Lecture-12 Sec6_14 Modal Analysis

TWO-DEGREE-OF-FREEDOM

SYSTEMS FORCED VIBRATION RESPONSE

USING MODAL ANALYSIS

MEMB343 Mechanical Vibrations

LEARNING OBJECTIVES

Upon completion of this lecture, you should be able to:

Understand the underlying principle of modal analysis.

Transform the physical coordinates of a system to its

principal coordinates.

Determine the forced response of an undamped two-degree-

of-freedom system using modal analysis.


MODAL ANALYSIS CONCEPTS

The equations of motion of the two degrees of freedom

system shown above without damping is expressed in

matrix form as:

2

1

2

1

322

221

2

1

2

1

0

0

F

F

x

x

kkk

kkk

x

x

m

m

(1)


MODAL ANALYSIS CONCEPTS (cont.)

In solving Eq. (1) for the response for a particular set

of exciting forces, the major obstacle encountered is the

coupling between the equations.

Both coordinates and occur in each of the

equations.

The types of coupling could either be:

static (elastic) coupling (non-diagonal stiffness matrix), or

dynamic (inertial) coupling (non-diagonal mass matrix).

Both types of coupling may also occur together.

1x

x

2x


MODAL ANALYSIS CONCEPTS (cont.)

If the system of equations could be uncoupled, so that we

obtain diagonal mass and stiffness matrices, then each

equation would be similar to that of a single degree of

freedom system.

These equations could then be solved independent of each

other.

The process of deriving the equations of motion into an

independent set of equations is known as modal analysis.


PRINCIPAL COORDINATES

The coordinate transformation that is required is one that

decouples the system inertially and elastically

simultaneously.

This transformation yield diagonal mass and stiffness

matrices.

It has been proven that if the mass or the stiffness matrix is

post and pre-multiplied by a mode shape and its transpose

respectively, the result is some scalar constant.

Thus with the use of a matrix whose columns are the

mode shape vectors, we already have at our disposal the

necessary coordinate transformation.


PRINCIPAL COORDINATES (cont.)

The coordinates are transformed to by the

equation:

is referred to as the modal matrix and is called the

principal coordinates, normal coordinates or modal

coordinates.

nnnnu

u

u

u

u

u

u

u

u

2

1

2

2

1

1

2

1

x

x

(3)

(2)



Eq. (1) can be rewritten as:

Substituting Eq. (2) into Eq. (4) yields:

Pre-multiplying Eq. (5) by the transpose of the modal

matrix we obtain:

Fxkxm

T

(4)

Fkm (5)

Fkm TTT (6)



The post and pre-multiplication of the mass matrix by the

mode shapes and their transpose results in a matrix .

The diagonal elements of are some constants while

all the off-diagonal terms are zero.

Similarly for the stiffness matrix

and are diagonal matrices.

DT

Mm

DM

(7)

(8) DT

Kk

DK DM

DM



Therefore Eq. (6) can be rewritten as:

Eq. (9) represents equations of the form:

is the ith column of the modal matrix i.e. the ith mode

shape.

and are the ith modal mass and the ith modal

stiffness, respectively.

iT

iiiii FFKM

n

(9) FKM TDD

iKiM

(10)

i



Eq. (10) is the equation of motion for the single degree of

freedom system shown in the above figure.

Since , Eq. (10) can be written as:

i

T

i

T

i

i

iiii

m

F

M

F

2

iii MK2

(11)



Once the solution (time responses) of Eq. (10) for all is

obtained, the solution in terms of the original coordinates

can be obtained by transforming back, i.e. substituting

in Eq. (2).

n

x

x


EXAMPLE 1

Consider the free vibration analysis for the system above

where = 5 kg, = 10 kg, = 2 N/m and = 4

N/m.

Substituting these values into the equations of motion, we

have:

042210

02225

212

211

xxx

xxx

1m 2m 21 kk 3k


EXAMPLE 1 (cont.)

