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 case

Again, the following relation is valid:

)1(22rr

r

rr i

m

k

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

5.100

010

005.

][M

311

131

113

][K

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

669800

033520

00950

][ 2r

142.493.635.

318.782.536.

318.1218.464.

][

Proportional Structural Damping

Assume proportional structural damping as:

6,,1,      05.0 jkd jj

)05.1(669800

0)05.1(33520

00)05.1(950

][ 2

i

i

i

r

)0(142.)0(493.)0(635.

)0(318.)180(782.)0(536.

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

][

Non-Proportional Structural Damping

Assume non-proportional structural damping as:

6,,2,      0

3.0 11

jd

kd

j

)078.1(669000

0)042.1(33540

00)067.1(957

][ 2

i

i

i

r

)1.3(142.)3.1(492.)0(636.

)7.6(316.)181(784.)0(537.

)181(321.1)173(217.)5.5(463.

][

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 case

The 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 XKsCsM

This 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 : H

q}{

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

MDOF Systems with viscous damping- general case

}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)

MDOF Systems with viscous damping- general case

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 case

Inserting these two into equations (1) and (2):

r

r

rHr

rHr

r

r

r

rHr

rHr

rr

m

k

M

K

m

c

M

C

}]{[}{

}]{[}{

}]{[}{

}]{[}{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

}{

N

x

xu

We write the equation of motion as:

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

FRF Characteristics (Viscous Damping)

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

M

Ku

M

MC

Which cab be simplified to:

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

FRF Characteristics (Viscous Damping)

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:

Nra

bs

r

rr 2,,1        

FRF Characteristics (Viscous Damping)

Let’s express the forcing vector as:

0}{ 12

FP N

Now using the previous series expansion:

N

r rr

rTr

Nsia

P

Xi

X 2

112)(

}}{{}{

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

)(

}}{{}{

)(

}}{{}{**

*

112 rsia

P

sia

P

Xi

X

r

rHr

N

r rr

rTr

N

FRF Characteristics (Viscous Damping)

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

aG

GS

GGR

}){/(}{

}Re{2}{

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