4 - Mechanical Systems - University of Massachusetts Lowell

1 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Chapter 4

Mechanical Systems

Peter AvitabileMechanical Engineering DepartmentUniversity of Massachusetts Lowell

jω

σ

UNSTABLESTABLE

TIME

FRF

FRF

TIME

FRF

TIME

FRF

TIME

TIME

TIME

> 1.0ζ

= 1.0ζ

= 0.7ζ

= 0.3ζ

= 0.1ζ = 0ζ


Mechanical Systems-Translational Mass Element

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

dtdpf =

Where:mvmomentump

forcef==

=

Considering the mass to be constant:

madtdvmfmdvfdt

dt)mv(df ==⇒=⇒=


Mechanical Systems – Translational Mass Element

Displacement, velocity, and acceleration are all related by time derivatives as:

2

2

dtxd

dtda =ν

=

xva &&& ==


Mechanical Systems – Rotational Mass Element

ω=ω

=α &dtd

dtd

dtd 2θ

=ω

=α

Centroidal mass moment of inertia – Ic (not to be confused with I – area moment of inertia used in strength of materials)

Angular acceleration

where: ω = angular velocityθ = angular displacement

Then:


Mechanical Systems - Translational Spring Element

( )21relk xxkkxf −==

A linear spring is considered to have no mass described by:

(Torsional spring follows the same relationship)

fk fk

x1 x2

k


Translational Spring Element

Hardening Spring

Bi-Linear

Linear

Softening Spring

Gap Cubicf f f

x x x

k

k = lb/in= N/m


Damper Element

relc cf ν=

Viscous (fluid), Coulomb (dry friction), and structural damping (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 ν−ν=

fc fcv1 v2

c


Equivalence - Springs in Parallel

Both springs see the same displacement

k2

x

f

k1

fx

keq

f2=k2x

f

f2

f1=k1x

f1

f=f1 + f2

keqx=k1x + k2x=(k1 + k2)x

21eq kkk +=∴


Equivalence - Springs in Series

Both springs see the same force but different displacements

But

∴

k2

x1

f

k1

x2

f=keq

x

keq

δ f=k1(x1-x2)=k1

k1

k2

f2=k2x2=k2

1δ

2δf

21 δ+δ=δ

2

2

1

1

eq kf

kf

kf

+= 21eq k1

k1

k1

+=

21

21eq kk

kkk+

=21 fff ==


Translational SystemsNewton’s 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

kxxc&

m )t(f

xxx &&& →→+


Translational Systems – Newton’s 2nd LawAssume 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 2nn =ω+ζω+ &&&

mkm2ccc

2n

nc

c

=ω

ω=

=ζ - damping ratio

– critical damping

ωn– natural frequency


D’Alembert’s Principle – The Fictitious ForceThe mass times acceleration is sometimes describedas a ‘fictitious force’, ‘reverse effective force’ or ‘‘apparent force’

( )∑ =−+ 0maF

Initially developed since it looks like a classical force balance – but often confuses many students.

DO NOT USE D’ALEMBERT!!!!!USE NEWTON’S SECOND LAW


Example – Pendulum Problem

Mass at end of massless string

l

mg

θ

2mJ l=

+x

θcosmgmg

θsinmg

T

FBD

sinθmgθJ l−=&& 0sinθmgθm 2 =+ ll &&OR

Then

0sinθgθ =+l

&&0θgθθ =+→

l&&for small .

Nat’l freq. lg

nω =


Example – Differential Equation about Equilibrium

y

x

+ ∑ = yy maFymmgky &&=+−

( ) xmmgxxk st &&=++−but stkxmg =

0kxxm =+∴ &&

Therefore, the equations can be written about the equilibrium point and the effect of gravity makes no difference.

