fundamental of vibration

7/23/2019 Fundamental of Vibration

1/17

EMC 4923

Mechanical Vibration

Outcome 3

Mechanical vibrations of two and multiple degree of freedom

systems

esson !

Learning outcomes

Analysis of a two degree of freedom system (FBD, equation of

motion, initial conditions, solution)


2/17

Vibrations of two degree of freedom systems

Systems that require two independent coordinates to describe their motion arecalled two degree of freedom systems !n general"umber of Degree of Freedom of the system # "umber of the masses in the

system $ number of possible typesof motion of each mass

!n such case, we get

%wo coupled di&erential equations'

%wo natural frequencies for harmonic solutions for each coordinate

!f we gie suitable initial ecitation, the system ibrates at one of the naturalfre"uencies

During free ibration at one of the natural frequencies, the amplitudes of the

two degrees of freedom (coordinates) are related in a speci*ed manner and thecon*guration is called a normal mode, principle mode, or natural mode ofvibrationA mode is a description of motion %here are arious +inds of modes, many withmodifying phrases, such as the st mode, -ndmode, a principle mode or acoupled mode . all describing a particular type of motionAt natural frequency, a ibrating system moes in a principle mode %he motionin which only one coordinate changes is called /a principle mode of motion0%he coordinates used to describe motion also describe the mode !f theamplitude of one mass is one unit of displacement, the mode is said to besimply in a normal or principle mode !t means that all parts of a system hae

same harmonic motion, with maimum displacements at identical times andmaimum elocities at still other identical times%he number of principle modes that eist, will correspond to the number ofdegrees of freedom

1 %hus a two degree of freedom system has two normal modes of ibrationcorresponding to two natural fre"uencies

1 !f we gie an arbitrary initial ecitation to the system, the resulting freeibration will be a superposition of the two normal modes of ibration

2oweer, if the system ibrates under the action of an eternal harmonic force,the resulting forced harmonic ibration ta+es place at the frequency of theapplied force

E#amples of two degree of freedom systems


3/17

$!

m

m-

$3

$2

-

- 3

-

2x

m-

+4 -

+-(- . )

22xm

1x

m

+

+-(- . )

11xm

%ree vibration analysis of an undamped system

&etermine the natural fre"uencies and relativeamplitudes of the following spring'mass systems with twodegrees of freedom(

5onsider a two degree freedom system with two masses ' three springs asshown below

Steps

Draw Free Body Diagram for each mass showing all the forcesdue to spring sti&ness as well as inertia force

- 6rite down the equation of motion for each mass applyingconditions of dynamic equilibrium

4 Assume S27 and substitutex# $ sin(t+) and

)t(sinXx 2 +=

8 Sole the resulting equations for amplitude ratio and naturalfrequencies

&raw %)& of each mass separately

%or mass *m1+ %or mass *m2+

-


4/17

,rite E"uation of motion for each mass seperately

11xm

+ k1x1 k2(x2 x1) = 0

22xm

+k2(x2 x1) + k3x2= 011xm

+ k1x1 k2x2+ k2x1=022xm

+k2x2 k2x1+ k3x2= 0

11xm

+(k1+k2) x1 k2x2 =022xm

k2x1+ (k2+k3)x2= 0

6e are interested in +nowing whether m1and m2can oscillate harmonically

with the same frequency

with the same phase angle 9

with di&erent amplitudes X1 and X2

!f motion is assumed to be in a principle mode, both generali:ed coordinates will

hae a harmonic motion of the same frequency, Assuming that it is possible to hae harmonic motion of m1and m2, we ta+e thesolutions to the aboe equations as follow;

x1= X1sin (t+ )

)t(sinXx 2

11 +=

x2= X2sin (t+ )

)t(sinXx 2

22 +=

Substituting the aboe two solutions into the *rst two equations, we hae;

11xm

+(k1+k2) x1 k2x2=0

0)t(sinXk)t(sinX)kk()t(sinXm 221212

11 =+++++

{ } 0)t(sinXkX)kk(m 221212

1 =+++ EQ# 1

22xm k2x1+ (k2+k3)x2= 0

[ ] [ ] [ ] 0)t(sinX)kk()t(sinXk)t(sinXm 23212222 =+++++

0)t(sinX)kk(mXk 2322

212 =++++ EQ# 2

Since the aboe equations


5/17

which represents two simultaneous homogeneous algebraic equations in theun+nowns $and $-!n each equation, the fraction /$-?$0 which is the ratio of amplitudes can be

determined in terms of system constants and 2 Amplitude @atio also de*nesthe modes of ibration

