MDOF SYSTEMS WITH DAMPING General case -...

MDOF SYSTEMS WITH DAMPING

General case

Saeed Ziaei Rad

MDOF Systems with hysteretic damping- general case

}0{}]){[]([}]{[ xDiKxM Free vibration solution:

Assume a solution in the form of:tieXx }{}{

Here can be a complex number. The solution here is likethe undamped case. However, both eigenvalues and Eigenvector matrices are complex.The eigensolution has the orthogonal properties as:

][]][[]]; [[]][[][ rT

rT kiDKmM

The modal mass and stiffness parameters are complex.

MDOF Systems with hysteretic damping- general caseAgain, the following relation is valid:

)1(22rr

r

rr i

mk

A set of mass-normalized eigenvectors can be defined as:

rrr m }{)(}{ 2/1

What is the interpretation of complex mode shapes?The phase angle in undamped is either 0 or 180.Here the phase angle may take any value.

Numerical Example with structural damping

m1 m3m2

x1

x2

x3

k1 k3

k2k4 k5

k6

m1=0.5 Kgm2=1.0 Kgm3=1.5 Kgk1=k2=k3=k4=k5=k6=1000 N/m

Undamped

5.100010005.

][M

3 1 1[ K ] 1000 1 3 1

1 1 3

Using command [V,D]=eig(k,M) in MATLAB

66980003352000950

][ 2r

142.493.635.318.782.536.

318.1218.464.][

V =

0.4639 0.2181 -1.31810.5361 0.7819 0.31810.6351 -0.4932 0.1419

D =

1.0e+003 *

0.9503 0 00 3.3518 00 0 6.6979

>> V'*M*V

ans =1.0000 -0.0000 0.0000-0.0000 1.0000 -0.00000.0000 -0.0000 1.0000

>> V'*K*V

ans =1.0e+003 *

0.9503 -0.0000 -0.0000-0.0000 3.3518 -0.0000-0.0000 -0.0000 6.6979

Proportional Structural DampingAssume proportional structural damping as:

6,,1, 05.0 jkd jj

)05.1(6698000)05.1(3352000)05.1(950

][ 2

ii

i

r

)0(142.)0(493.)0(635.)0(318.)180(782.)0(536.

)180(318.1)180(218.)0(464.][

D=0.05*K

D =

150 -50 -50-50 150 -50-50 -50 150

KC=K+i*D

KC =

1.0e+003 *

3.0000 + 1.5000i -1.0000 - 0.5000i -1.0000 - 0.5000i-1.0000 - 0.5000i 3.0000 + 1.5000i -1.0000 - 0.5000i-1.0000 - 0.5000i -1.0000 - 0.5000i 3.0000 + 1.5000i

[V1,D1]=eig(KC,M)

V1 =

1.0000 - 0.0000i 0.7304 + 0.0000i -0.2789 - 0.0000i-0.2413 + 0.0000i 0.8442 + 0.0000i -1.0000 - 0.0000i-0.1077 + 0.0000i 1.0000 + 0.0000i 0.6307 - 0.0000i

D1 =

1.0e+003 *

6.6979 + 0.3349i 0 0 0 0.9503 + 0.0475i 0 0 0 3.3518 + 0.1676i

imag(D1)/real(D1)

ans =

0.0500 0 00 0.0500 00 0 0.0500

D1(2,2)

ans =

9.5025e+002 +4.7513e+001i

V1(:,2)

ans =

0.7304 + 0.0000i0.8442 + 0.0000i1.0000 + 0.0000i

V1(:,2)*.464/.7304

ans =

0.4640 + 0.0000i0.5363 + 0.0000i0.6353 + 0.0000i

>> Mr=V1'*M*V1

Mr =

0.5756 - 0.0000i -0.0000 - 0.0000i 0.0000 - 0.0000i-0.0000 + 0.0000i 2.4794 + 0.0000i -0.0000 + 0.0000i0.0000 + 0.0000i -0.0000 - 0.0000i 1.6356 + 0.0000i

>> Kr=V1'*KC*V1

Kr =

1.0e+003 *

3.8554 + 0.1928i 0.0000 - 0.0000i 0.0000 + 0.0000i-0.0000 + 0.0000i 2.3561 + 0.1178i -0.0000 0.0000 + 0.0000i -0.0000 - 0.0000i 5.4822 + 0.2741i

>> diag(Kr)./diag(Mr)

ans =

1.0e+003 *

6.6979 + 0.3349i0.9503 + 0.0475i3.3518 + 0.1676i

Non-Proportional Structural DampingAssume non-proportional structural damping as:

1 1

j

d 0.1kd 0 , j 2 , ,6

2r

957 ( 1 .067 i ) 0 0[ ] 0 3354( 1 .0042 i ) 0

0 0 6690( 1 .078 i )

)1.3(142.)3.1(492.)0(636.)7.6(316.)181(784.)0(537.)181(321.1)173(217.)5.5(463.

