Ch.04 Single DOF System - Free-Response Characteristics

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04. Single DOF Systems: Free

Response Characteristics

Vibrations 4.01 Single DOF Systems: Free Response Characteristics

HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

Chapter Objectives

Determine the solutions for a linear, single DOF system that isunderdamped, critically damped, overdamped, and undamped

Determine the response of single DOF systems to initialconditions and use the results to study the response to impact

and collision

Determine when a system is stable and how to use the root-locus diagram to obtain stability information

Obtain the conditions under which a machine tool chatters

Use different models for damping: viscous (Voigt), Maxwell,hysteretic

Examine systems with nonlinear stiffness and nonlineardamping



1.Introduction

The governing equation for all linear single dof systems

2

2+ 2

+

2 =()

The solution can be determined by using

time-domain methods

the Laplace transform method



(4.1)

2

2+ 2

+

2 =()

(4.1)

2.Free Responses of Undamped and Damped Systems

1.Introduction

- In the absence of forcing, = 0 , the single dof reduces to

2

2+ 2

+

2 = 0

- Free responses: the responses of a system to initial conditions

= 0 and/or (0) = 0- Four distinct types of solutions to Eq. (4.1)

Underdamped System 0 < < 1

Critically damped system = 1

Overdamped System > 1

Undamped System = 0



(4.2)


- Underdamped System 0 < < 1, <

= 0 +

0+0

or

= 0 + (4.4)

where

: damped natural frequency, = 1 2

0 = 02 +

0 + 0

2

= 1

00 + 0



(4.3)

0 : amplitude, (4.5)

0 : phase, (4.6)


- Critically Damped System = 1, = = 0

+ 0 +0 (4.7)

- Over Damped System > 1, >

=0

+

+0

where

= 2 1 (4.9)



(4.8)

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- Undamped System = 0, = 0

= 0() +0

()

and

= 0 ( +

) (4.11)

where

0 = 0

2 +0

2

= 1

00



(4.10)

(4.12)


- To simplify matters

assume that 0 = 0; 0 0

introducing the nondimensional time variable =

and study the differences in the response of the mass forfour regions describe four different types of systems

()

0/= ()

()

0/=

1

1 2 1 2

()

0/=

()

0/=

1

1 2 1 2



= 0

0 < < 1

= 1

> 1


- Ex.4.1 Free response of a microelectromechanical system

A microelectromechanical system has a mass of 0.40, astiffness of 0.08/, and a negligible damping coefficient. Thegravity loading is normal to the direction of motion of this mass

Determine the displacement response of this system when

no forcing acting on this system, and

the initial values 0 = 2, and 0 = 0




Solution

0 = = = 0, from Eq.(4.10): = 0

The initial displacement 0 = 2

The natural frequency

=

=

0.08

0.4 109= 14142.14/(2250.8)

The displacement response

= 2 14142.14 ()

The displacement is a cosine harmonic function that varies

periodically with time and has the period

=2

=

1

=

1

2250.8= 444.29



= 0, = 0 : = 0() +0

() (4.10)


- Ex.4.2 Free response of a car tire

A wide-base truck tire is characterized with a stiffness of =1.23 106/, an undamped natural frequency of = 30,and a damping coefficient of = 4400/. In the absence offorcing, determine the response of the system assuming non-

zero initial conditions, evaluate the damped natural frequency

of the system, and discuss the nature of the response

Solution

The governing equation of motion of the tire system

2

2+ 2

+

2 = 0

= 2 = 2 30 = 188.5/

=

2=2

=4400188.5

21.23106=0.337



< 1 underdamped

where,

2

2+ 2

+

2 = 0 = 0( + )


The displacement response of the tire system

= 0( + )

0 = 02 +

0 + 0

2

, = 1

00 + 0

The damped natural frequency

= 1 2 = 188.5 1 0.3372 = 177.5/

The damped sinusoid period

=2

=

2

177.5= 0.0354 = 35.4

The amplitude of the displacement response decreases

exponentially with time, with

() = 0



where,

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- Ex.4.3 Free response of a door

