dynamic systems mechanical systems

5/22/2018 Dynamic Systems Mechanical Systems

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1 Dr. Peter AvitabileModal Analysis & Controls Laboratory

22.451 Dynamic Systems Chapter 4

Mechanical Systems

Peter AvitabileMechanical Engineering DepartmentUniversity of Massachusetts Lowell

j

UNSTABLESTABLE

TIME

FRF

FRF

TIME

FRF

TIME

FRF

TIME

TIME

TIME

> 1.0

= 1.0

= 0.7

= 0.3

= 0.1 = 0


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Mechanical Systems-Translational Mass Element

Translation of a particle moving in space due to anapplied force is given by:

dt

dpf=

Where:

mvmomentump

forcef

==

=

Considering the mass to be constant:

madt

dvmfmdvfdt

dt

)mv(df ====


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Mechanical Systems Translational Mass Element

Displacement, velocity, and acceleration are allrelated by time derivatives as:

2

2

dt

xddtda ==

xva &&&==


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Mechanical Systems Rotational Mass Element

=

= &

dt

d

dt

d

dt

d 2=

=

Centroidal mass moment of inertia Ic (not to beconfused with I area moment of inertia used instrength of materials)

Angular acceleration

where: = angular velocity

= angular displacement

Then:


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Mechanical Systems - Translational Spring Element

( )21relk xxkkxf ==

A linear spring is considered to have no massdescribed by:

(Torsional spring follows the same relationship)

fk fk

x1 x2

k


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Translational Spring Element

Hardening Spring

Bi-Linear

Linear

Softening Spring

Gap Cubic

f f f

x x x

k

k = lb/in

= N/m


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Damper Element

relc cf =

Viscous (fluid), Coulomb (dry friction), and structuraldamping (hysteretic)

Viscous Dashpot

fcfa

vrel

m

Coulomb DamperIn order to have motion, the applied force must

overcome the static friction. As soon as sliding occurs,the dynamic friction becomes appropriate.

fc

c

x&

Linear ( )21c cf =

fc fc

v1 v2

c


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Equivalence - Springs in Parallel

Both springs see the same displacement

k2

x

f

k1

f

x

keq

f2=k2x

f

f2

f1=k1x

f1

f=f1 + f2

keqx=k

1x + k

2x

=(k1 + k2)x

21eq kkk +=


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Equivalence - Springs in Series

Both springs see the same force but different displacements

But

k2

x1

f

k1

x2

f=keq

x

keq

f=k1(x1-x2)=k1

k1

k2

f2=k2x2=k2

1

2f

21 +=

22

11

eq k

f

k

f

k

f

+= 21eq k

1

k

1

k

1+=

21

21eq

kk

kkk

+=

21 fff ==


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Translational Systems

Newtons Second Law- THE RIGHT WAY

=maF OR

= xx maF +

= yy maF +Note that this applies to the center of mass which is not

necessarily the center of gravity.

Free-Body Diagram & Sign Convention

kx

xc&m )t(f

xxx &&&+


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Translational Systems Newtons 2ndLaw

Assume spring and dashpot are stretched

= xx maF

or in standard input-output differential form

xmFF)t(f kc &&=

xmkxxc)t(f &&& =

OR

)t(fkxxcxm =++ &&&m)t(fx

mkx

mcx =++ &&&

)t(fxx2x 2

nn =++ &&&

m

k

m2c

c

c

2n

nc

c

=

=

= - damping ratio

critical damping

n natural frequency


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DAlemberts Principle The Fictitious Force

The mass times acceleration is sometimes describedas a fictitious force, reverse effective force or apparent force

( ) =+ 0maFInitially developed since it looks like a classical

force balance but often confuses many students.

DO NOT USE DALEMBERT!!!!!USE NEWTONS SECOND LAW


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Example Pendulum Problem

Mass at end of massless string

l

mg

2mJ l=

+x

cosmg

mg

sinmg

T

FBD

sinmgJ l=&& 0sinmgm 2 =+ ll &&ORThen

0sing

=+

l

&& 0

g =+

l&&for small .

Natl freq. lg

n =


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Example Differential Equation about Equilibrium

y

x

+ = yy maFymmgky &&=+

( ) xmmgxxk st &&=++but stkxmg=

0kxxm =+ &&

Therefore, the equationscan be written about the

equilibrium point and theeffect of gravity makesno difference.

