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 2mk k k

x1 x2

FBD m 2mkx1 k(x1-x2) kx2

EOM 1211 )( xmxxkkx &&=−−−

2221 2)( xmkxxxk &&=−−

In matrix form, EOM is ⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡00

22

200

2

1

2

1

xx

kkkk

xx

mm

&&

&&

EOM -2 (example)

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡00

22

200

2

1

2

1

xx

kkkk

xx

mm

&&

&&

x

EOM

M K Fx

)()()()( tttt FKxxCxM =++ &&&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

=21 xx constant

time

x1 x2 x1x2

No phase diff. between x1 and x2

)sin(11 φω += tAx)sin(22 φω += tAx

)(1

φω += tjeA)(

2φω += tjeA

oror

Response of 2DOF system (example-1)

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡00

22

200

2

1

2

1

xx

kkkk

xx

mm

&&

&&EOM

Synchronous motion

Sub. into EOM

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

ω−ω−

00

22

200

2

1

2

12

2

xx

kkkk

xx

mm

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

−−−−

00

222

2

12

2

AA

mkkkmkω

ω

0KxMx =+ω− )()(2 tt

0xMK =ω− )()( 2 t

022

22

2

=ω−−

−ω−mkk

kmk 0)det( 2 =ω− MK Characteristic equation (CHE)

)sin(11 φω += tAx)sin(22 φω += tAx

)(1

φω += tjeA)(

2φω += tjeA

oror

Response of 2DOF system (example-2)

022

22

2

=ω−−

−ω−mkk

kmk0

233

224 =⎟

⎠⎞

⎜⎝⎛+ω⎟

⎠⎞

⎜⎝⎛−ω

mk

mk

mk634.01 =ω

mk366.22 =ω

CHE

Solve the CHE Natural frequencies of the system;

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

ω−−−ω−

00

222

2

12

2

AA

mkkkmk

kmk

mkk

AA 2

22

1 222

ω−=

ω−=

1ω=ω

731.0)634.0(2

)1(

2

1 =−

=⎟⎟⎠

⎞⎜⎜⎝

⎛

mmkk

kAA

2ω=ω

73.2)366.2(2

)2(

2

1 −=−

=⎟⎟⎠

⎞⎜⎜⎝

⎛

mmkk

kAA

From

Response of 2DOF system (example-3)

1ω=ω

731.0)1(

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛AA 73.2

)2(

2

1 −=⎟⎟⎠

⎞⎜⎜⎝

⎛AA

⎭⎬⎫

⎩⎨⎧

=φ1731.0

)(1 x⎭⎬⎫

⎩⎨⎧−

=φ1

73.2)(2 x

Amp. ratio Amp. ratio

The first mode shape The second mode shape

0.731 1

-2.73

1

2ω=ω

same directionOpposite direction

Response of 2DOF system (example-4)

In general, the free vibration contains both modes simultaneously (vibrate at both frequencies simultaneously)

)sin(1

73.2)sin(

1732.0

2221112

1 ψ+ω⎭⎬⎫

⎩⎨⎧−

+ψ+ω⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

tctcxx

2121 ,,, ψψcc are constants (depended on initial conditions)

Initial conditions (1)

)sin(1

73.2)sin(

1732.0

2221112

1 ψ+ω⎭⎬⎫

⎩⎨⎧−

+ψ+ω⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

tctcxx

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

42

)0()0(

2

1

xx

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

00

)0()0(

2

1

xx&

&Initial conditions and

)cos(1

73.2)cos(

1732.0

222211112

1 ψ+ω⎭⎬⎫

⎩⎨⎧−

ω+ψ+ω⎭⎬⎫

⎩⎨⎧

ω=⎭⎬⎫

⎩⎨⎧

tctcxx&

&

Velocity response

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

42

)0()0(

2

1

xx

2211 sin1

73.2sin

1732.0

42

ψ⎭⎬⎫

⎩⎨⎧−

+ψ⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

cc

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

00

)0()0(

2

1

xx&

&222111 cos

173.2

cos1732.0

00

ψ⎭⎬⎫

⎩⎨⎧−

ω+ψ⎭⎬⎫

⎩⎨⎧

ω=⎭⎬⎫

⎩⎨⎧

cc

Initial conditions (2)

2211 sin1

73.2sin

1732.0

42

ψ⎭⎬⎫

⎩⎨⎧−

+ψ⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

cc

222111 cos1

73.2cos

1732.0

00

ψ⎭⎬⎫

⎩⎨⎧−

ω+ψ⎭⎬⎫

⎩⎨⎧

ω=⎭⎬⎫

⎩⎨⎧

cc

4 Eqs., 4 unknowns

Solve for four unknowns,732.31 =c ,268.02 =c 2/21 π=ψ=ψ

)2

sin(1

73.2268.0)

