GEOMETRIC ANALYSIS OF AXISYMMETRIC

DISK FORGING

A Thesis Presented to

The Faculty of the

Fritz J. and Dolores H. Russ College of Engineering and Technology

Ohio University

In Partial Fulfillment

of the Requirement for the Degree

Master of Science

by

Corey Bevan Raub

March, 2000

ACKNOWLEDGEMENT

The research that went into this project would not have been possible without the invaluable assistance of many persons. Their expertise and patience have given me the strength and motivation to remain committed through the completion of this work.

My thanks and appreciations go out to my advisor, Dr. Bhavin Mehta, who has helped lead me through my educational path and arranged for my funding on this project.

I would especially like to thank Daniel Allwine, whose devotion and sincere encouragement has given me the will power to bring this project to an end.

I would like to thank the other members of my thesis committee (Dr. Khairul Alam, and Dr. Thomas Scott) for their valuable comments and suggestions and to the Faculty and Staff of the College of Mechanical Engineering for their commitment to higher education.

Finally, I would like to thank my parents, Daniel and Linda Raub, for their love, support, and patience in the pursuit of my education. Without their continued support, none of this would have been possible. TABLE OF CONTENTS

Approval Page

Acknowledgments

List of Figures, Illustrations, and Graphs vii

1 INTRODUCTION

1.1 PROJECT DESCRIPTION AND SIGNIFICANCE

1.2 BACKGROUND

2 MANUFACTURING

2.1 FORGING

2.1.1 TYPES OF FORGING

2.1.2 MACHINERY

2.1.3 HEAT TREATMENTS

2.2 OTHER METAL FORMING PROCESSES

2.2.1 CASTING

2.2.2 EXTRUSION

2.2.3 ROLLING AND BENDING

2.2.4 MACHINING

3 GEOMETRIC ANALYSIS OF AXISSYMMETRIC DISK FORGING

3.1 AXISYMMETRIC DISKS

3.2 PROGRAM SIGNIFICANCE

3.3 PROGRAM CONTENTS

3.4 GEOMETRIC ANALYSIS 4 MUTUAL VOLUMETRIC DISCRETIZATION OF AXISSYMMETRIC DISK

FORGING

4.1 MUTUAL VOLUMETRIC DISCRETIZATION 38

4.2 DERIVATION OF DISK VOLUME CALCULATIONS 40

4.3 EXAMPLES OF VOLUMETRIC DISCRETIZATIONS 42

5 CONCLUSION 47

REFERENCES

APPENDIX A MATHEMATICAL DERIVATIONS

A. 1 DERIVATION OF THE PARALLEL LINE SEGMENT MINIMUM

DISTANCE FORMULA 54

A.2 DERIVATION OF THE PARTIAL DIFFERENTIATION MINIMUM

DISTANCE FORMULA 59

A.3 DERIVATION OF THE VOLUME FORMULAS USED TO COMPUTE

THE MUTUAL VOLUMETRIC DISCRETIZATION 62

APPENDIX B PROGRAM FILES

B. 1 PROGRAM 1 - PROFILE-FINAL.M

B.2 PROGRAM 2 - MlNIMlJM-OFFSET-PROF1LE.M

B.3 PROGRAM 3 - MUTUAL DISCRETIZATION GROUP

APPENDIX C SAMPLE OUTPUT AND RESULTS C.l MINIMUM DISTANCE OF SEGMENTS (PROGRAM 1)

C.2 MINIMUM DISTANCE OF SEGMENTS (PROGRAM 2)

C.3 MUTUAL VOLUMETRIC DISCRETIZATION

GLOSSARY

ABSTRACT vii

LIST OF FIGURES, ILLUSTRATIONS, AND GRAPHS

Figure 1.1 A young blacksmith uses a hammer and anvil.

Figure 2.1 Flow lines in a forged part.

Figure 2.2 Set of closed impression dies.

Figure 2.3 Smith forging hammer

Figure 2.4 Board drop hammer

Figure 2.5 Direct extrusion

Figure 2.6 Direct extrusion, (b) hollow and (c) semi-hollow cross-sections

Figure 2.7 Indirect extrusion, (top) solid, and (bottom) hollow cross-sections

Figure 2.8 Forward impact extrusion

Figure 2.9 Backward impact extrusion

Figure 2.10 Hydrostatic extrusion

Figure 2.1 1 Polymer Extrusion

Figure 2.1 2 Relative cost for manufacturing an aircraft part

Figure 3.1 Axisymmetric closed die forging

Figure 3.2 Configuration of a simple four ringed disk profile

Figure 3.3 Profile intersection illustration

Figure 3.4 Minimum offset illustration

Figure 3.5 Illustration of segment minimum distance orientation

Figure 3.6 Minimum distance coordinates for two parallel lines

Figure 4.1 Ring cross-section

Figure 4.2 Simple discretization example

Figure 4.3 A second simple discretization example ... Vlll

Figure 4.4 Multiple ring discretization example

Figure 4.5 The "reverse" discretization

Figure 4.6 Configuration of a simple four ringed disk profile

Figure Al.l Representation of the minimum distance between two parallel lines

Figure A3.1 Configuration of a simple four ringed disk profile

Figure A3.2 Ring cross section

Figure C1.l Diagram of Program 1 - Example 1

Figure C1.2 Diagram of Program 1 - Example 2

Figure C1.3 Diagram of Program 1 - Example 3

Figure C3.1 Diagram of Original Profiles - Example 1

Figure C3.2 Diagram of Discretized Profiles - Example 1

Figure C3.3 Diagram of Original Profiles - Example 2

Figure C3.4 Diagram of Discretized Profiles - Example 2 CHAPTER 1

--INTRODUCTION --

1 .l- PROJECT DESCRIPTION AND SIGNIFICANCE

This project is in conjunction with the Air Force Research Laboratory at Wright-

Patterson Air Force Base in Dayton, Ohio. The project deals with a "simulation and optimization based system for design of multi-stage material processes". Once finished, the system will be used as a manufacturing a001 that will "consider alternative manufacturing sequences and parameters." [25, pp. 2871 There will be uses for such a system in industry, as well as research facilities and academic institutions. In industry, this system will be beneficial in determining the best set of "manufacturing methods and parameters" for a given situation. This system will also aid in the education and training of an assortment of manufacturing processes. Researchers will be able to apply this system to newly defined methods or to help in the improvement of pre-existing methods of manufacturing analysis.

Material processes have been improved, during the past 20 years, due to advances in the manufacturing analysis tools. It is now possible to ~unextremely accurate process models with the advancement of computers and "powerful computational tools such as finite-element method (FEM)" that produce results acceptable by industrial analysts.

However, these tools require extensive knowledge of the software and of the subject being analyzed. For these reasons alone:

". . .New computational tools" need to be developed "that car1

directly assist the designer in making process design decisions such as

those involving selection of the sequence of manufacturing operations and

specification of processing conditions for each of these operations.

Material process design methods that consider alternative materials and

processes in order to optimize life-cycle cost should enable the production

of components that are lighter, stronger, environmentally compliant, and

less expensive than those achieved though conventional design

practices." [25, pp. 2871

This project is part of an ongoing development of a design environment that integrates models for materials and manufacturing processes, and allows for the selection and optimization of components such as those used in aircraft structures and engines. [25]

The forging of axisymrnetric turbine-engine disks similar to those used in aircraft engines is the focus of this effort. Such a system will "significantly improve productivity and reduce the costs of multistage materials processes such as hot-metal forging." [ll, pp. 11

It will also "allow the evaluation, with respect to quality, performance, and cost of alternate materials, processes, and process parameters for the affordable manufacturing of reliable components." [25, pp. 287-2881

This system is expected to serve as "a tool for the design of multistage materials processes that will allow consideration of alternate routes and parameters for manufacturing both military and commercial components." 124, pp. 11 The benefits of a system, such as this, that will help in the optimization of algorithms for manufacturing 5 are phenomenal. "Optimization algorithms that can vary the sequence and parameters or processing stages . . . will allow the evaluation of alternative manufacturing sequences and materials to obtain designs optimized for quality, perfomlance and cost." [25, pp.

2881 This system can also be applied to other areas of manufacturing; it is not limited to only the production of metal-formed products and such processes.

Current methods of engineering analysis are not directly associated with the idea of "design to cost" and optimization paradigms, and are more useful when considering the "analysis of complex material flow and the evaluation of design choices." A principle concept for this project on a whole was to develop a design tool that would make use of much simpler models. Several reasons were considered for this approach.

1. "Simpler models can predict the most important characteristics of the

material behavior during processing in a small fraction of the time taken

by a Finite Element Analysis (FEA) run."

2. "They can be connected together to form simulations of multistage

processes."

3. "These connected simulations can be controlled by an optimization

algorithm that can change parameters of the whole sequence of operations

according to predetermined optimality criteria in order to obtain the best

combination of processes and process parameters that satisfy desired

product quality, performance, and cost requirements." [25, pp. 2891

The approach taken by this type of design tool is primarily aimed at the

"conceptual design of processes with potential for becoming a routine tool for problem solving and training." As mentioned, this system can also be applied to other areas of 4 manufacturing. In time, other processes using different materials will be developed that would benefit from such a system. "Emphasis is . . . " being ". ,. placed on the ability to evaluate alternative manufacturing strategies instead . ." of ". . . highly sophisticated simulations of a single manufacturing operation." [25, pp. 2921 A design tool such as this will help "predict the cost and properties of a component or system early in the product- realization process, when most of the cost is decided." [24, pp. 11

1.2 - BACKGROUND

Forging is a metal forming process used to produce large quantities of identical parts, as in the manufacture of automobiles, and to improve the mechanical properties of the metal being forged, as in aerospace parts or military equipment. The design of forged parts is limited when undercuts or cored sections are required. All cavities must be comparatively straight and largest at the mouth, so that the forging die may be withdrawn. [34, pp. 1621 The products of forging may be tiny or massive and can be made of steel (automobile axles), brass (water valves), tungsten (rocket nozzles), aluminum (aircraft structural members), or any other metal. More than two thirds of forging in the United States is concentrated in four general areas: 30 percent in the aerospace industry, 20 percent in automotive and truck manufacture, 10 percent in off- highway vehicles, and 10 percent in military equipment. [7] This process is also used for coining, but with slow continuous pushes. [12, pp. 591

The forging metal forming process has been practiced since the Bronze Age.

Hammering metal by hand can be dated back over 4000 years ago (See Figure 1.1). The 5 purpose, as it still is today, was to change the shape andlor properties of metal into useful tools. Steel was hammered into shape and used mostly for carpentry and farming tools.

An ax made easy work of cutting down trees and metal knives were much more efficient

than stone cutting tools. Hunters used metal-pointed spears and arrows to catch prey.

Blacksmiths used a forge and anvil to create many usefbl instruments such as horseshoes,

nails, wagon tires, and chains. Militaries used forged weapons to equip their armies,

resulting in many territories being won and lost with the use and strength of these

weapons. [35,pp. 431

Inc. Atl fights Resenred.

Figure 1.1 A young blacksmith uses a hammer and anvil. [7] CHAPTER 2

MANUFACTURING

2.1 - FORGING

"Forging changes the size and shape, but not the volume, of a part. The change is made by force applied to the material so that it stretches beyond the yield point. The

force must be strong enough to make the material deform. It must not be so strong,

however, that it destroys the material. The yield point is reached when the material will

reform into a new shape. The point at which the material would be destroyed is called

the fracture point." [35, pp. 431

In forging, a block of metal is deformed under impact or pressure to form the

desired shape. Cold forging, in which the metal is not heated, is generally limited to

relatively soft metals. Most metals are hot forged; for example, steel is forged at

temperatures between 2,100°F and 2,300°F (1,l 50°C to 1,260°C). These temperatures

cause deformation, in which the grains of the metal elongate and assume a fibrous

structure of increased strength along the direction of flow. Normally this results in

metallurgical soundness and improved mechanical properties. Strength, toughness, and

general durability depend upon the way the grain is placed. Forgings are somewhat

stronger and more ductile along the grain structure than across it. The feature of greatest

importance is that along the grain structure there is a greater ability to resist shock, wear,

and impact than across the grain. [34] Material properties also depend on the heat-treating

process after forging. Slow cooling in air may normalize workpieces, or they can be

quenched in oil and then tempered or reheated to achieve the desired mechanical 7 properties and to relieve any internal stresses. [7] Good forging practice makes it possible to control the flow pattern (See Figure 2-1) resulting in maximum strength of the material and the least chances of fatigue failure. These characteristics of forging, as well as fewer flaws and hidden defects, make it more desirable than some other operations (i.e. casting) for products that will undergo high stresses.

Figure 2.1 Flow lines in a forged part. [26, pp. 1291

In forging, the dimensional tolerances that can be held vary based on the size of the workpiece. The process is capable of producing shapes of 0.5 to >50.0 cm in

thickness and 10 to 400 crn in diameter. [17, pp. 161 The tolerances vary from +_ '13~in. for small parts to & ?4 in. for large forgings. Tolerances of 0.010 in. have been held in some precision forgings, but the cost associated with such precision is only justified in exceptional cases, such as some aircraft work. [9, pp. 2561 2.1.1 - TYPES OF FORGING

Forging is divided into three main methods: hammer, press, and rolled types. The sole principle is that of shaping metal without changing the volume. The particular method depends on the size of the workpiece, design requirements, and production requirements. [31, pp. 1571 A billet, or other bar stock, is used in the operation and the forgings are produced at elevated temperatures. Forging consists of three distinct classifications.

