AA242B: MECHANICAL VIBRATIONS
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AA242B: MECHANICAL VIBRATIONS
Undamped Vibrations of n-DOF Systems
These slides are based on the recommended textbook: M. Géradin and D. Rixen, “Mechanical
Vibrations: Theory and Applications to Structural Dynamics,” Second Edition, Wiley, John &
Sons, Incorporated, ISBN-13:9780471975465
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AA242B: MECHANICAL VIBRATIONS
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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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AA242B: MECHANICAL VIBRATIONS
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Linear Vibrations
Equilibrium configuration
qs (t) = qs (0)
q̇s (t) =
0
s = 1, · · · , n
(1)
Recall the Lagrange equations of motion
d ∂T
∂T
∂V
∂D
−
+
−
−
+ Qs (t) = 0
dt ∂ q̇s
∂qs
∂qs
∂ q̇s
where T = T0 + T1 + T2
Recall the generalized gyroscopic forces
Fs = −
n
n
X
X
∂ 2 T1
∂T1
q̇r +
=
∂ q̇s ∂qr
∂qs
r =1
r =1
∂ 2 T1
∂ 2 T1
−
∂qs ∂ q̇r
∂qr ∂ q̇s
!
q̇r ,
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 ∂T2
∂T2
∂V ?
∂D
∂ ∂T1
−
= Qs (t) −
−
+ Fs −
dt ∂ q̇s
∂qs
∂qs
∂ q̇s
∂t ∂ q̇s
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AA242B: MECHANICAL VIBRATIONS
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Linear Vibrations
Recall that
T0 (q, t)
=
D
=
2
N
3
∂Uik
1 XX
(q, t)
mk
2 k=1 i=1
∂t
N Z v (q̇)
X
k
Ck fk (γ)dγ
k=1
(transport kinetic energy)
(dissipation function)
0
From the Lagrange equations of motion
d ∂T2
∂T2
∂V ?
∂D
∂ ∂T1
−
= Qs (t) −
−
+ Fs −
dt ∂ q̇s
∂qs
∂qs
∂ q̇s
∂t ∂ q̇s
it follows that an equilibrium configuration exists if and only if
∂V ?
0 = Qs (t) −
⇒ Qs (t) and V ? are function of q only and not
∂qs
explicitly dependent on time ⇒ Qs (t) = 0 and V ? = V ? (q)
At equilibrium
Qs (t) = 0
and
∂V ?
∂(V − T0 )
=
= 0, s = 1, · · · , n
∂qs
∂qs
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AA242B: MECHANICAL VIBRATIONS
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Linear Vibrations
Free-Vibrations About a Stable Equilibrium Position
Consider first a system that does not undergo a transport or overall
motion ⇒ T = T2 (q̇)
The equilibrium position is then given by
Qs (t) = 0
and
∂V
= 0,
∂qs
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 deviation
from equilibrium)
Since the potential energy is defined only up to a constant, choose
this constant so that V(qs = 0) = 0
Now, suppose that a certain energy E(0) is initially given to the
system in equilibrium
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AA242B: MECHANICAL VIBRATIONS
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Linear Vibrations
Free-Vibrations About a Stable Equilibrium Position
Definition: the equilibrium position (qs = 0, s = 1, · · · , n) is said to
be 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 relative
minimum
if E(0) is small enough, V(t) will be small enough ⇒ and therefore
deviations from the equilibrium position will be small enough
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AA242B: MECHANICAL VIBRATIONS
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Linear Vibrations
Free-Vibrations About a Stable Equilibrium Position
Linearization of T and V around an equilibrium position
∂V
(qs = 0,
= 0)
∂qs
actually, this means obtaining a quadratic form of T and V in q and
q̇ so that the corresponding generalized forces are linear
since qs (t) represent deviations from equilibrium, V can be expanded
as follows
n
n
n
X
1 X X ∂2V
∂V
|q=0 qs +
|q=0 qs qr + O(q3 )
V(q) = V(0) +
∂q
2
∂q
s
s ∂qr
s=1
s=1 r =1
=
V(0) +
n
n
1 X X ∂2V
|q=0 qs qr + O(q3 )
