Post on 17-Mar-2018
AA242B: MECHANICAL VIBRATIONS 1 / 29
AA242B: MECHANICAL VIBRATIONSApproximation of Continuous Systems by Displacement Methods
These slides are partially based on the recommended textbook: M. Geradin 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 2 / 29
Outline
1 The Rayleigh-Ritz Method
2 The Finite Element Method
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AA242B: MECHANICAL VIBRATIONS 3 / 29
The Rayleigh-Ritz Method
Choice of Approximation Functions
Rayleigh-Ritz approximation
ui (x1, x2, x3, t) =n∑
j=1
nij(x1, x2, x3)qj(t), i = 1, 2, 3
qj , j = 1, · · · , n are the generalized coordinatesvector form of “assumed modes”
nj(x1, x2, x3) =
n1j(x1, x2, x3)n2j(x1, x2, x3)n3j(x1, x2, x3)
u(x1, x2, x3, t) =n∑
j=1
nj(x1, x2, x3)qj(t)
admissibility conditions: C 0 continuity and satisfaction of theessential boundary conditions
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AA242B: MECHANICAL VIBRATIONS 4 / 29
The Rayleigh-Ritz Method
Discretization of the Displacement Variational Principle
Rayleigh-Ritz approximation in matrix form
u(x1, x2, x3, t) = N(x1, x2, x3)q(t)
N ∈ R3×n is the displacement interpolation matrix
[N(x1, x2, x3)]ij = nij(x1, x2, x3), i = 1, 2, 3, j = 1, · · · , n
q(t) =[q1 · · · qn
]Tis the vector of generalized coordinates
Strain componentsrecall the spatial differentiation operator
DT =
∂
∂x10 0 ∂
∂x20 ∂
∂x3
0 ∂∂x2
0 ∂∂x1
∂∂x3
0
0 0 ∂∂x3
0 ∂∂x2
∂∂x1
continuous strain vector
ε(x, t) = Du = DN(x1, x2, x3)q(t) = B(x1, x2, x3)q(t)
where (B(x1, x2, x3) = DN(x1, x2, x3)) ∈ R6×n is the straininterpolation matrix
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AA242B: MECHANICAL VIBRATIONS 5 / 29
The Rayleigh-Ritz Method
Discretization of the Displacement Variational Principle
Displacement variational principle
kinetic energy
u = Nq⇒ T =1
2
∫V
ρ (Nq)T NqdV =1
2qTMq
where M =
∫V
ρNTNdV is the symmetric positive definite mass
matrix of the discrete systemstrain energy
Vint =1
2
∫V
εTHεdV =1
2
∫V
(Bq)T HBqdV =1
2qTKq
where H is the Hooke matrix of elastic coefficients, and
K =
∫V
BTHBdV is the symmetric positive semi-definite stiffness
matrix of the discrete systemthe rigid body modes u are the solutions of Ku = 0 — they produceno strain energy
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AA242B: MECHANICAL VIBRATIONS 6 / 29
The Rayleigh-Ritz Method
Discretization of the Displacement Variational Principle
Displacement variational principle (continue)external potential energy
Vext = −∫Sσ
(Nq)T tdS −∫V
(Nq)T XdV = −qTg
where g(t) =
∫Sσ
NT t(t)dS +
∫V
NT X(t)dV is called the external
load factordiscretized displacement variational principle
δ
∫ t2
t1
{1
2qTMq−
(1
2qTKq− qTg
)}dt = 0
=⇒∫ t2
t1
{δqTMq−
(δqTKq− δqTg
)}dt = 0
=⇒[δqTMq
]t2t1−∫ t2
t1
δqT {Mq + Kq− g} dt = 0
recall that δq(t1) = δq(t2) = 0
=⇒ Mq + Kq = g(t)
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AA242B: MECHANICAL VIBRATIONS 7 / 29
The Rayleigh-Ritz Method
Computation of Eigensolutions by the Rayleigh-Ritz Method
Discretized eigenvalue problem
assume free vibrationsassume harmonic motion
Mq + Kq = 0 ⇒ Kqa = ω2Mqa
Theorem: Each eigenvalue ω2i resulting from the discretization of
the displacement variational principle by the Rayleigh-Ritz method isan upper bound to the corresponding exact eigenvalue ω2
i,e
ω2i,e ≤ ω2
i , i = 1, · · · , n
