ME 375 System Modeling and Analysis Section 2 ... Handouts/chapter_2... · 1 ME 375 System Modeling...

1

ME 375 System Modeling and Analysis

Section 2 – Translating and Rotating g gMechanical Systems

Spring 2009School of Mechanical Engineering

Douglas E. AdamsAssociate Professor

2

Motivational ExampleEffects of suspension on exhaust system vibration

2.1

In automotive systems, a change to one component can cause noise and vibration problems in other components. - Consumers lose confidence when noises arise and mechanical parts can fail when vibrations are excessive. These problems result in warranty losses, recalls, and significant financial setbacks.

Increasein K2

© 2009 D. E. AdamsME 375 – Translating and Rotating Mechanical Systems

N&V problemRoad

3

Key Points to RememberThree elements and free oscillations

2.2

There are three physical elements in mechanical systems:p y y

Inertia- characterized by MASS (M, I)- maintains motionDissipation- characterized by DAMPING (C B)

We use lumped elementsthat are ideal

We usually deal withequivalent mass, dampingand stiffness- characterized by DAMPING (C, B)

- eliminates motionElasticity- characterized by (K)- opposes (or restores) motion

and stiffness

Mass and stiffness areeasiest to estimate

Damping is never correct


Free oscillations only occur when systems contain both mass and stiffness

4

Mechanical ElementsMass, damping and stiffness characteristics (LINEAR)

2.3

Each lumped (linear) element has its own characteristic:

M

K f

Bx

Δx

F B

F M

Also called “C”

This is

Δx

F KxM

Note that theseare straight lineswith constant slopes(also massless!) 1/K is called “flexibility”

Newton’s2nd Law fora particle


(also massless!)

Damper / spring forces are determined by relative motion (Δ).Inertia force is determined by absolute acceleration.

1/K is called flexibility

5

Equivalent Mechanical ElementsModeling mechanical elements using first principles

2.4

In most practical situations, we must use models to develop our models Consider the following example:

P

develop our models. Consider the following example:

Kequiv=E,I

3

3 equiv

PL P EffortDisplacmentEI K Stiffness

δ= = = =


equiv ff

Where did the mass go? Assume: 1) thin beam, 2) low frequencyWhy? Why?

6

Mechanical ElementsMass, damping and stiffness characteristics (NONLINEAR)

2.5

Lumped (nonlinear) elements have their own characteristics:

M

K f

Bx

Δx

F B

e.g., Shocks

p

e.g., Relativity

2

2

1cv

mvp

−

=

Δx

F K

e.g., Susp. coils

Note that theseare not straightlines

xM


Nonlinear characteristics often provide desirabledynamic characteristics so we use them intentionally.

7

Practical Nonlinear Example - Elastomers2.6

Linear?

Characteristic inshear is differentthan either of the Linear? Different in

compression thantension; anasymmetric nonlinear spring

other characteristics;it starts off like thetension curve and thentransitions into thecompression curve

Opposite behavior


8

2.7

Overcoming Design Trade-Offs in ShuttleTrade-off in isolation and impact resistance in Thermal Protection Systems

In the space shuttle orbiter, the thermal protection system involves a design trade-off: the orbiter airframe must be isolated from a hostile des g t ade o t e o b te a a e ust be so ated o a ost eacoustic environment and must also be capable of sustaining foreign object impact loads. This trade-off has been overcome to a large extent using a highly nonlinear silicon isolation pad.


Orbiter airframe is isolated from external acoustic loads (140-165 dB) in the low strain region of the strain isolation pad.

Resistance to moderate energy impacts is achieved in the high strain region of the strain isolation pad.

10

Mechanical ElementsWhen are mass, damping and stiffness important to consider?

2.8

Stiffness is important to consider when STATIC loads are applied because springs sustain static loadsapplied because springs sustain static loads.Stiffness is important to consider when DYNAMIC loads are applied as well because stiffness is constantly working to pull systems back to their resting position.

Damping is important to consider when DYNAMIC loads are applied because damping is the only element that can stop motion especially when resonances occur.

Mass is important to consider when DYNAMIC loads are applied because mass resists sudden changes in motion.


In general, as frequency increases, mass is most important; as frequency decreases, stiffness is most important; and in the middle, damping is most important.

11

Element ContributionsWhen are mass, damping and stiffness important to consider?

