Mechanical System Design [1 ed.] 9788122429350, 9788122421149

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Copyright © 2007, New Age International (P) Ltd., Publishers Published by New Age International (P) Ltd., Publishers All rights reserved. No part of this ebook may be reproduced in any form, by photostat, microfilm, xerography, or any other means, or incorporated into any information retrieval system, electronic or mechanical, without the written permission of the publisher. All inquiries should be emailed to [email protected] ISBN (13) : 978-81-224-2935-0

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Preface

v

Preface This book is written according to the syllabus prescribed by the U.P. Technical University, Lucknow. The main objective of this book is to produce a good textbook from the student’s point of view. It will be very useful for postgraduate and undergraduate students. The underlying theme of the book has been to expose the reader to large number of mechanical systems and the techniques of the systems. The subject matter has been presented in a very systematic and logical manner. This book will satisfy both average and brilliant students. So far, no complete and elaborate text is available in “Mechanical System Design”. An honest attempt has been made to make the topic easy and to strike an appropriate balance between the breadth and depth of coverage of various topics. We have tried our best to incorporate all the possible elements in this text. This book would not have been completed without the guidance and support of Dr M.A. Faruqi, Director of Engineering, Azad Institute of Engineering and Technology, Lucknow and Dr M.M. Hasan, Dean and Director of Engineering, Integral University, Lucknow, who guided us at crucial stages. We are highly thankful to Prof. Z.R. Nomani, Administrator, AIET Lucknow and Dr M.I. Khan, Prof. and HOD, Mechanical Engineering, Integral University, Lucknow, who took a took a lot of interest in completing the text and gave their invaluable suggestions. We take this opportunity to express our gratitude to Mr L.N. Mishra, Manager, Mr Vikas Roy, Marketing Executive and other staff of M/S New Age International (P) Ltd., Publishers for publishing this book in a very short time. The advice and suggestions of our esteemed readers to improve the text are most welcome and will be highly appreciated. Prof. K.U. Siddiqui Er. Manoj Kumar Singh

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Contents

Contents Preface

v

1. Introduction Design of Systems 1.1 1.2 1.3 1.4 1.5 1.6 1.7

Introduction 1 Concept of Systems 2 General Model of a System 4 Application of a System 5 Elements/Components of a System 6 Classification of a System in Mechanical System Design 7 A Case Study of a Mechanical System Design 8 Exercise 1 11 A Case Study 12

2. Engineering Processes and the System Approach 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9

1

Introduction (System Approach) 13 Application of Systems Concepts in Engineering 13 Identification of Engineering Functions of Systems 15 The Characteristics of a System in “MSD” 16 Engineering Activities Matrix 21 Defining the Proposed Effort 23 Role of Engineer in “Mechanical System Design” 23 Engineering Problem Solving 24 Concurrent Engineering (CE) 25

13

Contents

2.10 A Case Study: Viscous Lubrication System in Wire Drawing 29 Exercise 2 30 A Case Study 31

3. Design and Problem Formulation 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9

Introduction 32 Nature of Engineering Problem 34 Needs Statement 34 Identification and Analysis of Need 35 Hierarchical Nature of System and Hierarchical Nature of Problem Environment 37 Problem Scope and Constraints 41 A Case Study: Heating Duct Insulation System 42 A Case Study: High-Speed Belt Drive System 45 Chain Drives 54 Exercise 3 57 A Case Study 58

4. System Theories 4.1 4.2 4.3 4.4 4.5 4.6

59

Introduction 59 System Analysis View Point 59 Black Box or Decision Process Approach 61 State Theory Approach 62 Component Integration Approach 63 A Case Study: Automobiles Instrumentation Panel System 63 Exercise 4 68 A Case Study 68

5. System Modelling 5.1 5.2 5.3 5.4 5.5 5.6 5.7

32

Introduction 69 Need for Modelling 70 Modelling Types and Purposes 71 Linear Graph Modelling Concepts 74 Mathematical Modelling Concepts 79 Mathematical Modelling and System Behavior 81 Modelling and Simulation 84

69

Contents

5.8

A Case Study: A Compound Bar System 86 Exercise 5 94 A Case Study 94

6. Linear Graph Analysis 6.1 6.2 6.3 6.4 6.5 6.6

Introduction 95 Graph Modelling and Analysis Process 97 Linear Graph Analysis – A Path Problem 100 Linear Graph Analysis – Network Flow Problem 102 Goal Programming 108 A Case Study: Material Handling System 113 Exercise 6 118 A Case Study 119

7. Optimization Concepts 7.1 7.2 7.3 7.4 7.5 7.6

120

Introduction 120 The Optimization Process 121 Motivations and Freedom of Choice 122 Goals and Objectives—Criteria 122 Methods of Optimization 124 A Case Study: Aluminum Extrusion System 126 Exercise 7 130 A Case Study 130

8. System Evaluation 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8

95

Introduction 131 Feasibility Assessment 132 Planning Horizon 133 Time Value of Money 135 Financial Analysis 136 Selection between Alternatives 140 A Case Study: Manufacture of a Maize Starch System 147 Analysis of the Case Study: Manufacture of Maize-Starch System 149 Exercise 8 152 A Case Study 153

131

Contents

9. Calculus Methods for Optimization 9.0 9.1 9.2 9.3 9.4 9.5

Introduction 154 Model with One Decision Variable Optimization 154 Model with Two Decision Variables with No Constraints 158 Model with Two Decision Variables with Equality Constraints 167 Model with Two Decision Variables with Inequality Constraints 181 A Case Study: Optimization of an Insulation-System 184 Exercise 9 194 A Case Study 194

10. Decision Analysis 10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.8 10.9 10.10 10.11 10.12 10.13 10.14

195

Introduction 195 Elements of a Decision Problem 195 Decision Model 197 The Scientific Approach to the Decision Process 198 Quantitative Methods in Decision-making 199 Decision Trees 200 Advantages and Limitations of Decision Tree Approach 202 Mathematical Programming 205 Probability of a Density Function 205 Expected Monetary Value (EMV) 206 Utility Value 207 Fundamental of a Probability 208 Decision Analysis Making Under Condition 216 A Case Study: Installation of a Machinery 222 Exercise 10 232 A Case Study 233

11. System Simulation 11.1 11.2 11.3 11.4 11.5 11.6 11.7

154

Introduction 234 Simulation Concept 236 Simulation Models—Iconic, Analog, and Analytical 238 Waiting Line Simulation or Queuing Theory 239 Simulation Process 239 Problem Definition 239 Input Model Construction 240

234

Contents

11.8 11.9 11.10 11.11 11.12 11.13

Limitations of Simulation Approach 240 Simulation Programs and Languages 242 Advantages and Disadvantages of Simulation 243 Monte Carlo Method 244 Queuing Model with Random Input or Poisson’s Arrival 247 A Case Study: Inventory Control in a Production Plant 253 Exercise 11 265 A Case Study 265

12. Application of Mechanical System Design to Control System 12.1 12.2 12.3 12.4 12.5 12.6 12.7

Input-Output Relationship 266 Transfer Function 269 Mechanical System 270 Rotational System 271 D’Alembert’s Principle 272 Analogous System 279 Mechanical Equivalent Network 285 Exercise 12 290

13. The Product Design Process 13.1 13.2 13.3 13.4 13.5 13.6 13.7 13.8

291

Introduction 291 Product Design Process 293 Considerations of a Good Design 294 Regulatory and Social Issues 295 Conceptual Design 296 Human Factors in Design/Ergonomics 297 A Case Study: Human Factor Considerations in Man-Machine Environment System Design 298 Protection Against Noise 301 Exercise 13 310

14. Computer System Concept 14.1 14.2 14.3 14.4

266

Introduction 311 Data Flow Diagram 312 Conversion of Manual to Computer-based Systems 315 Fundamentals of CAD/CAM 316

311

Contents

14.5 14.6 14.7

Application of Computers in Design 318 Computer Integrated Manufacturing 319 Simulation and Off-line Programming 321 Exercise 14 322

15. Bond Graph 15.1 15.2 15.3 15.4 15.5 15.6 15.7 15.8 15.9

323

Introduction 323 Power Variables of Bond Graphs 323 Bond Graph Standard Elements 324 Power Directions on the Bonds 330 Assigning Numbers to Bonds 332 Causality 332 Generation of System Equations 338 Activation 342 Bond Graph Modelling 345 Exercise 15 347

Question Papers Tutorial Sheet References Index

348 357 363 365

Introduction to Design of Systems

1

1 Introduction to Design of Systems 1.1

INTRODUCTION

The idea of system covers an extremely broad range of interacting entities clubbed together. These can be natural as well as man-made. Examples from nature could be a solar system etc. The human body itself is a complex organisation where the skeletal system, the circulatory system, the nervous system and the digestive system are easily recognisable. Some of the commonly encountered engineering systems could be of transportation such as rail, road and sea, or communication such as telephone and internet or economy and agriculture etc. A mere aggregation of entities does not qualify for it being classified as a system. Some form of interaction is a must. New properties may appear with interaction. For example, consider what Luciano de Crescendo wrote– “We are all angels with only one wing We can only fly while embracing each other”. Luciano de Crescendo

What is design? If you search the literature for an answer to this question you will find out as many definitions as there are designs. Perhaps the reason is that the process of design is a common human experience. Webster’s dictionary says that to design is “to fashion after a plan” but that leaves out the essential fact that to design is also to create something that has never been. Certainly a “mechanical systems design” professional practices design by this definition but so does an artist, a sculptor, a composer, a playwright or any other creative member of our society. For engineers, design establishes and defines solutions to pertinent problems not solved before or new solutions to problems, which have previously been solved in a different way. Design engineers provide a service function to society by meeting societal needs within available resources and technology. ‘MSD’ process begins with the reorganisation of an existing need or issue in the real world that calls for a facility or system of facilities to satisfy the needs or the issue. During the ‘Mechanical Systems Design’, the engineer is involved in a continuous and cyclic interaction with society in which new needs have been felt. This interaction requires determining what is needed, when it is needed and how it can be attained and involves the engineer in creative activities.

2

1.1.1

Mechanical System Design

A System Design Example

To illustrate a system design problem, we consider the problem of designing a small online computer system, that is, a computer that responds immediately to message it receives. The function of the system is described by the flowchart message being received over a communication channel at the rate of M messages a second. On the average, there are m characters in a message. They are received in a buffer that can hold a maximum of b characters. A fraction k of the messages need replies which gave an average of r characters. The same buffer is used for both receiving and sending messages. Processing a message when it has been received, takes about 2000 instructions, preparing a relay requires a program that uses about 10,000 instructions. The finite size of the buffer means that messages and replies must be broken into section of not more than b characters. Servicing each such section causes an interruption of the processing, requiring the execution of 1000 instructions to transfer data either in or out of the computer. Three computers are being considered as possible system components, a slow, medium, and a fast computer, which have instruction execution rate of 25,000, 50,000 and 7,00,000 instructions a second. In addition, four buffer sizes are being considered. There could be a one, two, five or ten character buffer. The design problem is to determine which of the possible twelve combinations for computer speed and buffer size will be capable of carrying out the processing. Given the prices the cheapest design can then be determined.

1.2

CONCEPT OF SYSTEMS

The word system is derived from the Greek word “systema” which means an organised relationship among functioning units or components. A system exists because it is designed to achieve one or more objectives. We come into daily contact with the telephone system, the transportation system, the accounting system, the railway system, the production system and for over four-decades, the computer system. Similarly we talk of business and its organisation as a system consisting of the interrelated departments (sub systems) such as production, sales, personnel and information systems, etc. Many engineering systems may be generalised as processing units which receive certain input and are urged to act upon them in some desirable fashion to produce outputs with a purpose. Inputs may be in the form of energy matters information etc. Processing units may be autonomous or controlled by men. The outputs may be in the form of products, services or information and the objective may be to optimise certain features of the outputs. A system may have animate or inanimate interacting elements. The system approach by itself does not provide a set of rules for solving specific problems.

1.2.1

Bond Graph

Bond graph is an explicit graphical tool for capturing the common energy structure of systems. It increases one’s insight into systems behavior in the vector form; they give concise description of complex systems. Moreover, the notations of causality provide a tool for formulation of system equation. In 1959, Prof. H.M. Paynter gave the revolutionary idea of portraying systems in terms of power bonds, connecting the elements of the physical system to the so-called junction structures, which were manifestations of the constraints. This power exchange portrayal of a system, is called Bond Graph.

Introduction to Design of Systems

3

To understand the idea of Henry M. Paynter, let a simple system made of two elements be considered. To make the system closer to our perceptive understanding of nature around, let the system belong to the realm of mechanics, i.e. in the realm of force, velocities, power system, modeling and simulation. “In nineteen fifty nine Henry, the paynter drew a line until that point everything was fine Ergs and bits then started playing Amps, Volts, Magnet, Pressure, Price, Market Heat, Entropy, Motion with great communication cooked in a potion went unifying No one ever since drew the final line.” Amlendu Mukherjee (Prof. IIT Kharagpur)

1.2.2

Level of a System (a) Transportation system

Residential street

Vehicle

Collect for road

Rural road

Interchange

Freeway

(b) Interchange system

N

Pavement

E

S Pavement

Pavement

W Pavement

I

Bridge N-W ramp S-W ramp

E-N ramp

W-N ramp

N-E ramp

(c) Pavement system

Surface Lanes

Embankment Shoulder

Guard rail

Fig. 1.1

Drainage ditch

Lights Signal

Level of a system

Landscaping Lane identification

4

Mechanical System Design

Systems can be either abstract or physical. An abstract system is an orderly arrangement of interdependent ideas and constructs. A physical system consists of a set of elements, which operate together to accomplish an objective. A physical system may be further eleborated by examples shown in Table 1.1.

Table 1.1 S. No.

Mechanical System

Description

1.

Transportation system

The personnel, machines and organisations, which transport goods.

2.

Weapon system

The equipment, procedures, and personnel, which make it possible to use a weapon.

3.

School system

The buildings, teachers administrators, and textbooks that function together to provide instruction for students.

4.

Computer system

The equipment, which functions together to accomplish computer processing.

6.

Clutch

A mechanism to make or break transmission of power

7.

Telephone system

A device for transmitting sound signals between two places.

8.

Xerox copier

A quick mechanism for producing facsimiles of given document.

The examples illustrate that a system is not a randomly assembled set of elements; it consists of elements which can be identified as belonging together because of a common purpose, goal or objective. Physical systems are more than conceptual constructs; they display activity or behavior. The parts interact to achieve an objective.

1.3

GENERAL MODEL OF A SYSTEM

A general model of mechanical system may be depicted with a black-box with an input and output. This is of course very simplified because a system may have several inputs and outputs (Fig. 1.2 and Fig. 1.3). This feature, helps to define and delineate a system, from its boundary, the environment that is outside the boundary. The system thus is inside the boundary. In some cases, it is fairly simple to define what part of the system is and what is not. In other cases, the person studying the system may define the boundaries with some objectives. Some examples of boundaries are shown in Table 1.2.

Input

Process

Fig. 1.2

Simple system model

Output

Introduction to Design of Systems

5

Input1

Output1

Input2

Output2

Input

Output

Process

Input

Fig. 1.3

Output

Systems with many inputs and outputs

Table 1.2 System

Boundary

Human

Skin, hair, nails, and all parts contained inside form the system, all things outside are environment.

Automobile

The automobile body plus tyres and all parts contained within form the system.

Production

Production machines, production inventory of work in process, production employees, production procedures, etc. form the system. The rest of the company is in the environment.

Each system is composed of subsystems which in turn are made up of other subsystems, each subsystem being delineated by its boundaries. The interconnections and interactions between the subsystems are termed as interfaces. Interfaces occur at the boundaries and take the form of inputs and outputs across it for a computer.

1.4

APPLICATION OF A SYSTEM (i) Each system is composed of subsystems, which in turn are made up of other subsystems, each subsystem being delineated by its boundaries. The interconnections and interactions between the subsystems are termed as interfaces. Interfaces occur at the boundary and take the form of inputs and outputs. (ii) A subsystem at the lowest level may not be defined to be processor. The input and output are defined but not the transformation process. This system is termed as a black box. (iii) A system receives input and produces output after transforming input. The process transformation is regulated. In many systems, it is through a feedback control. The feedback control requires measurement of the outputs i.e., a comparison of the output with a standard, specifying what the system should be producing and a control mechanism which can cause the system process to be altered so as to bring the output closer to the standard. (iv) Each system has a boundary. The system is inside the boundary and the environment is outside the boundary. (v) Design and development of complex, highly engineered industrial equipment system. (vi) Design and development of military equipments and weaponary system.

6

Mechanical System Design (vii) Management of operations. (viii) In deciding major policy alternatives. (ix) System engineering involves the analysis and synthesis of system. System engineering includes the following three major steps. 1. Designing 2. Complexing, and 3. Tightly engineered system.

1.4.1

Implications of Systems Concept

The study of systems concept has three basic implications: 1. Interrelationships and interdependence must exist among the components. 2. The objectives of the organization as a whole have a higher priority than the objectives of its sub-systems. For example, computerising personnel applications must conform to the organization’s policy on privacy, confidentiality and security, as well as making selected data (e.g., payroll) available to the accounting division on request. 3. A system must be designed to achieve a predetermined objective. (e.g. contact basis). The Barauni Refinery refines petrol and diesel which is being supplied to (BKPL) BarauniKanpur Pipe Line.

1.5

ELEMENTS/COMPONENTS OF A SYSTEM

A system has following elements or components: Super System

Input

Process

Output

Feedback and Control

Environment

Fig. 1.6

Elements or components of a system

(i) Input: Input includes machines, manpower, raw materials, money, time etc. (ii) Processor: It is the element of a system that involves the actual transformation of input into output. It is the operational component of system. Processors may modify the input totally or partially depending on the specifications of the output. This means that as the

Introduction to Design of Systems

7

output specifications change so does the processing. In some cases, input is also modified to enable the processor to enable the transformation. (iii) Output: These are information in the right format, conveyed at the right time and place to the right person. (vi) Control: It includes processing the feedback and taking the necessary action. In a computer system, the operating system and accompanying software influence the behavior of the system. In systems analysis, knowing the attitude of the individual who controls the area for which a computer is being considered, can make a difference between success and failure of the installation. Management support is required for securing control and supporting the objective to the proposed change. (v) Feedback: Feedback is the data about the performance of the system. Feedback measures output against a standard in some form of cybernetic procedure that includes communication and control. Output information is feedback to the input and to management (controller) for deliberation. After the output is compared against performance standards, changes can result in the input or processing and consequently, the output. Feedback may be positive or negative, routine or informational. In system analysis, feedback is implemented in two different ways. (a) Negative feedback generally provides the controller with information action. (b) Positive feedback reinforces the performance of the system. It is routine in nature.

1.6

CLASSIFICATION OF A SYSTEM IN MECHANICAL SYSTEMS DESIGN

Systems are generally classified on the basis of the following.

1.6.1

System by Loop

(a) Open loop system (b) Closed loop system (a) Open loop system: This system does not have any influence on outputs. Examples of open loop system are (i) Watch, (ii) Metal cutting machine (iii) Metal working machine etc., the watch by itself does not observe its accuracy and adjust itself. Lathe which is used for metal cutting does not adjust its speed, feed etc. According to open loop system it has no feedback. Input

Processing unit

Output

Fig. 1.7 Open loop system (b) Closed loop system: According to work, the closed loop system brings results from the past action of the system and thereby it consists of the structure action. This system consists of input, processing unit and output. This output is compared with the desired output and the difference between the actual and desired result provides control action through feed back to the input and the actual output equals the desired output. For example:

8

Mechanical System Design (i) A thermostat receives temperature information, decides to start the furnace and changes the temperature accordingly. (ii) In a manufacturing system, orders and inventory levels leads to manufacturing decisions, which fills orders and correct inventories. (iii) A manufacturing unit attracts competition until the profit margin is reduced to equilibrium with economic forces. Processing unit

Input

Output

Computer sensing system Feed back

Fig. 1.8

1.6.2

Closed loop system

System by Nature of Mechanism 1. Mechanistic system. 2. Quasi-mechanistic system. 1. Mechanistic system: A mechanistic system is one, which though is fully mechanized yet the choice of system composition remains in the hands of human being, e.g. dial telephone, guided missiles, space rockets. 2. Quasi-mechanistic system: In this system human beings carry out some of the mechanical function e.g. a fighter plane.

1.6.3

System by Nature of Environment 1. Physical system. 2. Conceptual system. 1. Physical system: A system in a physical form can be seen and felt. e.g. human respiratory system, traffic system. 2. Conceptual system: Conceptual system do not exist in physical form. They exist in the mind of person. e.g. system like hour, days, weeks, months and year is a conceptual system. The system concept does not provide a set of rules for solving all problems but it is a useful device for viewing any phenomena.

1.7

A CASE STUDY OF A MECHANICAL SYSTEM DESIGN

Introduction—a Case Study These days a case study is used more and more in a preventive way rather than as a therapy after a situation has broken down. Most frequent application of the case study approach is in the study of persons or situations, which have gone awry. A case study is based on assumptions.

Introduction to Design of Systems

9

The assumption of uniformity in the basic human nature in spite of the fact that human behavior may vary according to situations. The assumption of studying the nature history of the chapter concerned. The assumption of comprehensive study of the chapter concerned, e.g. l

l

l

l

l

l

l

l

l

1.7.1

a case study of viscous lubrication system in wire drawing is fully concerned with application of system concepts in engineering—identifications of engineering functions, system approach, engineering activities matrix, defining the proposed effort, role of engineer, engineering problem solving and concurrent engineering. a case study of heating duct insulation system and a case study of high speed belt drive system, correlated with problem formulation, nature of engineering problem, needs statement, hierarchical nature of system, hierarchical nature of problem, and problem scope and constraints. a case study automobile instrumentation panel system is correlated with system theories— system analysis view points, black box approach and state theory approach e.g. a case study compound bar system is based on system modeling, modeling types and purposes, linear graph modeling and mathematical modeling concept. a case study of material handling system is correlated with linear graph analysis, graph modeling and analysis process, path problem and network problem. case study of aluminum extrusion system is based on optimization concept, optimization process, motivation and freedom of choice, method of optimization—analytical, combinational and subjective optimization. a case study of manufacture of maize starch system is based on system evaluation feasibility assessment, planning horizon, time value of money and financial analysis. a case study of optimization of an insulation system is fully concerted with calculus methods for optimization model with one decision variable, model with two decision variables, model with equality constraint, and model with inequality constraint. a case study of installation of machinery is correlated with decision analysis elements of decision problem, decision model, probability of a density function, expected monetary value, utility value and Bayes’ theorem. a case study of an inventory control in a production-plant is based on system simulation, simulation concepts, simulation models, analog, analytical, waiting line simulation, simulation process, problem definition, input model construction, solution process limitation and simulation approach.

Steps In a Case Study 1. Determining present status: The first step is to gather descriptive information which will determine, as precisely as possible, the present status of the unit under investigation. Here the investigator after knowing the problem tries to find out the nature and extent of the problem. 2. Gathering background information: Once the researcher is able to achieve an accurate description of the present situation, he collects background data. Here the researcher collects information about and examines the circumstances leading to the current status. At this stage, the investigator compiles a reasonable list of the possible causes of the present situation. He formulates the hypotheses about the true nature of the situation by making use of symptoms which appear in the data by using the researcher’s past experiences with similar situations, and by using the knowledge of the principles of human behavior.

10

Mechanical System Design 3. Testing suggested hypothesis: At this step the researcher gathers specific evidences in relation to each of the hypotheses suggested from the background information just gathered. The individual’s behavior is usually determined by several factors. The researcher tries to locate the factors, which are influential and therefore, are important. He tries to eliminate those, which are not. 4. Instituting remedial action: The case studies are generally carried on to make an intensive examination of problem cases. Therefore, the researcher tries to find out how one or more of the hypothesized difficulties actually contributed to the original difficulties. This is accomplished by instituting some remedial or corrective program and then by examining as to what effect the change has brought about.

1.7.2

Application of a Case Study

Case studies are conducted and the information achieved through it are used by physicians, social workers, sociologists, anthropologists, psychologists and many other persons for pragmatic purposes, and for extending their knowledge. Many good examples of such studies are available for further sociological research, for example, case histories of deviants have been collected and analysed in an attempt to locate the social pressures that may have led them to stray from the natural path. A case study method has been used not only as main method of data collection but also been used to supplement other research method. For example, take the study of the interplay of attitudes and personality traits by Smith et al. (1964). He used case histories as a part of the data to trace out the significant events leading to changes in the subjects. The social pressures affecting research projects have been studied by Colarelli and Siegel through case study approach.

1.7.3

Advantage of a Case Study

(i) Being an exhaustive study of a social unit, the case study method enables us to understand fully the behavior pattern of the concerned unit. (ii) Through case study a researcher can obtain a real and enlightened record of personal experiences which would reveal man’s inner strivings, tensions and motivations that drive him to action along with the forces that direct him to adopt a certain pattern of behavior. (iii) This method enables the researcher to trace out the natural history of the social unit and its relationship with the social factors and the forces involved in its surrounding environment. (iv) It helps in formulating relevant hypotheses along with the data which may be helpful in testing them. A case study thus enables the generalized knowledge to get richer and richer. (v) The method facilitates intensive study of social unit which is generally not possible if we use either the observation method or the method of collecting information through schedules. (vi) Information collected under the case study method helps a lot to the researcher in the task of constructing the appropriate questionnaire or schedule for the said task which requires thorough knowledge of the concerned universe. (vii) The researcher can use one or more of the several research methods under the case study method depending upon the prevalent circumstances. In other words, the use of different methods such as depth interviews, questionnaires, documents, study reports of individuals, letters and the like is possible under a case study method. (viii) A case study method has proved beneficial in determining the nature of units to be studied along with the nature of the universe.

Introduction to Design of Systems

11

(ix) This method is a means to understand well the past of a social unit because of its emphasis of historical analysis. Besides, it is also a technique to suggest measures for improvement in the context of the present environment of the concerned social units. (x) A case study method enhances the experience of the researcher and this in turn increases his analysing ability and skill.

1.7.4

Limitations of a Case Study

(i) Case situations are seldom comparable and as such the information gathered in a case study is often not comparable. Since the subject under the case study tells history in its own words, logical concepts and units of scientific classification have to be read into it or out of it by the investigator. (ii) Read Bain does not consider the case data as significant scientific data since they do not provide knowledge of the “impersonal, universal, non-ethical, non-practical, repetitive aspects of phenomena.” Real information is often not collected because the subjectivity of the researcher does not enter in the collection of information in a case study. (iii) The danger of false generalisation is always there in a view of the fact that no set rules are followed in collection of the information and only few units are studied. (iv) It consumes more time and requires a lot of expenditure. More time is needed under a case study method since one studies the natural history cycles of social units too minutely. (v) The case data are often vitiated because the subject according to Read Bain, may write what he thinks the investigators wants; and the greater the report, the more subjective the whole process is. (vi) A case study method is based on several assumptions, which may not be very realistic at times, and as such usefulness of case data is always subject to doubt. (vii) A case study method can be used only in a limited sphere; it is not possible to use it in case of a big society. Sampling is also not possible under a case study method. (viii) Response of the investigator is an important limitation of the case study method. He often thinks that he has full knowledge of the unit and can himself answer about it. In case the same is not true, then consequences follow. In fact, this is more the fault of the researcher rather than that of the case method.

Exercise 1. 2. 3. 4.

1 . . .

Why should we adopt system based design? Explain in brief. Why should mechanical system be designed on system basis? Define mechanical system design. Give examples using your own numbers. Write notes on: (a) Mechanical System (b) Electrical System (c) Translational System (d) Rotational System (e) Design

12 5. 6. 7. 8. 9.

Mechanical System Design Define system concept and also classification of system. Briefly explain system engineering in Mechanical Systems Design. What is designing? How can you checklist for an engineering design problem? Explain system design in brief. What is system concept? Write down application of systems and types of systems.

A Case Study 1. 2. 3. 4. 5. 6. 7.

Define case study. Why do we use case study? Briefly explain in the language of MSD. Write down the advantages of case study. What is the applications of case study? Briefly explain advantages and limitations of case study. What is the role of case study? Write down a suitable step of a case study in the field of mechanical engineering. Also write necessary condition of case study. 8. What is the statement problem of a case study? 9. A case study is used for various departments as a Mechanical, Electrical, Electronics and Computer Engg., etc. What is the basic concept of a case study on the basis of mechanical system design?

Engineering Processes and the System Approach

13

2 Engineering Processes and the System Approach 2.1

INTRODUCTION (SYSTEM APPROACH)

The system approach is such a technique and represents a broad based systematic approach to problems that may be interdisciplinary. It is particularly useful when problems are complex and affected by many factors, and it entails the creation of a problem model that corresponds as closely as possible in some sense to reality. There are several reasons why the system concept and approach have become more and more important in our life, as follows: 1. Recently all organisations and entities such as Manufacturing, Management, Economics, Politics, International affairs, etc. have tended to become large and global; hence, it has become necessary to consider everything systematically in connection with its surroundings to achieve its functions and objectives. 2. The recent progress of computers and information networks has enhanced the ability to gather, store, process, and transmit a large variety and quantity of data/information globally in less time than previously. This has contributed significantly to the solution of complex problems. In the area of manufacturing, Computer-Integrated Manufacturing (CIM) systems now play a role in automating the flow of information. 3. Optimisation techniques, such as operations research, management science, and system engineering and simulation techniques have been developed with the aid of these soft sciences or technologies. System thinking and optimisation and hence rational and logical decisionmaking to provide optimum solutions for large-scale systems/problems, have become possible.

2.2

APPLICATION OF SYSTEMS CONCEPTS IN ENGINEERING

A system concept of engineering gives a systematic application approach to get the task accomplishment more efficiently, effectively and economically.

14

Mechanical System Design

System approach is an organised approach for complex equipment design and the same can be completed in a much shorter duration with comparatively less efforts. System concept furnishes a true reference which tell, how to manage the jobs or how to analyse complex phenomena under different environments. System approach is more common in the field of physical science and engineering because it is comparatively easier to build a model of such system engineering primarily concerned with informative systems. The system concept does not provide a set of rules for solving all problems but it is a useful device for viewing any phenomena. The system concept of engineering involves the analysis and synthesis of system. It includes the designing of complex and highly engineered system in different environment system under approach. The term system is used in such a wide variety of ways that it is difficult to provide a definition broad enough to cover the many uses and at the same time concise enough to serve a useful purpose. We begin therefore with a simple definition of a system and expand upon it by introducing some of the terms that are commonly used when discussing a system. A system is defined as an aggregation or assemblage of objects joined in some regular interaction or interdependence. While this definition is broad enough to include static systems, the principal interest will be in dynamic systems where the interactions cause changes over time. As an example of a conceptually simple system, consider an aircraft flying under the control of an autopilot. A gyroscope in the autopilot detects the difference between the actual heading and the desired heading. It sends a signal to move the control surfaces. In response to the control surface movement the airframe steers forward the desired heading. As a second example, consider a factory that makes and assembles parts into a product. Two major components of the system are fabrication department making the parts and the assembly department producing the products. A purchasing department maintains a supply of raw materials and a shipping department dispatches the finished products. A product control department receives orders and assigns work to the other departments. In looking at these systems, we see that there are certain distinct objects, each of which possesses properties of interest. There are also certain interactions occurring in the system that cause changes in the system. The term entity will be used to denote an object of interest in a system; the term attribute will denote a property of an entity. There can of course be many attributes to a given entity. Any process that causes changes in the system will be called an activity. The term state of the system will be used to mean a description to all the entities, attributes, and activities, as they exist at one point in time. The progress of the system is studied by following the changes in the state of the system. In the description of the aircraft system, the entities of the system are the airframe, the control surfaces, and the gyroscope. Their attributes are such factors as speed, control surfaces angle and the gyroscope setting. The activities are the driving of the control surfaces and the response of the airframe to the control surface movement. In the factory system, the entities are the departments, orders, parts and products, the activities are the manufacturing processes of the departments, and attributes are such factors as the quantities for each order, type of part or number of machines in a department. In a Table 2.1 lists examples of what might be considered entities, attributes and activities for a number of other systems. If we consider the movement of cars as a traffic system, the individual cars are regarded as entities, each having as attributes as speed and distance traveled. Among the activities is the driving of a car. In the case of a bank system, the customers of the bank are entities with the

Engineering Processes and the System Approach

15

balances of their accounts and their credit statuses as attributes. A typical activity would be the action of making a deposit. Other examples are shown in a Table 2.1.

Table 2.1

Examples of Systems Traffic

Cars

Speed, distance

Driving

Bank

Customers

Balance

Deposting Credit status

Communications

Messages

Length

Transmitting priority

Supermarket

Customers

Shopping list

Checking out

The table does not show a complete list of all entities, attributes, and activities for the system. In fact a complete list cannot be made without knowing the purpose of the system description. Depending upon that purpose, various aspects of the system will be of interest and will determine what needs to be identified.

2.3

IDENTIFICATION OF ENGINEERING FUNCTIONS OF SYSTEMS

Following are the identifications of the engineering functions of systems: 1. There is some specific purpose or function that must be fulfilled or performed. 2. There are a number of components (at least two) that can be identified as necessary ingredients of the problem. Furthermore, each component has a variety of attributes that implicitly, physically, and behaviorally are necessary for its description. 3. The components are interrelated in some manner satisfying interface consistency between the components. 4. There are constraints that restrict the system’s behavior and the individual component response. In addition to identifying components, the interactions between the components must be defined. The components and the interactions define the structure of the system. The structure is extremely important because it identifies the manner in which the system behaves when some parts of the system are stimulated or modified. In essence, the structure is a key to determine system response to identification of the function. 5. In all design situations, the first task of the designer is to determine what exactly is needed. This identification of the need, simple as it may appear is a crucial first step in the design process, and which is the tripping point for many projects. 6. It frequently happens that a designer starts thinking of the solutions even before he has clearly identified the need and thus he is foredoomed to take the wrong track. His erroneous definition of the problem dictated by the broad pattern of the solution that he has pre-conceived, forces him to seek solutions of the sub-problems which result from the general pattern he has assumed. 7. Design decisions are decisions under uncertainty and insufficient information. The uncertainty arises out of two sources, (1) in which it is due to the reason that the additional information required depends upon the decision itself, and (2) in which it is due to statically random processes.

16

Mechanical System Design 8. Also gathering information costs time, money and effort—after some point the cost of additional information increases out of proportion to its worth. The point at which a designer stops gathering information and prefers taking a design under certainty is a decision in itself.

2.4

THE CHARACTERISTICS OF A SYSTEM IN “MSD”

Their are four types of a system on the basis of characteristics. 1. 2. 3. 4.

Linear and non-linear system Static and dynamic system Time invariant and time varying system and Distributed parameter and lumped parameter system

(1) Linear and Non-linear System If the input x and the output y of a system are related by a straight-line shown in Fig. (2.1) we may say that it is a linear system. y=mx

Y

y2 Output

y1

x1

x2

X

Input

Fig. 2.1 A spring obeying Hook’s law or a resistor obeying Ohm’s law are examples of linear system. For a linear system if an input x1 (t ) gives an output y1 (t ) i.e. x1 (t ) → y1 (t ) then k1 (t ) → ky1 (t ). This property is called homogeneity and is a property of all linear systems. Another condition for linearity is the property of superposition. This implies if x1 (t ) → y1 (t ) and x2 (t ) → y2 (t ) , then [ x1 (t ) + x2 (t )] → [ y1 (t ) + y2 (t )]

A system is called linear if it has homogeneity and superposition properties. Non-linear algebraic equations like y (t ) = x 3 (t ) or y (t ) = x (t ) clearly fail to satisfy the superposition property and hence cannot represent linear systems. Some of the non-linearities commonly encountered in engineering systems are:

(a) Spring Type Non-linearities Applied force ( f ) = k1 x (linear spring)

Engineering Processes and the System Approach

17

Applied force f = k1 x + k 2 x3 (hard and soft spring) k1 is +ve for hard spring k2 is –ve for soft spring Characteristics of linear, soft and hard spring are shown in Fig. 2.2 by displacement vs. force y Soft spring (k 2 ) Linear spring

Hard spring (k 1)

Displacement

x Force

Fig. 2.2

Displacement vs. force characteristics

Saturation is shown in Fig. 2.3

Output

Input

Fig. 2.3

Saturation

Transistors, operational amplifiers, magnetic and electromagnetic components all suffer from this type of non-linearity system.