2 s=1 r =1 ∂qs ∂qr
since the potential energy is defined only up to a constant, if this
constant is chosen so that V(0) = 0, then a second-order
approximation of V(q) is given by
V(q) =
n
n
1 X X ∂2V
|q=0 qs qr ,
2 s=1 r =1 ∂qs ∂qr
for q 6= 0
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AA242B: MECHANICAL VIBRATIONS
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Linear Vibrations
Free-Vibrations About a Stable Equilibrium Position
Stiffness matrix
let K = [ksr ] where ksr = krs =
=⇒ V(q) =
∂2V
|q=0
∂qs ∂qr
1 T
q Kq > 0,
2
for q 6= 0
K is symmetric positive (semi-) definite
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AA242B: MECHANICAL VIBRATIONS
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Linear Vibrations
Free-Vibrations About a Stable Equilibrium Position
Recall that
n
n
N
3
1 XXXX
∂Uik ∂Uik
T2 =
mk
q̇s q̇r
2 s=1 r =1
∂qs ∂qr
(relative kinetic energy)
k=1 i=1
T2 (q, q̇)
n
n
X
X
∂T2
∂T2
|q=0,q̇=0 qs +
|q=0,q̇=0 q̇s
∂qs
∂ q̇s
s=1
s=1
=
T2 (0, 0) +
+
n
n
n X
n
X
∂ 2 T2
1 X X ∂ 2 T2
|q=0,q̇=0 qs qr +
|q=0,q̇=0 qs q̇r
2 s=1 r =1 ∂qs ∂qr
∂q
s ∂ q̇r
s=1 r =1
+
n
n
1 X X ∂ 2 T2
3
3
|q=0,q̇=0 q̇s q̇r + O(q , q̇ )
2 s=1 r =1 ∂ q̇s ∂ q̇r
=
n
n
1 X X ∂ 2 T2
3
3
|q=0,q̇=0 q̇s q̇r + O(q , q̇ )
2 s=1 r =1 ∂ q̇s ∂ q̇r
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AA242B: MECHANICAL VIBRATIONS
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Linear Vibrations
Free-Vibrations About a Stable Equilibrium Position
Hence, a second-order approximation of T2 (q̇) is given by
T2 (q̇) =
where"
1 T
q̇ Mq̇ > 0,
2
for q̇ 6= 0
N
3
X
X
∂ 2 T2
∂Uik
∂Uik
M = msr = mrs =
|q=0 =
mk
|q=0
|q=0
∂ q̇s ∂ q̇r
∂qs
∂qr
i=1
k=1
is the mass matrix and is symmetric positive definite
#
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AA242B: MECHANICAL VIBRATIONS
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Linear Vibrations
Free-Vibrations About a Stable Equilibrium Position
Free-vibrations about a stable equilibrium position of a conservative
system that does not undergo a transport or overall motion
(T0 = T1 = 0)
∂T2
∂V
d ∂T2
−
=−
dt ∂ q̇s
∂qs
∂qs
=⇒
d
(Mq̇) − 0 = −Kq
dt
=⇒ Mq̈ + Kq = 0
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AA242B: MECHANICAL VIBRATIONS
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Linear Vibrations
Free-Vibrations About an Equilibrium Configuration
Consider next the more general case of a system in steady motion (a
transported system) whose equilibrium configuration defined by
∂V ?
∂(V − T0 )
=
= 0,
∂qs
∂qs
s = 1, · · · , n
corresponds
to the balance of forces
deriving from a potential
∂V
∂T0
and centrifugal forces
∂qs
∂qs
It is an equilibrium configuration in the sense that q̇s — which
represent here the generalized velocities relative to a steady motion
— are zero but the system is not idle
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AA242B: MECHANICAL VIBRATIONS
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Linear Vibrations
Free-Vibrations About an Equilibrium Configuration
Linearizations
effective potential energy V ? ⇒ effective stiffness matrix K?
1 T ?
q K q > 0,
for q 6= 0
2
2
∂ T0
?
?
where K = ksr = ksr −
|q=0
∂qs ∂qr
V ? (q) =
mutual kinetic energy
T1
=
n
n X
N X
3
X
X
∂Uik
∂Uik
∂T1
mk
q̇s =
q̇s
(q)
∂t
∂q
∂ q̇s
s
s=1
s=1
i=1
k=1
=
n
X
n
q̇s
s=1
n X
n
X
X ∂ 2 T1
∂T1
(0) +
|q=0 qr + O(q2 )
∂ q̇s
∂
q̇
s qr
r =1
!
∂ 2 T1
qr |q=0 q̇s qr ⇒ T1 (q, q̇) = q̇T Fq
∂
q̇
s
s=1 r =1
∂ 2 T1
where F = fsr =
|q=0
∂ q̇s ∂qr
≈
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Linear Vibrations
Free-Vibrations About an Equilibrium Configuration
Equations of free-vibration around an equilibrium configuration
