Mechanical Vibrations (10 ME 74) Chapter 5: Vibration measuring instruments

• The primary purpose of a vibration measuring

instrument is to give an output signal which closely

represents the vibration phenomenon.

• This phenomenon may be displacement, velocity or

acceleration of the vibrating system and accordingly,

these instruments are called vibrometers (vibration pick-

ups), velocity pick-ups or accelerometers.

• Vibration measuring devices having mass, spring

dashpot etc. are known as seismic instruments.1

Vibrometer

2

22 2

2 24

2 4 2 4

. . relative amplitude ratio is given by; (1- ) 2

1 , ( . . 3& ) & = ,

2

1 (Neglecting 1 compared to )(1- 2 ) 2

. In other words, the amplitude of re

Z rW K T

B r r

If r is high i e r above

Z r rr

B r r r r

Z B

lative motion ( ) is

equal to the amplitude of the vibrating body ( ).

Z

B

This results in output signal which is a true reproduction of input

quantity. Hence the device based on this works as a displacement

transducer or vibrometer.2

Vibrometer

3

Vibrometer

• Vibrometers are low natural frequency transducers used to

measure the amplitude of vibrations of a body vibrating with

high frequency.

• The natural frequency of vibrometers is small (1 to 5 Hz) and

hence it requires a heavy mass & a soft spring. This makes it

unsuitable for use in sophisticated applications. 4

Accelerometer

2

22 2

2 2

2 22 2

2 2

. . relative amplitude ratio is given by; (1- ) 2

1, & hence & (2 ) are neglegible,

1

As is a constant for the

n n

n

Z rW K T

B r r

If r is r r

Z r B Acceleration of vibrating bodyr Z r B

B

device, the amplitude of relative motion

( ) is proportional to the acceleration of the vibrating body ( ).Z B

Device based on this is an acceleration transducer or Accelerometer.

5

Accelerometer

• Accelerometers are high natural frequency transducers

used to measure the acceleration of a vibrating body which

vibrates with low frequency, such that frequency ratio r<<1

(0.25 & below).

• The natural frequency of accelerometers is high (above

100Hz) and hence it requires a light mass & a hard spring.

• Due to light weight, it is widely used in many applications.

• Also, by using electronic integrating devices, velocity &

displacement of the vibrating body can be calculated.

6

Numerical Problem 1

A seismic instrument with a natural frequency of 5

Hz is used to measure the vibration of a machine

operating at 120 rpm. The relative displacement of

the seismic mass as read from the instrument is

0.05 mm. Determine the amplitude of vibration of

the machine. Neglect damping.

7

2 2

22 2 2

12005 , 0.05 , 0, 1200 20

60

20= 4

5

=0(1 )(1 ) (2 )

n

n n

f Hz Z mm N rpm f Hz

f

f

Z r ras

B rr r

Z

The ratio of frequencies r

The ratio of amplitudes (relative motion) i

Data :

Solut

;

ion :

s

2

2

4 0.05 1.067 Hence 0.047 (4 1)

B mmB B

The amplitude of machine is 0.047 mm.

8

Numerical Problem 2

A vibrometer indicates 2 percent error in

measurement and its natural frequency is 5

Hz. If the lowest frequency that can be

measured is 40 Hz, find the value of damping

factor.

9

2

2 2 2

2

2 2 2

5 , 40 , 2%

40= 8

5

As error is 2%, 1.02(1 ) (2 )

8 1.02

(1 8 ) (2 8)

n

n n

f Hz f Hz Error

f

f

Z r Z

B Br r

The ratio of frequencies r

The ratio of amplitudes (relative motion) i

Data :

Solut

;

ion :

s

= 0.35

10

Numerical Problem 3

A commercial vibration pick-up has a natural

frequency of 5.75 Hz and a damping factor of 0.65.

What is the lowest frequency beyond which the

amplitude can be measured with in

(a) 1% error

(b) 2% error.

11

2

2 2 2

4 2

5.75 , 0.65

1.01

1.01 (1 ) (2 0.65 )

. . 0.02 0.31 1 0

3.30 2.14

In between 3.3 & 2.14, error is more than 1%.

The lowest va

nf Hz

Z

B

r

r r

i e r r

r and

r

When the e

Data :

Solution : rror is 1 %,

lue of beyond which the amplitude can be measured

within 1% error is But 3.3 5.75

. .

