4.0 Outline Free Vibration Natural Modes, Eigenvalue...

Ch. 4: Vibration of Multi-DOF System

4.0 Outline

Free VibrationNatural Modes, Eigenvalue ProblemsModal Analysis

4.0 Outline

4.1 Free Vibration

Free Undamped VibrationFor the undamped free vibration, the system will vibrate at the natural frequency. However, in M-DOF, thesystem not only vibrates at a certain natural frequencybut also with a certain natural displacementconfiguration. Moreover, there are as many naturalfrequencies and associated natural configurations asthe number of DOF of the system natural modesof vibrations.

4.1 Free Vibration


natural configurations

4.1 Free Vibration


( ) ( ) ( ) ( )( )

0 0

The equations of motion for undamped M-DOF systemcan be written as

, i.c. 0 and 0

where is the displacement vector, is the inertia matrix, is the stiffness matrix. For the ph

M t K t

t MK

+ = = =x x 0 x x x x

x

( )

ysical system, and are symmetric positive definite matrix.

To find the free vibration response, we assume the complex harmonic responseanalogous to the 1-DOF case, i.e. where and are con

i t

M K

t e ω

ω=x u

u

( )

( )

2

2 2

stant vector and scalar to be determined.Substitute the assumed solution into the equation

Eigenvalue-eigenvector problem:

or with unknowns and

i tM K e

K M K M

ωω

ω ω ω

− + =

− = =

u 0

u 0 u u u

4.1 Free Vibration

Ch. 4: Vibration of Multi-DOF System, the eigenvalue, is the natural frequency of the system., the eigenvector, is the mode shape of the system. tells the frequency of oscillation while dictates the displacement configuration.

That

ω

ωu

uis the system vibrates synchronously with the frequency

and the vibration forms a certain displacement pattern of whichits shape does not change throughout the motion; only the amplitude of the

co

may ω

( )nfiguration changes w.r.t. time in a harmonic fashion .

Hence it is necessary to determine and to solve for the free response.That is we need to solve the eigenproblem. This can be done by pen

i te ω

ω ucil and paper

for the system of DOF not greater than 3.

4.2 Natural Modes, Eigenvalue Problems


( )

( )

2

2

2

2

Eigenvalue-Eigenvector problem

For the system of equations to have nontrivial solution, must be singular. That is

det 0 algebraic equation in ; the system characteristic equat

K M

K M

K M

K M

ω

ω

ω

ω ω

=

− =

−

− = →

u u

u 0

2

ion

Substitute each back into the equation and solve for the accompanied .Note there are many solutions of because is singular. It iscommon to constrain the solution by setting the magnitud

K Mω

ω−

uu

e of to be one,and to have the positive value for the first element of the vector .

uu


Ch. 4: Vibration of Multi-DOF System2From the eigenproblem, we see that there are as many and as

the number of DOF, , of the system. Each natural frequency and the corresponding mode shape forms the .Bec

nnatural mode of vibration

ωωu

u

( ) ( ) ( )

( )

1 1

ause of linearity, the sum of the solutions is also a solution.Hence the general expression of the free undamped response can be written as

cos sin

cos

k k

n ni t i t

k k k k k k k kk k

k

t a e b e A t B t

t C

ω ω ω ω

ω

−

= =

= + = +

∴ =

∑ ∑x u u

x ( )

( ) ( )1

where the amplitudes and phase shifts are determined from i.c. 0 and 0 .Note that and are determined from the system parameters.

n

k k kk

k k

k k

t

C

φ

φω

=

+∑ u

x xu



Ex. Consider the simplified model of an automobile.Derive the equations of motion. Then calculatethe natural modes of the system and write anexpression for the free response. The parametershave the following values; m=1500 kg, IC=2000 kgm2,k1=36000 kg/m, k2=40000 kg/m, a=1.3 m, b=1.7 m.Determine the response if the system is subjectto the initial configuration x0=0.5 m.




