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AA242B: MECHANICAL VIBRATIONS 1 / 50

AA242B: MECHANICAL VIBRATIONSUndamped Vibrations of n-DOF Systems

These slides are based on the recommended textbook: M. Geradin and D. Rixen, “MechanicalVibrations: Theory and Applications to Structural Dynamics,” Second Edition, Wiley, John &

Sons, Incorporated, ISBN-13:9780471975465

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Outline

1 Linear Vibrations

2 Natural Vibration Modes

3 Orthogonality of Natural Vibration Modes

4 Modal Superposition Analysis

5 Spectral Expansions

6 Forced Harmonic Response

7 Response to External Loading

8 Mechanical Systems Excited Through Support Motion

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Linear Vibrations

Equilibrium configuration

qs(t) = qs(0)qs(t) = 0

s = 1, · · · , n (1)

Recall the Lagrange equations of motion

− d

dt

(∂T∂qs

)+∂T∂qs− ∂V∂qs− ∂D

∂qs+ Qs(t) = 0

where T = T0 + T1 + T2

Recall the generalized gyroscopic forces

Fs = −n∑

r=1

∂2T1

∂qs∂qrqr +

∂T1

∂qs=

n∑r=1

(∂2T1

∂qs∂qr−

∂2T1

∂qr∂qs

)qr , s = 1, · · · , n

Definition: the effective potential energy is defined as V? = V − T0

The Lagrange equations of motion can be re-written as

d

dt

(∂T2

∂qs

)− ∂T2

∂qs= Qs(t)− ∂V?

∂qs− ∂D

∂qs+ Fs −

∂

∂t

(∂T1

∂qs

)3 / 50


Linear Vibrations

Recall that

T0(q, t) =1

2

N∑k=1

3∑i=1

mk

(∂Uik

∂t(q, t)

)2

(transport kinetic energy)

D =N∑

k=1

∫ vk (q)

0

Ck fk (γ)dγ (dissipation function)

From the Lagrange equations of motion

d

dt

(∂T2

∂qs

)− ∂T2

∂qs= Qs(t)− ∂V?

∂qs− ∂D

∂qs+ Fs −

∂

∂t

(∂T1

∂qs

)it follows that an equilibrium configuration exists if and only if

0 = Qs(t)− ∂V?

∂qs⇒ Qs(t) and V? are function of q only and not

explicitly dependent on time ⇒ Qs(t) = 0 and V? = V?(q)

At equilibrium

Qs(t) = 0 and∂V?

∂qs=∂(V − T0)

∂qs= 0, s = 1, · · · , n

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Linear Vibrations

Free-Vibrations About a Stable Equilibrium Position

Consider first a system that does not undergo a transport or overallmotion ⇒ T = T2(q)

The equilibrium position is then given by

Qs(t) = 0 and∂V∂qs

= 0, s = 1, · · · , n

Assume next that this system is conservative ⇒ E = T + V = cst

Shift the origin of the generalized coordinates so that at equilibrium,qs = 0, s = 1, · · · , n (in which case the qs represent the deviationfrom equilibrium)

Since the potential energy is defined only up to a constant, choosethis constant so that V(qs = 0) = 0

Now, suppose that a certain energy E(0) is initially given to thesystem in equilibrium

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Linear Vibrations


Definition: the equilibrium position (qs = 0, s = 1, · · · , n) is said tobe stable if

∃ E? / ∀ E(0) < E?, T (t) ≤ E(0)

Consequences

T + V = E = cst = E(0)⇒ V(t) = E(0)− T (t)) ≥ 0

at a stable equilibrium position, the potential energy is at a relativeminimumif E(0) is small enough, V(t) will be small enough ⇒ and thereforedeviations from the equilibrium position will be small enough

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Linear Vibrations


Linearization of T and V around an equilibrium position

(qs = 0,∂V∂qs

= 0)

actually, this means obtaining a quadratic form of T and V in q andq so that the corresponding generalized forces are linearsince qs (t) represent deviations from equilibrium, V can be expandedas follows

