Mech Vibrations Complied P.dinesh

8/3/2019 Mech Vibrations Complied P.dinesh

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e-Notes by Prof. P.Dinesh, Sambram College, Bangalore

Mechanical Vibrations (ME-65)

Session 1 date: (27/2/07)

Additional references

Mechanical Vibrations,Grover. G.K . , Nemchand and Bros. Roorkee

Mechanical Vibrations, V.P.Singh, Dhanpat Rai and Co., Delhi.Vibrations and Noise for Engineers, K. Pujara, Dhanpat Rai and Sons,New Delhi.Mechanical Vibrations, J.B.K. Das and P.L. Srinivasamurthy, Sapna bookHouse, Bangalore.

.

Introduction:

Vibration is defined as a motion which repeats after equal interval of time and isalso a periodic motion.The swinging of a pendulum is a simple example ofvibration.Vibration occurs in all bodies which are having mass and elasticity. Theyare caused due to several reasons such as presence of unbalanced force in rotatingmachines, elastic nature of the system, external application of force or wind loadsand earthquakes. Vibrations are undesirable as they induce high stresses in systemcomponents leading to noise and failure, in such cases they are to be minimized ifnot totally eliminated.. The desirable effects are seen in musical instruments andcement compactors used in construction work.

From subject point of view the following notations and definitions are very

important:Periodic Motion:

It is a motion which repeats itself after equal intervals of time, e.g., theoscillations of simple pendulum

Time Period (T) :

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It is the time required for one complete cycle or to and fro motion.The unit isseconds.

Frequency (f or ) :It is the number of cycles per unit time. The unit are radians/sec. or Hz.

Amplitude (X or A) :It is the displacement of a vibrating body from its equilibrium position. It has

units of length in general.Natural Frequency (fn):

It is the frequency with which a body vibrates when subjected to an initialexternal disturbance and allowed to vibrate without external force being appliedsubsequently.

Fundamental Mode of Vibration:

A vibrating body may have more than one natural frequency and when it vibrateswith the lowest natural frequency ,it is the Fundamental mode of vibration.

Degrees of Freedom:It is the minimum number of coordinates required to describe the motion of

system. Typically in our discussions 1DOF system will have one mass, e.g., a springattached with one mass , 2 DOF system will have two masses and likewise we have3DOFsystem. A continuous system like a beam or plate consisting of infinite number ofparticles with mass, are systems with infinite number of DOF.

Simple Harmonic Motion (SHM):It is a periodic motion with acceleration always directed towards the equilibrium

position. It can also be defined as projection of motion of a particle along a circle withuniform angular velocity on the diameter of circle.Damping:

It is the resistance offered to the motion of a vibrating body by absorbing theenergy of vibrations. Such vibrations are termed as damped vibrations.Forced Vibrations:

It is the vibration of a body when subjected to an external force which is periodicin nature and vibrations occur as long as external force is present.

Resonance:It is said to occur in the system when the amplitude of vibrations are excessive

leading to failure. This occurs in forced vibrations when the frequency of externallyapplied force is same as that of natural frequency of the body.

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Linear and Non Linear Vibrations:When the vibrations are represented by linear differential equations and laws of

superposition are applicable for the system, we have Linear systems. Non linearvibrations

are experienced when large amplitudes are encountered and laws of superposition are notapplicable.

Longitudinal, Transverse and Torsional Vibrations:When the motion of mass of the system is parallel to the axis of the system, we

have Longitudinal vibrations. When the motion of mass is perpendicular to the systemaxis the vibrations are Transverse vibrations and when the mass twists and untwists

about the axis the vibrations are Torsional vibrations. Up and down motion of mass in aspring mass system represents Longitudinal vibrations. Vibration of a cantilever beamrepresents Transverse vibrations. The twisting and untwisting of a disc attached at the endof a shaft represents Torsional vibrations.

Vector representation of SHM:Any SHM can be represented as by the equation , x = A Sint ---(1) , where x is

the displacement , A is the amplitude , is the circular frequency and t is the time.Differentiating eqn.1 w.r.t. t we have velocity vector and differentiating eqn 1

twice we have the acceleration vector. If x1 and x2 are two displacement vectors withsame frequencies then the phase difference between them is given by .

