june 29, 2021 silent engineering
TRANSCRIPT
June 29, 2021
Silent Engineering(Lecture 3)
Forced Vibration Analysis of Plate and Formation of Sound Field
Due to Vibrating Source- Force Vibration and
Point sound source and wave equation -
Tokyo Institute of TechnologyDept. of Mechanical EngineeringSchool of Engineering
Prof. Nobuyuki Iwatsuki
１. Examples of Forced Vibration Based on Approximated Modal Analysis withRayleigh-Ritz Method
1.1 Vibration responses of peripherally clamped thin rectangular plate
Thickness: hDensity: ρYoung’s modulus: EPoison’s ratio: νPoint driving force: P(t)=P0cosωt
xa
b
O
y
Driving point
Response point
(xd,yd)
(xr,yr)
( )( ) ( )
)1(2tan
2
cos),(),,(
221
2222
Ni
tyxWQtyxw
i
iii
iiii
iii
～
=−
=
+−
−=
−
∑
ωωωωζφ
ωωζωω
φω
Vibration displacement:
(1)
),()( ddii yxWtPQ =Modal load for point excitation:
(2)Because the plate is excited by a force, P(t), at a certain point(xd,yd) , the distributed external force, p(x,y,t) can be represented as
)()()(),,( dd yyxxtPtyxp −−= δδwhere δ denotes Dirac’s delta function
),()(),( ddS iii yxWtPdsyxpWQ ∫ ==
Therefore Explained in the 1st lecture
Vibration velocity:( )
( ) ( )∑
+−
−−=
iiii
iii tyxWQtyxw2222 2
sin),(),,(ωωζωω
φωω (3)
( )( ) ( )
∑+−
−−=
iiii
iii tyxWQtyxw2222
2
2
cos),(),,(ωωζωω
φωω
Vibration acceleration:
(4)
where
−−−⋅
−−−=
⋅=
∑
∑
by
by
by
by
ax
ax
ax
ax
C
yWxWCyxW
jyjyjy
jyjy
ixixix
ixix
jiij
jyji
ixij
,,,
,,
,,,
,,
,
,,
,
sinsinhcoscosh
sinsinhcoscosh
)()(),(
λλα
λλ
λλα
λλ
(5)i λi
1 4.7302 7.8533 10.996: :
Coefficients, Cij, are set so that modal mass becomes unity.
Acceleration distribution, w(x,y,t), can be calculated.
The following frequency spectra of transfer functions are also calculated.
Compliance :)()()(
ωωω
jFjXjD = (6)
)()()(
ωωωjFjVjY = (7)Mobility :
)()()(
ωωωjFjAjG = (8)Accelerance :
Modal damping ratio, ζi, can be given as the experimentally measuredor estimated values.
FFT Analyzer
Modal Analysis SoftwareLX-cada
Accelerometer
ExciterForcesensor
Setup for Experimental Modal Analysis
Mode ζ
(0,0) 0.0222(1,0) 0.0149(2,0) 0.0242(0,1) 0.0541(1,1) 0.0248(3,0) 0.0142
Mode ζ
(2,1) 0.0120(4,0) 0.0130(3,1) 0.0095(0,2) 0.0100(1,2) 0.0211(4,1) 0.0109
Mode ζ
(5,0) 0.0254(2,2) 0.0148(3,2) 0.0115(5,1) 0.0120(6,0) 0.0491(4,2) 0.0106
The experimentally measured modal damping ratio
Frequency spectra of acceleranceof peripherally clamped rectangular thin plate
Driving point: (xd,yd)=(113.4,149.25)Response point: (xr,yr)=(1100.0,450.0)
(2,0)(1,0)
(0,0)
(0,1)(1,1)(2,1)
(4,0)(3,1)
(4,1)
(2,2) (3,2)
(5,1)
(4,2)
Driving point: (xd,yd)=(113.4,149.25)Response point: (xr,yr)=(663.4,374.75) Node
(0,0)(2,0)
(4,0)(2,2) (4,2)
(0,2)
1.2 Examples of calculation
Accelerance takes its peak values at natural frequency
In case where response point is located on nodal line, corresponding mode is not induced.