V(q) = V(0) +n∑

s=1

∂V∂qs|q=0 qs +

1

2

n∑s=1

n∑r=1

∂2V∂qs∂qr

|q=0 qsqr +O(q3)

= V(0) +1

2

n∑s=1

n∑r=1

∂2V∂qs∂qr

|q=0 qsqr +O(q3)

since the potential energy is defined only up to a constant, if thisconstant is chosen so that V(0) = 0, then a second-orderapproximation of V(q) is given by

V(q) =1

2

n∑s=1

n∑r=1

∂2V∂qs∂qr

|q=0 qsqr , for q 6= 0

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Linear Vibrations


Stiffness matrix

let K = [ksr ] where ksr = krs =∂2V∂qs∂qr

|q=0

=⇒ V(q) =1

2qT Kq > 0, for q 6= 0

K is symmetric positive (semi-) definite

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Linear Vibrations


Recall that

T2 =1

2

n∑s=1

n∑r=1

N∑k=1

3∑i=1

mk∂Uik

∂qs

∂Uik

∂qrqs qr (relative kinetic energy)

T2(q, q) = T2(0, 0) +n∑

s=1

∂T2

∂qs|q=0,q=0 qs +

n∑s=1

∂T2

∂qs|q=0,q=0 qs

+1

2

n∑s=1

n∑r=1

∂2T2

∂qs∂qr|q=0,q=0 qs qr +

n∑s=1

n∑r=1

∂2T2

∂qs∂qr|q=0,q=0 qs qr

+1

2

n∑s=1

n∑r=1

∂2T2

∂qs∂qr|q=0,q=0 qs qr +O(q3

, q3)

=1

2

n∑s=1

n∑r=1

∂2T2

∂qs∂qr|q=0,q=0 qs qr +O(q3

, q3)

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Linear Vibrations


Hence, a second-order approximation of T2(q) is given by

T2(q) =1

2qT Mq > 0, for q 6= 0

where

M =

[msr = mrs =

∂2T2

∂qs∂qr|q=0 =

N∑k=1

mk

3∑i=1

∂Uik

∂qs|q=0

∂Uik

∂qr|q=0

]is the mass matrix and is symmetric positive definite

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Linear Vibrations


Free-vibrations about a stable equilibrium position of a conservativesystem that does not undergo a transport or overall motion(T0 = T1 = 0)

d

dt

(∂T2

∂qs

)− ∂T2

∂qs= − ∂V

∂qs

=⇒ d

dt(Mq)− 0 = −Kq

=⇒ Mq + Kq = 0

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Linear Vibrations

Free-Vibrations About an Equilibrium Configuration

Consider next the more general case of a system in steady motion (atransported system) whose equilibrium configuration defined by

∂V?

∂qs=∂(V − T0)

∂qs= 0, s = 1, · · · , n

corresponds to the balance of forces deriving from a potential(∂V∂qs

)and centrifugal forces

(∂T0

∂qs

)It is an equilibrium configuration in the sense that qs — whichrepresent here the generalized velocities relative to a steady motion— are zero but the system is not idle

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Linear Vibrations


Linearizationseffective potential energy V? ⇒ effective stiffness matrix K?

