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Generalites Eq. des Vibs L Vibs L libres Eq. des Vibs T Vibs T libres Vibs L forcees Vibs T forcees Methodes approchees Systemes 2D
Vibrations & OndesVibrations des systemes continus
UPMC - Master Sciences de l’ingenieur
Septembre 2014
UPMC - Master SdI Vibrations & Ondes Vibrations des systemes continus sept. 14 1 / 112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Generalites
Vibratio
nsdansles
milieu
xela
stiques
1D
M1(T
C)→
vibrationsdes
structu
resa1dim
ension
:
x
Mϴ(x)
x
v(x)
x
u(x)
F
x
v(x)
F
Vibration
stran
sversesdes
cordes
Vibration
slon
gitudinales
des
pou
tres
Vibration
sdetorsion
des
arbres
Vibration
sdeflexion
des
pou
tres
Pou
rquoi
etudier
cessystem
essim
ples
?
Ilsmodelisen
tsim
plem
entlecom
portem
entdenom
breuses
structu
res
Ilsperm
ettentdecom
prendre
lesstru
ctures
pluscom
plexes
Ilsson
tles
constitu
ants
elementaires
des
structu
resen
elements
finis
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3/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Generalites
Differen
tstyp
esd’eq
uatio
ndumouvem
ent
Lim
itesdel’etu
de
Geom
etrie:
Pou
trerectilign
es,
Lon
gueurfinie
L
Section
constan
teS�
L
Materiau
:
Hom
ogene,
Lineaire
isotrope
Non
dissip
atif
Deu
xtyp
esd’eq
uatio
ndifferen
tielledumouvem
ent
Vibration
slon
gitudinales
Vibration
sdeTorsion
Vibration
sdes
Cord
es
1mem
eeq
uatio
nd’ordre
2:
l’Equation
ded’Alem
bert
∂2f(x,t)
∂x2
−1c2L
∂2f(x,t)
∂t2
=g(x,t)
Vibration
sdeflexion
Equation
d’ord
re4
∂4f(x,t)
∂x4
−1c2F
∂2f(x,t)
∂t2
=g(x,t)
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
longitudinalesDim
ensio
ns-Para
metres
-Hyp
otheses
Hyp
otheses
Pou
tredroite
:Lon
gueurL(�
e,l),
Section
constan
teS
Materiau
isotrope:ρ,E
,ν,non
dissip
atifDistrib
ution
deforces
f(x,t)
→Petites
pertu
rbation
s(gravite
non
priseen
compte)
Gran
deuretu
diee
:deplacem
entlon
gitudinal
local
u(x,t)
Meth
odes
pou
recrire
etresou
dre
lesequation
sdumou
vement:
Meth
odeloca
leavec
lePrin
cipefon
dam
ental
dela
dynam
ique
(PFD):adaptee
auxstru
ctures
simples.
Meth
odeenergetiq
ueou
variationnelle
:Theorem
edeHam
ilton+
variationsenergetiq
ues
Inclu
eles
condition
sauxlim
itesAdaptee
auxstru
ctures
complexes.
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
longitudinalesEquatio
ndumouvem
ent-Equilib
reLocal(1)
Onecrit
l’equilibre
dynam
iqued’unesection
delon
gueurdx:
Masse
:ρSdx
Deplacem
ent:u(x,t)
Acceleration
:∂2u
∂t2
Force
agau
che:−
F
Force
adroite
:F+
∂F
∂xdx
Force
exterieure
f(x,t)Onadon
c
ρSdx∂2u
∂t2=
−F+F+
∂F
∂xdx+f(x,t)d
x
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
longitudinalesEquatio
ndumouvem
ent-Equilib
reLocal(2)
ρSdx∂2u
∂t2
=��
��
−F+F+
∂F
∂xdx+f(x,t)d
x
⇔ρS� �dx
∂2u
∂t2
=∂F
∂x� �dx
+f(x,t)� �dx
⇔ρS∂2u
∂t2
=∂F
∂x+f(x,t)
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
longitudinaleslibres
Equatio
ndumouvem
ent=
Equatio
ndes
ondes
longitu
dinales
ρS∂2u
∂t2=
∂F
∂x+f(x,t)
Tractio
n/Compressio
npure
:
F=
σS=
EεS
=ES∂u
∂x
ρS∂2u
∂t2=
∂∂x(ES∂u
∂x)+
f(x,t)
Com
meE
etScon
stants
ρ∂2u
∂t2=
E∂2u
∂x2+
1Sf(x,t)
Onnote
c2=
Eρ⇔
c= √
Eρ
Finalem
ent:
∂2u
∂t2 −
c2∂2u
∂x2=
1ρSf(x,t)
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8/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
longitudinalesCelerite
des
ondes
longitu
dinales
[c]=
m/s
c=
celeriteduson
ouvitesse
des
ondes
longitu
dinales.
Ordres
degran
deur:
cacier
= √Eρ
= √2.10
11
8.103
=0.5.10
4=
5000m/s
Autres
valeurs
Materia
uc(m
/s)
Materia
uc(m
/s)
PVCmou
80
Glace
3200
Sab
lesec
10-300
Hetre
3300
Beto
n3100
Aluminium
5035
Plomb
1200
Verre
5300
PVCdur
1700
Acier
5600-5900
Gran
it6200
Perid
otite
17700
Rap
pel
:vitesse
duson
dan
sl’air
:343m/s,
dan
sl’eau
:1480m/s
1.Roch
emagmatiq
ueco
nstitu
antla
majeu
rpartie
dela
croute
terrestreUPMC
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
detorsion
dansles
arbresEquatio
ndumouvem
ent-Para
metres
x
xdx
L
ϴ(x,t)
Hyp
otheses
Pou
tredroite
:Lon
gueurL(�
e,l),
Section
constan
teS
Materiau
isotrope:ρ,E
,ν,G
(module
detorsion
),non
dissip
atif
→Petites
pertu
rbation
s+
gravitenon
priseen
compte
Gran
deuretu
diee
:deplacem
entangu
lairelocal
θ(x,t)
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
detorsion
dansles
arbresEquatio
ndumouvem
ent-Equilib
relocal
x
xdx
L
ϴ(x,t)
Onecrit
l’equilibre
dynam
iqued’unesection
d’ep
aisseurdx:
Mom
entd’in
ertie:ρIx d
x
Deplacem
ent:θ(x
,t)
Acceleration
:∂2θ
∂t2
Mom
entagau
che:−
M
Mom
entadroite
:M
+∂M∂xdx
Onadon
c
ρIx d
x∂2θ
∂t2=
−M
+M
+∂M∂xdx
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
detorsion
dansles
arbresEquatio
ndumouvem
ent-Equilib
relocal(2)
ρIx d
x∂2θ
∂t2=�
��
� �−M
+M
+∂M∂xdx
⇔ρIx � �dx
∂2θ
∂t2=
∂M∂x� �dx
⇔ρIx∂2θ
∂t2=
∂M∂x
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
detorsion
dansles
arbresEquatio
ndumouvem
ent=
Equatio
ndes
ondes
detorsio
n
ρIx∂2θ
∂t2=
∂M∂x
Torsio
npure
:
M=
GIx∂θ
∂x
ρIx∂2θ
∂t2=
∂∂x(G
Ix∂θ
∂x)
Com
meG
etIx
constan
ts:
ρ∂2θ
∂t2=
G∂2θ
∂x2
Onnote
c2=
Gρ⇔
c= √
Gρ
Finalem
ent:∂2θ
∂t2 −
c2∂2θ
∂x2=
0
Ordre
degrandeu
r:
cacier
= √Gρ
= √E
2(1
+ν)ρ
= √2.1011
2×(1
+0.3)×
8.103=
2000m/s
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
transversesdes
cordesEquatio
ndumouvem
ent-Para
metres
Hyp
otheses
surla
corde
Lon
gueurL
Materiau
:ρ(kg/m
)
Non
dissip
atif
Tension
constan
te:T
→Petites
pertu
rbation
s
Gran
deuretu
diee
:deplacem
ent
verticallocal
v(x,t)
Pente
locale
:α=
∂v
∂x
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14/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
transversesdes
cordesEquatio
ndumouvem
ent-Equilib
relocal
Bila
ndes
forces
Suivan
tx:
Tcosα(x
+dx)−
Tcosα(x)=
0
(cosα≈
1)⇔
0=
0
Suivan
ty:
Tsin
α(x
+dx)−
Tsin
α(x)=
ρdx∂2v
∂t2
⇔Tα(x
+dx)−
Tα(x)=
ρdx∂2v
∂t2
⇔T∂α
∂xdx=
ρdx∂2v
∂t2
⇔ρ∂2v
∂t2=
T∂2v
∂x2
∂2v
∂t2 −
c2∂2v
∂x2=
0avec
c= √
Tρ
Pourunecord
een
acier
dediametre
0.5
mm
tenduea10kg:cacier ≈
400m/s
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
longitudinaleslibres
Solutio
nlib
res=
Solutio
nsdel’eq
uatio
ndumouvem
entlorsq
uef(x,t)
=0
Aucunedistrib
ution
deforce
perm
anente
(mais
systemehors
d’eq
uilibre)
:
Equation
dumou
vementlibre
:
∂2u
∂t2 −
c2∂2u
∂x2=
0(1)
∃2typ
esdesolu
tions:
propagatives
statio
nnaires
Solu
tionstation
naire→
Hyp
othese
devaria
bles
separees
:
u(x,t)
=φ(t)X
(x)
alors(1)⇔∂2φ
∂t2X
−c2φ
∂2X
∂x2=
0
⇔φ(t)X
(x)−
c2φ
(t)X′′(x
)=
0
⇔φφ(t)
=c2X
′′
X(x)=
cte=
α(2)
Deuxequation
saresou
dre
:
l’unesurt,
l’autre
surx.