2sin(

1732.0

732.3 212

1 π+ω

⎭⎬⎫

⎩⎨⎧−

+π

+ω⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

ttxx

The response is

ttxx

212

1 cos268.0732.0

cos732.3732.2

ω⎭⎬⎫

⎩⎨⎧−

+ω⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

Initial conditions (3)

)sin(1

73.2)sin(

1732.0

2221112

1 ψ+ω⎭⎬⎫

⎩⎨⎧−

+ψ+ω⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

tctcxx

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

2464.1

)0()0(

2

1

xx

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

00

)0()0(

2

1

xx&

&(a) Initial conditions and

⎭⎬⎫

⎩⎨⎧−

=⎭⎬⎫

⎩⎨⎧

173.2

)0()0(

2

1

xx

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

00

)0()0(

2

1

xx&

&(b) Initial conditions and

Try to do

Summary (Free-undamped) (1)

0KxxM =+ )()( tt&&

The motion is synchronous: constant ω and φ

0KxMx =+ω− )()(2 tt

0xMK =ω− )()( 2 t

0)det( 2 =ω− MKCharacteristics equation

2nω Eigen value

nNnn ω

Eigen value problem

ωω ,,, 21 K N natural freq.

0xMK =ω− ini )( 2

ix Eigen vector

)sin( φ+ω= tAx )( φ+ω= tjeAor

N mode shapesNxxx ,,, 21 K

EOM1

2

3

4 5

Direct Method

Summary (Free-undamped) (2)

Free-undamped response

)sin()sin()sin()( 22221111 NNNN tAtAtAt φ+ω+φ+ω+φ+ω= xxxx K

∑=

φ+ω=N

iiiii tAt

1)sin()( xx

6

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

Direct Method

Example

1k2k

θx

2l1l

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

lb3220=W

ft5.41 =l lb/ft24001 =k

ft5.52 =l lb/ft26002 =k

ft4=r2rgWJC =

Forced harmonic vibration (1)

Example

EOMt

Fxx

kkkk

xx

mm

ω⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡sin

000 1

2

1

2221

1211

2

1

2

1

&&

&&

System is undamped, the solution can be assumed as

tXX

xx

ω⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡sin

2

1

2

1

Sub. into EOM⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

ω−ω−

01

2

12

22221

122

111 FXX

mkkkmk

[ ] ⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡ω

0)( 1

2

1 FXX

ZSimpler notation, [ ] ⎥⎦

⎤⎢⎣

⎡ω=⎥

⎦

⎤⎢⎣

⎡ −

0)( 11

2

1 FZ

XX

Forced harmonic vibration (2)

[ ] [ ]⎥⎦

⎤⎢⎣

⎡ωω

=⎥⎦

⎤⎢⎣

⎡ω=⎥

⎦

⎤⎢⎣

⎡ −

0)()(adj

0)( 111

2

1 FZ

ZFZ

XX

))(()( 222

22121 ω−ωω−ω=ω mmZWhere

ω1 and ω2 are natural frequencies

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

ω−−−ω−

ω=⎥

⎦

⎤⎢⎣

⎡0)(

1 12

11121

122

222

2

1 Fmkkkmk

ZXX

The amplitudes are))((

)(22

222

121

12

2221 ω−ωω−ω

ω−=

mmFmkX

))(( 222

22121

1212 ω−ωω−ω

−=

mmFkX

Forced harmonic vibration (3)

))(()2(

222

221

21

2

1 ω−ωω−ωω−

=m

FmkX

))(( 222

221

21

2 ω−ωω−ω=

mkFX

m mk k k

x1 x2

F1sinωt

tF

xx

kkkk

xx

mm

ω⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡sin

022

00 1

2

1

2

1

&&

&&

mk

mk 3, 21 =ω=ω

EOM

Force response of a 2 DOF system

0 1 2 3

012

3

45

-1

-2-3

-4

-5

FXk

1ωω

1

2

FkX

1

1

FkX

1ω=ω 2ω=ω

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

1k 2k

2l1l

mg

θ

xRef.

)( 11 θ− lxk)( 22 θ− lxk

1k 2k1l mg

θ1x

Ref.

11xk)( 12 θ+ lxk

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡θ⎥

⎦

⎤⎢⎣

⎡+−−+

+⎥⎦

⎤⎢⎣

⎡θ⎥

⎦

⎤⎢⎣

⎡00

00

222

2111122

112221 xlklklklklklkkkx

Jm

&&

&&⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡θ⎥

⎦

⎤⎢⎣

⎡ ++⎥

⎦

⎤⎢⎣

⎡θ⎥

⎦

⎤⎢⎣

⎡001

222

2211

11

1 xlklklkkkx

Jmlmlm

&&

&&

Concept of modal analysis

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡θ⎥

⎦

⎤⎢⎣

⎡+−−+

+⎥⎦

⎤⎢⎣

⎡θ⎥

⎦

⎤⎢⎣

⎡0

)(0

0222

2111122

112221 tFxlklklklklklkkkx

Jm

&&

&&)()()( ttt FKxxM =+&&

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

ωω

+⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡)()(

00

1001

2

1

2

12

2

21

2

1

tNtN

rr

rr

n

n

&&

&& )()()( ttt NΛrr =+&&

EOM in modal coordinate (Independent SDOF equations)

