beam elements. This is a crude model for a dynamic analysis,
and the resulting frequencies may not be very accurate. The consistent mass model of
Example 10.1, with the stiffness matrix of Eq. (11), Chapter 5, Example 5.5, leads to the
natural frequencies
ω
=
183
.
8 [159
.
5] rad
/
sec
or
f
1
=
29
.
3 [25
.
4] Hz
1
ω
=
711
.
6 [381
.
2]
or
f
2
=
113
.
26 [60
.
7]
2
ω
=
.
.
=
.
.
1415
6 [503
8]
or
f
3
225
30 [80
2]
3
(1)
ω
=
.
.
=
.
.
1587
3 [1332
9]
or
f
4
252
62 [212
1]
4
ω
=
2505
.
2 [1963
.
8]
or
f
5
=
398
.
72 [312
.
5]
5
ω
6
=
3686
.
0 [2252
.
3]
or
f
6
=
586
.
65 [358
.
5]
Shown in square brackets are the natural frequencies found using the lumped mass model of
Example 10.3 with rotary inertia. It is clear that the frequencies calculated using the lumped
mass model are lower than those obtained using the consistent mass model, especially for
the higher modes. If more elements are included in the model, the consistent and lumped
mass results are closer to each other.
If the masses associated with the rotational DOF are condensed out, the consistent
[lumped] mass model gives the natural frequencies
ω
=
184
.
4 [159
.
8] rad
/
sec
or
f
1
=
29
.
4 [25
.
4] Hz
1
ω
=
1555
.
6 [1279
.
4]
or
f
2
=
247
.
59 [203
.
6]
2
(2)
ω
=
2400
.
0 [1919
.
0]
or
f
3
=
381
.
98 [305
.
4]
3
ω
=
.
.
=
.
.
3244
3 [2178
4]
or
f
4
516
34 [346
7]
4
Apart from the lowest natural frequency, it appears as though condensing out the rotary
DOF has a major effect on the natural frequencies. However, it should be noted that the
second and third frequencies of (1) do not appear in (2), as these frequencies correspond to
rotary DOF which have been condensed out.
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