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Unit - IV
Chapter - 4
Fundamental of cnc and part
programming
Syllabus : Introduction to NC systems and CNC - Machine axis and Co-ordinate
system- CNC machine tools- Principle of operation CNC- Construction
features including structure- Drives and CNC controllers- 2D and 3D
machining on CNC- Introduction of Part Programming, types - Detailed
Manual part programming on Lathe & Milling machines using G codes and
M codes- Cutting Cycles, Loops, Sub program and Macros- Introduction of
CAM package.
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Section No.
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Topic Name
Page No.
4.1
Introduction
4.2
Numerical Control
4.3
Classification of NC System
4.4
Advantages of NC System
4.5
Disadvantages of NC System
4.6
Applications of NC System
4.7
Types of Numerical Control System
4.8
Conventional Numerical Control (NC)
4.9
Direct Numerical Control (DNC)
4 - 13
4.10
Computerized Numerical Control (CNC)
4 - 14
4.11
Constructional Features of CNC Machines
4 - 15
4.12
Advantages and Disadvantages of CNC Machines
4 - 28
4.13
Comparison between NC, CNC and DNC System
4 - 29
4.14
Adaptive Control System (ACS)
4 - 30
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4.15
Machining Centre
4 - 31
4.16
Program Reader
4 - 34
4.17
New Trends in Tool Materials
4 - 34
4.18
Tool Inserts
4 - 35
4.19
Work Holding in CNC Machines
4 - 36
4.20
Axis Nomenclature for CNC Machines
4 - 36
4.21
Part Programming
4 - 39
4.22
Procedure to Write a Part Program
4 - 49
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4.23
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Part Programming for Lathe
4 - 50
Part Programming for Milling and Drilling
4 - 67
Subroutine
4 - 90
4.26
Canned Cycle
4 - 93
4.27
Automatically Programmed Tools (APT)
4 - 96
4.28
Micromachining
4 - 99
4.29
Part Programming Using APT
4.30
Introduction of CAM Package
4.24
4.25
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Two Marks Questions with Answers
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Fundamental of CNC and Part Programming
4.1 Introduction :
·
When any machine tool is manually operated, the operator controls the relative
movements of the workpiece and tool.
·
The accuracy of these movements is controlled by reference to some form of
measuring device fitted on the machine slide or lead screw.
·
The operator has to perform functions like starting and stopping the machine,
turning the coolant on and off, etc.
·
With manual control accuracy of final workpiece, quality and time required to
manufacture depends on the skill, concentration and experience of the operator.
·
When many batches of identical parts are required, it is preferable to use jigs,
fixtures and templates.
·
Automatic machine tools are also used in order to minimize errors and variable
quality of manual operation.
·
So, to avoid human errors, minimize production cost and due to many other
reasons, NC i.e. Numerical Control machines comes into the picture.
·
On a numerically controlled machine tool the decisions which govern the
operation of the machine are made by a series of numbers in binary code which
are interpreted by an electrical system.
·
The electronic system converts these numerical commands into the physical
movement of the machine elements.
·
Now a days, these NC machines are modified into different machines as follows :
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Special purpose CNC machine tool with vertical and horizontal machining
centre.
m
Flexible Manufacturing System (FMS).
m
Gear cutting machines.
m
Electro-discharge machines with CNC.
m
Co-ordinate Measuring Machines (CMM).
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4.2 Numerical Control :
·
"Numerical control is a programmable automation in which actions are controlled
by means of coded numbers, letters and other symbols."
·
The numerical data which is required for producing a part is maintained on
punched tape.
·
This data is arranged in the form of blocks of information.
·
The block contains cutting speed, feed, dimensional information and contour
form.
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·
For preparing this data, part programmer is required which should have the
knowledge of tools, cutting fluids, use of machinability data and process
engineering.
·
Fig. 4.2.1 (a) shows the block diagram for the procedure of production through NC.
·
Fig. 4.2.1 (b) shows the block diagram for NC machine tool system.
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Fig. 4.2.1 (a) : The procedure of production through numerical control
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Fig. 4.2.1 (b) : Elements of a NC system
4.2.1 Basic Elements of NC System
A Numerical Control machine consists of following elements :
1. Machine Control Unit (MCU),
2. Machine tool and NC tooling,
3. Part program and drawings. (Refer Fig. 4.2.1(b)).
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1. Machine Control Unit (MCU) :
·
It is the heart of NC machine tool system and consists of many sub-units inside
it.
·
The first sub-unit is tape reader which receives the coded data from punched
tape.
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·
The tape reader reads this data and passes data to the buffer storage through the
decoding circuits.
·
The buffer storage stores the received information, till it is required and transfers
it to the required area.
·
This unit is also called as Data Processing Unit (DPU).
·
MCU also consists of sub-units like control unit, decoding circuits, feed control
units, etc.
·
Almost all the operations like tool movements, tool change, speed and feed
change and many others can be controlled by MCU.
2. Machine tool and NC tooling :
·
It is the manufacturing arm of NC machine tool system.
·
It receives the raw material and performs different operations like turning,
milling, drilling, grinding, etc.
·
For performing these operations, it should receive the information from the MCU.
·
As per the information, the desired shape and size is modified.
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3. Part program and drawings :
·
NC machine operates as per coded information, which is Input Data for the
machine.
·
The feeding of this data may be manually or automatic.
·
The manual feeding of input data includes operator and hence chances of error
increases.
·
Hence, data is fed by automatic means and for this purpose punched tape is
mostly used.
·
As punched tape is most widely used hence, it becomes a standard and due to
standardisation, similar tape punchers and tape readers are used in all the systems.
·
Punched tape uses a binary coded decimal system for containing operating
information of NC tool.
·
Punched tape have eight vertical columns (channels) numbering from 1 to 8 and
one feeding holes column between them.
·
Channel 1 to 3 is on one side and 4 to 8 on another side of the punched tape.
·
Also, it carries horizontal rows, which represent a code number, letter code or a
word.
·
The instructions are marked on the tape in the form of holes in binary codes
format, which is decoded in MCU and electric pulse is generated.
·
These pulses are fed further to the servo systems and mechanisms.
·
Generally, these tapes are manufactured by paper, which may be oiled or unoiled.
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Fig. 4.2.2 : Main features of a standard EIA Tape
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·
Punched tapes are cheap but they tear easily. Hence, laminated tapes are also
used.
·
Fig. 4.2.2 shows standard EIA (Electronics Industries Association) punched tape.
·
The other input media instead of punched tapes are :
m Punched cards
m
Magnetic tapes
m
Diskettes, etc.
Now a days, to enter the program, instead of these media, magnetic cassettes,
floppy discs, compact disc (CD) are used.
·
·
In many machines, MCU carries a keyboard also. This is used directly to
manipulate and feed the part program.
·
Due to this method, there is saving in machining time, hence now a days it is the
most popular type of input media to feed part program.
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4.3 Classification of NC System :
The classification of NC systems can be done in the following ways :
1. According to tool positioning or modes of programming :
a) Absolute system
b) Incremental system
2. According to motion control system :
a) Point to point system
b) Straight line or straight cut system
c) Continuous or contouring path system
3. According to servo control system :
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a) Open loop system
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b) Closed loop system
4. According to the types of feedback devices :
a) Analog transducer
b) Digital transducer
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4.3.1 According to Tool Positioning
a) Absolute system :
·
In this system, all the positions are indicated from a reference point, which is a
fixed zero point or set point.
·
Fig. 4.3.1 (a) shows that all positions are marked with a set point.
·
In Fig. 4.3.1 (a) point 'A' is a set point.
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Fig. 4.3.1 (a) : Absolute system
b) Incremental system :
·
In this method, the tool positions are indicated with respect to previous point.
·
Fig. 4.3.1 (b) shows an example of this system.
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Fig. 4.3.1 (b) : Incremental system
·
The main disadvantage of this system is that if an error occurs into the
dimensions of any location, all the locations marked after that will carry the same
error.
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4.3.1.1 Comparison of Absolute and Incremental System
Sr. No.
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Absolute positioning system
1.
In this system, all the positions are
indicated from a reference point,
which is a fixed zero point or set
point.
2.
The coordinates of each point are
independent of each other.
3.
If an error occurs into the dimensions
of any location, then the error will be
restricted to that location only.
4.
A reference point is a must here.
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Incremental positioning system
In this method, the tool positions are
indicated with respect to previous
point.
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The coordinates of each point are
dependent on each other.
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The main disadvantage of this system
is that if an error occurs into the
dimensions of any location, all the
locations marked after that will carry
the same error.
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Reference point is not needed here.
4.3.2 According to Motion Control System
a) Point to Point (PTP) system :
·
In this system, tool is accurately located at some specified position.
·
Fig. 4.3.2 (a) shows path of tool movement for drilling number of holes.
·
The spindle is first brought to the starting point, then moved to the next location
i.e. hole 1 along the marked path.
·
On that location, drilling operation is performed and then tool moves to next
location.
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Fig. 4.3.2 (a) : Point to point system
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b) Straight line or straight cut system :
·
In this system, the cutting tool can be moved along a straight line only, which is
parallel to the principal axes of motion.
·
It is not possible to combine the motion of axes. Hence, the tool motion is only
along the X-axis, Y-axis and Z-axis.
·
Due to this, angular cuts cannot be produced. (Refer Fig. 4.3.2 (b)).
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Fig. 4.3.2 (b) : Straight cut system
c) Continuous or Contouring path system :
·
In this, there is relative motion between the tool and workpiece, during the whole
operation.
·
Due to this relative motion, different curves and profiles can be cut.
·
Actually, it is a combination of PTP and straight cut system.
·
Fig. 4.3.2 (c) shows an example of continuous path system for a component on
NC-milling machine.
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Fig. 4.3.2 (c) : Continuous path system
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4.3.3 According to Servo Control System
Servo control is a group of electrical, mechanical, hydraulic and pneumatic devices,
which are used to control the slide position of NC machine tool.
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a) Open loop system :
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It is a simpler and cheaper system.
·
It involves feeding of tape, interpretation of information by tape reader, storing
the data in buffer storage.
·
After storing, it converts into electrical signal and send this signal to the control
unit.
·
The control unit is connected to servo control which controls the slide movement.
Refer Fig. 4.3.3 (a) which shows the block diagram of open loop NC system.
·
In open loop system, there is no feedback, to ensure whether the obtained slide
movement is same as desired or not and if not, what error is present.
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Amplifier
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Fig. 4.3.3 (a) : Open loop NC system
b) Closed loop system :
·
This is almost similar to open loop system, only carries an additional feed back
device.
·
This device is nothing but a transducer and accompanied by a comparator.
·
As this is similar to open loop system, the motion is same upto servo control.
·
The transducer fed back the slide displacement corresponding to the applied
signals, as shown in Fig. 4.3.3 (b).
·
The comparator compares the obtained slide motion with applied slide motion and
error, if any, is fed back to control unit through an amplifier.
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·
Then control unit sends correct commands to servo motor (control) and cycle
continues.
Fig. 4.3.3 (b) : Closed loop NC system
4.3.3.1 Comparison of Open Loop and Closed Loop System
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Sr. No.
Open loop system
Closed loop system
It is a system that involves feeding of
tape, interpretation of information by
tape reader, storing the data in buffer
storage, converting it into electrical
signal and sending this signal to the
control unit.
It is a system that carries an
additional feedback device along with
a transducer, accompanied by a
comparator.
2.
It is a simpler and cheaper system.
It is more complicated and costly than
the open loop system.
3.
Feedback device is absent.
4.
With no feedback device, chance of
error is always present.
1.
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Feedback device is present.
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As the comparator compares the
obtained slide motion, chance of error
is greatly reduced.
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4.3.4 According to Feedback Devices
A comparing mechanism is always needed to compare actual slide position with
applied slide position to ensure accuracy. Feedback devices are units which convey the
actual slide position to the control unit, so that comparison is easily done.
a) Analog transducers :
·
It produces a variable electrical voltage, which varies with rotational speed of the
shaft.
·
This voltage can be easily measured and converted into linear distances to
indicate machine tool position.
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b) Digital transducers :
·
It converts the rotary motion of machine screw into countable electrical pulses.
·
The number of electrical pulses indicates linear distance moved by the machine
table corresponding to the lead screw rotation.
4.4 Advantages of NC System :
·
High productivity : Due to less set up and lead time, productivity is higher.
·
Less scrap : As human errors are eliminated, accurate components are machined
hence, scrap is reduced.
·
Reduced jigs and fixtures : Work and tool positioning is done by NC tape,
hence less requirement of jigs and fixtures.
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·
High quality : Due to higher accuracy of NC systems, the quality of products is
easily controlled.
·
Flexibility in design :
produced at faster rate.
·
Utilization of manpower : In NC system, there is greater utilization of
manpower because, after setting a component, operator can perform other
operations.
·
Reduction in the inventory.
·
Safety to the operator and machine tool.
·
There is a greater flexibility in the manufacturing.
·
Less floor space is required.
·
As no jigs and fixtures are required, hence tooling cost is low.
·
Skilled operator is not required.
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In NC system complicated profiles can be easily
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4.5 Disadvantages of NC System :
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High initial cost : Initial investment is high.
·
High maintenance cost : Maintenance is costly and complicated.
·
Costly control system : Control systems are also costly.
·
Skilled operator : For part programming well trained and highly skilled operator
is required.
·
Unemployment : As only one operator is required, there is increase in
unemployment.
4.6 Applications of NC System :
·
NC system is used where 100 % inspection is required.
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·
NC system is suitable for machining of parts where frequent changes in design
occurs.
·
Repetitive production of precise parts in small and medium size production can
be done by NC system.
·
When accuracy requirement is high, NC system is suitable.
·
When high amount of material is to be removed, NC system is preferred.
·
For complex machining operations also, NC system is required.
4.7 Types of Numerical Control System :
With the same basic elements and same principle, NC machines can operate on
different systems of numerical control. The common types of NC systems used in machine
tools are :
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1. Conventional Numerical Control (NC)
2. Direct Numerical Control (DNC)
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3. Computerized Numerical Control (CNC)
4.8 Conventional Numerical Control (NC) :
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·
It is a hard wired NC system having IC's (Integrated Circuits) which are
permanently wired and arranged on circuit boards.
·
It is a hardware based system, and it is difficult to change features of MCU.
·
In conventional NC system, there are chances of mistakes while punching of tape.
·
There are no provisions for speed change, feed change hence, not suitable for
large production.
·
It is simple and cheaper than other NC systems.
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4.9 Direct Numerical Control (DNC) :
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·
It is a manufacturing system in which a number of machines are controlled by a
central computer through a direct connection of telecommunication lines and in
real time.
·
Instead of using a tape reader as in NC machines, the part program is transmitted
to the machine directly from computer memory. One computer can control more
than 200 separate machines.
·
The computer used for DNC system is designed in such a way that, on demand it
will provide instructions to each machine tool.
·
The Direct Numeric Control system consists of four components :
1) Central computer
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2) Bulk memory, which stores the NC part programs
3) Telecommunication lines
·
4) Machine tools. (Refer Fig. 4.9.1)
There are two types of Direct Numeric Control system, these are :
m Behind the Tape Reader system (BTR)
m
Special machine control unit
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Fig. 4.9.1 : Direct Numerical Control system (DNC)
Advantages :
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·
Control of more than one NC machine.
·
Elimination of punched tape and tape reader.
·
Convenient storage of NC part programs in computer files.
·
Greater computational capability and flexibility.
·
The data for tools and cutters can be centrally maintained and updated.
·
The data related to manufacturing can be effectively collected and hence,
inventory can be better controlled.
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Disadvantages :
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·
The crucial disadvantage of Direct Numerical Control system is that, if the
central computer goes down, all machines become inactive.
·
Initial cost is too high.
4.10 Computerized Numerical Control (CNC) :
·
In CNC, there is absence of hard-wired logic systems.
·
The functions of hard-wired are performed by the software program of the
computer. Hence called as software based system.
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·
A separate computer is attached with each machine tool, with stored
programmable logic, hence termed as self contained NC system.
·
The computer which is used, is known as mini-computer.
·
The main change in CNC is the hardware of NC is replaced with the software to
the maximum possible extent.
·
The program is entered into the computer early through the tape, but now a days
through a keyboard.
·
The program is stored in computer memory, which can be recalled whenever
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required.
·
Program can be easily edited and modified as per the requirement.
·
An extra feature in this system is the diagnostic software which enables easy
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trouble shooting, if CNC system fails.
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The cause can be easily detected and rectified through this software. Refer block
diagram of CNC system as shown in Fig. 4.10.1.
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Fig. 4.10.1 : CNC system
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4.11 Constructional Features of CNC Machines :
A CNC machine, which is now a days very popular, consists of following features :
1. Machine structure
2. Drives
3. Actuation system
4. Slideways for machines
5. Automatic Tool Changer (ATC) 6. Automatic pallet changer
7. Transducers/Control system
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4.11.1 Machine Structure
The CNC machine tool structure consists of following main parts :
a) Bed
b) Table
c) Column
a) Bed :
·
The bed of CNC machine is generally made of high quality cast iron with heavy
ribbing to provide high stiffness and low weight.
·
The cast iron structure provides the necessary damping, to reduce the vibrations
produced due to high speed, large material removal rates and heavy duty
machining.
·
Another area of consideration is the design from chip disposal point of view.
Fig. 4.11.1 shows a slant bed structure used in turning centres
·
These allows the chips to fall off from the cutting zone. It also provides the
operator easier and better access to the workpiece and tooling.
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Slant bed
Fig. 4.11.1 : Slant bed construction
b) Table :
·
The table is mounted on the bed which provides the machining centre with the
z-axis (linear movement).
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c) Column :
·
Column is mounted on the saddle and designed with high torsional strength, to
prevent distortion and deflection while machining.
4.11.2 Drives
· There are two basic applications where drives are used in CNC machines.
1) Spindle drives
2) Feed drives
1) Spindle drives :
·
Spindle drives are used to provide the main spindle power for cutting.
·
As large material removal rates are used in CNC, large power motors are used
for spindle drives.
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·
Also, the speed required during operations is infinitely variable.
·
Hence to provide such a speed control for infinitely variable speed DC motors
are used.
·
The speed control for DC motors can be achieved by varying the voltage
infinitely.
·
The use of AC motors are preffered in the generation of currents in CNC
machine tools. This is achieved by developments in the frequency converter.
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2) Feed drives :
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·
CNC machines are provided with independent axis drive to provide the feed
movements for the slides.
·
In order to obtain fast response and positional accuracy a special type of motor
called servomotor is used to power the slides.
·
Following are the feed drives that are used in the CNC machine tools :
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i) DC servomotors
ii) Brushless DC servomotors
iii) AC servomotor
iv) Stepper motor
v) Linear motor
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i) DC servomotor :
· DC servomotors are characterized by high overload capacity, excellent dynamic
response and low moment of inertia.
·
These are made up of permanent magnet type and have high acceleration torque.
·
The speed control in DC servomotors is achieved by flux control method, voltage
control method and rheostatic control.
ii) Brushless DC servomotor :
· In brushless DC servomotor, the motoring action achieved by electrical
commutation rather than by mechanical commutation.
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·
The flux created by current carrying conductor should rotate around the inside of
the stator.
·
The speed in this motor is proportional to the frequency of the applied voltage
and number of poles.
iii) AC servomotor :
· AC servomotors are used for low power servomechanisms and have constant
acceleration to maximum speed.
·
These motors are highly reliable and have high frequency response.
·
The speed control is achieved by controlling the applied controlled voltage.
