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18 April 2005
Dynamics of Non-viscouslyDamped Distributed Parameter
SystemsS Adhikari, Y Lei and M I Friswell
Department of Aerospace Engineering, University of Bristol, Bristol, U.K.
URL: http://www.aer.bris.ac.uk/contact/academic/adhikari/home.html
Non-viscously Damped Systems – p.1/35
18 April 2005
Outline of the Presentation
Introduction
Models of damping
Equation of motion
Outline of the solution method
Incorporation of boundary conditions
Numerical examples & results
Conclusions & future works
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Introduction (1)
Modelling and analysis of damping properties arenot as advanced as mass and stiffness properties.The reasons:
by contrast with inertia and stiffness forces, it isnot in general clear which variables are relevantto determine the damping forces
the spatial location of the damping sources aregenerally unclear - often the structural joints aremore responsible for the energy dissipationthan the (solid) material
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18 April 2005
Introduction (2)
the functional form of the damping model isdifficult to establish experimentally, and finally
even if one manages to address the previousissues, what parameters should be used in achosen model is still very much an openproblem
The ‘solution’ over the past 100 years:
Use viscous damping model
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18 April 2005
Viscous Damping Model
Introduced by Lord Rayleigh in 1877
instantaneous generalized velocities are theonly relevant variables that determine damping
However,
viscous damping is not the only damping modelwithin the scope of linear analysis.
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18 April 2005
Non-viscous Damping Model
Any causal model which makes the energydissipation functional non-negative is a possiblecandidate for a damping model
non-viscous damping models in general havemore parameters and therefore are more likelyto have a better match with experimentalmeasurements
Question:
What non-viscous damping model should beused?
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Equation of Motion
ρ(r)u(r, t) + L1u(r, t) + L2u(r, t) = p(r, t) (1)
specified in some domain D with homogeneouslinear boundary condition of the form
Mu(r, t) = 0; r ∈ Γ
specified on some boundary surface Γ.u(r, t): displacement variableρ(r): mass distribution of the systemp(r, t): distributed time-varying forcing functionL2: spatial self-adjoint stiffness operatorM: linear operator acting on the boundary
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18 April 2005
The Damping Operator
The damping operator L1 can be written in the form
L1u(r, t) =
∫
D
∫ t
−∞
C1(r, ξ, t − τ)u(ξ, τ) dτ dξ (2)
where C1(r, ξ, t) is the kernel function.
The velocities u(ξ, τ) at different time instantsand spatial locations are coupled through thekernel function
Eq. (1) together with the damping operator (2)represents a partial integro-differential equation
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18 April 2005
The Damping Operator
Any function that makes the energy dissipationfunction
F(t) =1
2
∫
D
{∫
D
∫ t
−∞
C1(r, ξ, t − τ)
u(ξ, τ)dτ dξ} u(r, t) dr (3)
non-negative can be used as a kernel function.The main assumption:
the damping kernel function C1(r, ξ, t) isseparable in space and time
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18 April 2005
Viscous Damping
The kernel function is a delta function in both spaceand time:
C1(r, ξ, t − τ) = C(r)δ(r − ξ)δ(t − τ) (4)