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
longitudinaleslibres
Solutio
nsta
tionnaire
gen
erale
enmodes
libres
∂2u
∂t2 −
c2∂2u
∂x2=
0
avecu(x,t)
=φ(t)X
(x)
⇔φφ(t)
=c2X
′′
X(x)=
α
d’ou
uneequation
surle
temps:
φ(t)−
αφ(t)
=0
(3)
etuneequation
dedeform
ation:
X′′(x
)−αc2X(x)=
0(4)
Solu
tionstation
naire
pou
r(3)
→α=
−ω2(<
0)
Lasolu
tions’ecrit
:
φ(t)
=Acos
ωt+Bsin
ωt
Dans(4)
onnote
γ=
ωc
X′′(x
)+
γ2X
(x)=
0
Lasolu
tions’ecrit
:
X(x)=
Ccos
γx+
Dsin
γx
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17/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
longitudinaleslibres
Form
egen
erale
dela
solutio
nsta
tionnaire
enmodes
libres
∂2u
∂t2 −
c2∂2u
∂x2=
0
Mou
vementvibratoire
longitu
dinal
libre:
u(x,t)
=(A
cosωt+Bsin
ωt)(C
cosγx+Dsin
γx)
Mou
vementharm
oniquedepulsation
ω
Conditio
nsinitia
les(C
.I.)dumou
vement→
(A,B
)
L’am
plitu
dedumou
vementdes
sectionsdependdela
position
dans
lapou
tre,mais
Ilfau
tidentifi
erles
pulsation
sωpossib
les
Conditio
nsauxlim
ites(C
.L.)dusystem
e→ωet
γ.
Les
C.L.determ
inentles
modes
propres(ou
natu
rels)dela
pou
tre
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
longitudinaleslibres
Solutio
nssta
tionnaires
particu
lieres-Conditio
nsauxlim
ites
2cas
simples
decon
dition
sauxlim
itespeuven
tetre
consid
eres:
Bord
libre
:aucuneff
ortal’extrem
ite
F=
0⇔ES∂u
∂x=
0
mou
vementindeterm
ine(u
=?)
Bord
Libre⇔
∂u
∂x=
0,∀t
Enca
stremen
t:
Section
extremebloquee
:u=
0
Effort
indeterm
ine(F
=?)
Enca
stremen
t⇔u=
0,∀t
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
longitudinaleslibres
Modes
propres
(ounaturels)
3com
binaison
spossib
lesdes
condition
sauxlim
ites:
Pou
treencastree
-encastree
Pou
treencastree
-libre
Pou
trelibre
-libre
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
longitudinaleslibres
Casdela
poutre
encastree-lib
re(1)
u(x,t)
=(A
cosωt+
Bsin
ωt)(C
cosγx+Dsin
γx)=
φ(t)X
(x)
Encastrem
enten
x=
0
u(0,t)
=0,∀
t⇔X(0)
=0
⇔C
=0
⇔X(x)=
Dsin
γx
Libre
enx=
L
⇔∂u
∂x(L,t)
=0,∀⇔
X′(L
)=
0
⇔cos
γL=
0⇔γL=
(2n+1)
π2
Valeu
rspossib
lesdeγ:
γn=
(2n+1)
π2L
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
longitudinaleslibres
Casdela
poutre
encastree-lib
re(2)
Deform
eespossib
les:
Xn (x
)=
sinγn x
Ceson
tles
modes
propres
Xn (x
)=
sin ((2n
+1)
πx
2L )avec
necessairem
ent:
γ=
ωc⇔
ωn=
cγn
⇒Pulsa
tionspropres
ωn:
ωn=
(2n+1)
πc
2L
⇔ωn=
(2n+1)
π2L √Eρ
Freq
uen
cespropres
(Hz)
:
fn=
2n+1
4L √Eρ
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
22/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
longitudinaleslibres
Casdela
poutre
encastree-lib
re(3)
Expression
dumodedevib
ratio
nlongitu
dinale
aωn:
un (x
,t)=
(A′ncos
ωn t
+B
′nsin
ωn t)
sin ((2n
+1)
πx
2L )
⇔un (x
,t)=
Uncos(ω
n t+ϕn )
sin ((2n
+1)
πx
2L )Repon
selibre
complete
=Com
bi.lin.des
solution
spossib
les:
u(x,t)
=∞∑n=0
un (x
,t)=
∞∑n=0
Uncos(ω
n t+
ϕn )
sin ((2n
+1)
πx
2L )
Con
dition
sinitiales
(u(x,0),u
(x,0))→
(Un ,ϕ
n )ou
(A′n ,B
′n )
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
23/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
longitudinaleslibres
Modes
longitu
dinauxdela
poutre
encastree-lib
re
X0 (x
)=
sin (πx
2L )
X1 (x
)=
sin (3π
x
2L )
X2 (x
)=
sin (5π
x
2L )
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
24/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
longitudinaleslibres
Resu
ltatpourles
3co
mbinaiso
nsdeC.L.
C.L.
Poutre
L-L
Poutre
E-L
Poutre
E-E
Modes
propres
cosnπxL
sin(2n
+1)
πx
2L
sinnπxL
Pulsatio
nspro
presnπL √
Eρ(2n+1)π
2L √
EρnπL √
Eρ
facier (L
=1m)
f1=
2500Hz
f0=
1250Hz
f1=
2500Hz
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
25/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
longitudinaleslibres
Casdela
poutre
encastree
auxdeu
xbouts
X0 (x
)=
sinπxL
X1 (x
)=
sin2π
x
L
X2 (x
)=
sin3π
x
L
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
26/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
longitudinaleslibres
Modes
longitu
dinauxdela
poutre
libre-lib
re
X1 (x
)=
cosπxL
X2 (x
)=
cos2π
x
L
X3 (x
)=
cos3π
x
L
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
27/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Sea
nce
5
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
28/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
longitudinaleslibres
Orth
ogonalite
des
modes
propres
longitu
dinaux
Onpeutecrire
lasolu
tionlibre
commeunecom
position
des
modes
propres:
u(x,t)
=∞∑n=0
un (x
,t)=
∞∑n=0
φn (t)X
n (x)
L’eq
uation
dumou
vementlibre
estverifi
eepou
rchaquemode
∂2u
n
∂t2
−c2∂2u
n
∂x2
=0⇔
−ω2n X
n −c2X
′′n=
0
Onpeutecrire
pou
rles
modes
net
m:
−ω2n X
n=
c2X
′′n(×
Xm)
et−
ω2mXm=
c2X
′′m(×
Xn )
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
29/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
longitudinaleslibres
Orth
ogonalite
des
modes
propres
longitu
dinaux(2)
Onadon
c:
−ω2n X
n=
c2X
′′n(×
Xm)
et−
ω2mXm=
c2X
′′m(×
Xn )
Soit
enmultip
liantresp
ectivementpar
Xmet
Xn ,
puisintegran
tsurla
longu
eurdela
barre
:
−ω2n ∫
L
0
Xn X
mdx=
c2 ∫
L
0
X′′nXmdx
−ω2m ∫
L
0
XmXn dx=
c2 ∫
L
0
X′′mXn dx
Enintegran
tpar
partie
etpou
rdes
extremites
encastrees
oulibres
onob
tient:
−ω2n ∫
L
0
Xn X
mdx=�
���
[X′n X
m] L0 −
c2 ∫
L
0
X′n X
′mdx
−ω2m ∫
L
0
XmXn dx=�
���
[X′mXn ] L0 −
c2 ∫
L
0
X′mX
′n dx
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
30/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
longitudinaleslibres
Orth
ogonalite
des
modes
propres
longitu
dinaux(3)
Onadon
c:
ω2n ∫
L
0
Xn X
mdx=
c2 ∫
L
0
X′n X
′mdx
etω2m ∫
L
0
XmXn dx=
c2 ∫
L
0
X′mX
′n dx
Ensou
strayantles
2equation
s:
(ω2n −
ω2m) ∫
L
0
Xn X
mdx=
0
avecω2n =
ω2m,on
obtien
tles
Rela
tionsd’orth
ogonalite
:
∫L
0
Xn X
mdx=
0et
∫L
0
X′n X
′mdx=
0
Les
modes
propres
etleu
rsderivees
sontorth
ogonaux
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