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Session 2 date : (28/2/07)

Principle of Superposition:When two SHM of same frequencies are added the resulting motion is also a

harmonic motion. Consider two harmonic motions x1 = A1Sint and x2 = A2 Sin(t +) . Then if x is the resultant displacement , x = x1 + x2. The resultant amplitude x = ASin (t +), where A is the resultant amplitude and is acting at an angle w.r.t vector x1.

The above addition of SHMs can also be done graphically.

**********

Sample Problems:

1) Add the following harmonics analytically and check the solution graphicallyx1= 3Sin (t + 30) , x2 = 4 Cos(t +10)

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Solution:Given : x1= 3Sin (t + 30) , x2 = 4 Cos(t +10)

Analytical method:We know that, x = x1 + x2 = A Sin (t+ )

Make x1 and x2 to have same Sin terms always , i.e., x2 = 4 cos(t +10 + 90) =4Sin (t +100 )Hence, A Sin (t+ )= 3 Sin (t + 30) + 4 Sin (t +100 )Expanding LHS and RHS

ASin t Cos + A cos t Sin =3 Sin t cos 30

+ 3Cos tSin30

+ 4 Sin t cos 100

+ 4 Sin t sin 100

A Sin t Cos + A cost Sin = Sin t(1.094) + Cos t (5.44)Comparing the coefficients of A Cos and A Sin in the above equationACos = 1.094, ASin = 5.44 , tan = ASin / ACos = 5.44/1.904Therefore , = 70.7 and A = 1.094/Cos 70.7 = 5.76 .

Graphical Method.:

Draw ox the reference line. With respect to ox, draw oa equal to 3 units in length atan angle of 30 to ox and ob equal to 4 units at an angle of 100 to ox. Complete thevectorpolygon by drawing lines parallel to oa and ob to intersect at point c. Measure ocwhich should be equal to A and the angle oc makes with ox will be equal to . Allangles measured in anticlockwise direction.

O

xxx

a

b

c

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2) Repeat the above problem given , x1 = 2Cos(t + 0.5) and x2 = 5Sin(t + 1.0) .The angles are in radians.(Hint: In the above problem the angles are to be converted to degrees. Ans. A =

6.195, = 73.49

)

3) Add the following harmonic motions analytically or graphically.x1 = 10 Cos(t + /4) and x2 = 8 Sin(t + /6) .

4) A body is subjected to 2 harmonic motionsx1 = 15sin(t + /6) , x2 = 8 cos(t + /6), what harmonic is to be given to the

body to it to equilibrium.Solution :Let the harmonic to be given to the two harmonics to make it to be in equilibrium beAsin (t + )Therefore, Asin (t + )+ x1 + x2 = 0Hence, Asint cos +A cost sin +15sint cos/6+15cost sin/6+8 costcos/6+8 sint sin/6 = 0

sint (A cos +8.99038)+ cost(A sin +14.4282) = 0Therefore, A cos = -8.99038

A sin = - 14.4282Therefore, tan = A sin / A cos = 14.4282/8.99038 , = 58.062From , A cos = -8.99038, substituting for = 58.062, A = 17.00Therefore, the motion is x= 17sin(t + 58.062)

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Beats Phenomenon:Consider two harmonics x1 and x2 of slightly different frequencies and the A cos

resulting motion will not be a SHM. Due to existence of different frequencies thephase difference of the two vectors keeps on changing and shifting w.r.t time. Thetwo harmonics when in phase have their resultant amplitude to be sum of individualamplitudes and when they are out of phase the resultant amplitude is difference ofindividual amplitudes. This phenomenon of varying of resultant amplitude is called asBeats and this occurs at a frequency given by the difference of the individualfrequencies of the two vectors.

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Fourier Theorem:

Any periodic motion can be represented in terms of sine and cosine terms calledas Fourier series. The process of obtaining the Fourier series of a periodic motion iscalled Harmonic analysis, i.e.,

F(t) a periodic function can be represented asF(t) = a0 + a1cost + a2cos2t + a3cos3t +.an cosnt

+ b1sint + b2sin2t + b3sin3t +.. bn sinnt

Resultant amplitude variation during one cycle w

time

Ampli

tude

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The constants a0,a1,a2,,,,,and b1,b2,b3. etc., are obtained using the followingformulae:

a0 = (/2) F(t), in the limits 0 to 2/

an=(/) F(t)cos(nt) dt, in the limits 0 to 2/bn= =(/) F(t)sin(nt) dt, in the limits 0 to 2/

4) Represent the above periodic motion using harmonic series

Sol: Mathematically for one complete cycle we have the eqn for AB asx(t) = -20t +2 for 0


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= (4/n)Thus, the harmonic series is , x(t) = 4/ (1/n) sin10nt , for n =1,2,..

