Feedback Control Systems ( FCS )

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Feedback Control Systems (FCS)

Dr. Imtiaz Hussainemail: [email protected]

URL :http://imtiazhussainkalwar.weebly.com/

Lecture-6-7-8Mathematical Modelling of Mechanical Systems

mailto:[email protected]

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Outline of this Lecture

•Part-I: Translational Mechanical System

•Part-II: Rotational Mechanical System

•Part-III: Mechanical Linkages

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Basic Types of Mechanical Systems

• Translational– Linear Motion

• Rotational– Rotational Motion

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TRANSLATIONAL MECHANICAL SYSTEMSPart-I

Basic Elements of Translational Mechanical Systems

Translational Spring

i)

Translational Massii)

Translational Damperiii)


i)

Circuit Symbols

Translational Spring• A translational spring is a mechanical element that

can be deformed by an external force such that the deformation is directly proportional to the force applied to it.


Translational Spring• If F is the applied force

• Then is the deformation if

• Or is the deformation.

• The equation of motion is given as

• Where is stiffness of spring expressed in N/m

2x1x

02 x1x

)( 21 xx

)( 21 xxkF

k

F

F

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Translational Spring• Given two springs with spring constant k1 and k2, obtain

the equivalent spring constant keq for the two springs connected in:

(1) Parallel (2) Series

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(1) ParallelFxkxk 21

Fxkk )( 21

Fxkeq

• The two springs have same displacement therefore:

21 kkkeq

• If n springs are connected in parallel then:

neq kkkk 21

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(2) SeriesFxkxk 2211

• The forces on two springs are same, F, however displacements are different therefore:

11 k

Fx 2

2 kFx

• Since the total displacement is , and we have 21 xxx xkF eq

2121 k

FkF

kFxxxeq

11


• Then we can obtain

21

21

21

111

kkkk

kk

keq

21 kF

kF

kF

eq

• If n springs are connected in series then:

n

neq kkk

kkkk

21

21

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Translational Spring• Exercise: Obtain the equivalent stiffness for the following

spring networks.

3k

i)

ii) 3k

Translational Mass

Translational Massii)

• Translational Mass is an inertia element.

• A mechanical system without mass does not exist.

• If a force F is applied to a mass and it is displaced to x meters then the relation b/w force and displacements is given by Newton’s law.

M)(tF

)(tx

xMF

Translational Damper

Translational Damperiii)

• When the viscosity or drag is not negligible in a system, we often model them with the damping force.

• All the materials exhibit the property of damping to some extent.

• If damping in the system is not enough then extra elements (e.g. Dashpot) are added to increase damping.

Common Uses of DashpotsDoor Stoppers Vehicle Suspension

Bridge Suspension Flyover Suspension


xCF

• Where C is damping coefficient (N/ms-1).

)( 21 xxCF


• Translational Dampers in series and parallel.

21 CCCeq 21

21

CCCC

Ceq

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Modelling a simple Translational System• Example-1: Consider a simple horizontal spring-mass system on a

frictionless surface, as shown in figure below.

or kxxm

0 kxxm

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Example-2• Consider the following system (friction is negligible)

• Free Body Diagram

MF

kfMf

k

Fx

M

• Where and are force applied by the spring and inertial force respectively.

kf Mf

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Example-2

• Then the differential equation of the system is:

kxxMF

• Taking the Laplace Transform of both sides and ignoring initial conditions we get

MF

kfMf

Mk ffF

)()()( skXsXMssF 2

21

)()()( skXsXMssF 2

• The transfer function of the system is

kMssFsX

2

1)()(

• if

12000

1000

Nmk

kgM

20010

2 ssF

sX .)()(

Example-2

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• The pole-zero map of the system is

20010

2 ssF

sX .)()(

Example-2

-1 -0.5 0 0.5 1-40

-30

-20

-10

0

10

20

30

40Pole-Zero Map

Real Axis

Imag

inar

y A

xis

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Example-3• Consider the following system

• Free Body Diagram

k

Fx

M

C

MF

kf

MfCf

CMk fffF

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Example-3

Differential equation of the system is:

kxxCxMF

Taking the Laplace Transform of both sides and ignoring Initial conditions we get

)()()()( skXsCsXsXMssF 2

kCsMssFsX

2

1)()(

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Example-3

kCsMssFsX

2

1)()(

• if

1

1

1000

2000

1000

msNC

Nmk

kgM

/

10000010

2

sssFsX .)()(

-1 -0.5 0 0.5 1-2

-1.5

-1

-0.5

0

0.5

1

1.5

2Pole-Zero Map

Real Axis

Imag

inar

y Ax

is

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• Free Body Diagram (same as example-3)

MF

kf

MfBf

BMk fffF kBsMssF

sX

21

)()(

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• Mechanical Network

kF

2x

M1x B

↑ M

k

BF

1x 2x

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Example-5

• Mechanical Network

↑ M

k

BF

1x 2x

)( 21 xxkF

At node 1x

At node 2x

22120 xBxMxxk )(

Example-6• Find the transfer function X2(s)/F(s) of the following system.

1M 2M

k

B

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Example-7

k

)(tf

2x

1M4B3B

2M

1x

1B 2B

↑ M1k 1B)(tf

1x 2x3B

2B M2 4B

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Example-8• Find the transfer function of the mechanical translational

system given in Figure-1.

Free Body Diagram

Figure-1

M

)(tf

kf

Mf

Bf

BMk ffftf )(kBsMssF

sX

21

)()(

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Example-9• Restaurant plate dispenser

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Example-10• Find the transfer function X2(s)/F(s) of the following system.

