Embed Size (px)
Citation preview
1 Basic Terms
Time period (T)Time period is the time taken by a particle to complete
one oscillation.
Frequency(ν) Number of oscillations complete per unit time by a particle.
ν =1
T(1)
Angular frequency(ω)It is the quantity obtained when the frequency is
multiplied by a factor 2π
ω =2π
T(2)
Phase(φ) The phase of a vibrating particle at any instant gives the state of
the particle as regards its position and direction of motion at that instant.Eg.
x (t)=Acos(ωt+φ), φ=ωt+φ0. φ0 is the initial phase or phase of particle at
t=0.
2 Basics of vibration
Any motion which repeats itself after certain interval of time is called vi-
bration.A motion which repeats itself after regular interval of time is called
periodic motion.Every vibration/ oscillation is periodic,but every periodic
motion is need not to be oscillatory/vibration.
2.1 Simple Harmonic Motion (SHM)
Simple harmonic motion is the simplest form of vibration.Eg. x (t)=Acos(ωt+φ),
where A is the amplitude,ω is the angular frequency and φ is the phase dif-
ference.SHM is not any periodic motion but one in which displacement is a
sinusoidal function of time.
2
3 Degree of Freedom
The minimum number of independent co-ordinates required to determine
completely the positions of all parts of a system at any instant of time defines
the degree of freedom of the system. It can be defined in both angular and
cartesian co-ordinate.
Figure 1: single degree of freedom system
4 Elementary Parts of Vibrating system
A vibratory system constitutes of kinetic energy and potential energy. Spring
stores potential energy and kinetic energy is associated with mass. Damper
is the element by which energy is taken away from the system. Vibration of
a system involves transfer of PE to KE and vice versa. If damping is present
some amount of energy is dissipated from the system after each cycle.
4.1 Spring element
Spring is a mechanical link that is generally assumed to be massless and low
damping property. Force is developed in spring when there is relative motion
between two ends of spring (deformation/displacement).The spring force is
proportional to the displacement/ deformation. The relation of spring force
3
to the displacement is given by the equation
F = kx (3)
where F is the spring force,x is the displacement and k is the spring constant.
If we plot a graph F Vs x, the result will be straight line. The work done (U)
in deforming spring is equal to the strain or potential energy stored in that
spring,and it is given by
U =1
2kx2 (4)
Generally springs are non-linear in nature.It exhibits linear nature up to
certain limit of deformation/displacement.Any vibrating system can be con-
verted to an equivalent spring mass system.Eg. A cantilever beam with end
mass m can be converted to a spring mass system has mass m and stiff-
ness/spring constant as
k =3EI
l3(5)
where E=Young’s modulus ,I=Moment of inertia of cross section of beam,l=Length
of beam
4.2 Mass or Inertia Element
Mass is assumed to be a rigid body which can lose or gain kinetic energy
when the velocity of the body changes.By Newton’s second law force applied
on a mass is the product of mass and its acceleration.Work is equal to the
force multiplied by the displacement in the direction of force applied and the
work done on the mass is stored in the form of mass’s kinetic energy.
KE =1
2mv2 (6)
where v is the velocity.
4
4.3 Damping Elements
The mechanism by which the vibrational energy is gradually converted into
heat or sound is known as damping .Damper is assumed to have neither
mass nor elasticity.The damping force exists only if there is relative veloc-
ity between two ends of the damper.Eg. viscous damping,coulomb damp-
ing,material damping.
5 Classification of vibration
5.1 Free and Forced vibration
Free Vibration If a system, after an initial disturbance is left to vibrates
its own such type of vibration is called free vibration. Eg. Motion of a simple
pendulum.
Forced Vibration If a system is subjected to an external force (external
force is applied to the system at particular interval of time) the resulting
vibration is called forced vibration. When the external forcing frequency
coincides with the natural frequency of the system resonance will occurs,
which leads to the failure of the system.
Natural frequency is the frequency of the system under free vibration.
5.2 Damped and Undamped Vibration
Undamped Vibration If no energy is dissipated from the system due
to friction or any resistance during oscillation is called Undamped vibra-
tion.Damped Vibration If energy is dissipated from the system (mainly
in the form of heat),such type of vibrations are called damped vibration.In
damped vibration the amplitude of vibration gradually decreases.Damping
5
is classified into viscous damping (fluid friction),coulomb damping (solid
friction),material damping (hysteretic damping).
5.3 Linear and Non-linear Vibration
Linear vibration If all the basic components of the vibrating system (the
spring, the mass and the damper) behaves linearly,the vibration is called
linear vibration. If vibration is linear it satisfy principle of superposition.