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AA242B: MECHANICAL VIBRATIONS 8 / 29
The Rayleigh-Ritz Method
Computation of Eigensolutions by the Rayleigh-Ritz Method
Eigenmodesonce the eigenvalues ω2
i are determined, the associated eigenmodesqai are obtained from the solution of
Kqai− ω2
i Mqai= 0
the corresponding approximate eigenmodes u(i)1 are given by
u(i) = N(x1, x2, x3)qai
orthogonality conditions
M and K symmetric⇒
{qTaiMqaj
= δij
qTaiKqaj
= δijω2j
assuming the eigenvectors are normalized with respect to Msubstitute the expressions for M and K in the above orthogonalityconditions
=⇒∫VρuT
(i)u(j)dV = δij∫V
(Du(i)
)TH(Du(j)
)dV = δijω
2i
1In this chapter, the subscript (i) is used to denote the i-th mode of u instead ofthe subscript i to avoid confusion with the i-th direction of this vector
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AA242B: MECHANICAL VIBRATIONS 9 / 29
The Rayleigh-Ritz Method
Computation of the Response to External Loading by the Rayleigh-Ritz Method
Response to external loading
modal superposition
q =n∑
s=1
ηs(t)qas
substitute in dynamic equations of equilibrium, premultiply by qTar ,
and exploit orthogonality conditions of the discrete eigenmodes
=⇒ ηr + ω2r ηr = qT
ar g = φr , r = 1, · · · , n
reconstruct the continuous solution as follows
u =n∑
s=1
ηs(t)uas
recall that g =∫SNT tdS +
∫VNT XdV
=⇒ φr = qTar g =
∫S
(Nqar )T tdS +
∫V
(Nqar )T XdV =
∫S
uTar tdS +
∫V
uTar XdV
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AA242B: MECHANICAL VIBRATIONS 10 / 29
The Rayleigh-Ritz Method
The Case of Prestressed Structures
Structure subjected to an initial stress field σ0ij
recall the additional term in the strain energy of the system
Vg =
∫V
σ0ijε
(2)ij dV
where ε(2)ij =
1
2
∂um∂xi
∂um∂xj
=1
2
∂nmk
∂xi
∂nml
∂xjqkql
=⇒ Vg =1
2qTKgq
where
[Kg ]kl =
∫V
σ0ij∂nmk
∂xi
∂nml
∂xjdV
dynamic equations of equilibrium
Mq + (K + Kg ) q = g(t)
associated eigenvalue problem
(K + Kg ) qa = ω2Mqa
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AA242B: MECHANICAL VIBRATIONS 11 / 29
The Rayleigh-Ritz Method
The Case of Prestressed Structures
Structure subjected to an initial stress field σ0ij (continue)
geometric stiffness driven by a parameter λ
σ0ij = λσ0?
ij ⇒ Kg = λK?g
parameterized (λ) eigenvalue problem(K + λK?
g
)q = ω2Mq
static case (ω = 0) (K + λiK
?g
)qi = 0
=⇒ eigenproblem where the eigenvalues λi correspond to criticalloadsthe eigenvalue λ1 with the smallest module yields the prestress statein which the system buckles
=⇒ σ0ijcr = λ1σ
0?ij
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AA242B: MECHANICAL VIBRATIONS 12 / 29
The Rayleigh-Ritz Method
The Case of Prestressed Structures
Application of Rayleigh-Ritz method to a continuous systemThe clamped-free uniform bar
u(x) =x
lq1 +
x2
l2q2
=⇒ q =
q1
q2
M = ml
13
14
14
15
K =EA
l
1 1
1 43
=⇒
ω21 = 2.49 EA
ml2> ω2
1,e = 2.46 EAml2
ω22 = 32.18 EA
ml2> ω2
2,e = 22.21 EAml2
u(x) =x
lq1 +
x2
l2q2 +
x3
l3q3
=⇒
ω2
1 = 2.47 EAml2
> ω21,e = 2.46 EA
ml2
ω22 = 23.4 EA
ml2> ω2
2,e = 22.21 EAml2
ω23 = 109.2 EA
ml2> ω2
3,e = 61.68 EAml2
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AA242B: MECHANICAL VIBRATIONS 13 / 29
The Finite Element Method
Finite Element Approximation
Finite Element approximation
particular application of the Rayleigh-Ritz methodthe structural domain is subdivided into a finite number of“elements” defined by the connectivity of “nodes”the assumed modes are local “shape” functions“local” refers to an element of the structural domain