31/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
longitudinaleslibres
Para
metres
modaux
Onaaussi
:
−ω2n ∫
L
0
X2ndx=
c2 ∫
L
0
X′′nXn dx
Quidon
nepar
integration
par
parties
avecles
C.L.:
ω2n ∫
L
0
X2ndx=
c2 ∫
L
0
X′2ndx
⇔ω2n=
ES ∫
L0X
′2ndx
ρS ∫
L0X
2ndx
=kn
mn
Onidentifi
ela
masse
modale
mnet
laraideu
rmodale
kn:
mn=
ρS ∫
L
0
X2ndx
etkn=
ES ∫
L
0
X′2ndx
Onpeutnorm
aliserles
modes
detelle
sorteque:
mn=
1alors
kn=
ω2n
UPMC
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SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
32/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Identification
descaracteristiques
modales
longitudinales
Onpeutecrire
lesenergies
modales
pou
ridentifi
erles
masses
etraid
eurs
modales
:Energie
cinetiq
uemodale
:
Tn= ∫
L
0
12ρS(u
n (x,t))
2dx=
12φ2n ρS ∫
L
0
X2n(x)dx=
12m
n φ2n
Onidentifi
ela
masse
modale
:
mn=
ρS ∫
L
0
X2n(x)dx
Energie
poten
tiellemodale
:
Un= ∫
L
0
12ES (
∂un
∂x(x,t) )
2
dx=
12φ2n E
S ∫L
0
X′2n(x)dx=
12kn φ
2n
Onidentifi
ela
raideurmodale
:
kn=
ES ∫
L
0
X′2n(x)dx
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
33/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
longitudinaleslibres
Exp
ressiondansla
base
modale
Equatio
ndumouvem
entlib
re:
u(x,t)−
c2u ′′(x
,t)=
0
avecu(x,t)
= ∑nφn (t)X
n (x)
⇔ ∑n
φn X
n −c2φ
n X′′n=
0
⇔ ∑n
φn ∫
L
0
XmXn dx−
c2φ
n ∫L
0
XmX
′′ndx=
0
Orth
ogonalite
des
modes
:
⇔φn ∫
L
0
X2ndx−
c2φ
n ∫L
0
Xn X
′′ndx=
0
(n=
0,...,∞)
Avec
lesrelation
ssuivan
tessur
lesmodes
propres:
∫L
0
X2ndx=
mn
ρS
∫L
0
Xn X
′′ndx=
−kn
ES
Onob
tient:
mn φ
n+kn φ
n=
0
ou
φn+
ω2n φ
n=
0
(n=
0,...,∞)
Chaquemodepropre
secom
porte
commeunsystem
ea1DDL.
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
34/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
longitudinaleslibres
Rep
onse
libre
enfonctio
ndes
conditio
nsinitia
les(1)
Onpeutecrire
unevibration
librequelcon
quecom
meunecom
binaison
lineaire
des
modes
propressou
sla
forme:
u(x,t)
= ∑n
(Ancos
ωn t
+Bnsin
ωn t)X
n (x)
Les
constan
tesAnet
Bnson
tdeterm
inees
avecles
condition
sinitiales
{u(x,0)
=u0 (x
)
u(x,0)
=u0 (x
)
⇔ ⎧⎪⎪⎨⎪⎪⎩u(x,0)
= ∑n
An X
n=
u0 (x
)
u(x,0)
= ∑n
ωn B
n Xn=
u0 (x
)
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
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us
sept.
14
35/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
longitudinaleslibres
Rep
onse
libre
enfonctio
ndes
conditio
nsinitia
les(2)
Onpeuttirer
avantage
des
proprietesdes
modes
propres:
⎧⎪⎪⎨⎪⎪⎩∑n
An X
n=
u0 (x
)
∑n
ωn B
n Xn=
u0 (x
)⇔ ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩
∑n
An ∫
L
0
XmXn dx= ∫
L
0
Xmu0 (x
)dx
∑n
ωn B
n ∫L
0
XmXn dx= ∫
L
0
Xmu0 (x
)dx
avecl’orth
ogonalite
des
modes
:
⎧⎪⎪⎪⎨⎪⎪⎪⎩An ∫
L
0
X2ndx= ∫
L
0
Xn u
0 (x)dx
ωn B
n ∫L
0
X2ndx= ∫
L
0
Xn u
0 (x)dx
⇔ ⎧⎪⎪⎪⎨⎪⎪⎪⎩An=
ρS
mn ∫
L
0
Xn u
0 (x)dx
Bn=
ρS
ωn m
n ∫L
0
Xn u
0 (x)dx
Les
integrales
peuven
tetre
diffi
cilesaevalu
erpou
rdes
Xncom
pliques
→Evalu
ationnumeriq
uedes
Anet
Bn
UPMC
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Vibratio
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Vibratio
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us
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36/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
enflexion
Equatio
nlocale
dumouvem
entpourunepoutre
-Para
metres
Poutre
droite
:
Section
constan
teS=
largeur×
hauteu
r=
l×e,
Lon
gueurL(�
e,l),
Mom
entquadratiq
uedesection
I⊥=
le3
12
Materiau
isotrope:ρ,E
,ν,non
dissip
atif
Varia
bles
:
Distrib
ution
deforces
transverses
f(x,t)
Petites
pertu
rbation
s+
gravitenon
priseen
compte
Gran
deuretu
diee
:deplacem
enttran
sverselocal
(flech
e)v(x,t)
UPMC
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SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
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us
sept.
14
37/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
enflexion
Equatio
nlocale
dumouvem
entdela
poutre
-Hyp
otheses
Hyp
othese
d’Euler-B
ernou
lli:L’in
ertiederotation
des
sectionest
negligee.
Con
sequence
:les
sectionsdroites
restent⊥
ala
ligneneutre.
Rotation
des
sections:θ(x
)=
∂v
∂x
UPMC
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Vibratio
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Ondes
Vibratio
nsdes
systemes
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us
sept.
14
38/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
enflexion
Equatio
nlocale
dumouvem
entdela
poutre
-Equilib
reLocal(1)
Onecrit
l’equilibre
dynam
iqued’unesection
d’ep
aisseurdx:
Masse
:ρSdx
Deplacem
ent(Flech
e):v(x,t)
Acceleration
:∂2v
∂t2
Effort
tranchantagau
che:T
Effort
tranchantadroite
:− (T
+∂T
∂xdx )
Mtflech
issantagau
che:−
M
Mtflech
issantadroite
:M
+∂M∂xdx
Force
exterieure
f(x,t)
UPMC
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Vibratio
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Ondes
Vibratio
nsdes
systemes
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us
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14
39/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
enflexion
Equatio
nlocale
dumouvem
entdela
poutre
-Equilib
reLocal(2)
Bila
ndes
Resu
ltantes
ρSdx∂2v
∂t2=
T−T
−∂T∂xdx+f(x,t)d
x
⇔ρS∂2v
∂t2=
−∂T∂x
+f(x,t)
Bila
ndes
Momen
ts
0=
−M
+M
+∂M∂xdx−
Tdx
⇔0=
∂M∂x
−T
d’apres
latheorie
des
pou
tres:
M=
EI∂θ
∂x=
EI∂2v
∂x2
d’ou
T=
∂∂x (
EI∂2v
∂x2 )
etEIetan
tcon
stant:
ρS∂2v
∂t2+EI∂4v
∂x4=
f(x,t)
UPMC
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Vibratio
nsdes
systemes
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us
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40/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
enflexion
Equatio
ndes
ondes
transverses
dansunepoutre
ρS∂2v
∂t2+
EI∂4v
∂x4=
f(x,t)⇔
∂4v
∂x4+
1c2
∂2v
∂t2=
1EIf(x,t)
Onnote
:
c= √
EI
ρS
cn’est
pas
unevitesse
!