Mathematical techniques are well developed to solve such problems.
Non-linear vibration If all the basic components of a vibrating system be-
haves non-linearly then such vibration is called non-linear vibrations.Mathematical
techniques are not well developed to solve non-linear problems.
5.4 Deterministic and Random Vibration
If the magnitude of force or motion of a vibrating system is known at any
given time is called deterministic vibration.In some cases the magnitude
of force or motion can’t be predicted such vibration is called random vi-
bration.
6 Undamped free vibration(Translational sys-
tem)
6.1 Equation of motion(EOM) and Natural Frequency
The equation of motion (EOM) of a single degree of freedom (SDoF) un-
damped system can be obtained using following methods:-
6
1. Newton’s method (Newtonian approach)
2. Energy method
3. Rayleigh’s method
6.1.1 Newton’s method
This method is based on Newton’s second law of motion,force acting on a
body is proportional to its acceleration. In vertical configuration the position
of mass after elongation of the spring by a distance δ due to gravitational
pull is taken as the static equilibrium position.In horizontal configuration
deflection of the spring due to gravitational pull can be neglected and also
we are assuming that mass is moving on a frictionless roller.
Figure 2: spring-mass system in horizontal position
Let x be the displacement,then dxdt
= x is the velocity,d2xdt2
= x is the acceler-
ation.
Restoring force F = ma = mx (restoring force is opposite to the direction of
displacement)Then,
mx = −kx ⇒ mx + kx = 0.This is the equation of motion of a SDoF un-
damped system.This is the equation of a SHM.The solution of the equation
is x(t) = Acosωnt+Bsinωnt (By fourier’s theorem any periodic function can
be expressed as a superposition of sine and cosine function).After substitut-
ing the solution in EOM we get,
7
−ω2n + k
m= 0 → k
m= ω2
n
ωn =√
km
rad/s is the natural frequency of the system.
fn = 12π
√km
Hz
6.1.2 Energy Method (Lagaranges method)
We know that the energy associated with a spring mass system is partially
kinetic energy and partially potential energy.According to this method sum
of energies associated with the system is constant.In this case we assume that
the system is conservative,ie;no energy is dissipated from the system.
Kinetic energy + Potential energy = Const.
we know that,kinetic energy = 12mv2,where v is the velocity.If x is the
displacement,dxdt
= x is the velocity.Then,kinetic energy = 12mx2
Potential energy = 12kx2
12mx2 + 1
2kx2 = Const.
ddt
( 12mx2 + 1
2kx2)=0
mx+ kx=0
Equation of motion obtained is same as we obtained in the above method,therefore
equation for natural frequency is same as above.Rayleigh’s method
8
6.1.3 Rayleigh’s Energy Method
This method is an extension of energy method. the method is based on the
principle that total energy of the vibrating system is equal to the maximum
potential energy of the system.At any moment total energy is either kinetic
or potential energy or sum of both kinetic and potential energy.Rayleigh’s
energy method,gives the natural frequency of the system directly.
Let x = Asinωnt
KEmax = 12mx2 = 1
2m(ωnA)2
PEmax = 12kx2 = 1
2kA2
KEmax = PEmax
12m(ωnA)2 = 1
2kA2
mω2n = k → ω2
n = km
ωn =√
kmrad/s
7 Equivalent stiffness of spring combinations
Springs in parallel Consider the springs shown in figure, assume a dis-
placement x is given to the mass m
The deflection of individual spring is equal to the deflection of the sys-
tem.So, k1x+ k2x = kex → ke = k1 + k2 where,
9
Figure 3: Springs in parallel
ke = equivalent stiffness of the system, k1, k2 = stiffness of springs, x =
deflection of the system.
Springs in series Consider the springs shown in figure, assume a
displacement x is given to the mass m
Figure 4: Springs in series
The total deflection of the system is the sum of deflection of individual
springs.So, x = x1 + x2 + x3 + .....
10
Forceke
= Forcek1
+Forcek2
+.......
or 1ke
= 1k1
+ 1k2
+.......
8 Undamped Torsional Vibration
If a rigid body oscillates about a specific reference axis,the resulting motion
is called torsional vibration.In this case,the displacement of the body is mea-
sured in terms of an angular co-ordinate.In torsional vibration,the restoring
moment may be due to the torsion of an elastic member or the unbalanced
moment of a force /couple.
