AA242B: MECHANICAL VIBRATIONS - Damped Vibrations of n ...

AA242B: MECHANICAL VIBRATIONS

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AA242B: MECHANICAL VIBRATIONS Damped Vibrations of n-DOF Systems

These slides are based on the recommended textbook: M. G´ eradin and D. Rixen, “Mechanical Vibrations: Theory and Applications to Structural Dynamics,” Second Edition, Wiley, John & Sons, Incorporated, ISBN-13:9780471975465

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Outline

1 Modal Superposition

2 Space-State Formulation & Analysis of Viscous Damped Systems

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Modal Superposition Damped Oscillations in Terms of Undamped Natural Modes

Assume that the damping mechanism can be described by a viscous, quadratic, dissipation function in the generalized velocities D=

1 T q˙ Cq˙ ≥ 0 2

where C is a symmetric1 and non-negative damping matrix Lagrange’s equations =⇒ M¨ q + Cq˙ + Kq = p(t) Consider again the eigenpairs (ωi2 , qai ), i = 1, · · · , n of the undamped system Look for a solution of the damped equations of dynamic equilibrium of the form n X q= yi (t)qai i=1 1 Symmetry

is required to obtain a conservative system (or obtain that D is a homogeneous function of degree 2 in q). ˙ 3 / 29


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M¨ q + Cq˙ + Kq = p(t);

q=

n X

yi (t)qai = Qy(t)

i=1 T

=⇒ QT MQ¨ y + QT CQy˙ + Q KQy = QT p(t) =⇒ I¨ y + QT CQy˙ + Ω2 y = QT p(t) In general, QT CQ = [βij ] is a full matrix Hence, n X yï + βij y˙j + ωi2 yi = qT ai p(t),

i = 1, · · · , n

j=1

where βij = qT ai Cqaj The above equation shows that unless some assumptions are introduced, the method of modal superposition is not that interesting for solving the dynamic equations of equilibrium, because the resulting modal equations are coupled in the presence of general damping 4 / 29


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However, if a small number of modes m n suffices to compute an accurate solution, the modal superposition technique can still be interesting because in this case, the size of the modal equations is much smaller than that of the original equations (reduced-order modeling) C −→ QT C Q n×n m×n n×n n×m | {z } m×m

=⇒ yï +

mn X

βij y˙j + ωi2 yi = qT ai p(t),

i = 1, · · · , m

j=1

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It will be shown that if the structure is lightly damped, a diagonal matrix QT CQ is a consistent even though not a physical assumption consider M¨ q + Cq˙ + Kq = 0 search for a solution of the form q = za e λt =⇒ (λ2k M + λk C + K)zak = 0 without damping, one would have λk = ±iωk and zak = qak if the system is lightly damped, it can be assumed that λk and zak differ only slightly from ωk and qak , respectively λk = iωk + ∆λ

zak = qak + ∆z

substituting in the characteristic equation and neglecting the second-order terms gives (K − ωk2 M)∆z + (i2ωk M + C)qak ∆λ + iωk C(qak + ∆z) ≈ 0 the light-damping assumption allows one to neglect the terms C∆λ and C∆z =⇒ (K − ωk2 M)∆z + iωk (C + 2∆λM)qak ≈ 0 6 / 29


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It will be shown that if the structure is lightly damped, a diagonal matrix QT CQ is a consistent even though not a physical assumption (continue) =⇒ (K − ωk2 M)∆z + iωk (C + 2∆λM)qak ≈ 0 2 T =⇒ qT ak (K − ωk M)∆z + qak iωk (C + 2∆λM)qak ≈ 0

=⇒ qT ak (C + 2∆λM)qak ≈ 0 1 =⇒ ∆λ ≈ − βkk 2 1 =⇒ λk ≈ − βkk + iωk 2 the first-order correction involves only the diagonal damping terms βkk = qTak Cqak , and thus the influence of the non-diagonal terms is only second-order 7 / 29