(2) Static and Dynamic System Consider a resistive network as shown below:

V (t ) = KV (t ) R2 + R3 The output at any instant depends upon the input at that instant. Such systems are static or memoryless.

In this system: i (t ) =

18

Mechanical System Design R2 Output i (t ) Input v (t )

R3

R1

Fig. 2.4

Resistive network

Dynamic system models are given by differential equations or difference equations. In such equations, time is the independent variable and the output is a function of time even when the input is constant. In such systems the output depends not only upon the input but the initial conditions also. Presence of energy storage elements e.g. spring capacitor or inductor in an electrical circuit makes this system dynamic. A sudden change in the input of the system variables in a dynamic system will not change instantaneously.

(a) Friction Viscous Characteristics For friction viscous, f = Kv where K = Coefficient of friction viscous; v = velocity of viscous friction/s liquid friction Viscous friction arises in situations like motional solids, through gases or liquids, relative motion of well-lubricated surface, etc. Viscous friction Sliding friction

f

V

Fig. 2.5

Friction viscous

(b) Hysteresis Characteristics Flux

Current

Fig. 2.6

Hysteresis

Engineering Processes and the System Approach

19

When the increasing portion of the input-output relationship follows different paths, we get hysteresis type of non-linearity. Similar type of non-linearity also arises in case of backlash of gear trains, relays and electromechanical devices.

(c) Dead Zone Characteristics This type of non-linearity is constructed in relays, motors and other types of actuators. Output

Input

Dead Zone

Fig. 2.7

Dead zone

(3) Time Invariant and Time Varying System If we define parameters of the system as the quantities which depend only on the properties of the system elements and not on its variables, these parameters appear as coefficients of independent variable and its derivatives in the differential equation model of dynamic systems. L

d 2i di i +R + =0 2 dt c dt

Parameters R, L, C, appear as coefficients if these parameters can be functions of independent variable (t) without affecting the property of linearity. If these parameters find the system, it is of time invariant type. When one or more parameters are functions of dependent variable time, the system is called a time varying system. Equation of motion of a rocket in flight is a good example of time variant system. M (t ) = Mass after time variant ‘t’ = ( Mo − kt )

Equation of motion M (t )

d2x dx +D =F 2 dt dt

M (t) = vertical displacement D = coefficient of friction between rocket body and atmosphere F = thrust developed by the rocket.

(a) Continuous Time Function In a system, if the variables of the system are continuous functions of time, such a system is called continuous time system.

20

Mechanical System Design x (t )

t

Fig. 2.8

(a) Continuous time function

(b) Discrete Time Discrete time functions arise in modeling of instrumentation and control systems employing multiplexing, computer control or digital signal processing. If the value of a signal is defined only as a set of instants or it, changes only at specified instants, the signal is called the discrete time signal. x (t, n)

t1

Fig. 2.8

t2

t3

tn

t4

(b) Discrete time function

(4) Distributed Parameter and Lumped Parameter System Let us consider a thermal system, like heating an iron slab. Heating is from one end, there is no dissipation of heat from the sides and heat flows in one direction only. Temperature T = T(t, x) is a function of two independent variables of t (time) and x (distance). Differential equation relating the input heat rate to output variable T(t, x) will be a partial differential equation. If we treat T = T(x, t) as continuously distributed in space, the system will be called as distributed parameter system.

Heat Source

Fig. 2.9

Iron Slab

Distributed parameter and lumped parameter system

Engineering Processes and the System Approach

21

On approximation if we divide the region of heat flow into a finite number of distinct regions and assume that the temperature of each region is the same over the whole region, in that case we have less number of temperature variables as function of only time. Such a system is called the lumped parameter system.

2.5

ENGINEERING ACTIVITIES MATRIX

Matrix in engineering activities is used when an organisation structure has to handle a variety of projects, ranging from small to large and when a pure project structure is superimposed on a functional structure, the result is a matrix structure. In other words, the engineering in activities matrix is a project matrix plus (+) a functional engineering activities, the project structure provides an horizontal lateral dimension to the traditional vertical orientation of the organisation structures. Organization Structures

Engg. Engg/group Engg/group

Production Proj I Prod/group Proj II Prod/group

Purchase Purchase/group purchase/group

Finance Finance/group Finance/group

Personnel Personnel/group Personnel/group

Fig. 2.10 The project terms are composed of persons drawn from the functional department for the duration of the project. When their assignment is over they return to their respective departments. During the project duration, such persons have two bosses, one from the functional department and second of the concerned project. Let us consider an engineering matrix of second order

a The symbol  1  a2

b1   b2  y  a1   b1    and   a  2  b2 

Functional Structure (direction of columns)

and

[a1 b1] [a2 b2 ]

x Project structure (Direction of Rows)

Fig. 2.11

Engg activity matrix functional structure vs project structure

22

Mechanical System Design

Consisting of 22 numbers called elements arranged in two rows and two columns, is called a matrix of second order. The functional structure is the direction of 1st and 2nd columns i.e., in the

a  b  vertical direction of  1  and  1  respectively of engineering matrix. While the project structure is the  a2  b2  direction of 1st and 2nd rows i.e., in the horizontal direction of [ a1 b1 ] and [ a2 b2 ] respectively of engineering matrix.

2.5.1

Matrix Organisation

The pure matrix organisation evolves from above setup when arrangement of sharing authority between project manager and functional manager is formalised. In this structure, different projects (rows of matrix) borrow resources from functional areas (columns). Senior management decides whether the project manager has little, equal or more authority than the functional manager with whom they negotiate for resources. The personnel working on a project have a responsibility to their functional superior as well as project manager, which means that the authority is shared between project manager and functional manager. The project manager integrates the contribution of personnel in various functional departments towards realisation of project objectives. The matrix form of organisation is incongruent with the traditional origination theory since there is dual sub-ordination, responsibility and authority are not commensurate, and the hierarchical principle is ignored. The matrix form of organisation involves greater organisational complexity and creates inherently conflictive situation, yet it is effective for simultaneous pursuit of twin objectives: efficient utilisation of resources and effective attainment of project objectives. Some of the advantages of pure matrix organisation are as follows. It enables project control over all resources, including cost and personnel. Policies can be set up independently provided that they do not contradict company policies. Authority to commit company resources by scheduling rests with the project manager. Rapid responses are possible to changes, conflicts and needs. Each person can be shown a career path even at the end of project. Key people can be shared thereby minimizing the costs. Strong technical base can be developed with knowledge being available for all projects on an equal footing. Better balance is possible between time, cost and performance. Rapid development of specialists and generalists occurs. Authority and responsibility are shared. Some of the disadvantages of a pure matrix organisation are as follows. (a) It enables multidimensional informational flow and work flow. (b) Reporting to multiple managers with continuously changing priorities. (c) Management goals may differ from project goals.

Advantage 1. It effectively focussed resources on single project permitting buffer planning and control to meet the deadline. 2. It is more flexible than a traditional functional hierarchy. 3. Service of specialists are better utilised as more emphasis is placed on the authority of knowledge than rank of the individuals in the hierarchy.

Limitation 1. It provides the principle of unity of command as a person works under two or more bosses. They give rise to conflicts.

Engineering Processes and the System Approach

23

2. Organizational relationship are more complex and or it creates problem of co-ordination. 3. Once the person is drawn temporarily from the different departments, the boss decides to have the line authority. 4. Project group is heterogeneous and due to which morale of the personnel may be low.

2.6

DEFINING THE PROPOSED EFFORT

Proposed effort determining the specific form of the end product, its size, shape, properties. Proposed effort defining the specific emphasis or character of the planning effort that is relevant to the situation, i.e., how much investigation, how many feasibility studies. Engineering proposed effort is often concerned with the detailed specification of the facility components and their interrelationships with one another. It requires consideration of the physical laws of nature and the properties of materials and equipment. In many cases, however, the end product of proposed effort process may be a specialized plan such as a transportation plan, a community development scheme, or the construction plan of the building.

2.7

ROLE OF ENGINEER IN “MECHANICAL SYSTEM DESIGN”

The engineer’s checklist for solving design problems are usually a compilation of ideas, associations and questions, which help and spur the ideation. In a sense, these act as reminders that there may be more than one way of looking at things. For engineering design situations a useful checklist would be one which is designed to incorporate the ways an engineer looks for design solutions. A lot of research into the patterns of creativity is being done at present. Some of things that act as catalysis in innovative designing are the following: (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k)

Memories of past design. Competitors’ product. Deliberate doodling and daydreaming. Self-association. Analogies. Word-association. Science friction. Deliberate distortion. Trying to describe the process. Deflection of problem. Use of formal proposals.

An engineer is expected to exhibit a variety of different talents and backgrounds. They include: 1. Ability to work in a team. 2. Greater ability to communicate and ‘sell’ one’s ideas—orally, electronically and on paper—not only among fellow employees, but to the supplier and the customers. 3. More manufacturing experience or atleast ability to work with and communicate with manufacturing persons.

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Mechanical System Design 4. Greater flexibility, performing more and different tasks. 5. Greater computer-aided design and manufacturing (CAD/CAM) background and some experience in solid modeling. 6. Greater creativity. 7. Increased problem-solving or project experience. 8. Awareness of environmental concern and knowledge of related laws. 9. Ability to keep up with rapidly changing technology. 10. Advisor/consultant: Available to others for interpretation of data, viewing. 11. Advocate activity: Promote activity a process or approach. 12. Analyst: Separate a whole into parts and examine them to explore for insight and characteristics. 13. Boundary spanner: Bridge the information/in length gap between engineering. 14. Motivator: Provide structure and skill availability to a group included. 15. Decision making: Select a preference from among many alternatives for topic of concern. 16. Designer: Produce the solution specification planner. 17. Expert: Provide a high level of knowledge, skill and experience of specific topic. 18. Co-ordinator and Integrator: Make a relationship between engineer and worker for co-ordinator and integrator skill. 19. Innovator/Inventor: Seek to produce a creative or advanced technology solution. 20. Measure: Obtain data and facts about existing condition. 21. Project managers: Operate, supervise and evaluate projects. 22. Trainer/Educator: In the skills and knowledge of concerned field of engineering.

It is observed from the above list of skills, the design engineer job is much more difficult than the engineers in production or marketing department. It is however more challenging and creative, giving much more satisfaction and rewards.

2.8

ENGINEERING PROBLEM SOLVING

In approaching and solving a problem, one of the initial challenges the engineer faces is to understand the nature of the problem, the environment in which it exists, and the response phenomena that are associated with the problem and the environment. The nature of the problem can provide an indication of the intrinsic factors that are involved in creating the problem. Thus the solution of the problem must also consider these same factors. The environments consists of the setting that contains or surrounds the problems.

2.8.1

Problem Area and its Environment 1. An interruption in a free way traffic in a city have an impact not only on the use of that freeway but the city’s entire transportation system. 2. Failure to receive material. 3. Failure of machine parts. 4. Motivation of a sum part of foundation.

Engineering Processes and the System Approach

2.9

25

CONCURRENT ENGINEERING (CE)

Concurrent Engineering has been recognised as a variable approach in which simultaneous design of a product and all its related processes in a manufacturing system, are taken into consideration ensuring required matching of the product’s structural with functional requirements and associated manufacturing implications. It is also known as Parallel or Simultaneous Engineering. It helps in early launch of product by visible product lead-time reduction. The central notation is that the team responsible for conceptualising the product does so correctly thereby dramatically reduces the changes to be made later. It strives to do right job the first time. Also the team manages parallel processing to reduce the delays and waste. If the design team makes decisions without having adequate knowledge of manufacturing process, corrections are needed in downstream process, which are expensive and time-consuming. Concurrent design is simultaneous planning of product and process of producing it. Cross-functional teams of experts from all relevant departments including marketing, materials, design and manufacturing simultaneously manages the entire development process. So the design team includes product design engineer, marketing manager or product marketing manager, production engineer, design engineer, testing engineer, materials engineer, quality control specialist, industrial engineer, assembly engineer and suppliers representative, etc. Prior to 1980’s, team formation was not a preferred idea and many designers worked in isolation. The role of manufacturing was to build what the designer conceived, improve the manufacture and assembly of the product. The reasons for difference between drawing and manufacturing parts varied because: (i) Drawing incomplete. (ii) Parts could not be made as specified. Modern Approach to Product Design (iii) Drawing was ambiguous. (iv) Parts could not be assembled if manufactured as drawn.

Traditional and CE Product Development Cycles Traditional and CE Product Development Cycles are shown in Fig. 2.12.

Traditional View of Design Refer Fig. 2.13.

Concept of Concurrent Engineering Refer Fig. 2.14

Some of the Goals of Concurrent Design Process (i) (ii) (iii) (iv) (v)

From start include all domains of expertise as active participants in design effort. Resist from making irreversible decisions before they are made. Perform continuous optimisation of product and process. Focus on component design for manufacturing assembly. Integrate manufacturing process design and product design that best match need requirement.

26

Mechanical System Design User

Market analysis R and D

Process planning

Design

Manufacturing

A sense of Engineering change order Users Concurrent design of product and processes

Market Analyses Rxd

Manufacturing

Advanced Product Modelers

Life-Cycle

Manufacturing

Process planning

Assemblability

Testability CE

Engineering Process Reliability and Maintenance

Fig. 2.12

Cost

Ergononics

Product development cycle employing concurrent engineering wheel. (CE)

Engineering Processes and the System Approach

27 Market’s needs

Product performance specification

Product design Production system technology

Production system specification

Investment design methods

Production system design

Production cost model

Fig. 2.13

Traditional view of design Inspection

Marketing

Manufacturing

Serviceability

Design Coordinator

Sales

Packaging

Assembly

Function

Fig. 2.14

2.9.1

Concept of concurrent engineering

Advantages of Concurrent Design

The application of C.E. approach results in reducing product design and development time and speeds the product to market. Recent example is that of Reliance Industries where involvement of design team reduced per-market time from 120 days to 14 days. This approach is also named as Business Process Re-engineering (BPR). C.E. helps to reduce cost, improve efficiency, reduction in throughout time of product development process.

28

Table 2.2

Mechanical System Design

Productivity and quality ranking for automobiles manufactures By Fundamental of CE shown below the Productivity and Quality Ranking for Automobile Manufactures S. No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Company Suzuki Toyota Daihatsu Honda Mitsubishi Mazda Isuzu Nissan Fuji (Suburu) Ford Chrysler Peugeot General Motors Volkswagon

Vehicles/Employee per Year

Quality Problems Per 100 Cars

70.4 61.0 57.0 56.2 50.4 42.0 41.7 39.5 38.7 20.0 18.0 13.3 12.5 11.2

0 117 – 112 – 133 – 111 – 149 176 – 169 –

UPTU 2005 Q. (1) (f) Explain a system design where environment and safety is of prime consideration. Solution: A system is often affected by changes occurring outside the system. Some system activities may also produce changes that do not react on the system. Such changes occurring outside the system are said to occur in the system environment. An important step in modelling system is to decide upon the boundary between the system and its environment. The decision may depend upon the purpose of the study. In the case of the factory system, for example, the factors controlling the arrival of order may be considered to be outside the influence of the factory and therefore part of the environment. However, if the effect of supply on demand is to be considered, there will be a relationship between factory output and arrival of orders, and this relationship must be considered an activity of the system. Similarly, in the case of a bank system, there may be limit on the maximum interest rate that can be paid. For the study of a single bank, this would be regarded as constraint imposed by the environment. In a study of the effects of monetary laws on the banking industry, however, the setting of the limit would be an activity of the system. The term endogenous is used to describe activities occurring within the system and the term exogenous is used to describe activities occuring in the environment that affect the system. A system for which there is no exogenous activity is said to be a closed system in contrast to an open system, which does have exogenous activities.

Engineering Processes and the System Approach

2.10

29

A CASE STUDY: VISCOUS LUBRICATION SYSTEM IN WIRE DRAWING

The process of wire drawing is an important one in the metal working industry. Drawn wire is used to produce nails, coat hangers, electrical wire, and a thousand other products. The process itself is simply that of making a longer, thinner wire from a shorter, fatter one. Usually the lengths involved are so long that the process can be considered essentially continuous, speeds vary about 100 m/second for small softwire to about 1m/second for a large, hard wire. Lubrication is very important to the wiredrawing process as it is to all metal-working process. In wire drawing, lubrication conventionally is done by allowing the wire to pass through a soap box just prior to its passing through the die. The soapbox contains a dry, powdered mixture of various hydrocarbons and often contains the equivalent of ordinary laundry soap. Companies, and even individual machine operators, have their own mixtures. Lubrication in wire drawing is very much an art today. Wire-drawing is a major expense in producing drawn wire, not only because of the cost of the dies, but also because of the production lost during frequent die changes. The kind of lubrication, which results when a soapbox is used, is called boundary lubrication. It is characterized by the presence of “slippery” stuff (e.g. soap) between the moving surfaces. The slipperiness is caused by chain molecules in the lubricant, which orient themselves, normal to the surfaces. Since such molecules are weak in shear, little resistance is offered to the sliding motion. Unfortunately, it is difficult to get a very thick film of such lubricants between the surfaces and hence the separation of the surfaces is not large. A close-up of the situation might look schematically as shown in Fig. 2.15. In boundary lubrication, surface-to-surface contact is not totally prevented, because the separation of the surfaces of the same magnitude as the surface roughness. (a) Stationary surface

Velocity

Fig. 2.15

(b) Stationary surface

Velocity (v )

The difference between boundary lubrication and full-film lubrication

In a viscous lubrication system in wire drawing operation, in addition to the work load and power required, the maximum possible reduction without any tearing failure of the workpiece is an important parameter. In the analysis that we give here, we shall determine these quantities. Since the viscous lubrication system in wire drawing operation is mostly performed with rods and wires, we shall assume the workpiece to be cylindrical, as shown in Fig. 2.16. A typical drawing die consists of four regions, (i) (ii) (iii) (iv)

A bell-shaped entrance zone for proper guidance of the workpiece A conical working zone A straight and short cylindrical zone for adding stability to the operation and A bell-shaped exit zone.

30

Mechanical System Design Lubrication

Die

r d1

α

x

O x

F dr

dx Job

Fig. 2.16 l l l l l

Drawing of cylindrical rod

The final size of the product is determined by the diameter of the stabilizing zone, dr The other importance die dimension being the half-cone angle, α Sometimes, a back tension is provided to keep the input workpiece straight, Fb The work load, i.e., the viscous lubrication system in wire drawing force, F A die can handle jobs having a different initial diameter, di

which, in turn, determines the length of the job-die interface. The degree of a viscous lubrication in a wire drawing operation is normally expressed in terms of the reduction, D factor in the cross-sectional area. Thus D =

=

Ai − Ar Ai di2 − d r2 di2

d  =1−  r   di 

2

where Ai and Ar being the initial and the final cross-sectional area of the workpiece.

Exercise 1. 2. 3. 4. 5.

2 . . .

What are the characteristics of a system? Give their importance in system design. UPTU 2005 What are the attributes of a system? Explain with suitable example. UPTU 2005 Explain engineering activities matrix. In what way it helps in system design? UPTU 2005 Why is the system approach becoming popular in the study of engineering problems? UPTU 2004 Explain the basic concept of concurrent engineering. UPTU 2004 What are major advantages of implementing the concurrent engineering for product design and production? 6. Describe a basic framework to be used for analyzing an engineering system. Give a list of optimization techniques used for the analysis of the system. UPTU 2004 7. Discuss the basic problems concerning systems. Illustrate how these problems can be made use of in case of an input–output system. UPTU 2004

Engineering Processes and the System Approach

31

8. Explain the system design where environment and safety is of prime consideration. 9. Write a short notes on: (a) Linear and non-linear system (b) Static and dynamic system (c) Friction viscous (d) Hysteresis (e) Dead zone (f) Time invariying system (g) Continuous time and discrete time system (h) Lumped parameter and distributed parameter system. 10. What is the role of engineer in mechanical system design? 11. What is the application of system concept of engineering in mechanical system design? 12. Explain brief identification of engineering of function and system approach engineering. 13. Defining the proposed effort, how can you solve the engineering problem in mechanical system design? 14. Give the basic aspects of concurrent engineering. How does it help in product design and development? Illustrate with suitable example. UPTU 2005 15. What do you mean by holistic approach?

A Case Study 1. Briefly explain viscous lubrication system in wire drawing. 2. Write the assumption of viscous lubrication of wire drawing.

32

Mechanical System Design

3 Design and Problem Formulation 3.1

INTRODUCTION

Design can be defined in any of the following ways: 1. The process by which we generate ideas in response to a certain need, and then carry out their transformation optimally into a reality. 2. It is a process of selectively applying the total spectrum of science and technology to the optional attainment of end results, which serves a valuable purpose. 3. The process of creating and innovating an optimum idea that is to be put to the service of mankind and society. 4. To suggest or outline ways to put together man-made things, or to suggest modifications in man-made things to satisfy optimally (under given constraints) some specified human need. 5. It is an iterative decision-making activity to produce the plans by which resources are converted optimally, into systems or devices to meet human needs. Given unlimited resources (such as time, money, human skill, technology), one can always aim for an ideal product. However, a designer always operates in a limited resource environment constraints. He has to contemplate only the available technology, the available skill of labour, and complete his work within the stipulated time and budgeted money. So, he can only produce the best possible one, within these given constraints. This process of creating the best product under the given constraints is known as optimization. Hence, optimization is always a part of designing, as can be seen from the definitions. Engineering product design being practical in nature, is concerned only with what is feasible. Considerations of physical reliability, cost effectiveness, financial feasibility and utility are a necessary requirement. Flow chart that show the difference between the process of Scientific Investigation and Design is shown below. Scientific investigation starts as a result of the curiosity the person has about his surrounding. Hence a scientist studies the work as it is. The outcome of his work is the accumulation of scientific knowledge and a better understanding of our world.

Design and Problem Formulation

3.1.1

33

Design

But design begins with the realization of unfulfilled needs of the society and ends with satisfying them. Thus a designer creates the world that has never been. The outcome of designer’s work is a better world for us. However, the knowledge of science and technology is required for design (See Figure). Curiosity

Need

Inputs Existing knowledge Technical Facilities Hypothesis Analysis

Scientific Enquiry

Better Understanding of world around

Engineering Design Process

Better world for us

Science Engg. Science & tech Ergonomics Aesthetics Sociology Psychology Economics Ecology Design

Scientific Investigation

Fig. 3.1

3.1.2

Design

Distinction between Engineering and Design

Engineering is a profession with high degree of interaction with the society, the core section of which is Designing. Engineering is concerned with keeping the material world of man adequately supplied. This calls for the following sequence of activities:

Engineering Research 1. 2. 3. 4.

Development Design Production Sales engineering

As we go down this sequence, the abstraction, which is larger at the beginning, gradually decreases, and particularity, which is smaller at the beginning, increases. As we can see, the designing, being in the middle of the sequence, implies equal importance both for abstraction and particularity and hence is the most challenging activity.

3.1.3

Problem Formulation

Introduction Engineers traditionally provide a service function to society by meeting societal needs within available resources and technology. Planning encompasses certain activities that specify how the product will be achieved.

34

Mechanical System Design

An engineer may be required to consider the following aspects in relation to a particular facility: profit, cost, marketability, quality, reliability, performance, life, simplicity and elegance, as well as political and social acceptability. The engineer must define a model that represents his concept of the problem. Based on this problem model, the engineer must develop an analysis and design procedure that enables him to define the problem and select the best solution to it. Finally, he must be concerned with the difficulties that are associated with the actual implementation of the conceived solution. A facility with an engineer involved is in reality part of some large entity.

3.2

NATURE OF ENGINEERING PROBLEM

It is not difficult to trace in general terms the various roles of engineering in the industrial and automation revolution. Civil Engineering checks up highway, bridges, and railways and builds port and cities. Next the development part of complex industrial society required increasing nature of Mechanical Engineering. Now the need for better controls, automation, the great increase in communication problem and many new materials have increased the need for electrical engineers. Projecting a little, one can predict a rise in the physical science, a degree of social concern and knowledge not previously required in engineering. 1. 2. 3. 4.

Food: High production low cost compact food manufacturing systems are needed. Water: Is a problem in advanced and developing societies, more fresh water is needed. Air Pollution: In metropolitan cities, creates a pollution problem. Communications and Transportation: Several problems in communications and transportation in a city by not following traffic rule and regulation. 5. Recreation and old age: Obviously recreation is a problem in old age.

In approaching and solving a problem, one of the initial challenges the engineer faces is to understand the nature of the problem, the environment in which it exists and the response phenomena that are associated with the problem and the environment. The nature of the problem can provide an indication of the intrinsic factors that are involved in creating the problem. Thus the solution of the problem must also consider these same factors. The environment consists of the setting that contains or surround the problem.

3.2.1

Problem area and its environment 1. An interruption in freeway traffic in a city can have an impact not only on the use of that freeway, but also on the city’s entire transportation system. By the same token, freeway improvement could affect the entire transportation system. 2. Failure of a machine part can cause a malfunction or limit the operation of that machine such that production is greatly reduced or stopped. 3. Failure to receive material that is required, can delay construction of a facility and result in the temporary unemployment of construction workers. 4. Modification of some part of a foundation can have widespread implications on the loads the building can sustain and therefore, limit the use of the building.

3.3

NEEDS STATEMENT

In all design situations the first task of the designer is to determine what exactly is needed. This identification of the need, simple as it may appear, is the crucial first step in the design process, and which is the

Design and Problem Formulation

35

tripping point for many project. It frequently happens that a designer starts thinking of the solutions even before he has clearly identified the need and thus he is foredoomed to take the wrong track. His erroneous definition of the problem dictated by the broad pattern of the solution that he has preconceived, forces him to seek solutions of the sub-problems which result from the general pattern he has assumed. Design decisions are decisions under uncertainty and insufficient information. The uncertainty arises out of two sources, one in which it is due to the reason that the additional information required depends upon the decision itself and two in which it is due to statistically random process. Also, gathering information costs time, money and effort and after some point the cost of additional information increases far out of proportion of its worth. The point at which a designer stops gathering information and prefers taking a decision under certainty is a decision in itself. Hence, the engineer may be required to consider the following aspects in relation to particular facility: profit, cost, marketability, quality, reliability, performance, life, simplicity, safety, and elegance. The basic need of design and planning process that the engineer utilizes in solving society’s problem is shown in figure given below: Investigation of problem Problem model

Real world environment Research data gathering feasibility studies

Social need

Design and analysis

Implementation project planning

Fig. 3.2

3.4

Optimum plan and design system understanding

Need of design and planning process

IDENTIFICATION AND ANALYSIS OF NEED

The beginning of any design process is the recognition of a need. The sources from which need perception results are as varied as the needs themselves. The designer may recognize a need all on his own or his employer or customer may communicate it to him. He may perceive the need of better and safer hand tools while trying to repair a faucet in his bathroom, or he may get all set to eradicate mice from the country after reading a newspaper report that almost as much grain is eaten by mice as by men in his country. Some needs may just be the perception of a lazy man who wants a robot to fetch him a glass of cold water while he relaxes in a comfortable chair. Recognizing a need is in itself a creative process. It takes organized thinking and a disciplined mind, forever on the lookout for challenging situations, to perceive most needs. Many of the great designers had an uncanny sense of recognizing the needs of society much before society itself realized them. Thomas Edison was one such designer who went through great lengths in developing the electric lamp

36

Mechanical System Design

much before any significant segment of society desired it. In the case of many of the modern household appliances the product was first developed on the hunch that the public would appreciate its need and only then was it released as a labor-saving miracle. Clearly, such anticipation and projection of the unexpressed and latent needs of society calls for acute powers of observation and inference, and promises ample material rewards in societies, which recognize individual initiatives. Many of the design problems encountered by a practicing engineer are however those that are communicated to him by his employers or customers. The need may be a basic need like the need for a faster, cheaper and lighter vehicle, or it may be one of a purely technical nature, like the need to protect a spacecraft from the intense heat generated by atmospheric friction during its re-entry at very high speeds from outer space into the earth’s atmosphere. It is the second kind of need that one encounters most often. An engineer may be asked to devise a system to reduce the level of poisonous gases in an automobile exhaust or he may be asked to design an engine cooling system for tractors operating in the polar regions where ordinary coolants (e.g. water) freeze. There may be a technical need to prevent a nuclear reactor from running amok and exploding like a bomb or a need for a system for the safe handling and manipulation of nuclear fuel.

3.4.1

Realization of Need

The problem is said to exist whenever there is a difference between the situation demands and the situation that exists. The realization of need for a design solution is the beginning of the design process. For example, if your mother finds it difficult to separate dosa from a hot plate, she might perceive a need for non-stick hot plate. When the man did not want to operate a device by reaching to it, he realized a need for a remote control mechanism. A person listening to news about loss of property due to earthquake may perceive a need for quake-resistant building structures. What is required for need for quake resistant building structures? What is required for need realization is organized thinking and disciplined mind that is always on the lookout for challenging situations? Some people are able to realize the need for a device or method, which others would have overlooked. Thomas Edison was one such designer who went through great lengths in developing the electric lamp much before any significant segment of society desired it. Thus some great designers have ability to recognize the needs of the society much before the society itself realizes them. The designer may recognize a need all on his own or, it may be communicated to him by his employer or customer. Today, it is the latter type that is common for practicing designers. The need communicated could be either basic (such as the need for fresh air, or cheaper and faster vehicle, or quake resistant building structures) or purely technical (such as designing a ceramic engine to reduce heat loss, to design a parallel processor system, designing an engine cooling system for tractors operating in polar regions where normal coolants like water freeze) in nature.

3.4.2

Preliminary Need Statement

The designer gives a more specific form to the need realized, by writing a preliminary need statement. That is, the designer determines what exactly is needed. Although this job sounds simple, it should be remembered that getting fooled is also equally simple. Often, the designers identify wrong a need and take up a wrong problem to solve, wasting not only their time, but also their sponsor’s time. This is because, the designer starts thinking in terms of solutions while stating the need and is foredoomed to take a wrong track. For example, consider the problem of loss of life due to earthquake. Given this news, the designer is tempted to define the problem as designing quake-resistant building structures and starts thinking of

Design and Problem Formulation

37

appropriate materials, method of construction etc. His solution area becomes small and the finally chosen design would be a relatively better building structure. However, if he thinks carefully, the problem can be stated as minimizing the loss of life due to earthquake. Now, the solution can be searched from a broader area which may include prior warning systems, use of dampeners, mobile residences, etc. All this is in addition to the earlier solution area. For example, consider the problem of grain being destroyed by mice in the country. Generally, when we look at this problem, we never think beyond a device that kills mice and tend to state the need as “A fool proof system for extermination of mice.” With this need statement, the designer has a limited area to search the solution from, i.e., he has to choose only among the killing machines. However, have we been successful in our war against mice for thousands of years? The answer is “No”. Thus, instead of regarding the problem as that of killing mice, it is much better to regard it as “a system to protect the grain from mice.” By this, we are broadening the possible solution space, i.e. we can now choose from devices that divert mice from area than kill them and prevent eating, etc. In fact, to define the problem as killing mice, is defining the solution and not really the problem. Thus, thinking in terms of solution even before the need is stated, forces the designer to seek solutions for an unnecessary bunch of sub problems, which could have been bypassed altogether as tried with mind that is open and not pre-conceived. Thus, the preliminary need statement should be as general as possible, which at the same time defines the essential nature of the problem. However, we should not make the statement natural that it doesn’t give us any real information. For example, in the earthquake problem considering evacuating the people to the safe area does not merit the designer’s attention, and in rice and grain problem, considering changing of the variety grain which is not palatable to mice, is far from the problem we are interested in. Below are given some of the preliminary need statements, and the products mentioned against them are the design solutions: Clutch: “A mechanism to make or break transmission of power”. “A device, which can transmit power when desired”. Solar Cooker: “A device that uses inexhaustible source of energy to cook food”. Satellite: “A device which receives and transmits electro signals between two places, regardless of the geographic location.” Xerox copier: “A quick mechanism for producing facsimiles of given document.” Identification and Analysis of Need Electric Iron: “An electric device for removing wrinkles from clothes”. Telephone: “ A device for transmitting sound signals between two places”. Paper Knife: “A simple device for slitting paper”.

3.5

HIERARCHICAL NATURE OF SYSTEM AND HIERARCHICAL NATURE OF PROBLEM ENVIRONMENT

Prior to developing a definitive statement of a problem, the engineer must first recognize and understand the environment in which the problem is embedded. This is an essential first step in approaching a problem irrespective of its size or complexity, because ultimately the implementation of any solution that he may conceive and propose is contingent on how well he understands the total problem environment. With respect to problem, the environment includes the systems that are involved in and influence the problem. As problem complexity increases, the engineer encounters greater difficulty in approaching

38

Mechanical System Design

and defining a problem. The system approach is such a technique and represents a broad based systematic approach to problems that may be interdisciplinary. It is particularly useful when problems are complex and affected by many factors. All views of the problem under consideration should be analyzed. Thus a clearer understanding of constraints, alternatives and consequences that are associated with the problem may be obtained. Problem environment and the components that are constrained in the system should be identified. This step is depicted graphically in Fig. 3.3. Statement describing problem environment and its limits

A

B

D

C General System Component Involved in problem

Fig. 3.3

Problem environment and component identification

The engineer must recognize that the problem may involve or be related to other systems. Thus in defining the problem, these other systems must be directly connected in reality to a component of some large system. Thus, for a comprehensive view of the problem, the engineer also must define the components and structure of the large system. The relationship of the system under investigation to a large system shown in Fig. 3.4. A microscopic investigation of the system under study can be made by sub-dividing each of the components into component parts shown in Fig. 3.5. Thus, a hierarchical system structure can be determined that permits analysis of the systems at the various levels. This hierarchical structure is shown in Fig. 3.6. Which co-positively depicts the system components that were shown in Figs 3.4 and 3.5. Figure 3.7 shows the problem environment and system components with structure. Figure 3.8 co-positively depicts the system components that were shown in Figs 3.3, 3.4, 3.5, 3.6 and 3.7. This hierarchical structure permits the analysis of the system in terms not only of the system under consideration, but also in terms of both higher and lower level systems. Furthermore, this structure provides the framework for analyzing the comprehensive aspects of the problems as well as the technical details that must be addressed and considered determining the hierarchical system structure. Each level of the system must be examined and goals and objectives defined. For each level of system, a specific purpose or function can be stated that indicates the needs that must be met by that system. In essence, these goals should reflect the stated purpose and function. Normally, the purpose and function can be ascertained by defining the role of a system in relation to the next higher-level system. The system goal is usually related to the objective of the next higher-level system, as indicated in the Fig. 3.8.

Design and Problem Formulation

39

System under investigation Other component large system

Fig. 3.4

System under investigation a component of a large system System which constitutes component of system under investigation

System under investigation

Fig. 3.5

Microanalysis of system component

Fig. 3.6

Hierarchical system structure

40

Mechanical System Design

B A D

C

Fig. 3.7 Problem environment and system components with structure Level A Goal

Level B

Goal

Objective

Objective

Objective

Criterion

Criterion

Criterion Goal

Objective Level C

Criterion

Fig. 3.8

System goal as defined by role of system in relation to next higher system

The objectives reflect ways in which the goal can be achieved for a particular system, the components can indicate the different variables that can be modified in order to fulfill the goal. Various problem statements or definitions can be generated depending on how many and which system levels and components are incorporated into the engineer’s view of the problem. For example in Fig. 3.9, different problem statements based upon components are included.

