theory of vibrations - saran - soil dynamics and machine foundation

8/4/2019 Theory of Vibrations - Saran - Soil Dynamics and Machine Foundation

http://slidepdf.com/reader/full/theory-of-vibrations-saran-soil-dynamics-and-machine-foundation 1/52



14 s.u /JyruuIfics& Machine Foundations

The forms of vibration mainly depend on the mass, stiffness distribution and end conditions of the

system.

To study the response of a vibratory system, in many cases it is satisfactory to reduce it to an idealized

system of lumped parameters. In this regard, the simplest model consists of mass, spring and dashpot

This chapter is framed to provide the basic concepts and dynamic analysis of such systems. Actual field problems which can be idealized to mass-spring-dashpot systems, have also been included.

2.2 DEFINITIONS

2.2.1 Vibrations: If the motion of the body is oscillatory in character, it is called vibration.. -, -

2.2.2 Degrees of Freedom: The number of independent co-ordinates which are required to define the

position of a system during vibration, is called degrees of freedom (Fig. 2.2).

~D:

m

(a) One degree of freedom (b) Two degrees offreedom

Z2

. .~

KI

Z,

.- -,

~ P@;J-PBBPZ)

- J.., .

(c) Three degrees of freedom' . (d) Six degrees 'offreedon~ (e) Infinite degrees offreedom-' , , . .: .',n ,-, t ~ "'_~

Fig. 2.2' :'Systems with different degrees of freedom



Theory of Vibrations 15

2.2.3 Periodic Motion: If motion repeats itself at regular intervals of time, it is called periodic motion.

2.2.4 Free Vibration: If a system vibrates without an external force, then it is said to undergo free

vibrations. Such vibrations can be caused by setting the system in motion initially and allowing it to move~~~~~. .

2.2.5 Natural Frequency: This is the property of the system and corresponds to the number of freeoscillations made by the system in unit time.

2.2.6 Forced Vibrations: Vibrations that are developed by externally applied exciting forces are called

forced vibrations. These vibrations occur at the frequency of the externally applied exciting force.

2.2.7 Forcing Frequency: This refers to the periodicity of the external forces which acts on the system

during forced vibrations. This is also termed as operating frequency.

2.2.8 Frequency Ratio: The ratio of the forcing frequency and natural frequency of the system is re-

ferred as frequency ratio.

2.2.9 Amplitude of Motion: The maximum displacement of a vibrating body from the mean position isamplitudeof motion. . ,

2.2.10 Time Period: Time taken to complete one cycle of vibration is known as time period.

2.2.11 Resonance: A system having n degrees of freedom has n natural frequencies. If the frequel}cyof excitation coincides with anyone of the natural frequencies of the system, the condition of resonance

occurs. The amplitudes of motion are very excessive at resonance.

2.2.12 Damping: All vibration systems offer resistance to motion due to their own inherent properties.

This resistance is called damping force and it depends on the condition of vibration, material and typeof the system..If the force of damping is constant, it is t&med Coulomb damping. If the damping forceis proportional to the velocity, it is termed viscous damping. If the damping in a system is free from itsmaterial property and is contributed by the geometry of the system, it is called geometrical or radiation

damping.

2.3 HARMONIC MOTION

Harmonic motion is the simplest form of vibratory motion. It may be described mathematically by thefollowing equation:

Z = A sin (rot - 0) ...(2.1)

N

L T:2!!-r- Go)

Timq.t

'. .,

c

Fig. 2.3 : Quantities describing harmonic motion



;. :'~f,t;,\r.j'~~!.

16 Soil Dynamics & Machine Foundations

The Eq. (2.1) is plotted as function of time in Fig. 2.3. The various terms of this equation are asfollows:

Z = Displacement of the rotating mass at any time t A = Displacement amplitude from the mean position, sometimes referred as single amplitude. The

distance 2 A representsthe peak-to-peak displacementamplitude, sometimesreferred to as double

amplitude, and is the quantity most often measured from vibration records.

ro = Circular frequency in radians per unit time. Because the motion repeats itself after 21tradians,

the'frequency of oscillation in terms of cycles per unit time will be ro/21t.It is denoted by f

8 = Phase angle. It is required to specify the time relationship between two quantities having the

same frequency when their peak values ha'ving like sign do not occur simultaneously. In Eq.

(2.1) the phase angle is a reference to the time origin.

More commonly, the phase angle is used as a reference to another quantity having the same fre-

quency. For example, at some reference point in a harmonically vibrating system, the motion may be

expressedby

ZI = AI sin rot

Motion at any other point in the system might be expressed as

Z, = A, sin ( rot -'e, )I I I

1t ~ 8 ~ - 1t.

...(2.2)

...(2.3)

with

For positive values of 8 the motion at point i reaches its peak within one half cycle after the peak

motion occurs at point 1. The angle 8 is then called phase lag. For negative values of 8 the peak motion

at i occurs within one half cycle ahead of motion at 1, and 8 is called as phase lead.

The time period, T is given by1 21t

T=-=- f roThe velocity and acceleration of motion are obtained from the derivatives of Eq. (2.1.).

dZ .Velocity = - = Z = roA cos (rot - 8)dt

= roA sin (rot - 8 + ~ )2

d Z .. 2Acceleration = -r = Z = ro A sin (rot - 8)

dt

= ro2A (sin rot - e + 1t)

Equations (2.5) and (2.6) show that both velocity and acceleration are also harmonic and can be

represented by vectors roA and ol A; which rotate at the same speed as A, i.e. ro rad/unit time. These,however, lead the displacement and acceleration vectors by 1tI2and 1trespectively. In Fig. 2.4 vector representation of harmonic displacement, velocity and acceleration is presented considering the dis-

placementas the referencequantity(8 = 0).

...(2.4)

...(2.5)

...(2.6)

, .J (., ..~4",t-t

",C.. .,.,~;<r'l!\"k..',", . ,~Ii<i"



Theory of Vibrations

N

z,z,z

..+'C~

E~v0a.UI

0

oN

...>-

+'

v0

~>

c0-0....

c:,I

c:,I

v0

.et

Fig. 2.4: Vector representation of harmonic displacement. velocity and acceleration

17

TimtZ,t

Ti mtZ,t

Timcz,t

When two harmonic motions having little different frequencies are superimposed. a non harmonic

motion as shown in Fig. 2.5 occurs. It appears to be harmonic except for a gradual increase and decrease

in amplitude. The displacement of such a vibration is given by:

Z = AI sin (0011- 91) + A2 sin (0021- 92)

N D,

-

2A max2Am\n

./.,/.. .,/ -

+'

Cc:,I

E,~

v0

a.III

c--- """'- ---,--- '-'"

.'J' ,.,

~T, b ~

" :' 3! j, ;I ',: ' ,' "

. ~~ 'i; 'P1>1Flg;'2.5':Motion containi.ng a beat

...(2.7)

TimtZ (t)



;;" C," 'i'j{':-;,':::;;;~,.


The dashed curve (Fig. 2.5), representing the envelop of the vibration amplitudes oscillates at afrequency, called the beat frequency, which corresponds to the difference in the two source frequencies:

I 1<01-<021 fb = Tb = 21t ...(2.8)

The frequency of the combined oscillations is the average of the frequencies of the two components

and is given by

f = i = (2~)(0) 1;1t0) 2 ) ...(2.9)

The maximum and minimum amplitudes of motion are the sum and difference of the amplitudes of

the two sources respectively.

Zmax = AI + A2 ...(2.10a)

"Zmin = IAI - A21 ,...(2.10b)

If the drive systemsof two machines designed to operate at the same speed are not synchronized, theymay result vibrations having the beat frequency.

2.4 VIBRATIONS OF A SINGLE DEGREE FREEDOM SYSTEM1

The simplest model to repre~ent a single degree of freedom system consisting of a rigid mass m supported

by a spring and dashpot is shown in Fig. 2. 6a. The motion of the mass m is specified by one co-ordinate

Z. Damping in this system is represented by the dashpot, and the resulting damping force is proportional

to the velocity. The system is sabject to an external time dependent force F (t).

----

Z - Djsplac(Zment

Z - V(Zlocity

Z - Ac c(zl(Zration

c

KZ+ Cl +1

..mz L - - -'-

Z-, -

m m

f F(t)

(I) Spring-mlss-dashpot system (b) Frcc-body diagram

~c ~ Pl8- 2.6, SI..,. ..,...' .,....... .-.:.., . . . .

~, ,." ~ ,~.,_."..~'~--"'"



>-":, ,;;[; /, '1\", ", ;,., "',c,...'" "-": ,,' ,.'r:,/'; ~:.: "'1~F"",';. ,

Theory of Vibrations . 19

, ,

Figure 2.6 (b) shows the free body diagram offue mass m at allYinstant dunng the course~fvibra-'tions. The forces acting on the mass m are:

(i) Exciting force, F (t): It is the externally applied force that causes the motion of the system.(ii) Restoring force, F,.: It is the force exertedby the spring on the mass emutends to restore the mass

, to its original position.For a linear system,restoringforce is equiJ.'to K . Z, where Kis thespring constant and indicates the stiffness. This force always acts towards the equilibrium posi-tion of the system.

(iii) Damping force, Fi The damping force is considered directly proportional to the velocity and

given by C . Z where C is called the coefficient of viscous damping; this force always opposesthe motion.

In some problems in which the damping is not viscous, the concept of viscous damping is still

used by defining an equivalent viscous damping which is obtained so that the total the energydissipated per cycle is same as for the actual damping during a steady state of motion.

(iv) Inertia force, F.: It is due to the acceleration of the mass and is given by mZ.According to De-l,

-Alemberfs principle, a body which is not in static equilibrium by virtue of some accelerationwhich it possess, can be brought to static equilibrium by' introduculg on it an inertia force. Thisforce actsthrough the centre of gravity of the body in the direction opposite to that of accelera-tion. " '

The equilibrium of mass m gives

mZ + CZ + KZ = F (t)

which is the equation of motion of the system. ,

2.4.1 Undamped Free Vibrations. For undamped free vibrations, the damping force and the exciting

force are equal to zero. Therefore the'"equation of motion of the system becomes .."

m Z + KZ = 0: '

, .::(2.11)

...(2.12a)

or ..

( K )Z+mZ=O ...(2.12b)

The solution of this equation can be obtained by substituting"

Z = A I cos con t + Az sin cont

where AI and Az are both constant~ and conis undamped natural frequency.

