Chapter 2 Free Vibration of Single Degree of Freedom

Free vibration of single degree

of freedom (SDOF)Chapter 2

Introduction

• A system is said to undergo free vibration when it oscillates only under an initial disturbance with no external forces acting after the initial disturbance

Introduction - SDOF

• One coordinate (x) is sufficient to specify the position of the mass at any time

• There is no external force applied to the mass

• Since there is no element that causes dissipation of energy during the motion of the mass, the amplitude of motion remains constant with time, undamped system

Introduction - SDOF

• If the amplitude of the free vibration diminished gradually over time due to the resistance the resistance offered by the surrounding medium, the system are said to be damped

• Examples: oscillations of the pendulum of a grandfather clock, the vertical oscillatory motion felt by a bicyclist after hitting a road bump, and the swing of a child on a swing under an initial push

Free Vibration of an Undamped

Translation System• Equation of Motion using Newton’s Second Law

▫ Select a suitable coordinate to describe the position of the mass or rigid body

▫ Determine the static equilibrium configuration of the system and measure the displacement of the mass or rigid body

▫ Draw the free body diagram of the mass or rigid body when a positive displacement and velocity are given

▫ Apply Newton’s second law of motion

FV of an undamped …

• Newton’s second law

• Applied to a undamped SDOF system

xmtF

)(

JtM )( For rigid body undergoing rotational motion


xmkxtF

)(

0 kxxm


• Equation of Motion using other methods

▫ D’Alembert’s Principle

▫ Principle of Virtual Displacements

▫ Principle of Conservation of Energy

0 kxxm

• Spring-Mass System in Vertical Position

0 kxxm

stkmgW

Wxkxm st

FV of an undamped …Solution!

• The solution can be found assuming,

substituting

stCetx )(

0)()(2

2

stst CekCedt

dm 02 kmsC

nim

ks

21

21

m

kn

characteristic equation eigenvalues

02 kms


• The general solution,

titi nn eCeCtx

21)(

tite ti sincos using,

tAtAtx nn sincos)( 21

A and C can be determine from the initial conditions


• The initial conditions at t =0,

• The solution is

01)( xAtx

tx

txtx n

n

n

sincos)( 00

02)( xAtx n

FV of an undamped …Harmonic

• Previous equations are harmonic function of time

• The motion is symmetric about the equilibrium position of the mass

• The velocity is maximum and the acceleration is zero at the equilibrium position

• At the extreme displacement the velocity is zero and the acceleration is a maximum

• The quantity is the natural frequency n


sin2 AA cos1 AA

amplitudex

xAAAn

2/12

02

0

2

2

2

1

21

anglephasex

x

A

A

n

0

01

1

21 tantan

tAtAtx nn sincos)( 21


2/12

02

00

n

xxA

0

01

0 tanx

x n

tAtx ncos)(

Substituting, the solution can be written

using the relation,001 cosAA 002 sinAA

00 sin)( tAtx n

• Obtain the free response of

a) in the form

b) in the form

Initial condition are x(0) = 0.05 m and x(0) = -0.3 m/s

Example - Harmonic

)(1282 tfxx

tAtAx nn cossin 21

tAx nsin

Example - solution

ttttxa 8cos05.08sin0375.08cos05.08sin8

3.0)

062.08

3.005.0)

2

22

Ab

806.0062.0

05.0sin 605.0

)8(062.0

3.0cos

rad214.20927.927.)333.1(tan605.0

806.0tan 11

mtx 214.28sin062.0


2/1

m

kn

1)Note the following aspect,

stst

mgWk

2/1

st

n

g

2/1

2

1

st

n

gf

2/1

21

gf

st

n

n


tAtx ncos)(

2cossin)(

tAtAtx nnnn

tAtAtx nnnn coscos)( 22

2)