Design and Problem Formulation

41 Problem statement

Problem statement

1

1

2

3

4

2

3

4

Problem statement

Fig. 3.9

3.6

Problem statement Problem statement based on component 2, 3, 4,

Problem statement based upon component

PROBLEM SCOPE AND CONSTRAINTS

Problem scope and constraints are explained by an example. Example (1) A freeway interchange may be considered as a transportation system. Problem scope Constraints System purpose: To allow the orderly flow of traffic at the intersection of two highways. System components: Two highways, overpass, ramps, traffic signals, traffic signs, traffic, etc. System structure: Physical layouts of interchange, traffic laws. System constraints: Traffic volume, human reaction time, traffic regulations, etc. Example (2) A city building is an engineering system. System purpose: To provide a place for shops, offices, and apartments. System components: The physical structure, the floors, the elevators, heating and lighting, etc. System structure: Use of space. Physical layout, etc. System constraints: Floor-space requirements, safety features, cost of construction, surrounding environments, etc. Example (3) A sewage disposal system in a city: System purpose: To dispose of refuse from city buildings. System components: The buildings, sewage pipes, processing plans, outlet for processed sewage, etc. System structure: Sewage flow, processing, engineering layout, etc. System constraints: Pollution levels, pollution density and distribution etc.

42

Mechanical System Design

Question: Discuss the basic problems concerning systems. Illustrate how these problems can be made use in case of an input-output system. UPTU 2004 Answer: Basic problem concerning system For the system shown in Fig. 3.10 transformation on inputs ‘I’ produces output ‘O’. This inputoutput relationship is expressed as follows Eq. 3.1 TI → O

(3.1)

T is called the operator Inputs (I )

Transformation

Output (O)

Environment

Fig. 3.10

Environment basic problem concerning system

Basic Problem Defined 1. System analysis: Classify contents of TI and O (Investigate the process, raw material and products in case of manufacturing system). 2. (a) System operation: Given T and I, find ‘O’ (design the products to be made using the given raw material and process). (b) System inversion: Given T and O find ‘I’ (determine the raw material to make a certain product using a certain process). (c) System synthesis: Given I and O, determine a suitable ‘T’ (Given a raw material and product to be made) establish a suitable production process. 3. System optimization: Pick I, O or T so that a specific criterion is optimised. 4. System design: This involves contacting new useful system static structure and operation procedure.

3.7 3.7.1

A CASE STUDY: HEATING DUCT INSULATION SYSTEM Statement of the Problem

In a rapidly growing city, a new building code was recently inserted in the section on warm-air heating; the code requires that contractors cover all warm-air ducts with 1/32 centimetre asbestos insulating paper. One contractor does not do so however because he claims adding the insulation increases, rather than decreases, the heat loss from these ducts. A special meeting has been called to consider the problem, and an engineer has been hired by the city to comment on the contractor’s contention. Now before going on, the student, imagining himself to be the consulting engineer, should define the problem for himself.

3.7.2

Definition of the Problem

The engineer in this situation is faced with a rather vague problem situation, though it is not an extreme case. He decides quite readily that what he must do is determine the heat loss from the duct under a set of typical conditions, with and without the insulation. But what are “typical” conditions, if indeed they exist at all? Perhaps this question can be avoided if he derives a general expression for the heat loss from

Design and Problem Formulation

43

a duct first. Then it will be a simple matter to insert whatever numbers the city and/or the contractor feel would be typical. The engineer still must prescribe some specific things, for example, he decides to work with a horizontal round duct, but notes that he could easily also repeat his work for a rectangular duct if necessary. It seems doubtful that the results will be qualitatively different. The engineer states his problem as follows in Fig. 3.11.

D

Air velocity, force (F) Air temperature

Material x m , km , tm

Surrounding temperature to insulation x m , k a , rd

Fig. 3.11

Heating duct insulation system

Determine an expression for the heat loss to the surroundings per metre of duct in terms of V, D, ts, to, xa, xm, ∈a and ∈m with and without insulation. Now notice that the engineer has accomplished. He has gone from a general assignment (“Comment” on the problem) to a specific problem, which can be solved by engineering principles. Later, of course, when he has his equations derived and his numerical examples worked, he will have to go back and make his evaluations and generalizations.

3.7.3

Making a Guess

For both term and experience, try now to make a guess about the outcome. Doing so is a good way to gain engineering judgment. Guessing can also be an aid to understanding the problem because it forces one to consider only the most essential factors. In this case, for example, the engineer realized that adding the insulation will increase the thermal resistance of the duct wall and will decrease the convection loss because the outer surface temperature will be lower. However, ducts are made of galvanized sheet metal or aluminum, both of which have low emissivity, which will tend to increase the radiation loss, which effect will dominate? Make a guess before going on.

3.7.2

Model Formulation

The engineer has done a little model formulation already, because he has decided to consider a round duct. This illustrates the fact that it is impossible to keep these stages of problem solving from overlapping, reversing order, etc. depending on the problem. In this problem there are several simplifying assumptions which do not change the numerical results significantly but which reduce the amount of work necessary by a great deal.

44

Mechanical System Design

r2

r1

Insulation

l

Fig. 3.12

Heating duct insulation of a cylindrical system

r2 r1 We know that thermal resistance, (R) of a cylindrical system = , 2π kl ln

where r2 and r1 are the outside and inside radii respectively. Let the length of the cylinder be l and thermal conductivity be k Model formulation ln x = ( x − 1) −

x=

if

x −1=

then

⇒ x −1=

r2 and r2 = r1 + xa , r1 r2 −1 r1 r1 + xa −1 r1

⇒ x −1=1+ ⇒ x −1=

1 1 ( x − 1) 2 + ( x − 1)3 ... 2 3

xa −1 r1

xa r1

In another problem

ln

x r2 r2 ≅ −1= a r1 r1 r1 r2 r1 xa ≅ R= 2π kl 2π kr1l ln

∴ Thermal resistance,

Design and Problem Formulation

3.8

45

A CASE STUDY: HIGH-SPEED BELT DRIVE SYSTEM

3.8.1

Statement of the Problem

A company that builds conventional drills and drill processes has designed a new, very high-speed, lowpower drill.

Specifications Flat leather belt density = 0.035 gm/cm3

σ working = 275 psi Pulley diameter = 150 mm Angle of contact θ = 180° = π radian Speed = 3000 rpm Power = 0.5 H.P. Assumptions coefficient of friction µ = 0.20 Derivation of equations:

T1 = e µθ T2 V = ⇒ V =

π dN 60 π × .15 × 3000 = 23.56 m/sec. 60

T − T  HP =  1 2  v  75 

⇒ T1 − T2 =

HP × 75 v

⇒ T1 − T2 =

0.5 × 75 23.54

⇒ T1 − T2 = 1.5

=

...(1)

T1 = e.20π T2

T1 = 1.87 T2

...(2)

46

Mechanical System Design Solving equation (1) and (2) we get; T1 − T2 = 1.5

T1 = 1.874 T2

...(1) ...(2)

T1 = 1.87 T2 1.87 T2 − T2 = 1.5

⇒ T2 =

1.5 0.87

⇒ T2 = 1.71 kgf T1 = 1.5 + T2

= 1.5 + 1.71 = 3.21 kgf

A=

T1 3.21 = 6 275

= 0.011 mm 2 This is the communication of old belt design result. Old design result: A flat leather belt and pulley drive in the prototype of the new drill has repeatedly failed. The belt was designed on the same basis used successfully by the company on past machines, and the report on this design is shown in statement of the problem. An engineer has been assigned the job of recommending corrective action and of finding out why the regular companies design procedure has failed in the case.

3.8.2

Definition of the Problem

The engineer in this project must convert his “find the trouble” assignment into an analysable problem. What could be the trouble? To avoid some obvious error, he checks the figures given in the design report and finds that the arithmetic there is okay. He also checks the specifications given and finds them to be standard and conservative. No previous difficulties have been encountered with design procedure. Because the high-speed involved in the new drill is a new feature, he expects that it may affect the belt design in some way not accounted for in the analysis. But how? This he does not know (Do you? Think about it!). But he decides to start at the beginning and rederive the basic equations making sure not to make any unwarranted assumptions. He is especially interested in the origin of the equation,

T1 = e µθ which appears in the report. T2

Design and Problem Formulation

47

His problem definition then is this: Analyse the mechanics of a high-speed belt on a pulley to determine the relationship among T1 , T2 , µ , θ , r , w and horsepower. Model Formulation: Here, T1 = slack tension depends upon movement T2 = tight tension movement µ = coefficient of friction θ = angle of contact r = radius w = speed in rpm v = velocity After drawing Fig. 3.13, the engineer very nearly has his model. He assumes that angle of contact is 1800 and that µ is constant. He neglects bending stresses in the belt and any area changes in the belting due to the varying tension. He also decides to neglect the effects to centrifugal force on the belt. rdθ

rdθ θ

dθ

µdN

T

T + dT

T2

T1

dN

Fig. 3.13 Radial force balance

dN = T sin

dθ dθ + (T + dT ) sin 2 2

⇒ dN = T sin

dθ dθ dθ + T sin + dT sin 2 2 2

dθ dθ  dθ  ⇒ dN = T  sin + sin  + dT sin 2 2  2 

sin

dθ dθ = 2 2

48

Mechanical System Design

dN = Tdθ

...(1)

Tangential force balance

µ dN + (T + dT ) cos

⇒ µ dN + T cos

dθ dθ = T cos 2 2

dθ dθ dθ + dT cos = T cos 2 2 2

⇒ µ dN + dT cos

dθ =0 2

dθ   = 1  cos 2   ⇒ µ dN = − dT

...(2)

Application of Physical Principles we get dN = Td θ

...(1)

µ dN = − dT

...(2)

Dividing (1) by (2) we get:

dN −Tdθ = µ dN dT ⇒−

dT = T . dθ µ

⇒−

dT = µ dθ T

⇒

dT = − µ dθ T T2

⇒

∫T

dT

=µ

T1

∫ dθ

⇒ [loge T ]T2 = − µθ T

1

⇒ log e T2 − log e T1 = −µθ ⇒ − (log e T1 − log e T2 ) = − µθ ⇒ log e T1 − log e T2 = µθ

Design and Problem Formulation

49

[log a N = x ⇒ N = a x ] ⇒ log e

T1 = µθ T2 T1 = e µθ T2

3.8.3

Evaluation and Generalization

This is the equation used in the design that has failed so it is certain that either the bending energy or the centrifugal force, both neglect here, becomes important at the new, lighter speed. Now, since bending energy is directly proportional to velocity the centrifugal-force component between, is proportional to v2 .

Decision To relay the assumption that allow centrifugal force and include it into analysis.

3.8.4

Model Formulation

Same as before, only don’t neglect centrifugal effect. Application of physical principles: The power equation and the tangential force balance remain the same. The radial balance now is

dN +

wv 2 dθ dθ ⋅ r dθ = T sin + (T + dT ) sin 2 2 gr

⇒ dN +

wv 2 dθ dθ dθ dθ dθ = T sin + T sin + T sin + dT sin 2 2 2 2 g  dθ dθ   sin 2 = 2   

⇒ dN +

wv 2 dθ   dθ dθ = T sin + sin 2 2  g 

⇒ dN +

wv 2 dθ   dθ = T 2sin 2  g 

⇒ dN +

wv 2 dθ dθ = T ⋅ 2 2 g  wv 2  ⇒ dN = dθ  T −  g  

dθ    dT sin 2 = 0  

50

Mechanical System Design

 wv 2  ⇒ dN =  T −  dθ g  

...(1)

Tangential force balance equation,

µ dN + (T + dT ) cos ⇒ µ dN + dT cos

dθ dθ = T cos 2 2

dθ dθ   = 0  cos = 1 2 2  

⇒ µ dN = − dT

Putting the values of dN in eqn. (2)

 wv 2  ⇒ µ T −  dθ = − dT g   ⇒

dT = − µ dθ  wv 2  T −  g  

⇒ integrating we get, T2

⇒

∫

∫

dT = − µ dθ 2 wv T1 T − g T

2 2   wv   ⇒ log e  T −   = −µθ g     T1

   wv 2  wv 2   ⇒  log e  T2 −  − log e  T1 −   = − µθ g  g     

   wv 2  wv 2   ⇒ = − log e  T1 −  − loge  T2 −   = − µθ g  g       wv 2   T1 −  g  ⇒ loge  = µθ  wv 2   T2 − g   

...(2)

Design and Problem Formulation

51

wv 2 g ⇒ = e µθ 2 wv T2 − g T1 −

This result shows clearly that unless the form

wv 2 is small compared with T’s, the centrifugal g

effects are not negligible.

3.8.5

Evaluation, Generalization and Communication

The engineer has learned that of high speeds, centrifugal force is important in belt design and he also knows how to determine what ‘High’ means If

wv 2 is not small compared with T2 , the speed is high. g

A slightly thicker belt should correct the failure. Question 1: A leather belt is required to transmit 0.2 k/w from a pulley 150 mm diameter, running speed 3,000 rpm. The angle embraced is 180° and coefficient of friction between the belt and the pulley is 0.2. If the safe working stress for the leather belt 198 × 10−3 kgf/mm2, density of leather 1.08 × 10−3 kg/ mm3. Find the (i) tension in tight and slack side of belt in kgf. (ii) Velocity in m/sec (iii) And minimum belt area in (mm)2 Solution: Given that P = 0.2 kw = 200 watt Diameter of pulley = 150 mm = 0.15 m Angle of contact θ = 180° = π radian Coefficient of friction = 0.2 T1 = tension in the tight side of the belt in kgf. T2 = tension in the slack side of the belt in kgf.

Velocity (v) =

π dN 60

π × 0.15 × 3000 60 v = 23.56 m/s. =

And power transmitted, = P = (T1 − T2 ) v

52

Mechanical System Design For a given belt, value of P will be maximum when the product (T1 – T2) v is maximum.

T1 − T2 = ⇒ T1 − T2 =

P v 200 23.56

⇒ T1 − T2 = 8.488 kgf

...(1)

We know that

T1 = e µθ T2 ⇒

T1 = e 0.2 × π T2

⇒

T1 = 1.874 T2

⇒ T1 = 1.874 T2

Putting the value of T1 in Equation (1) we get; ⇒ (1.874 T2 − T2 ) = 8.488 ⇒ 0.874 T2 = 8.488 ⇒ T2 =

8.488 0.874

T2 = 9.711 kgf T1 = 1.874 × 9.711 T1 = 18.1996 kgf

For the minimum belt area

Minimum belt area

=

T1 σ

=

18.1996 kgf 198 × 10−3 kgf /mm 2

= 91.91 (mm) 2

So the required answer is (i) Tension in tight side T1 = 18.1996 kgf and Tension in slack side T2 = 9.711 kgf (ii) Velocity = 23.56 m/s (iii) Minimum belt area = 9191 (mm) 2

...(2)

Design and Problem Formulation

53

3.8.5.1 Ratio of tensions (i) Flat belt:

T1 = e µθ T2

(ii) v-belt or rope

T1 = e µ cos ec α T2

Where, 2α = angle of groove.

3.8.5.2 Angle of contact Open belt. θ = 180 − 2α where

α = sin −1

[ r1 − r2 ] d

α = sin −1

[ r1 + r2 ] d

Cross belt θ = 180 + 2α where Centrifugal stress in belt or rope

Fe =

wv 2 = mv 2 g

where

v = speed of belt or rope. w = weight of belt or rope per unit length. Hence centrifugal stress is independent of the radius of curvature of the path.

Power transmitted by a Belt or Rope If the effective tension and speed are known, then

P=

(T1 − T2 )v HP 75

where T1 and T2 are tension in the tight and slack side of belt in kgf, and v is the speed in m/s. For a given belt, value of P will be maximum when the product (T1 − T2 ) v is maximum.

3.8.5.3 Effect of centrifugal tension on the power transmitted Horse power transmitted, P =

where

(T1 − T2 )v 1  v   = T1  1 −    75 k    75 

k=

T1 T2

To determine maximum stress in the belt, total tension on tight side

54

Mechanical System Design T = T1 + Tc

1  v   P = (T − Tc ) 1 −   k  75   If T is assumed to be constant, the horse power will first increase to maximum then decrease to zero as the speed is increased. But Tc =

wv 2 , so that P is maximum when g

 wv 2  T −  v is a maximum. g  

Differentiating with respect to v and equating to zero, we get

T =

v=

3wv 2 = 0 or T = 3Tc g Tg 3ω

Also P = 0, when Tc = T, i.e., when wv =

3.9

T1g w

CHAIN DRIVES

Chains are used for the transmission of power over comparatively long distances of upto 8 m approximately. Main criterion of operating ability of a chain drive is the wear resistance of the joint. A very small number of teeth in the sprocket unfavorably affects chain service life and intensifies noise.

3.9.1

Design of Transmission Roller Chains 1. Transmission ratio or velocity ratio of the chain drive

i=

Z 2 n1 = Z1 n2

where, Z1 = number of teeth on sprocket pinion (smaller) Z 2 = number of teeth on sprocket wheel (larger) n1 = number of rotation of pinion in rpm n2 = speed of rotation of rear in rpm

Preferred transmission ratios are 1, 1.12, 1.25, 1.4, 1.6, 1.8, 2.0, 2.25, 3.15, 4.0, 4.5, 5, 5.6, 6.3 and 7.1 2. Average chain speed, v =

pZ1 n1 m/s 60 × 1000

Design and Problem Formulation

55

where, P = pitch of chain, mm n1 = speed of the smaller sprocket, rpm Z1 = number of teeth on the smaller sprocket

Normal velocity of chain ranges from 2.5 to 7.5 m/s 3. Number of teeth: Minimum recommended number of teeth on the small sprocket is 17 and maximum number of teeth on the bigger sprocket can be 114. Minimum number of teeth on the sprocket,

Z min = = where,

4d r + 5 for pitches of 2.54 mm p 4d r + 4 for pitches of 3.175 to 5.785 mm p

d r = roller pin diameter

Question 2. A leather belt is required to transmit 7.5 kW from a pulley 1.2 m diameter, running 250 r.p.m. The angle embraced is 165° and the coefficient of friction between the belt and the pulley is 0.3. If the safe working stress for the leather belt is 1.5 MPa, density of leather 1 mg/m3 and thickness of belt 10 mm, find the width of the belt taking centrifugal tension into account. Ans. Given, P = 7.5 kW = 7500, d = 1.2 m, N = 250 r.p.m.

θ = 165° = 165 ×

π = 2.88 rad , µ = 0.3, σ = 1.5 = 1.5 × 106 N/m 2 180

P = 1mg/m3 = 1000 kg/m3 , t = 10 mm = 0.01m Let

b = width of belt in meters T1 = tension in the tight side of the belt in N and T2 = tension in the slack side of the belt in N

Velocity of the belt, v =

π dN 250 = π × 1.2 × = 15.75 m/s 60 60

and power transmitted, P = (T1 − T2 ) v value of P will be maximum when the product (T1 – T2)v is maximum.

T1 − T2 = Again we know that 2.3 log

P 7500 = = 477.4 N v 15.75

T1 = µθ = 0.3 × 2.88 = 0.864 T2

...(i)

56

Mechanical System Design

log

T1 0.864 = = 0.3756 2.3 T2 T1 = 2.375 T2

or

...(ii)

From equations (i) and (ii), we get T1 = 824.6 N and T2 = 347.2 N Now mass of the belt per meter length, m = area × length × density = b t . l. p = b × 0.01 × 1 × 1000 = 10 b kg

And, centrifugal tension, Tc = mv 2

= 10b (15.71) 2 = 2468 b N Maximum tension in the belt, T = σ bt =1.5 × 106 × b × 0.01 = 15000 bN We know that, T = T1 + Tc or

15000 b = 824.6 + 2468 b b = 65.8 mm The width of the belt taking centrifugal tension into account is 65.8 mm Ans.

Question 3. Following data is given for a rope pulley transmitting 24 kW: Diameter of pulley = 40 mm, Speed = 110 r.p.m., Angle of groove = 45°, Angle of lap on smaller pulley =160°, Coefficient of friction = 0.28, Number of ropes = 10, Mass is kg/m length of ropes = 53C2 and working tension is limited to 122C 2 kN, where C is girth of rope in meter. The initial tension in each rope will be Solution: Given, P = 24 kW = 24 × 103 W; d = 400 mm = 0.4 m N = 110 r.p.m; 2 β = 45° or β = 22.50

m = 530C2 kg/m, T = 122C 2 kN

π = 2.8 rad , µ = 0.28, n = 10, 180 Initial tension: Power transmitted per rope, = 160° = 160 ×

P=

Total power transmitted 24 = = 2.4 kW = 2400 W Number of ropes 10

And velocity of the rope, v =

π dN π × 0.4 × 110 = = 2.3m/s 60 60

Let T1 = tension in the tight side of the rope, and T2 = tension in the slack side of the rope

Design and Problem Formulation

57

We know that, power transmitted per rope 2400 = (T1 − T2 ) v = (T1 − T2 ) 2.3

T1 − T2 =

2400 = 1043.5 N 2.3

...(i)

T  Again we know that 2.3 log  1  = µ cosec β = 0.28 × 2.8 × cosec 22.5° = 2.05  T2  T  log  1  = 2.05 = 0.8913  T2  T1 = 7.786 T2

or

...(ii)

Solving equation (i) and (ii), we get T1 = 1197.3 N , T2 = 153.8 N 1197.3 + 1538 + Initial tension in each rope, To = T1 T2 = = 675.55 N 2 2

Initial tension in each rope To = 675.55 N Diameter of each rope Let d1 = diameter of each rope Centrifugal tension, Tc = mv 2 = 53 C 2 (2.3) 2 = 280.4 C 2 N and working tension (T) or or

= 122 × 103 C 2 = T1 + Tc = 1197.3 + 280.4 C 2

C 2 = 9.836 × 10−3 C = 0.0992 m =99.2 mm Girth (i.e. circumference) of rope C = 0.0992 m C = π d1 = 99.2 mm

⇒ d1 =

Exercise

99.2 = 31.57 mm π

3 . . .

1. Prepare a check list for carrying out the need analysis of the design of a product. Give suitable explanations wherever necessary. 2. For each of the following products, give at least two need statements of different generality, for which the listed product is a solution. (i) Electric iron (ii) Oil pipeline (iii) Solar cooker (iv) Telephone

58

3. 4. 5. 6. 7.

8.

9. 10. 11.

Mechanical System Design (v) Envelope (vi) Handkerchief Explain nature of engineering problems in system design. What is needs statement? Briefly explain by line diagram in the basic of goal, objective, criteria. In problem formulation give brief idea of hierarchical nature of systems, hierarchical nature of problem environment with neat and clean figure. What is difference between problem scope and constraint? Where should it be applicable? Write a note on: (a) Need statement (b) Nature of engineering problem (c) Identification and analysis need (d) Hierarchical nature of problem environment in engineering. (e) Basic problem concerning system. (f) Problem formulation. Write at least one need statement, for each of the given product: (i) Clutch (ii) Solar cooker (iii) Satellite (iv) Xerox (v) Electric iron (vi) Telephone (vii) Paper knife (viii) Envelope (ix) Handkerchief (x) Oil pipeline. Describe the steps involved in nature of engineering problem and needs statement in mechanical system design. Explain a system design where environment and safety is of prime consideration. Explain a system design where problem scope and constraints are prime consideration of problem formulation.

A Case Study Heating Duct Insulation System 1. 2. 3. 4. 5.

Briefly explain a case study of heating duct insulation system. Explain the definition of the problem of heating duct insulation system. How can you formulate a mode by heating duct insulation system? What is the assumption of heating duct insulation system? Define heating duct insulation system.

High Speed Belt Drive System 1. 2. 3. 4. 5. 6. 7.

Briefly explain a case study of high speed belt drive system. Write down application of physical principle of high speed belt drive system. What is difference between evaluation and generation on the basis of case study ‘High speed belt drive’? What is difference between flat belt and V belt? Briefly explain application and computation of high speed belt drive. Write a short note of high speed belt drive. Briefly explain a case study of high speed belt drive of evaluation, generation and communication.

System Theories

59

4 System Theories 4.1

INTRODUCTION

System Analysis consists of a continuing dialogue between the policy maker and the analyst in which the former asks for alternative solution to his problem, makes decision to exclude some, and makes value judgments and policy decision, while the analyst attempts to classify the conceptual framework in which decision must be made, to define alternative possible objectives and criteria and to explore in as clear as possible the costs and effectiveness of alternatives considerations of action. The analyst is helping someone else to make decision. System analysis emphasises basic economic concepts, mostly the simple concepts of marginal product and margin cost; the system analysis approach has developed a variety of techniques for analysis of complex problem of decision-making in such a way as to make calculation several judgment.

4.1.1

The Elements of System Analysis

The following elements are identified in any analysis: 1. The objective: The first and one of the most important tasks of system analysis is to discover what the objectives of the decision maker are or should be trying to attain through the options open to him, and how to measure the extent to which they are attained. 2. The alternatives: The alternatives are the means by which it is hoped the objectives can be attained. 3. The costs: Costs are to be measured not only in money but also in terms of opportunity that they preclude.

4.2

SYSTEM ANALYSIS VIEW POINT

It involves the study and construction of each part of the system, both as an individual and its relation to the whole. In order to design, modify and improve the system, it is mounted whether on an abstract system like inventory control procedure or physical system such as power from mechanical system design. It involves a continuous cycle of following:

60

Mechanical System Design 1. Defining objectives: It is the conceptual phase in which the objectives are classified and defined with a view to select a policy regarding action or decision making for solving the problem. 2. Promising alternatives: Promising alternative programming which can possibly achieve the objectives. 3. Model building programming: Alternatively programmes can be shaped into models. A model is an analog of reality. It represents the system qualitatively or quantitatively. A model may be in the physical form or it may be mathematical representative of the system. It is always cheaper and convenient to forecast the consequence and to test the performance of a system from its model rather getting the same information after going through the complicated process of actively fabricating the system being considered. Depending upon the type of problem, there can be an overall process model, performance model, reliability model, time model or cost model. 4. Criterion: The individual models are evaluated in terms of the criterion specified before the attention may be effectiveness, cost, performance or cost-gain performance etc. 5. Performance-wise alternatives: From the performance of models, various alternative programs are tested in order of performance. 6. Verification: The most promising alternatives are tested in experiments and their good points are verified.

4.2.1

Techniques in System Analysis (Cobb-Douglas Models)

It depends upon the mathematical equation to exhibit the behavior of the system components and the effect may exercise on one another. Some such techniques are: (a) Operations research (b) CPM and PERT Supply (output) Labour Production

Financing

Investment

(Input)

Machinery

Fig. 4.1

Position of corporate model

To demonstrate a system analysis study, we will consider a small part of the corporate model developed earlier. Looking at just the financial and the production models and ignoring the influence of the economies that might affect results, gives the simple system illustrated in Fig. 4.1. The system involves four variables denoted by the following symbols: K - capital investment L - labour M - machinery S - supply With regards to production, economists have found that the output of an enterprise is often related to the investment in labour and machinery by an equation of the general form,

System Theories

61

S = f  La M b  Here f is a constant. Models of this form are known as Cobb-Douglas models. They imply that a given percentage increase in the applied resources produces a proportional percentage increase in output. We will assume the exponents and the production function, when, a = 1 and b = 1 then the form S = f [LM ] The financial model assumes that there are possible substitutions between labour and machinery. In the present case, we will assume a linear relationship as follows: K = eL + M The coefficient e is a constant. It implies that one unit of investment in labour is equivalent to investing e units in a machinery. Both equations depend upon L and M. The financial model shows how a given amount of investment K can be divided between L and M. The production model shows how the output, or supply S, depends upon the amount invested in L and M. It should be possible to find an assignment of L and M that maximizes supply for a given investment.

4.3

BLACK BOX OR DECISION PROCESS APPROACH

We have given a simple definition of a system as a set of interacting objects. The objects might be considered the basic entities of the system, but usually the description of system can be made at many levels of detail. It is customary to describe a system as consisting of interacting subsystems. Any subsystem might itself be considered a system consisting of subsystems at a still lower level of detail and so on. A system study must begin by deciding on the level of subsystem detail to be used. The block-building principle mentioned, help to organise a system description by isolating subsystems and identifying their inputs and outputs. Each subsystem at a given level of detail is described as a block, giving relationships between the inputs and outputs. The relationships should be such that they are sufficient to determine the outputs from the inputs when the subsystem stands by itself, that is to say, there should be no need to use an endogenous variable with itself in the block. The term “black box “borrowed from the engineering field, is often used to describe an element of this nature which given an output in response to an input without any need or ability to know how the transformation is made. Each subsystem has its own inputs and outputs and standing by itself its response can be derived from the relationship that defines the subsystem. The interactions that occur in a system arise from the fact that the outputs of some subsystems become the inputs of others as they become the endogenous variables of the system. In the same way that a system breaks down into subsystems so also a model of a system breaks into sub models. When describing systems in terms of a blocks the terms block subsystem or sub model tend to be used interchangeably, as will occur in the subsequent discussion. In fact when concentrating on a particular part of a system or its model, we will often drop the prefix “sub” and talk about a part of a system or model as being itself a system or model. The context should make it quite clear what is intended. The system provides a meaningful framework for describing and understanding the features and problems of the subject. System is defined as a set of elements arranged in an orderly manner to accomplish an objective.

62

Mechanical System Design Input

Process

Output

Fig. 4.2 A system may have single input and multiple outputs or may have serveral input and outputs. In any system, the inputs are transferred into the output by a process. We say that the process is transparent to us when we are also to understand the system. But if the process of input transformation is not visible and understandable, then we say that the system is a black box and it is not transparent.

Input

Output

Fig. 4.3 Black box system Black box approach also called decision analysis criterion and is also called decision process approach.

4.4

STATE THEORY APPROACH

The system variables that are used to design or describe the system state (status) at any time, are called state variables. A particular system state is then identified by a set of instantaneous values of the state variables. If the values of one or more of the state variables is changed so that a new configuration can be recognised then a new state exists. A transformation function defines the new state in terms of the old state and the changes in the values of the state variables. State concepts are demonstrated in the following examples:

 S1  Si =    S2  Si +1 = Si + S1 a  i1 =  1   a2  where Si = amount of the checking accounts i = month i + 1 = financial state beginning S i + 1 = beginning amount i1 = input during month

System Theories

4.5

63

COMPONENT INTEGRATION APPROACH

There are a number of components (at least two) that can be identified as necessary integration of the problem. Furthermore, each component has a variety of attributes that implicitly, physically, and behaviorally are necessary for its description. The component are integrated in some manner satisfying interface consistency between the components.

System Components Let us consider a freeway interchange which may be considered as a transportation system. Two highways, overpass, ramps, traffic signals, etc. are system components.

Example of Component Integration Approach Let us consider: A city building is an engineering system. The physical structure, the floors, the elevators, heating and lighting etc.

System Components E.g. Sewage disposal system in a city. The buildings, sewage pipes, processing plants, outlet for processed sewage, etc. are system components. In addition to identifying component integration approach the interaction between the components must be defined. The components and the interactions define the structure of the system. In the conduct of his work the engineer may be concerned with view of the problem under component integration approach.

4.6 4.6.1

A CASE STUDY: AUTOMOBILES INSTRUMENTATION PANEL SYSTEM Introduction

The World of Automobiles 1. Dipping of the headlights through manual switch control by the driver is a normal feature on automobiles. But in Cadillac, Buick and some other American vehicles, the headlight dips automatically depending upon the intensity of light emanating from the oncoming vehicles. This is achieved by the use of light-sensitive ‘photo-electric cells’ used within the headlight assembly. 2. Early automobiles employed acetylene gas lamps that glared with brilliant white flame. Kerosene lamps were also used since the car speeds were low and only a few vehicles were plying on the roads. 3. A two-beam headlamp system is universally adopted now on all classes of automobiles. This system was introduced for the first time in 1940 on some American car models, although the developmental work for non-dazzle light beam started as early as 1918. 4. On an average, the power consumed by the lighting system of car at night is about 75% of its total electrical energy.

64

Mechanical System Design 5. History of air-conditioning of automobiles dates back to few buses in late 1930s, but real strides came in late 1950s. 6. In India, various auto electric products are marketed under the following brand names. 7. Bulbs: Lumen, Auto pal, Lucas-TVS, KOITO (Japan), Bosch, BPF, Mega, Vijay, Franco. 8. Stop and tail lamps: MICO, Auto pal, Altar, Moonlight, Brita, and Mind. 9. Cables and harnesses: AMP, UBC, JMC, Mind. 10. Fog lamps and Halogen bulbs: MICO, Auto pal, Aware, Oakes, Hi-Brite 11. Horns and Contacts: MICO, Loud Padmini, Toyota, Mind, Jalwa, Autosafe, Shankh, Him co, Hella, Lucas, Union, Shark (Spartan), Motility, Acme, Prestolite. 12. Electronic flashers: Loud Padding, Healed. 13. Switches: Lumen, Pavna Padmini, S-Kay, Minda, Tokai Rika (Japan), Prestige. 14. Wiper blades, arms and accessories: Cam flex, MICO, Wipro, Syndicate, 15. Wiper motors: Far, Globe 16. Car heaters: Subros. 17. Air-conditioners: Sanden (Japan), Sanden Vices, Suborns, Nippon Denso (Japan). 18. Automotive instruments: Pricol, Denso (Japan), Nippon Seiki (Japan), Alert, razki. 19. Central locking systems: Loud Padding, Auto Cop 20. Power windows: Optional (Italy) 21. World leaders in various fields are Denso Corporation of Japan in Air-Conditioning. Fukoku of Japan in Wiper blade profiling.

4.6.2

Terminology

Since we shall deal with the light rays, bulbs and the wiring circuits in this chapter, it is imperative to understand the meaning of some commonly appearing terms. These terms are briefly explained below with reference to auto electricals.

Earthing It is essential to provide both the flow and the return paths between the source of electrical energy and the device to which it is supplying the energy. It envisages that two wires must be used for this purpose. Out of these two wires, one incorporates a controlling switch while the other is earthed to complete the circuit. In automobiles, the chassis itself acts as the return wire. All unstitched wires may be incorporated into one, and one terminal of each electrical device may be connected to it to serve as earthing.

Short-circuiting It is an unwanted occurrence in which some or all of the current escapes into the chassis when the insulation of a cable fails due to its damage. The unused current in the circuit generally overheats the cable and may cause fire in it.

Lens It is a combination of large number of prisms which helps in either concentrating the light rays or in speeding them out. A convex lens concentrates the light rays but a concave lens diverges them. Use of these characteristics is made in the working of sealed-beam light units.

System Theories

65

Reflection On striking a flat and polished surface, a traveling light ray returns back at the same angle at which it strikes the surface. This is called reflection. Use of this phenomenon is made in the working of curved reflector of headlight due to which the radiated rays from bulb are concentrated into a strong beam.

Refraction The phenomenon due to which a straight-moving light rays are bent, when they transit from one to another medium, is called refraction. This is a desired property in headlamp’s working to get a proper beam of light in front of the vehicle.

4.6.3

Wiring Systems

The auto electric circuits are not merely the arrangement of wires and electrical loads in any zigzag or network form, rather they are planned systems and follow certain rules and basic principles of electricity. Accordingly, the wiring systems may be one of the following three styles: 1. Insulated-return or double-wire system 2. Earth-return or single-wire system 3. Balanced-load or three-wire system Insulated cable

Switch

–

12 V battery

+ Load units

Insulated cable (a)

Insulated cable

12 V battery

Switch

– +

Load units

Earthed to metal parts of engine and chassis (b)

Fig. 4.6

Wire circuiting systems (a) double-wire insulated return type, and (b) single wire earth return type

66

Mechanical System Design Insulated cable

12 v

24 V battery

Neutral cable

Insulated cable (c)

Fig. 4.6 (Contd.) (c) three-wire balance load type In the first system the current flows from and into each load unit via twin insulated cables. It is similar to domestic wiring system. In the earth-return system, a single insulated cable carries current from the battery to various load units. The metal members of the engine and the chassis form a common return path for the current and also act as earthing.

