NOISE NOI SE & VI VIB B RA RATIO TION N
Topic op ic 6 : Vibra ibr ation tio n Cont Contro roll
Ir Dr Zainal Fitri B Zainal Abidin Sept 2016
Chapter Out Outli line ne • Introduction • Vibration Nomograph and Vibration Criteria • Reduction of Vibration at the source • Balancing of Rotating Machines • Whirling of Rotating Shafts • Balancing of Reciprocating Engines • Control of Vibration • Control of Natural Frequencies • Introduction of Damping • Vibration Isolation • Vibration Absorbers © 2011 Mechanical Vibrations Fifth Edition in SI Units
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Chapter Out Outli line ne • Introduction • Vibration Nomograph and Vibration Criteria • Reduction of Vibration at the source • Balancing of Rotating Machines • Whirling of Rotating Shafts • Balancing of Reciprocating Engines • Control of Vibration • Control of Natural Frequencies • Introduction of Damping • Vibration Isolation • Vibration Absorbers © 2011 Mechanical Vibrations Fifth Edition in SI Units
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Introduction • Vibration leads to wear of machinery and
discomfort of humans, thus we want to eliminate vibration • Designer must compromise between acceptable
amount of vibration and manufacturing cost • We shall consider various techniques of
vibration control in this chapter. © 2011 Mechanical Vibrations Fifth Edition in SI Units
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Chapter Outline • Introduction • Vibration Nomograph and Vibration Criteria • Reduction of Vibration at the source • Balancing of Rotating Machines • Whirling of Rotating Shafts • Balancing of Reciprocating Engines • Control of Vibration • Control of Natural Frequencies • Introduction of Damping • Vibration Isolation • Vibration Absorbers © 2011 Mechanical Vibrations Fifth Edition in SI Units
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Vibration Nomograph and Vibration Criteria • Vibration nomograph displays the variations of displacement,
velocity and acceleration amplitudes wrt frequency of vibration • Harmonic motion: x • Velocity:v •
(t ) = X sin ω t
(t ) = x (t ) = ω X cos ω t = 2π fX cos ω t
(t ) = −ω X sin ω t = −4π f X sin ω t Acceleration: a (t ) = x 2
• Amplitude of velocity: vmax
= 2π fX
• Amplitude of acceleration: amax © 2011 Mechanical Vibrations Fifth Edition in SI Units
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2
2
(9.4)
= −4π 2 f 2 X = −2π fvmax
(9.5)
Vibration Nomograph and Vibration Criteria
• Taking log of Eq. 9.3 and Eq. 9.4:
ln vmax = ln (2π f ) + ln X ln vmax = − ln amax − ln (2π f ) • When X is constant, ln vmax varies linearly with ln(2π f) • When amax is constant, ln vmax varies linearly with ln(2π f ) • This is shown as a nomograph in the next slide. • Every pt on the nomograph denotes a specific sinusoidal vibration.
Vibration Nomograph and Vibration Criteria
Vibration Nomograph and Vibration Criteria
Ranges of Vibration
Vibration Nomograph and Vibration Criteria
Frequency sensitivity of different parts of human body
Vibration Nomogr aph and Vibration Criteria
• Vibration severity of machinery is defined in
terms of the root mean square (rms) value of vibration velocity. (ISO 2372) • Vibration severity of whole building vibration
(ISO DP 4866) • Vibration limits for human (ISO 2631)
Vibration Nomograph and Vibration Criteria
•Example 9.1 Helicopter Seat Vibration Reduction The seat of a helicopter, with the pilot, weights 1000N and is found to have a static deflection of 10 mm under self-weight. The vibration of the rotor is transmitted to the base of the seat as harmonic motion with frequency 4 Hz and amplitude 0.2 mm. •a) What is the level of vibration felt by the pilot? •b) How can the seat be redesigned to reduce the effect of vibration? © 2011 Mechanical Vibrations Fifth Edition in SI Units
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Vibration Nomograph and Vibration Criteria
•Example 9.1 Helicopter Seat Vibration Reductio n Solution
Mass = m = 1000/9.81 = 101.9368 kg Stiffness = k = W/δ st = 1000/0.01 = 105N/m Natural frequency = ωn =
k m
•Frequency ratio = r =
=
105 101.9368
4.9849
ω =
ω
n
=
= 31.3209 rad/s = 4.9849 Hz
1.2462
4.0
•Amplitude of vibration felt by pilot: X = ±
Y
1 − r 2
Y is the amplitude of base displacement •where © 2011 Mechanical Vibrations 12 Fifth Edition in SI Units
Vibration Nomograph and Vibration Criteria
•Example 9.1 Helicopter Seat Vibration Reduction Solution 0.2 X = vmax
= 0.3616 mm
1 − 1.2462 = 2π fX = 2π (5)(0.3616) = 9.0887 mm/s 2
amax = (2π f ) X = 228.4074 mm/s2 = 0.2284 m/s 2 2
•At 4 Hz, the amplitude of 0.3616 mm may not cause much discomfort. •However the velocity and acceleration at 4 Hz are not acceptable for a comfortable ride. •Try to bring
amax down
to 0.01 m/s 2
Vibration Nomograph and Vibration Criteria
•Example 9.1 Helicopter Seat Vibration Reduction Solution 2 2 2 amax = 10 mm/s = −(2π f ) X = −(8π ) X = 0.01583 mm X
=
0.01583
Y
0.2
ω n =
ω
=±
=
1 1 − r
8π
2
or r = 3.6923
= 6.8068 rad/s =
k
3.6923 3.6923 m m = 101.9368 kg ∴ k = 4722.9837 N/m
•Either use softer material for seat or increase mass of seat.