[c]=
m3
s2
UPMC
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Vibratio
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Vibratio
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systemes
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us
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
libresen
flexion
Vibratio
nslib
resd’unepoutre
lorsquef(x,t)
=0
∂4v
∂x4+
1c2
∂2v
∂t2=
0(1)
Solu
tionstation
naire
:
v(x,t)
=φ(t)X
(x)
(1)⇔c2φ
X′′′′+
φX
=0
⇔c2X
′′′′
X(x)=
−φφ(t)
=−ω2
(2)
Equatio
nen
temps
(2)⇔φ(t)
+ω2φ
(t)=
0(3)
Equatio
nsurla
defo
rmatio
n
(2)⇔X
′′′′(x)−
γ4X
(x)=
0(4)
avec
γ4=
ω2
c2
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Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
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14
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
libresen
flexion
Exp
ressiongen
erale
des
vibratio
nslib
resd’unepoutre
(1)
v(x,t)
=φ(t)X
(x)
Surle
temps:φ(t)
+ω2φ
(t)=
0⇔φ(t)
=Acos
ωt+Bsin
ωt
Surl’esp
ace:X
′′′′(x)−
γ4X
(x)=
0⇔X(x)=
X0 e
rt
Equatio
ncaracteristiq
ue:r4−
γ4=
0⇔r2=
±γ2
4racin
espossib
les:r=
−γ,
+γ,
−iγ,
+iγ
Lasolutio
ngenerale
estC.L.des
solutio
nspossib
les
X(x)=
Ceγx+
De −
γx+
Geiγ
x+
He −
iγx
⇔X(x)=
C1cos
γx+C2sin
γx+C3cosh
γx+C4sin
hγx
Onpeutaussi
ecrire:
⇔X(x)=
D1 (co
sγx+
cosh
γx)+
D2 (co
sγx−
cosh
γx)
+D
3 (sinγx+
sinhγx)+
D4 (sin
γx−
sinhγx)
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
libresen
flexion
Exp
ressiongen
erale
des
vibratio
nslib
resd’unepoutre
(2)
v(x,t)
=φ(t)X
(x)
Dependance
temporelle
:
φ(t)
=Acos
ωt+Bsin
ωt
Dependance
spatiale
X(x)=
C1cos
γx+C2sin
γx+
C3cosh
γx+C4sin
hγx
Com
mentidentifi
erles
constan
tesincon
nues?
Con
dition
sauxlim
ites→(C
1 ,C2 ,C
3 ,C4 )
et(X
n ,ωn )
Con
dition
sinitiales→
(A,B
)Finalem
entle
mou
vementlibre
peuts’ecrire
commeunecom
binaison
lineaire
des
modes
propres.((D
e)composition
modale).
v(x,t)
= ∑n
Xn (A
ncos
ωn t
+Bnsin
ωn t)
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Vibratio
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
libresen
flexion
Orth
ogonalite
des
modes
propres
Xn:Modepropre
d’ord
ren
γ4n:Valeu
rpropre
associee
γ4n= (
ωn
c )2
Ona∀n,
X′′′′n
−γ4n X
n=
0(5)
(5)⇔ {X
′′′′n
=γ4n X
n
X′′′′m
=γ4mXm
⇔ {∫L0XmX
′′′′n
dx=
γ4n ∫
L0XmXn dx
∫L0Xn X
′′′′m
dx=
γ4m ∫
L0Xn X
mdx
Dou
ble
integration
par
parties
+CL:
⇔ {�
���
[XmX
′′′n] L0 −
��
� �[X
′mX
′′n] L0+ ∫
L0X
′′mX
′′ndx=
γ4n ∫
L0X
mX
n dx
����
[Xn X
′′′m] L0 −
��
� �[X
′n X′′m] L0+ ∫
L0X
′′nX
′′mdx=
γ4m ∫
L0X
n Xmdx
⇔(γ
4n −γ4m) ∫
L
0
Xn X
mdx=
0
D’ou
3relation
sd’orth
ogonalite
des
modes
quandm
=n:
∫L
0
Xn X
mdx=
0
∫L
0
X′′nX
′′mdx=
0
∫L
0
Xn X
′′′′m
dx=
0
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
libresen
flexion
Conditio
nsauxlim
itespourunepoutre
(1)
Pou
rdeterm
iner
lescon
stantes
dela
fonction
deform
edumodepropre,
onpeutcon
siderer
3cas
simples
avecchacu
n2param
etrescon
nus:
Bord
libre
:Effort
tranchantnulau
bord
T=
0⇔EI∂3v
∂x3=
0
Mom
entdeflexion
nulau
bord
M=
0⇔EI∂2v
∂x2=
0
Deplacem
ents
indeterm
ines
v=?et
∂v
∂x=?
Bord
Libre⇔
∂2v
∂x2=
0et
∂3v
∂x3=
0,∀
t
Appuisim
ple
:Flech
enulle
aubord
v=
0
Mom
entdeflexion
nulau
bord
M=
0⇔EI∂2v
∂x2=
0
Indeterm
ines
:θet
T∂v
∂x=?et
∂3v
∂x3=?
Appuisim
ple⇔
v=
0et
∂2v
∂x2=
0,∀
t
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
libresen
flexion
Conditio
nsauxlim
itespourunepoutre
(2)
Enca
stremen
t:
Flech
enulle
aubord
v=
0
Rotation
nulle
aubord
θ=
0⇔∂v
∂x=
0
Efforts
indeterm
ines
T=?et
M=?
Enca
stremen
t⇔v=
0et
∂v
∂x=
0,∀
t
Reca
pitu
latif
C.L.
Bord
Libre
Appuisim
ple
Encastrem
ent
v→X
?0
0
θ→X
′?
?0
M→
X′′
00
?
T→
X′′′
0?
?
UPMC
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
libresen
flexion
Differen
tesco
mbinaiso
nspossib
lesdes
conditio
nsauxlim
itespourunepoutre
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
libresen
flexion
Modes
propres
dela
poutre
simplem
entappuyee
(1)
Deplacem
entX(x)
etmom
entflech
issantX”(x)
nulsen
x=
0et
x=
L
X(x)=
D1 (co
sγx+
cosh
γx)+
D2 (co
sγx−
cosh
γx)
+D
3 (sinγx+
sinhγx)+
D4 (sin
γx−
sinhγx)
X′′(x
)=
γ2(D
1 (−cosγx+
cosh
γx)+
D2 (−
cosγx−
cosh
γx)
+D
3 (−sin
γx+
sinhγx)+
D4 (−
sinγx−
sinhγx))
X(0)=
0⇒D
1=
0
X′′(0
)=
0⇒D
2=
0
X(L)=
0et
X′′(L
)=
0
⇒D
3=
D4et
sinγL=
0
d’oules
valeurs
possib
lesdeγet
des
modes
etpulsatio
nspro
pres:
γn=
nπL
⇔Xn (x
)=
sinnπx
Lωn=
γ2n c⇔
ωn= (
nπL )
2 √EI
ρS
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
libresen
flexion
Modes
propres
dela
poutre
simplem
entappuyee
(2)
Xn (x
)=
sinnπx
Lωn= (
nπL )
2 √EI
ρS
X1 (x
)=
sinπxL
X2 (x
)=
sin2π
x
L
X3 (x
)=
sin3π
x
L
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
libresen
flexion
Modes
propres
dela
poutre
libre-lib
re(1)
X′′(0)
=0⇒
D2=
0et
X′′′(0)
=0⇒
D4=
0
⇒X
=D
1 (cosγx+cosh
γx)+
D3 (sin
γx+sin
hγx)
X′′(L
)=
0et
X′′′(L
)=
0don
nentle
systeme
(S) {
D1 (−
cosγL+cosh
γL)+
D3 (−
sinγL+sin
hγL)=
0D
1 (sinγL+sin
hγL)+
D3 (−
cosγL+cosh
γL)=
0
Ledeterm
inantdoit
etrenul.Cequidon
nel’eq
uation
:
det(S
)=
0⇔cos
γLcosh
γL=
1
Les
racines
decette
equation
sonttrou
veesnumeriq
uem
ent(ou
graphiquem
ent)
:
γ1 L
γ2 L
γ3 L
γ4 L
γ5 L
4.73
7.85
11
14.13
17.28
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
libresen
flexion
Modes
propres
dela
poutre
libre-lib
re(2)
cosγLcosh
γL=
1→
γ0 L,γ
1 L,γ
2 L,γ
3 L,γ
4 L,γ
5 L,....
Les
frequences
propresson
tob
tenues
encalcu
lant:
ωn=
cγ2n ⇔
fn=
ωn
2π=
12π (γn LL )
2 √EI
ρS
Ensubstitu
antles
racines
successivem
entdansle
systemepreced
ent,
onob
tientles
rapports (
D1
D3 )
nquideterm
inentla
formedes
modes
propres.
→Pas
d’expression
generale
exactedes
modes
propres
→solu
tionnumeriq
ue(cf
codeMatlab
)
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Vibratio
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
libresen
flexion
Modes
propres
dela
poutre
libre-lib
re(3)
X1 (x
)
X2 (x
)
X3 (x
)
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
libresen
flexion
Modes
propres
dela
poutre
bi-en
castree
(1)
X(0)
=0⇒
D1=
0et
X′(0)
=0⇒
D3=
0
⇒X
=D
2 (cosγx−
coshγx)+
D4 (sin
γx−
sinhγx)
X(L)=
0et
X′(L
)=
0:
(S) {
D2 (cos
γL−
coshγL)+
D4 (sin
γL−
sinhγL)
D2 (−
sinγL−
sinhγL)+
D4 (cos
γL−
coshγL)
det(S
)=
0⇔cos
γLcosh
γL=
1
Onadon
cles
mem
esracin
eset
mem
esfreq
uences
propres.
