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Slide 10.1
Bolton, Mechatronics PowerPoints , 4th Edition, © Pearson Education Limited 2008
Basic system Models• Objectives:
• Devise Models from basic buildingblocks of mechanical, electrical, fluid
and thermal systems• Recognize analogies between
mechanical, electrical, fluid and thermal
systems
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Slide 10.2
Bolton, Mechatronics PowerPoints , 4th Edition, © Pearson Education Limited 2008
• Mathematical Models
• Mechanical system building blocks – Rotational systems
– Building up a mechanical system
Electrical system building blocks- Building up a model for electrical systems
- Electrical and mechanical analogyies
Fluid system building blocksThermal system building blocks
Basic system Models
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Slide 10.3
Bolton, Mechatronics PowerPoints , 4th Edition, © Pearson Education Limited 2008
Mathematical Models
• In order to understand the behavior ofsystems, mathematical models are needed.
Such a model is created using equations andcan be used to enable predictions to bemade of the behavior of a system underspecific conditions.
• The basics for any mathematical model isprovided by the fundamental physical lawsthat govern the behavior of the system.
• This chapter deals with basic buildingblocks and how to combine such blocks tobuild a mathematical system model.
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Slide 10.4
Bolton, Mechatronics PowerPoints , 4th Edition, © Pearson Education Limited 2008
Figure 10.1 Mechanical systems: (a) spring, (b) dashpot, (c) mass
Mechanical system building blocksThe models used to represent mechanical systems have the basic building blocks of:
Springs: represent the stiffness of a system
Dashpots: dashpots are the forces opposing motion, i.e. friction or damping
Masses: the inertia or resistance to acceleration
All these building blocks can be considered to have a force as an inputand a displacement as an output
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Slide 10.5
Bolton, Mechatronics PowerPoints , 4th Edition, © Pearson Education Limited 2008
• The stiffness of aspring is described by:
F=k.x
The object applying theforce to stretch the
spring is also acted onby a force (Newton’sthird law), this force
will be in the oppositedirection and equal insize to the force usedto stretch the spring
Mech. sys blocks: Spring
k is thestiffnessconstant
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Slide 10.6
Bolton, Mechatronics PowerPoints , 4th Edition, © Pearson Education Limited 2008
c : speed of the body
It is a type of forces when wepush an object through a fluid
or move an object againstfriction forces.
Thus the relation between thedisplacement x of the piston,i.e. the output and the forceas input is a relationshipdepending on the rate ofchange of the output
Mech. sys blocks: Dashpots
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Slide 10.7
Bolton, Mechatronics PowerPoints , 4th Edition, © Pearson Education Limited 2008
• F=ma
m: mass, a: acceleration
Mech. sys blocks: Masses
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Slide 10.8
Bolton, Mechatronics PowerPoints , 4th Edition, © Pearson Education Limited 2008
Energy in basic mechanical blocks• The spring when stretched stores energy, the
energy being released when the spring springsback to its original length.
The energy stored when there is an extensionx is:E= kx2 /2=
Energy stored in the mass when its moving with avelocity v, its called kinetic energy, and releasedwhen it stops moving:
E=mv2 /2
No stored energy in dashpot, it dissipates
energy=cv2
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Slide 10.9
Bolton, Mechatronics PowerPoints , 4th Edition, © Pearson Education Limited 2008
Basic Blocks or Rotational System• For rotational system, the equivalent three building blocks are:
a Torsion spring, a rotary damper, and the moment of inertia
With such building blocks, the inputs are torque and the
outputs angle rotated
With a torsional spring
With a rotary damper a disc is rotated in a fluid and
the resistive torque T is:
The moment of inertia has the property that the greaterthe moment of inertia I, the greater the torque neededto produce an angular acceleration
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Slide 10.10
Bolton, Mechatronics PowerPoints , 4th Edition, © Pearson Education Limited 2008
• The stored energy in rotary system:
• For torsional spring:
• Energy stored in mass rotating is :
• The power dissipated by rotary damper
when rotating with angular velocity ωωωω is:
Energy in rotary system
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Slide 10.11