3) If the initial displacement is zero,

tx

tAx

tx n

n

n

n

sin

2cos)( 00

If the initial velocity is zero,

txtx ncos)( 0

4)The response of a SDOF system can be represent in the displacement ( ) – velocity ( ), known as the state space or phase plane


xx

tAtx ncos)(

tAtx nn sin)(

A

x

A

y

12

2

2

2

A

y

A

x

• The column of the water tank shown is 300 ft. high and is made of reinforced concrete with a tubular cross section of inner diameter 8 ft. and outer diameter 10 ft. The tank weighs 6 x 105 lb with water. By neglecting the mass of the column assuming the Young’s modulus of reinforce concrete as 4 x 106 psi, determine the following:

Example - Harmonic

• the natural frequency and the natural time period of transverse vibration of the water tank

• the vibration response of the water tank due to an initial transverse displacement of 10 in.

• the maximum values of the velocity and acceleration experienced by the water tank

Example

Example – solution

3

3

l

EIPk

a) The stiffness of the beam,

l= 3600 in. , E = 4 x 106 psi, 444444

0 10600961206464

inddI i

inlbl

EIk /1545

3600

)10600)(104(333

46

3

sec/9977.0106

4.38615455

radm

kn

sec2977.62

n

n


b) Using the initial displacement of x0 =0 and the initial velocity =0,

00 sin)( tAtx n

inxx

xAn

100

2/12

02

00

20tan 01

0

nx

tttx 9977.0cos102

9977.0sin10)(


c) The velocity and acceleration can be determine by differentiating,

29977.0cos9977.010)(

ttx

sec/977.99977.0100max inAx n

29977.0sin9977.010)(

2 ttx

222

0max sec/9540.99977.010 inAx n

• A simply supported beam of square cross section 5 mm x 5 mm and length 1 m, carrying a mass of 2.3 kg at the middle, is found have a natural frequency of transverse vibration of 30 rad/s. Determine the Young’s modulus of elasticity of the beam.

Example - Harmonic


3

192

l

EIk

41033 105208.010510512

1mI

m

kn

I

lmE n

192

32

2

3

192nm

l

EIk

29

10

32

/100132.207)105208.0(192

0.10.303.2mNE

• An industrial press is mounted on a rubber pad to isolated it from its foundation . If the rubber pad is compressed 5 mm by the self-weight of the press, find the natural frequency of the system

Example - Harmonic

mst

3105

Hzradm

g

st

n 0497.7sec/2945.44105

81.92/1

3

2/1

• An air-conditioning chiller unit weighing 2,000 lb is to be supported by four air springs. Design the air springs such that the natural frequency of vibration of the unit lies between 5 rad/s and 10 rad/s

Example - Harmonic


sradn /5.7

4.386

2000m

m

keq

n

inlbmk neq /1491.2915.74.386

2000 22

inlbkinlbk /78.72/1491.2914

Example

• An electrical switch gear is supported by a crane through a steel cable of length 4 m and diameter 0.01 m. If the natural time period of the axial vibration of the switch gear is found to be 0.01 s, find the mass of the switch gear


mN /)10(064.4)10(07.201.044

1 6112

l

AEk stiffness cable

nn

nf

211.0

m

kn

20

1.0

2

kgk

mn

53.1029)20(

)10(0644.42

6

2

Example

• A bungee jumper weighing 160 lb ties one end of a elastic rope of length 200 ft and stiffness 10 lb/in to bridge and the other to himself and jumps from the bridge. Assuming the bridge to be rigid, determine the vibratory motion of the jumper about his static equilibrium position


sec/88.361,1

)12(200)4.386(22mgh

jumper theofvelocity

2

21

in

ghvormv

inlbkinch

lbm /10,

sec

4.386

160 2

88.362,1)0(,0)0(

:position mequilibriu about the

00 txxtxx

)sin()(:jumper theof response 00 tAtx n


ink

mxxxx

nn

12.2774.386

160

10

1361A

where,

00

2

2

00

21

0tan

where,

01

0

n

n

x

x

Free Vibration of an Undamped

Torsional System• If a rigid body oscillate about a specific reference

axis, the resulting motion is called torsionalvibration

• The displacement of the body is measured in terms of a angular coordinate

• The restoring moment may be due to the torsion of an elastic member or to the unbalanced moment of a force or couple

