PDF Response of MDOF systems

Response of MDOF systems

Degree of freedom (DOF): The minimum number of

independent coordinates required to determine completely the positions of all parts of a system at any instant of time.

Two DOF systems Three DOF systems

The normal mode analysis (EOM-1)

Example: Response of 2 DOF system

m 2m

k k k

x₁ x₂

FBD ^{kx1 m k(x1-x2) 2m kx2}

EOM

− kx

₁

− k ( x

₁

− x

₂

) = m x &&

₁

2 2

2

1

) 2

( x x kx m x

k − − = &&

In matrix form, EOM is

⎥

⎦

⎢ ⎤

⎣

= ⎡

⎥ ⎦

⎢ ⎤

⎣

⎥ ⎡

⎦

⎢ ⎤

⎣

⎡

− + −

⎥ ⎦

⎢ ⎤

⎣

⎥ ⎡

⎦

⎢ ⎤

⎣

⎡

0 0 2

2 2

0

0

2 1 2

1

x x k

k

k k

x x m m

&&

&&

EOM -2 (example)

⎥ ⎦

⎢ ⎤

⎣

= ⎡

⎥ ⎦

⎢ ⎤

⎣

⎥ ⎡

⎦

⎢ ⎤

⎣

⎡

− + −

⎥ ⎦

⎢ ⎤

⎣

⎥ ⎡

⎦

⎢ ⎤

⎣

⎡

0 0 2

2 2

0

0

2 1 2

1

x x k

k

k k

x x m

m

&&

&&

x ¨ EOM

M K x F

) ( )

( )

( )

( t C x t Kx t F t x

M && + & + =

In general form

M is the inertia of mass matrix (n x n) C is the damping matrix (n x n)

K is the stiffness matrix (n x n)

F is the external force vector (n x 1)

x is the position vector (n x 1)

Synchronous motion

From observations, free vibration of undamped MDOF system is a synchronous motion.

• All coordinates pass the equilibrium points at the same time

• All coordinates reach extreme positions at the same time

• Relative shape does not change with time

2

=

1

x

x

constant

time

x₁ x₂ x₁

x₂

No phase diff. between x₁ and x₂

)

1

sin(

1

= A ω t + φ

x

)

2

sin(

2

= A ω t + φ

x

) ( 1

φ ω +

= A e

^{j t}

) ( 2

φ ω +

= A e

^{j t}

or or

Response of 2DOF system (example-1)

⎥ ⎦

⎢ ⎤

⎣

= ⎡

⎥ ⎦

⎢ ⎤

⎣

⎥ ⎡

⎦

⎢ ⎤

⎣

⎡

− + −

⎥ ⎦

⎢ ⎤

⎣

⎥ ⎡

⎦

⎢ ⎤

⎣

⎡

0 0 2

2 2

0

0

2 1 2

1

x x k

k

k k

x x m m

&&

&&

EOM

Synchronous motion

Sub. into EOM

⎥ ⎦

⎢ ⎤

⎣

= ⎡

⎥ ⎦

⎢ ⎤

⎣

⎥ ⎡

⎦

⎢ ⎤

⎣

⎡

− + −

⎥ ⎦

⎢ ⎤

⎣

⎥ ⎡

⎦

⎢ ⎤

⎣

⎡

ω

− ω

−

0 0 2

2 2

0

0

2 1 2

1 2

2

x x k

k

k k

x x m m

⎥ ⎦

⎢ ⎤

⎣

= ⎡

⎥ ⎦

⎢ ⎤

⎣

⎥ ⎡

⎦

⎢ ⎤

⎣

⎡

−

−

−

−

0 0 2

2 2

2 1 2

2

A A m

k k

k m

k

ω ω

0 Kx

Mx + =

ω

−

²

( t ) ( t )

0 x

M

K − ω ) ( ) =

(

²

t

2 0 2

2

2

2

=

ω

−

−

− ω

−

m k

k

k m

k det( K − ω

²

M ) = 0 Characteristic equation (CHE)

)

1

sin(

1

= A ω t + φ

x

)

2

sin(

2

= A ω t + φ

x

) ( 1

φ ω +

= A e

^{j t}

) ( 2

φ ω +

= A e

^{j t}

or or

Response of 2DOF system (example-2)

2 0 2

2

2 2

ω =

−

−

− ω

−

m k

k

k m

k 0

2 3 3

2 2

4

⎟ =

⎠

⎜ ⎞

⎝ + ⎛

⎟ ω

⎠

⎜ ⎞

⎝

− ⎛

ω m

k m

k

m 634 k .

