محاضرات كادكام 2013 · the arithmetic logic unit or input/output section for...

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CAD/CAM LECTUERS

4th Class-Electromechanical Eng. Dept.

BY

Assistant Professor

Dr. Farag Mahel Mohammed

2013-2014

REFERENCE 1. Computer Aided Design and Manufacturing, C.B. Besant, 1986.

2. CAD/CAM, Mc Mahan and Browne, 1998.

3. CAD/CAM in practice, Medland, 1988.

4. Computer Aided Manufacture, Chang and Richard, 2006.

5. CAD/CAM Principles and Applications, Rao, 2010.

6. Computer Aided manufacturing, S. Vishal, 2013.

7. Fundamental of Computer Aided Design, Goyal, 2013.

1

CHAPTER ONE

INTRODUCTION 1.1 HISTORICAL BACKGROUND

In 1730 the industrial revolution starts:

Manual power

Animal

Steam

Higher production rates

Higher in comes

Markets large demand for machines- especially automobiles.

In 1950 introduce computer:

Production under computer control using CNC machine high

accuracy and high mass production.

The manual and animal power used reduced and increase using steam power

2

1.2 THE DESIGN PROCESS The process of designing something is characterized as an iterative

procedure, which consists of six identifiable steps or phases:

1. Recognition of need.

2. Definition of problem.

3. Synthesis.

4. Analysis and optimization.

5. Evaluation.

6. Presentation.

Recognition of need involves the realization by someone that a

problem exists for which some corrective action should be taken. This might be

the identification of some defect in a current machine design by an engineer or the

perception of anew product marketing opportunity by a sales person.

Definition of the problem involves a thorough specification of the

item to be designed. This specification includes physical and functional

characteristics, cost, quality, and operating performance. Synthesis and analysis

are closely related and highly iterative in the design process.

A certain component or subsystem of the over all system is

conceptualized by the designer, subjected to analysis improve through this

analysis procedure and redesigned. The process is repeated until the design has

been optimized with in the constraints imposed on the designer. The components

and subsystems are synthesized into the final overall system in a similar iterative

manner.

Specification

Initial Design

3

Evaluation is concerned with measuring the specifications established in the

problem definition phase. This evaluation often requires the fabrication and

testing of a prototype model to assess operating performance, quality, reliability

and other criteria. The final phase in the design process is the presentation of the

design. This includes documentation of the design by means of drawings, material

specifications, assembly lists, and so on .Essentially, the documentation requires

that a design data base be created.

Engineering design has traditionally been accomplished on drawing

boards with the design being documented in the form of detailed engineering

drawing. Mechanical design includes the drawing of the complete product as well

as its components and subassemblies, and the tools and fixture required to

manufacture the product figure (1.1) illustrates the basic steps in the design

process indicating its iterative nature.

Electrical design is concerned with the preparation of circuit diagrams,

specification of electronic components and soon, similar manual documentation is

required in other engineering design fields (structural design, aircraft design,

chemical engineering design etc.). In each engineering discipline, the approach

has traditionally been to synthesize a preliminary design manually and then to

subject that design to same form of analysis. The analysis may involve

sophisticated engineering calculations or it may involve a very subjective

judgment of the aesthete appeal possessed by the design .the analysis procedure

identifies certain improvements that can be made in the design. As stated

previously, the process is iterative process in that it is time consuming many

engineering labor hours are required to complete the design project.

4

Figure (1.1): The general design process.

Recognition of need

Problem definition

Synthesis

Analysis and optimization

Evaluate Design

Presentation Automated drafting

Engineering analysis

Geometric modeling

Satisfy Specification

No

Yes

5

1.3 THE PRODUCT CYCLE AND CAD/CAM

CAD/CAM is a term which means computer aided design and

computer-aided manufacturing. It is the technologies concerned with the use of

digital computers perform certain functions in design and production. This

technology is moving in the direction of greater integration of design and

manufacturing, two activities which have traditionally been treated as distinct and

separate functions in a production firm.

CAD (Computer Aided Design): can be defined as the use of computer systems

as assist in the creation, modification, analysis, or optimization of a design

CAM (Computer Aided Manufacturing): can be defined as the use of computer

systems to plan, mange, and control the operations of a manufacturing

plant through either direct or indirect computer interface with the plants

production resources.

CAD/CAM: is concerned with the application of computers to the manufacture

of engineering components starting from the drawing office, to the

production department, to the machine assembly shops, to the quality

control department, right to the finished parts store.

6

Figure (1.2): The product cycle.

Factory Specific

Customer

Design

Process Planning

Scheduling

Stock control

Inspection Fabrication

Test

Sales

Customer

7

From this diagram it appears that each of these activities needs to altered

or modified according to the feedback from other activities which are interacting

with it. The fast and efficient means are needed for:

1. Producing product design (geometry, loads, appearance).

2. Predicting product behavior (simulation).

3. Process planning (material, tool, and machine sequence).

4. Scheduling of machine.

5. Product fabrication (machining, forming, assembly).

6. Product inspection (in process, past process).

7. Material handling (raw material, semi finish, finish tools).

8. Marketing (cost, sale, price sale, service after sale).

9. Information handling and exchange.

The CAD include (1 and 2) while CAM include (3 to 9).

Computer is a power full and fast tool for performing, computing, is

the right means to use for improving the efficiency of production activities. The

main benefits of computer in industry include:

1. To increase the productivity of the designer, This is accomplished by

helping the designer to visualize the product and it is component

subassemblies and parts: and by reducing the time required in synthesizing,

analyzing, and documenting the design. This productivity improvement

translates not only into lower design cost but also into shorter project

completion times.

8

2. To improve the quality of design, A CAD system permits a more thorough

engineering analysis and a larger number of design alternatives can be

investigated. Design errors are also reduced through the greater accuracy

provided by the system. These factors lead to a better design.

3. To improve documentations, use of a CAD system provides better

engineering drawings, more standardization in the drawings, better

documentation of the design, or fewer drawing errors, and greater legibility.

4- To create a data base for manufacturing, In the process of creating the

documentation for the product design (geometries and dimension of the

product and it is components, material specifications for components, bill of

materials, etc…..), much of the required data base to manufacture the product

is also created.

9

COMPUTER INTEGRATED MANUFACTURE (CIM) Is that all the individual functions of design and manufacture are

computerized and these function are tied together through a central data base

system, that is shred among all department and activities.

Computer-integrated manufacturing (CIM) is manufacturing supported by

computers. It is the total integration of Computer Aided Design / Manufacturing

and also business operations and databases. This term has generally been replaced

by the wide field of PLM – Product Lifecycle Management. Some components of

CIM are: CAD, CAP (Computer Aided Planning), CAQ (Computer Aided Quality

Assurance), CAM (Computer aided manufacturing).

Figure (1.3): CIM diagram.

10

1.5 BENEFITS OF CAD/CAM

1. Reduce the number of steps involved in the design process.

2. Make each design step much easier and less tedious for the designer to

perform.

3. Make better decisions and will reduce the possibility of having errors.

4. The designer arrives at an optimal solution.

5. Scheduling of components and tools through manufacturing is much easier.

6. Reduce delivery time of products.

7. The working capital required by company is reduced.

8. All information is stored in the computer in stead of on paper.

1.6 APPLICATION OF CAD/CAM 1. Study of molecular structure in chemistry.

2. Medical research.

3. Aircraft flight simulation.

4. Structure design in aircraft.

5. Ship building.

6. Automobile industries.

7. Town planning.

8. Integrated circuits and printed circuit board design.

9. Mesh data preparation for finite element analysis.

11

CHAPTER TWO

COMPUTER SYSTEM

2.1 DIGITAL COMPUTER SYSTEM

Computers are now in common use in both scientific and commercial

fields. The digital computer is a major and central component of CAD/CAM

systems, so it is essential to be familiar with the technology of the digital

computer and the principle on which it works.

The modern digital computer is an electronic machine that can

perform mathematical logical calculations and data processing functions in

accordance with a predetermined program of instructions.

The computer system consist of the hardware and software to perform

the specialized design function required by the particular user firm, the hardware

typically includes the computer, one or more graphics display terminals,

keyboard, and other peripheral equipment.

The software consists of the computer programs and instructions to

implement computer graphics on the system plus application program to facilitate

the engineering functions of the user company. Examples of these application

programs include stress-strain analysis of components, dynamic response of

mechanism, heat transfer calculations, and numerical control pant programming.

