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Feedback Control Systems (FCS)
Lecture-3-4-5
Introduction Mathematical Modeling
Mathematical Modeling of Mechanical Systems
Dr. Imtiaz Hussain
email: [email protected]
URL :http://imtiazhussainkalwar.weebly.com/
1
Lecture Outline
• Introduction to Modeling
– Ways to Study System
– Modeling Classification
• Mathematical Modeling of Mechanical Systems
– Translational Mechanical Systems
– Rotational Mechanical Systems
– Mechanical Linkages
2
Model
• A model is a simplified representation or
abstraction of reality.
• Reality is generally too complex to copy exactly.
• Much of the complexity is actually irrelevant in
problem solving.
3
What is Mathematical Model?
A set of mathematical equations (e.g., differential eqs.) that
describes the input-output behavior of a system.
What is a model used for?
• Simulation
• Prediction/Forecasting
• Prognostics/Diagnostics
• Design/Performance Evaluation
• Control System Design
Ways to Study a System
System
Experiment with a
model of the System
Experiment with actual
System
Mathematical Model
Physical Model
Analytical Solution
Simulation
Frequency Domain
Time Domain
Hybrid Domain
5
Mathematical Models
• Black box
• Gray box
• White box
6
Black Box Model
• When only input and output are known.
• Internal dynamics are either too complex or
unknown.
Input
Output
• Easy to Model
7
Grey Box Model
• When input and output and some information
about the internal dynamics of the system are
known.
u(t)
y(t)
y[u(t), t]
• Easier than white box Modelling.
8
White Box Model
• When input and output and internal dynamics
of the system are known.
u(t)
dy(t )
du(t ) d 2 y(t )
3

dt
dt
dt 2
y(t)
• One should have complete knowledge of the
system to derive a white box model.
9
MATHEMATICAL MODELING OF
MECHANICAL SYSTEMS
10
Basic Types of Mechanical Systems
• Translational
– Linear Motion
• Rotational
– Rotational Motion
11
Basic Elements of Translational Mechanical Systems
Translational Spring
i)
Translational Mass
ii)
Translational Damper
iii)
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
i)
Circuit Symbols
Translational Spring
Translational Spring
• If F is the applied force
x2
x1
• Then x1 is the deformation if x2  0
• Or ( x1  x2 ) is the deformation.
• The equation of motion is given as
F  k ( x1  x2 )
• Where k is stiffness of spring expressed in N/m
F
F
Translational Mass
• Translational Mass is an inertia
element.
Translational Mass
ii)
• 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.
F  Mx
x(t )
F (t )
M
Translational Damper
• Damper opposes the rate of
change of motion.
• 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.
Translational Damper
iii)
Common Uses of Dashpots
Door Stoppers
Bridge Suspension
Vehicle Suspension
Flyover Suspension
Translational Damper
F  Cx
F  C( x1  x 2 )
• Where C is damping coefficient (N/ms-1).
Example-1
• Consider the following system (friction is negligible)
k
x
M
F
• Free Body Diagram
fk
F
M
fM
• Where f k and f M are force applied by the spring and
inertial force respectively.
19
Example-1
fk
F
M
fM
F  fk  fM
• Then the differential equation of the system is:
F  kx  Mx
• Taking the Laplace Transform of both sides and ignoring
initial conditions we get
F ( s )  Ms2 X ( s )  kX ( s )
20
Example-1
F ( s )  Ms2 X ( s )  kX ( s )
• The transfer function of the system is
X (s)
1

F(s)
Ms2  k
• if
M  1000kg
k  2000Nm 1
X (s)
0.001
 2
F(s)
s 2
21
Example-2
X (s)
0.001
 2
F(s)
s 2
• The pole-zero map of the system is
Pole-Zero Map
Imaginary Axis
 2
0
− 2
-1
-0.5
0
Real Axis
0.5
1
22
Example-2
• Consider the following system
k
F
x
M
C
• Free Body Diagram
fk
F
M
fC
fM
F  f k  f M  fC
23
Example-3
Differential equation of the system is:
F  Mx  Cx  kx
Taking the Laplace Transform of both sides and ignoring
Initial conditions we get
F ( s )  Ms2 X ( s )  CsX ( s )  kX ( s )
X ( s)
1