1

= 0

ω m

366 k .

2

= 2 ω

CHE

Solve the CHE Natural frequencies

of the system

;

⎥ ⎦

⎢ ⎤

⎣

= ⎡

⎥ ⎦

⎢ ⎤

⎣

⎥ ⎡

⎦

⎢ ⎤

⎣

⎡

ω

−

−

− ω

−

0 0 2

2 2

2 1 2

2

A A m

k k

k m

k

k

m k

m k

k A

A

²

2 2

1

2 2

2

ω

= − ω

= − ω

1

= ω

731 . 0 )

634 . 0 ( 2

) 1 (

2

1

=

−

⎟⎟ =

⎠

⎜⎜ ⎞

⎝

⎛

m m k k

k A

A

ω

2

= ω

73 . 2 )

366 . 2 ( 2

) 2 (

2

1

= −

−

⎟⎟ =

⎠

⎜⎜ ⎞

⎝

⎛

m m k k

k A

A

From

Response of 2DOF system (example-3)

ω

1

= ω

731 . 0

) 1 (

2

1

⎟⎟ =

⎠

⎜⎜ ⎞

⎝

⎛ A

A 2 . 73

) 2 (

2

1

⎟⎟ = −

⎠

⎜⎜ ⎞

⎝

⎛ A

A

⎭ ⎬

⎫

⎩ ⎨

= ⎧

φ 1

731 .

) 0

1

( x

⎭ ⎬

⎫

⎩ ⎨

= ⎧−

φ 1

73 . ) 2

2

( x

Amp. ratio Amp. ratio

The first mode shape The second mode shape

0.731 1

-2.73 1

ω

2

= ω

same direction Opposite

direction

Response of 2DOF system (example-4)

In general, the free vibration contains both modes

simultaneously (vibrate at both frequencies simultaneously)

) 1 sin(

73 . ) 2

1 sin(

732 . 0

2 2

2 1

1 1

2

1

ω + ψ

⎭ ⎬

⎫

⎩ ⎨ + ⎧−

ψ +

⎭ ω

⎬ ⎫

⎩ ⎨

= ⎧

⎭ ⎬

⎫

⎩ ⎨

⎧ c t c t

x x

2 1

2

1

, c , ψ , ψ

c are constants (depended on initial conditions)

Initial conditions (1)

) 1 sin(

73 . ) 2

1 sin(

732 . 0

2 2

2 1

1 1

2

1

ω + ψ

⎭ ⎬

⎫

⎩ ⎨ + ⎧−

ψ +

⎭ ω

⎬ ⎫

⎩ ⎨

= ⎧

⎭ ⎬

⎫

⎩ ⎨

⎧ c t c t

x x

⎭ ⎬

⎫

⎩ ⎨

= ⎧

⎭ ⎬

⎫

⎩ ⎨

⎧

4 2 )

0 (

) 0 (

2 1

x x

⎭ ⎬

⎫

⎩ ⎨

= ⎧

⎭ ⎬

⎫

⎩ ⎨

⎧

0 0 )

0 (

) 0 (

2 1

x x

&

&

Initial conditions and

) 1 cos(

73 . ) 2

1 cos(

732 . 0

2 2

2 2 1

1 1

1 2

1

ω + ψ

⎭ ⎬

⎫

⎩ ⎨ ω ⎧−

+ ψ +

⎭ ω

⎬ ⎫

⎩ ⎨ ω ⎧

⎭ =

⎬ ⎫

⎩ ⎨

⎧ c t c t

x x

&

&

Velocity response

⎭ ⎬

⎫

⎩ ⎨

= ⎧

⎭ ⎬

⎫

⎩ ⎨

⎧

4 2 )