1. Hammer Forging (Flat Die) - Preferred method for individual forgings. The

shaping of a metal, or other material, by an instantaneous application of

pressure to a relatively small area. A hammer or ram, delivering intermittent

blows to the section to be forged, applies this pressure. The hammer is

dropped from its maximum height, usually raised by steam or air pressure.

(See Figure 2.2 and 2.3 for examples of hammer forging equipment)

2. Press Forging - This process is similar to kneading, where a slow continuous

pressure is applied to the area to be forged. The pressure will extend deep into

the material and can be completed either cold or hot. A cold press forging is

used on a thin, annealed material, and a hot press forging is done on large

work such as armor plating, locomotives and heavy machinery.

3. Die Forging - Open and closed die operations can be used in forging. In

open-die forging the dies are either flat or rounded. Large forgings can be

formed by successive applications of force on different parts of the material. 9 Hydraulic presses and forging machines are both employed in closed die

forging. In closed-die forging (See Figure 2.2) the metal is trapped in

recessed impressions, which are machined into the top and bottom dies. As

the dies press together, the material is forced to fill the impressions. Flash, or

excess metal, is squeezed out between the dies. Closed-die forging can

produce parts with more complex shapes than open-die forging.

Copyright O 1994. 1995, 1996, 1997 The Learning Company. Inc. All hghts Resaved.

Figure 2.2 Set of closed impression dies. [7]

In all three cases, sufficient reduction must be performed to produce a high quality product, both sound and dense in structure, and to cause adequate grain flow. [34, pp. 411 An excessive amount of reduction will result in poor traverse mechanical properties. Parts will commonly require multiple intermediate reductions to ensure the proper grain flow.

Some forging methods have definite advantages versus others. Hammer forging can produce a wide variety of shapes and sizes and, if sufficiently reduced, can create a high degree of grain refinement at the same time. The disadvantage to this process is that 10 finish machining is often required, as close dimensional tolerances cannot be obtained.

Press Forging is more economical than hammer forging (except when dealing with low production numbers), and closer tolerances can be obtained. A greater proportion of the work done is transmitted to the workpiece, differing from that of the hammer forging operation, where much of the work is absorbed by the machine and foundation. This method can also be used to produce larger forgings, as there is no limitation in the size of the machine.

Die forging is the best method, as far as tolerances that can be met, and also results in a finished part that is completely filled out and is produced with the least amount of flashing. The final shape and the improvement in metallurgical properties are dependent on the skill of the operator. Closer dimensional tolerances can be held with closed die forgings than with open die forgings and the operator requires less skill. [9]

2.1.2 - MACHINERY

The type of machinery to be used depends on the shape, size, material, and number of pieces to be made. Forging hammers apply force by the impact of a large ram.

This may be a drop hammer, or weight falling under the force of gravity, or it may be a power hammer, driven by steam or compressed air. 'Two types of power hammers are:

the smith forging hammer [26, pp 1301 and the drop hammer [26, pp. 1311 (See Figures

2.3 and 2.4). Heavy workpieces could be processed using a smith-forging hammer, and

smaller forgings are die formed in drop hammers. The largest hammers can provide a

total force as high as 80,000 pounds. -. Figure 2.3 Smith forging hammer Figure 2.4 Board drop hammer

Smith forging hammers are typically steam or air-operated, consisting of a power actuated ram supported by a heavy cast iron frame. The final product is a result of the ram being powered into the dies containing the workpiece. A drop hammer differs in that the anvil is an integrated part of the hammer base. It is necessary for the alignment between the forging die elements used. [26, pp. 1321 This method "is advantageous in that the physical properties of the metal are improved by the severe mechanical working, the operation is rapid, many complicated parts can be forged to shape, a minimum amount of machning is necessary, and internal defects are eliminated." [5, pp. 2011 The disadvantages are the cost of machinery and dies, which dernands a high quantity of parts to be manufactured in order for the process to be cost effective.

A forging press consists of a hydraulic press, which exerts a force capable of pressing steel or a metal alloy into the shape of the forging die. These machines can be 12 positioned horizontally or vertically. This method can be used to form car wheels, gears, bushings, and other such parts.

"Mechanical presses have a motor-driven flywheel that stores

energy to drive a ram--much lighter than a hammer--through a crank or

other mechanical device. The ram in a press moves more slowly than a

hammer and squeezes the workpiece. The largest mechanical presses have

a total force of 12,000 tons and cannot forge as large or complicated parts

as the larger hammers.

Hydraulic presses, in whch high-pressure fluid produced by

hydraulic pumps drives a ram, are about 100 times slower than hammers.

They are used for large or complex die forgings and for extrusion. Presses

with a total force of 50,000 tons have been developed in the United States

primarily for the forging of large airplane components. Even larger

hydraulic presses, up to 78,000 tons, have been introduced in Europe." [7]

2.1.3 - HEAT TREATMENTS

Materials can be improved before or after manufacturing by different heat

treatment processes. Forging is usually performed to hot metals, allowing for smoather

flow and easier deformation. Steel is heated to varying temperatures, usually between

1700°F to 2000°F but can reach as high as 2400°F, depending on the carbon content.

Depending on the amount of work required to the piece, it may be necessary to reheat the

piece one or more times. [23, pp. 91 The temperature of the metal when completely 13 forged is called the finishing temperature. After forging, the material must be cooled uniformly and protected from moisture or cold air. This is done by placing the material into "dry ashes, lime or mica dust in order to retard the rate of cooling." [20, pp. 4111

Preheating of materials is done to help prevent cracking or distortion of the material. This is done by placing the metal in a series of furnaces of increasing temperatures instead of throwing it directly into the furnace used to heat the metal "for

forging, annealing, normalizing or hardening". [20, pp. 421 Another way to achieve this is

to start in a cold haceand slowly bring it to temperature.

If necessary, other operations should follow forging as soon as possible such as

annealing whenever machining is required. Annealing is the heating and then cooling of

metal to make the metal less brittle, or more malleable and ductile. This will soften the

steel that was previously hardened and reduce internal stresses. Amealing is done by

heating the metal to a temperature beyond the critical temperature and holding it there for

a period of time. The metal is then cooled with the furnace and not removed until the

furnace is cold. It can also be cooled to a temperature withln the furnace that is known to

be below the lower critical temperature, at which the annealing is complete. Slower

cooling rates are required as carbon content increases in the metal.

Another method of heat treatment is normalizing. This is done to improve the

crystalline structure of the steel, thus obtaining superior properties. Heating the forged

part just beyond the critical temperature and then allowing it to air-cool completes

normalizing. This allows the grain-size to be refined and, if not held at that temperature

too long, will result in a newly formed crystalline structure. The internal stresses, if any, 14 will be relieved, hardened steels will be softened, overheated steels will have a more

favorable, normal fine-grained structure, and structural distortion will be removed.

Hardening of steels can also be done after forging. The workpiece is heated

slowly, to obtain the finest grain-sizes, to its hardening temperature - much higher than

annealing temperatures. The metal is kept at this temperature only until "uniform heat

distribution and completion of the thermal transformation." [20, pp. 481 Prolonged

exposure at these elevated temperatures will result in increased grain growtli and surface

decarbonization, if no protection from oxidation is provided. Oxidation can be avoided

by surrounding the metal with some material that will use up the oxygen that is present in

the furnace. Once the metal has been uniformly heated to temperature, it is removed

from the furnace and placed directly into a quenching tank. This rapidly cools the metal

and the metal retains its new qualities. 2.2 - OTHER METAL FORMING PROCESSES

Forging is just one of many types of metal forming. Some other methods include casting, extrusion, rolling, machining, etc.

2.2.1 - CASTING

Casting is a forming process that involves a liquid or viscous material being poured into a prepared mold or form. The material is given time to solidify and the

object results in the shape of the mold. Castings are often cheaper than forgings,

depending on the quantity, type of material, and cost of patterns as compared to the cost

of the dies for forging. [9].

2.2.2 - EXTRUSION

Extrusion is a compressive metal forming process in which a billet (a plastic

material) is forced through a die orifice, thus reducing its cross-sectional shape to that of

the die orifice. This process was developed around 1800 in England. During this period,

England was the dominant leader in the Industrial Revolution. The invention of the

extrusion process consisted of the first hydraulic press, used with lead, and was improved

by the Germans in 1890 with the development of the horizontal extrusion press, used

with a variety of metals. 16 Steel and other metals may also be formed into shapes by forcing, or extruding,

the hot or cold metal into dies. Jet-engine rings, watchcases, and turbine blades are produced by the extrusion method. Tremendous pressure is exerted by machines to force

the steel into the dies. [7]

Rods and bars can also be formed by extrusion. Here the metal is trapped in a

cylinder and forced out through an opening in a die, much like toothpaste from a tube.

More complex cross sections, like Z-shaped or H-shaped bars, are readily formed by

extrusion. [7]

This process is not limited only to metals. Extrusion can be applied to polymers,

ceramics, and other composite materials. It is used with polymers, particularly for

thermoplastics and elastomers, to produce tubing, pipes, hose, structural shapes (such as

window and door molding), sheet and film, continuous filaments, and coated electrical

wire and cable. [13] The end product, or extrudate, is cut into desired lengths to meet

specific requirements. This is also the case for ceramics. Long uniform cross sections

are formed and cut to desired lengths. These sections are widely used as hollow bricks,

shaped tiles, drain pipes, tubes and insulators. Some are also used as starting slugs for

other ceramic processing methods, such as jiggering and plastic pressing. Jiggering is a

manufacturing process used to produce products similar to those on a pottery wheel.

Products such as bowls and plates are mass-produced by compressing ceramic slugs. 1131

There are various methods of extrusion. Four particular cases are direct, indirect,

impact, and hydrostatic extrusion. Direct extrusion, or forward extrusion, involves a

metal billet being inserted into a container and a ram is used to force the billet through a

reduced cross-section. This process leaves a small portion of the billet in the container, 17 which cannot be pressed through the die. This remaining material, or butt, is cut fi-om the extrudate. Although this process is simple, high frictional forces are generated between the container walls and the surface of the billet. This results in a very high ram force required to complete the operation. Complex geometries are possible with this application resulting in hollow and semi-hollow cross-sections. This type of extrusion is shown in Figures 2.5 and 2.6.

Work Blllet /

F+ d Final Work ) /I 1 t

~le 1container

Figure 2.5 Direct extrusion

Work Billet

QB)

Figure 2.6 Direct extrusion, (b) hollow and (c) semi-hollow cross-sections

Indirect Extrusion, or backward extrusion, is unique in that the billet moves relative to the die and not the container. A compressive force pushes the die and hollow 18 ram into the billet. The material is extruded through the ram. This results in no fhctional forces on the container walls, since the billet is not moving relative to the container.

Hence, indirect extrusion has much lower ram forces. A drawback to this operation is the

lower strength of the ram, and the awkwardness of supporting the extrudate after it exits the die. Cross-sections Similar to those created by direct extrusion can be produced by

indirect extrusion. Figure 2.7 illustrates indirect extrusions.

Impact extrusion is another method of extrusion. Ths operation is performed at

higher speeds and shorter strokes. This process produces individual parts by impacting a

slug of material and deforming it around the die, or, if using backwards impact extrusion,

around the punch. Since this process is performed at high speeds, material can undergo

large reductions in area and there is the possibility of high production rates. [33] An

example of both forward impact extrusion and backward impact extrusion is shown in

Figures 2.8 and 2.9.

Figure 2.7 Indirect extrusion, (top) solid, and (bottom) hollow cross-sections Die,

Starting Blank Extruded Part

Figure 2.8 Forward impact extrusion

Starting Blank

Figure 2.9 Backward impact extrusion

Hydrostatic extrusion is very similar to direct extrusion. However, to minimize the amount of friction between the billet and container, the billet is surrounded with a fluid. The ram moving forward pressurizes this fluid and the resulting ram force is significantly lower than with direct extrusion. The pressure on the surface of the billet actually increases the materials ductility. This property allows for brittle materials to be 2 0 extruded and ductile materials to undergo high reductions. One drawback with this operation is that the billet must be originally tapered to create a seal with the die, allowing the container to be pressurized. Figure 2.10 illustrates this procedure.

Container I

Work Billet 1

Figure 2-10. Hydrostatic extrusion

The extrusion of polymers differs from the extrusion of metals. The "billet" material is not in the form of bar stock, but rather in a pellet or powder form. This substance is added to a machine that utilizes a screw-type action to assist in mixing and pushes the plastic material through the die opening. Figure 2.1 1 shows a representative piece of equipment.

EXTRUSION MOLDING (cutawlay view} .. I ,. nozzle or die

I- motor conveyor - Copyright (c) 1994, 1995 Compton's NewMedia, Inc. AU Rights Reserved

Figure 2.1 1 Polymer extrusion [7] The polymer is fed onto the rotating screw, heated, and mixed. The screw is made up of three sections:

1.Feed Section - the stock is moved from the hopper and preheated.

2.Compression Section - the polymer is transformed into a liquid

consistency, air entrapped among the pellets is extracted fi-om the melt, and the

material is compressed.