Figure 5: Torsional vibration of disc
consider the above figure,a disc has polar mass moment of inertia J0
mounted at one end of solid circular shaft and the other end is fixed.The
angular rotation of the disc about the axis of shaft be θ;θ also represents the
shaft’s angle of twist.From the theory of torsion of circular shafts,
Mt = GIol
,where
11
Mt = Torque that produces twist θ
G = Shear modulus, l = length of the shaft, Io = Polar moment of inertia
of the shaft
Polar moment of inertia Io = πd4
32, d = diameter of the shaft.
If the disc rotated by an angle θ from its equilibrium position,the shaft
provides a restoring torque of magnitude Mt. thus the shaft acts as torsional
spring with torsional stiffness
Kt = Mt
θ
Consider the free body diagram,the equation of motion of the system by
Newton’s method is Joθ +Ktθ = 0.
9 Damped free vibrations
9.1 Types of damping
1. Viscous damping
Viscous damping is the most commonly used damping mechanism.When
a system vibrates in a fluid medium such as air,gas and oil, the resis-
tance offered by the fluid to the moving body causes energy to be dissi-
pated.In viscous damping,damping force is proportional to the velocity
of vibrating body.
2. Coulomb damping / Dry friction damping
12
In coulomb damping the damping force is constant in magnitude but
opposite in direction to that of motion of vibrating body. It occurs
due to friction between rubbing surfaces that are either dry or have
insufficient lubrication.
3. Material or Hysteresis damping
When materials are deformed,energy is absorbed and dissipated by the
material.This effect is due to friction between internal planes.
4. Others like structural damping,slip or interfacial damping
9.2 Free Viscously damped system
In this system we added a damper to the existing spring mass system.The
damper dissipates energy from the system,based on the value of damping
co-efficient the nature of system’s response varies.
Figure 6: Damped free vibration
m = mass of the system , c = damping co-efficient, k = spring constant.
By Newton’s law the equation of motion of the system is
mx+ cx+ kx = 0
13
Let x(t) = est, substitute the value of x(t) in equation of motion.The equa-
tion of motion becomes ms2est + csest + kest = 0
ms2 +cs+k = 0,now the equation of motion is in the form of a quadratic
equation.Since it is a quadratic equation it has two roots.The two roots of
the equation is :-
s1, s2 = −c2m±
√( c
2m)2 − k
m
x(t) = A1es1t + A2e
s1t
x(t) = A1e−c2m
+√
( c2m
)2− km + A2e
−c2m
−√
( c2m
)2− km
9.2.1 Critical damping(cc)
It can be simply defined as the value of damping co-efficient (c) at which
the terms inside the radical sign become zero.It is represented by cc.
[ c2m
]2 = km
= [ cc2m
]2 = km
cc = 2m√
km
cc = 2√km , critical damping co-efficient.
9.2.2 damping factor (ζ)
It is the ratio of damping coefficient to critical damping co-efficient.
14
ζ = ccc
Again consider the the equation of motion
x(t) = A1e−c2m
+√
( c2m
)2− km + A2e
−c2m
−√
( c2m
)2− km
from the above equation consider the term c2m
c2m
= ccc
cc2m
c2m
= ζωn Now, x(t) = A1e(−ζ+√ζ2−1)ωnt+A2e
(−ζ−√ζ2−1)ωnt
Based on the values of damping ratio/ factor damped vibrations are classified
into :-
1. Over damped system, ζ > 1
2. Critically damped system, ζ = 1
3. Under damped system, ζ < 1
9.2.3 Over damped system
• At time t=0, the displacement of the system x(t) = A1 + A2
• Even though ζ > 1, the co-efficient of A1andA2 having negative ex-
ponential terms.Which indicates the amplitude of vibration/ the dis-
placement x(t) decreases as time increases
• In this case the response of the system is an aperiodic motion.the system
doesn’t vibrates.The response approach equilibrium position as time
tends to ∞.(Once the system get disturbed it will take infinite time to
come back to its equilibrium position). The values of A1andA2 can be
find by applying initial conditions x(0)andx(0)
15
Figure 7: Response of over damped system
9.2.4 Critically damped system ζ = 1
• c2m
=√
km
.Since the roots are same the solution of equation of motion
is x(t) = (A1 + A2t)e−ωnt
• The values ofA1andA2 can be find out from initial conditions x(0)andx(0).
x(0) =A1
x(t) = −ωn(A1 + A2t)e−ωnt + A2, x(0) = −ωnA1 + A2
⇒ x(0) = −x(0)ωn + A2
A2 = x(0) + x(0)ωn
• x(t) = [x(0) + (x(0) + x(0)ωnt)]e−ωnt.
• In this case also we can see negative exponential term in the displace-
ment of the system.It indicates the displacement is decreasing as the
time increases and finally it reaches the equilibrium position.the re-
sponse of the system is aperiodic(there is no vibration/oscillation).