Chapter Outline • Introduction • Vibration Nomograph and Vibration Criteria • Reduction of Vibration at the source • Balancing of Rotating Machines • Whirling of Rotating Shafts • Balancing of Reciprocating Engines • Control of Vibration • Control of Natural Frequencies • Introduction of Damping • Vibration Isolation • Vibration Absorbers © 2011 Mechanical Vibrations Fifth Edition in SI Units
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Control of Vibration • Some import methods to control vibrations: Control
ωn and avoid resonance under external excitations.
Introduce
damping mechanism to prevent excessive response of system
Use
vibration isolators to reduce transmission of excitation forces from one part of the machine to another
Add
an auxiliary mass neutralizer or vibration absorber to reduce response of system
© 2011 Mechanical Vibrations Fifth Edition in SI Units
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Chapter Outline • Introduction • Vibration Nomograph and Vibration Criteria • Reduction of Vibration at the source • Balancing of Rotating Machines • Whirling of Rotating Shafts • Balancing of Reciprocating Engines • Control of Vibration • Control of Natural Frequencies • Introduction of Damping • Vibration Isolation • Vibration Absorbers © 2011 Mechanical Vibrations Fifth Edition in SI Units
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Control of Natural Frequencies • Resonance Large displacements large strains and
stresses failure of system • Often the excitation frequency cannot be controlled. • Hence must control natural frequency by varying mass m
or stiffness k to avoid resonance. • Practically mass cannot be changed easily. • Hence we change stiffness k by altering the material or
number and location of bearings.
Chapter Outline • Introduction • Vibration Nomograph and Vibration Criteria • Reduction of Vibration at the source • Balancing of Rotating Machines • Whirling of Rotating Shafts • Balancing of Reciprocating Engines • Control of Vibration • Control of Natural Frequencies • Introduction of Damping • Vibration Isolation • Vibration Absorbers © 2011 Mechanical Vibrations Fifth Edition in SI Units
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Introduction of Damping • System may be required to operate over a range of speed, hence
cannot avoid resonance • Can use material with high internal damping to control the response. • Can also use bolted or riveted joints to increase damping. • Bolted or riveted joints permit slip between surfaces and dissipate
more energy compared to welded joints. • However they also reduce stiffness of structure, produce debris and
cause fretting corrosion.
Introduction of Damping
m x + k (1 + iη ) x = F 0e ω
i t
where loss factor η =
=
(∆W / 2π )
W Energy dissipated during 1 cycle of harmonic displacement/radian
Maximum strain energy in cycle
F 0 k η
=
F 0 aE η
, a = constant
Intro ntroduct duction ion of Da Dampin mping g • Viscoelastic materials have larger values of η and are
used to provide internal damping. • Disadvantage is their properties change with temperature,
frequency and strain. • Sandwich viscoelastic material between elastic layers –
Constrained layer damping • Material with largest η will be subjected to the smallest
stresses.