12

There are three basic hardware components of a general purpose

digital computer as shown in figure (2.1):

Figure (2.1): Computer System.

1. Central processing unit (CPU):The central processing unit is often

considered to consist of two subsections that:

a. Control unit: the control unit coordinates the operations of all the other

components. It control the input and output of information between the

computer and the outside world through the input/output section,

synchronizes the transfer of signals between the various sections of the

computer and commands the other section in the performance of their

function.

Controller

Memory

Arithmetic Logic unit

Mass storage unit

I/O unit Output Input

CPU

13

b. Arithmetic Logic unit: the arithmetic logic unit carries out the arithmetic

and logic manipulations of data. It adds, subtracts, multiplies, divides and

compares number according to programmed instructions.

2. Memory: the memory of the computer is the storage unit. The data stored

in this section are arranged in the form of words which can be transferred to

the arithmetic logic unit or input/output section for processing. In general

the memory classified into main and auxiliary memory.

3. Input/Output section: the input/output provides the means for the

computer to communicate with the external world. This communication is

accomplished through peripheral equipment such as printers, monitors,

keyboard, mouse…etc.

2.2 DATA REPRESENTATION

Information is handled within the computer by electrical components such

as transistors, integrated circuits, semi conductors and wires, all of which can only

indicate two states or conditions.

The binary number system is thus particularly suitable for

mathematically representation the two states possible. The binary number system

is based on the number (two) and involves only two digits zero (0) and one (1).

The meaning of successive digits in the binary system is based on the number (2)

raised to successive powers.

14

The first digit is 20

The second digit is 21

The third digit is 22

And so forth

Binary Decimal

0000 0

0001 1

0010 2

0011 3

0100 4

0101 5

0110 6

0111 7

1000 8

1001 9

Example-1: Convert the binary number 11010011 into decimal one.

11010011=1x20 +1x21 +0x22 +0x23 +1x24 +0x25 +1x26 +1x27 =211

A part from the decimal and binary system, the octal and hexadecimal

number system is in common use in computers. The octal number system is based

on the number eight and involves the eight digits, zero (0) to seven (7), where as

the hexadecimal number system is based on the number sixteen and involves the

sixteen digits, zero (0) to nine (9) and A to F which represent 10 to 15.

15

Decimal Binary Octal Hexadecimal

0 0 0 0

1 0001 1 1

2 0010 2 2

3 0011 3 3

4 0100 4 4

5 0101 5 5

6 0110 6 6

7 0111 7 7

8 1000 10 8

9 1001 11 9

10 1010 12 A

11 1011 13 B

12 1100 14 C

13 1101 15 D

14 1110 16 E

15 1111 17 F

16

Example-2: Convert the binary number 10111 into decimal, octal and

hexadecimal numbers.

1. To decimal number:

10111=1x20 +1x21 +1x22 +0x23 +1x24 = 23.

2. To octal number: (Split into groups of three binary digits) 010111=

27

3. To hexadecimal number: (Split into groups of four binary digits)

00010111= 17

HW

1. Determine the binary, octal and hexadecimal numbers equivalent to

decimal numbers (25, 90, and 1990).

2. Determine the binary, octal and decimal numbers equivalent to

hexadecimal numbers (A53C, 3D5).

3. Convert the Octal number (57011) into hexadecimal number.

17

2.3 PROGRAMMING LANGUAGE In general there are three basic categories of computer programming

language:

1. Machine Language. Low level Language.

2. Assembly Language.

3. High level language such as:

FORTRAN, BASIC, PASCAL and COBOL………etc.

Example-3: using high level language (BASIC) to draw a line from point (X1,Y1)

to point (X2,Y2).

10 SCREEN 0: CLS

20 INPUT “from point”; X1,Y1

30 INPUT “from point”; X2,Y2

40 SCREEN 1: CLS

50 FOR Y= Y1 To Y2

60 A= ((X2-X1)*(Y-Y1))/(Y2-Y1)

70 X= A+X1

80 PSET (X,Y)

90 NEXT Y

18

Example-4: Draw a circle according to it is center (150,100) and radius R using

BASIC Language.

10 SCREEN 0 : CLS

20 INPUT “Radius”; R

30 SCREEN 1 : CLS

40 PI = 3.141569

50 FOR TH = 0 TO 360

60 X = 150 + R*COS(TH*PI/180)

70 Y = 100 + R*SIN(TH*PI/180)

80 PSET (X,Y)

90 NEXT TH

19

CHAPTER THREE

GEOMETRICAL TRANSFORMATIONS

3.1 INTRODUCTION Many of the editing features involve transformations of the graphics

elements or cell composed of the elements or even the entire model. In this

chapter we begin with a brief mathematical review of matrix algebra and then

discuss the mathematic of these transformations. Two dimensional

transformations are considered, first to illustrate concepts. And deal with three

dimensions. There are several common transformations used in computer graphics

such as:

1. Scaling.

2. Reflection.

3. Rotation.

4. Translation.

3.2 MATRICES A matrix is a rectangular array of numbers (which can be viewed as a

row of vectors) which is extensively used in computer graphics since it gives us

very compact notations. A general matrix will be represented by an upper case

letter:

=

333231

232221

131211

aaaaaaaaa

A

20

Figure (3.1): Matrix multiplication.

The element on the ith row and jth column is denoted by aij. Note that

we start indexing at 1, where as C indexes arrays from 0 - beware! A matrix is

said to be of dimension n by m (written n x m) if it has n rows and m columns.

Matrix multiplication is more complex. Given two matrices, A and B if we want

to multiply B by A (that is form AB) then if A is (n x m), B must be (m x p). This

produces a result, C = AB, which is (n x p), with elements cij =∑=

m

1kkjik ba , that is

the i; jth element of C is the ith row of A dot product with the jth column of B.

Note that matrix multiplication is not commutative, indeed in this case we cannot

multiply BA, since the sizes are wrong..

Matrix multiplication distributes over addition, that is A(B + C) = AB

+ AC, and there is an identity matrix for multiplication, denoted I, which is square

and has ones on the diagonal with zeros everywhere else. The transpose of a

matrix, A, which is either denoted AT or A/ is obtained by swapping the rows and

columns of the matrix. Thus:

=′⇒

=

2313

2212

2111

232221

131211

aaaaaa

Aaaaaaa

A

If we consider a (n x1) matrix (that is a column vector, s) then it

transpose s/ is a (1 x n) matrix (which we would call a row vector).

21

3.3 Mathematical elements in 2-D graphics This section considers some of the transformations that are applied to 2-D

graphics primitives and objects. This section uses many of the results that were

shown above for matrices.

1. Scaling.

The scaling of an element is used to enlarge it or reduce its size, scaling is

the simple stretching of the object, generally about the origin. Given a point r and

a scaling matrix S, where:

=

′′=′

YX

].S[YX

]P].[S[]P[

=

y

x

S00S

S

Figure (3.2): 2-D Scaling.

Where Sx is the x-axis scaling and Sy is the y-axis scaling the location of the new

point can be written r`= rS. If Sx = Sy = S the scaling is said to be uniform and r` =

rS, otherwise the scaling is called differential. An example is shown in Figure

(3.2).

Example-1: Apply scaling by a factor 2.For the line defined by the points A(1,1)

and B(3,2).

=

4262

2131

2002

The line scaled to A′(2,2) and B′(6,4).

22

Commands in Auto CAD:

Scale ↵ or

Select objects: “use mouse left click sign the objects” ↵

Specify base point:

Specify scale factor or [Reference]: ….↵

2. Reflection or mirror:

Reflection or mirror is a transformation, which allows a copy of the object

to be displayed while the object is reflected about a line or plane.

=

′′=′

YX

].F[YX

]P].[F[]P[

a. about Y-axis

−=

=

′′

1001

F

YX

].F[YX

b. about X-axis.

−

=

=

′′

1001

F

YX

].F[YX

c. about origin

−

−=

=

′′

1001

F

YX

].F[YX

Figure (3.3): 2-D Reflection

23


Mirror ↵ or


Specify first point of mirror line:

Specify second point of mirror line: ↵

3. Rotation

In this transformation the points of the objects are rotate about the origin

by an angle θ. The final position and orientation of a geometric entity is described

by the angle of rotation and the base point about which the rotation, the general

formula is given by:

=

′′

YX

].R[YX

From figure (3.4), the general position is:

X=rcosα

Y=rsinα

The new position is specified by:

−=

′′

+=+=

+=′−=

−=+=′

YX

.cossinsincos

YX

cosYsinXsincosrcossinr

)(sinrYsinYcosX

sinsinrcoscosr)rcos(X

θθθθ

θθαθαθ

θαθθ

αθαθθα

Where,

−=

θθθθ

cossinsincos

]R[

Figure (3.4): 2-D Rotation.