·
Following are some methods to control the speed :
m Supply frequency control
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m
Supply voltage control
m
Controlling number of poles
m
Adding external resistance in rotor circuit
m
Cascade control
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iv) Stepper motor :
· Stepper motors convert the input pulses into a precisely defined increment in the
shaft position.
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·
Each size or step angle is determined by construction of motor and type of drive.
·
These motors are suitable for position control systems in plottors, disk drives,
machine tools, robotics, etc.
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· The speed and position is controlled by input pulses.
v) Linear motor :
· Linear motors are widely used in high performance CNC machine tools.
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·
These motors give higher positional accuracy at higher feeds and speeds.
·
Also, linear motors have higher acceleration and deceleration rate.
·
The maximum speed of linear motor is limited by the bus voltage and speed of
control electronics.
4.11.3 Actuation System
An important element of actuation system is recirculating ball screw.
Recirculating ball screw :
·
In this, the sliding friction is replaced by rolling friction.
·
It consists of screw with circular form threads and nut assembly with internal
helical ball groove, to allow a continuous flow of steel balls.
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·
Either the nut or the screw rotates and causes the rolling of balls through a
helical path, as shown in Fig. 4.11.2.
Fine pitch worm gear
Ball bearings
Recirculation channel
Nut (shown cutaway)
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Sector gear
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Fig. 4.11.2 : Re-circulating ball screw
Circular form
threads
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Nut
Fig. 4.11.3 : Recirculating ball screws
·
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Ball screw
Balls
·
·
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A return tube deflects the balls and recirculates them.
The rigidity of the drive system can be improved by preloading the ball screw
and nut assembly.
Also, precise positioning can be achieved by the application of preloading. Here
axial displacement is eliminated and hence reduce the backlash.
·
Preloading can be achieved by fitting the balls in the component tightly while
assembly.
·
One of the method used for preloading the ball screws is keeping the spacer
between two nuts as shown in Fig. 4.11.4.
·
Spacer provided between nut A and nut B avoid the axial movement of nut and
hence the balls are tightly fitted in the gap.
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Spacer
Nut B
Fa
Nut A
FA
FB
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Fig. 4.11.4 : Preloading of the recirculating ball screw and nut arrangement
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Comparison of Conventional Screw and Re-circulating Ball Screw
Table 4.11.1 : Conventional screw and Re-circulating ball screw
Sr.
No.
Parameter
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Conventional power screw Re-circulating ball screw
(Square, trapezoidal)
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Advantages of recirculating ball screw
1.
Efficiency
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There is sliding friction,
hence high input torque is
required to overcome
friction. Thus efficiency of
screw is as low as 40 %.
There is rolling friction,
hence low input torque is
required to overcome
friction. Thus efficiency of
screw is as high as 90 %.
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2.
Load carrying capacity
It has lower load carrying
capacity as compared to
recirculating ball screw.
It has high load carrying
capacity as compared to
conventional power screw.
For the same load carrying
capacity, recirculating ball
screw is more compact and
light weight.
3.
'Stick-slip' phenomenon
In conventional power
screw, 'stick-slip'
phenomenon is observed.
This is due to difference
between the value of
coefficient of static friction
and coefficient of sliding
friction.
The operation of
recirculating ball screw is
smooth and free from any
'stick-slip' phenomenon.
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4.
Compensation for wear
and tear
It requires periodic
adjustment to compensate
for wear on the surfaces of
the screw and nut.
It is virtually wear-free due
to presence of lubricating
film between the contacting
surfaces.
Limitations of recirculating ball screw
5.
Cost
It is low at initial cost.
The initial cost of
recirculating ball screw is
very high.
6.
Special operating
environment
It can be operated in any
environment with
satisfactory life.
It requires high degree of
cleanliness and restricted
entry of foreign particles.
It can be easily lubricated
by grease.
It requires a continuous thin
film of lubricant between
the balls and grooves in the
nut and screw.
Due to high friction between
thread surfaces, these
screws are self-locking.
Due to negligible friction
between balls and thread
surfaces, these screws are
over-hauling. Hence special
brake is required to hold
the load in its place.
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7.
Lubrication
8.
Self-locking and
over-hauling
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Applications of Recirculating Ball Screw
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Re-circulating ball screw is used in high speed applications such as :
(i) Automobile steering gears
(ii) Power actuators
(iii) Hospital bed adjusters
(iv) Machine tool controls
(v) X-Y recorders of CNC machines.
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4.11.4 Slideways for Machines
· Precise positioning and repeatability of machine tool slides are the major
functional requirements of CNC machine.
·
To eliminate stick slip, there are different slideway systems such as rolling
friction slideway and slideways with low friction PTFE (Poly Tetra Fluoro
Ethylene).
·
These slides have low wear, good vibration damping, easy machinability, low
coefficient of friction and low price.
·
The plastic coated slideway have static coefficient of friction, which is less than
dynamic coefficient of friction.
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Fundamental of CNC and Part Programming
·
With increase in speed, dynamic coefficient of friction increases upto a certain
value and then remains constant.
·
Following are the techniques used to meet requirements in CNC machine tool
slideways :
i) Hydrostatic slideways
ii) Linear bearings with balls
iii) Rollers or Needles
iv) Surface coatings
i) Hydrostatic slideways :
·
Fig. 4.11.5 shows the working of hydrostatic slideways.
·
The carriage or slides in contact with the slideway are provided with the small
pockets or cavities.
·
Air or oil is pumped into small pockets and then flows out from pockets between
slide and slideways.
·
These slideways provides almost frictionless condition for slide movement. The
slideways should be kept clean for the efficient operation.
·
The only disadvantage is that, it requires very large surface area to provide
proper support.
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Oil pocket
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Oil
inlet
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Oil inlet
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Fig. 4.11.5 : Hydrostatic slideway
ii) Linear bearings with balls :
· In this technique, the sliding friction is replaced with rolling friction.
·
Fig. 4.11.6 shows the linear ball bearing with recirculating balls.
·
These ball bearings are designed to give frictionless movement for shaft rotation
as well as over varying strokes of length with high linear precision.
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Ball
Shaft
Cage
Fig. 4.11.6 : Linear ball bearing
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iii) Rollers or Needle bearing :
· Rollers or needle bearings are also called as tachoway and are provided for the
movement along a flat plane.
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·
Linear roller bearings provides unlimited linear movement.
·
Rollers in the roller bearings are guided between shoulders of the supporting
element with very close tolerance. Refer Fig. 4.11.7.
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Rollers
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Fig. 4.11.7 : Linear roller bearing
·
The guiding element prevents the falling of rollers from the shoulders and also
recirculate them with ease.
·
While using the rollers the bed should be machined accurately and surface in
contact should be hardened.
iv) Surface coatings :
· In this technique, the guiding surface is coated with low friction material such as
polytetrafluroethylene (PTFE).
·
Sometimes, replaceable strips of low friction material are used on guide ways.
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4.11.5 Automatic Tool Changer (ATC)
· The complicated jobs can be machined on NC and CNC machines and for that
they require different types of tools.
·
For changing and resetting the tool, more time is required.
·
For this purpose, tools can be automatically changed with the help of Automatic
Tool Changer i.e. ATC.
·
By using ATC, a complete job can be machined in one pass, on one machine,
with one program only.
·
Hence, productivity and repeatability of manufacturing increases.
·
As per part program, the machining centre select a desired tool and machining is
done.
·
Generally tools are stored in drum type or chain type magazine as shown in
Fig. 4.11.8 (a) and (b).
·
Fig. 4.11.8 (c) shows a 180º type of rotation tool changer mechanism.
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(c)
Fig. 4.11.8 : Automatic tool changer
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·
Generally, machining centres with 16 to 24 tools are used, but now a days
machining centres with 160-200 tools are used.
·
MCU receives tool change command and send spindle to its fixed tool change
co-ordinates.
At that time, tool magazine is indexed to the proper position and tool changer
rotates.
This tool changer engages the tool in the spindle and tool in the magazine at that
time.
Both the tools are removed by tool changer from their places and turns by 180º
to swap both tools.
Thus old tool is returned to magazine and new in the spindle. Hence, completes
the tool changing operation.
·
·
·
·
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4.11.6 Automatic Pallet Changer (APC)
· Machine downtime because of loading-unloading, clamping-releasing, etc. can be
minimized with the help of automatic workpiece loader/unloader system.
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·
In this system, the workpieces are mounted on the pallet and the pallets are
moved around the machine in a logical manner. This system is called as pallet
changing system.
·
According to logical movement of pallet, the system can be linear or rotary.
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Linear pallet changer system :
A typical linear pallet changer system is as shown in Fig. 4.11.9.
·
In this system, the table moves in a linear motion, hence called as linear pallet
changer system.
·
The workpiece can move in two ways.
i) Linear motion
i)
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·
ii) Inverted U-path.
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Linear motion :
· In Fig. 4.11.9 (a) the workpiece is on left side track waiting for completion of
machining operation of earlier workpiece.
·
In Fig. 4.11.9 (b) after completion of earlier workpiece, it moves onto the
unloading table and the next component is ready to move onto the machining
table.
·
In Fig. 4.11.9 (c) the next component moves onto the machining table and the
system continuous.
ii) Inverted U-path :
·
In Fig. 4.11.9 (d) the table is in linear motion but the component is rotated in an
inverted U-path to move onto the machining table and then moves linearly.
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Table 1
Table 2
Setting
station 1
(a)
(b)
(c)
Machine
spindle
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ATC
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Pallets
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(d) Inverted U-path
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Fig. 4.11.9 : Linear pallet changer system
Rotary pallet changer system :
·
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Rotary pallet changer system, is shown in Fig. 4.11.10 which is same as linear
pallet changer system except that the table is rotated for the movement of the
workpieces.
Machine
spindle
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Pallets
Fig. 4.11.10 : Rotary pallet changer system
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·
For the maching purpose of workpieces, the table is moved in rotary motion with
the help of indexing mechanism.
4.11.7 Transducers/Control Elements
· The control unit should indicate the current status and position of various
machine tool elements.
·
The control unit part, for allowing manual control and programming of machine,
may be housed on machine structure itself.
·
To monitor the position of slides, linear and rotary transducers are used.
4.11.8 Feedback Devices
· In closed loop control system, feedback devices are required for proper position
control of slides or drives, for holding workpieces and for controlling tool
motions.
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·
The feedback devices can be either of rotary or linear form.
·
There are two types of rotary transducers which are resolvers and encoders that
can be connected directly to the ball screw.
·
Linear transducers have a portion attached to the structure and their other part is
fixed to the slide which moves over the stationary part.
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Encoder (Rotary transducer)
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·
Encoders are numerical devices which indicates output in digital form directly
and widely used as position and motion sensors.
·
It consists of a glass disc with accurately etched lines at regular intervals. Refer
Fig. 4.11.11.
·
The glass disc rotates between the light source and photodiodes.
Light source
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Photo diode
Glass disc
Fig. 4.11.11 : Encoder
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·
The lines make and break this photoelectric beam which generates a pulse signal
and this signal is amplified to give a square wave output.
·
Number of signals generated per revolution depends on the number of lines on
the disc.
Linear transducer :
·
The principle of linear transducer is similar to rotary transducer except that, the
signal directly translates into linear displacements of the slide.
·
In linear transducer instead of disc, glass scale with line grating is used, which
have line graduations. Refer Fig. 4.11.12.
Light source
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Slide motio
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Grated
glass
scale
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Photocells
Fig. 4.11.12 : Linear transducer
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·
The relative movement between glass scale (fixed to slide) and photocells (fixed
to guide) generates an electric pulse.
·
The amount of pulses produced for a given resolution of the gratings decide the
magnitude of the travel.
4.12 Advantages and Disadvantages of CNC Machines :
Advantages
Advantages of CNC machines are similar to NC machine. Some additional advantages
due to additional feature in CNC machine over conventional machines are as follows :
· Program storage : As computer is available, hence multiple programs can be
stored in the machine.
·
Reliability of system : As the data is directly entered with the help of computer,
no need to use punched tape. It also improves reliability of the system.
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·
Online part programming : The part program can be done online with editing,
if required.
·
Flexibility of system : The system is too flexible, as new systems can be added
at low costs.
·
Metric conversions : Part program, which is written in inches, can be easily
converted into millimeters i.e. metric conversion is easy.
·
Interpolations : In NC system, there is interpolation for straight and circular
path, but in CNC it is available for helical, parabolic and cubic curves also.
·
Expanded tool compensations : For the purpose of tool offset and tool wear,
tool compensation is provided.
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Disadvantages
Disadvantages of CNC machines are similar to NC machines.
· High initial cost : Initial investment is high.
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·
High maintenance cost : Maintenance is costly and complicated.
·
Costly control system : Control systems are also costly.
·
Skilled operator : For part programming well trained and highly skilled operator
is required.
·
Unemployment : As only one operator is required, there is increase in
unemployment.
·
Computer problem : If there is any problem with computer, then the whole
machine will get stop.
·
Costly software : The software required for the operation of CNC machines is
also costly.
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4.13 Comparison between NC, CNC and DNC System :
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Sr. No.
Parameters
NC
CNC
1.
Flexibility
Less
High
2.
Tape editing
Not possible
at site
Is possible
Is possible
3.
Productivity
Less
High
Highest
4.
Number of programs
stored
Only one at
a time
5.
Number of operations
done at a time
One
One
Multiple
6.
Initial cost
Low
High
Highest
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DNC
High
Multiple programs Multiple programs
can be stored
can be stored.
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Fundamental of CNC and Part Programming
4.14 Adaptive Control System (ACS) :
·
Adaptive control system is the logical extension or improvement over the NC and
CNC systems.
·
In conventional systems, the total production time is increased due to the
non-productive time such as time required for workpiece handling, setup, tool
changing time, operators delay, lead time between order processing and
production, etc.
·
Hence NC, CNC systems find major application in the areas where reduction in
non productive time is the primary requirement.
·
The adaptive control system automatically determines the process variables such
as cutting speed, feed, depth of cut during the machining process and make
changes in the prescribed limit as per the requirement. Refer Fig. 4.14.1.
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·
Also, it makes optimal use of machine capability to reduce the non-productive
time.
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Input
Program
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Speed and feed
NC system
Command
signals
corrections
Position
feedback
NC machine
tool
Adaptive
controller
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Process constraints
Strategy
Performance
index
Measured
data
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Sensors
Fig. 4.14.1 : The adaptive control system
·
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Following are the two common systems of adaptive control :
1. Adaptive Control with Optimisation (ACO)
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2. Adaptive Control with Constraints (ACC)
1. Adaptive Control with Optimisation (ACO) :
·
In this system, the overall performance of the process is indicated by performance
index (PI) or merit figure and it is given by,
Material Removal Rate (MRR)
PI =
Tool Wear Rate (TWR)
·
Sensors mounted on the machine tool measures the various parameters such as
tool wear, cutting temperature and torque, machine vibration, etc.
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·
These sensors fed this data to adaptive controller along with the process
constraints such as speed, feed rate, etc.
·
This performance index is then compared with the set value and maintain it by
continuously changing the process variables.
2. Adaptive Control with Constraints (ACC)
·
In this system the various constraints such as torque, cutting forces, motor power,
tool wear, cutting temperature, etc. are specified.
·
When the process is in progress, the ACC system manipulate spindle speed, feed
rates to maintain the specified constraints in prescribed limit.
·
These constraints are measured with the help of sensors and transducers and
compare with the set value.
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4.15 Machining Centre :
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·
Generally, each and every machine tool is designed to perform basically one type
of operation. But, during the manufacturing process most of the components
require various operations on their different surfaces.
·
For example, consider any workpiece on which various machining operations like
milling, drilling, boring, threading, etc are to be performed.
·
The conventional method to perform these operations is move the workpiece from
one machine tool to another machine tool until all the operations are completed.
But, this method will take more time and may be some errors.
·
For this purpose Computerised Numerical Control (CNC) machine tool is used
which is designed to perform various cutting operations on different surfaces of
workpiece.
·
CNC machines are available in the form of Lathe (CNC-Lathe), Milling
(CNC-Machining center), EDM (CNC-EDM), etc.
·
Machining centre or CNC milling machine is capable of performing milling,
drilling, boring, counter-boring, threading and so many operations.
·
In these machines, the workpiece does not have to be moved to another machine
tool for other operations.
·
Some of the machining centres are provided with two work tables called as
pallets.
·
When the workpiece on one pallet is being machined, the operator set the
workpiece on the free pallet. After machining, the pallet changer moves the pallet
of finished workpiece away from the operator and the other pallet comes with the
new workpiece for machining.
·
In machining centres, generally the workpiece is stationary and the tool is moving
(rotating).
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Fundamental of CNC and Part Programming
·
Machining centres are classified as follows :
i) Horizontal machining centre
ii) Vertical machining centre.
4.15.1 Horizontal Machining Centre (HMC)
· HMC carries horizontal machining
spindle (head) which can slide along Column
the horizontal guideways. Refer
Fig. 4.15.1.
·
In these machines, software is used to
move the tool or workpiece.
·
These machines are used for machining
of large workpieces on its all the
surfaces.
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·
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HMC's are very heavy in construction.
Fig. 4.15.1 shows the principal parts of
HMC, which are described as follows :
i) Bed : It is a heavy structure which
supports the complete machine and
carries guide-ways over its top surface.
It is generally made of cast iron.
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ii) Saddle : It is mounted
guide-ways on the bed and
column over it. Generally,
X-axis movement to the
centre.
+
–
Z
+
Table
–
+
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over the
also carries
it provides
machining
Spindle
Y
–
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X
Bed
Saddle
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Fig. 4.15.1 : Horizontal machining centre
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iii) Table : It is mounted over the guide-ways provided on the saddle. It is made of
cast iron. For mounting the work holding devices, T-slots are provided on the
table. It provides Z-axis movement to the machining centre.
vi) Column : It is mounted over the saddle. It provides Y movement to the machining
centre. The column can be of fixed type or travelling type.
v) Automatic Tool Changer (ATC) : It is used to change the tool from the machine
spindle. It is placed closed to the spindle and enables the tool change rapidly.
vi) Spindle and servo system : Spindle is mounted on the headstock and it provides
Z-axis movement to the machining centre. Servo system consists of servo motors
and feedback system. It provides accurate and rapid movement along all the axes.
4.15.2 Vertical Machining Centre (VMC)
·
It carries a vertical machining spindle (head) which can slide along the vertical
guide-ways provided on the column. Refer Fig. 4.15.2.
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·
·
The vertical head can be tilted
(swivelled) in either direction.
Column
These machines are suitable for
machining flat surfaces with deep
cavities like the manufacturing of
moulds, dies, etc.
·
These machines are also very heavy
in construction. Fig. 4.15.2 shows
the principal parts of VMC which
are almost similar to HMC.
i) Bed : It is a heavy structure which
supports the complete machine and
carries guide-ways over its top
surface. It is generally made of cast
iron.
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Spindle
+
Z
–
+
–
Table
X
–
Y
+
ii) Saddle : It is mounted over the
guide-ways on the bed and also
carries column over it. It provides
Bed
Y-axis movement to the machining
Saddle
centre.
iii) Table : It is mounted over the
guide-ways provided on the saddle.
Fig. 4.15.2 Vertical machining centre
It is made of cast iron. For
mounting the work holding devices,
T-slots are provided on the table. It provides X-axis movement to the machining
centre.
iv) Column : It is mounted over the saddle. It provides Z-axis movement to the
machining centre.
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v) Automatic Tool Changer (ATC) : It is used to change the tool from the machine
spindle, rapidly.
vi) Spindle and servo system : Spindle is mounted on the headstock and it provides
Z-axis movement to the machining centre. Servo system consists of servo motors
and feedback system. It provides accurate and rapid movement along all the axes.