5) A periodic motion is represented by a saw tooth wave form, the amplitude is 0 at t=0and rises to 10 cm, at t=0.3, it then drops down to zero at t = 0.3 ,and remainszero for next 0.2 seconds and one cycle is completed. The next cycle again starts at t=0.5secs. Represent the above cycle in form of a harmonic series.

Further problems refer the texts given in the additional references.

*************

10

0.3 0.2

Sec.

X,

cms

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Mechanical Vibrations (ME 65)


Undamped freee vibrations

Single degree of freedom System

This consists of a single spring attached with a single mass. The Various ways inwhich the equation of motion is obtained are :

a) Newtons Method b) Energy Method and c) Rayleigh Method

Newtons MethodWhen a mass m is attached to a spring it deflects by and the system is under

equilibrium as mg = weight = k, where k is the spring stiffness, defined as force per unitlength. If now the mass m is given a displacement x in the downward direction and thesystem is allowed to vibrate, we have the following forces acting on the system: thespring force, k(x+) acting in the upward direction, inertia forcemxacting in the upward direction and force mg acting in the direction of displacement x

downwards. The equation of motion is written taking equilibrium of forces as:

mx = - k(x+) + mg= -kx-k+mg= -kx-k+k

Or mx + kx = 0, which is the governing differential equation for a single degrr offreedom system. Rewriting the equation of motion as

m

k

Displ. x

m

mg

k(x+)

mx

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x+ (k/m) x = 0, we have the quantity (k/m)1/2

as the natural frequency of the systemn .

Energy Method:In this method the concept of total energy of the system, which is the

sumof Kinetic energy (T) and Potential energy(V) , is made use of which remainsconstant always for any configuration of system while it is vibratingFor a single DOF system of spring and mass, the kinetic energy is givenby (1/2)mx andthe potential energy stored in the system is (1/2)kx2 . As the total energy of the systemremains constant, we have T+V = 0 or d(T+V)/dt = 0. Differentiating we have thegoverning differential equation as mx + kx = 0, and the natural frequency is given by

n = (k/m)

1/2

.Rayleighs Method:

In this method the max kinetic energy of the system is equated to themaximum potential energy. For SHM the max. kinetic energy is at the mean positionwhich is equated to the potential energy. If A is the amplitude of vibration and n is thenatural frequency the max. kinetic energy is given by (1/2)m(n A)2 and max. potentialenergy is (1/2)kA2 . Equating the two equations and simplifying we have again n =(k/m)1/2.

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Session 5 date:(7/3/07)SPRINGS IN SERIES AND SPRINGS IN PARALLEL

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Consider figure (b) where the springs are in series. When the mass is subjected to a forceF, the displacement of mass m is equal to deflections of sprimgs 1 & 2. Hence we canwrite, the displacement of the equivalent spring as,

= 1 + 2

Where 1 deflection of spring 1.and 2 deflection of spring 2.

Hence we can write

F/Ke = F/K1 + F/K2 , where Ke= equivalent spring stiffness

1/Ke = 1/K1 + 1/K2

Considering fig (a) where springs are in parallel when the mass is subjected to aforce F we have the total spring force equal to sum of individual spring forces.

Hence, we can write the total force in the equivalent spring as

Ke .x = K1.x + K2.x

Therefore Ke = K1 + K2Therefore the equivalent spring stiffness for springs in parallel is equal to

Ke = K1 + K2

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Problem:1) Obtain an equivalent spring mass system and expression for n. for 3 springs in series

and in parallel configuration

i) for series spring combination.

1/Ke = 1/K1 + 1/K2 + 1/K3

Therefore Ke = K1.K2.K3K1K2 + K2K3 + K1K3

ii) for parallel spring combination.

