Mechanical Vibrations (10 ME 74) Chapter 5:
Vibration measuring measuring instruments •
T he
primary
purpose
of
a
vibration
measuring
ins instrument is to give an output signa gnal which closely represents the vibration phenomenon. •
Thi This phenome men non may be disp isplacement, veloc locity or acce accele lerratio ation n of the the vibr vibrat atin ing g syst system em an and d acco accord rdin ingl gly y, these instruments are called vibrometers (vibration pickups), velocity pick-ups or accelerometers.
•
Vibration
measuring
devices
having
mass,
spring
dashpot etc. are known as seismic instruments . 1
Vibrometer Z W .K .T relative amplitude ratio is given by; B If r is high, (i.e. r Z
3 & above) & =
r2
B
(1 - 2r 2 r 4 ) 2r 2 Z
r 2 r4
1 2
r 2
2 2
(1 - r )
2 r
2
,
1 (Neglecting 1 compared to r 4 )
B. In other words, the amplitude of re lative motion ( Z ) is is
equa equall to the ampl amplitu itude de of of the the vibr vibrati ating ng bod body y ( B). This results in output signal si gnal which is a true reproduction reproduction of input quantity. Hence the device based on this works work s as a displacement transducer transducer or vibrometer. 2
Vibrometer Z W .K .T relative amplitude ratio is given by; B If r is high, (i.e. r Z
3 & above) & =
r2
B
(1 - 2r 2 r 4 ) 2r 2 Z
r 2 r4
1 2
r 2
2 2
(1 - r )
2 r
2
,
1 (Neglecting 1 compared to r 4 )
B. In other words, the amplitude of re lative motion ( Z ) is is
equa equall to the ampl amplitu itude de of of the the vibr vibrati ating ng bod body y ( B). This results in output signal si gnal which is a true reproduction reproduction of input quantity. Hence the device based on this works work s as a displacement transducer 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 4
Accelerometer Z W .K .T relative amplitude ratio is given by; B If r is Z B
r2 1
r
r 2
2 2
(1- r )
2 r
2
1, & hence r 2 & (2 r )2 are neglegible, 2
Z
2
r B
2 B
2 n
Acceleration of vibrating body n2
As n is a constant for the device, the amplitude of relative motion ( Z ) is proportional to the acceleration of the vibrating body (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
D ata :
1200 f n 5 Hz, Z 0.05 mm , 0, N 1200 rpm f 20 Hz 60 Solut i on : The r atio of fr equenci es r
n
=
f f n
20 5
4
Th e r atio of ampli tudes (r el ative motion) i s ;
Z B
Z B
r2 (1 r 2 ) 2 (2 r ) 2
42 (4 1) 2
0.05 B
r 2
(1 r ) 2
as =0
1.067 Hence B 0.047 mm
The ampli tude of machine i s 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
D ata : f n
5 Hz, f
40 Hz, Error 2%
Solut i on : The r ati o of fr equenci es r
n
=
f
40
f n
5
8
The r ati o of ampli tudes (r elati ve moti on) i s ;
Z B
r2
2 2 2 r r (1 ) (2 )
1.02
As error is 2%,
82 2 2
(1 8 )
(2 8)
2
Z B
1.02
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
D ata : f n 5.75 Hz , 0.65 Solution : Wh en th e e r r or is 1%,
1.01
Z B
1.01
r 2 (1 r 2 ) 2 (2 0.65 r ) 2
i.e. 0.02 r 4 0.31r 2 1 0 r 3.30 and 2.14
In between r 3.3 & 2.14, error is more than 1%. The lowest value of r beyond which the amplitude can be measured within 1% error is r = 3.30. But r
f f n
3.3
f 5.75
f 19 H z
i.e. the lowest frequency beyond which the amplitude can be measured within 1% error is 19 H z. 12
Solution : Wh en th e er r or i s 2%,
1.02
r 2 (1 r 2 ) 2 (2 0.65 r ) 2
Z B
1.02
,
0.04 r 4 0.31r 2 1 0, r becomes imaginary. Hence take
Z
0.98 i.e. 0.04 r 4 0.31r 2 1 0
B r 1.565 (only possible value ) But r
f f n
1.565
f 5.75
f 9 H z
i.e. the lowest frequency beyond which the amplitude can be measured within 2% error is 9 H z. 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. Undeflected position O
A
y e
B
S G
Deflected position
15
Critical speed of shaft with a single rotor (without damping) Undeflected position O
A
y e
B
S G
Deflected position
The shaft being of neglegible weight, carries a disc of mass m. O 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 erfectly mounted without any eccentricity, O, S & G will coincide. But it is only an ideal situation. Let; k be the transverse stiffness of the shaft, y be the dynamic amplitude of vibration the shaft, ( radial distance OS )
e be the eccentricity of disc from shaft axis, ( radial distance SG )
be the angular velocity of the shaft.