Advantages of machining centre :
·
·
·
·
Machining centres have high metal removal rate (MRR) capability.
Machining centres are highly versatile.
It increases productivity.
It consists of automatic tool changer (ATC) and automatic pallet changer system
(APC), hence it is more flexible and economical then the conventional machines.
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·
·
As various operations can be performed at one place, handling of the workpiece
and time of machining is minimum.
Due to less handling of the workpiece, errors are minimum.
It can machine the components with closed tolerances.
Complicated components are also machined very easily.
·
Faster cutting speed, heavier cutting depths and feeds can be obtained.
·
·
4.16 Program Reader :
It is a device used to read the coded instructions from the program of instructions.
They are classified on the basis of programming input medium.
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1) Punched tapereader :
·
When a punched tape is passed through a tapereader, the electric connections are
either close or open depending on whether there is a hole punched at a particular
track or not.
·
The coded instructions on tape are transformed and utilized for various machine
tool functions.
·
The commonly used tapereaders are :
i) Pneumatic
ii) Photo electrical
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2) Card readers :
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iii) Mechanical.
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·
It reads the information punched into a card, converting the presence or absence
of hole into an electric signal representing a binary 0 or 1.
·
It operates at speeds ranging from 12 to 1000 cards per minute.
4.17 New Trends in Tool Materials :
·
·
·
·
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Trends in the manufacturing industries leads to the new trends in the tool
material.
Now-a-days changes in the various workpiece materials and manufacturing
processes affects the material to be used for tooling.
As the industries continually looking for new manufacturing materials that are
lighter and stronger, the tool makers must develop the tools that can easily
machine these materials keeping the highest possible rate of productivity.
New trends in the tool material involves the various combinations of tool material
compositions, coatings and tool geometries.
New tool materials :
·
Refer section 6.4 for various tool materials used for tooling.
·
Along with these materials, polycrystalline diamond (PCD) cutting tools and
polycrystalline boron nitrides (PCBN) cutting tools are presently dominate in
turning, milling, drilling of various alloys.
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·
Cermets cutting tools, comprised of titanium carbonitride (TiCN), are hard and
chemically stable, leading to high wear resistance.
·
Cermet tools are also effective in dry machining.
·
Development in ceramic tool technology enables these tools to move into new
areas of applications. For example, silicon nitride tools offer improved fracture
resistance as compare to other ceramic materials.
Coatings :
·
Coatings for tool inserts can be classified as Chemical Vapour Deposition (CVD)
coatings and Physical Vapour Deposition (PVD) coatings.
·
These coatings includes titanium carbonited, titanium aluminium nitride which
offers high hardness, increased toughness and wear resistance.
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·
Recent development in coatings includes soft coatings that are used in dry
machining.
·
Diamond coated tools are used for machining hard materials.
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4.18 Tool Inserts :
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·
Inserts are removable cutting tips used in CNC tooling for high cutting speeds
and feeds.
·
Tool inserts are usually indexable. They can be rotated, flipped or changed
without disturbing the overall geometry of the tooling system.
·
These inserts are available in different shapes with varying geometry.
·
Inserts may be in the form of triangle, square, round, rhombus, diamond shapes
with different angles.
·
Coated and uncoated cemented carbides are most commonly used cutting inserts
in the market.
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Coated Inserts :
·
·
·
·
·
·
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Coated inserts are used while working with ferrous materials such as steel, cast
iron, iron, stainless steel, etc.
For machining super alloys, it is best to use coated inserts.
Also it is beneficial to use coated inserts for titanium alloys.
Both the coating material and coating processes are considered while selecting
inserts for a particular operation.
Physical or chemical vapour deposition coatings are used for carbide tool
materials to improve productivity and tool life.
Coating materials : TiN, TiC, Al2O3 , TiCN, TiAlN, etc.
Uncoated Inserts :
·
Uncoated inserts are used while machining soft materials due to its sharp and
uncoated cutting edge.
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Fundamental of CNC and Part Programming
·
It is ideal for non-ferrous materials such as aluminium, brasses, bronzes and
many composites and wood.
·
Uncoated inserts avoid formation of built up edge.
4.19 Work Holding in CNC Machines :
·
CNC machines performed number of operations using variety of tools, on
different faces of the workpiece to accomplish finished component in single
setting.
·
This requires that the workpiece should be operated from different sides without
repositioning it.
·
Hence, the work holding devices in the CNC machine tool has to bear
multidirectional cutting forces.
·
For very simple components conventional work holding devices such as chucks,
vices are used but for complex shapes of workpiece it becomes necessary to use
modular fixturing for work holding purpose.
·
Grid plate modular fixturing facilitates precise and exact positioning of the
component.
·
Also, this fixture in conjunction with a rotory table will allow to be used as an
indexing fixture.
·
It provides clamping of more workpieces in a single fixture.
·
This type of work holding reduces the clamping and unclamping time.
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Requirements of work holding devices
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·
Work holding devices should restrict the linear and rotory movement of the
workpiece.
·
It should allow quick loading and unloading of the workpiece.
·
Holding force as well as cutting force should not distort or deflect the workpiece.
·
It should ensure the proper loading of the workpiece.
·
It should allow the workpiece to be operated on different faces in single setting.
·
It should have provision for easy chip removal.
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4.20 Axis Nomenclature for CNC Machines :
·
A program in CNC system, specify the various axis about which motion is
required.
·
For this purpose, a standard axis system is considered due to which relative tool
position with respect to work must be obtained.
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·
A right hand rule for machine tool axis is as shown in Fig. 4.20.2 (a).
·
Machine tool co-ordinate axis is defined for providing a means of locating a tool
in relation to the workpiece.
·
According to the machine, a point can be located by several methods. Generally,
that point is at the origin.
·
In NC machine, the origin is defined in two ways which are fixed zero and
floating zero.
·
In a fixed zero method, the origin is always predefined. It is generally at the
lowermost left hand corner of the worktable. Refer Fig. 4.20.1 (a).
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(0, 0)
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(0, 0)
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(a) Fixed zero
(b) Floating zero
Fig. 4.20.1 : Fixed zero and floating zero
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·
All other points on the worktable are defined from this point.
·
But in modern NC machines, floating zero concept is provided which allows the
operator to define his origin.
·
This makes it convenient to develop programs of symmetrical components by
providing the origin at the point of symmetry in the workpiece.
Refer Fig. 4.20.1 (b).
·
This setting of the zero is done manually by
the operator, by positioning the tool about
the point at which origin is to be defined
and by pressing the zero button at that
point.
·
Fig. 4.20.2 (b) shows the axis nomenclature
for different machines.
·
The motion of three axes i.e. X, Y and Z
are specified as follows :
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Fig. 4.20.2 (a) : Machine tool
co-ordinate axis
m
Z-axis is always spindle axis or parallel to spindle axis.
m
X-axis is always horizontal axis and parallel to the surface of the work.
m
Y-axis is perpendicular to both X and Z-axis.
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Fundamental of CNC and Part Programming
(Drilling)
(Turning)
(Milling)
+Z
+Z
+X
+Y
+Y
+X
+Z
+X
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+Z
+Z
+Y
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+X
+X
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+Z
+Y
+X
+Y
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Fig. 4.20.2 (b) : X, Y, Z motions
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Example 4.20.1 : Explain with neat sketch 2, 21/ 2 and 3 axes of CNC machines.
Solution. : i) 2 axes of CNC Machines :
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If the machine tool controls 2 axes at the same time, then it is called as a 2-axes CNC
machine.
Z Axis control plane
Tool
path
+Z
+Y
XY Plane
+X
Fig. 4.20.3 2-Axes Machine tool
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Z Axis control plane
ii) 2 1 2 axes of CNC Machines :
+Z
If the tool can be controlled to
follow an inclined Z-axis control
plane, the machine tool is called as
2 1 2 axes CNC machine.
+Y
Tool
path
XY Plane
+X
Fig. 4.20.4 2 1 2 axes of machine tool
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iii) 3 Axes machine tool :
+Z
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If the machine tool can control
X, Y and Z axes simultaneously at
the same time, then the machine tool
is called as 3 axes machine tool.
Tool
path
+Y
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+X
Fig. 4.20.5 Axes machine tool
4.21 Part Programming :
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·
It is a set of instructions which instruct machine tool about processing steps to be
performed to manufacture the component.
·
The various techniques used for generating CNC instructions are as follows :
m Manual CNC part programming.
·
m
Computer assisted part programming.
m
CAD-CAM based programming.
m
Modelling based programming.
m
Automatically programmed tools (APT).
Instructions given by part program carry dimensional and non-dimensional data,
which is written in specific format.
4.21.1 Manual Part Programming
· The program for machining any type of workpiece not only varies from person to
person but also varies from machine to machine.
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·
Various controllers use different syntax during instructing the machine tools.
·
But, there are similarities between the codes and by understanding the basics, one
can easily adapt to other controllers with minor changes.
·
It is necessary that, the programmer understands the different processes involved
by carefully studying the drawings, fixtures and machine tools.
·
A typical block diagram of this process is shown in Fig. 4.21.1.
Machine tool
MCU
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Tape
Punch
Program
Programmer
Drawing
Fig. 4.21.1 : Manual part programming
·
It is a simplest method of part programming, in which program is written
manually on a paper.
·
The program is nothing but an instruction set for machine tool which defines the
tool position relative to the workpiece.
·
After verifying the program, corresponding punch tape is prepared.
·
Each line of program is called as block, which consists of an operation number
word, data word, etc.
·
The format of each block is as follows :
N…. G…. X…. Y…. Z…. U…. V….…. W…. F…. S…. T…. M….;
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Above each letter signifies a particular operation, which is as follows :
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a) N is a Sequence Number and used for giving the number to the lines. It is
generally, minimum three digit number. It is written as N001, N002, N003,
etc.
b) G is a Preparatory Function which changes the control mode of the
machine and called as G-codes. Generally, G-codes are followed by two
digit number. It is written as G01, G02, etc. Some common G-codes are
tabulated in the Table 4.21.1.
c) 'X, Y, Z and U, V, W' represents co-ordinate positions of tools. For two axes
system, only two letters are specified. In case of multiple axes (Milling
machine) other additional letters i.e. 'U, V, W' are specified. 'X, Y, Z' can be
positive or negative according to dimension. Generally 'X, Y, Z' are called as
dimensional data.
d) F is the Feed rate function, which defines feed rate of operation. For
example, F100, it means feed rate is 100 mm/min. If it is specified once,
then no need to specify again. It continues unless and until another value is
specified.
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e) S is a Cutting Speed Function which specifies spindle speed. For example,
S2000 means spindle is rotating at a speed of 2000 rpm.
f) T is a Tool Change Function. Generally, all the CNC machines are having
'ATC'. For programming, each tool is associated with an index number. For
example, T-04 it means tool number 4 is in ready position.
g) M is a Miscellaneous Function which is generally called as M-codes. By
specifying M-codes, other auxiliary operations are performed. Some common
M-codes are given in the Table 4.21.2.
h) ; is called as End of Block (EOB) which is written after each and every
line.
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G code
G00
G01
G02
Function
G code
Function
Positioning (rapid
traverse)
G42
Tool nose
compensation right
Linear interpolation
(feed)
G63
Tapping mode
Circular interpolation
clockwise
G65
Macro calling
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G03
Circular interpolation
anti-clockwise
G04
Dwell
G10
Data setting
G17
XPYP plane selection
G69
G18
ZP XP plane selection
G70
Inch data input
G19
YP ZP plane selection
G71
Metric data input
G20
Outer diameter /
internal diameter
cutting cycle
G72
Finishing cycle
G21
Thread cutting cycle
G73
Stock removal in
turning
G22
Stored stroke limit
function ON
G74
Stock removal in
facing
G23
Stored stroke limit
function OFF
G75
Pattern repeating
G24
End face turning cycle
G76
Peck drilling on Z-axis
G25
Spindle speed
fluctuation detect OFF
G77
Grooving on X-axis
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G66
G67
G68
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Macro modal call
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Macro modal call
cancel
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Mirror image for
double turret ON
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Mirror image for
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G26
Spindle speed
fluctuation detect ON
G78
Multiple threading
cycle
G27
Reference point return
check
G81
Drilling cycle on milling
G28
Return to reference
point
G84
Tapping cycle
G30
2 , 3 , 4 reference
point return
th
G88
Boring manual dwell
G31
Skip cutting
G90
Absolute programming
G33
Thread cutting
G91
Incremental
programming
G34
Variable-lead thread
cutting
G92
Co-ordinate system
setting, maximum
spindle speed setting
Automatic tool
compensation X
G94
Per minute feed
Automatic tool
compensation Z
G95
Per revolution feed
G40
Tool nose radius
compensation cancel
G96
Constant surface
speed control
G41
Tool nose radius
compensation left
G97
Constant surface
speed control cancel
nd
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G36
G37
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Table 4.21.1 : List of G-codes
M Code
Function
M Code
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Function
M00
Program stop
M18
Turret reverse rotation
M01
Optional stop
M22
Chip conveyor forward
M02
Program stop-reset
M23
Chip conveyor reverse
M03
Spindle normal rotation
M24
Chip conveyor stop
M04
Spindle reverse rotation
M30
Program stop-Reset and
rewind
M05
Spindle stop
M31
Tailstock base unclamp
M06
Tool change
M32
Tailstock base clamp
M08
Coolant ON
M40
Spindle neutral gear
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M09
Coolant OFF
M41
Spindle low gear
M10
Chuck clamp
M42
Spindle high gear
M11
Chuck unclamp
M51
Air blow ON
M12
Tail stock quill OUT
M52
Air blow OFF
M13
Tailstock quill IN
M70
Tool presetter arm down
M17
Turret forward rotation
Table 4.21.2 : List of M-codes
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4.21.2 Preparatory Functions
· Preparatory function changes the control mode of the machine.
i)
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·
G-codes are generally followed by two digit number. Refer Table 4.21.1.
·
Out of Table 4.21.1, few G-codes are required in almost all programs and hence
discussed below :
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G00 (Rapid traverse function) :
·
It is the positioning function and enables rapid movement for tool positioning.
·
This code is used during a typical situation in a machining operation where, the
tool is to be positioned near the cutting surface in smallest possible time, without
any machining.
·
This code remains valid till it is cancelled by another G code.
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ii) G01 (Linear interpolation function) :
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·
Any machining during straight or taper lines is done using this function G01.
·
The feed rate at which the cutting tool is required to move is also specified by
using G01.
·
The use of G00 and G01 is explained with the help of Fig. 4.21.2 and
corresponding instruction blocks :
N001
G00
X10
Y40;
(From origin (0,0) the tool moves at rapid feed rate to position X = 10 and Y = 40
i.e. point P)
N002
G01
X30
Y10
F200;
(In linear interpolation the tool moves to point Q i.e. X = 30 and Y = 10 with a
feed of 200 mm/min).
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Y
P
40
30
20
Q
10
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X
0,0
10
20
30
40
Fig. 4.21.2 : Linear interpolation
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iii) G02/G03 (Circular interpolation function) :
·
If the cutting tool is required to move along an arc then circular interpolation,
functions are used. It permits the cutting tool to move along an arc of circle in
clockwise or counter-clockwise direction.
·
When the circular interpolation is to be used, it is necessary to determine the
plane in which the arc is positioned. For this purpose plane selection codes G17,
G18 or G19 are used.
·
To define the movement of cutting tool along the circle, find the end point
coordinates and the respective radius vectors in a given plane (write the values of
I J K i.e. coordinates of centre of arc with respect to starting point).
·
Sometimes, I J represents the coordinates of the centre of the arc. But generally
this method is not used.
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G02 (Clockwise circular interpolation)
·
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For clockwise circular interpolation G02 code is used. The use of G02 is
explained with the help of Fig. 4.21.3 and corresponding instruction blocks :
N001
G02
X50
Y30
I–10
J–30;
(Clockwise circular interpolation from A to B, where I J represents the value centre
of arc with respect to A). Refer Fig. 4.21.3 (a).
· If direct radius method is used then,
N001 G02 X50 Y30 R31.62;
(Clockwise circular interpolation from A to B, where R= 30 2 + 10 2 = 31.62 ).
·
For Fig. 4.21.3 (b) we can write,
N002 G02 X30 Y20 I–40 J20;
(clockwise circular interpolation from A to B).
· If direct radius method is used then,
N002 G02 X30 Y20 R44.72;
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40 2 + 20 2 = 44.72 ).
(Clockwise circular interpolation from A to B, where R =
Y
Y
70
60
60
A
50
20
I = – 10
10
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(0, 0)
30
B
20
A
40
J = 30
30
R
50
R
40
J = 20
10 20 30 40 50 60
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B
10
X
(0, 0)
I = – 40
10 20 30 40 50 60 70
(a)
X
(b)
Fig. 4.21.3 : Clockwise circular interpolation
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G03 (Anticlockwise circular interpolation)
·
For anticlockwise circular interpolation G03 code is used. The use of G03 is
explained with the help of Fig. 4.21.4 and corresponding instruction blocks :
N001 G03 X30 Y50 I–30 J–10;
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(Anticlockwise circular interpolation from B to A). Refer Fig. 4.21.4 (a).
· If direct radius method is used then
N001 G03 X30 Y50 R31.62;
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(Anticlockwise circular interpolation from B to A, where R = 30 2 + 10 2 = 31.62 ).
Y
Y
70
60
60
A
50
40
R
J = 10
(0, 0)
20
I = – 30
10 20 30 40 50 60
J = 50
30
B
20
10
A
50
40
30
R
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10
X
(a)
(0, 0)
B
I = – 10
10 20 30 40 50 60 70
X
(b)
Fig. 4.21.4 : Anticlockwise circular interpolation
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·
For Fig. 4.21.4 (b) we can write,
N002 G03 X60 Y50 I–10
J50;
(Anticlockwise circular interpolation from B to A).
· If direct radius method is used then,
N002 G03 X60 Y50 R 50.99;
(Anticlockwise circular interpolation from B to A, where R = 50 2 + 10 2 = 50.99 ).
Note : For XY plane I J values are specified, similarly for YZ plane J K values are
specified.
iv) G43/G49 (Tool length compensation)
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·
We know that, tools used for machining can vary in length.
·
This method must be employed to compensate for these varied lengths.
·
·
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There are two methods to account of tool lengths.
i) Premeasuring the tools.
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ii) Tool length compensation using CNC controller's.
In the first method, the tool length is measured and known length can then be
added in the program's Z-axis dimensions to account for the tool. This is known
as presetting the tool.
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·
In a CNC machine setup, a programmable tool register is provided which is a
memory location in the computer where the tool length may be stored.
·
When a particular tool is called, the required information for the tool offset is
called from the tool register.
Tool 3
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Tool offset
Tool offset
Tool 2
Tool offset
Tool 1
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Fig. 4.21.5 : Tool length offset
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·
The MCU then shifts the Z-axis by the amount stored in the shift register.
Generally, the values of the offset are entered by the operator at the time of
programming. Refer Fig. 4.21.5 for tool length offset.
·
Cutter length compensation is given by G43 and cancelled by G49.
v) G40/G41/G42 (Cutter radius compensation)
·
During profile cutting operations, an allowance is provided to the cutter radius in
the programmed co-ordinates.
·
Consider a component as shown in Fig. 4.21.6 in which the tool path centre line
is decided by the spindle axis centre-line, whereas the workpiece edge is offset
from it by the cutter radius.
