Prof.johnson July16 APSS2010

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USC ViterbiSchool of Engineering

Example 2DOF System

Unforced Case — free responses

Review SDOF: modal free responses of is

where the ai & bi , or c i & " i , depend on and

k 1!=!4

k 2!=!2

2

1

˙q i +" i

2q i = 0

q i (t ) =a i sin" i t +b i cos" i t = c i sin(" i t +# i )

q i (0) !!!!!!! q i (0)

For example, with initial conditions:

q 1 (0) =1= a

1sin0+ b

1cos0 =c

1sin("

1)

q 1 (0) = 0 =a 1# 1cos0 $b 1# 1sin0 = c 1# 1cos(" 1)

q 2(0) = 0 = a 2 sin0+ b 2 cos0 = c 2 sin(" 2)

q 2(0) =1=a 2#

2cos0 $b 2#

2sin0 = c 2#

2cos("

2)

then: a1=0, b1=1, c 1=1, " 1=–! /2 and a 2=1/2, b 2=0, c 2=1/2, " 2=0.

q 1 (t ) = cost

q 2(t ) = 0.5sin2t

More generally, the responses in the original coordinates are:

u(t ) = "i a i sin#

i t +b

i cos#

i t [ ]

i

$ = c i "i sin(#

i t +%

i )

i

$

= ai sin"

i t +bi cos"

i t [ ]i

# = ci sin("

i t +$ i )

i

#


Example 2DOF System

Example Free Responses

Displaced in shape of: 2nd mode1st mode combination


c 2!=!0.1

Example 2DOF System

Unforced Case with Damping

Let’s add damping. (What is damping???)

k 1!=!4

k 2!=!2

2

1

k 1!=!4 k 2!=!1m 1!=

!2 m 2

!=

!1

c 1!=!0.2

c 2!=!0.1

c 1!=!0.2

Already know " =

1 1

2 #1

$

% &

'

( ) . Let u ="q and premultiply by " T

M ˙u+Cu+Ku =0

C =

c 1+c

2"c

2

"c 2

c 2

#

$ %

&

' ( =

0.3 "0.1

"0.1 0.1

#

$ %

&

' (

1 2

1 "1

#

$ %

&

' (

2 0

0 1

#

$ %

&

' (

1 1

2 "1

#

$ %

&

' (

˙q 1

˙q 2

) * +

, - . +

1 2

1 "1

#

$ %

&

' (

0.3 "0.1

"0.1 0.1

#

$ %

&

' (

1 1

2 "1

#

$ %

&

' ( q

1

q 2

) * +

, - . +

1 2

1 "1

#

$ %

&

' (

6 "2

"2 2

#

$ %

&

' (

1 1

2 "1

#

$ %

&

' ( q

1

q 2

) * +

, - .

=

6 0

0 3

#

$ %

&

' (

˙q 1

˙q 2

) * +

, - . +

0.3 0

0 0.6

#

$ %

&

' ( q

1

q 2

) * +

, - . +

6 0

0 12

#

$ %

&

' ( q

1

q 2

) * +

, - . =

0

0

) * +

, - .

"TM" ˙q+"TC"q+"TK"q =0

"

˙q 1 +

0.05q 1 +

1q 1 = 0

˙q 2+0.20q

2+ 4q

2= 0

˙q i +2" i # i

˙q i +# i q i = 0

" 1

= 1, # 1=0.025 =2.5%

" 2 =

2, # 2 =

0.050 =5.0%


c 2!=!0.1

Example 2DOF System

Unforced Case with Damping — free response

So modal damped free responses are

where the ai & bi , or c i & " i , depend on and , and.

k 1!=!4

k 2!=!2

2

1

k 1!=!4 k 2!=!1m 1!=

!2 m 2

!=

!1

c 1!=!0.2

c 2!=!0.1

c 1!=!0.2

q i (t ) =a i e "# i $ i t sin$ i

dt +b i e "# i $ i t cos$ i

dt = c i e "# i $ i t sin($ i

dt +% i )

q i (0) !!!!!!! q i (0)

