Hugh Hunt, Trinity College, Cambridge *

Hugh Hunt, Trinity College, Cambridge www.hughhunt.co.uk

*


Some Counterintuitive Problems in Vibration

Hugh Hunt

Cambridge University Engineering Department


Something is counter-intuitive if:

• it requires advanced/specialist knowledge

• it is obscure or difficult to observe

• it doesn’t fit with our experience

• we’ve never noticed it before

• we believed what our teachers said


VIBRATION

“Common sense will carry one a long waybut no ordinary mortal is endowed with an inborn

instinct for vibrations”.

“Vibrations are too rapid for our sense of sight … common sense applied to these phenomena is too common to be other than a source of danger”.

Professor Charles Inglis, FRS from his “James Forrest” Lecture, Inst Civil Engineers, 1944


“design process”

concept

iteration

product

vibration problemsVibrationConsultant


Important concepts

• StiffnessFrequency =

• Massmk

m

k c

The mkc model

• Nodal points• Vibration modes• Non-linearity• Damping


Helmholtz ResonatorNeck plug of mass m

Contained air of stiffness k

mk

m

k

V1

Smallervolumeof air:

stiffnessincreased

V2

Wallsmade

flexible: stiffness

decreased

V2 V2

Water recreates rigid enclosure:

stiffnessincreased



The tips of the tuning fork move on thearcs of circles and centrifugal inertia forces are generated, twice per cycle.

Suppose tip amplitude is 0.2mm, oscillating frequency is 440Hz, moving mass is 20% of the fork mass, thenthe 880Hz component of tip force Fis about 10% of the weight of the fork.

F

P

Tuning Fork:“P” is a nodal point, so why do we get more sound when “P” is put on a table?


mode 1 mode 2 mode 3 mode 4

AXIALVIBRATION

cLn

fn 2

E

c where

1n 2n 3n 4n


mode 1 mode 2 mode 3 mode 4

EULERBENDINGVIBRATION

cAI

L

af n

n 2 2

E

c where

4.221 a 7.612 a 1213 a 2004 a


A vibrating beam marked out with the nodal points is very useful. The location of the nodal points are:

Position of nodal points for a beam of L=1000mm (measured in mm from one end)

mode 1: 224 776 mode 2: 132 500 868 mode 3: 94 356 644 906 mode 4: 73 277 500 723 927 mode 5: 60 226 409 591 774 940 mode 6: 51 192 346 500 654 808 949 mode 7: 44 166 300 433 567 700 834 956 mode 8: 39 147 265 382 500 618 735 853 961

See the Appendix for details of how to derive these


Axisymmetric bodies

Turbocharger blade vibration

Questions: 1. Do the blades fatigue less

rapidly if they are perfectly tuned, or is it better to mistune them?

2. Can vibration measurements made on a rotor be used to estimate its fatigue life?


Constrained-layer damping 1. Works by introducing damping

material in places where shear strain is large

2. Material selection is important(i) not too rubbery(ii) not too glassy

- just right!

3. Temperature dependent

4. Effective over wide range of frequencies

5. Compromises strengthfrequency

ampl

itude


Tuned absorber 1. Works by attaching a resonant

element, with just the right amount of damping

2. Works at one frequency only

3. Material selection again is important owing to temperature dependence of damping

frequency

ampl

itude


m

k c

The mkc model has great virtues:

- simple

- huge range of application

- “intuitive” … with a bit of thought


Appendix

Nodes of a Vibrating Beam


zy

Equation of motion:

For vibration, assume y(x,t)=Y(x)cos(t), so

This has general solution

Boundary condition for a fee end at z=0:

02

2

4

4

t

ym

z

yEI

mass per unit length m

flexural rigidity EI, length L

EI

mY

dz

Yd 244

4

4

with ,0

Free vibration of a beam

zDzCzBzAzY sinhcoshsincos)(

0and00

3

3

0

2

2

zz

dz

Yd

dz

Yd


so

i.e. C=A and D=B

Boundary condition for a free end at z=L:

so

and

or, in matrix form,

0and0 DBCA

0and03

3

2

2

LzLz

dz

Yd

dz

Yd

0sinhcoshsincos LBLALBLA

0coshsinhcossin LBLALBLA

0coshcossinhsin

sinhsincoshcos

B

A

LLLL

LLLL


For a non-trivial solution, the determinant must be zero, so

0coshcossinhsin

sinhsincoshcos

LLLL

LLLL

0)sinh)(sinsinh(sin)cosh(cos 2 LLLLLL

0coshcos22 LL L

L

cosh

1cos

L0 2

32

25

27

29

1

Exact solutions for L: 4.730 7.853 10.996 14.137


From aL the frequencies of free vibration are found using

aj= 22.37, 61.67, 120.90, 199.86, ...

or aj

The corresponding mode shapes are obtained by substituting j intothe matrix equation to find the ratio between A and B

so that

The location of nodal points is then found by looking for where Y(z)=0

EI

m 24

4

2

mL

EIa jj

...,,,, 2

292

272

252

23

0)sinh(sin)cosh(cos BLLALL

)sinh)(sincosh(cos

)cosh)(cossinh(sin)(

zzLL

zzLLzY

jjjj

jjjj


The location of the nodal points needs to be computed numerically, and the values are:

Position of nodal points for a beam of L=1000mm (measured in mm from one end)

mode 1: 224 776 mode 2: 132 500 868 mode 3: 94 356 644 906 mode 4: 73 277 500 723 927 mode 5: 60 226 409 591 774 940 mode 6: 51 192 346 500 654 808 949 mode 7: 44 166 300 433 567 700 834 956 mode 8: 39 147 265 382 500 618 735 853 961

Hugh Hunt, Trinity College, Cambridge *

Documents

trinity college

axial vibration slide

thought slide

end mode

stiffness increased

vibrating beam slide

euler bending vibration

vibration measurements