24

Note that positive θ implies an anti-clockwise rotation. It is a simple

exercise to show, using simple trigonometry, that R is indeed a matrix which

rotates points by θ.

Example-2: Rotate the line

2142 about the origin by 30º CCW.

=

−732.2866.1464.2232.1

2142

30cos30sin30sin30cos


Rotate ↵ or


Specify base point:

Specify angle rotation or [Reference]: …..↵

4. Translation

Involves moving the geometric entity from one location to another, the

new entity is parallel at all the points to the old entity. The general formula in

matrix notation is:

)t,t(TYX

].T[YX

yx=

+=

′′

Where tx is the unit translates in X-axis and ty is the unit translates in Y-axis.

Figure (3.5): 2-D Translation.

Before translation After translation

25

Example-3: Translate the line defined by (4,5) and (3,7) by 1 unit in X-direction

and 2 unit in Y-direction.

+=

′′

YX

].T[YX

=

+

9745

7534

2211


Move ↵ or


Specify base point of displacement:

Specify second point of displacement or<use first reference as displacement>:↵

5. Concatenation of transformation

Many a times it becomes necessary to combine the individual transformations as

shown above in order to achieve the required results. In such cases, the combined

transformation matrix can be obtained by multiplying the respective

transformation matrices. However, care must be taken to see that the order of the

matrix multiplication be done in the same as that of the transformations as

follows:

[T]=[Tn][Tn-1][Tn-2]……….[T3][T2][T1].

26

3.4 HOMOGENEOUS COORDINATES Representing 2D coordinates in terms of vectors with 2 components

turns out to be rather awkward when it come to the sort of manipulation that needs

to be carried out for computer graphics.

Homogeneous coordinates allow us to treat all transformations in the

same way, as matrix multiplications. The consequence is that our 2-vectors

become extended to 3-vectors, with a resulting increase in storage and processing.

Homogeneous coordinates mean that we represent a point (x; y) by the extended

triple (x; y; w). In general w should be non-zero. The normalized homogeneous

coordinates are given by (x/w; y/w; 1) where (x/w; y/w) are the Cartesian

coordinates of the point. Note in homogeneous coordinates (x; y; w) is the same

as (x/w; y/w; 1) as is (ax; ay; aw) where a can be any real number. Points with

w = 0 are called points at infinity, and are not frequently used.

Vector triples usually represent points in 3D space, however here we are

using them to represent points in 2D space, so what is going on. Well we are using

a bit of mathematical trickery to make life easy for ourselves. If you like then you

can think of 2D space corresponding to plane w = 1.

Now in homogeneous coordinates the transformations can be given as:

=

′′

1YX

1000S000S

1YX

Scaling Y

X

27

=

′′

θθθ−θ

=

′′

−

−=

′′

−=

′′

−

−=

′′

−

1YX

100t10t01

1YX

nTranslatio

1YX

1000cossin0sincos

1YX

Rotation

1YX

100010001

1YX

origintheabout

1YX

100010001

1YX

axisXabout

1YX

100010001

1YX

axisYabout

flectionRe

Y

X

28

3.5 2-D ROTATION ABOUT AN ARBITRARY POINT Q(q1,q2) If we wanted to rotate an object about any arbitrary point (Q), by angle

(θ) this can easily be achieved by:

1. Translate the object by (-Q),

−−

=100q10q01

]T[ 2

1

1

2. Rotate object by angle (θ),

θθθ−θ

=1000cossin0sincos

]T[ 2

3. Translate the object back to the original position by (Q).

=

100q10q01

]3T[ 2

1

One of the big advantages of homogeneous coordinates is that transformations can

be very easily combined. All that is required is multiplication of the

transformation matrices. This makes otherwise complex transformations very easy

to compute. For instance if we wanted to rotate an object about some point, Q, this

can easily be achieved by:

]T[]T][T[]T[ 123=

−−

θθθ−θ

=

100q10q01

1000cossin0sincos

100q10q01

]T[ 2

1

2

1

θ−θ−θθθ+θ−θ−θ

=100

sinq)cos1(qcossinsinq)cos1(qsincos

]T[ 12

21

Figure (3.6): Rotation about an

arbitrary point

29

3.6 REFLECTION ABOUT AN ARBITRARY AXIS (Y =a+bX) The transformations given earlier for reflection are about origin or about

the coordinate’s axes. However sometimes it may be necessary to get the

reflection about an arbitrary line as shown in figure (3.7). To derive the necessary

transformation matrix, the following complex procedure is required:

1. Translate the mirror line along the

Y-axis such that the line passes

through the origin (o):

−=100a10

001]T[ 1

2. Rotate the mirror line such that it

coincide with the X-axis:

θθ−θθ

=1000cossin0sincos

]T[ 2

3. Mirror the object through the X-axis:

−=

100010001

]T[ 3

4. Rotate the mirror line back to the original angle with the X-axis:

θθθ−θ

=1000cossin0sincos

]T[ 4

Figure (3.7): Reflection about an

arbitrary axis.

30

5. Translate the mirror line along the Y-axis back to the original position:

=

100a10001

]T[ 5

The required transformation matrix is given by:

[T]=[T5][T4][T3][T2][T1]

+θθ−θθ−θθ

=

−

θθ−θθ

−

θθθ−θ

=

100)12(cosa2cos2sin

2sina2sin2cos]T[

100a10

001

1000cossin0sincos

100010001

1000cossin0sincos

100a10001

]T[

31

Example-4: Given the line (5,7) and (9,9)

a- Translate the line through (-6,3.3).

b- Rotate the line through 35° about the origin.

c- Rotate the line about its end point (5,7) by 40° CW.

Solution:

a-

=

′′

1YX

100t10t01

1YX

Y

X

−=

−=

′′

113.123.10

31

119795

1003.310601

1YX

b-

θθθ−θ

=

′′

1YX

1000cossin0sincos

1YX

=

−=

′′

11535.126.821.2081.0

119795


1YX

C-1. Translate the object by (-Q),

−−

=

−−

=100710501

100q10q01

]T[ 2

1

1

2. Rotate object by angle (θ),

−−−−−

=

θθθ−θ

=100040cos40sin040sin40cos

1000cossin0sincos

]T[ 2

32

3. Translate the object back to the original position by (Q).

=

=

100710501

100q10q01

]3T[ 2

1

Thus:

=

−−

−−−−−

=

′′

1196.5735.95

119795

100710501


100710501

1YX

Example-5: Reflect the triangle (20,40), (50,50) and (30,60) about the arbitrary

axis Y= 15- (15/10)X.

Solution:

1. Translate the mirror line along the Y-axis such that the line passes through

the origin (o):

−=

1001510001

]T[ 1

2. Rotate the mirror line such that it coincide with the X-axis:

−−−−−

=

θθ−θθ

=

−=−=θ −

10003.56cos3.56sin03.56sin3.56cos

1000cossin0sincos

]T[

3.56)10/15(tan

2

1 o

33

3. Mirror the object through the X-axis:

−=

100010001

]T[ 3

4. Rotate the mirror line back to the original angle with the X-axis:

−−−−−

=

θθθ−θ

=10003.56cos3.56sin03.56sin3.56cos

1000cossin0sincos

]T[ 4

5. Translate the mirror line along the Y-axis back to the original position:

=

=

1001510001

100a10001

]T[ 5

The required transformation matrix is given by:

[T]=[T5][T4][T3][T2][T1]

−

−−−−−

−

−−−−−

=

1001510001

10003.56cos3.56sin03.56sin3.56cos

100010001

10003.56cos3.56sin03.56sin3.56cos

1001510001

]T[

+−−−−

−−−−=

′′

+−−−−

−−−−=

111605040305020

100)1)3.56(2(cos15)3.56(2cos)3.56(2sin

)3.56(2sin15)3.56(2sin)3.56(2cos

1YX

100)1)3.56(2(cos15)3.56(2cos)3.56(2sin

)3.56(2sin15)3.56(2sin)3.56(2cos]T[

−

−−−=

′′

11144.46.172.6

87.5253.5175.30

1YX

34

Example-6: Two points, A and B, constituting a portion of a two-dimensional

shape are moved to points C and D, respectively, resulting in

transformation the original shape. List the required transformation

matrices, in proper order, that have to be applied to all the points of

the shape. The coordinates of the points are A(2,2), B(5,5), C(5,2)

and D(7,2 + 2√3).