Chapter Out Outli line ne • Introduction • Vibration Nomograph and Vibration Criteria • Reduction of Vibration at the source • Balancing of Rotating Machines • Whirling of Rotating Shafts • Balancing of Reciprocating Engines • Control of Vibration • Control of Natural Frequencies • Introduction of Damping • Vibration Isolation • Vibration Absorbers © 2011 Mechanical Vibrations Fifth Edition in SI Units
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Vib ibra rati tion on Iso Isola lati tion on • Insert isolator between vibrating mass and vibration source to reduce
response • Passive isolators: springs, cork, felt etc. • E.g. Mounting of high-speed punch press
Vibration Isolation
• Exhaust Hangers Design
• Engine mounting
Bad example Exhaust Hangers
Vibration Isolation • Active isolator comprised of servomechanism with sensor,
signal processor and actuator. • Effectiveness given in terms of transmissibility Tr which is
the ratio of amplitude of the transmitted force to that of the exciting force • 2 types of isolation situations:
Protect base of vibrating machine against large unbalanced or impulsive forces
Protect system against motion of its foundation
Vibration Isolation • Protect base of vibrating machine against large unbalanced or
impulsive forces
F i (t ) = kx(t ) + c x (t ) • Protect system against motion of its foundation
F i (t ) = m x(t ) = k [ x(t ) − y (t )] + c[ x (t ) − y (t )]
Vibration Isolation • Vibration Isolation System wit h Rigid Found ation
•
•
Resilient member placed between vibrating machine and rigid foundation Member is modeled as a spring k and a dashpot c as shown:
Vibration Isolation • Vibration Isolation System wit h Rigid Found ation
•
Reduction of force transmitted to foundation:
•
Equation of motion:
m x + c x + kx = F 0 cos ω t
•
Steady state solution: x(t ) = X cos(ω t − φ ) F 0 ω c and φ = tan −1 where X = 2 2 2 2 2 ω k m − (k − mω ) + ω c
•
Force Ft transmitted to the foundation:
•
F t (t ) = kx (t ) + c x (t ) = kX cos(ω t − φ ) − cω X sin (ω t − φ ) Magnitude of total transmitted force FT:
•
F 0 k + ω c 2
F T =
(kx ) + (c x ) = X 2
2
k + ω c = 2
2 2
2 2
(k − mω )
2 2
+ ω 2 c 2
Vibration Isolation • Vibration Isolation System wit h Rigid Found ation
•
Reduction of force transmitted to foundation:
• •
Transmissibility T r =
F T F 0
=
k 2 + ω 2 c 2
(k − mω )
2 2
1 + (2ζ r )
+ ω 2c 2
2
=
(1 − r ) + (2ζ r ) 2 2
2
where r =
ω ω n
• •
Following graphs shows the variation of Tr with r .
Vibration Isolation • Vibration Isolation System wit h Rigid Found ation
• •
Reduction of force transmitted to foundation:
Vibration Isolation • Vibration Isolation System wit h Rigid Found ation
•
Reduction of force transmitted to mass:
m z + c z + kz = − m y where z = x − y
• •
Displacement transmissibility
T d = •
Td is
X Y
1 + (2ζ r )
2
=
(1 − r ) + (2ζ r ) 2 2
2
also the ratio of the maximum steady-state accelerations of the mass and the base.
Vibration Isolation • Isolation of Source of Vibration from Surroundin gs
T r =
• • By defining
r =
1 r − 1 2
ω ω n
N =
30
π
=
for r > 2 and small ζ
2π N δ st 60
g
=
2 − R 1 − R
where R = 1 − T r
2 − R g 2 − R = 29.9092 δ st 1 − R δ st (1 − R )
Vibration Isolation • Isolation of Source of Vibration from
Surroundings (Isolation efficiency) •
Vibration Isolation •Example 9.4 •Spring Support for Exhaust Fan •An exhaust fan, rotating at 1000rpm, is to be supported by 4 springs, each having a stiffness of K. If only 10% of the unbalanced force of the fan is to be transmitted to the base, what should the value of K? Assume the mass of the exhaust fan to be 40kg.
Vibration Isolation •Example 9.4 •Spring Support f or Exhaust Fan •Solution •Transmissibility = 0.1 2
ω 1 + 2ζ ω n 0.1 = 2 ω 2 ω 2 1 − + 2ζ ω n ω n
•Forcing frequency
ω =
1000 × 2π 60
= 104.72 rad/s
Vibration Isolation •Example 9.4 •Spring Suppor t for Exhaust Fan •Solution •Natural frequency: •Assuming ζ =0,
ω n =
0.1 =
k m
=
4 K 40
=
K
3.1623
±1 104.72 × 3.1623 2 1 − K
•To avoid imaginary values, 331.1561 K
= 3.3166 or K = 9969.6365 N/m
Vibration Isolation •Example 9.8 •Isolation from Vibrating Base •A vibrating system is to be isolated from its supporting base. Find the required damping ratio that must be achieve by the isolator to limit the transmissibility at resonance to Tr =4. Assume the system to have a single degree of freedom.
