dynamics and vibration, wahab,2008-ch6

Dynamics & Vibration

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© John Wiley & Sons, Ltd

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A harmonic motion is a periodic motion that varies in a sinusoidal way.

Displacement

Velocity

Acceleration

Simple harmonic motion

tAx nsin

tAdt

dxx nn cos.

tAdt

xdx nn sin2

2

2..

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0 2 4 6 8 10 12 14

Time (sec)D

isp

lace

men

t (m

)

Chapter 6: Free Vibration Of A Single Degree Of Freedom System (page 266)


A cycle of vibration: is a measure of the motion of a vibrating system, from equilibrium (x=0) to a maximum, then to equilibrium again, then to a minimum and back to equilibrium.

Amplitude of vibration: is the maximum displacement of a vibrating system from its equilibrium position.

A period of oscillation: is the time taken to complete one cycle

A frequency of oscillation: is the number of cycles per unit time and is calculated as the inverse of the period of oscillation.

Definition

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0 2 4 6 8 10 12 14

Time (sec)

Dis

pla

cem

ent

(m)


A natural frequency of a system: is the frequency at which the system is left to vibrate freely without the action of any external forces.

A phase angle of two harmonic motions: is the shift of the maximum values between two harmonic motions.

A Single Degree Of freedom (SDOF) system is the simplest vibrating system, which in general consists of a mass, a spring and a damper

Definition

x

k

c

f

m


Free body diagram

Undamped free vibrationEquation of motion (page 268)

+x

kxm

x

k

m

..

xmFx ..

xmkx

Applying Newton’s 2nd Law

Rearrange

0..

kxxm 02..

xx n

Where the angular frequency of the system in rad/s is

m

kn


Free body diagram

Undamped free vibrationEquation of motion

Applying Newton’s 2nd Law

From static equilibrium

Substituting in the equation of motion

x

k

m

mg

)( oxk

+x

m

mg

ok

+x

m

Static equilibrium

..

)( xmxkmg o

0 okmg o is the static deflection.

0....

kxxmxmkx

Which is the same equation of motion as before


Undamped free vibrationEquation of motion using Energy method (page 270)

Total Energy in the general position

The total energy becomes

The static deflection is

Which is again the same equation of motion as before

x

k

m

k

m

(a) Equilibrium position (b) General

position

Datum

222.

2

1)(

2

1

2

1oo kxkmgxxmSEPEKE

22.

2

1

2

1kxxmSEPEKE

kmg

o

Conservation of energy

.333222111 constSEPEKESEPEKESEPEKE

02

1

2

1 22.

kxxm

dt

d00

...... kxxmxxkxxm


The solution of the equation of motion is:

Time response (page 271)

)0( 2..

xx n

tBtAx nn sincos

Where A and B are constants

oxx oxx..

Initial conditions: at t=0,

tBtAx nnnn cossin

.

AxBAx oo 0sin0cos

n

o

nonnox

BBxBAx

.

..

0cos0sin

tx

txx nn

o

no

sincos

.

Which gives a final form of:


The solution of the equation of motion is:

Time response)0( 2

..

xx n

tBtAx nn sincos

Where A and B are constants

oxx oxx..

Initial conditions: at t=0,

tBtAx nnnn cossin

.

AxBAx oo 0sin0cos

n

o

nonnox

BBxBAx

.

..

0cos0sin

tx

txx nn

o

no

sincos

.

Which gives a final form of: )sin( tCx nOr

2.

2

n

o

o

xxC

o

no

n

o

o

x

x

x

x.

1.

1 tantan

Where


Time response

C is the maximum amplitude of vibration and

is the phase angle between the initial displacement and initial velocity

.

-1.5

-1

-0.5

0

0.5

1

1.5

0 0.2 0.4 0.6 0.8 1 1.2 1.4

Time (sec)

Am

plitu

de (m

m)

nn

2

nn

2 = period of motion in seconds (s)

2

1 n

nnf = natural frequency in Hz

(1 Hz = 1 cycle per second)


Example 6.1: Undamped free vibration: wind turbine (page 274)

.

Wind turbine

m

k54m

Tower cross-section

0.3m0.5m

The rotor and hub mass is 15 tons

The tower is made of steel that has a Young’s modulus of 200 GPa.

Ignoring the tower own weight, determine:

a) Natural Frequency

b) The time response due to initial transverse displacement c) The maximum values of velocity and acceleration.

Solution:

The moment of inertia I is:

434444 m1067035.2)3.05.0(64

)(64

io DDI


Example 6.1: Undamped free vibration: wind turbine

.

Wind turbine

m

k54m

Tower cross-section

0.3m0.5m

N/m1.1017554

1067035.210200333

39

3

L

EIk

The stiffness k of a cantilever beam is:

3

3

L

EI

u

Fk

rad/s 8236.01015

1.101753

m

kn

Hz131.02

nnf

The time response is: )sin( tCx n

m1.001.0 22

2.