More generally, the responses in the original coordinates are:

u(t ) = e "# i $ i t %

i a i sin$

i

dt +b

i cos$

i

dt [ ]

i

& = c i %i e "# i $ i t sin($

i

dt +

i

&

= e "# i $ i t a

i sin$

i

dt +b

i cos$

i

dt [ ]

i

& = ci e "# i $ i t sin($

i

dt +'

i

i

&

" i d=" i [1#$ i

2]1/2


General MDOF Systems

Equation of motion: M˙

u+C˙

u+Ku =Ef

First use the undamped, unforced equation to find natural

frequencies and mode shapes:(M"

2#K)$ =0 or ("

2I#M

#1K)$ =0

The mode shapes have the properties that

"i TM" j =M i # ij =

M i , i = j

0, i $ j

% & '

"i TK" j =K i # ij =

K i , i = j

0, i $ j

% & '

The fact that !i TM! j = 0 and !i

TK! j = 0 for i ! j means the

mode shapes are orthogonal . Note also that K i /M i = ! i 2.

Note: some systems have (M! 2 – K) be less than rank (n – 1), whichimplies there are multiple modes at one or more frequencies.

Note: some systems (e.g., satellites) may have ! = 0 be a solution;

this implies one or more rigid body modes in which the wholestructure moves or rotates, possibly without internal deformation.


General MDOF Systems

Forced Harmonic Response

Forced response of a linear system is the sum of the freeresponse from any non-zero initial conditions and theeffects of the external forcing.

Let us assume the initial conditions are zero and consider a harmonic excitation.

M ˙u +Cu+Ku =Ef =Ef0sin" t

Then the response is also harmonic. This is most easily

demonstrated using a SDOF example …




Damping Models

In the previous damped 2DOF example, we used specific

damping coefficients in the model: c 1 = 0.2, c 2 = 0.1.

k 1!=!4

k 2!=!2

2

1c 2!=!0.1

c 1!=!0.2

2 0

0 1

"

# $

%

& '

˙u 1

˙u 2

( ) *

+ , - +

0.3 .0.1

.0.1 0.1

"

# $

%

& ' u 1

u 2

( ) *

+ , - +

4 +2 .2

.2 2

"

# $

%

& ' u 1

u 2

( ) *

+ , - =

0

0

( ) *

+ , -

" TC" =

1 2

1 #1

$

% &

'

( )

0.4 #0.2

#0.2 0.2

$

% &

'

( ) 1 1

2 #1

$

% &

'

( ) =

0.4 #0.2

#0.2 1

$

% &

'

( )

"

˙q 1 +

1

15q 1 #

1

30q 2 +

1q 1

= 0

˙q 2+

1

3q 2#

1

15q 1+ 4q

2= 0

" 1

= 1, # 1=2.5%

" 2=2, #

2=5.0%

We found that the damping #TC# diagonalized.

However, a different choice (c 1 = 0.2, c 2 = 0.2) may not:

0.2

c 2!=!0.1

k 1!=!4 k 2!=!1m 1!=!2 m 2!=!1

c 1!=!0.2 0.2

This is what is called a non-proportionally

damped or non-classically damped system.

The equations

are still coupled!

We will see later how to solve systems like this.(! 1=1.00037, ! 2=1.99926, & 1=3.3313%, & 2=8.3368%)


Damping Models

Special Cases of Proportional Damping: Modal Damping

We often do not know the damping coefficients (while mass

and stiffness can be computed or easily measured, dampingis much harder to quantify).

Damping ratios & i in individual modes are easier to estimate(e.g., excite the structure in a particular mode, and estimate & from the decay e –&! t ; repeat for other modes).

So, one way to construct the full damping matrix C is byassuming a modal decomposition

" TC" =

! 02# i $ i M i

0 !

%

&

' ' '

(

)

* * *

C =" # T! 0

2$ i % i M i

0 !

&

'

( ( (

)

*

+ + + " #1 =M"

! 02$ i % i

0 !