Vibration Isolation •Example 9.8 •Isolation fro m Vibrating Base •Solution •Setting ω=ωn, T r =
1 + (2ζ )
2
2ζ
or ζ =
1 2 T − 1 2 r
=
1 2 15
= 0.1291
Vibration Isolation •Example 9.11 •Isolation Under Shock •An electronic instrument of mass 20kg is subjected to a shock in the form of a step velocity of 2m/s. If the maximum allowable values of deflection (due to clearance limit) and acceleration are specified as 20mm and 25g respectively, determine the spring constant of an undamped shock isolator.
Vibration Isolation •Example 9.11 •Isolation Under Shock •Solution •Magnitude of velocity of mass: xmax = X ω n •Magnitude of acceleration of mass: xmax = X ω n displacement amplitude
X =
xmax
ω n
< 0.02 or ω n >
xmax
=
X
2 0.02
= 100 rad/s
2 2 X ω n ≤ 25(9.81) = 245.25m/s or ω n ≤
100 rad/s ≤ ω n ≤ 110.7362 rad/s
where X is the
xmax X
=
245.25 0.02
= 110.7362 rad/s
Vibration Isolation •Example 9.11 •Isolation Under Shock •Solution •Selecting the value of
ωn
as 105.3681,
2 5 k = mω n = 20(105.3681) = 2.2205 ×10 N/m 2
Vibration Isolation • Active Vibration Control
•
An active vibration isolation system is shown below.
Vibration Isolation • Active Vibration Control
•
System maintains a constant distant between vibrating mass and referee
•
Depending on the types of sensor, signal processor and actuator used, the system can be electromechanical, electrofluidic, electromagnetic, piezoelectric or fluidic.
Chapter Outline • Introduction • Vibration Nomograph and Vibration Criteria • Reduction of Vibration at the source • Balancing of Rotating Machines • Whirling of Rotating Shafts • Balancing of Reciprocating Engines • Control of Vibration • Control of Natural Frequencies • Introduction of Damping • Vibration Isolation • Vibration Absorbers © 2011 Mechanical Vibrations Fifth Edition in SI Units
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Vibration Absorbers • When the excitation freq coincides with the
ωn, the system may experience excessive vibration. • Dynamic vibration absorber is another
spring mass system designed to shift ωn of the resulting system away from the excitation freq.
Vibration Absorbers • Undamped Dynamic Vibration Absorber
m1 x1 + k 1 x1 + k 2 ( x1 − x2 ) = F 0 sin ω t m2 x2 + k 2 ( x2 − x1 ) = 0
Vibration Absorbers •
Undamped Dynamic Vibration Absorber
•
x j (t ) = X j sin ω t , j = 1,2
•
Assuming
•
Amplitude of masses:
(k
2 ) m F 0 − ω 2 2 X 1 = (k 1 + k 2 − m1ω 2 )(k 2 − m2ω 2 ) − k 22
X 2 =
(k + k 1
•
We want to reduce
X1.
− m1ω 2 )(k 2 − m2ω 2 ) − k 22
Thus set numerator of
ω = 2
2
k 2 F 0
k 2 m2
, ω ≈ ω = 2
2 1
k 1 m1
X1 to
zero.
Vibration Absorbers • Undamped Dynamic Vibration Absorber
• X1 and X2 can
be rewritten as: 2
ω 1 − ω 2 X 1 = δ st k ω 2 ω 2 k 1 + 2 − 1 − − 2 k 1 ω 2 ω 2 k 1 X 2
δ st
=
1
k 1 + 2 k 1
ω − ω 2
2
ω 2 k 1 − − 2 ω 2 k 1
Vibration Absorbers • Undamped Dynamic Vibration Ab sorb er
• •
2 peaks correspond to 2 ωn of composite sys.
