Mechanical Engineers Handbook Episode 7 potx

The equation of motion for the linear displacement, x, is or The equation of motion for the angular displacement, y, is or Consider the single degree of freedom model with viscous dampin

for a free vibration with linear damping, and Fig 2.7b represents a model for

a free vibration with torsional damping

The equation of motion for the linear displacement, x, is

or

The equation of motion for the angular displacement, y, is

or

Consider the single degree of freedom model with viscous damping shown

in Fig 2.8 The differential equation of motion is

o ˆpk=m

is the undamped natural (angular) frequency, and

a ˆ c=2m

is the damping ratio

The characteristic equation in r for Eq (2.38) is

Case 1 a2ÿ o2< 0 ) the roots r1 and r2 are complex conjugate(r1; r22 C, where C is the set of complex numbers).

Case 2 a2ÿ o2> 0 ) the roots r1 and r2 are real and distinct(r1; r22 R; r16ˆ r2, where R is the set of real numbers)

Case 3 a2ÿ o2 ˆ 0 ) the roots r1and r2are real and identical, r1ˆ r2

In this case, the damping coef®cient is called the critical dampingcoef®cient, c ˆ ccr, and

a ˆ o )2mccr ˆ



km

r

The expression of the critical damping coef®cient is

ccr ˆ 2pkmˆ 2mo: …2:42†One can classify the vibrations with respect to critical damping coef®cient asfollows:

Case 1 c < ccr) complex conjugate roots (low damping and tory motion)

oscilla-Case 2 c > ccr) real and distinct roots (great damping and aperiodicmotion)

Case 3 c ˆ ccr) real and identical roots (great damping and aperiodicmotion)

Case 1: Complex Conjugate Roots a2ÿ o2 < 0The term

b ˆpo2ÿ a2

…2:43†

is the quasicircular frequency

The roots of Eq (2.39) are

The solution of the differential equation is

x ˆ eÿat…C1sin bt ‡ C2cos bt†; …2:45†

or

where C1, C2 (A and j) are constants

If we use the initial condition

the constants are

C2ˆ x0; C1 ˆv0‡ axb 0; …2:49†or

 2s

Equation (2.56) shows that the displacement measured at equal time intervals

of one quasiperiod decreases in geometric progression To characterize thisdecay, the logarithmic decrement d is introduced

Case 2: Real and Distinct Roots a2ÿ o2> 0The roots of the characteristic equation are negative

Case 3: Real Identical Roots a2ÿ o2 ˆ 0The roots of the characteristic equation are

The solution of the differential equation, namely the law of motion, in thiscase becomes

x ˆ eÿat…C1t ‡ C2† ˆC1t ‡ C2

When t ! 1, one obtains the undeterminate x ˆ 1=1 The l'Hospital rule

is applied in this case:

x ˆ limt!1

namely, the motion stops aperiodically

The diagrams of motion are similar to the previous case, with the sameboundary conditions Critical damping represents the limit of periodicmotion; hence, the displaced body is restored to equilibrium in the shortestpossible time, and without oscillation Many devices, particularly electricalinstruments, are critically damped to take advantage of this property

2.4 Forced Undamped Vibrations

2.4.1 RESPONSE OF AN UNDAMPED SYSTEM TO A SIMPLE HARMONIC EXCITING FORCE WITH CONSTANT AMPLITUDE

Common sources of harmonic excitation imbalance in rotating machines, themotion of the machine itself, or forces produced by reciprocating machines.These excitations may be undesirable for equipment whose operation may

be disturbed or for the safety of the structure if large vibration amplitudesdevelop Resonance is to be avoided in most cases, and to prevent largeamplitudes from developing, dampers and absorbers are often used.General Case

An elastic system is excited by a harmonic force of the form

where F0is the amplitude of the forced vibration and p is the forced angularfrequencies The differential equation of motion for the mechanical model inFig 2.12 is

The following notation is used:

The constant C is determined using Eqs (2.67), (2.69), and (2.70) for o 6ˆ p:

ÿCp2sin pt ‡ o2C sin pt ˆ q sin pt; …2:71†

or

C oÿ 2ÿ p2

ˆ q ) C ˆ…o2ÿ pq 2†: …2:72†

Introducing in Eq (2.68) the obtained value of C, one can get

x ˆ C1sin ot ‡ C2cos ot ‡o2qÿ p2sin pt; …2:73†

which may be written as

Equation (2.77) is a combined motion of two vibrations: one with the naturalfrequency o and one with the forced angular frequency p The resultant

is a nonharmonic vibration (Fig 2.13), here for o ˆ 1 rad=s, q ˆ 1 N=kg,

p ˆ 0:1 rad=s The amplitude is

A ˆo2qÿ p2ˆ

F0m

o2ÿ p2ˆ

F0m

mk

1 ÿ op 2

ˆF0k

where xst ˆ F0=k is the static deformation of the elastic system, under themaximum value F0, and

jA0j ˆ 1

1 ÿ op 2 …2:79†

is a magni®cation factor In Fig 2.14 is shown the magni®cation factorfunction of p=o

From Fig 2.14 one can notice:

j Point a ( p ˆ 0, A0ˆ 1) and x ˆ xst The system vibrates in phase withforce

j Point b ( p ! 1, A0! 0), which corresponds to great values ofangular frequency p The in¯uence of forced force is practically null

j Point c ( p ˆ o, A0! 1) This phenomenon called resonance and isvery important in engineering applications

The curve in Fig 2.14 is called a curve of resonance

ResonanceWhen the frequency of perturbation p is equal to the natural angularfrequency o, the resonance phenomenon appears The resonance is char-acteristic through increasing amplitude to in®nity In Eq (2.77) for o ˆ p, thelimit case is obtained, limp!ox ˆ 1
0 Using l'Hospital's rule one cancalculate the limit:

limp!ox ˆ q lim

Figure 2.14

Curve of

resonance

factor e ˆ o ÿ p, and in this case p=o  1 and p ‡ o  2o The vibrationbecomes

x ˆo2qÿ p2‰sin pt ÿ sin otŠ

ˆ…o ‡ p†…o ÿ p†q 2 sin…p ÿ o†t2 cos…p ‡ o†t2

ˆ2oe2q sin…ÿet†2 cos2ot2 ˆ ÿoeq sin 2et cos ot: …2:81†The amplitude is in this case is

f…t† ˆ ÿoeq sin 2et : …2:82†The vibration diagram is shown in Fig 2.16 for q ˆ 2 N=kg, e ˆ 0:12 rad=sand o ˆ 0:8 rad=s

2.4.2 RESPONSE OF AN UNDAMPED SYSTEM TO A CENTRIFUGAL EXCITING FORCE

Unbalance in rotating machines is a common source of vibration excitation.Frequently, the excited (perturbation) harmonic force came from an

unbalanced mass that is in a rotating motion that generates a centrifugalforce For this case the model is depicted in Fig 2.17 The unbalanced mass

m0is connected to the mass m1with a massless crank of lengths r The crankand the mass m0rotate with a constant angular frequency p The centrifugalforce is

 2

1 ÿ op 2

ˆ po

2.4.3 RESPONSE OF AN UNDAMPED SYSTEM TO A NONHARMONIC PERIODIC EXCITING FORCE

One may consider the nonharmonic periodic exciting force F (t) Thefunction F (t) obeys the Dirichlet conditions The exciting force F (t) can bedeveloped in Fourier series:

Fpˆ F …t† ˆ a0‡ a1cos pt ‡ a2cos 2pt ‡


‡ b1sin pt ‡ b2sin 2pt ‡

…2:86†For n terms,

…2:89†or

x ˆ C1sin ot ‡ C2cos ot ‡Pn

iˆ0

Fim‰o2ÿ …ip†2Šsin…ipt ‡ ji†: …2:90†

The resonance appears for the ®rst harmonic (the fundamental harmonic)and for superior harmonics (o ˆ p, o ˆ 2p, o ˆ 3p, , o ˆ np)

2.4.4 RESPONSE OF AN UNDAMPED SYSTEM TO AN ARBITRARY EXCITING FORCE

For this general case the exciting force is an arbitrary force Fig (2.19) Thedifferential equation of motion is

The vibration in this case is

x ˆ C1sin ot ‡ C2cos ot ‡mo1

…t

0F …t† sin‰o…t ÿ t†Šdt; …2:92†

where t is presented in Fig 2.19 and C1, C2are constants The integral in Eq

(2.92) is called the Duhamel integral

2.5 Forced Damped Vibrations

2.5.1 RESPONSE OF A DAMPED SYSTEM TO A SIMPLE HARMONIC EXCITING FORCE WITH CONSTANT AMPLITUDE

The mechanical model is shown in Fig 2.20 The differential equation ofmotion is

m x ‡ c _x ‡ kx ˆ F0sin pt: …2:93†

The following notation is used:

c2mˆ 2a;

where x1represent the solution of the differential homogenous equation, and

x2 is a particular solution of the differential nonhomogeneous equation Inthis case x1represents the natural vibration of the system and x2is the forcedvibration of the system The solution of the free damped system is

x1ˆ eÿat…C1sin ot ‡ C2cos ot†; …2:100†or

x1ˆ B1eÿatsin…ot ‡ j†; …2:011†where

_x2ˆ D1p cos pt ÿ D2p sin pt

x2ˆ ÿD1p2sin pt ÿ D2p2cos pt:



…2:104†Introducing Eq (2.103) and Eq (2.104) in Eq (2.95) one can obtain

ÿD1p2sin pt ÿ D2p2cos pt

‡ 2a‰D1p cos pt ÿ D2p sin ptŠ

‡ o2D1sin pt ‡ D2cos ptˆ q sin pt: …2:105†Using the identi®cation method, the linear system is obtained

D1…o2ÿ p2† ÿ 2D2ap ˆ q2D1ap ‡ D2…o2ÿ p2† ˆ 0:



…2:106†Solving this system, one ®nds

Therefore the forced vibration x2 has the expression

In conclusion, the motion of the system is represented by the equation

x ˆ B1eÿatsin…ot ‡ j† ‡ B2sin…pt ÿ f†: …2:111†

The graphic of the vibration is shown in Fig 2.21 for x0ˆ 0 m, v0ˆ 0:2 m=s,

o ˆ 5 rad=s, q ˆ 1 N=kg, p ˆ 0:3 rad=s, a ˆ 0:1 sÿ1.According to Eq (2.110) the amplitude of the vibration is

B2 ˆ D2

1 ‡ D2 2

q

where xst represents the static deformation

The amplitude of forced vibration can be written as

where

a

oˆ

c2m

km

In Fig 2.22 the magni®cation factor A1 is plotted versus the ratio p=o.Different curves for different values of the ratio c=ccr are obtained In thecase of c=ccr ˆ 0 (system without damping),

The maximum of A1 is obtained from the relation

 2s

 2ˆ

2cccr

po

1 ÿ po

 2; …2:119†

and

f ˆ arctan

2cccr

po

1 ÿ po

The variation of delay versus frequency ratio is plotted in Fig 2.23 From Fig

2.23, for p=o < 1 is obtained f ˆ 0, namely, the vibration is in phase withthe perturbation force For p=o > 1 is obtained f ˆ p, that is, the vibration

and perturbation forces are in opposition of phase For the case c=ccr 6ˆ 0(systems with damping), for p=o < 1 the phase angle between the vibrationand perturbation forces is 0 < f < p=2, and for p=o > 1 the phase angle isp=2 < f < p For p ˆ o and for any parameter c=ccr, the phase angle is thesame.

Using d'Alambert's principle for a general system with one DOF, weobtain the differential equation

ÿm x ÿ c _x ÿ kx ‡ F0sin pt ˆ 0; …2:121†where m x is the inertia force, c x is the damping force, kx is the spring force,and F0sin pt is the perturbation force Equation (2.121) is another form of Eq.(2.93)

To represent Eq (2.121) with rotating vectors (Fig 2.24), the equation ofmotion is considered as

In Fig 2.24 the vector A represents the amplitude of the excited vibration.The angular frequency of all rotative vectors is p A difference of phase fbetween the vector A and perturbation force is noticed The four vectorsillustrated in Fig 2.24 are in equilibrium and the projection on the vector Aand its normal directions yields

Figure 2.24

Dampedvibration

is rotating with angular velocity p ˆ const (Fig 2.25) The mass m0producesthe perturbation centrifugal force

 2p2

o2s

2.5.3 RESPONSE OF A DAMPED SYSTEM TO AN ARBITRARY EXCITING FORCE

In a general case, the perturbation force is a arbitrary function of time F (t).The differential equation of motion is (Fig 2.27)

The general solution of the homogenous differential equation x1ing to Eq (2.132) is calculated with Eqs (2.45) or (2.46), and it is the naturalvibration of the system The particular solution x2 of Eq (2.132) is deter-mined with the help of the conservation of momentum theorem Thevariation of perturbation force on mass unit q1 is shown in Fig 2.28 For

an arbitrary time one can consider an elementary impulse q1dt, as sented in Fig 2.28 For the direction of motion one can write

displacement dx The effect of continuous action of perturbation force intime interval …0; t† is obtained through integration:

0F …t†eÿa…tÿt†sin b…t ÿ t†dt: …2:141†

In Eq (2.141) the ®rst term represents the effect of the initial displacement x0and of the initial speed v0, and the second term is the effect of theperturbation force F (t ) If the damping is neglected, a ˆ 0 and b ˆ o, theequation of motion is

x ˆvo0sin ot ‡ x0cos ot ‡mo1

…t

0F …t† sin o…t ÿ t†dt: …2:142†

If we use the identity

sin…ot ÿ ot† ˆ sin ot cos ot ÿ cos ot sin ot; …2:143†

and Eq (2.142) can be rewritten as

x ˆvo0sin ot ‡ x0cos ot ‡ A sin ot ‡ B cos ot: …2:146†If

one can obtain the results from section 2.4.1

Equation (2.142) can be used in case of a mass system acted on by aseries of discontinuous impulses, which produce jumps in speed Dv0, Dv1,

Dv2; , at time moments t ˆ 0, t ˆ t0, t ˆ t00 For x0 ˆ 0 one can obtain theequation of motion on the (0, t) interval:

x ˆo1‰Dv0sin ot ‡ Dv1sin o…t ÿ t0† ‡ Dv2sin o…t ÿ t00† ‡


Š: …2:148†

2.6 Mechanical ImpedanceFor the mechanical system shown in Fig 2.29a, the spring force is

where x is the displacement and k is the spring constant The case of a forcedvibration with viscous damping is considered in Fig 2.29b The differentialequation of motion is

where Z is mechanical impedance

In this way, the study of the forced vibration is reduced to a staticproblem, and

z ˆeZiptF0 ˆÿmp2e‡ k ‡ icpipt F0: …2:156†

Equation (2.158) is the same as Eq (2.112).

In the case of n mechanical impedances Z1, Z2; , Zn in parallel, onecan write

Z ˆ Z1‡ Z2‡


‡ Zn; …2:159†or

1Zi

Machine Directly on a Foundation

In this case (Fig 2.30), the perturbation force is transmitted to the foundation.The transmissibility coef®cient is t ˆ 1 and the machine is not isolated

Machine on a Foundation with Elastic ElementsThe machine is connected to a foundation through elastic elements with theequivalent elastic constant k (Fig 2.31) The force is transmitted to thefoundation through elastic elements; therefore, the transmitted force is theelastic force kx In Fig 2.32 is depicted the variation of magni®cation factorversus frequency ratio.

In the case of undamped forced vibration, the particular solution is

xpˆ A sin pt ˆo2qÿ p2sin pt; …2:164†

where

A ˆo2ÿ pq 2ˆ

F0m

o2ÿ p2ˆ

F0m

mk

The maximum transmitted force is kA, and the transmissibility coef®cient is

ˆ A

xst ˆ A0: …2:168†

The diagram in Fig 2.32 represents the variation of the transmissibilitycoef®cient The case of a machine directly on a foundation (rigid joint)gives a particular case, o ! 1, point a on the diagram in Fig 2.32 Also, if

p <p2o one can obtain jtj > 1, i.e., the force transmitted to the foundation

is greater than the perturbation force For good operation it is necessary thatjtj > 1 As a result in calculus one will take the negative values; thus,

1 ÿ po

where (p=o†2> 2

In conclusion, in the case of a machine on a foundation with elasticelements, it is recommended that p=o >p2 The dangerous situation iswhen p=o ˆ 1

Machine on a Foundation with an Elastic Element and a DamperThe machine is settled on a foundation with the help of an elastic elementwith the elastic constant k, and a damper with the viscous dampingcoef®cient c (Fig 2.33) The transmitted force is not in the same phase with

the perturbation force In this case, the transmitted force is



1 ‡ 4 cccr

 2

p2

o2

vuu

From Eq (2.175) one can observe that the transmissibility coef®cient t doesnot depend on the amplitude of perturbation force Equation (2.175) isplotted in Fig 2.34, which shows the variation of transmissibility coef®cient

as a function of the ratio c=ccr If t ˆ 1, the perturbation force is transmittedintegral to the foundation For the case a=o ˆ 0 one may obtain

jtj ˆ 1 ˆ ... phenomenon appears The resonance is char-acteristic through increasing amplitude to in®nity In Eq (2 .77 ) for o ˆ p, thelimit case is obtained, limp!ox ˆ
Using l''Hospital''s rule one cancalculate... that is, the vibration