Solution:

To move points A and B to C and D, respectively, four steps are involved:

1. Translation the line AB from the location A to the origin, and the

transformation matrix is:

−−

=100210201

]T[ 1

2. Rotation the line AB about Z-axis by angle and the angle is calculated>

The angle between AB and the X-axis is:

o452525tan 1

AB =−−

=α −

The angle between CD and the X-axis is:

o0657

2)322(tan 1CD =

−−+

=β −

The angle between the line AB and CD is 60º -45º =15º, rotate the line AB

about the Z-axis by 15º, the transformation matrix is:

−=


]T[ 2

3. Scale the line AB so that the line is the same in length as line CD:

35

42)464.3(CD

242.433AB22

22

=+=

=+=

The scaling factor is CD/AB=4/4.242=0.942

=

1000942.0000942.0

]T[ 3

4. Translation the line AB to the location of C so that A and B will coincide

with C and D respectively. The transformation matrix is:

=

100210501

]T[ 4

The equivalent transformation matrix is:

+=

−

−=

′′

−

−=

−−

−

=

113222

75

115252

10031.091.024.0

67.324.091.0

1

10031.091.024.0

67.324.091.0

100210201


1000942.0000942.0

100210501

][

YX

T

36

3.7 Mathematical elements in 3-D graphics

1. 3-D Translation: To translate the point (X,Y,Z) to new point )Z,Y,X( ′′′

through (tx, ty, tz) we use:

=

′′′

1ZYX

1000t100t010t001

1ZYX

z

y

x

2. 3-D Reflection: An object is reflected through a plane by manipulating the

diagonal elements in 3-D matrix.

=

′′′

1ZYX

]F[

1ZYX

−

=

1000010000100001

F

Reflection through YZ plane (around X-axis)

−

=

1000010000100001

F Reflection through XZ plane (around Y-axis)

−=

1000010000100001

F Reflection through XY plane (around Z-axis)

3. 3-D Scaling: The diagonal terms of the general (4×4) transformation matrix

provide scaling:

37

=

′′′

1ZYX

10000S0000S0000S

1ZYX

z

y

x

3-D Rotation:

=

′′′

1ZYX

]R[

1ZYX

The rotation transformation matrix about x-axis:

θθθ−θ

=θ

10000cossin00sincos00001

)(R x

The rotation transformation matrix about y-axis:

θθ−

θθ

=θ


)(R y

The rotation transformation matrix about z-axis:

θθθ−θ

=θ

1000010000cossin00sincos

)(R z

38

Example-7: For a rectangular object of points (2,2,6), (8,2,6), (8,8,6), (2,8,6),

(2,2,3), (8,2,3), (8,8,3) and (2,8,3):

a- Change the scale by 3,2,1 in x,y,z respectively.

b- Reflect the object through XY plane.

c- Rotate the object around Z-axis by 20°.

Solution

a-

=

′′′

1ZYX

10000S0000S0000S

1ZYX

z

y

x

=

=

′′′

1111111133336666

161644161644624246624246

11111111333366668822882228822882

1000010000200003

1ZYX

b-

=

′′′

1ZYX

]F[

1ZYX

−−−−−−−−=

−=

′′′

1111111133336666

161644161644624246624246

11111111333366668822882228822882

1000010000100001

1ZYX

c-

=

′′′

1ZYX

]R[

1ZYX

39

−−

=

−

=

′′′

11111111333366662.825.1061.456.22.825.1061.456.285.078.483.619.185.078.483.619.1

11111111333366668822882228822882

100001000020cos20sin0020sin20cos

1ZYX

40

3.8 Window to viewport transformation In general the objects and primitives represented in the application model

will be stored in world coordinates that is their size, shape, position, etc. will be

given in terms of logical units for whatever the object represent (e.g. mm, cm, m,

km or light years). Thus to display the appropriate images on the screen (or other

device) it is necessary to map from world coordinates to screen or device

coordinates. This transformation is known as the window to viewport

transformation, the mapping from the world coordinate window to the viewport

(which is given in screen coordinates).

Figure (3.8): The window to viewport transformation.

In general the window to viewport transformation will involve a scaling and

translation as in Figure (3.8), where the scaling in non-uniform (the vertical axis

has been stretched). Non uniform scaling result from the world coordinates

window and viewport having different aspect ratios. In Figure the screen window

(that is the viewport) covers only part of the screen.

The transformation is generally achieved by a translation in world

coordinates, a scaling to viewport coordinates and another translation in viewport

coordinates, which are generally composed to give a single transformation matrix.

Often the clipping of visible elements will be carried out at the same time as the

41

transformation is applied. Typically the region will be clipped in world

coordinates and then transformed.

Figure (3.9): The procedure used to transform from world coordinate window to viewport.

The required transformations are shown in Figure (3.9). If the world

coordinate window has dimensions (xwmin; ywmin) and (xwmax; ywmax), while the

viewport (or screen coordinate window) has dimensions xVmin; yVmin) and (xVmax;

yVmax), then the transformation will be given by: a translation;

−−

=100

Y10X01

T minw

minw

1

a scaling:

==

1000S000S

ST Y

X

2

and finally another translation;

=

100Y10X01

T minV

minV

3

These can be combined to yield the transformation matrix:

42

minwmaxw

minVmaxVy

minwmaxw

minVmaxVx

minwyminVy

minwxminVx

wv

minw

minw

Y

X

minV

minV

wv

YYYYS

XXXXS

Where100

YSYS0XSX0S

M

100Y10X01

1000S000S

100Y10X01

M

−−

=

−−

=

−−

=

−−

=

Example: Estimate the transformation matrix required to transform the (2,2) and

(4,3) from the world coordinates to the screen coordinates at (0.5,1) and

(3,2.25).

Solution:

−−

=

−−

=

−−

=

=−

−=

−−

=

=−

−=

−−

=

1005.125.10

2025.1M

1002*25.1125.102*25.15.0025.1

M

100YSYS0XSX0S

M

25.123

125.2YYYYS

25.1245.03

XXXXS

wv

wv

minwyminVy

minwxminVx

wv

minwmaxw

minVmaxVy

minwmaxw

minVmaxVx

43

Clipping:

Clipping is a very important element in the displaying of graphical images.

This helps in discarding the part of the geometry outside the viewing window,

such that all the transformation that are to be carried out of zooming and panning

of the image on the screen are applied only on the necessary geometry.

In order to carry out the clipping operation for lines, it is necessary to know

the lines are completely inside the clipping rectangle, completely outside the

rectangle or partially inside the rectangle as shown in figure (3.10).

Figure (3.10): The clipping operation

To know whether a line is completely inside or outside the clipping

rectangle, the end points of the line can be compared with the clipping boundaries.

For example, the line P1P2 is completely inside the clipping rectangle, similarly

line P3P4 and P9P10 are completely outside the clipping rectangle.

In Cohen- Sutherland clipping algorithm of 2-D, the lines are classified as

to whether they are in, out or partially in the window by doing an edge test. The

end points of the line are classified as to where they are with reference to the

window by means of a 4-digit binary code. The code is given as TBRL. The code

is identified as follows:

44

T=1 if the point is above the top of the window.

=0 otherwise.

B= 1 if the point is above the bottom of the window.

=0 otherwise.

R= 1 if the point is above the right of the window.

=0 otherwise.

L= 1 if the point is above the left of the window.

=0 otherwise.

1001 1000 1010

0001 0000 0010

0101 0100 0110

The 4-digit coding of the line end points for clipping

- The line is completely inside the window if both the end points are equal to

(0000).

- The line is completely outside the window if both the end points are not

equal to (0000) and a 1 in the same bit position for both ends.

45

HW

1. Reflect an object defined by the points (1,2), (3,5), (7,8) and (4,4) about an

arbitrary axis basses through the points (4,4) and (9,6).

2. An object is defined by the points (1,2), (6,4), (8,7) and (3,5), perform the following transformations on it:

a. Reflect the object about an arbitrary axis defined by the points (1,2) and (6,4).

b. Scale the object by a factors of (0.5 in X-direction and 2 in Y- direction).

3. An object is defined by the points (5,7), (4,8) and (1,-2). Perform the

following transformation on it:

a. Rotate the object about the point (2,-1) by 35º CW.

b. Apply scaling on the object by 2 in X-direction.

4. Reflect the object defined by the points (2,3), (5,5), (5,7) and (2,8), about

the axis Y=3-X.