Automobile instrumentation panel system It is very essential that the driver is continuously informed of proper working of the engine, different systems and the components of the auto vehicle. To serve this purpose the most convenient place is dashboard (facia) panel of the vehicle. The instruments fitted on this panel are called dashboard instruments. The architecture of an instrument panel needs to be stylish, efficient and convenient. Depending upon the number of instruments accommodated on it the instrument panel may be small or big since the bodies of instruments are of circular, rectangular, elliptical shape etc. Figure 4.7 shows various automotives instruments used on the dashboard/instrument panels of some vehicles. The scooter instruments panel incorporates an inbuilt ‘quartz clock’ which is energized with a 1.5volt, heavy duty, open torch cell. The dashboard of super-luxury cars are equipped with trip computer switches also to programme an on-board computer.

Instruments and their function 1. Speedometer: It shows instant speed of the vehicle in kilometer per hour (Kmph) throughout its running. 2. Odometer: It records the distance traveled by the vehicle in kilometer ever since the vehicle is in operation. 3. Tachometer: It shows the engine speed in r.p.m. It is provided with a red zone mark, which implies that the engine operation in this speed range may cause serious damage to it.

System Theories

67

4. Ammeter: It is used to indicate the charging current generated by the dynamo and the discharging current from the battery. Automobile ammeters are usually of centre zero style in which right side of scale shows charging current and the left side depicts discharging current. 5. Trip computer: Display several information such as: (a) Average speed of the vehicle. (b) Average fuel consumption in liters/100 km. (c) Total running time of the vehicle. (d) Distance traveled from start of a particular trip. (e) Total quantity of fuel consumed after computer was reset. (f) The distance after which the fuel tank will be empty. (g) The time of arrival at a particular destination of the journey. (h) Over speeding of the vehicle than the preset speed. (i) Given by the speedometer, odometer and tachometer. 6. In car temp. sensor: It senses the temperature inside the car component and accordingly sets the heater/cooler, or air-conditioning system automatically. 7. Fuel sensor: It keeps vigil on fuel quantity and informs it through a gauge. 8. Sunlight sensor: It senses the intensity of sunlight and informs the temp. sensor inside the car. 9. Pressure sensor: It senses the pressure of oil in lubrication system and informs to the gauge for display. 10. Temperature gauge: Shows the temperature of engine cooling water when the ignition is switched on. 11. Lubrication oil pressure gauge: It displays the oil pressure if the pressure is lower than the stipulated value, a red bulb indication appears on dash board. 12. Fuel level gauge: Fuel gauge shows the level of the fuel in fuel tank when the ignition is switched on.

A

1

2

3

4

5

6

7

8

(a)

Fig. 4.7

9

10

11 12

BCD

S R

E FG H

I

N

(b)

J

Q

K L

PO N

Dashboard of LML Supremo Scooter and a Car

M

68

Mechanical System Design 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

Quartz clock Fuel guage Fuel level indicator Head light Signal on indicator Neutral indicator Speedometer Speed indicator needle Odometer Spot light indicator On indicator Head light low-beam indicatore

Exercise

4 . . .

1. Explain the black box approach of system analysis and design. UPTU 2005 2. Explain the decision process approach for system analysis. 3. Write a short note of (a) State theory (b) Black box approach (c) Component integration approach (d) Decision process approach. UPTU 2005 4. Explain briefly system analysis view-point. 5. What is component integration approach? Also explain decision process approach on the basis of system theories.

A Case Study 1. 2. 3. 4. 5. 6.

What is dashboard system of instrumentation panel system? Write down the instrument names and their functions. Draw a neat and clean figure of LML supremo dashboard and explain. Why do we use automobile instrumentation panel system? What is the role of instrument & their function? Write a short note on: (a) Trip computer (b) In car temp. sensor (c) Temp. range (d) Fuel level range (e) Ammeter (f ) Tachometer

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69

5 System Modelling 5.1

INTRODUCTION

Models are extensively used by engineers as aids in the description analysis and design phases of problem solving to facilitate the communication of ideas and as a means of storing information for future reference and use. A model is a representation of some object or condition that exists or of an issue under consideration. It contains information expressed in certain specific forms about the objective models and requires interpretation according to certain prescribe rule. Often the modelling medium is the symbolism of mathematics and requires the logic and understanding of mathematics for its interpretation. In many cases, mathematical models require the support of both descriptive and schematic models to define the nature of the model and to aid in interpreting the symbols and variables used in the model. A mathematical model can thus be considered to be made up of four basic component models namely: 1. A descriptive model of the problem and purpose. 2. A symbolic definition model describing all the model symbols, variables, parameters and constants used in the mathematical model itself. 3. A schematic model illustrating the interpretation of the system basis and interrelationships among the system components. 4. The mathematical model itself. To study a system, it is sometimes possible to experiment with the system itself. The objective of many system studies however is to predict how a system will perform before it is built. Clearly, it is not feasible to experiment with a system while it is in this hypothetical form. An alternative that is sometimes used is to construct a number of prototypes and test them but this can be very expensive and time consuming. Even with an existing system, it is likely to be impossible or impractical to experiment with the actual system, for example it is not feasible to study economic systems by arbitrarily changing the supply and demand of goods. Consequently, system studies are generally conducted with a model of the system. For the purpose of most studies, it is not necessary to consider all the details of a system; so a model is not only a substitute for a system, it is also a simplification of the system.

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We define a model as the body of information about a system gathered for the purpose of studying the system. Since the purpose of the study will determine the nature of the information that is gathered, there is no unique model of a system. Difficult models of the same system will be produced by different analysis interested in different aspects of the system or by the same analyst as his understanding of the system changes. The task of drawing a model of a system may be divided broadly into two subtasks: Establishing the model structure and supplying the data. Establishing the structure determines the system boundary and identifies the critical attributes, and activities of the system. The data provide the values the attributes can have and define the relationships involved in the activities. The two jobs of creating a structure and providing the data are defined as parts of one task rather than as two separate tasks, because they are usually so intimately related that neither can be done without the other. Assumptions about the system direct the gathering of data and analysis of the data confirms or refutes the assumptions. Quite often, the data gathered will disclose an unsuspected relationship that changes the model structure. Essentially the same description is rewritten in a table to identify the entities.

Table 5.1

Elements of supermarket model Shopper

No. of Items

Arrive at

Basket

Availability

Shop queue check out

Counter

Number occupancy

Return basket and leave

Shoppers needing several items of shopping arrive at a supermarket. They get a basket, if one is available, carry out their shopping, and arrange to check out at one of the several counters. After checking out they return the basket and leave.

5.2

NEED FOR MODELLING

A model is a simplified representation of an operation or a process in which only the basic aspects of the most important features of a typical problem under investigation are considered. The objective of models is to provide a means for analysis of the behavior of the system for the purpose of improving its performance. For each system or component, a need model can be developed as shown in Figure 5.1. The essential features of a need model is that it portrays a disparity between what is wanted and what is available. The resolution of a need implies a real location of resources so that what is wanted is made possible.

5.2.1

Availability and Use of Resources

These need models reflect the issues that must be resolved in each component and the manner in which they can be resolved. Thus for the problem statement, a model emerges that represents how the problem as defined in the problem statement is to be analysed.

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71

Defining problem and modelling the problem statement are basically only part of the overall design and planning process. A model is a conceived image of reality; it normally will be a simplification of reality. For each system or component, a need model can be developed as shown in Fig. 5.1. Need model

Criterion

Model to determine what is wanted

How to measure demand What is wanted Objective Need

Gap between what is wanted and what is available Trade of model required

What is available

Objective Model to determine what is available

Criterion

Fig. 5.1 Need model What is available is an indication of the resources that are actually and potentially available for the problems solution. What is wanted is an unconstrained statement of an ultimate state of the problems solution. The disparity is an indication that a limited resource problem exists that can be solved with the aid of a decision model. In order to fully develop the need model, the engineer is required to formulate a descriptive model of what is available and what is wanted. Having determined a model of the problem definition and its structure the engineer must acquire necessary data and carry out the analysis and design according to the framework and issues that he has formulated. Although these activities lead to a technical solution of the problem, the task of implementing this solution must be achieved if the design and planning efforts are to be successful.

5.3

MODELLING TYPES AND PURPOSES

Types of model in mechanical system design

72

Mechanical System Design 1. Model by function (a) Descriptive model (b) Predictive model (c) Normalising model 2. Model by structure (a) Iconic models (b) Analog models (c) Symbolic models 3. Model by nature of the environment (a) Deterministic model (b) Probabilistic model 4. Model by the extent of generality: (a) Specific models (b) General models

1. Model by Function (a) Descriptive model: It describe some aspects of a situation based on observation, survey, question results or the available data. The result of an opinion will represent a descriptive model. (b) Predictive model: These can answer ‘questions made by prediction regarding certain events’ for example, based on survey results technique method obtained to explain and predict the election result before all the votes are a duly counted. (c) Normalising model: When a predictive model has been successful it can be used to prescribe a source of a action in linear programming because it prescribes what the manager ought to do.

2. Model by Structure (a) Iconic model: Iconic models are pictorial representation of a real system and have the appearance of real thing. An iconic model is said to be “scaled down” or “scaled up”. According to the dimension there model are smaller or greater than those of the real item e.g. city maps have blue print, globe and biology structure of a cell for reading purpose. An iconic model is difficult to be manipulated. (b) Analog model: Analog models are more abstract than the iconic ones for there is no lookalike correspondence between these models and real system. They are built by utilising a set of properties to represent the components. e.g. A network of pipes through which water is running could be used as parallel for understanding the distribution electric current, graphs and maps in various colours and analog model in which different atoms corresponds to difficult characteristics. A flow process chart is an analog model, which represents the order of occurrence of various events to make a product. (c) Symbolic model: Symbolic model are also called mathematical model. Symbolic model are most abstract in notation. They employ a set of mathematical system to represent the components and their inbetween relationship of the real system.

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3. Model by Nature of Environment (a) Deterministic model: In deterministic models all parameters and technical relationship are assumed to be known with certainty when the decision is to be made. (b) Probabilistic model: Linear programming and break even models in which at least one parameter or decision variable is a random variable are called probabilistic or stochastic models.

4. Model by the Extent of Generality (a) Specific model: When a model presents a system at same specific time, it is known as specific models. (b) General model: These models are used to explain alternative strategies. These models do not yield any optimum solution to the problem, but give a solution to a problem depending and based on the past experience. Question: Discuss different types of models.

(UPTU 2004)

Ans: Different types of models Models

Physical

Static

Mathematical

Dynamic

Numerical

Dynamic

Static

Analytical

Numerical

System simulation

Fig. 5.2

Types of models

Question: Describe the steps involved in modelling a mechanical system. (UPTU 2004) Ans: The steps involved in modeling a mechanical system Model is based on a representation of the operation of the system. To the user of a model, the model is either a mathematical formula or a computer program that when supplied with the numerical values of various parameters, will make a numerical prediction of the system performance. Model can be involved for a variety of steps:

74

Mechanical System Design 1. Understanding the model is used to explain why and how: It helps in developing an insight. It explains why the system behaves in certain way. 2. Learning: A model may be intended to teach about factors that determine system performance. A model may omit certain features of the real system and focus on those aspects, which are crucial for effective operation of the system. 3. Improvement: A model may help improve the system design and operation. 4. Optimisation model: An optimization model determines the optimal combination of system parameters. 5. Decision making: A model can help in decision making in designing or operation of the system.

Question: Model building is the essence of system design. Discuss.

(UPTU 2005)

Ans: Model building is the essence of system design To illustrate a system design problem, we consider the problem of designing a small part of online computer system, that is, a computer that responds immediately to messages it receives. The function of the system is described by the flowchart of messages are being received over a communication channel at the rate of message a second on the average there are x characters in a message. They are received in a buffer that can hold a maximum of b characters. A fraction k of the messages need replies which have an average of r characters. The same buffer is used for both receiving and sending messages. Processing a message when it has been received takes about 2000 instructions. Preparing a reply requires a program that uses about 10,000 instructions. The finite size of the buffer means that messages and replies must be broken into sections of not more than b characters. Servicing each such section causes an interruption of the processing, requiring the execution of 1,000 instructions to transfer data either in or out of the computer. Three computers are being considered as possible system components: A slow, a medium, and a fast computer, which have instruction execution rates of 25,000, 50,000 and 100,000 instructions a second. In addition four buffer sizes are being considered. There could be a one, two, five or ten character buffer. Given the process the cheapest design can then be determined.

5.4

LINEAR GRAPH MODELLING CONCEPTS

A network system may be graphically represented by a set of points, together with a set of lines connecting some of the points. A simple form is one in which system parameters or components are represented by points (called nodes). The interrelationships or connections that exist in the system between the parameters or components are represented by lines (called branches) joining the relevant graphical nodes. The graph itself is called a linear graph model of the system. Linear graphs are not drawn to scale and the relative positions of the system parameters or components. Therefore, linear graphs should not be confused with geometric graphs and the physical properties of lines and points. Linear graph theory is an abstract concept and is called linear because of its concern with the connectivity of lines and nodes. The primary purpose of a linear graph is to present a concise, graphical representation of the interrelationships of the system that are relevant to the system problem at hand. It thus provides a conceptual way of portraying the environment or system in which a problem is embedded.

System Modelling

5.4.1

75

Linear Programming Model

The non-negativity restrictions x1 ≥ 0 and x2 ≥ 0 imply that values of the variables x1 and x2 can lie only in the first quadrant ( x1 , x2 , plane). This is shown by shaded area of Fig. 5.3. Other quadrants do not satisfy the non-negativity restrictions and hence the point ( x1 , x2 ) can not lie in them. Therefore a number of alternatives are eliminated. The effect of the remaining constraints can now be added to figure constraints with their inequality sign changed into equality sign. The direction in which each constraint holds good is then determined from the direction of the inequality and is indicated by an arrow on its associated straight line. The constraint conditions define the boundary of the region containing feasible solution. Use finding the optimum solution x1 + x2 ≤ 450 2 x1 + x2 ≤ 600 Z max = 3 x1 + 4 x2

Max Z = 1

X2

0

Fig. 5.3

5.4.2

X1

Linear programming graphical method has been solved by simplex method

Graphical Method for Linear Programming Problem Solution

Procedure: For solving the linear programming problem by graphical solution the following procedure is adopted. Step.1: Identify the problem, the decision variables, the objective function and the constraints restrictions. Step.2: Draw a graph that indicates all the constraints/restrictions and identify the feasible region (solution space). Step.3: The feasible region obtained in step 2 may be bounded or unbounded. Compute the co-ordinates of all the corner points of the feasible region.

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Mechanical System Design

Step.4: Select the corner point obtained in step 3 that optimises (maximises or minimises) the objective function of the optimum solution. Step.5: Interpret the results.

Linear Programming Problem Solved by Graphical Method: Linear programming problems involving two decision variables can easily be solved by graphical method. Also solve by simplex method Problem Maximize Z = 4 x1 + 3 x2 Subject to constraints 2 x1 + x2 ≤ 1000 x1 + x2 ≤ 800 x1 ≤ 400 x2 ≤ 700 x1 ≥ 0 , x2 ≥ 0

Solution by Graphical Method

1000

2x1 + x 2 = 1000

x 2 = 700

800

T 700

S x1 = 400

R

600

x2 400

x1 + x 2 = 800

Q

200

P 0

200

400

500 600 x1

Fig. 5.4

800

1000

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Table 5.2 Extreme point

( x1 , x2 )

z = 4 x1 + 3 x2

O P Q R S T

(0, 0) (400, 0) (400, 200) (200, 600) (100, 700) (0, 700)

0 1600 2200 2600 2500 2100

Z Maximum = 2600 at a extreme point R Ans. Example 2: A manufacturer wishes to determine how to produce two products X and Y so as to realize the maximum total profit from the sale of the products. Both products are made in two process A and B. It takes 7 hours in process A and 4 hours in process B to manufacture 100 units of product X. It requires 6 hours in process A and 2 hours in process B to manufacture 100 units of product Y. Process A can handle 84 hours of work and process B can take 32 hours of work in the scheduled period. If the profit is Rs 11 per 100 units of product X and is Rs 4 per 100 units of product Y how many products should be manufactured to realize the maximum profit? Assume that whatever is produced is sold and set up time for the process is negligible. Solution: Formulation of the problem Let x1 = number of units (in hundreds) of product X to be manufactured x2 = number of units (in hundreds) of product Y to be manufactured Z = Total incremental profit of the firm The objective function is: Maximize,

Z = 11x1 + 4 x2

...(1)

Here, x1 ≥ 0; x2 ≥ 0 Constraints: availability of time on process A and B expressed as 7 x1 + 6 x2 ≤ 84 4 x1 + 2 x2 ≤ 32

Thus, the problem becomes: Maximize Z = 11x1 + 4 x2 Subjected to

7 x1 + 6 x2 ≤ 84 4 x1 + 2 x2 ≤ 32

x1 ≥ 0 , x2 ≥ 0 The problem can be solved in 2 ways: 1. Graphical method 2. Simplex method For solving this problem graphically the constraints (inequalities) are converted temporarily into equations.

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Mechanical System Design Graphical Method: 7 x1 + 6 x2 = 84

4 x1 + 2 x2 = 32 Any one of the points 1, 2, 3, and 0 respectively (0, 14), (2.4, 11.2), (8, 0) and (0, 0) may give optimum solution. From point (1); z = 11x1 + 4x2 = 11 × 0 + 4 × 14 z = 56

16 14 1 12

Region of feasible solution

2

10 Product Y (x 2 )

4x 1 + 2x 2 = 32

8 7x 1 + 6x 2 = 84

6 4 2 3 0

4

2

4

6

8 10 12 14 16 18 Product X ( x1)

Fig. 5.5 For point (2);

z = 11 × 2.4 + 4 × 11.2 = 71.2 For point (3); z = 11 × 8 + 4 × 0 = 88 For point (4); z = 11 × 0 + 4 × 0 =0 Maximum profit occurs at point 3 and is Rs 88

Example 3: Find the minimum value of z min = 20x1 + 10x2 z = x1 + x2 ≤ 40 3 x1 + x2 ≥ 30 4 x1 + 3 x2 ≥ 60 x1 , x2 > 0

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79

Solution: x1

3x1 + x 2 = 30

P

F

4x1 + 3x 2 = 60

x1 + 2x 2 = 40

Feasible region

G

C

x2

A

Fig. 5.6 The co-ordinates of the extreme points are: C = (15, 0), A = (40, 0), F = (4, 18), G = (6, 12)

Table 5.3 Extreme point

x1 , x2

z = 20 x1 + 10 x2

A F G C

(40, 0) (4, 18) (6, 12) (15, 0)

800 260 240 300

Clearly the minimum value of Z occurs at point G, hence the optimum solution is x1 = 6 and x2 = 12 with minimum Z = 240.

5.5

MATHEMATICAL MODELLING CONCEPTS

All system models must portray in some manner the number and behavior of its components, the system structure of the component interaction and system conditions introduced by the exposure of the system to physical laws and requirements imposed by organisational considerations. In addition the relationship and use of the model by the modeler in the design process may require that the model exhibit decision aspects. The essential difference between the mathematical and graphical modelling of systems is that the mathematical model must provide explicit structural statements in its formulation, whereas the graphical model analysis can build on an existing graph structure consequently. The mathematical model and the individual equations, it uses often exhibit the mathematical characteristics of the system graph structure. The use of mathematical model is usually more convenient than a graphical analysis when analysing large systems.

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Mechanical System Design

The entire example developed explicitly utilize and exhibit graphical structural properties in some segments of the models. Thus, if the connectivity of branches and nodes is used to define the structural interaction of the components, then the branch node incidence matrix A or its transposed AT will appear in the mathematical formulation. All equations expressing compatibility between branch and node quantities will exhibit the following form: Branch vector = A (Node vector) For example, in the compound bar problem, the relationship between elongations and joint displacements is expressed as u = Au′ In the leveling network problem, the relationship between elevation difference and elevation is expressed as (V + L) = AH Similarly, all equations involving equilibrium or continuity in which branch quantities are summed up at a node will exhibit the form: Node vector = AT (branch vector) For example, in the compound bar problem, equilibrium of branch and nodal forces at the joints is expressed as

P′ = AT P For the survey problem the distribution of errors in the measurements yields

AT V = 0 For the highway problem, the continuity of flow along highways and through intersections gives

AT φ = φ ′ Although the examples exhibit this modelling similarity, they represent the solution of engineering problems that differ in complexity. In the case of the compound bars, the components and structures are well defined. The constraints must conform to the laws of science that apply to the problem. However the purpose of the system is not really defined in terms of what objectives or goals the system must fulfill. The second example, involving a surveying network focuses on an observational and measurement problem. The formulation requires an explicit recognition of the fact that measurements are always inexact and an implied error behavior is built into the model. The highway problem presented in the third example represents a system with an expressed purpose that must be fulfilled. It also illustrates that the exact form of the equations in the model will depend on how the problem is stated. If the objective is to determine the maximum network capacity, one model will be appropriate, but if the objective is to minimize cost of network improvement, a model with slightly different equations becomes appropriate. In reality, the problem could be expanded to involve other considerations, such as construction schedule and roadway maintenance. Thus in this case, all the problem components have not been fully defined at this point. The consideration of these other aspects involves the management of the system.

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Note that regardless of the level of complexity of the problem, mathematical modeling can be utilised. In planning and designing engineering systems, mathematical modeling can serve as a valuable analytical tool.

5.6

MATHEMATICAL MODELLING AND SYSTEM BEHAVIOR

One of the important activities of an engineer is the application of his knowledge and creative skill in designing new systems and redesigning existing ones to prove their performance. However before he can start designing systems, the engineer must have a good understanding of the system behavior under three conditions. 1. Automobile suspension system 2. A biomedical system, and 3. A liquid level system in mechanical system design.

1. Automobile Suspension System This damps out vibrations and jerks produced in a vehicle due to uneven road surface. The weight of the passenger and car body is transferred to the axle through a spring and a shock absorber. M

Car body and passenger

x 2 (t )

Shock absorber

D

K

x1(t ) Road surface

Fig. 5.7

Automobile suspension system

K = spring constant D = coefficient of damping M = mass of passenger + car body The system variables are displacement (x), velocity (v) and acceleration (a). The input variable is x1 (t) and the output variable is the vertical motion x2(t) of the mass M. The mathematical model should relate to input and outputs. The physical law governing the behavior of this system is D’ Alembert’s principle. It states that algebraic sum of externally applied forces ( f ) and the forces resisting the motion in any given direction = 0. In this case, there is no external applied force. The force opposing the motion are Ma = M

dv d 2x = M 2 (by mass) dt dt

82

Mechanical System Design Kx = spring DV = Ddx/dt (dash port) Using D’ Alembert’s principle M

d 2x dx +D + Kx = 0 dt dt 2

(1)

Equation (1), along with appropriate initial condition is the mathematical model of the system in mechanical systems design.

2. A Biomedical System When a drug is injected into the body, it suddenly raises the concentration of that drug in the blood. In due course, some part of the drug is passed out of the blood stream and the remaining part is converted into other chemicals. As a result the drug concentration gradually decreases. A mathematical model to compute the drug concentration at any time after the injection is desirable. Variables of interest are: Drug concentration c(t) mg/litre Drug flow rate qi(t) mg/sec Drug outflow rate qo(t) The volume outflow rate is generally a known constant = K Therefore qo = Kc As the volume of the blood in a body = constant = v Thus v and K are parameters of the system The continuity equation for the system is

 Vc  qi = qo + d    dt  ⇒ V Subject for

dc + qo = qi dt qo = Kc V

dc = Kc = qi dt

This is the required mathematical model in mechanical systems design.

3. A Liquid Level System Consider the liquid level system as shown: The inflow and outflow for the tank is controlled by inlet and outlet valves. Suppose the inlet valve is suddenly opened, we are interested in how the liquid level in the tank will change with time. System variable: Input and output flow rates and liquid level in the tank. Parameters: Valve, resistance (R) and area of x-section the tank.

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83

H (Turbulent flow) R If we assume that change in head is small around the operating point P (Q1 , H1 ) , the incremental resistance is constant around this point. Q=

Q + qi X

Inlet valve

C

H + hi

R

Q + qo

Outlet valve

Fig. 5.8

Liquid level system

Q

Q1

R

R

h

q

H

Fig. 5.9 i.e.

h =R Q

or,

q0 = h/r

Input = output + accumulation or, inflow rate = outflow rate + accumulation Qi = q0 + Adh/dt Qi = h/R + Adh/dt

H

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Mechanical System Design

5.7

MODELLING AND SIMULATION

5.7.1

The Role of Models in Engineering Design

A model is an idealisation of part of the real world that helps in the analysis of a problem. You have employed models in much of your education, and especially in the study of engineering you have learned to use and construct models such as the free body diagram, electric circuit diagram, and the control volume in a thermodynamic system which illustrates some of the common types of conceptual models. A model may be either descriptive or predictive. Descriptive model enables us to understand realworld system or phenomenon; an example is a sectioned model of an aircraft gas turbine. Such a model serves a device for communicating ideas and information. However it does not help us to predict the behavior of the system. A predictive model is used primarily in engineering design because it helps us to both understand and predict the performance of the system. We also can classify models as follows: (1) static or dynamic (2) deterministic or probabilistic and (3) iconic, analog, and symbolic. A static model is one whose properties do not change with time; a model in which time-varying effects are considered is dynamic. In the deterministic-probabilistic class of models there is differentiation between models that predict what will happen. A deterministic model describes the behavior of a system in which the outcome of an event occurs with certainty. In many real-world situations the outcome of an event is not known with certainty, and these must be treated with probabilistic models. An iconic model is one that looks like the real things. Examples are a scale model of an aircraft for wind tunnel test and an enlarged model of polymer molecule. Iconic models are used primarily to describe the static characteristics of system, and combined with the length scale ratio to express the ratio of velocities at equivalent positions in similar systems. Typical kinematics similarity ratios are: Acceleration:

ar =

Velocity:

Vr =

am Lm Tm−2 = = Lr Tr−2 ap L p T p−2

...(i)

vm Lm Tm−1 = = Lr Tr−1 vp L p T p−1 or

...(ii)

Qm L3m Tm−1 = 3 −1 = L3r Tr−1 ...(iii) Volume Flow Rate: Qp L pT p In dynamic similarity the forces acting at corresponding times and on corresponding locations in the model and the prototype are in a fixed ratio. In fluid-flow situations the forces arise from inertia, viscosity, gravity, pressure, vibration, centrifugal force, or surface tension. In a system in which the forces produce fluid motion, e.g., flow under gravity, dynamic similarity automatically ensures kinematics similarity. In situations, in which movement is produced mechanically, it is possible to obtain kinematics similarity without satisfying dynamic similarity. Some dynamics similarity ratios are: Qr =

Inertial force Fi = mass × acceleration

= ρ L3

L = ρ L2 v 2 T2

…(iv)

System Modelling

Viscous force

85

Fµ = τ L2 = µ

du L2 v L2 =µ = µ Lv dy L

…(v)

Dimensionless groups involved with dynamic similarity usually are formulated by taking the ratio of the inertia force to the other fluid forces. Thus,

Inertial force F ρ L2 v 2 ρ Lv = i = = = N Re Viscous force Fµ µ Lv µ

…(vi)

The resulting dimensionless group is the familiar Reynolds number. N Re often is used in similitude considerations of a fluid system. We can write Eq. (vi) as N Re =

where v =

Lv Lm vm L p v p = = v Vm vp

…(vii)

µ is the kinematics viscosity and L is the linear dimension, usually taken as the diameter of ρ

the pipe.

v=

L p Vm Lm V p

vp =

1 Vm vp s Vp

…(viii)

From Eq. (viii) we can see that if we wish to build a 1 to 10 scale model (s = 0.1) the velocity in the model must be increased by a factor of 10 to maintain dynamic similarity. If the higher velocity is difficult to attain, then we may be able to maintain similitude by changing the fluid. At 60° F the kinematic velocity of water is 0.0435 ft2/h compared with 0.568 ft2/h for air at the same temperature. This gives a ratio Vm /Vp of about 1:13. Or since v of gases increases with temperature, a model employing cold air would result in a 1 to 20 scale factor to simulate air in a furnace at 2500° F. Thermal similarity requires that the temperature profiles in the model and the prototype must be geometrically similar at corresponding times. In addition, when system involve bulk movement of material from one point to another, thermal similarity requires that kinematic similarity must also be obtained. Chemical similarity requires that the rate of a chemical reaction at any location in the model must be proportional to the rate of the same reaction at the corresponding location in the prototype. This requires both thermal and kinematic similarity plus a concentration profile within the model that corresponds to the one in the prototype. Chemical simulation usually is difficult to achieve because it requires proportionality of time, temperature and concentration between the model and the prototype. Since the rate of chemical reaction usually is very sensitive to temperature, chemical models often are operated at the same temperature and concentration conditions as the prototype. Chemical similarity is achieved when the temperatures and concentration are the same for the model and the prototype at corresponding location and times. Similitude and dimensional analysis are necessary requirements for physical modeling. The use of dimensional analysis provides a way of reducing the number of variables needed to describe a system. It also allows data from scale models to be properly interpreted with respect to the design. The use of dimensional analysis can reveal strength and weakness in existing designs and thus point the way to improve new designs.

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5.8 5.8.1

Mechanical System Design

A CASE STUDY: COMPOUND BAR SYSTEM Introduction

Models are extentively used by engineers: as aids in the description, analysis, and design phases of problem solving; to facilitate the communication of ideas; and as a means of storing information for future reference and use. A model is a representation of some objects or conditions that exists, or of an issue under consideration. It contains information expressed in certain specific forms about the object it models and requires interpretation according to certain predefined rules. Often the modeling medium is the symbolism of mathematics and requires the logic and understanding of mathematics for its interpretation. In many cases, mathematical models require the support of both descriptive and schematic models to define the nature of the model and to aid in interpreting the symbols and variables used in the model. A mathematical model can thus be considered to be made up of four basic component models, namely: 1. A descriptive model of the problem and purpose; 2. A symbolic definition model describing all the model symbols, variables, parameters, and constants used in the mathematical model itself; 3. A schematic model illustrating the interpretation of the symbols and interrelationships among the system components; 4. The mathematical model itself. As indicated in Fig. 5.10, the first three component models define the manner in which the final mathematical model (component 4) is to be interpreted. In the mathematical modeling of engineering systems, each component of the system may require modeling components as defined above and as illustrated in Fig. 5.10. Descriptive problem and purpose of model Modelling Symbolic definition model

Schematic model

Mathematical model

Modelling procedure

Model itself

Fig. 5.10 Basic component of a mathematical model

The Fundamental Use of Models As an illustration of the modeling process and the various kinds of models used in problem solving, consider the following example:

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Model 1: Descriptive Model of Problem and Purpose A straight uniform bar of specified length is to be made from a light elastic material. The bar is to hang vertically from a support and carry a specific axial load at its free bottom end within a specific deflection range. This model is a statement of the modeler’s view and evaluation of the problem that exists for him in reality. It formulates the model purpose and system constrains; its definition is a critical step since it provides the requirements and objectives for the final model.

Model 2: The Symbolic Definition Model. Let

A B L

be the connection point of the bar to the support. be the bottom end point of the bar be the length of the bar in meter. be the cross-sectional area of the bar in square metre. As E be the Young’s modulus of the material in newton per square metre. P be the axial force along the bar member in Newton be the bar member stress in N/m2 σ be the bar member strain ∈ be the bar member elongation in meter ∆L be the applied axial load at end B1 and the reaction load at A1 respectively, in newtons PB′′ PA′ be the displacements of the points B and A, respectively, in metre u B′ ′ u ′A be the maximum permissible movement of the applied load at B in metre. ∆ Model 2 is the interface between the modeler’s knowledge of the real problem and the requirements of a specific technological model. These symbols, in fact, represent the system components and their attributes that are relevant to the problem.

Model 3: The Schematic Model PA′ N

A

As m2 Component 2

E (N/m )

System components Joint A Joint B Bar AB

uB′ (metre) B PB′ (N )

Fig. 5.11

Attributes ′ PA′ ⋅u A ′ ′ PB ⋅uB

P , σ , ∈, ∆L

88

Mechanical System Design

Model 4: Mathematical Model of the Bar Behavior Bar assumed weightless P = constant

mathematical logic and derivation i.e., program segment for deriving

σ = P/ As ∈=

∆L = σ /E = P/ As E L

∴ ∆L =

PL As E

Mathematical model Mathematical model

Model 4 is a mathematical model for the technological description of the bar behavior. Its selection depends on the level of technology available and known to the modeler and the relevant depth considered necessary to meet the problem requirements. The mathematical model identifies the five parameters that describe the system response and structure of the bar. If any four parameters are known, the fifth can be found directly from the model. It can be used to study the bar response to the design problem model described later.

Model 5: Mathematical Model of the System Component Interaction ∆L = u B′ − u A′ u A′ = 0 P = PB′ PA′ = PB′ Model 5 defines the system structure relating the component bar parameters and the node parameters. It normally requires a unique creative effort on the modeler’s part, since a specific technological formulation for the problem may not exist.

Model 6: Mathematical Model of the System Response From model 4; ∆L =

PL where PB′ is the system input As E

 L  L From model 5; ∆L = u B′  is the system mode of response; ,  As E  As E P = PB′ u B′ is the system response and system output

PB′ L As E Model 6 is a statement indicating how the modeler intends to read the system. It focuses on the input-output description of the system and represents the solution process for the systems model. ∴ uB′ =

Model 7: Mathematical Model of the Design Problem uB′ =

PB′ L ≤∆ As E

System Modelling

89 ∴ As E ≥ =

PB′ L = a constant ∆

Model is a mathematical model exposing the design variables As and E. The modeler can now establish a design criterion to test whether a particular design is suitable or not. Since the design criterion is bounded on one side only, the designer must establish his own value system for departures of AsE from the constant PB′ L/∆. The total mathematical model for the system design problem is summarized in Fig. 5.12. The seven model components portray in various ways such things as the embodiment of model purpose, scope of investigation, reality constraints on condition and choices, and the modeler’s bias and knowledge of technology. They do not, of course, model the design process and selection of specific attributes for the model components. The model itself can be checked for internal consistency; for level of accuracy relevant to the real world problem, with problem identification, needs and resources; and for the impact of the modeler’s bias. Model 1, for example, prescribes that the structure be a bar (not a truss or combination of structural elements), that it be uniform, straight, and be made from a light elastic material. The inclusion of all (or any one) of these descriptions severely limits the design choices potentially available and, in effect, plays a dominant role in the specific forms of models 2 to 7. The requirement for a light material (the validity of which should be considered) tentatively supports the assumption of a weightless bar in model 4 and the vague selection criterion imposed in model 7. Existing technology permits model 4 to be updated to include the effect of the self-weight of the bar. Whether this is done or not reflects a decision made by the modeler. If the material selection aspect of the problem is to be elaborated, at least models 1 and 7 are affected and require replacement by new or updated versions. Descriptive Model 1

Schematic

Symbolic

Model 3

Model 2

Mathematical Model

Model 4 - Bar behaviour Model 5 - System structure Model 6 - System response Model 7 - Design problem

Fig. 5.12 Mathematical model of system design problem

90

Mechanical System Design

Similarly, an investigation can be directed to the necessity for the imposition of constraint such as straightness and uniformity on the bar. If the load is applied dynamically or suddenly, model 4 may not be accurate or suitable; in which case, models 1, 4, 6 and 7 must be changed. Suppose that the cost of the structure and the time to manufacture and erect the structure enter into consideration and affect the model’s demands purpose. The requirement for internal consistency in the models demands that if costs enter into model 1, a consideration of costs must appear in at least one of the following models. The specific form of its inclusion will expose the modeling and design rationale of the modeler, which can then be examined and critiqued. The development of useful and valid models requires a high level of creative ability, technical and professional knowledge, and an understanding of the forces, constraints, and values at work in the environment in which the problem exists.

