제어응용연구실 1 Modeling of Mechanical systems Ⅰ. 제어응용연구실 2 CONTENTS ▶ Equations of Mechanical System ▶ Modeling of Mechanical System Elements

제어응용연구실 1

Modeling of Mechanical Modeling of Mechanical systems Ⅰsystems Ⅰ


CONTENTSCONTENTS

▶ ▶ Equations of Mechanical SystemEquations of Mechanical System

▶ ▶ Modeling of Mechanical System ElementsModeling of Mechanical System Elements


IntroductionIntroduction

PointPoint

1. 선형 / 비선형 모형과 시변 / 시불변 모형의 형태와 물리적인 의 미를 학습

2. Translational motion, Rotation Motion 의 운동 해석


Modeling of Modeling of Mechanical System ElementsMechanical System Elements


Translational MotionTranslational Motion

Modeling of Mechanical System ElementsModeling of Mechanical System Elements

• MassMass

Ma forces

: An element that stores the kinetic energy of translational motion

g

wm ▶ ( g is the gravitational acceleration constant )

2222 in./s386ft/s2.32s/cm981s/m81.9 g

※ ※ Mass Mass is analogous to inductance inductance of electric networks



※ ※ Conversion TableConversion Table


※ ※ Conversion TableConversion Table




Force : 1 N = 0.2248 lb (force) = 3.5969 oz (force)

Mass : 1kg = 1000g = 2.2046 lb (mass) = 35.274 oz (mass) = 0.06852 slug

Distance : 1 m = 3.2808 ft = 39.37 in. 1 in. = 25.4 mm 1 ft = 0.3048 m

dt

tdvM

dt

tydMtMatf

)()()()(

2

2

Figure. Force-mass system

M

y (t)

f (t)

The force equationThe force equation

―

―

―



• Linear SpringLinear Spring

( K : the spring constant, or simply stiffness )( K : the spring constant, or simply stiffness )

Figure. Force-spring system

)()( tKytf The force equationThe force equation

: An element that stores potential energy

※ ※ Linear Spring Linear Spring is analogous to a capacitor capacitor in electric networks



• Friction Friction

― Viscous Friction

Figure. Dashpot for viscous friction

dt

tdyBtf

)()(

The force equationThe force equation

: A retarding force that is a linear relationship between the applied force and velocity.

f

y

BSlope( B : Viscous frictional coefficient )( B : Viscous frictional coefficient )

0



― Static Friction

: A retarding force that tends to prevent motion from beginning.

f

y0

sF

0)()(

ysFtf

― Coulomb Friction

: A retarding force that has a constant amplitude with respect to the change of velocity.

f

y0

cF

dttdy

dttdyFtf c /)(

/)()(

cF



Rotational MotionRotational Motion

J torques

• InertiaInertia

: An element that stores the kinetic energy of rotational motion

Figure. Torque-inertia system

2

2 )()()()(

dt

tdJ

dt

tdJtJtT

221 MrJ

For instance, the inertia of a circular disk or shaft about its geometric axis is given byFor instance, the inertia of a circular disk or shaft about its geometric axis is given by

The torque equationThe torque equation



22

22

222

25-

cm-g 980sec-cm-g 1

in-oz 386 sec-in.-oz 1

ft-lb 32.2sec-in.-oz 192 sec-ft-lb 1

sec-in.-oz 10 1.417 cm-g 1 : Inertia

ft-lb 0.00521 in.-oz 1

in.-oz 192 ft -lb 1

in.-oz 0.0139 cm-g 1 : Torque

deg/sec 6 rpm 1

rad/sec 1047.060

2 rpm 1 : Velocity Angular

deg 3.57180 rad 1 :nt DisplacemeAngular

―

―

―

―



• Torsional SpringTorsional Spring

Figure. Torque-torsional spring system

The torque equationThe torque equation

)()( tKtT

• Friction for Rotational MotionFriction for Rotational Motion

dt

tdBtT

)()( : friction Viscous

0)()( : friction Static

sFtT

dttd

dttdFtT c /)(

/)()( : friction Coulomb



Conversion ( Conversion ( Between Translational and Rotational MotionsBetween Translational and Rotational Motions ))

Figure. Rotary-to-linear motion Control system ( lead screw)

2

2

L

g

WJ

• A load may be controlled to move along a straight line through a rotary motor-and-screw assembly.


Figure. Rotary-to-linear motion Control system ( rack and pinion)


Figure. Rotary-to-linear motion Control system ( belt and pulley)

• A similar situation in which a rack -and-pinion is used as a mechanical linkage.

• The control of a mass through a pulley by a rotary motor, such as the control of a printwheel in an electric typewriter.