0 (

) 0 (

2 1

x x

2 2

1

1

sin

1 73 . sin 2

1 732 . 0 4

2 ψ

⎭ ⎬

⎫

⎩ ⎨ + ⎧−

⎭ ψ

⎬ ⎫

⎩ ⎨

= ⎧

⎭ ⎬

⎫

⎩ ⎨

⎧ c c

⎭ ⎬

⎫

⎩ ⎨

= ⎧

⎭ ⎬

⎫

⎩ ⎨

⎧

0 0 )

0 (

) 0 (

2 1

x x

&

&

2 2

2 1

1

1

cos

1 73 . cos 2

1 732 . 0 0

0 ψ

⎭ ⎬

⎫

⎩ ⎨ ω ⎧−

+

⎭ ψ

⎬ ⎫

⎩ ⎨ ω ⎧

⎭ =

⎬ ⎫

⎩ ⎨

⎧ c c

Initial conditions (2)

2 2

1

1

sin

1 73 . sin 2

1 732 . 0 4

2 ψ

⎭ ⎬

⎫

⎩ ⎨ + ⎧−

⎭ ψ

⎬ ⎫

⎩ ⎨

= ⎧

⎭ ⎬

⎫

⎩ ⎨

⎧ c c

2 2

2 1

1

1

cos

1 73 . cos 2

1 732 . 0 0

0 ψ

⎭ ⎬

⎫

⎩ ⎨ ω ⎧−

+

⎭ ψ

⎬ ⎫

⎩ ⎨ ω ⎧

⎭ =

⎬ ⎫

⎩ ⎨

⎧ c c

4 Eqs.,

4 unknowns

Solve for four unknowns

, 732 .

1

= 3

c c

₂

= 0 . 268 , ψ

₁

= ψ

₂

= π / 2

2 ) 1 sin(

73 . 268 2

. 0 2 )

1 sin(

732 . 732 0 .

3

_{1 2}

2

1

ω + π

⎭ ⎬

⎫

⎩ ⎨ + ⎧−

+ π

⎭ ω

⎬ ⎫

⎩ ⎨

= ⎧

⎭ ⎬

⎫

⎩ ⎨

⎧ t t

x x

The response is

t x t

x

2 1

2

1

cos

268 . 0

732 . cos 0

732 . 3

732 .

2 ω

⎭ ⎬

⎫

⎩ ⎨ + ⎧−

⎭ ω

⎬ ⎫

⎩ ⎨

= ⎧

⎭ ⎬

⎫

⎩ ⎨

⎧

Initial conditions (3)

) 1 sin(

73 . ) 2

1 sin(

732 . 0

2 2

2 1

1 1

2

1

ω + ψ

⎭ ⎬

⎫

⎩ ⎨ + ⎧−

ψ +

⎭ ω

⎬ ⎫

⎩ ⎨

= ⎧

⎭ ⎬

⎫

⎩ ⎨

⎧ c t c t

x x

⎭ ⎬

⎫

⎩ ⎨

= ⎧

⎭ ⎬

⎫

⎩ ⎨

⎧

2 464 . 1 )

0 (

) 0 (

2 1

x x

⎭ ⎬

⎫

⎩ ⎨

= ⎧

⎭ ⎬

⎫

⎩ ⎨

⎧

0 0 )

0 (

) 0 (

2 1

x x

&

&

(a) Initial conditions and

⎭ ⎬

⎫

⎩ ⎨

= ⎧−

⎭ ⎬

⎫

⎩ ⎨

⎧

1 73 . 2 )

0 (

) 0 (

2 1

x x

⎭ ⎬

⎫

⎩ ⎨

= ⎧

⎭ ⎬

⎫

⎩ ⎨

⎧

0 0 )

0 (

) 0 (

2 1

x x

&

&

(b) Initial conditions and

Try to do

Summary (Free-undamped) (1)

0 Kx

x

M && ( t ) + ( t ) =

The motion is synchronous:

constant ω and φ

0 Kx

Mx + =

ω

−

²

( t ) ( t ) 0 x

M

K − ω ) ( ) =

(

²

t

0 )

det( K − ω

²

M =

Characteristics equation

2

ω

n Eigen value

nN n

n

ω

Eigen value problem

ω

ω

₁

,

₂

, K ,

^N natural freq.