the spatial delta function means that thedamping force is ‘locally reacting’ and the timedelta function implies that the force dependsonly on the instantaneous value of the motion
in general this represents the non-proportionalviscous damping model
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18 April 2005
Viscoelastic Damping
The kernel function is a delta function in space butdepends on the past time histories:
C1(r, ξ, t − τ) = C(r)g(t − τ)δ(r − ξ) (5)
Represents a locally reacting viscoelasticdamping model where the damping forcedepends on the past velocity time historiesthrough a convolution integral over the kernelfunction g(t)
g(t) is known as retardation function, heredityfunction or relaxation function
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18 April 2005
Non-local Viscous Damping
The kernel function is a delta function in time butdepends on the spatial distribution of the velocities:
C1(r, ξ, t − τ) = C(r)c(r − ξ)δ(t − τ) (6)
velocities at different points can affect thedamping force at a given point via a convolutionintegral
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18 April 2005
Non-local Viscoelastic Damping
This is the most general form of damping model
the only assumption is that the kernel functionis separable in space and time:
C1(r, ξ, t − τ) = C(r)c(r − ξ)g(t − τ) (7)
all the previous three damping models can beidentified as special cases of this model
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Parametrization of Models (1)
Plausible functional form of the kernel functions inspace and time is requiredRequirement:For a physically realistic model of damping
ℜ
[
G(ω)
∫
D
∫
D
C(r)c(r − ξ)U ∗(ξ, ω)U(r, ω) dξ dr
]
≥ 0
for all ω
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Non-viscous Damping Functions
Damping functions (in Laplace domain) Author, Year
G(s) =
Pnk=1
aks
s + bk
Biot (1955, 1958)
G(s) =E1sα − E0bsβ
1 + bsβBagley and Torvik (1983)
0 < α < 1, 0 < β < 1
sG(s) = G∞
"
1 +
Pk αk
s2 + 2ζkωks
s2 + 2ζkωks + ω2
k
#Golla and Hughes (1985)
and McTavish and Hughes (1993)
G(s) = 1 +
Pnk=1
∆ks
s + βk
Lesieutre and Mingori (1990)
G(s) = c1 − e−st0
st0Adhikari (1998)
G(s) = c1 + 2(st0/π)2 − e−st0
1 + 2(st0/π)2Adhikari (1998)
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18 April 2005
Parametrization of Models (2)
g(t) = g∞ µ exp(−µt) so that G(ω) = g∞ µ
iω+µ
c(r − ξ) = α2exp(−α|r − ξ|) and
C(r), g∞, µ and α are all positive
The damping force:
∫
D
∫ t
−∞
C(r) g∞ µ exp(−µ{t − τ})
α
2exp(−α|r − ξ|)u(ξ, τ) dξ dτ
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18 April 2005
Special Cases
if α → ∞, µ → ∞ one obtains the standardviscous model in (4)
if α → ∞ and µ is finite one obtains the localnon-viscous model in (5)
if α is finite but µ → ∞ one obtains thenon-local viscous damping model in (6)
if both α and µ are finite one obtains thenon-local viscoelastic damping model in (7)
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18 April 2005
Damped Euler-Bernoulli Beam������x L x R
x 1 x 2
x 0
Homogeneous Euler-Bernoulli beam withnon-viscous damping
Objectives:
To obtain eigenvalues and eigenvectors of thesystem
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18 April 2005
Equation of Motion (1)
Part within the damping patch:
EI∂4w(x, t)
∂x4+ ρA
∂2w(x, t)
∂t2+
∫ x2
x1
∫ t
−∞
α
2exp (−α |x − ξ|)
g∞µ exp (−µ(t − τ))∂w(ξ, t)
∂t
∣
∣
∣
∣
t=τ
dξdτ = f(x, t)
(8)
when x ∈ [x1, x2]
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18 April 2005
Equation of Motion (2)
Part outside the non-viscous damping patch:
EI∂4w(x, t)
∂x4+ρA
∂2w(x, t)
∂t2+C0
∂w(x, t)
∂t= f(x, t) (9)
when x ∈ (xL, x1) ∪ (x2, xR).