V?(q) =1

2qT K?q > 0, for q 6= 0

where K? =

[k?sr = ksr −

(∂2T0

∂qs∂qr

)|q=0

]mutual kinetic energy

T1 =n∑

s=1

N∑k=1

3∑i=1

∂Uik

∂tmk

∂Uik

∂qsqs =

n∑s=1

qs∂T1

∂qs(q)

=n∑

s=1

qs

(∂T1

∂qs(0) +

n∑r=1

∂2T1

∂qsqr|q=0 qr +O(q2)

)

≈n∑

s=1

n∑r=1

∂2T1

∂qsqr |q=0 qsqr ⇒ T1(q, q) = qT Fq

where F =

[fsr =

∂2T1

∂qs∂qr|q=0

]13 / 50


Linear Vibrations


Equations of free-vibration around an equilibrium configuration

the equilibrium configuration generated by a steady motion remainsstable as long as V? = V − T0 ≥ 0this corresponds to the fact that K? remains positive definitein the neighborhood of such a configuration, the equations of motion(for a conservative system undergoing transport or overall motion)are

d

dt

(∂T2

∂qs

)− ∂T2

∂qs︸︷︷︸0

= Qs (t)︸︷︷︸0

−∂V?

∂qs− ∂D

∂qs︸︷︷︸0

+Fs −∂

∂t

(∂T1

∂qs

)︸︷︷︸

0

where Fs =n∑

r=1

(∂2T1

∂qs∂qr− ∂2T1

∂qr∂qs

)qr = (FT − F)q

Mq + Gq + K?q = 0

where G = F− FT = −GT is the gyroscopic coupling matrix

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Linear Vibrations


Example

X = (a + x) cos Ωt Y = (a + x) sin Ωt v 2 = X 2 + Y 2 = (a + x)2Ω2 + x2

V =1

2kx2 T0 =

1

2Ω2m(a + x)2 T1 = 0 T2 =

1

2mx2 V? =

1

2kx2 −

1

2Ω2m(a + x)2

equilibrium configuration

∂V?

∂x= 0 =⇒ kx − Ω2m(a + x) = 0 =⇒ xeq =

Ω2ma

k − Ω2m

the system becomes unstable for Ω2 =k

m

k? =∂2V?

∂x2= k − Ω2m⇒ system is unstable for Ω2 ≥ k

m

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Natural Vibration Modes

Free-vibration equations: Mq + Kq = 0

q(t) = qaeiωt ⇒ (K− ω2M)qa = 0⇒ det (K− ω2M) = 0

If the system has n degrees of freedom (dofs), M and K are n × nmatrices ⇒ n eigenpairs (ω2

i ,qai )

Rigid body mode(s): ω2j = 0⇒ Kqaj = 0

For a rigid body mode, V(qaj ) =1

2qT

ajKqaj = 0

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Orthogonality of Natural Vibration Modes

Distinct Frequencies

Consider two distinct eigenpairs (ω2i ,qai ) and (ω2

j ,qaj )

qTaj

Kqai = qTajω2

i Mqai (2)

qTai

Kqaj = qTaiω2

j Mqaj (3)

Because M and K are symmetric

(2)− (3)T ⇒ 0 = (ω2i − ω2

j )qaj

T Mqai

since ω2i 6= ω2

j ⇒ qaj

T Mqai = 0 and qaj

T Kqai = 0

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Physical interpretation of the orthogonality conditions

qaj

T Mqai = 0⇒ qaj

T(ω2

i Mqai

)= (ω2

i Mqai )T qaj = 0

which implies that the virtual work produced by the inertia forces ofmode i during a virtual displacement prescribed by mode j is zero

qaj

T Kqai = 0⇒ (Kqai )T qaj = 0

which implies that the virtual work produced by the elastic forces ofmode i during a virtual displacement prescribed by mode j is zero

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Rayleigh quotient

Kqai = ω2i Mqai ⇒ qai

T Kqai = ω2i qai

T Mqai ⇒ ω2i =

qaiT Kqai

qaiT Mqai

=γi

µi

γi = generalized stiffness coefficient of mode i (measures thecontribution of mode i to the elastic deformation energy)

µi = generalized mass coefficient of mode i (measures thecontribution of mode i to the kinetic energy)

Since the amplitude of qai is determined up to a factor only ⇒ γi

and µi are determined up to a constant factor only

Mass normalization

qaj

T Mqai = δij

qaj

T Kqai = ω2i δij

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Degeneracy Theorem

What happens if a multiple circular frequency is encountered?