0 x

M

K − ω

_ni

)

_i

=

(

²

x

i Eigen vector

) sin( ω + φ

= A t

x

^or

= A e

^j(ωt+φ)

N mode shapes

x

N

x

x

₁

,

₂

, K , 1

EOM

2

3

4 5

Direct Method

Summary (Free-undamped) (2)

Free-undamped response

) sin(

) sin(

) sin(

)

( t = x

₁

A

₁

ω

₁

t + φ

₁

+ x

₂

A

₂

ω

₂

t + φ

₂

+ x

_N

A

_N

ω

_N

t + φ

_N

x K

∑

=

φ + ω

=

^N

i

i i

i

i

A t

t

1

) sin(

)

( x

x 6

where A and φ are from initial condition x(0) and v(0)

Direct Method

Example

k1

k2

x θ

l2

l1

Determine the normal modes of vibration of an automobile simulated by simplified 2-dof system with the following numerical values

lb

= 3220 W

ft 5 .

1

= 4

l k

₁

= 2400 lb/ft

ft 5 .

2

= 5

l k

₂

= 2600 lb/ft

ft

= 4

2

r

g r

J

_C

= W

Forced harmonic vibration (1)

Example

EOM

F t

x x k

k

k k

x x m

m ⎥ ω

⎦

⎢ ⎤

⎣

= ⎡

⎥ ⎦

⎢ ⎤

⎣

⎥ ⎡

⎦

⎢ ⎤

⎣ + ⎡

⎥ ⎦

⎢ ⎤

⎣

⎥ ⎡

⎦

⎢ ⎤

⎣

⎡ sin

0 0

0

₁

2 1 22

21

12 11

2 1 2

1

&&

&&

System is undamped, the solution can be assumed as

X t X x

x ⎥ ω

⎦

⎢ ⎤

⎣

= ⎡

⎥ ⎦

⎢ ⎤

⎣

⎡ sin

2 1 2

1

Sub. into EOM

⎥ ⎦

⎢ ⎤

⎣

= ⎡

⎥ ⎦

⎢ ⎤

⎣

⎥ ⎡

⎦

⎢ ⎤

⎣

⎡

ω

− ω

−

0

1 2

1 2

2 22

21

12 2

1

11

F

X X m

k k

k m

k

[ ] ⎥

⎦

⎢ ⎤

⎣

= ⎡

⎥ ⎦

⎢ ⎤

⎣ ω ⎡

) 0

(

¹

2

1

F

X Z X

Simpler notation,

[ ] ⎥

⎦

⎢ ⎤

⎣ ω ⎡

⎥ =

⎦

⎢ ⎤

⎣

⎡

₋

) 0

(

^{1 1}

2

1

F

X Z

X

Forced harmonic vibration (2)

[ ] [ ]

⎥ ⎦

⎢ ⎤

⎣

⎡ ω

= ω

⎥ ⎦

⎢ ⎤

⎣ ω ⎡

⎥ =

⎦

⎢ ⎤

⎣

⎡

−

0 )

(

) ( adj ) 0

(

^{1 1 1}

2

1

F

Z F Z

X Z X

) )(

( )

( ω = m

₁

m

₂

ω

₁²

− ω

²

ω

²₂

− ω

²

Where

Z

ω₁ and ω₂ are natural frequencies

⎥ ⎦

⎢ ⎤

⎣

⎥ ⎡

⎦

⎢ ⎤

⎣

⎡

ω

−

−

− ω

−

= ω

⎥ ⎦

⎢ ⎤

⎣

⎡

0 )

(

1

₁

2 1 11

21

12 2

2 22

2

1

F

m k

k

k m

k X Z

X

The amplitudes are

) )(

(

) (

2 2

2 2

2 1 2 1

1 2 2 22

1

ω − ω ω − ω

ω

= −

m m

F m

X k

) )(

(

₁^{2 2 2}₂ ²

2 1

1 21

2

ω − ω ω − ω

= −

m m

F

X k

Forced harmonic vibration (3)