2.9

Consider the following scenario of an aircraft landing gear undergoing a hard landingundergoing a hard landing.

Spring supportsMass of wheelis dominant at

Dampingi d i t


weight of aircraftafter landing during taxi

is dominant at instant tire patch impacts runway and wheel moves quickly

is dominant when tire patchimpacts runway and wheel resonates

12

Mechanical Interconnection LawsCompatibility and continuity in mechanical systems

2.10

Interconnections are described with two mechanical laws:N t /E l L ( tibilit )Newton/Euler Laws (compatibility)

jiij

CMext

ff

vMdtdFrr

rr

−=

=∑ 2nd law

3rd lawPhysics

Kinematics

Kinematics (continuity)

break!t can' Connection

0)( =∑loop

iMotion


Free body diagrams (FBDs) are a book keeping mechanism for mechanical systems to implement interconnection laws.

13

Vertical Single Degree of Freedom (SDOF) System

Gravity is an additional static force/undeformed length

2.11

K

fB

xg

M

3. Define coordinates1. Framework for model

- Define the problem- What to model?- Assumptions- Draw a good picture

(Kinematics)- Absolute or relative- Positive or negative

xuoxs

Deformedlength

4. Draw FBD(s)

2. Determine # of DOFS (system order)

sconstraintKinematic

#sCoordinate

## −=orient) & (locate

DOFs

EngineeringJudgment

SystemThis will beI t t

M

xxd

Absolute

Undeformedlength

xu

( ) KxuxK =−xB &Judgment+ System Important

later

Mg

M

( ) uo KxuxK =−

)(tf

xB

Directions for internal forces? They oppose motion.


14

d

Vertical Single Degree of Freedom (SDOF) System

Gravity is an additional static force/undeformed length5. Apply Newton’s (or D’Alembert’s) Law

t h DOF

2.12

From the

( )

( ) ( ) ( ))(

)(

)()( )(

2

2

MgtfKxxBxM

MgtfuuxKuxdtdBux

dtdM

MgtfuxKxBxMtfMguxKxB

xMxMdtdF

uuu

oououou

o

o

+=++

+=−+++++

+=−++++−−−=

==∑

&&&

&&&

&

&&&to each DOF(Kinetics/physics)

x=xu+uo

From the

undeformedposition

( ) ( ) ( )( )

)()(][

)(2

2

tfKxxBxMtfMguxKKxxBxM

MgtfuxxKxxdtdBxx

dtdM

OR

ddd

osddd

osdsdsd

=++=−−+++

+=−+++++

&&&

&&&

Static FBD

x=xd+xs

deformed(static equil.)position

)(tfKxxBxM ddd ++

6. Eliminate kinematic constraints7. Put model in appropriate form and solve for response8. Interpret results and validate


15

Vertical single degree of freedom (SDOF) system

What if we move the base instead of forcing the mass?

41

2.13

2

MMg

( ) uo KxuxK =−xB &

4K

fB

xg

M

xb1

112sconstraint

Kinematic#

sCoordinate##

=−=

−=orient) & (locate

DOFs

( ) ( )

bddd

bo

ob

xMKxxBxMMgxMKuKxxBxM

MguxKxBxxdtdM

&&&&&

&&&&&

&

−=++

+−=−++

+−−−=+2

2

5

xuoxs

Deformedlength

3 xb

Mxd

Relative

Undeformedlength

xu


16

Series and Parallel ElementsSeries and parallel combinations

Parallel combinations FF 21

2.14

Series combinations

( ) ( )ji MotionMotionMotionSame

Δ=Δelements across

( )KK

KKFK

FFx

FKKF

KF

KFxx

211211

11

22

2

2

1

121

+=

+=

=

=⇔=

Series combinations

( ) ( )ji ForceForceForceSame

=elements across

eqeq KKK 1

( ) ( )( )( )02012

21

2011

021220111

21

llxKK

Klx

lxxKlxKFF

+−+

=−

−−=−=

ji

jiSERIESjiPARALLEL EE

EEEEEE

+=+= ,

( )( )

( )( )02012

0201221

21

llxK

llxKK

KKF

equiv

i

+−=

+−+

=⇒


17

Two degree of freedom system (MDOF)Think about relative motion and how springs anddampers oppose motion

2.15

2

M1 M2

f1

C2C1

K1 K2

x1

f2

x21

No gravity inthis problem.Springs initiallyundeformed.