2

n

o

o

xxC


Example 6.1: Undamped free vibration: wind turbine

.

20

1.0tantan 1

.1

n

n

o

o

x

x

)2

8236.0sin(1.0

tx

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0 2 4 6 8 10 12 14

Time (sec)

Dis

pla

ce

me

nt

(m)

)2

8236.0cos(8236.01.0)cos(. ttCx nn

Velocity

Maximum velocity

m/s082.08236.01.0max

.

nCx Acceleration

)2

8236.0sin(8236.01.0)sin( 22.. ttCx nn

Maximum acceleration 222

max

..

m/s068.08236.01.0 nCx


Example 6.2: Undamped free vibration : lift (page 277)

Natural frequency of the cage should not exceed 2.5 Hz x

E=200GPa

m=1100kg

L=50mA lift cage has a mass of 1100 kg

Starts its motion with an initial velocity of 3 m/s

.

Cable made of steel

Determine the minimum cross-section area of the cable and the amplitude of vibration

Solution:

A

P ELo

The stiffness k is

AA

L

EAPk

L

E

A

P

o

o 99

10450

10200

)(o

Pk


Example 6.2: Undamped free vibration : lift

x

E=200GPa

m=1100kg

L=50m

.

69

1085.671100

10425.2

A

A

m

kn m2=67.85mm2

)sin( tCx n

time response

m19.025.2

30

22

2.2

n

oo

xxC

the maximum amplitude


A pendulum (page 280)

mg

o

sinmg cosmg

L

+x

+y

The moment equivalent Newton’s 2nd law of motion ..

oo IM oI is the mass moment of inertia about ‘o’

..

Is the angular acceleration..

0sin ImgL

Rearranging and assuming small oscillation angle

0..

mgLIo Or 02..

n

Where

on I

mgL is the angular frequency in rad/s.


A pendulum

mg

o

sinmg cosmg

L

+x

+y

The mass moment of inertia is 2

go mRI

where gR is the radius of Gyration

For a simple pendulum LRg

L

g

L

gLn

2

22gg

nR

gL

mR

mgL


A hinged pin support at B

A mass m is attached at C

Write the equation of motion for the bell crank for small oscillation angle

Solution:

k

l2

l1

m

A

B

C

kxo

l2

l1

mg

A

Rx

C

Ry

B

Static equilibrium

From static equilibrium

012 lklmgM oB

1

2

l

l

k

mgo

Example 6.5: Undamped free vibration: bell crank (page 282)


Example 6.5: Undamped free vibration: bell crank

l2

l1

A C

Vibrating system

l2

l1

mg

A

Rx

C

Ry

B

Free body diagram

)( 1 olk

From the free body diagram ..

112 )( BoB IllklmgM

..

11

212 )( BIl

l

l

k

mglklmg

Using 1

2

l

l

k

mgo

02

1..

BI

kl

Rearranging

2

2mlIB Where

m

k

l

l

l

l

m

knn

2

1

2

2

1..2

0

22

21

..

ml

kl


A rotor on a fixed shaft (page 283)

k,G,IP

L

T,

J

The moment equation equivalent to Newton’s 2nd law of motion ..

JT T

..

Where is the sum of the torques

is the torsional angular acceleration.

J is the mass moment of inertia of the rotor (or disc)

L

GITk p

From static torsional analysis, the stiffness k is given by

Where G is the shear modulus of the shaft material and Ip is the polar moment of inertia of the shaft.


A rotor on a fixed shaft

k,G,IP

L

T,

J

Where

The sum of the torques is ..

Jk

0..

kJ

Or

02..

n

J

kn is the angular frequency in rad/s.

LJ

GI pn

L

GIk pUsing gives


Example 6.6: Undamped free vibration:torsional vibration (page 284)

4m

BrakeRotor

The rotor has a mass of 12 tons and a radius of gyration of 18m

A uniform hollow shaft of inner and outer diameters 0.2 m and 0.4 m

A brake is installed at a distance of 4m behind the rotor

Determine the period of free torsional vibration of the rotor when the break is stopping the rotor rotation

Solution:

The polar moment of inertia of the shaft is

434444 m10356.2)2.04.0(32

)(32

iop DDI


Example 6.6: Undamped free vibration:torsional vibration

4m

BrakeRotor

The rotor mass moment of inertia J is

38880001812000 22 mRJ kg.m2

The angular frequency is

415.338880004

10356.21077 39

LJ

GI pn rad/s

The period of free vibration is:

84.1415.3

22 n

n s


Viscous Damped Free VibrationEquation of motion (page 286)

The damping force produced by a viscous damper is equal to

.

xc

where c is the viscous damping coefficient in N.s/m.