3.Metering Section - the melt is homogenized and sufficient pressure is

developed to pump it through the die opening.

As the melt reaches the die, it must pass through a screen pack, a series of wire meshes supported by a stiff plate (called a breaker plate) containing small axial holes. The screen pack functions to:

1.Filter the contaminants and hard lumps fi-om the melt,

2.Build pressure in the metering section, and

3.Straighten the flow of the polymer melt and remove its memory of the

circular motion imposed by the screw. [13]

This melted polymer is finally forced through a tapered dielnozzle and the extrudate is formed.

Extrusion is beneficial for many reasons. It can be used to create a variety of shapes, the grain structure and strength properties are enhanced, close tolerances can be held, continuous lengths are produced, and as is often the case, there is only minimal waste. Extrusion performed at high speeds is even more beneficial, resulting in good surface finish, product straightness, good flow of metal, materials maintain mechanical properties, and close dimensional tolerances are held. [6] The major limitation with extrusion is that the final cross-sectional area is the same throughout the entire length of product. This may require that additional machining be done to meet design criteria. Product defects are another concern with extrusion, as with any metal forming operation.

Some common defects are:

1. Centerburst - a defect that propagates from an internal crack in the

billet as a result in tensile stresses along the centerline of the

workpiece.

2. Piping - the formation of a sink in the end of a billet.

3. Surface cracking - this is a result of high workpiece temperatures,

extremely high extrusion rates, and high friction.

Extrusion is an excellent way of producing identical products of equal cross-sectional area. The advancement of technology will only better the process. Finite clement packages already allow simulations to replace expensive testing. Extrusion is often cheaper than forging for larger quantities of equal parts. This is primarily due to a forged part requiring more machining than an identical extruded part. There are, however, reduced tooling costs associated with forging compared to that of extrusion. Therefore, for fewer parts, the tooling costs for forging prove to be less expensive than that of extrusion. [9] See figure 2-12 for a relative cost analysis comparison. METHOD (1) MACHINED FROM BARSTOCK (2) FORGED AND EXTRUDED (3) COLD EXTRUDED AND MACHINED

NUMBER OF PIECES

Figure 2.12 Relative cost for manufacturing an aircraft part [9, pp. 2621

2.2.3 - ROLLING AND BENDING

Rolling and similarly bending are both used to shape, or curve, flat sheet metal and straight extrusions. Metals are used based on their modulus of elasticity, E. This is a measure of how much a material can be stretched and deformed without craclung.

During the deformation process a solid material can experience three phases changes: the elastic phase, the plastic phase, and the rupture phase. The elastic phase is the period when the material will return to its original shape after a load is removed. The plastic phase is when a material that is acted upon by a force will retain its newly formed

shape when unloaded. The rupture phase is when a material reaches its failure strength

and cracks. This phase is rarely pursued in metal forming. [12, pp. 501 24 Exerting a force and a counter force to a given material completes this metal forming process. The bending or reshaping of the material is done by a third pressure point, called a fulcrum or die, which controls how the material will flow through a machine. This method of metal forming is merely a reshaping operation and is often used in conjunction with a previously extruded object, such as metal tubing or flat bar stock.

2.2.4 - MACHINING

Machining is a finishing operation. This can be used to complete a cast part, such as adding a smooth surface finish to the faces on vise jaws, or to finish an extruded object, such as turning threads on a screw. Milling machines and lathes are the two most common pieces of equipment used for metal machining. Other equipment, includes grinding machines, electrical discharge machines (EDM), electro-chemical machines

(ECM), and many others. "Machining is one of the most expensive metal fabrication processes owing to the substantial setup time and expertise required of machine tool operators. Machined parts should be specified only when their primary attribute- precision-is required." [12, pp. 671 CHAPTER 3

GEOMETRIC ANALYSIS OF AXISYMMETRIC DISK FORGING

3.1 - AXISYMMETRIC DISKS

An axisymmetric forging model was the primary focus of this project. This type of disk forging is carried out in a closed die forging press, as seen in the Figure 3.1.

Figure 3.1 Axisyrnmetric closed die forging 2 6 For this project, no disks with curved surfaces can be analyzed. All disks should be considered as being composed as a series of concentric rings each with quadrilateral

cross-section. The "outer" vertices (in a radial sense) of a particular ring are used as the

"inner" vertices (again in a radial sense) of the subsequent ring. The inner and outer

boundaries for each ring must be vertical. The top and bottom boundaries may have any

slope. Graphically, Figure 3.2 shows a half cross-section of a simple disk (four rings

with a hole in the center, axially symmetric about the z-axis).

Figure 3.2 Configuration of a simple four ringed disk profile 3.2 - PROGRAM SIGNIFICANCE

This program analyzes the minimum offset between two axisymmetric profiles to determine whether or not the profiles developed by the system optimizer feasible and within tolerance. Rules associated with the intersection of such profiles are the topic of further research. The discussed techniques are valid for any axisynmetric profiles and are applicable to all types of forging.

The program deals with "a new process design method for controlling microstructures and mechanical properties through the optimization of preform and die shapes. The approach involves the determination of optimal shape trajectories for material deformation and the use of these ideal trajectories in the design of perform and die shapes." Models such as the one presented here will be combined to perform the

"simulation of a metalforming process similar to those used for the manufacturing of turbine disks." This software package will help "compare and evaluate.. .different manufacturing alternatives" for the production of such parts. "The comparison will be based on the total materials and processing cost" of such parts. [25, pp. 2901

3.3 - PROGRAM CONTENTS

One of the primary goals of this project was to develop a program to determine line segment intersection between two cross-section profiles (See Figure 3.3). Figure 3.3 Profile intersection illustration

A second objective was to determine the minimum offset between the two profiles (See

Figure 3.4).

Figure 3.4 Minimum offset illustration

Matlab0 was used to develop the software necessary to complete these objectives. A program was first developed to determine the desired objectives between two line segments. Coordinates for the line segments were selected at random to reduce errors only associated with particular cases (i.e. two parallel lines, two perpendicular lines, etc.).

Data files containing the profile coordinates to be used by the system model will come ffom parameterized model geometries or from the software directly as changes are made by the optimizer.

The solution that was developed involves a combination of a partial derivative minimum distance method and geometric analysis to determine whether the lines 2 9 intersect. This was expanded on to account for two profiles of n-segments each. Two programs were developed that solve the problem.

The first program plots the two profiles and shows the minimum distance by plotting a line between the two profiles at the minimum offset. It does this by comparing all segments of the outer profile with each segment of the inner profile to determine the offset distances. A matrix of offset values is generated along with the coordinates of the respective minimum distance segment. The program outputs the absolute minimum

offset value or terminates if the profiles intersect at any point.

The second program produces much simpler results since it only outputs the

absolute minimum offset. There is no plotting of the profiles. Values for the mini~num

offsets are still stored in matrix form, but the coordinates of the respective minimum

distance segments are neglected. This program runs slightly faster and is preferred if

only the output is desired. This program also terminates if the profiles intersect at any

point.

3.4 - GEOMETRIC ANALYSIS

The first attempt was to solve this problem using only geometric analysis of the

two profiles. The first program is computed this way and the second is a combination of

this and also partial differentiation to help find the minima. These programs apply to 2-D

cases only. The code for these programs can be seen in Appendix B, sections B.l and

B.2. 30 The first step in solving this problem was to determine the slope of the segments.

This is calculated as:

where m is the slope, and x and y are the coordinates of the line segment endpinis. The y-intercept also needed to be calculated in order to determine line intersection. This if evaluated as:

b=y-mex, (3.02) where b is the y-intercept. Flags were set to determine if the segments were vertical or horizontal. This was done to simplify the minimum distance calculation m-d to test if the segments intersect or were perpendicular. If the segments were both vertical the minimum distance is simply the difference between the x-coordinates. If the segments were both horizontal the minimum distance is simply the difference between the y- coordinates.

Any two lines that are not parallel intersect at one point. The method used to compute line intersection was to check if the segments contained this collinear point of intersection. This was done by determining the point at which the two lines would intersect and then running a check to see if the segments contained that point. The following shows how this was derived.

y,= m, +b,,

y2= m2.x2 +b2, 3 1 where y, and y~ are the equations of the lines containing segments 1 and 2, respec,tively.

At the point of intersection, the x-coordinates and y-coordinates of the two equations are equal, or

m, ex, +b, =m2 ex2 +b2. (3.05)

m, *x+bl =m2ax+b2. (3.06)

Solving the above equation for x and simplifying gives the x-coordinate of the point of intersection (xint),

Substituting this value into the equations of the lines, (3.06), gives the vaiue of the y- coordinate of the point of intersection (yint),

The coordinates of the point of intersection (xintand yint)were then checked to see if they were contained within both line segments. If the point of intersection was contained within the range of coordinates of both segments 1 and 2, then the segments intersect. If the segments do not intersect, a minimum distance between the two segments was then calculated.

The minimum distance between any two segments that do not intersect occurs at the endpoints of at least one of the segments. The programs construct line segments representing the minimum distance. This minimum distance, defined in Figure 3.5, is either the distance between the endpoint of one segment to the endpoint of a second 32 segment (endpoint distances - 2,5,7,8), or the distance from the endpoints of one segment and orthogonal to the second segment (orthogonal distances - 1,3,4,6).

- .. -

. .. . PUINT ElF ., INTERSCCTIOPI 7 . . 'i. . i ,ORTHOGONAL DILTA$-EC'- , .

,

/ . / . - - .

Figure 3.5 Illustration of segment minimum distance orientation

The first program computes each of the eight minimum distance possibilities.

(For thls project, the minimum distance had to occur such that the endpoints of the minimum distance segment were contained within the endpoints of the line segments.)

These are then checked to determine which satisfies the stated criteria for particular set of segments. The first check determined if the endpoints of the four orthogonal minimum distance segments fell between the endpoints of the two line segments. If this condition was satisfied, then the respective minimum distance values were stored and compared with each other to determine an absolute minimum. This value was compared to the absolute minimum of the endpoint minima and a value for the overall minimum distance of the two segments was stored. An example is explained below: 33 As seen in Figure 3.5, the distance segment, labeled 1, would not satisfy the criteria explained because its "orthogonal endpoint" is not within the range of segment coordinates on that line. Therefore, the actual minimum distance for this case would be one of the next two distance segments, 2 or 3, both of which satisfy the mentioned criteria.

The distances were calculated using the standard equation for finding the distance between any two points in a plane.

where d is the distance. The coordinates of the orthogonal minimum distance segments were found using the known equations of the lines along with characteristics of perpendicular lines (i.e. mL=-llm).

This calculation was simplified when cases arose where at least one segment was either vertical or horizontal. If both segments are vertical, the minimum distance is simply the difference between the x-coordinates,

d = lx, - x, 1. (3.10)

If both segments are horizontal, the minimum distance is the difference between the y- coordinates, or

A special case arose if the segments were collinear and did not intersect. In this case the minimum distance is the difference between the nearest endpoints. y-axis

,. / ,/ ,. ,r

.' .' C ,/ / / , x-axis-

Figure 3.6 Minimum distance coordinates for two parallel lines

If both segments are parallel and neither vertical nor horizontal, the minimum distance is found using a slightly different method. The following method only applies if the segments are arranged similar to those in Figure 3.6. If the endpoints of the minimum distance segment are not contained within segments 1 and 2, then the minimum distance is calculated using the standard distance equation. The minimum distance, for cases similar to Figure 3.6, is calculated as the distance from one endpoint of either of the two segments and perpendicular to the other segment. The coordinates of the minimum distance segment used in deriving this formula were from point (xll, yll) of segment 1 to a new point (xo, yo) of segment 2.

These new coordinates are defined as: 3 5 where bZ is the y-intercept of line 2 and m is the slope of both lines. Using these coordinates in the distance equation,

d = ,/(XI, -xd2 +(YI, 9 yields the equation used to find the minimum distance, or

See Appendix A. 1 for a complete derivation of this distance calculation.

Once the minimum distance was found for the first segment, the program continued this same operation for each segment of the outer profile with respect to each segment of the inner profile (See Figure 3.4 for an example). The program outputs the minimum offset value and a graph of the two profiles showing where the minimum offset occurs. All of the minimum distance values are stored in a matrix along with the coordinates where the minimum distances occur.

The second program differs only slightly in that it computes the minimum distance between each set of line segments using a combination of the methods used in the first program and a newly derived partial differentiation method. This method works by evaluating the distances between a fixed point on one segment and the entire range of coordinates of the second segment. The minimum distance occurs when the partial derivative of the square of the distance equation, with respect to the x-coordinates, is equal to zero. The following is a summary of the derivation of this partial differentiation method. For a complete derivation, see Appendix A.2.

For two arbitrary lines 11 and l2 given by lI 3 y, = m, x, + b,, and

the expression for the distance between two arbitrary points (one on each line) is

This expression can be re-written as:

(1 +m; + (1 + m:)x; -2(l+ m,m2)xlx,... D=[ + 2m, (b, - b2)x1- 2m2(b, - b2)x2+ (b, - b2Y

The minimum distance (Dmin)can also be represented by the following expression:

DL,, = min {D' }, xt15x1 ~XIZ

where xll, xlz are the (ordered) x-coordinates of the endpoints of sl, and ~21,~22 are the

(ordered) x-coordinates of the endpoints of s2. [2] For two arbitrary segments, sl and sz

(one on each line), the minimum distance between the segments occurs when the partial derivative of the equation (3.19) is equal to zero. The following are the results from taking this partial derivative.