5.8.2

A Case Study: e.g. A Compound Bar System Model

If a single equation can be considered as a model of a condition or component, then a set of equations intuitively represents a model of a system. In fact, many system models take the form of a set of equations, either algebraic, functional, deferential, or integral. In some cases, each individual equation introduces a new condition initiated by the consideration of another component in the system model. In other cases, each equation introduces a new constraint on the relationships among the set of system variables. Great insight into mathematical models can be gained if the various equations are grouped into sets in which each set of equations focuses on a specific type of system constraint; i.e., component behavior, system structure, system equilibrium. As an example, consider a compound bar system in structural design.

5.8.2.1 Statement of the Problem The compound bar system shown in Fig. 5.13 consists of four bar components, 1, 2, 3 and 4 connected to three short horizontal members, A, B, and C, in which member A is a fixed support. The construction is such that these horizontal members can be considered as rigid cross pieces so that the vertical bar members are constrained to axial deformations only; i.e., the horizontal members remain horizontal and can be treated as individual joints in the structure. The system can be loaded with joint forces PB′ and PC′ at B and C, developing the reaction force PA′ . The joint forces PB′ and PC′ can be considered as system inputs. As a result of the applied system loading, the bars will be loaded, producing axial forces P1, P2, ... P4, and the bars will deform by the amounts u1, u2, ..., u4. This internal loading and deformation of the system will cause joints B and C to displace amounts u B′ , and uC′ . Since joint A is a fixed support, the displacement u A′ is zero. The joint displacement u B′ and uC′ can be considered as measurements of the system response (i.e., output) to the system loading input. The problem is to develop a mathematical behavior model that can be used to analyze the system response under various combinations of loading at B and C. This is realized if a set of equations can be developed to express the system response parameters u B′ and uC′ as functions of the system input parameters PB′ and PC′ .

System Modelling

91 PA′ O

A 1

2

(P1, u1)

(P2 , u2 )

4

(P4 , u4 )

B

′ uB

PB′

(P3 , u3 ) C

′ uC

PC′

Fig. 5.13

Compound bar system

System Constraints The following constraints exist in the structure: 1. Component behavior: The response of a bar member when subjected to an axial load is governed by a physical law, commonly referred to as an equation of state; 2. System compatibility: Geometric relationships exist between the elongation of the bar members and the physical displacements of the joints; 3. System equilibrium: The components are in stable equilibrium under static loading.

1. Component Behavior It has already been shown that the following relationship exists between the axial force Pi acting on a bar and the corresponding elongation ui

EA  Pi =  i i  ui = Ki ui ...(1)  Li  In structural engineering, Ki is called the stiffness coefficient of bar i and is expressed in units of bar force per unit displacement. The stiffness coefficient Ki thus reflects the “stiffness” of bar i and is a function of the material, the cross-sectional area and its length. One each equation can be used to describe the behavior response of each of the four members in the given structure. Thus, P1 = K1u1 P2 = K 2u 2 P3 = K 3u3

...(2)

92

Mechanical System Design P4 = K 4u 4

In matrix notation, equation 2 becomes:

 P1   K1     P2  =  0  P3   0     P4   0

0

0

K2 0 0

0 K3 0

0   u1    0   u2  0   u3    K 4   u4 

P(4 × 1) = K (4 × 4) µ(4 × 1)

i.e., Where P(4 × 1)

u(4 × 1)

...(3) ...(4)

is the column matrix of component axial forces; is the column matrix of component axial deformations;

K(4 × 4) is a square diagonal matrix of component stiffnesses. Its diagonal form indicates (and establishes through matrix multiplication with U (4 × 1) ) the independence of the four equations.

2. System Compatibility Connecting the bars together to form the system structure requires the development of mathematical expressions to enforce the geometric compatibility between member elongations and the joint displacements. System compatibility is insured if the individual members continue to be connected to the joints whatever the joint displacements. This implies that the members must elongate just enough to suit the joint displacements. Therefore:

u1 = − u ′A + uB′ + 0 u2 = − u ′A + u B′ + 0 u3 = 0 − uB′ + uC′

…(5)

u4 = − u ′A + 0 + uC′ Equations 3–5 state that member exogamous depend on relative joint displacements. In matrix notation,

 u1   −1 +1 0   − u A′        u2  =  −1 +1 0   uB′   u3   0 −1 +1  uC′         u4   −1 0 +1  0 

…(6)

3. System Equilibrium Assuming that the applied loads P2′ and PC′ are static, the system is in equilibrium if the algebraic sum of the forces acting on any joint in the structure is equal to zero. Any direction (up or down) can be assumed to be positive since a reversal of sign does not affect the equation, but a positive sense must be maintained during the formation of the equations. Usually the same positive direction is used for all joints in the structure.

System Modelling

93

Figure 5.15 illustrates the forces acting on the present structure. Bar axial forces are shown as tensile (i.e., positive) forces, and joint forces as downward (i.e., positive). The following equilibrium equations can be derived: At joint A

PA′ + P1 + P2 + P4 = 0

At joint B

PB′ − P1 − P2 + P3 = 0

…(7)

At joint C PC′ − P3 − P4 = 0 Thus, rearranging terms to separate the applied joint loads (cause) from the bar component member loads (effect) results in:

PA′ = − P1 − P2 − 0.P3 − P4 PB′ = + P1 + P2 − P3 + 0.P4 PC′ = 0.P1 + 0.P2 + P3 + P4

…(8)

PA′ A

P1

P2

B

P4

PB′

P3

C

PC′

Fig. 5.15

Member forces and joint loads

In matrix notation

 P1   PA′   −1 −1 0 −1        P2   PB′  =  +1 +1 −1 0   P   P′   0 0 +1 +1  3   C   P   4

...(9)

94

Mechanical System Design

Exercise 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

5 . . .

Discuss different types of models, used commonly, for mechanical systems design. Illustrate your answer suitably. Describe the steps involved in modeling a mechanical system. UPTU 2004 Write in brief about iconic, analogue and mathematical models. UPTU 2004 Model building is the essence of system design. Discuss. UPTU 2005 What is mathematical model of real situation? Discuss the graphical model in system design. UPTU 2005 In mechanical system design what are the suitable steps need for modeling? UPTU 2005 In system modeling design system briefly explain modeling types and purposes. Write a note on (a) Linear Graph Modeling Concept (b) Mathematical Modeling Concept and (c) Need Statement Why do we use linear graph modeling in Mechanical system design? In new approach how can you explain about linear graph modeling concept? What is mathematical modeling concept? Write in suitable steps. Explain mathematical as well as linear graph modeling in the form of system modeling.

A Case Study 1. Briefly explain a case study of compound bar system model. 2. Write down in suitable step the statement problem of the compound bar system model. 3. Write a short note on: (a) Compound behavior (b) Compound capability (c) System equilibrium and (d) System constraints 4. Draw a figure of compound bar system model and explain it.

Linear Graph Analysis

95

6 Linear Graph Analysis 6.1

INTRODUCTION

A common feature of many engineering systems is that they are composed of a number of components physically interwoven in the form of a network. Typical examples are the inter state highway system, a city sewage collection and disposal system; a local telephone system; the national transportation system; a city water supply system of dams, pipelines, and reservoirs; and the steel frame of a skyscraper. In addition, there are many engineering decision problems and organisational systems that although they don’t have the physical appearance of networks, can be usefully interpreted as such e.g. the flow of decisions and authority within an industrial firm can be described by a network system. Similarly, the schedule of job in a construction project may be viewed as a network of activates. The flow of cash among the agencies within a state government or among the departments within a construction company may be considered as network of cash flow. A common feature of these decision and organisational examples is that they are viewed in such a way that the inherently logical or procedural aspects are used as a means of identifying components and structuring them together into a network system. These networks can be depicted and analysed through the use of linear graphs; and in this respect, linear graphs are valuable aid to the engineer in seeking solutions to problems that involve network system. Alternative routings for the manufacture of a product consists of 4 stages as shown in the network diagram below. The arrows represent operations with defined work controls and operation time (cost), Further a set of successive arrows from the start point 1 (source) to the end point 9 (sink) indicates the route of the production process for producing a finished product. Selecting a route with least time or cost is the optimum routing analysis. The optimum routing problem has finite number of combinations; a solution for this type of combinatorial problem can always be obtained by calculating the total production times for all possible routes by trial and error. Complete enumeration requires substantial computational effort, when the problem involves large number of arrows. Such problems can be solved for network technique or dynamic program.

96

6.1.1

Mechanical System Design

Solution by Network Technique

In the figure given below we have a set of nodes (vertices) and a set of arcs. Arc Eij connects nodes Ni and Nj . If there exists a sequence of nodes and arcs such that N1 , E12 , N 2 , E23 ... N K − 1 , EK − 1 and N k , this is known as a chain or path from the node N1 to N k . When N1 = N k we get a directed cycle. An acyclic directed network has no loops or cycles. Shortest, Path Algorithm Step 1

Set the evaluation for node h, eh = 0

Step 2

Continue to label the remaining nodes (chosen in ascending order) according to the formula ej = min (ej + dij) dij = Cap of the path ij i = 1, h, h + 1 ... j – 1

Step 3 When the node k has been labelled ek is the value of shortest path from node h to k. The path is determined by tracing backward from node n to all such nodes such that ej + dij = ej ( j = k, k – 1 ... (h + 1) (h) 4

3

3 7 2 4 2

2

6

3 1

5

9

5 4

4

3

3 3

4

5 1st stage

2nd stage

Turning

Planning

Fig. 6.1

3

6

4 3rd stage

4th stage

Drilling

Finishing

Linear graph analysis

Solution: e1 = 0 e2 = min (e j + dij ) = 6(1 → 2) e3 = 4 (1 → 3) e4 = 9 (2 → 4)

8

Linear Graph Analysis

97 e5 = min [6 + 2, 4 + 3] e5 = 7 (3 → 5) e6 = 9 (3 → 6) e7 = 12 (4 → 7, 6 → 7) e8 = min i = 3, 4, 5, 6[ei + d i8] = 8 (3 → 8)

e9 = min [e7 + d 79 , e8 + d89 ] = 11(8 → 9) Min production time = 11 hours.

Optimal route = 1 → 3 → 8 → 9

The Machine Problem can be Solved Using Backward Calculus Method To begin f (9) = 0 f (8) = min [T8i + fj ] = 3(8 → 9) f (7) = 2 (7 → 8)

Next

f (6) = min [3 + 2, 4 + 3] = 5 (6 → 7)

Similarly f (5) = 7 (5 → 8) f (4) = 5 (4 → 7)

Further,

f3 = min [T3 j + f (1)] j = 5, 6, 8 = 7 (3 → 8)

f (2) = min[T2 j + f (1)]

and

j = 4, 5, 6

= 8(2 → 4) Finally

f (1) = min [Ti + f (1)]

j = 2, 3 = 11(1 → 3)

Optimal route 1 → 3 → 8 → 9 , Total flow time = 11 hours.

6.2

GRAPH MODELLING AND ANALYSIS PROCESS

A network system may be graphically represented by a set of points, together with a set of lines connecting some of the points. A simple form is one in which system parameters or components are

98

Mechanical System Design

represented by points (called nodes). The interrelationships or connections that exist in the system between the parameters or components are represented by lines (called branches) joining the relevant graphical nodes. The graph itself is called a linear graph model of the system. Linear graphs are not drawn to scale, and the relative positions of the nodes and branches do not necessarily represent the actual relative positions of the system parameters or components. Therefore, linear graphs should not be confused with geometric graphs and the physical properties of lines and points. Linear graph theory is an abstract concept, and is called linear because of its concern with the connectivity of lines and nodes. The primary purpose of a linear graph is to present a concise graphical representation of the interrelationship of the system that is relevant to the system problem at hand. It thus provides a conceptual way of portraying the environment or system in which a problem is embedded. The modelling capability of linear graphs is shown in the following examples: Example: (1) (a) Consider the problem of a construction contractor who is choosing a route to move heavy construction equipment from Charbag to Engineering College, Lucknow. He has no preference between two-lane or four-lane highways but he does want to avoid traveling on secondary road because of load limitations. He wishes to find the route that requires the shortest travel time. After studying the highway map and consulting the highway authorities, he has narrowed his choices to three, as indicated in Fig. 6.2. (b) is a linear graph model of the three feasible routes. Charbag

C 80

57

Polytechnic

60

P

Haz 45

60 Hazratganj 66

Chi

M

47

90 35

Munsipulia 60

Chinhat

Tedhipulia 74

T 20

Engineering College

E (1) (a) Highway Network Map

Fig. 6.2

(1) (b) Linear graph model

Linear graph representation of a highway network

The nodes denote the cities and towns, and the branches, the highway connections. The numbers along the branches indicate the driving time in minutes. From this graph, he easily arrives at conclusion that his best route is CHARBAG-HAZRATGANJ-MUNSIPULIA-TEDHIPULIA-ENGINEERING COLLEGE which takes 160 minutes. The second best choice is the CHARBAG-POLYTECHNIC ENGINEERING COLLEGE, which requires 170 minutes. The route going through CHINHAT requires 200 minutes.

Linear Graph Analysis

99

A linear graph is constructed with the specific purpose of helping to solve a system problem. Therefore, the choice of parameters to be represented in the graph must depend on the nature of the problem and the system objective. Suppose that an engineer is preparing a plan for improving the highway network in the same area as is illustrated in Fig. 6.2. His graph model may well include the following additional features. Feasible route for new highways, existing highways for which additional lanes can be added, towns that require better highway connections because of industrial developments, recreational sites, etc. He may also choose to use several different symbols among the nodes to differentiate towns, cities, and recreational sites, as well as different symbols among the branches to differentiate the different classes of highway. Example 2: Consider the problem encountered by a contractor who is performing some work in a river flats area that has been subjected to high water conditions and occasional destructive flooding in past. There is a four month period during which he has no use for the equipment either on this job or on others. He can keep the equipment on this job in the river flats, or else move it out, store it, and then move it back at a total cost of Rs 1,800. If he keeps it in the river flats, he has the option of building a platform for the equipment at a cost of Rs 500, which will protect it against high water, but not against a destructive flood. The damage that would be caused by high water amounts to Rs 10,000, if there is no platform. A destructive flood would entail a loss of Rs 60,000, regardless of whether or not he builds a platform. The probability of high water in the four-month period is 0.25; the probability of a destructive flood is 0.02. The contractor has to choose from three possible options: 1. Move equipment 2. Leave equipment in the area and build a protective platform and 3. Leave equipment in the area unprotected. Figure 6.3 is a linear graph model of this decision problem. This type of linear graph model is called a decision tree. A branch represents either an alternative available to the decision-maker or a possible outcome that can result from an uncontrollable chance event. A node denotes either the occurrence of an event when outcome is determined by chance, or a final condition. The decision nodes, denoted by rectangles, are those at which the decision-maker must choose one course of action from among the alternatives then available to him. The chance nodes, indicated by circles, are those at which the decision-maker has absolutely no control. The actual specific outcome from a chance node, of course, depends on nature but in the prior situation for the contractor, is modeled in Fig. 6.3. As a probability outcome using past observations of nature the numbers on the chance outcome branches indicate the probability that these outcome would occur. The final outcomes, shown as triangular nodes at the tips of the tree, show the cost to the decision-maker for each possible combination of events. The preceding examples of the modeling of physical, organisational and sequential decision system problem illustrate the use of linear graphs as an aid in modeling complex problems. The actual modeling of problems using linear graphs requires creativity and an intimate knowledge of the problem on the part of the investigator. The process of applying graphs to systems problems might be described as follows:

100

Mechanical System Design – Rs 1800

nt

– Rs 500

e ipm

EMV

rma

l (.7

3)

qu

E ve Mo

– Rs 500

)

Do

No

.25 er (

at hw

no

Hig

e ov tm eq

Flood (.02

en pm

ui ild Bu

)

orm latf

– Rs 60500

p

t

No Pl

0

)

r fo

at

.73 al (

m

rm

No

High Water (.25)

– Rs 10000

Decision Node d

oo

Fl

Chance Node

2)

(.0

Outcome Node – Rs 60000

Fig. 6.3

Graph model of a contractor’s decision problem

1. As with other systems problems, the first step is to identify the problem and the systems components, structures, and attributes that are involved. 2. The problem must be represented as a model according to some representational form and convention. In linear graph modeling, the model is limited to the use of nodes and branches as the representational forms. In addition, attributes must be assigned to the branches and nodes. 3. The problem must be posed in terms of a graphical structural property that exists in the model. 4. A linear graph analysis procedure must be established that will enable the graphical property to be processed through the linear graph model. 5. The graphical solution must be transferred back into the systems structure of the actual problem. In the remaining sections of this chapter the various aspects of the application and use of linear graph models and analysis are presented.

6.3

LINEAR GRAPH ANALYSIS, A PATH PROBLEM

Graphical properties relate to the many different structural relationships that can be imagined as existing between the various nodes and branches of a linear graph. The development of these properties permits graph theory to be applied to many engineering problems.

Linear Graph Analysis

101

As an illustration, several examples that depict graphical properties as related to specific problems are presented in this and the following section. Consider again, the contractor’s problem involving the movement of heavy equipment over a road network between two towns. See example of graph modelling and analysis process, the contractor is interested in routes and is confined to travel along a sequence of roads passing through towns. In addition although he knows that the roads carry two-way traffic, he only wants to go from Charbag to Engineering College so that a directed one-way route concept results and Fig. 6.2 should be changed to portray a directed graph. A directed graph is one in which at least one branch is assigned a unique direction. An arrowhead usually shows the branch direction. This discussion raises the need for two graphical properties describing routes namely, chains and paths. A chain a sequence of alternating nodes and branches between two terminal nodes where successive branches have one node in common. A path is a chain in a directed graph where, in addition, all the branches in the path point in the same direction. The linear graph analysis problem then reduces to enumerating path directed form node C to E in the directed graph of Fig. 6.2(b). Because no one-way highway segment exists the individual chains and path are identical according to the contractor’s criterion of minimum travel time. Example 3: A linear graph represented the scheduling problem in assembling a prefabricated house. Each arrow models a job and is labeled with its estimated duration in hours. The directions of the arrows indicate the order in which the jobs must be performed. For example, frames cannot be erected until flooring is completed, but can be started before completing electrical wiring and plumbing. Thus, the model clearly shows the order or precedence for all the tasks and the length of time needed to complete each task. Using such a diagram, the total project duration and the amount of extra or slack time available for each task can be computed. In this example, the shortest possible time the project can be completed is 66 hours, or 8¼ days. Such a diagram is called the CPM (Critical Path Method) diagram and is being used extensively for planning and controlling construction progress. Solution: Consider the CPM linear graph model of Fig. 6.4. It is a directed graph since technological and organizational conditions establish the various conditions before an activity can start. Because all the activities must be completed in specific directed sequences, it is necessary to transverse all the seven paths joining the start mode A and the end node G. Consequently, if the problem concerns determining the minimum project duration, the graphical structural property forming the basis of the analysis is again the path. The total time to complete each path is as follows: Path A-B-C-G A-B-C-D-E-F-G A-B-C-E-F-G A-B-C-H-F-G A-B-C-H-F-J-G A-B-C-I-F-G A-B-C-I-F-J-G Thus, the longest path takes 66 hours.

Time required for completion 24 hours 44 hours 40 hours 66 hours 62 hours 61 hours 57 hours

102

Mechanical System Design Landscaping 8

Site clearing (8) A

B

Foundation

Flooring

(8)

(2) C

Frames

Roofing

(3) D

F

Painting 20

(3)

G

E

Electrical wiring (30)

H

J

Plumbing

Install utilities (16)

(25) I

Fig. 6.4

Linear graph model of a construction plan

This is the minimum time required to complete the project. If there is a delay in completing any of the jobs along this critical path, the entire project will be delayed. On the other hand, if the duration of the project is to be shortened, one or more of the jobs along this critical path must be shortened. Therefore this critical path controls the duration of the project. A similar method of analysis can be used for the problem in Fig. 6.3. The contractor has the choice of three possible decision paths: 1. Move equipment; 2. Do not move equipment, but build a protective platform; 3. Do not move equipment and build no protective platform; The consequence of the first path is a definite loss of Rs 1800. However, the consequence of the other two paths depends on chance. Without a definite knowledge of future events, the contractor must estimate what he can expect to gain or lose in taking path 2 or 3. These estimates provide a single measure of value for the paths with uncertain outcomes. Taking into consideration his available capital and his willingness to assume risk, he must decide on which path he will take.

6.4

LINEAR GRAPH ANALYSIS NETWORK FLOW PROBLEM

A large class of engineering networks can be viewed as capacitated flow problems in which one or more commodities (e.g., traffic, water, information, cash etc.) can be considered as flowing through the network whose branches have various constraints and flow capacities. Consider for example, a two-lane, two-way highway network connecting a metropolitan area with a residential Delhi town, as shown in Fig. 6.5 (a) Along each road branch is a vector containing the relevant attributes of the road it represents: road identification number, mileage, and flow capacity in one direction in vehicles per hour (vph). Thus, the road between nodes 1 and 3 is identified as road 1, which is 15 miles long and has a maximum capacity of 700 vph in the direction from node 1 to node 3.

Linear Graph Analysis

103

The two-way capacity would be twice this value. The engineer is interested in determining the maximum flow or capacity, of the network during the morning and evening peak traffic hours. Assuming that the morning and evening flows follow identical characteristics, but in opposite directions, the study can be confined to either the morning or evening conditions; e.g., only the flow from node 5 to node 1 or the flow from node1 to node 5 needs to be studied, not both considering the evening peak traffic. Each path joining node 1 to node 5 represents one possible route of travel. The first step is to assign directions to the flow in as many branches as possible. The direction of flow in roads 1, 2, 3, 4, 5, and 7 must be from node 1 to 5 and can therefore, be assigned. However, the flow in road 6 can be from node 2 to node 4 or from 4 to node 2, and there may be no a priori way of knowing the correct direction to assign to the flow in this branch. Therefore, the problem must be analysed for both possible conditions. For the first case, assume that the flow is from node 2 to node 4 and thus, the direction of flow in each branch is shown in the directed graph of Fig. 6.4(b). a = Road number b = Mileage c = Flow capacity in vehicles per hour

3

(4:15:800)

(1:15:700) (5:12:1000)

(2:14:900) Metropolitan Area

5

4

1

Delhi Town

(6:2:400) (7:12:700) 2

(3:14:1000) (b)

(a)

Fig. 6.5

(c)

(a) Highway network

3 (4:15:800)

(1:15:700)

(5:12:1000)

(2:14:900)

5

4

1

(3:14:1000)

(7:12:700)

2

Fig. 6.5 (b) Directed graph model If the engineer is interested in determining the flow of vehicles between two intersections that are connected by a highway segment, he can place a pneumatic tube across the roadway and connect it to

104

Mechanical System Design

a counting device. The road tube may be placed any where along the highway segment, the only requirement being that the path between the two intersections is traversed. Fig. 6.6(a) depicts this situation in terms of a linear graph.

1

(b) Traffic counter for each highway segment

(a) Traffic counter for one highway segment

3 1:15:700

4:15:800

2:14:900

1

4

5:12:1000

6:2:4000 3:14:1000

7:12:700 2 (d) One possible placement of traffic counter to count traffic flow from node 1 to node 5 (proper cut-set)

(c) Traffic counter for total of all highway segments leaving intersection

3 1:15:700

1

4:15:800

2:14:900

4

5:12:1000

6:2:4000

3:14:1000

7:12:700 2 (e) Another possible proper cut-set

Fig. 6.6

5

5

Linear Graph Analysis

105

= 700 + 1000 + 1000 = 2700 The engineer is interested in obtaining the value of total maximum flow, therefore instead of two counting devices and two tubes he really wants to determine maximum flow counted. Question 1: A truck delivers concrete from the readymix plant to construction site. The network in Fig. 6.7 represents the available routes between the plant and the site. The distance from node to node is given along the route lines in km. What is the best route from plant to site? UPTU 2005 2

2

4 5

3 1 Plant

3

7

6 6

1

4

6

4 3

5

1

Fig. 6.7 Solution: The least distance travelled from plant to site is f (6) = 0 f (5) = 4 f (4) = min (5, 3 + 4) = min (5, 7)

= 5(4 → 6) f (3) = min (1 + 4, 6) = min (5, 6) = 5 (3 → 5 → 6) f (2) = min (1 + 1 + 4, 2 + 5, 7 + 4) = min (6, 7, 11) = 6 (2 → 3 → 5 → 6) f (1) = min (3 + 1 + 1 + 4, 4 + 1 + 4) = min (9, 9) = 9 (1 → 2 → 3 → 5 → 6) or

(1) → (3) → (5) → (6)

Site

106

Mechanical System Design 6

2 3

4

1

1

5 1

3

6 Or 5

1

1

3

4

Fig. 6.8

4

Two alternatives route from site to plant

The best route is 1 → 2 → 3 → 5 → 6 or 1 → 3 → 5 → 6 The least distance travelled plant to site is 9 km Ans. Question 2: For the network shown in Fig. 6.9, find the shortest path from node 1 to node 8. What is the longest path? The figures (values) adjacent to the arcs denote their lengths. UPTU 2004 3 5

Start

3

4

2

1

7

2

4

8

1 2 3

2

4

4 1 5

4

6

Fig. 6.9 Q (3)(a) Shortest Path f (8) = 0 minimum f (7) = min (4, 3 + 4) = 4 (7 → 8) f (6) = 4 (6 → 8)

f (4) = min (1 + 4, 1 + 3 + 4) = min (5, 8) = 5 ((4) → (7) → (8)) f (5) = min (1 + 1 + 4, 4 + 4, 1 + 1 + 3 + 4) = min (6, 8, 9)

End

Linear Graph Analysis

107 = ((5) → (4) → (7) → (8)) f (3) = min (4, 3 + 4, 3 + 3 + 4) = min (4, 7, 10) = 4 (3 → 8) f (1) = min (2 + 1 + 4, 2 + 6, 5 + 4, 5 +3 + 4, 5 + 3 + 3 + 4) = min (7, 8, 9, 12, 15) = 7 ((1) → (4) → (7) → (8))

So, node 1 to node 8 the shortest path is 7

1

1 2

Fig. 6.10

4

4 8

Shortest path

Total time taken in shortest path is 7 minutes. Longest path f (8) = 0 f (7) = 4 (7 → 8) f (6) = max (3 + 4, 4) = 7 ((6) → (7) → (8)) f (4) = max (1 + 3 + 4, 1 + 4) = max (8, 5) =8 f (5) = max (1 + 1 + 3 + 4, 4 + 4, 1 + 1 + 4) = max (9, 8, 6) = 9 ((5) → (4) → (7) → (6) → (8)) f (3) = max (3 + 3 + 4, 3 + 4, 4) = max (10, 7, 4) = 10 ((3) → (7) → (6) → (8)) f (1) = max (5 + 3 + 3 + 4, 5 + 3 + 4, 5 + 4, 6 + 2, 2 + 1 + 4) = max (15, 12, 9, 8, 7) = 15 ((1) → (3) → (7) → (6) → (8)) So, node 1 to node 8 the longest path is

108

Mechanical System Design

5

3 3

1 7 3

4

8

6

Total time taken in longest path is 15 minutes. Ans.

6.5 GOAL PROGRAMMING “Success could be achieved easily when one has a definite goal.”

Introduction Linear programming basically is the technique applicable only when there is a single goal (objective function) such as maximising the profit or minimizing the cost or loss. There are situations where the system may have multiple (possible conflicting) goals. For example, a firm may have a set of goals, such as employment stability, high product quality, maximisation of profit, minimizing overtime or cost, etc. In such situations, we need a different technique that seek a compromise solution based on the relative importance of each objective. This technique is known as Goal Programming. Goal programming technique starts with the most important goal and continues until the achievement of a less important goal. Whether the goals are attainable or not, the objective function is stated in such a manner that the optimization means: “as close as possible to the indicated goals.” Illustration: Consider the following Linear Programming Procedure: Maximize z = x1 + x2 subject to the constraints: 3 x1 + 2 x2 ≤ 12 , x1 + x2 ≥ 8, − x1 + x2 ≥ 4, 5 x1 ≤ 10 and x1 ≥ 0, x2 ≥ 0

Feasible region of the problem is given as under: Although the two shaded areas may be considered to represent the feasible region, yet there is no area of feasible solution. This is so because we cannot find any point ( x1 , x2 ) to lie on both these shaded regions. Thus problem cannot be solved by the useful linear programming procedure. Let us assume that all the constraints of the problem represent some management goals, say for example, overtime operation, utilization time of machines, minimum requirement quality, fulfilment of promised order, etc. If profit be fixed arbitrarily high (say Rs. 10,000) then the objective function z may also be treated as a management goal to acquire the specified profit. Thus there shall be 5 management goals in the problem and the real objective may be changed from profit maximization to goal attainment. As stated above, goals may not always be attainable. The objective of management may be rather a general approach of attaining the goals as closely as possible. This may be achieved by changing the objective function to minimization of

z = d1 + d2 + d3 + d4 + d5

Linear Graph Analysis

109 x2 − x1 + x 2 = 4

x1 + x 2 = 8

5x 1 = 10

3x1 + 2x 2 = 12

x1

0

Fig. 6.11 where d1 = x1 + x2 − 10, 000 , d 2 = 3 x1 + 2 x2 − 12 d 3 = 5 x1 − 10 , d 4 = x1 + x2 − 8 d 5 = − x1 + x2 − 4 This is the case when all the goals are considered equally important. But in most of the business situations, one goal is more important to the management than the others.

6.5.1

Simplex Method for Goal Programming Problem

The major steps of the simplex method for the linear goal programming problem are: Step 1. Identify the decision variables of the key decision and formulate the given problem as linear goal programming problem. Step 2. Determine the initial basic feasible solution and set up initial simplex table. Compute zj and zj – cj values separately for each of the ranked goals p1 p2 ... and enter at the bottom of the simplex table. These are shown from bottom to top i.e.; first priority goal ( p1 ) is shown at the bottom and least priority goal at the top. Step 3. Examine zj – cj values in the p1 row first. If all (zj – cj) < 0 at the highest priority levels, then the optimum solution has been obtained. If at least one (zj – cj) > 0 at a certain priority level and there is no negative entry at the higher unachieved priority levels in the same column, then the current solution is not optimum. Step 4. In the target values of each goal in the solution if column (xB) is zero, the current solution is optimum.

110

Mechanical System Design

Step 5. Examine the positive values of (zj – cj) of the highest priority ( p1 ) and choose the largest of these. The column corresponding to this value becomes the key column. Otherwise move to the next higher priority ( p2 ) and select the largest positive value of (zj – cj) for determining the key column. Step 6. Determine the key row and key number (leading element) in the same way as in the simplex method. Step 7. Any positive value in the (zj – cj) row which has negative (zj – cj) under any lower priority rows are ignored. It is because deviations from the highest priority goal would be increased with the entry of this variable in the basis. Question 3: A textile company produces two types of materials, A and B. The material A is produced according to direct orders from furniture manufacturers. The material B is distributed to retail fabric stores. The average production rates for the materials A and B are identical i.e. 1,000 meters/hour. By running two shifts the operational capacity of the plant is 80 hours per week. The marketing department reports that the maximum estimated sales for the following week is 70,000 meters of material A and 45,000 meters of material B. According to the accounting department the profit from a meter of material A is Rs 2.50 and from a meter of material B is Rs 1.50. The management of the company decides that a stable employment level is a primary goal for the firm. Therefore, whenever there is demand exceeding normal production capacity, the management simply expands production capacity by providing overtime. However, the management feels that overtime operation of the plant of more than 10 hours per week should be avoided because of the accelerating costs. The management has the following goals in order of their importance. P1 : Avoid any underutilization of production capacity. P2 : Limit the overtime allowed for plant operation to10 hours per week. P3 : Achieve the sales target of 70,000 meters of material A and 45,000 meters of material B. P4 : Minimize operation of the plant as much as possible. Formulate this problem as a goal-programming problem.

Solution → Formulation: → Production hours constraint: Regular production hours available to the firm are limited to 80 hours. The management may allow some extra hours of work (overtime) if necessary. Thus the production hours constraint could be set up as follows:

x1 + x2 + d1− − d1+ = 80 where, x1 = number of hours used for producing the material A. x2 = number of hours used for producing the material B. d1− = amount of underutilization of variable normal production hours. d1+ = amount of extra hours (overtime) beyond normal production hours allowed by the management. Sales constraint: Since the maximum sales for the following week is 70,000 meters of material A and 45,000 meters of B, the sales constraint is x1 + d 2− = 70,000

Linear Graph Analysis and

111

x3 + d3− = 45, 000

where d2− and d3− are the underachievement of sales target of materials A an B respectively. Overtime constraint: The overtime constraint is d1+ + d4− − d4+ = 10 where d4− = amount of underutilized hours between the actual overtime allowed and 10 hours of overtime, and d4+ = amount of overtime in excess of the allowed 10 hours. Both negative and positive derivations are included from the allowed 10 hours of overtime, because the actual overtime can be less than, equal to, or even more than 10 hours. Substituting the value of d1+ in the production hours constraint, we get

x1 + x2 + d 4− d4− = 90 In this constraint, underutilization ( d1− ) is meaningless and hence been ignored. Achievement function: Since sales goals for materials are equally important, the profit contribution ratio (5:3) between these two will be considered as different weights. The achievement function, therefore, is: minimize z = p1 d1− + p2 d 4− + 5 p3 d 2− + 3 p3 d3− + p4 d1+ The complete goal programming problem, therefore, is as under: minimize z = p1 d1− + 5 p3 d 2− + 3 p3 d3− + p2 d 4+ + p4 d1+ subject to the constraints:

x1 + x2 + d1− d1+ = 80 x1 + x2 + d 4− − d4+ = 90 x1 + d2− = 70 and x2 + d3− = 45 x1 , x2 , d1− , d1+ d2− , d3− , d4− , d4+ , ≥ 0 Question 4: Suppose 2 products are to be produced with a single unit of A requiring 2.4 mins of moulding time and 5.0 min. of assembly time. The profit for product A is 0.6 per unit. A single unit of product B requires 3.0 min. of moulding time and 2.5 mins of finishing time. The unit profit for B is 0.7. Suppose the production manager is to determine the number of units of each product to be produced per week in consideration of the following equally ranked goals: (i) A profit target of Rs 350/- per week should be met. (ii) Overtime should not exceed 20% of the available production time in moulding department and should be avoided, if possible. (iii) Available production time in assembly and finishing departments should be fully utilized but not exceeded. Assume: (i) Available production time in moulding department is equal to 1200 mins/week. (ii) Available time in assembly shop = 1500 mins/week. (iii) Available time in finishing shop = 600 min/week. Formulate the production problem in goal programming model form.