l

GIM o

t

32

4dIo

l

Gd

l

GIMk ot

t32

4

• Equation of Motion▫ The equation of the angular

motion of the disc about its axis can be derived by using Newton’s second laws

00 tkJ

0J

ktn

t

nk

J02 02

1

J

kf t

n

• Important aspects of this system

▫ If the cross section of the shaft supporting the disc is not circular, an appropriate torsionalspring constant is to be used

▫ The polar mass moment of inertia of a disc is given by

▫ The general solution

g

WDDhJ

832

24

0

tAtAt nn sincos)( 21

01 A

nA /02

Example

• The figure shows a spacecraft with four solar panel. Each panel has the dimension of 5 ft. x 3 ft. x 1 ft. with a density of 0.1 lb/in3 , and is connected to the body by aluminum rods of length 12 in. and diameter 1 in. Determine the natural frequency of vibration each panel about the axis of the connecting rod


5820.1386.4

0.283)112)(12)(3(5panel a of massm

98.17036112

5820.1

12

m axis- xabout the panel theof inertia momentof mass

22

22

0

baJ

444

0 098175.0132

32

rod of inertia momentofpolar ind I

radinlbl

GIkt /101089.3

12

098175.0108.3 48

0

sec/4841.13

21

0

radJ

ktn

Example

• Find the equation of motion of the uniform rigid bar OA of length l and mass m shown in the figure. Also find its natural frequency


llkaakkJ t 210

:motion ofEquation

2

2

2

03

1

212

1 where ml

lmmlJ

03

1 2

2

2

1

2 lkakkml t

21

2

2

2

2

13

ml

lkakktn

Rayleigh’s Energy Method

• Uses the energy method to find the natural frequencies of a single degree of freedom systems

• The principle of conservation of energy, in the context of an undamped vibrating system, can be restated as

• Subscript 1 denote the time when the mass is passing through its static equilibrium position (U1=0)

2211 UTUT

Rayleigh’s Energy Method

• Subscript 2 indicate the time corresponding to the maximum displacement of the mass (T2=0)

• If the system is undergoing harmonic motion, then T1 and U2 denote the maximum values

21 00 UT

maxmax UT

Example

• Find the natural frequency of the transverse vibration of the water tank considered in the first example by including the mass of the column

Example

32

3

max

2

32

36

xlxl

y

xlEI

Pxxy

The maximum kinetic energy of the beam,

dxxyl

mT

l2

0

max2

1

Example

32

3

max 32

xlxl

yxy

dxxlxl

y

l

mT

l

0

232

2

3

maxmax 3

22

2

max

7

6

2

maxmax

140

33

2

1

35

33

42yml

l

y

l

mT

Example

2

maxmax2

1ymT eq

mmeq140

33

eqeff mMM

meq denotes the equivalent mass of the cantilever at the free end, its maximum kinetic energy

The total effective mass acting at the end, M is the mass of the water tank

mM

k

M

k

eff

n

140

33

Free Vibration with Viscous Damping

• The viscous damping force F is proportional to the velocity,

• c is the damping constant or coefficient of viscous damping

• The negative sign indicates that the damping force is opposite to the direction of velocity

xcF

Free Vibration with Viscous Damping

kxxcxm

0 kxxcxm

FV with Viscous Damping - Solution

0 kxxcxm

02 kcsms

stCetx

m

k

m

c

m

c

m

mkccs

22

2,1222

4

tstseCeCtx 21

21 The general solution

FV with Viscous Damping• Critical Damping Constant and Damping Ratio

▫ The critical damping is defined as the damping constant for which the radical becomes zero

▫ The damping ratio is defined as the ratio of the damping ratio to the critical damping constant