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Modal Superposition Damped Oscillations in Terms of Undamped Natural Modes It will be shown that if the structure is lightly damped, a diagonal matrix QT CQ is a consistent even though not a physical assumption (continue)

it is also possible to obtain the eigenmode correction as follows ∆z =

n X αi qai i=1 i6=k

∀l 6= k,

qTal (K − ωk2 M)

n X αi qai + qTal iωk (C + 2∆λM)qak ≈ 0 i=1 i6=k

=⇒ αl =

iωk βkl ωk2 − ωl2

=⇒

zak = qak + i

n X ωk βkj qa 2 ω − ωj2 j j=1 k j6=k

note that ∆z is pure imaginary ⇒ a free-vibration of the damped system is no longer a synchronous motion of the whole system qk = zak e λk t = (qak + i=(∆z)) e (iωk +∆λ)t =⇒ dof qi is no longer in phase with another dof qj ! 8 / 29


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In summary, the lightly-damped assumption simplifies the modal equations to yï + βi y˙i + ωi2 yi = qT ai p(t),

i = 1, · · · , n

where βi ≡ βii = qT ai Cqai in which case the modal damping coefficient is defined as ξi =

βi 2ωi

=⇒ yï + 2ξi ωi y˙i + ωi2 yi = qT ai p(t),

i = 1, · · · , n

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A simple method for constructing a diagonal damping matrix is the so-called Rayleigh proportional damping C = aK + bM 2 =⇒ βi = qT ai Cqai = aωi + b

=⇒ ξi =

1 b (aωi + ) 2 ωi

Whenever possible, the modal coefficients ξi should be determined from experimental vibration testing

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Modal Superposition Forced Harmonic Response in the Lightly Damped Case

M¨ q + Cq˙ + Kq = fa e iωt After a certain amount of time, the homogeneous response is damped out ⇒ response can be limited to the forced term (particular solution) q = za e iωt =⇒ (K − ω 2 M + iωC)za = fa Develop the solution in the form za =

n P

yi qai

i=1

=⇒ (K − ω 2 M + iωC)

n X

yi qai = fa

i=1

Pre-multiply by qT aj =⇒ yj =

qT aj fa ωj2

− ω 2 + i2ξj ωωj 11 / 29


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Modal Superposition Forced Harmonic Response in the Lightly Damped Case

Note: (K − ω 2 M + iωC)−1 is the dynamic influence (or admittance) matrix n P yj qaj with Since za = (K − ω 2 M + iωC)−1 fa and za = j=1

yj =

qT aj fa

, it follows that the spectral expansion of ωj2 − ω 2 + i2ξj ωωj this matrix in terms of the undamped modes is given by (K − ω 2 M + iωC)−1 =

n X

qaj qT aj

j=1

ωj2 − ω 2 + i2ξj ωωj

Observe that if ω → 0, the right hand-side converges to K−1 , which is consistent with the limit when ω → 0 of the left hand-side

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Modal Superposition Forced Harmonic Response and Force Appropriation Testing

Vibration test ⇒ modal characteristics (natural modes and frequencies) Procedure: force the system to vibrate along successive modes by tuning the frequency and relative force amplitude of the excitation Phase lag quadrature criterion damped system, harmonic vibration test (K − ω 2 M + iωC)za = fa with za = qaj and ω = ωj =⇒ (K − ωj2 M + iωj C)qaj = faj where faj is the appropriate excitation to achieve za = qaj and ω = ωj since (qaj , ωj2 ) satisfies (K − ωj2 M)qaj = 0, it follows that faj = iωj Cqaj which shows that the excitation force is in phase with the dissipation forces and has a 90deg phase lag with respect to the response za = qaj is the converse true? 13 / 29


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Suppose all the exciting forces are synchronous and the response at every point of the structure is in phase quadrature with the excitation The phase relationship between response and excitation may be expressed by assuming that za is a real vector and fa an imaginary one (K−ω 2 M+iωC)za = fa ⇒ (K−ω 2 M)za = 0

and

fa = iωCza

showing that the only admissible solution for ω and za are the eigensolutions of the associated conservative system Hence, the converse is true and the 90deg phase idea is one way to measure qak and ωk2