Theorem: to a multiple root ω2p of the system

Kx = ω2Mx

corresponds a number of linearly independent eigenvectors qaiequal to the root multiplicity

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Modal Superposition Analysis

n-dof system: M ∈ Rn×n, K ∈ Rn, and q ∈ Rn

Coupled system of ordinary differential equations Mq + Kq = 0q(0) = q0

q(0) = q0

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Natural vibration modes (eigenmodes)

Kqai = ω2i Mqai , i = 1, · · · , n

Q = [qa1 qa2 · · · qan ]⇒

QT KQ = Ω2

QT MQ = I

Ω2 =

ω21

. . .

ω2n

Truncated eigenbasis

Qr =[qa1

qa2· · · qar

], r << n ⇒

QT

r KQr = Ω2r (reduced stiffness matrix)

QTr MQr = Ir (reduced mass matrix)

Ω2r =

ω2

1

. . .

ω2r

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Modal superposition: q = Qr y =r∑

i=1

yi qai where

y = [y1 y2 · · · yr ]T and yi is called the modal displacement

Substitute in Mq + Kq = 0

=⇒MQr y + KQr y = 0

=⇒ QTr MQr y + QT

r KQr y = 0

=⇒ Ir y + Ω2r y = 0

Uncoupled differential equations (modal equations)

yi + ω2i yi = 0, i = 1, · · · r

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yi + ω2i yi = 0, i = 1, · · · r

Case 1: ω2i = 0 (rigid body mode)

yi = ai t + bi

Case 2: ω2j 6= 0

yj = cj cosωj t + dj sinωj t

General case: rb rigid body modes

q =

rb∑i=1

(ai t + bi )qai +

r−rb∑j=1

(cj cosωj t + dj sinωj t)qaj

Initial conditions

q(0) = q0 = Qy(0) and q(0) = q0 = Qy(0)

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q0 = Qy(0) and q0 = Qy(0)

=⇒ qTai

Mq0 = qTai

MQy(0)

From the orthogonality properties of the natural mode shapes (eigenvectors) it follows that

qTai

Mq0 = qTai

MQy(0) = qTai

Mqaiyi (0) = yi (0)⇒ yi (0) = qT

aiMq(0)

Case 1: ω2i = 0 (rigid body mode) ⇒ ai × 0 + bi = qT

aiMq(0)⇒ bi = qT

aiMq(0)

Case 2: ω2j 6= 0 ⇒ cj × 1 + dj × 0 = qT

ajMq(0)⇒ cj = qT

ajMq(0)

Case 1: ω2i = 0 (rigid body mode) ⇒ ai = qT

aiMq(0)

Case 2: ω2j 6= 0 ⇒ dj = qT

ajMq(0)

Thus, the general solution is

q(t) =

rb∑i=1

([qT

aiMq(0)

]t + qT

aiMq(0)

)qai

+

r−rb∑j=1

(qT

ajMq(0) cosωj t + qT

ajMq(0)

sinωj t

ωj

)qaj

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Spectral Expansions

∀x ∈ Rn, x =

n∑s=1

αs qas ⇒ qTaj

Mx =n∑

s=1

αs qTaj

Mqas = αj

=⇒ ∀x ∈ Rn, x =

n∑s=1

(qT

asMx)

qas =n∑

s=1

qas

(qT

asMx)

=n∑

s=1

(qas qT

as

)Mx =

(n∑

s=1

(qas qTas

)M

)x

⇒n∑

s=1

(qas qT

as

)M = I

This is the same result as QT MQ = I

A given load p can be expanded in terms of the inertia forcesgenerated by the eigenmodes, Mqaj as follows

p =n∑

j=1

βj Mqaj ⇒ qTai

p =n∑

j=1

βj qTai

Mqaj = βi

=⇒ βi = qTai

p = modal participation factor→ p =n∑

j=1

(qT

ajp)