Bolton, Mechatronics PowerPoints , 4th Edition, © Pearson Education Limited 2008Table 10.1 Mechanical building blocks
Summary of Mechanical building blocks
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Slide 10.12
Bolton, Mechatronics PowerPoints , 4th Edition, © Pearson Education Limited 2008
Figure 10.2 (a) Spring–dashpot–mass, (b) system, (c) free-body diagram
Building up a mechanical systemMany systems can be considered to be a mass, a springand dashpot combined in the way shown below
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Slide 10.13
Bolton, Mechatronics PowerPoints , 4th Edition, © Pearson Education Limited 2008
Building up a mechanical system• The net forced applied to
the mass m is F-kx-cv
V: is the velocity with whichthe piston (mass) ismoving
The net fore is the force
applied to the mass to
cause it to accelerate thus:
net force applied to mass
=ma =
F kxdt
dxcdt
xd mor
dt
xd m
dt
dxckxF
=++
=−−
2
2
2
2
2nd order differential equationdescribes the relationshipbetween the input of force F tothe system and the output ofdisplacement x
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Slide 10.14
Bolton, Mechatronics PowerPoints , 4th Edition, © Pearson Education Limited 2008
Figure 10.3 Model for (a) a machine mounted on the ground, (b) the chassis of a car
as a result of a wheel moving along a road, (c) the driver of a car as it is drivenalong a road
Example of mechanical systems
The model in b can beused for the study of thebehavior that could be
expected of the vehiclewhen driven over a roughroad and hence as a basisfor the design of thevehicle suspension model
The model in C can beused as a part of a largermodel to predict how the
driver might feel whendriving along a road
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Slide 10.15
Bolton, Mechatronics PowerPoints , 4th Edition, © Pearson Education Limited 2008
Figure 10.4 Example
Analysis of mechanical systems
The analysis of such systems is carried out by drawing afree-body diagram for each mass in the system, thereafterthe system equations can be derived
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Slide 10.16
Bolton, Mechatronics PowerPoints , 4th Edition, © Pearson Education Limited 2008
• Procedure to obtain the differential
equation relating the inputs to the outputsfor a mechanical system consisting of anumber of components can be written as
follows
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Slide 10.17
Bolton, Mechatronics PowerPoints , 4th Edition, © Pearson Education Limited 2008
Figure 10.5 Mass–spring system
Example: derive the differential
equations for the system in Figure
Consider the free body diagram
For the mass m2 we can write
For the free body diagram ofmass m1 we can write
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Slide 10.18
Bolton, Mechatronics PowerPoints , 4th Edition, © Pearson Education Limited 2008
Figure 10.6 Rotating a mass on the end of a shaft: (a) physical situation,(b) building block model
Rotary system analysisThe same analysis procedures can also be applied to rotary system,
so just one rotational mass block and just the torque acting on thebody are considered
Spring
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Slide 10.19
Bolton, Mechatronics PowerPoints , 4th Edition, © Pearson Education Limited 2008
Figure 10.7 Electrical building blocks
Electrical system building blocks
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Slide 10.20
Bolton, Mechatronics PowerPoints , 4th Edition, © Pearson Education Limited 2008
Table 10.2 Electrical building blocks
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Slide 10.21
Bolton, Mechatronics PowerPoints , 4th Edition, © Pearson Education Limited 2008
Figure 10.8 Resistor–capacitor system
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Slide 10.22
Bolton, Mechatronics PowerPoints , 4th Edition, © Pearson Education Limited 2008
Figure 10.9 Resistor–inductor–capacitor system
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Slide 10.23
Bolton, Mechatronics PowerPoints , 4th Edition, © Pearson Education Limited 2008
Figure 10.10 Resistor–inductor system
Slid 10 24
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Slide 10.24
Bolton, Mechatronics PowerPoints , 4th Edition, © Pearson Education Limited 2008
Electrical System Model
Resistor–capacitor–inductor system
Slide 10 25
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Slide 10.25
Bolton, Mechatronics PowerPoints , 4th Edition, © Pearson Education Limited 2008Figure 10.12 Analogous systems
Electrical and Mechanical Analogy
F I
Velocity Volt
C dashpot 1/R
Spring inductor
Mass capacitor
Slide 10 26
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Slide 10.26
Bolton, Mechatronics PowerPoints , 4th Edition, © Pearson Education Limited 2008