02

2

m

k

m

cc

nc mkmm

kmc 222

ccc /

FV with Viscous Damping• Critical Damping Constant and Damping Ratio

▫ the solution,

nc

c m

c

c

c

m

c

22

ns 12

2,1

tt nn

eCeCtx

1

2

1

1

22

FV with Viscous Damping▫ The nature of the roots and hence the behavior of

the solution depends upon the magnitude of damping;

▫ Case 1: Underdamped

s vibrationundamped toleads,0

nis 2

1 1

overdamped critical, d,underdampe :cases three,0

mkmcorccor c /2/1

nis 2

2 1

expressed becan roots theand negative is1 condition, for this 2

FV with Viscous Damping

titi nn

eCeCtx

22 1

2

1

1

tCtCetx nn

tn 2'

2

2'

1 1sin1cos

Case 1: Underdamped

n

n xxCandxC

2

00'

20

'

1

1

For the initial condition, 00 )0()0( xtxandxtx

t

xxtxetx n

n

nn

tn

2

2

002

0 1sin1

1cos

the solution,

nd 21

The frequency of damped vibration is


2'

2

2'

10 CCXX

'

2

'

1

1 /tan CC

The constants are,

'1'

2

1

0 /tan CC

tXetx n

tn 21sin

0

2

0 1cos

teXtx n

tn

the solution can be expressed as ,

FV with Viscous Damping▫ Case 2: Critical damped

nc

m

css

221

mkmcorccor c /2/1

equal, are and roots two thecase In this 21 ss

the solution, tnetCCtx

21)(

For the initial condition,00 )0()0( xtxandxtx

01 xC 002 xxC n

t

nnetxxxtx

000)(

zero odiminish t eventually lmotion wil the,0 as 0

Since . isequation by the drepresentemotion the

te

aperiodic

tn

FV with Viscous Damping▫ Case 3: Overdamped

mkmcorccor c /2/1

distint, and real are and roots two thecase In this 21 ss

the solution,

tt nn

eCeCtx

1

2

1

1

22

)(

012

1 ns 012

2 ns

FV with Viscous Damping▫ Case 3: Overdamped

For the initial condition,00 )0()0( xtxandxtx

12

1

2

0

2

0

1

n

n xxC

12

1

2

0

2

0

2

n

n xxC

mely with tiexponental diminishesmotion the, negative are roots the

Since . isequation by the drepresentemotion the aperiodic


• The nature of the roots with varying values of damping can be shown in a complex plane. The semicircle represent the locus of the roots for different values of damping ratio in the range of 0 to 1


• A critically damped system will have the smallest damping required for aperiodic motion: hence the mass returns to the position of rest in the shortest possible time without overshooting

• The figure represent the phase plane or state space of a damped system


• Logarithmic Decrement▫ Represent the rate at which the amplitude of a free

damped vibration decreases. It is defined as the natural logarithm of the ratio of any successive amplitude

n

n

n

eteX

teX

x

x

n

t

n

t

020

010

2

1

cos

cos2

1

m

c

x

x

dn

ndn2

2

1

2

1

2ln

222

1

damping smallfor

22 22

or


• Energy Dissipated in Viscous Damping▫ The rate of change of energy with time is given

▫ The negative sign denotes that energy dissipate with time. Assuming a simple harmonic motion

2

2 velocity force

dt

dxccvFv

dt

dW

tdcXdtdt

dxcW ddd

t

n

22

0

2

2/2

0cos

2XcW d

FV with Viscous Damping• Energy Dissipated in Viscous Damping

▫ The fraction of the total energy of the vibrating system that is dissipated in each cycle of motion is called the specific damping capacity