][

D=[0.1*K(1,1) 0 00 0 00 0 0]

D =

300 0 00 0 00 0 0

KC=K+i*D

KC =

1.0e+003 *

3.0000 + 0.3000i -1.0000 -1.0000 -1.0000 3.0000 -1.0000 -1.0000 -1.0000 3.0000

M =

0.5000 0 00 1.0000 00 0 1.5000

>> [V2,D2]=eig(KC,M)

V2 =

0.9242 - 0.0758i 0.6577 - 0.0012i -0.2683 + 0.0341i-0.2170 + 0.0472i 0.7597 + 0.0725i -0.9795 - 0.0205i-0.0989 + 0.0151i 0.8993 + 0.1007i 0.6144 - 0.0130i

D2 =

1.0e+003 *

6.6899 + 0.5219i 0 0 0 0.9565 + 0.0640i 0 0 0 3.3536 + 0.0140i

>> imag(D2)/real(D2)

ans =

0.0780 0 00 0.0669 00 0 0.0042

>> abs(V2(:,2))

ans =

0.65770.76310.9049

>> abs(V2(:,2))*.463/.6577

ans =

0.46300.53720.6370

>> angle(V2(:,2))*180/pi

ans =

-0.10695.44986.3887

abs(V2(:,1))*1.321/.9273

ans =

1.32100.31640.1424

>> angle(V2(:,1))*180/pi

ans =

-4.6881167.7416171.3307

>> V2'*M*V2ans =

0.4943 - 0.0000i 0.0115 - 0.0624i -0.0051 + 0.0442i0.0115 + 0.0624i 2.0270 - 0.0000i -0.0070 - 0.0440i-0.0051 - 0.0442i -0.0070 + 0.0440i 1.5629 + 0.0000i

>> V2'*KC*V2ans =

1.0e+003 *3.3066 + 0.2580i 0.0150 - 0.0590i -0.0177 + 0.1483i0.0442 + 0.4238i 1.9387 + 0.1298i -0.0227 - 0.1475i-0.0111 - 0.2987i -0.0095 + 0.0416i 5.2415 + 0.0220i

>> conj(V2')*M*V2ans =

0.4834 - 0.0950i -0.0000 - 0.0000i -0.0000 - 0.0000i-0.0000 - 0.0000i 1.9860 + 0.3810i -0.0000 - 0.0000i-0.0000 - 0.0000i 0.0000 - 0.0000i 1.5604 + 0.0070i

>> conj(V2')*KC*V2ans =

1.0e+003 *3.2835 - 0.3831i 0.0000 - 0.0000i -0.0000 - 0.0000i0.0000 - 0.0000i 1.8752 + 0.4915i 0.0000 - 0.0000i-0.0000 - 0.0000i 0.0000 - 0.0000i 5.2330 + 0.0454i

>>diag(conj(V2')*KC*V2)./diag(conj(V2')*M*V2)

ans =

1.0e+003 *

6.6899 + 0.5219i0.9565 + 0.0640i3.3536 + 0.0140i

Non-Proportional Structural Damping

Each mode has a different damping factor.All eigenvectors arguments for undamped and proportional damp cases are either 0 or 180.All eigenvectors arguments for non-proportional case are within 10 degree of 0 or 180 (the modes are almost real).

Exercise: Repeat the problem withm1=1Kg, m2=0.95 Kg, m3=1.05 Kgk1=k2=k3=k4=k5=k6=1000 N/m

FRF Characteristics (Hysteretic Damping)

titi eFeXMDiK }{}]){[][]([ 2 Again, one can write:

The receptance matrix can be found as:T

rMDiKH ]][][[])[][]([)( 2212 FRF elements can be extracted:

N

r rrr

krjrjk i

H1

222)(

or

N

r rrr

jkrjk i

AH

1222)(

Modal Constant

MDOF Systems with viscous damping- general caseThe general equation of motion for this case can be written as:

}{}]{[}]{[}]{[ fxKxCxM Consider the zero excitation to determine the natural frequencies and mode shapes of the system:

steXx }{}{ This leads to:

}0{}]){[][]([ 2 XKsCsMThis is a complex eigenproblem. In this case, there are2N eigenvalues but they are in complex conjugate pairs.

