actuators & mechanisms - mechatronics engineering...

Mechatronics (2)

4th Year- Mechatronics Major

Prof. Dr. Magdy M. Abdelhameed

Course Code: MDP 454, Course Name: Mechatronics (2), Second Semester 2014

Actuators & Mechanisms


Mechatronics (2)



Lectures

Power/Energy

Conversion

(Electrical Motors)Power/Energy

Transmission

(Gears,

Belt Drives,

Power Screws)Transmission

Support

(Bearings)

Joints

(Fasteners,

Connectors)

Structural

Support

(Frames

Shafts

Axles

Spindles)

Tools

Stress Analysis,

Failure Theories

Dynamics, Statics, Etc….

Mechatronics (2)



Electric Motors

Principles and Applications

Mechatronics (2)



Power/Energy Converters

• Rotary

Electrical Input -> Mechanical Rotary Motion/Torque

=DC Motor

=AC Motor

=Stepper Motor

Combustion -> Mechanical Rotary Motion /Torque

=Gasoline Engine

=Gas Turbine

Mechatronics (2)



• Linear

Electrical Input -> Mechanical Linear Motion/Torque

=Lead screw linear actuators

=Solenoids

Pressure -> Mechanical Linear Motion/Torque

=Hydraulic Pumps

=Hydraulic Actuators

=Pneumatic Actuators

=Compressors

Power/Energy Converters

Mechatronics (2)



Motor Actuators:

• Types

• Theories

• Applications

Motor Actuators

Mechatronics (2)



DC Motors

Just as the rotor reaches alignment, the brushes move across the commutator contacts and energize the next winding. In the animation the commutator contacts are brown and the brushes are dark grey. A yellow spark shows when the brushes switch to the next winding.

Mechatronics (2)



DC Motor Applications

•Automobiles:

–Windshield Wipers

–Door locks

–Window lifts

–Antenna retractor

–Seat adjust

–Mirror adjust

–Anti-lock Braking System

•-Cordless hand drill

•-Electric lawnmower

•-Fans

•-Toys

•-Electric toothbrush

•-Servo Motor

Mechatronics (2)



Brushless DC Motors

• A brushless dc motor has a rotor with permanent magnets and a stator with windings. It is essentially a dc motor turned inside out. The control electronics replace the function of the commutator and energize the proper winding.

Mechatronics (2)



Brushless DC Motor Applications

• Medical: centrifuges, orthoscopic surgical tools, respirators, dental surgical tools, and organ transport pump systems

• Model airplanes, cars, boats, helicopters

• Microscopes

• Tape drives and winders

• Artificial heart

Mechatronics (2)



Full Stepper Motor

This animation demonstrates the principle for a stepper motor using full step

commutation. The rotor of a permanent magnet stepper motor consists of permanent

magnets and the stator has two pairs of windings. Just as the rotor aligns with one of

the stator poles, the second phase is energized. The two phases alternate on and off

and also reverse polarity. There are four steps. One phase lags the other phase by

one step. This is equivalent to one forth of an electrical cycle or 90°.

Mechatronics (2)



Half Stepper Motor

•This animation shows the stepping pattern for a half-step stepper motor. The

commutation sequence for a half-step stepper motor has eight steps instead of four.

The main difference is that the second phase is turned on before the first phase is

turned off. Thus, sometimes both phases are energized at the same time. During the

half-steps the rotor is held in between the two full-step positions. A half-step motor

has twice the resolution of a full step motor. It is very popular for this reason.

Mechatronics (2)



Stepper Motors

• This stepper motor is very simplified. The rotor of a real stepper motor usually has many poles. The animation has only ten poles, however a real stepper motor might have a hundred. These are formed using a single magnet mounted inline with the rotor axis and two pole pieces with many teeth. The teeth are staggered to produce many poles. The stator poles of a real stepper motor also has many teeth. The teeth are arranged so that the two phases are still 90° out of phase. This stepper motor uses permanent magnets. Some stepper motors do not have magnets and instead use the basic principles of a switched reluctance motor. The stator is similar but the rotor is composed of a iron laminates.

