Mech Vibrations notes

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 book House, Bangalore.

.

Introduction:

Vibration is defined as a motion which repeats after equal interval of time and is also a periodic motion.The swinging of a pendulum is a simple example of vibration.Vibration occurs in all bodies which are having mass and elasticity. They are caused due to several reasons such as presence of unbalanced force in rotating machines, elastic nature of the system, external application of force or wind loads and earthquakes. Vibrations are undesirable as they induce high stresses in system components leading to noise and failure, in such cases they are to be minimized if not totally eliminated.. The desirable effects are seen in musical instruments and cement 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., the

oscillations of simple pendulum

Time Period (T) :

1

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

Frequency (f or ω) : 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 initial external disturbance and allowed to vibrate without external force being applied subsequently.

Fundamental Mode of Vibration: A vibrating body may have more than one natural frequency and when it vibrates

with 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 spring attached with one mass , 2 DOF system will have two masses and likewise we have 3DOFsystem. A continuous system like a beam or plate consisting of infinite number of particles 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 with uniform angular velocity on the diameter of circle.Damping:

It is the resistance offered to the motion of a vibrating body by absorbing the energy 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 periodic in 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 externally applied force is same as that of natural frequency of the body.

2

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 linear vibrations

are experienced when large amplitudes are encountered and laws of superposition are not applicable.

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 system axis 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 a spring mass system represents Longitudinal vibrations. Vibration of a cantilever beam represents Transverse vibrations. The twisting and untwisting of a disc attached at the end of a shaft represents Torsional vibrations.

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

the displacement , A is the amplitude , ω 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 with same frequencies then the phase difference between them is given by φ.

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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 = A1Sinωt and x2 = A2 Sin(ωt + φ) . Then if x is the resultant displacement , x = x1 + x2. The resultant amplitude x = A Sin (ωt +θ), where A is the resultant amplitude and is acting at an angle θ w.r.t vector x1. The above addition of SHMs can also be done graphically.

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Sample Problems:

1) Add the following harmonics analytically and check the solution graphically x1= 3Sin (ωt + 30◦) , x2 = 4 Cos(ωt +10◦)

3

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

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

Make x1 and x2 to have same Sin terms always , i.e., x2 = 4 cos(ωt +10◦ + 90◦) = 4Sin (ωt +100◦ )Hence, A Sin (ωt+ θ)= 3 Sin (ωt + 30◦) + 4 Sin (ωt +100◦ )Expanding LHS and RHSASin ωt Cos θ + A cos ωt Sin θ =3 Sin ωt cos 30◦ + 3Cos ωtSin30◦ + 4 Sin ωt cos 100◦

+ 4 Sin ωt sin 100◦

A Sin ωt Cos θ + A cosωt Sin θ = Sin ωt(1.094) + Cos ωt (5.44)Comparing the coefficients of A Cos θ and A Sin θ in the above equationACos θ = 1.094, ASin θ = 5.44 , tan θ = ASin θ/ ACos θ = 5.44/1.904Therefore , θ = 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 at an angle of 30◦ to ox and ob equal to 4 units at an angle of 100◦ to ox. Complete the vector polygon by drawing lines parallel to oa and ob to intersect at point c. Measure oc which should be equal to A and the angle oc makes with ox will be equal to θ. All angles measured in anticlockwise direction.

O

xxx

a

b

c

4

2) Repeat the above problem given , x1 = 2Cos(ωt + 0.5) and x2 = 5Sin(ωt + 1.0) . The angles are in radians.(Hint: In the above problem the angles are to be converted to degrees. Ans. A = 6.195, θ = 73.49 ◦ )

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

4) A body is subjected to 2 harmonic motions x1 = 15sin(ωt + π/6) , x2 = 8 cos(ωt + π/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 be Asin (ωt + φ)Therefore, Asin (ωt + φ)+ x1 + x2 = 0Hence, Asinωt cos φ+A cosωt sin φ+15sinωt cosπ/6+15cosωt sinπ/6+8 cosωt cosπ/6+8 sinωt sinπ/6 = 0 sinωt (A cos φ+8.99038)+ cosωt(A sin φ+14.4282) = 0Therefore, A cos φ = -8.99038 A sin φ= - 14.4282Therefore, tan φ = A sin φ / A cos φ = 14.4282/8.99038 , φ = 58.062˚From , A cos φ = -8.99038, substituting for φ = 58.062˚, A = 17.00Therefore, the motion is x= 17sin(ωt + 58.062˚)