{ } 0XkX)kk(m 221212

1 =++

2

121

2

2

1

mkkk

XX

+=

{ } 0X)kk(mXk 2322

212 =+++

2

2

232

2

1

k

mkk

X

X +=

@

2

232

2

2

2121

1

2

mkkk

kmkk

XX

+=+=

quating the aboe two equations, $-' $can be eliminated

2

2

121

2

232

2

k

mkk

mkk

k

+=

+

Soling the aboe equation gies

( ) ( )[ ] ( )

( ) ( )0

mm

kkkkkk

m

kk

m

kk

0kkkkkkmkkmkkmm

21

1332212

2

32

1

214

133221

2

132221

4

21

=+++

+++

=++++++

Alternately, soling by using determinants, for a nonC:ero solution of $and $-,the determinant of coecients of $and $- must be :ero

det[m12+ (k1+k2) k2k

2 m

2

2+(k2+k3 )]=0m

1m

2

4[ (k1+k2 )m2+(k2+k3 )m1 ]2+[ (k1+k2 ) (k2+k3 )k22 ]=0

%he aboe equation is called the fre"uency or characteristic e"uationbecause solution of this equation yields the frequencies of the characteristicalues of the system %he roots of the aboe equation are gien by;

8


6/17

1

2,

2

2=1

2 [(k1+k

2 )m2+ (k2+k3)m1m

1m

2 ] 12 {[(k1+k2 )m2+(k2+k3 )m1m

1m

2 ]

2

4[(k1+k2) (k2+k3 )k22

m1m

2 ]}

1 /2

%his shows that it is possible for the system to hae a nonC:ero harmonicsolution of the form

x1= X1sin (t+ )

x2= X2sin (t+ )

when=

1 and=

2 gien by the rule stated aboe

6e shall denote the alues of $and $-corresponding to

1 as X1(1)

and

X2(1)

and those corresponding to

2 as X1(2)

and X2(2)

( )

( ) 2

1232

2

2

2

1121

1

1

1

2

mkk

k

k

mkk

X

X

+

=+

=

( )

( ) 22232

2

2

2

2121

2

1

2

2

mkk

k

k

mkk

X

X

+=

+=

( ) ( )0

mm

kkkkkk

m

kk

m

kk

21

1332212

2

32

1

214 =++

+

++

+

%re"uency or characteristice"uation

%he roots (

2

2,1

) of the aboe equation which is quadratic can be determinedusing the formula

0cbxaxequationquadratictoapplicablea2

ac4bbx 2

2

2,1 =++

=

( ) ( )

21

133221

2

32

1

21

mm

kkkkkkc

m

kk

m

kkb

1a

where

++=

++

+=

=

E#ample-

E


7/17

Find the natural frequencies and mode shapes of a spring

mass system, which is constrained to moe in the ertical

direction

.olution

m1=m

2=m;k

1=k

2=k

3=k

%he equation of the natural frequencies is

( ) ( )

0m

k3

m

k4

0m

kkk

m

k2

m

k2

0mm

kkkkkk

m

kk

m

kk

2

224

2

22224

21

1332212

2

32

1

214

=+

=++

+

+

=++

+

++

+

sing quadratic formula to *nd the roots (

2

2,1

) for the aboe equation which

*nally gies the natural frequencies 1& 2

a # , b # . 4k/m ' c # 3k2/m2

m

k

m

k3

m

k,

m

k3

m

k2

m

k4

2

1

2

m

k12

m

k16

m

k4

2

m

k3*1*4

m

k4

m

k4

2

1

2

2

2

2

2

22

2

2,1

=

=

=

=

=

=

=

G


8/17

%he amplitude ratios are gien by;

( )

( )

1

m

k3mk2

k

k

m

k3mk2

m

k3when

mkk

k

k

mkk

X

X 22

1232

2

2

2

1121

1

1

1

2

=

=

=

=+

=+

=

( )

( )

1

m

kmk2

k

k

m

kmk2

m

kwhen

mkk

k

k

mkk

X

X 22

2232

2

2

2

2121

2

1

2

2

=

=

=

=+

=+

=

Hhysically the aboe results meanx2 'x1are in phase and equal for the *rstmode, whateer their absolute alues!n the second mode, they are equal but out of phase

%he natural modes are gien by

I


9/17

$!

m

m-

$3

$2

-

- 3

ercis Hroblem

Determine the two natural frequencies and two modes of ibration for the

system shown;

m1= m, m2= 2m

k1= k, k2= k3= 2k

J


10/17

$!

m

m-

$2

-

- 3

$!

m,

m-

$2

),

)-

)-3),

Answer:

( )

( )

( )

( ) 2

1

X

X,1

X

X

m

k2,

m

k

2

1

2

2

1

1

1

2

21

==

==

/ssignment 0roblem ! (To be submitted in the next class session)

Determine the natural frequencies and relatie amplitudes of the following

springCmass systems with two degrees of freedom

Show all the wor+ . should include the following;

i FBD>s of the two massesii quation of motion of each mass using equation of dynamic equilibriumiii Deriation of amplitude ratios (7odes of ibration)i Deriation of frequency equaion

/ssignment 0roblem 2A large car manufacturer company analy:ed the problem of a car by ta+ing an

entire car apart By weighing each section, the following alues of equialentmasses and spring constant were found;

m1, ale mass # JK +g

m2, body mass # GIK +g

Springs, equialent k1# E4J "?mm

%ires, equialent k2 # 8EE "?mm

Determine the two natural frequencies and two modes of ibration foraboe car!