5. A cube of 10 unit length has one of its corners at the origin (0,0,0) and the

three edges along the three principle axes. If the cube is to be rotated about

the Z-axis by an angle of 30º CCW direction. Calculate the new position of

the cube.

6. Using Cohen- Sutherland clipping algorithm to test whether the line shown

inside or outside the window.

46

CHAPTER FOUR

FINITE ELEMENT METHOD 4.1 INTRODUCTION

The finite element method has developed simultaneously with the

increasing of high speed electronic digital computer and with the growing

emphasis on numerical method for engineering analysis. Although the method

was originally developed for structural analysis, the general nature of the theory

on which it is based has also made possible its successful application for solutions

of problems in other field of engineering.

The analytical solution only for desired unknown quantity at any location in

the body. Analytical solutions can be obtained only for a certain simplified

situations. For problems involves complex material properties and boundary

conditions using Finite Element Method. The process of selecting only a certain

number of discrete points in the body can be termed discretizations.

The Finite Element Method consists of five essential states:

1. Definition of the Finite Element Method mesh.

2. Selecting a displacement model.

3. Formulate the discrete equation.

4. Solving the stiffness equation.

5. Determining element stresses and strains.

47

4.2 TYPES OF FINITE ELEMENT

48

4. 3 SPRING ELEMENT

50

Example-1

52

Example-2: For the spring system with arbitrary numbered nodes and elements, as

shown. Find the global stiffness matrix.

Solution:

First construct the following:

Element Node (1) Node (2)

1 4 2

2 2 3

3 3 5

4 2 1

[ ]

[ ]

[ ]

−

−=

−

−=

−

−=

33

333

53

22

222

32

11

111

24

KKKK

K

uuKKKK

K

uuKKKK

K

uu

53

[ ]

[ ]

−−

−+−−−++−

−

=

−

−=

33

11

3322

124214

44

54321

44

444

12

K0K000K0K0K0KKK00KKKKKK000KK

K

uuuuuKKKK

K

uu

54

6.4 APPLICATION TO THE UN AXIALLY STRESSED BAR

SUBJECTED TO VARIOUS END CONDITIONS.

1. Defining the finite element mesh.

2. Selecting the displacement model.

We will assume

ξ+=ξ 10e aau

[ ] )1......(..........aa

1u1

0e

ξ=ξ

l10e2

0e1

aau

au

+=

=

)2........(aa

101

uu

1

0

e2

e1

=

l

=

l101

A

From equations (1) and (2)

[ ]

−=−

ll

1101

A 1

[ ][ ]

ξ=∴ −

ξ e2

e11e

uu

A1u

55

e2

e1

e uu1ull

ξ+

ξ

−=ξ

ξ

ξ

−=ξ e2

e1e

uu

1ull

( )[ ] e0uN)(u ξ=ξ

Where :-

( )[ ]=ξN Shape function.

=e0u Displacement vector.

3. Generate the stiffness equilibrium equation.

i- Find the eU : on the basis of equation (1) , the strain eε "strain within the

element " .

[ ] )3(..............................uBuddN

ddu e

0eo

e =

ξ

=ξ

=ε

where

[ ]

−=

ll

11B

Hooks law is σ =E.ε, or in matrix form

[ ] [ ] D ε=σ

[ ] )4.........(..........dvolD21dvolE

21U t2e εε=⋅ε⋅= ∫∫

Substitute from equation (3) in equation (4) get:

[ ] [ ] [ ] ξ⋅Α⋅= ∫ duBDBu21U e

0t

0

e0

el

( [ ] [ ] [ ] ) e0

0

tte0 udBDBu

21

ξΑ⋅= ∫l

56

⇒ [ ] e0

te0

e uu21U Κ=

where : [Κ] = element stiffness matrix.

[Κ] = )5....(........................................11

11

−

−

ΕΑl

ii- To find total energy stored U:

∑=e

e )6...(............................................................UU

it is convenient to express this equation in matrix formation:

=

−

n0

30

20

10

u

u

u

u

U ,[ ][ ]

[ ][ ]

[ ]

)7.........(

k0000k0000k0000k

K

n

3

2

1

=

⇒ [ ] )8........(........................................uku21U t=

iii- compatibility

for compatibility

111 uu =

212 uu =

221 uu =

322 uu =

and so on

57

⇒ )9...(..............................

u

uuu

10000000

..........00100

..........00010

..........00010

..........00001

u

u

uu

u

u

n

3

2

1

n2

n1

22

21

12

11

=

[ ] ucu =

where [C ] = connection compatibility

from equation (9) into (8) get:

=

−

uckcu21U

tt

[ ] uku21 t=

where [Κ] = system of global or assembly

[ ] [ ] [ ] )10.......(..................................................ckck t

=

−

iv- potential energy of applied loads

As a first case, suppose concentrated axial forces are applied at the nodes (1,2 ,

…., n+1) then:

1n1n332211 xu....xuxuxu ++−−−=Ω , where Χ is applied load .

or [ ] )11.....(..............................pu t−=Ω

v- total potential energy :-

[ ] [ ]puuku21UV tt −=Ω+=

and

58

( [ ] [ ] ) [ ]puukuuku21V ttt δ−δ+δ=δ

since [Κ] symmetric

( [ ] [ ] ) )12..(..............................pukuV t −δ=δ

But for equilibrium δν=0 , and tuδ arbitrary

∴[Κ]υ=Ρ

Example-3: Using finite element method to formulate the equilibrium equation

and find the displacement and stresses of the bar subjected to various

end conditions shown.

Solution:

1. Element stiffness matrices:

[ ]

[ ]

[ ]

−

−=

−

−=

−

−=

1111

L4EA7K

1111

L4EA5K

1111EAK

2

1

l

2. Assembly of [K]

[ ]

+=

222

221

212

211

122

121

112

111

kk0kkkk0kk

K

59

[ ]

[ ]

−

−−

−

=

−

−+−

−

=

L4EA7

L4EA70

L4EA7

L4EA12

L4EA5

0L4

EA5L4

EA5

K

L4EA7

L4EA70

L4EA7

L4EA7

L4EA5

L4EA5

0L4

EA5L4

EA5

K

3. Equilibrium equation:

−

−−

−

=

=

3

2

1

47

470

47

412

45

04

54

5

0

][

uuu

LEA

LEA

LEA

LEA

LEA

LEA

LEA

P

R

uKP

As the left hand side clamped so (u1=0) and therefore:

−

−=

3

2

uu

77712

L4EA

P0

EAPL4u7u

1249

P]u7u7[L4

EA

andu127u

0]u7u12[L4

EA

33

32

32

32

=+−

=+−

=⇒

=−

60

EAPL8.0u

EAPL371.1u

EAPL4u

1235

2

3

3

=

=

=

4. Stress in element:

[ ]

ε=σ

−

==ε

E

anduL1

L1uB e

0e0

For element (1)

[ ] [ ]

[ ]A

P8.0

EAPL8.0

011

LE

uu

11LEu11

LEE

2

11011

=

−=

−=−=ε=σ

For element (2)

[ ]

[ ]A

P57.0

EAPL371.1

EAPL8.0

11LE

uu

11LEE

3

222

=

−=

−=ε=σ

The exact solution

AP667.0

A5.1Pcentertheat

A2P

AP

2

1

==σ

=σ

=σ

61

Example-4: Find the stresses in the two bar assembly which is loaded with force

P, and constrained at the two ends, as shown in the figure.

Solution:

Use two 1-D bar elements.

62

Example-5: Determined the support reaction forces at the two ends of the bar

shown, if P=60 KN, E=20 GN/m2, A=250 mm2, L=150 mm, and

Δ=1.2mm.

S0lution:

Δo=PL/EA=(60*103*150)/(20*103*250)=1.8mm

Since Δo=1.8> Δ=1.2, so the contact of the bar

with the wall on the right will occur.

[ ]

[ ]

[ ]

−

−=

−

−=

−

−=

1111

150250*10*20K

1111

150250*10*20K

1111

LEAK

32

31

[ ]

−−−

−=

110121

011

150250*10*20K

3

63

−−−

−=

3

2

131

uuu

110121

011

150250*10*20

3FPF

Since left side is fixed so u1=0 and get:

mm5.12.1250*10*2010*60*150

21u

)u2(150

250*10*20P

u1112

150250*10*20

FP

3

3

2

2

3

23

3

=

+=

∆−=

∆

−

−=

To find the support reaction forces:

KN10)2.15.1(150

250*10*20)uu(150

250*10*20F

KN50)5.10(150

250*10*20)uu(150

250*10*20F

3

32

3

3

3

21

3

1

−=+−=+−=

−=−=−=

64

H.W

1. Consider a two degree of freedom bar elements as shown in figure. Using finite

element method to formulate the equilibrium equation of it. If the cross

sectional area is 12 mm2 and E=200 GN/m2.