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

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

5

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Fourier Theorem:

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

F(t) a periodic function can be represented asF(t) = a0 + a1cosωt + a2cos2ωt + a3cos3ωt +…….an cosnωt + b1sinωt + b2sin2ωt + b3sin3ωt +…….. bn sinnωt

Resultant amplitude variation during one cycle w

time

Amplitude

6

The constants a0,a1,a2,,,,,and b1,b2,b3…. etc., are obtained using the following formulae: a0 = (ω/2π) ∫ F(t), in the limits 0 to 2π/ω

an=(ω/π) ∫ F(t)cos(nωt) dt, in the limits 0 to 2π/ω bn= =(ω/π) ∫ F(t)sin(nωt) dt, in the limits 0 to 2π/ω

t.2

2cm

2cm

A

B

4) Represent the above periodic motion using harmonic seriesSol: Mathematically for one complete cycle we have the eqn for AB as x(t) = -20t +2 for 0<t<0.2T=0.2, ω= (2π/T) = 10π 0.2 a0=(ω/2π) ∫ x(t)dt 0

0.2 a0= (10π/2π) ∫ (-20t +2 ) dt = 0 0

0.2 an = (ω/π) ∫ x(t)cos(nωt)dt 0 = 0 0.2 bn = (ω/π) ∫ x(t)sin(nωt)dt 0

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= (4/πn)Thus, the harmonic series is , x(t) = 4/π∑ (1/n) sin10πnt , for n =1,2,…..

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

Further problems refer the texts given in the additional references.

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10

0.3 0.2

Sec.

X,cms

8

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 in which the equation of motion is obtained are :

a) Newton’s Method b) Energy Method and c) Rayleigh Method

Newton’s Method When a mass m is attached to a spring it deflects by δ and the system is under

equilibrium as mg = weight = kδ, where k is the spring stiffness, defined as force per unit length. If now the mass m is given a displacement x in the downward direction and the system is allowed to vibrate, we have the following forces acting on the system: the spring force, k(x+δ) acting in the upward direction, inertia force mx”acting 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+δ) + mg = -kx-kδ+mg = -kx-kδ+kδ

m

k

Displ. x

m

mg

k(x+δ)

mx”

9

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

x”+ (k/m) x = 0, we have the quantity (k/m)1/2 as the natural frequency of the system ωn .

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 remains constant 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” and the potential energy stored in the system is (1/2)kx2 . As the total energy of the system remains constant, we have T+V = 0 or d(T+V)/dt = 0. Differentiating we have the governing differential equation as mx” + kx = 0, and the natural frequency is given by ωn = (k/m)1/2.

Rayleigh’s Method: In this method the max kinetic energy of the system is equated to the

maximum potential energy. For SHM the max. kinetic energy is at the mean position which is equated to the potential energy. If A is the amplitude of vibration and ωn is the natural frequency the max. kinetic energy is given by (1/2)m(ωn A)2 and max. potential energy is (1/2)kA2 . Equating the two equations and simplifying we have again ωn = (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 force ‘F’, the displacement of mass ‘m’ is equal to deflections of sprimgs 1 & 2. Hence we can write, the displacement of the equivalent spring as,

б = б1 + б2 Where б1 – deflection of spring 1.and б2 – 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 a force 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 + K2

Therefore 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 ωn. 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.K3

K1K2 + 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

ωn = √(K e / m) , Therefore ωn =

Therefore ωn =

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

13

Ke1 = K1 + K2 = 20 N/mm

Ke = Ke1 + K3Ke1 + K3

= 20 (5)

25= 4 N/mm

= 4000 N/mωn = √ (Ke / m) Therefore ωn = 18.97 rad/s

Frequency = fn = ωn /2π= 3.012 Hz

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NATURAL FREQUENCY OF A SPRING : Considering mass of spring

Consider a spring mass system as shown in the figure where the mass is displaced by ‘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.E considering the mass of the spring.