Substitut ing Eq. (2.13) in Eq. (2.12), we get? ,

-(j)~ (AI cos (j)i + Az sin (j)nt j+(~)(AI ~os oont + Az sin:oo~t) = 0

"

~'co =:1: -, " n m

,-,' , .

The values of constants A I and A2 are obtained by supstituting proper boundary conditions. We maynave the following two boundary conditions: ' " ' '

'" . ~

(i) At time t = 0, displacement Z = Zo' and

(ii) At time 1 = '0, velocity Z = V0

Substituting the first boundary condition in Eq. (2.13)

...(2.B)

"

,,'

...(2.14)or

Now,

. "/, ; "'"..',', "' Z:. """':.!'",I;'j,d",. ,',:'}.., :';"h' ,,',", , " :!':"'" '"

',,' ,-"""".."Ar-r;"'O:iI'i),+.'nji;~:J}'i"..ql.d")Jiti..j}iJ'iI.J'!,';~"; >is:.:,,,, '

':,; 'z ,=: -:' AI" 00,; si~ cont + A2 C1)n'~os cont "

C.' ...(2.15 ):; j

, ...(2.16)



20Soil Dymunics & Machine Fo"ndations

Substituting the second boundary condition.in Eq. (2.16)V.

A =--2..2 ~ n ...(2.17)

Hence . Vo2 = 20 cos oont + - sin oont

. con...(2.18)

...(2.19) Now let.

and

20 = Az cos 9V

--2.. = A sin 9co Zn ...(2.20)

where

Substitution of Eqs. (2.19) and (2.20) into Eq. (2.18) yields

2 = Az cos (oont - 9)

9 = tan-I(~

)con20

...(2.21 )

...(2.22)

( )

2

2 VoAz = ,/20 + -

. con

The displacement of mass given by Eq. (2.21) can be represented graphically as shown inFig. 2.7. It may be noted that

...(2.23)

c+)

~ One cycle

Acceleration /.-0,

\ % ." y, '" 0'

1'," /. 3

e\ 2~\ TI.r /2 "IT.~, 9 2lT +9 /

\ / / '\ /" / 0 '. 0 /~/ , V' ., / '- -'" "-- , -A

Z 0

0 isplacement "

+Az

:N..

oN

..

N Time,t

"1'/

velocity

(-)

Fig, 2.7 : Plot of displacement. velocity and acceleration for the free vibration of a mass-spring systemI>



'reory ilf Jl"l6iatiOns 21

At time t equal to Displacement Z is ..

0

8

Az cos 8

Az(J)n

1t +8L- 0

0)n

1I+8

0)

3-1t+82 .

-AZ

0(J)n

21t + 8

O)nAZ

It is evident from Fig. 2.7 that nature of foundation displacement is sinusoidal. The magnitude of

maximum displacement is Az. The time required for the motion to repeat itself is the period of vibration,T and is therefore given by. .

T = 21tO)n

...(2.24)

The natural frequency of oscillation, 1" is given by

J. =1- =~ =...!.. (K n T 21t 21t v-;;

...(2.25)

Now mg W- =-=0

K K st .

Where g = Acceleration due to gravity, 9.81 mIs2

W = Weight of mass m

°st = staticdeflectionof the springTherefore

...(2.26)

- I rg In - 21t Vfut

Eq. (2.27) shows that the natural frequency is a function of static deflection. The relation of In andOs!given by Eq. (2.27) gives a curve as shown in Fig. 2.8.

The nature of variation of the velocity and acceleration of the mass is also shown in Fig. 2.7.

...(2.27)

I,

- .

,.,....~ ~.~.n

I



22 Soil Dynamics & Machine Foundiuions . :~40

30

0-0 2 4 6

. 6stat (mm)

8 10

Fig. 2.8 : Relationship between natural frequency and static deflection

2.4.2 Free Vibrations With Viscous Damping. For damped free vibration system (i.e., the excitation

force Fo sin (J)t on the system is zero), the differential equation of motion can be written as

mZ + Cl + KZ = 0 ...(2.28)

where C is the damping constant or force per unit velocity. The solution of Eq. (2.28) may be written as'),.t . .

Z = A e ...(2.29)

where A and A are arbitrary constants. By substituting the value of Z given by Eq. (2.29) in Eq. (2.28),we get

m A A2it + C A A It + K A it = 0

2

(C

)K

or A + ni A + m = 0

By solving Eq. (2.30)

C FC

)2 K

. A,1,2 = - 2m :i: V~~) -;;The completesolutionof Eq.(2.28)is given by

.

Z - A Alt A ' A2t '

- le + 2e

The physical significance of this solution depends upon the relative magnitudes 'of (K/m), which determines whether the exponents are real or complex quantities.

...(2.30)

...(2.31 )

...(2.32)2

(C/2m) and

Case I : (~

)2 > K 2m m

The roots AI and A2are real and negative. The motion of the system is not oscillatory but is an

exponential subsiden~~(Fig. 2. 9). Because.of the relatively large damping, so much energy is dissipated

'-..----

,.... N 20:I:-

c.....

10



Theory of Vi!'rations ,

23

by the damping force that there is sufficient kinetic energy left t~ carry the mass and pass the equilibrium

position. Physically this means a relatively large damping and the system is said to be over damped,

z 2C > 4 km ... -"

Tim(l,t

Fig. 2.9 : Free vibrations ofviscously overdamped system

Case 11 : (~ )2 = K 2m m ,-

The roots Al and Az are equal and negative. Since the equality must be fulfilled, the solution is

given by

Z = (AI.+ Az t) le = (AI + Az t) e-Ct/Zm ...(2,33)

In this case also, there is no vibratory motion. It is similar to oyer damped case except that it is

possible for the sign to change once as shown in Fig. 2010.This,case is of little importance in itself; itassumes greater significance as a measure of the damping capacity of the system. " '

z

c2=l"kmTime,t

.,

Fig. 2.to': Free vibrations of a vlscouslycritically damped system

(~ ) = K. C = C2m m' c

Then Cc "=,2 ~Km". ...(2.35)

The system in this conditioonis known as ~ritically damped system anaC ~ is known as critical damp-

ing constant.' The ratio of the actual damping constant to the critical damping constant. is. defined as

damping ratio:

When ...(2.34)

'Now

C~=- Cc

C - C Cc - C 2JK"m_c:'fK

2m - Cc . 2m - Cc' 2m - Cc'Vm

By substitutingthis valueof' 2: ' as ~(On in Eq. (2.31), ~~ ~et"

..

,..(2.36)

...(2.37)

.".~.....- ",



, -.. ".' r , . [.',


AI, 2 =(_;:!:~;2-1) COn...(2.38)

Case III : (~ )2 < K 2m m

The roots Al and Al are complex and are given by .

AI,2 = [-;:!:i~I-;2 ]COn

The complete solution of Eq. (2.28) is given by

- (-~+j~I-~2 )O>i (-~-j~l-e )(J)"IZ - A I e + Az e

r:-:2 r ,or Z = e-~o>"t A j"I-~2 0>,,1+ A e-;~I-~- O),i'

the Eq. (2.41) can be written as I e z

...(2.39)

...(2.40)

...(2.41 )

Z = e-~O)",[Cl sin( (J)n~ t)+Cz cos( (J)1I~t)]...(2.42)

or Z = e-~O)II' [Cl sin(J)ndt+CZ COS(J)ndt] ...(2.43 )

where wild = (01/ ~ 1- ;z = Damped natural frequency.

The motion of the system is oscillatory (Fig. 2.11) and the amplitude of vibration goes on decreasing

in an exponential fashion.

z2

C < 41<m

Fig. 2.11 : Free vibrations of a viscously underdamped system

As a convenient measure of damping, we may compute the ratio of amplitudes of the successive

cycles of vibration

or

-0> f,1Z e "

---L = -0> f,(t+Zn/o>"cI)Z2 e n

ZI 0> f,.21t/o)ncl - = en

Zz r:2ZI Znl;!"l-f,-- =eZz

ZI - 21t;

loge 22.- ~

...(2.44 a)

or ...(2.44 b)

or ...(2.44 c)

...(2.44 d)



~ }, ",inF1': j' /," " 't~.. ' .~t .o,~; ',.

Tlreory of Vtb",tiolU ,--:)

. .Natural logarithm of ratio of two successive peak amplitudes {i,e, log, (~)} is called as logarith-mk. ,decrement. .

1 Z\ r:-:2

or ~= 2x loge~ ' As for small valuesof~, V1- ~- :: 1 ...(2.44 e)

tbus, damping of a system can be obtained from a free vibration record by knowing the successiveamplitudes which are one cycle apart. .'

If the damping is very small, it may be convenient to measure the differences in peak amplitudes for a number of cycles, say n.

In such a case, if Z" is the peak amplitudes of the n,h cycle, then

Zo Zl Z2 Zn-I 0- = - = - = . . . = - = e where~=2x ~Z\ Z2 ZJ Zn

Zo, =[

Zo] [

~] [

Z2

]..

[Z"-I

]= eno

Zn Z, Z2 Z) Z"Therefore,

or

Z1I 0 ..

~ = - oge Zn n

Z1I 0..

}: = - oge.z~ 2xn n

Hence ...(2.441)

...(2.44 g)

Therefore, a system is

over damped if ~ > 1;

critically damped if ~ = 1 andunder damped if ~ < 1.

2.4.3 Forced Vibrations Of Single Degree Freedom Syst~m. In many cases of vibrations caused by

rotating parts of machines, th~ systems are subjected to periodic exciting forces. Let us consider the case

of a single degree freedom sys~.:mwhich is acted upon by a steady state sinusoidal exciting force having

magnitude F and frequency 0>(i.e. F(t) = Fosin rot). For this case the equation of motion (Eq. 2.11) can be written as :

.. .

111Z + C Z + K Z = Fo Sin ro t ...(2.45)

Eq-;(2.45) is a linear, non-homogeneous, second order differential equation. The solution of this

equation consists of two parts namely (i) complementary function, and (ii) particular integral. The

complementary function is obtained by considering no forcing function. Therefore the equation of motionin this case will be :

.. .m Z, + C Z, + K Z, = 0 ...(2.46)

The solution of Eq. (2.46) has already been obtained in the previous st?ctioIland is given by,

ZI = e-O>/"(C\sinrondt+C2cosrondt) ...(2.47)

Here ZI represents the displacement of mass m at any instant t when vibrating without any forcingfunction. .

The particular integral is obtained by rewriting Eq. (2.45) as

m 2:2+ C 22 + K Z2 = Fo sin rot

where Z2 = displacement of mass m a~~nYinstant t when vibrating with forcing function.