Vibration Absorbers • Undamped Dynamic Vibration Ab sorb er
=−
k 1
δ st = −
F 0
•
At X1=0, ω= ω1, X 2
•
Size of absorber can be found from: k 2 X 2
•
•
k 2
k 2
= m2ω 2 X 2 = − F 0
Absorber introduces 2 resonant frequencies Ω1 and Ω2, at which the amplitudes are infinite. Values of Ω1 and Ω2 can be found by noting
k 2 k 1
=
k 2 m2 m1 m2 m1 k 1
=
m2 ω 2
m1 ω 1
2
Vibration Absorbers •
•
Undamped Dynamic Vibration Absorber
Setting denominator of
X 1
δ st
ω ω 2
4
2
ω 2 ω − ω 1 ω 2
2
= 0,
m ω 2 1 + 1 + 2 2 + 1 = 0 m1 ω 1
• •
2 roots of the equation: 2 m2 ω 2 Ω1 1 + 1 + ω 2 m1 ω 1 = 2 Ω 2 ω 2 2
2
2 m ω 2 ω 1 + 1 + 2 2 − 4 2 m1 ω 1 ω 1
ω 2 2 ω 1
2
Vibration Absorbers •Example 9.15 •Vibration Absorber for Diesel Engine •A diesel engine, weighing 3000 N, is supported on a pedestal mount. It has been observed that the engine induces vibration into the surrounding area through its pedestal at an operating speed of 6000 rpm. Determine the parameters of the vibration absorber that 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 auxiliary mass is to be limited to 2 mm.
Vibration Absorbers •Example 9.15 •Vibration Absorb er for Diesel Engin e •Solution •We have f =
6000 60
= 100Hz or ω = 628.32 rad/s
•Amplitude of motion of auxiliary mass is equal and opposite to that of the exciting force. 2 F 0 = m2ω X 2
250 = m2 (628.32) (0.002) 2
m2 = 0.31665 kg k 2 = ω 2 m2 = (628.32) (0.31665) = 125009 N/m 2
Vibration Absorbers • Damped Dynamic Vibration Abso rber
•
Amplitude of machine can be reduced by adding a damped vibration absorber as shown.
Vibration Absorbers • Damped Dynamic Vibration Abso rber
•
Equations of motion m1 x1 + k 1 x1 + k 2 ( x1 − x2 ) + c2 ( x1 − x2 ) = F 0 sin ω t m2 x2 + k 2 ( x2 − x1 ) + c2 ( x 2 − x1 ) = 0
•
Assume solution: x j (t ) = X j e
iω t
, j = 1,2
• Steady-state solutions:
( ) X = [(k − m ω )(k − m ω ) − m k ω ]+ ic ω (k − m ω − m ω ) F 0 k 2 − m2ω + ic2ω 2
1
2
1
X 2 =
1
2
2
2
X 1 (k 2 + ic2ω )
(k − m ω + ic ω ) 2
2
2
2
2
2 2
2
2
1
1
2
2
Vibration Absorbers • Damped Dynamic Vibration Absorber µ = m2 / m1 = Mass ratio = Absorber mass/main mass δ st = F 0 / k 1 = Static deflection of the system ω a2 = k 2 / m2 = Square of natural frequency of absorber 2 ω n = k 1 / m1 = Square of natural frequency of main mass
f = ω a / ω n = Ratio of natural frequencies g = ω / ω n = Forced frequency ratio cc = 2m2ω n = Critcial damping constant
ζ = c2 / cc = Damping ratio
Vibration Absorbers • Damped Dynamic Vibration Absorber X 1
δ st X 2
δ st
(2ζ g )2 + (g 2 − f 2 )
2
=
=
(2ζ g ) (g − 1 + µ g 2
2
) + [µ f g − (g
2 2
2
2
2
− 1)(g − f 2
2
)]
2
(2ζ g )2 + f 4 (2ζ g ) (g − 1 + µ g 2
2
) + [µ f g − (g
2 2
2
2
2
− 1)(g − f 2
2
)]
2
Vibration Absorbers • Damped Dynamic Vibration Abso rber
•
If c2=ζ =0, resonance occurs at 2 undamped resonant frequencies
•
If ζ =∞, m2 and m1 are clamped together and system behaves as 1-DOF system. Resonance occurs at
g= •
ω ω n
=
1 1 + µ
= 0.9759
All curves intersect at pt A and B which can be located by 2 2 2 1 2 + + µ f f f + g 4 − 2 g 2 =0 2 + µ 2 + µ
Vibration Absorbers • Damped Dynamic Vibration Abso rber
•
The most efficient absorber (tuned vibration absorber) is one where pts A and B coincides.
•
Make curve horizontal at either A or B.
•
Set slope =0 at A and B:
µ
µ + 2
µ 3 − ζ 2 =
8(1 + µ )
3
µ
µ + 2
µ 3 + ζ 2 =
for point A
8(1 + µ )
3
for point A
Vibration Absorbers • Damped Dynamic Vibration Absor ber
•
Average value of ζ 2 used in design: 2 = ζ optimal
3µ 8(1 + µ )
3
X 1 X 1 2 = = 1+ µ δ st optimal δ st max