2. Consider a two degree of freedom bar element as shown in figure. Using finite

element method to formulate the equilibrium equation of it, and then estimate

the stress distributions. If Esteel=200 GN/m2, ECopper= 110GN/m2 and EAL= 120

GN/m2.

65

CHAPTER FIVE

GEOMETRIC MODELING 5.1 INTRODUCTION

Geometric models in CAD/CAM can be based on a wire frame, surface

models or solid models.

5.2 WIRE FRAME MODEL

The wire frame model is represented by tables defining edges and points

of the shapes. The coordinates (x,y,z) of each point are stored in the points-table.

This model can not be used to determine surface area or volume of an object.

Mathematic description:

– A list of curve equations;

– Coordinates of points; and

– Connectivity information for shape curves and points.

Advantages:

1. Require simple user input;

2. Easy for users to develop systems by themselves.

• Disadvantages:

1. Ambiguous in shape;

2. No information about inside and outside boundary surface.

5.3 SURFACE MODEL

The surface model is represented by tables of edges and points as a wire

frame plus a table of faces. A record of faces stores edges from which the face is

consist.

66

This model can be used to determine surface area only but can not used to

determine volume or moment of inertia.

5.4 SOLID MODEL

The solid model is represented in either of the following ways:

1. Boundary representation of solid model: it has information on the faces,

edges and vertices plus topological information which defines the

relationships between the faces, edges and vertices. So the solid is

represented as closed space in 3-D space.

2. Constructive solid geometry: using Boolean operations, the solid object is

made from two intersecting solid primitives. The operations consist of

67

union, intersection and difference or subtraction. The basic 3-D primitives

are:

a- Block (box)

b- Cylinder

c- Sphere

d- Cone

e- Torus

f- Wedge

c- Sphere

d- Cone

a- block

b- cylinder

68

In which the Boolean operations are:

a. Union: combines two volumes into a single solid.

b. Intersect: keeps only the volume common to both solid objects.

c. Subtract: the volume of one solid object from the other solid.

e- torus

f- wedge

73

CHAPTER SIX

SYSTEM DESIGN AND MANUFACTURE

6.1 MANUFACTURING PRODUCTION CYCLE The manufacturing production cycle consist of many steps that:

1. Processing operation:

a. Basic process (metal casting, plastic molding).

b. Secondary process (milling, turning, drilling…).

c. Process to enhance physical properties (heat treatment, surface

treatment).

d. Finishing operation (polishing, painting).

2. Assembly operation:

a. Mechanical fastening (screw, rivet, nut…….).

b. Joining process (welding, soldering ……).

3. Material handling and storage:

a. Transporting between successive operations.

b. Transporting to final stage.

4. Control:

a. Inter stage control.

b. Final stage control.

74

Example-1: List the main operations required to produce the shape shown.

1. Cutting

2. Milling

3. Drilling

4. Grinding

5. Final control final storage assembly.

6.2 METHOD OF WORKPIECE TRANSPORT

1. Continuous transfer mechanism:

• Easy to design.

• High rate of production.

2. Intermitted transfer mechanism:

• Greater flexibility.

6.2.1 TYPE OF TRANSFER MECHANISM

There are two types of transfer mechanism that linear and rotary

mechanism:

1. Linear transfer mechanism:

a. Walking beam.

Piston

Table

Input-Linear movement. Output- Linear movement.

. . . .

I.C I.C I.C

75

b. Conveyor.

2. Rotary transfer mechanism:

a. Rack and pinion.

b. Geneva mechanism:

Indexing table

Rack Piston . Pinion

. +

Input-Rotary. Output-Linear.

76

timeCycle1RateoductionPr

360BAngleCAngle180BA

18090B21A

21

timeDwelltimeIndextimeCycle

=

=+

=+

=++

+=

o

o

o

Example-2: An index table driven by a Geneva mechanism has six stations with a

driver speed of 12rpm calculate: Index time, Dwell time, Cycle time

and production rate.

Solution:

.hr/721

36009.4

1timeCycle

1rateoductionPr

sec99.433.366.1timeDwelltimeIndextimeCycle

.sec33.3min0555.0121

360240timeDwell

CtimeDwell

.sec66.1min0277.0121

360120timeIndex

BtimeIndex240120360C

360BC12060180B

180BA

606

360AAngle

===

=+=+=

==×=

α

==×=

α=−=

=+

=−=

=+

==

o

o

o

o

o

77

6.3 AN AUTOMATION BLOCK BUILDING The basic component of automation as primarily belonging to one of the

following:

1. Sensors: sensors are first link between the typical automation system and

conventional process. The most sensors type are:

• Manual switches.

• Proximity switches: Some switch do not require physical contact or

light radiation to sense an object. Such switch is called proximity

switches.

• Photo electric sensors: In wider use than proximity switches are

sensors that are sensitive to light radiation. Two approaches for

employing photoelectric are in use:

a. The first approach merely uses a photocell to detect the presence of light

radiating naturally from some object in the process.

b. The second approach to photoelectric employs a beam of light emitted by

an artificial light source. The principal purpose of this approach is to

detect the presence or absence of objects in the path of the beam

• Infrared Sensors: Sometimes it is useful to detect electromagnetic

radiation outside the visible range. Infrared sensors respond to

radiation in the range of wave length just beyond the visible spectrum

at the red end. Infrared sensors are very useful when used with

artificial beams to detect the presence or absence of objects, even

more than Photo electric systems.

• Fiber optical: Convenient supplement to Photo electric or infrared

sensor systems are fiber optic light tubes, which are flexible pipes of

glass or plastic that can be used to bend light beams around corners.

78

Fiber optics are so efficient that it becomes worthwhile for the

telephone industry to convert communications circuits from

electrical signals to modulated light signals for transmission via

fiber optics and subsequent reconversion at the receiving end.

• Laser: In automated systems, the laser is useful in providing very

long, precise light beams. The precision of the beams makes them

excellent for detecting tiny objects that are capable of breaking the

beam at large and varying distances. Such precision also makes the

laser a good tool for dimensional measurement.

79

2. Analyzer: information is sensed by automated system it must be registed and

analyzed and then decision must be made by the system as to what action

should be taken. The analyzers type are:

• Computers: Digital computers are primary means of analyzing

automation system inputs. Computers are extremely versatile in that

the ways they can be programmed to manipulate data are limitless.

• Counters: It is frequently useful for an automated system to

determine how many of various items are present or pass through an

automated system. The counter can be mechanical, solid state

electronic counters.

• Timer: If pieces clock pulses are available, a counter that counts

these pulses becomes a timer basically a clock. An industrial timer is

more similar to an alarm clock than an ordinary clock. When the

elapsed time becomes equal to a preset value, an output signal is

generated.

• Bar Code Readers: is an analyzing system that incorporates a

conventional photoelectric or laser scanner along with timers and

counters.

• Optical Encoder: The capability of rapidly scanning. It’s a series of

bars make possible additional automation opportunities when light and

dark bars are placed concentric rings on a disk. A portion of such a

disk that can be rigidly attached to a shaft and housed in an assembly

consisting of optical sensors for each ring. The assembly is called an

optical encoder and is useful for automatically detecting shaft rotation.

80

Beam of light passes through slots in a disk and are detected by a light

sensor. When the disk is rotated, a pulsed output is produced by the

sensor, with the number of pulses being proportional to the angle

through which the disc rotates. Angular rotation of disc (shaft rotation)

can be determined by the number of pulses produced since some

reference position.

A= Angular value of ring (i).

mi= 0 If the ring is white.

mi= 1 If the ring is dark.

n= Total number of rings.

Example-3: An absolute optical encoder disk has eight rings and eight LED

sensors and turn to provide eight bit output. Suppose the output

pattern is [10010110], what is the angular position of the shaft?

