Mechanical Vibration of Continuous media.pdf

8/16/2019 Mechanical Vibration of Continuous media.pdf

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MECH300GDr. YK Lee

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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MECH300GD YK L


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ρ π ρ π β EI

l l EI l

f nn 242 515.3

21

2)( ==


Continuous theory: β 1l = 1.875104

MECH300GDr YK Lee


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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

MECH300GDr YK Lee


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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

MECH300GDr YK Lee


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Example of String VibrationMusical instruments

MECH300GDr YK Lee


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String Excitation Sources: Plucked String(harp, guitar, mandolin)

http://www.mandolincafe.com/archives/briefhistory.html

MECH300GDr. YK Lee


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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 )

MECH300GDr. YK Lee


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String Excitation Sources: Struck String (piano)

MECH300GDr. YK Lee


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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

MECH300GDr. YK Lee


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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

MECH300GDr. YK Lee


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Frequency of Piano Keys

http://www.vibrationdata.com/piano.htm

Middle C = 261.63 Hz

MECH300GDr. YK Lee


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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

MECH300GDr. YK Lee


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Biochemo-optomechanical chip

http://web.mac.com/majumdargroup/iWeb/Site/Biosensing.html

MECH300GDr. YK Lee


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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

MECH300GDr. YK Lee


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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


MECH300GDr. YK Lee


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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



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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

MECH300GDr. YK Lee


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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

MECH300GDr. YK Lee


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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


Freq equation

Mechanical Vibration of Continuous media.pdf

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