( ) ( )( ) ( )

( ) ( )( ) ( )

( )( )

1 2

1 2 1 2

2 1

2 21 2 1 2

1 2 1 22 2

1 2 1 2

0

00

x

C C C

C

C

F k x a k x b F mx

mx k k x k a k b F

M I k x b b k x a a Fc I

I k a k b x k a k b Fc

m k k k a k bx x FI k a k b k a k b Fc

θ θ

θ

θ θ θ θ

θ θ

θ θ

⎡ ⎤= − − − + + =⎣ ⎦+ + − − =

⎡ ⎤= − + × + − × + =⎣ ⎦

− − + + =

+ − −⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤+ =⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢− − +⎣ ⎦ ⎣ ⎦ ⎣⎣ ⎦ ⎣ ⎦

∑

∑

( )

( )

2

2

To calculate the natural modes, we must determine and from eigenproblem.1500 0 76000 21200

0 2000 21200 176440

is determined from the characteristic equation det 0

det det

n

n

M K

K M

K M

ω

ω ω

ω

⎥⎦

⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

− =

− =

u

( )

2

2

6 4 2

2

76000 1500 2120021200 176440 2000

3 10 138.887 4320 0

47.0296, 91.8571 6.8578, 9.5842 rad/s

ωω

ω ω

ω ω

⎡ ⎤−⎢ ⎥−⎣ ⎦

= × − + =

= =



( )2

1

41

2

42

is determined from substituting into and solve for .

For 6.8578,0.5456 2.12 1

1 10 2.12 8.2381 0.257341

For 9.5842,6.1786 2.12 1

1 10 2.12 0.7274 2.914

K Mω ω

ω

ω

−

=

⎡ ⎤ ⎡ ⎤× = =⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

=

−⎡ ⎤× = =⎢ ⎥−⎣ ⎦

u u = 0 u

u 0 u

u 0 u

( ) ( )

1 1

2 2

1 1

417

1The first natural mode is 6.8578 and

0.257341

1The second natural mode is 9.5842 and

2.914417Therefore the general free response is

1cos 6.8578

0.257341x t C t

ω

ω

φ

⎡ ⎤⎢ ⎥⎣ ⎦

⎡ ⎤= = ⎢ ⎥−⎣ ⎦

⎡ ⎤= = ⎢ ⎥

⎣ ⎦

⎡ ⎤= + +⎢ ⎥−⎣ ⎦

u

u

( )2 2

1cos 9.5842

2.914417C t φ

⎡ ⎤+ ⎢ ⎥

⎣ ⎦



( ) ( ) ( )

( ) ( ) ( )

( )

1 1 2 2

1 1 2 2

1

Therefore the general free response is1 1

cos 6.8578 cos 9.58420.257341 2.914417

1 16.8578 sin 6.8578 9.5842 sin 9.5842

0.257341 2.914417

0.50 co

0

x t C t C t

x t C t C t

x C

φ φ

φ φ

⎡ ⎤ ⎡ ⎤= + + +⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎡ ⎤= − + − +⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦⎡ ⎤

= =⎢ ⎥⎣ ⎦

( )

1 2 2

1 1 2 2

1 1 2 2

1 1 2 2

1 2 1 2

1 1s cos

0.257341 2.914417

0 1 10 6.8578 sin 9.5842 sin

0 0.257341 2.914417cos 0.4594, cos 0.0406sin 0, sin 0

0, 0, 0.4594, 0.0406

C

x C C

C CC C

C C

φ φ

φ φ

φ φφ φ

φ φ

⎡ ⎤ ⎡ ⎤+⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎡ ⎤ ⎡ ⎤= = − −⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦

= == =

∴ = = = =

∴ ( )0.4594cos 6.8578 0.0406cos9.58420.1182cos 6.8578 0.1183cos9.5842

not a harmonic response; combination of two oscillations

t tx t

t t+⎡ ⎤

= ⎢ ⎥− +⎣ ⎦



Two natural modes

4.3 Modal Analysis

Modal analysis is the procedure for solving the simulta-neous system of ODEs. The system is transformed into a set of independent ODEs of which its solution is readily determined. The actual solution are formed bycombining the basis together; transform the solutionback to the original system.

4.3 Modal Analysis


4.3 Modal Analysis

Ch. 4: Vibration of Multi-DOF SystemUndamped System

Consider the linear transformation where is a constant nonsingular square transformation matrix.Substituting this transformation into the equation of motionto change

M K

U Uη

=

=

q + q Q

q

' '

' '

the system coordinate from to : and premultiply both sides by

Note that , , and are the mass and stiffness matrix andgeneralized force vector in new

T

T T T

MU KU UU MU U KU UM K

M K

η

η η

η η

η η

+ =

+ =

+ =

qQ

QN

N

' '

' '

coordinate system .