the equilibrium configuration generated by a steady motion remains
stable as long as V ? = V − T0 ≥ 0
this corresponds to the fact that K? remains positive definite
in the neighborhood of such a configuration, the equations of motion
(for a conservative system undergoing transport or overall motion)
are
∂V ?
∂D
∂ ∂T1
d ∂T2
∂T2
= Qs (t) −
−
+Fs −
−
dt ∂ q̇s
∂qs
∂ q̇s
∂t ∂ q̇s
| {z } ∂qs
|{z}
|{z}
|
{z
}
0
0
where Fs =
n X
r =1
0
∂ 2 T1
∂ 2 T1
−
∂qs ∂ q̇r
∂qr ∂ q̇s
0
q̇r = (FT − F)q̇
Mq̈ + Gq̇ + K? q = 0
where G = F − FT = −GT is the gyroscopic coupling matrix
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AA242B: MECHANICAL VIBRATIONS
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Linear Vibrations
Free-Vibrations About an Equilibrium Configuration
Example
X = (a + x) cos Ωt
V =
1 2
kx
2
T0 =
Y = (a + x) sin Ωt
1 2
2
Ω m(a + x)
2
T1 = 0
2
2
2
2
2
v = Ẋ + Ẏ = (a + x) Ω + ẋ
T2 =
1
2
mẋ
2
V
?
=
2
1 2
1 2
2
kx − Ω m(a + x)
2
2
equilibrium configuration
∂V ?
Ω2 ma
= 0 =⇒ kx − Ω2 m(a + x) = 0 =⇒ xeq =
∂x
k − Ω2 m
the system becomes unstable for Ω2 =
k? =
k
m
∂2V ?
k
= k − Ω2 m ⇒ system is unstable for Ω2 ≥
∂x 2
m
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AA242B: MECHANICAL VIBRATIONS
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Natural Vibration Modes
Free-vibration equations: Mq̈ + Kq = 0
q(t) = qa e iωt ⇒ (K − ω 2 M)qa = 0 ⇒ det (K − ω 2 M) = 0
If the system has n degrees of freedom (dofs), M and K are n × n
matrices ⇒ n eigenpairs (ωi2 , qai )
Rigid body mode(s): ωj2 = 0 ⇒ Kqaj = 0
1
For a rigid body mode, V(qaj ) = qT
Kqaj = 0
2 aj
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AA242B: MECHANICAL VIBRATIONS
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Orthogonality of Natural Vibration Modes
Distinct Frequencies
Consider two distinct eigenpairs (ωi2 , qai ) and (ωj2 , qaj )
T 2
qT
aj Kqai = qaj ωi Mqai
(2)
qT
ai Kqaj
(3)
=
2
qT
ai ωj Mqaj
Because M and K are symmetric
(2) − (3)T ⇒ 0 = (ωi2 − ωj2 )qaj T Mqai
since ωi2 6= ωj2 ⇒ qaj T Mqai = 0
and qaj T Kqai = 0
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Orthogonality of Natural Vibration Modes
Distinct Frequencies
Physical interpretation of the orthogonality conditions
qaj T Mqai = 0 ⇒ qaj T ωi2 Mqai = (ωi2 Mqai )T qaj = 0
which implies that the virtual work produced by the inertia forces of
mode 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 of
mode i during a virtual displacement prescribed by mode j is zero
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AA242B: MECHANICAL VIBRATIONS
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Orthogonality of Natural Vibration Modes
Distinct Frequencies
Rayleigh quotient
Kqai = ωi2 Mqai ⇒ qai T Kqai = ωi2 qai T Mqai ⇒ ωi2 =
qai T Kqai
γi
=
qai T Mqai
µi
γi = generalized stiffness coefficient of mode i (measures the
contribution of mode i to the elastic deformation energy)
µi = generalized mass coefficient of mode i (measures the
contribution 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
T
qaj Kqai
=
δij
=
ωi2 δij
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Orthogonality of Natural Vibration Modes
Degeneracy Theorem
What happens if a multiple circular frequency is encountered?
Theorem: to a multiple root ωp2 of the system
Kx = ω 2 Mx
corresponds a number of linearly independent eigenvectors {qai }
equal to the root multiplicity
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AA242B: MECHANICAL VIBRATIONS
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Modal Superposition Analysis
n-dof system: M ∈ Rn×n , K ∈ Rn , and q ∈ Rn
Coupled system of ordinary differential equations