A door undergoes rotational motions about the vertical

axis pointing in the direction. The governing equationof motion of this system

+ + = 0

where, = 202

= 48/

= 28.8/

Determine the response of this system when the door

is opened with an initial velocity of 0 = 4/ fromthe initial position (0) = 0, then plot this response asa function of time and discuss its motion




Solution

Rewritten the governing equation of motion

+

+

= + 2 + 2 = 0

=

=

28.8

20= 1.2/

=

2=

48

2 20 1.2= 1.0

the system is critically damped

The displacement response

= 0 + 0 + (0)

= 41.2



where,


Displacement time history of the door

= 41.2



= 0 + , = 12,0 = 0

2 +0+0

2, =

1 00+0


2.Initial Velocity

- The amplitude and phase with the initial condition 0 = 0, 0

0 = 02+

0+0

2

=0

, = 1

00+0

= 0

=0

= =0

12 ( )

= =0

12 (2)



= 1 12 = 112

with

(4.13)

(4.14)

(4.15)


- Time histories of displacement, velocity, and acceleration of a

system with prescribed initial velocity 0



0 = 0, 0 0: =0

, = 1 2, = 1 1 2 = 1

12


Extrema of Displacement Response / = = =0

= 0

1 2 = 0

= + , = 0,1,2, (4.17)

= / =0

=0

/ 12 +

= (1)0

+ /, = 0,1,2,

/ occurs when = 0, or



(4.16)

. =

=

1 2

(4.18)

(4.19)

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0 =0, 0 0: =0

12 (), = 12,=

1 12 = 112


Extrema of Velocity Response / = = =0

= =0

12 (2) = 0

= 2 + , = 0,1,2, (4.20)

= /

=0

1 2

=0

1 2/ 1

2 +

= (1)+10 2+ /, = 0,1,2,



(4.21)

=0

, = =0

12 ( )


Force Transmitted to Fixed Surface

The dynamic component of the force transmitted to the base of a

single dof system

= + (4.22)

=0

2 +

The reaction force acting on the base at = 0

(0) =20

=20

when the mass of a single dof system is subjected to aninitial velocity, the force is instantaneously transmitted to the

base



(4.23)

(4.24)


State-Space Plot and Energy Dissipation

State-space plot of a single degree-of-freedom system with a prescribed initial velocity 0




- Ex.4.4 Impact of a vehicle bumper

Consider a vehicle of mass that is travelling at a constant velocity0

Model of a car bumper colliding with a stationary barrier






Time history of velocity of mass

Equivalent impact configuration

and rebound immediately

after impact with the floor


- Ex.4.5 Impact of a container housing a single dof system

Consider the effects of dropping onto the

floor a system that resides inside a

container that has a coefficient of

restitution with respect to the floor

The magnitude of the velocity at the time

of impact with the floor

0 = 2

At the instant = 0+ after impact, thecontainer bounces upwards with a velocity

whose magnitude is 0 The container and the single dof system

can be modeled as a single dof system

with a moving base



dropped from a height

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Define the relative displacement

= () 2

2+

+ =

2

2

The container is decelerating during the rebound upwards: =

2

2+

+ = ()

The initial conditions

0 = 0 0 = 0

0 = 0 0 =0 0 = 1+ 0 = 1+ 2

=

1 2 1 2 + 1

1

12 12+



() : unit step function

= (1 + )2

=


The corresponding velocity

=

1 2 1 2

+1

1 2 1 2



Normalized maximum relative

displacement of a system

inside a container that is

dropped from a height as afunction of coefficient of

restitution of the container and

the damping ratio of the single

degree-of-freedom system


- Ex.4.6 Collision of two viscoelastic bodies

Using the single dof model to analyze the impact (collision)

Solution

The relative acceleration between and = = = = (a)

The magnitude of the contact force acting on each mass

=

+



(b)


From the free-body diagram of each mass during impact

=

, =

Since there are no external forces acting on the system at the

time of impact, the systems linear momentum is conserved

+ = 0 + = 0

=

, =

, =

+

From (a,b,c,f,g) the governing equation

= = +

+ 2 + 2 = 0 (h)