&

'

( ( (

)

*

+ + + " #1


Damping Models


So, the equation of motion is: M ˙u+M"!

2# i $ i

!

%

& ' '

(

) * * " +1u+Ku =0

Decouple method 1: let u = #q and premultiply by #T:

" TM" ˙q+" TM"!

2# i $ i

!

%

& ' '

(

) * * " +1"q+" TK"q =0

!

M i

!

%

& ' '

(

) * * ˙q+

!

M i

!

%

& ' '

(

) * *

!

2# i $ i

!

%

& ' '

(

) * * q+

!

K i

!

%

& ' '

(

) * * q =0

!

M i

!

%

& ' '

(

) * * ˙q+

!

M i

!

%

& ' '

(

) * *

!

2# i $ i

!

%

& ' '

(

) * * q+

!

K i

!

%

& ' '

(

) * * q =0

˙q+

!

2# i $ i

!

%

& '

'

(

) *

* ˙q+

!

$ i

2

!

%

& '

'

(

) *

* q=

0

˙q i +2# i $ i q i +$ i 2q i =0, i =1,...,n


Damping Models


So, the equation of motion is: M ˙u+M"!

2# i $ i

!

%

& ' '

(

) * * " +1u+Ku

Decouple method 2: let u = #q and premultiply by # –1M –1:

" #1M#1M" ˙q+" #1M#1M"!

2$ i % i

!

&

' ( (

)

* + + " #1"q+" #1M#1K"q =

" #1" ˙q+" #1"!

2$ i % i

!

&

' ( (

)

* + + " #1"q+" #1M#1K"q =0

Let " =#$1M

$1K#

% M#" =K# % #T

M#" =#TK#

% ! M i

!

&

' (

(

)

* +

+ " =

!

K i

!

&

' (

(

)

* +

+

% " =

!

K i M i

!

&

' ( (

)

* + + =

!

, i 2

&

' ( (

˙q+

!

2" i # i

!

$

% & &

'

( ) ) q+

!

# i 2

!

$

% & &

'

( ) ) q =0

˙q i +2" i # i q i +# i 2q i = 0, i =1,...,n


Damping Models

Special Cases of Proportional Damping: Rayleigh Damping

Rayleigh damping (1877) is given by C = ' M + ( K.Since both M and K diagonalize with #, it is easy to see

that #TC# = ' #TM# + ( #TK# is diagonal.

The result: 2& i ! i = ' + (! i 2 or & i = (' /! i + (! i )/2.

Since there are only two parameters, ' and ( , we can

choose the damping of two modes; all other modal

damping ratios are given by the above equation.

If we know (or choose) & r and & s then' = 2! r ! s(& r ! s – & s! r ) / (! s

2 – ! r 2)

( = 2(& s! s – & r ! r ) / (! s2 – ! r

2)& i = (' /! i + (! i )/2


Damping Models

Conditions for Proportional Damping

It can be shown that the system is classically damped if either of the following are true:• C = M F(M –1K) + K G(K –1M)

• C = F(KM –1)!M + G(MK –1) K for some matrix functions F(") and G(").

Note: should add references here.




Laplace and Fourier Transforms

Time history p(t ) is related to its

Laplace transform P (s) and Fourier Transform P ( j ! ) as

P (s ) = p (t )e "st dt "#

#

$

p (t ) =1

2"

P (s )e st ds

#$

$

%

P ( j " ) = p (t )e # j " t dt

#$

$

%

p (t ) =1

2" P ( j # )e j # t d #

$%

%

&

Laplace transform P (s) may also be denoted L{ p(t )}

and Fourier Transform P ( j ! ) by F { p(t )}.