UPTU 2004

112

Mechanical System Design

Ans.: Production minutes constraint: Regular production minutes available to the firm are limited to 350 minutes. The management allow some extra minutes of work (overtime) if necessary. Thus the production hours constraint could be set up as follows:

x1 + x2 + d1− − d1+ = 350 where

x1 = number of minutes used for moulding of A x2 = number of minutes used for moulding of B

d1− = number of minutes uses for assembling of A d1+ = number of minutes uses for finishing of B Since the available production time in moulding department of A and B are x3 = 1200 x4 = 1200

d 2− = 1500 (Available time in assembly) d2+ = 600 (Available time in finishing) Overtime constraint is

d3− − d3+ = 240 d3− = amount of overtime allowed d3+ = overtime is not allowed x1 + x2 + d1− + d3− − d1+ − d3+ = 350 + 240 x1 + x2 + d1− + d3− − d1+ − d3+ = 590 z = p1 d1− + p2 d3+ + 6 p3 d3− + 7 p3 d 2− + p3 d2+ The complete goal programming problem, therefore, is as under: Minimize

Minimize z = p1 d1− + p2 d3+ + 6 p3 d3− + 7 p3 d2− + p3 d 2+ + subject to the constraints.

x1 + x2 + d1− − d1+ = 350 x1 + x2 + d1− + d3− − d1+ − d3+ = 590 d 2− = 1500 d2+ = 600 and

d3− − d3+ = 240 x1 , x2 , d1− , d2− , d3− , d3+ ≥ 0

Linear Graph Analysis

6.6

113

A CASE STUDY: MATERIAL HANDLING SYSTEM

6.6.1

Definition of Material Handling

“Material handling includes all movements of materials in manufacturing situation. It is an art and science involving the moving, packing and storing of substances in any form.” American Society of Mechanical Engineers (ASME) “Material handling involves the movements of materials, manually or mechanically in batches or one item at a time within the plant. The movement may be horizontal, vertical or the combination of horizontal and vertical.” “Material handling embraces the basic operations in connection with the movement of bulk, package and individual products in a semisolid or solid state by means of gravity, manually or power actuated equipment and within the limits of and individual producing, fabricating, processing or servicing establishments.” HAYNES Material handling embraces the movement, handling and storage of everything within and around the establishment—raw materials, components and supporting goods in stores and between operations, regardless of shape, size, form, nature or weight. H.K. Compton

Starting from the time, the raw materials enters the factory gate and goes out of the factory gate in the shape of finished product, it is handled at so many stages, so material handling is the integral part of the manufacturing process.

6.6.2

Characteristics/Features of Materials Handling 1. It is the handling of raw materials, semi-finished materials, finished goods, packing materials, tools, jigs and fixtures through production as well in storage area. 2. It covers all producing, manufacturing, servicing and fabricating establishments. 3. It is the science and art of moving materials from one place to another place, machine to machine, from one processing department to another and further moved to final assembly section. 4. Movement of materials may be vertical or horizontal or both. 5. The materials handled may be solid, fluid or gaseous. 6. All material handling is by transport and all transport is material handling in the form of manual and mechanical devices.

6.6.3

Objectives of Materials Handling

The overall objectives of materials handling is to reduce production cost. This general objective can be subdivided into more specific goals, such as 1. Increased equipment and space utilisation 2. Reduced costs 3. Increased capacity 4. Improved working conditions 5. Improved customer service.

114

6.6.4

Mechanical System Design

Functions of Material Handling

The basic function of the material handling is to choose most appropriate material handling equipment which is safe and can fulfill material handling requirements at the minimum possible overall cost and to choose production machinery and assist in plant layout so as to eliminate, as far as possible, the need of material handling. In general, the functions of good materials handling system includes: (a) Using the principles of containerization, unit load or cartelization, aim at moving optimum number of pieces in one unit. (b) Safe, standard, efficient, effective, appropriate, flexible and proper sized materials handling equipment should be selected. (c) Employ mechanical aid in place of manual labour in order to speed up the materials movements. (d) Minimize the movement involved in a production operation. (e) Changes in sequences of production operations may be suggested in order to minimize back tracking and duplicate handling. (f) Handling equipment arrangement should minimize distance moved by products and at the same time handling equipment should not interfere with the production line. (g) Minimize the distances moved, by adopting shortest routes. (h) Utilize gravity for assisting material movements wherever possible. (i) Design containers, packages, drums etc. to economic handling and to reduce change to the materials in transit. (j) Material handling equipments should periodically be resorted to check ups, repairs and maintenance.

6.6.5

Materials Handling Principles

The basic guidelines or principles for making material handling effective and efficient are: 1. To determine combination of handling activities, all operations must be analysed. This will result in simplification and possibly, reduction in handling cost. 2. The selection of equipment must be made after careful consideration of the cost of moving, and economics can be measured by studying the cost of operation of handling in each move. The physical state of the material is a determining factor in the selection process. 3. It may be desirable to have specialised equipment, but the first cost of operation, maintenance, and repair are generally more than those of standardised equipment. Present worth and life span value should be evaluated. 4. Material handling equipment should be selected in such a manner as to afford flexibility and be capable of performing multiple operations, and also should be standardised. This will result in reduction of cost of operations, maintenance and repair and also in storage. 5. Simultaneously with other planning activities, selection and procurement of material handling equipment should be conducted in advance to take care of all aspects of handling and storage particularly of standardized equipment and combining methods. 6. Movement must be studied in detail to reduce ‘back tracking’ of materials. Length of moves must also be studied as to afford better utilisation of men and equipment.

Linear Graph Analysis

115

7. Rated capacity should be carefully examined and never exceeded, as overloading causes under wear, entails excessive maintenance and repairing cost. It also creates potential accident hazards, violating safely first principle in material handling.

6.6.6

Material Handling Equipments

1. Wheel barrow 2. Two wheel hand truck or sack truck 3. Three or four wheel hand trolley 4. Shelf trolley. 5. Lift trucks 6. Trolley trucks 7. Large mobile job cranes 8. Movable ramps 9. Fixed ramp 10. Detachable ramps 11. Screw conveyor 12. Small mobile job cranes 13. Forklift truck 14. Gently cranes 15. Platform truck 16. Roller conveyor 17. Industrial tractor. 18. Chute 19. Tripod stand Many engineering systems can be modeled and analysed like the material handling system using linear graph models and concepts. Linear graph analysis requires that the problem be expressed in graphical form, that the problem be identified with a graphical property, and that an analysis procedure be developed that focuses on this specific graphical property. The problem can be often be identified as a path problem or a cut-set problem. If a problem formulation requires an analysis that is inconvenient it may be possible to use dual graph concepts to transform the original graph model and graphical property into an equivalent problem using a different graph model and graphical property.

6.6.7

Factor Considerations 1. Selection of region, tax benefits, natural resources 2. Selection of capacity (Proximity of raw material) Labour market, transport 3. Selection of actual site 4. Raw material availability 5. Market proximity 6. Transportation

116

Mechanical System Design

1. Wheel Barrow

2. Two wheel hand truck or sack truck

3. Three or four wheel hand trolley

5. Lift trucks

Fig. 6.12

4. Shelf trolley

6. Trolley trucks

Material handling equipments

Linear Graph Analysis

117

8. Movable ramps

9. Fixed ramps

7. Large mobile job cranes

Material In 10. Detachable

Shaft

Material In

11. A Screw conveyor

12. Small mobile job cranes

Fig. 6.13

13. Forklift truck (Counter balance type)

Small mobile job cranes

118

Mechanical System Design

15. Platform truck 14. Gently cranes

17. Industrial tractor 16. Roller conveyor

19. Tripod stand

18. Chute

Fig. 6.13 (Contd) 7. 8. 9. 10. 11.

Energy Availability of labour Climate Water Finance and business facilities.

Exercise

6 . . .

1. Briefly explain linear graph analysis. 2. What is graph modeling and analysis process?

Linear Graph Analysis 3. 4. 5. 6.

119

What is difference between network flow and path flow problem? Briefly explain network flow problem with suitable example. Briefly explain path flow problem with suitable example. What is Goal Programming?

A Case Study 1. 2. 3. 4. 5. 6. 7.

Explain a case study of material handling system. What is the characteristics/features of material handling? What is the objective of material handling system? Write down function of material handling system. Write down principle of material handling. Write down names of material handling equipments. What is the main factor considerations in the study of material handling system?

120

Mechanical System Design

7 Optimization Concepts 7.1

INTRODUCTION

It is common experience that for almost any favorable feature in a design there is an unfavorable one. Any good thing in life is attained at the cost of some other thing. With the advent of automobiles, what we gained through increased mobility for work and pleasure, we lost in increased pollution, increased deaths on roads and in large portion of the inhabitable space in our cities taken up by facilities related to automobiles, such as roads, petrol stations, parking garages, etc. Industrialization leads to more prosperity but also leads to increased dehumanization and the poisoning of the environment. Faster means of communication create a kind of global village with events happening in any part of the world affecting all others almost instantaneously. This permits fighting of plagues across national borders (eradication of small pox and international AIDS control are two such examples), prediction of global weather and providing immediate aid to distress-stricken regions of the world. But such rapid means of communication also foster international gangs of terrorists and other such anti-social activities. All these examples show that there are positive and negative effects of almost any development. In fact one of the things that make a designer’s task difficult is that he has to strike a balance between these opposing effects. Another way to look at this problem is to consider the requirements that a designer has to design, for any design will always have conflicting requirements. Thus the requirements for a transport engineer that a road should have a high capacity for traffic (measured in vehicles carried per hour) and that there should be maximum safety appear to be conflicting. If he increases the speed limit he increases the flow rate but puts at risk the safety. If he tries to increase the safety by increasing the spacing between vehicles, the traffic capacity is adversely affected. There are many other situations where we come across such conflicting requirements of performance. An insecticide, which is very efficient for mosquitoes, cannot be very safe for people. A soap that cleans stains, which ordinarily would be difficult to remove, must necessarily do more damage to the fabric. The safety fuse used in the domestic distribution systems to protect against the slightest of overloads also has a great tendency to blow off accidentally. An aeroplane, which is more stable and thus easier to control has less manipulability. This means that while designing one must try to strike a balance between conflicting requirements. Thus in determining the best speed for a freeway to carry the maximum number of vehicles, one must balance the benefits of larger speeds with the penalties of a greater distance between vehicles which these larger speeds require or in deciding on the potency of the insecticide to be used. One should

Optimization Concepts

121

balance the savings resulting from the higher efficiency with the additional cost of preventive measures which need to be taken to ensure that the insecticide will not harm people. The process of arriving at such balances is known as optimization. Take another example, suppose we have to transmit a given amount of electrical power through a high voltage transmission line. As a result of the resistance of the transmission line there is a loss of energy due to ohmic heating. This heating is inversely proportional to the resistance of the transmission line and represents loss. To reduce the loss, we can reduce the resistance of the line. This requires thicker conductors, which means spending more money on the installation of the line. The use of heavier lines will also require strengthening of the towers and insulators. Thus we see that increasing the thickness of the conductors saves money due to reduced ohmic losses on the one hand, while increasing the capital costs on the other. The capital costs can be converted into an annual cost by the rate of depreciation and interest. The optimization problem in this situation is to determine the sizes of the conductor, which give the minimum cost (Fig 10.1). In addition to these requirements, there may be constraints on the choice of conductor size. One of the constraints may be due to the tensile strength of the conductor materials which may impose an upper limit on the weight of the conductor between the two towers. Similarly, the limits on the strength of insulator and other hardware material may also act as constraints on the size of the conductors. There is another class of problems where the process of optimization is called for. In this class of problems the number of design parameters whose values are to be fixed for a complete description of the problem is higher than the number of equation available to fix them. In such a simulation a large number of solution are possible, each satisfying the basic constraints of the problem and giving the desired functional performance. However, all of them are not equally acceptable.

7.2

THE OPTIMIZATION PROCESS

In most engineering problems, many feasible solutions exist. It is the professional responsibility of the engineer to seek out the best possible solution according to the goals and objectives for the total system. The process by which the best solution is determined is called ‘optimization’. Optimization is an essential and necessary phase of the design and planning process. After the problem environment has been carefully defined and initial solutions have been found, system design models are developed to describe the design characteristics and interactions of the system requirements. These models may be graphical, mathematical or physical. They provide a vehicle to analyse the behavior and response of the design when it is expected to satisfy a wide range of environmental conditions. As a result of the understanding gained from these analyses, design modifications may be made or completely new designs may be needed. This cyclic processor design and analysis is reiterated until a set of most promising solutions to the system problems emerges. These solutions must then be evaluated and ranked according to how well they fulfill the goals and objectives that have been established for the proposed system. During the ranking and evaluation phase, new understanding may be gained and new system constraints may be discovered. It is then necessary to return again to the design and analysis cycle. Thus, the optimization process is a cyclic process that consists of a continuous interplay of design, analysis and ranking of alternative solutions.

122

7.3

Mechanical System Design

MOTIVATIONS AND FREEDOM OF CHOICE

For optimization to be possible, the engineer must have the desire or at least the incentive to find the best solution. The motivation comes from the need to efficiently utilize materials and resources to improve the quality of life. The degree of success in optimization is often governed by the extent to which the designer is motivated to seek an optimum solution. In addition, optimization is possible only if a freedom of choice exists. The problem statement may be restrictive that there is no opportunity for choice or selection; i.e. there may be only one feasible solution to the problem as stated. As the problem constraints are relaxed the number of feasible solution increases and task of optimization becomes correspondingly more complex. It is important, therefore, that during the problem definition phase, the system constraints imposed must be truly representative of the goals and objectives of the problem. Unnecessary system constraints may disqualify an otherwise attainable optimum solution from further consideration.

7.4

GOALS AND OBJECTIVES—CRITERIA

The purpose of the optimization process is to determine the system that allows the decision-maker to most nearly achieve his stated goals and objectives. The degree to which he achieves these goals and objectives must be measured by a specific set of criteria. Therefore, it is absolutely essential that the goals, objectives and criteria of the problem be clearly defined during problem definition. The goals must truly reflect the ultimate purposes of those who have valid direct or indirect interest in the problem. For example, consider again the problem of designing a High-Raise Building (HRB). Among the many different interested parties, the most important ones are the owner, the tenants, and the general public, which is represented by the city administration. The primary goal of the owners may be to obtain a profit large enough to justify the venture; the tenants may be primarily interested in acquiring an attractive living or business environment for a reasonable cost; and the city administration may be primarily interested in improving the economic and social well-being of the city. These goals are not mutually independent. The tenants may or may not already have been identified during the design of the HRB, but it is apparent that if the goals of the owners are to be achieved, the interests of the prospective tenants must be taken into serious consideration. Similarly, the well being of the city strongly affects the lives of tenants and, therefore, the future prospects of the HRB. The problem objectives specify which characteristics of the system are to be optimized in order to achieve the goals. Objectives are commonly related to such factors as cost, profit, weight, speed, time, artistic features, colour, shapes, forms, and efficiency. In order to achieve the HRB owner’s goals of obtaining a large profit, some obvious objectives are: 1. To minimize the capital, maintenance and operational cost, and 2. To maximize the total income from the HRB. For the tenant’s goals the objectives may be: (a) Maximize the attractiveness of the living and/or business environment, both within and around the HRB. (b) To maximize the safety of the building, and (c) To maximize the convenience of entering and leaving the building.

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For the city administration, the objectives may be to maximize the economic gain of the city while minimizing financial and operational burdens imposed on the city by the HRB. Again these goals are not mutually exclusive, but confict with, as well as complement, one another. Finally, the criteria are the set of parameters used to measure how optimum a solution is with respect to the goals and objectives. For the owner’s objectives of minimizing cost the criteria may be expressed in terms of the capital investment in rupees, the interest rate for loans, inflation rate, city taxes, maintenance cost, and operational cost. For maximizing income the criteria may include rental rates in rupees, occupancy rate, tenants income levels and expected life of the HRB.

Table 7.1 Possible goals, objectives, and criteria for an HRB project Goals 1. Maximize profit from investment

Objectives 1. Minimize cost 2. Maximize income

2. Maximize tenant satisfaction

3. Maximize safety

4. Provide attractive living and/or business environment for tenants

5. HRB to be architecturally attractive 3. Contribute to the well-being of the city

Criteria Capital investment in rupees; operational cost, maintenance cost; interest rate; city taxes. Rental rates; occupancy rate, tenant’s income levels; life of HRB. Structural safety factor against natural hazards such as storms, tornadoes, earthquakes, etc. fire hazards and fire escapes, city building codes; design structural loading; emergency procedures. Functional needs; scenic views; shopping and recreational facilities; personal security; transportation of goods and people; noise; proximity to employment; playgrounds and parks, churches and schools; prestige and pride; police protection Conformity to surrounding buildings and environments; modern, traditional, or classic; degree of inventiveness; colour; shape.

6. Maximize economic gain for the city.

Creation of new industry and employment; improve housing environment.

7. Minimize financial and operational burden on the city

Increased demand for water and power supply, sewage disposal and processing facilities; impact on neighborhood traffic; demands for parks and recreational facilities; impacts on local schools.

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7.5

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METHODS OF OPTIMIZATION

The optimization methods that may be utilized for a given problem depend on the degree of order that can be established in the problem. The order reflects how well the system structure and behavior characteristics are known and how well the system constraints and criteria can be quantified. The optimization methods can be broadly grouped into three approaches: 1. Analytical 2. Combinatorial and 3. Subjective.

7.5.1

Analytical Approach

This approach is applicable only to a totally ordered system problem. A total order implies complete knowledge of structure and behavior and either completely automates the optimization processor or enables both the problem structure and the criterion preference function to be analytically stated. In such a case, the optimization process can be supported by a well-defined set of criteria that may be used to judge the quality of the design. The criteria should be included in the statement of systems objectives and should define the technical, political, and operational goals of the system. The system variables are quantified and each variable is assigned a system value according to a common measure, such as a monetary value. Thus, if x1 , x2 , x3 , x4 , ... and xn , are the design variables in the problem and ai is the value of one unit of xi, to the total system, then the total value of the system may be expressed in a mathematical function as follows;

c=

n

∑a

i

xi

…(1)

i =1

The objective of optimization then, is to find the set of xi’s such that the total system value c is maximized or minimized. Equation (1) represents the simplest form of a criteria function. It is a linear equation involving only first-degree terms of the design variables. In general cases, the total system value c may appear in any functional form depending on the nature of the problem, i.e. c = F ( x1 , x2 , x3 , ... xn )

…(2)

There may also be more than one criteria function for a given system problem. Thus, if c1, c2 and c3 are three independent measures of a system, then c1 = F1 ( x1 , x2 , x3 , ... xn ) c2 = F2 ( x1 , x2 , x3 , ... xn ) c3 = F3 ( x1 , x2 , x3 , ... xn ) And the system design must be optimized with respect to all three criteria functions. If it is not possible to optimize simultaneously all three criteria functions, then an order of priorities must well be specified to resolve the conflicts. Thus, the optimization model for a totally ordered system problem takes the following general format:

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Maximize or minimize the following criteria functions: c1 = F1 ( x1 , x2 , x3 , ... xn ) c2 = F2 ( x1 , x2 , x3 , ... xn ) ... ck = Fk ( x1 , x2 , x3 , ... xn ) subject to the following system constraints:

Lower Bounds G1 ( x1 , x2 , x3 , ... xn ) ≥ a1 G2 ( x1 , x2 , x3 , ... xn ) ≥ a2

…(3)

... Gm ( x1 , x2 , x3 , ... xn ) ≥ am

Upper Bounds H1 ( x1 , x2 , x3 , ... xn ) 0 ≤ b1 H 2 ( x1 , x2 , x3 , ... xn ) 0 ≤ b2

…(4)

... H r ( x1 , x2 , x3 , ... xn ) 0 ≤ br

Equalities P1 ( x1 , x2 , x3 , ... xn ) = C1 P2 ( x1 , x2 , x3 , ... xn ) = C 2

…(5)

... Ps ( x1 , x2 , x3 , ... xn ) = Cs

Approximations Q1 ( x1 , x2 , x3 , ... xn ) ; d1 Q2 ( x1 , x2 , x3 , ... xn ) ; d 2

…(6)

... Qt ( x1 , x2 , x3 , ... xn ) ; d t

Such an analytical approach is called Mathematical Programming.

7.5.2

Combinatorial Approach

As the degree of order possessed in the problem is weakened, the optimization process may reduce to purely evaluating a multiple selection of alternatives and selecting the best. Suppose that there are n design variables in the problem and that they are denoted as x1 , x2 , x3 ... and xn . The approach is to

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identify the most probable range of values for each of these variables and then analyze all the possible combinations. If there are ten discrete probable values for each parameter, a complete combinatorial analysis will involve 10 n cases. It is obvious that a strict application of such an approach becomes impracticable even for simple systems involving only new variables. For example, for n = 4, there will already be about 10 4 cases for analysis. However, technical procedures together with past experience often can be used to eliminate a large percentage of the feasible combinations and leave only a few for detailed study and analysis. This approach is used extensively in decision analysis. All the possibilities are analyzed and compared to arrive at an optimum decision polices.

7.5.3

Subjective Approach

In complex problem situations, it may be difficult or even impossible to establish specific models or system orders, in which case purely subjective methods become necessary. The subjective approach is the most important and often is the deciding method in optimization. Intangible factors such as social values, political influences, and psychological effects are extremely difficult to quantify and measure. For example, a safe highway cannot be designed without considering the response characteristics and behavior of the drives.

7.6

A CASE STUDY: ALUMINIUM EXTRUSION SYSTEM

Aluminium is a non-ferrous metal. Non-ferrous metals are those which do not contain iron as the base material. The most commonly used non-ferrous metals in workshop are aluminium, copper, lead, tin, nickel and zinc. They also form very useful alloys amongst themselves, known as a non-ferrous alloys, which posses very significant characteristics like high resistance to corrosion, conductivity of heat and electricity, lightness in weight and also being non-magnetic. These properties enable these metals and alloys to be preferred over iron, steel and their alloys where these characteristics stand as the primary considerations. Non-ferrous metals and alloys can also be cast and machined without any appreciable amount of difficulty, but they are more expensive as compared to the ferrous products. However, apart from the cost factor, there are some inherent disadvantages associated with non-ferrous metals, when compared with ferrous metals, such high shrinkage, hot shortness and lower strength at elevated tempratures. Aluminium ore is found as a hydrated aluminium oxide, called bauxite. The impurities present in it are oxides of iron, silicon and titanium. The first process, therefore, is to separate aluminium oxide from these impurities. For this purpose, bauxite is fused in an electric furance and carbon is added to reduce the impurities, which form a sludge and can be removed. As a result of this refining, pure aluminium oxide is separated from the impurities. Then an electrolytic bath is used to reduce aluminium from its oxide. As the electrolytic process proceeds the oxygen escapes through the bath and molten aluminium collects at the bottom (cathode), from where it is periodically tapped off. This mineral is mainly available in our country in Bihar, Mahrashtra, Madhya Pradesh, Karnataka and Tamilnadu.

7.6.1

Properties and uses of Aluminium 1. High electrical conductivity: used for heavy conductors and busbar work. 2. High heat conductivity: used in various domestic utensils and other heat conducting appliances.

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3. Good resistance to corrosion: used in manufacture of containers for chemical industry and window frames etc. 4. It can be readily worked, extruded, rolled, drawn and forged. 5. It has high ductility and is extremely light in weight. Widely used in aircraft industry. 6. Its corrosion resistance can be considerably increased by anodising. 7. It becomes hard by cold working and, therefore, needs frequent annealing. 8. Its low tensile strength can be sufficiently improved by adding 3 to 4 percent copper.

7.6.2

Alloys of Aluminium

Alloys of aluminium are light in weight compared to steel, brass, nickel and copper. These resist corrosion and have good electrical and thermal conductivities. These can readily accept a wide range of surface finishes and can be fabricated by all common processes. These lose part of their strength at elevated temperature. At sub zero temperatures, however, their strength increases without loss of ductility and thus these are best suited for low-temperature applications. Aluminium alloys can be classified according to the method of shaping them viz. wrought products and cast aluminium alloys. These are further classed as to whether they respond to heat-treatment of strengthening-type or not. Based on the sequences of basic treatments used to produce the various tempers, all forms of wrought and cast aluminium and aluminium alloys could be classified as: 1. Fabricated 2. Annealed 3. Strain-hardened 4. Solution heat-treated 5. Thermally treated to produce stable tempers 1. Fabricated: Products which acquire some temper from shaping processes not having any special control over the amount of strain-hardening or thermal treatment. 2. Annealed, recrystallised (wrought products only): Wrought products subjected to softest temper. 3. Strain-hardened (wrought-products only): Products whose strength is increased by strainhardening with or without supplementary thermal treatments to produce partial softening. 4. Solution heat-treated: This refers to unstable temper possible to alloys which spontaneously age at room temperature after solution heat treatment. 5. Thermally treated to produce stable tempers: Products which are themally treated with or without supplementary strain-hardening, to produce stable tempers.

7.6.2.1 Wrought Aluminium Alloys Strength greater than that of pure aluminium can be achieved by addition of other elements. Alloys can be strengthened by heat treating in some cases. Alloys which can’t be strengthened by heat treating are known as non-heat treatable alloys. The initial strength of these alloys depends upon the hardening effects of elements such as Mn, Si, Fe and Mg, singly or in various combinations. These alloys are work-hardenable and further strengthening is achieved by various degrees of cold working. Alloy containing appreciable amount of magnesium when supplied in strain-hardened temperature are usually given a final elevated temperature treatment called stabilizing. Heat treatable alloys show increasing solid

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solubility in aluminium with increasing temperature and as such it is possible to subject them to thermal treatments which will impart pronounced strengthening. In the first step (solution heat-treatment), the solution element is put in solid solution at elevated temperature of about 490 °C. Then it is rapidly quenched (usually in water) which momentarily freezes the structure and for a short time renders the alloy very workable. Since alloy in this stage is very soft and ductile, it can be easily worked. It is at this stage that some fabricators retain this more workable structure by storing the alloys at below freezing temperature until they are ready to form them.

7.6.2.2 Casting Aluminum Alloy These alloys contain large quantities of Si or Mg to produce a lower melting point alloy. Only a few alloys are suitable for die casting and sand casting. Some of these alloys naturally age after solution treatment, but others have to be given precipitation treatment.

7.6.2.3 Clad Aluminum Alloys Heat treatable alloys containing copper or zinc are major alloying elements are found to be less resistant to corrosion attack than the majority non-heat and treatable alloys. To increase corrosion resistance in sheet and plate form these are often clad (up to 2.5 to 5% of total thickness on either side) with high purity aluminium, a low magnesium-silicon alloy, or an alloy containing 1% zinc. Cladding not only protects the composite due to its own inherently galvanic effect which further protects the core metal.

7.6.2.4 Duraluminium It is a very important alloy of aluminium. Its composition is as given below: Copper → 3.5 to 4.5% Magnesium → 0.4 to 0.7% Manganese → 0.4 to 0.7% Iron or Silicon → not more than 0.7 % Aluminum → Rest This alloy has got high machinability. It can be heat treated to increase its tensile strength. By heat treatment, tensile strength of the alloy can be raised up to 4300 kg/cm2 without affecting ductility of the alloy. Duraluminium is as strong as steel but has only about one-third of its weight. So, it is used to fabricate parts of aircraft and automobiles. Aluminium, silicon alloys have excellent casting qualities and resistance to corrosion. The alloys are not hot-short and are easy to cast in thin or thick sections, but these are difficult to machine. Aluminium– magnesium alloys are superior to practically all other aluminum casting alloys with respect to resistance to corrosion and machinability. At the same time these have high mechanical strength and ductility. Aluminum alloys for pressure die-casting must possess considerable fluidity and be free from hotshortness.

7.6.3

Indian Standard Specifications on Aluminum and its Alloys

Bureau of Indian Standards (BIS) have brought out the following IS specification on aluminium metal and its alloys:

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Table 7.2 S.No.

Standard No.

1 2 3 4 5 6 7 8 9 10 11

7.6.4

IS IS IS IS IS IS IS IS IS IS IS

617 733 734 736 737 738 739 740 1284 6751 7793

Year 1975 1975 1975 1974 1974 1977 1977 1977 1975 1972 1975

Specification Casting for general engineering Wrought bars, rods and sections Wrought forging stock Wrought plates Wrought sheet and strip Wrought drawn tube Wrought wires Wrought rivet stock Wrought bolt and screw stock Castings and strips for bearings Wrought I.C. engine pistons.

Effects of Alloying Elements on Aluminimum

Table 7.3

Effects of alloying elements on aluminium

Alloying elements

Effects

1.

Iron

Impurity in aluminum ores. Small percentages increase the strength and hardness of some alloys and reduces hot-cracking tendencies in castings. Iron reduces pick up tendency in die-casting cavities.

2.

Manganese

It is used in combination with iron to improve castability. It alters the nature of the inter metallic compounds. Increases ductility and impact strength and reduces shrinkage.

3.

Silicon

Increases fluidity in casting and welding alloys and reduces solidification and hot cracking tendencies. Improves corrosion resistance. Addition in excess of 13% make the alloy extremely difficult to machine.

4.

Copper

Improves elevated temperature properties and machinability. Increase strength up to about 12%; higher concentrations cause brittleness.

5.

Magnesium

It decreases cast ability and improves strength by solid solution strengthening alloys with over about 6% will precipitation harden.

6.

Zinc

Lowers cast ability. High percentage promotes hot cracking and high shrinkage. Percentages over 10% produces tendencies for stress corrosion cracking, promotes very high strength in combination with other elements.

Designation of Aluminium and Alloys Bureau of Indian Standard (BIS) have brought out (Indian Standards) IS: number 6051 in year 1971 to describe the designation for aluminum and its alloys:

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Table 7.4

Elements number designations Elements 1. 2. 3. 4. 5. 6. 7. 8. 9.

Exercise

Unalloyed aluminum (wrought and cast form) Copper Manganese Silicon Magnesium Magnesium Silicate (Mg2Si) Zinc Other elements (such as Ni, Ti, Cr, Pb, Bi, etc.) Unassigned

Group Number 1 2 3 4 5 6 7 8 9

7 . . .

1. Explain the difference between goal and objective. Give example for each. 2. Formulate maximization of Cobb-Douglas type utility function under budget limitation C, if unit capital and unit labour cost are Pk and Pl respectively. 3. What is optimization concept? 4. Briefly write about optimization process. 5. What is motivation and freedom of choice? 6. Distinguish between goal, objective and criteria. 7. Briefly explain method of optimization (i) analytical (ii) combinatorial (iii) subjective. 8. Write a short note of (a) Analytical optimization (b) Combinatorial optimization and (c) Subjective optimization. 9. Briefly explain goal, objective and criteria in the technique of optimization concept. 10. What is the difference between motivation and freedom of choice? 11. Explain optimization process and what do you mean by motivation and freedom of choice in the form of optimization concept.

A Case Study 1. 2. 3. 4. 5.

Write down properties and use of Aluminium. Write down alloys of Aluminium. What is the effect of alloying elements on the basis of Aluminium extrusion system? Explain designation of Aluminium and alloys. Explain different process of Aluminium extrusion system.

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8 System Evaluation 8.1

INTRODUCTION

All proposed engineered systems usually have to pass engineering, economic, financial, political, and social feasibility tests. These tests are supplied to ensure that the system will be able to perform its intended function, that the benefits to be gained from the system are greater than the cost of implementing and operating the system, that the owner will be able to pay for the system, and that the individuals who will be affected by the system will respond favorably. The purpose and scope of the planning are major factors in determining the planning horizon to be used in a study. There is considerable uncertainty associated with many of the benefits and costs of any system under study. The longer the period of analysis used, the more uncertain many of the costs and benefits become. Sometimes techniques, such as simulation (Chapter 11), are used to gain an understanding of the possible time profile of cash flows associated with the system. The value of money varies with time because of its ability to grow if property invested or to earn interest if placed in a savings account. Therefore, discounting factors must be used to convert costs and benefits to the same time period so that they can be compared. Along with the comparision of benefits and costs comes the desire to rank alternatives according to their desirability. If all benefits and costs associated with the system can be expressed in terms of rupees, then criteria such as the net present worth, annual equivalent value, benefit-cost ratio, and internal rate of return can be used to rank the desirability of alternatives. The particular criterion to be used will depend on the objective of owner. If the costs can be expressed in terms of rupees, but the benefits cannot, then it may be possible to rank alternatives according to the cost required to provide the same level of benefits. This criterion is usually satisfactory only for single purpose systems. The economic ranking criteria were originally developed for single purpose systems that were independent of one another, and their use further assumes an unlimited capital available for investment. The ranking of interrelated systems under conditions of limited capital becomes a much more complex problem, as shown by Masse (1962). When the benefits and costs cannot all be expressed in the terms of rupees or any other common units, then the final selection must be made on the basis of the decision-maker’s subjective judgment. This is usually the type of problem encountered in reality. Hence, we see that evaluation techniques are

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helpful in ranking alternatives, but the final selection depends on the values and objective of the decisionmaker, who is often the owner (individual, company, or government agency). A financial analysis determines if money can be made available for system implementation after the owner has determined the alternative that he prefers. If the proposed system passes the above feasibility tests, then the engineer must begin to plan the sequence and duration of activities necessary to construct the system.

8.2

FEASIBILITY ASSESSMENT

Before a new plant or business unit is started a thorough survey should be undertaken to acertain whether the venture gives sufficient promise of profitable returns to justify the expenditure of time and investment in its promotion and operation. Survey: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

8.2.1

Analyse the product and survey the market. Make an economic survey. Determine the major objectives and scope of the operation, design the product Determine the volume and output size of the plant. Select the location. Decide whether to buy or make the parts. Develop the manufacturing technology and equipment. Select a plant type building and layout. Determine the capital needs, profitability, and Financial plan.

Product Analysis of Feasibility Assessment

Determine the marketability of the product, its physical nature and commercial possibilities i.e. potential demand. If the potential demand looks promising an intensive market survey is taken to estimate the actual sales in immediate future. If the market does not look promising expensive sales plan and production analysis need not be done. Well-established procedures (exist for initiating and developing) engineered systems. The owner (government, company, or individual) of the proposed system employs or contracts with a professional engineer to whom he explains his problem as he sees it, usually in terms of a proposed system. Then the engineer, in co-operation with the owner, should review the owner’s goals and objectives to determine if the proposed system is compatible with these goals and objectives or whether a different system is really what the owner wants. After this step, the engineer evaluates the alternative designs applicable to the proposed system. The goal and objectives will depend on the problem of owner and evaluation procedures will usually be required to pass tests for engineering, economic, financial, political and social feasibility. Engineering feasibility requires that the proposed system be capable of performing its intended function. Design analysis procedures as described in standard engineering texts can be used to indicate the ability of a proposed system to perform its intended function. In addition, the construction or implementation of the system must be possible. A proposed system is economically feasible if the total value of the benefits that result from the system exceed the costs that result from the system. Economic

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feasibility depends on engineering feasibility because a system must be capable of producing the required output in order to produce benefits. The owner must have sufficient funds to pay for system installation and operation before the proposed system is considered to be financially feasible. Financial feasibility may or may not be related to economic feasibility. An owner may be able and willing to pay for a system in order to fulfill non economic goals. It may also be that an economically feasible project is financially non-feasible because the owner is not able to raise enough money to implement the system. Political and the social feasibility is assured if the required political approval can be secured and if the potential users of the system will respond favorably to system construction. Every system is subject to review at different stages of planning. A private company has executive officers or a broad of directors that reviews proposed systems. Public systems are subject to public hearing and review by commities of elected representatives. Usually political support is gained after evidence of engineering and economic feasibility has been presented. However political pressure may be quite strong for specific system if it is adversely affected and often oppose economically feasible system. Constraints of economic factors have not received sufficient emphasis. Political and social feasibility can best be sustained by active participation of representatives from all interested groups in planning and designing a proposed system.

8.3

PLANNING HORIZON

Planning is necessarily connected with the future. How far the planning penetrates into the future, and the planning horizon depends upon the scope of the problem. Top management’s concern for overall objective is a panasonic problem. But once the long-range objectives are settled, the tactics required to implement are a series of short-range operations. For example, a firm considering the introduction of a new product must consider demand factors over a span of years. A decision to go ahead with the introduction leads to facility, supply, distribution and similar problems over the shorter period needed to establish the product. Each problem area then explodes into detailed action plans for the next period, perhaps a few months or a year. Finally, at operating level, scheduling decisions are in terms of weeks, days and hours (see Fig. 8.1.) The planning horizon is the most distant future time considered in the engineering economic study. The planning horizon to be used for a specific study depends on the purpose of the planning and the scope or a real extent for which the planning is done. If the purpose of the planning is to develop a framework plan that is to serve as a guideline for development, a time of 50 years or more in the future may be chosen as the planning horizon. The framework plan will indicate the types of systems needed to achieve the desired results from the development. Framework plan for a river basin region might include the number of reservoirs and other types of water resources development, along with the purposes that each of these elements should serve in order to sustain the projected economic and population growth of the region for the next 50 years. A planning horizon of 20 to 30 years in the future may be chosen for a study concerning investment decisions. The size and type of water treatment plant that a city will invest in may be based on existing water treatment methods and the projected economic and population growth planning horizons. The construction of a large dam may require that a concrete mixing plant be constructed at the dam site to provide concrete for the dam. The size and type of mixing plant to be used may be determined from a study with

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a planning horizon which coincides with the projected completion date of the dam, some 1 to 5 years in the future. In some cases the planning horizon for a major plant stables several projects, in which case the salvage value of the plant at the end of particular project may influence both its economic and financial feasibility.