+x

kx

.

xc

Free body diagram

From the free body diagram, Newton’s 2nd law is: ...

xmxckx

Rearranging

0...

kxxcxm Or 02 2...

xxx nn

Wherem

kn and is the damping ratio and is equal to:

nm

c

2

Defining the critical damping coefficient, cc, as nc mc 2

becomescc

c

x

k

m

c


Time response (page 287)

+x

kx

.

xc

Free body diagram

The solution of the equation of motion is: tAex

Differentiating twice with respect to time and substitute in the equation of motion

02 22 nn

The roots are

)1( ),1( 22

21 nn

By adding the two solutions together

tt eAeAx 2121

Or tt nn eAeAx )1(2

)1(1

22

x

k

m

c


Time response

tt nn eAeAx )1(2

)1(1

22 There are three categories of damped motion that can be defined a) Overdamped system 1

b) Critically damped system 1

c) Underdamped system 1

x

k

m

c


Overdamped system (page 287)

1

1 2 and are real negative numbers

The time response will be in the form atAex

where a is a positive real number

tt nn eAeAx )1(2

)1(1

22

oxx oxx..

Applying initial conditions at t=0, and

21 AAxo

22

12

.

)1()1( AAx nno


Overdamped system

Solving for A1 and A2 and substituting in the displacement time response equation

t

n

noot

n

noonn e

xxe

xxx

)1(

2

2.

)1(

2

2.

22

12

)1(

12

)1(

Displacement time response of an overdamped system

0

0.2

0.4

0.6

0.8

1

1.2

0 2 4 6 8 10 12

Time (sec)

Am

plit

ud

e (

mm

)

25

20The higher the damping ratio , the longer the time to decay

12

For heavily overdamped system, is much larger than 1,

)2

1(2

2

..

t

n

o

n

oo

nexx

xx

The displacement time response becomes


Critically damped system (page 289)

The displacement time response is the form

0

0.2

0.4

0.6

0.8

1

1.2

0 1 2 3 4 5 6

Time (sec)

Am

plit

ud

e (

mm

)

1

n 21

tnetAAx )( 21

oxx oxx..


1Axo

12

.

AAx no


tonoo

netxxxx ))((.


Underdamped system (page 290)

1 012 and 22 11 i where 1i

The time response will be in the form baba eee Using the relationship

ttiti nnn eeAeAx

22 12

11

Defining the damped natural frequency d21 nd

The time response becomes

ttiti ndd eeAeAx 21

Using Euler formula xixe ix sincos

tdddd

netitAtitAx )sin(cos)sin(cos 21

tt nn eAeAx )1(2

)1(1

22


Underdamped system

oxx oxx..


Or tdd

netAtAx sincos 43

Where A3 and A4 are new constants

Another form for the equation

)sin( tDex dtn

Where 24

23 AAD

4

31tanA

A

3Axo

34

.

AAx ndo


Underdamped system


td

d

ono

donet

xxtxx

sincos

.

The period of motion in seconds is defined as

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6

Time (s)

Dis

pla

cem

ent

(mm

)

1x2x

dd

2

)sin( tDex dtn

dd

2

The ratio of two successive amplitudes

))(sin(

)sin(

1)(

1

2

1

1

1

ddt

dt

tDe

tDe

x

xdn

n

dnex

x 2

1


Underdamped system

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6

Time (s)

Dis

pla

cem

ent

(mm

)

1x2x

dd

2

)sin( tDex dtn

Defining a logarithmic decrement

dnx

x 2

1ln

dd

2Using and

21 nd

22)2(


Example 6.7: Damped free vibration I (page 293)

x

k

m

c

Mass = 300 kg

Damped period of vibration is 1.2s

The ratio between the two maximum amplitudes in two successive cycles is

15

1

1

2 x

x

Determine:a)The logarithmic decrement.b)The damping ratio.c)The damped angular frequency.d)The undamped angular frequency.e)The critical damping constant.f)The damping coefficient.g)The stiffness.


Example 6.7: Damped free vibration I

x

k

m

c

a) The logarithmic decrement

708.215lnln2

1 x

x

b) The damping ratio

3958.0708.2)2(

708.2

)2( 2222

c) The damped angular frequency.

236.5667.12.1

22.1

2

dd

d rad/s

d) The undamped angular frequency

7.53958.01

667.1

1 22

d

n rad/s



x

k

m

c

f) The damping coefficient.

e) The critical damping constant.

N.s/m34207.530022 nc mc

N.s/m64.135334203958.0 ccc

g) The stiffness k

N/m97477.5300 22 nn mkm

k



0 1 2 3 4 5 6

Time

Dis

pla

cem

ent

1x

2x3x

Damped natural frequency is 1 Hz

The ratio between its maximum amplitude in the first cycle and that after two cycles is

81

3x

x

Find the damping ratio

Solution:

))2(sin(

)sin(

1)2(

1

3

1

1

1

ddt

dt

tDe

tDe

x

xdn

n

dnex

x 2

3

1 8ln2ln3

1 dnx

x

As dn 03972.12

8lnthus

The damping ratio is 163.003972.1)2(

03972.1

)2( 2222


dynamics and vibration, wahab,2008-ch6

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period of motion

periodic motion

ltda harmonic motion

maximum amplitude of

maximum displacement

maximum values of velocity