Free Body Diagram

M1

1kf

1Mf

Bf

M2

)(tF

1kf

2Mf

Bf2kf

2k

BMkk fffftF 221

)(

BMk fff 11

0

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Example-11

1k

)(tu

3x

1M

4B3B

2M

2x

2B 5B

2k 3k

1x

1B

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Example-12: Automobile Suspension

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Automobile Suspension

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Automobile Suspension

).()()( 10 eq ioioo xxkxxbxm

2 eq. iiooo kxxbkxxbxm

Taking Laplace Transform of the equation (2)

)()()()()( skXsbsXskXsbsXsXms iiooo 2

kbsmskbs

sXsX

i

o

2)()(

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Example-13: Train Suspension

Car BodyBogie-2

Bogie

Frame

Bogie-1

WheelsetsPrimary

Suspension

Secondary

Suspension

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Example: Train Suspension

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ROTATIONAL MECHANICAL SYSTEMSPart-I

Basic Elements of Rotational Mechanical Systems

Rotational Spring

)( 21 kT

21


Rotational Damper

21

)( 21 CT

T

C


Moment of Inertia

JT

TJ

Example-1

1

T 1J

1k1B

2k

2J2

3

↑ J1

1k

T

1 31B

J2

2

2k

Example-2

↑ J1

1k

1BT

1 32B

3B J2 4B

2

1

T 1J

1k

3B

2B4B

1B

2J2

3

Example-3

1T

1J

1k

2B 2J

22k

Example-4

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MECHANICAL LINKAGESPart-III

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Gear• Gear is a toothed machine part, such

as a wheel or cylinder, that meshes with another toothed part to transmit motion or to change speed or direction.

Fundamental Properties• The two gears turn in opposite directions: one clockwise and

the other counterclockwise.

• Two gears revolve at different speeds when number of teeth on each gear are different.

Gearing Up and Down • Gearing up is able to convert torque to

velocity.

• The more velocity gained, the more torque sacrifice.

• The ratio is exactly the same: if you get three times your original angular velocity, you reduce the resulting torque to one third.

• This conversion is symmetric: we can also convert velocity to torque at the same ratio.

• The price of the conversion is power loss due to friction.

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Why Gearing is necessary?

• A typical DC motor operates at speeds that are far too

high to be useful, and at torques that are far too low.

• Gear reduction is the standard method by which a

motor is made useful.

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Gear Trains

Gear Ratio• You can calculate the gear ratio by using

the number of teeth of the driver divided by the number of teeth of the follower.

• We gear up when we increase velocity and decrease torque. Ratio: 3:1

• We gear down when we increase torque and reduce velocity. Ratio: 1:3

Gear Ratio = # teeth input gear / # teeth output gear = torque in / torque out = speed out / speed in

FollowerDriver

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Example of Gear Trains• A most commonly used example of gear trains is the gears of

an automobile.

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Mathematical Modelling of Gear Trains• Gears increase or reduce angular velocity (while

simultaneously decreasing or increasing torque, such that energy is conserved).

2211 NN

1N Number of Teeth of Driving Gear

1 Angular Movement of Driving Gear

2N Number of Teeth of Following Gear

2 Angular Movement of Following Gear

Energy of Driving Gear = Energy of Following Gear

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Mathematical Modelling of Gear Trains• In the system below, a torque, τa, is applied to gear 1 (with

number of teeth N1, moment of inertia J1 and a rotational friction B1).

• It, in turn, is connected to gear 2 (with number of teeth N2, moment of inertia J2 and a rotational friction B2).

• The angle θ1 is defined positive clockwise, θ2 is defined positive clockwise. The torque acts in the direction of θ1.

• Assume that TL is the load torque applied by the load connected to Gear-2.

B1

B2

N1

N2

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Mathematical Modelling of Gear Trains• For Gear-1

• For Gear-2

• Since

• therefore

B1

B2

N1

N2

2211 NN

11111 TBJa Eq (1)

LTBJT 22222 Eq (2)

12

12

NN

Eq (3)

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Mathematical Modelling of Gear Trains• Gear Ratio is calculated as

• Put this value in eq (1)

• Put T2 from eq (2)

• Substitute θ2 from eq (3)

B1

B2

N1

N2

22

11

1

2

1

2 TNN

TNN

TT

22

11111 TNN

BJa

)( La TBJNN

BJ 22222

11111

)( La TNN

NN

BNN

JNN

BJ2

12

2

121

2

12

2

11111

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Mathematical Modelling of Gear Trains

• After simplification

)( La TNN

NN

BNN

JNN

BJ2

12

2

121

2

12

2

11111

La TNN

BNN

BJNN

J2

112

2

2

11112

2

2

111

La TNN

BNN

BJNN

J2

112

2

2

1112

2

2

11

2

2

2

11 J

NN

JJeq

2

2

2

11 B

NN

BBeq

Leqeqa TNN

BJ2

111

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Mathematical Modelling of Gear Trains

• For three gears connected together

3

2

4

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2

12

2

2

11 J

NN

NN

JNN

JJ eq

3

2

4

32

2

12

2

2

11 B

NN

NN

BNN

BBeq

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Home Work• Drive Jeq and Beq and relation between applied

torque τa and load torque TL for three gears connected together.

J1 J2 J3

13

2

τa

1N2N

3N

1B2B

3B

LT

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END OF LECTURES-6-7-8

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Feedback Control Systems ( FCS )