Source: Dynamic Systems – Vu & Esfandiari


Systems with Displacement Input

In terms of natural frequency and damping ratio

∑ = xx maF+

m

ck

x

y

m x

)xy(c && −)xy(k −

mck

x

y

FBD

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

The force exerted can be found to be

yy2xx2x 2nn

2nn ω+ζω=ω+ζω+ &&&&

( ) ( )xycxyk)t(f && −+−=


Transfer Function and State Space

m

c

k

x

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

002 XmmsXmX(s)s)x(m &&& −−=L

0)()( cXsscXxc −=&L)()( skXkx =L)())(( sftf =L

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


Transfer Function and State Space

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

Grouping and rearranging:

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

kcsms1

)s(F)s(X)s(H 2 ++==

Sometimes written with

kcsms)s(b 2 ++=


The frequency response function is the system transfer function evaluated along

Frequency Response Function - SDOF

ω= js

kcsms1)s(h 2 ++

=

Source: Vibrant Technology

Recall:

The complex valued function defines the surface shown


Polynomial Form

Pole-Zero Form

Partial Fraction Form

Exponential Form

kcsms1)s(h 2 ++

=

)ps)(ps(m/1)s(h *

11 −−=

)ps(a

)ps(a)s(h *

1

*1

1

1

−+

−=

tsinem1)t(h d

t

d

ωω

= ζω−

SDOF – Transfer Function


Polynomial Form

Pole-Zero Form

Partial Fraction Form

kcjωmω1)h(jω 2 ++−

=

)pj)(pj(m/1)j(h *

11 −ω−ω=ω

)pj(a

)pj(a)j(h *

1

*1

1

1

−ω+

−ω=ω

SDOF – Frequency Response Function


Transfer Function approach is used extensively in design 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 excitation3. T.F. contains necessary units but does not provide

physical structure of system4. If T.F. is known, then response can be evaluated

due to various inputs5. If T.F. is unknown, it can be established

experimentally by measuring output response due to known measured inputs


S-plane Plots

jω

σ

UNSTABLESTABLE

TIME

FRF

FRF

TIME

FRF

TIME

FRF

TIME

TIME

TIME

> 1.0ζ

= 1.0ζ

= 0.7ζ

= 0.3ζ

= 0.1ζ = 0ζ


Experimental Determination of Damping Ratio

Determine decay of amplitudex1 at t1 and again at n cycleslater xn at t1 + (n-1)T

Then

x1 x2 xn

t1 t2 tn

T - period

( )T

TTt

t

2

1 n

nn

n

ee1

ee

xx ζω

ζω−+ζω−

ζω−===

OR

( )( ) T1n

T1nn

1 n

ne

e1

xx ζω−

−ζω− ==


Log Decrement

Txxln

1n1

xxln n

n

1

2

1 ζω=

−

=

For damping < 10%

2dn

1

22

ζ−

πζ=

ωπ

ζω=

2n

1

1

2xxln

1n1

ζ−

πζ=

−

∴

πζ≅ 2xxln2

1

Note: This damping ratio formulation is applicable to any 2nd

order system of this form


Estimate of Response Time

whose time constant T of the exponential is

( )φ−ωζ−

= ζω− tcose1

X)t(x dt

20 n

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

The exponential response envelope ist

20 ne

1

X ζω−

ζ−

σ=

ζω11

n


Estimate of Response TimeThe response of the second-order system in terms of the settling time is

σ=

ζω==

44T4tn

s

which will cause 2% of the initial value



State Space RepresentationThe ‘state’ of the system can be described in terms of the displacement and velocity as

=

xx

xx

2

1&

=

=)velocity(x)displ(x

xxX2

1&

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

Then )t(fm1x

mcx

mkx +−−= &&&

OR

um1x

mcx

mkx 212 +−−=&


State Space RepresentationSo that the state space representation is

u0xx10

xx

m1

2

1

mc

mk

2

1

+

−−

=

&

&State Equation

Output Equation [ ] u0

xx01y2

1 ⋅+

=


Lagrange’s Equations

nciiiiQ

qV

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 /



nciii

QqL

qL

dtd

=∂∂

−

∂∂&

iq&

One standard form of Lagrange’s Equation

where L = T-V

We can then write

( ) ( )nci

ii

QqVT

qVT

dtd

=∂−∂

−

∂−∂&

(Note V is not a function of )



2c

2

2

I21mv

21TD2

mv21TD1

ω+=⇒

=⇒

Kinetic energy for a particle

If the mass is not located at a point (such as a particle), then a more complicated form of these equations is necessary