1%

n

r

f fr f

f

i e the lowest frequency beyond which the amplitude can be measured

within error is

r = 3.30. 19 Hz

19 Hz.12

2

2 2 2

4 2

4 2

1.02

1.02 , (1 ) (2 0.65 )

0.04 0.31 1 0, .

Hence take 0.98 . . 0.04 0.31 1 0

1.565 ( )

But n

Z

B

r

r r

r r r becomes imaginary

Zi e r r

B

r only possible value

fr

f

When the error is 2%, Solution :

1.565 5.75

. .

2%

ff

i e the lowest frequency beyond which the amplitude can be measured

within error is

9 Hz

9 Hz.

13

Critical speed of shafts (Whirling or whipping speed)

• Critical speed of a rotating shaft is the speed at which

the shaft starts to vibrate violently in the transverse

direction.

• At these speeds, large amount of force is transmitted

to the foundation or bearings & cause failure of the

same.

• Hence it is very important to determine the critical

speed range and avoid such speeds either by keeping

the operating speed well below or above that value by

quickly crossing over the critical speed.14

Critical speed of shaft with a single rotor (without damping)

• Critical speed may occur because of eccentric mounting

of the rotor, non uniform distribution of rotor material,

bending of shaft, etc.

• Let us consider a shaft rotating horizontally between

bearing A & B as shown in fig.

G

S

O

e

yA B

Undeflected position

Deflected position

15

Critical speed of shaft with a single rotor (without damping)

G

S

O

e

yA B

Undeflected position

Deflected position

The shaft being of neglegible weight, carries a disc of mass .

is the point on axis of rotation, S is the coincident point on the shaft axis,

G is the center of gravity of the disc. If the disc is p

m

O

erfectly mounted without

any eccentricity, O, S & G will coincide. But it is only an ideal situation.

Let; be the transverse stiffness of the shaft,

be the dynamic amplitude of vibration the

k

y shaft, ( )

be the eccentricity of disc from shaft axis, ( )

be the angular velocity of the shaft.

radial distance OS

e radial distance SG

16

Critical speed of shaft with a single rotor (without damping)

G

S

O

e

yA B

Undeflected position

Deflected position

Considering equilibrium of the shaft, the centrifugal force acts

radially outwards through the center of gravity G of the disc &

the restoring force due to stiffness acts radially inwards through S.

i 2

22 2

2

. . ( ) 0. Rearranging the terms,

( )( )

e m y e ky

y mk m y m e

e k m

17

Critical speed of shaft with a single rotor (without damping)

22

222

2 2 2

2

2

2

as common factor in the denominator,

But ,

1 1 1

where 1

When 1, i.e. , amplitude

n

n

n

n

n

Taking k

m

ky m yk m

e em mk

k k

y rr frequencyratio

e r

r

becomes infinite. This frequency of the shaft

is called critical frequency & it is equal to natural frequency of transverse vibrations.

60Hence, critical speed (or whirling speed) N 60

2

nc

y

f

, where;

, being the static deflection of the shaft under the weight of mass .

n

n

rpm

gm

18

Important notes for calculation of static deflection of beams

22

222

2 2 2

2

2

2

as common factor in the denominator,

But ,

1 1 1

where 1

When 1, i.e. , amplitude

n

n

n

n

n

Taking k

m

ky m yk m

e em mk

k k

y rr frequencyratio

e r

r

becomes infinite. This frequency of the shaft

is called critical frequency & it is equal to natural frequency of transverse vibrations.

60Hence, critical speed (or whirling speed) N 60

2

nc

y

f

, where;

, being the static deflection of the shaft under the weight of mass .

n

n

rpm

gm

19

Case (i): When =n (r =1) Forcing frequency coincides with the natural frequency of

transverse vibration of the shaft. y/e – approaches infinity i.e., the deflection of geometric

centre of the disc tends to infinity.

The disk has a tendency to fly out, if the damping is insufficient. There will be severe

vibrations of the shaft thereby producing huge bearings reactions.

Case (ii): < n, r < 1

y/e = is positive. The deflection y and eccentricity ‘e’ are in the same sense. This condition

of disc is referred as “Heavy side outwards” i.e.,. The disc rotates with heavy side outside.

Thus S will lie between O and G. Positive sign indicates that y is in phase with centrifugal

force.