~.;Y,

...(2.48)



8111;1'

"

26"--H,' " "

Soil Dynamics & Mac/line Foundations

; The,solution of Eq. (2A8).,\~ gi'{en by'. " ,,,', " . ~ ',", ,,' '" t,; "

, 22 = AI sin 00t + A2cos 00't

where AI and A2 are two, arbitrary constants.Substituting Eq. (2.49) in ~q. (2.48) ',' -

m (- At 002sin 00t - A2002cos (0t) + C (AI 00cos 00t - A2 ID;in 00t) + K (AI sin 00t + A2cos 00t) ,

,",,' ""'='Fosin'ro,t':. : , ,..'(2.50:

Considering .sine and Cosine functions in Eq. (2.50) separately, , ' ' , ,'.,

2 " " ,; , , ' (,. ' ',' ", ,"~ .

(- m AI 00 + KAt - CA2 00)sin 00t = Fo sin 00t

(- m A2 002+ KA2 + CA! 00).I::osffi t,~ O. 'J"

From Eg. (2.51 a), .,

Al(~ - o}) - A2(~

and from Eg. (2,51 b)

A{~-W )+A2(~-w2) =0

Solving Egs, (2,52 a) and (2.52 b), we get

(K-moo2) FoAt-

- 2

(K - mm2) +C2m2

and A2 = - CmFo

" ". '<c, "',, >, (K-~~2)2+c"2m2,"."

By substitutingJhe values 'of Al and A~"inEq::2'.49;:'" ,

' """

, :' ," ','

, "',.."

...(2.49)

"... ~ ,'" . . .:', "

",'

. ...(2,51 a)

,..(2.? 1 b)

.)

F. 0,

m = Q. 0

: ',n",(2,52 a)

...{2.52 b)

0.. '.)2.53 a)

"

...(2,53 b).

t '

~, ':22':°2" ~ 2 {(K'-mm2)sinmi-cwcosmt}

(K - m m ) + Cm"

'...(2.54)

let, ' tan e = 'C 0)

K-l~cd2

-, ' .,.(2.55)

By substituting Eg. (2.55) in Eg. (2.54), one can obtain

22 = Fo . sin(mt - e)

~(K~';'m2t+'c2 cii1' :" :,., "Eg. (2.56) may be written as t . . " , ,

22 = . '.ccFo!K , . 'sin{~t -"'er

~(I:!12)+(2Tl~)~.. "I , -"~; . '". W ,',

11 = Frequency ratio =;- "',,;, ".,:.' ,n

-,' ':Ci"","'G'}""::"',',_:;,~'-',~ = Dam p ing ratio = -

c.:;=

,

' ';JKi{i . '.., 0"'2' 'Km 'c

...(2.56)

,,' ,

, ,'!" ..~

. " ...(2.57)

Where,,'

""""'~!i1I"1!",?' ~"" "".",..1d."" "~~",...""."..",""."""",."""""""",-,,i""'-~-' .



Theory of Vibrations. . ,'..,

~J'(,,~\'k

27

The complete solution is obtained by adding the compJimentaryfunction and the particular integral.

Since the 'coriipliIne~tarYfti~l(!tioh:lsan'expJnenii~nf'decayin~ function,:iit'will die out'soon and themotion will be des~ribed by only the p~uticula:rmtegral(Fig. 2:i 1)'.:The syStemwill vibrate harmonically

,with the same frequency as the forcing and the pe~ ap11'1!tu4~.,is,g~ven byF. /K '.

Az = 0 " oi

. . ", ..., ~(l~ 1]2)2+(21]~)2 ",,"', ;;"

N.., ,~ '~','--' '

"I

+'

C

~

E~u

0a.UI ....

....,.,..

Transi~nt , ,

0

~211"

(;) ~ N

N

..+'C

~

E~u0a.UI

0

Time.t

. "

-,

'h ..'

:'1 i." ..;.

. . / ' -;. ..i , ,- #" "

'.'. -' ,.;" ~,: ,~~#"st'~c;!:t>~tatcz

4;i;" ':~~.~.",;..' , :~~.::'---'- ~,'

"

"

N\

'I\" .

+'C~

Ec:.Iu0

a.III

Time,t

.,'. " .q , ','" .),

0< , " "

. "! ".' '.', ,', .

Co mpl~t~ solu tion

Fig. 2.12 Superpos,ition of transient and steady state vibrations

i, - ~ rr"'- "'-T--'1iiiiii[-'

...(2,58)

~-




Differentiating Eq. (2.59) with respect to 11and equating to zero, it can be shown that resonance will

occur at a frequency ratio given by

11 = ~1-2~2

which is approximately equal to unity for small values of ~.

or ffind = ffin ~1-2~2

...(2.60 a)

...(2.60 b)

where ffind = Damped resonant frequency

°30

~ =0-5

180°

150°

<D120°

c:.J

C'I

C

0

c:.J 90°' r =0.707\11

0.J:a..

600

00 '1.0 2.0 3.0

FrczquQ ncy rat io# -rz.

Fig. 2.14: Phase Jagversus frequency ratio for different amounts of damping, . ' . , : " ..,' ~ .

;':J '



<'. , ~ ,";~,'~',{f.(;"..


'By substituting Eq. (2.60) in Eq. (2.59), the maximum value of magnification factor is obtained. It

is given by ,

I

J.lmax = 2E,~I-E/

I

- 2E,

Assuming a damping of 5% in a structure, its amplitude at resonance will be 10 times the static

deflection. This indicates that systems will be subjected to very large amplitudes at resonance whichshould be avoided.

...(2.61 )

(For small values of;) ...(2.62)

. '. --- - -

The phase angle e given by Eq. (2.55) indicates the phase difference between the motion and the

exciting force: It can}e.-writt~.n~s .

-I

(

211E,

)e = Tan

_ 1- 2'" . T\

Variation of e with respect to 11is shown in Fig. 2.14

...(2.63 )

2.4.3.1 Rotating mass type excitation. Machines with unbalanced rotating masses develop alternating

force as shown in Fig. 2.15 a. Since horizontal forces on the foundation at any instant cancel, the net

vibrating force on the foundation is vertical and equal to 2 me ero2 sin rot, where me is the mass of eachrotating element, placed at eccentricity e from the centre <,>f rotating shaft and ro is the angular frequency

of masses. Fig. 2.15b shows such a system mounted on elastic supports with dashpot representing viscous

damping. .

12m~. ~~

Forc~ gczn~rat~d

(a) Rotating masstype excitation (b) Mass-spring-das~ot system

Fig. 2.15 : Single degree freedom system with rotating mass type excitation

The equation of motion can be written as-. 2

m Z + C Z + K Z = 2 111ero sin rot .. . - e. .

where m is the mass 'of foundation including 2 me' Equations (2.64) imd (2.48) are similar, except that

2 Ill" ero2 appears in Eq. (2.64) in place of Fo' The solution of Eq. (2.64) may therefore be written as,

...(2.64)



'.'; :~,(i:', '.'),' '<'F '" ". ,." ',', , ". .(\~.tJi;"":


where

Since

or

.---.

';Iile

31

Z = A; sin (0)t + 0') .

(2mee/ m)'T}2

Az = I 2 2

(1-T}2) +(2~T})2

F=2m .eO)0 e

F (J) 2 0)2

K = 2 me . e K = 2 me . e (mro~)

...(2.65)

..(2.66)

= (2 me :}T}2

e = Tan-I(

2T}\)1-11

...(2.67)

3.0 0.10

2.0

5.0

1.0

0-0 4.01..0 ' ,2.0, '. 3.0

Frequency rati o. 1) .

(a) Az 1(2m~elm) versus Irequency rauo 11

0° -0 4.0 5.01.0 2.0 3.0

Frqquqncy ratio., - 't(b) Phase angle versus frequencYT&tio11

Fig. 2.16 : Response,oh system with rotating unbalance

180°.0.05

'(D

I 0.25'...

0.50aI

C7Ic:0

90°aIU\

0.s::.a..




The Eq. (2.66) can be expressed in non-dimensional form as given below:

A~ 1"\2-

/ = ~ ...(2.68)(2mee m) (1-1"\2)2+(21"\1;)2

The value of A= /(2me elm) is plotted against frequency ratio 1"\in Fig. 2.16 a. The curves are similar

in shape to those in Fig. 2.13 except that these starts from origin. The variation of phase angle e with 11

is shown in Fig. 2.16 b. Differentiating Eq. (2.68) with respect to 11and equating to zero. it can be shown

that resonance will occur at a frequency ratio given by1

1"\=-

F-2e

...(2.69 a)

0011or ro =-

nd .JI=21;2 .

By substituting Eq. (2.69 a) in Eq. (2.68), we get

(

Az

)- l'

2meelm max - 21;~1-1;2

~ 2\ for small damping

...(2.69 b)

...(2.70)

...(2.71 )

2.5 VIBRATION ISOLATION

In case a machine is rigidly fastened to the foundation, the force will be transmitted directly to the

foundation and may cause objectionable vibrations. It is desirable to isolate the machine from the foun-

'dation through a suitably designed mounting system in such a way that the transmitted force is reduced.

For example, the inertial force developed in a reciprocating engine or unbalanced forces produced in any

other rotating machinery should be isolated from the foundation so that the adjoining structure is not set

into heavy vibrations. Another example may be the isolation of delicate instruments from their supports

which may be subjected to certain vibrations. In either case the effectiveness of isolation may be mea-

sured in terms of the force or motion transmitted to the foundation. The first type is known as force

isolation and the second type as motion isolation.

2.5.1 Force Isolation. Figure 2.17 s~ows a machine of mass m supported on the foundationby means of an isolator having an equivalent stiffness K and damping coefficient C. The machine is excited with

unbalanced vertical force of magnitude 2 me eci sin 00t . The equation .ofmotion of the ~achine can bewritten as:

w~ere

... 2m Z + CZ + KZ = 2 me eoo sin 00t

The steady state motion of the mass of machine can be worked out as

22m eoo / K .

