3_CSA_System_Modelling_Mechanical.pdf

8/13/2019 3_CSA_System_Modelling_Mechanical.pdf

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System Modeling

Chapter 2a


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echanical Systems2


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Mechanical Systems

In a translationalor linearmotion system, the

variables involved include displacement, velocity,

accelerationand force.

Displacementrefers to the positional displacement of

a system component with reference to a referencepoint.

It is commonly denoted x(t). It is a function of time, t,

because the displacement could change from time to

time.

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Mechanical Systems

Velocityis the rate of change of displacement,

commonly denoted v(t).

Since it is the rate of change of displacement, we can

express it as

Accelerationis the rate of change of velocity,

commonly denoted a(t). Mathematically, we could

express it as

dt

tdx

tv

dt

txd

dt

tdvta

2

4


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Mass

Components of a linear motion system can be

modeled into three categories of elements: mass,

linear spring, and friction.

All physical things have mass. It is the property that

stores the kinetic energy of translational motion. It isoften denoted as M, and its unit is kg.

M

5


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Reaction Forces - Mass

The reaction force in a mass is always in the opposite

direction of the displacement of the mass.

The magnitude of the reaction force in a mass moved

a displacement ofx(t) is calculated with the well-

known force equation.

M

Displacement,x(t)

Force, 2

2

dt

txdMMaf

M

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Dashpot

Linear motion, is of course subject to friction, which

hinders its motion.

Frictionis the retarding force directly proportional to

the linear velocity. It is often represented by the

symbol of a dashpot. The coefficient of viscosity, or viscous frictional

coefficient, is denoted B, and its unit is Nm-1s-1.

B

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Reaction Forces - Dashpot

A dashpot can be extended, or compressed.

x(t)

Compress

A pair of

outward

forces,

extend

A pair of

inward

forces,

x(t)

dttdxBfD

dttdxBfD

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Spring

Linear springis used to model any component in thesystem that has elasticity.

The coefficient of stiffnessis denoted as k, and it has

a unit of Nm-1.

k

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Reaction Forces - Spring

A spring can be extended, or compressed.

x(t)

compress

A pair of

outward

forces,

x(t)

extend

A pair of

inward

forces,

tkxfS tkxfS

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Mechanical Systems

Let us assume that we have a mechanical system suchas the one below:

How do we determine the transfer function thatrelates the output displacement,x(t), to the input

force,f(t)?

MInput force,f(t)

Displacement,x(t)

k

B

Fixed

point

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Mechanical Systems

Since one end of the spring-and-dashpot section is

connected to a fixed point, the compression of the

section is equal to the movement of the other end,

which isx(t).

MInput force,f(t)

Displacement,x(t)

k

B

Compressed byx(t)

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Mechanical Systems

Cut across parallel sections of dashpots and springsto divide the system into several stand-aloneequilibrium of forces.

Reaction

force, r1(t)

Reaction

force, r2(t)kx(t)

kB

Section B

dt

tdxB

BM

Input force,

f(t)

k

kx(t)

Section A

2

2

dt

txdM

dt

tdx

B

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Mechanical Systems

BM

Input force,

f(t)

k

kx(t)

Section A

2

2

dt

txdM

dt

tdx

B

15

tkxdt

tdxB

dt

txdMtf

2

2


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Mechanical Systems

Reaction

force, r1(t)

Reaction

force, r2(t)kx(t)

kB

Section B

dt

tdxB

16

tkxdt

tdxBtrtr 21


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Mechanical Systems

The last step involves Laplacetransforming the

equations of forces obtained from Step 5 into the

s-domainto form the required transfer function.

[f(t)] =F(s) and [x(t)] =X(s)

skXtkx

0xssXBdt

tdxB

0

2

2

2

0tdt

tdxsxsXsM

dt

txdM

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Mechanical Systems

Assuming initial conditions to be zero

and

This would reduce the last two equations to

00 x

00

tdt

tdx

sXMs

dt

txdM 2

2

2

sBsXdttdx

B

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Mechanical Systems

So

The transfer function is

tkx

dt

tdxB

dt

txdMtf

2

2

skXsBsXsXMssF 2

sXkBsMssF 2

kBsMssFsX

2

1

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Exercise 1

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Determine the transfer functions,

and

21

f(t)

x2(t)

k1

k2

B

x1(t)

M1

M2

sFsX

1

sFsX2


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22

f(t)

x2(t)

k1 B

x1(t)

M1

M2

2

21

1

dt

txdM

dt

tdx

B

1

2

22

2

dt

txdM

k1x

1(t)

k2x2(t)

k1x2(t)

dt

tdxB

2

k1x2(t)

dt

tdxB

2


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23

f(t)

x1(t)

2

21

1

dt

txdM

dt

tdx

B

1k1x1

k1x2

txtxktxtxdtd

Bdt

txdMtf 2112121

2

1

dt

tdxB

2


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24

x2(t)

M2

dt

tdxB

1

2

22

2

dt

txdM

k1x1(t)

k2x2(t)

dt

tdxB

2

dt

tdxB

2

k1x2(t)

txtxktxtxdtdBtxk

dttxdM 21121222

2

2

2


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Laplace Table

sXMsdt

txdM 2

2

2

sBsXdttdx

B

26

skXtkx


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27

sXsXksXsXBssXsMsF 2112112

1

1212

11 kBssXkBssMsXsF

}{ 2112121

2

1 txtxktxtxdt

dB

dt

txdMtfL

txtxktxtxdtd

Bdt

txd

Mtf 2112121

2

1

sFsX

1


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28

11122

22 kBssXkBsksMsX

122

2

112

kBsksM

kBssXsX

sXsXksXsXBssXksXsM 211212222

2

txtxktxtxdt

dBtxkdt

txdM 211212222

2

2

}{ 211212222

2

2 txtxktxtxdt

dBtxk

dt

txdML


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29

112221

11

2

11 kBskBsksM

kBs

sXkBssMsXsF

12

2

2

2

112

2

21

2

11

kBsksM

kBskBsksMkBssMsXsF

12

2

2

2

112

2

21

2

11

kBsksM

kBskBsksMkBssMsXsF

12

2

2

2

2

211

2

12

2

2

2

11

kBsksM

ksMkBskBssMksMsMsXsF


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30

12

2

2

2

2

2

2

112

2

2

2

11

kBsksMksMsMkBsksMsMsXsF

2222112222112

2

21

ksMsMkBsksMsMkBsksM

sFsX


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31

sXsXksXsXBssXksXsM 211212222

2

11122

22 kBssXkBsksMsX

1

122221

kBskBsksMsXsX

sFsX

2


32/32

32

21121122112

kBskBssMkBsksM

kBs

sF

sX

1212

1

1

12

2

12 kBssXkBssM

kBskBsksMsXsF

1

2

11

2

112

2

1

2

kBs

kBskBssMkBsksM

sXsF

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3_CSA_System_Modelling_Mechanical.pdf