Source: Dynamic Systems Vu & Esfandiari


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Systems with Displacement Input

In terms of naturalfrequency and damping

ratio

= xx maF+

m

ck

x

y

m x

)xy(c &&)xy(k

m

ck

x

y

FBD

( ) ( ) xmxycxyk &&&& =+ kyyckxxcxm +=++ &&&&OR

The force exerted canbe found to be

yy2xx2x 2

nn

2

nn

+=++ &&&&

( ) ( )xycxyk)t(f &&+=


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Transfer Function and State Space

m

c

k

x

f(t) )t(fkxxcxm =++ &&&

00

2 XmmsXmX(s)s)x(m &&& =L

0)()( cXsscXxc =&L

)()( skXkx =L

)())(( sftf =L

)s(f)s(kXcX)s(scXXmmsX)s(mXs 0002 =++ &


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Transfer Function and State Space

( ) 002 XmXcms)s(f)s(Xkcsms &+++=++

Grouping and rearranging:

Assume initial conditions are zero and rearrangingterms to obtain OUT/IN formThen:

kcsms1

)s(F)s(X)s(H

2 ++==

Sometimes written with

kcsms)s(b 2 ++=


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The frequency response function is the systemtransfer function evaluated along

Frequency Response Function - SDOF

= js

kcsms

1)s(h

2 ++=

Source: Vibrant Technology

Recall:

The complex valuedfunction defines the

surface shown


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Polynomial Form

Pole-Zero Form

Partial Fraction Form

Exponential Form

kcsms

1)s(h

2 ++=

)ps)(ps(

m/1)s(h

*

11 =

)ps(

a

)ps(

a)s(h

*

1

*

1

1

1

+

=

tsinem

1)t(h d

t

d

=

SDOF Transfer Function


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Polynomial Form

Pole-Zero Form

Partial Fraction Form

kcjm

1)h(j

2 ++=

)pj)(pj(

m/1)j(h

*11

=

)pj(

a

)pj(

a)j(h

*1

*1

1

1

+

=

SDOF Frequency Response Function


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Transfer Function approach is used extensively indesign but is limited to linear, time-invariant systems.

SDOF Transfer Function

1. T.F. method to express output relative to input2. T.F. system property independent of the nature

of excitation

3. T.F. contains necessary units but does not providephysical structure of system

4. If T.F. is known, then response can be evaluateddue to various inputs

5. If T.F. is unknown, it can be establishedexperimentally by measuring output response due toknown measured inputs


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S-plane Plots

j

UNSTABLESTABLE

TIME

FRF

FRF

TIME

FRF

TIME

FRF

TIME

TIME

TIME

> 1.0

= 1.0

= 0.7

= 0.3

= 0.1 = 0


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Experimental Determination of Damping Ratio

Determine decay of amplitudex1at t1and again at n cycleslater xnat t1+ (n-1)T

Then

x1 x2 xn

t1 t2 tn

T - period

( )T

TTt

t

2

1 n

nn

n

ee

1

e

e

x

x +

===

OR

( )( ) T1n

T1n

n

1 n

n

e

e

1

x

x ==


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Log Decrement

Tx

xln1n

1

x

xln n

n

1

2

1 =

=

For damping < 10%

2d

n122

==

2n

1

1

2xxln

1n1

=

2xxln2

1

Note: This dampingratio formulation isapplicable to any 2nd

order system of this

form


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Estimate of Response Time

whose time constant T of the exponential is

( )

= tcose1

X)t(x

d

t

2

0 n

The response of a mechanical system due to aninitial displacement is given as:

The exponential response envelope is

t

2

0 ne1

X

=

11

n


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Estimate of Response Time

The response of the second-order system in termsof the settling time is

===

44

T4t ns

which will cause 2%

of the initial value



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State Space Representation

The state of the system can be described interms of the displacement and velocity as

=

x

x

x

x

2

1

&

=

= )velocity(x)displ(x

x

x

X 2

1

&

u = f (force) and y = x (measured by sensor)

Then )t(fm

1x

m

cx

m

kx += &&&

ORu

m

1x

m

cx

m

kx 212 +=&


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State Space Representation

So that the state space representation is

u0

x

x10

x

x

m1

2

1

m

c

m

k

2

1

+

=

&

&State

Equation

OutputEquation [ ] u0x

x01y2

1 +

=


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Lagranges Equations

nciiii

QqV

qT

qT

dtd =

+

T Kinetic energyV Potential energyQnci non-conservative generalized forcesqi

independent generalized coordinatesn total # independent generalized coordinates

Kinetic Energy is a function of

Potential energy is the sum of elastic potential Veand gravitational potential VgPotential Energy is a function of