Ke= K1 + K2 + K3

1/Ke = 1/K1 + 1/K2 + 1/K3

Ke= K1 + K2 + K3

Natural frequency

K1

K2

K3

m

12

Ke


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n = (Ke / m) , Therefore n =m

KKK 321 ++

Therefore n =m

KKK )3/12/11/1( ++

2). Obtain the natural frequency of the system

Solution:Given m = 109 N

K1 = 10N/mmK2 = 10N/mmK3 = 5N/mm

The spring equivalent when parallel springs are added, we have

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Ke1 = K1 + K2= 20 N/mm

Ke = Ke1 + K3Ke1 + K3

= 20 (5)

25= 4 N/mm

= 4000 N/mn = (Ke / m)Therefore n = 18.97 rad/s

Frequency = fn = n /2= 3.012 Hz

***********

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Session 6 date:(9/3/07)

NATURAL FREQUENCY OF A SPRING: Considering mass of spring

Consider a spring mass system as shown in the figure where the mass is displacedby x. dy is a small elemental spring length at a distance of y from the fixed end.L be the length of the spring. Let x and x be the velocity and acceleration of mass.

The total K.E of the system is the sum of K.E of the mass m and K.Econsidering the mass of the spring.

The velocity of the spring element at a distance of y from the fixed end is xy/LWe can write the K.E of the spring element dy as ( )(Rho)(dy)(xy/L)2

Where Rho is the mass densityAbove expression is of the form KE = mv2.The K.E for the entire spring considering the mass of the spring becomesL( )(Rho)(dy)(xy/L)2

0L

= (Rho) x2 / 2l2 (y3/3)0

= 1/6 (Rho)x2L

= 1/6 Ms.x2 where, Ms = (Rho)L = Mass spring

Therefore the entire K.E of the system

K.E = 1/2mx2 + 1/6 Ms.x2

kL

y

dy

xm

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The potential energy of the systemP.E = Kx2

Therefore the total energy of the system

= K.E + P.E = constant

1/2mx2 + 1/6 Ms.x2 + Kx2 = constantDifferentiating the above expression w.r.t time we get

Mxx + 1/3 Msxx+ Kxx = 0

Mx + 1/3 Msx+ Kx = 0

x {m + 1/3 Ms} + Kx = 0

From the above expression,Therefore , n = (K/(m+1/3(Ms))1/2 , rad/s

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Session 7 date :(16/3/07)

Problems

3 ) Determine the equation of motion and natural frequency of the system shown

Solution:It is assumed that:

The string is inextensible The friction between string and disc is neglected.

The disc is given an angular displacement , due to which the mass m isdisplaced by x, from the figure, we have;

x = 2r .

Also, the vertical displacement of centre O is

X1 = r

m

x

r

k

M

O

x1

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By making use of the energy principle, we have the total energy of the system is constantat any given instant of time.K.E. System = 1/2mx2 +1/2M2 +1/2I2

= 1/2 m(2r)2+1/2M(r)2+1/2(1/2Mr2)()2

Simplifying= r2 ()2(2m+(3/4)M)

Similarly, P.E. of the system is the strain energy stored in the spring due to displacementof centre x1, i.e, (1/2)k(x1)2 , which is (1/2)k(r2)()2

According to Energy Method, d (KE+PE) = 0dt

Differentiating the sum of KE and PE

(2m+(3/4)M)2+k = 0Or+ k/(4m+(3/2)m) = 0 , which is in the form + n2 = 0

orn2 = k/(4m+(3/2)m), i.e.

n = k/(4m+(3/2)m)1/2

and the natural frequency inHz, ,

fn = (k/(4m+(3/2)m)1/2)/2University Problems for practice

1) A homogenus cylinder of mass m and radius r is suspended by a spring and aninextensible cord as shown. Obtain the equation of motion and find the naturalfrequency of vibration of the cylinder.

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Answer: +(8k/3m) = 0, fn = (8k/3m)/2, Hz.

2) A simple pendulum is as shown in fig. Determine the natural frequency of thesystem if the mass of the rod m r is not negligible .

Answer: + ((m +(mr/2))/((m+(mr/3))(g/L) = 0fn = (((m +(mr/2))/((m+(mr/3))(g/L))/2, Hz

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3) A circular cylinder of mass m and mass moment of inertia I is connected by aspring of stiffness k as shown. If it is free to roll without slipping, determine thenatural frequency.

r

k

m

L

m

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Answer: fn = ((2k/3m))/2 , Hz.***************

4) The Mass of an uniform rod is negligible compared to the mass attached toit. For small oscillations, calculate the natural frequency of the system.

**

a

Lk

m

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Mechanical Vibrations (ME65)


Forced Vibrations

Forced vibrations are those whose amplitudes are maintained by application ofexternal forces.Ringing of electric bell or machine tool vibrations are examples of forcedvibrations.The external force maintaining the vibrations are called external excitation andare random, periodic or impulsive in nature.