Mqaj

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Spectral Expansions

Recall thatn∑

s=1

(qas qT

as

)M = I

Hence, ∀A ∈ Rn, A =n∑

s=1Aqas qT

asM and A =

n∑s=1

qas qTas

MA

A = M⇒ M =n∑

s=1Mqas qT

asM =

n∑s=1

Mqas (Mqas )T (because M is symmetric)

A = K⇒ K =n∑

s=1Kqas qT

asM =

n∑s=1ω2

s Mqas qTas

M =n∑

s=1ω2

s Mqas (Mqas )T

A = M−1 ⇒ M−1 =n∑

s=1qas qT

asMM−1 ⇒ M−1 =

n∑s=1

qas qTas

A = K−1 ⇒ K−1 =n∑

s=1qas qT

asMK−1 =

n∑s=1

qas (Mqas )T K−1 ⇒ K−1 =n∑

s=1

qas qTas

ω2s

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Forced Harmonic Response

Mq + Kq = sa cosωtq(0) = q0

q(0) = q0

Solution can be decomposed asq = qH (homogeneous) + qP (particular)

qp = qa cosωt ⇒ qa = (K− ω2M)−1sa where (K− ω2M)−1 iscalled the admittance or dynamic influence matrix

The forced response is the part of the response that is synchronousto the excitation — that is, qp

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Rigid body modes:

uai

rbi=1

∀qa ∈ Rn, qa =rb∑

i=1

αi uai+

n−rb∑j=1

βj qaj

=⇒ sa = (K− ω2M)qa =

rb∑i=1

αi (K− ω2M)uai+

n−rb∑j=1

βj (K− ω2M)qaj

=⇒ sa = −rb∑

i=1

αiω2Muai

+

n−rb∑j=1

βj (ω2j − ω

2)Mqaj

Premultiply by uTaj⇒ αj = −

uTaj

sa

ω2and premultiply by qT

ai⇒ βi =

qTai

sa

(ω2i − ω2)

=⇒ qa = −rb∑

i=1

uTai

sa

ω2uai

+

n−rb∑j=1

qTaj

sa

(ω2j − ω2)

qaj=

− rb∑i=1

uaiuT

ai

ω2+

n−rb∑j=1

qajqT

aj

(ω2j − ω2)

sa

Since qa = (K− ω2M)−1sa

=⇒ (K− ω2M)−1 = −1

ω2

rb∑i=1

uaiuT

ai+

n−rb∑j=1

qajqT

aj

ω2j − ω2

(4)

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Which excitation sam will generate a harmonic response with anamplitude corresponding to qam ?

qam = (K− ω2M)−1sam ⇒ sam = (K− ω2M)qam

=⇒ sam = Kqam − ω2Mqam = (ω2m − ω2)Mqam

At resonance (ω2 = ω2m) sam = 0⇒ no force is needed to maintain

qam once it is reached

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The inverse of an admittance is an impedance

Z(ω2) = (K− ω2M)

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Application: substructuring (or domain decomposition)

the dynamical behavior of a substructure is described by its harmonicresponse when forces are applied onto its interface boundariesa subsystem is typically described by K and M, has n1 free dofs q1,and is connected to the rest of the system by n2 = n − n1 boundarydofs q2 where the reaction forces are denoted here by g2(

Z11 Z12

Z21 Z22

)(q1

q2

)=

(0g2

)=⇒ Z11q1 + Z12q2 = 0⇒ q1 = −Z−1

11 Z12q2

=⇒ (Z22 − Z21Z−111 Z12)q2 = Z?22q2 = g2

Z?22 is the “reduced” impedance (reduced to the boundary)