Z = r e 2 .sm(oot-8)

(1-'12). +(21"\1;)2

8 = Tan-I

[2'11;2

]1-.1"\

...(2.72)

...(2.73)

...(2.74)



':;0:"<\',:,-;' '-' ,'t "t" "',"',' '1

Theory of Jlibrlltions;;::

33

--'-~, '

~~~ Ma chine

K 2'

cK 2

Iso lata r

Foundation Ground surfac~

Fig. 2.17 : Machine-isolator-roundation systcm

The only force which can be applied to the foundation is the spring force KZ and the damping force. '

C Z; hence the total force tqmsmitted to the foundation during steady state forced vibration is

Ft = KZ + CZ

Substituting Eq. (2.73) in Eq. (2.75), we get2

2m em .F = e .sm(mt- e)+

t~ 2 2(1-1l) +(211~) ,

..,(2.75)

C.2me em2 /K '00 cos(mt-e)2 2

(1-112) +(211~)

...(2.76)Equation (2.76) can be written as:

2 Jl+(211~)2F = 2me em , .sin(m t - P)t 2 2

. (1-112) +(211~)

where p is the phase difference between the exciting force and the force transmitted to the foundation and

is given by , ' i

- P ~-e --=~-~~-l[

coo

]'" r K,, ", J -

...(2.77)

...(2.78)~

;"": ','"o/,,'n1.."n" ',' i J " ".



34Soil Dynamics & Machine. Foundations

Since the force 2 m e ol is the force which would be transmitted if springs were infinitely rigid, ae .

measure of the effectiveness of the i~olation mounting system is given by

. ' Ft ~1+(211~)2

IlT =2 = ~ ...(2.79)2 meem (1-112)2 + (211~)2

IlT is called the transmissibility of the system. A plot of IlTversus 11for different values of ~is shown

in Fig. 2.18. It will be noted from the figure that for any frequency ratio greater than 12, the force

transmitted to the foundation will be less than the exciting force. However in this case, the presence of

damping reduces the effectiveness of the isolation system as the curves for damped case are above the

undamped ones for 11>12. A certain amount of damping, however, is essential to maintain stability

under transient conditions and to prevent excessive amplitudes when the vibrations pass through reso-

nance during the starting or stopping of the machine. Therefore, for the vibration isolation system to be

effective 11should be greater than 12.

4.5

4.0

1.0

. ~ =0

f =0.125

~ =0

~ =0.125

0 -

0

~ =0.5

~ =1.0

~ =2.0

~ : 0.125'

I ~ =0

1.0 2.0 3.0

Frczquczncyr(:itio I 1'\.

Fig. 2.18: Transmissibility (J.1-r)versus freqeuncy ratio (Tt>

1- 3.0=<.

»....-.-oD

U\U\.-

2-0EIIIC0...I-



,f "c"'" f ','" ,,': ,r:,,<"f:' ';:£'4/ l',,' tt~~7T{~:,:~";';:s:",.:';


, . . : ';' , ' ' '

2.5.2. Motion Isolation. In many situations, it would be necessary to isolate structure or mechanical

systems from vibrations transmitted from the neighboring machines. Again we require a suitable mount-

ing system so that least vibrations are transmitted to the system due to the vibrating base. We consider asystem mounted through a spring and dashpot and attached to the surface which vibrates harmonically

with frequency (I)and amplitude Y0 as shown in Fig. 2.19.

Machina

z " .

Foundation. , ,

Iso lator

v = Yo Sin GJt

Vi brating ground.

d u (l to n (l i9 h bo u r in 9machines

Fig. 2.19: Motion isolation system

Let Z be the absolute displacement of mass; the equation of motion of the system can be written as:

m Z + C (2 - Y) + K (Z -:- Y) = 0 ...(2.80)

or m Z + K 2 .+ K Z = C Y .+ K Y = C (I) Y0 cos (I) 1 .+ K Y0 sin (I) 1

m Z'+ C 2.+ KZ = Yo ~K2.+(Cro)2 sin (rol'+ ex) ...(2.81)or

h T-I CO)

ere ex =, an K l:he solution of Eq. (2.81) will give the maximum amplitude as:

Z - . ~ 1. +( 21l ~) 2 ,

max - Vo'~,. , " (l'-1l2)2+(21l~l" , . ) . ., ' - '. , ~

...(2.8?)

...(2.83)

" " . ,, ' " ,



18E:.

36 SoU Dynamics & Machine Foundations

The effectiveness of the mounting system (transmissibility) is given by

- Zmax ~ ~1 + (2T\~)2

~T- -y; - ~(1-T\2)2+(2T\~)2

...(2.84)

Equation (2.84) is the same expression as Eq. (2.79) obtained earlier. Transmissibilityof such system

can also be studied from the response curves shown in fig. 2.18. It is again noted that for the vibration

isolation to be effective, it must be designed in such a way that T\> .fi.

2.5.3. Materials Used In Vibration Isolation. Materials used for vibration isolation are rubber, felt.

cork and metallic springs. The effectiveness of each depends on the operating conditions.

1.5.3.1. Rubber. Rubber is loaded in compression or in shear, the latter mode gives higher flexibility.

With loading greater than about 0.6 N per sq mm, it undergoes much faster deterioration. Its damping

and stiffness properties vary widely with applied load, temperature, shape factor, excitation frequency

and the amplitude of vibration. The maximum temperature upto which rubber can be used satisfactorily

is about 65°c. It must not be used in presence of oil which attacks rubber. It is found very s' ".,ble for high

frequency vibrations.

2.5.3.2.Felt. Felt is used in compressfun only and is capable of taking extremely high loads. It has very

high damping and so is suitable in the range of low frequency ratio. It is mainly used in conjunction with

metallic springs to reduce noise transmission.

2.5.3.3. Cork. Cork is very useful for accoustic isolation and is also used in small pads placed under-

neath a large concrete block. For satisfactory working it must be loaded from 10 to 25 N/sq mm. It is not

affected by oil products or moderate temperature changes. However, its properties change with the fre-

quency of excitation.

1.5.3.4. Metallic springs. Metallic springs are not affected by the operating conditions or the environ-

ments. They are quite consistent in their behaviour and can be accurately designed for any desired

conditions. They have high sound transmissibility which can be reduced by loading felt in conjunction -

with it. It has negligible damping and so is suitable for working in the range of high frequency ratio.

2.6 THEORY OF VIBRATION MEASURING INSTRUMENTS

The purpose of a vibrationmeasuring instrument is to give an output signal which represents, as closely

as possible, the vibrationphenomenon. This phenomenon may be displaceme~t, velocity or accelerationof the vibrating system and accordingly the instrument which reproduces signals proportional to theseare called vibrometers. velometers or accelerometers.

There are essentially two basic systems of vibration measurement. One method is known as the

directly connected system in which motions can be measured from a reference surface which is fixed.

More often such a referencesurface is not available. The second system, known as "Seismic system" doesnot require a fixed reference surface and therefore is commonly used for vibration measurement.

Figure 2.20 shows a Vibration measuring instrument which is used to measure anyof the vibration

phenomena. It consists of a frame in which the mass ~ is supported by means of a spring K and dashpotC. The frame is mounted on a vibrating body and vibrates al~ng with it. The system reduces to a springmass dashpot system having base on support excitation as discussed in Art. 2.5.2 illustrating motionisolation.



. .(~

Thtory of VibratiOns 37

zm

cK

y = Yo Sin '->t

Fig.2.20 : Vibration measuring instrument

Let the surface S of the structure be vibrating harmonically with an unknown amplitude Y0 and anunknown frequency (0. The output of the instrument will depend upon the relative motion between the

mass and the structure, since it is this relative motion which is detected and amplified. let 2 be the

absolute displacement of the mass, then the output of the instrument will be proportional to X = 2 - Y.The equation of motion of the system can be written as

m Z + C (Z - Y) + K (2 - Y) = 0

Subtracting m Yfrom both sides,... .. 2

m X + C X + K X = - m Y = m Y0(0 sin (0t The solution can be written as

.. .( 2.8S) ,

..:.(2.86)

where

2

X = ~ TJ Yo sin «(0 t- e)(1- TJ2)2+ (2TJ~)2

(0.11 = - = frequency ratio

(On

~ = damping ,ratio

1

(2 TJ~

)and e = tan- 1- TJ2

Equation(2.S7)ca~ be rewrittenas:. ,

X = 1)2.J! Y0 sin «(0 t - 8)

...(2.S7)

(2.,SS)

where;°1 ">;(1..;>' ,

J! = ~1- TJ2)2+ (2T1~l

ill



38. Soil Dynamics & Machine Foundlltions- -." '-" .' . -. .

2.6.1. Displacement Pickup. The instrument will read the displacement of the structure directly if

1121.1= I and 8 = O.The variation o{Tl~ with-~'aiid'~-isshown in Fig-.2.21. The variation of8 with 1'\

is already given in Fig. 2.14. It is seen"'tnatwneifff is" large, 1'\21.1is approximately equal to 1 and 8 is

approximately equal to 180°. Therefore to design a displacement pickup, 1'\should be large which means

that the natural frequency of the instrument itself'shou~d be low compared to the frequency to be mea-

sured. Or in other words, the instrument should have a soft spring and heavy mass. The instrument is

sensitive, flimsy and can be used in a weak vibration environment. The instrument can not be used for

measurement of strong vibrations.

3.0

,- -t

I \I \0

I

I

. .

-- . -

- -"

2 0,

1 .0

0 -0 1.0 2.0 3.0 4.0 5.0

FrequClncy ratio, '1.

Fig. 2.21 : Response of a vibration measuring instrument to a vibrating base

2.6.2. Acceleration Pickup (Accelerometer). Equation (2.88) can be rewritten as. I 2

X = 2 1.1 Yoro sin (rot - 8)(J,)n . '

The output of the instrument will be proportional to the acceleration of the structure if J.1is constant.

Figure 2.13 shows the variation of J.1with 1'\and;. It is seen that J.1is approximately equal to unity for

small values of 1'\.Therefore to design an acceleration pickt!p, 11should be small which means that th~'

...(2.89)

" ",.~'n'7""~:"'"""'"

.'1::' 'lr~-r\f

........--.



<~


natural frequency of the instrument itself should be high compared to the frequency to be measured. In

other words, the instrument should have a stiff spring and small mass. The instrument is less sensitive

and suitable for the measurement of strong motion. The instrument size is small.

2.6.3 Velocity Pickup. Equation (2.88) can be rewritten as

1X = - TIJ!Y0(J)sin «(J)1- 0)

COn

The output of the instrument will be proportional to velocity of the structure if ~ 111lis a constant.O)n

At 11 = 1, Eq. (2.90) can be written as

1 1 . .. .1X = -

2 ~ Yo(J) sm (0)1- 0) .: atTl = 1, f.1= - ...(2.91)con ':1 .. . . 2~

Since O)nand ~are constant, the instrument will measure the velocity at 11= 1.

It may be noted that the same instrument can be used to measure displacement, acceleration andvelocity in different frequency ranges. . ,

X a Y if TI» 1 Displacement pickup (Vibrometer~)

...(2.90)

..

X a Y if 11 « lAcceleration pickup (Accelerometers)

X a Y if 11= 1 Velocity pickup (Velometers) .

Displacement and velocity pickups have the disadvantage of having rather a large size if motions

having small frequency of vibration are to be measured. Calibration of these pickups is not simple. Fur-

ther. corrections have to be made in the observations as the response is not flat in the starting regions.

From the point of view of small size, flat frequency response, sturdiness and ease of calibration, accelera-

tion pickups are to be favoured. They are relatively less sensitive and this disadvantage can easily be

overcome by high gain electronic instrumentation..

. .

2.6.4. Design of Acceleration Pickup. The relative displacem~nt between, the mass an~ the supportwould be a measure of the support acceleration ifTl is less than 0.75 an4 ~is of the order of 0.6 to 0.7.

Of the various methods of measurement of relative displacement, .electrical gauging,:in 'whIch {he me-

chanical quantity is converted into an equivalent electrical quantity is best suited for a~.<;elerationpick-

ups. Electrical gauging offers the possibility of high magnification of ~e signals which are usually weak

because the spring is stiff and the displacements are small. The mechanica,l quantity alters either the

resistance, or capacitance or the inductance of the circuit which consequently alters the current in thecircuit. '

2.7 VIBRATION OF MULTIPLE DEGREE FREEDOM SYSTEMS

In the preceding sections, vibrations of systems having single degree of freedom havebeen discussed. In

many engineering problems, one may come across the systems which may have more than one degree of

freedom. Two degrees freedom cases arise when the foundation of the system is yielding thus adding

another degree of freedom or a spring'mass system is attached to the main system to reduce its vibrations.

In systems when there are a number of masses con~ected with each other, even if each mass is con-

strained to have one degree of freedom, the system as a whole h~s as many degrees of freedom as there

are masses. Such an idealization is done for carrying out dyn,!mic analysis of multistoreyed buildings.Q T



40

(a) Four storeyedframe

(b) Idealisation (c) First mode

-- .

Soil Dynamics & Machine Foundations

(d) Second mode (e) Third mode (t) Fourth mode