 Mq̈ + Kq = 0
q(0) = q0

q̇(0) = q̇0
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AA242B: MECHANICAL VIBRATIONS
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Modal Superposition Analysis
Natural vibration modes (eigenmodes)
Kqai = ωi2 Mqai ,
Q = [qa1
i = 1, · · · , n
T
Q KQ =
qa2 · · · qan ] ⇒
QT MQ =
 2

ω1


..
Ω2 = 

.
2
ωn
Ω2
I
Truncated eigenbasis
Q r = qa 1
qa 2
···
qa r ,
QT
r KQr
QT
r MQr
r << n ⇒


2
Ωr = 

ω12
=
=
Ω2r
Ir
(reduced stiffness matrix)
(reduced mass matrix)

..
.
ωr2



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AA242B: MECHANICAL VIBRATIONS
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Modal Superposition Analysis
Modal superposition: q = Qr y =
r
P
yi qai where
i=1
y = [y1
y2
···
T
yr ] and yi is called the modal displacement
Substitute in Mq̈ + Kq = 0
=⇒ MQr ÿ + KQr y = 0
T
=⇒ QT
r MQr ÿ + Qr KQr y = 0
=⇒ Ir ÿ + Ω2r y = 0
Uncoupled differential equations (modal equations)
ÿi + ωi2 yi = 0,
i = 1, · · · r
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Modal Superposition Analysis
ÿi + ωi2 yi = 0,
i = 1, · · · r
Case 1: ωi2 = 0 (rigid body mode)
yi = ai t + bi
Case 2: ωj2 6= 0
yj = cj cos ωj t + dj sin ωj t
General case: rb rigid body modes
q=
rb
rX
−rb
X
(ai t + bi )qai +
(cj cos ωj t + dj sin ωj t)qaj
i=1
j=1
Initial conditions
q(0) = q0 = Qy(0)
and
q̇(0) = q̇0 = Qẏ(0)
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AA242B: MECHANICAL VIBRATIONS
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Modal Superposition Analysis
q0 = Qy(0)
and
T
i
q̇0 = Qẏ(0)
T
i
=⇒ qa Mq0 = qa MQy(0)
From the orthogonality properties of the natural mode shapes (eigenvectors) it follows that
T
i
T
i
T
i
T
i
qa Mq0 = qa MQy(0) = qa Mqai yi (0) = yi (0) ⇒ yi (0) = qa Mq(0)
T
Case 1: ωi2 = 0 (rigid body mode) ⇒ ai × 0 + bi = qT
a Mq(0) ⇒ bi = qa Mq(0)
i
i
T
Case 2: ωj2 6= 0 ⇒ cj × 1 + dj × 0 = qT
a Mq(0) ⇒ cj = qa Mq(0)
j
j
Case 1: ωi2 = 0 (rigid body mode) ⇒ ai = qT
a Mq̇(0)
i
Case 2: ωj2 6= 0 ⇒ dj = qT
a Mq̇(0)
j
Thus, the general solution is
q(t) =
rb h
r −rb i
X
X
sin ωj t
T
T
T
T
qa Mq̇(0) t + qa Mq(0) qai +
qa Mq(0) cos ωj t + qa Mq̇(0)
qa j
i
i
j
j
ωj
i=1
j=1
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Spectral Expansions
n
∀x ∈ R ,
x=
n
X
T
j
αs qas ⇒ qa Mx =
s=1
n
=⇒ ∀x ∈ R ,
x=
n
X
T
j
αs qa Mqas = αj
s=1
n n
n X
X
X
T
T
T
qas Mx qas =
qas qas Mx =
qas qas Mx =
n
X
T
(qas qas )M
s=1
s=1
s=1
⇒
n X
s=1
T
qas qas
!
x
M=I
s=1
This is the same result as QT MQ = I
A given load p can be expanded in terms of the inertia forces
generated by the eigenmodes, Mqaj as follows
p=
n
X
j=1
βj Mqaj ⇒ qT
ai p =
n
X
βj qT
ai Mqaj = βi
j=1
=⇒ βi = qT
ai p = modal participation factor → p =
n X
qT
p
Mqaj
aj
j=1
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Spectral Expansions
Recall that
n P
qa s q T
M=I
as
s=1
Hence, ∀A ∈ Rn ,
n
P
A=
s=1
A=M⇒M=
n
P
s=1
A=K⇒K=
n
P
s=1
Aqas qT
as M and A =
Mqas qT
as M =
Kqas qT
as M =
A = M−1 ⇒ M−1 =
n
P
s=1
A = K−1 ⇒ K−1 =
n
P
s=1
n
P
s=1
n
P
s=1
n
P
s=1
Mqas (Mqas )T
ωs2 Mqas qT
as M =
−1
⇒ M
qas qT
as MM
−1
=
qas qT
as MK
−1
=
qas qT
as MA
(because M is symmetric)
n
P
s=1
n
X
ωs2 Mqas (Mqas )T
T
qas qas
s=1
n
P
s=1
−1
qas (Mqas )T K−1 ⇒ K
=
n
X
qas qT
as
s=1
ωs2
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Forced Harmonic Response

