Mechanical Vibration of Continuous media.pdf
Transcript of Mechanical Vibration of Continuous media.pdf
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Mechanical Vibration
Vibration of Continuous media
Textbook: Chap 8
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Vibration of Continuous media
Extending previous section to∞ DOF
Textbook: Chap 8
(Distributed Parameter Systems)
Strings , rods and beams :distributed mass and stiffness
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• It is difficult (or time-consuming ) to identifydiscrete masses, dampers and springs incontinuous systems
• They are systems of infinite degree of freedom .
• If they are modeled as discrete systems, theequations are ordinary differential equations(ODE)
• If they are modeled as continuous systems, theequations are partial differential equations (PDE) ,which are more accurate but harder to solve
Introduction
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Introduction• The frequency equation of a continuous
system (for natural frequencies) is atranscendental equation
• Yields infinite number of natural
frequencies and natural modes• Need to apply boundary conditions to find
the natural frequencies, unlike discretesystems.
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The string/cable equation
• Start by considering auniform string stretchedbetween two fixedboundaries
• Assume constant, axialtension in string
• Let a distributed force f ( x,t ) act along thestring
f (x,t)
τ
x
y
http://www.kettering.edu/~drussell/Demos/HangChain/HangChain.html
http://www.kettering.edu/~drussell/Demos/string/Fixed.html
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Examine a small element of string
xt x f t
t xw x F y
Δ++−=Δ=∑
),(sinsin
),(
2211
2
2
θ τ θ τ ∂
∂ ρ
• Force balance on an infinitesimal element
• Now linearize the sine with the small angleapproximate sin x = tan x = slope of the string=
θ1θ2
τ2
τ1x1 x2 = x 1 +Δx
w (x ,t )
f (x ,t ) m × a y
x
xw
∂∂
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2
2
2
2
2
1
2
2
),(),(
),(
),(),(
),(
)(
:about/of seriesTaylortheRecall
),(
),(
),(),(
1
112
12
t t xw
t x f x
t xw x
xt
t xw xt x f x
xt xw
x
xO xw
x x xw
xw
x xw
xt
t xw
xt x f x
t xw
x
t xw
x
x x x
x x
∂ ∂
ρ ∂
∂ τ
∂ ∂
∂ ∂
ρ ∂
∂ τ
∂ ∂
∂ ∂
τ ∂ ∂
∂ ∂
τ ∂ ∂
τ
∂ τ∂
∂
∂
ρ ∂
∂
τ ∂
∂
τ
=+⎟ ⎠ ⎞
⎜⎝ ⎛
⇒Δ=Δ+Δ⎟
⎠
⎞⎜
⎝
⎛
+Δ+⎟ ⎠ ⎞
⎜⎝ ⎛
Δ+⎟ ⎠ ⎞
⎜⎝ ⎛
=⎟ ⎠ ⎞
⎜⎝ ⎛
Δ=Δ+⎟ ⎠ ⎞
⎜⎝ ⎛
−⎟ ⎠ ⎞
⎜⎝ ⎛
K
EOM Linearization xt x f Δ+− ),(sinsin 1122 θ τ θ τ
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0 ,0),(),0(
0at)()0,( ),()0,(
,),(),(
00
2
22
2
2
>==
===
==
t t wt w
t xw xw xw xw
c x
t xwc
t t xw
t
l
&
ρ τ
∂ ∂
∂ ∂
Since τ is constant , and for no external force the equationof motion becomes:
Second order in time and second order in space , therefore4 constants of integration. Two from initial conditions:
And two from boundary conditions:
, wave speed
Free Vibration Equation for String or Cable
cf. Eq. (8.9)
Wave equation
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Physical quantities
• Deflection is w( x,t ) in the y-direction• The slope of the string is w x( x,t )• The restoring force is τw
xx( x,t )
• The velocity is wt ( x,t )• The acceleration is wtt ( x,t )
at any point x along the string at time t
Note that the above applies to cables as well as strings
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The solution
∑∞=
+=
+=+=
1
)sin()cos()sin()sin(),(
)sin()cos()sin()sin(
sincossinsin),(
nnn
nn
nnnnnnn
xnct nd xnct nct xw
xn
ct n
d xn
ct n
c
xct d xct ct xw
llll
llll
π π π π
π π π π σ σ σ σ
The derivation of the solution to wave equation is neglected here
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Mode shape (Normal mode, Eigenfunction)
⎟ ⎠ ⎞
⎜⎝ ⎛ =
⇒=
===
∀=
=
∫
t c
xt xw
d
ndx xn
xd
nc
n x xw
n
n
ll
Klll
l
l
π π
π π
π
cos)sin(),(
1
3,2 ,0)sin()sin(2
,0
1)=(ioneigenfunctfirsttheis which,sin)(
1
0
0
= Called the 1st mode of vibrationOr 1st harmonic or 1st normal mode→vibration in the first mode shape
l
π ω
ncn = = N-th frequency
)sin()cos()sin()sin(),( xn
ct n
d xn
ct n
ct xw nnn llllπ π π π +=
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Plots of mode shapes
0 0.5 1 1.5 2
1
0.5
0.5
1
X ,1 x
X ,2 x
X ,3 x
x
sinn
2 x
⎝
⎞
⎠
nodes
n=1
n=2
n=3
See the animation: http://www.kettering.edu/~drussell/Demos/string/Fixed.html
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Example: Piano wireL=1.4 m, τ=11.1x10 4 N, m =110 gramsCompute the first natural frequency.