The x-coordinate of the minimum distance segment on segment 2 is found by taking the partial derivative of (3.19) with respect to x2,

where xl is the x-coordinate of either endpoint of segment 1 [i.e. xl 1 or x12].

The x-coordinate of the minimum distance segment on segment 1 is found by taking the partial derivative of (3.19) with respect to XI, where x2 is the x-coordinate of either endpoint of segment 2 [i.e. ~21or x~~].Substituting the x-coordinate values of (3.22) and (3.21) into equations (3.16) and (3.17) and solving for yl and y2 results in the y-coordinate values. The minimum distances are then computed using the standard equation for distance between two points, (3.18).

The second program continues to find the minimum distances for all of the profile segments as done in program 1. The output is not as detailed as program one in that only the absolute minimum offset of the two profiles is output &om the program. No graphs are generated; this was done to speed up the computation. The minimum distances for all of the outer profile segments with respect to the inner profile segments are stored in a matrix [MI, which can be viewed if additional offset information is desired. 3 8 CHAPTER 4

MUTUAL VOLUMETRIC DISCRETIZATION OF AXISWMETRIC

FORGED DISKS

4.1 - MUTUAL VOLUMETRIC DISCRETIZATION

Forging changes the size and shape, but not the volume, of a part. Mutual volumetric discretization is only applicable if the disks in question contain the same volume. The purpose of this program was to develop a software package that could be used as a crude strain estimator. Volumetric discretization is performed in order to redefine the ring structure of multiple profiles containing the same volume.

The program works by discretizing the profiles into individual sections by adding discreet rings as a means of "volume tracking". After mutually discretizing the various profiles, each profile will have new rings added that pertain to the ring volumes of the other profiles (i.e. all profiles will contain the same number of rings and corresponding rings will have equal volumes). Strain calculations can be computed by comparing the material flow between the various rings of the profiles.

Several constraints limited by the system model are that the slab model assumes complete die fill for all profiles, the profiles rnust be of equal volume, and that the profiles are two-dimensional. Changes being made to the software will allow for a variation in the profile volumes off 5%. Rings are added to the profiles based on the 39 areas that the designer wants to analyze. An example of a simple quadrilateral ring is illustrated in Figure 4.1.

Figure 4.1 Ring cross-section

All disks are analyzed by looking at only the half cross-sections, defined radially from the origin. The rings are positioned such that its inner and outer radii define the

"width" and the respective z-coordinates define the height of the rings. The taper of the rings is defined by the angles, a and P. An example of a simple disk profile data file is shown in Table 4.1. This profile would have five rings and is seen to have no taper.

Table 4.1 Example of disk geometry file

Radius Lower Z-Coordinate Upper Z-Coordinate [OY -10, 10; 10, -10, 10; 15, -10, 10; 20, -10, 10; 25, -10, 10; 30, -10, 101; 4.2 - DERIVATION OF DISK VOLUME CALCULATIONS

Volumetric discretization can be more easily understood once the information pertaining to the volume calculations and disk profiles is presented. A complete derivation can be seen in Appendix A.3. The following outline is excerpted from unpublished work by Daniel Allwine [3].

The thickness of the ring at a given radius, r, is given by the following function,

h(r)=hl +(r-r,)(tana+tanp), (rl 5rSr2) (4.01) where hl is the height, a is the upper ring taper, P is the lower ring taper, rl is the inner ring radius, and r2 is the outer ring radius. The height at the inner radius is defined as,

hl = zI2-z,,. (4.02)

The ring taper angles are defined as tangent functions of the coordinates shown in Figure

4.1, or respectively as,

222 - 212 a = tan-'( ), and (4.03) r2 -5

The volume of a specific ring is given by

Performing the integration yields which can be reduced to the following form when the upper and lower ring taper is zero,

V = dl(r; - r12). (4.07)

Substituting the coordinate forms of the taper angles into (4.06) yields the following form

of the volume equation as an explicit function of the vertex coordinates,

(4.08)

The volume occupied withn a specific radius, r, is given by:

where a is defined as

a = tana + tanp. (4.1 0)

Solving equation (4.09) for r results in the amount a particular ring will be filled radially by a given volume of material. This assumes that the given volume is smaller than the

total volume occupied by the ring. Therefore the following equations are formed,

r3+ a2r2+a, = 0 (4.1 1)

where

let 1 1 12 P=--ao -- a3 and q=--a2 2 27 9

Further, let

s1 = [P+~m]'3and s, = [p-Jmji3 (4.14) The roots of the cubic equation (4.1 1) are then given by

The real root that falls between the inner radius, rl, and the outer radius, rzl is the desired solution.

4.3 - EXAMPLES OF VOLUMETRIC DISCRETIZATIONS

In the Figure 4.2, profile 1 is discretized with respect to profile 2. The newly formed profile, profile 112, is the result of the first profile, containing one ring, being volumetrically discretized with respect to the second profile, composed of two rings.

This can only be done since the profiles 1 and 2 are of the same volume. Rings, or partitions, are added to profile 1, such that it may "match" the volume distribution of profile 2. The final result has shape of profile 1, all the original partitions of profile 1, as well as the new partitions required to "match" the volume (ring-wise) of profile 2. [3] 4.2 Simple discretization example

In the Figure 4.2, profile 1 is discretized with respect to profile 2. The newly formed profile, profile 112, is the result of the first profile, containing one ring, being volumetrically discretized with respect to the second profile, composed of two rings.

This can only be done since the profiles 1 and 2 are of the same volume. Rings, or partitions, are added to profile 1, such that it may "match" the volume distribution of profile 2. The final result has shape of profile 1, all the original partitions of profile 1, as well as the new partitions required to "match" the volume (ring-wise) of profile 2. [3]

UProfile - I ,aProfile - 2

Profile - 112

4.3 A second simple discretization example

Figure 4.3 shows an example similar to the previous, except that the shape of the second profile is different than the first profile. However, the profiles still contain the 44 same volume of material. In Figure 4.3, the profile 112 is the result of profile 1 being volumetrically discretized with profile 2. The result is a new profile with the shape of profile 1, all the original partitions of profile 1, as well as the newly formed partitions that are placed to "match" the volume distribution of profile 2. 7

Profile - 1 Proflle - 2

Profile - 112

4.4 Multiple ring discretization example

In Figure 4.4, the discretization is between two unlike profiles of equal volume, as in the previous example. The same technique is used as before, the only exception being that there are now more rings. The figure clearly shows in this situation how the discretization of the two profiles is performed. The original partition of profile 1 remains in the new profile 112 and new rings are added to make up for the volume distribution from profile 2.

In the next example, the procedure is reversed in that profile 2 is being discretized with respect to profile 1. It is seen, in Figure 4.5, that the procedure is done the same way as in the previous examples. The difference is that the final profile, profile 211, has the shape of profile 2 and all the original partitions of profile two, as well as the new ring partition due to the volume distribution of profile 1. Profile - 1 Profile -2

I

Profile - 211 L

4.5 The "reverse" discretization

The difference between the method of Figure 4.4 and Figure 4.5 demonstrates that there is a defined direction associated with volumetric discretization. These two examples show that the mutual discretization of the profiles is not commutative, i.e. the profiles 112 and 21 1 are not the same. However, the rings of the two profiles 112 and 2)1 would contain the same volume, hence they form a mutually discretized pair.

These concepts were used in creating a program to perform the mutual discretization of up to N-disks. The previous examples were simplified to explain the process, whereas the program will also discretize profiles that have inclined surfaces.

All disks should be considered as being composed as a series of concentric rings each with quadrilateral cross-section. The "outer" vertices (in a radial sense) of a particular ring are used as the "inner" vertices (again in a radial sense) of the subsequent ring. The inner and outer boundaries for each ring must be vertical. The top and bottom boundaries may have any slope. Graphically, Figure 4.6 shows a half cross-section of a simple disk (four rings with a hole in the center). Figure 4.6 Configuration of a simple four ringed disk profile

See the program code in Appendix B.3 and the output in Appendix C.3 for examples using this process. CHAPTER 5

CONCLUSION

The models developed for this project have been coded in Matlab@ and will be converted into C programming language and used in co~ljunctionwit11 AML (Adaptive

Modeling Language). The purpose for this project was to produce reliable results in a more timely and cost effective manner. This project is a small part of an "ongoing development of a design environment that integrates models for materials and processes and allows selection and optimization of materials and manufacturing processes for components such as those used in aircraft structures and engines." Such a system will

"allow the evaluation, with respect to quality, performance, and cost of alternate materials, processes, and process parameters for the affordable manufacturing of reliable components." [25, pp. 2921

These models could greatly aid in the ability to run the simulation and optimization based design of multi-stage manufacturing processes. The ability to compute the minimum offset and the mutual volumetric discretizations of axisymmetric disk profiles, similar to aircraft turbine engine disks, are two tools being added to work currently being developed. Extensive reviews of all possible scenarios were considered in the development of these tools to ensure high reliability and accurate results.

The limitations of these tools are that the disks must only be composed of flat

surfaces. Disks with curved surfaces could be analyzed by this package but

modifications would have to be made. Currently, curved surfaces are dealt with by 48 breaking down the curves into small line segments to closely represent the contour of the curved surface.

Results obtained from the two programs can be found in sections C.1, C.2, and

C.3 of Appendix C. The code for the programs can be seen in Appendix Byin sections

B.1, B.2, and B.3. Axisyrnmetric profiles, similar to those of aircraft turbine engine disks, were used for the analysis in all the examples. The following are descriptions of each of the examples used:

Appendix C. 1

EXAMPLE 1 - This example shows two ten-segment non-intersecting profiles and the relative absolute minimum distance for these profiles. The minimum distance matrix is included for this example, whlch displays the distances from each outer profile segment to each inner profile segment.'

EXAMPLE 2 - Similar to example one except that the minimum distance matrix is not included. The figure shows the two non-intersecting profiles and the minimum distance between the profiles.

EXAMPLE 3 - This is an example of two intersecting profiles and the output that results. No minimum distance is shown because the profiles intersect.

Appendix C.2

The results in this section are identical to the results in Appendix C.l except that no plots are generated using this program. This method executes slightly faster and has the same precision. 4 9 Appendix C.3

This section shows the results of the Mutual Discretization program. The results shown are explained below.

EXAMPLE 1 - This example shows the mutual discretization between four preforms and a final forged shape. The first figure shows the original profiles and the second figure shows the discretized profiles. The coordinate arrays containing the geometric data for the newly discretized profiles are also shown. These data files are the essential output of this program and are what would be used in computing the strain using the strain estimator of the optimizer.

EXLUMPLE 2 - This example shows the mutual discretization between two preforms and the final forged shape. The figures associated with the discretization of these profiles are shown. The geometric data files are not shown for this example. 5 0 REFERENCES

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pp 15-17. 54

APPENDIX A.1

Derivation of the parallel line segment minimum distance formula

/ . I # , fl .- ,#

/'segment1

.-/' / .' # # .' 1 x-axis < .-.' ,,

Figure Al.1. Representation of the minimum distance between two parallel lines

The two segments can be represented by the foilowing equations:

where m is the slope of the two parallel lines, b2 is the y-intercept of line 2, and bl is the y-intercept of line 1. The equation of the minimum distmce line is:

where mo is the slope and bo is the y-intercept. Since the minim*umdistance line is perpendicular to the parallel line segments, the slope can be written as,

Such that, Considering that the minimum distance line intersects line segments 1 and 2 at points

(xl 1, yl and (xo, yo), respectively, the following equations can be formed.

yo =m*x0 +b2

Solving for bo in equation (A1.06) gives,

1 bo = y,, +-*xll. (A1.08) m

Substituting equation (A1.08) into equation (A1.05) yields,

Setting equations (A1.07) and (A1.09) equal to each other and solving gives the value for xo.

Substituting this value for xo into equation (A1.09) and solving for yo, yields *m yo ;--*[ 1 y,, *m+x,,m2 +1-b, 1 m

y,, em *(mi +l)-(y,, *m+x,, -b2 *m)+xIl*(m2 +1) Yo = m *(m2+I)

y,, em3+b2 *m+x,, em2 Yo = m*(m2+l)

Therefore, the coordinate set (xo, yo) is represented as

yI1*m+x,, -b2 em y,, em2+x,, *m+b2 ( m2 +l > m2 +l

The minimum distance between the parallel line segments can be represented as

the distance from points (xl 1, yl l) to (xo, yo), or

To simplifL this expression, expand the squares and substitute the formulas for the

coordinates (xo, yo) from (A1.19). Ths approach is shown below.