) )(

(

) 2

(

2 2

2 2 2

1 2

1 2

1 ω −ω ω −ω

ω

= − m

F m

X k

) )(

( ₁^{2 2 2}₂ ²

2

1

2 = ω −ω ω −ω

m X kF

m m

k k k

x₁ x₂

F₁sinωt

F t x

x k k

k k

x x m

m ⎥ ω

⎦

⎢ ⎤

⎣

= ⎡

⎥⎦

⎢ ⎤

⎣

⎥⎡

⎦

⎢ ⎤

⎣

⎡

− + −

⎥⎦

⎢ ⎤

⎣

⎥⎡

⎦

⎢ ⎤

⎣

⎡ sin

0 2

2 0

0 ₁

2 1 2

1

&&

&&

m k m

k 3

, ₂

1 = ω =

ω

EOM

Force response of a 2 DOF system

0 1 2 3

0 1 2 3 4 5

-1 -2 -3 -4 -5

F Xk

ω1

ω

1 2

F k X

1 1

F k X

ω1

=

ω ω= ω₂

Same direction

Opposite direction

Solving methods

Modal analysis

• is a method for solving for both transient and steady state responses of free and forced MDOF systems through

analytical approaches.

• Uses the orthogonality property of the modes to

“decouple” the EOM breaking EOM into independent SDOF equations, which can be solved for response separately.

Introduction

Coordinate coupling

k1 k₂

l2

l1

mg

θ x

Ref.

)

( ₁

1 x−l θ k

)

( ₂

2 x−l θ k

k1 l¹ mg k₂

θ

x1 Ref.

1 1x k

) ( ₁

2 x +lθ k

⎥⎦

⎢ ⎤

⎣

= ⎡

⎥⎦

⎢ ⎤

⎣

⎡

⎥ θ

⎦

⎢ ⎤

⎣

⎡

+

−

− + +

⎥⎦

⎢ ⎤

⎣

⎡

⎥ θ

⎦

⎢ ⎤

⎣

⎡

0 0 0

0

2 2 2 2 1 1 1 1 2 2

1 1 2 2 2

1 x

l k l k l k l k

l k l k k

k x

J m

&&

&&

⎥⎦

⎢ ⎤

⎣

= ⎡

⎥⎦

⎢ ⎤

⎣

⎡

⎥ θ

⎦

⎢ ⎤

⎣ + ⎡ +

⎥⎦

⎢ ⎤

⎣

⎡

⎥ θ

⎦

⎢ ⎤

⎣

⎡

0

1 0

2 2 2

2 2

1 1

1 1

1 x

l k l

k

l k k

k x

J ml

ml m

&&

&&

Concept of modal analysis

⎥⎦

⎢ ⎤

⎣

= ⎡

⎥⎦

⎢ ⎤

⎣

⎡

⎥ θ

⎦

⎢ ⎤

⎣

⎡

+

−

− + +

⎥⎦

⎢ ⎤

⎣

⎡

⎥ θ

⎦

⎢ ⎤

⎣

⎡

0 ) ( 0

0

2 2 2 2 1 1 1 1 2 2

1 1 2 2 2

1 x F t

l k l k l k l k

l k l k k

x k J m

&&

&&

) ( )

( )

( t Kx t F t x

M && + =

⎥⎦

⎢ ⎤

⎣

= ⎡

⎥⎦

⎢ ⎤

⎣

⎥⎡

⎦

⎢ ⎤

⎣

⎡

ω + ω

⎥⎦

⎢ ⎤

⎣

⎥⎡

⎦

⎢ ⎤

⎣

⎡

) (

) ( 0

0 1

0 0 1

2 1 2

1 2

2 2

1 2

1

t N

t N r

r r

r

n n

&&

&&

) ( )

( )

( t Λr t N t r && + =

EOM in modal coordinate (Independent SDOF equations) EOM in physical coordinate (Coordinates are coupled)

Solve for r (t )

Transform r(t) back to x(t)

Orthogonality

x = eigen vector (vector of mode shape)

ii i

T

i

Mx = M

x

j

i

i

T

j

Mx = 0 , ≠

x x

^T_j

Kx

_i

= 0 , i ≠ j

ii i

T

i

Kx = K

x

If M and K are symmetric and then x

_i

and x

_j

are said to be “orthogonal” to each other.