=110 g per 1.4 m = 0.0786 kg/m
ω 1 = c
l=
1.4τ
=
1.411.1 ×10 4 N0.0786 kg/m
=2666.69 rad/s or 424 Hz
Ref: Piano has 88 keys.
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Bending vibrations of a beam
2
2 ),()(),(
aboutinertia
of momentareasect.-cross)(
modulusYoungs
)(stiffness bending
xt xw
x EI t x M
z
x I
E
x EI
∂ ∂ =
==
=Next sum forces in the y - direction (up, down)
Sum moments about the point Q
Use the moment given from
stenght of materials
Assume sides do not bend
(no shear deformation)
f (x ,t)
w (x ,t)
x
dx A(x)= h 1h
2
h 1
h 2
M (x ,t )+M x (x ,t )dx
M (x ,t )
V (x ,t )
V (x ,t )+V x (x ,t )dx
f (x ,t )
w (x ,t )
x x +dx
·Q
(Transverse vibration)
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Summing forces and moments
0)(2
),(),(),(
),(
02
),(
),(),(),(
),(),(
),()(),(),(),(),(
2
2
2
=⎥⎦
⎤⎢⎣
⎡ ++⎥⎦
⎤⎢⎣
⎡ +⇒
=+
⎥⎦
⎤⎢⎣
⎡ ++−⎟ ⎠ ⎞
⎜⎝ ⎛ +
=+−⎟ ⎠ ⎞⎜⎝ ⎛ +
dxt x f
xt xV
dxt xV dx x
t x M
dxdxt x f
dxdx x
t xV t xV t x M dx
xt x M
t x M
t t xwdx x Adxt x f t xV dx
xt xV t xV
∂ ∂
∂ ∂
∂ ∂
∂ ∂
∂ ∂ ρ
∂ ∂
0
m × a
I × α → 0
F
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A EI c
xt xwc
t t xw
t x f x
t xw x EI xt
t xw x A
t t xw
dx x Adxt x f dx x
t x M
xt x M
t xV
ρ ∂ ∂
∂ ∂ ρ
∂ ∂
∂ ∂
∂ ∂ ρ
∂ ∂
ρ ∂
∂
∂ ∂
==+
=⎥⎦
⎤⎢⎣
⎡+
=+−
−=⇒
,0),(),(
),(),()(),()(
),()(),(
),(
),(),(
4
42
2
2
2
2
2
2
2
2
2
2
2
2
Substitute into force balance equation yields:
Dividing by dx and substituting for M yields
Assume constant stiffness to get:
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Spatial equation (BVP)
xa xa xa xa x X
Ae x X EI
A
c
x X c
x X
x
β β β β
ω ρ ω β
ω
σ
coshsinhcossin)(
:getto)(Let
Define
.0)()(
4321
224
2
+++==
=⎟ ⎠
⎞⎜
⎝
⎛ =
=⎟ ⎠ ⎞⎜
⎝ ⎛ −′′′′
Apply boundary conditions to get 3constants and the characteristic equation
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Solve numerically ( fsolve in Matlab) to obtain solution totranscendental equation
4)14(
5
493361.16351768.13
210176.10068583.7926602.3
54
321
π β
β β
β β β
+=
⇒>==
===
n
n
nl
Kll
lll
⎥⎦
⎤⎢⎣
⎡ +−−−−= x x x xa x X nnnn
nn
nnnn
llllll
ll β β β β
β β β β coscosh)sin(sinh
sinsinhcoscosh)()( 4
Eigenvalues & eigenfunctions
ll β β tanhtan =
EI A
c
224 ω ρ ω β =⎟
⎠ ⎞
⎜⎝ ⎛ =
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Mode shapes
X ,n x .cosh n cos n
sinh n sin nsinh .n x sin .n x cosh .n x cos .n x
0 0.2 0.4 0.6 0.8 1
2
1.5
1
0.5
0.5
1
1.5
X ,3.926602 x
X ,7.068583 x
X ,10.210176 x
x Mode 1
Mode 2Mode 3Note zero slope
Non zero slope
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Natural Freq of the Euler-Bournoulli Beam
42)(
l EI
l nn ρ β ω =
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Rayleigh ’s Method
• Consider the beam shown below.
