Harmonic Excitation Applications of Vibratory

Systems Subha Kumpaty

ASME Milwaukee Section

Webinar 3/27/21

Introduction

• Vibrations in mechanical systems are typically an undesirable phenomena (transmit forces, generate noise)

• Tacoma Narrows Bridge- 1940

• Suspension systems are used to isolation a body from vibrations and/or damp out transmitted vibrations

• Vibration isolation systems:• Reduce the transmission of vibrations from a machine to the

environment

• Reduce the transmission of vibrations from the environment to the machine

Response to Harmonic Excitation0 cosmx cx kx F t+ + =

( ) cos( )x t X t = −

0

2 2 2( ) ( )

FX

k m c =

− +

1

2tan

c

k m

−=−

2220

21

1

+

−

=

nn

F

Xk

1

2

2

tan

1

n

n

−

=

−

3

Response to Harmonic Excitation0 cosmx cx kx F t+ + =

( ) cos( )x t X t = −

0

2 2 2( ) ( )

FX

k m c =

− +

1

2tan

c

k m

−=−

2220

21

1

+

−

=

nn

F

Xk

1

2

2

tan

1

n

n

−

=

−

4

Plate Supporting a Pump: From SS Rao text

A reciprocating pump, weighing 68 kg, is mounted at the middle of a steel plate ofthickness 1 cm, width 50 cm, and length 250 cm. clamped along two edges as shownin Fig. During operation of the pump, the plate is subjected to a harmonic force, F(t)= 220 cos (62.832t) N. if E=200 GPa, Find the amplitude of vibration of the plate.

Solution

• The plate can be modeled as fixed – fixed beam has the following stiffness:

• Natural frequency

Wn = sqrt (k/m) = 38.8 rad/s ; Frequency ratio is 1.619

• The maximum amplitude (X) is found as:

( )( )

( )( )( )

mNx

xxkSo

mxxxbhIBut

l

EIk

/ 82.400,10210250

10667.4110200192 ,

10667.41101105012

1

12

1

192

32

99

493223

3

==

===

=

−

−

−−−

( )mm

mk

FX o 32487.1

832.626882.400,102

2202

−=−

=−

=

-ve means that the response is out of

phase with excitation

8

2( ) ( sin )

dM m x m x e t cx kx

dt− + + = − −

Unbalance in rotating

machines is of a common

source of vibration

excitation. The

unbalance is represented

by an eccentric mass m

with eccentricity e that is

rotating with angular

velocity .

2 sinMx cx kx me t + + =

Rotating Unbalance

9

1

2

2

tan

1

n

n

−

=

−

2

2 22

1 2

n

n n

MX

me

=

− +

Rotating Unbalance contd.

10

1 DOF Lab: Free Vibration

2 2 2 0Mp bq kr + + =

Mp

q

b k

r

Constants

p = 15 in = 38 cm

q = 7.5 in = 19 cm

r = 26 in = 66 cm

M =16.1 lbm = 7.3 kg

22 0n n + + =

( ) ( sin sin )nt

d dt e A t B t −

= +

11

Lab I(a): Free Vibration

2

dd

=

d = 160 ms

2 24

+=

.

Avg. = 0.18755

2139.287d rad

n s

−= =

Given M and the geometry, k and b can be calculated.

12

Lab Forced Vibration (Rotating Unbalance)

Rotating unbalance

m = 0.0614 lbm =

0.1351 kg

e = 2.25 in. = 5.715

cm

2

2

2 2 2(1 ) (2 )

n

me

MX

=

− +

13

Forced Vibration (Rotating Unbalance)

Frequency Response

0

1

2

3

4

5

6

7

8

9

0.7 0.8 0.9 1.0 1.1 1.2Wbar

Theoretical Data Experimental Data

Vibration Isolation

Vibration isolation system:

a) instrument with vibration mounts,

b) representative mechanical system.

Goal: reduce the transmission of base motion xb(t) to the instrument

(a) (b)

15

( ) ( )mx c x y k x y= − − − −

In many cases, the dynamical system is excited by the motion

of the support point. We let y be the harmonic displacement of

the support point and measure the displacement x of the mass m

for an inertial reference. EOM becomes

Support Motion

16

OR

2 sinmz cz kz my mY t + + = =

The steady-state solution of the previous section can be

used where Z replaces X and mY2 replaces me2

17

2 2

2 2 2

( )

( ) ( )

X k c

Y k m c

+=

− +

3

2 2tan

( ) ( )

mc

k k m c

=

− +

2

22 2

21

21

n

n n

X

Y

+

=

− +

Support Motion contd.