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Component
profile
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Tool
path
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Fig. 4.21.6 : Cutter diameter compensation
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·
As per the cutter radius, the programmer will have to generate new co-ordinate
positions.
·
Usually, the CNC machines have a built-in feature called as cutter diameter
compensation which allows the user to input the cutter diameter compensation for
each tool.
·
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Cutter compensation is accomplished by using following G codes :
(a) Cutter diameter compensation Left (G41) : When G41 command is given, the
tool will compensate to the left of the programmed surface when seen in the
direction of the tool movement. Refer Fig. 4.21.7.
G 41
G 42
Fig. 4.21.7 : Cutter diameter compensation to the LEFT and RIGHT
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(b) Cutter diameter compensation right (G42) : When G42 command is given, the
tool will compensate to the right of the programmed surface when seen in the
direction of the tool movement. Refer Fig. 4.21.7.
(c) Cutter diameter compensation cancel (G40) : G40 command cancels any
compensation as applied previously. The tool will change from a compensated
position to an uncompensated position.
vi) G90/G91 (Absolute and Incremental programming)
G90 (Absolute programming system)
· In this system, all the positions are indicated from a reference point, which is a
fixed zero point or set point.
·
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Fig. 4.21.8 (a) shows that all positions are marked with a set point.
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D
C
B
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Fig. 4.21.8 (a) : Absolute system
·
In Fig. 4.21.8 (a) point 'A' is a set point.
N001 G90 G01 X10 Y5 F200;
N002 X20 Y10;
N003 X30 Y15;
X
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(Linear interpolation from A to B in absolute mode).
(Linear interpolation and tool moves from B to C).
(Linear interpolation and tool moves from C to D).
G91 (Incremental programming system)
· In this method, the tool positions are indicated with respect to previous point.
·
Fig. 4.1.82 (b) shows an example of this system.
·
The main disadvantage of this system is that if an error occurs into the
dimensions of any location, all the locations marked after that will carry the same
error.
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Y
D
C
B
A
X
Fig. 4.21.8 (b) : Incremental system
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N001 G91 G01 X10 Y5 F200;
(Linear interpolation from A to B in
N002 X10 Y5;
incremental mode).
(Linear interpolation from B to C).
N003 X10 Y5;
[Coordinates of C are given with respect to B].
(Linear interpolation from C to D).
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[Coordinates of D are given with respect to C].
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4.22 Procedure to Write a Part Program :
The general procedure to write a part program for any component is as follows :
· Study the given part or component carefully.
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·
Assume required data like feed, speed, cutter diameter, etc. and decide the path to
be followed by the cutter.
·
Mark the tool path on a separate dimensionless drawing and give the number to
various points along the path.
·
Write the coordinates of all the points on a separate table or on the drawn
drawing.
·
The first four blocks and last two blocks are almost same in each program.
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Note :
i) G codes, M codes and other data remains active or continue, unless and until it is
cancelled by other codes and data. Hence there is no need to write this data again
and again on each line. Even if it is written, it is not wrong.
ii) Number of blocks can be written as N001, N01, N1, N10, N100 in any of the way.
iii)We can write more than one G codes or M codes in one block.
iv)We can write any number as a program number.
v) The initial position (position 1 which is not reference position) can be assumed any
where near the workpiece surface.
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4.23 Part Programming for Lathe :
0
50
Example 4.23.1 : Write a manual part program to finish the stepped shaft in f 40 mm
section as shown in Fig. 4.23.1. Assume spindle speed as 350 rpm and feed rate
0.4 mm/rev.
40
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Fig. 4.23.1
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Solution : Given data :
S = 350 rpm
and
F = 0.4 mm/rev.
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Assume that the machine by default is in diameter programming mode.
(Home position)
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3
2
50
+X
1 +Z
40
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0
4
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Fig. 4.23.1 (a)
Refer Fig. 4.23.1 (a).
Program
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Description
N001 G90 G71;
(Absolute programming mode and data input is in metric
mode)
N002 G95 G28 U0 W0;
(Feed in mm/rev. and go to home position)
N003 M06 T0101;
(Tool change, tool no. 01 with offset value 01)
N004 S350 M03;
(Spindle speed 350 rpm and in clockwise direction)
N005 G00 X0 Z0 M08;
(Rapid tool positioning at point 1 and coolant ON)
N006 G01 X40 F0.4;
(Finish the face till point 2 at 0.4 mm/rev.)
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N007 Z – 40;
(Finish the turn upto point 3)
N008 X52;
(Finish cut face till point 4)
N009 G00 G28 U0 W0 M09;
(Rapid traverse to home position and coolant OFF)
N0010 M05 M30;
(Spindle stop and program stop).
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Example 4.23.2 : Write a manual part program for turning and facing the barstock to
the required dimensions as shown in Fig. 4.23.2. Assume one roughing and one finish
face cut; two roughing and two finish turning cut. Assume spindle speed 450 rpm and
feed 0.5 mm/rev.
50
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Fig. 4.23.2
Solution : Given data :
S = 450 rpm
and
F = 0.5 mm/rev.
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Assume that the machine by default is in diameter programming mode. Refer
Fig. 4.23.2 (a).
1296
8
2
1
52
50
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1
4
7
10
5
50
69
72
65
11
+X
32
+Z
Fig. 4.23.2 (a)
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Program
Description
N001 G90 G71;
(Absolute programming mode and data input is in metric
mode)
N002 G95 G28 U0 W0;
(Feed in mm/rev. and go to home position)
N003 M06 T0101;
(Tool change, tool no. 01 with offset value 01)
N004 S450 M03;
(Spindle speed is 450 rpm and in clockwise direction)
N005 G00 X75 Z2 M08;
(Rapid traverse to point 1 and coolant ON)
N006 G01 X0 F0.5;
(Rough cut face till point 2 at a feed of 0.5 mm/rev.)
N007 Z0;
(Advance or depth of cut upto point 3)
N008 X60;
(Finish cut face till point 4)
ww
N009 Z – 48;
N0010 X70;
w.E
N0011 G00 X60 Z0;
N0012 G01 X55;
N0013 Z – 49;
(Rough turning upto point 5)
(Rough turning upto point 6)
asy
E
(Rapid traverse to point 4)
(Rough turning upto point 7)
ngi
(Rough turning upto point 8)
N0014 X70;
(Rough turning upto point 9)
N0015 G00 X60 Z0;
(Rapid traverse to point 4)
N0016 G01 X50;
(Rough turning upto point 10)
N0017 Z – 50;
(Rough turning upto point 11)
N0018 X70;
(Rough turning till point 12)
N0019 G00 G28 U0 W0 M09;
(Rapid traverse to home position and coolant OFF)
N0020 M05 M30;
(Spindle stop and program stop)
nee
rin
g.n
et
Explanation :
The tool cycles are as follows :
1-2
(Rough facing)
6-4
(Rapid traverse)
2-3
(Feed for finishing)
4-7-8-9
(Again rough turning of shaft and
shoulder)
3-4
(Finish facing)
9-4
(Rapid traverse)
4-5-6
(Rough turning of shaft and
shoulder)
4-10-11-12
(Finish turning of shaft and
shoulder)
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Fundamental of CNC and Part Programming
Example 4.23.3 : Write a manual part program for finishing a forged component as
shown in Fig. 4.23.3. Assume speed 300 rpm and feed 0.4 mm/rev. Assume 1 mm
material is to be removed radially from external diameter.
 50
 100
R5
ww
w.E
Solution : Given data :
S = 300 rpm
and
20
30
30
asy
E
Fig. 4.23.3
F = 0.4 mm/rev.
ngi
Assume that the machine by default is in diameter programming mode.
Refer Fig. 4.23.3 (a).
6
25
5
3 R5
2
4
1
50
60
100
nee
30
(Home position)
rin
+X
+Z
g.n
et
60
80
Fig. 4.23.3 (a)
Program
Description
N001 G90 G71;
(Absolute programming mode and data input is in metric
mode)
N002 G95 G28 U0 W0;
(Feed in mm/rev. and go to home position)
N003 M06 T0101;
(Tool change, tool no. 01 with offset value 01)
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N004 S300 M03;
(Spindle speed is 300 rpm and in clockwise direction)
N005 G00 X50 Z2 M08;
(Rapid traverse to point 1 and coolant ON)
N006 G01 Z – 25 F0.4;
(Finish turning upto point 2 with a feed of 0.4 mm/rev.)
N007 G02 X60 Z – 30 R5;
(Clockwise interpolation upto point 3 with a radius of 5)
N008 G01 Z – 60;
(Finish turning upto point 4)
N009 X100 Z – 80;
(Finish turning upto point 5)
N0010 G00 X102;
(Rapid traverse to point 6)
N0011 G28 U0 W0 M09;
(Return to home position and coolant OFF)
N0012 M05 M30;
(Spindle stop and program stop)
ww
Example 4.23.4 : Write a part program for the component as shown Fig. 4.23.4.
Assume that spindle speed 500 rpm and feed is 0.3 mm/rev.
30
20
ngi
R10
Fig. 4.23.4
Solution : Given data :
20
asy
E
40
20
60
w.E
nee
rin
S = 500 rpm and Feed = 0.3 mm/rev.
g.n
et
Assume that the machine by default is in diameter programming mode.
Refer Fig. 4.23.4 (a).
6
(Home position)
R10
5
4
40
60
8
7
3
2 1
20
+Z
20
+X
40
50
70
Fig. 4.23.4 (a)
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Program
Description
N001 G90 G71;
(Absolute programming mode and data input is in metric
mode)
N002 G95 G28 U0 W0;
(Feed in mm/rev. and go to home position)
N003 M06 T0101;
(Tool change, tool no. 01 with offset value 01)
N004 S500 M03;
(Spindle speed 500 rpm and in clockwise direction)
N005 G00 X0 Z2 M08;
(Rapid traverse to point 1 and coolant ON)
N006 G01 Z0 F0.3;
(Linear interpolation upto point 2 with a feed of
0.3 mm/rev.)
ww
N007 X20;
(Finish turn upto point 3)
w.E
N008 X40; Z – 20;
(Finish turning upto point 4)
N009 Z – 40;
(Finish turning upto point 5)
asy
E
N0010 G02 X60 Z – 50 R10;
(Circular interpolation upto point 6 with a radius of 10)
N0011 G01 Z – 70;
(Finish turning upto point 7)
N0012 G00 X65;
ngi
(Rapid traverse to point 8)
nee
N0013 G28 U0 W0 M09;
(Return to home position and coolant OFF)
N0014 M05 M30;
(Spindle stop and program stop)
rin
g.n
et
Example 4.23.5 : Write a manual part program to finish the following component
(Refer Fig. 4.23.5). Assume spindle speed 600 rpm and feed 0.45 mm/rev.
Ø 30
R15
Fig. 4.23.5
Solution : Given data :
S = 600 rpm and F = 0.45 mm/rev.
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Assume that the machine by default is in diameter programming mode.
Refer Fig. 4.23.5 (a).
(Home
position)
4
3
+X
R15
Ø 30
2
1
+Z
ww
w.E
Program
N001 G90 G71;
Fig. 4.23.5 (a)
asy
E
Description
ngi
(Absolute programming mode and data input is in metric
mode)
nee
N002 G95 G28 U0 W0;
(Feed in mm/rev. and go to home position)
N003 M06 T0101;
(Tool change, Tool no. 01 with offset value 01)
N004 S600 M03;
(Spindle speed 600 rpm and in clockwise direction)
N005 G00 X0 Z2 M08;
(Rapid traverse to point 1 and coolant ON)
N006 G01 Z0 F0.45;
(Linear interpolation upto point 2 with a feed of
0.45 mm/rev.)
N007 G03 X30 Z – 15 R15;
(Anticlockwise interpolation upto point 3 with a radius of
15)
N008 G00 X32;
(Rapid traverse upto point 4)
N009 G28 U0 W0 M09;
(Return to home position and coolant OFF)
N0010 M05 M30;
(Spindle stop and program stop)
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rin
g.n
et
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Fundamental of CNC and Part Programming
Example 4.23.6 : Fig. 4.23.6 shows the finished size of a round bar. The original
diameter of the bar was 28 mm. Make a part program for facing, parting and
reduction of diameter. Take feed = 200 mm/min, spindle speed = 640 rpm and
depth of cut = 2 mm per cut.
ww
Ø 28
Ø 24
Ø 20
X
40
60
w.E
Solution : Given data :
Z
Fig. 4.23.6
asy
E
ngi
Feed F = 200 mm/min, Spindle speed S = 640 rpm, Depth of cut D = 2 mm
It is assumed that the machine by default is in diameter programming mode. Refer
nee
Fig. 4.23.6 (a) for turning and Fig. 4.23.6 (b) for parting.
12
rin
7
6
11
5
10 4
9
3
8
X
2
1
Home position
g.n
et
Z
40
60
Fig. 4.23.6 (a)
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Ø 28
Ø 24
Ø 20
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Fundamental of CNC and Part Programming
Parting Tool
Fig. 4.23.6 (b)
Program
Description
N001 G90 G71;
(Absolute programming mode and data input is in metric
mode)
ww
N002 M06 T0101;
(Tool change, tool No.1 with offset value 01)
N003 M03 S640 G94;
(Spindle speed 640 rpm and in clockwise direction, Feed in
mm/min)
N004 G00 X0 Z5 M08;
(Rapid traverse to point 1 and coolant ON)
w.E
asy
E
N005 G73 P6 Q11 U2 W1 D2 (See explanation at the end)
F200;
N006 G01 X0 Z0;
(Linear interpolation upto point 2)
ngi
N007 X20;
(Linear interpolation upto point 3)
N008 Z – 40;
(Linear interpolation upto point 4)
nee
N009 X24;
(Linear interpolation upto point 5)
N0010 Z – 60;
(Linear interpolation upto point 6)
N0011 X30;
(Linear interpolation upto point 7)
N0012 G72 P6 Q11;
(See explanation at the end)
N0013 G28 U0 W0;
(Return to home position)
N0014 M06 T0202;
(Tool change for parting with offset value 02)
N0015 G00 X – 30 Z – 62;
(Rapid traverse to parting position)
N0016 G01 X0 F50;
(Linear interpolation for parting with Feed of 50 mm/min)
N0017 G28 U0 W0;
(Return to home position)
N0018 M03 M09 M30;
(Coolant OFF, spindle stop and program stop).
rin
g.n
et
Line N005 : G73 P6 Q11 U2 W1 D2 F200;
G73 –
Rough facing and rough turning
P6 –
Indicates sequence number at which the machining starts
Q11 –
Indicates sequence number at which the machining ends
U2 –
Leave 2 mm stock on X axis for machining
W1 –
Leave 1 mm stock on Z axis for finishing
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Fundamental of CNC and Part Programming
D2 –
Depth of cut for each roughing pass is 2 mm
F200 –
Feed is 200 mm/min
Line N0012 : G72 P6 Q11;
G72 –
Finish facing and turning
P6 –
Indicates sequence number at which finish pass starts
Q11 –
Indicates sequence number at which finish pass ends
Example 4.23.7 : Write a part program for that part shown in Fig. 4.23.7.
5
ww
 20
w.E
(0, 0)
 10
E
D
10 mm
A
5 mm
asy
E
C
–x
B
40
–z
20
ngi
+z
+x
Fig. 4.23.7
Solution : Given data :
Assume, S = 500 rpm
and
F = 0.3 mm/rev
Program
nee
rin
Description
g.n
et
N001
G90 G71;
(Absolute programming mode and data input is in metric
mode)
N002
G95 G28 U0 W0;
(Feed in mm/rev. and go to home position)
N003
M06 T0101;
(Tool change, tool no. 01 with offset value 01)
N004
S500 M03;
(Spindle speed 500 rpm and in clockwise direction)
N005
G00 X0 Z0 M08;
(Rapid traverse to point A and coolant ON)
N006
G01 X5 Z – 25 F0.3;
(Linear interpolation upto point D with a feed of
0.3 mm/rev.)
N009
Z – 60;
(Finish turn upto point E)
N008
G00 X15;
(Rapid traverse to point C)
N009
X15 Z – 20;
(Rapid traverse to point B)
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N010
G28 U0 W0 M09;
(Return to home position and coolant OFF)
N011
M05 M30;
(Spindle stop and program stop)
Example 4.23.8 : A 110 mm long cylindrical rod of f 75 mm is to be turned into a
component as shown in Fig. 4.23.8, using a CNC lathe. Write a CNC program for
manufacturing this component.
asy
E
Ø 30
w.E
Ø 55
Ø 70
ww
Ø 44
31
R12
20
45
67
ngi
15º
95
Fig. 4.23.8
Solution : Given data :
nee
rin
Assume spindle speed 500 rpm and feed 0.15 mm/rev. Refer Fig. 4.23.8 (a).
9
8
6
7
31
5
4
3
2
1
Ø 30
Ø 55
Ø 44
R12
Ø 70
g.n
et
20
45
67
15º
95
Fig. 4.23.8 (a)
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Fundamental of CNC and Part Programming
Program
Description
N001
G90 G71;
(Absolute programming mode and data input is in metric
mode)
N002
G95 G28 U0 W0;
(Feed in mm/rev and go to home position)
N003
M06 T0101;
(Tool change, tool no. 01 with offset value 01)
N004
S500 M03;
(Spindle speed 500 rpm and in clockwise direction)
N005
G74 P006 Q015 D2 R1;
(Canned cycle starts from line 006 to 015 with depth of
cut 2 mm and relief 1 mm)
N006
G00 X0 Z0 F0.15 M08;
(Rapid traverse to origin, coolant ON and feed is
0.15 mm/rev)
ww
N007
G01 X28 Z0;
(Linear interpolation upto point 1)
N008
X30 Z – 1;
(Linear interpolation upto point 2)
N009
X30 Z – 20;
(Linear interpolation upto point 3)
N010
G03 X44 Z – 31 R12;
(Circular interpolation upto point 4)
N011
G01 X44 Z – 45;
(Linear interpolation upto point 5)
N011
X55 Z – 45;
N012
X55 Z – 67;
N013
X70 Z – 69;
N014
X70 Z – 95;
(Linear interpolation upto point 9)
N015
G72 P006 Q014;
(Finish facing and turning from line 006 to 014)
N016
G28 U0 W0 M09;
(Return to home position and coolant OFF)
N017
M05 M30;
(Spindle stop and program stop)
w.E
asy
E
(Linear interpolation upto point 6)
ngi
(Linear interpolation upto point 7)
(Linear interpolation upto point 8)
nee
rin
g.n
et
Example 4.23.9 : Write a manual part program to turn the component shown on a
CNC Lathe from 75 mm bar stock. The following data may be assumed :
i) There will be two rough turnings and one finish turning. The first cut is with a depth
of 3 mm for a length of 58 mm, the second with a depth of 3 mm for a length of
59 mm and the third with a depth of 1.5 mm for the full length of 60 mm.
ii) The shoulder of the work-piece is also machined during each cut.
iii) The spindle speed is 400 rpm and the feed rate is 0.5 mm/rev.
Make a free-hand sketch showing relevant points of tool positions for each of the three
turning operations and then write the manual part program. State also what each line
of the program does.
Note : If the exact G-codes and M-codes are not known, the student can use his/her
own code-numbers, but the function of such codes must be clearly stated.
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Fundamental of CNC and Part Programming
 60
+Z
 75
+X
60
Fig. 4.23.9
Solution : Given data :
ww
S = 400 rpm, F = 0.5 mm/rev
Assume that the machine is in diameter programming mode.
w.E
Also, assuming one rough and one finish facing cut.