• Kinetic energy• Assuming w( x,t )=W ( x)cos ω t , maximum
KE:
( )∫∫ == l l
dx x AwdmwT 0
20
2
21
21 ρ &&
( ) ( )∫= l
dx xW x AT 02
2
max 2 ρ ω
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Rayleigh ’s Method• Ignoring work done by shear forces, potential
energy:
• Thus
• Max value of w( x,t ) is W ( x). Hence
( ) ( ) ( ) xwt x xw x EI t x M Md V l
∂∂=∂∂== ∫ θ θ and ,, where21 22
0
∫∫ ⎟⎟
⎠ ⎞
⎜
⎜
⎝ ⎛
∂∂
=∂∂⎟
⎟
⎠ ⎞
⎜
⎜
⎝ ⎛
∂∂
=
l l
dx x
w
EI dx x
w
x
w
EI V 0
2
2
2
0 2
2
2
2
2
1
2
1
( ) ( )∫ ⎟⎟ ⎠ ⎞
⎜⎜
⎝ ⎛
∂∂=
l dx
x xW
x EI V 0
2
2
2
max 21
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Rayleigh ’s Method• Equating T max to V max , Rayleigh ’s quotient
• For a stepped beam,
( ) ( ) ( )
( )( )∫∫ ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
== l
l
dx xW A
dxdx
xW d x EI
R
0
2
0
2
2
2
2
ρ ω ω
( )...
21
21
0
220
21
0
2
2
2
220
2
2
2
11
2
++
+⎟⎟
⎠
⎞⎜⎜
⎝
⎛ +⎟⎟
⎠
⎞⎜⎜
⎝
⎛
== ∫∫∫∫
l l
l l
dxW AdxW A
dx
dx
W d I E dx
dx
W d I E
R ρ ρ
ω ω
L
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Rayleigh ’s Method for the Cantilever Beam
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Rayleigh ’s Method for the Cantilever Beam
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Rayleigh ’s Method for the Cantilever Beam
MECH300GD YK L
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ρ π ρ π β EI
l l EI l
f nn 242 515.3
21
2)( ==
Rayleigh ’s Method for the Cantilever Beam
Continuous theory: β 1l = 1.875104
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Example 8.12• Find the fundamental frequency of transverse
vibration of the nonuniform cantilever beamshown below, using the deflection shapeW ( x)=(1- x / l)2
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Solution
• Cross sectional area
• Moment of inertia
• Rayleigh’s quotient
• The exact value of ω is known to be
( )l
hx x A =
( )3
121
⎟ ⎠ ⎞
⎜⎝ ⎛ =
l hx
x I
( ) 42
4
2
0
4
0
2
23
33
2 5811.1or5.2
1
212
l
Eh
l
Eh
dxl
xl
hx
dxl l
xh E
Rl
l
ρ ω
ρ ρ
ω ω ==⎟ ⎠ ⎞⎜
⎝ ⎛ −⎟
⎠ ⎞⎜
⎝ ⎛
⎟ ⎠ ⎞
⎜⎝ ⎛ ⎟⎟ ⎠ ⎞
⎜⎜
⎝ ⎛
==
∫
∫
4
2
5343.1l
Eh ρ
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Example of String VibrationMusical instruments
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String Excitation Sources: Plucked String(harp, guitar, mandolin)
http://www.mandolincafe.com/archives/briefhistory.html
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- Plucking a string provides it with an initial energydisplacement (potential energy)
- The shape of the string before its release completely definesthe harmonic signature of the resulting motion
- A string plucked at 1/n-th the distance from one end will nothave energy at the n-th partial and its integer multiples
- The strength of excitation of the n-th vibrational mode isinversely proportional to the square of the mode number
String Excitation Sources: Plucked String
⎭
⎬⎫
⎩
⎨⎧ +−= L
L
ct
L
x
L
ct
L
xht xw
π π π π
π
3cos
3sin
9
1cossin
8),( 2
L x
Ex 8.1, p.508
hw( x,t )
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String Excitation Sources: Struck String (piano)
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- A struck string is given an initial velocity distribution (kineticenergy)
- A string struck at 1/n-th the distance from one end will nothave energy at the n-th partial and its integer multiples
- The harmonic amplitudes in the vibration spectrum of a
struck string fall off less rapidly with frequency than those ofplucked strings.