 Mq̈ + Kq =
q(0) =

q̇(0) =
sa cos ωt
q0
q̇0
Solution can be decomposed as
q = qH (homogeneous) + qP (particular)
qp = qa cos ωt ⇒ qa = (K − ω 2 M)−1 sa where (K − ω 2 M)−1 is
called the admittance or dynamic influence matrix
The forced response is the part of the response that is synchronous
to the excitation — that is, qp
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AA242B: MECHANICAL VIBRATIONS
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Forced Harmonic Response
Rigid body modes:
∀qa ∈ Rn ,
qa =
uai
rb
P
i=1
r
b
i=1
αi uai +
n−rb
P
β j qa j
j=1
2
=⇒ sa = (K − ω M)qa =
rb
X
n−rb
2
αi (K − ω M)uai +
i=1
=⇒ sa = −
rb
X
j
=⇒ qa = −
rb
X
uT
a sa
i
i=1
ω2
ua i +
uT
a sa
j
ω2
2
βj (K − ω M)qaj
j=1
n−rb
2
αi ω Muai +
i=1
Premultiply by uT
a ⇒ αj = −
X
X
2
2
βj (ωj − ω )Mqaj
j=1
and premultiply by qT
a ⇒ βi =
i

n−rb
X
qT
a sa
j=1
(ωj2 − ω 2 )
j
qaj
= −
rb
X
ua i uT
a
i
i=1
ω2
qT
a sa
i
(ωi2
+
− ω2 )
n−rb
X
qaj qT
a
j=1
(ωj2 − ω 2 )
j

 sa
Since qa = (K − ω 2 M)−1 sa
2
=⇒ (K − ω M)
−1
=−
rb
n−rb q qT
X
aj a
1 X
j
T
ua i ua +
2 − ω2
2
i
ω i=1
ω
j
j=1
(4)
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AA242B: MECHANICAL VIBRATIONS
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Forced Harmonic Response
Which excitation sam will generate a harmonic response with an
amplitude corresponding to qam ?
qam
=
(K − ω 2 M)−1 sam ⇒ sam = (K − ω 2 M)qam
=⇒ sam
=
2
Kqam − ω 2 Mqam = (ωm
− ω 2 )Mqam
2
At resonance (ω 2 = ωm
) sam = 0 ⇒ no force is needed to maintain
qam once it is reached
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AA242B: MECHANICAL VIBRATIONS
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Forced Harmonic Response
The inverse of an admittance is an impedance
Z(ω 2 ) = (K − ω 2 M)
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AA242B: MECHANICAL VIBRATIONS
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Forced Harmonic Response
Application: substructuring (or domain decomposition)
the dynamical behavior of a substructure is described by its harmonic
response when forces are applied onto its interface boundaries
a 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 boundary
dofs q2 where the reaction forces are denoted here by g2
Z11 Z12
q1
0
=
Z21 Z22
q2
g2
=⇒ Z11 q1 + Z12 q2 = 0 ⇒ q1 = −Z−1
11 Z12 q2
?
=⇒ (Z22 − Z21 Z−1
11 Z12 )q2 = Z22 q2 = g2
?
Z22
is the “reduced” impedance (reduced to the boundary)
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AA242B: MECHANICAL VIBRATIONS
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Forced Harmonic Response
Spectral expansion of Z?
22
2
look at Z−1
11 and let (ω̃i , q̃ai ) denote the n1 eigenpairs of the associated dynamical
n1
q̃aj q̃T
X
aj
−1
subsystem: from (4), it follows that Z11
=
ω̃j2 − ω 2
j=1
apply twice the relation
ω2
1
1
= 2 + 2 2
ω̃j2 − ω 2
ω̃j
ω̃j (ω̃j − ω 2 )
−1
−1
=⇒ Z11 = K11 +ω
2
n1
X
q̃aj q̃T
a
j=1
ω̃j2 (ω̃j2 − ω 2 )
j
−1
= K11 +ω
2
n1 q̃ q̃T
X
aj a
j
j=1
+ω
ω̃j4
4
n1
X
q̃aj q̃T
a
j=1
ω̃j4 (ω̃j2 − ω 2 )
j
owing to the M11 -orthonormality of the modes and the spectral expansion of K−1
11 ,
the above expression can further be written as
−1
Z11
?
=⇒ Z22
−1
2
−1
−1
=
K11 + ω K11 M11 K11 + ω
=
K22 − K21 K11 K12
−
n1
X
q̃aj q̃T
a
j
ω̃j4 (ω̃j2
j=1
− ω2 )
−1
−1
2
−1
−1
−1
ω [M22 − M21 K11 K12 − K21 K11 M12 + K21 K11 M11 K11 K12 ]
n1
−
4
ω
4
X [(K21 −
i=1
ω̃i2 M21 )q̃ai ][(K21 −
ω̃i4 (ω̃i2 − ω 2 )
ω̃i2 M21 )q̃ai )]T
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AA242B: MECHANICAL VIBRATIONS
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Forced Harmonic Response
?
Z22
=
K22 − K21 K−1
11 K12
−1
−1
−1
−ω 2 [M22 − M21 K−1
11 K12 − K21 K11 M12 + K21 K11 M11 K11 K12 ]
n1
X
[(K21 − ω̃i2 M21 )q̃ai ][(K21 − ω̃i2 M21 )q̃ai )]T
−ω 4
ω̃i4 (ω̃i2 − ω 2 )
i=1
The first term K22 − K21 K−1
11 K12 represents the stiffness of the
statically condensed system
The second term
−1
−1
−1
M22 − M21 K−1
11 K12 − K21 K11 M12 + K21 K11 M11 K11 K12 represents
the mass of the subsystem statically condensed on the boundary
The last term represents the contribution of the subsystem
eigenmodes since it is generated by q̃ai q̃T
ai
(K21 − ω̃i2 M21 )q̃ai is the dynamic reaction on the boundary
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AA242B: MECHANICAL VIBRATIONS
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Response to External Loading