5
11
62
60
ngi
1
4
7
10
nee
rin
+X
32
75
8
asy
E
60
63
69
1296
2
1
g.n
et
+Z
Fig. 4.23.9 (a)
Program
Description
N001 G90 G71;
(Absolute programming mode and data input is in metric
mode)
N002 G95 G28 U0 W0;
(Feed in mm/rev. and go to home position)
N003 M06 T0101;
(Tool change, tool no. 01 with offset value 01)
N004 S400 M03;
(Spindle speed is 400 rpm and in clockwise direction)
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N005 G00 X80 Z2 M08;
(Rapid traverse to point 1 and coolant ON)
N006 G01 X0 F0.5;
(Rough cut face till point 2 at a feed of 0.5 mm/rev.)
N007 Z0;
(Advance or depth of cut upto point 3)
N008 X69;
(Finish cut face till point 4)
N009 Z – 58;
(Rough turning upto point 5)
N0010 X80;
(Rough turning of shoulder upto point 6)
N0011 G00 X69 Z0;
(Rapid traverse to point 4)
N0012 G01 X63;
(Rough turning upto point 7)
N0013 Z – 59;
(Rough turning upto point 8)
ww
N0014 X80;
w.E
(Rough turning upto point 9)
N0015 G00 X69 Z0;
N0016 G01 X60;
N0017 Z – 60;
N0018 X80;
(Rapid traverse to point 4)
asy
E
(Rapid traverse to point 10)
(Finish turning upto point 11)
ngi
(Finish turning of shoulder till point 12)
N0019 G00 G28 U0 W0 M09;
(Rapid traverse to home position and coolant OFF)
N0020 M05 M30;
(Spindle stop and program stop)
Explanation :
nee
The tool cycles are as follows :
rin
g.n
et
1-2
(Rough facing)
6-4
(Rapid traverse)
2-3
(Feed for finishing)
4-7-8-9
(Again rough turning of shaft and
shoulder)
3-4
(Finish facing)
9-4
(Rapid traverse)
4-5-6
(Rough turning of shaft and
shoulder)
4-10-11-12
(Finish turning of shaft and
shoulder)
Example 4.23.10 : Write a complete part program using G and M codes for the job
shown in Fig. 4.23.10. Assume suitable speed and feed for machining.
Billet size - Diameter : 60 mm and Length : 90 mm.
Thread : Major Diameter, D0 = 20 mm, Minor Diameter,
Dc = 17 mm and Pitch : 2.5 mm, Groove : Width = 5 mm and depth = 2.5 mm.
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R10
 60
 40
 20
Thread
Groove
ww
w.E
20
Solution :
10
9
20
5
10
20
asy
E
Fig. 4.23.10
8
7
ngi
6
5
3
4
X
nee
2
rin
1
g.n
et
Z
Fig. 4.23.10 (a)
Program
Description
N01 G90 G71 ;
Absolute programming mode and data input is in metric mode.
N02 G95 G21 ;
Feed in mm/rev ; Metric mode
Facing
N03 M42 T01 ;
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N04 S200 M03 M07 ;
Spindle speed 200 rpm, clockwise ON, coolant ON
N05 G00 X0 Z3 ;
Rapid to point 1
N06 G01 Z0 F0.5 ;
Facing operation begins on the stock
N07 X62 ;
Facing ends at point 2
Turning
N08 G71 U2 R1 ;
See explanation below
N09 P010 Q015 W0
F0.5 ;
See explanation below
N010 G00 X20 Z0 ;
Rapid to point 2
ww
N011 G01 Z–25 ;
Turning begins till point 6
N012 G02 X40 Z–35 ;
Circular clockwise interpolation upto point 7
N013 G01 X40 Z–45 ;
Linear interpolation upto point 8
N014 G01 X60 Z–65 ;
Linear interpolation upto point 9
w.E
N015 G01 X60 Z–85 ;
Finishing
asy
E
Linear interpolation upto point 10
ngi
N016 M42 T02 ;
High gear, tool No. 02
N017 G70 P010 Q015 ;
See explanation below
N018 G28 U0 V0 W0
M09 ;
Return to home position, coolant off
N019 M05 ;
Spindle off
Grooving
nee
rin
g.n
et
N020 M42 T03 ;
High gear, tool no. 03
N021 G95 G21 G96 ;
Feed per revolution, metric mode, constant surface speed
N022 S400 M03 M07 ;
Spindle speed 400 rpm, clockwise ON coolant ON
N023 G00 X30 Z0 ;
Rapid to point 2
N024 Z–20 ;
Rapid to point 3
N025 X30 ;
Rapid to point 3
N026 G75 R0.5 ;
See explanation
N027 X15 Z–25 P1000
Q500 F0.15 ;
See explanation
N028 X30 ;
Rapid to point 6
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Fundamental of CNC and Part Programming
N029 Z0 X30 ;
Rapid to point 2
Threading Tool
N030 M42 T04 ;
High gear, tool no. 0.4
N031 S400 M03 M07 ;
Spindle speed 400 rpm, clockwise ON, coolant ON
N032 G00 X20 Z3 ;
Rapid to point 2
N033 G76 P020060
Q300 R100 ;
See explanation below
N034 X17 Z–22 P2000
Q300 F2.5 ;
See explanation below
N035 G28 U0 V0 W0
M09 ;
Return to home position, coolant off
N036 M05 M30 ;
Spindle off and program stop
ww
Explanation :
N08
Sequence
Number
N09
Sequence
Number
w.E
G71
asy
E
Stock removal
in turning
P010
R1
Depth of each roughing
pass in 2 mm
ngi
Q015
Sequence
Number at
beginning
U2
Sequence
number at end
Tool escape of
1 mm
W0
F0.5
Facing operation is done
and there is no stock left
on 2 axis
Feed rate is
0.5 mm/ rev
nee
N017
G70
P010
Sequence
Number
G code for finish
face and turn
Sequence number
at beginning
rin
g.n
et
Q015
Sequence number at end
N026
G75
R0.5
Sequence
Number
G code for
Grove cutting
Tool escape
of 0.5 mm
N027
X14
Z–23
P1000
Q500
F2.5
Sequence
Number
Finished
Grooved
diameter is
15 mm
End point of
the groove
Incremental
depth of cut on
X axis
1000/1000 = 1
mm
Tool
advance
on Z axis
500/1000 =
0.5 mm
Feed rate of
0.15 mm/ rev
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N033
G76
P020060
Q300
R100
Sequence
Number
Multiple
threading
cycle
Double finish pass
indicated by 02, No
chamfer (00) and
angle of tool 60º
Minimum cutting
depth,
300/1000 = 0.3 mm
Finishing allowance,
100/1000 = 0.1 mm
N034
X17
Z–22
P2000
Q300
F2.5
Sequence
Number
Minor
diameter is
17 mm
Thread
length
Height of
thread =
2000/1000 =
2 mm
Depth of
initial cut is
300/1000 =
0.3 mm
Pitch of
threads is
2.5 mm
Example 4.23.11 : Write CNC part program for the component shown in Fig. 4.23.11
Mention the assumption made.
ww
130.00
asy
E
120.00
65.00
45.00
80.00
100.00
R8
Chramfer 3  45º
nee
54.56
ngi
32.00
R10

w.E
150.00
rin
g.n
et
M20 = 2.5 Thread
root diameter 16.75
Grove 3 DEEP
Billet diameter 110 mm
Billet length 200 mm
29.00
All dimensions in mm
32.00
Fig. 4.23.11
Solution : Refer example 4.23.10.
4.24 Part Programming for Milling and Drilling :
Example 4.24.1 : Write a part program to machine a workpiece as shown in
Fig. 4.24.1 (a). Assume cutter diameter as 10 mm end mill type, depth of workpiece
10 mm and feed rate 200 mm/min. Take spindle speed 500 r.p.m.
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Y
Y
20
P6(0, 50)
P5(20, 50)
30
P3(50, 20)
P4(20, 20)
20
ww
X
50
w.E
Z
X
P1(0, 0)
P2(50, 0)
P0(–10, –10)
Z
asy
E
X
X
– 10
10
(a) Workpiece
ngi
Fig. 4.24.1
Solution : For the tool path refer Fig. 4.24.1 (b).
Programs
(b) Tool path
nee
rin
Description
g.n
et
101;
Program number.
N01 G28 U0 V0 W0;
Return to machine reference position.
N02 G90 G71 G94;
Absolute programming mode, metric (mm) data input and
feed in mm/min.
N03 G17 M06 T01;
Selection of XY plane, tool change and select tool no.01.
N04 G41 M03 S500;
Cutter radius compensation to left, spindle start in clockwise
direction and spindle speed is 500 r.p.m.
N05 G00 X–10 Y–10 Z5
M08;
Rapid travel to position (–10, – 10) and cutter is 5 mm
above the workpiece surface, coolant ON.
N06 G01 X0 Y0 Z–10 F200;
Linear interpolation and move tool 10 mm downward along
Z-axis with feed 200 mm/min to position (0, 0).
N07 X50 Y0;
Linear interpolation and tool move to position (50, 0)
N08 X50 Y20;
Linear interpolation and tool move to position (50, 20)
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N09 X20 Y20;
Linear interpolation and tool move to position (20, 20)
N10 X20 Y50;
Linear interpolation and tool move to position (20, 50)
N11 X0 Y50;
Linear interpolation and tool move to position (0, 50)
N12 X0 Y0;
Linear interpolation and tool move to position (0, 0)
N13 G00 Z5;
Rapid travel 5 mm above the workpiece surface.
N14 G28 U0 V0 W0;
Return to machine reference position.
N15 G40 M05;
Cutter radius compensation cancel and spindle stop.
N16 M09 M30;
Coolant off, program end and tape rewind.
ww
Example 4.24.2 : Write a part program to machine a workpiece as shown in
Fig. 4.24.2 (a). Assume cutter diameter as 10 mm end mill type, depth of workpiece
10 mm and feed rate 150 mm/min. Take spindle speed 600 r.p.m.
Y
w.E
Y
asy
E
ngi
R20
40
50
P3(50, 40)
P6(0, 40)
X
nee
I=0
and J = 20
P1(0, 0)
P0(–10, –10)
Z
rin
X
g.n
et
P2(50, 0)
Z
X
X
10
– 10
(a) Workpiece
(b) Tool path
Fig. 4.24.2
Solution : For the tool path refer Fig. 4.24.2 (b).
Program
Description
102;
Program number
N01 G28 U0 V0 W0;
Return to machine reference position.
N02 G90 G71 G94;
Absolute programming mode, metric (mm) data input and
feed in mm/min.
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N03 G17 M06 T01;
Selection of XY plane, tool change and select tool no. 01.
N04 G41 M03 S600;
Cutter radius compensation to left, spindle ON with speed
600 r.p.m.
N05 G00 X–10 Y–10 Z5 M08;
Rapid travel to position (–10, –10) and cutter is 5 mm
above the workpiece surface, coolant ON.
N06 G01 X0 Y0 Z–10 F150;
Linear interpolation and move tool 10 mm downward along
Z-axis with feed 150 mm/min.
N07 X50 Y0;
Linear interpolation and tool move to position (50, 0).
N08 G03 X50 Y40 I0 J20;
Anticlockwise circular interpolation and tool move to position
(50, 40).
N09 G01 X0 Y40;
Linear interpolation and tool move to position (0, 40)
N10 X0 Y0;
Linear interpolation and tool move to position (0, 0).
N11 G00 Z5;
Rapid travel 5 mm above the workpiece surface.
ww
w.E
N12 G28 U0 V0 W0;
Return to machine reference position.
N13 G40 M05;
Cutter radius compensation cancel and spindle stop.
N16 M09 M30;
asy
E
Coolant off, program end and tape rewind.
Example 4.24.3 : Write a part program to machine a workpiece as shown in
Fig. 4.24.3 (a). Assume cutter diameter as 10 mm end mill type, depth of workpiece
10 mm and feed rate 100 mm/min. Take spindle speed 800 r.p.m.
ngi
Y
P10(0, 70)
Y
R15
nee
P9(20, 70) P8(50, 70)
P7(70, 70)
rin
20
g.n
et
R15
P6(70, 50)
P5(70, 20)
20
R15
20
20
X
X
P1(0, 0)
P2(20, 0)
P3(50, 0)
P4(70, 0)
P0(–10, –10)
(a) Workpiece
(b) Tool path
Fig. 4.24.3
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Solution : For tool path refer Fig. 4.24.3 (b).
Program
Description
103;
Program number.
N01 G28 U0 V0 W0;
Return to machine reference position.
N02 G90 G71 G94;
Absolute programming mode, metric (mm) data input and
feed in mm/min.
N03 G17 M06 T01;
Selection of XY plane, tool change and select tool no. 01.
N04 G41 M03 S800;
Cutter radius compensation to left, spandle ON with speed
800 r.p.m.
ww
N05 G00 X–10 Y–10 Z5 M08;
w.E
N06 G01 X0 Y0 Z–10 F100;
N07 X20 Y0;
N08 G02 X50 Y0 I15 J0;
N09 G01 X70 Y0;
Rapid travel to position (–10, –10) and cutter is 5 mm
above the workpiece surface, coolant ON.
Linear interpolation and move tool 10 mm downward along
Z-axis with feed 100 mm/min to position (0, 0).
asy
E
Linear interpolation and tool move to position (20, 0).
Clockwise circular interpolation and tool move to position
(50, 0).
ngi
Linear interpolation and tool move to position (70, 0).
nee
N10 X70 Y20;
Linear interpolation and tool move to position (70, 20).
N11 G02 X70 Y50 I0 J15;
Clockwise circular interpolation and tool move to position
(70, 50).
N12 G01 X70 Y70;
Linear interpolation and tool move to position (70, 70).
N13 X50 Y70;
Linear interpolation and tool move to position (50, 70).
N14 G02 X20 Y70 I–15 J0;
Clockwise circular interpolation and tool move to position
(20, 70).
N15 G01 X0 Y70;
Linear interpolation and tool move to position (0, 70).
N16 X0 Y0;
Linear interpolation and tool move to position (0, 0).
N17 G00 Z5;
Rapid travel to 5 mm above the workpiece surface.
N18 G28 U0 V0 W0;
Return to machine reference position.
N19 G40 M05;
Cutter radius compensation cancel and spindle stop.
N20 M09 M30;
Coolant off, program end and tape rewind.
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g.n
et
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Fundamental of CNC and Part Programming
Example 4.24.4 : Write the program for the job as shown in Fig. 4.24.4 (a). Use the
following machining data :
Speed = 800 r.p.m., feed = 10 mm/min,
Depth of cut = 3 mm, Thickness of job = 3 mm.
Y
P10(12, 165)
160
P9(75, 165)
P8(75, 145)
P7(148, 145)
75
R12
20
P6(160, 133)
R12
P11(0, 153)
45
P5(160, 100)
R20
ww
(0, 0)
110
P4(130, 100)
w.E
(a) Workpiece
100
P3(110, 80)
asy
E
X
P1(0, 0)
ngi
Fig. 4.24.4
Solution : For tool path refer Fig. 4.24.4 (b).
Program
P2(110, 0)
P0(–10, –10)
(b) Tool path
nee
rin
Description
g.n
et
104;
Program number.
N01 G28 U0 V0 W0;
Return to machine reference position.
N02 G90 G71 G94;
Absolute programming mode, metric (mm) data input
and feed in mm/min.
N03 G17 M06 T01;
Selection of XY plane, tool change and select tool
no.01.
N04 G41 M03 S800;
Cutter radius compensation to left, spindle ON with
speed 800 r.p.m.
N05 G00 X–10, Y–10 Z5 M08;
Rapid travel to position (–10, –10) and cutter is
5 mm above the workpiece surface, coolant ON.
N06 G01 X0 Y0 Z–3 F10;
Linear interpolation and move tool 3 mm downward
along Z-axis with feed 10 mm/min to position (0, 0).
N07 X110 Y0;
Linear interpolation and tool move to position (110,0).
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N08 X110 Y80;
Linear interpolation and tool move to position
(110, 80).
N09 G02 X130 Y100 I20 J0;
Clockwise circular interpolation and tool move to
position (130, 100).
N10 G01 X160 Y100;
Linear interpolation and tool move to position
(160, 100).
N11 X160 Y133;
Linear interpolation and tool move to position
(160, 133).
N12 G03 X148 Y145 I–12 J0;
Anticlockwise circular interpolation and tool move to
position (148, 145).
N13 G01 X75 Y145;
Linear interpolation and tool move to position
(75, 145).
N14 X75 Y165;
Linear interpolation and tool move to position
(75, 165).
ww
w.E
N15 X12 Y165;
Linear interpolation and tool move to position
(12, 145).
asy
E
N16 G03 X0 Y153 I0 J–12;
Anticlockwise circular interpolation and tool move to
position (0, 153).
N17 G01 X0 Y0;
Linear interpolation and tool move to position (0, 0).
N18 G00 Z5;
ngi
Rapid travel to 5 mm above the workpiece surface.
nee
N19 G28 U0 V0 W0;
Return to machine reference position.
N20 G40 M05;
Cutter radius compensation cancel and spindle stop.
N21 M09 M30;
Coolant off, program end and tape rewind.
rin
g.n
et
Example 4.24.5 : Write the program for the job as shown in Fig. 4.24.5 (a).
P3(200, 100)
P4(0, 100)
C
80
(0, 0)
100
80
A
120
P6(80, 20)
P5(0, 20)
B
P0(0, 0)
P1(80, 0)
P2(200, 0)
Fig. 4.24.5
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Fundamental of CNC and Part Programming
Solution : For tool path refer Fig. 4.24.5 (b).
Assume speed 500 r.p.m., thickness of job 5 mm and feed rate as 20 mm/min.
Program
Description
105;
Program number.
N01 G28 U0 V0 W0;
Return to machine reference position.
N02 G90 G71 G94;
Absolute programming mode, metric (mm) data input
and feed in mm/min.
N03 G17 M06 T01;
Selection of XY plane, tool change and select tool
no.01.
N04 G41 M03 S500;
Cutter radius compensation to left, spindle ON with
speed 500 r.p.m.
N05 G00 X0, Y0 Z5 M08;
Rapid travel to position (0, 0) and cutter is 5 mm
above the workpiece surface, coolant ON.
ww
w.E
N06 X80 Y0 Z–5 F20;
N07 G01 X200 Y0;
asy
E
Rapid travel to position (80, 0) and move tool 5 mm
downward along Z-axis with feed 20 mm/min.
ngi
Linear interpolation and tool move to position (200,0).
nee
N08 X200 Y100;
Linear interpolation and tool move to position
(200, 100).
N09 X0 Y100;
Linear interpolation and tool move to position (0, 100).
N10 X0 Y20;
Linear interpolation and tool move to position (0, 20).
N11 X80 Y20;
Linear interpolation and tool move to position (80, 20).
N12 X80 Y0;
Linear interpolation and tool move to position (80, 0).
N13 G00 Z5;
Rapid travel to 5 mm above the workpiece surface.
N14 G28 U0 V0 W0;
Return to machine reference position.
N15 G40 M05;
Cutter radius compensation cancel and spindle stop.
N16 M09 M30;
Coolant off, program end and tape rewind.
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et
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Fundamental of CNC and Part Programming
Example 4.24.6 : Write a part program for the workpiece as shown in Fig. 4.24.6.
Assume cutter diameter 10 mm, depth of workpiece 10 mm, spindle speed 600 r.p.m.
and feed 15 mm/min.