2 yll *m+xlI-b2 em y,, om+x,, -b2 em X,, -2 ex1, m2+l m2+l ..... [ )+[ (A1.22) 2 yI1em2 +xI1 *m+b2 yI1em2 +xI1 *m+b2 +Yll -2*ylIe )+( m2 +I j' 2 2 ~@XII~YII*~2.~11 i 2ox,,~b2*m y,12em2 XI, - m2 +1 m2 +1 m2 +1 + (m2 + l)i -...- 2 + 20y,, ox,, em - 2*y,, *b2am2 - 2*x,, *b2em x,, (mi + l)i (m2+lp (m2+l)i +(m2+l)i'""

20y,, ex,, .m+-2.~,,

...a. I+ m2 +1

20y,,ex,, em3+xl1 2 em2+b2 2 (m2 + ly 2 2 XI, +y,, ..... 2 2 -4*x,, oy,,*m+2*x1, -2*x,, *b2*m+2*yI, *m2+2*yll 06, - ..... (Al..26) m2 +1 + ylI2 m2 (mi +I)+2 y,, x,, m (m2+I)+ xIl2 (m2 +I)+h2' (m2+ 1) (m2 + 1r

y,,2 em2+2*yll*xll*~+x~~ 2 +b2 2 m2 +12

Equation (A1.31) is the final simplified expression for the minimum distance between two parallel lines. 59 APPENDIX A.2

Derivation of the partial differentiation minimum distance formula

This is a derivation of the minimum distance calculation from crny given point on one line segment to any arbitrary point on a second line segment.

For two arbitrary lines 1, and l2 given by

1, 3 y2 = m2x2+b2, the expression for the distance between two arbitrary points (one on each line) is

Substituting the expressions in (A2.01) and (A2.02) into (A2.03) gives

This expression when expanded results in the following:

This expression is written in this form to help simplify the partial differentiation. For two arbitrary segments, sl and s2 (one on each line), the minimum distance between the segments is Dmi,= min (D). XI IS+l SXl2 where xll, x12 are the (ordered) x-coordinates of the endpoints of slyand ~21,x22 are the

(ordered) x-coordinates of the endpoints of s2.

Distance cannot be represented as a negative number, hence, only the positive roots of the above expression are of interest. Equation (A2.07) can be re-written as

DL, = min (D') XI IS*l

The distance expression is then

The minimum distance from a fixed point on segment 1 [i.e. (x11, yll) or (~12,

y12)]to any segment on segment 2 occurs when the partial derivative of D~ with respect

to x2 is zero, or

0 = 2x, (1 + mi)- 2(1+ m,m2)xl- 2m2(b, - 4). (A2.10)

Substituting the known value for xl into this expression and solving for x2 will yield the

x-coordinate of the minimum distance segment on segment 2. This can be shown as

where xl would be the known value from either endpoint of segment 1.

The minimum distance from a fixed point on segment 2 [i.e. (~21,y21) or (~22,

y22)]to any segment on segment 1 occurs when the partial derivative of 0' with respect

to xl is zero, or 6 1 0 = 2x, (1 + m:)- 2(l+ mlrn2)x,+ 2rn, (b, - 4). (A2.12)

Substituting the known value for x2 into this expression and solving for xl will yield the x-coordinate of the minimum distance segment on segment 1. This can be shown as

where x2 would be the known value from either endpoint of segment 2.

The x-coordinate values from expressions (A2.11) and (A2.13) can be substituted into equations (A2.01) and (A2.02) to get the appropriate y-coordinate values (yl and y2, respectively) of the minimum distance segment endpoints. The minimum distance is then calculated using equation (A2.03). APPENDIX A.3

DERIVATION OF THE VOLUME FORMULAS USED TO

COMPUTE THE MUTUAL VOLUMETRIC DISCRETIZATION

Figure A3-1. Configuration of a simple four ringed disk profile

The above figure is an example of a four-iinged disc profile; similar to one that could be analyzed using this type of analysis. The following is a derivation of the formulas presented briefly in Chapter 4.

The thickness of the ring (See Figure A3-2) at a given radius, r, is given by the following function, 63 where hl is the height, a is the upper ring taper, P is the lower ring taper, rl is the inner ring radius, and r2 is the outer ring radius. The height, hl, is defined as,

The ring taper angles are defined as tangent hnctions of the coordinates shcwn in Figure

A3.2, or respectively as,

Figure A3-2. Ring Cross-Section

The volume of a specific ring is given by 64 Performing the integration yields the following

v = I*r [h; + (r - ")('ma + tan ~))drd@. (A3.06)

2 (h, -rl(tana+ tanp))(r; -r;)+-(tana+tan~)(r: -r, (A3.12) 3

This can be reduced to the following form when the upper and lower ring taper is zero

[i.e. a = -P or (tan a + tan P = O)],

v = dzl(r; -r12). (,43.13)

Substituting the coordinate form of the taper angles (A3.14) into (A3.12) yields the

following form of the volume equation as an explicit hnction of the vertex coordinates, V = n (h, - rl (211 + 222 - (212 + 221) 2 2 (211 +22,)-(2 4-2 ) [ r2-r1 )Ti -r~I+?( r2 - '-1 r - 11 )I.

The volume occupied within a specific radius, r, is given by:

2 (h, -rla)(r2-r12)+-a(r3 -r,3) , 3 I where a is defined as

a = tana+ tanp.

Equation (A3.16)can be rewritten in the following,

Solving (A3.21) for r, results in the amount a particular ring will be filled radially by a given volume of material. This assumes that the given volume is smaller than the total volume occupied by the ring. Before solving for r, let the following expressions be introduced:

r 3 +r2 a, +ao =O. (A3.22) where Further, let

1 p = --2a0 --a2 13 and q=--a:1 27 9

s, = [p + ,/=I" and s2= [p - ,/mji3(A3.27/28) The roots of the cubic equation (A3.22) are then given by

The real root that falls between rl, the inner radius, and r2, the outer radius, is the desired solution. APPENDIX B.1

PROGRAM 1 - PROFILE- F1NAL.M

Programmer: Corey Raub and Dan Allwine Filename: profile-fina1.m Compiler: MatlebB Input: Two profile -y Output: Profile plots with minimum offset shown, and the offset value

[rowMAT,columnMAT]=size(MAT); [rowFUX,columnRIX]=size(RIX); MAT(:,columnMAT+ l)=MAT(:, I); RIX(:,columnRIX+l)=RIX(:,1); jrowMAT,columnMAT]=size(MAT); [rowRIX,columnRIX]=size(RE); N 1=columnMAT- 1; N2=columnRIX- 1; N=N I *N2; M=zeros([S,N]); figure(1); plot(MAT(1 ,:),MAT(2,:),'r1,RIX(I ,:),RIX(2,:),'b'); hold on; minimum~distance=l000; % initialize the minimum distance between the profiles for i=l :(columnMAT-1) for j= 1:(columnW-1) if minimum-distance>eps subMAT1=MAT(:,i); subMAT2=MAT(:,i+l); xl l=subMATl(l);xl2=subMAT2(1); y 1l=subMAT 1(2);y 12=subMAT2(2); subRIX1=RE(: j); subRIX2=RE(: j+l); x21=subRIXl (l);x22=subRIX2(1); y2 l=subRIX1(2);y22=subRIX2(2);

% Sort the endpoints so that x-coord of first endpoint is <= x-coord of second endpoint if(xll>xl2) % SwapP11 and PI2 temp-xl 1; x11=x12; xl2=temp; temp=yll; y11=y12; yl2=temp; end if(x21>x22) % Swap P21 and P22 temp-x2 1; x21=x22; x22=temp; temp=y21; y21=y22; y22=temp; end

% Check to see whether each segment is valid (ie: Does the segment have length?) if (((x 1 1=x 12) & (y 11=y 12)) ( ((x2 1=x22) & (~21722))) error(['Invalid Profile(s) (One or more segment has no length)! !'I) end

% Set flags that indicate whether a given segment is exactly vertical or horizontal seg 1vert=O; segl horiz=O; seg2vert=O; seg2hori~O; % Initialize slope variables ml=O; m2=0;

% Compute slope and y-intercept for first line if (xl l=xl2) seglvert=l; % Segment 1 is vertical (slope is infinite, y-intercept is not defined) else ml=(yl2-yl l)/(x12-XI 1); if (ml=O) seglhori~l;% Segment 1 is horizontal (slope is zero) end bl=yll-ml*xll; end

% Compute slope and y-intercept for second line if (x2 1=x22) seg2veel; % Segment 2 is vertical (slope is infinite, y-intercept is not defined) else m2=(y22-y2 1)/(x22-x21); if (m2=0) seg2horiz=l; % Segment 2 is horizontal (slope is zero) end b2=y21-m2*x21; end

% Determine whether lines are parallel if (segIvert*seg2vert) % Both lines are vertical plot(x1 l ,y 11,'bo',x l2,y l 2,'bo1,x2 1,y2 1,'kot,x22,y22,'ko'); minimum-distance=abs(x2 1-x 11); aa=[xll,x21]; bb=[yll,yl I]; elseif (seglhoriz&seg2vert) % Segment one is horizonta! and contains an endpoint that lies on segment two % which is vertical if ((XI1=~211~12=~21))&((y11<=y21&y11>=y22)l(y11>=y21&y11<=y22)) error(['Invalid Profile(s) Segments intersect!!']) end elseif (seg2horiz&seglvert) % Segment two is horizontal and contains an endpoint that lies on segment one % which is vertical if (x21=x111x22=xl l)&((y21<=yl 1&y21>=y12)l(y21>=yl1&y21<=y12)) error(['Invalid Profile(s) Segments intersect! !'I) end elseif ((abs(m1-m2) < eps)&-(seglhoriz*seg2horiz)) % Lines are parallel but not vertical plot(x1 l ,y 11,'bo1,x12,y 12,'bo1,x21 ,y21 ,'koq,x22,y22,'ko'); minimum-distance=abs((m2*x 1I -y l l+b2)/sqrt(m2"2+1)); m2_ortho=- l/m2; bl l-ortho=yl I -m2_ortho*xlI; x 11-ortho=@2-b 11-ortho)/(m2-ortho-m2); y l 1-ortho=rn2*xl l_ortho+b2; aa=[xl 1-ortho,xl I]; bb=[yl 1-ortho,yll]; elseif (segl horiz*seg2horiz) % Segments are both horizontal plot(x1 l ,y 11,'bo9,xl 2,y 12,'bo1,x21,y21 ,'ko',x22,y22,'koq); if (y 11=y21)&(x12

% Determine if the segments intersect if . . . (x l l Xp&x l2>Xp)l(x2 1Xp&x22>Xp)

% Determine the minimum distance between the two segments % Existing slopes of lines 1 and 2 are ml, and m2 respectively % Accounts for cases where the segments are vertical or horizontal

% Compute perpendicular line from (x I l ,y 11) to line 2 and point of intersection on line 2 if seg2vert % Segment 2 is vertical (m2 = undefined) m2_0rtho=O; xll_ortho=x21; yll-ortho=yll; elseif seg2horiz % Segment 2 is horizontal (-l/m2 = undefined) xl l-ortho=xll; y l 1-ortho=y21; % Because Segment 2 is horizontal else % Segment 2 is neither horizontal nor vertical m2_ortho=-lIm2; bl l-ortho=y 11-m2_ortho*xI 1; x 1 1-orth~(b2-b11-ortho)/(m2-ortho-m2); y 11-ortho==*xlI-ortho+b2; end

% Compute perpendicular line from (x 12,y 12) to line 2 and point of intersection on line 2 if seg2vert % Segment 2 is vertical (m2 = undefined) rn2_ortho=O; x 12_ortho=x21; y 12-ortho=y 12; elseif seg2horiz % Segment 2 is horizontal (-llm2 = undefined) x 12-ortho=x 12; y l2_ortho=y21; % Because Segment 2 is horizontal else % Segment 2 is neither horizontal nor vertical m2_ortho=- l Im2; b 12~ortho=yl2-m2~ortho*x12; x 12-ortho=@2-b 12-ortho)/(m2-ortho-m2); y 12-ortho==*x 12_ortho+b2; end

% Compute perpendicular line from (x21 ,y21) to line 1 and point of intersection on line 1 if seglvert % Segment 1 is vertical (ml = undefined) ml-ortho=O; x2l-ortho=x11; y2l_ortho=y21; elseif seg 1horiz % Segment 1 is horizontal (- l /m 1 = undefined) x2 1-ortho=x2 1; y2l-ortho=yll; % Because Segment 1 is horizontal else % Segment 1 is neither horizontal nor vertical m l -ortho=- 1/ml ; b2l~ortho=y21-ml~ortho*x21; x2 1-ortho=(bl -b21~ortho)/(ml~ortho-ml); y2l-orthml *x2l-ortho+bl; end

% Compute perpendicular line from (x22,y22) to line 1 and point of intersection on line 1 if seglvert % Segment 1 is vertical (ml = undefined) ml-ortho=O; x22-ortho=x 11 ; y22_ortho=y22; elseif seglhoriz % Segment 1 is horizontal (-l/ml = undefined) x22_ortho=x22; y22-orth~yl l; % Because Segment 1 is horizontal else % Segment 1 is neither horizontal nor vertical ml-ortho=- l /ml ; b22-ortho=y22-ml-ortho*x22; x22_ortho=@l -b22-ortho)/(ml-ortho-ml); y22-ortho=ml *x22_ortho+b 1; end