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Spectral expansion of Z?22

look at Z−111 and let (ω2

i , qai) denote the n1 eigenpairs of the associated dynamical

subsystem: from (4), it follows that Z−111 =

n1∑j=1

qajqT

aj

ω2j − ω2

apply twice the relation

1

ω2j − ω2

=1

ω2j

+ω2

ω2j (ω2

j − ω2)

=⇒ Z−111 = K−1

11 +ω2n1∑

j=1

qajqT

aj

ω2j (ω2

j − ω2)= K−1

11 +ω2n1∑

j=1

qajqT

aj

ω4j

+ω4n1∑

j=1

qajqT

aj

ω4j (ω2

j − ω2)

owing to the M11-orthonormality of the modes and the spectral expansion of K−111 ,

the above expression can further be written as

Z−111 = K−1

11 + ω2K−1

11 M11K−111 + ω

4n1∑

j=1

qajqT

aj

ω4j (ω2

j − ω2)

=⇒ Z?22 = K22 − K21K−111 K12

− ω2[M22 −M21K−1

11 K12 − K21K−111 M12 + K21K−1

11 M11K−111 K12]

− ω4

n1∑i=1

[(K21 − ω2i M21)qai

][(K21 − ω2i M21)qai

)]T

ω4i (ω2

i − ω2)

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Z?22 = K22 −K21K−111 K12

−ω2[M22 −M21K−111 K12 −K21K−1

11 M12 + K21K−111 M11K−1

11 K12]

−ω4n1∑

i=1

[(K21 − ω2i M21)qai ][(K21 − ω2

i M21)qai )]T

ω4i (ω2

i − ω2)

The first term K22 −K21K−111 K12 represents the stiffness of the

statically condensed system

The second termM22 −M21K−1

11 K12 −K21K−111 M12 + K21K−1

11 M11K−111 K12 represents

the mass of the subsystem statically condensed on the boundary

The last term represents the contribution of the subsystemeigenmodes since it is generated by qai q

Tai

(K21 − ω2i M21)qai is the dynamic reaction on the boundary

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Response to External Loading

Mq + Kq = p(t)q(0) = q0

q(0) = q0

General approach

consider the simpler case where there is no rigid body mode ⇒eigenmodes (qai , ω

2i ), ω2

i 6= 0, i = 1, · · · , n

modal superposition: q = Qy =n∑

i=1

yi qai

substitute in equations of dynamic equilibrium

=⇒MQy + KQy = p(t)⇒ QT MQy + QT KQy = QT p(t)

modal equations yi + ω2i yi = qT

aip(t), i = 1, · · · , n

yi (t) depend on two constants that can be obtained from the initialconditionsq(0) = Qy(0), q(0) = Qy(0) ⇒

orthogonality conditionsyi (0), yi (0)

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Response to an impulsive force

spring-mass system: m, k, ω2 =k

mimpulsive force f (t): force whose amplitude could be infinitely largebut which acts for a very short duration of time

magnitude of impulse: I =

∫ τ+ε

τ

f (t)dt

impulsive force = Iδ(t) where δ is the “delta” function

centered at t = 0 and satisfying

∫ ε

0

δ(t)dt = 1

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Response to an impulsive force (continue)

dynamic equilibrium

m

∫ τ+ε

τ

du

dtdt + k

∫ τ+ε

τ

udt =

∫ τ+ε

τ

f (t)dt = I

assume that at t = τ , the system is at rest (u(τ) = 0 and u(τ) = 0)

u = A/ε⇒∫ τ+ε

τ

udt = A and

∫ τ+ε

τ

udt = Aε2/6

hence

m

∫ τ+ε

τ

du

dtdt = I ⇒ m∆u = I ⇒ ∆u = u(τ + ε)− u(τ) =

I

m⇒ u(τ + ε) =

I

m∫ τ+ε

τ

udt = 0⇒ ∆u = 0⇒ u(τ + ε)− u(τ) = 0⇒ u(τ + ε) = 0

the above equations provide initial conditions for the free-vibrationsthat start at the end of the impulsive load