P8(0, 60)
P7(35, 60)
35
P5(60, 50)
R10
R15
P6(45, 50)
20
ww
P9(0, 30)
30
w.E
20
15
P3(35, 30) P4(60, 30)
25
P1(0, 0)
asy
E
P2(20, 0)
P0(–10, –10)
(b) Tool path
(a) Workpiece
ngi
Fig. 4.24.6
Solution : For tool path refer Fig. 4.24.6 (b).
Program
nee
Description
rin
107;
Program number.
N01 G28 U0 V0 W0;
Return to machine reference position.
N02 G90 G71 G94;
Absolute programming mode, metric (mm) data
input and feed in mm/min.
N03 G17 M06 T01;
Selection of XY plane, tool change and select
tool no.01.
N04 G41 M03 S600;
Cutter radius compensation to left, spindle ON
with speed 600 r.p.m.
N05 G00 X–10, Y–10 Z5 M08;
Rapid travel to position (–10, –10) and cutter is
5 mm above the workpiece surface, coolant ON.
N06 G01 X0 Y0 Z–10 F15;
Linear interpolation and move tool 10 mm
downward along Z-axis with feed 15 mm/min to
position (0, 0).
N07 X20 Y0;
Linear interpolation and tool move to position
(20, 0).
N08 X35 Y30;
Linear interpolation and tool move to position
(35, 30).
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et
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Fundamental of CNC and Part Programming
N09 X60 Y30;
Linear interpolation and tool move to position
(60, 30).
N10 X60 Y50;
Linear interpolation and tool move to position
(60, 50).
N11 X45 Y50;
Linear interpolation and tool move to position
(45, 50).
N12 G03 X35 Y60 I–10 J0;
Anticlockwise circular interpolation and tool move
to position (35, 60).
N13 G01 X0 Y60;
Linear interpolation and tool move to position
(0, 60).
N14 G03 X0 Y30 I0 J–15;
Anticlockwise circular interpolation and tool move
to position (0, 30).
N15 G01 X0 Y0;
Linear interpolation and tool move to position
(0, 0).
ww
N16 G00 Z5;
w.E
N17 G28 U0 V0 W0;
N18 G40 M05;
Rapid travel 5 mm above the workpiece surface.
Return to machine reference position.
asy
E
Cutter radius compensation cancel and spindle
stop.
ngi
N19 M09 M30;
Coolant off, program end and tape rewind.
nee
Example 4.24.7 : Write a part program for the component as shown in Fig. 4.24.7 (a).
Take feed 30 mm/min; spindle speed 1000 r.p.m. and thickness of component as 10 mm.
Use end mill type cutter of diameter 10 mm.
P6(20, 90)
Y
Y
20
55
rin
P5(75, 90)
55
40
15
20
R15
15
g.n
et
P4(90, 75)
P7(20, 35)
P3(90, 35)
40
P1(35, 20)
X
P2(75, 20)
X
P0(0, 0)
(a) Workpiece
(b) Tool path
Fig. 4.24.7
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Solution : For tool path refer Fig. 4.24.7 (b).
Program
Description
108;
Program number.
N01 G28 U0 V0 W0;
Return to machine reference position.
N02 G90 G71 G94;
Absolute programming mode, metric (mm) data input
and feed in mm/min.
N03 G17 M06 T01;
Selection of XY plane, tool change and select tool
no.01.
N04 G41 M03 S1000;
Cutter radius compensation to left, spindle ON with
speed 1000 r.p.m.
N05 G00 X0, Y0 Z5 M08;
Rapid travel to position (0, 0) and cutter is 5 mm
above the workpiece surface, coolant ON.
N06 X35 Y20 Z–10 F30;
Rapid travel to position (35, 20) and cutter is 10 mm
downward along Z-axis with feed 30 mm/min.
ww
w.E
N07 G01 X75 Y20;
N08 G02 X90 Y35 I15 J0;
asy
E
Linear interpolation and tool move to position
(75, 20).
ngi
Clockwise circular interpolation and tool move to
position (90, 35).
nee
N09 G01 X90 Y75;
Linear interpolation and tool move to position
(90, 75).
N10 X75 Y90;
Linear interpolation and tool move to position
(75, 90).
N11 X20 Y90;
Linear interpolation and tool move to position
(20, 90).
N12 X20 Y35;
Linear interpolation and tool move to position
(20, 35).
N13 X35 Y20;
Linear interpolation and tool move to position
(35, 20).
N14 G00 X0 Y0 Z5;
Rapid travel to position (0, 0) and cutter is 5 mm
above the surface of workpiece.
N15 G28 U0 V0 W0;
Return to machine reference position.
N16 G40 M05;
Cutter radius compensation cancel and spindle stop.
N17 M09 M30;
Coolant off, program end and tape rewind.
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et
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Fundamental of CNC and Part Programming
Example 4.24.8 : Prepare a part program for the given job as shown in
Fig. 4.24.8 (a). By using following data :
Speed = 1000 r.p.m., Feed = 10 mm/min, Depth of cut = 3 mm
Tool position from the surface of the workpiece is 10 mm above. Thickness of
job = 3 mm.
P5(10, 65)
10
55
P6(0, 55)
10
55
ww
P4(65, 65)
55
w.E
(0, 0)
asy
E
R10
R10
P3(65, 10)
ngi
P1(0, 0)
P2(55, 0)
nee
P0(–10, –10)
(a) Workpiece
Fig. 4.24.8
Solution : For tool path refer Fig. 4.24.8 (b).
Program
(b) Tool path
rin
Description
g.n
et
109;
Program number.
N01 G28 U0 V0 W0;
Return to machine reference position.
N02 G90 G71 G94;
Absolute programming mode, metric (mm) data input and
feed in mm/min.
N03 G17 M06 T01;
Selection of XY plane, tool change and select tool no.01.
N04 G41 M03 S1000;
Cutter radius compensation to left, spindle ON with speed
1000 r.p.m.
N05 G00 X–10 Y–10 Z10
M08;
Rapid travel to position (–10, –10) and cutter is 10 mm
above the workpiece surface, coolant ON.
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N06 G01 X0 Y0 Z–13 F10;
Linear interpolation and move tool 13 mm downward along
Z-axis with feed 10 mm/min to position (0, 0).
N07 X55 Y0;
Linear interpolation and tool move to position (55, 0).
N08 G03 X65 Y10 I0 J10;
Anticlockwise circular interpolation and tool move to position
(65, 10).
N09 G01 X65 Y65;
Linear interpolation and tool move to position (65, 65).
N10 X10 Y65;
Linear interpolation and tool move to position (10, 65).
N11 X0 Y55;
Linear interpolation and tool move to position (0, 55).
N12 X0 Y0;
Linear interpolation and tool move to position (0, 0).
ww
w.E
N13 G00 Z10;
Rapid travel 10 mm above the surface of workpiece.
N14 G28 U0 V0 W0;
Return to machine reference position.
N15 G40 M05;
N16 M09 M30;
asy
E
Cutter radius compensation cancel and spindle stop.
Coolant off, program end and tape rewind.
ngi
nee
Example 4.24.9 : Write a manual program for drilling holes as shown in
Fig. 4.24.9. The spindle is initially at the lower left corner of the workpiece. Take
spindle speed 1200 rpm.
60
10 Drill (3 Holes)
rin
30
12
g.n
et
#1
65
55
#2
35
25
#3
75
Fig. 4.24.9
Solution : Given data :
S = 1200 rpm,
Refer Fig. 4.24.9 (a).
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Fundamental of CNC and Part Programming
#1
#2
#3
+Y
+X
(Home position)
Fig. 4.24.9 (a)
ww
Program
w.E
Description
N001 G90 G71;
N002 M06 T0101;
N003 S1200
M03;
(Absolute programming mode and data input is in metric
mode)
asy
E
(Tool change, Tool no.1 with offset value 01)
(Spindle speed 1200 rpm and in clockwise direction)
ngi
N004 G00 X12 Y55 M08;
(Rapid traverse to point 1 and coolant ON)
N005 M00;
(Auto-stop for drilling)
N006 X30 Y35;
(Rapid traverse to point 2)
N007 M00;
(Auto-stop for drilling)
N008 X60 Y25;
(Rapid traverse to point 3)
N009 M00;
(Auto-stop for drilling)
N0010 X – 40 Y – 30;
(Rapid traverse to any arbitrary point)
N0011 M09 M05 M30;
(Coolant OFF, spindle stop and program stop).
nee
rin
g.n
et
Example 4.24.10 : Write a manual part program for drilling two holes of 8 mm
diameter and two holes of 6 mm diameter. Take spindle speed as 1200 rpm for 8 mm
diameter hole and 1600 rpm for 6 mm diameter hole. Refer Fig. 4.24.10.
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Fundamental of CNC and Part Programming
40
20
8, Drill (2 Holes)
#1
6, Drill (2 Holes)
20
30
#4
40
#3
#2
10
60
75
R2.5
90
100
ww
w.E
Solution : Given data :
Fig. 4.24.10
asy
E
S = 1200 rpm (for 8 mm diameter)
S = 1600 rpm (for 6 mm diameter). Refer Fig. 4.24.10 (a).
ngi
#1
nee
#3
#2
rin
#4
+Y
+X
(Home position)
g.n
et
Fig. 4.24.10 (a)
Program
Description
N001 G90 G71;
(Absolute programming mode and data input is in metric
mode)
N002 M06 T0101;
(Tool change, tool no.1 with offset value 01)
N003 S1200 M03;
(Spindle speed 1200 rpm and in clockwise direction)
N004 G00 X20 Y60 M08;
(Rapid traverse to point 1 and coolant ON)
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N005 M00;
(Auto-stop for drilling)
N006 Y30;
(Rapid traverse to point 2)
N007 M00;
(Auto-stop for drilling)
N008 X – 10 Y0;
(Rapid traverse to tool change position)
N009 M00;
(Change of drill tool from 8 mm diameter to
6 mm diameter)
N0010 G00 G90 G71;
(Rapid traverse, absolute and metric mode)
N0011 S1600 M03;
(Spindle speed 1600 rpm and in clockwise direction)
ww
N0012 X90 Y30;
(Rapid traverse to point 3)
N0013 M00;
(Auto-stop for drilling)
N0014 Y10;
N0015 M00;
w.E
N0016 X – 10 Y0;
N0017 M09 M05 M30;
(Rapid traverse to point 4)
(Auto-stop for drilling)
asy
E
(Rapid traverse to tool change position)
(Coolant OFF, spindle stop and program stop).
ngi
nee
Example 4.24.11 : Write a manual part program for drilling and milling the
machine component as shown in Fig. 4.24.11. Assume 10 mm diameter milling
cutter. Take spindle speed 2000 rpm (for milling) and 3000 rpm (for drilling). Feed
1400 mm/min.
rin
15
R5 Typ.
g.n
et
15
30
60
8 Typ.
30
50
15
60
80
90
Fig. 4.24.11
Solution : Given data :
D = 10 mm, S = 2000 rpm (for milling), S = 3000 rpm (for drilling),
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F = 1400 mm/min. Refer Fig. 4.24.11 (a).
#6
#5
#10
#8
#7
#9
#3
#4
Home
Position
#1
ww
#2
Fig. 4.24.11 (a)
w.E
It is assumed that the machine (0, 0) is at the left lower most corner of the workpiece.
Program
N001 G90 G71 G94;
asy
E
Description
(Absolute programming mode, data input is in mertic
mode and feed in mm/min)
ngi
N002 G00 X – 5, Y – 5 S2000
M03;
(Rapid traverse to point 1 and spindle speed is
2000 rpm in clockwise direction)
N003 M00;
(Allows the operator to clamp the spindle)
N004 G01 X65 F1400 M08;
(Machining upto point 2 with a feed of 1400 mm/min
and coolant ON)
N005 Y25;
(Machining upto point 3)
N006 X95;
(Machining upto point 4)
N007 Y65;
(Machining upto point 5)
N008 X25;
(Machining upto point 6)
N009 Y35;
(Machining upto point 7)
N0010 X – 5;
(Machining upto point 8)
N0011 Y – 5;
(Machining upto point 1)
N0012 M00;
(Auto-stop to raise the spindle)
N0013 G00 X – 10 Y0;
(Rapid traverse to tool change position)
N0014 M00;
(Change to drill tool of 8 mm diameter)
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N0015 G00 G90 G71;
(Rapid traverse, absoulte and metric mode)
N0016 X15 Y15 S3000 M03;
(Spindle speed 3000 rpm in clockwise direction and
locate point 9)
N0017 M00;
(Auto stop for drilling)
N0018 X80 Y45;
(Locate point 10)
N0019 M00;
(Auto-stop for drilling)
N0020 X – 10 Y0;
(Rapid traverse to tool change position)
N0021 M09 M05 M30;
(Coolant OFF, spindle stop and program stop)
ww
Example 4.24.12 : Write the part program for the work piece shown in Fig. 4.24.12.
Material : Aluminium, Work piece size : 100 mm ´ 80 mm ´ 15 mm.
w.E
y
20
asy
E
20
R20
ngi
60
nee
R20
5
5
80
Fig. 4.24.12
rin
x
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et
Solution : Given data :
Workpiece size = 100 mm ´ 80 mm ´ 15 mm.
Workpiece material = Aluminium
Assuming the machining zero at the origin and milling cutter is selected of f 20 mm.
Refer Fig. 4.24.12 (a).
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Y
2
7
3
8
9
6
10
ww
5
w.E
Home position
11
4
X
1
asy
E
Fig. 4.24.12 (a)
As, cutter diameter D = 20 mm
ngi
nee
Generally for aluminium, cutting speed V = 60 m/min
p DN
m/min
But,
V =
1000
1000 V 1000 ´ 60
= 954.929 » 1000 rpm
\ Spindle speed, N =
=
pD
p ´ 20
Assuming
rin
Feed F = 1400 mm/min
Program
Description
g.n
et
N001 G90 G71 G94;
(Absolute programming mode,data input is in metric mode
and feed in mm/min)
N002 M06 T0101;
(Tool change, Tool no. 1 with offset value 01)
N003 S1000
(Spindle speed 1000 rpm and in clockwise direction)
M03;
N004 G00 X10 Y–10 Z–15
M08;
(Rapid traverse to point 1 and coolant ON)
N005 G41 D20;
(Tool compensation on RIGHT and cutter diameter is
20 mm)
N006 G01 Y70;
(Linear interpolation upto point 2)
N007 X90;
(Linear interpolation upto point 3)
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N008 Y10;
(Linear interpolation upto point 4)
N009 X10;
(Linear interpolation upto point 5)
N0010 Y50;
(Linear interpolation upto point 6)
N0011 X30 Y70;
(Linear interpolation upto point 7)
N0012 X70;
(Linear interpolation upto point 8)
N0013 G02 X90 Y50 R20;
(Clockwise interpolation upto point 9 and radius is 20 mm)
N0014 G01 Y30;
(Linear interpolation upto point 10)
N0015 G03 X70 Y10 R20;
(Anticlockwise interpolation upto point 11 and radius is
20 mm)
ww
N0016 G00 Z15;
(Rapidly move spindle upwards)
N0017 G28 U0 W0 Z0
(Return to home position)
N0018 M03 M09 M30;
(Coolant OFF, spindle stop and program stop)
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asy
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Example 4.24.13 : Write a program (manual part program) to drill five holes in the
locations shown in Fig. 4.24.13 and pause at each location where a hole should be
drilled.
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3
19
2
100
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4
32
5
1
25
25
Start
point
25
6
25
rin
g.n
et
44
100
Drill 5 holes-12.5 DIA
Fig. 4.24.13
Solution : Given data :
S = 300 rpm, F = 50 mm/min, Thickness of plate = 10 mm
Refer Fig. 4.24.13 (a).
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Y
3
19
4
2
100
32
5
1
ww
25
X
w.E
25
25
Home position
6
44
100
asy
E
Start
point
25
Fig. 4.24.13 (a)
ngi
Program
nee
Description
rin
N001 G90 G71 G94;
(Absolute programming mode and data input is in metric
mode and feed in mm/min)
N002 M06 T0101;
(Tool change, tool no. 01 with offset value 01)
N003 M03 S300;
(Spindle speed is 300 rpm and in clockwise direction)
N004 G00 X25 Y25 Z10;
(Rapid traverse to start point and tool is above the plate)
N005 M08 X25 Y25;
(Coolant ON and Rapid traverse to point 1)
N006 G04 X2;
(Stoppage of axis motion i.e. pause for 2 sec.)
N007 G01 Z–10 F50;
(Linear interpolation, drill the hole with a feed of 50 mm/min.)
N008 G01 Z10;
(Move spindle upwards)
N009 G00 X31 Y57 Z10;
(Rapid traverse of point 2 and tool is above the plate)
N0010 G04 X2;
(Stoppage of axis motion i.e. pause for 2 sec.)
N0011 G01 Z – 10 F50;
(Linear interpolation, drill the hole with a feed of 50 mm/min)
N0012 G01 Z10;
(Move spindle upwards)
N0013 G00 X31 Y76 Z10;
(Rapid traverse to point 3 and tool is above the plate)
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N0014 G04 X2;
(Pause for 2 sec)
N0015 G01 Z – 10 F50;
(Linear interpolation, drill the hole with a feed of 50 mm/min)
N0016 G01 Z10;
(Move spindle upwards)
N0017 G00 X75 Y76 Z10;
(Rapid traverse to point 4 and tool is above the plate)
N0018 G04 X2;
(Pause for 2 sec)
N0019 G01 Z – 10 F50;
(Linear interpolation, drill the hole with a feed of 50 mm/min)
N0020 G01 Z10;
(Move spindle upwards)
N0021 G00 X75 Y25 Z10;
(Rapid traverse to point 5 and tool is above the plate)
N0022 G04 X2;
(Pause for 2 sec)
N0023 G01 Z–10 F50;
(Linear interpolation, drill the hole with a feed of 50 mm/min)
ww
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N0024 G01 Z10;
(Move spindle upwards)
N0025 G28 U0 W0;
(Return to home position)
N0026 M03 M09 M30;
asy
E
(Coolant OFF, spindle stop and program stop)
Example 4.24.14 : Write a part program for drilling holes in the part shown in
Fig. 4.24.14. The plate thickness is 20 mm.
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2
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3
rin
90
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et
1
60
4
25
A
15
25
+Y
50
+z
80
95
+X
20
–z
Fig. 4.24.14
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Solution : Given data :
Assume, S = 300 rpm
and
F = 50 mm/min, Thickness of plate = 20 mm,
Also, assume size of hole = 10 mm
Refer Fig. 4.24.14 (a).
+Y
3
ww
2
w.E
90
60
1
asy
E
4
25
A
25
50
Home
position
15
ngi
80
95
Fig. 4.24.14 (a)
Program
+X
nee
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Description
g.n
et
N001
G90 G71 G94;
(Absolute programming mode and data input is in metric
mode and feed in mm / min)
N002
M06 T0101;
(Tool change, tool no. 1 with offset value 01)
N003
M03 S300;
(Spindle speed 300 rpm and in clockwise direction)
N004
G00 X25 Y25 Z2 M08;
(Rapid traverse to point 1 and coolant ON and tool is
2 mm above the plate)
N005
G01 Z – 20;
(Tool moves 20 mm inside the workpiece)
N006
G00 Z2;
(Rapid traverse of tool 2 mm above the plate)
N007
G00 X50 Y60;
(Rapid traverse to point 2)
N008
G01 Z – 20;
(Tool moves 20 mm inside the workpiece)
N009
G00 Z2;
(Rapid traverse of tool 2 mm above the plate)
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N010
G00 X80 Y90;
(Rapid traverse of tool to point 3)
N011
G01 Z – 20;
(Tool moves 20 mm inside the workpiece)
N012
G00 Z2;
(Rapid traverse of tool 2 mm above the plate)
N013
G00 X95 Y15;
(Rapid traverse of tool to point 4)
N014
G01 Z – 20;
(Tool moves 20 mm inside the workpiece)
N015
G00 Z2;
(Rapid traverse of tool 2 mm above the plate)
N016
G00 X – 10 Y – 10;
(Rapid traverse of tool to home position)
N017
M09 M05 M30;
(Coolant OFF, spindle stop and program stop)
ww
4.25 Subroutine :
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·
Subroutine is also called as subprograms.