We would like these new and matrix to be diagonal so thatthe new system is actually -independent harmonic equations andcan be solved readily: , 1, 2, , .j j j j j

M Kn

m k N j n

η

η η+ = = …

4.3 Modal Analysis

Ch. 4: Vibration of Multi-DOF System( )

' '

If is the modal matrix collection of modal vectors of the system ,

and will be diagonal. Coordinates are called modal coordinates.

and the new system of equations are called modal equations.

I

j

U

M K η

'

21

2' 2

f is chosen so , the matrix is said to beorthonormal with respect to and . The modal coordinates

are called normal coordinates. The consequence is that

0 00 0

00 0 0

T

j

T

U U MU M I UM K

U KU K

η

ωω

= =

= = Ω =

2

spectral matrix

This special always exists because is always symmetric positive definite.The orthogonality property is fundamental to modal analysis because it makesthe response t

n

U Mω

⎡ ⎤⎢ ⎥⎢ ⎥ =⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

o be representable as a linear combination of natural modes.Note that needs not be a symmetric matrix.U

4.3 Modal Analysis

Ch. 4: Vibration of Multi-DOF System( ) ( )Free response of the undamped system with i.c. 0 and 0

By the defined transformation and the orthonormality property and

and Therefore the initial condit

T T

T T

M K

UU MU I U KU

U M U K

η

η η

=

=

= = Ω

∴ Ω =

q + q 0 q q

q

= q q

( ) ( ) ( ) ( )( )

( ) ( ) ( )

ions can be transformed into modal coordinates0 0 and 0 0

They will be used to determine the constant of integration of contribution of each mode to the response

cos 0 cos

T T

r r r r r r

U M U M

t

t C t

η η

η

η ω φ η ω= + =

= q = q

( )

( ) ( ) ( )

( )1

0sin

00 cos sin , 1, 2, ,

Hence the response is

rr

rTrT

r r r rr

n

r rr

t t

Mt M t t r n

U t

ηω

ω

η ω ωω

η η=

+

∴ = + =

= =∑

u qu q

q u

…


Ex. Consider the system of three disks on a shaft withthe parameters shown. For simplicity, let theequivalent torsional spring constant between twodisks be the same, i.e. k=GJi/Li. Also all diskinertia, I, are the same. Calculate the motion ofthree disks subject to i.c.

4.3 Modal Analysis

( ) [ ] ( ) [ ]0 1 1 1 and 0 1 0 1T TΘ = Θ =


4.3 Modal Analysis

( )

1 1

2 2

3 3

2

2 2

First develop the equations of motion free vibration

0 0 0 00 0 2 00 0 0 0

Eigenvalue

0det 2

0

M K

I k kI k k k

I k k

k I kK M k k I k

k k

θ θθ θθ θ

ωω ω

Θ+ Θ =

⎡ ⎤ −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥+ − − =⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦

− −− = − − −

− −

0

2

0

30, ,r

I

k kI I

ω

ω

=

=


4.3 Modal Analysis

21

22

23

1

2

3

Eigenvector

0 02 0

0 0

1 1 0.51 , 0 , 11 1 0.5

Normalize the eigenvectors or for each mod

r r

r r

r r

r

r

r

T

k I kk k I k

k k I

M I

ω θω θ

ω θ

θθθ

⎡ ⎤− − ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − − =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − ⎣ ⎦ ⎣ ⎦⎣ ⎦⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Θ Θ =

1

2

3

e 11 1 0.5

1 1 11 , 0 , 13 2 1.51 1 0.5

Tr r

r

r

r

M

I I I

θθθ

Θ Θ =

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦


4.3 Modal Analysis

( ) ( ) ( ) ( )

1 1

2 2

3 3

Transformation matrix

1 1.5 0.5 21 1 0 2 , 3

1 1.5 0.5 2

Transform the initial conditions

43 3

0 0 0 , 0 0 00 2

3

T T

U UI

II

U M U M

I

θ ηθ ηθ η

η η

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= − =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− ⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

