Mechatronics Engineering II

Instructor: Prof. Dr. A. Halim Bassiuny

email: bassiuny@yahoo.com

Assistant: Eng. M. Samy

Eng. Magdy Abdalla

Department of Mechanical Engineering (Mechatronics) HU

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Mechatronics Engineering II
Topics
1-4 1- Modeling and simulation of mixed dynamic systems
Approximate time allocation

5 2- Principles of process monitoring

6 3- Digital signal processing techniques
7 4- Actuator Sizing
8 -- Mid Term Exam
9-10 5- Fourier transform and FFT – smoothening windows
11 6- Computer based signal processing (Matlab &
Labview)
12 7- Monitoring and control applications
13 8- Advanced monitoring and control experiments
14 9- Term Project Discussion
Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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Mechatronics Engineering II
Lecture 1

Motion Transmission
“Modeling”
Rotational elements
Gears
Rack and Pinion
Lead Screw and Ball Screw
Error Detectors
Actuator Sizing

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Introduction
Single axis

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 Introduction

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Review
• Mechanical Motion:
– Mechanisms are devices that have been designed to make
jobs easier.
– Mechanisms involve some kind of motion and force
– They must have some kind of input and produce some
kind of output.
– If we connect mechanisms together we can build
mechanical systems
• Motions Types:
Linear Rotary Intermittent Oscillating Reciprocating

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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 Remember
TRANSLATIONAL MECHANICAL COMPONENTS

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Examples

Example 1:

Solution
Analysis of forces applied to the mass on both t- and s-domain:

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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 Examples
Solution Cont.:
ODE Model (t-domain):

Ms2 +Fvs +k)X(s) = F(s)

Taking f(t) as the input signal (the cause) and x(t) as the output
response signal (the effect), we have the transfer function model
G(s) (s-domain):

[Sum of impedance] X(s) = [Sum of applied forces]

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Examples
Example 2: Find the transfer function X2(s)/F(s)

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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 Examples
Solution Ex2:
Analysis of forces applied to the mass 1 on s-domain:

Forces on M1 due Forces on M1 due

only to motion of M1 only to motion of M2

all forces on M1

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Examples
Solution 2 Cont.

Analysis of forces applied to the mass 2 on s-domain:

a. Forces on M2 due only to

motion of M2;
b. forces on M2 due only to
motion of M1;
c. all forces on M2

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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 Examples
Solution 2 cont.

ODE Model:

After applying Laplace Transformation:

and reorganizing the common factors:

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Examples
Taking f(t) as the input signal and x2(t) as the output signal, we can
solve the transfer function model G(s) from the set of equations:

Similarly, we can derive the transfer function model from f(t) to x1(t):

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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 Summary

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

ROTATIONAL MECHANICAL COMPONENTS

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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 Example 1
• Find the transfer function 2(s) / T(s) for the rotational
system given below

a. Physical system b. schematic

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Example 1: Solution

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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 Example 1: Solution

a. Torques on J2 due only b. torques on J2 due only c. final free-body diagram

to the motion of J2; to the motion of J1 for J2

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Example 1: Solution

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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ROTATIONAL MECHANICAL SYSTEM
Summary

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Quiz
• Write, but not solve, the equation of motion for the
mechanical system given below.

Solution

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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 1- Gears

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

1- Gears
• Generalizing the results, we can make the following
statement : Rotational mechanical impedances can be
reflected through gear trains by multiplying the mechanical
impedances by the ratio

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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 1- Gears
In order to eliminate gears with the large radii,
a gear train is used to implement large gear
ratios by cascading smaller gear ratios.

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Reflected Inertia
Reflected inertia at
the input shaft

N = Gear ratio

Important for
actuator sizing

Characteristics of motion transmission mechanisms ?

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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 Example
Given the rotational system shown in the Figure, find the
transfer function, G(s) = 6(s)/ 1 (s).