),( tqT i

),( tqV i

dtdq /


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Lagranges Equations

nci

ii

Q

q

L

q

L

dt

d=

&

iq&

One standard form of Lagranges Equation

where L = T-V

We can then write

( ) ( )nci

ii

Qq

VT

q

VT

dt

d=

&

(Note V is not a function of )


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Lagranges Equations

2c

2

2

I2

1mv2

1TD2

mv2

1TD1

+=

=

Kinetic energy for a particle

If the mass is not located at apoint (such as a particle), then

a more complicated form ofthese equations is necessary

Potential Energy of anelastic element is

Potential Energy ofa mass is

2e k

2

1V =

mghVg=


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Lagranges Equations

Non-conservative Forces are those that cannot bederived from a potential function (i.e., external forces,frictional forces)Generalized Forces are given by Virtual Work.

To determine Qj, obtain , then let all except

L++== 2211ii qQqQqQWW 0qi= jq

Thus

j

j

q

WQ

=

0qj=

ij


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Lagranges Equations

Non-conservative Forces are then:L++== 22nc11ncincinc qQqQqQW

Then

i

nc

nci q

WQ

=

0qi=

ij


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Example using Lagrange Equation

Use Lagrange EQ to obtaindifferential equation forSDOF system

Only one independent generalized coordinate exists: q=x

22

c xm2

1

mv2

1

T &

==

m

c

k

x

f(t)

Kinetic Energy

Potential Energy

Non-conservativeForces

( )0Vkx2

1VV g

2e ===

applied)t(f

edissipativxc &


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Non-Conservative Forces[ ] xxc)t(fWnc = &

xQW ncnc =

xc)t(fx

WQ ncnc &=

=

Lagrange Equation

ncQ

x

V

x

T

x

T

dt

d=

+

&


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Lagrange Equation ncQxV

xT

xT

dtd =

+

&

Where

Then substitutingxmx

xm

xT &

&

&

&=

=

)2

1( 2

xmdt

)

x

T(d

&&&

=

0x

)xm2

1(

x

T2

=

=

&

kx

x

)kx2

1(

x

V2

=

=

OR

xc)t(fkxxm &&& =+

)t(fkxxcxm =++ &&&


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Now lets repeat this with the Lagrange function

22 kx2

1xm

2

1VTL == &

xmt

)

x

L(

&&&

=

xmx

)kxxm21(

x

L22

&

&

&

&=

=

)t(fkxxcxm =++ &&&

ncQx

L

t

)x

L(

=

&

xm)xL( && =


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Example Translational Mechanical System

m

k

xst

xy

g

Undeformed position

Dynamic position

Static equilibrium

Solution



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Example - Two DOF Systems

Consider

m1

f1(t)

x2x1

f2(t)

m2

c

k

FBD (assume x1> x2)

1121211 xm)xx(k)xx(c)t(f)1(maF &&&& ==

2221212 xm)xx(k)xx(c)t(f)2(maF &&&& =++=

221

211 f

)xx(c

)xx(kf

&&m2m1

STATEASSUMPTIONS!!!


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Rearranging Terms)t(fkxkxxcxcxm 1212111 =++ &&&&

=

+

+

2

1

2

1

2

1

2

1

2

1

ff

xx

kkkk

xx

cccc

xx

m00m

&

&

&&

&&

)t(fkxkxxcxcxm 2212122 =++ &&&&


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Consider

FBD (assume x2> x1)1112212111 xm)xx(k)xx(cxk)t(f)1(maF &&&& =++=

m1

f1(t)

x2x1

f2(t)

m2

c

k2

k1

22122122 xm)xx(k)xx(c)t(f)2(maF &&&& ==

)t(f)xx(k

)xx(c

xk

)t(f2

122

12

11

1

&&m2m1


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Rearranging terms

)t(fxkx)kk(xcxcxm 1221212111 =+++ &&&&

=

++

+

2

1

2

1

22

221

2

1

2

1

2

1

f

f

x

x

kk

k)kk(

x

x

cc

cc

x

x

m0

0m

&

&

&&

&&

)t(fxkxkxcxcxm 222122122 =++ &&&&


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Rotational Systems

A rotational system follows the same equationsdeveloped for translation

= 00 IM

dmrIorJ 20 =

Newtons Second Law

.accelangularw

intofmomentmassI

appliedmomentsM

0

0

&

Mass moment of inertia of rigid body about axis


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Rotational Systems

Torsion spring stiffness similar to translation

)(KT 12Tk =

Dashpot similar to translation

Right hand rule convention determines +/-

)(KT RELTk =

)(CT 12DD = &&

)(CT RELDD =

usedoftenBorCD

RADLBFT /


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Rotational Systems

FBD

OUTJ

IN

J DT

ST


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Rotational Systems - Equations

Most systems we will treat will be 2D or planarsystems. Modeling of general 3D bodies is more

complex and beyond the scope of this course.