Fig. 2.22 : A four storeyed frame with mode shapes

Figure 2.22 a shows the frame work of a four storeyed. building. It is usual to lump the masses at the

floor levels and the lumped mass has a value corresponding to weight of the floor, part of the supporting

system (columns) above and below the floor and effective live load. The restoring forces are provided by

the supporting systems. Figure 2.22b shows such an idealization and it gives a four degrees of freedom

system. In free vibration a system having four degrees of freedom has four natural frequencies and the

vibration of the any point in the system, in general, is a combination of four harmonics of these four

natural frequencies respectively. Under certain conditions, any point in the system may execute har-

monic vibrations at any of the four natural frequencies, and these are known as the principal modes of

vibration. Figure 2.22<.to 2.221'show the four modes of vibration. If all the masses vibrate in phase (Fig.

2.22c), the mode is termed the first or lowest or fundamental mode of vibration and the frequency asso-

ciated with this mode would be the lowest in magnitude compared to other modes. If all adjacent masses

vibrate out of phase with each other (Fig. 2.22£), the mode is termed the highest mode of vibration andthe frequency associated with this mode would be highest in magnitude compared to other modes.

2.7.1. Two Degrees of Freedom Systems.

2.7.1.1. Undampedfree vibration: Figure 2.23 shows a mass-spring system with two degrees of free-

dom. Let Z\ be the displaceuent of mass ml and Z2 the displacement of mass m2' The equations of motion

of the system can be written:

In t Z\ + K t ZI + K2 (Z\ - Z2) = 0

1nl Z2 + K3 Zl + K2 (Z2 - Z,) = 0

The solution of Eqs. (2.92) and (2.93) will be of the following form:

ZI=A\sinro,,( .

Z2 = A2 sin ro" t

Substitution of Eqs. (2.94) and (2.95), into Eqs. (2.92) and (2.93), yields:

(KI + K2 - ml C1)~)Al - K2A2 = 0

- K2 Al + (~ + KJ - m2 C1);)A2 = 0

...(2.92)

...(2.93)

...(2.94 )

...(2.9S)

...(2.96)

...(2.97)

, ,



I .

~heory of Vibrations

~!'\Wm

IL

%L

,

3KT%%P%S

Z2

Jz~-

Fig. 2.23 : Free vibration of a two degrees freedom system

For nontrivial solutions of oon in Eqs. (2.96) and (2.97),

2Kt +K2 -mt ron -K2

21 = 0- K2 K2 + K3 - ~ ron

.- f2 9;!)

00: _ [

Kt + K2 + K2 + K3

]O)~+ K) K2 + K2 K3+ K3K) -= 0

m) ~ ml~

Equation (2.99) is quadratic in ro2, and the roots of this equation are:n

ro~= .!.

[(K) +K2:t- K2 ~K3

)+

{(

KI +K2 K2 +K3

)2 + 4 K~

}

1/2

]2 m) "'2 m) ~ mj ~

...(2.100)

From Eq. (2.100), two valuesofnatura!~!!e9~~ncies oon)and oon2can be obtained. con)is correspond-ing to the fIrst mode and COn2is of the second mode.

or

. ,

...(2.99)

- 11iiii



42

..,

Soil Dynamics & Machine Foundatiofls

The general equation of motion of the two masses can now be written as

Z = A ( I) sin (0 t + A (2) sin ( 0 tI I nl I n2

and Z = A(I) si n (0 t + A(2) si n (0 t2 2 n I 2 n2

The superscripts in A represent the mode.

The relative values of amplitudes AI and A2 for the two modes can be obtained using Eqs. (2.96) and(2.97). Thus

(0 2Al - K2 - K2 +KJ -"'2 ffinJ

(i)- 2- K A2 KI+K2-mlffinl 2

(2)' 2A I - K2 - K2 + KJ - "'2 ffin2

(2) - 2 - K A2 KJ + K2 - m) ffin2 2

2.7.1.2. Undamped forced vibrations. Consider the system shown in Fig. 2.24 with excitation force Ft

sin (0 t acting on mass ml. In this case, equations of motion will be:

ml Zt + Kt Zt + K2 (Zl - Z2) = Fo sin (0t

1n2 Z2 + KJ Z2 + K 2 ( Z2 - Z\) = 0

F0 sin G.)tl

...(2.101)

...(2.102)

...(2.103j

...(2.10{

(2.105)

...(2.106

21

22

.

Fig. 2.24 : Forced vibration of a two degrees freedom system

.. ',",y



leory of Vibrations 43

For steady state, the solutions will be as

21 = Al sin 00t

2z = Azsin 00t

Substituting Eqs. (2.107) and (2.108) in Eqs. (2.105) and (2.106), we getZ(KI + Kz - ml 00) AI - KzAz = Fo

z- KzA I + (Kz + K3- mz 00) Az = 0

Solving for AI and Az from the above two equations, we getz

(Kz +K3 -rnz co ) FoA =I

[

4

(

KI+Kz Kz+K3

)

z KIKz+KzK3+K3KI

]ml rnz co - + co +

ml rnz mlrnz

K3Fo -A =z

[

4

(

KI+Kz Kz+K3

)

Z' KIKz+KzK3+K3KI

"

mlrnz co - + co +

ml rnz m\rnz The above t\VOequations give steady state amplitude of vibration of the ~wo masses respectively, as

a function of 00. The denominator of the two equations is same. It may be noted that:

(i) The expression inside the bracket of the denominator of Eqs. (2.1110) and (2.111b) IS of the

same type as the expression of natural frequency given by Eq. (2,99). Therefore at 00= oolll andCl) = Cl)nZ values of A laud Az will be infinite as the denominator will become zero.

(ii) The numerator of the expression for Al becomes zero when

/K2 +KJ)Cl) = rnz ...(2.112)

Thus it makes the mass ml motionless at this frequency. No such stationary condition exists for

0 mass ml' The fact that the mass which is being excited can have zero amplitude of vibration under

certain conditions by coupling it to another spring -mass system forms the principle of dynamic vibrationabsorbers which will be discussed in Art. 2.8. '

...(2.107)

...(2.108)

...(2.109)

...(2.110)

...(2.111 a)

and ...(2.11Ib)

2.7.2. System With n Degrees of Freedom.

2.7.2.1. Undamped free vibrations: Consider a system shown in Fig. 2.25 having n-degree of freedom.

If Z\' 2z, Z3 ... 2n are the displacements of the respective masses at any instant, then equations of motionare:

rn, 2( + K( Z\ + Kz ( ZI - Zz) = 0

mz 2z - Kz(Z( - 2z) + K3 (2z - 23) = 0

...(2.113)

...(2.114)

m3 23 - K3(2z - 23) + K4 (23 - 24) ;: 0 ...(2.115)

'. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

mn 2n - Kn (2n - I -'- 2n) = 0 ...(2.116)

".,'~'.,-..~,,-",'" "' . ,c-"" '; '; . "', ..; . :--';co,,---, - ... n



181:.J

44Soil Dynamics & Machine Foundations

Z1

Z2

Z3

Kn-1

Zn-1

Zn

Fig. 2.25: Undamped free vibrations of a multi-degree freedom system



"'.'~..2,,~,.."t;,~.>, :"""',;,n rA';".!n~";'F."'" "'.,.,


The solution of Eqs. (2.113) to (2.116) will be of the follow:"'~ IO':n:

ZI = Al sin cont

Z2 = A2 sin cont Z) = AJ sin cont

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

Zn = An sin cont

Substitution of Eqs. (2.117) to (2.120) into Eqs. (2.113) to (2.116), yields:

[(KI+K2)-mIID~] AI-K2~ =0

-K2A1 + [(K2+K))-~ID~] A2-KJAJ =0

- KJ Az + [(KJ + K4) - mJ ID~] A) - K4 A4 = 0

, . . , . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

- Kn An - I + (Kn -mnID~) An = 0

For nontrivial solutions of oonin Eqs. (2.121) to (2.124),

[ (KI + Kz) - ml ID;]-K2

0

-Kz

[(Kz+KJ)-~ID;] '"

-K)

0

0

0

0

0

0 =0

0 2'" -Kn (Kn-mnIDn)0

- ~

45

...(2.117)

...(2.118)

...(2.119)

...(2.120)

...(2.121)

...(2.122)

...(2.123)

...(2.124)

(2.125) .

Equations (2.125) is of nthdegree in CI);and therefore gives n values of con correspondingto n natural

frequencies. The mode shapes can be obtained from Eq. (2.121) to (2.124) by using, at one time, one of

the various values of conas obt1incd from Eq. (2.125).

When the numht.'Tvi degreeS of freedom exceeds three, the problem of forming the frequency equa-

tion and s01";~jgit for determ41ation of frequencies and mode sh<1pesbecomes tedius. Numerical tech-. J'iG.~esare invariably resorted to in such cases.. ,

Holzer's numerical technique is a convenient method of solving the problem for the system idealizedas sho~ in Fig. 2.26. By sUI11II1iPgJfotcesat free end, .

'-, . --.'O".::.,:-=C°'-:"~",_,- ;: "

~-::--===.::.~-~:---==E'~--,'::.._~



If

46Soil Dynamics & Machine Foundations

..m1 Z1

m2 Zz

m3 Z3

.

K.J

m.11-

~_1. Kj-1(Zj-Zi-1)

K.11-

m.I

mi+1

Fig. 2.26: An idealised multi-degree freedom system

Inertia force at a level below mass mi - I;-1 ..= ".

I m. Z. L...J= } }

Spring force at that level corresponding to the difference of adjoining masses= K. I (Z. - z. 1)1- 1 1-

Equating Eqs. (2.126) and (2.127) .

i-I ..

Lj=lmjZi = Ki-I (Zi- Zi-I)

Putting Zi = Ai sin (()t in Eq. (2.128), we get

...(2.126)

...(2.127)

...(2.128)

I'i~'t m i (- Ai U)~ sin U)n t) = Ki - I (Ai sin (Unt - Ai - I sin (Unt)

.'

'2

U)n "i-I AAi = Ai-I - K £...j=lmj ji-I

Equation (2.129) gives a relationship b~tween any two succ~sive amplitudes. Starting with any

arbitraryvalueof AI' amplitudeof all othermassescanbe deterinined.A plot of An+ 1 versus (0~ would

have the shape as shown in Fig. 2.27. Finally An + I should worked out to zero' ~ue to fIXityat the base.

The intersection of the curve with (0~ axis would give various val~~s.pfQ);.~ode $ape.can be obtained.. .. -. .

by substituting the correct value of (O~in Eq. (2.129).

...(2.129)or

iI

m1.J

K1

Iml

K2




1.0

~

t

' ..-J +J c:~«~1:

0

(..)2n.

-1.0

2wn1

2""nz

---Fig.-~7: Residual a~a flinction of frequency in Holzer method

2.7.2.2. Forced vibration. Let an undamped n degree of freedom system be subjectedto forced vibration,

and Fj(t) represents the for~e on mass mr The equation of motion for the mass mj will ben

m. Z. + I K.. Z. = F. ( t)I I =1 IJ ) ,

where i = 1,2,3, , n

...(2.130)

The amplitude of vibration of a mass is the algebraic sum of the amplitudes of vibration in various

modes. The individual modal response would be some fraction of the total response with the sum of

fractions being equal to unity. If the factors by which the modes of vibration are multiplied are repre-

sented by the coordinates d, then for mass mj'

- (1) (2) (r) (n)Z. - A. d 1 + A. d 2 + ... + A. d +... + A. d '" 1 r ,n

Equation (2.131) can be written as

...(2.131 )

n

Z. = I A~r) d -' r=l ' r

Substituting Eq. (2.132) in Eq. (2.130)

...(2.132)

n (r) .. n n (r)- Im. A. d + I I K.. A. d - F. ( t ) I ,r 'I 1 r ,r=1 r=1 j=l . .

Under free vibrations, it can be shown

n (r) - 2 (r)~ Kij Aj - oonr mj Aj1=1

...(2.133)

...(2.134)

Substituting Eq. (2.134) in Eq. (2.133), we get

n (r) .. n 2 (r) -ImjAj dr + I oonrmjAj dr - Fj(t)

r=1 r=l ..(2.135)

or n (r)'. 2 -ImjAj (dr+oonr.dr) -Fj(t)

r=l ...(2.136)

l' -.-'T .~ ""}'- -")J'!!I~'



~,


Since the left hand side is a summation involving different modes of vibration, the right hand side

should also be expressed as a summation of equivalent force contribution in corresponding modes.

Let F; (t) be expanded as:

Fj (t) = i mj A~r) fr (t)r=1

where fr (t) is the modal force and given by

...(2.137 a) -"J

n

Ili (t) .A~r)

fr(t) = i~1 Z

Lm[(A~r»)];=1

Substituting Eq. (2.137 a) in Eq. (2.136), we get

.. zdr + O}nrd,. = fr (t)

Equation (2.138) is a single degree freedom equation and its solution can be written as

1 I

dr = - J fr Ct) sinO}nr(t-1:) dt where 0 < 1:< I0}nr 0

It is observed that the co-ordinate d, uncouples the n degree of freedom system into n systems of

single degree of freedom. The d's are termed as normal co-ordinates and this approach is known as

normal mode theory. Therefore the total solution is expressed as a sum of contribution of individualmodes.

(2.137 b)

(2.138)

..(2.139)

2.8 UNDAMPED DYNAMIC VIBRATION ABSORBER

A system on which a steady oscillatory force is acting may vibrate excessively, especially when close to

resonance. Such excessive vibrations can be eliminated by coupling a properly designed spring masssytem to the main system. This forms the principle of undamped dynamic vibration absorber where the

excitation is finally transmitted to the auxiliary system, bringing the main system to rest.

Let the combination of K and M be the schematic representation of the main system under consid-

eration with the force F0 sin CJ}tacting on it. A spring - mass (auxiliary) absorber system is attached to themain system as shown in Fig. 2.28. The equations of motion of the complete system can be written as:

MZ1 + KZI + Ko (ZI-ZZ) = Fa sin rot ...(2.140)

...(2.141)moZZ+Ka(ZI-ZZ) =0

The forced vibration solution will be of the form

ZI = Al sin rot

~ = Azsin rot

Substitution of Eqs. (2.142) and (2.143) in Eqs. (2.140) and (2.141) yields

Al (-M0}2 R 3 R Ka)-KaAZ = Fa

-KnAI + Az (-mnO}Z+ Kn) = 0

...(2.142)

...(2.143)

...(2.144)

(2.145)

PPV

PV VV PV UO PV PO




Subtituting:

Z2

ma

Absor ber

syst <zm

Ka

Z1

M

Main

syst<zm

Fig. 2.28 : Vibration absorber

F ZSl = -2... = Staticdeflectionof main systemK

2 Karo = - = Natural freqeuncy of the absorber

na ma

K .

ro~ = M = Natural frequency of mam system

m Ilm = -.Jl.. = Mass ratio = Absorber massIMain massM

The Eqs. (2.144) and (2.145) can be written as

(1

K{/ <i

)

K{/A I+--- -A 2 - =Z

K <.02 K SIn

A2 =

(1:~:

)$*

and

Solving Eqs. (2.146) and (2.147) for At and A2' we get

Cil 1-- 2

~t. =

(

L 2

)

'

(

; -roK ' no - 2

)

'K Si ro ro'

, <, n' " "--, ~-- -~, " .,".. ld"2. 1+ K" 2 K'

- rona ron

, J

.

.49

...(2.146)

...(2.147)

...(2.148)

I ".

i , ". .", .-, " -~t 'u'" ,.' ." ',,



50 SoU Dynamics & Machine Foundlltions