·
When a similar machining operation is to be performed repeatedly then the
general programming method will not be used because,
¡ It is very lengthy.
·
asy
E
¡
It is tedious.
¡
It is more time consuming.
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¡ It consumes more space in the computer memory.
In such case, subroutine method is used. It is a time saving technique.
rin
·
It is an independent program similar to general program and stored in the
computer memory under separate program number.
·
It can be called anywhere in the main program and for any number of times.
·
To call the subroutine in the main program, the miscellaneous code M98 is used.
The instruction block for subroutine can be written as follows :
N10 M98 P50 L1;
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et
where,
M98 indicates a call to subroutine,
P50 indicates the program number (here 50),
·
L1 indicates to call subroutine only one time.
After execution of subroutine return back to the main program and continue it.
·
To end the subroutine and return back to the main program M99 code is used.
·
It is important to note that, the main program is written in absolute programming
mode (G90) and subroutine is written in incremental programming mode (G91).
·
Hence, use code G91 at the start of subroutine and G90 before use of M99.
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Example 4.25.1 : Write a part program using subroutine for milling a square pocket
of 30 ´ 30 mm and 5 mm depth. Refer Fig.4.25.1. Take diameter of cutter as 5 mm,
speed 500 r.p.m. and feed 100 mm/min.
Solution : Fig. 4.25.1 (b) shows the enlarged view of pocket and corresponding tool path.
From Fig. 4.25.1 (a) coordinates of point P1 = (15, 30) and of point P5 = (60, 30). In this
case we will first write a subroutine program which is then called in the main program.
15
30
ww
PA(0, 20)
PB(20, 0)
5
1
10
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B
D
C
30
2
asy
E
15
P1
A
P5
10
ngi
(a) Workpiece
Fig. 4.25.1
PD(–20, 0)
PC(0, –20)
Incremental
mode
25
(b) Enlarged view
of pocket
nee
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Subroutine program
Program
g.n
et
Description
50;
Subroutine program number.
N01 G91;
Set to incremental mode.
N02 G01 Z–10 F100;
Linear interpolation and cutter moves 5 mm
downward along Z-axis with feed 100 mm/min.
(Total depth = 5 + 5 = 10)
N03 Y20;
Cutter moves 20 mm in Y direction i.e.
position A.
N04 X20;
Cutter moves 20 mm in X direction i.e.
position B.
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N05 Y–20;
Cutter moves 20 mm in Y direction i.e.
position C.
N06 X–20;
Cutter moves 20 mm in X direction i.e.
position D.
N07 G00 Z10;
Rapidly move the cutter 10 mm in
Z-direction.
N08 G90 M99;
Set absolute programming mode and end of
subroutine program.
Main program
ww
51;
Program
w.E
Description
Main program number.
N10 G28 U0 V0 W0;
N15 G90 G71 G94;
N20 G17 M06 T01;
Return to machine reference position.
asy
E
Absolute programming mode, metric (mm)
data input and feed in mm/min.
Selection of XY plane, tool change and tool
no. 01 is selected.
ngi
N25 G41 M03 S500;
nee
Cutter radius compensation to left, spindle
ON with speed 500 r.p.m.
rin
N30 G00 X15 Y30 Z5 M08;
Rapid travel to position P1 (15, 30) and
cutter is 5 mm above the surface of
workpiece, coolant ON.
N35 M98 P50 L1;
Call for subroutine, program number 50 and
for only one time.
N40 G00 X60 Y30;
Rapid travel to position P5 (60, 30).
N45 M98 P50 L1;
Call for subroutine, program number 50 and
only once.
N50 G28 U0 V0 W0;
Return to machine reference position.
N55 G40 M05;
Cutter radius compensation cancel and
spindle stop.
N60 M09 M30;
Coolant off, program end and tape rewind.
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Fundamental of CNC and Part Programming
4.26 Canned Cycle :
·
Canned cycle is also called as multiple-repetitive cycle.
·
It is a set of instructions stored in the computer memory which is used to
perform a fixed sequence of operation.
·
It is commonly used for repetitive operation where material is to be cut in
number of passes.
·
The main advantage of canned cycle is that, it reduces length of the program
hence memory space and the complexity of the program.
·
In this cycle, the final position is mentioned in the instruction block and the
cutter path is automatically plotted by the controller itself.
ww
·
Canned cycle can be called and cancelled by using G-codes (preparatory codes).
·
According to the shape of workpiece various G-codes are used for canned cycle.
For example : G74 for slot or rectangular pocket milling and G77 for circular
pocket milling.
·
Canned cycle can be cancelled by using G80 code.
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4.26.1 Comparison between Subroutine and Canned Cycle
Give the comparison between subroutine and canned cycle.
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Sr. No.
Subroutine
Canned cycle
1.
It is a separately written program which
is called in the main program.
It is not a separately written program. It
is the part of the program.
2.
It is used when multiple passes are
required at different places.
It is used when multiple passes are
required at the same place.
3.
It is called and executed by using
miscellaneous function (M-codes).
It is written by using preparatory
function (G-codes).
4.
It is a separate program hence separate
program number is given.
It is not a separate program hence no
need to give any program number.
5.
It is given in every block of instruction
till the operation is completed.
In this cycle, directly final point is given
in the instruction block.
6.
The path of cutter for every point is
mentioned by the programmer.
The path of cutter for every pass is
automatically generated.
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4.26.2 Slot Milling (G74)
· Slot milling is commonly used for producing keyway in shafts as shown in
Fig. 4.26.1.
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·
For this purpose G74 code is used which is written
in the following format :
N05 X I Y I Z I ;
N06 G74 X F Y F Z F K S F;
where,
End mill
cutter
X I Y I Z I be the initial position of the tool
centre point of the slot.
X F Y F Z F be the final position of the tool
centre point of the slot.
K be the peck depth.
S be the peck feed.
aft
Sh
Fig. 4.26.1 : Keyway milling
ww
·
Keyway
(closed)
F be the feed.
For example : Write a program to produce a 10 mm deep slot by using a cutter
of 10 mm diameter. Take peck feed 50 mm/min, feed 100 mm/min and peck
depth 2 mm. Refer Fig. 4.26.2 (a).
w.E
P1
Start
asy
E
Peck depth K = 2 mm
P2
ngi
10
40
50
nee
End
rin
(b) Tool path
(a) Workpiece
Fig. 4.26.2
g.n
et
Solution : Fig. 4.26.2 (b) shows the tool path or peck depth. The starting blocks and end
blocks are similar to all programs hence only the block of slot milling is explained as
follows :
Program
Description
N10 G00 X10 Y20 Z5;
Rapid travel to initial position i.e. P1 (10, 20)
and cutter is 5 mm above the workpiece
surface.
N15 G74 X60 Y20 Z–10 K2 S50 F100;
Slot milling cycle and cutter moves to
P2 (60, 20) and 10 mm downward along
Z-axis with peck feed 50 mm/min and feed
100 mm/min.
N20 G80;
Canned cycle cancel.
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4.26.3 Rectangular Pocket Milling (G75)
· It is used for producing rectangular or square pockets in the component.
·
Initially the tool will move by a distance equal to crossover towards the outer
surface and then it will move in clockwise direction.
·
At the end of one cycle the tool will again move outward by the distance equal
to crossover and take a cut.
·
For this cycle instruction block is written in the following format :
N10 G75 X Y Z I K S F ;
where,
XYZ be the width, thickness and depth of pocket respectively.
ww
2
times the cutter diameter).
3
For example : Write a part program to mill 30 ´ 20 mm pocket for 10 mm depth.
Take cutter diameter 10 mm, peck feed 30 mm/min and feed 100 mm/min. Refer
Fig. 4.26.3 (a).
I be the crossover (generally
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·
R5
40
asy
E
Crossover
ngi
10
20
60
(a) Workpiece
nee
Start
End
rin
g.n
et
(b) Tool path for pocket
Fig. 4.26.3
Solution : Fig. 4.26.3 (b) shows the tool path for pocket. The starting blocks and end
blocks are similar to all programs, hence only the block of pocket milling is explained as
follows :
Program
Description
N10 G00 X30 Y20 Z5;
Rapid travel to initial position i.e. centre of
pocket and cutter is 5 mm above the
workpiece surface.
N15 G75 X20 Y10 Z–10 I6 K2 S30 F100;
Rectangular pocket milling cycle with all
values.
N20 G80;
Canned cycle cancel.
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Fundamental of CNC and Part Programming
4.27 Automatically Programmed Tools (APT) :
·
APT is a language of computer assisted part programming for CNC machine
tools. In this case, the programmer gives instruction to the computer in the form
of programming language.
·
These instructions are stored in source file. Language processor converts these
statements into a center line data file. This file has cutter centerline locations
along with spindle, cutter diameter comensation ON and OFF.
·
Postprocessor is used over here to convert CL (cutter location) file into tape
commands needed by NC control.
·
APT is widely used language as it can generate complex geometries in all axis
simultaneously.
ww
Elements involved in computer assisted part programming are shown in Fig. 4.27.1.
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Drawing
Source file
Translation
asy
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CL Data
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NCtapecomnds
Machine tool
APT
·
MCU
Post processor
nee
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NC Tape commands
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Fig. 4.27.1 : Elements of computer assisted part programming
4.27.1 Structure of APT
·
APT consists of different types of statements made of letters, numbers and
punctuation marks.
·
Punctuation marks used in APT are given in the following table :
Symbol
Name
/
Slash
,
Comma
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Description
It divides a statement. Major words are at the left
of slash while minor words, symbols modifying the
words are on the right side. example : Go/To, L6
It is used as a separator between various
elements. Normally, it is on right side of slash.
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·
·
=
Equal to
It is used to assign an entity to a symbolic name.
Example : SP = POINT / 80, 30, 50
( )
Parenthesis
These are used to enclose the nested definitions.
$
Dollar
It is placed at the end of line. This indicates that
the statement continues in next line.
$ $
Double Dollar
Any statement after this sign is comment. It is not
a part of program
Words in the statement contain maximum six letters. It can contain alphabets or
numbers but no special characters.
Keywords are the commands in APT to perform a certain function only.
Keywords are of following two types :
ww
i) Major word : This defines type of statement.
w.E
ii) Minor word : This defines various parameters.
·
·
·
Symbols are the words used for geometrical definitions and numerical values.
They must be defined before using in the program.
Lables are the words used to refer a statement so that control can be passed to it.
They can be defined with all characters.
Mathematical operations are represented in APT as follows :
Symbol
asy
E
Operation
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Symbol
–
nee
Operation
+
Addition
*
Multiplication
* *
Exponential
ABS
SQRT
Square root
SIN
Sin of angle
COS
Cosine of angle
TAN
Tangent of angle
EXP
Value of e to the power
LOG
Natural log
/
Subtraction
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Division
Absolute value
Table 4.27.1 : Mathematical Operations in APT
·
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et
An APT program consists of four major statements :
i)
Geometry statement
ii) Motion statement
iii) Post processor statement
iv) Auxiliary statement
i) Geometry statement
·
Geometry of a given component can be defined by the number of elements.
A geometry statement in APT is written as follows :
<Geometry name> = Major word/ <definition>
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·
Geometry name is a major word such as POINT, LINE, CIRCLE, PLANE,
etc.
· <definition> is the statement defining geometry of the major word.
· Following are some examples of geometry statements.
a) Point : In 3D system point is defined by X, Y and Z co-ordinates. It is given
as,
P1 = POINT/5, 70, 0
b) Line : Lines are considered to have infinite length and do not have any
specific direction. It is represented as follows,
<symbol> = LINE / point 1, point 2
L1 = LINE / 50, 60, 90, 94
C) Circle : Circle is considered as an infinitely long cylinder kept perpendicular
to XY plane. It's radius value can not be negative. It is represented as,
<symbol> = CIRCLE / X1, Y1 radius
C 1 = CIRCLE / 61, 62, 30
ii) Motion statement :
· This statement represents motion control of the machine. This defines tool
path as per geometry. This is written as follows :
· Major word/ Minor word, scalar
· In 3D system, path of cutting tool is defined by three intersecting surfaces,
they are Drive Surface (DS), Part Surface (PS) and Check Surface (CS).
· Tool moves along intersection of part and drive surface. Tool is stopped by
check surface.
· They are divided in the following three subgroups :
ww
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asy
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a) Setup commands
b) Point to point motion command
c) Continuous path motion command
rin
g.n
et
a) Setup command : In this case, end point of a motion is starting point of
subsequent motion. FROM command is used to show starting point of cutter
for first motion.
FROM/ X1, Y1
b) Point to point motion command : In this type of command motion between
two points are given in operations like drilling, milling etc. The following
statements are used :
· GODLA : This specifies relative movement along the specified axis.
GODLA/dx, dy, dz.
· GOTO : This is an absolute movement statement. It moves spindle to a
specific point from a current position.
GOTO/ x, y
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c) Continuous path motion commands : In milling or turning, these are used to
specify the continuous path to generate different surfaces.
iii) Post processor statement :
· In this command, machine tool functions and their settings are specified.
This statement converts CL data to machine tool co-ordinate. This can be
stated as follows for cooling fluid. This command tells us about when to
start or stop cutting fluid.
ì ON ü
ïï OFF ïï
COOLNT / í
ý
ï MIST ï
ïî FLOOD ïþ
ww
iv) Auxiliary statement
· This statement control output of the program. This also makes computer to
accept a part program and make it readable. For example,
PARTNO / <literal string> : PARTNO (This statment is used to identify part
program).
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4.28 Micromachining :
·
·
·
·
·
·
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Micromachining is a advanced machining process which is used to fabricate the
components having dimensions in micrometers.
It usually involves chemical etching process on a very fine scale.
The following are the examples of components manufactured by micromachining :
m Probes and sensors
m
Measuring devices
m Microactuators
m
Microgimbals
Micromachining has important role in fabrication of microelectronic devices such
as Printed Circuit Boards (PCB).
PCB technology is now-a-days foundation for information systems,
telecommunications, automotive controls, robotics, aerospace, military weapons,
etc.
The most commonly semiconductor material used is silicon for micromachined
parts.
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4.28.1 Wafer Machining
· Wafer machining is the primary process used in manufacturing of
microelectronics devices.
· After purifying the silicon used for fabrication, a single crystal slilicon is
obtained through the process known as Czochralski process.
· This process utilizes a seed crystal that is dipped into a silicon melt and then
slowly pulling out while being rotated.
· This results in silicon crystal of 150 mm - 300 mm in diameter and over 1 m in
length.
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·
This crystal is then sliced into individual wafers by using a inner diameter blade
known as wafer machining. (Refer Fig. 4.28.1)
Step 1 :
Silicon crystal boule
Step 4 :
Grind outer diameter
between live centers
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·
·
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Step 2 :
Slice off ends
Step 3 :
Drill center bore
Step 5 :
Grind orientation flat
Step 6 :
Slice with diamond-coated
wire saw
Fig. 4.28.1 : Wafer Machining
In this method, rotating blade with inner diameter is used for cutting and then the
wafers are cut to a thickness of required microns (upto 05
. ´ 10 3 mm).
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This thickness provides necessary physical and mechanical support for
temperature absorption and fabrication.
· Finally, these wafers are cleaned and polished to get surface without damage.
The fabrication of whole microelectronic device takes place over this wafer surface.
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4.29 Part Programming using APT :
Example 4.29.1 : Write an APT
program for drilling holes on a
component as shown in Fig. 4.29.1 (a)
The component is 10 mm thick.
The post processor statement is
MACHIN/MM. Assume spindle speed
as 950 rpm and feed as 0.5 mm/rev.
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(25, 30)
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(50, 40)
(45, 20)
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SP (70, 35)
6,4 Holes
(15, 10)
through
Z
10
X
Fig. 4.29.1 (a)
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Solution : Tool path for the given part is shown in Fig. 4.29.1 (b). By using point to
point programming method, the program is written as follows :
(50, 40)
SP (70, 35)
(25, 30)
P4
P3
(45, 20)
(15, 10)
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P2
P1
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Fig. 4.29.1 (b)
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PARTNO / EXAMPLE 1
MACHIN/MM, 1
PRINT/ON
CLPRINT/ON
$ $ PRINT GEOMETRY
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$$ PRINT CLDATA
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SP = POINT/70, 35, 40
$$ (Starting point co-ordinates 70, 35, 40)
P1 = POINT/15, 10, 5
$$ (Co-ordinates of point 1 (P1) 15, 10, 5)
P2 = POINT/45, 20, 5
$$ (Co-ordinates of point 2 (P2) 45, 20, 5)
P3 = POINT/25, 30, 5
$$ (Co-ordinates of point 3 (P3) 25, 30, 5)
P4 = POINT/50, 40, 5
$$ (Co-ordinates of point 4 (P4) 50, 40, 5)
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$$ DRILL DIA 6 MM
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SPINDL/950, CLW
$$ (Spindle speed is 950 rpm clockwise)
SPINDL/ON
$$ (Spindle ON)
COOLNT/MIST
$$ (Coolant ON and its mist type)
FROM/SP
RAPID, GOTO/P1
$$ (From starting point go to point P1)
$$ DRILLING WITH FEED RATE 0.5 MM/REV
CYCLE/DRILL, – 15, MMPR, 0.5, 2
$$ (Drill 15 mm below origin with feed
of 0.5 mm/rev and 2 mm offset)
GOTO/P2
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CYCLE/DRILL, – 15, MMPR, 0.5, 2
$$ (Drill 15 mm below origin with feed of
0.5 mm/rev and 2 mm offset)
GOTO/P3
CYCLE/DRILL, – 15, MMPR, 0.5,2
$$ (Drill 15 mm below origin with feed of
0.5 mm/rev and 2 mm offset)
GOTO/P4
CYCLE/DRILL, – 15, MMPR, 0.5, 2
$$ (Drill 15 mm below origin with feed of
0.5 mm/rev and 2 mm offset)
CYCLE / OFF
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COOLNT / OFF
$$ (cycle and coolant OFF)
RAPID, GODLTA/15
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RAPID, GOTO / SP
REWIND
FINI
$$ (Go to starting point)
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$$ (Program finish)
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Example 4.29.2 : Write an APT program for a component shown in Fig.4.29.2 (a). It
is 20 mm thick. Post processor statement is MACHIN/MMPOST, 3. The mill diameter is
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60
10 mm.