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Example: Solution
Find the equivalent
Inertia

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 Example: Solution
Equivalent Inertia

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Example: Solution

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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 Assignment I

Assignment I Due Date: week 4

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

2- Belt and Pulley

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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 Example

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Ball Screw

Convert rotary to linear motion

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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 Ball Screw
• For each turn of the screw, the nut advances a distance
equal to the lead L
• If n is the number of continuous threads or starts on the
screw and p is the pitch, then L=np
• The reflected Jeff inertia seen at the input shaft due to a
translational mass on the nut ml (kg)
ml
J eff 
(2 . p ) 2
• The equivalent torque Teff inertia seen at the input shaft
due to an external force F at the nut is:
F
Teff 
2 . p
Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Example

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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 Rack and Pinion
• The pinion meshes with a gear rack.
• If the wheel drives the rack, the device converts
rotary to linear motion.
• Conversely, if the rack is the driver (for
instance, if a cylinder piston rod is connected to
the rack), then the device converts linear to
rotary motion.
V is the linear rack velocity (mm/sec)
ω is the angular pinion velocity (rad/sec) V  R
R is the pinion pitch radius (mm).
• The equivalent moment of inertia of this mass as reflected back to
the pinion shaft is given by mR2 2
V  
J eq  J l  mR 2  J l  m 
 
Jl
Equivalent linear inertia as felt by the pinion driving the rack meq  m 
R2
Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Tachometers and Error Detectors

• Tachometers: has the property that its output voltage vo(t) is
proportional to the angular velocity (t) of the motor, i.e.,

vo (t )  K t w(t )
• Error Detectors: Special components whose output is the difference
of two signals. Such components are very useful in control systems.
• Operational Amplifier with Resistors: The output voltage vo(t) of the
amplifier is
Rf Rf
vo (t )  v1 (t )  v 2 (t )
R1 R2

Hence, for R1 = R2 = Rf , the voltage vo will be vo (t )  v1 (t )  v 2 (t )

• Potentiometer: The error voltage eo(t) is proportional to the voltage

drop between the two points, V1 and V2, i.e: eo(t)= Kp(V1 - V2)

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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 Modeling of Electrical Systems
Remember

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Modeling of Electrical Systems

Simplify

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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Complex Circuits via Mesh Analysis
• Replace passive element values with their
impedances.
• Replace all sources and time variables with their
Laplace transform.
• Assume a transform current and a current direction
in each mesh.
• Write Kirchhoffs voltage law around each mesh.
• Solve the simultaneous equations for the output.
• Form the transfer function

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Complex Circuits via Mesh Analysis

Example: Given the network shown in Figure, find the
transfer function I2(s)/V(s).
1- convert the network
into Laplace transforms
for impedances and
circuit variables

2- Summing voltages
around each mesh through
which the assumed
currents, I\(s) and I2(s),
flow.

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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Complex Circuits via Mesh Analysis

• Around Mesh 1, where I1(s) flows:

• Around Mesh 2, where I2 (s) flows:

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Complex Circuits via Mesh Analysis

• Combining terms, Eqs. 1 and 2 become simultaneous
equations in I\ (s) and I2(s)):

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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Complex Circuits via Mesh Analysis

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Complex Circuits via Mesh Analysis

• Example: Write, but do not
solve, the mesh equations
for the network shown in
Figure

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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 Electromechanical System
• A motor: Electromechanical system yields a mechanical
output (displacement) generated by an electrical input
(voltage)
• Example: armature-controlled dc servomotor

Drive the transfer

function of the armature-
controlled dc servomotor

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Electromechanical System
• Fixed field: magnetic field is developed by stationary
permanent magnets
• Armature: current ia, flows through rotating circuit
passes through this magnetic field at right angles and
feels a force, F = B L ia .
– B is the magnetic field strength and
– L is the length of the conductor.

The resulting torque turns

the rotor, the rotating
member of the motor.

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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 Electromechanical System
• A conductor moving at right angles to a magnetic field
generates a voltage at the terminals of the conductor
equal to e = BLv,
– e is the voltage and v is the velocity of the conductor
normal to the magnetic field.
– B is the magnetic field strength and
– L is the length of the conductor.
Since the current-carrying
armature is rotating in a
magnetic field, its voltage
is proportional to speed.

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Electromechanical System
• vb back emf;
• Kb is a constant of proportionality “back emf constant”

writing a loop equation.