MDOF Systems with viscous damping- general case

Nrss

rr

rr ,,1 }, {}{

, *

*

It is customary to express each eigenvalues as:

)1( 2rrrr is

Next, consider the following equation:}0{}]){[][][( 2 rrr KCsMs

Then, pre-multiply by : Hq}{

}0{}]){[][][(}{ 2 rrrHq KCsMs *


}0{}]){[][][( 2 qqq KCsMs A similar expression can be written for : q}{

This can be transposed-conjugated and then multiply by r}{

}0{}]){[][][(}{ 2 rqqHq KCsMs **

Subtract equation * from **, to get:

}0{}]{[}){(}]{[}){( 22 rHqqrr

Hqqr CssMss

This leads to the first orthogonality equations:

}0{}]{[}{}]{[}){( rHqr

Hqqr CMss (1)


Next, multiply equation (*) by and (**) by : qs rs

}0{}]{[}{}]{[}{ rHqr

Hqqr KMss (2)

Equations (1) and (2) are the orthogonality conditions:If we use the fact that the modes are pair, then

*

2

}{}{

)1(

rq

rrrq is

MDOF Systems with viscous damping- general caseInserting these two into equations (1) and (2):

r

r

rHr

rHr

r

r

r

rHr

rHr

rr

mk

MK

mc

MC

}]{[}{}]{[}{

}]{[}{}]{[}{2

2

Where , , are modal mass, stiffness and damping. rm rk rc

FRF Characteristics (Viscous Damping)

The response solution is:

}{])[][]([}{ 12 FMCiKX We are seeking to a similar series expansion similar to theundamped case.To do this, we define a new vector {u}:

12

}{

Nx

xu

We write the equation of motion as:

1122 }0{}]{0:[}{]:[ NNNN uKuMC


This is N equations and 2N unknowns. We add an identityEquation as:

}0{}]{:0[}]{0:[ uMuM Now, we combine these two equations to get:

}0{}{0

0}{

0

u

MK

uM

MC

Which cab be simplified to:

}0{}]{[}]{[ uBuA 3


Equation (3) is in a standard eigenvalue form. Assuming a trial solution in the form of

NrBAs rr 2,,1} 0{}]){[][(

steUu }{}{

The orthogonality properties cab be stated as:

][]][[][][]][[][

rT

rT

bB

aA

With the usual characteristics:

Nrabs

r

rr 2,,1


Let’s express the forcing vector as:

0}{ 12

FP N

Now using the previous series expansion:

N

r rr

rTr

N siaP

XiX 2

112 )(}}{{}{

And because the eigenvalues and vectors occur in complex conjugate pair:

)(}}{{}{

)(}}{{}{

**

*

112 rsia

Psia

PXi

X

r

rHr

N

r rr

rTr

N


Now the receptance frequency response functionResulting from a single force and response parameter

jkH

kF jX

)1(()1(()(

2*

**

12

rrrr

N

r rrrrr

krjrjk

iaiaH

r

krjr

or

N

r rr

jkrrjkrjk i

SiRH

r122 2

))(/()(

Where:

rrkrkr

krkr

rkrkrrkr

aGGS

GGR

}){/(}{}Re{2}{

)1}Im{}Re{(2}{ 2

M =

1 0 0 00 1 0 00 0 1 00 0 0 1

K =

2000 -1000 0 0-1000 2000 -1000 0

0 -1000 2000 -10000 0 -1000 1000

C =

20 0 0 00 5 0 00 0 0 00 0 0 0

A=[C MM zeros(4,4)];

B=[K zeros(4,4)zeros(4,4) -M];

[V1,D1]=eig(B,-A);

D11=diag(D1)

D11 =

-2.4697 +58.3513i-2.4697 -58.3513i-4.6931 +47.9909i-4.6931 -47.9909i-4.3513 +31.7082i-4.3513 -31.7082i-0.9859 +11.0507i-0.9859 -11.0507i

>> [V1(:,1) V1(:,2)]

ans =

0.0055 - 0.0066i 0.0055 + 0.0066i-0.0022 + 0.0144i -0.0022 - 0.0144i-0.0024 - 0.0137i -0.0024 + 0.0137i0.0017 + 0.0055i 0.0017 - 0.0055i0.3695 + 0.3393i 0.3695 - 0.3393i

-0.8333 - 0.1667i -0.8333 + 0.1667i0.8070 - 0.1070i 0.8070 + 0.1070i

-0.3263 + 0.0838i -0.3263 - 0.0838i

>> [V1(1:4,1) (V1(5:8,1)/V1(5,1))*V1(1,1)]

ans =

0.0055 - 0.0066i 0.0055 - 0.0066i-0.0022 + 0.0144i -0.0022 + 0.0144i-0.0024 - 0.0137i -0.0024 - 0.0137i0.0017 + 0.0055i 0.0017 + 0.0055i

MDOF SYSTEMS WITH DAMPING General case -...

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