2 =

, =

2



(c)

(d,e)

(f,g)

(i)


- Ex.4.7 Vibratory system employing a Maxwell model

The governing equations of

motion and solution for response

2

2++1 =

1 =

Define

=

, ,

1

+ 1 + = /

= = 2

2 =



Maxwell Model

Kelvin-Voigt Model

(b,c)

(a)

(e)

(d)

+ 1 + = /, = = 2 (d)


In the limiting case, when (1 ), (d) canbe used study a vibratory system with a Maxwell

model as well as a Kevin-Voigt model

(d) 2 + 1+ = ()

+ 2 = 0 (f)

+ 0 + (0)

The response

=() + 2

23 + 2 + 2 1 + +

=()

23 + 2 + 2 1 + +



(h)

(g)Maxwell Model

Kelvin-Voigt Model

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Force transmitted to the fixed support

The reaction force on the base is seen to be

= 1 + =

+

in terms of the nondimensional quantities

= 2 +

=() + 2 1 +

23 + 2 + 2 1 + +

= 0, 0 = 0,(0)

=

(0)

= 0

= 0/ (m)

0/=

+ 2 1 +

23 + 2 + 2 1 + +



(i)

(j)

(k)

(l)

(n)


0/=

1 + 2

2 + 2 + 1

Reaction force of the system for = 0.15



(o)

2

2++1 = , 1 =

(a)


- Ex.4.8 Vibratory system with Maxwell model revisited

Now consider the case where the support consists only of a

Maxwell element; that is, = 0

2

2+1 = , 1 =

Define12 = 1/, 1

+ = ()/1 = 21

21 =11

The Laplace transforms of Eqs (c)

2 + 1 = 1()

1 + 21 = 0 (e)



(b)

(a)

(c)

(d)


1

1+ 0 + (0)

The response

=1() + 21

212 + + 21

=1()

212 + + 21

2 =1

21=

11

When the spring with stiffness = 0, the reaction force on the base

= 1 =

,()

1= 1 = 21

()

1=

211()

212 + + 21



(f)

(g)

(h)

(I,j)

(k)


= 0, 0 = 0,(0)

= 1

(0)

= 0

= 0/1 (m)

0/1=

21212 + + 21

=1

2 + 2 + 1



(n)

(l)

Maxwell element: Displacement response

of the mass for = 0.15.Maxwell element: Reaction force of the

system for = 0.15 and = 0.12

= 0 + , = 12,0 = 0

2 +0+0

2, =

1 00+0


3.Initial Displacement

- The amplitude and phase with the initial condition 0, 0 = 0

0 =0

1 2, =

11 2

=

=0

1 2 +

= = 0

12

= =0

2

12 ()



(4.28)

(4.29)

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Time histories of displacement, velocity, and acceleration of a

system with a prescribed initial

0 =0

1 2

= = 1

1 2

= 0 +

= 0

= 02 ()




State-space plot of single degree-of-freedom system with

prescribed initial displacement

0 =0

1 2

= = 1

1 2

= 0 +

= 0

= 02 ( )




Logarithmic Decrement

Consider the displacement response of a single dof system

subjected to an initial

displacement

Logarithmic decrement

()

( + )

where

2

=2

1 2



Quantities used in the definition of the logarithmic decrement


Let = + , = 1,2, (4.32)

then, by definition01

=12

=23

= =1

=

0

=01

12

23

1

=

The logarithmic decrement in terms of two amplitudes

measured cycles apart

=1

0

=1

()

( + ), = 1,2,

From Eq.s (4.28), (4.30) and (4.34)

=2

1 2



(4.33)

(4.34)

(4.35)


From a measurement of the amplitudes 0 and , one can

obtain the damping ratio

= 1/ 1 (2/)2 (4.36)



Curve fit to a set of sampled data from the response of a system with prescribed initial displacement


- Ex.4.9 Estimate of damping ratio using the logarithmicdecrement

It is found from a plot of the response of a single dof system to

an initial displacement that

at time the amplitude is 40% of its initial value

two periods later the amplitude is 10% of its initial value

Determine an estimate of the damping ratio

Solution

From Eq. (4.34)