=⇒ u(t) =I

mωsinωt

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Response to an impulsive force (continue)

linear system ⇒ superposition principle

du =f (τ)dτ

mωsinω(t − τ)⇒ u(t) =

1

mω

∫ t

0

f (τ) sinω(t − τ)dτ

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Time-integration of the normal equations

let pi (t) = qTai

p(t) = i-th modal participation factormodal or normal equation: yi + ω2

i yi = pi (t)yi (t) = yH

i (t) + yPi (t), yH

i = Ai cosωi t + Bi sinωi tthe particular solution yP

i (t) depends on the form of pi (t)however, the general form of a particular solution that satisfies therest initial conditions is given by the Duhamel’s integral

yPi (t) =

1

ωi

∫ t

0

pi (τ) sinωi (t − τ)dτ

complete solution

yi (t) = Ai cosωi t + Bi sinωi t +1

ωi

∫ t

0

pi (t) sinωi (t − τ)dτ

Ai = yi (0),Bi =yi (0)

ωiand yi (0) and yi (0) can be determined from

the initial conditions q(0) and q(0) and the orthogonality conditions

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Response truncation and mode displacement method

computational efficiency ⇒ q = Qr y =r∑

i=1

yi qai , r n

what is the effect of modal truncation?consider the case where p(t) = g

static loaddistribution

× φ(t)amplification

factor

for a system initially at rest (q(0) = 0 and q(0) = 0)yi (0) = qT

aiMq(0) = 0 and yi (0) = qT

aiMq(0) = 0 ⇒ Ai = Bi = 0

=⇒ yi (t) =1

ωi

∫ t

0

pi (τ) sinωi (t−τ)dτ =qT

aig

ωi

∫ t

0

φ(τ) sinωi (t − τ)dτ

=⇒ q(t) =r∑

i=1

[qai q

Tai

gspatial factor

] 1

ωi

∫ t

0


temporal factor

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Response truncation and mode displacement method (continue)

general solution for restrained structure initially at rest

=⇒ q(t) =r∑

i=1

[qai q

Tai

gspatial factor

] 1

ωi

∫ t

0


temporal factor θi (t)

truncated response is accurate if neglected terms are small, which istrue if:

qTai

g is small for i = r + 1, · · · , n ⇒ g is well approximated in therange of Qr

1

ωj

∫ t

0φ(τ) sinωj (t − τ)dτ is small for j > r , which depends on the

frequency content of φ(t)

φ(t) = 1 ⇒ θi (t) =1− cosωi t

ωi→ 0 for large circular frequencies

φ(t) = sinωt ⇒ θi (t) =ωi sinωt − ω sinωi t

ωi (ω2i − ω2)

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Mode acceleration method

Mq + Kq = p(t)⇒ Kq = p(t)−Mq

apply truncated modal representation to the acceleration

q(t) = Qr y(t)⇒ q(t) = Qr y(t)

=⇒ Kq = p(t)−r∑

i=1

Mqai yi =⇒ q = K−1p(t)−r∑

i=1

qai

ω2i

yi

recall thatyi + ω2

i yi = qTai

p(t) (5)

and that for a system initially at rest

yi (t) =1

ωi

∫ t

0

qTai

p(τ) sinωi (t − τ)dτ (6)

(5) and (6) ⇒ yi (t) = qTai

(p(t)− ωi

∫ t

0p(τ) sinωi (t − τ)dτ

)42 / 50



Mode acceleration method (continue)

substitute in q(t) = K−1p(t)−r∑

i=1

qai

ω2i

yi

=⇒ q(t) =r∑

i=1

qai qTai

ωi

∫ t

0

p(τ) sinωi (t−τ)dτ+

(K−1 −

r∑i=1

qai qTai

ω2i

)p(t)

recall the spectral expansion K−1 =n∑

i=1

qai qTai

ω2i

=⇒ q(t) =r∑

i=1

qai qTai

ωi

∫ t

0

p(τ) sinωi (t − τ)dτ +

(n∑

i=r+1

qai qTai

ω2i

)p(t)

which shows that the mode acceleration method complements thetruncated mode displacement solution with the missing terms usingthe modal expansion of the static response