Mais
lesform
esdes
modes
sontdifferen
tes:γn L→ (
D2
D4 )
n
Freq
uences
propresωn=
cγ2n ⇔
fn=
ωn
2π=
12π (
γn LL )
2 √EI
ρS
UPMC
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Vibratio
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Vibratio
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
libresen
flexion
Modes
propres
dela
poutre
bi-en
castree
(2)
X1 (x
)
X2 (x
)
X3 (x
)
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
libresen
flexion
Modes
propres
dela
poutre
encastree-lib
re(1)
X(0)
=0⇒
D1=
0et
X′(0)
=0⇒
D3=
0
⇒X
=D
2 (cosγx−
coshγx)+
D4 (sin
γx−
sinhγx)
X′′(L
)=
0et
X′′′(L
)=
0⇔
cosγLcosh
γL=
−1
Racin
eset
modes
propres
γ1 L
γ2 L
γ3 L
γ4 L
γ5 L
γ6 L
1.88
4.69
7.86
11
14.14
17.28
→ (D2
D4 )
n
Freq
uences
propres
ωn=
cγ2n ⇔
fn=
ωn
2π=
12π (γn LL )
2 √EI
ρS
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Vibratio
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Vibratio
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
libresen
flexion
Modes
propres
dela
poutre
encastree-lib
re(2)
Reso
lutio
ngraphiquedecos
xcosh
x=
±1
01
23
45
67
89
1011
1213
1415
−10
−8
−6
−4
−2 0 2 4 6 8 10
x
cos(x)* cosh(x)1−
1
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
libresen
flexion
Modes
propres
dela
poutre
encastree-lib
re(3)
X1 (x
)
X2 (x
)
X3 (x
)
UPMC
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Vibratio
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Vibratio
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
libresen
flexion
Modes
propres
dela
poutre
encastree-a
ppuyee
(1)
X(0)
=0⇒
D1=
0et
X′(0)
=0⇒
D3=
0
⇒X
=D
2 (cosγx−
coshγx)+
D4 (sin
γx−
sinhγx)
X(L)=
0et
X′′(L
)=
0⇔
tanγL=
tanhγL
Racin
eset
modes
propres
γ1 L
γ2 L
γ3 L
γ4 L
γ5 L
3.93
7.07
10.21
13.35
16.49
→ (D2
D4 )
n
Freq
uences
propres
ωn=
cγ2n ⇔
fn=
ωn
2π=
12π (γn LL )
2 √EI
ρS
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Vibratio
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
libresen
flexion
Modes
propres
dela
poutre
encastree-a
ppuyee
(2)
01
23
45
67
89
1011
1213
1415
−10
−8
−6
−4
−2 0 2 4 6 8 10
x
tan(x)tanh(x)
UPMC
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Vibratio
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
libresen
flexion
Modes
propres
dela
poutre
encastree-a
ppuyee
(3)
X1 (x
)
X2 (x
)
X3 (x
)
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Com
paraisondes
frequencespropres
Let
T
12
34
50
1000
2000
3000
4000
5000
6000
7000
8000
9000
10000
11000
12000
13000
Ordre des m
odes
Fréquence (Hz)
Modes longitudinaux (A
cier : L = 1m
)
LLEL
EE
12
34
50 20 40 60 80
100
120
140
160
180
200
220
Ordre des m
odes
Fréquence (Hz)
Modes transversaux (A
cier : L = 1m
, e = 3m
m)
AA
EA
EL
LL
UPMC
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Identification
descaracteristiques
modales
enflexion
Onpeutecrire
lesenergies
associees
achaquemodedevibration
(ouenergies
modales)
enrep
onse
libreou
forceepou
rretrou
verles
masses
etraid
eurs
modales
:
Energie
cinetiq
ue:
Tn= ∫
L
0
12ρS(v
n (x,t))
2dx=
12φ2n ∫
L
0
ρSX
2n(x)dx=
12m
n φ2n
onidentifi
ela
masse
modale
:m
n= ∫
L0ρSX
2n(x)dx
Energie
poten
tielle:
Un= ∫
L
0
12EI (
∂2v
n
∂x2(x,t) )
2
dx=
12φ2n ∫
L
0
EIX
′′2n(x)dx=
12kn φ
2n
onidentifi
ela
raideurmodale
:kn= ∫
L0EIX
′′2n(x)dx
UPMC
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Vibratio
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Ondes
Vibratio
nsdes
systemes
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us
sept.
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
libresen
flexion
Rep
onse
libre
d’unepoutre
aunedeform
atio
net/
ouim
pulsio
ninitia
le
Oncon
sidere
lasolu
tionecrite
par
decom
position
modale
:
v(x,t)
= ∑n
(Ancos
ωn t
+Bnsin
ωn t)X
n (x)
Anet
Bnadeterm
iner
avecles
condition
sinitiales
don
nees
:
⎧⎪⎪⎨⎪⎪⎩v(x,0)
= ∑n
An X
n=
v0 (x
)
v(x,0)
= ∑n
ωn B
n Xn=
v0 (x
)⇔ ⎧⎪⎪⎪⎨⎪⎪⎪⎩
An=
ρS
mn ∫
L
0
Xn v
0 (x)dx
Bn=
ρS
ωn m
n ∫L
0
Xn v
0 (x)dx
Integrales
diffi
cilesaevalu
erpou
rdes
Xncom
pliques
→Evalu
ationnumeriq
uedes
Anet
Bn
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
libresen
flexion
Exp
ressiondansla
base
modale
Les
modes
propresverifi
entles
relationssuivan
tes:
∫L
0
X2ndx=
αn
∫L
0
Xn X
′′′′n
dx=
γ4n α
n
∫L
0
X′′2ndx=
γ4n α
n
αncon
stante
arbitraire,
determ
inee
par
norm
alisation.
Equatio
ndumouvem
entlib
re:
ρSv(x,t)
+EIv ′′′′(x
,t)=
0
avecv(x,t)
= ∑nφn (t)X
n (x)
⇔ ∑n
ρSφn X
n+
EIφ
n X′′′′n
=0
⇔ ∑n (
ρSφn ∫
L0XmXn dx
+EIφ
n ∫L0XmX
′′′′n
dx )
=0
⇔ρSφn ∫
L
0
X2ndx
︸ ︷︷︸α
n
+EIφ
n ∫L
0
Xn X
′′′′n
dx
︸︷︷
︸γ4nα
n
=0
(n=
0,...∞)
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
libresen
flexion
Para
metres
modaux
L’eq
uation
dumou
vementdevelop
pee
enbase
modale
don
ne:
ρSv(x,t)
+EIv ′′′′(x
,t)=
0⇔ρSφn α
n+
EIφ
n αn γ
4n=
0
⇔m
n φn+
kn φ
n=
0(n
=0,...,∞
)
avecla
masse
modale
:
mn=
ρS ∫
L
0
X2ndx
=ρSαn
etla
raideu
rmodale
:
kn=
EI ∫
L
0
X′′2ndx
etaussi
kn=
EI ∫
L
0
X′′′′n
Xn dx
ChaquemodeXnse
comporte
comme
unsystem
ea1d
dl:
φn+ω2n φ
n=
0
avecla
pulsa
tionpropre
ω2n=
EI ∫
L0X
′′2ndx
ρS ∫
L0X
2ndx
=kn
mn
UPMC
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Vibratio
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Vibratio
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Rapp
elsur
lesmodes
propreslongitudinaux
Rela
tionsd’orth
ogonalite
&Para
metres
modaux
Orth
ogonalite∀
m=
n∀m
=n
∫L
0
Xn X
m=
0
∫L
0
X′n X
′m=
0
Para
metre
modaux
Masse
modale
:
mn=
ρS ∫
L
0
X2n(x)dx
Raid
eurmodale
:
kn=
ES ∫
L
0
X′2n(x)dx
UPMC
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
longitudinalesforcees
Rep
onse
auneexcita
tionperm
anen
telongitu
dinale
(1)
∂2u
∂t2 −
c2∂2u
∂x2=
1ρSf(x,t)
=g(x,t)
Onsuppose
larep
onse
combinaison
lineaire
des
modes
propres:
u(x,t)
= ∑n
φn (t)X
n (x)
oules
φn (t)
sontles
contrib
ution
sincon
nues
des
modes.