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AA242B: MECHANICAL VIBRATIONS 14 / 29
The Finite Element Method
Finite Element Approximation
Interpolation of the displacement field
u(x1, x2, x3, t) =n∑
j=1
nj(x1, x2, x3)qj(t)
let (x(l)1 , x
(l)2 , x
(l)3 ), l = 1, · · · , n denote the coordinates of the nodes
of the elementsthe generalized coordinates ql are the “nodal” values of thedisplacement field
ql(t) = ui(x
(l)1 , x
(l)2 , x
(l)3 , t
)=⇒ nij(x
(l)1 , x
(l)2 , x
(l)3 ) = δjl , i = 1, 2, 3
indeed
ui (x(l)1 , x
(l)2 , x
(l)3 , t) =
n∑j=1
nij(x(l)1 , x
(l)2 , x
(l)3 )qj(t) =
n∑j=1
δjlqj(t) = ql(t)
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AA242B: MECHANICAL VIBRATIONS 15 / 29
The Finite Element Method
Finite Element Approximation
u(x1, x2, x3, t) =n∑
j=1
nj(x1, x2, x3)qj(t)
nj are continuous functions — usually, polynomials
global viewpoint: nj is attached to node j and has support in theelements attached to this node — it vanishes elsewhere
local viewpoint: inside each element e, the displacement field isapproximated by a linear combination of a small number of shapefunctions ϕe
i equal to the number of nodes i attached to element e
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AA242B: MECHANICAL VIBRATIONS 16 / 29
The Finite Element Method
Finite Element Approximation
Examplesfinite element approximation of a scalar field in one dimension
nj(x (l))
= δjl
⇒ u(x (l), t
)=
n∑j=1
nj(x (l))qj(t) = ql(t)
finite element approximation of a vector field in one dimension
n1j
(x (l))
= δj(3l−2), n2j(x(l)) = δj(3l−1), n3j(x
(l)) = δj(3l)
⇒
u1
(x (l), t
)=
n∑j=1
n1j
(x (l))qj(t) = q3l−2(t)
u2
(x (l), t
)=
n∑j=1
n2j
(x (l))qj(t) = q3l−1(t)
u3
(x (l), t
)=
n∑j=1
n3j
(x (l))qj(t) = q3l(t)
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AA242B: MECHANICAL VIBRATIONS 17 / 29
The Finite Element Method
Finite Element Approximation
Partition of unity
u(x1, x2, x3, t) =n∑
j=1
nj(x1, x2, x3)qj(t)
the chosen shape functions nj must satisfy
n∑j=1
nij(x1, x2, x3) = 1, ∀i = 1, 2, 3
in order to preserve a constant displacement field (rigid body mode)
for example, in one dimension
qj (t) = u(t), ∀j = 1, · · · , n
⇒ u(x , t) =n∑
j=1
nj (x)qj (t) = u(t)n∑
j=1
nj (x) = u(t)
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AA242B: MECHANICAL VIBRATIONS 18 / 29
The Finite Element Method
Generation of P1 Elements
P1 family of elements
in one dimension
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AA242B: MECHANICAL VIBRATIONS 19 / 29
The Finite Element Method
Generation of P1 Elements
Restriction of a global function to an element
consider the domain V as a union of elements
V =N⋃
e=1
Ve
restriction of the global function nj(x1, x2, x3) to an element e
ϕej (x1, x2, x3) = ϕj(x1, x2, x3) = nj(x1, x2, x3)|Ve
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AA242B: MECHANICAL VIBRATIONS 20 / 29
The Finite Element Method
Generation of P1 Elements
P1 family of elementsshape functions in one dimension
Ne(x) =[ϕ1(x) ϕ2(x)
]=[
1− x
h
x
h
]
displacement field
u(x , t) = ϕ1(x)u1(t) + ϕ2(x)u2(t) = Ne(x)qe(t)
with qe(t) =[q1(t) q2(t)
]T20 / 29
AA242B: MECHANICAL VIBRATIONS 21 / 29
The Finite Element Method
Element-Level Matrices
Element energies
element-level kinetic energy
Te =1
2
∫Ve
ρe uT udV