The velocity of the spring element at a distance of ‘y’ from the fixed end is x’y/LWe can write the K.E of the spring element ‘dy’ as (½ )(Rho)(dy)(x’y/L)2

Where Rho is the mass densityAbove expression is of the form KE = ½ mv2.The K.E for the entire spring considering the mass of the spring becomesL∫(½ )(Rho)(dy)(x’y/L)2

0 L

= (Rho) x’2 / 2l2 ∫ (y3/3) 0 = 1/6 (Rho)x’2L

= 1/6 Ms.x’2 where, Ms = (Rho)L = Mass spring

Therefore the entire K.E of the system

k

L

y

dy

x m

15

K.E = 1/2mx’2 + 1/6 Ms.x’2

The potential energy of the system P.E = ½ Kx2

Therefore the total energy of the system

= K.E + P.E = constant

1/2mx’2 + 1/6 Ms.x’2 + ½ Kx2 = constantDifferentiating the above expression w.r.t time we get

Mx’x” + 1/3 Msx’x”+ Kxx’ = 0

Mx” + 1/3 Msx”+ Kx = 0

x” {m + 1/3 Ms} + Kx = 0

From the above expression,Therefore , ωn = (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 ‘θ’, due to which the mass ‘m’ is displaced by ‘x’, from the figure, we have;

x = 2r θ.

Also, the vertical displacement of centre ‘O’ is

X1 = r θ

m

x

r

k

M

θO

x1

17

By making use of the energy principle, we have the total energy of the system is constant at any given instant of time.K.E. System = 1/2mx’2 +1/2Mθ’2 +1/2Iθ”2

= 1/2 m(2rθ’)2+1/2M(rθ’)2+1/2(1/2Mr2)(θ’)2

Simplifying = r2 (θ’)2(2m+(3/4)M)

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

According to Energy Method, d (KE+PE) = 0 dt Differentiating the sum of KE and PE

(2m+(3/4)M)2θ”+kθ = 0 Or θ”+ k/(4m+(3/2)m) = 0 , which is in the form θ”+ ωn

2 = 0 orωn

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

ωn = k/(4m+(3/2)m)1/2

and the natural frequency inHz, ,

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

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

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

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

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

*************3) A circular cylinder of mass m and mass moment of inertia I is connected by a

spring of stiffness k as shown. If it is free to roll without slipping, determine the natural frequency.

r

k

m

L

m

19

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

4) The Mass of an uniform rod is negligible compared to the mass attached to it. 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 of external forces.Ringing of electric bell or machine tool vibrations are examples of forced vibrations.The external force maintaining the vibrations are called external excitation and are random, periodic or impulsive in nature.

Basic sources of excitation are external or inherent to the system. Machine subsystems are heated unevenly during operation and give rise to uneven deformation leading to generation of unbalanced force,Resonance of system produces large amplitudes leading to unbalanced forces.Similarly, defective assembly , bending and distortion of components, bearing defects leading to misalignment, uneven distribution of mass in rotating components lead to creation of unbalanced forces causing a system to vibrate 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 and

Fsinωt is the external excitation . x is the displacement of mass in the direction shown. The equation of motion is written as

mx”+cx’+ kx = Fsinωt ----(i)

m

kx Cx’ mx”

x

Fsi

nωt

21

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 of equation mx”+cx’+ kx = 0, which is written as Ae-ξωnt sin(ωd t + φ). The particular integral x(pi) is assumed to be in the form x = X sin(ωt - φ), thus we have dx/dt = x’ as ωXsin( ωt-φ+π/2) and (dx/dt)2 = x” = ω2 Xsin( ωt-φ+π), substituting the values of x’ amd x” in eqn. (i), we have,

m(ω2 Xsin( ωt-φ+π) + c(ωXsin( ωt-φ+π/2)) + k(X sin(ωt - φ)) = Fsinωt Rearranging ,

Fsinωt - kX sin(ωt - φ) - cωXsin( ωt-φ+π/2) - mω2 Xsin( ωt-φ+π) = 0

Where,Fsinωt is the external forceDiplacement x lags the external forcekX is the spring force lagging F by φcωX is the damping force lagging F by (φ+π/2)mω2 X is the inertia force, lagging F by (φ+π), The vector diagram of these is as shown below:

From the geometry of diagram, we have, F2 = (kx – mω2 X) 2 + (cωX)2, simplifying X = F/((k– mω2 ) 2 + (cω)2)Therefore, the total solution can be written as, x = x(cf) + x(pi)

= Ae-ξωnt sin(ωd t + φ) + F sin(ωt-φ) (/((k– mω2 ) 2 + (cω)2) ----(ii)

x

kx

cωX

mω2 X

F

reference

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The eqn(ii) is total response which consists of two parts, first being the transient part, the first term in RHS , which dies out with time and the second part the x(pi), is the steady 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-(ω 2/ωn

2 )2 + (2ξω/ ωn )2 , (F/k) is called the Zero frequency deflection which is the deflection of spring mass under a steady force.

The phase angle, φ = tan-1 ((2ξω/ ωn )/ 1-(ω 2/ωn 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 external excitations exsist, and this makes the study of steady state vibrations to be important for study and analysis.Magnification factor M.F. is one parameter in study of forced vibrations which is defined as the ratio of amplitude of steady state response X to Xst the zero frequency deflection or the static response under steady load F.

The M.F . is given by, M.F. = X/Xst = 1/(√(1-(ω 2/ωn 2 )2 + (2ξω/ ωn )2 )

This M.F. depends upon the frequency ratio ω/ωn and the damping factor ξ . From the plots of M.F. versus frequency ratio and phase angle,φ versus frequency ratio also called frequency response curves following observations can be made: (Refer any standard text for detailed curves)

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

MF

ω/ωn ω/ωn

φ

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iv) Maximum amplitude occurs at left of resonancev) For small values of frequency ratio, the inertia and damping forces are small

resulting 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.The

impressed force balances the damping force.vii) For large values of frequency ratio, inertia force increases to a large value and

damping and spring forces are small.viii) The frequency at which the maximum amplitude occurs is obtained by using

the relation ωp = ωn √(1 - 2ξ 2 ), where ωp is the frequency at which maximum amplitude occurs.

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

Solution by complex algebra:

Let the equation of motion be written as

mx”+cx’+kx = Feiωt

the response of which is x = X ei(ωt – φ) . substituting the expressions for x’ and x” into the equation of motion and simplifying , we have (-mω2 +icω +k) X ei(ωt – φ) = Feiωt , from which , Xe -iφ =(F/(k - mω2 ) + icω), from which using x = X ei(ωt – φ) , the real part of x is given by Re(Feiωt /(k - mω2 ) +icω).Introducing the complexfrquency response H(ω) as ratio of output Xe –iφ to input F i.e, H(ω) = Xe –iφ / F = (1/(k - mω2 ) +icω) = X/Xst = 1/√((k - mω2 ) 2 +(cω) 2 )The phase angle, φ = tan -1 (cω/(k – mω2 ))

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Rotating and Reciprocating unbalance

The figure shows a rotating equipment rotating at a speed of ω rad./sec. Let mo be the unbalance mass rotating with its CG at a distance of e from centre.This unbalanced mass gives rise to a centrifugal force , equal to mo ω2 e .Let m be the total mass of equipment inclusive of mo and at any instant of time mo make an angle of ωt. The equation of motion for this system can be written considering the effective mass ‘m-mo’ and the unbalanced mass ‘mo’.

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

mo ω2 e

a

x

a = esinωte = eccentricity

kc

m

x

mo

25

(m-mo)x” + (mo)d2(x + esinωt) /dt2 = - Kx – Cx’

Ie, mx” - mox” +mox”+ mod{ωe cosωt}/dt = -Kx –Cx’

mx” - mo ω2e sinωt = -Kx –Cx’

mx” + Cx’ + Kx = moω2sinωt

The above equation is similar to

mx” + Cx’ + Kx = Fsinωt

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

X = moω 2 e/K

Therefore __X__ = (w/wn) 2 (moe/m)

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Φ = tan-1 {2ξ(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 –ξωnt (sinωd t + Φ2 ) + (moew 2 /k)/ (√(1-(ω 2/ωn 2 )2 + (2ξω/ ωn )2 )

The following points are concluded for unbalanced system:

Damping factor plays an important role in controlling the amplitudes during

resonance. 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 = 1/2ξ

Also, (X ) resonance = moe/ 2mξ

From the plot of (X / emo m) v/s ω/ωn , it iserved that at low speeds, because the inertia force is small, all the curves start from zero and at resonance (X / moe/m )= 1/2ξ and the amplitude of such vibrations can be controlled by the damping provided in the system. 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 of excessive amplitudes and due to the unbalance forces being setup, the foundations can be damaged. Hence there is a need to eliminate or reduce the vibrations being transmitted to the foundations, springs, dampers, etc. are placed between the machines and the foundations 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 in terms of force or motion transmitted to the foundation. The requirements of these isolating 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 in of 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 usually preferred for light loads & high frequency oscillation. Felt pals are used for low frequency ratios. Many small sized felt pads are used instead of a single large pad. Cork can 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 reduced by 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 the foundation (Ftr ) is vector sum of the spring force (kX) and damping force (cωX). We can write,Ftr = X √(K2 + c2ω2 ), substituting for X as X = F/((k– mω2 ) 2 + (cω)2), we have Ftr equal to,Ftr = F (√(K2 + c2ω2 ) / ((k– mω2 ) 2 + (cω)2)Transmissibility is defined as the ratio of force transmitted to the foundation to the force impressed on the system i.e.,

kc

m

x

Fsinωt

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Tr = ε = Ftr / F = √(1 + (cω/k)2 / (√(1-(ω 2/ωn 2 )2 + (2ξω/ ωn )2 )

The angle of lag of the transmitted force is ,

(φ – α) = tan – 1 (( 2ξω/ ωn) / 1-(ω 2/ωn 2)) - tan – 1(2ξω/ ωn )

Plot of Tr versus ω/ ωn (refer a text book) for various values of ξ , is called the transmissibility curve . From the plot it is seen that all curves start from 1 and transmissibility Tr is always desired to be less than 1, as it ensures that transmitted force to 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 beyond this 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 √2 is damping control zone and beyond that is mass control zone.

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FORCED VIBRATION DUE TO EXCITATION OF SUPPORT:

Frame

y

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m

Z

KC

x

VIBRATING BODY/ BASE

Figure shows a basic sesmic instrument used for measuring vibrations. When the system is excited by the vibrations of the base, the mass ‘m’ is subjected to a displacement ‘x’. If we consider ‘y’ be the motion of the base, then the absolute amplitude of mass ‘m’ is the displacement ‘x’. If ‘Z’ is considered as the displacement of mass ‘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 = Ysinωt

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 – Kysinωt – Cωycosωt = 0

mx” + Cx’ + Kx = y {Kysinωt + Cωcosωt = 0}

= y(√K2 + (cω)2 ).sin(ωt + α) (1)

Where α = tan-1 {cω/K} = tan-1 {2ξω/ωn}

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

The PI is x = X sin(ωt + α - φ (a)

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(a) is similar to x = X sin (ωt – φ), where X is the steady state amplitude.

X = y (√K 2 + (cω) 2 ) (√(K - (cω2)2 + (cω)2)

Therefore X/y = (√1+ (2ξω/ωn)2 ) / (√(1 – (ω/ωn)2)2 + (2ξω/ωn)2 ) (b)

Φ = (tan-1 (2ξω/ωn2 / 1 – (ω/ωn)2 ))

α = (tan-1 {2ξω/ωn})

(Φ - α ) = (tan-1 (2ξω/ωn2 / 1 – (ω/ωn)2 )) - (tan-1 {2ξω/ωn}) ---- ©

Equations (a), (b) and (c) completely define the motion of the mass due to the support or base 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) = 0 my”+mz”+cz’+kz = - my” mz” + cz’ + kz = -m(-ω2 y sinωt) i.e., mz” + cz’ + kz = mω2 y sinωt similar to eqn b, we havez/y = (ω/ωn)2 / (√(1 – (ω/ωn)2)2 + (2ξω/ωn)2 )The expressions for Φ – α 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 = πcωx2 , 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 the sharpness of resonance. It also gives the side band of frequencies ω1 and ω2 on either side of the resonance by which resonance can be avoided during operation. The expression of Q is given as follows: Q = ωn / (ω2 - ω1) = 1/2ξ

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