Case (iii): When > c, r > 1

y/e = negative, the deflection y and the eccentricity e are in opposite sense. This condition of

the disc is referred as “Heavy side inwards”. G falls between O and S. Negative sign

indicates that y is out of phase with centrifugal force.

Discussion on critical speeds (without damping)

20

ye

O SGy e

O S G

< n, r < 1, heavy side outside >n, r > 1, heavy side inside

21

Numerical Problem 1(Critical speeds with out damping)

A rotor has a mass of 12 Kg and is mounted midway on a

horizontal shaft of 24 mm simply supported at the ends by

two bearings. The bearings are 1 m apart. The shaft rotates

at 1200 rpm. The mass center of the rotor is 0.11 mm away

from the geometric center of the rotor due to certain

manufacturing errors. Determine the amplitude of steady

state vibrations and dynamic force transmitted to the

bearings if E = 200 GN/m2.22

3 3

49

12 ,  1 , 24 , 200 , 0.11 , 1200

Static deflection for a simply supported beam with a central point load;

12 9.81 1

48 0.02448 200 10

64

m kgs l m d mm E GPa e mm N rpm

mgl

EI

-

Data :

7

Solu

.53

ion :

× 10

t

4

2 2

9.81

7.53 10

2 2 1200

60 60

125.66

114.14

12 114.14

n

n

g

N

m

4m

Natural frequency = 114.14 rad / sec

Operating frequency 125.66 rad / sec

Ratio of frequencies r = 1.1

Stiffness of spring k = = 156335 N / m

23

2 2

2 2

0 (undamped).

(1.1) 0.11

1 (1 1.1 )

sign indicates the displacement is out of phase with centrifugal force.

156335 (0.6

r ey

r

ve

ky

Damping ratio =

Amplitude of shaft, = 0.634 mm

Dynamic load on shaft334 10 )

(12 9.81) 99.12

216.84

2 2

Dead load Dynamic load

W

99.12 N

Total load on shaft, W =

W 216.84 N

Load on each bearing 108.42 N

24

Numerical Problem 2

A shaft of 14 mm and the length 1.2 m is held in long

bearings. It carries a rotor of 16 Kgs at its midspan. The

eccentricity of the mass center of the rotor from the shaft

center is 0.4 mm. The shaft is made of steel for which E =

200 GN/m2 and permissible stress is 70 MPa.

Determine the critical speed of the shaft and range of

speed over which it is unsafe to run the shaft when;

(a) the shaft is horizontal

(b) the shaft if vertical.

25

3

16 ,  1.2 , 14 , 200 , 0.4 , 70

Static deflection of a beam with both ends fixed with a central point load;

1

m kgs l m d mm E GPa e mm MPa

mgl

Shaft is supported in long be

D

arings fixed ends

ata :

Solution :

3

49

3

2 2

16 9.81 1.2

92 0.014192 200 10

64

9.81

3.75 10

16 (25.6)

60 51.1851.18 / sec

2

n

n

n

EI

g

m

rad

-33.75 × 10 m

Natural frequency = 51.18 rad / sec

Stiffness of spring k = = 41910 N / m

Critical speed

489 rpm

26

3 3

70 ,

32 32. . 70

14

Bending moment on shaft 18857.4 -

1200For a shaft with fixed ends, M = 18857.4

8 8

b b

b bb

Given the bending stress MPa

M Mi e

d

M N mm

W l W

Additional

Deflection due to dynamic load :

3.75Additional deflection 125.7

16 9.81

b

b

bending load on shaft W

Wmg

125.7 N

3 mm

27

2 2

2 2

( )

3 3 .

(1 ) 0.4 (1 )

Taking +ve sign, 0.94 0.94,

Hen

c

static deflection due to disc weight neglected

y r rAmplitude y mm

e r r

Nr

N

(i) When the shaft is vertical :

Range of unsafe speeds of operation :

ce operating speed 0.94 489

Taking ve sign, 1.074 1.074,

Hence operating speed 1.074 489

c

N

Nr

N

N

459 rpm.

525 rpm.

unsafe speed range is from 459 rpm & 525 rpm, as stress exceeds 70 MPa

28

2 2

2 2

( )

6.75 (3 3.75)=6.75 .

(1 ) 0.4 (1 )

Taking +ve sign, 0.972 0.972, 0.972 489c

static deflection due to disc weight considered

y r rAmplitude y mm

e r r

Nr N

N

(ii) When the shaft is horizontal :

475 rp

Taking ve sign, 1.03 1.03, 1.074 489c

Nr N

N

m.