The torque developed by

the motor is proportional
to the armature current

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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 Electromechanical System

To find Tm in terms of m , we are

to separate the input and output
variables

La, is <<<<< Ra

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Assignment I

Assignment I due: Monday 2.04.2018

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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 Actuator Sizing
• The size of the actuator refers to its power capacity and must be
large enough to be able to move the axis under given inertial and
load force/torque conditions.
• Undersized actuator: The axis will not be able to deliver the
desired motion, i.e. velocity and acceleration.
• Oversized actuator: Additional cost and the motion axis will have
slower bandwidth (the larger power the slower the bandwidth)
• Sizing of an electrical actuator requires the determination of:
– Maximum torque (peak torque) Tmax
– Rated torque (RMS or continuous)
– Maximum speed required
– Position accuracy
– Gear mechanism parameters: Gear ratio, its inertial and resistive load,
stiffness, backlash characteristics

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Actuator Sizing
• For a given application, the load motion requirements specify
the desired position accuracy x and maximum speed xmax.
• The acceptable gear ratio range is specified by the accuracy
and maximum speed requirements:

 
 N   ma x
x x max
• The actuator needs to generate torque in order to move two
different categories of inertia and load
– Load inertia and force/torque (including gear mechanism)
– Inertia of actuator itself and any resistive force
• For rotary actuator, the total toque is given by:

JT =total moment of inertia reflected at the
TT  J T . motor shaft
Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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 Actuator Sizing
• Actuator sizing requires the following
– Determine the net inertial JT
– Determine the net load torques Jl includes all inertia
reflected on the
– Specify the desired motion profile
motor shaft
Net inertia J T  J m  J l ,eff
Ball screw inertia includes all inertia components; coupling Jc, ball
screw inertia Jbs and moving load W inertia
W /g
J l ,eff  J c  J bs 
(2p ) 2
The sum of all external torques Tl include friction Tf, gravity Tg, process
torque Ta, and torque due to nonlinearity Tnl Tl  T f  Tg  Ta  Tnl
1
the reflected torque on the motor side is Tl ,eff  Tl
N
Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Actuator Sizing
Motion profile
• The most commonly used motion profile is the trapezoidal velocity
profile shown in the figure. The typical motion includes:
– constant acceleration period
Speed
– constant speed period Tcyc
– constant deceleration period
– dwell period (zero speed)

Time

Ta Tr Td Tdw

From motion profile, the required torque as a function of time is calculated

as: 
Tm (t )  J T  (t )  Tl ,eff
Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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 Actuator Sizing
Thus the maximum torque and the RMS torque can be determined:
t cyc
1
Tmax  max(Tm (t )) , Trms 
t cyc T
0
m (t ) 2 dt

Example: A rotary motion axis is driven by an electric servo motor.

The load is a solid steel cylinder d=75mm, l=50 mm and =7.8x10-6
kg/mm3. The desired motion of the load is periodic as shown in the
figure. The total distance to be traveled is 1/4 revolution. The period
of motion is 250msc, and the dwell period is 100msec. The remaining
period is equally divided between constant acceleration, constant
speed and constant deceleration. If the rotary load is directly
connected to the gear shaft, determine motor size of this application

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Actuator Sizing
Solution
Speed
Tcyc
The load inertia JL
1 2 1
Jl  mr   .  . r 2 . l . r 2
2 2
From motion profile Time

1 
1 Ta Tr Td Tdw
 a   . ta  
2 42
 1 π
θ  2 . . (rad) / 0.05  80 / 16 [rad/s]  2400 / 16[ rev / min]  150rpm
4 2

 θ 5π   
θa    100π rad/ sec 2 , θ r  0.0 θ d  100 θ dw  0.0
ta 0.05

Take the motor inertia equal to the load inertia

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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 Actuator Sizing
• The required torque to move the load through the cycle is

Ta  (Tm  J l ) θ a 
Tr  0.0

Td  ( J m  J l ) θ a
Tdw  0.0  Tmax  Ta

1 2 2 2 2
Trms  (Ta .t a  Ta t r  Td .t d  Tdw .t dw
t cyc
Conclusion: A motor with inertia of … , maximum speed
capability of …, peak and RMS torque rating of …, … would
be sufficient to do the required task

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Assignment

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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 Assignment II

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Assignment II

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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 Assignment II

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Assignment II

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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 Assignment II

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Assignment II

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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 Assignment III

Simulation of a Brushed DC Motor

Using LabVIE

Two Weeks
Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

Assignment

Simulation of a Brushed DC Motor

Using LabVIE
Two Weeks

Mechatronics Engineering II Prof. A. Halim M. Bassiuny Mechatronics Division

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