% Calculate the distance of the orthogonal line segments min-dist-S 1P 1-S2=sqrt((x 11-ortho-x 11)"2+(y 1 1-ortho-y 1 1)"2); min-dist-S 1P2-S2=sqrt((x 12-ortho-x 12)"2+(y 12-ortho-y 12)"2); min-dist-S2P 1-S 1=sqrt((x2 1-ortho-x2 1)"2+(y2 1-ortho-y2 1)"2); min-distPS2P2-S 1=sqrt((x22-ortho-x22)"2+(y22-ortho-y22)"2); % Calculate the distance between the segment endpoints P11-P21=sqrt((xll-x21)"2+(y 11-y2 1)"2); P 1l -P22=sqrt((x 11-x22)"2+(y 1 l -y22)"2); P 12-P2 1=sqrt((x 12-x2 1)"2+(y 12-y2 1)"2); P 12-P22=sqrt((x 12-x22)"2+(y12-y22)"2);

% Plot the LINES to prove the logical statements are correct figure(] ); plot(x11 ,y 11 ,'bol,x l2,y 12,'bo1,x21 ,y21 ,'ko',x22,y22,'ko'); hold on;

% Determine the true minimum distance between the segments % Using letters to simplify the logical statements a-min-dist-S 1P 1-S2; bm- in-dist-S 1P2-S2; c=min-dist-S2P 1--S 1; d=rnin-dist-S2P2-S 1; e=PI 1-P2 1; f=P11-P22; g=P 12-P2 1; h=P 12-P22;

% Determine the minimum of the endpoint distances if (&=f)&(e<=g)&(e<=h) temp-min 1=e; ==[XI I ,x2l];bb=[y 1l,y21]; % coordinates of the minimum distance line elseif (f<=e)&(f<=g)&(f<=h) temp-min l =f; aa=[xl 1,x22];bb=[y 11,y22]; % coordinates of the minimum distance line elseif (g<=e)&(g<=f)&(g<=h) temp-minl =g; aa=[xl2,~2l];bb=[yl2,y21]; % coordinates of the minimum distance line else temp-min 1=h; aa=[xl2,~22];bb=[y12,y22]; % coordinates of the minimum distance line end

% Determine the minimum distance of all valid orthogonal intersection lines temp_min2=2*ternp_minl; % Sets the default orthogonal minimum to twice % that of the endpoint minimum - must have a % default ~fthe following statements are not % satisfied % Checking if orthogonal line from linel-point1 falls between line2- % pointsl&2 if... ((x2 1<=x 1 1-ortho&xl l~ortho~=x22)&-~(x21=x1 :-ortho&xl l_ortho=x22)) temp_min2=a; cc=[xl I-ortho,xl I];dd=[yll-ortho,yll]; % coordinates of the minimum distance line end % Checking if orthogonal line from linel -point2 falls between line2-pointsl&2 if ((x2 1<=x 12-ortho&x 12_ortho<=x22)&-(x2 I =x l2-onho&x 12_ortho==x22)) if b<=temp_min2 temp_min2=b; cc=[x 12-ortho,x 12];dd=[y 12-ortho,y 121; % coordinates of the minimum distance line end end % Checking if orthogonal line from line2-point 1 falls between linel -points1 &2 if ((x I l <=x2 l_ortho&x2 1-ortho<=x 12)&,-(x I 1=x2 l_ortho&x2 1rtho=x 12)) if c<=temp_min2 temp_min2=c; cc=[x2 l-ortho,x21];dd=[y2l_ortho,y2 11; % coordinates of the minimum distance line end end % Checking if orthogonal line from line2-point2 falls between line 1-points 1 &2 if ((x 11<=x22-0rtho&x22-0rtho<=x 12)&-(X 11=~22-orth0&~22_0rth0=~ 12)) if d<=temp_min2 temp_min2=d; cc=[x22_ortho,x22];dd=[y22_ortho,y22]; % coordinates of the minimum distance line end end

if temp_minl

% If the segments intersect - ignore all plots

else- - disp('The minimum distance is equal to zero at the poini of intersection'); error(['The profiles intersect! !'I) end end M(l ,(i- l)*NZ+jF-minimum-distance; M(2:3,(i- 1)*N2+j)=aat; M(4:5,(i-1)*N2+j)=bb1;

else % If minimum-distance

PROGRAM 2 - MINIMUM- OFFSET - PROF1LE.M

Programmer: Corey Raub and Dan Allwine Filename: minimum_offsetqrofile.m Compiler: MatlebB Input: Two profile coordinate arrays Output: The absolute minimum offset between the two profiles

[rowMAT,columnMAT]=size(MAT); [rowRIX,columnRIX]=size(RU(); MAT(: ,columnMAT+ l)=MAT(:,I); RIX(:,columnRIX+1)=RD[(:,1); [rowMAT,colurnnMAT]=size(MAT); [rowRIX,columrlRIX]=size(RlX); N 1=columnMAT- 1; N2=columnRD[- 1; N=N 1*N2; ~M=zeros([l,N]); % minimum distances matrix

Minimum~Distance=l000;% initialize the minimum distance between the profiles

for i= 1:(columnMAT- 1 ) for j=l :(columnRIX-1) if Minimum-Distanc~eps subMAT 1=MAT(: ,i); subMAT2=MAT(:,i+l); x l l=subMATl(l);xl2=subMAT2(1); y 1 l=subMAT1(2);y 12=subMAT2(2); subRU(l=RU((: j); subRD(2=RU((:j+l ); x2 1=subRIXl (l);x22=subRK2(1); y2 l=subRCX1(2);y22=subRIX2(2);

% Sort the endpoints so that x-coord of first endpoint is <= x-coord of second endpoint if(x1 lrxl2) % Swap PI1 and P12 temp=xl 1; XIl=x12; xl2=temp; temp==11; y11=y12; yl2=temp; end if(x2 1>x22) % Swap P2 1 and P22 temp=x2 1; x2 1=x22; x22=temp; temp=y21; y2 1=y22; y22=temp; end

% Check to see whether each segment is valid (ie: Does the segment have length?) if (((xlI=x12) & (y11--y12)) 1 ((~211x22) & (~21~~22))) error(['Invalid Segment(s) (Segment has no length)! !'I) end

% Set flags that indicate whether a given segment is exactly vertical or horizontal seglvert-0; seglhoriz=O; % Initialize statements as false seg2vert=O; seg2horiz-0;

% Initialize slope variables ml=0; m2=0;

% Compute slope and y-intercept for first line if (xll=xl2) seglvert=l; % Segment 1 is vertical (slope is infinite, y-intercept is not defined) else ml=(yl2-yl I)/(x12-x11); if (ml=O) segl horiz=l; % Segment I is horizontal (slope is zero) end bl=yll-ml*xll; end

% Compute slope and y-intercept for second line if (x21=x22) seg2vert=l; % Segment 2 is vertical (slope is infinite, y-intercept is not defined) else m2=(y22-y2 1)/(x22-x21); if (m2=0) seg2horiz-1; % Segment 2 is horizontal (slope is zero) end b2=y21 -m2*x21; end

% Determine whether lines are parallel if (segIvert*seg2vert) % Both lines are vertical if (xl l=x21)&(y12y21) % Segments overlap - ERROR Minimum-Distance=O; else Minimum-Distance=abs(x2 1-x 1 1); end elseif (segl horiz*seg2horiz) % Both lines are horizontal if (y 11 =y2 1)&(x 12x21) % Segments overlap - ERROR Minimum-Distance=O; else % segments are parallel Minimum-Distance=abs(y2l -y 1I); end elseif (seg 1horiz&seg2vert) if (XI1=~21(~12=~2l)&((yl l<=y2l&yll>=y22)#yl I>=y21&yll<=y22)) % Segment one is horizontal and contains an endpoint that lies on segment two which is vertical Minimum-Distance=O; end elseif (seg2horiz&segl vert) if (x21=x1 I)x22=x1 l)&((y21<=yl 1&y21>=y12)l(y21>=yl1&y21<=y12)) % Segment two is horizontal and contains an endpoint that lies on segment one which is vertical Minimum-Distance=O; end elseif (ml *m2-=0)&(abs(rnl-m2)=0) % Lines are parallel but not vertical Minimum-Distance=abs((rn2*x 1 I -y l l+b2)lsqrt(rn2"2+1)); else % The lines are not parallel (note that one may still be vertical) if seglvert % Segment 1 is vertical (equation is 'x=constar,t') Xint=x 11 ; Yint==m2*Xint+b2; elseif seg2vert % Segment 2 is vertical (equation is 'x=constantl) Xint=x2 1 ; Yint=ml *Xint+bl ; else % Neither segment is vertical % Comp~itepoint of intersection for the two lines Xint=(b2-bl)/(ml -m2); Yint=ml *Xint+b 1; end Xp-Xint; YpzYint; % Xp and Yp are the coordinates of the line-intersection point

% Determine if the segments intersect (Do the segments contain the point of intersection?) If (x 1 1<=Xp&x 12<=Xp)l(xl l >=Xp&x l2r=Xp)l(x2 I ~=Xp&x22~=xp)~(x21>=xp&x2>=x?)

% Determine the minimum distance between the two segmenis % Existing slopes of lines 1 and 2 are ml, and rn2 respectively % Accounts for cases where the segments are vertical

if (seg 1vertlseg2vert) if seg2vert % Segment 2 is vertical (m2 = kndefined) xl 1-ortho=x21; % Compute perpendicular line from (x I 1,y 11) to line 2 and point of intersection on line 2 yll-orthwyll; xl2-ortho=x21; % Compute perpendicular line from (x21 ,y21) to line 1 and point of intersection on line 1 y l2_ortho=y 12; else % Segment 2 is not vertical, Determine the coordinates of minimun line m2_ortho=- l Im2; bl 1-ortho=yl I-m2_ortho*xll; xll~ortho=@2-blI~ortho)/(m2~ortho-m2); y 1 1-ortho==*xl l_ortho+b2; bl2-orthwy 12-m2-0rtho*xl2; x 12-ortho=(b2-b 12-ortho)/(m2-ortho-m2); y 12-ortho=m2*x 12_ortho+b2; end if seglvert % Segment 1 is vertical (ml = undefined) x2l-ortho=xl l; %Compute perpendicular line from (x21,y21) to line 1 and point of intersection on line 1 y2l-ortho=y21; x22-ortho=x1 1; % Compute perpendicular line from (x22,y22) to line 1 and point of intersecrion on line 1 y22-ortho=y22; else % Segment 1 is not vertical, Determine the coordinates of minimum line ml-ortho=-l/ml; b21~ortho=y21-ml~ortho*x2l; x2l-ortho=@l -b2 1-ortho)/(ml-ortho-ml); y2 1-orthorn- 1*x2 1-ortho+bl ; b22-ortho=y22-ml-ortho*x22; x22-ortho=@ 1-b22_ortho)/(ml -ortho-ml ); y22-ortho=ml *x22-ortho+b 1; end min-dist-S 1P 1-S2=sqrt((~ll~ortho-x11)"2+0-y11)"2); % Compute the minimum orthogonal distances min-dist-S 1P2-S2=sqrt((x 12-ortho-x 12)"2+(y 12-ortho-y 12)"2); % Compute the minimum orthogonal distances min-dist-S2P 1-S 1=sqrt((x2 1-ortho-x2 1)"2+(y2 1-ortho-y2 1)"2); % Compute the minimum orthogonal distances min-dist-S2P2-S 1=sqrt((x22-ortho-x22)"2+(y22~ortho-y22)"2); % Compute the minimum orthogonal distances % Determine the minimum distance of all valid orthogonal intersection lines temp-min=1000; % Sets a default value for the orthogonal minimums

% Checking if orthogonal line from linel-point1 falls between line2-pointsl&2 if ((x2 1<=x 1I-orthokx 11-ortho<=x22)&-(x2 1==XI I -ortho&x 11 _ortho==x22)) temp-min=a; end % Checking if orthogonal line from linel-point2 falls between line2-pointslSr2 if ((x2 1<=x 12-ortho&x 12-ortho<=x22)&-(~2 I ==x 12-prtho&x12-ortho--=x22)) if b<=temp-min temp-min=b; end end % Checking if orthogonal line from line2-point1 falls between linel-pointsl&2 if ((x 1I <=x2 1-ortho&x2 1-ortho<=x l2)&-(x 1I -x2 1-0rthohx2 l -ortho=xl2)) if c<=temp-min temp-min=c; end end % Checking if orthogonal line from line2-point2 falls between linel-pointsl&2 if ((XI1 ~=x22-ortho&x22-ortho<=x 12)&-(x 1 1=x22-ortho&x22-ortho=x 12)) if d<=temp-min temp-min=d; end end minl=temp-min; % This will be the default 1000 if the above checks are not satisfied % As a result the minimum will be one of the endpoint minimums else

% Determine the minimum distance of the segments using the partial derivatives method as described % below: % DA2=(x1-x2)"2+(yl-y2)"2 %DA2=(l+ml"2)*x 1A2+(l+m2A2)*x2"2-2*(It-ml*mZ)*xl *x2+2*ml *@l -b2)*x1-2*rr~2*(bl-b2)*~2+(bl-h2)~2 % Take Partials of DA2with respect to xl and x2 to find minimums