Basic sources of excitation are external or inherent to the system. Machinesubsystems are heated unevenly during operation and give rise to uneven deformationleading to generation of unbalanced force,Resonance of system produces largeamplitudes leading to unbalanced forces.Similarly, defective assembly , bending anddistortion of components, bearing defects leading to misalignment, uneven distribution ofmass in rotating components lead to creation of unbalanced forces causing a system tovibrate forcibly.

Forced vibration of damped single degree of freedom system

kx is the spring force, cx is the damping force and mx is the inertia force andFsint is the external excitation . x is the displacement of mass in the direction shown.The equation of motion is written as

mx+cx+ kx = Fsint ----(i)

m

kx Cx mx

x

Fs

in

t

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The solution of above equation is in 2 parts, i) Complimentary function (cf) and ii)particular integral(pi). The total solution x = x(cf) + x(pi). The x(cf) is the solution ofequation mx+cx+ kx = 0, which is written as Ae-nt sin(d t + ). The particular integralx(pi) is assumed to be in the form x = X sin(t - ), thus we have dx/dt = x asXsin( t-+/2) and (dx/dt)2 = x = 2 Xsin( t-+), substituting the values of x amd

x in eqn. (i), we have,m(2 Xsin( t-+) + c(Xsin( t-+/2)) + k(X sin(t - )) = Fsint

Rearranging ,

Fsint - kX sin(t - ) - cXsin( t-+/2) - m2 Xsin( t-+) = 0

Where,Fsint is the external forceDiplacement x lags the external forcekX is the spring force lagging F by

cX is the damping force lagging F by (+/2)m2 X is the inertia force, lagging F by (+),The vector diagram of these is as shown below:

From the geometry of diagram, we have,F2 = (kx m2 X) 2 + (cX)2, simplifying

X = F/((k m2 ) 2 + (c)2)Therefore, the total solution can be written as,

x = x(cf) + x(pi)

= Ae-nt sin(d t + ) + F sin(t-) (/((k m2 ) 2 + (c)2) ----(ii)

x

kx

cX

m2 X

F

reference

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The eqn(ii) is total response which consists of two parts, first being the transientpart, the first term in RHS , which dies out with time and the second part the x(pi), is thesteady state vibration which does not die with time.The expressions for amplitude X in dimensionless form and phase angle are as follows:X = (F/k)/(1-( 2/n 2 )2 + (2/ n )2 , (F/k) is called the Zero frequency deflection which

is the deflection of spring mass under a steady force.

The phase angle, = tan-1 ((2/ n )/ 1-( 2/n 2))

Magnification Factor:

In a vibrating system the transient vibrations die out after passage of time and the

steady state vibration continues with constant amplitude as long as the externalexcitations exsist, and this makes the study of steady state vibrations to be important forstudy and analysis.Magnification factor M.F. is one parameter in study of forced vibrations which is definedas the ratio of amplitude of steady state response X to Xst the zero frequency deflection orthe static response under steady load F.

The M.F . is given by, M.F. = X/Xst = 1/((1-( 2/n 2 )2 + (2/ n )2 )

This M.F. depends upon the frequency ratio /n and the damping factor .From the plots of M.F. versus frequency ratio and phase angle, versus frequency ratioalso called frequency response curves following observations can be made: (Refer anystandard text for detailed curves)

i) Phase angle is 90 at resonanceii) M.F. is infinity at resonance and = 0iii) For all frequencies th MF reduces with damping

MF

/n

/n

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iv) Maximum amplitude occurs at left of resonancev) For small values of frequency ratio, the inertia and damping forces are smallresulting in small phase angles.Impressed force is nearly equal to spring force.

vi) For frequency ratio of 1, the inertia force is balanced by the spring force.Theimpressed force balances the damping force.

vii) For large values of frequency ratio, inertia force increases to a large value anddamping and spring forces are small.

viii) The frequency at which the maximum amplitude occurs is obtained by usingthe relation p = n (1 - 2 2 ), where p is the frequency at which maximumamplitude occurs.

ix) Condition for resonance , (MF)resonance = (Xr/Xst ) = (1/2)

Solution by complex algebra:

Let the equation of motion be written as

mx+cx+kx = Feit

the response of which is x = X ei(t ) . substituting the expressions for x and x into theequation of motion and simplifying , we have (-m2 +ic +k) X ei(t ) = Feit , fromwhich ,Xe -i =(F/(k - m2 ) + ic), from which using x = X ei(t ) , the real part of x is given by

Re(Feit /(k - m2 )+ic).Introducing the complexfrquency response H() as ratio of output Xei to input F

i.e, H() = Xe i / F = (1/(k - m2 )+ic)= X/Xst = 1/((k - m2 ) 2 +(c) 2 )

The phase angle, = tan -1 (c/(k m2 ))

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Session 9 date: (27/3/07)Rotating and Reciprocating unbalance

The figure shows a rotating equipment rotating at a speed of rad./sec. Let mobethe unbalance mass rotating with its CG at a distance of e from centre.This unbalancedmass gives rise to a centrifugal force , equal to mo 2 e .Let m be the total mass ofequipment inclusive of mo and at any instant of time mo make an angle of t.

The equation of motion for this system can be written considering the effectivemass m-mo and the unbalanced mass mo.

Referring figure as shown below, we have the effective displacement of mo is sumof x and esint. Hence we can write the equation of motion in the vertical direction as

mo2 e

ax

a = esinte = eccentricity

kc

m

x

mo

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(m-mo)x + (mo)d2

(x + esint) /dt2

= - Kx Cx

Ie, mx - mox +mox+ mod{e cost}/dt = -Kx Cx

mx - mo 2e sint = -Kx Cx

mx + Cx + Kx = mo2sint

The above equation is similar to

mx + Cx + Kx = Fsint

Hence for an under damped system, we get the expression for steady state amplitude as

X = mo2e/K222

n )/2())w/w(1( nwlyw+

Therefore __X__ = (w/wn)2(moe/m) 222n )/2())w/w(1( nwlyw+

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= tan-1

{2(w/wn) / (1 - (w/wn )2

)Same analysis is extended to reciprocating masses where exciting force becomesmoew2sinwt where mo = Unbalanced mass of reciprocating masses.

The complete solution for the unbalanced system is

x = A2e nt (sind t + 2 ) + (moew2 /k)/ ((1-( 2/n 2 )2 + (2/ n )2 )

The following points are concluded for unbalanced system:

Damping factor plays an important role in controlling the amplitudes duringresonance. For low values of frequency ratio, X tends to 0. For low values of frequency ratio (w/wn), X tends to 0. At high speeds of operation, damping effects are negligible. The peak amplitudes occur to right of resonance unlike for balanced systems. At resonance, w = wn ie: X / moe/m = 1/2

Also, (X ) resonance = moe/ 2m

From the plot of (X / emo m) v/s /n , it iserved that at low speeds, because the

inertia force is small, all the curves start from zero and at resonance (X / moe/m )= 1/2and the amplitude of such vibrations can be controlled by the damping provided in thesystem. For very large frequency ratio, (X / moe/m ) tends to one.

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Session 10 date : (28/3/07)

VIBRATION ISOLATION AND TRANSMISSION:

Vibration Isolation:

High speed machines and engines due to unbalance give rise to vibrations ofexcessive amplitudes and due to the unbalance forces being setup, the foundations can bedamaged. Hence there is a need to eliminate or reduce the vibrations being transmitted tothe foundations, springs, dampers, etc. are placed between the machines and thefoundations to reduce the vibrations or minimize then. These elements isolate the

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vibrations by absorbing the vibration energy. This isolation of vibrations is expressed interms of force or motion transmitted to the foundation. The requirements of theseisolating elements are that there should be no connection between the vibrating system &the foundation & it is to be ensured that in case of failure of isolators the system is still inof position on the foundation. Rubber acts effectively as an isolator during shear loading.The sound transmitted by it is also low. Heat and oil affect the rubber and it is usuallypreferred for light loads & high frequency oscillation. Felt pals are used for lowfrequency ratios. Many small sized felt pads are used instead of a single large pad. Corkcan be used for compressive loads.

Helical & leaf springs of metal are used as isolators for high frequency ratios.

They are not affected by air, water or oil. The sound transmitted by them can be reducedby covering them with pads of felt, rubber or cork.