~~~

( ro')( ~..~' )

K

1- OO~a 1+ i - oo~ -~lithe natural frequency oonaof the absorber is chosen equal to 00 i.e. frequency of the excitatipn force,

it is evident from Eq. (2.148) that Al = 0 indicating that the main mass does not vibrate at all. Further Eq. (2.149) gives

...(2.149)

Az - -K Zst - Ka

or Az Ka = - K Zst ...(2.150)

Thus the absorber system vibrate in such a way that its spring force at all instmts is equal and

opposite to F0 sin 00 t. Hence, there is no net force acting on main mass M and the same therefore doesnot vibrate.

The addition of a vibration absorber to a main system is not much meaningful unless the mainsystem is operating at resonance or at least near it. Under these conditions, 00= oon' But for the absorber to be effective, 00should be equal to 00 .

na

Therefore, for the effectiveness of the absorber at the operating frequency corresponding to the natu-

ral frequency of the main system alone, we have

or oona = ,oon ...(2.151 a)

Ko = K

ma M...(2.151 b)

K mor -!L=-!!..=Ii ...(2.151 e)

K M t"'m

When the condition enumerated in Eqs. (2.151) is fulfilled, the absorber is known as a tunedabsorber. .

For a tuned absorber, Eqs. (2.148) and (2.149) become:

(

002

J

1- -

~,: ~ ( 00') ( OO~a00' J1- --y- 1+Jlm- --y- - Jlm

OOna OOna

...(2.152)

~: = ( 00') ( I 00' )1- --y- 1+ Jlm- --y- - Jlm

OO"a OOna

The denominators of Eqs. (2.152) and (2.153) are identical. At a value of 00when these denomina-

tors are zero the two masses have infinite amplitudes of vibration. Let when-00= oonl' the denominators

becomes zero. For this condition the expression for the denominators can be written as

..(2.153)

- .. .~ ..., ~- .-.--~.-




(

OOnt

)

4-<2+llm)(

OOnt

)2 +1 =0

OOna <Ona...{2.154)

The Eq. (2.154) is quadratic in <0;1' and therefore there are two values of oonl for which the denomi-

nators ofEqs. (2.152) and (2.153) become zero. These two frequencies are the natural frequencies of the

3ystem. Solution of Eq. (2.154) gives:

(:J ~ (1+~2m)~J~m+~4~

1.6

...(2.155)

1.2

_1.4

_

1

°c c

331.0

0.8

0.60 0.2 0.4 0.6 0.8

Mass ratio }.Im

Fig. 2.29: Natural frequency ratio versus mass ratio

The relationships of Eq. (2.155) is plotted in Fig. 2.29. From this plot, it is evident that greater the

mass ratio, greater is the spread between the two resonant frequencies. The frequencyresponse curve for

themainsystemisshownin Fig.2.30fora valueof ~'"=0.2. The dotted curves shown actually mean thatthe amplitude is negative or its phase difference with respect to the exciting force is 1800. It can be

noticed from this figure that by attaching a vibration absorber {oona= (On)to the main system vibrating

at resonance reduces its vibration to zero. Now if the exciting frequency is absolutely constant, the

system will work efficiently. Any change in the exciting frequency will shift the operating point from the

optimum point and the main system response will no longer be zero. It may be noted that by adding the

vibration absorber, we have introduced two resonant points instead of one in the original system. Now

I --:::w;-' -Hr.