% D1=2*(l+mlA2)*x1-2*(l+ml*m2)*x2+2*ml *@I-b2); % D 1=0 when a minimum occurs xl~minl=(2*(l+ml*m2)*x21-2*ml*@1-b2))l(2*(l+m1A2)); if XI l~xl~minl&xl2~xl~minl yl-minl=mI *xl-minl+bl; min1 =sqrt((x 1-min 1-x2 1)"2+(y 1-minl -y2 1)"2); else min 1=2*sqrt((xll-x21)"2+(y 1I-y21)"2); % Set min to twice endpoint minimum end xl-min2=(2*(l+ml *m2)*x22-2*ml *@I-b2))l(2*(l+mlA2)); if x 1Ix 1-min2 y 1-min2=ml *x l_min2+b 1; min2=sqrt((x l -min2-x22)"2+(y 1-min2-y22)"2); else rnin2=2*sqrt((x11-~21)"2+(yl1-y21)"2); % Set min to twice endpoint minimum end

% D2=2*(l+m2A2)*x2-2*(l+ml*m2)*x1-2*m2*(bl -b2); % D2=0 when a minimum occurs x2-min 1=(2*(l+ml *m2)*x 1 1+2*m2*(bl-b2))/(2*(1+rn2"2)); if x2 1~2_rnin2 y2_min2=m2*~2_min2+b2; min4=sqrt((x2-min2-x 12)"2+(y2-min2-y 12)"2); else min4=2*sqrt((x 1 1-x2 1)"2+(y 1 1-y2 1)"2); % Set min to twice endpoint minimum end minqartial=[min 1,min2,min3,min4]; min 1=min(mingartial); end

% Calculate the endpoint distances - min 1-endpoint=sqrt((x 11 -x2 1)"2+(y 11 -y2 1)"2); min2-endpoint=sqrt((x 12-x2 1)"2+(y 12-y21)"2); min3_endpoint=sqrt((x 11-x22)"2+(y 11 -y22)"2); rnin4_endpoint=sqrt((x12-x22)"2+(y 12-y22)"2); min-endpoint=[min 1~endpoint,min2~endpoint,min3~endpoint,min4~endpoint]; min2=min(min_endpoint);

% Determine the minimum distance % Endpoint minimums are compared to the partial derivatives minimum method or the orthogonal YO minimum method if minlcmin2 Minimum-Distance=minl ; % Minimum occurs within the segments else Minimum_Distance=min2; % Minimum occurs at the endpoints end % If the segments intersect else error(['Invalid Prof les(s) -- Segments intersect!!']) end end M(l ,(i- l)*N2+j)=Minimum-Distance;

else % If minimum-distance

PROGRAM 3 - MUTUAL DISCRETIZATION GROUP

Programmer: Corey Raub and Dan Allwine Filename: minimumumoffsetqrofile.m Compiler: MatlebB Output: A graph of the original disk profiles and one of the mutually discretized profiles, and a cell array of the newly discretized profile coordinates

Mutual Discretization Group contains the following files: (1) profile2.m, (2) rundisc2, (3) plotdisc2, (4) profilemap, (5) provmap, (6)volcomp, (7) volcompr, (8) indandsumvols, (9) volmapr ...... Profile2.m - Contains the disk geometry coordinates function [p]=profile2

P5=CO1-5,5;7.5,-5,5;15,-5,5;20, -10,10;30, -10t10;35,-5t5;42.5t-5f 5;50t- 5151 ; dis kvolume=volcomp (P5)

% Calculations to determine intermdediate disk sizes rl=30;hl=1/2*diskvolume/(pi*rl"2); h2=12.5;r2=sqrt(diskvolume/(2*h2*pi)); h3=10;r3=sqrt(diskvolume/(2*h3*pi)); r4=45;h4=1/2*di~kvolume/(pi*r4~2);

% Intermediate disk profiles Pl=[O, -hl,hl;l/S*rl,-hl,h1;2/5*rl,-hl,hl;3/5*rlf -hlth1;4/.5*r11 - hlthl;rlt -hl, hl] ; P2=[0,-h2lh2;1/7*r2,-h2,h2;2/7*r2,-h2,h2;3/7*r2f-h2lh2;4/7*r2f- h2,h2;5/7*r2,-h2,h2;6/7*rZI -h2,h2;r2! -h2,h2]; P3=[0, -h3,h3;1/4+r3# -h3, h3;1/2*13, -h3, h3;3/4kr3, -h3!h3;r3, -h3#h31 ; P4=[0,-h4,h4;1/8*r4,-h4,h4;1/6*r4,-h4,h4;1/3*r4,-h4th4;2/5*r41- h4th4;5/8*r4t-h4,h4;2/3*r4t-h4Ih4;6/7*r4f-h4fh4;r4t-h4fh4~; Rundisc2.m - runs the Mutual Volumetric Discretization for N-disk profiles

function [p]=rundisc2 8 function rundisc2 (pl,p2 ) clc; clear; clf;

[p]=profile2; [m,n] =size (p);

% Plot the profiles with original rings

figure (1); for i=2:n plotdisc2(p{i-ll,p{i}); % using curly brackets {I since dealing with cell arrays hold on; end title('0riginal Disk Profiles');

% Begin Mutual Volumetric Discretization process % "dummy" is an array that is not used other than to gather the correct new pl ("pln") coordinate array

[pln,dummyl=profilemap(p~l~,p{2}) ; % Disc 1 discretized w.r.t disc 2; yields disc 112

for i=3:n [pln,dummy] =profilemap (pln,p{i} ) ; % Discretizing disc 112 w.r.t discs 3:n to create "Master Disc1' end p{ll=pln; % Set the first profile equal to the "Master Disc"

% Now discretizing original disc profiles (2:n) w.r.t the "Master Disc"

for i=2:n Ip{i} ,dummyl=profilemap(p{i},pin) ; % Discretizing discs 2:n w.r.t the "Master Disc" to place new rings % stops after this step for disc 2, does not rediscretize disc 3 end

% Plot the newly discretized profiles to show the Mutual Volumetric Discretization

figure (3); for i=2:n plotdisc2(p{i-l),p~i~); hold on; end axis auto; title('Mutua1 Volumetric Discretized Profiles');

% Output to Matlab screen num-of-disks -and - rings=zeros(2,n); for i=l:n [f,s]=size(p{i)) ; num-of-disks and rings (1,i) =i; num,-of-disksrandIrings (2,i) =f ; end num- of - disks-and-rings

...... plotdisc2.m - plots the disk profiles function plotdisc2 (pl,p2) % function plotdisc2 (pl,p2) [n,dummy] =size (pl); [m,dummy] =size (p2); for i=l:n-1, x(1, i)=pl (it1) ; x(2, i)=pl (it1); x(3,ij=pl (itl,1) ; x(4, i)=pl (i+l,1); y(lli)=pl(i,2); y(2,i)=pl (it3) ; y(3,i)=pl (i+l,3) ; y(4,i)=pl(i+l12); end u = x; v = y; clear x y for i=l:m-1, x~l,i)=p2~ill); x (2,i)=p2 (it1) ; x(3, i)=p2(i+l,1) ; x (4,i) =p2 (i+l,1) ; y(l,i)=p2(i,2); y (2,i)=p2 (it3) ; y (3,i)=p2 (i+1,3); y(4,i)=pZ(i+l,Z); end fill (u,v, 10,x, y, 20) ; profi1emap.m-maps partitions in Standard Disc Profile Arrays pl and p2 such that the discs are "mutually" discretized in a volumetric sense. function [new-pl, new-p2l=profilemap(pl,p2) 8 function [new-pl, new-p2 I =profilemap (~142) % % This m-function maps partitions in Standard Disc Profile Arrays pl and p2 such that % the discs are "mutually" discretized in a volumetric sense. % % INPUTS: % pl,p2 = Standard Disc Profile Arrays consisting of N, and M rings respectively. % The format of each array representing an (N-1)-ring disc is: % [ rO , z0 low, z0 high ; % rl , zl~low,zlIhigh ; % % rn , zN-low, zN-high ] % % Note that row-0 represents a "virtual ring", which is used to specify the 9 inner radius and thickness of the innermost ring/disc. 9 0 % OUTPUTS: % new-pl, new p2 = New Standard Disc Profile Arrays each consisting of K rings, where- % where K <= ( M + N )

% Austral Engineering & Software, Inc. % 19-JUN-1999 -- Daniel A. Allwine

errorcode = 0;

if (abs( (volcomp(pl) -voIcomp (p2)) /volcomp (pl) < 0.05) [new~p2,errorcode]=provmap(pl,p2); if (errorcode == -1) new-pl = [-1 -1 -1:-1 -1 -11; newp2 = new-pl; save profmap.mat; return; end [new-pl,errorcode]=provmap(newWp2,pl); if (errorcode == -1) new- pl = [-2 -2 -2;-2 -2 -21; new-p2 = new-pl; errorcode = -2; save profmap.mat; return; end else 82 disp(['Hey! These profiles do not contain the same volume! !'I) disp(['Profile-1 Volume = ',num2str(volcomp(pl))I); disp(['Profile-2 Volume = ',num2str(vo~comp(p2))I); new-pl = [-3 -3 -3;-3 -3 -31; new-p2 = new- pl; errorcode = -3; save profmap.mat; return; end if (size(new-pl) -= size (new-p2) ) new- pl = [-4 -4 -4;-4 -4 -41; new- p2 = new-pl; errorcode = -4; save profmap.mat; return; end save profmap.mat; return;

pr0vmap.m - adds rings to a profile pz with respect to profile p,, such that the they will match in a volumetric sense

function [new-p2, errorcodel =promap (pl,p2) % function [new-p2]=promap (pl,p2) % % This m-function adds rings to profile-2 (p2) based on the volume distribution of % profile-1 (pl). In other words, partitions are created in p2 that "line up" (in % a volumetric sense) with the partitions in pl. The original partitions in p2 % are not altered. % % INPUTS: % pl,p2 = Standard disc profile arrays consisting of n, and m rings respectively. % The format of each array representing an (n-1)-ring disc is: % [ rO , zO-low, z0 high ; % rl , zl-low, zl1high ; % % rn , zn-low, zn-high ] 9- % Note that row-0 represents a "virtual ring", which is used to specify the % inner radius and thickness of the innermost ring/disc. % % OUTPUT: % new- p2 = Newly partitioned profile p2.

8 Austral Engineering % Software, Inc. % 19-JUN-1999 -- Daniel Allwine errorcode = 0; [nl,dummy] =size (pl); [n2,dummy] =size (p2); tvoll=volcomp (pl); tvol2=volcomp (p2); vi=indandsumvols(pl); %Compute the individual and "summed" volumes for P1 v2=indandsumvols(p2); %Compute the individual and "summed" volumes for ~2 if abs((tvol1-tvol2)/tvo11)>0.05 % Simple notification of volume discrepancy. More is needed! disp(llHey! These profiles do not contain the same volume! ! '1) errorcode = -1; else for i=l:nl-1, %Loop through the rings of pl k = min(find(vl(i,2) <= v2(:,2))); % Finds ring in p2 where Rl,i "fills to" if isempty (k) break end if (k==l) deltav = vl (it2) ; else deltav = vl (i,2) - v2 (k-1,2); % Finds amount of "spill-~ver" volume in ring k of p2 end rl = p2(k,l) ; r2 = pZ(kt1,l); 211 = p2 (k,2) ; 212 = p2 (k,3) ; 221 = p2 (ktl,2) ; 222 = p2 (ktl,3) ; hl = 212-211; alfa = atan2(z22-~12,r2-r1); beta = atan2 (211-z21,r2-rl) ; new r = volmapr(deltav,rl,r2,hl,alfa,beta); if ?(abs (new- r - r2) > le-38) & (abs(new-r - rl) > le-08) ) new-zl = 211 - (new-r - rl)*tan(heta); new-22 = 212 t (new- r - rl)*tan(alfa); p2 = [p2(1:k,:) ; [new-r new-21 new-z21 ; p2(k+l:n2,:)]; %Insert new ring [n2,dmy]=size (p2); v2=indandsurnvols(p2); %Compute the individual and "summed" volumes for updated p2 end

end % end of loop on i end new-p2=p2; vo1comp.m - computes the volume of a particular disk profile function vol=volcomp(p) % function vol=volcomp(p) % % This m-function computes the volume an entire disc described by the Standard Disc Profile Array (p) % % INPUTS: % p = Standard Disc Profile Array % % OUTPUT: % vol = Volume of entire disc %

% Austral Engineering % Software, Inc. % 19-JUN-1999 -- Daniel Allwine vol=O; [m,dummy] =size (p); for n=2 :m, rl=p (n-1,1) ; zll=p (n-1,2); z12=p (n-1,3); r2=p (n,1) ; z21=p (n,2) ; z22=p (n,3) ; a= ( (211-212)+ (222-221)) / (r2-rl); if abs (a)

% Austral Engineering % Software, Inc. % 19-JUN-1999 -- Daniel Ailwine

[n,dummy] =size (p); v=zeros (n-1,2); for i=l:n-1, %Compute the individual and "sumrr,ed" volumes for updated p2a v (i,1) =volcompr (p,i) ; v(i,2)=sum(v(l:i,l)); end

vo1rnapr.m - determines the new radius in a profile pz where a new ring will be placed function [r2]=volmapr (v,rl, r2,h1, alfa,beta) % function r2=volmapr (v,rl, hl, alfa,beta) % % This m-function computes the radius within a given ring that will define % the specified input volume. The ring is defined by inner-radius, inner- % thickness, and top/bottom taper angles (in radians). % % INPUTS: % v = The required volume 86 3 rl = The inner radius of the given ring