▫ Another quantity used to compare damping capacity of engineering materials is called loss coefficient and is define as the ratio of energy dissipated per radian and the total energy

constant422

22

22

21

2

m

c

Xm

Xc

W

W

dd

d

W

W

W

W

2

2/tcoefficien loss


• Torsional Systems with Viscous Damping

tcT

d,underdampe for the case;linear in the as

00 tt kcJ

is,motion ofequation the

nd 21

0J

ktn

00 22 Jk

c

J

c

c

c

t

t

n

t

tc

t

Example- FV viscous damping• The human leg has a measured natural

frequency of around 20 Hz when in its rigid (knee-locked) position in the longitudinal direction ( i.e., along the length of the bone) with damping ratio of ξ = 0.224. Calculate the response of the tip of the leg bone to an initial velocity of v0 = 0.6m/s and zero initial displacement ( this would correspond to the vibration induced while landing on your feet, with your knee locked from a height of 18 mm. What is the maximum acceleration experience by the leg assuming no damping?

Example- FV viscous damping• Highway crash barrier are design to absorb a

vehicle’s kinetic energy without bringing the vehicle to such an abrupt stop that the occupants are injured. Knowledge of the barrier’s materials provide the spring constant k and the damping coefficient c; the mass m is the vehicle mass. For this application , t=0 denotes the time at which the moving vehicle contacts the barrier at x=0; thus v(0) is the speed of the vehicle at the time of contact and x(0) =0. The applied force is zero. Most of the barrier’s resistance is due to the term cv, and it stops resisting after the vehicle comes to rest; so the barrier does reverse the vehicle motion. Cont……

Example- FV viscous damping• A particular barrier’s construction gives

k=18000 N/m and c = 20000 N s/m. A vehicle 1800 kg vehicle strikes the barrier at 22 m/s. Determine how long it takes for the vehicle to come to rest, how far the vehicle compress the barrier, and the maximum deceleration of the vehicle

Example- FV viscous damping• An underdamped shock absorber is to be design

for a motorcycle of mass 200 kg. When the shock absorber is subjected to an initial vertical velocity due to a road bump, the resulting displacement-time curve is to be as shown. Find the necessary stiffness and damping constants of the shock absorber if the damped period is to be 2 s and the amplitude x1 is to be reduced to one-fourth in one half cycle (i.e., x1.5 = x1/4 ). Also find the minimum initial velocity that leads to a maximum displacement of 250 mm

Example- Solution• Approach: We use the equation for the logarithm

decrement in term of the damping ratio, equation for the damped period of vibration, time corresponding to maximum displacement for an underdamped system, and envelop passing through the maximum points of an underdamped system

22

1

1

27726.2)16ln(ln

x

x

16/4/,4/ 15.1215.1 xxxxx

4037.0

Example - Solution

21

222

nd

d

sradn /4338.34037.012

2

2

msNmc nc /54.13734338.320022

msNcc c /4981.55454.13734037.0

mNmk n /2652.23584338.320022

Example - Solution

2

1 1sin td

sec3678.0

9149.0sin 1

1

t

tnXex

21

The displacement of the mass attain its maximum value at time t,

9149.04037.01sinsin2

11 ttd

The envelop passing through the maximum points,

Example - Solution

tnXex

21

The envelop passing through the maximum points,

3678.0433.44037.024037.0125.0 Xe

mX 4550.0

The velocity of the mass can be obtained by differentiating,

tXetx d

tn sin

Example - Solution

The velocity of the mass,

ttXetx nddn

tn cossin

2

0 10 nd XXxtx

sm /4294.1

4037.01)4338.3)(4550.0(2

Example

• The maximum permissible recoil distance of a gun is specified as 0.5 m. If the initial recoil velocity is to be 8 m/s and 10 m/s, find the mass of the gun and the spring stiffness of the recoil mechanism. Assume that a critically damped dashpot is used in the recoil mechanism and the mass of the gun has to be at least 500 kg

Example - Solution

1000 Eetxxxtxt

nn

20

2

00 Etxtxxetx nn

tn

Let tm = time at which x=xmax and v=0 occur. Here x0 = 0 and v0 =0 initial recoil velocity. By setting v(t)=0,