52

!IIZ

Soil Dynamics & Machine Foundations

if the variation of the exciting frequency is such that the operating point shifts near one of the new

resonant points, then amplitudes will be excessive. Thus depending upon the variation of the exciting

frequencies the spread between the two resonant frequencies has to be decided to remain reasonably away

from the resonant points. After deciding the spread between the resonant frequencies, a proper value of

!J.mcan be chosen from the curve of Fig. 2.29. Undamped dynamic vibration absorbers are not suitable for

varying forcing frequency excitation. To make the vibration absorber effective over an extended range of

frequencies of the disturbing force, it is advantageous to introduce a damping device in the absorber

system. Such an absorber system is called a damped dynamic vibration absorber.

8

6

...

<I~ 4,

z

1.0

00

}Jm=0.20

\

'---2.0 2.51.50.5 1.0

G.)nl

TO)

The Eq. (2.154) can also be written as

Fig. 2.30 : Response versus frequency of a vibration absorber

~m J~)-lr ( )

2

(J) nl

(J)//(/

...(2.156)




t ILLUSTRATIVE EXAMPLEs!

Example 2.1The motion of a particle is representedby the equationz = 20 sin rot. Show the relative positions and

magnitudes of the displacement, velocity and acceleration vectors at time t = 0, and ro= 2.0 rad/s and0.5 rad/s.

Solution:Z = 20 sin rot

Z = 20 ro cos rot = 20 ro sin lrot + ~ )- 2 2Z =- 20 ro sin rot = 20 ro sin (w t + 1t)

The magnitudes of displacement, velocity and acceleration vectors are 10, 10 ro and 10 ro2 respec-

tively. The phase difference is such that the velocity vector leads the displacement vector by 1t/2 and the

acceleration vector leads the velocity vector by another 1t/2. Figures 2.31 a and 2.31 b show the three

vectors for ro= 2.0 and O.?Orad/s respectively. I 20 (V el.)

40 (Accln.)

10(oispl.)

(a) G..)= 2.0 rod/sec

s(Vel.)

. ~/z ".2.S(Acc!n.) 10(OlspL)

( b) CV = 0.5 rod I s e c

Fig. 2.31 : Vector diagram (Example 2.1)

21t 21tTime period = - = - = 1t s

ill 2for ro= 2.0 rad/s

21t 21tTime perIod = - = - = 41ts

ill (0.5) for ro = 0.5 rad/s

Example 2.2 A body performs,simultaneously,the motions

Zl (mm) = 20 sin 8.0 t

Z2 (mm) = 21 sin 8.5 t

Determine the maximwn an~ minimum )~1itude of the combuled motion, and the time period of the

periodic motion.

~~tm~f~dJ "T'f:ifrir" """"-";"~'.'.<f:;-:F;;':.;;,:Jijj~ .,.-,'!.,!,,~,~~

'111""



54 Soil Dynamics & Machine 'Foundations

Solution:

Z = 21 + 20 = 41 mmmax

Z . = 21 - 20 = 1 mmmm

The beat frequency is given by

8.5-8.0 0.5

1= 21t = 2 1t = 0.0795 H z, a nd

T = 2.. - 21tI - 0.5 = 41t = 12.57 s

Example 2.3

A mass of 20 kg when suspended from a spring, causes a static deflection of 20 mm. Find the natural

frequency of the system.

Solution:

Stiffness of the spring, K = W~\t

20 x 9.81 ::::104 N/mK = 20xl0-J

1[K Natural frequency,In = 21t V;

~ 1 ~1O421t 20 = 3.6 Hz.

KZ

Example 2.4

For the system shown in Fig. 2.32, determine the naturalfrequency of the system if

K1

Kt = 1000 N/m

Kz = 500 N/m

KJ = 2000 N/m

K4 = Ks = 750 N/m

Mass of the body = 5 kg

Solution:

Let Ket and Ke2represent respectively the effective stiffnesses

of the top three springs and the lower two springs, then

K3

1 1 1 1- = -+-+-Kel Kt Kz KJ

K4, KS

=~+~+~- .1000 500 2000-0.0035 ". 'Fig. 2.32:"MuHpriags system

~.....- -~_'".,-~._........._-_....-



.- ..... ... -Theory of Vibratimrs 55

Kel = 285.7 N/m

Ke2 = K4 + Ks = 750 + 750 = 1500 N/m

Now Kel and Ke2are two springs in parallel, therefore effective stiffness,

Ke = Kel + Ke2:;:: 285.7 + 1500 = 1785.7 N/m

f. ~ ~. /K = ~~1785.7 = 3.0 Hzn 21tV;; 21t 5.0Example 2.5

A vibrating system consists of a mass of 5 kg, a spring stiffnessof 5 N/mm and a dashpot with a damping

coefficient of 0.1 N-s/m. Determine (i) damping ratio and (ii) logarithmic decrement.

Solution:

(i) Cc = 2 ~km = 2J5x 10-3 x 5 = 0.319 N-s/m

J: C 0.1

~=C=0.319=0.313c

(in 27t~ - 27t.x0.313 = 2.07

Lograthimic decrement = ~ 1- ~2 - ~ 1- 0.3132

, ,

Zlog ::,..l = 2.07

eZ 2

~ = 7.92Z2

Therefore the free amplitude in the next cycle decreases by 7.92 times.

Example 2.6

A mass attached to a spring of stiffness of 5 N/mm has a viscous damping device. When the mass wasdisplaced and released, the period of vibration was found to be 2.0 s, and the ratio 01 the consecutive

amplitudes was 10/3. Determine the amplitude and phase angle when a force F = 3 sin 4 t acts on the

system. The unit of the force is Newton. .

Solution:

i.e.

or,

(ii)

. 27t~ ZI 10 .

1- ~2 = loge Z2 = loge 3 = 1.2

~ = 0.195

TII = 2.0 S

21t 21t

(lJ n = T = '2 = 3.14 radls(lJ = 4.0 rad/s

T1 ='~'= 4.0 = 1.273. ID 3.14n

.' i.:' F 3.0, Fo = 3.0N; AsI = -9..= - = 0.6mm.

. '.' .,', .;..",fi..,;'! "."K. 5.0 t:<,'..'""

(i)

..

'.

----., -=., -,.;".,'~ -""'~-.~ ~~,,~,..-'>,:;<..."".","..;g""~,,;,;~,,,,:.u ,.~~



56 Soil Dynamics & Machine' Foundations

From Eq. (2.58),A

Az = ~ 2

t . 2; Asl = Static Deflection

;

(1-11) +(2~11). " . . , '

= '. 0.6 , ~Q.755llll11

~(1-1.2732)2 +(2 x.0.195 x 1.273)2

T-I

(

211~

)T

-1'

(

2 x 1.273 x 0.195

)e = an --r = an 2 = 141.4°1-11 1-1.273

Example 2.7

Show that, in frequency - dependent excitation the damping factor ~is given by the followingexpres-S1On:

): -.: ..!.

(

12 ~ 11

J~ - 2 2/n

Where 11 and 12 frequencies at which the amplitUdeis 1/.J2 times the peak amplitude.Solution:

In a forced vibration test, the system is excited with constant force of excitation and varying frequencies.

A response curve as shown in Fig. 2.33 is obtained.

0.09

0.05

Amox = 0.084

0.08

E

E 0.07"C:I

"'0::J-

c. 0.06E

et

0.0410

. . n, ,

Fig. 2.33: Determination of viscous damping in forced vibrations by Bandwidth method

-.------

II

II I

I I IIII I

f1 I fn Ifl

14 18 22 24

F r (Zq u (Zn c y .,0f (Zxci to t ion, Hz



"

Th~o,.,of v~,!s57

At resonance, 11= 1 ana A~I Zst = 1/2~(for small values of ~). If the frequency ratio is T\when

amplitude of motion is 1/..[i times the peak amplitude, then fr~m Eq. 2.59, we get

I 1 1

.J2' 2~ = 4(1 :"112)2+4~2 ~2

or 114-2112(1 -2~2) + (1- 8~J) = 0

or 11~,2= ~[2(1-2~2):t~4(1-2~2)2_4(1-8~2)]

;, (1 - 2~2):f:2~~1+~2

11~-11i = 4~~1-~2 = 4~

1

-2

= Il- 112= (

12- 11

) (

12+ I.

)112 111 In2 In In

(

I - J;

)

1 + f = 2 2 . s ince 2 . =2

In In

Now [for small values of~]

Also

~ = !( Iz - 11

)2 In

This methodfor determiningviscousdamping is knownas the band width method.

Example 2.8 .

A machine of mass 100 kg is supported on springs of total stiffness of 784 N/mm. The machine produces

an unbalanced disturbing force of 392 N at a speed 50 c/s. Assuming a damping factor of 0.20, determine(i) the amplitude of motion due to unbalance,

(ii) the transmissibility, and

(iii) the transmitted force.

Therefore

'Solution:

( i), 184 x 103

(J)II = ~KI m = ..'/ " ==87.7 rad/s, V 100

, 00= 21t x 50 = 314 rad/s

Now

00 - ,314 = 3.58,'n =;., ~ 87.7" 'UJln; ,

, Fi) '- 392 '= 0.5 'innl

" ~st ~ ~ - , ;784 ,

, , ' ;. : :'?it .'. - 2

.Az ==4(~.~2l'+(2T\~)

-

= ,0.5

'~(1-'3582)2 +(2 x 3.58 x 0.2)2 = 0.042 mm

" " .'

~'\, ",', ',','. ':;' cc:

, .

..- - - .~ "'-; ~~~,~"F.---



58 Soil Dynamics & Machine Foulldations

~1+(211~)2

(ii) Transmissibility JlT = ~(1-n2)2 +(211~)2

- ~1+(2X3.58XO.2)2

- ~(1-3.582)2 +(2 x 3.58 x 0.2)2

==0.1467

(hi) Force transmitted = 392 x 0.1467 = 57.5 N.