Frequency spectra of acceleranceof peripherally clamped rectangular thin plate
Driving point: (xd,yd)=(663.4,374.75)Response point: (xr,yr)=(1100.0,450.0)
Driving point: (xd,yd)=(113.4.4,374.75)Response point: (xr,yr)=(1100.0,450.0)
NodeNode
In case where driving point is located on nodal line, corresponding mode is not induced.
2. Sound Generation Mechanism2.1 Fundamental equation of sound wave
Sound pressure applying to infinitesimal element of medium
Sound pressure : p(x,y,z)Density of medium : ρDisplacement of medium ; d(ξ,η,ζ)
By propagating sound to the element, density, ρ, pressure at a point (x,y,z) and volume,V, were changed as ρ0, P0 , V0 → ρ, P, V
The sound pressure at the point can be defined as
)(),,,( 0 tPPtzyxp −= (6)The volume is calculated as
zyxzyx
zyxzyx
V
∆⋅∆⋅∆
∂∂
+∂∂
+∂∂
+≅
∆⋅∆⋅∆
∂∂
+
∂∂
+
∂∂
+=
ζηξ
ζηξ
1
111
(7)
The ratio of volume change is given as
dd •∇==∂∂
+∂∂
+∂∂
=−
=∆ div0
0
zyxVVV ζηξ (8)
Since mass of the elements should be kept, the following equation holds.VV ρρ =00 (9)
Therefore )1(0 ∆+= ρρ (10))1(0 s+= ρρ (11)
:∆ dilatation
s: condensation
∆−=s (12)
( )
ζηξ ,,u =
By differentiating Eq.(8) with respect to time, we obtain
ud divdiv =∂∂
=∂∆∂
ttwhere particle velocity is
(13)
(14)From Eqs.(10) and (13), we obtain
ud divdiv0 ==∂∂ ρρ
t(15)
Equation of continuity
Equation of motion in x-direction can be expressed as
( )
zyxxp
zyxxppzyx
xpp
tzyx
∆⋅∆⋅∆∂∂
−=
∆⋅∆
∆
∂∂
+−∆⋅∆
∆
∂∂
−=∂∂
∆⋅∆⋅∆21
21ξρ
Namely
xp
t ∂∂
−=∂∂ξρ (16)
Also in y- and z- direction, we obtain
p-pt
∇=−=∂∂ graduρ (17)
In the infinitesimal element, the following equation holds
tts
t ∂∂
≅∂∂
+=∂∂ uuu
00 )1( ρρρ (18)
2.2 Wave equation
∆−== KKspBy using bulk modulus, K,
(20)
By differentiating Eq.(19) with respect to time, we obtain
udivKzyx
K
zyxtK
tK
tp
−=
∂∂
+∂∂
+∂∂
−=
∂∂
+∂∂
+∂∂
∂∂
−=∂∆∂
−=∂∂
ζηξ
ζηξ
(19)
By differentiating Eq.(20) with respect to time, we obtain
pKzp
yp
xpKpK
tK
tp
2
2
2
2
2
2
2
2
2
grad1-divdiv
∇=
∂∂
+∂∂
+∂∂
=
−=
∂∂
−=∂∂
ρ
ρρu
(21)
pctp 222
2
∇=∂∂
Wave equation
By setting sound speed as
ρKc = (22)
(23)
The wave equation of particle velocity is also given as
uu divgrad2
2
•=∂∂
ρK
t(24)
Wave equation with respect to particle velocity
2.3 Velocity potential
By using velocity potential, φ, we obtain
tp
ztytxt ∂∂
=∂∂
=∂∂
∂∂
=∂∂
∂∂
=∂∂ φρφζφηφξ
,,, (25)
The wave equation can be rewritten as
∂∂
+∂∂
+∂∂
=∇=∂∂
2
2
2
2
2
2222
2
2
zyxcc
tφφφφφ (26)
Wave equation with velocity potential
2.4 Solution of wave equationPlane sound wave:
2
22
2
2
xc
t ∂∂
=∂∂ φφ
(27)
Wave equation:
Solution:
)()(
)cos()cos(),(kxtjkxtj BeAe