Potential Energy of an elastic element is

Potential Energy of a mass is

2e k21V δ=

mghVg =


Lagrange’s EquationsNon-conservative Forces are those that cannot be derived 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+δ+δ=δ=δ ∑ 2211ii qQqQqQW

Wδ 0qi =δ jqδThus

jj q

WQδδ

=0q j =δij ≠


Lagrange’s EquationsNon-conservative Forces are then:

L+δ+δ=δ=δ ∑ 22nc11ncincinc qQqQqQW

Then

i

ncnci q

WQδδ

=0qi =δij ≠


Example using Lagrange EquationUse Lagrange EQ to obtain differential equation for SDOF system

Only one independent generalized coordinate exists: q=x

22c xm

21mv

21T &==

m

c

k

x

f(t)

Kinetic Energy

Potential Energy

Non-conservative Forces

( )0Vkx21VV g

2e ===

applied)t(f −

edissipativxc −&


Example using Lagrange EquationNon-Conservative Forces

[ ] xxc)t(fWnc δ−=δ &

xQW ncnc δ=δ

xc)t(fxWQ nc

nc &−=δ

δ=

Lagrange Equation

ncQxV

xT

xT

dtd

=∂∂

+∂∂

−

∂∂&


Example using Lagrange Equation

Lagrange Equation ncQxV

xT

xT

dtd

=∂∂

+∂∂

−

∂∂&

WhereThen substituting

xmx

xm

xT

&&

&

&=

∂

∂=

∂∂ )

21( 2

xmdt

)xT(d

&&& =∂∂

0x

)xm21(

xT

2

=∂

∂=

∂∂ &

kxx

)kx21(

xV

2

=∂

∂=

∂∂

OR

xc)t(fkxxm &&& −=+

)t(fkxxcxm =++ &&&


Example using Lagrange EquationNow let’s repeat this with the Lagrange function

22 kx21xm

21VTL −=−= &

xmt

)xL(

&&& =∂∂∂

∂

xmx

)kxxm21(

xL

22

&&

&

&=

∂

−∂=

∂∂

)t(fkxxcxm =++∴ &&&

ncQxL

t

)xL(

=∂∂

−∂∂∂

∂&

xm)xL( &&

=∂∂


Example – Translational Mechanical System

m

k

xst

xy

g

Undeformed position

Dynamic positionStatic equilibrium

Solution



Example - Two DOF Systems

Consider

m1

f1(t)

x2x1

f2(t)

m2c

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(k

f →→−←→−←

→ &&m2m1

STATE ASSUMPTIONS!!!



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 =+−+− &&&&



Consider

FBD (assume x2 > x1)

1112212111 xm)xx(k)xx(cxk)t(f)1(maF &&&& =−+−+−=∑

m1

f1(t)

x2x1

f2(t)

m2c

k2

k1

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

)t(f)xx(k)xx(c

xk)t(f

2122

12

11

1 →←−→←−→

←→ &&

m2m1



Rearranging terms

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

=

−

−++

−

−+

2

1

2

1

22

221

2

1

2

1

2

1

ff

xx

kkk)kk(

xx

cccc

xx

m00m

&

&

&&

&&

)t(fxkxkxcxcxm 222122122 =+−+− &&&&


Rotational Systems

A rotational system follows the same equations developed for translation

α=∑ 00 IM

dmrIorJ 20 ∫=

Newton’s Second Law

.accelangularwintofmomentmassI

appliedmomentsM

0

0

&−α−−

Mass moment of inertia of rigid body about axis


Rotational Systems

Torsion spring stiffness similar to translation

)(KT 12Tk θ−θ=

Dashpot similar to translation

Right hand rule convention determines +/-

)(KT RELTk θ=

)(CT 12DD θ−θ= &&

)(CT RELDD θ=

usedoftenBorCD

RADLBFT /−


Rotational Systems

FBD

OUTθJ

INθ

J DTST


Rotational Systems - Equations

Most systems we will treat will be 2D or planar systems. Modeling of general 3D bodies is more complex and beyond the scope of this course.