Level of authority

Time

No. of Employers

Long range Profit planning Shorter range planning resources planning Co-ordination scheduler Operation schedulers

Fig. 8.1 Level of authority vs. no. of employers The scope or area extent of a proposed system will be a factor in determining the planning horizon to be used in a study. The planning horizon for a rural frame to market road will usually be shorter than for an inter-state highway system. Traditionally three different periods have been considered in planning studies: the physical life of the system, the economic life of the system, and the period of analysis. Physical life of a system ends when it can no longer physically perform its intended function. The physical life of a building does not end if the building is converted from a hotel to a museum. Its physical life ends when it can no longer provide shelter or support the loads sustained in the use of building. The economic life of a system ends when the incremental benefit from continuing operation of the system for one more time period no longer exceeds the incremental costs of continuing operation for one more time period. This point usually occurs when the annual operation, maintenance, and repair (OMR) costs equal or exceed annual benefits from the system. Since a program of regular maintenance and periodic replacement or worn parts may extend the physical life of a system almost indefinitely, the economic life is usually shorter than the physical life. The period of time over which the system consequences are considered to affect the system benefits and costs is refered to as the period of analysis. Because of the uncertainty associated with

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future benefits and costs problem solving, it is necessary to confine studies to a certain extension in space and time. This confinement within specific limitations is required because of the uncertainty associated with future benefits and costs and because of our inability to trace and evaluate the consequences of the system beyond the immediate vicinity of the system. Regardless of the period of analysis used, the salvage value of the system should be considered in the analysis. The salvage value is the worth of the system at the end of the period of analysis. For some systems, this worth will be positive, indicating that the system still has value or can be sold and thus can be thought of as a benefit that occurs at the end of the period of analysis. For other systems, this worth may be negative indicating that it is a liability that must be disposed of and thus can be thought of as a cost, that occurs at the end of the period of analysis. When alternative schemes of development are being considered for the same purpose, all alternatives must be evaluated for the same period of analysis. If some component requires periodic replacement, then its costs are usually assumed to be repeated at the end of each component life until the total period of analysis is completed. However, this assumption should not be made without considering the effects of inflation, the development of new production techniques through technological advances, and the changing nature of demand for the system outputs with time. Uncertainty about any of these factors tends to favor alternatives with short lives.

8.4

TIME VALUE OF MONEY

The value of money is not fixed. It varies with time because of its ability to grow, if profitably invested or to earn interest, if placed in a savings account. A rupee today, will be equivalent to Rupees 1.05, one year from now. If an investment opportunity is available that will earn 5% interest per year. If a person is given the choice of receiving Rs 1 today or Rs 1 after one year from now, he would choose the Rs 1 today because the Rs 1 could be invested to earn interest. This concept of equivalence is important in economic studies that justify proposed system because it allows values that occur at different points in time to be compared on a common time basis. Narrowly defined, in a borrowing–lending situation, interest may be defined as money paid for the use of borrowed money. The rate of interest is the ratio, expressed as a percentage of the interest payable at the end of a period of time, usually a year or less, and the money owned at the beginning of the year. Thus, if Rs 5 is payable annually on debt of Rs 100, the interest is Rs 5 and interest rate is Rs 5/100 = .05 or 5 percent per year. Although the interest rate may be loosely defined as any expression of the time value of money, a more precise definition distinguishes between interest rate and discount rate. Interest is the fee for borrowing money. A discount rate is the expression of the time value of money used in the equivalence computations. The discount rate should reflect the opportunity cost, which is equal to the return that would have been realized had money been invested in other alternatives available to the individual. It may or may not be equal to the interest rate. If the individual can earn 8% interest on his money from some other source, if the system is not constructed, then he should receive 8% for the discount rate even though he can borrow money from a bank at 6%. Interest is said to be compounded if interest is earned on interest. If interest is based only on the original amount for each period , the interest is said to be simple interest. Most economic studies are performed at compound interest wherein values are adjusted forward or backward in time using a compounding growth or discount factor.

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If an individual behaves rationally, he would value the opportunity to receive a specific amount of money now equally with the opportunity to have the same amount at some future date. Most individuals value the opportunity to receive money now higher than waiting for one or more years to receive the same amount. This phenomenon is referred to as an individual’s time value of money. Thus, an individual’s preference for possession of a given amount of cash now, rather than the same amount at some future time, is called, “time value for money”. Three reasons may be attributed to the individual’s preference for money. (i) Risk: We live under risk or uncertainty. As an individual is not certain about future cash receipts he prefers receiving cash now. (ii) Preference for consumption: Most people have subjective preference for present consumption over future consumption of goods and services either because of the urgency of their present wants or because of the risk of not being in a position to enjoy future consumption that may be caused by illness or death, or because of inflation. As money is the means by which individuals acquire most goods and services, they may prefer to have money now. (iii) Investment opportunities: Most individuals prefer present cash to future cash because of the available investment opportunities to which they can put present cash to earn additional cash. Time value of money or time preference for money is one of the central ideas in finance. Individuals as well as business organisations, frequently encounter the situations involving cash receipts or disbursements over several periods of time. When this happens time value of money becomes important and some time vital consideration in decision-making.

8.5

FINANCIAL ANALYSIS

A system may be economically feasible but not financially feasible in addition to the requirement that the benefits to the owner must be greater than his cost; there must be a plan for paying for the system. Few large systems are ever paid for from cash on hand. Most large systems are financed by loans that have to be repaid.

8.5.1

Bonds

Loans are often obtained through the sale of bonds. A bond is an instrument setting forth the conditions under which money is loaned. It consists of a pledge by borrower to repay a specified principal at a stand time and to pay a stated interest rate on the principal in the main time.

8.5.2

Repaying the Loan

A city must be providing a means of repaying the loan. The money for the repayment may come partly from revenues, received as result of the sale of hydroelectric power and water and partly from tax revenues. The time profile of the annual payments, A as shown in Fig. 8.2 must be equivalent to a final payment, F at the end of the period. The time profile of the annual payment = A Final Payment = F, Interest rate = I The value of the last annual payment at the end of year 50 is A. The value of payment made at the end of year 49 has increased into A (1 + i) at the end of the 50 year period. Similarly, the payment made

System Evaluation

137

at the end of year 48 has increased to A (1 + i)2. Since the final sum, F, is equal to future value of all annual payments. A

A

A

1

2

3

Fig. 8.2

A

A

A

A

A

A

45

46

47

48

49

50

Time profit of payments to sinking fund

F = A + A (1 + i ) + A (1 + i)2 + A (1 + i)3 + ... + A (1 + i) n −1 F = A[1 + (1 + i ) + (1 + i)

2+

(1 + i) + ... + (1 + i ) 3

n −1

where n is the number of time period. This equation can be simplified to

F=A

[(1 + i) n − 1] i

  i A= F  n  (1 + i ) − 1    i Annual payment A = F   n  (1 + i ) − 1 Some important formulae are given below: 1. Compound value of lump sum =

F = P (1 + i ) n

2. Compound value of an unity

F=

3. Sinking fund

A=F

1 (1 + i )n − 1

4. Present value of a lump sum

P=F

1 (1 + i ) n

5. Capital recovery

A=P

i (1 + i ) n (1 + i ) n − 1

n

6. NPV (Net present value)

A (1 + i )n − 1 i

∑ (1 + r ) i =1

F

n

−C

]

...(1)

138

Mechanical System Design

 F  F F NPV =  + + + ... − C 1 2 3 (1 + r ) (1 + r )  (1 + r ) 

8.5.3

Present Value

Present value of a given sum of money, which is done at the end of a specified period of time, is that principal, when invested at interest will accumulate to the given sum in specified time P=

A (1 + i ) n

…(1)

where, A = the given sum n = number of conversion periods i = rate of interest per conversion period Problem 1.: Find the present value of Rs 10,000 due in 5 years at 8% interest compounded semi annually. P=

Answer:

A (1 + i ) n

8 % 2 = 4% η = 5× 2 = 10 years i=

P= =

10,000 (1 + 0.04)10 10,000 1.48

= 6755.64 Ans.

8.5.4

Annuity

An annuity is a series of periodical payments ordinarily of a fixed amount, payable regularly at the end of fixed period for a number of periods. For example, annuity of Rs 200 payable monthly for 10 years, means: l l l

Payment period or internal is 1 month. The term is 10 years. The sum of payment made in one year = Rs 2400

8.5.4.1 Amount of an Annuity It is the sum of the compound amounts, which would be obtained if each payment when due were kept at interest until the end of the specified term.

System Evaluation

139

 (1 + i ) n − 1  Amount of Annuity = R   i   where

R = Rate of periodic payment.

Illustration to Understand Amount of Annuity Problem 2: Find the amount of annuity for Rs 200 payable monthly for 10 years at 8% (Here n = 120 and i = 0.08/12 = 0.0067). We know that Amount of annuity

Answer: Amount of annuity

 (1 + i) n − 1  = R  i    (1 + 0.0067)120 − 1  = 200   0.0067    (1.0067)120 − 1  = 200   0.0067    (2.2284 − 1)  = 200    0.0067 

=

200(1.2284) 0.0067

= Rs 36, 668.66 Ans. Problem 3: Determine rate of periodic payment (R), the amount of annuity so that in 20 years one can get Rs 1, 00,000; payments to be made quarterly; the interest rate is 8 % compounded. Answer: Given that n = 4 × 20 = 80 years i=

0.08 4

i = 0.02 We know that Amount of annuity

 (1 + i) n − 1  = R  i    (1 + i) n − 1  1,00, 000 = R   i   R=

1,00, 000 × 0.02 (1 + 0.02)80 − 1

140

Mechanical System Design

R=

1,00, 000 × 0.02 4.8754 − 1

R = Rs 516.075 Ans

8.6

SELECTION BETWEEN ALTERNATIVES

Present Worth Method Question 1: The data regarding the initial cost, operating costs etc. of two equipments A and B is as given below. Find out the economical machine for selection by using present worth method.

Initial cost in Rs Operating cost per year Rs Life of equipment in years Interest rate 10%

Equipment A

Equipment B

10,000 1,000 4

15,000 800 5

Solution: Equipment A (i) Present worth of cost of equipment A is Rs. 10,000. (ii) Present worth of operating cost in first year

 1  P = F 2  (1 + i )    1 = 1000  1  (1 + 0.10)  = 1000 × 0.909 = Rs 909 In second year =

1000 (1 + 0.10) 2

= 1000 × 0.8264 = Rs 826 In third year =

1000 (1 + 0.10)3

= 1000 × 0.7513 = Rs 751.30 In fourth year =

1000 (1 + 0.10) 4

System Evaluation

141

= 1000 × 0.6209 = Rs 620.90 Therefore, expenses converted in terms of present worth = Present worth of cost + present worth of operating cost = Rs 10,000 + Rs 909 + Rs 826 + Rs 751.30 + Rs 620.90 = Rs 13,107.20

…(1)

Equipment B (i) Present worth of cost of equipment B = Rs 15, 000 (ii) Present worth of operating cost in first year

 1  P = F 2  (1 + i )  =

800 (1 + 0.10)1

= 800 × 0.909 = Rs 727.20 Present worth of operating cost in second year =

800 (1 + 0.10) 2

= 800 × 0.8264 = Rs 661.12 Present worth of operating cost in third year =

800 (1 + 0.10)3

= 800 × 0.7513 = Rs 601.04 Present worth of operating cost in fourth year =

800 (1 + 0.10) 4

= 800 × 0.6209 = Rs 496.72 Therefore, total expenses converted in terms of present worth = Rs 15,000 + Rs 727.20 + Rs 661.12 + Rs 601.04 + Rs 496.72 = Rs 17486.08 from equation (1) and (2), it is clear that equipment A should be selected.

…(2)

142

Mechanical System Design

Question 2: A company is considering to purchase a machine for a job. Two alternative machines are available in the market: one a standard machine and other a semi-automatic one. The comparative data of them are given below: Item Purchase price Estimated economical life (year) Salvage value at the end of 4 years Labour cost per year Maintenance cost during first year Increase in maintenance cost per year after the first year

M/c A

M/c B

Rs 20,000 4 Rs 4000 Rs 12,000 Rs 1000 Nil

Rs 30,000 4 Rs 10,000 Rs 6000 Rs 200 Rs 200

All the costs incurred during a year can be assumed to occur at the end of the year. The prevailing interest rate for money is 10%. Using the present value method, determine which machine is to be preferred. Solution: Machine A (i) Purchase price of machine = Rs 20,000 (ii) Present worth of labour cost First year P = F

1 (1 + i)

= 12,000 ×

1 (1 + 0.10)1

= Rs 10,909.09 Second year = 12,000 ×

1 (1 + 0.10)2

= Rs 9917.35 Third year = 12,000 ×

1 (1 + 0.10)3

= Rs 9015.78 Fourth year = 12,000 ×

1 (1 + 0.10) 4

= Rs 8156.16 Present worth of total labour cost = Rs 10,909.09 + Rs 9917.35 + Rs 9015.78 + Rs 8156.16 = Rs 38,112.38 (iii) Present worth of maintenance cost

System Evaluation

143

  1 First year P = 1000  1  (1 + 0.10)  = Rs 909.09 1 (1 + 0.10) 2

Second year = 1,000 ×

= Rs 826.45 1 (1 + 0.10)3

Third year = 1,000 ×

= Rs 751.31 Fourth year = 1,000 ×

1 (1 + 0.10) 4

= Rs 683.01 Total present worth of maintenance cost = Rs 909.09 + Rs 826.45 + Rs 751.31 + Rs 683.01 = Rs 3169.86 (iv) Present worth of salvage value at the end of 4 years = 4000 ×

1 (1 + 0.10) 4

= Rs 2732. 05 Therefore, total present worth of machine A = Rs 20,000 + Rs 38, 112.38 + Rs 3169.86 + Rs 2732.05 = Rs 64,014.29 Machine B (i) Purchase price of machine = Rs 30,000 (ii) Present worth of labour cost First year = 6, 000 ×

1 (1 + 0.10)1

= Rs 5454.55 Second year = 6, 000 ×

1 (1 + 0.10) 2

= Rs 4958.68 Third year = 6, 000 ×

1 (1 + 0.10)3

…(1)

144

Mechanical System Design = Rs 4507.86 Fourth year = 6, 000 ×

1 (1 + 0.10) 4

= Rs 4098.10 Therefore, total present worth of labour cost = Rs 5454.55 + Rs 4958.68 + Rs 4507.86 + Rs 4098.10 = Rs 19,019.22 (iii) Present worth of maintenance cost

  1 First year = 2000  1  (1 + 0.10)  = Rs 1818.19

  1 Second year = 2000  2   (1 + 0.10)  = Rs 1818.18 Third year =

2000 (1 + 0.10)3

= Rs 1803.15 Fourth year =

2000 (1 + 0.10) 4

= Rs 1775.83 Total present worth of maintenance cost = Rs 1818.19 + Rs 1818.18 + Rs 1803.15 + Rs 1775.83 = Rs 7215.35 (iv) Present worth of salvage value at the end of 4 years = 10,000 ×

1 (1 + 0.10)4

= Rs 6830.13 ∴ Total present worth of machine B = Rs 30,000 + Rs 19019.22 + Rs 7215.35 + Rs 6830.13 = Rs 630647 From equation (1) and (2) we get; The present worth of M/c B is less, therefore M/c B should be preferred. Ans.

…(2)

Question 3: Suppose that 2 robot alternatives are being considered for a certain industrial application. The cost and life data of the robots are given below: UPTU 2004

System Evaluation

Investment Life Salvage value Annual disturbances

145 Robot type 1

Robot type 2

Rs 1,50,000 3 years 0 Rs 80,000

Rs 1,25,000 5 years 0 Rs 50,000

Which alternative should be selected? Use MARR of 20%. Assume a salvage value for the robot type 2 at the end of 3 years as Rs 50,000. Solution:

P: Investment Life in years S: Salvage value A: Annual disbursement

Robot type 1

Robot type 2

Rs 1,50,000 3 years 0 Rs 80,000

Rs 1,25,000 5 years 0 Rs 50,000

Robot Type 1 (i) Given that, investment = Rs 1,50,000 Present worth of cost of robot type 1 is = Rs 1,50,000 (ii) Present worth of Annual disbursement in first year

 1  P = F n  (1 + i )    1 = 80,000  1  (1 + 0.20)  = Rs 66666.66   1 Annual disbursement in second year = 80,000  2   (1 + 0.20)  = Rs 55555.5

  1 Annual disbursement in third year = 80, 000  3  (1 + 0.20)  = Rs 46296.29 Total annual disbursement = Rs 66666.66 + Rs 55555.5 + Rs 46296.29 = Rs 168518.50 Total present worth of robot type 1

146

Mechanical System Design = Rs 1,50,000 + Rs 168518.50 = Rs 318518.5

...(1)

Robot Type 2 (i) Investment = Rs 1,25,000 Present worth cost of equipment of robot type 2 is Rs 1,25,000 (ii) Annual disbursement in first year

P=F

1 (1 + i)

= 50,000

1 (1 + 0.20)1

= Rs 41666.66 Annual disbursement in second year = 50,000

1 (1 + 0.20) 2

= Rs 34722.22 Annual disbursement in third year = 50,000

1 (1 + 0.20)3

= Rs 28935.18 Total annual disbursement = Rs 41666.66 + Rs 34722.22 + Rs 28935.18 = Rs 105324.06 (iii) Present worth of salvage value at the end of third year = 50,000

1 (1 + 0.20)3

= Rs 28935.18 Total present worth of cost of type 2 = Rs 1,25,000 + Rs 105324.06 + Rs 28935.18 = Rs 259259.06 From equation (1) and (2) it is clear that Robot type 2 should be selected. Ans

...(2)

Question 4: In a company there are two machines M 1 and M 2 . The former being replaced after every three years while the latter is replaced in six years. Given below are yearly costs of both the machines. UPTU 2005 Year

1

2

3

4

5

6

M1

1000

200

400

1000

200

400

M2

1700

100

200

300

400

500

System Evaluation

147

Taking the value of money as 10% find out which machine should be purchased. Solution: Since the money carries the rate of interest, the present worth of the money to be spent over in a period of one year is 1 1 P ⇒ = = n F (1 + i ) (1 + 0.10)1 = 0.9091 ∴ The total discounted cost (present worth) of A for 3 years is = 1000 + 200 × (0.9091) + 400 (0.9091)2 = Rs 1512 approx. Again, the total discount cost of B for six years is = 1,700 + 100 × (0.9091) + 200 × (0.9091)2 + 300 × (0.9091)3 + 400 × (0.9091)4 + 500 × (0.9091)5 = Rs 2,765 Average yearly cost of machine A = Rs 1,512/3 = Rs 504 Average yearly cost of machine B = Rs 2765/6 = Rs 461 This shows that the apparent advantage is with machine B. But, this comparision is unfair since the period for which the costs are considered are different. So if we consider 6 years period for machine A also, then the total discount cost of A wil be = 1,000 + 200 × 0.9091 + 400 × (0.9091)2 + 1,000 × (0.9091)3 + 200 × (0.9091)4 + 400 × (0.9091)5 = Rs 2467 Hence machine A should be purchased.

8.7 8.7.1

A CASE STUDY: MANUFACTURE OF A MAIZE-STARCH SYSTEM Place of Manufacture of Maize-Starch System in Mechanical Systems Design

Introduction: When a designer proposes a design, he does so with the confidence that it will meet the need. A designer is a visionary and a dreamer, but he is of no use if he is not practical minded. One of the first tests of practically of any design is that it should be economically viable. A design is of no use if it does not make money for the manufacturer or does not give enough benefits to its user to justify its cost. A design is not worthwhile if it is not properly implemented, the production factory set up and the product released in the market. Money is required to implement a design. A project starts costing money from the day it is conceived as an idea. The first set of costs relates to design and development expenses and includes all costs including the prototype fabrication and testing stage. Development costs are incurred on a continual basis even after the production has started. These are on account of the product improvements which any self-respecting manufacturer continuously attempts. Manufacturing costs are incurred after initial design and development are completed. One set of the manufacturing costs consists of direct costs and the other of what are incidental to the establishment. Direct costs include the cost of labour, raw materials, etc. The indirect costs are termed as overhead

148

Mechanical System Design

costs and include such charges as cost of plant, cost of power, water and other utilities, cost of watch and ward staff, cost of personnel management and workers welfare, costs of fire and other insurance, etc. All these costs go into making up the total cost of the product. Sales revenue must cover this cost and then some more to reward the entrepreneur for his labour and the risk he has taken. The designer strives to maximize this profit, either by reducing the costs or by making the product so attractive that the market sales price goes up. It is important to differentiate between these two broad categories. The first type of costs are termed as fixed costs as opposed to the second which are termed as variable costs. At times, it is not very clear whether cost is fixed or variable. Overhead costs are thus seen to be essentially fixed costs. The capital tied up in goods in transit etc. is thus a variable cost. Cost of Product

Development Costs

Overhead Costs

Marketing Costs

Agents commission

Advertising

Distribution

Transportation

Tied up capital

Packaging

Direct labour

Bought components

Raw materials

Labour welfare

Accounts office

Taxes

Utilities

8.7.2

Interest on loans

Insurance

Supervision

Special Tools

Machinery

Factory space

Research

Protoype costs

Design office space

Designer’s fee or salary

Fig. 8.3

Production Costs

Some major costs heads

Designing for Profit in Manufacture of a Maize-Starch System

The aim of the designer is to design for maximum profits, subject to professional ethics and all legal and safety constraints. If we assume that the sales price for a given design is dictated by the market, the only way to maximize profit is to reduce costs. We may reduce either fixed costs or variable costs, or hopefully, both. It happens often that there is a trade-off involved between the two costs. We may be able to reduce fixed costs only at the cost of increased variable costs, or vice-versa. To give a simple example, suppose you have to drill holes in a wall for driving in some nails. You may either purchase a hand drill which costs Rs 20 and spend 3 minutes drilling each hole or you may purchase an electric drill costing Rs 200 and spend one minute drilling each hole. In this case the time spent on each hole is the variable cost and the cost of the drill is the fixed cost. It is obvious that the electric drill should be purchased only if a very large number of holes have to be drilled. A hand drill would be sufficient if only a few were needed. A designer, particularly when he is designing for production, constantly faces such problems. It is clear that if he can reduce the variable costs, event at increased fixed costs, he may make more profit if he sells ‘enough’. But the question, at what level of production does it pay to incur more expensive fixed costs but less variable costs methods, must be answered. This is done by resorting to what may be termed as the location of the indifference point.

System Evaluation

8.8

149

ANALYSIS OF THE CASE STUDY: MANUFACTURE OF MAIZE-STARCH SYSTEM

Maize-starch is an important input material used by textile mills, pepper mills, food and other industries. The annual demand of starch in the state of Uttar Pradesh is estimated to be above 25,000 tonnes and at present there is no starch factory in the state. Maize is abundantly grown in the state and the annual production is about 1,00,000 tonnes. The industrial separation of pure starch from maize is achieved by the standard wet-milling process. There are four main steps in this process: steeping, grinding, germ and fibre separation and product recovery. Figure 8.4 gives the basic process flow chart. Maize, either from storage or as received from the market, is cleaned and transferred to large steeping vats. Here maize is soaked in a hot aqueous solution containing dissolved sulphur dioxide. The steeping solution removes soluble components and loosens the starch granules from the protein matrix in which they are embedded. The steep water is drawn off, concentrated in multi-effect evaporators to about 50 percent solid content. The steep concentrate thus obtained is a profitable by-product which is sold to cattlefeed manufactures. Cleaned maize Determination by cross grinding Fine grinding

Steeping vat

Germs Wash press and dry

Seive Step water

Pump

Slit screen

Oil expeller Maize tank Maize oil

Oil care

Fibre Gluten and starch Centrifugal separation

Kernels Gluten

Fig. 8.4

Maize starch

Basic Process Flow Diagram for the Manufacture of Maize Starch

The steeped maize is conveyed to determinating mills where the softened kernels are cracked to release the germs. Germs are separated, washed and dried. Maize oil is then recovered from germs by mechanical expellers, and this also is a valuable by-product. A second grinding of the kernels now release the starch granules from the fibre. Screens, filters and washers are used to separate starch from fibre. Protein also comes out with starch and is finally separated by centrifugal separators. The starch is further purified by washing. It is then dewatered, centrifuged and dried. To carry out an economic analysis we have assumed a set-up using about 16,000 tonnes of maize to produce about 10,000 tonnes of starch per year.

150 Equipment Costs For maize receiving and cleaning For steeping stage For coarse milling For fine milling and fibre washing For gluten and starch separation For dewatering and drying Oil expeller and germ-drier Miscellaneous equipment

Mechanical System Design Rs

Rs

1,27,000 2,00,000 1,25,000 3,03,000 20,50,000 9,15,000 2,30,000 25,35,000

Total sum equal to

64,85,000

Erection charges, (including costs of pipings, valves etc.) Engineering design costs (roughly 5%) Land and buildings … Preoperative expenses to be during the incurred/construction period

8,00,000 3,65,000 16,50,000 3,00,000

Total capital investment in plant Working capital (funds needed for payment of wages, raw materials, fuel, etc. for the period of a few months depending upon the trade credit terms. This is roughly equivalent to the costs of product which at anytime is in production or in the distribution network but for which payment is yet to be received). For a product like maize-starch 2-month production is typical, costs of maize:2,000 tonnes at Rs 1,600 per tonne sulphur: 10 tonnes at Rs 1,000 per tonne Credit for finished product Salary and wages Utilities Total capital requirement is the total of plant cost and working capital

96,00,000 32,00,000

10,000 40,00,000 1,40,000 1,50,000 75,00,000 1,71,00,000

For setting up the plant, we must have this much finances available with us. These could be raised from the savings of the promoters, bank loans, credit from other financial organisations, or from floating a public issue of stock. Unless financial recources required are available or promised, the project should not be launched, at least on this scale.

System Evaluation

151

We next evaluate the economic feasibility of the project by calculating the cost of production and sales revenue.

Fixed Costs Plant cost: depreciation at 15 percent on equipment at 10 per cent on buildings

11,48,000 1,71,000 13,19,000

Interest on borrowed capital Insurance against fire etc. Administrative and sales costs

18,00,000 4,00,000 13,00,000

Total Cost: Sum of Fixed Costs

48,19,000

Variable Cost (for 10,000 tonnes starch production) Raw Material Maize: 16,000 tonnes at Rs 1,600 per tonne Sulphur: 80 tonnes at Rs 1,000 per tonne

2,56,00,000 80,000

Total sum

2,56,80,000

Wages Utilities Power, water, fuel, etc. Repairs and maintenance Packaging and freight

18,00,000 12,20,000 3,00,000 10,00,000

Total Cost: Sum of Variable Costs

3,00,00,000

Total Cost: Sum of fixed costs + Sum of variable costs

3,48,19,000

Sales revenue (for 16,000 tonnes maize consumption a year, see Table 8.1)

4,07,60,000

Table 8.1 S. No. 1. 2. 3. 4.

Total Sales Revenue from 16,000 Tonnes of Maize Product Starch Maize Oil Oil cake Maize gluten

Per cent recovery

Production (tonne)

Rate/ tonne

Revenue (Rs)

63.0 2.5 3.5 10.0

10,800 400 560 1,600

3,200 7,000 1,000 800

3,45,60,000 28,00,000 5,60,000 12,80,000 (Contd)

152

Mechanical System Design

S. No. 5. 6.

Product Maize bran Steep concentrate (Losses about 4 per cent)

Per cent recovery

Production (tonne)

Rate/ tonne

11.0 6.0

1,700 960

450 800

Total

Revenue (Rs) 7,92,000 7,69,000 4,07,60,000

Conclusion

Re ve nu e

40

6

Cost of revenue (10 rupees)

Thus, if the plant operates at full capacity utilization, it will make a profit of Rs 58,41,000 a year. To calculate the break-even point, note that the sales revenue per-percentage point capacity utilization is Rs 4,07,800. Fixed costs are Rs 48,19,000 and variable costs are Rs 3,00,000 per percentage point capacity utilization. We can plot cost and revenue curves against percentage utilization as shown in Fig. 8.5. From the graph, it is seen that the break-even capacity utilization is roughly 45 per cent. This is a low enough break-even point and permits a comfortable hedge against unforeseen developments like power shortage or even against reasonable changes in the cost structures of inputs or outputs.

20 Fixed costs

10

Break-even capacity utilization 25

Fig.8.5

Exercise 1. 2. 3. 4. 5.

s

st

Co

30

50 75 100 Capacity utilization, percent

Break-even analysis of maize-starch plant

8 . . .

Explain the various factors, which are considered, while making a feasibility analysis of a system. Discuss the element of feasibility analysis. UPTU 2005 Discuss the system evaluation. UPTU 2004 Write down about planning horizon with a suitable example. Write notes on: (a) Time value of money (b) Financial analysis

System Evaluation

6. 7. 8. 9.

153

(c) Planning horizon (d) Feasibility assessment. (e) System evaluation. What is difference between time value of money and financial analysis? Discuss about time value of money. What is the main role of financial analysis in system evaluation? Explain the planning horizon. What factors are considered while planning as horizon of system?

A Case Study 1. 2. 3. 4.

Explain a case study of manufacture of a maize-starch system. Draw a neat flow diagram for the manufacture of maize starch. Write down the suitable steps of manufacturing of maize starch system. Briefly explain about production of maize starch in our country and U.P. How much production contains of maize starch. 5. You are a small company making curtain rods of standard sizes, 2 m in length. There are three materials A, B and C which you can use. Each material cost for a different process and machine for manufacturing and their cost data may be summarized as below:

Item Raw material Rs/m Equipment cost Rs/year Labour cost Rs/rod

Material A

Material B

Material C

2.20 5,000.00 0.50

2.50 3,000.00 0.60

2.60 4,000.00 0.20

Draw the total cost versus the yearly production volume curve for each of the three materials. If a sales volume of 10,000 rods per year is anticipated, which material should be used? 6. You are a traveling salesman carrying your own samples around. You may either travel by a taxi in which case the cost will be reimbursed by your company or you may buy and use your own car in which case the company will give a traveling allowance at the rate of Rs 1.00 per km traveled. The expenditures involved with a car are: Depreciation Tax and Insurance Maintenance (Plus) Fuel and Oil

Rs 2,000 per year 600 per year 500 per year 0.20 per Km. 0.60 per Km.

Plot the per kilometer costs and compute the distance traveled per month which will make owning your car an attractive proposition.

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Mechanical System Design

9 Calculus Method for Optimization 9.0

INTRODUCTION

The calculus methods of optimization are useful in finding the optimum solution of continuous and differentiable functions. These methods are analytical and make use of the techniques of differential calculus in locating the optimum points. Since some of the practical problems involve objective functions that are not continuous and/or differentiable, the classical optimization techniques have limited scope in practical applications. However, a study of the calculus methods of optimization forms a basis for developing most of the numerical techniques of optimization presented in subsequent chapters. In this chapter, we present the necessary and sufficient conditions in locating the optimum solution of a one decision variable function, a two-variable function with no constraints, and a two-variable function with equality and inequality constraints.

9.1

MODEL WITH ONE-DECISION VARIABLE OPTIMIZATION

A function of one variable f ( x) is said to have a relative or local minimum at x = x* if f ( x*) ≤ f ( x* + h) for all sufficiently small positive and negative values of h. Similarly, a point x* is called a relative or local maximum if f ( x*) ≥ f ( x* + h) for all values of h sufficiently close to zero. A function f ( x) is said to have a global or absolute minimum at x* if f ( x*) ≤ f ( x) for all x, and not just for all x close to x*, in the domain over which f ( x) is defined. Similarly, a point x* will be a global maximum of f ( x) if f ( x*) ≥ f ( x) for all x in the domain. Figure 9.1 shows the difference between the local and global optimum points. A1 A2 A3 = Relative maxima = Global maximum A2 = Relative maxima B1 B2 = Global minimum B1 A one-decision variable optimization problem is one in which the value of x = x* is to be found in the interval [a, b] such that x* minimizes f ( x). The following two theorems provide the necessary and sufficient conditions for the relative minimum of a function of a one-decision variable optimization.

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155

A1, A2, A3 = Relative maxima A2 = Global maximum B1, B2 = Relative minima f (x)

B1 = Global minimum

f (x)

A2

Relative minimum is also global minimum

A3 A1 B2

B1 a

x

b

a

b

x

Fig. 9.1 Relative and global minima Theorem 9.1.1: Necessary Condition. If a function f ( x) is defined in the interval a ≤ x ≤ b, and has a relative minimum at x = x*, where a ≤ x * ≤ b and if the derivative df ( x)/dx = f ′( x) exists as a finite number at x = x*, then f ′( x*) = 0 . Proof: It is given that f ( x* + h) − f ( x*) (9.1) h exists as a definite number, which we want to prove to be zero. Since x* is a relative minimum, we have f ′( x*) = lim

h →o

f ( x) ≤ f ( x* + h) for all values of h sufficiently close to zero. Hence f ( x* + h) − f ( x*) ≥ 0 if h > 0 h f ( x* + h ) − f ( x*) ≤ 0 if h < 0 h Thus, Eq. (9.1) gives the limit as h tends to zero through positive values as f ′( x*) ≥ 0 while it gives the limit as h tends to zero through negative values as f ′( x*) ≤ 0 The only way to satisfy both Eqs. (9.2) and (9.3) is to have f ′( x*) = 0

(9.2) (9.3) (9.4)

This proves the theorem. Notes: 1. This theorem can be proved even if x* is a relative maximum. 2. The theorem does not say what happens if a minimum or maximum occurs at a point x* where the derivative fails to exist. For example, in Fig. 9.2.

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lim

h →o

f ( x* + h ) − f ( x*) = m + (positive) or m − (negative) h

depending on whether h approaches zero through positive or negative values, respectively. Unless the numbers m+ and m− are equal, the derivative f ′( x*) does not exist. If f ′( x*) does not exist, the theorem is not applicable. f (x) Negative slope m −

Positive slope m+

m+

m− f ( x *)

x

x*

Fig. 9.2

Derivative undefined at x* f (x)

Stationary point, f ′(x ) = 0

0

x

Fig. 9.3 Stationary (inflection) point 3. The theorem does not say what happens if a minimum or maximum occurs at an endpoint of the interval of definition of the function. In this case

lim

h →o

f ( x* + h) − f ( x*) h

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157

exists for positive values of h only, or for negative values of h only, and hence the derivative is not defined at the endpoints. 4. The theorem does not say that the function necessarily will have a minimum or maximum at every point where the derivative is zero. For example, the derivative f ′( x ) = 0 at x = 0 for the function shown in Fig. 9.3. However, this point is neither a minimum nor a maximum. In general, a point x* at which f ′( x ) = 0 is called a stationary point. If the function f ( x) possesses continuous derivatives of every order that come in question, in the neighborhood of x = x*, the following theorem provides the sufficient condition for the minimum or maximum value of the function. Theorem 9.1.2: Sufficient Condition Let f ′( x) = f ′′( x*) = ... = f ( n − 1) ( x*) = 0, but f ( n ) ( x*) ≠ 0. Then f ( x*) is (i) a minimum value of f ( x) if f ( n ) ( x*) > 0 and n is even; (ii) a maximum value of f ( x) if f ( n ) ( x*) < 0 and n is even; (iii) neither a maximum nor a minimum if n is odd. Proof: Applying Taylor’s theorem with remainder after n terms, we have

f ( x* + h) = f ( x*) + hf ′( x*) + for

h2 hn − 1 hn ( n) f ′′( x*) + ... + f ( n−1) ( x*) + f ( x* + θ h) 2! ( n − 1)! n!