F ( s ) Ms2  Cs  k
24
Example-3
X ( s)
1

F ( s ) Ms2  Cs  k
• if
Pole-Zero Map
2
1.5
M  1000kg
k  2000Nm 1
C  1000N / ms
1
Imaginary Axis
1
0.5
0
-0.5
-1
X ( s)
0.001
 2
F ( s ) s  s  1000
-1.5
-2
-1
-0.5
0
0.5
1
Real Axis
25
Example-4
• Consider the following system
x2
x1 k
B
F
M
• Mechanical Network
x1
F
↑
k
x2
M
B
26
Example-4
• Mechanical Network
x1
F
At node
↑
k
x2
M
B
x1
F  k ( x1  x2 )
At node
x2
2
0  k ( x2  x1 )  Mx2  Bx
27
Example-5
• Find the transfer function X2(s)/F(s) of the following system.
k
M1
B
M2
Example-6
x1
x2
B3
k
f (t )
B4
M1
M2
B1
B3
x1
f (t ) ↑
k
M1
B2
B1
B2
x2
M2
B4
29
Example-7
• Find the transfer function of the mechanical translational
system given in Figure-1.
Free Body Diagram
fk
fB
Figure-1
M
f (t )
f (t )  f k  f M  f B
fM
X ( s)
1

F ( s ) Ms2  Bs  k
30
Example-8
• Restaurant plate dispenser
31
Example-9
• Find the transfer function X2(s)/F(s) of the following system.
Free Body Diagram
f k1 f k f B
2
M2
k2
F (t ) f M 2
f k1
fB
M1
f M1
F(t )  f k1  f k2  f M 2  f B
0  f k1  f M1  f B
32
Example-10
x1
u (t )
x2
k1
B1
x3
B4
B3
M1
k2
B2
M2
k3
B5
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Basic Elements of Rotational Mechanical Systems
Rotational Spring
2
1
T  k (1   2 )
Basic Elements of Rotational Mechanical Systems
Rotational Damper
C
2
1
T  C(1  2 )
T
Basic Elements of Rotational Mechanical Systems
Moment of Inertia

J
T  J
T
Example-11
1
k1
B1
2
3
T
J1
1
T
↑
k1
2
J1
k2
J2
B1
3
J2
k2
Example-12
1
k1
B2
2
T
3
J2
J1
B3
B1
1
T
↑
k1
2
J1
B4
B2
B1
B3
3
J2
B4
Example-13
1
k1
J1
B2
2
T
J2
k2
Example-14
MECHANICAL LINKAGES
41
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.
42
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.
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.
45
Gear Trains
46
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
Driver
Follower
• We gear down when we increase torque
and reduce velocity.
Ratio: 1:3
  ℎ   
   
  =
=
=
  ℎ     
 
Example of Gear Trains
• A most commonly used example of gear trains is the gears of
an automobile.
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Mathematical Modeling of Gear Trains
• Gears increase or descrease angular velocity (while
simultaneously decreasing or increasing torque, such
that energy is conserved).
Energy of Driving Gear = Energy of Following Gear
N11
N1
1
N2
2

N 2 2
Number of Teeth of Driving Gear
Angular Movement of Driving Gear
Number of Teeth of Following Gear
Angular Movement of Following Gear
49
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.
N1
N2
B1
B2
50
Mathematical Modelling of Gear Trains
• For Gear-1
 a  J11  B11  T1
Eq (1)
• For Gear-2
T2  J 22  B22  TL
Eq (2)
N1
N2
B1
• Since
B2
N11  N 2 2
• therefore
N1
2 
1
N2
Eq (3)
51
Mathematical Modelling of Gear Trains
• Gear Ratio is calculated as
T2
N2
N1

 T1 
T2
T1
N1
N2
N2
N1
• Put this value in eq (1)
B1
N
 a  J 11  B11  1 T2
N2
B2
• Put T2 from eq (2)
N1



 a  J 11  B11 
( J 22  B22  TL )
N2
• Substitute θ2 from eq (3)
N1
N1 
N1 
N1



 a  J 11  B11 
(J2
 1  B2
2 
TL )
N2
N2
N2
N 2 52
Mathematical Modelling of Gear Trains
N1
N1 
N1 
N1



 a  J 11  B11 
(J2
 1  B2
2 
TL )
N2
N2
N2
N2
• After simplification
2
2
 N1 
 N1 
N





 J 21  B11  
 B21  1 TL
 a  J11  
N2
 N2 
 N2 
2
2




 N1 


N
N1
1







 J 2 1  B1  
 B2 1 
 a  J1  
TL
N2




 N2 
 N2 




2
J eq
N 
 J1   1  J 2
 N2 
Beq
N
 B1   1
 N2
N1



 a  J eq1  Beq1 
TL
N2
2

 B2

53
Mathematical Modelling of Gear Trains
• For three gears connected together
J eq
Beq
 N1
 J1  
 N2
 N1
 B1  
 N2
2
2
2
2

 N1 
 J 2  


 N2 

 N1 
 B2  


 N2 
 N3

 N4
 N3

 N4
2

 J 3

2

 B3

54
Example-15
• Drive Jeq and Beq and relation between applied
torque τa and load torque TL for three gears
connected together.
1
2
3
N1
N2
J1
τa
J2
B2
B1
N3
J3
TL
B3
55
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END OF LECTURES-3-4-5
56

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