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