AA242B: MECHANICAL VIBRATIONS 1 / 50 AA242B: MECHANICAL VIBRATIONS Undamped 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 1 / 50
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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
1 / 50
AA242B: MECHANICAL VIBRATIONS 2 / 50
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
2 / 50
AA242B: MECHANICAL VIBRATIONS 3 / 50
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
AA242B: MECHANICAL VIBRATIONS 4 / 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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AA242B: MECHANICAL VIBRATIONS 5 / 50
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
5 / 50
AA242B: MECHANICAL VIBRATIONS 6 / 50
Linear Vibrations
Free-Vibrations About a Stable Equilibrium Position
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
6 / 50
AA242B: MECHANICAL VIBRATIONS 7 / 50
Linear Vibrations
Free-Vibrations About a Stable Equilibrium Position
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
7 / 50
AA242B: MECHANICAL VIBRATIONS 8 / 50
Linear Vibrations
Free-Vibrations About a Stable Equilibrium Position
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
8 / 50
AA242B: MECHANICAL VIBRATIONS 9 / 50
Linear Vibrations
Free-Vibrations About a Stable Equilibrium Position
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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AA242B: MECHANICAL VIBRATIONS 10 / 50
Linear Vibrations
Free-Vibrations About a Stable Equilibrium Position
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
10 / 50
AA242B: MECHANICAL VIBRATIONS 11 / 50
Linear Vibrations
Free-Vibrations About a Stable Equilibrium Position
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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AA242B: MECHANICAL VIBRATIONS 12 / 50
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
12 / 50
AA242B: MECHANICAL VIBRATIONS 13 / 50
Linear Vibrations
Free-Vibrations About an Equilibrium Configuration
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
AA242B: MECHANICAL VIBRATIONS 14 / 50
Linear Vibrations
Free-Vibrations About an Equilibrium Configuration
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
14 / 50
AA242B: MECHANICAL VIBRATIONS 15 / 50
Linear Vibrations
Free-Vibrations About an Equilibrium Configuration
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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AA242B: MECHANICAL VIBRATIONS 16 / 50
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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AA242B: MECHANICAL VIBRATIONS 17 / 50
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
17 / 50
AA242B: MECHANICAL VIBRATIONS 18 / 50
Orthogonality of Natural Vibration Modes
Distinct Frequencies
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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AA242B: MECHANICAL VIBRATIONS 19 / 50
Orthogonality of Natural Vibration Modes
Distinct Frequencies
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
19 / 50
AA242B: MECHANICAL VIBRATIONS 20 / 50
Orthogonality of Natural Vibration Modes
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
20 / 50
AA242B: MECHANICAL VIBRATIONS 21 / 50
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
21 / 50
AA242B: MECHANICAL VIBRATIONS 22 / 50
Modal Superposition Analysis
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
22 / 50
AA242B: MECHANICAL VIBRATIONS 23 / 50
Modal Superposition Analysis
Modal superposition: q = Qr y =r∑
i=1
yi qai where
y = [y1 y2 · · · yr ]T and yi is called the modal displacement
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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AA242B: MECHANICAL VIBRATIONS 33 / 50
Forced Harmonic Response
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)
33 / 50
AA242B: MECHANICAL VIBRATIONS 34 / 50
Forced Harmonic Response
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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AA242B: MECHANICAL VIBRATIONS 35 / 50
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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AA242B: MECHANICAL VIBRATIONS 36 / 50
Response to External Loading
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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AA242B: MECHANICAL VIBRATIONS 37 / 50
Response to External Loading
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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AA242B: MECHANICAL VIBRATIONS 38 / 50
Response to External Loading
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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AA242B: MECHANICAL VIBRATIONS 39 / 50
Response to External Loading
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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AA242B: MECHANICAL VIBRATIONS 40 / 50
Response to External Loading
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
φ(τ) sinωi (t − τ)dτ
temporal factor
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AA242B: MECHANICAL VIBRATIONS 41 / 50
Response to External Loading
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
φ(τ) sinωi (t − τ)dτ
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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AA242B: MECHANICAL VIBRATIONS 42 / 50
Response to External Loading
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
AA242B: MECHANICAL VIBRATIONS 43 / 50
Response to External Loading
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
43 / 50
AA242B: MECHANICAL VIBRATIONS 44 / 50
Response to External Loading
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
44 / 50
AA242B: MECHANICAL VIBRATIONS 45 / 50
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
45 / 50
AA242B: MECHANICAL VIBRATIONS 46 / 50
Mechanical Systems Excited Through Support Motion
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