Note: L p (t ){ } = p (t )e "st dt "#

#

$ = pe "st "#

#+ p (t )se "st dt

"#

#

$ = s L p (t ){ }

(assuming p(t )$0 as t $#)integration by parts

Similarly: F p (t ){ } = j " F p (t ){ }


Transfer Functions

SDOF

Consider single degree of freedom system m u +c u +ku =f

ms 2U (s )+csU (s )+kU (s ) =F (s )

(ms 2+cs +k )U (s ) =F (s )

U (s )

F (s )=

1

ms 2+cs +k

Laplace transform both sides:

Or, rearranging:

U ( j " )

F ( j " )=

1

m ( j " )2+cj "

=

1

k #m " 2+cj

Could do similar using Fourier transform:

If f (t ) = F 0 sin! t , then

u (t ) = F 0

1

k "m # 2+cj #

sin(# t +$ )

" = angle1

k #m $ 2+cj $

%

& '

(

) * = tan

#1 c

m $ 2 #k

These are calledtransfer function


Transfer Functions

SDOF Example

For example:

k !=!8

2c !=!0.2

2˙u +0.2u +8u =f

10−1

100

101

10−2

10−1

100

Frequency [rads/sec]

M a g n i t u d e

10−1

100

101

−150

−100

−50

0


P h a s e [ d

e g r e e s ]

10−1

100

101

−2

0

2


R e a l P a r t

10−1

100

101

−3

−2

−1

0


I m a g i n a r y P a r t

U (s )

F (s )=

1/2

s 2+0.1s + 4 Graph with s = j ! :


Transfer Functions

SDOF Example — Poles

For example:

k !=!8

2c !=!0.2

2˙u +0.2u +8u =f

U (s )

F (s )=

1/2

s 2+0.1s + 4

Roots of the denominator polynomial are called the poles

of the system:

s poles ="0.1± 0.12 " 4

2= "0.05 ± j 3.9975 # "0.05 ±1.9994

R

Im


Transfer Functions

SDOF Example — Poles and Zeros

Some TFs have a numerator that is also a polynomial in s.For example, the transfer function from ground accelerationto the absolute acceleration of a SDOF system:

k !=!8

2c !=!0.2

2˙u +0.2u +8u = "2˙v g

L ˙v (t ){ }

L ˙v g(t ){ }=

L ˙u + ˙v g{ }L ˙v g(t ){ }

=

L "0.1u " 4u { }

L ˙v g(t ){ }

= ("0.1s " 4)L u { }

L ˙v g(t ){ }=

0.1s + 4

s 2+0.1s + 4

Roots of the numerator polynomial are

called the zeros of the system:

s root =

"40

Re

Im


Transfer Functions

MDOF

For a MDOF system, must be careful to handle matricescorrectly in determining the transfer function:

M ˙u+Cu+Ku =Ef

Ms 2+Cs +K[ ]U(s ) =EF(s )

U(s ) = Ms 2+Cs +K[ ]

"1

EF(s )

Now, the transfer function [Ms2 + Cs + K] –1E is a matrix,

each element of which is a scalar transfer function.




State-Space Formulation

Unforced Response — Mode Shapes

If the damping term does not decouple, then the state-

space approach must be used and

must be solved directly to get the complex eigenvalues and

complex eigenvectors. Once the the eigenvalues ) i ,) i * are

found, then ! i = |) i | and & i = –Re{) i }/(2! i ).

If some eigenvalues are purely real, then no oscillator

mode corresponds to that eigenvalue.

(" 2I+ " M

#1C +M

#1K)$

d=0


k 3!=!a

1c 3!=!c

k 2!=!a

1c 2!=!c


Visualization of Non-Proportional Damping

Let’s look at an example of a

3DOF structure with classicaldamping (-c = 0). The natural

frequencies & damping are:! 1 = 1 , & 1 = 3.14%;! 2 = 2.80, & 2 = 8.80%;

! 3 = 4.05, & 3 = 12.72%.