TRANSMISSIBILITY:

In a spring mass dashpot system subjected to harmonically varying external force,the spring and dashpot become the vibration isolators and the spring force and damping

force are the forces between the mass and foundation. Thus the force transmitted to thefoundation (Ftr) is vector sum of the spring force (kX) and damping force (cX). Wecan write,Ftr= X (K2 + c22 ), substituting for X as X = F/((k m2 ) 2 + (c)2), we have Ftrequalto,Ftr = F ((K2 + c22 ) / ((k m2 ) 2 + (c)2)Transmissibility is defined as the ratio of force transmitted to the foundation to the forceimpressed on the system i.e.,

k c

m

x

Fsint

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Tr= = Ftr/ F = (1 + (c/k)2 / ((1-( 2/n 2 )2 + (2/ n )2 )

The angle of lag of the transmitted force is ,

( ) = tan 1 (( 2/ n) / 1-( 2/n 2)) - tan 1(2/ n )

Plot of Tr versus / n (refer a text book) for various values of , is called thetransmissibility curve . From the plot it is seen that all curves start from 1 andtransmissibility Tris always desired to be less than 1, as it ensures that transmitted forceto the foundation is minimum and better isolation is achieved. The operating values of

frequency ratio to achieve this effect should be greater than 2 and the region beyondthis value of frequency ratio is called mass control zone where isolation is most effective.In the plot the frequency ratio values upto 0.6 are spring control zone and from 0.6 to 2is damping control zone and beyond that is mass control zone.

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Session 11 date:(30/3/07)

FORCED VIBRATION DUE TO EXCITATION OF SUPPORT:

Frame

y

29

m

Z

KC

x

VIBRATING BODY/


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Figure shows a basic sesmic instrument used for measuring vibrations. When thesystem is excited by the vibrations of the base, the mass m is subjected to adisplacement x. If we consider y be the motion of the base, then the absoluteamplitude of mass m is the displacement x. If Z is considered as the displacement ofmass m w.r.t the frame, then we have a relative motion of m w.r.t the frame.

Absolute amplitude: (neglect z)

Let the displacement of base be y viz: a sinusoidal motion, given byy = Ysint

For such a system the equation for motion can be written as

x

mx

K(x-y) C(x-y)

mx + K(x-y) + C(x-y) = 0i.e, mx + Cx + Kx Ky - Cy = 0Substituting for y and y, we get

mx + Cx + Kx Kysint Cycost = 0

mx + Cx + Kx = y {Kysint + Ccost = 0}

= y(K

2

+ (c)

2

).sin(t + )

(1)Where = tan-1 {c/K} = tan-1 {2/n}

The solution of (1) consists of CF and PI.

The PI is x = X sin(t + - (a)

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(a) is similar to x = X sin (t ), where X is the steady state amplitude.X = y (K2 + (c)2 )

((K - (c2)2 + (c)2)

Therefore X/y = (1+ (2/n)2 ) / ((1 (/n)2)2 + (2/n)2 ) (b)

= (tan-1 (2/n2 / 1 (/n)2 ))

= (tan-1 {2/n})

( - ) = (tan-1

(2/n2

/ 1 (/n)2

)) - (tan-1

{2/n}) ---- Equations (a), (b) and (c) completely define the motion of the mass due to the support orbase excitation. The ratio X/y is called the displacement transmissibility

Relative Amplitude:If the displacement of the mass is considered relative to the frame and if this

relative displacement is called z , then we have,z = x y

or, x = y + zsubstituting this value of x in the equation of motion,

m(y + z)+c(y+z)+K(y+z-y) = 0my+mz+cz+kz = - mymz + cz + kz = -m(-2 y sint)

i.e., mz + cz + kz = m2 y sintsimilar to eqn b, we have

z/y = (/n)2 / ((1 (/n)2)2 + (2/n)2 )The expressions for is same as given above for absolute amplitude.

Energy dissipated by Damping.

When a syatem undergoes steady state forced vibration swith viscous damping, energy

gets absorbed by the dashpot . The energy dissipated or workdone per cycle is given by,Energy dissipated/ cycle = cx2 , where x is the amplitude of steady state vibrations.The power required for vibrating the system can be obtained by the relation

Power = Energy dissipated/ cycle / Sec. , Watts.

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Sharpness of Resonance:In forced vibration, quantity Q is related to damping which becomes a measure of thesharpness of resonance. It also gives the side band of frequencies 1 and 2 on either sideof the resonance by which resonance can be avoided during operation. The expression ofQ is given as follows:

Q = n / (2 - 1) = 1/2

Mech Vibrations Complied P.dinesh

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