(2int;C mathematical model will be as shown by FBD)


11/17

E#ercise !Find the natural frequencies and mode shapes of a springCmass system that is constrained to moe in a erticaldirection (m#4+g, +#-KKK "?m)Find the solution for the following set of initial conditionsx

1(0 )=0.01m; x

1(0 )=0

x2(0 )=0.02m; x

2(0 )=0.1m /s

K


12/17

Answer;

1=15.958

rad

s ;

2=41.778

rad

s ;

First mode; x1(1) (t)=X

1

(1 )cos (15.958 t+ 1 ) M x2

(1) ( t)=1.618X1

(1)cos (15.958 t+ 1) M

Second mode; x1(2) ( t)=X

1

(2 )cos (41.778 t+ 2 ) M x2

(2) ( t)=0.618X1

(2)cos (41.778t+ 2 ) M

@esponses; x1(t)=0.012cos (15.958 t0.235 )0.002 cos (15.958 t0.56 ) M

x2 ( t)=0.019cos (41.778 t0.235 )0.0012cos (41.778 t0.56 ) M

E#ercise 2

An oerhead traeling crane can be modeled as indicated in the *gure %hegirder has a span L#8K m, an area moment of inertia ! # KK- m8, and a

modulus of elasticity # -KG K"?m-, a sti&ness +#48

EI

L3 %he trolley

has the mass m# KKK +g, the load being lifted has a mass m-# EKKK +g, andthe cable through the mass m-is lifted has a sti&ness + # 4 K

E "?m


13/17

a) Determine the natural frequencies and themode shapes of the system

b) For the initial conditionsx

1(0 )=0m; x

1(0 )=0

M

x2(0 )=0.02m; x

2(0 )=0.1m/s

*nd the responses of the system

Answer;

1=7.389

rad

s ;

2=58.27

rad

s ;

First mode;

x1

(1) (t)=X1

(1 )cos (7.389 t+ 1 )

M

x2

(1) ( t)=11.118X1

(1)cos (7.389 t+ 1 )

M

Second mode; x1(2) ( t)=X

1

(2 )cos (58.27 t+ 2 ) M x2

(2) ( t)=0.018 X1

(2)cos (58.27 t+ 2 ) M

@esponses;x

1(t)=0.0021cos (7.389t0.595 )0.0018cos (7.389 t0.086 )m

M

x2( t)=0.024cos (58.27 t0.595)+0.000032cos (58.27 t0.086 ) mM

-


14/17

E#ercise !Find the natural frequencies and mode shapes of a springCmass system that is constrained to moe in a erticaldirection (m#4+g, +#-KKK "?m)Find the solution for the following set of initial conditionsx

1(0 )=0.01m; x

1(0 )=0

x2(0 )=0.02m; x

2(0 )=0.1m /s

4


15/17

First mode; x1(1) (t)=X

1

(1 )cos (15.958 t+ 1 ) M x2

(1) ( t)=1.618 X1

(1)cos (15.958 t+ 1) M

Second mode; x1(2)

( t)=X1(2 )cos

(41.778

t+ 2 ) M x2(2)

( t)=0.618

X1(2)cos

(41.778

t+ 2 ) M

@esponses; x1 (t)=0.012cos (15.958 t0.235 )0.002 cos (15.958 t0.56 ) M

x2( t)=0.019cos (41.778 t0.235 )0.0012cos (41.778 t0.56 ) M

8


16/17

E#ercise 2

An oerhead traeling crane can be modeled as indicated in the *gure %hegirder has a span L#8K m, an area moment of inertia ! # KK- m8, and a

modulus of elasticity # -KG K"?m-, a sti&ness

+#48

EI

L3 %he trolley has the mass m# KKK +g,

the load being lifted has a mass m-# EKKK +g, andthe cable through the mass m-is lifted has a sti&ness+ # 4 KE "?m

c) Determine the natural frequencies and the modeshapes of the system

d) For the initial conditionsx

1(0 )=0m; x

1(0 )=0

M

x2(0 )=0.02m; x

2(0 )=0.1m/s

*nd the responses of the system

E


17/17

@esponses; x1 (t)=0.0021cos (7.389t0.595 )0.0018cos (7.389 t0.086 )m M

x2 ( t)=0.024cos (58.27 t0.595)+0.000032cos (58.27 t0.086 ) mM

fundamental of vibration

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