A non-classically dampedversion with additional

damping in the first floor (-c = 15c ) has natural

frequencies & damping:! 1 = 1.15, & 1 = 23.6%;! 2 = 2.71, & 2 = 85.9%;

! 3 = 3.64, & 3 = 15.7%.

k 1!=!a

1

c 1!=!c !+!-c

c !=!0.01a

a !=!199.3233(so first natural

frequency is 1!Hz)

"0.328 0.737 "0.591

"0.591 0.328 0.737

"0.737 "0.591 "0.328

0.065± 2.06 j "1.142!12.9 j 1.913±1

0.117± 3.71 j "0.508! 5.8 j "2.385!1

0.146± 4.63 j 0.916±10.4 j 1.061±

#

$

% % % % % % %

"0.233! 0.19 j "0.545! 3.99 j 0.003± 0

"0.557! 0.08 j "0.977! 0.93 j 0.537! 0

"0.737 "0.591 "0.328

"0.966±1.97 j "26.85± 63.2 j 3.749! 0

0.409± 4.04 j 6.213± 22.1 j "3.575!1

1.260± 5.17 j 8.655± 5.2 j 1.174± 7

#

$

% % % % % % %



Visualization of Non-Proportional Damping

ProportionallyDamped:

Non-Proportionally

Damped:



Unforced Free Response

The unforced free response of the state-space system

could be found through modal decomposition, but there is aneasier way. To compute the response at time t , break the time

up into r smaller steps. Given the definition of a derivative:

˙" =A"

"(1

r t ) # "(0

r t )+ " 1

r t # "(0)+A"(0) 1

r t = [I+A t

r ]"(0)

The same can be used to approximate ( after each t /r .

"(2r t ) # [I+A t

r ]"(1

r $t ) = [I+A t

r ]2"(0)

!

"(r

r t ) # [I+A t

r ]"(r %1

r $t ) = [I+A t

r ]r "(0)

To eliminate the approximation, let r $#.

"(t ) = [I+ (At ) + 1

2!(At )

2+

1

3!(At )

3+!]"(0)

"(t ) =e At "(0)



Unforced Free Response — Notes about eAt

eAt is a matrix; its elements are not the exponential of theelements of At . The computation of eAt can be computedthrough the power series formulation, though that is not

computationally efficient.

An efficient computation uses the fact that the eigenvectors of

a power of a matrix are the same as those of the matrix itself.

If A+ = +., where + is the eigenmatrix and . is a diagonalmatrix of eigenvalues, then it can be shown that An+ = +.n.

Thus, e At

= I+ (At ) + 1

2!(At )

2+

1

3!(At )

3+!

="#1" +"#1$"t +1

2!"#1$"2

t 2+

1

3!"#1$3"t

3+!

="#1[I+ ($t ) + 1

2!($t )

2+

1

3!($t )

3+!]"

= "#1

"

e % i t

"

&

'

( ( (

)

*

+ + +

"



Impulse Response

The state impulse response matrix is given by

H" =AH" +B# (t ), H" (0$) = 0

Hy =CyH" +Dy# (t )

Integrating from t = 0 – to t = 0+, will give the initial conditions

that make a free response equal to the impulse response.

H"(0

+

) #H"(0#) =AH"

avg(0

+

#0#)+B

0 0

H" (0+

) =B

For t > 0, the impulse response is just an unforced free

response (which was solved several slides ago), so

H"(t ) =e

At H

"(0

+

)

H" (t ) =e At B





Forced Response

The forced response is, then, the combination of the initial

condition free response and the effects of the force

"(t ) =e At "(0)+ e

A(t #$ )Bf($ )d $

0

t

%

Add Example!USC ViterbiSchool of Engineering

Discrete-Time State-Space System

We’ve already seen the free response of the unforced state-

space system . So, ((t + -t ) = eA-t ((t ).

The effect of the forced response of in [t , t + -t ]

is the superposition of the free response with the impulse

response of f during the time step.

˙" =A"

Defining ((k ) = ((k -t ), then the discrete-time state space form

"(k +1) = Ad"(k )+B

df(k )

y(k ) =Cy"(k )+Dyf(k )

˙" =A" +Bf

"(t +

#t )=e

A#t

"(t )+ e

A(t +#t $% )

t

t +#t

& Bf

(%

)d %

In the interval [t , t + -t ), if f($ ) is constant (a zero-order hold )

"(t +#t ) =e A#t

"(t ) + e A(#t $% )

0

#t

& d %

'

( )

*

+ , Bf(t )

Ad

Bd

Ad Bd

0 I

"

# $

%

& ' =e

A

0

"

# $

Note: one can also assume a first-o

(linear) hold on f to determine Bd.


