13 - vibration suppression - absorber

Week 13

1

Mainly based on the text book by D.J. Inman, with some

addition from the text book by S.G. Kelly

Week 13

2

Dynamic Vibration Absorber

A harmonic disturbance to a single DOF system can cause large

amplitude vibration when the excitation frequency is near to the

system natural frequency.

The vibration amplitude can be reduced by adding a degree of

freedom, such that the natural frequencies of the 2-DOF system

are away from the excitation frequency.

A “vibration absorber” is a second spring-mass system added to

this primary mass

Designed to absorb the input disturbance from the primary mass

Motion of primary mass becomes minimum

Motion of absorber mass becomes substantial

Also termed as a “Tuned Mass Damper”

Week 13

3

Applications

Reciprocating machines

Structures excited by earthquakes

Transmission lines or telephone

lines excited by wind blowing

A tuned mass damper beneath the

platform of the Millennium Bridge, UK.

A tuned mass damper on

transmission lines.

Week 13

4

Applications

Tuned mass damper in the

Taipei 101 skyscraper

Suspended from Level 92 to Level 87. It

weighs 660 metric tons and has a

diameter of 5.5 meters. It can reduce 30

to 40 percent of the building's movement

when it's hit by strong winds.

Week 13

5

Undamped Vibration Absorber

Vibration absorber is applied to a machine which operation

frequency meets its resonance frequency.

Absorber mass

Primary mass

Vibration absorber is often used with machines that run at constant

speed or systems with constant excited frequency because the

combined system has narrow operating bandwidth we will see later

Week 13

6


Initially, the primary system experiences resonance

Add absorber system as indicated

The system now has 2 DOF see the equation…

Week 13

7


The equations of motion become:

To solve the EoM, assume harmonic

solutions corresponding to

synchronous motions of both mass,

Insert to EoM and sort out…

Week 13

8


We get:

The form of the response magnitude suggests a design condition

allowing the motion of the primary mass to become zero.

Applying Cramer’s Rule, we solve:

Choose ka and ma

to make X zero

All the system

response goes to

the absorber motion

Week 13

9


Choose the absorber mass and stiffness from:

= 0

This causes the primary mass to be fixed and the absorber mass to

oscillate at:

The magnitude of the force acting on the absorber mass is:

As in the case of the isolator,

static deflection, rattle space and

force magnitudes need to be checked.

Same magnitude but

opposite to disturbance,

hence zero force on

primary mass.

Week 13

10

Pitfalls in Absorber Design

The success of the vibration absorber depends on several factors:

The frequency of the harmonic excitation ω must be well known

This frequency should not deviate much from its constant value

If ω shifts much, it could end up exciting a system natural frequency

(resonance)

Damping, which always exists to some degree, spoils the

absorption.

The primary mass will not have zero displacement

Only desirable if the excitation frequency range is too wide

The absorber spring stiffness ka must be capable of withstanding

the full force of the excitation and the corresponding deflection.

Avoiding resonance can be quantified by examining:

Week 13

11

Robustness to Resonance

Original natural frequency of primary system

without the absorber attached

Natural frequency of absorber system before it

is attached to primary mass

Mass ratio (absorber to primary mass)

Stiffness ratio

Frequency ratio

From the displacement equation of the primary mass:

We define a normalized displacement of the primary mass:

Week 13

12


ra

a

, rp

p

Xk

F0

1 ra

2

1 2 rp2 1 ra

2 2

Week 13

13


If ω drifts to 0.781ωa or

1.28ωa the combined

system will experience

resonance and fail.

If ω drifts such that |Xk/F0|

> 1, the force transmitted

to the primary mass is

amplified, absorber is not

improving the system.

ma and ka are chosen

such that ω/ωa is within

the band-width, |Xk/F0| < 1

The shaded area is the useful operating bandwidth for the

absorber design (0.908ωa < ω < 1.118ωa)

Week 13

14

Robustness to Driving Frequency Shifts

Further design consideration by checking μ and β, which specifies

the mass and stiffness of the absorber system.

From

we get a characteristic equation by setting the matrix coefficient to

zero and assuming ω as the system natural frequency.

Dependence of system frequency on mass ratio μ and frequency ratio β

Week 13

15

Robustness to Driving Frequency Shifts

As increases, system

natural frequencies n split

farther apart for fixed and

farther apart from driving

frequency .

If is too small, system will

not tolerate much fluctuation

in driving frequency.

Rule of thumb 0.05 < < 0.25

Very large large ma stress and fatigue problems

Week 13

16

Example: Absorber design

Based on Example 5.3.1 from the text book of D.J. Inman.