kxtBkxtAtx−− +=
++−=ωω
ωωφ
(28)
where
λπω 2
==c
k wavenumber (29)
Spherical sound wave:Sound wave radiating from spherical sound source
Wave equation:
∂∂
+∂∂
+∂∂
=∂∂
2
2
2
2
2
22
2
2
zyxc
tφφφφ
Spherical coordinate system
By substituting coordinate transform as
θϕθϕθ
cossinsincossin
rzryrx
===
(30)
into the wave equation, we obtain
∂∂
+
∂∂
∂∂
+∂∂
+∂∂
=∂∂
2
2
2222
22
2
2
sin1sin
sin12
ϕφ
θθφθ
θθφφφ
rrrrrc
t (31)
Since velocity potential is a function of the radius, we obtain
2
22
2
2 )()(rrc
tr
∂∂
=∂
∂ φφ(32)
Solution:
)(
)cos(krtjAe
krtAr−=
−=ω
ωφConsider only divergent wave
)(
)cos(
krtjerA
krtrA
−=
−=
ω
ωφTherefore
(33)
Sound pressure and particle velocity can be given as
)(
)(
1 krtjr
krtj
erAjk
rr
erAj
tp
−
−
+=
∂∂
−=
=∂∂
=
ω
ω
φξ
ωρφρ
(34)
(35)
Sound wave from pulsating sphere:Point sound source
Let’s consider the harmonic vibration in radial direction on sphericalsound source with radius r0.
tjr e ωξξ 0
=
Since particle velocity on spherical surface is equal to the harmonic vibration velocity,the following equation holds.
tjkrtjr ee
rAjk
rωω ξξ 0
)(
00
01 =
+= − (36)
Therefore
00
20 1
0
jkrerA
jkr
+= ξ (37)
If
10 <<kr
02
00
0
44
ξππ
rA
AA
=
=
where
(38)
)(0
)(0
)(0
41
4
4
krtjr
krtj
krtj
er
Ajkrr
er
Ajt
p
er
A
−
−
−
+=
∂∂
−=
=∂∂
=
=
ω
ω
ω
πφξ
πωρφρ
πφ
Therefore, velocity potential, sound pressure and particle velocity can be represented as
(39)
(40)
(41)
Important equations for point sound source
Point sound source on a plane:
Low of mirrors
Low of mirrors
Since a rigid plane does not vibratethe particle velocity on the reflectionpoint on the plane should be zero and there is no change of velocitypotential based on the low of mirrors.
Point sound source of mirror image
Actual point sound source
Reflection point
Rigid plane
Observationpoint
r1
r2
Therefore the velocity potential at an observation point P can be given as
)(
2
0)(
1
0 21
44krtjkrtj e
rAe
rA −− += ωω
ππφ (42)
Therefore let’s consider two point sound sources with same sound intensity and phase just at both side of the rigid plane.
Observationpoint
Two point sound sources at both side of rigid plane
The following conditions are takeninto account.
θ
θ
sin2
sin2
2
1
drr
drr
+≅
−≅r>>d
(43)
By substituting Eqs.(43) into Eq.(42), we obtain
+=
−− θθω
πφ
sin2
1
sin2
1
)(0 114
kdjkdjkrtj er
er
eA (44)
rrrr ≅≅ 21 ,By assuming that ,
)sin2
cos(2
)(0 θπ
φ ω kder
A krtj −= (45)Directivity
Let assume that d→0
)(0
2krtje
rA −= ω
πφ (46)
Sound radiation from vibrating plate:Let’s assume that many point sound source are located on the vibrating plate and radiate sound.