504 rpm.

Hence it is unsafe to operate the shaft between

475 rpm & 504 rpm, as the stress exceeds 70 MPa.

29

Critical speed of shaft with a single rotor (with damping)

When the damping force is present, the center of gravity G

will not be in line with O & S. The centrifugal force acts

through G at an angle to the vertical as shown.

Phase angle b/n amplitude and y

eccentrcity is .e

G

S

OA B

c y

ky

x

m x

y

e

G'

30

G

S

OA B

c y

ky

x

m x

y

e

G'

2

2

2

Resolving the forces horizontally, sin

From the fig, sin sin . Substituting,

(

Resolving the forces vertically, cos

From the fig, cos ( cos ).

Substituting, (

m x c y

x e

m x ky

x y e

m

2mω e)sin = cωy (i)

2

cos )

( )

y e ky

2mω e) cos = y(k - m (ii)

31

G

S

OA B

c y

ky

x

m x

y

e

G'

2 2 2 2 2 2 2 2

2

2 2 2

2 2

Squaring & adding (i) & (ii), we get,

( ) (sin cos ) ( ) ( )

( ) ( )

( ) ( ) . .

m e y k m c

m

k m c

Dynamic force on the bearings ky c y i e

Phase a

2

2 2 2

2 2

D

y r

e (1 - r ) + (2ζr)

(i) F = y (k + (cω)

(ii

N

)

ote :

& , ngle between amplitude y eccentricity e

-1

2

2ζr= tan

1 - r32

Case (i): When =n (r =1) and 900

Forcing frequency coincides with the natural frequency of transverse vibration of the shaft.

y/e – becomes maximum. i.e., the deflection of geometric center of the disc tends to infinity

in absence of damping. It is called critical speed.

Case (ii): < n, r < 1, and 0 < 900

y/e = is positive. The deflection y and eccentricity ‘e’ are in the same sense. This condition

of disc is referred as “Heavy side outwards” i.e.,. The disc rotates with heavy side outside.

Case (iii): When > n, r > 1, and 900 < 1800

y/e = negative, the deflection y and the eccentricity e are in opposite sense. This condition of

the disc is referred as “Heavy side inwards”.

Discussion on critical speeds (without damping)

33

Numerical Problem 1(Critical speeds with damping)

A disc of mass 5 kg is mounted midway between two

simple bearing supports which are 480 mm apart, on a

horizontal steel shaft 9 mm in diameter. The CG of the disc

is displaced by 3 mm from its geometric center. Equivalent

viscous damping at the center of the disc is 49 Ns/m. If the

shaft rotates at 760 rpm, determine the maximum stress in

the shaft. Also compare it with the dead load stress in the

shaft. Take E= 200 GPa.34

3 3

49

4 ,  480 ,   9 , 200 , 3  

49 / , 760

Static deflection for a simply supported beam

4 9.81 0.48

48 0.00948 200 10

64

m kgs l mm d mm E GPa e mm

c Ns m N rpm

mgl

EI

-3

Data :

Solution :

1.4 × 10 m

3

2 2

9.81

1.4 10

2 2 760 79.6

60 60 83.6

4 83.6

49

2 2 4 83.6

n

n

n

g

N

m

c

m

Natural frequency = 83.6 rad / sec

Operating frequency 79.6 rad / sec r = 0.952

Stiffness of spring k = = 27956 N / m

Damping ratio = =

2 2

2 2 2 2 2 2

(0.952) 3

(1 ) (2 ) (1 0.952 ) (2 0.0733 0.952)

r ey

r r

0.0733

Amplitude = 16.17 mm

35

2 2 3 2 2

3 3

( ) 16.17 10 27956 (49 79.6)

(4 9.81) 456.43

496 0.4832 32

32 4 4

d y k c

Dead load Dynamic load

Wl

M

d d

Dynamic load on shaft :

F 456.43 N

Total load on shaft, W =

W 496 N

Max bending stress 3

3 3

(0.009)

(4 9.81)

39.24 0.4832

32 ' 4'

(0.009)

Max bending stress under dynamic conditi

Weight of disc mg

M

d

831.64 MPa

Bending stress considering only disc weight (dead load) :

39.24 N

65.8 MPa

on 831.64

Max bending stress under static condition 65.8 12.64

36