A radial saw base has a mass of 73.16 kg and is driven by a motor

that turns the saw’s blade. The motor runs at constant speed and

produces a 13-N force at 180 cycle/min due to a small unbalance in

the motor. The manufacturer wants a vibration absorber designed to

drive the table oscillation to zero, simply by retrofitting the absorber

onto the base.

Design the absorber assuming

that the stiffness provided by

the table legs is 2600 N/m. The

absorber must fit inside the

table base and hence has a

maximum deflection of 0.2 cm.

Week 13

17

Example: Absorber design

Given: F0 = 13 N, = 180 cpm

m = 73.16 kg, k = 2600 N/m, Xa < 0.002 m

To meet the deflection requirement, choose the stiffness first.

The absorber is design so that

Then all of the force is absorbed by ma. Hence:

Check:

Week 13

18

Example: Absorber bandwidth

Based on Example 5.3.2 from the text book of D.J. Inman.

Continuing Ex. 5.3.1, compute the bandwidth of the absorber design.

We calculate the boundaries for the operating range, i.e.

For

After some manipulation and sorting (which you should try), we get:

1180.10.25= with ,1 aa

22222

0

1111

p

a

app

a

aF

Xk

Week 13

19


For

After some manipulation and sorting (which you should try), we get:

with

22222

0

111

1

p

a

app

a

a

F

Xk

1382.1 , 3929.025.0 ,16.73

2600 ,

29.18

6500 22 a

pa

0212

2242

ap

a

ap

a

Week 13

20


We found four values:

Observing the form of the plot, we can conclude that the driving

frequency is allowed to vary between

Or, with a = 18.85, the operating range is

In this range, the driving frequency will not cause resonance and the

absorber will still reduce the vibration of the primary mass.

1382.1 , 1180.1 , 3929.0 , 1180.1a

aa 1180.13929.0

rad/s 0821.214089.7

Week 13

21

Damping in Vibration Absorbers

Damping is often present in devices

It has the potential to destroy the

ability of the absorber to fully protect

the primary mass, X ≠ 0

So, amplitude of the primary mass at

operating point increase with

increasing damping

ca

However:

Damping can reduce the resonance amplitude of the system.

Damping also improve the effective bandwidth of operation.

Week 13

22


ca

Assuming harmonic solutions:

The amplitude of the motion of the primary mass becomes: Cannot be zero

Week 13

23


In terms of the dimensionless ratio, we write the dimensionless

displacement of the primary mass as:

Where: is the static deflection of mp

is the mixed “damping ratio”

is the ratio of the driving frequency to the

primary natural frequency

and

Week 13

24


As damping increases,

the absorber fails, but the

resonance goes away.

Notice that for fixed β

and μ, the curves of

various ζ passes through

two fixed points.

Region of absorption

Week 13

The Quest for Damped Absorber Design

Three parameters effect making the amplitude small: ζ , β , μ

Previously we saw the plot for μ = 0.25, β = 1 and ζ = 0

|Xk/F0| = 0 at r = 1, infinite at r = 0.782 and 1.281

Operating bandwidth 0.897 ≤ r ≤ 1.103

0.25, 1.0,min 0.4

Addition of damping reduces the

amplitude and widens the bandwidth, but

amplitude does not go to zero.

Note how ζ = 0.1 gives a lower amplitude

over a wider range than ζ = 0.4

25

Week 13

26


0.25, 1.0,min 0.4

0.25, 0.8,min 0.27These curves show that

just increasing the

damping does not result

in the smallest

amplitude. The μ and

also matter.

Choosing the best ζ , β ,

and μ brings us to the

question of optimization.

Notice that the

two fixed points

shift location.

Week 13

27


Since it is not possible to eliminate steady-state motion of the

primary system when damping is present, a damped vibration

absorber must be designed to reduce the peak at the lower

resonant frequency and to widen the effective operating range.

The two peaks should have approximately the same

magnitudes

The value of should lead to |Xk/F0| having the same value at

the two fixed points

The value of should make the two fixed points near the peaks

The optimum design of the damped vibration absorber requires:

1

1

18

3opt

Week 13

28


Example: For μ = 0.2, the optimum design is = 0.833 and = 0.25

Notice that the

magnitude |Xk/F0| > 1,

but the damped vibration

absorber has reduced

the steady state

amplitude to an

acceptable level over the

entire operating range.

Week 13

29

Try Yourself

A diesel engine, weighting 3000 N, is supported on a pedestal

mount. The engine induces vibration through its pedestal mount at

an operating speed of 6000 rpm. Determine the parameters of the

vibration absorber what will reduce the vibration when mounted on

the pedestal. The magnitude of the exciting force is 250 N, and

the amplitude of motion of the absorber mass is to limited to 2

mm.

13 - vibration suppression - absorber

Documents