Appropriate boundary conditions must besatisfied at x = xL and at x = xR
relevant continuity conditions at the internalpoints x1 and x2 must be satisfied
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Outline of the Solution Method
Transform the equations into Laplace domain
differentiate with respect to the spatial variableto eliminate the spatial correlation terms(possible due to the exponential assumption)
express the BCs corresponding to the higherorder derivatives in terms of the known BCs
repeat the process for all three segments
merge the solutions from the three segments bymatching the displacements and theirderivatives at the interfaces
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18 April 2005
Eigensolutions of the Beam
The eigenvalues λj are the roots of
det[
M(s) exp(
Φ(s)(xL − x1))
T (x1, s)
+N(s) exp(
Φ(s)(xR − x2))
T (x2, s)]
= 0
The corresponding mode shapes are
ψj(x) =
exp(
Φ(λj)(x − x1))
T(x1, λj)u0(λj), xL ≤ x ≤ x1
T(x, λj)u0(λj), x1 ≤ x ≤ x2
exp(
Φ(λj)(x − x2))
T(x2, λj)u0(λj), x2 ≤ x ≤ xR
Non-viscously Damped Systems – p.23/35
18 April 2005
Boundary Conditions
The matrices M(s) and N(s) depend on the boundaryconditions:
Clamped-clamped (C-C):
M(s) =
I2×2 O2×2
O2×2 O2×2
, N(s) =
O2×2 O2×2
I2×2 O2×2
Free-Free (F-F):
M(s) =
O2×2 I2×2
O2×2 O2×2
, N(s) =
O2×2 O2×2
O2×2 I2×2
Non-viscously Damped Systems – p.24/35
18 April 2005
Example 1: The System
Part 1 Part 2
Damped beam with step variation in the system propertiesand pinned boundary conditions (Friswell and Lees, 2001)
Parameters Part 1 Part 2
Li 1m 2m
ρAi 10 kg/m 20 kg/m
ci 0 Ns/m2 10 Ns/m2
EIi 100 Nm2 100 Nm2
Non-viscously Damped Systems – p.25/35
18 April 2005
Example 1: Results
Proposed method Friswell and Lees (2001)-2.2552 ± 1.2711i -2.2552 ± 1.2711i-1.7936 ± 10.903i -1.7936 ± 10.903i-1.5741 ± 24.863i -1.5741 ± 24.863i-1.7876 ± 43.165i -1.7876 ± 43.165i-1.8781 ± 68.118i -1.8781 ± 68.118i-1.6984 ± 99.327i -1.6984 ± 99.327i-1.6775 ± 133.66i -1.6775 ± 133.66i-1.8549 ± 174.00i -1.8549 ± 174.00i
The first eight eigenvalues of the beam
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18 April 2005
Example 2: The System (1)
g ( t ) , c ( x )
C wR
K θ R
M wR
C θ R
K wR
x L x 1
x R x 2
C w2 g ( t ) C w1 g ( t )
Euler-Bernoulli beam with complex boundary conditions andmiddle supports
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18 April 2005
Example 2: The System (2)
The numerical values (in SI units) of the systemparameters are as follows:
Case 1: Local viscous damping:L=1, EI=1, m=16, MwR=4, KwR=8, CwR=4,g∞=1.6,g(t)=δ(t), c(x) = δ(x),CθR = KθR = Cw1 = Cw2=0
Case 2: Non-local non-viscous damping:L=1, EI=1, m=16, MwR=4, KwR=8, CwR=4, g∞=16,g(t) = µ exp (−µt),c(x) = α exp (−α |x|),KθR = 8, CθR = Cw1 = Cw2=4
Non-viscously Damped Systems – p.28/35
18 April 2005
Example 2: Results
jλj
Proposed method Yang and Wu (1997)
1 -0.2705 ± 1.1451i -0.2705 ± 1.1451i
2 -0.1357 ± 4.4930i -0.1357 ± 4.4930i
3 -0.0896 ± 13.2586i -0.0896 ± 13.2586i
4 -0.0727 ± 26.8877i -0.0727 ± 26.8877i
5 -0.0647 ± 45.4297i -0.0647 ± 45.4297i
6 -0.0602 ± 68.8941i -0.0602 ± 68.8941i
7 -0.0575 ± 97.2862i -0.0575 ± 97.2862i