Example 2.9

The rotor of a motor having mass 2 kg was running at a constant speed of 30 c/s with an eccentricity of

160 mm. The motor was mounted on an isolatorwith damping factor of 0.25. Determine the stiffness of the isolator spring such that 15% of the unbalanced force is transmitted to the foundation.Also ~eterminethe magnitude of the transmitted force. '.

Solution:

(i) Maximum force generated by the motOi

= 2 me eo:? = 7. x 2.0 x 0.16 x (21t x 30/ = 22716 N

= 22.72 kN

:ii)Force transmitted

Jl = = 0 15T unbalancedforce .

. ~1 +4112~2 .

I.e. ~ . = 0.15(1- 112)2+ (211~)2

or 1 + 4112x (0.25)2 = (0.15)2 [(1 -r 112)2+ (211 x 0.25)2]

or 114 - 12.84112 - 43.44 = 0

It gives(0

11= 3.95 i.e. ;- = 3.95"

(0 601t

(0" = .JK/ m = 3.95 = 3.95 = 47.7 rad/s

K = m (47.7)2 =:=2.0 x (47.7)2 = 4639 N/m

(iii) Force transmitted to the foundation

Therefore

= 0.15 x 22.72 = 3.4 kN.

Example 2.10

A seismic instrument with a natural frequency of 6Hz is used tomeasure the vibration of a machine

running at 12.9rpm. The in~trumentgives the reading for th~r~lative displacement of the seismic massas 0.05 mm. Determine the amplitudes of displacement: velo'city and acceleration of the vibrating ma-

chine. Neglect damping.



~;,:,~ .;,~

'/reory 01 v.ibrations59

';olution :

(i) CJJ = 6 Hz = 37.7 rad/sn

120 x 21t(J) = 120 rpm = =' 12 57 rad/s

60 .

'1 = 12.57 = 0.33337.7

1~ = - for ~= 0

1- ,,2 ,

1= = 1.125

1- (0.333)2

(ii) For displacement pickup, Eq. (2.88) gives2

X=,,~yo0.05 = (0.333)2 x 1.125 x Yo

or Yo = 0.40 mm

(iii) For velocity pickup, Eqo(2.91) gives'

or

1X = - 11~ (Y 00)

00 0n

0.333 x 1.125x (Yo 00)0.05 = (37.7)

or o. 0 (Y 000) = velocity = 5.03 ,mm/s

(iv) For acceleration pickup, Eq. (2.89) gives

X ='4 (YOOO2), OOn ' ,

I.e,

0.05 = 1.125 (Y 002 )(37.7)2 0

(Y() 002) = Acceleration = (37.7)2 x 0.05 = 63.17 mm/S21.125

or

Example 2.11

Determine the natural frequencies and mode shapes of the system represented by a mathematical mod~1

shown in Fig. 2.34 a.,/

. , co:" '1,,';

,- ~:::."'O'~,""""", ,~>=j"~



, i

J

60 Soil Dynamics & Machi'Je FOUlrilatiolls

j

+1

+ 1

(a) Two degrees freedom system (b) First mode (c) Second mode

Fig. 2.34 : Two degrees freedom system with mode shapes

Solution:

(i) The system shown in Fig. 2.34a is a two degree freedom system. The solution of such a system

has already been described in Art. 2.7. '

(ii) The two natural frequencies of the system can be obtained using Eq. (2.100) by putting KI = K,

Kz = 2 K and KJ = K, and m I = m2 = m. By doing this, we get

(02 = .!.

[(

3K + 3K )

_

{

4 ~ (2 K)2

}

\l2

)

= K III 2 m m ,m2 m

(02 = .!.r-l6K + 4K

]= 5.K

112 2 m, m , m,

Hence, COni= .JK/ m and CiJ~2?:, [sK/.m , " .,' '

(iii) The relative values of amplitudes Al and ~ 'for the two modes can be obtainedusing Eqs.(2.103) and (2.104). . . '

,}~~r .

,,"-;, ,:;. ..

' '-'--'

.~."



eory of Vtbtalions_/ 6i

A (1) K . 2K -1-= 2 = =+1A(l) K + K -m o:l K+2K-m x K/m2 1 2 1 nl

A~2) 2K

A (2) = K + 2 K - m x 5 K / m = - 12

. The mode shapes are shown in Fig. 2.34 band 2.34 c.

B:xample2.12

Determinethe natural frequencies and mode shapes of the system represented by the mathematical moqelshown in Fie. 2.35 a.

0.761

1.0

(a) Three degree freedom system (b) First mode (c) Second mode (d) Third mode

Fig. 2.35 : Three degrees freedom system with mod-e shapesSolution:

(i) Equations of motion for the three masses can be written as

m 21 + K 21 + 2 K (21 - 22) = 0

m 7..2 + 2 K (22 - 21) + K (22 - Z3) = 0

m 23'+ K (23 - 22) = 0For steadystate, the solutionswit be as

2t = At sin ront

~ =~ sin ron t

23 = A3 sin ront

...(2.157 a)

.~:(2.157 b)

...(2.157 c)

- '-,

.-"

" -;"-. ; -:-

.

...(2.158 a)

...(2.158 b)

...(2.1"58 c)

.1'~;-.

' -~ "" ~" ~- =' ~ : :: -. "' :: :: "" 'c .: :- - . :" _._-

~

.~- ~ --

'W;;i'i~!~:t'jj



'11,"

62 Soil Dynamics & Machine Follndations

substituting Eqs. (2.158) in Eqs. (2.157), we get

., .(3 K - 111oo~)Al - 2 K A2 = 0

PK3=T +(3K-moo~)A2-KA3'=O

- K A2 + (K - 111 co;) A3 = 0

For nontrivial solutions of 00/1in Eqs. (2.159)

...(2.159a),

...(2.159 b)

...(2.159 c)

...(2.160)

.," . IIIW-Pllttll1~ A -= !l Eg . (2.160) become as

~ K

or

...(2.161 a)

...(2.161 b)

Eg. (2.161 b) is cubic in A. The values of A are worked out as

A( = 0.238; A2= 1.637; and A) = 5.129

00/1( = .J0.238Kl m; CO/l2= .J1.637 Kl m; and 00/13= ~5.129 Kl mTherefore,

(ii) Egs. (2.159) in terms of A can be written as

(3 - A) Al - 2 A2 = 0

- 2 A I + (3 - A) A2 - A) = 0

- A2 + (1 - A) A3 = 0

A = 0.238

...(2.162a)

...(2.162 b)

...(2.162 c)For r mode:

Eg. (2.162 a) gives

" AI

(3 - 0.238) AI - 2 A2 = 0 or A = 0.7242

Eg. (2,162 h) gives

- 2 x 0.724 A2 -t-(3 - 0.238) Az - AJ = 0

A2 '

A = 0.7613

, ,.'

or

-

"

,' .1 " iI; -' ", (.. '

.,03K-11l00 -2K

/I

-2K "'K 2-K 1=0

-' -moo/l

0 -K 2 K-mcon

3-A -2 0

-2 3-A-1 I =0

0 -1 I-A

3 2f. - 7 A + lOA - 2 = 0



:," 'h, , .\


Assuming Al = a, A2 = 1.381 a and A) = 1.815 a

:. Al : A2 : A) = a : 1.381 a : 1.815 a

= 1: 1.381: 1.815

;::;0.551 : 0.761 : 1

Similarly,

For II mode; A. = 1.637

Al : A2 : A) = - 0.933: - 0.635 : 1, and

For III mode: A.= 5.129

Al : A2 : A) = 3.891 : - 4.14 : 1

The mode shapes are plotted in Figs. 2.35 b, 2.35 c and 2.35 d,

Example 2.13A small reciprocating machine weighs 50 kg and runs at a constant sp~~d of 6000 rpm. After it was

installed, it was found that the forcing frequency is very close to th~}latural frequency of the system.

What dynamic absorber should be added if the nearest natural frequency of the system should be at least

20 percent from the forcing frequency.

Solution:

(i) 21t N 2 T t x 6000 = 628 rad/sill = W "" 60

At the time of installation of machine,

Forcing frequency ==Natural frequency of system

~ = 6282K = m x 628

= 50 x 6282 = 201 x 105 N/m

Therefore,

or

(ii) Aner adding the vibration absorber to the system, the natural frequency becomes (1 :i:0.2) 628i.e. 753.6 rad/s or 502.4 rad/s

For tuned absorber:

ma Ka

M = K = JIM

Now from Eq. (2.156)

J(:,:J-If

JInl - ~)l;;:)

illnl = 0.80)//(/



.


2}

2

{(0.8) -1 = 0.2025- 2J-lm- 0.8

and when (Onl = 1.2(Ona

{(1.2)2 -1}2J-l = 2 = 0.134

m 1.2

Adopting the higher value of J-lm

Ka = 0.2025 x 201 x 105 = 40.7 x 105 N/m

ma = 0.2025 x 50 = 10.12 kg

PRACTICE PROBLEMS

2.1 A single degree (mass-spring-dashpot) system is subjected to a frequency dependent oscillatory

force (m eo (02sin (0 f). Proceeding from fundamentals, derive the expression of the amplitudeof the system.

2.2 'Presence of damping reduces the effectiveness of the isolation system'. Is this statement true?

If yes, explain with neat sketches. .

2.3 Give two methods of determining 'damping factor' of,a single degree freedom system.

2.4 Starting from fundamentals~explain the principles involved in the design of (i) Displacement

pickup, (if) Velocitypickup,and (iif) Accelerationpickup.Illustrateyour answerwithneatsketches.

2.5 Describe the principles involved in a 'tuned dynamic vibration absorber'. Illustrateyour answer with neat sketches. Discuss clearly its limitations.

2.6 A mass of 25 kg when suspended from a spring, causes a static deflection of 25 mm. Find the

natural frequency of the system. Ans. (20 rad/s)

2.7 A spring mas system (K\, m) has a natural frequency of f\. ~fa second spring of stiffness K2 is

attached in series with the first spring, the natural frequency becomes f\/2. Determine K2 in

terms ofK\. Ans. (K/3)

2.8 A mass of 5 kg is attached to the lower end of a spring whose upper end is fixed. The nawral

period of this system is 0.40s. Determine the natural period when a mass of 2.5 kg is attached

to the mid point of this spring with the upper and lower ends fixed. Ans. (0.14 sec)

2.9 Determine the differential equation of motion of the system shown in Fig. 2.36. The moment of

inertia of weight W about the point 0 is Jo' Show that th~ system becomes unstable when:

K.ab >-

- W

..

theory of vibrations - saran - soil dynamics and machine foundation

Documents