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The Phase Quadrature Criterion the structure vibrates according to one of the eigenmodes of the associated conservative system if and only if all degrees of freedom π vibrate synchronously and have a phase lag of with respect to the 2 excitation

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Space-State Formulation & Analysis of Viscous Damped Systems

Damped equations of motion M¨ q + Cq˙ + Kq = p(t) where C is also symmetric Above equation of dynamic equilibrium can be written in first-order form as q˙ K 0 p(t) C M q + = ¨ 0 −M 0 q M 0 q˙ {z } | {z } | {z } | {z } | {z } | B

r˙

A

r

s(t)

=⇒ B˙r + Ar = s(t) Above first-order form is called the space-state form of the equation of dynamic equilibrium 16 / 29


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Space-State Formulation & Analysis of Viscous Damped Systems The Homogeneous Case

B˙r + Ar = 0 r = ya e λt ⇒ (λBya + Aya )e λt = 0 ⇒ λBya + Aya = 0 ⇒ Aya = −λBya

(EVP)

A and B are 2n × 2n matrices if M, C, and K are n × n matrices ⇒ 2n eigensolutions (λk , yak ) Orthogonality relationships since M, C, and K are symmetric, then A and B are symmetric recall that the eigenvectors of a symmetric pencil (A, B) are orthogonal with respect to A and B  T  yaj Byai = bi δij =⇒  T yaj Ayai = ai δij 17 / 29


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Conjugate eigensolutions ¯ k , y¯a ) if (λk , yak ) is an eigensolution, the complex conjugate pair (λ k is also an eigensolution of Aya = −λBya proof λk = µk + iνk yak = uak + ivak =⇒ A(uak + ivak ) + (µk + iνk )B(uak + ivak ) = 0 =⇒

Auak + µk Buak − νk Bvak Avak + νk Buak + µk Bvak

= =

0 0

(real part) (imaginary part)

(1)

note that equations (1) above are invariant with respect to the simultaneous changes νk → −νk and vk → −vk , which implies that ¯ k = µk − iνk ), (¯ (λ yak = uak − ivak ) is also a solution of the eigenvalue problem Aya = −λBya modal superposition delivers a real-valued solution of a real-valued problem

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Stability of the general solution ˙ T recall thatM¨ q + Cq˙ +Kq = 0 ⇐⇒ B˙r + Ar = 0 where r = [q q] q za za r= = e λt = ya e λt ⇒ ya = q˙ λza λza hence,the 2n eigenmodes of B˙r + Ar = 0 are of the form zak yak = where zak is a complex mode of the damped λ k z ak system, and λk is a complex eigenvalue of either system above and therefore solution of (λ2k M + λk C + K)zak = 0,

k = 1, · · · n

pre-multiply the above equation by ¯zTak and assume that the system is lightly damped =⇒ λ2k mk + λk ck + kk = 0 where mk = ¯zTak Mzak , ck = ¯zTak Czak , and kk = ¯zTak Kzak stability condition µk = 1 =⇒

   λi  

r

q = −ωi ξi ∓ ξi2 − 1 < 0 q q n P −ω ξ + ξ2 −1 −ω ξ − ξ2 −1 i i = + Bi e i i )yai (Ai e i i i=1

where the constants Ai and Bi are determined from the initial conditions and orthogonality relationships of the eigenvectors yai which are in this case real-valued

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Space-State Formulation & Analysis of Viscous Damped Systems The Non-Homogeneous Case In the following section, the system is assumed to be underdamped and therefore ωkd is real and λk is complex Governing system of equations T B˙r + Ar = s(t); s(t) = pT (t) 0T 2n n n P P P Modal superposition: r = ηk (t)yak = ηk (t)yak + ηk (t)¯ yak i=1