EOM in physical coordinate (Coordinates are coupled)

Solve for )(tr

Transform r(t) back to x(t)

Orthogonality

x = eigen vector (vector of mode shape)

iiiTi M=Mxx

jiiTj ≠= ,0Mxx jii

Tj ≠= ,0Kxx

iiiTi K=Kxx

If M and K are symmetric and then xi and xj are said to be “orthogonal” to each other.

njni ω≠ω

Normalization

u = normalized eigen vector (respect to mass matrix)

1=iTi Muu

jiiTj ≠= ,0Muu

constant is , CC ii xu

0xMK =ω− )()( 2 tFrom eigen value problem

=

0uMK =ω− )()( 2 t

or iii MuKu 2ω=

22ii

Tiii

Ti ω=ω= MuuKuu

Modal matrix

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

[ ]nuuuU K21=

Then

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

==

100

010001

K

MOMM

K

K

IMUUT

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

ω

ωω

=

2

22

21

00

0000

nN

n

n

T

K

MOMM

K

K

KUU

Λ (Spectral matrix)

Modal analysis (undamped systems)-1

1. Draw FBD, apply Newton’s law to obtain EOM2. Solve for natural frequencies through CHE3. Determine mode shapes through EVP4. Construct modal matrix (normalized)

Procedures)()()( ttt FKxxM =+&&

0xMK =ω− )()( 2 t0)det( 2 =ω− MK

[ ]nuuuU K21=

IMUU =T

ΛKUU =T

5. Perform a coordinate transformation )()( tt Urx =

)()()( ttt FKxxM =+&& )()()( ttt FKUrrMU =+&&

)()()( ttt TTT FUKUrUrMUU =+&&

)()()( ttt TFUΛrr =+&&

Modal analysis (undamped systems)-2

)()()( ttt TFUΛrr =+&&

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

ω

ωω

+

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

)()()()(

)()()()(

)(

)()(

00

0000

)(

)()(

4

3

2

1

4

3

2

1

21

22221

11211

2

1

2

22

21

2

1

tNtNtNtN

tFtFtFtF

uuu

uuuuuu

tr

trtr

tr

trtr T

NNNN

N

N

NnN

n

n

N K

MOMM

K

K

M

K

MOMM

K

K

&&

M

&&

&&

Independent SDOF equations, can be solve for r(t)

6. Transform the initial conditions to modal coordinates

)()( tt Urx =

IMUU =T

)0()0( Urx =

)0()0( MUrUMxU TT =

)0()0( MxUr T=

From

and )0()0( xMUr && T=

Modal analysis (undamped systems)-3

7. Find the response in modal coordinates8. Transform the response in modal coordinate

back to that in original coordinate )()( tt Urx =

K=)(tr)(tr

)(tx

Example (Modal analysis) -1

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡00

33327

1009

2

1

2

1

xx

xx&&

&&

EOM

⎥⎦

⎤⎢⎣

⎡=

01

0x ⎥⎦

⎤⎢⎣

⎡=

00

0vInitial conditions

Example (Modal analysis) -2

2dof string-bead system

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡00

2112

00

2

1

2

1

xx

aT

xx

mm

&&

&&

EOM

⎥⎦

⎤⎢⎣

⎡=

00

0x ⎥⎦

⎤⎢⎣

⎡=

10

0vInitial 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. 0=ωn

Rigid-body modes

Compute the solution of the system. Let m1 = 1 kg, m2 = 4 kg and k = 400 N/m.

Initial condition ⎥⎦

⎤⎢⎣

⎡=

001.0

0x ⎥⎦

⎤⎢⎣

⎡=

00

0v

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)

)()()()( tttt FKxxCxM =++ &&&

CKMKCM 11 −− =

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

KMC β+α=

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 )()()()( tttt FKxxCxM =++ &&&

)()()()()( tttt FKxxKMxM =+β+α+ &&&

)()( tt Urx = )()()()()( tttt FKUrrUKMrMU =+β+α+ &&&

TU )()()()()( TTTT tttt FUKUrUrUKMUrMUU =+β+α+ &&&

)()()()()()( T ttttt NFUΛrrΛIr ==+β+α+ &&&

)()()(2)( 1121111 tNtrtrtr nn =ω+ζω+ &&&

M2112 nn βω+α=ζω

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.

/radkg.m10 2321 === JJJ

N.m/rad10321 == kkN.m.s/rad2=c

• Zero initial conditions• Applied moment M(t) is a unit

impulse function

Modal analysis on damped systems (Ex.)