Assuming solutions of the form , we have:

We can express the above equation as:

06102

0245

2

2

1

21

2

tj

tj

eXX

eXX

0

0

6102

245

2

1

2

2

X

X

tj

tj

eXtx

eXtx

22

11


EXAMPLE 1 (cont.)

Equating the determinant of the dynamic matrix to zero:

The characteristic equation therefore becomes:

The roots of the characteristics equation are given by:

06102

245det

2

2

0207050 24

502

205047070 222,1


EXAMPLE 1 (cont.)

Thus and and the two natural frequencies

are given by:

Substitution of and into either one of the equations

of motion will give the two natural mode shapes.

5/221

5/21

122

12

1 2

0

0

6102

245

2

1

2

2

X

X


EXAMPLE 1 (cont.)

The mode shape for is given by:

The mode shape for the natural frequency is:

is arbitrary.

1

1

1X

X

1

1

022 21 XX

21 XX

1X

0

0

65/2102

245/25

2

1

2

2

X

X


EXAMPLE 1 (cont.)

Similarly the mode shape for is given by:

The mode shape for the natural frequency is:

2/1

1

2X

X

2

2

02 21 XX

2

12

XX

0

0

61102

2415

2

1

2

2

X

X


EXAMPLE 1 (cont.)

For an arbitrary deflection the two mode shapes

would be:

5/21

11 X

1

11

2/1

12 12


EXAMPLE 1 (cont.)


EXAMPLE 1 (cont.)

The system thus can vibrate freely with simple harmonic

motion when started in the correct way at one of two

possible frequencies.

The masses move either in phase or 180 degrees out of

phase with each other.

Since the masses reach their maximum displacements

simultaneously, the nodal points are clearly defined.


EXAMPLE 2

Consider the forced vibration response analysis for the

system above where = 5 kg, = 10 kg, = 2

N/m and = 4 N/m.

Substituting into Eq. (1) and neglecting damping, we

obtain:

2

1

2

1

2

1

62

24

100

05

F

F

x

x

x

x

1m 2m 21 kk

3k


EXAMPLE 2 (cont.)

The two natural frequencies and natural mode shapes were

previously obtained as:

Thus the modal matrix using the natural mode shapes is:

5/21

12

1

11

2/1

12

2/11

11


EXAMPLE 2 (cont.)

The coordinates are transformed by the equation:

Substitution of the above equation into the equations of

motion and pre-multiplying by yields the equations of

motion in principal coordinates: T

2

1

2

1

2

1

62

24

100

05

F

FTTT

x

x

2

1

2

1

2/11

11

x

x


EXAMPLE 2 (cont.)

The products and are calculated to

be:

mT kT

2/150

015

2/11

11

100

05

2/11

11 m

T

2/10

06

2/11

11

62

24

2/11

11 k

T


EXAMPLE 2 (cont.)

Substituting these products into the equations of motion in

principal coordinates gives:

Thus the equations of motion in are:

2

1

2

1

2

1

2/11

11

2/150

06

2/150

015

F

F

2/2/152/15

615

2122

2111

FF

FF


EXAMPLE 2 (cont.)

The original set of equations of motion are shown to be

uncoupled.

In other words the two degree of freedom system is broken

down to two single degree of freedom systems shown in

the above figure.


EXAMPLE 2 (cont.)

Once the time response for and have been

determined, they can be substituted into the coordinate

transformation equation to give the time response in terms

of the original coordinates .

The preceding equation illustrates a very important

principle in vibration, namely that any systems motion in

physical coordinates can be written as the sum of its

motion in each principle mode in some proportion and

relative phase.

tttx

tttx

212

211

2

1

1 2

x


EXAMPLE 2 (cont.)

In general for an degree of freedom system:

nn

nn

n

nnn

t

u

u

u

t

u

u

u

t

u

u

u

x

x

x

cos......coscos2

1

22

2

2

1

211

1

2

1

1

2

1

n


EXAMPLE 2 (cont.)

If the two degrees of freedom system discussed above is

given arbitrary starting conditions, the resulting motion

would be the sum of the two principal modes in some

proportion and would look as shown in the above figure.