^{ni nj}

ω

≠

ω

Normalization

u = normalized eigen vector (respect to mass matrix)

= 1

i T

i

Mu

u

j

i

i

T

j

Mu = 0 , ≠ u

constant is

, C

C

_i

i

x

u

0 x

M

K − ω ) ( ) =

(

²

t

From eigen value problem

=

0 u

M

K − ω ) ( ) =

(

²

t

or Ku

i

= ω

i²

Mu

i

2 2

i i

T i i i

T

i

Ku = ω u Mu = ω

u

Modal matrix

Modal matrix is the matrix that its columns are the mode shape of the system

[ u u u

_n

]

U =

1 2

K Then

⎥ ⎥

⎥ ⎥

⎦

⎤

⎢ ⎢

⎢ ⎢

⎣

⎡

=

=

1 0

0

0 1

0

0 0

1

K M O M

M

K K I

MU U

^T

⎥ ⎥

⎥ ⎥

⎥

⎦

⎤

⎢ ⎢

⎢ ⎢

⎢

⎣

⎡

ω ω

ω

=

2 2

2 2

1

0 0

0 0

0 0

nN n

n T

K

M O

M M

K K KU

U

Λ (Spectral matrix)

Modal analysis (undamped systems)-1

1. Draw FBD, apply Newton’s law to obtain EOM 2. Solve for natural frequencies through CHE 3. Determine mode shapes through EVP

4. Construct modal matrix (normalized)

Procedures

) ( )

( )

( t Kx t F t x

M && + =

0 x

M

K − ω ) ( ) =

(

²

t

0 ) det( K − ω

²

M =

[ u u u

_n

]

U =

1 2

K I

MU U

^T

=

Λ KU

U

^T

=

5. Perform a coordinate transformation

x ( t ) = Ur ( t ) )

( )

( )

( t Kx t F t x

M && + = MU r && ( t ) + KUr ( t ) = F ( t )

) ( )

( )

( t

^T

t

^T

t

T

MU r U KUr U F

U && + =

) ( )

( )

( t Λr t U

^T

F t

r && + =

Modal analysis (undamped systems)-2

) ( )

( )

( t Λr t U

^T

F t

r && + =

⎥⎥

⎥⎥

⎦

⎤

⎢⎢

⎢⎢

⎣

⎡

=

⎥⎥

⎥⎥

⎦

⎤

⎢⎢

⎢⎢

⎣

⎡

⎥⎥

⎥⎥

⎦

⎤

⎢⎢

⎢⎢

⎣

⎡

=

⎥⎥

⎥⎥

⎦

⎤

⎢⎢

⎢⎢

⎣

⎡

⎥⎥

⎥⎥

⎥

⎦

⎤

⎢⎢

⎢⎢

⎢

⎣

⎡

ω ω

ω +

⎥⎥

⎥⎥

⎦

⎤

⎢⎢

⎢⎢

⎣

⎡

) (

) (

) (

) (

) (

) (

) (

) (

) (

) (

) (

0 0

0 0

0 0

) (

) (

) (

4 3 2 1

4 3 2 1

2 1

2 22

21

1 12

11 2

1

2 2

2 2

1 2

1

t N

t N

t N

t N

t F

t F

t F

t F

u u

u

u u

u

u u

u

t r

t r

t r

t r

t r

t

r ^T

NN N

N

N N

nN N n

n

N K

M O M

M

K K

M K

M O M

M

K K

&&

M

&&

&&

Independent SDOF equations, can be solve for r(t) 6. Transform the initial conditions to modal coordinates

) ( )

( t Ur t x =

I MU U

^T

=

) 0 ( )

0

( Ur

x =

) 0 ( )

0

( U MUr

Mx

U

^T

=

^T

) 0 ( )

0

( U Mx

r =

^T

From

and

r & ( 0 ) = U

^T

M x & ( 0 )

Modal analysis (undamped systems)-3

7. Find the response in modal coordinates

8. Transform the response in modal coordinate

back to that in original coordinate

x ( t ) = Ur ( t )