16
Critical speed of shaft with a single rotor (without damping) Undeflected position O
A
y
B
S
e
G
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.e. m 2 ( y e) ky
0. Rearranging the terms,
(k m 2 ) y m 2e
y e
m 2 (k m 2 ) 17
Critical speed of shaft with a single rotor (without damping) Taking k as common factor in the denominator,
m 2 k y m 2 e m 2 m 2 k 1 1 k k y e
r 2
1 r 2
2 2 n y 2 But k m n , e 2 1 2 n
where r frequencyratio
n
When r 1, i.e. n , amplitude y becomes infinite. This frequency of the shaft is called critical frequency & it is equal to natural frequency of transverse vibrations.
60 n Hence, critical speed (or whirling speed) N c 60 f n rpm, where; 2 n
g
, being the static deflection of the shaft under the weight of mass m. 18
Important notes for calculation of static deflection of beams Taking k as common factor in the denominator,
m 2 k y m 2 e m 2 m 2 k 1 1 k k y e
r 2 2 1 r
2 2 n y 2 But k m n , e 2 1 2 n
where r frequencyratio
n
When r 1, i.e. n , amplitude y becomes infinite. This frequency of the shaft is called critical frequency & it is equal to natural frequency of transverse vibrations.
60 n Hence, critical speed (or whirling speed) Nc 60 f n rpm, where; 2 n
g
, being the static deflection of the shaft under the weight of mass m.
19
Discussion on critical speeds (without damping) 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. 20
O
G S y e
O
G y
<
n,
r < 1, heavy side outside
>
n,
S e
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
D ata :
m 12 kgs, l 1 m, d 24 mm, E 200 GPa, e 0.11 mm, N 1200 rpm Solu t i on :
Static deflection for a simply supported beam with a central point load;
mgl 3 48 EI
12 9.8113
0.0244 48 200 10 64
7 .53 ×10 - 4 m
9
Natur al fr equency n = Oper ating fr equency
g
9.81
4
7.53 10 2 N 2 1200
60 125.66
60
114.14 r ad / sec 125.66 r ad / sec
1.1 114.14 2 2 Stiffness of spring k = m n = 12 114.14 156335 N / m Ratio of frequencies r =
23
Dampin g r atio = 0 (undamped). Amplitude of shaft, y =
r 2e 2
1 r
(1.1) 2 0.11 2
(1 1.1 )
0.634
mm
ve sign indicates the displacement is out of phase with centrifugal force.
Dynami c load on shaft ky
156335 (0.634 10 3 ) 99.12 N
Total load on shaft, W = Dead load W
Dynamic load
(12 9.81) 99.12 216.84 N
L oad on each bearing
W 2
216.84 2
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
D ata :
m 16 kgs, l 1.2 m, d 14 mm, E 200 GPa , e 0.4 mm , 70MPa Solution : Shaf t i s supported i n long be ar ings f ixed ends
Static deflection of a beam with both ends fixed with a central point load;
mgl 3 192 EI
16 9.811.23
0.0144 192 200 10 64
Natur al fr equency n =
3.75 ×10 -3 m
9
g
9.81 3.75 10
3
51.18 r ad / sec
Stiffness of spr ing k = m n2 = 16 (25.6) 2 41910 N / m
60 51.18 489 r pm 2
Cr iti cal speed n 51.18 rad / sec
26
Deflection due to dynami c load :
Given the bending stress 70 MPa, i.e.