 Mq̈ + Kq =
q(0) =

q̇(0) =
p(t)
q0
q̇0
General approach
consider the simpler case where there is no rigid body mode ⇒
eigenmodes (qai , ωi2 ), ωi2 6= 0, i = 1, · · · , n
n
P
modal superposition: q = Qy =
yi qai
i=1
substitute in equations of dynamic equilibrium
=⇒ MQÿ + KQy = p(t) ⇒ QT MQÿ + QT KQy = QT p(t)
modal equations ÿi + ωi2 yi = qTai p(t),
i = 1, · · · , n
yi (t) depend on two constants that can be obtained from the initial
conditions
q(0) = Qy(0), q̇(0) = Qẏ(0)
⇒
yi (0), ẏi (0)
orthogonality conditions
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AA242B: MECHANICAL VIBRATIONS
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Response to External Loading
Response to an impulsive force
k
m
impulsive force f (t): force whose amplitude could be infinitely large
but which acts for a very short
Z duration of time
spring-mass system: m, k, ω 2 =
τ +
magnitude of impulse: I =
f (t)dt
τ
impulsive force = I δ(t) where δ isZthe “delta” function
centered at t = 0 and satisfying
δ(t)dt = 1
0
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AA242B: MECHANICAL VIBRATIONS
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Response to External Loading
Response to an impulsive force (continue)
dynamic equilibrium
Z τ +
Z τ +
Z τ +
d u̇
m
dt + k
udt =
f (t)dt = I
dt
τ
τ
τ
assume thatZat t = τ , the system isZat rest (u(τ ) = 0 and u̇(τ ) = 0)
τ +
τ +
üdt = A and
udt = A2 /6
ü = A/ ⇒
τ
hence
τ +
Z
m
τ
d u̇
dt
dt
τ
I
I
⇒ u̇(τ + ) =
m
m
=
I ⇒ m∆u̇ = I ⇒ ∆u̇ = u̇(τ + ) − u̇(τ ) =
=
0 ⇒ ∆u = 0 ⇒ u(τ + ) − u(τ ) = 0 ⇒ u(τ + ) = 0
τ +
Z
u̇dt
τ
the above equations provide initial conditions for the free-vibrations
that start at the end of the impulsive load
=⇒ u(t) =
I
sin ωt
mω
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AA242B: MECHANICAL VIBRATIONS
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Response to External Loading
Response to an impulsive force (continue)
linear system ⇒ superposition principle
du =
f (τ )dτ
1
sin ω(t − τ ) ⇒ u(t) =
mω
mω
Z
t
f (τ ) sin ω(t − τ )dτ
0
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AA242B: MECHANICAL VIBRATIONS
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Response to External Loading
Time-integration of the normal equations
let pi (t) = qTai p(t) = i-th modal participation factor
modal or normal equation: ÿi + ωi2 yi = pi (t)
yi (t) = yiH (t) + yiP (t), yiH = Ai cos ωi t + Bi sin ωi t
the particular solution yiP (t) depends on the form of pi (t)
however, the general form of a particular solution that satisfies the
rest initial conditions is given by the Duhamel’s integral
Z t
1
yiP (t) =
pi (τ ) sin ωi (t − τ )dτ
ωi 0
complete solution
yi (t) = Ai cos ωi t + Bi sin ωi t +
1
ωi
Z
t
pi (t) sin ωi (t − τ )dτ
0
ẏi (0)
and yi (0) and ẏi (0) can be determined from
ωi
the initial conditions q(0) and q̇(0) and the orthogonality conditions
Ai = yi (0), Bi =
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AA242B: MECHANICAL VIBRATIONS
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Response to External Loading
Response truncation and mode displacement method
computational efficiency ⇒ q = Qr y =
r
P
r n
yi qai ,
i=1
what is the effect of modal truncation?
consider the case where p(t) =
g
×
static load
φ(t)
amplification
distribution
factor
for a system initially at rest (q(0) = 0 and q̇(0) = 0)
yi (0) = qTai Mq(0) = 0 and ẏi (0) = qTai Mq̇(0) = 0 ⇒ Ai = Bi = 0
=⇒ yi (t) =
1
ωi
t
Z
=⇒ q(t) =
pi (τ ) sin ωi (t−τ )dτ =
0
r
X
i=1
"
#
qai qTai g
spatial factor
qTai g
ωi
t
Z
φ(τ ) sin ωi (t − τ )dτ
0