Onsubstitu
ela
composition
modale
dansl’eq
uation
dumou
vement:
∑n
φn X
n −c2 ∑
n
φn X
′′n=
g(x,t)
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
longitudinalesforcees
Rep
onse
auneexcita
tionperm
anen
telongitu
dinale
(2)
∑n
φn X
n −c2 ∑
n
φn X
′′n=
g(x,t)
Onmultip
liepar
Xmet
onintegre
de0aL:
∑n
φn ∫
L
0
XmXn dx−
c2 ∑
n
φn ∫
L
0
XmX
′′ndx= ∫
L
0
Xmg(x,t)d
x
Integration
par
parties
etC.L
pou
rle
secondterm
e:
∑n
φn ∫
L
0
XmXn dx+
c2 ∑
n
φn ∫
L
0
X′mX
′n dx= ∫
L
0
Xmg(x,t)d
x
avecl’orth
ogonalite
des
modes
:
φn ∫
L
0
X2ndx+
c2φ
n ∫L
0
X′2ndx= ∫
L
0
Xn g
(x,t)d
x
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Vibratio
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
longitudinalesforcees
Rep
onse
auneexcita
tionperm
anen
telongitu
dinale
(3)
Com
position
modale
intro
duite
dansl’eq
uation
dumou
vementforce
:
φn ∫
L
0
X2ndx+
c2φ
n ∫L
0
X′2ndx= ∫
L
0
Xn g
(x,t)d
x
Onidentifi
eles
param
etresmodaux:
⇔φnm
n
ρS
+φnkn
ρS
=1ρS ∫
L
0
Xn f(x,t)d
x
⇔m
n φn+kn φ
n= ∫
L
0
Xn f(x,t)d
x
D’ou
l’equation
dela
repon
seforcee
pou
rchaquemode:
⇔φn+ω2n φ
n=
1mn ∫
L
0
Xn f(x,t)d
x
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
longitudinalesforcees
Rep
onse
auneexcita
tionharm
oniquelongitu
dinale
(1)
Onadon
cl’eq
uation
dela
repon
seforcee
pou
rchaquemode:
φn+
ω2n φ
n=
1mn ∫
L
0
Xn f(x,t)d
x
L’excitation
harm
oniquerep
arties’ecrit
:
f(x,t)
=F(x)cos
Ωt
Larep
onse
harm
oniques’ecrit
:
φ(t)
=Φcos
Ωt
etφ(t)
=−Ω
2Φcos
Ωt
Etl’eq
uation
differen
tielledevien
t:
Φ(ω
2n −Ω
2)=
1mn ∫
L
0
Xn F
(x)dx
D’ou
l’amplitu
dedela
repon
seen
fonction
dela
frequence
d’excitation
:
Φ(Ω
)=
1mn
1
ω2n −
Ω2 ∫
L
0
Xn F
(x)dx
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Cas
particulierde
l’excitationpar
vibrationdu
support
Poutre
E-L
-Excita
tionquelco
nque
Lesupport
estan
imed’undeplacem
entvariab
leus (t)
Deplacem
enttotal
des
sectionsdela
poutre
:u(x,t)
+us (t)
Equatio
ndu
mouvem
ent:
∂2(u
+us )
∂t2
−c2∂2(u
+us )
∂x2
=0
⇔(u
+us )−
c2u ′′
=0⇔
u−c2u ′′
=−us
Onutilise
lesmodes
propres
dela
poutre
encastree
libre:
u(x,t)
= ∑n
φn (t)X
n (x)
etX
n=
sin(2n+
1)π
2L
x
Pouridentifi
erles
φn (t),
onreso
utpourtoutus (t)
:
φn+
ω2n φ
n=
−us ρS
mn
∫L
0
sin(2n+
1)πx
2Ldx⇔
φn+
ω2n φ
n=
−2Lus ρS
(2n+
1)π
mn
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Vibratio
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us
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Cas
particulierde
l’excitationpar
vibrationdu
support
Poutre
E-L
-Vibratio
nharm
oniquedusupport
(1)
Pou
runmou
vementdusupport
onapou
rchaquemode:
φn+ω2n φ
n=
−2L
us ρS
(2n+1)π
mn
Sile
mou
vementdusupport
estharm
oniqueon
a:
us (t)
=U0cos
Ωt
etus (t)
=−Ω
2U0cos
Ωt
Larep
onse
estharm
oniquedemem
efreq
uence
φn (t)
=Φ
ncos
Ωt
etφn (t)
=−Ω
2Φncos
Ωt
d’ou(ω
2n −Ω
2)Φn=
2LU0 Ω
2ρS
(2n+1)π
mn ⇔
Φn (Ω
)=
2LU0 ρS
(2n+1)π
mn
Ω2
ω2n −
Ω2
avec
mn=
ρS ∫
L
0
X2ndx=
ρS ∫
L
0 (sin
(2n+1)π
x
2L
)2
dx=
ρSL
2
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Excitation
parvibration
longitudinaledu
support
Poutre
E-L
-Vibratio
nharm
oniquedusupport
(2)
Larep
onse
dechaquemodeau
mou
vementharm
oniquedusupport
est:
φn (t)
=Φ
n (Ω)cos
Ωt=
4U0
(2n+1)π
Ω2
ω2n −
Ω2cos
Ωt
Lemou
vementtotal
dela
pou
treen
vibrationlon
gitudinale
est:
u(x,t)
= ∑n
φn (t)X
n (x)⇔
u(x,t)
=4U
0
π ∑n
1
2n+1
Ω2
ω2n −
Ω2sin (
(2n+1)π
x
2L
)cos
Ωt
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Vibratio
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Excitation
parvibration
longitudinaledu
support
Poutre
E-L
-Vibratio
nharm
oniquedusupport
(3)
u(x,t)
=4U
0
π ∑n
1
2n+1
Ω2
ω2n −
Ω2sin (
(2n+1)π
x
2L
)cos
Ωt
02500
50007500
1000012500
15000−
50
−40
−30
−20
−10 0 10 20 30 40 50
Frequence (H
z)
Amplitude (dB)
Mvt harm
onique du support − R
éponse en fréquence − A
cier − L =
1m
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Rapp
elssur
lesmodes
propresen
flexion
Orth
ogonalite
&Para
metres
modaux
Orth
ogonalite
∫L
0
Xn X
mdx=
0
∫L
0
X′′nX
′′mdx=
0
∫L
0
Xn X
′′′′m
dx=
0
Para
metres
modaux
Masse
modale
:
mn=
ρS ∫
L
0
X2ndx
Raid
eurmodale
:
kn=
EI ∫
L
0
X′′2ndx
etaussi
kn=
EI ∫
L
0
X′′′′n
Xn dx
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibration
forceesen
flexion
Rep
onse
perm
anen
teauneexcita
tionquelco
nque
Equation
dumou
vement:
∂4v
∂x4+
1c2
∂2v
∂t2=
1EIf(x,t)
avecv(x,t)
= ∑n
φn (t)X
n (x)
⇔ ∑n
X′′′′n
φn+
1c2Xn φ
n=
1EIf(x,t)
Orth
ogonalite
des
modes
propres:
⇔ ∑n
φn ∫
L
0
XmX
′′′′n
dx+
1c2φn ∫
L
0
XmXn dx=
1EI ∫
L
0
Xmf(x,t)d
x
⇔m
n
c2ρSφn +
kn
EIφn=
1EI ∫
L
0
Xn f(x,t)d
x⇔φn+ω2n φ
n=
1mn ∫
L
0
Xn f(x,t)d
x
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibration
forceesen
flexion
Rep
onse
auneexcita
tionharm
oniqueponctu
elle(1)
Cas
general
:distrib
ution
quelcon
que:f(x,t)
=F(x)cos
Ωt
φn+
ω2n φ
n= ∫
L0Xn F
(x)dx
mn
cosΩt⇔
φn (t)
= ∫L0Xn F
(x)dx
mn (ω
2n −Ω
2)cos
Ωt
Force
harm
oniquepon
ctuelle
appliquee
enx0:
f(x,t)
=F0 δ(x−
x0 )cosΩt
Lacon
tribution
dela
forceal’excitation
dechaquemodes’ecrit
:∫
L
0
Xn (x
)f(x)dx= ∫
L
0
Xn (x
)F0 δ(x−
x0 )d
x=
F0 X
n (x0 )
Finalem
ent:
v(x,t)
= ∑n
Xn (x
)φn (t)
=F0 ∑
n
Xn (x
0 )Xn (x
)
mn (ω
2n −Ω
2)cos
Ωt
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibration
forceesen
flexion
Rep
onse
auneexcita
tionharm
oniqueponctu
elle(2)
Force
harm
oniquepon
ctuelle
appliquee
enx0:
f(x,t)
=F0 δ(x−
x0 )cos
Ωt
v(x,t)
=F0 ∑
n
Xn (x
0 )Xn (x
)
mn (ω
2n −Ω
2)cos
Ωt
Lacon
tribution
dumodeXnau
mou
vementforce
dependdesa
valeurau
poin
td’ap
plication
dela
force:Xn (x
0 )Par
conseq
uent:
F0aunnoeud:Xn (x
0 )=
0→con
tribution
nulle
F0aunven
tre:Xn (x
0 )=
max→
contrib
ution
max
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
79/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibration
forceesen
flexion
Rep
onse
auneexcita
tionharm
oniqueponctu
elle(3)
Casdela
poutre
appuyee
Xn=
sinnπx
Lm
n=
ρS ∫
L
0 (sin
nπx
L )2dx=
ρSL
2ωn= (
nπL )
2 √EI
ρS
v(x,t)
=2F0
ρSL
∞∑n=1
sinnπx0
Lsin
nπx
L
ω2n −
Ω2
cosΩt
F0en
L/2
v(x,t)
=2F0
ρSL
∞∑n=1
sinnπ2sin
nπx
L
ω2n −
Ω2
cosΩt
=2F0
ρSL
∞∑p=0
(−1)psin
(2p+1)π
xL
ω22p+1 −
Ω2
cosΩt
v(x,t)
=2F0
ρSL (
sinπxL
ω21 −
Ω2 −
sin3πx
L
ω23 −
Ω2+
sin5πx
L
ω25 −
Ω2 −
sin7πx
L
ω27 −
Ω2... )
cosΩt
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
80/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibration
forceesen
flexion
Rep
onse
auneexcita
tionharm
oniqueponctu
elle(4)
Casdela
poutre
appuyee
v(x,t)
=2F
0
ρSL
∞∑n=1
sinnπx0
Lsin
nπx
L
ω2n −
Ω2
cosΩt
010
2030
4050
60−
80
−70
−60
−50
−40
−30
−20
−10 0 10 20 30
Frequence (H
z)
Amplitude (dB)
Acier 5m
m −
Poutre A
ppuyée Flexion −
Réponse en fréquence à une force ponctuelle
Xo qcq
Xo =
L/2
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
81/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Metho
desappro
cheesGen
eralitesP
ourdes
systemes
contin
usdeform
eeven
tuellem
entcom
plexe.