=

=+= Idt

dJTTJM

20

2

DS

)(KT OUTINTS =

OUTD BT = &

OUTOUTOUTINT JB)(K = &&&

INTOUTTOUTOUT KKBJ =++ &&&


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Example - SDOF Torsional System

A torsional system: (a) physical system, (b) FBD

&BK

)t(T

J

K

B

)t(T J

)a( )b(


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Example - SDOF Torsional System

This sign convention is simpler and useful for the given angle .Thus,

Consider a single-degree-of-freedom (SDOF) torsionalsystem. The system consists of a shaft of torsionalstiffness K, a disk of mass-moment of inertia J, and atorsional damper B. Derive the differential equation.

Solution. Applying the moment equation about the mass centeralong the longitudinal axis.

+=+ cc IM

= &&&JBK)t(T

)t(TKBJ =++ &&&The differential equation in the input-output form is


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Example - Two DOF Torsional System



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Example - Two DOF Torsional System



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Example - Rigid Body in Planar Motion

As before

= 00 IM+= &&)ml0(mgsinL 2

Point mass on string Moment method

0sin

2

=+ mgLmL &&

0sin =+ Lg

&&

L

mg

0

sinL


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Thin uniform rod of mass m and length l is a pendulum

Then

= 00 IM

&&

0sin2 Img

L

=

FBD

0sin2

lmgI0 =+&&

0

mg

Lg


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Linearization: For small

where 2c0 mdII +=

or

sin

02

0 =+ mgL

I &&

02 0

=+ I

mgL&&

2

2

)L

m(Ic+=

2

03

1mLI =

2

12

1mLIc=



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Mixed Translation and Rotation

Pulley system

Rotation of pulley

Newtons second law for mass m

and

Txm =&&

Icof pulley, mass m of radius r,Tension in string

m x

r

k

cI

(everything measured from equilibrium so no mg term)

For small angle

or

The natural frequency is

kxrTrJ =&&

=rx kxrrxmJ = &&&&then

0kr)mrJ( 22 =++ &&

0mrJ

kr2

2

=+

+&&

2

2

n mrJ

kr

+=


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Example Cart-Pendulum Problem

Consider the pendulum systemshown attached to a horizontalcart

This is a mixed problem. First solve the pendulum and

then the cart translation.

Cart moves horizontally onfrictionless surface. Mass oninextensible string

The general moment about point P (where the string isattached to the cart mass) is needed to sum the forces

for Newtons Second Law.

M

c L

k

m


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pp/cp axmrIMp +=

yP

xPP

L

mg

sinL

xap &&=

Lr p/c1

=

1C



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The cross product term is

Using the parallel axis theorem, the mass of thependulum at a distance L gives

The moment about P due to the mass on the pendulum is== &&2

p

2

p mLImLI

= sinLmgM

sinararpc/ppc/p

=== cosxL)90sin(xL &&&&

The general moment equation becomes

+= cosxLmLmsinLmg 2 &&&&


E l d l l


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Now the translational equation is evaluated. Only horizontalis considered.

M )t(fkx

L

msinL

xc& xapx &&=

2L& &&L

sinL 2& cosL&&

centripetaltangential


E l P d l P bl


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For the cart, Newtons Second Law

xckx)t(fmaF &=

The acceleration of the pendulum mass

sinLcosLx 2&&&&& +(cart) (tangential) (centripetal)

xM)sinLcosLx(m 2 &&&&&

&& ++=

0sinmgLcosxmLmL2 =++ &&&&

( ) )t(fkxxcsinLcosLmxmMx 2 =++++ &&&&&&


E l C P d l P bl


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For small motion, theequations can be linearized

0mgLxmLmL2 =++ &&&&

or in matrix forms as

0;sin;1cos 2

&

( ) )t(fkxxcmLxmM =++++ &&&&&

=

+

+

+ )t(f

0

xk0

0mgL

xc0

00

x)mM(mL

mLmL2

&

&

&&

&&



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Chapter 4 - Review Slide

k2

xf

k1

21eq kkk +=

k2

x1

f

k1

x2

21

21eq

kk

kkk

+=

2x

xln

2

1x1 x2 xn

t1 t2 tn

T

dynamic systems mechanical systems

Documents

dynamic systems chapter

dynamic friction

center of mass

translational systemsnewtons

equivalence springs

torsional spring

linear spring

parallelboth springs