• The response of the critically damped system is similar to over damped
system, but the difference is that, critically damped system approaches
equilibrium position at the shortest possible time.
16
• Shock absorber, recoil spring in gun,door closer etc are some examples
which we can see the application of critical damping.
Figure 8: response of critically damped system
9.2.5 Under damped system ζ < 1
• In this case the damping factor is less than unity.The roots of equation
of motion are complex roots.
x(t) = A1e(−ζ+i√
1−ζ2)ωnt + A1e(−ζ−i√
1−ζ2)ωnt
• Here we can see negative exponential along with complex terms in
the displacement of the system.It shows that the system undergoes
harmonic oscillationbut decaying along with the time.
We know that eiθ = cosθ + isinθ
• The above solution of x(t) can be written as
x(t) = [A1cos(√
1− ζ2.ωnt)+iA1sin(√
1− ζ2.ωnt)]+[A2cos(√
1− ζ2.ωnt)−
iA2cos(√
1− ζ2.ωnt)]
• Let (A1 + A2) = C1 and i(A1 − A2) = C2
x(t) = e−ζωnt[C1cos(√
1− ζ2)ωnt+ C2sin(√
1− ζ2)ωnt]
• Therefore x(t) = C3cos(ωnt+ φ0) or C4sin(ωnt+ φ0) ,C3, C4, andφ are
constants which can be find from initial conditions
17
Figure 9: Response of an under damped system
• The term√
1− ζ2ωn = ωd is the damped natural frequency.
9.2.6 Logarithmic decrement
It is defined as the natural logarithm of the ratio of two successive amplitudes
on the same side of the mean line.Let amplitude of two successive peaks be x1
and x2.Then, logarithmic decrement δ = ln(x1x2
),damped time period td = 2πωd
by solving x(t) we get δ = 2πζ√1−ζ2
.If ζ << 1,δ = 2πζ
9.2.7 Relationship between vibrational energy and logarithmic
decrement
We know that δ = ln(x1x2
) ⇒ x1x2
= eδ ⇒ x2x1
= e−δ
e−δ can be written as e−δ = 1 − δ + δ2
2!− ................ Let E1 be the energy
associated with the amplitude x1 is E1 = 12kx2
1 and energy associated with
x2 is E2 = 12kx2
2.
Now,the energy being dissipated form the system per cycle and its ratio with
maximum energy in that cycle.
∆EE1
= E1−E2
E1= 1− E1
E2
18
= 1-( 12kx2
2 /12kx2
1 )
= 1- x22 / x
21
e−δ = x2 / x1 → x22 / x
21 = e−2δ
∆EE1
= 1− e−2δ
→ 1− e−2δ = 1− (1− 2δ + 2δ2
2!− .........)
higher order terms can be neglected then,
∆EE1
= 2δ.
The ratio of energy dissipated during the cycle to the maximum energy in
that cycle is equal to the twice of logarithmic decrement.
9.2.8 Energy dissipation in viscous damping
For viscously damped system force F can be expressed as F = cx, where c
is the damping co-efficient.
Work done = Force * displacement.
= F * dx → c ∗ dxdtdx
The rate of change of work per cycle,
∆E =∫ 2πωdc∗( dx
dt)2dt
0
10 Forced vibration problems
1) A machine having mass m=100kg,spring stiffness k = 7.84×105N − m
has a unbalanced rotating force of 392N at speed of 3000 rpm. Assuming
damping factor is 0.2.Determine amplitude of vibration,Transmissibility ra-
tio and transmitted force.
Given data:-
19
m=100kg, k= 7.84×105N −m, F=392N, N=3000 rpm,ζ=0.2
ω=2π×300060
= 314.16 rad/sec
ωn=√
km
=√
7.84×105
100= 88.54 rad/sec
r= ωωn
= 3.5
A= F/k√(1−r)2+(2ζr)2
= 4.41×10−5m
Transmissibility ratio (TR)=
√1+(2ζr)2√
(1−r2)2+(2ζr)2= 0.151
Force transmitted FT , TR = FTF⇒ FT = TR× F = 0.151× 392 = 59N
2) A machine of 100kg mass has a 20kg rotor with 0.5mm eccentric-
ity,stiffness 85 × 103N/m,damping factor is 0.02, N=600rpm. Determine
dynamic amplitude,phase angle,force transmitted.