3

1

0

0

00

0 Ex

x

xx

xt

nnnn

m

Example - Solution

n

t

m

exetxx mn

1

00max

7178.25.0max0 nn exx

smx /10 when, 0

sec/3575.77178.2*5.0/10 radn

kg, 500 isgun of masswhen

mNmk n /403.066,275003575.722

FV with Coulomb Damping

• In vibrating structure, whenever the components slide relative to each other, dry friction damping appears internally

• Coulomb’s law of dry friction states that, when two bodies are in contact, the force required to produce sliding is proportional to the normal force acting in the plane of contact

mgWNF


• Equation of Motion

▫ Case 1

When x is positive and dx/dt is positive or when x is negative and dx/dt is positive (half cycle)

Nkxxm



▫ Case 1

This is a second-order nonhomogeneous DFQ. The solution is

k

NtAtAtx nn

sincos 21

k

NxA

301 02 A

k

Nxxt

4/2/ position, extreme At the 02nn

0)0(/2)0( 0 txandkNxtx



▫ Case 2

When x is positive and dx/dt is negative or when x is negative and dx/dt is negative (half cycle)

Nkxxm

FV with Coulomb Damping▫ Case 2

The solution is

k

NtAtAtx nn

sincos 43

k

NxA

03 04 A

k

Nxx

2/ tposition, extreme At the 01n

0)0()0( 0 txandxtx


k

Nk

Nx

r

2

cycles ofnumber

0

kNxn /at stopsmotion

FV with Coulomb Damping▫ The previous solutions can be expressed as a

single equation

▫ Where sgn(y) is called the signum function and it is define as

0)sgn( kxxmgxm

10

11

11

yfor

yfor

yfor


• The equation of motion is nonlinear with Coulomb damping, while it is linear with viscous damping

• The natural frequency of the system is unaltered with the addition of Coulomb damping, while it is reduced with addition of viscous damping

• The motion is periodic with Coulomb damping, while it can be nonperiodic in a viscouly damped (overdamped) system


• The system comes to rest some time with Coulomb damping, whereas the motion theoretically continues forever with viscous and hysteresis damping

• The amplitude reduces linearly with Coulomb damping, whereas it reduces exponentially with viscous damping


• In each successive cycle, the amplitude of motion is reduced by the amount 4μN/k, so the amplitude at the end of any two consecutive cycles are related:

The slope of the enveloping straight lines is

kNXX mm /41

k

N

k

N n

n

224

Example

• A metal block, placed on a rough surface, is attached to a spring and is given an initial displacement of 10 cm from its equilibrium position. It is found that the natural time period of motion is 1 s and the amplitude reduces by 5 cm in each cycle. Find (a) the kinetic coefficient between the metal block and the surface and (b) the number of cycles of motion executed by the block before it stops.


• Torsional Systems with Coulomb Damping▫ If a constant frictional torque acts on a torsional

system, the equation governing the angular oscillations can be derived similar to that of the linear,

torquedampingconstant thedenoted T

TkJ t 0

0J

ktn

t

rk

Tr

2

amplitude, the

0

TkJ t 0

k

Tk

T

r2

ceased,motion

0

FV with Hysteretic Damping

• The damping caused by the friction between the internal planes that slip or slide as the material deforms is called hysteretic damping

• This causes a loop to be formed in the stress-strain of force-displacement curve

• The energy loss in one loading and unloading cycle is equal to the area enclosed by the loop

• The energy loss per cycle is independent of the frequency but proportional to the square of the amplitude


constant damping hysteretic theish and ct coefficien Damping

hc

,dissipatedenergy The 2hXW

relation,nt displaceme-force The

xihkF

stiffness,complex

ikihk 1


decrement, logarithm hysteresis The

1lnln1j

j

X

X

12

2

1j

j

X

X


ratio damping viscousequivalent The

k

heq

2

is,constant damping equivalent The

hkmkmkcc eqceq

22

isfrequency The

m

k

k

heq

22

Chapter 2 Free Vibration of Single Degree of Freedom

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Transcript of Chapter 2 Free Vibration of Single Degree of Freedom