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R5
50
70
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50
Fig. 4.29.3 (a)
Solution : The tool path for the above geometry is shown in Fig. 4.29.2(b)
PARTNO/EXAMPLE 2
MACHIN / MMPOST, 3
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3
4
60
L7
L6
1
L3
L2
5
6
95
2
L5
L4
Y
L1
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SP
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PRINT / ON
CLPRINT / ON
X
7
8
50
70
Fig. 4.29.2 (b)
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SP = POINT / 0, 0, 40
L1 = LINE/XAXIS
50
L2 = LINE / Y AXIS
$$ (Starting point co–ordinates 0, 0, 40)
$$ (Line L1 is along X - axis)
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$$ (Line L2 is along Y - axis)
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L3 = LINE/PARLEL, L2, X LARGE, 50
$$ (Line L3 parallel to L2 at 50 mm)
L4 = LINE/PARLEL, L3, X LARGE 70
$$ (Line L4 parallel to L3 at 70 mm)
L5 = LINE/PARLEL, L4, X LARGE, 50
$$ (Line L5 parallel to L4 at 50 mm)
L6 = LINE/PARLEL, L1, Y LARGE, 95
$$ (Line L6 parallel to L1 at 95 mm)
L7 = LINE/PARLEL, L6, Y LARGE, 60
$$ (Line L7 parallel to L6 at 60 mm)
$$ DEFINE PLANE 2 MM ABOVE THE PART
PL1 = PLANE/0, 0, 1, 2
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$$ (Select plane)
$$ = SETUP STATEMENTS
CUTTER/10
$$ (Cutter diameter is 10 mm)
SPINDL/800, CLW
$$ (Spindle speed 800 rpm, clockwise)
SPINDL/ON
FEDRAT/MMPM, 240
COOLNT/ON
$$ MOTION STATEMENTS
$$ (Spindle ON)
$$ (Feed rate is 240 mm/min)
$$ (Coolant ON)
RAPID, GOTO, SP
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$$ (Goto starting point rapidly)
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GO/TO, L6, TO, PL1, TO, L3
$$ (Go from L6 to PL1 and then from PL1 to L3)
GODLTA/ – 18
GOLFT/L6, PAST, L2
$$ (Tool moves from 1 to 2)
GORGT/L2, PAST, L7
$$ (Tool moves from 2 to 3)
GORGT/L7, PAST, L5
$$ (Tool moves from 3 to 4)
GORGT/L5, PAST, L6
$$ (Tool moves from 4 to 5)
GORGT/L6, TO, L4
$$ (Tool moves from 5 to 6)
GOLFT/L4, PAST, L1
$$ (Tool moves from 6 to 7)
GORGT/L1, PAST, L3
$$ (Tool moves from 7 to 8)
GORGT/L3, TO, L6
RAPID, GODLTA/18
$$ (Tool moves from 8 to 1)
RAPID, GOTO/SP
$$ (Go to starting point rapidly)
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COOLNT/OFF
SPINDL/OFF
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$$ (Coolant OFF)
$$ (Spindle OFF)
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REWIND
FINI
$$ (Program finish)
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Example 4.29.3 : Write an APT program for a part as shown in Fig. 4.29.3(a).
Thickness of plate is 10 mm. End mill diameter is 10 mm. Assume feed rate as 0.3
mm/rev and spindle speed as 1000 rpm.
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50
100
R20
R20
150
Fig. 4.29.3 (a)
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Solution : The tool path for given part is shown in Fig. 4.29.3 (b)
5
C2
R20
6
100
L3
4
L4
P0
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Y
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C1
X
R20
L1
1
Z
3
2
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150
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Fig. 4.29.3 (b)
PART NO/ EXAMPLE 3
MACHINE/MMPOST, 3
X
10
SP
50
L2
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PRINT/ON
CLPRINT/ON
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$$ PRINT GEOMETRY
SP = POINT / –10, –10, 50
$$ (Starting point co-ordinates –10, – 10, 50)
L1 = LINE/XAXIS
$$ (Line L1 along X-axis)
C1 = CIRCLE/150, 0, 20
$$ (Co-ordinates of center of circle C1)
L4 = LINE/YAXIS
$$ (Line L2 along Y-axis)
L2 = LINE/PARLEL, L4, XLARGE, 100 $$ (Line L2 parallel to L4 at 100 mm)
C2 = CIRCLE/20, 100, 20
$$ (Co-ordinates of center of circle C2)
P0 = POINT/150, 50
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$$ (Co-ordinates of point P0)
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L3 = LINE/P0, RIGHT, TANTO, C2
$$ (Line L3 tangent to C2)
$$ DEFINE PLANE 15 MM Below THE PART SURFACE
PL1 = PLANE/0, 0, 1, – 15
$$ (Select plane)
$$ SETUP STATEMENTS
CUTTER/10
$$ (Diameter of cutter is 10 mm)
SPINDL/1000, CLW
$$ (Spindle speed 1000 rpm, clockwise)
FEDRAT/MMPR, 0, 3
$$ (Feedrate 0.3 mm/rev)
SPINDL/ON
$$ (Spindle ON)
COOLNT/ON
$$ (Coolant ON)
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$$ MOTION STATEMENTS
RAPID, GOTO/SP
$$ (Rapid Go to starting point)
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GO/TO, L1, TO, PL1, TO, L4
$$ (Go from L1 to PL1 and then from PL1 to L4)
$$ CUTTER POSITION 2
GOFWD/L1, PAST, 1, INTOF, C1
$$ (CUTTER POSITION 3)
GOLFT/C1, PAST, 1, INT OF, L2
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$$ (Cutter moves from 1 to 2)
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$$ (Cutter moves from 2 to 3)
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GOLFT/L2, PAST, L3
$$ (Cutter moves from 3 to 4)
GOLFT/L3, TANTO, C2
$$ (Cutter moves from 4 to 5)
GOFWD/C2, TANTO, L4
$$ (Cutter moves from 5 to 6)
GOFWD/L4, PAST, L1
$$ (Cutter moves from 6 to 1)
RAPID, GODLTA/50
COOLNT/OFF
$$ (Coolant off)
SPINDL/OFF
$$ (Spindle off)
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RAPID, GOTO/SP
FINI
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$$ (Program finish)
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4.30 Introduction of CAM Package :
Computer Aided Manufacturing (CAM) is an application technology that uses
computer software and machinery to facilitate and automate the processes. It often use the
computer aided design, real time controls and robotics.
Thus it helps in reducing waste and energy for enchanced manufacturing and
production efficiency via increased speed, raw material and precise tooling accuracy.
The CAM systems can be linked with the CAD softwares such as Auto CAD, Solid
works, Catia and many more to produce the given product by using different CAM
software such as
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i) Master cam
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ii) Virtual Gibbs
iii) Surf cam
iv) Edge cam
v) Smart cam
vi) Alpha cam
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By using any of the CAM software it helps us to develop the CNC program for the
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given product. The most commonly used software in industries as well as in education
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institutes is master cam. It provides an easy assess with the popular CAD system which
helps take care of the investments in the industries.
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The software incorperates a comprehension cutting tool database for cutting process
parameter. It used the actual tools from the various tool manufactures, which helps us to
manufacture as per the industrial requirements. Master cam also contains many inbuilt tool
path modules for different tool path generation thus it helps to manufacture the parts
easily.
Thus by using the CAM software one can manufacture the given part in a specific
time with more accuracy.
Review Questions
1. What is mean by 'Numerical control' ?
2. What are the types of NC system ?
3. Write note on punch tape.
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4. What are the advantages, limitations and applications of NC systems over
conventional system ?
5. What is machining centre ? Briefly describe the horizontal and vertical
machining centre.
6. What are the elements of NC systems ?
7. Explain NC motion control system.
8. Differentiate between open loop and closed loop system.
9. Write note on CNC.
10. What are the constructional features of CNC machines ?
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11. Explain automatic tool changer.
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12. Write a short note on DNC.
13. Explain briefly NC part programming.
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14. Write note on G codes and M codes.
15. Explain following codes:
a) G03
d) M08
b) G90
e) M30
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c) G96
f) T05
16. Explain axis nomenclature of NC system.
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17. Write a manual part program for the following components. Assume spindle
speed 500 rpm and feed 0.5 mm/rev.
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100
R6
R5
50
20
30
30
22
10
30
15
Fig. 4.1 (a)
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6
10
5
Fig. 4.1 (b)
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80
70
+X
R9
20
24
18
26
R4.5
+Z
19
21
49
51
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Fig. 4.1 (c)
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+X
R20
50
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Fig. 4.1 (d)
10
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2×45º
R5
+Z
2×45º
50
36
40
R40
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7
15
22
42
47
Fig. 4.1 (e)
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18. What is NC machine ?
19. Name the advantages of NC machines.
20. Mention the different components of DNC.
21. Draw the simple configuration of CNC.
22. Write the main difference between CNC and DNC.
23. Write the advantages of CNC.
24. Mention the requirements of spindles in CNC.
25. Write the requirements of sideways.
26. What are capabilities of MCU ?
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27. Classify machining centres.
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28. List the data needed essentially to make a part program.
29. Name the methods of creatings part programming.
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30. What is absolute and incremental programming ?
31. Enlist the important steps followed in preparing a part program.
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32. Explain the meaning of following G codes :
i) G94
iii) G90
ii) G03
iv) G70
33. Explain the meaning of following M codes :
i) M03
iii) M09
ii) M05
iv) M30
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34. What is recirculating ball screw ?
35. Explain any two applications of NC system.
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36. What is the difference between open loop and closed loop system ?
37. What is mean by continuous path system ?
38. What do you mean by straight line system ?
39. Classify NC system.
40. What is pallet changer system ?
41. Explain Automatic Tool Changer.
42. Explain axis nomenclature for milling machine.
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Part A : Two Marks Questions with Answers :
Q.1
What is numerical control ?
Ans. : Numerical control is a programmable automation in which actions are controlled by
means of coded numbers, letters and other symbols.
Q.2
Name the basic elements of NC system.
Ans. : A numerical control machine consists of following elements :
i) Machine Control Unit (MCU)
ii) Machine tool and NC tooling
iii) Part program and drawings.
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Q.3
How NC system is classified ?
Ans. : NC system is classified as,
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i) According to tool positioning or modes of programming
a. Absolute system
b. Incremental system
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ii) According to motion control system
a. Point to point system
b. Straight line or straight cut system
c. Continous or contouring path system
iii)According to servo control
a. Open loop
b. Closed loop
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iv) According to type of feedback device
a. Ananlog transducer
Q.4
b. Digital transducer.
Explain absolute and incremental system.
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Ans. : Absolute system : In this system, all the positions are indicated from a reference point,
which is a fixed zero point or set point.
Incremental system : In this system, the tool positions are indicated with repsect to
previous point.
Q.5
State any four advantages of NC system.
Ans. : i)
High productivity.
ii) Less scrap.
iii) Flexibility in design.
iv) Reduction in inventory.
v) Less floor space required.
vi) Skilled operator not required.
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Q.6
Mention some disadvantages of NC system.
Ans. : i)
High initial cost.
ii) High maintainance cost.
iii) Costly control system.
iv) Uneployment.
Q.7
What are the applications of NC system ?
Ans. : i)
NC system is used where 100 % inspection is required.
ii) Where frequent changes in design occur.
iii) Where repetitive production is required.
iv) For complex machining operations.
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Q.8
Mention the types of NC systems.
Ans. : The common types of NC systems used in machine tools are
i)
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Coventional Numerical Control (NC)
ii) Computerized Numerical Control (CNC)
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iii) Direct Numerical Control (DNC)
Q.9
What is the purpose of Automatic Tool Changer (ATC).
Ans. :
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i) For complicated job on NC and CNC machines different types of tools are
required; for changing and resetting the tool, more time is required.
ii) For this purpose, tools can automatically be changed with the help of automatic
tool changer (ATC), hence productivity and repeatability of manufacturing
increases.
Q.10 What is Direct Numerical Control (DNC).
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Ans. : DNC is a manufacturing system in which a number of machines are controlled by a
central computer through a direct connection of telecommunication lines and in real time.
Q.11
What are G codes and M codes ?
Ans. : i) G is a preparatory function which changes the control mode of the machine.
e.g. G01 - Linear interpolation
G04 - Dwell
ii) M is a miscellaneous function which is generally called as M codes.
e.g. - M00 - Program stop
M06 - Tool change.
Q.12
Explain fixed zero and floating zero method.
Ans. :
i) Fixed zero : In this method, the origin is always predefined. It is generally at
the lowermost left hand corner of the worktable.
ii) Floating zero : Floating zero concept is provided which allows the operator to define his
origin.
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Q.13
List the commonly used co-ordinate systems of CNC machine tools.
Ans. : i) Absolute co-ordinate system.
ii) Incremental co-ordinate system.
Q.14
What is point-to-point (PTP) system ?
Ans. : In this system, tool is accurately located at some specified position. The spindle is brought
to the starting point, then moved to the next location i.e. from point 1 to point then point 3 etc.
On that location, operation is performed and then tool moves to next location.
Q.15
What are G-codes and M-codes ? Give examples.
Ans. : G-codes or preparatory function, changes the control mode of the machine. G-codes are
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followed by two digit number. It is written as G 01, G 02, etc.
M-codes or miscellaneous function, controls other auxiliary operations. It is also followed by two
digit number. It is written as M 03, M 04, etc.
Q.16
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What is the difference between incremental and absolute system ?
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Ans. : In absolute system, all the co-ordinates are indicated from a referance point which is a
fixed zero or set point. In incremental system, all the co-ordinates or tool positions are indicated
with respect to previous point.
Q.17
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What is the role of computer for CNC machine tool ?
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Ans. : The program is entered into the computer directly through keyboard. The program is
stored in computer memory, which can be recalled whenever required. Programs can be easily
edited and modified as per the requirement. These features makes the system flexible.
Q.18
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Differentiate between fixed zero and floating zero in CNC terminology.)
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Ans. : In a fixed zero method, the origin is always predefined. It is generally at the lowermost
left hand corner of the worktable. All other points are defined from this point.
In a floating zero concept, operator can define his origin, which makes it convenient to develope
programs of symmetrical components by providing the origin at the point of symmetry in the
workpiece.
Q.19
Name the various elements of CNC machines.
Ans. : i) Mini computer
ii) Machine tool and CNC tooling
iii) Part program and drawings.
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Q.20
What are the classifications of NC machines ? What are the types of motion
control system used in NC machines ?
Ans. :
1. According to the tool positioning
a) Absolute system
b) Incremental system
2. According to the motion control system
a) Point to point system
b) Straight line system
c) Continuous path system.
3. According to servo control system
a) Open loop system
b) Closed loop system
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Q.21
Define NC.
Ans. : NC is a programmable automation in which various actions are controlled by means of
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coded numbers, letters and other symbols.
Q.22
Mention the major elements of NC machines.
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Ans. : i) Machine Control Unit (MCU)
ii) Machine tool and NC tooling
iii) Part program and drawings.
Q.23
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Compare closed loop NC system with open loop NC system.
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Ans. : In open loop system, there is no feedback, to ensure whether the obtained slide movement
is same as desired or not and if not, what error is present.
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In closed loop system, there is a feedback device to compare the slide movement. It is nothing
but a transducer.
Q.24
Show the axes of a CNC horizontal boring machine ?
Ans. :
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+Z
+Y
+X
Q.25
What are the basic assumptions made while programming in APT language?
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Ans. : In APT programming, it is assumed that the workpiece remains stationary and cutting tool
does all the movements.
Q.26
What is mean by APT language ?
Ans. : Automatic Programming of Tools (APT) is the oldest and one of the most powerful NC
processor languges. It is generally used on large capability computers and can perform the
mathematics required for complex curves using four or five axis contouring techniques.
Q.27
Compare a closed loop NC system with open loop NC system.
Ans. : Open loop system involves feeding of tape, interpretation of information by tape reader,
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storing the data in buffer storage and it converts into electrical signal and send this signal to the
control unit.
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Closed loop system is almost similar to open loop system only it carries an additional feed back
device which is nothing but a transducer and accompanied by a comparator.
Q.28
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What is a preparatory function ? How is it important in CNC Programming ?
Ans. : G is the preparatory function which changes the control mode of the machine and it is
called as G-codes. Generally, they are followed by two digit number. It is written as G01, G02
etc.
Q.29
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Distinguish between point to point and continuous path systems.
Ans. : Point to point system : In this system, tool is accurately located at some specified
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position. Fig. 4.2 (a) shows path of tool movement for drilling number of holes.
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Fig. 4.2 (a) : Point to point system (PTP)
Continuous path system : In this, there is relative motion between the tool and workpiece,
during the whole operation. Due to this relative motion, different curves and profiles can be
cut. Actually, it is a combination of PTP and straight cut system.
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Fig. 4.2 (b) : Continuous path system
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Q.30
What do you mean by machining centre with respect to NC machines ?
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Ans. : It is a versatile NC machine tool, which can perform different operations in a single
setting and can automatically change tool under the programmable control. Several operations
like milling, boring, drilling, reaming, tapping, counter-boring, etc. can be performed in a single
set up, in any sequence.
Q.31
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What is meant by 'tool magazine' in a CNC machine ?
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Ans. :
i) Tool magazine is used in the automatic tool changer for storing the different tools.
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ii) Tool magazines are of two types such as drum type magazine and chain type magazine
which can store 60 or more tools.
Q.32
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What is the function of subroutine in NC part programming ?
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Ans. : i) A subroutine is a sequence of operations which is seperately defined and stored by
the user.
ii) Subroutine can be called at any point in the main part programes.
Q.33
With reference to CNC manual part programming, state what is linear
interpolation.
Ans. : Any machining during straight or taper lines is done using linear interpolation function
G01. The feed rate at which the cutting tool is required to move is also specified by linear
interpolation.
Q.34
Mention the advantages of stepping motor.
Ans. : Advantages :
i) Stepper motors offer precise rototation control.
ii) Stepper motor exhibit excellent positional accuracy and errors are non cummulative.
iii) Stepper motor has long maintenance free life and hence it is cost effective.
iv) These are directly compatible with digital control technique.
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Fundamental of CNC and Part Programming
Q.35
State the differences between CNC and DNC.
Ans. :
Sr. No.
Q.36
Ans. :
Parameter
1
Productivity
2
3
CNC
DNC
High
Highest
Number of operations done at a time
One
Multiple
Initial cost
High
Highest
What do you understand by 'canned cycle' in manual part programming ?
· Canned cycle is a fixed sequence of particular operations.
· This set of instructions are permanently stored in the control system and can be
called and used by a single command in the part programming.
For example : Canned cycle for drilling, tapping, boring, etc.
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Q.37
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Define CNC and DNC.
Ans. : DNC (Direct Numerical Control) (Refer Two Marks Q.10 of Chapter - 4)
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CNC (Computerised Numerical Control) : CNC is a manufacturing system in
which a seperate compuer is attached to each machine tool, with stored programmable
logic, with the absence of hard-wired logic systems.
Q.38
What is adaptive control ?
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Ans. : Adaptive control is the control technique which automatically determines the process
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variables like feed, cutting speed etc. during machining and makes changes in the prescribed
limit as per requirement.
Q.39
How are various functions timed in NC machines ?
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Ans. : The various functions in NC machines are timed by using punched tape. It uses a binary
coded decimal system for containing operating information of NC tool.
Q.40
Distinguish a fixed zero and floating zero.
Ans. :
Fixed Zero
Floating Zero
1) In this method, the origin is
always predefined.
1) In this method, the setting
of zero is done manually.
2)
2)
(0, 0)
(0, 0)
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(a) Fixed zero
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(b) Floating zero
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Fundamental of CNC and Part Programming
Q.41
State the functions of the following G and M codes :
G00
G03
M06
M03
Ans. :
G00 = Positioning (Rapid traverse)
G03 = Circular interpolation anticlockwise
M06 = Tool change
M03 = Spindle normal rotation
Q.42
Define "micromachining" with the help of an example.
Ans. :
·
Micromachining is a advanced machining process which is used to fabricate the
components having dimensions in micrometers.
·
It usually involves chemical etching process on a very fine scale.
·
The following are the examples of components manufactured by micromachining :
m Probes and sensors
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Measuring devices
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Microactuators
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Microgimbals
Fundamental of CNC and Part Programming ends ...
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