Matlab MDOF Notes Part 1
B.P. Wang, MAE UTA, March 25,2010
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2.5 Distributed Parameter Systems
Fig. 2.6 Schematic
representation of a clamped
beam under transversal
vibration
41
x
0
dx
y
Fig. 2.7 Forces and
moments acting on the
infinitesimally small portion
of the clamped beam under
transversal vibration
x
0
T
M
M
2.5 Distributed Parameter Systems
45
and substituting this back into (2.84) we will get the solution in the following form
[4, 10]:
a(x) = C1 cosh x + C2 sinh x + C3 cos x + C4 sin x
(2.87)
Constants C1 , C2 , C3 and C4 are integration constants and can b
42
2 Basics of Vibration Dynamics
J
M
3 y(x, t)
=T
t 2 x
x
(2.66)
2.5.1.2 Simplifying the Equation of Motion
It is possible to collect these two equations into one, by expressing T from the second
equation (2.66) and substituting it back into the first o
2.5 Distributed Parameter Systems
43
y + c2 y iv = 0
(2.72)
This equation expresses the free transversal vibration of a beam, neglecting the
dynamic effects of the longitudinal forces and rotational inertia. Clearly, there is a
lot of simplification assum
40
2 Basics of Vibration Dynamics
2.5 Distributed Parameter Systems
In practice, the vibration of continuously distributed parameter systems is solved and
analyzed through the finite element method. As in the case of other fields of science, in
FEM vibrat
44
2 Basics of Vibration Dynamics
y(x, t) = a(x)V (t)
(2.77)
meaning that the position at a given place and time is composed of a combination of
function a(x), which is only dependent on the horizontal position and V (t) which
is only dependent on time. W
2.4 Multiple Degree of Freedom Systems
39
while multiplying the stiffness matrix by from both sides we get
T
Ks [1 2 3 . . . N ]
T Ks = 1 2 3 . . . N
T
T
1 Ks 1 1 Ks 2 . . . 1T Ks N
T Ks 1 T Ks 2 . . . T Ks N
2
2
2
=
=
.
.
.
. .
.
.
(2.54)
T K T K . .
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2 Basics of Vibration Dynamics
The solution to the eigenvalue problem expressed by our physical vibrating system is a set of N eigenvalues ranging from 12 , 22 . . . 2N , where N is the number
of degrees of freedom. These eigenvalues have a very well-d
2.4 Multiple Degree of Freedom Systems
37
The solution of such systems is in fact very similar to the solution of SDOF
systems. To illustrate this, let us consider a case without damping and with no outside
force. Removing these effects from the equation
2.2 Free Vibration with Damping
Fig. 2.3 Free vibration of a
point mass with viscous
damping
31
0
qt
Spring: k
Fk
kq t
Fb
b
bq t
Mass: m
We can improve our previous model by adding this damping force to our system.
Let us have the same vibrating mass m co
36
2 Basics of Vibration Dynamics
0
k1
b1
0
q1 t , f e1 t
q2 t
f e2 t
k2
Mass: m1
b2
k3
Mass: m2
b3
Fig. 2.5 Multiple degrees of freedom system: connected set of two masses
moving masses illustrated in Fig. 2.5 is sliding on a frictionless surface. Now in
34
2 Basics of Vibration Dynamics
Fig. 2.4 Forced vibration of
a point mass with viscous
damping
0
qt
Spring: k
Fk
kq t
Fb
b
bq t
Mass: m
m q(t)
+ bq(t)
+ kq(t) = f e (t)
f e (t)
(2.34)
This is a second order ordinary differential equation, just like in
2.3 Forced Vibration of a Point Mass
35
possible to transform our ODE into a decoupled form, which is referred to as the
state-space representation.
The solution of the equation of motion consists of two parts. The transient response
describes the passing
2.2 Free Vibration with Damping
The second term is the damped natural frequency d or
d = d2 n2
33
(2.29)
Depending on the magnitude of the damped natural frequency d we may have
overdamping, critical damping or underdamping. Overdamping is the case, when
30
2 Basics of Vibration Dynamics
Substituting the trial solution in (2.11) to the original equation of motion and differentiating results the same expressions for n , however can assume both negative
and positive values. The undamped natural frequency n
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2 Basics of Vibration Dynamics
where critical damping is denoted by bc = 2 km. Instead of expressing the simplified differential equation in the terms of the damping coefficient d as in Eq. (2.19)
we may express it using proportional damping and get [4
46
2 Basics of Vibration Dynamics
n = (n l)
2
EJ
Al 4
(2.98)
The equations given by (2.89) will not make it possible to compute the integrating
constants, though it is possible to compute their ratios and substitute it back into (2.87)
to get an (x). The