1

ωi

t
Z
0

φ(τ ) sin ωi (t − τ )dτ 
temporal factor
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AA242B: MECHANICAL VIBRATIONS
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Response to External Loading
Response truncation and mode displacement method (continue)
general solution for restrained structure initially at rest


#
"
Z t
r
X
1


qai qTai g 
=⇒ q(t) =
φ(τ ) sin ωi (t − τ )dτ 
ωi 0
i=1 spatial factor
temporal factor θi (t)
truncated response is accurate if neglected terms are small, which is
true if:
qT
ai g is small for i = r + 1, · · · , n ⇒ g is well approximated in the
range
Z tof Qr
1
φ(τ ) sin ωj (t − τ )dτ is small for j > r , which depends on the
ωj 0
frequency content of φ(t)
φ(t) = 1
⇒
φ(t) = sin ωt
⇒
1 − cos ωi t
→ 0 for large circular frequencies
ωi
ωi sin ωt − ω sin ωi t
θi (t) =
ωi (ωi2 − ω 2 )
θi (t) =
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AA242B: MECHANICAL VIBRATIONS
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Response to External Loading
Mode acceleration method
Mq̈ + Kq = p(t) ⇒ Kq = p(t) − Mq̈
apply truncated modal representation to the acceleration
q(t) = Qr y(t) ⇒ q̈(t) = Qr ÿ(t)
=⇒ Kq = p(t) −
r
X
Mqai ÿi =⇒ q = K−1 p(t) −
i=1
r
X
qai
ÿi
ωi2
i=1
recall that
ÿi + ωi2 yi = qTai p(t)
and that for a system initially at rest
Z t
1
yi (t) =
qTai p(τ ) sin ωi (t − τ )dτ
ωi 0
Rt
(5) and (6) ⇒ ÿi (t) = qTai p(t) − ωi 0 p(τ ) sin ωi (t − τ )dτ
(5)
(6)
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AA242B: MECHANICAL VIBRATIONS
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Response to External Loading
Mode acceleration method (continue)
substitute in q(t) = K−1 p(t) −
r
X
qai
ÿi
ωi2
i=1
!
Z
r
r
X
X
qai qTai
qai qTai t
−1
p(t)
p(τ ) sin ωi (t−τ )dτ + K −
=⇒ q(t) =
ωi
ωi2
0
i=1
i=1
recall the spectral expansion K−1 =
n q qT
P
ai ai
2
i=1 ωi
Z
r
X
qai qTai t
p(τ ) sin ωi (t − τ )dτ +
=⇒ q(t) =
ωi
0
i=1
n
X
qai qTai
ωi2
i=r +1
!
p(t)
which shows that the mode acceleration method complements the
truncated mode displacement solution with the missing terms using
the modal expansion of the static response
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AA242B: MECHANICAL VIBRATIONS
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Response to External Loading
Direct time-integration methods for