Approxim
ationdes
frequences
propres?
Approxim
ationdela
formedes
modes
propres?
Hyp
otheses
raisonnables
pou
rla
deform
ationdusystem
e:Les
deplacem
ents
verifientles
condition
sauxlim
itesgeom
etriques.
Oncalcu
leles
energies
cinetiq
ues
etpoten
tielles
Equation
sdeLagran
ge→Equation
sdumou
vement
uneseu
lefon
ctiondeform
e:Meth
odedeRayleig
h
Plusieu
rsfon
ctionsdeform
e:Meth
odedeRitz
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
82/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Metho
desappro
cheesExp
ressionsgen
erales
des
energ
iescin
etiqueet
poten
tielle
Pou
rles
vibrationslon
gitudinales
:
T=
12 ∫L
0
ρS (
∂u(x,t)
∂t
)2
dx
U=
12 ∫L
0
ES (
∂u(x,t)
∂x
)2
dx
Pou
rles
vibrationsdeflexion
:
T=
12 ∫L
0
ρS (
∂v(x,t)
∂t
)2
dx
U=
12 ∫L
0
EI (
∂2v(x,t)
∂x2
)2
dx
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
83/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Metho
dede
Rayleigh
Exem
ple
:poutre
encastree
libre
enflexio
n(1)
Approxim
atio
ndela
defo
rmee
enflexio
n:
v(x,t)
= [3 (
xL )2− (
xL )3 ]
φ(t)
→φ(t)?
C.L.geo
metriq
ues
:v(0,t)
=0
∂v(0,t)/∂
x=
0
Energ
ies:
T=
12φ2(t) ∫
L
0
ρS [
3 (xL )
2− (xL )
3 ]2
dx
U=
12φ2(t) ∫
L
0
EI [
6L2 −
6xL3 ]
2
dx
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
84/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Metho
dede
Rayleigh
Exem
ple
:poutre
encastree
libre
enflexio
n(2)
v(x,t)
= [3 (xL )
2− (xL )
3 ]φ(t)
Energ
ies:
T=
0.471ρSLφ2(t)
U=
6EI
L3φ2(t)
Eq.deLagrange
→Equatio
ndumvt
:
ddt
∂T∂φ
+∂U∂φ
=0
⇔0.943
ρSLφ+12
EI
L3φ=
0
Approxim
atio
ndela
Pulsa
tionpropre
ωRayleig
h= √
12EI
0.943ρSL4=
3.567
L2 √
EI
ρS
Onrap
pelle
lava
leurvra
ie:
ωexa
cte=
(γ1 L)2
L2 √
EI
ρS
=3.534
L2 √
EI
ρS
Erreu
rrela
tiveΔωω<
2%.
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
85/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Metho
dede
Rayleigh-R
itzPrin
cipe
Pou
rob
tenirdes
approxim
ationsdes
premieres
frequences
propres:
Com
biner
plusieu
rsfon
ctionsdeform
erealistes
Ces
fonction
sdoiven
tverifi
erles
CLgeom
etriques
Leurcon
tribution
relativeφn (t)
estadeterm
iner
Processu
s
Calcu
lerles
energies
Deriver
lesequation
sdeLagran
ge
Identifi
erles
matrices
d’in
ertieet
deraid
eur
Diagon
aliser→Freq
uences
propresappro
chees
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
86/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Metho
dede
Rayleigh-R
itzExem
ple
:vib
ratio
nslongitu
dinales
d’unepoutre
E-L
(1)
Hyp
othese
dedefo
rmee
=com
binaison
de4fon
ctionsrealistes.
u(x,t)
=xLφ1 (t)
+ (xL )
2
φ2 (t)
+ (xL )
3
φ3 (t)
+ (xL )
4
φ4 (t)
φn (t)?
Rem
arques
:
Les
fonction
sverifi
entles
condition
sauxlim
itesgeom
etriques
Elles
neson
tpas
orthogon
ales
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
87/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Metho
dede
Rayleigh-R
itzExem
ple
:vib
ratio
nslongitu
dinales
d’unepoutre
E-L
(2)
Approxim
atio
ndela
defo
rmee
:
u(x,t)
=xLφ1 (t)
+ (xL )
2
φ2 (t)
+ (xL )
3
φ3 (t)
+ (xL )
4
φ4 (t)
Energ
ies:
T=
12 ∫L
0
ρS (
∂u(x,t)
∂t
)2
dx
=12ρS ∫
L
0 (φ1 (t)
xL+φ2 (t) (
xL )2
+φ3 (t) (
xL )3
+φ4 (t) (
xL )4 )
2
dx
U=
12 ∫L
0
ES (
∂u(x,t)
∂x
)2
dx
=12ρS ∫
L
0 (φ1 (t)
1L+
φ2 (t)
2xL2+φ3 (t)
3x2
L3
+φ4 (t)
4x3
L4 )
2
dx
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
88/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Metho
dede
Rayleigh-R
itzExem
ple
:vib
ratio
nslongitu
dinales
d’unepoutre
E-L
(3)
Matrices
d’in
ertieet
deraideu
r:
M=
ρSL ⎛⎜⎜⎝
1/31/4
1/5
1/61/4
1/51/6
1/71/5
1/6
1/7
1/81/6
1/71/8
1/9 ⎞⎟⎟⎠K
=ESL ⎛⎜⎜⎝
11
11
14/3
3/28/5
13/2
9/52
18/5
216/7 ⎞⎟⎟⎠
Equatio
nmatricielle
dumouvem
ent:
ρSL ⎛⎜⎜⎝
1/3
1/4
1/5
1/6
1/4
1/5
1/6
1/7
1/5
1/6
1/7
1/8
1/6
1/7
1/8
1/9 ⎞⎟⎟⎠ ⎛⎜⎜⎝
φ1
φ2
φ3
φ4 ⎞⎟⎟⎠
+ESL ⎛⎜⎜⎝
11
11
14/3
3/2
8/5
13/2
9/5
21
8/5
216/7 ⎞⎟⎟⎠ ⎛⎜⎜⎝
φ1
φ2
φ3
φ4 ⎞⎟⎟⎠
=0
→Mise
enœuvre
des
techniques
pou
rsystem
esdiscrets
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
89/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Metho
dede
Rayleigh-R
itzExem
ple
:vib
ratio
nslongitu
dinales
d’unepoutre
E-L
(3)
Diagonalisa
tion→
Freq
uen
ceset
vecteurs
propres
approch
es
ω1Ritz=
1.571
L √Eρ
X1Ritz=
xL+0.028 (
xL )2−
0.5 (xL )
3
+0.11 (
xL )4
acom
parer
ala
solutio
nexa
cte:
ω1vra
i=
π2L √Eρ
=1.571
L √Eρ
(Δω1
ω1
=0 )
X1vra
i=
sinπx
2L
etω2Ritz=
4.724
L √Eρ
X2Ritz=
xL+0.69 (
xL )2−
2.56 (xL )
3
+2.06 (
xL )4
acom
parer
ala
solutio
nexa
cte:
ω2vra
i=
3π2L √
Eρ=
4.712
L √Eρ
(Δω2
ω2
=0.25% )
X2vra
i=
sin3πx
2L
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
90/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Metho
dede
Rayleigh-R
itzExem
ple
:vib
ratio
nslongitu
dinales
d’unepoutre
E-L
(3)
Compara
isondes
modes
vrais
etdeleu
rapproxim
atio
n
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
91/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
deflexion
desplaques
Cara
cteristiques
dela
plaque
Form
equelcon
que
Dim
ension
caracteristiquea
Epaisseu
rh�
a
Materiau
hom
ogeneet
isotrope:
ρ,E
,ν
Les
norm
alesau
plan
median
restentperp
endicu
lairesau
plan
median
deform
e.(B
ernou
illi)
Gran
deuraidentifi
er:
w(x,y
,t)//z
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
92/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
deflexion
desplaques
Equatio
nlocale
dumouvem
entlib
re
Form
egen
erale
D∇2∇
2w+
ρh∂2w
∂t2=
0
Encoord
onnees
cartesiennes
:
ρh∂2w
∂t2+
D (∂4w
∂x4+2
∂4w
∂x2∂
y2+
∂4w
∂y4 )
=0
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
93/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
deflexion
desplaques
Solutio
ndel’eq
uatio
ndumouvem
entlib
re
ρh∂2w
∂t2+D (
∂4w
∂x4+2
∂4w
∂x2∂
y2+
∂4w
∂y4 )
=0
Solutio
navaria
bles
separees
w(x,y
,t)=
φ(t)X
(x)Y
(y)
Don
ne:
φ(t)
+ω2φ
(t)=
0⇔φ(t)
=Acos
ωt+Bsin
ωt
et
−ω2ρhXY
+D (
∂4X
∂x4Y
+X∂4Y
∂y4+2∂2X
∂x2
∂2Y
∂y2 )
=0
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
94/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
deflexion
desplaques
Solutio
ndel’eq
uatio
ndumouvem
entlib
re
−ω2ρhXY
+D (
∂4X
∂x4Y
+X∂4Y
∂y4+2∂2X
∂x2
∂2Y
∂y2 )
=0
Don
nedes
solution
sdela
forme:
X(x)=
ax e
αx+bx e −
αx
Y(y)=
ay e
βx+by e −
βx+
cy e
γy+dy e −
γy
avecles
relations:
α2+β2=
−(α
2+γ2)
=ω √
ρhD
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
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us
sept.
