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Problems and Solutions for Section 4.4 (4.44 through 4.55)
4.44
A vibration model of the drive train of a vehicle is illustrated as the threedegree
offreedom system of Figure P4.44. Calculate the undamped free response [i.e.
M
(
t
) =
F
(
t
) = 0,
c
1
=
c
2
= 0] for the initial condition
x
(0) =
0
,
!
x
(0) = [0 0 1]
T
.
Assume that the hub stiffness is 10,000 N/m and that the axle/suspension is
20,000 N/m. Assume the rotational element
J
is modeled as a translational mass
of 75 kg.
Solution:
Let
k
1
= hub stiffness and
k
2
= axle and suspension stiffness.
The equation of motion is
75
0
0
0
100
0
0
0
3000
!
"
#
#
#
$
%
!!
x
+
10,000
1
’
1
0
’
1
3
’
2
0
’
2
2
!
"
#
#
#
$
%
x
=
0
x
0
( )
=
0
and
!
x
0
( )
=
0 0 1
!
"
$
%
T
m/s
Calculate eigenvalues and eigenvectors:
M
!
1/2
=
0.1155
0
0
0
0.1
0
0
0
0.0183
"
#
$
$
$
%
’
’
’
!
K
=
M
!
1/2
KM
!
1/2
=
133.33
!
115.47
0
!
115.47
300
!
36.515
0
!
36.515
6.6667
"
#
$
$
$
%
’
’
’
det
!
K
!
"
I
( )
=
3
!
440
2
+
28,222
=
0
1
=
0
#
1
=
0 rad/s
2
=
77.951
2
=
8.8290 rad/s
3
=
362.05
3
=
19.028 rad/s
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1
=
0.1537
0.1775
0.9721
!
"
#
#
#
$
%
,
v
2
=
’
0.8803
’
0.4222
0.2163
!
"
#
#
#
$
%
,
v
3
=
0.4488
’
0.8890
0.0913
!
"
#
#
#
$
%
Use the mode summation method to find the solution.
Transform the initial conditions:
q
0
( )
=
M
!
1/2
x
0
( )
=
0
,
!
q
0
( )
=
M
1/2
!
x
0
( )
=
0
0
54.7723
"
#
$
%
T
The solution is given by:
q
t
( )
=
c
1
+
c
4
t
( )
v
1
+
c
2
sin
!
2
t
+
"
2
( )
v
2
+
c
3
sin
3
t
+
3
( )
v
3
where
i
=
tan
"
1
#
i
v
i
T
q
0
( )
v
i
T
q
0
( )
$
%
’
(
)
i
=
2,3
c
i
=
v
i
T
!
q
0
( )
i
cos
i
=
2,3
Thus,
2
=
3
=
0,
c
2
"
1.3417, and
c
3
=
0.2629
So,
q
0
( )
=
c
1
v
1
+
c
i
sin
i
v
i
i
=
2
3
"
!
q
0
( )
=
c
4
v
1
+
i
c
i
cos
i
v
i
i
=
2
3
"
Premultiply by
v
1
T
;
v
1
T
q
0
( )
=
0
=
c
1
v
1
T
!
q
0
( )
=
53.2414
=
c
4
So,
q
t
( )
=
53.2414
t
v
1
+
1.3417sin 8.8290
t
( )
v
2
+
0.2629sin 19.028
t
( )
v
3
Change to
q(
t
):
x
t
( )
=
M
!
1/2
q
t
( )
x
t
( )
=
0.9449
t
1
1
1
"
#
$
$
$
%
’
’
’
+
!
0.1364
!
0.05665
0.005298
"
#
$
$
$
%
’
’
’
sin8.8290
t
+
0.01363
!
0.02337
0.0004385
"
#
$
$
$
%
’
’
’
sin19.028
t
m
4.45
Calculate the natural frequencies and normalized mode shapes of
4
0
0
0
2
0
0
0
1
!
"
#
#
#
$
%
!!
x
+
4
’
1
0
’
1
2
’
1
0
’
1
1
!
"
#
#
#
$
%
x
=
0
Solution:
Given the indicated mass and stiffness matrix, calculate eigenvalues:
M
!
1/2
=
0.5
0
0
0
0.7071
0
0
0
1
"
#
$
$
$
%
’
’
’
(
!
K
=
M
!
1/2
KM
!
1/2
=
1
!
0.3536
0
!
0.3536
1
!
0.7071
0
!
0.7071
1
"
#
$
$
$
%
’
’
’
det
!
K
!
"
I
( )
=
3
!
3
2
+
2.375
!
0.375
=
0
1
=
0.2094,
2
=
1,
3
=
1.7906
The natural frequencies are:
!
1
=
0.4576 rad/s
2
=
1 rad/s
3
=
1.3381 rad/s
The corresponding eigenvectors are:
v
1
=
!
0.3162
!
0.7071
!
0.6325
"
#
$
$
$
%
’
’
’
v
2
=
0.8944
0
!
0.4472
"
#
$
$
$
%
’
’
’
v
3
=
0.3162
!
0.7071
0.6325
"
#
$
$
$
%
’
’
’
The relationship between eigenvectors and mode shapes is
u
=
M
!
1/2
v
The mode shapes are:
u
1
=
!
0.1581
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This note was uploaded on 03/31/2009 for the course MECHANICAL MAE351 taught by Professor J.g.lee during the Spring '09 term at Korea Advanced Institute of Science and Technology.
 Spring '09
 J.G.Lee

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