= K ) r (t )

r (t )

x (t

Example (Modal analysis) -1

⎥ ⎦

⎢ ⎤

⎣

= ⎡

⎥ ⎦

⎢ ⎤

⎣

⎥ ⎡

⎦

⎢ ⎤

⎣

⎡

− + −

⎥ ⎦

⎢ ⎤

⎣

⎥ ⎡

⎦

⎢ ⎤

⎣

⎡

0 0 3

3

3 27

1 0

0 9

2 1 2

1

x x x

x

&&

&&

EOM

⎥ ⎦

⎢ ⎤

⎣

= ⎡ 0 1

x

0

⎥

⎦

⎢ ⎤

⎣

= ⎡ 0 0 v

0

Initial conditions

Example (Modal analysis) -2

2dof string-bead system

⎥ ⎦

⎢ ⎤

⎣

= ⎡

⎥ ⎦

⎢ ⎤

⎣

⎥ ⎡

⎦

⎢ ⎤

⎣

⎡

− + −

⎥ ⎦

⎢ ⎤

⎣

⎥ ⎡

⎦

⎢ ⎤

⎣

⎡

0 0 2

1

1 2

0

0

2 1 2

1

x x a

T x

x m m

&&

&&

EOM

⎥ ⎦

⎢ ⎤

⎣

= ⎡ 0 0

x

0

⎥

⎦

⎢ ⎤

⎣

= ⎡ 1 0 v

0

Initial conditions

Example (Modal analysis) -2_2

Rigid body mode

• Rigid body mode is the mode that the system moves as a rigid body.

• The system moves as a whole without any relative motion among masses.

• There is no oscillation. ω

_n

= 0

Rigid-body modes

Compute the solution of the system.

Let m₁ = 1 kg, m₂ = 4 kg and k = 400 N/m.

Initial condition

⎥

⎦

⎢ ⎤

⎣

= ⎡

0 01 . 0

x

0

⎥

⎦

⎢ ⎤

⎣

= ⎡

0

0

v

0

More than two degrees of freedom

Calculate the solution of the n-degree-of-freedom system in the figure for n = 3 by modal analysis. Use the values m₁ = m₂ = m₃ = 4 kg and k₁ = k₂ = k₄ = 4 N/m, and the initial condition x₁(0) = 1 m with all other initial

displacements and velocities zero.

Modal analysis on damped systems (1)

) ( )

( )

( )

(t Cx t Kx t F t

x

M && + & + =

C KM K

CM⁻¹ = ⁻¹

The original modal analysis can be applied to MDOF damped system if and only if

K M

C = α +β

α _β

However, there are subsets of the above systems where C can be written as a linear combination of M and K.

and are constants. Such system is called “proportionally damped.”

Necessary and sufficient condition

Sufficient but not necessary condition Such system is called “classically damped”.

EOM

Modal analysis on damped systems (2)

For proportionally damped Mx&&(t) +Cx&(t) + Kx(t) = F(t) ) ( )

( )

( ) (

)

(t M K x t Kx t F t

x

M && + α +β & + =

) ( )

(t Ur t

x = MUr&&(t) +(αM +βK)Ur&(t) +KUr(t) = F(t)

UT ^UTMUr&&^{(t) +UT(αM +βK)Ur}&^{(t)+UTKUr(t) = UTF(t)}

) ( )

( )

( )

( ) (

)

(t I Λ r t Λr t U^TF t N t

r&& + α +β & + = =

) ( )

( )

( 2

)

( _{1 1} ²_{1 1 1}

1 t r t r t N t

r&& + ζω_n & +ω_n =

M

2 1

2ζω_n1 = α +βω_n Let

Premultiply by

Thus, the system when it is written in modal coordinates r(t) can be decoupled into k sets of SDOF equations

where

Modal analysis on damped systems (Ex.)

A belt-driven lathe

•bearings are modeled as providing viscous damping

•shafts provide stiffness

•belt drive provides and applied torque.

/rad kg.m

10

²

3 2

1

= J = J =

J

N.m/rad 10

³

2

1

= k =

k

N.m.s/rad

= 2 c

• Zero initial conditions

• Applied moment M(t) is a unit impulse function

Modal analysis on damped systems (Ex.)