⎡ ⎤⎢ ⎥⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥= Θ = = Θ =⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦


4.3 Modal Analysis

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( )

( )

21 1 1 1 1

22 2 2 2 2

23 3 3 3 3

1

2

3

Modal equations

40, 0 3 , 03

0, 0 0, 0 0

20, 0 0, 03

43 rigid body motion with no acceleration3

0

2 / 3 3 2 3sin sin33 /

II

I

It I t

t

I k I kt t tI k Ik I

η ω η η η

η ω η η η

η ω η η η

η

η

η

+ = = =

+ = = =

+ = = =

= +

=

= =


4.3 Modal Analysis

1 1

2 2

3 3

Motion

2 1 31 sin3 3 32 2 31 sin3 3 32 1 31 sin3 3 3

I kt tk II kU t tk II kt tk I

θ ηθ ηθ η

⎡ ⎤+ +⎢ ⎥

⎢ ⎥⎡ ⎤ ⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥= = + −⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎢ ⎥⎢ ⎥+ +⎢ ⎥⎣ ⎦

4.3 Modal Analysis

Ch. 4: Vibration of Multi-DOF SystemDamped System

Recall we can decouple if we transform .Do the same thing as before with the damped system. Let :

and premultiply with T

T T

M C K

M K UU

MU CU KU UU MU U CU

ηη

η η η

η

+ =

+ = ==

+ + =

+

q + q q Q

q q Q qq

Q

If is normalized, and . Define .

For the new system of equations to be decoupled, must bea diagonal matrix. We would like to find the condition tha

T T

T T T

T

T

U KU UU U MU I U KU UU CU

U CU

η η

η η η

+ =

= = Ω =

+ +Ω =

QN Q

N

t makes a diagonal matrix.TU CU

4.3 Modal Analysis


( ) ( )

Let as .Therefore as ; eigenproblems.In other words, , has the same eigenvectors as ,but with different eigenvalues, and , respectively.We can write thes

T T T TU CU U MU U KU U MUCU MU KU MU

C M U K M

= Λ = Λ = Ω = Ω= Λ = Ω

Λ Ω1

1

1 1

1 1

e equations as and .Or and .Multiply each pair of equations together:

and But they are equal, which requires .Hence, fo

T T

T T

T T T T

M CU U U K U MU C U M M KU U

U KM CU U MU U CM KU U MUKM C CM K

−

−

− −

− −

= Λ = Ω

= Λ = Ω

= Ω Λ = ΩΛ = Λ Ω = ΛΩ

=1 1r the transformed system to be decoupled, .KM C CM K− −=

4.3 Modal Analysis


( )

A special case of the above condition is that ,which is called proportionally damped. Substitute into the equationsand apply the orthonormality property, we have

T T T T

C M K

U MU U M K U U KU U

α β

η α β η η

η

= +

+ + + = Q

( ) ( )( ) ( ) ( ) ( )

( ) ( )

nd

2

decoupled 2 order ODE

with the new i.c. 0 0 and 0 0

We can write each equation in the standard form2 with i.c. 0 and 0 , 1, 2, ,

which we hav

T T

r r r r r r r r r

I n

U M U M

N r n

α β η η

η η

η ζ ω η ω η η η

+ + Ω +Ω = ⇐ −

= =

+ + = =

N

q q

…

( ) ( ) ( )

( ) ( )0

e learned that one form of its solution is

1cos sin

Finally, the response in the physical coordinates are .

r r r r

tt

r r dr r r drdr

t C e t N t e d

t U t

ζ ω ζ ω τη ω φ τ ω τ τω

η

− −= + + −

=

∫q


Ex. The effect of bearing lubrication is considered in thisproblem. It can be reasonably modeled asviscous friction with the damping constant c.Determine the response of the system if thesecond disk is subject to a constant load τu(t)and the system initial conditions are

4.3 Modal Analysis

( ) [ ] ( ) [ ]0 1 1 1 and 0 1 0 1T TΘ = Θ =


4.3 Modal Analysis

1 1 1

2 2 2

3 3 3

First develop the equations of motion

0 0 0 0 0 00 0 0 0 2 ( )0 0 0 0 0 0

subject to

M C K

I c k kI c k k k u t

I c k k

θ θ θθ θ θ τθ θ θ

Θ+ Θ+ Θ =

⎡ ⎤ ⎡ ⎤ −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥+ + − − =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦

Q

( ) [ ] ( ) [ ]initial conditions 0 1 1 1 and 0 1 0 1 .

