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Transcript of Modelling of rotational mechanical system
1
Feedback Control Systems (FCS)
Dr. Imtiaz Hussainemail: [email protected]
URL :http://imtiazhussainkalwar.weebly.com/
Lecture-6-7-8Mathematical Modelling of Mechanical Systems
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Outline of this Lecture
•Part-I: Translational Mechanical System
•Part-II: Rotational Mechanical System
•Part-III: Mechanical Linkages
3
Basic Types of Mechanical Systems
• Translational– Linear Motion
• Rotational– Rotational Motion
4
TRANSLATIONAL MECHANICAL SYSTEMSPart-I
Basic Elements of Translational Mechanical Systems
Translational Spring
i)
Translational Massii)
Translational Damperiii)
Translational Spring
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
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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Translational Spring
(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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Translational Spring
(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
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Translational Spring
• 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
Translational Damper
xCF
• Where C is damping coefficient (N/ms-1).
)( 21 xxCF
Translational Damper
• 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
20
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
22
• 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
26
Example-4• Consider the following system
• Free Body Diagram (same as example-3)
MF
kf
MfBf
BMk fffF kBsMssF
sX
21
)()(
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Example-5• Consider the following system
• 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
34
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
37
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
Basic Elements of Rotational Mechanical Systems
Rotational Damper
21
)( 21 CT
T
C
Basic Elements of Rotational Mechanical Systems
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
48
MECHANICAL LINKAGESPart-III
49
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.
52
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
32
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
63
END OF LECTURES-6-7-8
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