Solution:

Encoder ring inner Angular value Observed pattern Computed value 1 180 1 180 2 90 0 0 3 45 0 0 4 22.5 1 22.5 5 11.25 0 0 6 5.625 1 5.625 7 2.8125 1 2.8125 8 1.40625 0 0

210.9375

∑=

=n

1iiiAmA

81

3. Actuators: The real world condition is sensed of analyzed some thing may

need to be done about it (Automated system are closing the loop by taking

physical action automatically without operator intervention). Some of the

actuators used are:

• Cylinder actuators: are defined by their ability to exert a linear force and

hold it at any specified position indefinitely. The cylinder actuators

divided into hydraulic and pneumatic actuators.

• Solenoids.

• Relay.

4. Drivers: It takes some action upon the process at the command of computer

or other analyzed and this action is taken by continuous movement typified by

rotation. There are many type of drivers used such as:

• Motors.

• DC servomotors.

• Stepper motors.

• Kinematic linkage.

• Walking beams.

• Geneva.

82

Sensors Analyzer Driver Automation components

Actuators

General manufacturing process

Operator Piece Process machine

83

CHAPTER SEVEN FUNDAMENTAL OF NUMERICAL CONTROL

7.1 INTRODUCTION The (NC) machine is one type of the programmable machine, controlled the

machine tool by a special program for each work piece required. The program

consists of (numbers, letters and etc) precise instructions about the methodology

of manufacture as well as the movement, for example, what tool to be used, at

what speed, at what feed to move from which point to which point in what path,

and then saved on the punched tape. The punched tape read by the control device.

When replaced the work piece required to product the program must be replaced

also, these style made the (NC) machine suitable for low and mid production.

7.2 BASIC COMPONENT OF (NC) SYSTEM The general structure of the operation of a typical numerical control system

is shown in figure below. The (NC) machine must be consisting of the following

components:

1. Program of instruction.

2. Control unit.

3. Machine tool.

The part program consist of instructions written in the numerical codes

constitute the basic operations to be carried out in machining of the part. These

instructions are then entered into an input medium such as a paper tap. The

program is then read by paper tape reader. The controller translates these

numerical codes into the machine actuation details, which are then used to control

the individual machine functions such as the movement of the axes.

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7.3 CLASSIFICATION OF NUMERICAL CONTROL

Numerical control (NC) machines can be classified as follows:

1. Control loops.

2. Motion control.

3. Power drives.

4. Hardwired (NC) and softwired computer numerical control (CNC).

5. Positioning systems.

1. CONTROL LOOPS OF (NC) SYSTEM

The control loops of (NC) system are designed to control the position and

velocity of the machine tool axis. There are two types of control systems in the

programmable machines; open loop and closed loop. The overall accuracy of the

machines is determined by the type of control loop used.

The open loop control system doesn’t positioning feedback to the control

unit. The movement pulses are sent out by the control unit and they are received

Input -punched tape -data carrier

Sensor Control unit -punch tape reader

Read data buffer signal CM -Table -Spindle -Motors -Control -Data display

85

by a special type of servomotor called a stepper motor. The number of pulses that

the control sends to the stepper motor controls the amount of the rotation of the

motor. The stepper motor then proceeds with the next movement command. Since

this control system only counts pulses and can’t identify discrepancies in

positioning, the control has no way of knowing that the tool not reaches the proper

location. The machine will continue this inaccuracy until somebody finds the

error. The advantage of the open loop control system is that it is less expensive,

since it doesn’t require the additional hardware and electronics needed for

positioning feedback. The disadvantage is difficulty of detecting a positioning

error.

In the closed loop control system the electronic movement pulses are sent

from the control to the servomotor, enabling the motor to rotate with each pulse.

The movement are detected and counted by a feedback deviance called

transducer.

With each step of movement, a transducer sends a signal back to the

control, which compares the current position of the driven axis with the

programmed position. When the number of pulses sent and received match, the

control starts sending out pulses for the next movement.

Drive unit Stepping motor

Machine table

Drive unit Stepping motor y-axis y-axis

x-axis x-axis

Command pulses

86

Example-1: Calculate the displacement in accuracy of (NC) system without

velocity feed back for the following motion:

N09 G01 X82.55 Y44.45 F100

Take the smallest programmable increment is (0.002) mm/pulses.

Solution:

Number of required pulses:

pulses22225002.0

45.44Y

pulses41275002.055.82X

pulses

pulses

==

==

Time intervals:

.sec/pulses33.83367.26

22225

.sec67.26min4445.0100

45.44Y

.sec/pulses33.83353.49

41275

.sec53.49min8255.0100

55.82X

time

time

=⇒

===

=⇒

===

Open loop NC machine Closed loop NC machine

87

X-error:

mm033.05129.8255.82mm5129.82002.041258

pulses4125883353.49

=−=×

=×

Y-error:

mm0178.04322.4445.44mm4322.44002.011.2216

pulses11.221683367.26

=−=×

=×

2. MOTION CONTROL OF (NC) SYSTEM

There are three types used to control the motion in (NC) system:

1. Point to point positioning control.

2. Straight-cut positioning control.

3. Continuous (contouring) path (NC) system.

The function of PTP motion control system is to move the machine table or

spindle to a specified position to perform the machining operation at that point.

The path taken to reach the specific point is not defined by the programmer in this

system. Because this movement is nonmachining, as shown in figure below there

are some paths that may be taken between the two point P and Q. a PTP-NC

machine is able to perform simple milling operations if the machine is equipped

with feed control mechanism. With most PTP machines, the only directions that

are accurately controlled are straight lines parallel to the machine axes (right and

left, forward and backward).

In continuous control system, the machine control two or more axes

simultaneously. The tool contacts the workpiece and the desired shapes are made

as shown.

88

PTP motion control Continuous path control

89

Example-2: Define the path of three holes in figure below.

The tool path may be Sequential (parallel to the axes) or Simultaneous (by 45˚) thus:

Path of 3-holes

Tool path motion

Sequential Simultaneous From To

X25Y35 0-a-1

0-c-1 0-b-1 0 1

X50Y-20 1-d-2

1-f-2 1-e-2 1 2

X20Y30 2-h-3

2-g-3 2-i-3 2 3

Y g 3 10 C 1 f i

20 b

d e 2 h 15

a 0 25 50 20 X

90

Mathematics for programming coordinate system

1. Cartesian coordinate system:

Point PT[x,y,z]

Line L[PT1,PT2]

Plane PL[PT1,PT2,PT3]

Example-3: using Cartesian coordinate to locate the point in the figure shown.

2. Polar coordinate system: Point [radius, angle] PT1[R,A] PT1[8.6,39˚] PT2[R,A,Z] PT2[8.6,39˚,30]

0 1 -10 4 -20 6 20 20

Z Y 2 35 3 5 X

PT0=[0,0,0]

PT1=[20,0,0]

PT2=[20,35,0]

PT3=[20,35,-10]

PT4=[20,0,-10]

PT5=[40,35,-10]

PT6=[40,0,-30]

Z-axis Y-Z Y-axis Plane X-Z Plane X-Y Plane X-axis

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3. POWER DRIVES

Most modern NC machines are use DC or AC servo motor to drive the

axes and the spindle. Their small size, ease of control, and low cost are

advantages of servo motor. In large machines hydraulic drives are used. Also

pneumatic drives are rarely used in NC positioning system.

4. NC and CNC

The difference between NC and CNC lies in the controller technology.

The NC machines developed in the early days had the total control system

developed using the hardware. This is sometimes called hardwired numerical

control. These are characterised by a part program input media such as

magnetic or paper tape. These had very little part program memory.

The new control systems are termed as computer numerical control

(CNC) which are characterised by the availability of computer and enhanced of

memory in the controller as shown in the figure.

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CNC machines have many advantages over conventional machines.

Some of them are:

1. There is a possibility of performing multiple operations on the same

machine in one setup.

2. Because of the possibility of simultaneous multi-axis tool movement,

special profile tools are not necessary to cut unusual part shapes.

3. The scrap is significantly reduced because of the precision of the CNC

machine and lesser operator impact.

4. It is easy to incorporate part design changes when CAD/CAM systems

are used.

5. It is easy to perform quality assurance by spot-check instead of

checking all parts.

6. Production is significantly increased.

CNC machines also have some disadvantages:

1. They are quite expensive.

2. They have to be programmed, set up, operated, and maintained highly

skilled personnel.

5. POSITIONING SYSTEMS

In general there are two methods to define the tool location:

1. Absolute: Tool location always defined in relation to zero point.

2. Increment: The next tool position must be defined with reference to the

previous tool location.

Note: In increment of the polar coordinate the angle only vary.

93

Modern NC machines also allow the user to choose the types of

positioning systems through software. In an absolute positioning system NC

machine, the coordinate origin also can be reset. Machines that allow their

coordinate origin to be reset are called floating zero machines; other called

fixed zero machines.