32 M b 3
d
70
32 M b
143
Bending moment on shaft M 18857.4 N - mm For a shaft with fixed ends, M b =
Wbl 8
18857.4
W b 1200 8
Additional bending load on shaft W b 125.7 N 3.75 Additional deflection W b 125.7 3 mm 16 9.81 mg 27
Range of un safe speeds of oper ation : (i) When the shaft is ver ti cal : ( static deflection due to disc weight neglected )
Amplitude y 3 mm.
y e
r2
(1 r
Taking +ve sign, r 0.94
N N c
2
3
)
0.4
r 2 2
(1 r
)
0.94,
Hence operating speed N 0.94 489 459 rpm. Taking
ve
sign, r 1.074
N N c
1.074,
Hence operating speed N 1.074 489 unsafe
525 rpm.
speed r ange is f r om 459 rpm & 525 r pm, as str ess exceeds 70 M Pa 28
(ii ) When the shaf t i s hor izontal : ( static deflection due to disc weight considered )
Amplitude y (3 3.75)=6.75 mm.
Taking +ve sign, r
Taking
ve
0.972
sign, r 1.03
N N c N
N c
y e
r2
(1 r
0.972,
1.03,
2
)
N
6.75 0.4
r 2 2
(1 r
)
m. 0.972 489 475 r p
N 1.074 489 504 rpm.
H ence i t i s unsafe to oper ate the shaft between 475 r pm & 504 r pm, as th e str ess exceeds 70 M Pa.
29
Critical speed of shaft with a single rotor (with damping)
O
B
A y
ky S
c y G'
e
x G
m
x
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 y and eccentrcity e is . 30
O
B
A ky
y
S
c y
e
G'
Resolving the forces horizontally, m 2 x sin From the fig, x sin
(m ω2 e)sin = cωy
x G
c y
m
x
e sin . Substituting, (i)
Resolving the forces vertically, m 2 x cos ky From the fig, x cos
( y e cos ).
Substituting, m 2 ( y e cos ) ky
(m ω2 e) cos = y(k - m
2
)
(ii)
31
O
B
A ky
y
S
c y
e
G'
x G
x
m
Squaring & adding (i) & (ii), we get, ( m 2 e) 2 (sin 2 cos 2 ) y 2 (k m 2 ) 2 (c ) 2
y
e
m 2 ( k m 2 ) 2 (c ) 2
r
2
2
2
2
(1 - r ) + (2ζr)
N ote : (i) Dynamic force on the bearings
(ky ) 2 (c y ) 2 i.e. F D = y (k 2 + (cω)2
2 ζr (i i ) Phase angle between amplitude y & eccentricity e, = tan 2 1 - r 32 -1
Discussion on critical speeds (without damping) Case (i): When
=
n (r
=1) and
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 <
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 <
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”. 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
Data : m 4 kgs, l 480 mm, d 9 mm, E 200 GPa, e 3 mm
c 49 Ns / m, N 760 rpm Solution : Static deflection for a simply supported beam
mgl 3 48 EI
4 9.81 0.483
1.4 ×10 -3 m
0.0094 48 200 10 64 9
g
Natural fr equency n = Operating fr equency
9.81
2 N 60
1.4 10
3
2 760 60
83.6 rad / sec 79.6 79.6 r ad / sec r = 0.952 83.6
Stiffness of spr ing k = m n = 4 83.6 27956 N / m 2
Damping ratio = Amplitude y =
c
2
49
= 0.0733 2m n 2 4 83.6
r 2e (1 r ) (2 r ) 2 2
2
(0.952) 2 3 (1 0.952 ) (2 0.0733 0.952) 2 2
2
16.17 mm 35