14
95/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
deflexion
desplaques
Conditio
nsauxbord
s(1)
Norm
aleau
bord
:n=
(nx ,n
y ),
Tangen
teau
bord
s=
(−ny ,n
x )
Deplacem
ent:w
Rotation
(gradien
t): (
∂w
∂nx ,
∂w
∂ny )
Mom
entFlech
issant:
M=
D [(∂2w
∂x2+
ν∂2w
∂y2 )
n2x+ (
∂2w
∂y2+
ν∂2w
∂x2 )
n2y+
2(1−
ν)
∂2w
∂x∂y2nx n
y ]
Effort
tranchant
T=
∂M∂n
+D
∂∂s [
2(1
+ν) (
∂2w
∂x2−
∂2w
∂y2 )
nx n
y+
2(1−
ν)∂2w
∂x∂y(n
2x −n2y) ]
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
contin
us
sept.
14
96/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Conditions
auxbords
(2)
C.L.
Bord
Libre
Appuisim
ple
Encastrem
ent
w?
00
∂w
∂n
??
0
M0
0?
T0
??
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
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us
sept.
14
97/112
Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
deflexion
desplaques
Deform
atio
nset
contra
intes
Deform
ationsdansle
petit
element(dx,dy,dz):
εx=
−z∂2w
∂x2
εy=
−z∂2w
∂y2
εxy
=−2z
∂2w
∂x∂y
Con
traintes
resultan
tes:
σx=
E
1−ν2(ε
x+
νεy )
=Ez
1−ν2 (
∂2w
∂x2+ν∂2w
∂y2 )
σy=
E
1−ν2(ε
y+
νεx )
=Ez
1−ν2 (
∂2w
∂y2+
ν∂2w
∂x2 )
σxy
=Gεxy
=−
Ez
1+
ν
∂2w
∂x∂y
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
deflexion
desplaques
Energ
ies
Energie
dedeform
ationdupetit
element(dx,dy,dz):
dU
=12(ε
x σx+εy σ
y+εxy σ
xy )dxdydz
Energie
dedeform
ationdela
plaq
ue:
U=
12D ∫
a
0 ∫b
0 [(∂2w
∂x2 )
2
+ (∂2w
∂y2 )
2
+2ν∂2w
∂x2
∂2w
∂x2
+2(1−
ν) (
∂2w
∂x∂y )
2 ]dxd
y
ouD
=Eh3
12(1−
ν2)
:Rigid
itedeflexion
dela
plaq
ue
Energie
cinetiq
uedela
plaq
ue:
T=
12ρh ∫
a
0 ∫b
0
w2dxdy
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
deflexion
desplaques
Casdela
plaquerecta
ngulaire
appuyee
auxbord
s(1)
Seule
configu
rationpou
rlaq
uelle∃
unesolu
tionanalytiq
ue
Avec
lescon
dition
sauxlim
ites,on
obtien
t:
Xm(x)=
sinmπx
Lx
Yn (y
)=
sinnπy
Ly
ωmn=
π2 (
m2
L2x+
n2
L2y ) √
Dρh
Finalem
ent
w(x,y
,t)=
∞∑m=1
∞∑n=1
Amncos(ω
mn t
+φmn )
sinmπx
Lx
sinnπy
Ly
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Vibrations
deflexion
desplaques
Casdela
plaquerecta
ngulaire
appuyee
auxbord
s(2)
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Exem
plede
mise
enœuvre
deselem
entsfinis
Etudedes
vibratio
nsd’ungroupemotopropulseu
r
Diverses
configu
rations
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Exem
plede
mise
enœuvre
deselem
entsfinis
Etudedes
vibratio
nsd’ungroupemotopropulseu
r
Diverses
configu
rations
UPMC
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SdI
Vibratio
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Exem
plede
mise
enœuvre
deselem
entsfinis
Etudedes
vibratio
nsd’ungroupemotopropulseu
r
Com
paraison
des
modes
EF/M
esure
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Exem
plede
mise
enœuvre
deselem
entsfinis
Etudedela
restauratio
nd’uncla
vecinduXVII
eme
UPMC
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Exem
plede
mise
enœuvre
deselem
entsfinis
Etudedela
restauratio
nd’uncla
vecinduXVII
eme
Spectre
acoustiq
uedela
table
d’harm
onie
UPMC
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SdI
Vibratio
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Exem
plede
mise
enœuvre
deselem
entsfinis
Etudedela
restauratio
nd’uncla
vecinduXVII
eme
Com
paraison
des
modes
EF/M
esure
UPMC
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SdI
Vibratio
ns&
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Vibratio
nsdes
systemes
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us
sept.
14
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Exem
plede
mise
enœuvre
deselem
entsfinis
Etudedela
restauratio
nd’uncla
vecinduXVII
eme
Com
paraison
des
modes
EF/M
esure
UPMC
-Master
SdI
Vibratio
ns&
Ondes
Vibratio
nsdes
systemes
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us
sept.
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Analyse
modale
experimentale
Etudedes
vibratio
nsd’unetable
d’harm
onie
devio
lon
Presen
tationdes
mesu
res
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Analyse
modale
experimentale
Etudedes
vibratio
nsd’unetable
d’harm
onie
devio
lon
Identifi
cationdes
param
etresdynam
iques
UPMC
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SdI
Vibratio
ns&
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Vibratio
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systemes
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us
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14
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Analyse
modale
experimentale
Etudedes
vibratio
nsd’unetable
d’harm
onie
devio
lon
Com
paraison
Analyse
modale
/EF
UPMC
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SdI
Vibratio
ns&
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Gen
eralites
Eq.des
VibsL
VibsLlib
resEq.des
VibsT
VibsT
libres
VibsLforcees
VibsT
forceesMeth
odes
approch
eesSystem
es2D
Cequ’il
fautretenir
dessystem
escontinus
Etab
liret
connaitre
lesequation
sdes
ondes
Let
T
Etab
liret
connaitre
leursolu
tiongen
erale(Separation
des
variables)
Interpreter
etEcrire
lescon
dition
sauxlim
ites
Endeduire
lesfreq
uences
etmodes
propres
Con
naitre
lesrelation
sd’orth
ogonalite
des
modes
Exprim
erla
solution
libreen
fonction
des
modes
propres
Determ
iner
lasolu
tionlibre
enfon
ctiondes
CI
Exprim
erla
solution
forceeen
fonction
des
modes
propres
Interpreter
toutcela
physiq
uem
ent
Con
naitre
lesmeth
odes
deRayleigh
etRayleigh
-Ritz
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