=1

0

=1

2

0.4

0.1= 0.693

Then, from Eq.(4.36)

=1

1 (2/)2=

1

1 (2/0.693)2= 0.11



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= 0 + , = 12,0 = 0

2 +0+0

2, =

1 00+0


4.Initial Displacement and Initial Velocity

- The amplitude and phase with the initial condition 0, 0

= 0 ( + ) (4.37)

= = 0 ( + ) (4.39)

= = 0 ( 2)

0 = 02 +

0 + 0

2

= 0 1 + + 2

1 2

= 1

00 + 0

= 11 2

+

= 1 12 = 112



where,

(4.38)

(4.15)


Displacement response of a system with prescribed initial

displacement and prescribed initial velocity




- Ex.4.10 Inverse problem: information from a state-space plot

From the given graph, determine the following: (a) the value of

the damping ratio and (b) the time = at whichthe maximum displacement occurs




Solution

at = 0: 0 = 0, 0 = 1.60 at = 0: = 1.80



From the graph


a.Determine

Along the line = 0

0.950 + 0.50

The logarithmic decrement

= 0.9500.50

= 1.90 = 0.642

The damping factor

=1

1 (2/)2=

1

1 (2/0.642)2= 0.10



= 0 ( + ) (4.37)


b.Determine

=(0)

(0)=

1.600

= 1.6

0 = 0 1 + + 2

1 2= 0 1 +

1.6 + 0.1 2

1 0.12= 1.9760

= 1

12

+= 1

10.12

0.1+1.6= 10.5853= 0.53

Eq. 3(4.7): 1.80 = 1.9760.1 ( 1 0.12 + 0.53)

0.91 = 0.1 (0.995 + 0.53)

Therefore

= 0.945



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3.Stability of Single Degree-of-Freedom System

- A linear single dof system is considered stable if, for all

selections of finite initial conditions and finite forcing functions

, > 0

where has a finite value

- This is a boundedness condition, which requires the system

response () be bounded for bounded system inputs

- If this is not the case, then the system is considered unstable



2

2+ 2

+

2 =()

(4.1)


- Instability of Unforced System

(4.1) = 0

+2 0 + 0

where = 2 + 2 + 2

= 2 + (/) + /= ( 1)( 2) (4.41)

1,2 =1

2

2 4

1

()=

1

( 1)( 2)=

1

1 2

1

1

1

2

()=

1

( 1)( 2)=

1

2 1

2 2

1

1



(4.40)

= /(2)

= /

(4.42)

(4.43)

(4.44)

note


System response

= 1 ()

= 1 0

+2 0 + 0

= 0 1

+ 2 0 + 0

11

=(0)

2 11

2 2

1

1

+2 0 + 0

1 21

1

1

1

2

=(0)

212

211 +

2 0 + 0

121 2



= 0

+

2 0 + 0

,

1

()=

1

12

1

1

1

2,

()=

1

21

22

1

11,2 =

1

2

2 4 , = /(2) , = / (4.42)


System is table if a finite value 0

2 12

2 11 +

2 0 + 0

1 21 2

a finite value

1,2 remains finite as > 0, or

1,2 0 (4.45)

0, 0 (from (4.42))

The system is stable if 0 0

The system is unstable if < 0 < 0




Root locus diagram




- Ex.4.11 Instability of inverted pendulum

The inverted pendulum that was examined in Ex.3.11 is a system

that can be unstable, depending on the values of the parameters

=2

51

2 +112 +

1

322

2

= 12

= 12 11 2

22

For this system > 0, > 0, and > 0 if

12 11 +2

22



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Asymptotic Stability

- Consider the system

2

2+ 2

+

2 = 0

The equilibrium position = 0 of this system is said to beasymptotically stable if

()

that is, the equilibrium position is approached as time increases

- Since the governing equation is an equation with constant

coefficients, a solution to this equation can be written in the form

= (4.48)

: a constant

: an unknown quantity



(4.46)

(4.47)