Given data, m = 100kg, k = 85 × 103N/m, m0 = 20kg, ζ = 0.02, N =
600rpm
Soln:-
ω = 2πN60
= 2π×60060
=62.83 rad/s
ωn = km
= 85×103
100= 29.15 rad/s
r = ωωn
= 2.155
20
mAm0e
= r2√(1−r2)2+(2ζr)2
A = r2m0e
m√
(1−r2)2+(2ζr)2
= 2.1552×20×0.005
100√
(1−2.155)2+(2×0.02×2.155= 1.27× 10−4m
Phase angle, tanφ = 2ζr1−r2 , φ = tan−1[2×0.02×2.155
1−2.1552] = -1.33
Transmission ratio (TR) = FTF
=
√1+(2ζr)2√
(1−r2)2+(2ζr)2= 0.29
FT = F × 0.29 = 11.8N ,where F = m0ω2e
3) A seismic instrument with frequency of 6 Hz is used to measure vi-
bration of a machine at 120 rpm.The relative displacement of seismic mass
is 0.05 mm. determine amplitude of vibration of machine. Neglect damping.
Solution:-
ζ = 0
ωn = 2πfn = 2π × 6 = 37.69 rad/s
ω = 2πN60
= 2π×12060
= 12.56 rad/s
Z = 0.05mm = 0.05× 10−3 m
r = ωωn
= 0.33
21
ZB
= r2√(1−r2)2+(2ζr)2
= 0.122
0.05×10−3
B= 0.122 → B = 4.09× 10−4m
4) A vibrating system have total mass m = 25kg,at speed of 1000 rpm,the
system and eccentric mass have a phase difference of 90o and corresponding
amplitude is 1.5 cm. The eccentric unbalanced force m0 = 1kg at radius
of rotation of 4cm. determine natural frequency of the system,damping fac-
tor,amplitude at 1500rpm,phase angle at 1500 rpm.
Given data:-
m = 25kg , e = 4cm = 0.04m, A = 1.5cm = 0.015 m, m0 = 1kg, φ = 90o, N
= 1000rpm
Solution:-
ω = ωn = 2πN60
= 104.71 rad/s
at ω = ωn, r = 1 then,
mAm0e
= 1√(2ζr)2
= 12ζ
25×1.51×4
= 12ζζ = 0.053
ω at 1500 rpm = 2π×150060
= 157.07rad/s
frequency ratio r = ωωn
= 1.5
A =m0em
r2√(1−r2)2+(2ζr)2
= 2.82×10−3m
22
phase angle φ, tanφ = 2ζr1−r2 = 2×0.053×1.5
1−1.52= 172o
11 Vibration measuring instruments
instruments used to measure the displacement ,velocity or acceleration of a
vibrating body is called vibration measuring instruments.Instruments having
spring,mass and dashpot (damper) is called seismic instruments.The vibra-
tion measuring instruments are classified into Vibrometer and Accelerometer.
Vibrometer measures the displacement of the vibrating body and it have
low natural frequency.Accelerometer measures the acceleration of the body
and it have high natural frequency.We have equation,
ZB
= r2√(1−r2)2+(2ζr)2
,When the value of ’r’ is very high (more than 3)the
equation can be written as
ZB
= r2√(1−r2)2
' 1, Z' B The relative amplitude Z is equal to the am-
plitude of the vibrating body B on the screen.Even though Z and B are not
in same phase,but B being a single harmonic (motion which displacement
is directly proportional to restoring force)which makes the output quantity
same as the input quantity.
Vibrometer
• It is known as low frequency transducer and it is used to measure high
frequency ω of vibrating body
23
Figure 10: Vibrometer
• The natural frequency of the instrument is low,the value of r is high r
= ωωn
• Since the natural frequency of the instrument is low, the mass of the
instrument is high
• It is used rarely,the range of natural frequency of the instrument is 1
Hz to 5 Hz
Accelerometer
• Used to measure acceleration of a vibrating body
• The natural frequency of the instrument is high compared to the fre-
quency of vibrating body.So it is light in construction.
• It is widely used in vibration measurement
• An accelerometer composed of internal mass compressed in contact
with relatively a stiff force measuring load cell(usually piezoelectric
24
Figure 11: Accelerometer
crystal)by relatively soft preloaded spring.For an accelerometer system
damping is negligible.
6) Discuss all condition of transmissibility of vibration.How will you
model transmissibility? Why is vibration isolation is important.
hint- define transmissibility- transmissibility happens in externally ex-
cited system-draw base excitation-derive the equation for A/B ratio-define
isolation-two types of isolation-materials used- application
25
![Notes [2012]](https://img.pdfslide.tips/doc/110x75/55cf96b4550346d0338d4387/notes-2012.jpg)