% hl = The inner thickness of the given ring % alfa = The "top" taper angle (CCW = positive, origin = (rl, +h1/2)) % beta = The "bottom" taper angle (CW = positive, origin = (rl, - h1/2) ) % % OUTPUT: 9- r2 = Radius inside given ring that will define specified volume

% Austral Engineering % Software, Inc. % 19-JUN-1999 -- Daniel Allwine a=tan (alfa)+tan (beta); if (a==O) % We have a ring with uniform thickness (an easy problem) r2 = sqrt (v/pi/hl + rlA2); else % The thickness changes as a function of radius (a tougher problem) c=hl-rl*a; a2=3* (hl-rl*a)/ (2*a); a0= (3/(2*a) ) * (1/3*(rlA3) *a - (rlA2)*hl - v/pi) ;

r(l, l)= (sl + s2)-a2/3; r (2,1)= - (sl+s2)/2 - a2/3 + sqrt (-3)/2* (~l-~2); r (3,l)= - (slts2)/2 - a2/3 - sqrt (-3)/2* (~l-~2);

% The following line will compute volumes based on the determined radii % vlist=pi*((hl-rl*a)*(r."2-rlA2) + 2/3+a*(r."3-rlA3i);

for i=l:length(r) if (isreal(r (if 1) ) ) & (rl<=r(i, 1) &r (i,1) <=r2) r2=r (it1) ; break end end end APPENDIX C.l

PROGRAM 1 - PROFILE- F1NAL.M Example 1 :

Output Based on the following coordinates:

MINIMUM DISTANCE LINE IS GREEN 2s------I 1.5 1 "4I 1(3 "-"-\ -I 1

0.5 - 7m I 0 - 1 /' i -o.+,

-1 - +Yd/\ \ 1

-1.5 I 1 I I -/( 1 1.5 2 2% 3 3.5 4 44.5 Figure C1.l Diagram of Program 1 - Example 1 Matlab Output

Minimum- Distance =

0.1 1180339887499

An Example of the Minimum Dist.ance matrix is shown below. There are 100 column representing minimum distances of the two, ten-segment profiles. Each column represents a minimum distance segment. The rows of each column are the mininlum distances.

Columns 1 through 4

Columns 5 through 8

1.50000000000000 2.10085766799573 2.01556443707464 1.52069063257455

Columns 9 through 12

1.25000000000000 1.25000000000000 1.00000000000000 0.80498447189992

Columns 13 through 16

0.80498447189992 1.34164078649987 1.90065778087482 2.10085766799573

Columns 17 through 20

Columns 21 through 24

Columns 25 through 28

Columns 29 through 32 89 Columns 33 through 36

1.06066017177982 0.90138781886600 0.9013S781886600 1.34629120178363

Columns 37 through 40

2.65165042944955 2.70416345659799 3.00520382004283 3.0103986446380'7

Colms41 through 44

3.01039864469807 1.67630546142402 1.67630546142402 0.90138783886600

Columns 45 through 48

0.50000000000000 1.07186615714068 1.25000000000000 1.52069063257455

Columns 49 through 52

2.30488611432322 2.75000000000000 1.89010581714358 1.89010581714358

Columns 53 through 56

2.23606797749979 1.82002747232013 1.34629120178363 0.70710678118655

Columns 57 through 60

0.70710678118655 1.06066017177982 1.80277563773199 2.35849528301415

Columns 61 through 64

1.50000000000000 1.50000000000000 2.500000000000~70 2.26384628453435

Columns 65 through 68

2.26384628453435 0.50000000000000 0.50000000000000 0.75000000000000

Columns 69 through 72

1.03077640640442 1.03077640640442 0.50000000000000 0.78102496759067

Columns 73 through 76

1.74928556845359 2.12132034355964 2.12426457862480 0.89442719099992

Columns 77 through 80 Columns 81 through 84

0.50000000000000 0.78102496759067 1.50000000000000 2.12132034355964

Columns 85 through 88

2.12132034355964 1.34629120178363 1.03077640640442 0.25000000000000

Columns 89 through 92

0.11180339887499 0.11180339887499 0.11180339887499 1.67630546142402

Columns 93 through 96

1.67630546142402 2.50000000000000 3.00000000G00000 2.30488611432322

Columns 97 through 100

2.01556443707464 1.25000000000000 0.79056941504209 0.75000000000000 Example 2:

Output Based on the following coordinates:

Outer-Profile=[l,2,2.5,4,4.5,4.5,3.75,2.5,2,1;; % x-coordinates 1y1y2,2y1.5,-1,-1 -5,-1.5,-.5,-.5]; % y-coordinates Inner- Profile=[l.75,2.6,2.9,3.5,4,4,3.25,3,2.25,1.75; % x-coordinates 0.75,0.75,1.75,1.75,1.5,1,-0.25,0,.25,.5];% y-coordinates

MINIMUM DISTANCE LINE is GREEN 2 , I 7------7 *---7 \ I

i" 0.5 I

-0.

Figure C 1.2 Diagram of Program 1 - Example 2

Matlab Output Example 3:

Output Based on the following coordinates:

Outer-Profile=[l,2,2.5,4,4.5,4.5,3.75,2.5,2,1; % x-coordinates 1,1,2,2,1.5,-1,-1-5,-1.5,--5,-.5]; % y-coordinates Inner- Profile=[l.75,2.6,2.9,3.5,4,4,3.25,3,2.25,1.75; % x-coordinates 0.75,0.75,1.75,1.75,1.5,1,-1.75,0,.25,.5];% y-coordinates

Figure C1.3 Diagram of Program 1 - Example 3

Matlab Output

The minimum distance is equal to zero at the point of intersection ??? Error using => profile-final The profiles intersect! !

* As seen by this example - if the profiles intersect, the program is aborted APPENDIX C.2

PROGRAM 2 - MINIMUM- OFFSET - PROF1LE.M

NOTE: THE PROFILE COORDINATE ARRAYS ARE THE SAME USED FOR PROGRAM 1 - OUTPUT. ALL RESULTS ARE THE SAME

Example 1 :

Output Based on the following coordinates:

Outer~Profile=[1,2,2.5,4,4.5,4.5,3.75,2.5,2,1; % x-coordinates 1,1,2,2,1.5,-1,-1.5,-1.5,--5,--51; % y-coordinates Inner-Profile=[l.75,2.6,2.9,3.5,4,4,3.25,3,2.25,1.75; % x-coordinates 0,0,1,1,0.75,0.25,-1,-0.75,-0.5,-0.251;% y-coordinates

Matlab Output

Minimum- Distance =

An Example of the Minimum Distance follows on the next page. There are 200 column representing minimum distances of the two, ten-segment profiles. Each column represents a minimum distance segment. The rows of each column are the minimum distances.

Columns 1 through 4

Columns 5 through 8

Columns 9 through 12

Columns 13 through 16 Columns 17 through 20

2.01556443707464 1.52069063257455 1.27475487839820 1.03077640640442

Columns 21 through 24

2.00000000000000 1.00000000000000 1.00000000000000 1.00000000000000

Columns 25 through 28

1.00000000000000 1.75000000000000 2.75000000000000 2.51246890528022

Columns 29 through 32

2.37170824512628 2.37170824512628 2.40416305603426 1.48660687473185

Columns 33 through 36

1.06066017177982 0.90138781886600 0.90138781886600 1.34629120178363

Columns 37 through 40

Columns 41 through 44

Columns 45 through 48

0.50000000000000 1.07186615714068 1.25000000000000 1.52069063257455

Columns 49 through 52

2.30488611432322 2.75000000000000 1.89010581714358 1.89010581714358

Columns 53 through 56

2.23606797749979 1.82002747232013 1.34629120178363 0.707106781 18655

Columns 57 through 60

0.70710678118655 1.06066017177982 1.80277563773199 2.35849528301415 95 Columns 61 through 64

Columns 65 through 68

2.26384628453435 0.50000000000000 0.50000000000000 0.75000000000000

Columns 69 through 72

1.03077640640442 1.03077640640442 0.50000000000000 0.78102496759067

Columns 73 through 76

1.74928556845359 2.12132034355964 2.12426457562480 0.89442719099992

Columns 77 through 80

Columns 81 through 84

0.50000000000000 0.78102496759067 1.50000000000000 2.12132034355964

Columns 85 through 88

2.12132034355964 1.34629120178363 1.03077640640442 0.25000000000000

Columns 89 through 92

0.11180339887499 0.11180339887499 0.11180339887499 1.67630546142402

Columns 93 through 96

1.67630546142402 2.50000000000000 3.00000000000000 2.30488611432322

Columns 97 through 100

2.01556443707464 1.25000000000000 0.79056941504209 0.750G0000000000 Example 2:

Output Based on the following coordinates:

0uterProfile=[1,2,2.5,4,4.5,4.5,3.75,2.5,2,1;; % x-coordinates 1,1,2,2,1.5,-ly-l.5y-l.5~-.5y-.5]; % y-coordinates Inner- Profile=[l.75,2.6,2.9,3.5,4,4,3.25,3,2.25,1.75; % x-coordinates 0.75,0.75,1.75,1.75,1.5,1,-0.25,0,.25,.5];% y-coordinates

Matlab Output

Example 3:

Output Based on the following coordinates:

Outer-Profile=[l,2,2.5,4,4.5,4.5,3.75,2.5,2,1; % x-coordinates 1,1,2,2,1.5y-1y-1.5,-1.5y-.5y-.5]; % y-coordinates Inner-Profile=[l.75,2.6,2.9,3.5,4,4,3.25,3,2.25,1.75; % x-coordinates 0.75,0.75,1.75,1.75,1.5,1,-1.75,0,.25,.5];% y-coordinates

Matlab Output

??? Error using => minimum-offsetqrofile Invalid Profiles(s) -- Segments intersect! !

* As seen by this example - if the profiles intersect, the program is aborted APPENDIX C.3

PROGRAM 3 - MUTUAL DISCRETIZATION GROUP

Example 1 :

Output is based on the final profile geometry as follows:

This and four intermediate disk profiles were used to gather the following output.

This is a plot of the original disk profiles

Figure C3.1 Diagram of Original Profiles - Example 1 This is a plot of the discretized profiles

2 0 I I I I I I I I I 1

-20 1 I I I I I I I I 0 5 10 15 20 25 30 35 40 45 50 Figure C3.2 Diagram of Discretized Profiles - Example 1

Matlab Output diskvolume =

num- o f-disks-and-rings =

% This shows the number of disks % and their respective number of rings

profile-set =

Columns 1 through 4 [26x3 double] [26x3 double] [26x3 double] [26x3 double]

Columns 5 through 5 [26x3 double] 99 The array "profile-set" is a 1 x 5 cell array. The following are the discretized profile arrays with the added updated ring coordinates.

profile-set (2) profile-set (5) Example 2:

Output is based on the final profile geometry as follows:

This and three intermediate disk profiles were used to gather the following output.

Matlab Output diskvolume =

nurn- of - disks-and-rings =

profile-set =

[18x3 double] [18x3 double] [I 8x3 double] This is a plot of the original disk profiles

Figure C3.3 Diagram of Original Profiles - Exarnple 2

This is a plot of the discretized disk profiles

2 0

-20 o s 10 15 20 25 30 3s rO 45 SO Figure C3.4 Diagram of Discretized Profiles - Example 2 GLOSSARY

1. Anneal(ing) Heating and slowly cooling a metal part to produce a

crystalline structure that is softer and easier to work. To improve the

machinability of forgings made bmalloy steels

2. Axisymmetric An object that is symmetric about one or more of its cross-

sectional axes

3. Ductile Capable of being shaped. That which can be stretched,

drawn, or hammered thin without breaking; not brittle - referring to metals. Easily

molded; plastics; pliant.

4. Malleable Impressionable, able to be hammered, pounded, or pressed

into various shapes without breaking or returning to its original shape.

5. Normalize Preformed in order to obtain a homogeneous microstructure

with a well-refined grain. Relieves the strains set up by the forging operations.

Performed at -lOO°F higher than the critical temperature.

6. Quenching Heating and then cooling of steel parts in a controlled bath

(usually water, brine, or oil) to alter the hardness of the part.

7. Temper Heating metal and then cooling it in a controlled manner to

change its hardness

8. Traverse In a direction across the grain structure. ABSTRACT

This thesis focuses on the forging metal forming process, and newly found techniques of geometric analysis that pertain to this operation. This project is part of an

"ongoing development of a design environment that integrates models for materials and processes and allows selection and optimization of materials and manufacturing processes for components such as those used in aircraft structures and engines." The forging of axisyrnmetric turbine-engine disks similar to those used in aircraft engines is the focus of this project.

The objective of this project was to create low-fidelity models that deliver reasonable results in quick, cost-effective manner. Models were created using Matlab@ that can greatly aid in the ability to run the simulation and optimization based design of multi-stage manufacturing processes. This project adds to the above-mentioned work, the ability to compute the minimum offset and the mutual volumetric discretization of axisymmetric disk profiles.

These programs deal with "a new process design method for controlling microstructures and mechanical properties through the optimization of preform and die shapes." The models presented here will be used to perform the "simulation of a metalforming process similar to those used for the manufacturing of turbine disks." Such a system will "allow the evaluation, with respect to quality, performance, and cost of

alternate materials, processes, and process parameters for the affordable manufacturing of reliable components."