22 rg

WMrJ



Gear Trains, Levers, and Timing BeltsGear Trains, Levers, and Timing Belts

Figure. Gear train

2211

2211

1221

.3

.2

.1

TT

rr

NrNr

2

1

1

2

2

1

1

2

2

1 r

r

N

N

T

T



Figure. Gear train with friction and inertia

1. The torque equation for gear 2

2

22

222

22

22

)()()(

cFdt

tdB

dt

tdJtT

2. The torque equation on the side of gear 1

)()()(

)( 11

11

112

12

1 tTFdt

tdB

dt

tdJtT c

• Example of torque equationExample of torque equation



3. Converte

2

22

2

112

2

2

12

12

2

2

2

12

2

11

)()()()(

cFN

N

dt

tdB

N

N

dt

tdJ

N

NtT

N

NtT

2

22

2

12

2

1

22

12

2

2

1

22

12

2

2

1

: orquefriction t Coulomb : Torque

:locity Angular ve :t coefficienfriction Visous

:nt displacemeAngular : Inertia

cFN

NT

N

N

N

NB

N

N

N

NJ

N

N

※ The following quantities are obtained when reflecting from gear 2 to gear 1 :


Equations of Mechanical SystemEquations of Mechanical System




Figure. Mass-spring-friction system Figure. Free-body diagram

)()()(

)(2

2

tKydt

tdyB

dt

tydMtf

)(1

)()()(

2

2

tfM

tyM

K

dt

tdy

M

B

dt

tyd

• Example ofExample of Mechanical system Mechanical system


)()(

21 txdt

tdx

)(1

)()()(

212 tf

Mtx

M

Btx

M

K

dt

tdx

KBsMssF

sY

2

1

)(

)(






)()()()()(

)( 21121

21

2

11 tytyKdt

tdy

dt

tdyB

dt

tydMgMtf

)()()(

)()()()(

2222

2

222

2

221121

12 tyKdt

tydB

dt

tydMtytyK

dt

tdy

dt

tdyBgM

)()( 11 tytx

dt

tdx

dt

tdytx

)()()( 11

2

)()( 23 tytx

dt

tdx

dt

tdytx

)()()( 32

4


― Force equations for the system

― State variables


)()(

21 txdt

tdx

gtfM

txtxM

Btxtx

M

K

dt

tdx )(

1)]()([)]()([

)(

142

1

131

1

12

)()(

43 txdt

tdx

gtxBBM

txM

KKtx

M

Btx

M

K

dt

tdx

)()(

1)()()(

)(421

23

2

212

2

11

2

14

)()( 11 txty

)()( 32 txty


― State equations

― Output equations



― State diagram


s

gKMKMKMsBMBMBMsMM

sFKKsBBsM

gsY

1221111221112

21

21212

21

)(

)()()(

)(

s

gMMKsMMBsMMsF

KsBsY

)()(

)()( 2112112

21112

2121212

2121211

321211

421

)(])([

])([

KKsKBBKsBBMKKKM

sMBBBMsMM


― Applying the gain formula to the state diagram



uemotor torq)( tTm

tcoefficienfriction ousmotor viscmB

ntdisplaceme laod)( tLshaft theofconstant springK

velocitylaod)( tL

inertiamotor mJ

inertia loadLJ

citymotor velo)( tmntdisplacememotor m

▲

▲

▲

▲

▲

▲

▲

▲

▲

Figure. Motor-load system



)(1

)]()([)()(

2

2

tTJ

ttJ

K

dt

td

J

B

dt

tdm

mLm

m

m

m

mm

2

2 )()]()([

dt

tdJttK L

LLm

)(1

)]()([)()(

2

2

tTJ

ttJ

K

dt

td

J

B

dt

tdm

mLm

m

m

m

mm

)]()([)(

2

2

ttJ

K

dt

tdLm

L

L


― Torque equations for the system



)()()(

231 txtxdt

tdx

)()(

12 tx

J

K

dt

tdx

L

)(1

)()()(

313 tT

Jtx

J

Btx

J

K

dt

tdxm

mm

m

m

― State equations

― State diagram


])([ )(

)(

)(

)(23

23

KBsJJKsJBsJJs

KsJ

ssT

sX

sT

s

mLmLmLm

L

mm

m

])([ )(

)(

)(

)(23

2

KBsJJKsJBsJJs

K

ssT

sX

sT

s

mLmLmLmmm

m


― Applying the gain formula to the state diagram

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제어응용연구실 1 Modeling of Mechanical systems Ⅰ. 제어응용연구실 2 CONTENTS ▶ Equations of Mechanical System ▶ Modeling of Mechanical System Elements