Mechatronics (2)



More on Stepper Motors

• Note how the phases are driven so that the rotor takes half steps

Mechatronics (2)



• Animation shows how coils are energized for full steps


Mechatronics (2)



Full step sequence showing how

binary numbers can control the

motor

• Half step sequence

of binary control

numbers


Mechatronics (2)



•Film Drive

•Optical Scanner

•Printers

•ATM Machines

•Pump

•Blood Analyzer

•FAX Machines

•Thermostats

Stepper Motors Applications

Mechatronics (2)



•A switched reluctance or variable reluctance motor does not contain any permanent magnets. The stator is similar to a brushless dc motor. However, the rotor consists only of iron laminates. The iron rotor is attracted to the energized stator pole. The polarity of the stator pole does not matter. Torque is produced as a result of the attraction between the electromagnet and the iron rotor in the same way a magnet is attracted to a refrigerator door. An electrically quiet motor since it has no brushes.

Switched Reluctance Motor

Mechatronics (2)



• Motor scooters and other electric and hybrid vehicles

• Industrial fans, blowers, pumps, mixers, centrifuges,

machine tools

• Domestic appliances

Switched Reluctance Motor

Mechatronics (2)



A brushless ac motor is driven with ac sine wave voltages. The permanent magnet rotor rotates synchronous to the rotating magnetic field. The rotating magnetic field is illustrated using a red and green gradient. An actual simulation of the magnetic field would show a far more complex magnetic field.

Brushless AC Motor

Mechatronics (2)



• The stator windings of an ac induction motor are distributed around the stator to produce a roughly sinusoidal distribution. When three phase ac voltages are applied to the stator windings, a rotating magnetic field is produced. The rotor of an induction motor also consists of windings or more often a copper squirrel cage imbedded within iron laminates. Only the iron laminates are shown. An electric current is induced in the rotor bars which also produce a magnetic field.

AC Induction Motor

Mechatronics (2)



• Aircraft Window Polarizing Drives

• Antenna Positioning and Tuning Devices

• Audio/Video Recording Instruments

• Automated Inspection Equipment

• Automated Photo Developing Equipment

• Automated Photo Slide Trimming & Mounting Equipment

• Automatic Carton Marking & Dating Machines

• Automatic Dying and Textile Coloring Equipment

• Automatic Food Processing Equipment

• Automatic I.V. Dispensing Equipment

• Automatic Radio Station Identification Equipment

• Automotive

• Automotive Engine Pollution Analyzers

Huge List of applications

Mechatronics (2)



• Baseball Pitching Machine

• Blood Agitators

• Blood Cell Analyzer

• Warning Light Flashers

• Railroad Signal Equipment

• Remote Focusing Microscopes

• Resonator Drives for Vibraphones

• Silicone Wafer Production Equipment

• Solar Collector Devices

• Sonar Range Recorders and Simulators

• Steel Mill Process Scanners

• Tape Cleaning Equipment

• Tape Input for Automatic Typewriters


Mechatronics (2)



• Telescope Drives

• Ultrasonic Commercial Fish Detectors

• Ultrasonic Medical Diagnostic Equipment

• Voltage Regulators

• Water and Sewage Treatment Controls

• Weather Data Collection Machines

• Welding Machines

• X-Ray Equipment

• XY Plotters


Mechatronics (2)



Linear Stepping Motor

Mechatronics (2)



The duty cycle factor, labeled f on the curves is defined as:

where Ton is the on-time and Toff is the off-time.

Solenoid

Mechatronics (2)



• Electric Motors convert electrical power to mechanical

power.

Electric Motor

Electrical Power

I, V

Mechanical Power

F, v or T, w

I : Current

V: Voltage

F: Force

v: VelocityT: Torque

w: Angular Velocity

Electrical Power = I*V

Mechanical Power = F*v for linear motor

= T*w for rotary motor

Analysis of Electric Motors

Mechatronics (2)



Features of basic motor types

DC Motors: speed and rotational direction control via

voltage

= Easy to control torque via current

= low voltage

= linear torque-speed relations

= Quick response

AC Motors: smaller, reliable, and cheaper

= speed fixed by AC frequency

= low torque at low speed

= difficult to start

Electric Motors

Mechatronics (2)



R

Vini

iieind lF, v

x

x

x

x

Bi = Vin/R (Ohm’s Law)

B is

constant

When the switch is on, the wire will experience a sudden increase

of force from 0 to F.

v will increase.eind will be induced in the direction opposite to i.

i will decrease.

F will decrease.

v will decrease.

Eventually v reaches to a

constant velocity.

Principle and Analysis of DC Motor

Mechatronics (2)



At steady-state, apply Fload to slow down the wire.

The net force will decrease. F - Floadv will decrease.

i will increase.

v will increase.

F will increase.

eind will decrease.