- Light “hammers ” (mass much less than the mass of the
string) result in little spectral drop-off with frequency. Heavierhammers produce a drop-off roughly proportional to theinverse of the mode number
String Excitation Sources: Struck String
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String Excitation Sources: Bowed String(violin family: violin, viola, cello and ??)
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Musical Sound quality : Vibration Spectra
from Giancoli, Phys. For Scientists and Engineers
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Frequency of Piano Keys
http://www.vibrationdata.com/piano.htm
Middle C = 261.63 Hz
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SPM-like bio-sensing
1981 Scanning Probe Microscopy invented by
Binnig and Rohrer; cantilevers use to imageatomic structure, magnetic properties o fatoms and biological molecules
1986 Binnig and Rohrer awarded Nobel Prize
~1996 Cantilevers use as b iological sensorsbegins
http://www.sciencemag.org/cgi/content/full/288/5464/316
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Biochemo-optomechanical chip
http://web.mac.com/majumdargroup/iWeb/Site/Biosensing.html
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Start of construction = 23 Nov 1938
Opened for traffic = 1 Jul 1940
Collapse of bridge = 7 Nov 1940
L = 2800 ft = span between towersh = 232 ft = maximum sag of cables
b = 39 ft = wdith between cables
d = 17 in = diameter of cable
h/L = 0.0823 = 1/12 = sag-to-span ratio
b/L = 0.0139 = 1/72 = width-to-span ratio
Vibration of Tacoma Bridge as a Cable
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w f
= 4300 lb/ft = f loor weight/ft along the bridge
wg = 323 lb/ft = girder weight/cable/ft
wc = π / 4 (17/12)2* 0.082*490= 632 lb/ft of cable
wt = 4300/2 + 320+632 = 3105 lb/ft = total weight carried per cableρ= wt /g = 3105/32.2 = 96.4 lb*ft 2*s2
T 0= total tension at the tower = 13.82 ×10 6 lb
Vibration of Tacoma Bridge as a Cable
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cpm4
cpm3.95nHz0658.0
104.961.13
2800226
≈==
××
==
n
nT Ln
f n ρ
f 1 ~ 4cpm, f 2 ~ 8cpm:
consistent with Prof F.B. Farquharson’s report about Tacoma bridge
cpm = cycle per minute, 1 cpm = 60 Hz
Vibration of Tacoma Bridge as a Cable
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Torsional Vibration of Tacoma Bridge
The girder & the floor are both open sections →Torsional stiffness K g Kf very small
Consider a pair of cables spaced b ft apart & under tension T :For 3 consecutive stations: i-1, i, i+1
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Torsional Vibration of Tacoma Bridge
If the cross section at i has a small roation θ, yi = x φ = (b/2) θ , the vertical component of T is
x
Tb
x
bT T F θ
θ φ ===2
22
The torque of the cable is Fb = Tb 2θ /xTorsional stiffness = Torque per unit length of the cableTb2 = 13.11 ×10 6 ×39 2 = 19,900 10 6 lb*ft 2
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2
2
2
2
x K
t J
∂∂=
∂∂ θ θ
0sin = L K
J ω
π π π ω n L K J
i ,,2, K=
J
K
L f
12 =
s94.3
10900,19400,39
2800 32
=
×== − K J
Lτ
Torsional Vibration of Tacoma Bridge
Freq equation