Example-4: Locate the position of the points shown using absolute and

increment.

Y 1 3 5 14

2 4 8

10 10 10 10 10 X

Point Absolute Increment

1 (10,22) (10,22)

2 (20,8) (10,-14)

3 (30,22) (10,14)

4 (40,8) (10,-14)

5 (50,22) (10,14)

94

Example-5: Using absolute and increment methods to locate the position of the

points shown.

Example-6: Using absolute and increment methods to locate the tool position of

the octagonal object with radius of 30.

2 1

3 8 30 4 7 5 6


1 (30,45˚) (30,45˚)

2 (30,90˚) (30,45˚)

3 (30,135˚) (30,45˚)

4 (30,18˚) (30,45˚)

5 (30,225˚) (30,45˚)

6 (30,270˚) (30,45˚)

7 (30,315˚) (30,45˚)

8 (30,360˚) (30,45˚)

Angle=360/8=45˚

Y 75˚

5 60˚ 4 3 30˚ 2 1 10˚ 40 20 X


1 (60,10˚) (60,10˚)

2 (60,30˚) (60,20˚)

3 (40,60˚) (40,30˚)

4 (40,75˚) (40,15˚)

5 (60,75˚) (60, 0˚)

95

CHAPTER EIGHT CNC MACHINES PART PROGRAMMING

8.1 INTRODUCTION

Since its introduction, CNC technology has found many applications,

including lathes and turning centers, milling machines and machining

centers, punches, electrical discharge machines, flame cutters, grinders, and

testing and inspection equipment. The most complex CNC machine tools are

the turning center and the machining center.

When preparing a program for a particular operation the programmer

must select all cutting data using recommendations for conventional

machining. This includes proper selection of cutting speeds, feed rates, tools

and tool geometry, and so on. When the programmer has chosen all of the

necessary information property, the operator loads the program into the

machine and presses a button to start the cutting cycle. The CNC machines

moves automatically from one machining operation to another, changing the

cutting tool and applying the coolant. In a surprisingly short time the work

piece is machined according to the highest quality standards.

8.2 AUTOMATIC TOOL CHANGER

Most of the time, several different cutting tools are used to produce

a part. The tools must be replaced quickly for the next machining operation.

For this reason, the majority of CNC machine tools are equipped with

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automatic tool changer ATC. For the automatic tool changer to operate, it is

necessary to have the following:

1. A tool magazine.

2. The tool adaptor that has a provision for pick-up the tool change arm.

3. The ability in the control to perform the tool change function.

4. Tool change procedure.

The tool magazines to be used have to be considered in terms of the

following attributes:

• Storage capacity.

• Type and shape.

• Tool change procedure.

Storage capacity typically starts with about 12 and can go as high as 200

while 30 to 60 appears to be the most common capacity of the tool

magazines. The simplest type of tool magazine is turret. This method

combines tool storage with the tool change procedure, without need for the

tool change arm. The turret simply indexes to bring the tool into the position

of machining, since the spindle is combined with the tool turret. It is a

simple method and the time is normally more for actual tool change.

The next type of tool magazine found in most of the machine tools with

lower tooling requirements is the drum or disc type magazine. The drum

rotates for the purpose of tool change to bring the required tool to the tool

change arm. For storing large number of tools, a chain type tool magazine

provides the necessary flexibility. The chain allows for a very large variety

of arrangements.

The tool change time range from as low as 2 to a maximum of 10

seconds.

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8.3 COORDINATE SYSTEMS In NC system, each axis of motion is equipped with a separately controlled

driving source. The relative movement between tools and workpiece is

achieved by the motion of the machine tool slides. The three main axes of

motion are referred to as the X, Y, and Z.

The Z-axis is perpendicular to both the X and Y axes in order to create

a right hand coordinate system. A positive motion in the Z direction moves

the cutting tool away from the workpiece.

Z-axis

1. On a workpiece-rotating machine, such as a lathe, the Z axis is

parallel to the spindle, and the positive motion moves the tool away

from the workpiece.

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2. On a tool-rotating machine, such as a drilling or milling machine, the

Z axis is parallel to the tool axis, and the positive motion moves the

tool away from the workpiece.

3. On other machines, such as a press, a planning machine, or shearing

machine, the Z axis is perpendicular to the tool set, and the positive

motion increases the distance between the tool and the workpiece.

X-Axis

1. On a lathe, the X axis is the direction of tool movement, and the

positive motion moves the tool away from the workpiece.

2. On a horizontal milling machine, the X axis is parallel to the table.

3. On a vertical rnilling machine, the positive X axis points to the right

when the programmer is facing the machine.

The Y axis is the axis left in a standard Cartesian coordinate system.

102

8.4 AN INTRODUCTION TO PART PROGRAMMING

There are two types of formatting language that the (APT) and

(G-function). The G-function is the main one, while the APT consider is the

secondary language. The G-function codes are classified into:

1. G Codes (Preparatory functions).

2. M Codes (Miscellaneous functions).

To write any program three steps must be apply that:

1. Partitioning the drawing into main parts.

2. List the coordinates of the points.

3. Write the program steps.

Any line programming consist of some functions and written as:

N10 G01 X14 Z26 F3 S150 T12 M8 EOB Block number Function

Dimensional deformation

Feed rate

Spindle speed

Tool number

Miscellaneous function

End of block

103

Prepartory functions This is genoted by “G”. it is pre-set function associated with the movement of machine axes and the associated geometry. It has two digits e.g., G01, G42, and G90 as per ISO pspecifications. However, some of the current-day controllers accept up to 3 or 4 digits. Here, we will only discuss some of the regular functions. ISO has standardised a number of these preparatory functions, also popularly called G code. The standardised code are:

Miscellaneous function, M

105

Miscellaneous function, M

107

Example-1: Using G-function to write the appropriate program for machining

the shape shown, using 10mm/min feed and 500 rpm and tool

number (1).

Solution:

%

01003

N1 G71

N2 G18 M3

N3 M8 G90 S500 F10 T1

N5 G0 X0 Z0

N6 G1 X40 Z0

N7 G1 X40 Z30

N8 G2 X30 Z40 I-10 J0

N9 G1 X0 Z40

N10 G1 X0 Z7

N11 G3 X7 Z0 I0 J-7

N12 M5

N13 M9

N14 M30

108

Example-2: The component shown to be machined. Write a program without

and with using canned cycles to drill all the holes.

Solution:

Without using canned cycles:

%

01301

N1 G71

N3 G00 X25 Y35 Z2

N4 G01 Z-18 F125

N5 G00 Z2

N6 X55 Y55

N7 G01 Z-18 F125

N8 G00 Z2

N9 X75 Y70

N10 G01 Z-18 F125

N11 G00 Z2

N12 X0 Y0 Z30

With using canned cycles:

%

01302

N1 G71 G17

N2 G81 X25 Y35 Z-18 R2 F125

N3 X55 Y55

N4 X75 Y70

N5 G80 X0 Y0 Z30

All dimension in mm

109

Example-3: Write the G-code steps required to obtain the motion statement for

machining the part shown in figure. Using absolute coordinate

system, F20 mm/min and S400 rpm.

Solution:

%

01003

N01 G71

N02 G17

N03 G90 F20 S400

N04 G0 X30 Y50

N05 G1 X80 Y50

N06 G1 X80 Y82

N07 G1 X120 Y82

N08 G1 X120 Y50

N09 G3 X120 Y122 I0 J36

N10 G1 X30 Y122

N11 G1 X30 Y50

N12 G0 X0 Y0

N13 M30

110

Example-4: A complete part program using the ISO codes for the external

contour of the following component shown in the figure. If the stock

dimension is (105*60*5) mm, end mill size is15 mm, feed

100mm/min and 800 rpm speed.

Solution:

%

01313

N10 G71 G92 X0 Y0 Z50

N15 G90 T01 S800 M06

N20 G00 X-7.5 Y-7.5 Z2

N25 G01 Z-6 F100

N30 X-7.5 Y62.5

N35 X62.5 62.5

N40 X62.5 Y42.5

N45 G02 X62.5 Y12.5 R-25

N50 G01 X62.5 Y-7.5

N55 X42.5 Y-7.5

N60 G02 X30.5 Y21 R15

N65 G03 X24.5 Y21 R-10

N70 Go2 X12.5 Y-7.5 R15

N75 G01 X-7.5 Y-7.5

N80 G00 Z2

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