8 -0.0557 ± 130.6088i -0.0557 ± 130.6088i
9 -0.0546 ± 168.8632i -0.0546 ± 168.8632i
10 -0.0537 ± 212.0505i -0.0537 ± 212.0505i
First ten eigenvalues of the beam for Case 1
Non-viscously Damped Systems – p.29/35
18 April 2005
Example 2: Results
jλj
µ = ∞, α = ∞ µ = 100, α = 10 µ = 100, α = 0.1 µ = 1, α = 10
1 -0.67921 ± 1.1780i -0.65439 ± 1.1966i -0.56874 ± 1.2444i -0.52895 ± 1.4404i
2 -0.98559 ± 6.4492i -0.92155 ± 6.4835i -0.61797 ± 6.4677i -0.49096 ± 6.5182i
3 -1.3833 ± 16.611i -1.2245 ± 16.723i -1.0592 ± 16.703i -0.61363 ± 16.656i
4 -1.4263 ± 31.584i -1.2455 ± 31.754i -1.1706 ± 31.727i -0.73244 ± 31.620i
5 -1.0714 ± 51.442i -0.90384 ± 51.521i -0.83720 ± 51.483i -0.74997 ± 51.456i
6 -1.0353 ± 76.208i -0.79867 ± 76.276i -0.74770 ± 76.237i -0.71290 ± 76.207i
7 -1.3874 ± 105.87i -0.92919 ± 106.11i -0.90512 ± 106.08i -0.71677 ± 105.88i
8 -1.4200 ± 140.46i -0.91144 ± 140.68i -0.89997 ± 140.67i -0.75157 ± 140.47i
9 -1.0745 ± 179.97i -0.78527 ± 180.03i -0.77680 ± 180.01i -0.75400 ± 179.98i
10 -1.0541 ± 224.42i -0.74574 ± 224.45i -0.74022 ± 224.44i -0.73084 ± 224.42i
First ten eigenvalues of the beam for Case 2
Non-viscously Damped Systems – p.30/35
18 April 2005
Example 2: Results
−0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 0.5−0.1
−0.08
−0.06
−0.04
−0.02
0
0.02
x (m)
ℜ(ψ j(x
)) Mode 1
Mode 2
Mode 3
Mode 4
µ = ∞, α= ∞
µ = 100, α= 10
µ = 100, α= 0.0
µ = 1, α= 10
Real parts of the first four modes for Case 2
Non-viscously Damped Systems – p.31/35
18 April 2005
Example 2: Results
−0.5 0 0.5−5
0
5
10
15x 10
−3
x (m)
ℑ(ψ 1(x)
)
Mode 1
−0.5 0 0.5−2
0
2
4
6x 10
−3
x (m)
ℑ(ψ 2(x)
)
Mode 2
−0.5 0 0.5−4
−2
0
2
4x 10
−4
x (m)
ℑ(ψ 3(x)
)
Mode 3
−0.5 0 0.5−2
−1
0
1
2x 10
−3
x (m)
ℑ(ψ 4(x)
)
Mode 4
µ = ∞, α= ∞µ = 100, α= 10
µ = 100, α= 0.0
µ = 1, α= 10
Imaginary parts of the first four modes for Case 2
Non-viscously Damped Systems – p.32/35
18 April 2005
Conclusions (1)
A method to obtain the natural frequencies andmode-shapes of Euler-Bernoulli beams withgeneral linear damping models has beenproposed
it is assumed that the damping force at a givenpoint in the beam depends on the past historyof velocities at different points via convolutionintegrals over exponentially decaying kernelfunctions
Non-viscously Damped Systems – p.33/35
18 April 2005
Conclusions (2)
conventional viscous and viscoelastic dampingmodels can be obtained as special cases of thisgeneral linear damping model
the choice of damping models effects theimaginary parts of the complex modes
future work will discuss computational issues,forced vibration problems and experimentalidentification of non-viscous damping models
Non-viscously Damped Systems – p.34/35
18 April 2005
Open Problems
To what extent different damping models with‘correct’ sets of parameters influence thedynamics?
which aspects of dynamic behavior are wronglypredicted by an incorrect damping model?
how to choose a damping model (not theparameters!) for a given system?
Non-viscously Damped Systems – p.35/35
References
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35-1
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35-2