=⇒

i=1

2n X

Byak η˙ k (t) +

k=1

i=1

2n X

Ayak ηk = s(t)

k=1

Pre-multiply the above equation by yaT and recall the orthogonality conditions j

=⇒

T (ya Byaj )η˙ j j

T j

T j

+ (ya Ayaj )ηj = ya s

Recall that an eigenvector yaj satisfies T j

T j

ya Ayaj = −λj ya Byaj ⇒ λj = −

yaT Ayaj j

yaT Byaj j

=⇒ η˙ j − λj ηj =

yaT s j

yaT Byaj j | {z } Φj

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Space-State Formulation & Analysis of Viscous Damped Systems The Non-Homogeneous Case

η˙ j − λj ηj =

yaTj s yaTj Byaj | {z } Φj

Pre-multiply by e −λj t =⇒ e −λj t η˙ j − e −λj t λj ηj = e −λj t Φj d e −λj t ηj = e −λj t Φj dt Z t =⇒ ηj (t) = e λj t Φj (τ )e −λj τ dτ + ηj (0)e λj t | {z } 0 {z } homogeneous solution | =⇒

particular solution

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Space-State Formulation & Analysis of Viscous Damped Systems The Harmonic Case

s(t) = sa e

iωt

where

sa =

pT a

0T

T

⇒ Φj (t) =

yaT sa j

yaT Byaj

e

iωt

= Φaj e

iωt

j

After the transient response is damped out, the amplitude ηj (t) of the forced response becomes e

ηj (t) = amp

λj t

t

Z 0

Φaj e

(iω−λj )τ

dτ

=

e iωt iω − λj

! Φaj ,

?

?

∀t > t , t 6= 0

Recalling that λi = −ξi ωi ± iωid , it follows that the forced harmonic response of an underdamped system can be written as      n   ¯ yaT yaj yaT ya j ¯ X 1 1 j j iωt    + T s e r= d) d) a  yaT Byaj ¯ y B¯ y ξ ω + i(ω + ω ξ ω + i(ω − ω a j j j j a j j j j=1 j

j

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Space-State Formulation & Analysis of Viscous Damped Systems The Harmonic Case      n   ¯ yaj yaT yaj ¯ yaT X 1 1 j j iωt     r= + s e d) d) a T B¯  yaT Byaj ¯ y y ξ ω + i(ω + ω ξ ω + i(ω − ω a j j j j a j j j j=1 j

j

Recall that (summary for harmonic case) r=

q q˙

= ya e

iωt

=

za iωza

e

iωt

,

s = sa e

iωt

=

pa 0

e

iωt

And note that ya Byaj

=

T ¯ ya B¯ ya j

=

T j

j

T j

yaj ya sa

=

iωj zT a

¯ zT a

−iωj ¯ zT a

j

j

j

zaj T T = za Czaj + 2iωj za Mzaj = ρj j j iωj zaj ¯ zaj C M T T za M¯ zaj = ρ¯j =¯ za C¯ zaj − 2iωj ¯ j j −iωj ¯ zaj M 0 ! ¯ zT zaj ¯ aj T ¯ pa ya j ¯ ya s a = j −iωj ¯ zaj ¯ zT a

zT a

C M

j

zaj zT aj

!

iωj zaj zT a

pa ,

M 0

j

j

and za

=

2

(K − ω M + iωC)

−1

pa

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Space-State Formulation & Analysis of Viscous Damped Systems The Harmonic Case

It follows that   n   ¯ zaj zT zaj ¯ zT X aj aj 1 1 za = + p d) d) a  ρj ρ ¯ ξ ω + i(ω + ω ξ ω + i(ω − ω j j j j j j j j=1 and therefore the expansion of the admittance matrix in terms of the damped modes is given by

2

−1

(K − ω M + iωC)

=

n X k=1

(

1 ξk ωk + i(ω + ωkd )

!

T zT a za k

ρk

k

+

1 ξk ωk + i(ω − ωkd )

!

¯ zT zak ¯ a

)

k

ρ¯k

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