where Ve is the volume of element eelement-level strain energy
Vint,e =1
2
∫Ve
(Du)T H (Du) dV
element-level potential energy
Vext,e = −∫Sσ,e
uT tdS −∫Ve
uT XdV
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AA242B: MECHANICAL VIBRATIONS 22 / 29
The Finite Element Method
Element-Level Matrices
Element-level mass and stiffness matrices
starting from u = Neqe
consider the element-level kinetic energy
Te =1
2
∫Ve
ρe (Ne qe)T Ne qedV =1
2qTe Me qe
where the element-level mass matrix is defined as
Me =
∫Ve
ρeNTe NedV
consider the element-level strain energy
Vint,e =1
2
∫Ve
(DNeqe)T H (DNeqe) dV =1
2qTe Keqe
where the element-level stiffness matrix is defined as
Ke =
∫Ve
(DNe)T H (DNe) dV
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AA242B: MECHANICAL VIBRATIONS 23 / 29
The Finite Element Method
Element-Level Matrices
Generalized loadconsider the element-level potential energy
Vext,e = −∫Sσ,e
qTe N
Te tdS −
∫Ve
qTe N
Te XdV = −qT
e ge
where
ge =
∫Sσ,e
NTe tdS +
∫Ve
NTe XdV
is the generalized load
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AA242B: MECHANICAL VIBRATIONS 24 / 29
The Finite Element Method
Global Matrices
Global energies
global kinetic energy and mass matrix
T =N∑
e=1
Te =N∑
e=1
1
2
∫Ve
ρe uT udV =
N∑e=1
1
2qTe Me qe
global strain energy and stiffness matrix
Vint =N∑
e=1
Vint,e =N∑
e=1
1
2
∫Ve
(Du)T H (Du) dV =N∑
e=1
1
2qTe Keqe
global potential energy and generalized force vector
Vext =N∑
e=1
Vext,e = −N∑
e=1
∫Sσ,e
uT tdS −∫Ve
uT XdV = −N∑
e=1
qTe ge
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AA242B: MECHANICAL VIBRATIONS 25 / 29
The Finite Element Method
Global Matrices
Localization operator
consider the global vector of generalized displacements
q =[q1 q2 q3 · · · qn
]Tconsider an element e and its generalized displacements qe
qe = Leq
where Le is a localization operatorfor example, if e = (1, 2)
Le =
[1 0 0 · · · 00 1 0 · · · 0
]⇒ qe =
[q1 q2
]T
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AA242B: MECHANICAL VIBRATIONS 26 / 29
The Finite Element Method
Global Matrices
Global mass matrix
N∑e=1
1
2qTe Me qe =
1
2qT(
N∑e=1
LTe MeLe
)q =
1
2qTMq =⇒ M =
N∑e=1
LTe MeLe
Global stiffness matrix
N∑e=1
1
2qTe Keqe =
1
2qT(
N∑e=1
LTe KeLe
)q =
1
2qTKq =⇒ K =
N∑e=1
LTe KeLe
Global generalized force vector
N∑e=1
qeT ge = qT
(N∑
e=1
LTe ge
)= qT g =⇒ g =
N∑e=1
LTe ge
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AA242B: MECHANICAL VIBRATIONS 27 / 29
The Finite Element Method
Assembly Process
Scatter and gather
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AA242B: MECHANICAL VIBRATIONS 28 / 29
The Finite Element Method
Assembly Process
Assembly of the mass matrix
loop over all the elements e = 1, · · · ,Nadd the local contribution LT
e MeLe to the mass matrix of theassembled system
M =N∑
e=1
LTe MeLe
Assembly of the stiffness matrix
similarly
K =N∑
e=1
LTe KeLe
Assembly of a load vector
similarly
g =N∑
e=1
LTe ge
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AA242B: MECHANICAL VIBRATIONS 29 / 29
The Finite Element Method
Lumped Mass Matrix
Lumping of the mass matrix
consistent mass M = [mij ]M is usually sparsemass lumping
m11 m12 . . . m1n
m21
. . ....
.... . .
...mn1 . . . . . . mnn
︸ ︷︷ ︸
M
→
n∑i=1
m1i 0 . . . 0
0n∑
i=1
m2i
...
.... . . 0
0 . . . 0n∑
i=1
mni
︸ ︷︷ ︸
M(L)
M(L) is the lumped mass matrix
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