Distributed Parameter Systems

Euler-Bernoulli Beam Example

Continuous systems (e.g., beams, plates, shells, or more

complex structures) also have modes of vibration (generallyinfinitely many). Let’s look at an Euler-Bernoulli beam;

transverse displacement v ( x ,t ) must satisfy the partialdifferential equation of motion where (")/ = 0/0 x :

(EI " " v " " ) + # A ˙v =f (x ,t )

L

x

v (x ,t )

EI , * A

Substitute v ( x ,t ) = V ( x )T (t ) into

the unforced, undamped system.

(EI " " V " " ) T + # AV T = 0

(EI " " V " " )

# AV = $

˙T

T

LHS is a function of x , the RHS of t , so both must be constant.

=" 2

˙T +"

2T = 0

" " " " V # $ 4V =0

Assuming EI and * A are constant, let ) 4 = [* A! 2/EI ]1/2 so:

V (x ) =c 1sinh" x +c

2cosh" x +c

3sin" x +c

4cos" x

T (t ) = c sin(" t +# )



Euler-Bernoulli Beam Example — exact simply-supported

The coefficients c i depend on the boundary conditions:

• V = 0 and V / = 0 at a fixed (cantilevered) end

• V = 0 and V // = 0 at a simply supported end

• V // = 0 and V /// = 0 at a free end

1. Simply supported at both ends: c 2 = c 4 = 0

c 1sinh" L + c

3sin" L = 0

" 2c 1sinh" L # "

2c 3 sin" L = 0

$ sinh" L sin" L

" 2sinh" L #"

2sin" L

= 0

The non-trivial solution is ) i L = i ! where n = 1, 2, …; the

resulting eigenfunctions are V i ( x ) = C sin i ! x /L and thenatural frequencies are ! i = (i ! /L)2 [* A/EI ]1/2.

! 3!=!9! 1

! 2!=!4! 1! 1



Euler-Bernoulli Beam Example — exact cantilevered

cos" L cosh" L +1= 0

which has solutions ) L = 1.8751, 4.6941, 7.8548, 10.996, …;

larger solutions are approximately ! (1"+ 2n)/2.

! 3!=!17.55! 1

! 2!=!6.27! 1

! 1

2. Cantilevered at x = 0 and free at x = L: c 4 = –c 2, c 3 = –c 1 Using the remaining equations leads to

" 1=

3.516

L2

EI

# A

" 2=

22.03

L2

EI

# A

" 3=

61.70

L2

EI

# A

V i (x ) =C cosh"

i x # cos"

i x #

cosh" i L+cos" i L

sinh" i L+sin" i Lsinh"

i x # sin"

i x [ ]{ }



Euler-Bernoulli Beam Example — assumed modes method

Sometimes solving for the exact mode shapes from thepartial differential equations of motion is difficult.

An alternate approach is to assume a set of mode shapes

V i ( x ) that satisfy the boundary conditions. Then, Lagrange’s

equations can be used to determine the equations of motion.

For example, for an Euler-Bernoulli beam:

V =1

2EI ( " " v )2dx

0

L

# =

1

2v i

2(t ) EI " " V i

2(x )dx

0

L

# i

$

T =1

2% Av

2dx

0

L

# =

1

2v i

2(t ) % AV

i

2(x )dx

0

L

# i

$

f i (t ) = f (x ,t )V i (x )d0

L

"

v (x ,t ) = V i (x )v

i (t

i

"

Which leads to: M˙v +Kv = f

k ij = EI " " V i

(x ) " " V j (x )dx 0

L

#

m ij = " AV i (x )V j (x )dx 0

L

#

Prof.johnson July16 APSS2010

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