Again, v eventually reaches a

constant velocity. vnew < v

R

VinlF, v

x

x

x

x

B

Fload

i

ii

DC Motor under external load

Mechatronics (2)



v

Fload

Basic Trend: Fload v ; Fload v

Note: v is the steady-state velocity.

Velocity Vs. Force Relation – DC Motor

Mechatronics (2)



Motor Output Torque:

T

AAT

T

A

K

TiTiK

K

Ti

T0

0

0

for

0for 0

T: motor output torque

KT: torque constant (motor specific, based on the construction)

iA: armature current [Amp]

f: magnetic flux [webers] = [T m2]

T0: torque loss (e.g. frictional)

Basic Equations:

PM DC Motor: Theory of operation

Mechatronics (2)



TTA

K

TT

Ki 01

Current Line:

i

T

T0/KTf

Slope = 1/KTf


Mechatronics (2)



Electromotive Force (EMF):

nKe Eind

eind: induced voltage or EMF

KE: EMF constant (motor specific, based on construction)

n: rotational speed of the motor [rpm]


Mechatronics (2)



Armature Voltage (Counter EMF):

eind: induced voltage (voltage generated within motor)

Vin: input voltage

VA: actual voltage available in the armature

indinA eVV

eind works in opposite direction to input voltage Vin.


Mechatronics (2)



Armature Current

A

indinA

R

eVi

RA : armature resistance


Mechatronics (2)



Torque-Speed Curve (Different Motors)

Mechatronics (2)



-good speed regulation to

two times full-load torque

-speed/torque curve is very

steep, soft performance,

high no-load speed and

starting torque

-high starting torque and soft

speed characteristic, limited

no-load speed

-linear speed/torque curve,

current draw varies linearly

with torque

Types of DC Motors

Mechatronics (2)



• Constant

Torque (or force) is independent of speed

=Example - when weight has to be lifted or load moved

against dry “Coulomb” friction.

• Linear

Torque is proportional to speed.

=Example - stirring pancake batter with egg-beater

Often a consequence of laminar flows

• Quadratic

Torque is proportional to square of speed

=Examples - fans, air drag on aircraft automobiles.

Often a consequence of turbulent flows

Load Lines – Working load Vs. Speed

Mechatronics (2)



Operation occurs at intersection of motor’s operation line and

load line, i.e.

where load curve and torque-speed curve intersect

Constant

Load Lines

Mechatronics (2)



Two main operations regions:

I. Start Up Torque

Torque required to change speed in a motor.

=Normally used for starting a motor from a dead stop.

=Inertia and required response time is important.

II. Operating Torque

Torque required to operate motor at constant speed.

=Since motor is operating at a constant speed, inertia is

not as important as the driven load.

Operating Regions

Mechatronics (2)



Operating Regions

Mechatronics (2)



The figure shows the torque speed curve of a DC motor. Find the

speed and motor current for the following:

No-load and stall conditions

Lifting a 10-oz load with a 2 in radius pulley.

A motor driving a robot arm with a weight.

Operating Torque Example

Mechatronics (2)



•If voltage is applied to motor with

no load attached to shaft, motor

would turn at its no-load speed of

1000 rpm

•On the other hand, if the shaft was clamped so it could not turn,

motor would exert the stall torque of 100 in-oz on clamp and draw

260 mA of current

Operating Torque Example – Solution (a)

Mechatronics (2)



•Torque equals force times

distance

•Thus motor torque to lift the weight

is

T = 2 in. * 10 oz. = 20 in-oz

•From graph, at torque of 20 in-oz,

•Speed has declined to 800 rpm

•Current is up to 125 mA

Operating Torque Example – Solution (b)

Mechatronics (2)



•Motor attached to a 12-in. robot arm (weighing 10 oz)

•Apple (weighing 8 oz) rests on end of arm

•Torque due to arm and apple:

T = (6 in.*10 oz.) + (12 in.*8 oz.) = 156 in-oz

Operating Torque Example – Solution (c)

Mechatronics (2)



•From graph, torque of 156 in-oz

exceeds stall torque of 100 in-oz,

motor will not be able to lift load

Consider gear ratio of 5:1

Torque required of motor is then only 1/5

of original torque: Tnew = 156/5 = 31.2

in-oz.

Inserting a gear train between

motor and load might solve the

problem


Mechatronics (2)



•Motor will now rotate at 690 rpm and a require a current of 150 mA

•However, because of gear train, load will only rotate at 690/5 = 138

rpm


Mechatronics (2)



Thank You For Your Attention!

Questions?

actuators & mechanisms - mechatronics engineering...

Documents