lecture-11 wk 9 2dof coupling and semi-definite
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TWO-DEGREE-OF-FREEDOM
SYSTEMS COORDINATE COUPLING AND SEMI-
DEFINITE SYSTEMS
MEMB443 Mechanical Vibrations
LEARNING OBJECTIVES
Upon completion of this lecture, you should be able to:
Understand the concepts of static and dynamic couplings in
the equations of motion of a two-degree-of-freedom system.
Recognize that the choice of generalized coordinates
determines the type of coupling in the equations of motion.
Recognize that a system that undergoes rigid body motion,
known as semi-definite system, will have one or more
natural frequencies that is zero corresponding to rigid body
modes.
MEMB443 Mechanical Vibrations
COORDINATE COUPLING
The differential equations of motion for the two degrees of
freedom are in general coupled.
This means both coordinates appear in each equation.
In the most general case the two equations for the
undamped vibration system are expressed in matrix form
as:
0
0
2
1
2221
1211
2
1
2221
1211
x
x
kk
kk
x
x
mm
mm
MEMB443 Mechanical Vibrations
COORDINATE COUPLING (cont.)
Mass or dynamical coupling exists if the mass matrix is
non-diagonal.
Stiffness or static coupling exists if the stiffness is non-
diagonal.
It is possible to find a coordinate system which has neither
form of coupling.
The two equations are then decoupled and each equation
may be solved independently of the other.
MEMB443 Mechanical Vibrations
COORDINATE COUPLING (cont.)
Although it is always possible to decouple the equations of
motion for the undamped system, this is not always the
case for a damped system.
If , then the damping is said to be proportional
(proportional to the stiffness or mass matrix).
The system equations can therefore be decoupled.
0
0
2
1
2221
1211
2
1
2221
1211
2
1
2221
1211
x
x
kk
kk
x
x
cc
cc
x
x
mm
mm
02112 cc
MEMB443 Mechanical Vibrations
STATIC COUPLING
Choosing coordinates and , the system will have
static coupling.
If , the coupling disappears, and we obtain
uncoupled and vibrations.
0
0
0
02
22
2
111122
112221
x
lklklklk
lklkkkx
J
m
x
2211 lklk
x
MEMB443 Mechanical Vibrations
DYNAMIC COUPLING
There is some point C along the bar where a force applied
normal to the bar produces pure translation i.e. .
Choosing coordinates and eliminates static coupling
and introduces dynamic coupling.
0
0
0
02
42
2
31
21
cc
c
x
lklk
kkx
Jme
mem
4231 lklk
cx
MEMB443 Mechanical Vibrations
STATIC AND DYNAMIC COUPLING
If we choose at the end of the bar, the following
equations of motion will result:
Both static and dynamic coupling are now present.
0
01
2
22
2211
11
1
x
lklk
lkkkx
Jml
mlm
1xx
MEMB443 Mechanical Vibrations
SEMI-DEFINITE SYSTEMS
Semi-definite systems are also known as unrestrained or
degenerate systems.
Physically this implies that the system is supported in such
a manner that rigid-body motion is possible.
MEMB443 Mechanical Vibrations
SEMI-DEFINITE SYSTEMS (cont.)
For the spring-masses system, the equations of motion is
given by:
For free vibration, we assume the motion to be harmonic
of the form:
0
0
1222
2111
xxkxm
xxkxm
iii tXtx cos
2,1i
MEMB443 Mechanical Vibrations
SEMI-DEFINITE SYSTEMS (cont.)
Substitution of the solution into the equations of motion
yield:
Equating the determinant of the coefficients of and
to zero, we obtain the characteristic equation:
0
0
2
2
21
21
2
1
XkmkX
kXXkm
021
2
21
2 mmkmm
1X
2X
MEMB443 Mechanical Vibrations
SEMI-DEFINITE SYSTEMS (cont.)
The natural frequencies are therefore found to be:
It is seen that one of the natural frequencies of the system
is zero, which means that the system is not oscillating.
The system moves as a whole without any relative motion
between the two masses (rigid body translation).
01
21
212
mm
mmk
MEMB443 Mechanical Vibrations
EXAMPLE 1
Two identical circular cylinders, of radius r and mass m
each, are connected by a spring of stiffness k. Determine
the natural frequencies of oscillation of the system.
Assume no slipping occurs.
MEMB443 Mechanical Vibrations
EXAMPLE 1 (cont.)
Determine the equivalent mass of one of the cylinder for
translational motion.
Since both mass are identical the equivalent masses are the
same.
mm
m vr
vm rvm
m vJvm
eq
GG
Geq
GGGeq
2
3
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
2
22
222
MEMB443 Mechanical Vibrations
EXAMPLE 1 (cont.)
Write the equations of motion.
Assume the solutions given below and substitute these
solutions and their derivatives into the above equations.
02
3
02
3
212
211
xxkxm
xxkxm
tXtx
tXtx
cos
cos
22
11
MEMB443 Mechanical Vibrations
EXAMPLE 1 (cont.)
The following equations are then obtained.
The above equations in matrix form are:
0
0
2
32
3
2
1
2
2
X
X
kmk
kkm
02
3
02
3
12
2
21
2
kXXkm
kXXkm
MEMB443 Mechanical Vibrations
EXAMPLE 1 (cont.)
Equate the determinant of the dynamic matrix in the
previous equation to zero, and we obtain the characteristics
equation.
02
6
4
9
02
6
4
9
222
242
m km
m km
0
2
32
3
det2
2
kmk
kkm
MEMB443 Mechanical Vibrations
EXAMPLE 1 (cont.)
The roots of the characteristics equation give the natural
frequencies of this system.
rad/s 3
4
0
2
1
m
k
MEMB443 Mechanical Vibrations