2

2+ 2

+

2 = 0 (4.46)

() (4.47)

= (4.48)


- Upon substituting Eq.(4.48) into Eq. (4.46) and requiring that

0, we obtain

2 + 2 +2 = 1 2 = 0 (4.49)

Eq. (4.49) : the characteristic equation

1, 2 : characteristic roots or eigenvalues

1,2 = 2 1 (4.50)

Then the solution of (4.48)

= 11 +2

2 (4.51)

- If the real parts of the exponents 1 and 2 are negative, Eq. (4.47)is satisfied, and the equilibrium position is asymptotically stable



4.Machine Tool Chatter

- Consider a model of a turning operation on a lathe

When the cutting parameters such as spindle speed andwidth of cut are carefully chosen, the turning operation can

produce the desired surface finish on the work piece

However, this turning operation can become unstable forcertain values of spindle speed and width of cut. When these

undesirable conditions are present, the tool and work piece

system chatters, producing an undesirable surface finishand a shortening of tool life

Explore the loss of stability that leads to the onset of chatter




- For a rigid work piece and a flexible tool, the cutting force

acting on the tool due to the uncut material and the associated

damping can be modeled as shown in the figure

: the mass of the tool and tool holder

: the stiffness of the tool holders support structure

: the equivalent viscous damping of the structure

: the dynamic cutting force, the sum of the forces due tothe change in chip thickness and the change in the

penetration rate of the tool




: the overlap factor (0 1)1 : an experimentally determined dynamic coefficient

called the cutting stiffness

: the experimentally determined penetration ratecoefficient

: the rotational speed of either the tool or the work piecein revolutions per second




- Carrying out a force balance based on the figure, the tool

vibrations can be described by the following equation

2

2+

1

+

+ 1 +

1

1

1

= 0

=

2, =

, =

1

2, 2 =

: quality factor

: the nondimension time, =



(4.52)

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2

2+

1

+

+ 1 +

1

1

1

= 0 (4.52)


- A solution to Eq. (4.52) is of the form =

The characteristic equation

2 +1

+

+ 1 +

1

1 = 0

in general, = +

- For the system to be stable, the < 0, that is, < 0

The boundary between the stable and unstable regions

corresponds to = 0

- To find the stability boundary, let = and substitute thisvalue into the quasipolynomial Eq. (4.53)

1

+

+1

(/)

= 0,2 = 1+

1

1cos(/)



(4.53)

(4.54)


Stability chart for one set of parameters in turning = 1



5.Single Degree-of-Freedom System with Nonlinear

1.Nonlinear Stiffness

System with Hardening Cubic Spring

2

2+ 2

+ + 3 = 0, =



Comparison of the responses of linear (solid lines) and nonlinear (dashed lines) systems with prescribed initial displacement: (a) displacement and (b) phase portrait

(4.55)


System with Piecewise Linear Springs

Consider a second nonlinear system shown in the figure

The governing equation

2

2+ 2

+ + = 0

where

() = 0 1

() > 1

= , =

, =

, 2 =



Single dof system with

additional springs that are

not contacted until the

mass displaces a distance

in either direction


Response of the system with prescribed initial velocity0

= 10



Single dof system with

additional springs that are

not contacted until the

mass displaces a distance

in either direction

2

2+ 2

+ + () = 0


2.Nonlinear Damping

Compare the free responses of systems with

Linear viscous damping

2

2+ 2

+ = 0

Coulomb damping

2

2+

+ = 0

Fluid damping

2

2+

+ = 0

=

, =



where

(4.56)

(4.57)

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Comparisons of displacement responses for three different

damping models



(a) Displacement histories and (b) phase portraits for the free response of a system with

dry friction subjected to two different initial displacements: = 0.86


Nonlinear System Response Dependence on Initial Conditions

During the free oscillations, the system will come to a stop or

reach a rest state when / = 0 and



(4.57)

Ch.04 Single DOF System - Free-Response Characteristics

Documents

response of single dof

free responses of undamped

linear single dof systems22

overdamped system

damped systems1

different types of systems0

distinct types of solutions

different models