202sconstraint

Kinematic#

sCoordinate##

=−=


DOFs

4M1

11

11

xKxC &

)(tf

M2( )( )122

122

xxKxxC

−

− &&

)(tfM1 M2

(abs) x23

)(1 tf )(2 tf

( ) ( )( ) ( ) )(

)(

212212222

1122122111111

tfxxKxxCxMtfxxKxxCxKxCxM

+−−−−=+−+−+−−=

&&&&

&&&&&

5

(abs) x1Coordinates fromdeformed positions


18

Two degree of freedom system (MDOF)Different choices of coordinates lead to different kindsof coupling – but the dynamics are identical

2.16

M1 M2

f1

C2C1

K1 K2

x1

f2

x2

Static coupling (check)

( ) ( ))()(

21212222222

1222212112111

tfxKxCxKxCxMtfxKxCxKKxCCxM

=−−++=−−++++

&&&&&&&&

ABSOLUTE COORDINATES

RELATIVE COORDINATES

)()(

221221221212

1212212111111

tfxKxCxMxMtfxKxCxKxCxM

=+++=−−++

&&&&&

&&&&

Dynamic (inertial) coupling© 2009 D. E. AdamsME 375 – Translating and Rotating Mechanical Systems

19

Two DOF system – Matrix form of EOMCoupling mechanisms are important and we can usematrices to make them easier to understand

2.17

M1 M2

f1

C2C1

K1 K2

x1

f2

x2


Symmetric mass,damping, and stiffnessmatrices+static coupling

Response (output) vector

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡−

−++

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡−

−++

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡)()(

00

2

1

2

1

22

221

2

1

22

221

2

1

2

1

tftf

xx

KKKKK

xx

CCCCC

xx

MM

&

&

&&

&&


RELATIVE COORDINATES

Response (output) vector

Force (input) vector

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡ −+

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡ −+

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡)()(

000

2

1

21

1

2

21

21

1

2

21

21

1

22

1

tftf

xx

KKK

xx

CCC

xx

MMM

&

&

&&

&&

Non-symmetric parameter matrices+static and dynamic coupling


20

When Are Lumped Models Not Appropriate?Biological example of orb web spider and prey detection

When mass and stiffness properties are distributed throughout the s stem and the freq enc of interest is high then e m st se

2.18

the system and the frequency of interest is high, then we must use partial differential equations instead of ordinary differential equation.

Spider has

Orb web spider

Propagating waves sent to locate prey

plocated and ensnared prey using a beam-forming type approach

Prey

Actuators Sensors


Propagating waves sent to locate damage

21

Rotational element laws (linear)2.19

Each lumped (linear) element has its own characteristic

Δθ

T B

T I?

ICM

K τ

B θ

This isEuler’s

Δθ

T Kθ

Note that theseare straight lineswith constant slopes(also MASSLESS!)

Euler sLaw for arigid body


Damper and spring rotational torques are determined by relative motion (Δ) .Inertial torques are determined by absolute acceleration.

22

Interconnections are the defining features of systemsEuler’s Laws (compatibility)

Mechanical Interconnection LawsCompatibility and continuity in mechanical systems

2.20

jiij

CMext Idtd

ττ

ωτrr

rr

−=

=∑ 2nd law

3rd lawPhysics

Kinematics

(it’s a little morecomplicated thanthis with movingreference frames)

Kinematics (continuity)

break!t can'Connection

0)( =∑i

iMotion

Free body diagrams (FBDs) are even more important for rotational systems because reference points for moments are crucial


23

Rotational single degree of freedom (SDOF) LEVER

A simple example of rotational modeling technique

2.21


- Define the problem- What to model?- Assumptions (No mass?