Infinite rigid plane
Vibratingplate
)(0
)(0
2
2krtj
krtj
er
dS
er
Ad
−
−
=
=
ω
ω
πξπ
φ
Velocity potential at an observationpoint due to a point sound sourceof infinitesimal element onthe vibrating plate can begiven as
(47)Velocity potential at an observationpoint can then be given as
１∫ −=S
)(0
2dSe
rkrtj ωξ
πφ
(48)
∫
∫
−
−
=
=∂∂
=
S
)(0
S
)(0
2
2
dSer
dSer
jt
p
krtj
krtj
ω
ω
ξπρ
ξωπρφρ
Sound pressure at the observation point can then be given as
Sound pressure can be calculated with vibration acceleration distribution
on vibrating plate
(49)
2.5 An example of calculationLet’s consider a simple case where a peripherally simply supported thin rectangular plate vibrates with a certain vibration mode shape at its natural frequency. All vibration modes have the same maximum vibration amplitudeof 0.2mm.
Calculated in last week.
NnMmtb
yna
xmCtyxw mn ,...,2,1,,...,2,1,cossinsin),,( === ωππ
(50)
a=1326.8mm b=749.5mm h=4.5mm E=205GPaν=0.3ρ=7860kg/m3
dxdyd
tyxw
dSd
tyxwtzyxp
b aair
Sair
PPP
∫ ∫
∫
=
=
0 0
)',,(2
2)',,(),,,(
πρ
πρ
( ) ( )
air
PPP
vdttyyyxxd
/'
222
−=
+−+−=
Sound pressure at the observation point:
where
xy
z
O
dxdy
),,( PPP zyxObservation point P:
d
ab
(1,1)
(3,1) (5,1)
(1,3)(3,3)
(7,1)(5,3)
(2,1)
(1,2)
(2,2)
(4,1)(3,2)
(4,2)
(2,3)
(5,2)
(6,1)
(6,1)
(6,2)
(7,2)
(6,3)
SPL of the mode having oddnumber nodal lines is small.
Pa20,log20 10 µ== refref
RMS pppSPL
Sound pressure level:(50)
Observation point: )0.1000,75.474,4.763(),,( =OOO zyxA little bit far from the center of plate
Effect of mode shape:
＋
＋
－－
＋
＋
－－
＋＋
－－
＋
A vibration mode having odd number nodal lines
A vibration mode having even number nodal lines
Sound pressures formed by vibration area with opposite phases tend to offset each other.
ds
ds
P P
d
d ds
ds
d
dw‥
-w‥
w‥
w‥
Interference of two soundwaves with opposite phase minimize SPL.
Interference of two soundwaves with the same phase make SPL twice.
(7,1) mode at 323.51Hz, The observation point :(0.0, 0.0, zO)
DISTANCE zO m
Distance
SPL is affected by the distance between sound source and observation point.
Effect of distance:
z
xy
O
θ
ϕ
Rd
Observation point P:
)cos,sinsin,cossin(),,(
θϕθϕθ RRRzyx PPP
=
Vibrating rectangular
Acceleration w(x,y)‥
Sound pressure p
Effect of direction:
Azimuth
Zenith angle
θ=0° 10°
20°
30° 40°50°
60°
70°
80°
(7,1) mode at 323.51Hz, Radius of the observation point is 1m
SPL is affected by the dirction angle from sound source to observation point.
Please note that these results are obtained froma simple vibration such as only a certain mode shapeand that we will have to consider the response as thesummation of all mode of vibration and the responsecharacteristics which should be determined by excitingpoints.
3. Concluding remarksExamples of forced vibration analysis and soundgeneration mechanism are explained. (1)Acceleration distribution on vibrating plate can
be calculated based on modal analysis of the plate.
(2)Frequency spectrum of accelerance takes peakvalues at natural angular frequency.
(3)Vibration response is strongly affected by positions of excitation and response.
(4)Sound radiation can be calculated by solvingsound wave equation of velocity potential.
(5)Sound pressure radiating from a vibrating platecan be calculated with accelerationdistribution of vibrating plate.
(6)Sound pressure radiating from vibratingplate is strongly affected by mode shapeand distance and direction between sound source and observation point.
3. Concluding remarks