 Mq̈ + Kq
q(0)

q̇(0)
solving
= p(t)
= q0
= q̇0
will be covered towards the end of this course
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AA242B: MECHANICAL VIBRATIONS
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Mechanical Systems Excited Through Support Motion
The general case
q=
M11
M21
M12
M22
q1
q2
q̈1
q̈2
where q2 is prescribed
+
K11
K21
K12
K22
q1
q2
=
0
r2 (t)
The first equation gives
M11 q̈1 + K11 q1 = −K12 q2 − M12 q̈2 ⇒ q1 (t)
Substitute in second equation
=⇒ r2 (t) = K21 q1 + M21 q̈1 + K22 q2 + M22 q̈2
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AA242B: MECHANICAL VIBRATIONS
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Mechanical Systems Excited Through Support Motion
Quasi-static response of q1
0 + K11 q1 = −K12 q2 − 0
−1
=⇒ qqs
1 = −K11 K12 q2 = Sq2
Decompose q1 = qqs
1 + z1 = Sq2 + z1
q1
I
=⇒ q(t) =
=
q2
0
S
I
z1
q2
where z1 represents the sole dynamics part of the response
Substitute in the first dynamic equation and exploit above results
=⇒ M11 Sq̈2 + M11 z̈1 + K11 qqs
1 + K11 z1 = −K12 q2 − M12 q̈2
M11 z̈1 + K11 z1 = g1 (t)
=⇒
g1 (t) = −M11 q̈qs
1 − M12 q̈2 = −(M11 S + M12 )q̈2
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AA242B: MECHANICAL VIBRATIONS
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Mechanical Systems Excited Through Support Motion
Consider next the system fixed to the ground and solve the
corresponding EVP
e = [· · · q̃a · · · ]
K11 x = ω 2 M11 x ⇒ Q
i
e
=⇒ z1 (t) = Qη(t)
Solve M11 z̈1 + K11 z1 = g1 (t) for z1 (t) using the modal
superposition technique
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AA242B: MECHANICAL VIBRATIONS
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Mechanical Systems Excited Through Support Motion
Case of a global support acceleration q̈2 (t) = u2 φ(t), where
u = [u1 u2 ]T denotes a rigid body mode
decomposition of the solution into a rigid body motion and a relative
displacement y
u1
ÿ1
q̈1
=
φ(t) +
q̈ = q̈rb + ÿ ⇒
q̈2
u2
0
u1 = Su2 ⇒ g1 (t) = −(M11 Su2 + M12 u2 )φ(t)
=⇒ g1 (t) = −(M11 u1 + M12 u2 )φ(t)
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AA242B: MECHANICAL VIBRATIONS
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Mechanical Systems Excited Through Support Motion
Method of additional masses (approximate method)
suppose the system is subjected not to a specified q2 (t) but to an
imposed
0
f(t)
suppose that masses associated with q2 are increased to M22 + M?22
then
K11 K12
q1
0
q̈1
M11
M12
+
=
q̈2
K21 K22
q2
f(t)
M21 M22 + M?22
by elimination one obtains
q̈2 = (M22 + M?22 )−1 (f(t) − K22 q2 − K21 q1 − M21 q̈1 ) and
?
M11 q̈1 + K11 q1 = −K12 q2 − M12 (M22 + M22 )
−1
(f(t) − K22 q2 − K21 q1 − M21 q̈1 )
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AA242B: MECHANICAL VIBRATIONS
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Mechanical Systems Excited Through Support Motion
Method of additional masses (continue)
therefore
{M11
?
−1
−
M12 (M22 + M22 )
=
−K12 q2 − M12 (M22 + M22 )
M21 } q̈1 +
?
−1
n
o
? −1
K11 − M12 (M22 + M22 ) K21 q1
(f(t) − K22 q2 )
now if (M22 + M?22 )−1 → 0 and if f(t) = (M22 + M?22 )q̈2 , one obtains
M11 q̈1 +K11 q1 = −K12 q2 −M12 q̈2 =⇒ M11 q̈1 +M12 q̈2 +K11 q1 +K12 q2 = 0
which is the same equation as in the general case of excitation
through support motion
⇒ for M?22 very large, the response of the system with ground
motion is the same as the response of the system with added lumped
mass M?22 and the forcing function
0
0
≈
(M22 + M?22 )q̈2
M?22 q̈2
where q̈2 is given ⇒ apply same solution method as for problems of
response to external loading
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