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Direct time-integration methods for solving Mq + Kq = p(t)q(0) = q0

q(0) = q0

will be covered towards the end of this course

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Mechanical Systems Excited Through Support Motion

The general case

q =

(q1

q2

)where q2 is prescribed(

M11 M12

M21 M22

)(q1

q2

)+

(K11 K12

K21 K22

)(q1

q2

)=

(0

r2(t)

)The first equation gives

M11q1 + K11q1 = −K12q2 −M12q2 ⇒ q1(t)

Substitute in second equation

=⇒ r2(t) = K21q1 + M21q1 + K22q2 + M22q2

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Quasi-static response of q1

0 + K11q1 = −K12q2 − 0

=⇒ qqs1 = −K−1

11 K12q2 = Sq2

Decompose q1 = qqs1 + z1 = Sq2 + z1

=⇒ q(t) =

(q1

q2

)=

(I S0 I

)(z1

q2

)where z1 represents the sole dynamics part of the response

Substitute in the first dynamic equation and exploit above results

=⇒M11Sq2 + M11z1 + K11qqs1 + K11z1 = −K12q2 −M12q2

=⇒

M11z1 + K11z1 = g1(t)g1(t) = −M11qqs

1 −M12q2 = −(M11S + M12)q2

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Consider next the system fixed to the ground and solve thecorresponding EVP

K11x = ω2M11x⇒ Q = [· · · qai · · · ]

=⇒ z1(t) = Qη(t)

Solve M11z1 + K11z1 = g1(t) for z1(t) using the modalsuperposition technique

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Case of a global support acceleration q2(t) = u2φ(t), whereu = [u1 u2]T denotes a rigid body mode

decomposition of the solution into a rigid body motion and a relativedisplacement y

q = qrb + y⇒[

q1

q2

]=

[u1

u2

]φ(t) +

[y1

0

]u1 = Su2 ⇒ g1(t) = −(M11Su2 + M12u2)φ(t)

=⇒ g1(t) = −(M11u1 + M12u2)φ(t)

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Method of additional masses (approximate method)

suppose the system is subjected not to a specified q2(t) but to animposed (

0f(t)

)suppose that masses associated with q2 are increased to M22 + M?

22

then(M11 M12

M21 M22 + M?22

)(q1

q2

)+

(K11 K12

K21 K22

)(q1

q2

)=

(0

f(t)

)by elimination one obtainsq2 = (M22 + M?

22)−1 (f(t)−K22q2 −K21q1 −M21q1) and

M11q1 + K11q1 = −K12q2 −M12(M22 + M?22)−1 (f(t)− K22q2 − K21q1 −M21q1)

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Method of additional masses (continue)therefore

M11 − M12(M22 + M?22)−1M21 q1 +

K11 −M12(M22 + M?

22)−1K21

q1

= −K12q2 −M12(M22 + M?22)−1 (f(t)− K22q2)

now if (M22 + M?22)−1 → 0 and if f(t) = (M22 + M?

22)q2, one obtains

M11q1+K11q1 = −K12q2−M12q2 =⇒M11q1+M12q2+K11q1+K12q2 = 0

which is the same equation as in the general case of excitationthrough support motion⇒ for M?

22 very large, the response of the system with groundmotion is the same as the response of the system with added lumpedmass M?

22 and the forcing function(0

(M22 + M?22)q2

)≈(

0M?

22q2

)where q2 is given ⇒ apply same solution method as for problems ofresponse to external loading

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AA242B: MECHANICAL VIBRATIONS - Stanford University · PDF fileAA242B: MECHANICAL VIBRATIONS...

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