18

0 sinmx cx kx F t+ + =

Vibratory forces generated by machines and other causes are

often unavoidable; however, their effects on a dynamical

system can be minimized by proper isolator design. The force

to be isolated is transmitted through the spring and damper.

2 2 2( ) ( ) 1 (2 )nTF kX c X kX

= + = +

FT

Vibration Isolation

( )

2 2

22 2 2

( ω) 1 4ξ β( ω)

( ω)1 β 4ξ β

T TF j F

TR G jF j F

+= = = =

− +

2

( )( )

( )

TF s cs kG s

F s ms cs k

+= =

+ +

19

2

22 20

21

21

nT

n n

F X

F Y

+

= =

− +

The transmissibility TR, is the ratio of transmitted force

to that of the disturbed force.

=F0 sin t

FT

Isolation contd.

Transmissibility - Observations

r = / n

Notes:

1. TR depends on damping

ratio and frequency ratio r

2. When , TR when r = 1 (or, input is resonance)

3. When , TR = 1 for all

damping ratios. Therefore, the

transmitted amplitude is equal

to the base amplitude without

gain or loss if input frequency is

0 →

𝑟 = 2

2n =

Transmissibility - Observations

r = / n

Notes:4. When r is very small (stiff

spring), TR ~ 1 for all

5. When input frequency ratio

is , increasing

decreases TR and improves

vibration isolation.

6. When input frequency ratio

is , increasing

increases TR and degrades

vibration isolation.

𝑟 < 2

𝑟 > 2

TR < 1 occurs only when 𝑟 > 2

Transmissibility: Summary

r = / n

1. For minimum transmitted force or motion,

choose vibration isolation mounts with

undamped natural frequency (i.e.,

stiffness) much lower than the expected

input vibration frequency (therefore,

is high).

2. If possible, select mounts with light

damping for min TR (only for ).

𝑟 = 𝜔/𝜔𝑛

𝑟 > 2

3. Many times the disturbing input frequency

may not be constant but may be in a

known range. The isolator design must

consider a range of r

4. The designer may need to balance TR

requirements with other design

constraints (shock response, settling

time, etc).

23

The transmissibility ratio TR, is less than 1 when frequency ratio is

greater than 1.414.

To accomplish this, natural frequency must be smaller, that is have

a soft spring.

Isolation/ Transmissibility

24

Vibration MeasurementAnalyze detection of platform vibration through relative motion of mass

( ) ( )mx c x u k x u= − − − −

y x u= −

my cy ky mu+ + = −

2

2

( )( )

( )

Y s msG s

U s ms cs k

−= =

+ +

( ) ( )

2

2 22 2

( ω) ω( ω)

( ω)ω ω 2ξω ωn n

Y j YG j

U j U= = =

− +

ω >> ωn Y U displacement measurementthen

( )2

2

1ω

ωn

Y Uω << ωn acceleration measurementthen

25

Attach a secondary mass and spring making the system a 2-DOF

with two natural frequencies away from the natural frequency of the

primary mass-spring system

Dynamic Vibration Absorber

26

27

Two-dof system: Free Vibration

1 1 1 2 2 1

2 2 2 2 2

0 0

0 0

m x k k k x

m x k k x

+ − + = −

28

Two-dof system: Free Vibration

2 0K M− =

1 0.618 km

=

2 1.618 km

=

With equal masses and equal springs

{0.618

1.000}

𝐾 −𝑀𝜔2 = 0

{-1.618

1.000}

29

01 1 11 12 1

2 2 21 22 2

sin0

0 0

F tm x k k x

m x k k x

+ =

Start with 2-DOF system under harmonic excitation

21 111 1 12

2221 22 2

0

X Fk m k

Xk k m

− =

−

Expecting harmonic response, the amplitudes are given by

2

22 2 11 2 2

11 1 22 2 12 21

( )

( )( )

k m FX

k m k m k k

−=

− − −

21 12 2 2

11 1 22 2 12 21( )( )

k FX

k m k m k k

−=

− − −

The denominator=0 gives the natural frequencies

Dynamic Vibration Absorber

30

The secondary system is m-c-k (damped vibration absorber)

(Damped) Dynamic Vibration Absorber

31

Dynamic Vibration Absorber Concept: arresting

primary mass motion by tuning the secondary

system frequency to the excitation frequency

Questions?

kumpaty@msoe.edu