It is obvious that fits the pattern where and 0.

T T

cC C M KI

α β α β

Θ = Θ =

= + = =


4.3 Modal Analysis

( ) ( ) ( ) ( )

1 1

2 2

3 3

From the eigenproblem analysis,

30, ,

1 1.5 0.5 21 1 0 2 , 3

1 1.5 0.5 2

43 3

0 0 0 , 0 0 00 2

3

r

T T

k kI I

U UI

II

U M U M

I

ω

θ ηθ ηθ η

η η

=

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= − =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− ⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥= Θ = = Θ =⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦


4.3 Modal Analysis

( )

( ) ( ) ( )

( )

1 1 1 1

2 2 2 2

Transformed equations of motion

0 0 0 0 01 0 0 10 1 0 0 0 0 0 0

30 0 1 230 0 0 0

4, 0 3 and 033

0, 0

cI

u tc kI I I

c kI I

u tc III Ic kI I

τη η η

τη η η η

η η η η

⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥

⎡ ⎤⎢ ⎥⎡ ⎤ ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥+ + = ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ −⎣ ⎦ ⎣ ⎦⎢ ⎥⎢ ⎥

⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

+ = = =

+ + = ( )

( ) ( ) ( )

2

3 3 3 3 3

0 and 0 0

3 2 2, 0 0 and 03 3

c k Iu tI I I

η

η η η τ η η

= =

+ + = − = =


4.3 Modal Analysis

( )

( )

( )

1

1 2

2

3

Response of each modal coordinate cannot be determined in a conventional way. Use Laplace transform.

23 13 3 3

0 because of zero i.c. and no excitation

3

c tI

t

I I I I It I t ec c c

t

kI

η

τ τη

η

ω

−⎛ ⎞⎛ ⎞= + + − −⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠=

=

( ) ( )

( )

3 3 3 3

3 3

2 23 3 3 3

3 33 3 3 3 3 3

3

3 33

3 33 3

3

1, , 1 1222 3

2cos 1 cos sin3

1 2Apply the initial condition, and 3 2

2 2 1 2 2cos3 3 3 3

d

t td d d

d

d

td

d

c Ik cIIk

t C e t e t tk

IC

It e tk k k

ζ ω ζ ω

ζ ω

ζ ω ω ζ

ζ ωτη ω φ ω ωω

πφω

ζ ω ττ τη ωω

− −

−

= = − = −

⎡ ⎤⎛ ⎞= + − − +⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

= = −

= − + + + 3sin d tω⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦


4.3 Modal Analysis

( )

3 3

1 1 1

2 2 2

3 3 3

1 2

3 33

3

The motion of each disk is

1 1.5 0.5 21 1 0 23

1 1.5 0.5 2

21 13 3 3 3 3

1 cos3 3

c tI

td

d

UI

I It t ec c c Ik

e tIk

ζ ω

θ η ηθ η ηθ η η

τ τ τθ

ζ ω ττ ωω

−

−

⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = −⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦⎛ ⎞⎛ ⎞= + + − − −⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

+ +

( )

( ) ( )

3 3

3

2 2

3 33 3

3

3 1

1 sin33 3

2 21 13 3 3 3 3

2 2 1 cos sin33 3 3 3

d

c tI

td d

d

tIk

I It t ec c c Ik

e t tIk Ik

t t

ζ ω

ω

τ τ τθ

ζ ω ττ ω ωω

θ θ

−

−

⎡ ⎤⎛ ⎞+⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

⎛ ⎞⎛ ⎞= + + − − +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎡ ⎤⎛ ⎞− + +⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

=


4.3 Modal Analysis

3 3 3

Let 3, 300, 10,000.Hence 100, 0.5, 86.6025, 100Plot of the responses are as shown.

d

I c kω ζ ω τ= = =

= = = =

Damping made the motion linear, instead of parabola, function of time

4.0 Outline Free Vibration Natural Modes, Eigenvalue...

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