No deformation? No friction?) 4. Draw FBD(s)

(abs) θ

θ&B2. Determine # of DOFs

112sconstraint

Kinematic#

sCoordinate##

=−=


DOFs

5. Apply Newton’s (or D’Alembert’s) Lawto each DOF

1f

2fyR

xRa

b

O

θB

Locate CM and orient

C ld l d thi b

Parasiticdynamics

Desired staticrelationship ( )θθ

θθ&&&

&&&

BIb

fbaf

bfafBI

O

O

++=

+−−=1 12

21

Could only do this becausepoint O is not moving


24

Coupled translation-rotation (e.g., MEMS)Equivalent inertia and damping in T-R mechanical system

τrI

θ

2.22

grF

rI cmo ,−

rM

x

xBFxM rgrr −= &&& rB

gB

xrB

xr

Ir

F

rx

BrFI

gcmogr

ggrcmo

rgrr

τ

θ

θτθ

−−=

=

−−=

−

−

&&&

&&&

22

, so where

rxBxM effeff

τ=+ &&&

rx

rB

Bxr

IM gr

cmor

τ=⎟⎟

⎠

⎞⎜⎜⎝

⎛++⎟

⎠⎞

⎜⎝⎛ + − &&&

22


25

Complicated translational-rotationalmechanical system example

When there are absolute and relative motions…BE CAREFUL

2.23

M1 M2

M3,Icm3

M4,Icm4

B1

K1

M5,Icm5

K2

B2

K3


- Looks difficult – take the same approach- Model motion of all masses in the plane- No friction in wheels or pins, massless rod- Bodies 1 and 2 only translate

(Kinematics)- Absolute or relative- Positive or negative

4, cm4

55443321 ,,,,,,, θθθ xxxxxabs rel rel abs rel abs abs abs

2. Determine # of DOFS (What are the constraints?)

448sconstraint

Kinematic#

sCoordinate##

=−=


DOFs

abs rel rel abs rel abs abs abs


26


4. Draw FBDs (6 of them) (What’s the sixth one?)

Doesn’t rotate

2.24

M1

Doesn t rotate

MM3,Icm3

M2Doesn’t rotate

Doesn’t rotateM5,Icm5

Rotates/translates

Rotates/translates

Rotates/translatesRotates/translates

M4,Icm4


27


5. Apply Newton’s Laws (Just apply it where and when you need to)

2.25

( ) ( )

( ) ( ) :3 MASS

:2 MASS

:1 MASS

2

,31233,43312

2

3

2332222212

2

2

,1,312222111111

RxxKRxxdtdM

xxKxBxKxxdtdM

RRxBxKxBxKxM

xx

xRx

+−−=+

−+−−=+

−−++−−=

&

&&&&

( ) ( )

( )

equation) 1 MASSinto (subsitute :5 MASS

: forces, internal unknown thefind toneed We

cossin :4 INERTIA

2

,1,5525

5,5

,

312

2

44444442

444,

d

RRxr

IM

R

xxdtdrMgrMrMI

xRxRcm

qij

cm

−==⎟⎟⎠

⎞⎜⎜⎝

⎛+

+−−=+ θθθ

&&

&&

( )equations) 3 MASSand 1 MASSinto e(substitut :3 INERTIA

equation) 3 MASSinto e(substitut :4 MASS

3,3133,

,43443124

rRI

RrxxdtdM

xcm

x

=

−=++

θ

θ

&&


28


5. Apply Newton’s Laws (continued)

5,3

3,2222111111 x

IM

IxBxKxBxKxM cmcm &&&&&&&& ⎟⎟

⎞⎜⎜⎛

++−++−−= θ

2.26

( ) ( )

( ) ( ) ( )

( ) ( )2

33

3,23344314313

2332222212

525

533

2222111111

cossin xxrMgrMrMI

rI

xxKrxxMxxM

xxKxBxKxxM

xr

Mr

xBxKxBxKxM

cm

&&&&&&

&&&&&&&&&&&&

&&&&&

+−−=+

+−−++−=+

−+−−=+

⎟⎟⎠

⎜⎜⎝

++++=

θθθ

θθ

θ

( ) ( )314444444444, cossin xxrMgrMrMIcm +=+ θθθ

6. Eliminate kinematic constraints

7. Put model in appropriate form and solve

Let’s look at this term

⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

−+−=

!5!31

!5!3sin

42

53

θθθ

θθθθ L

7. Put model in appropriate form and solve

Note that the equations are nonlinear in θ4

8. Interpret results and validate

It’s like a stiffness,but it’s stiffer thannormal when θ getsbigger (NONLINEAR).


Date post:	28-May-2018
Category:	Documents
Upload:	vokhanh
View:	217 times
Download:	0 times

ME 375 System Modeling and Analysis Section 2 ... Handouts/chapter_2... · 1 ME 375 System Modeling...

Documents