α=θ

=+⇒α=∑ IdtdJTTJM 2

02

DS

)(KT OUTINTS θ−θ=

OUTD BT θ−= &

OUTOUTOUTINT JB)(K θ=θ−θ−θ &&&

INTOUTTOUTOUT KKBJ θ=θ+θ+θ &&&


Example - SDOF Torsional System

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

θ&BθK

θ

)t(TJ

K

B

θ

)t(T J

)a( )b(


Example - SDOF Torsional System

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

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

Solution. Applying the moment equation about the mass center along the longitudinal axis.

+→α=∑→+ cc IM

θ=θ−θ− &&& JBK)t(T

)t(TKBJ =θ+θ+θ &&&

The differential equation in the input-output form is


Example - Two DOF Torsional System



Example - Rigid Body in Planar Motion

As before

α=∑ 00 IM

θ+=θ− &&)ml0(mgsinL 2

Point mass on string – Moment method

0sin2 =+ θθ mgLmL &&

0sin =+ θθLg&&

L

mg

θ

0

θsinL



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

Then

α=∑ 00 IM

θθ &&0sin

2ImgL

=−

FBD

0sin2lmgI0 =θ+θ&&

0

mg

θLg

θ


Linearization: For small

where 2c0 mdII +=

or

θ⇒θ→θ sin

020 =+ θθ mgLI &&

02 0

=+ θθImgL&&

2

2)Lm(Ic +=

20 31mLI =

2

121 mLIc =



Mixed Translation and Rotation

Pulley system

Rotation of pulley

Newton’s second law for mass m

and

Txm −=&&

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

m x

r θ

kcI

(everything measured from equilibrium so no mg term)

For small angle

or

The natural frequency is

kxrTrJ −=θ&&

θ= rx kxrrxmJ −−=θ &&&&then

0kr)mrJ( 22 =θ+θ+ &&

0mrJkr

2

2=θ

++θ&&

2

2

n mrJkr+

=ω


Example – Cart-Pendulum Problem

Consider the pendulum system shown attached to a horizontal cart

This is a mixed problem. First solve the pendulum and then the cart translation.

Cart moves horizontally on frictionless surface. Mass on inextensible string

The general moment about point P (where the string is attached to the cart mass) is needed to sum the forces for Newton’s Second Law.

M

c L

θ

k

m


pp/cp axmrIMp +α=∑

yP

xPP

L

θ

mgθsinL

xap &&=

Lr p/c1 =

1C

θφ



The cross product term is

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

The moment about P due to the mass on the pendulum isθ=α⇒= &&2

p2

p mLImLI

θ−= sinLmgM

φsinarar pc/ppc/p =×θ=θ−= cosxL)90sin(xL &&&&

The general moment equation becomes

θ+θ=θ− cosxLmLmsinLmg 2 &&&&



Now the translational equation is evaluated. Only horizontal is considered.

M )t(fkx

Lθ

mθsinL

xc& xapx &&=

2θL& θ&&L

θθ sinL 2& θθcosL&&

centripetaltangential



For the cart, Newton’s Second Law

xckx)t(fmaF &−−⇒=∑

The acceleration of the pendulum masssinθθLcosθθLx 2&&&&& −+

(cart) (tangential) (centripetal)

xM)sinLcosLx(m 2 &&&&&&& +θθ−θθ+=

0sinmgLcosxmLmL2 =θ+θ+θ→θ &&&&

( ) ( ) )t(fkxxcsinLcosLmxmMx 2 =++θθ−θθ++→ &&&&&&



For small motion, the equations can be linearized

0mgLxmLmL2 =θ++θ &&&&

or in matrix forms as

( )0;sin;1cos 2 ≈θθ≈θ≈θ &

( ) )t(fkxxcmLxmM =++θ++ &&&&&

=

θ

+

θ

+

θ

+ )t(f0

xk00mgL

xc000

x)mM(mLmLmL2

&

&

&&

&&



Chapter 4 - Review Slide

k2

x

f

k1

21eq kkk +=

k2

x1

f

k1

x2

21

21eq kk

kkk+

=

πζ≅ 2xxln2

1x1 x2 xn

t1 t2 tn

T

4 - Mechanical Systems - University of Massachusetts Lowell

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Transcript of 4 - Mechanical Systems - University of Massachusetts Lowell