0 0 and negative for hk < 0. This means that X* cannot be an extreme point. The same conclusion can be obtained even if we assume that ∂ f ( X *) /∂ X k < 0. Since this conclusion is in contradiction with the original statement that X* is an extreme point, we may say that ∂ f /∂ xk = 0 at X = X *. Hence the theorem is proved. Theorem 9.2.2: Sufficient Condition. A sufficient condition for a stationary point X* to be an extreme point is that the matrix of second partial derivatives (Hessian matrix) of f (X) evaluated at X* is (i) positive definite when X* is a relative minimum point, and (ii) negative definite when X* is a relative maximum point. Proof: From Taylor’s theorem we can write n

f ( X * + h) = f ( X *) + ∑ hi i =1

1 n n ∂f ∂2 f ( X *) + ∑ ∑ hi h j ∂ xi ∂ xi ∂ x j 2! i =1 j =1

0 0 then f (x, y) has a minimum value. if rt – s 2 < 0, f ( x, y ) has no extreme value if rt – s 2 = 0 can is doubtful and need further integration. Example 4: Consider a function in two variables x and x2 given as

f ( x1 , x2 ) = x12 − 8 x2 + 2 x22 − 6 x1 + 30

6

50/3

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Mechanical System Design

Find the optimal values of x1 and x2 and verify the values obtained for maxima/minima. UPTU 2004 Solution:

f ( x1 , x2 ) = x12 − 8 x2 + 2 x22 − 6 x1 + 30 df = 2 x1 − 6 dx1 = 2( x1 − 3)

df = − 8 + 4 x2 dx2 = 4 ( x2 − 2)

df = 0 ⇒ 2( x1 − 3) = 0 dx1 = x1 = 3

df = 2 x1 − 6 dx1 = 2 ( x1 − 3)

df = 0 = 4 ( x2 − 2) = 0 dx = ( x2 − 2) d2 f =2 dx12 d2 f =4 dx22 f ( x1 , x2 ) = 33 − 8 × 2 + 2 × (2) 2 − 6 × 3 + 30 = 9 – 16 + 8 – 18 + 30 = –17 + 30 = 13

 2, 0  0, 4   J1 = 2 J2 = 8 So the values are obtained for minima.

Calculus Method for Optimization

9.3

167

MODEL WITH TWO DECISION VARIABLES WITH EQUALITY CONSTRAINTS

In this section, we consider the optimization of continuous functions subjected to equality constraints: Minimize f = f ( x) Subject to (9.16) g j ( x) = 0, j = 1, 2, ..., m where

 x1  x    x =  2 M  xn  Here m is less than or equal to n; otherwise (if m > n), the problem becomes over defined and, in general, there will be no solution. There are several methods available for the solution of this problem. The methods of direct substitution, constrained variation, and Lagrange multipliers are discussed in the following sections.

9.3.1

Solution by Direct Substitution

For a problem with n variables and m equality constraints, it is theoretically possible to solve simultaneously the m equality constraints and express any set of m variables in terms of the remaining n – m variables. When these expressions are substituted into the original objective function, there results a new objective function involving only n – m variables. The new objective function is not subjected to any constraint, and hence its optimum can be found by using the unconstrained optimization techniques discussed in Section 9.2. This method of direct substitution, although it appears to be simple in theory, is not convenient from practical point of view. The reason for this is that the constraint equations will be nonlinear for most of practical problems, and often, it becomes impossible to solve them and express any m variables in terms of the remaining n – m variables. However, the method of direct substitution might prove to be very simple and direct for solving simpler problems, as shown by the following example. Example 5: Find the dimensions of a box of largest volume that can be inscribed in a sphere of unit radius. Solution: Let the origin of the Cartesian coordinate system x1 , x2 , x3 be at the centre of the sphere and the sides of the box be 2 x1 , 2 x2 and 2 x3 . The volume of the box is given by f ( x1 , x2 , x3 ) = 8 x1 x2 x3 ( E1 ) Since the corners of the box lie on the surface of the sphere of unit radius, x1 , x2 , and x3 have to satisfy the constraint

x12 + x22 + x32 = 1

( E2 )

This problem has three design variables and one equality constraint. Hence the equality constraint can be used to eliminate any one of the design variables from the objective function. If we choose to eliminate x3 , Eq. ( E2 ) gives

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Mechanical System Design

(

x3 = 1 − x12 − x22

)

1/2

( E3 )

Thus the objective function becomes

(

f ( x1 , x2 ) = 8 x1 x2 1 − x12 − x22

)

1/2

( E4 )

which can be maximized as an unconstrained function in two variables. The necessary conditions for the maximum of f give

 ∂f = 8 x2  1 − x12 − x22 ∂ x1  

(

)

1/2

 =0 − 1/2  2 2  1 − x1 − x2 

(

x12

 1/2 x22 ∂f = 8 x1  1 − x12 − x22 − ∂ x2  1 − x12 − x22  Equations ( E5 ) and ( E6 ) can be simplified to obtain

(

)

(

)

 =0 1/2   

)

( E5 )

( E6 )

1 − 2 x12 − x22 = 0 1 − x12 − 2 x22 = 0 from which it follows that x1* = x2* = 1/ 3 and hence x3* = 1/ 3. This solution gives the maximum volume of the box as

f max =

8 3 3

To find whether the solution corresponds to a maximum or a minimum, we apply the sufficiency conditions to f ( x1 , x2 ) of Eq. ( E4 ). If A is taken as the base point ( x1* , x2* ), the variations in x1 and x2 leading to points B and C are called admissible variations. On the other hand, the variations in x1 and x2 representing point D are not admissible since point D does not lie on the constraint curve, g ( x1 , x2 ) = 0. Assuming that ∂ g /∂ x2 ≠ 0, It can be rewritten as

dx2 = −

∂ g /∂ x1 * * ( x1 , x2 ) dx1 ∂ g /∂ x2

(9.17)

This relation indicates that once the variation in x1 ( dx1 ) is chosen arbitrarily, the variation in x2 ( dx2 ) is decided automatically in order to have dx1 and dx2 as a set of admissible variations.

∂f ∂ g /∂ x1 ∂ f  * * df =  −  ( x1 , x2 ) dx1 = 0  ∂ x1 ∂ g /∂ x2 ∂ x2 

(9.18)

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169

The expression of the left-hand side is called the constrained variation of f. Note that Eq. (9.18) has to be satisfied for all values of dx1. Since dx1 can be chosen arbitrarily, Eq. (9.18) leads to

 ∂ f ∂g ∂ f ∂g  * * −   ( x1 , x2 ) = 0  ∂ x1 ∂ x2 ∂ x2 ∂ x1 

(9.19)

Equation (9.19) represents a necessary condition in order to have ( x1* , x*2 ) as an extreme point (minimum or maximum). Example 6: A beam of uniform rectangular cross section is to be cut from a log having a circular cross section of diameter 2a. The beam has to be used as a cantilever beam (the length is fixed) to carry a concentrated load at the free end. Find the dimensions of the beam that correspond to the maximum tensile (bending) stress carrying capacity. UPTU 2005 Solution: From elementary strength of materials, we know that the tensile stress induced in a rectangular beam (σ ) at any fiber located at a distance y from the neutral axis is given by

σ M = y I where M is the bending moment acting and I is the moment of inertia of the cross section about the x axis. If the width and depth of the rectangular beam shown in Fig. 9.6 are 2x and 2y, respectively, the maximum tensile stress induced is given by y

a x (Neutral axis)

2y

x 2 + y 2 = a2

2x

Fig. 9.6 σ max =

Cross section of the log

3 M M My y= = 2 1 I (2 x ) (2 y )3 4 xy 12

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Mechanical System Design

Thus for any specified bending moment, the beam is said to have maximum tensile stress carrying capacity if the maximum induced stress (σ max ) is a minimum. Hence, we need to minimize k / xy 2 or maximize kxy 2 , where k = 3M/4 and k = 1/k, subject to the constraint

x2 + y 2 = a 2 This problem has two variables and one constraint; hence Eq. (9.19) can be applied for finding the optimum solution. Since

f = kx −1 y −2

( E1 )

g = x2 + y 2 − a 2

( E2 )

We have

∂f = − kx −2 y −2 ∂x ∂f = − 2kx −1 y −3 ∂y ∂g = 2x ∂x ∂g = 2y ∂y Equation (9.19) gives

−kx −2 y −2 (2 y ) + 2kx −1 y −3 (2 x) = 0 at (x*, y*) that is, y* =

2 x*

Thus the beam of maximum tensile stress carrying capacity has a depth of The optimum values of x and y can be obtained from Eqs. ( E3 ) and ( E2 ) as

x* =

9.3.2

( E3 ) 2 times its breadth.

a a and y* = 3 3

Necessary Conditions for a General Problem

The procedure indicated above can be generalized to the case of a problem in n variables with m constraints. In this case, each constraint equation g j ( X ) = 0, j = 1, 2, ..., m, gives rise to a linear equation in the variations dxi , i = 1, 2 ..., n. Thus there will be in all m linear equations in n variations. Hence, any m variations can be expressed in terms of the remaining n – m variations. These expressions can be used to express the differential of the objective function, df, in terms of the n – m independent variations. By letting the coefficients of the independent variations vanish in the equation df = 0, one obtains the necessary conditions for the constrained optimum of the given function. These conditions can be expressed as

Calculus Method for Optimization

171

∂f ∂ xk ∂ g1 ∂ xk  f , g1 , g 2 , ..., g m  J  = ∂ g2  xk , x1 , x2 , x3 , ..., xm  ∂ xk M

∂f ∂ x1 ∂ g1 ∂ x1 ∂ g2 ∂ x1

∂f ∂ x2 ∂ g1 ∂ x2 ∂ g2 ∂ x2

∂ gm ∂ xk

∂ gm ∂ x1

∂ gm ∂ x2

∂f ∂ xm ∂ g1 K ∂ xm ∂ g2 = 0 K ∂ xm K

K

(9.20)

∂ gm ∂ xm

where k = m + 1, m + 2, ..., n. It is to be noted that the variations of the first m variables ( dx1 , dx2 , ..., dxm ) have been expressed in terms of the variations of the remaining n – m variables ( dxm + 1 , dxm + 2 , ..., dxm ) in deriving Eqs. (9.20). This implies that the following relation is satisfied:

 g , g , ..., g m  J 1 2 ≠0  x1 , x2 , ..., xm 

(9.21)

The n – m equations given by Eqs. (9.20) represent the necessary conditions for the extremum of f ( x) under the m equality constraints, g (X) = 0, j = 1, 2, ..., m.

9.3.2.1 Sufficient Condition for a General Problem By eliminating the first m variables, using the m equality constraints (this is possible, at least in theory), the objective function f can be made to depend only on the remaining variables, xm + 1 , xm + 2 , ..., xn . Then the Taylor’s series expansion of f, in terms of these variables, about the extreme point X* gives n ∂f  n n  ∂2 f 1 + ∑ ∑  f ( X * + dX ) = f ( X *) + ∑   dxi + i = m +1  ∂ xi  2! i =m +1 j =m +1  ∂ xi ∂ x j g

  dxi dx j g

(9.22)

where (∂ f /∂ xi ) g is used to denote the partial derivative of f with respect to xi (holding all the other variables xm + 1 , xm + 2 , ..., xi + 1 , xi + 2 , ..., xn constant) when x1 , x2 , ..., xm are allowed to change so that the constraints gi ( X * + dX ) = 0 , j = 1, 2, ..., m, are satisfied; the second derivative, (∂ f /dxi dx j ) g , used to denote a similar meaning. As an example, consider the problem of minimizing f ( X ) = f [ x1 , x2 , x3 ]

subject to the only constraint

gi ( X ) = x12 + x22 + x32 − 8 = 0 Since n = 3 and m = 1 in this problem, one can think of any of the m variables, say x1 , to be dependent and the remaining n – m variables, namely x2 , and x3 , to be independent. Here the constrained partial derivative [∂ f /dxi ] g , for example, means the rate of change of f with respect to x2 [holding the other independent variable x3 constant] and at the same time allowing x1 to change about X* so as to

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Mechanical System Design

satisfy the constraint g i [ x] = 0. In the present case, this means that dx1 has to be chosen to satisfy the relation

g1 ( X * + dX ) + g1 ( x*) +

∂ g1 ∂g ∂g ( X *) dx1 + 1 ( X *) dx2 + 1 ( X *) dx3 = 0 ∂ x1 ∂ x2 ∂ x3

that is,

2 x1* dx1 + 2 x2* dx2 = 0 since gi ( X *) = 0 at the optimum point and dx3 = 0 ( x3 is held constant). Notice that ( df /dxi ) g has to be zero for i = m + 1, m + 2, ..., n since the dxi appearing in Eq. (9.22) are all independent. Thus the necessary conditions for the existence of constrained optimum at x* can also be expressed as ∂f    = 0 , i = m + 1, m + 2, ..., n  ∂ xi  g

(9.23)

Of course, with little manipulation, one can show that Eq. (9.23) is nothing but Eq. (9.20) Further, as in the case of optimization of a multivariable function with no constraints, one can see that a sufficient condition for X* to be a constrained relative minimum (maximum) is that the quadratic form Q defined by n

Q= ∑

i = m +1

n  ∂2 f ∑  j = m +1  ∂ xi ∂ x j 

  dxi dx j g

(9.24)

is positive (negative) for all non vanishing variations ∂ xi . As in Theorem 9.4, the matrix

  ∂2 f    2    ∂ xm+1  g   ∂ 2 f     dxn ∂ xm +1  g

 ∂2 f   ∂ xm + 1 ∂ xm + 2    ∂2 f    ∂ xn ∂ xm + 2   g

   g

 ∂2 f ...   ∂ xm + 1 ∂ xn  ...

 ∂2 f  2  ∂ xn

  g

      g    

has to be positive (negative) definite to have Q positive (negative) for all choices of dxi . It is evident that computation of the constrained derivatives ( dx 2 f /dxi dx j ) g is a difficult task and may be prohibitive for problems with more than three constraints. Thus the method of constrained variation, although it appears to be simple in theory, is very difficult to apply since the necessary conditions themselves involve evaluation of determinants of order m + 1. This is the reason that the method of Lagrange multipliers, discussed in the following section, is more commonly used to solve a multivariable optimization problem with equality constraints.

9.3.3

Solution by the Method of Lagrange with a Two Decision Variables

The basic features of the Langrange multiplier method is given initially for a simple problem of two variables with one constraint. The extension of the method to a general problem of n variables with m constraints is given later.

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173

Problem with Two Variables and One Constraint. Consider the problem: Minimize f ( x1 , x2 )

(9.25)

subject to g ( x1 , x2 ) = 0

For this problem, the necessary condition for the existence of an extreme point at X = X* was found in ∂f ∂ f /∂ x2 ∂ g  −    ∂ x1 ∂ g /∂ x2 ∂ x1 

=0

(9.26)

( x1* , x2* )

By defining a quantity λ , called the Lagrange multiplier, as

 ∂ f /∂ x2  λ = −   ∂ g /∂ x2 

(9.27) ( x1* ,

x2* )

Equation (9.26) can be expressed as ∂f ∂g  +λ   ∂ x1   ∂ x1

( x1* , x2* )

 ∂f ∂g  +λ   ∂ x2   ∂ x2

( x1* , x2* )

=0

(9.28)

=0

(9.29)

and Eq. (9.27) can be written as

In addition, the constraint equation has to be satisfied at the extreme point, that is,

g ( x1 , x2 )

( x1* , x*2 )

=0

(9.30)

Thus, Eqs. (9.28) to (9.30) represent the necessary conditions for the point ( x1* , x*2 ) to be an extreme point. Notice that the partial derivative ( dg /dx2 ) ( x* , x* ) has to be nonzero to be able to define λ by Eq. 1 2 (9.27). This is because the variation dx2 was expressed in terms of dx1 in the derivation of Eq. (9.26) [see Eq. (9.17)]. On the other hand, if we choose to express dx1 in terms of dx2 , we would have obtained the requirement that (dg /dx1 )

( x1* , x2* )

be nonzero to define λ . Thus the derivation of the necessary

conditions by the method of Lagrange multipliers requires that at least one of the partial derivatives of g ( x1 , x2 ) be nonzero at an extreme point. The necessary conditions given by Eqs. (9.28) to (9.30) are more commonly generated by constructing a function L, known as the Lagrange function, as L( x1 , x2 λ ) = f ( x1 , x2 ) + λ g ( x1 , x2 )

(9.31)

By treating L as a function of the three variables x1 , x3 and λ , the necessary conditions for its extremum are given by

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Mechanical System Design

∂L ∂f ∂g ( x1 , x2 , λ ) = ( x1 , x2 ) + λ ( x1 , x2 ) = 0 ∂ x1 ∂ x1 ∂x1 ∂L ∂f ∂g ( x1 , x2 , λ ) = ( x1 , x2 ) + λ ( x1 , x2 ) = 0 ∂ x2 ∂ x2 ∂x2

(9.32)

∂L ( x1 , x2 , λ ) = g ( x1 , x2 ) = 0 ∂λ Equations (9.32) can be seen to be same as Eqs. (9.28) to (9.30). The sufficiency conditions are given later. Example 7: Find the solution of Example 9.7 using the Lagrange multiplier method: Minimize f ( x, y) = kx −1 y −2 subject to

g ( x, y ) = x 2 + y 2 − a 2 = 0 Solution: The Lagrange function is

L( x, y , λ ) = f ( x, y ) + λ g ( x, y ) = kx −1 y −2 + λ ( x 2 + y 2 − a 2 ) The necessary conditions for the minimum of f ( x, y) [Eqs. (9.38)] give

∂L = − kx −2 y −2 + 2πλ = 0 ∂x

( E1 )

∂L = − 2kx −1 y −3 + 2 yλ = 0 ∂y

( E2 )

∂L = x −2 + y −2 − a 2 = 0 ∂λ

( E3 )

Equations ( E1 ) and ( E2 ) yield 2λ =

2k k = 4 2 x y xy 3

from which the relation x* = (1/ 2) y* can be obtained. This relation, along with Eq. ( E3 ) , gives the optimum solution as

x* =

a a and y* = 2 3 3

9.3.3.1 Necessary Conditions for a General Problem The equations derived above can be extended to the case of a general problem with n variables and m equality constraints: Minimize f ( X ) Subject to

g j ( X ) = 0, j = 1, 2 ..., m

(9.33)

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175

The Lagrange function, L, in this case is defined by introducing one Lagrange multiplier λ j for each constraint g j ( X ) as L( x1 , x2 , ..., xn , λ1 , λ2 , ..., λm ) = f ( X ) + λ1 g1 ( X ) + λ2 g 2 ( X ) + ... + λm g m ( X )

(9.34)

By treating L as a function of the n + m unknowns, x1 , x2 , ..., xn , ..., λ1 , λ2 , ..., λm , the necessary conditions for the extremum of L, which also correspond to the solution of the original problem stated in Eq. (9.33), are given by m ∂g j ∂L ∂ f = + ∑ λj = 0, i = 1, 2, ..., n j =1 ∂ xi ∂ xi ∂ xi

∂L = g j ( x) = 0, i = 1, 2, ..., m ∂ λi

(9.35) (9.36)

Equations (9.35) and (9.36) represent n + m equations in terms of the n + m unknowns, xi and λ j . The solution of Eqs. (9.35) and (9.36) gives

 x1*  λ1*   *  *  x2  λ  X * =   and λ * =  2  ...    ...   x*  λ *   n  n The vector X* corresponds to the relative constrained minimum of f (X) (sufficient conditions are to be verified) while the vector λ * provides the sensitivity information, as discussed in the next subsection.

9.3.3.2 Sufficiency Conditions for a General Problem A sufficient condition or f (X) to have a constrained relative minimum at X* is given by the following theorem. Theorem 9.3.1: Sufficient Condition A sufficient condition or f (X) to have a relative minimum at X* is that the quadratic, Q, defined by

∂2 L dxi dx j (9.37) i =1 j =1 ∂ xi ∂ x j evaluated at X = X* must be positive definite for all values of dX for which the constraints are satisfied. Proof: The proof is similar to that of Theorem 9.4 n

n

Q=∑ ∑

Notes: 1. If n

∂2L ( X *, λ *) dxi dx j j =1 ∂ xi ∂ x j n

Q=∑ ∑ i =1

is negative for all choices of the admissible variations dxi , X* will be a constrained maximum of f (X).

176

Mechanical System Design 2. It has been shown by Hancock that a necessary condition for the quadratic form Q, defined by Eq. (9.37), to be positive (negative) definite for all admissible variations dX is that each root of the polynomial zi , defined by the following determinant equation, be positive (negative):

L11 − z L12 L21 L22 − z M

L13 L23

K K

K Lnn − z K g1n K g2n

Ln1

Ln 2

Ln3

g11 g 21 M

g12 g 22

g13 g 23

gm1

gm 2

gm3 K

L1n L2 n

g11 g12

g 21 K g m1 g 22 K g m 2

g1n 0 0

g 2 n K g mn =0 0 K 0 0 K 0

gmn

0

0

K

(9.38)

0

where Lij =

∂2L ( x*, λ *) ∂ xi ∂ x j

(9.39)

gij =

∂ gi ( x*) ∂xj

(9.40)

3. Equation (9.38), on expansion, leads to an (n – m) th-order polynomial in z. If some of the roots of this polynomial are positive while the others are negative, the point X* is not an extreme point. The application of the necessary and sufficient conditions in the Lagrange multiplier method is illustrated with the help of the following example. Example 8: Find the dimensions of a cylindrical tin (with top and bottom made up of sheet metal to UPTU 2004 maximize its volume such that the total surface area is equal to A0 = 24π ). Solution: If x1 and x2 denote the radius of the base and length of the tin, respectively, the problem can be stated as: Maximize f ( x1 , x2 ) = π x12 x2 subject to

2π x12 + 2π x1 x2 = A0 = 24π The Lagrange function is

L( x1 , x2 , λ ) = π x12 x2 + λ (2π x12 + 2π x1 x2 − A0 ) and the necessary conditions for the maximum of f give ∂L = 2π x1 x2 + 4πλ x1 + 2πλ x2 = 0 ∂ x1

( E1 )

∂L = π x12 + 2πλ x1 = 0 ∂ x2

( E2 )

Calculus Method for Optimization

177

∂L = 2π x12 + 2πλ x1 x2 − A0 = 0 ∂λ

( E3 )

Equation ( E1 ) and ( E2 ) lead to

λ=−

1 x1 x2 = − x1 2 x1 + x2 2

that is, x1 =

1 x2 2

( E4 )

and Eqs. ( E3 ) and ( E4 ) give the desired solution as 1/2

A  x1* =  0   6π 

1/2

 2A  , x2* =  0   3π 

1/2

 A  , and λ* = −  0   24π 

This gives the maximum value of f as

 A3 f * =  0  54π

1/2

  

If A0 = 24π , the optimum solution becomes

x1* = 2, x2* = 4, λ * − 1, and f * = 16π To see that this solution really corresponds to the maximum of f, we apply the sufficiency condition of Eq. (9.38). In this case L11 =

L12 =

L22 =

g11 =

g12 =

∂2L ∂ x12 x*, λ * ∂2L ∂ x1 ∂ x2

= L21 = 2π x1* + 2πλ * = 2π x *, λ *

∂2L ∂ x22

∂ g1 ∂ x2 ∂ g1 ∂ x2

= 2π x2* + 4πλ * = 4π

=0 x *, λ *

= 4π x1* + 2π x*2 = 16π x*, λ *

= 2π x1* = 4π x*, λ *

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Mechanical System Design

Thus Eq. (9.38) becomes

4π − z 2π 16π 2π 0 − z 4π = 0 16π 4π 0 that is,

272π 2 z + 192π 3 = 0 This gives z=−

12 π 7

Since the value of z is negative, the point ( x1* , x*2 ) corresponds to the maximum of f.

9.3.3.3 Interpretation of the Lagrange Two Decision Variable To find the physical meaning of the Lagrange multipliers, consider the following optimization problem involving only a single equality constraint: Minimize f ( X )

(9.41)

g ( x) = b or g ( x ) = b − g ( x) = 0

(9.42)

Subject to where b is a constant. The necessary conditions to be satisfied for the solution of the problem are

∂f ∂g +λ = 0, i = 1, 2, ..., n ∂ xi ∂ xi g = 0

(9.43) (9.44)

Let the solution of Eqs. (9.43) and (9.44) be given by X *, λ*, and f * = f ( X *). Suppose that we want to find the effect of a small relaxation or tightening of the constraint on the optimum value of the objective function (i.e., we want to find the effect of a small change in b on f *). For this we differentiate Eq. (9.42) to obtain db – dg = 0 or ∂g dxi i =1 ∂ xi n

db = dg = ∑

(9.45)

Equation (9.43) can be rewritten as

∂f ∂g ∂ f ∂g +λ = −λ =0 ∂ xi ∂ xi ∂ xi ∂ xi

(9.46)

∂ g ∂ f /∂ xi , i = 1, 2, ..., n = λ ∂ xi

(9.47)

or

Calculus Method for Optimization

179

Substituting Eq. (9.47), into Eq. (9.45) we obtain 1 ∂f ∂f dxi = i =1 λ ∂ xi λ

(9.48)

∂f dxi i =1 ∂ xi

(9.49)

df df * or λ * = db db

(9.50)

df * = λ * db

(9.51)

n

db = ∑

since n

df = ∑

From equation (9.48)

λ= or

Thus λ * denotes the sensitivity (or rate of change) of f with respect to b or the marginal or incremental change in f * with respect to b at x*. In other words, λ * indicates how tightly the constraint is binding at the optimum point. Depending on the value of λ * (positive, negative, or zero), the following physical meaning can be attributed to λ *. 1. λ * > 0. In this case, a unit decrease in b is positively valued since one gets a smaller minimum value of the objective function f. In fact, the decrease in f * will be exactly equal to λ * since df = λ * ( −1) = − λ * < 0. Hence λ * may be interpreted as the marginal gain (further reduction) in f * due to the tightening of the constraint. On the other hand, if b is increased by 1 unit, f will also increase to a new optimum level, with the amount of increase in f * being determined by the magnitude of λ * since df = λ * ( +1) > 0. In this case, λ * may be thought of as the marginal cost (increase) in f * due to the relaxation of the constraint. 2. λ * < 0. Here a unit increase in b is positively valued. This means that it decreases the optimum value of f. In this case the marginal gain (reduction) in f * due to a relaxation of the constraint by 1 unit is determined by the value of λ * as df * = λ * (+ 1) < 0. If b is decreased by 1 unit, the marginal cost (increase) in f * by the tightening of the constraint is df * = λ * (−1) > 0 since, in this case, the minimum value of the objective function increases. 3. λ * = 0. In this case, any incremental change in b has absolutely no effect on the optimum value of f and hence the constraint will not be binding. This means that the optimization of f subject to g = 0 leads to the same optimum point x* as with the unconstrained optimization of f. In economics and operations research, Lagrange multipliers are known as shadow prices of the constraints since they indicate the changes in optimal value of the objective function per unit change in the right-hand side of the equality constraints. Example 9: Find the maximum of the function f ( x) = 2 x1 + x2 + 10 subject to g ( X ) = x1 + 2 x22 − 3 using the Lagrange multiplier method. Also find the effect of changing the right-hand side of the constraint on the optimum value of f.

180

Mechanical System Design

Solution: The Lagrange function is given by

L( x, λ ) = 2 x1 + x2 + 10 + ... (3 − x1 − 2 x22 ) .

( E1 )

The necessary conditions for the solution of the problem are

∂L =2−λ =0 ∂ x1 ∂L = 1 − 4λ x2 = 0 ∂ x2 ∂L = 3 − x1 − 2 x22 = 0 ∂λ

( E2 )

The solution of Eqs. ( E2 ) is  x*   2.97  X* =  1  =   *  x2   0.13 

λ * = 2.0

( E3 )

The application of sufficient condition of Eq. (9.46) yields

L11 − z L21 g12

L12 L22 − z g12

g11 g12 = 0 0

0 0 −z −1 −z −1 0 −4λ − z −4 x2 = 0 −8 − z −0.52 = 0 0 0 −1 −4 x2 −1 −0.52 0.2704z + 8 + z = 0 z = –6.2972 Hence x* will be a maximum of f with f * = f ( x*) = 16.07. One procedure for finding the effect on f * of changes in the value of b (right-hand side of the constraint) would be to solve the problem all over with the new value of b. Another procedure would involve the use of the value of λ *. When the original constraint is tightened by 1 unit (i.e., db = –1), Eq. (9.51) gives

df * = λ * db = 2( −1) = − 2 Thus the new value of f * is f * + df * = 14.07. On the other hand, if we relax the original constraint by 2 units (i.e., db = 2), we obtain

df * = λ * db = 2(+2) = 4 and hence the new value of f * is f * + df * = 20.07.

Calculus Method for Optimization

9.4

181

MODEL WITH TWO DECISION VARIABLES WITH INEQUALITY CONSTRAINTS

This section is concerned with the solution of the following problem: Minimize f ( X ) Subject to g1 ( X ) ≤ 0, j = 1, 2, ..., m

(9.52)

The inequality constraints in Eq. (9.52) can be transformed to equality constraints by adding non2 negative slack variables, y j , as

g1 ( X ) + y 2j = 0, j = 1, 2, ..., m where the values of the slack variables are yet unknown. The problem now becomes

(9.53)

Minimize f ( X ) subject to

G1 ( X , Y ) = g ( x) + y 2j = 0, j = 1, 2, ..., m

(9.54)

 y1  y   2 where Y =   is the vector of slack variables. M   ym  This problem can be solved conveniently by the method of Lagrange multipliers. For this, we construct the Lagrange function L as m

L( X , Y , λ ) = f ( X ) + ∑ λ j G j ( X , Y ) j =1

(9.55)

 λ1  λ   2 where λ =   is the vector of Lagrange multipliers. The stationary points of the Lagrange function M  λm  can be found by solving the following equations (necessary conditions): m ∂g ∂L ∂f ( X , Y , λ) = ( X ) + ∑ λ j i ( X ) = 0, i = 1, 2, ..., n j =1 ∂ xi ∂ xi ∂ xi

(9.56)

∂L ( X , Y , λ ) = G j ( X , Y ) = g j ( X ) + y12 = 0, j = 1, 2, ..., m ∂ xi

(9.57)

∂L ( X , Y , λ ) = 2λ j y j = 0, j = 1, 2, ..., m ∂ yi

(9.58)

182

Mechanical System Design

It can be seen that Eqs. (9.56) to (9.58) represent (n + 2m) equations in the (n + 2m) unknowns, X, λ , and Y. The solution of Eqs. (9.56) to (9.58) thus gives the optimum solution vector X*, the Lagrange multiplier vector, λ*, and the slack variable vector, Y*. Equations (9.57) ensure that the constraints g j ( x) ≤ 0, j = 1, 2, ..., m, are satisfied, while Eqs. (9.58) imply that either λ j = 0 or y j = 0. If λ j = 0, it means that the jth constraint is inactive and hence can be ignored. On the other hand, if y j = 0, it means that the constraint is active ( g j = 0) at the optimum point. Consider the division of the constraints into two subsets, J1 and J 2 , where J1 + J 2 represent the total set of constraints. Let the set J1 and J 2 include the indices of all the inactive constraints. Thus for j ∈ J1 , y j = 0 (constraints are active), for j ∈ J 2 , λ j = 0 (constraints are inactive), and Eqs. (9.56) can be simplified

∂g j ∂f + ∑ λj = 0, i = 1, 2, ..., n j ∈ j1 ∂ xi ∂ xi

(9.59)

Similarly, Eqs. (9.57) can be written as

g j ( X ) = 0,

j ∈ J1

(9.60)

g j ( X ) + y 2j = 0,

j ∈ J2

(9.61)

Equations (9.59) to (9.61) represent n + p + (m – p) = n + m equations in the n + m unknowns, xi (i = 1, 2, ..., n), λ ( j ∈ J1 ) and y j ( j ∈ J 2 ), where P denotes the number of active constraints. Assuming that the first P constraints are active, Eqns. (9.59) can be expressed as

−

∂g p ∂f ∂g ∂g , i = 1, 2, ..., n = λ1 1 + λ2 2 + ... + λ p ∂ xi ∂ xi ∂ xi ∂ xi

(9.62)

These equations can be written collectively as

−∇f = λ1∇g1 + λ2 ∇g 2 + ... + λ p ∇g p

(9.63)

Those constraints that are satisfied with an equality sign, g j = 0, at the optimum point are called the active constraints, while those that are satisfied with a strict inequality sign, g j < 0, are termed inactive constraints. The symbol ∈ is used to denote the meaning “belongs to” or “element of.” In Eq. (9.63) ∇f and ∇g j are the gradients of the objective function and the jth constraint, respectively:

 ∂ g j /∂ x1   ∂ f /∂ x1    ∂ f /∂ x   ∂ g j /∂ x2  2 and ∇f =  ∇ = g    j M  M    ∂ f /∂ xn  ∂ g j /∂ xn    Equation (9.63) indicates that the negative of the gradient of the objective function can be expressed as a linear combination of the gradients of the active constraints at the optimum point.

Calculus Method for Optimization

183

Further, we can show that in the case of a minimization problem, the λ j values ( j ∈ J1 ) have to be positive. For simplicity of illustration, suppose that only two constraints are active (p = 2) at the optimum point. Then Eq. (9.63) reduces to −∇f = λ1∇g1 + λ2 ∇g 2

(9.64)

Let S be a feasible direction at the optimum point. By premultiplying both sides of Eqn. (9.64) by S T , we obtain

− S T ∇f = λ1 S T ∇g1 + λ2 S T ∇g2

(9.65)

where the superscript T denotes the transpose. Since S is a feasible direction, it should satisfy the relations

S T ∇g1 < 0

(9.66)

S T ∇g 2 < 0 A vector S is called a feasible direction from a point X if at least a small step can be taken along S that does not immediately leave the feasible region. Thus for problems with sufficiently smooth constraint surfaces, vector S satisfying the relation S T ∇g j < 0

g1 = 0

g1 = 0

g1 < 0 g2 < 0

g2 = 0

S

g1 > 0

(Linear constraint)

g2 > 0

g1 > 0

g2 > 0

∇g1

Angles greater ∇g 2 than 90°

∇g 2

(a)

(b)

g1 < 0 g2 < 0

g1 = 0

g2 = 0

S Concave constraint surface g1 > 0

g2 > 0

90° ∇g1

Angles greater than 90°

g2 = 0

S

Angle equal to 90°

Angles greater than 90° ∇g1

g1 < 0 g2 < 0

∇g 2

(c)

Fig. 9.7 Feasible direction S

184

Mechanical System Design

can be called a feasible direction. On the other hand, if the constraint is either linear or concave, as shown in Fig 9.7(b) and (c), any vector satisfying the relation

S T ∇g j ≤ 0 can be called a feasible direction. The geometric interpretation of a feasible direction is that the vector S makes an obtuse angle with all the constraint normals, except that for the linear or outward-curving (concave) constraints, the angle may go to as low as 90°. Thus if λ1 > 0 and λ2 > 0, the quantity S T ∇f can be seen always to be positive. As ∇f indicates

the gradient direction, along which the value of the function increases at the maximum rate. S T ∇f represents the component of the increment of f along the direction S. If S T ∇f > 0, the function value increases as we move along the direction S. Hence, if λ1 and λ2 are positive, we will not be able to find any direction in the feasible domain along which the function value can be decreased further. Since the point at which Eq. (9.62) is valid is assumed to be optimum, λ1 and λ2 have to be positive. This reasoning can be extended to cases where there are more than two constraints active. By proceeding in a similar manner, one can show that the function values have to be negative for a maximization problem.

9.5

A CASE STUDY: OPTIMIZATION OF AN INSULATION SYSTEM

Insulation is applied on metallic pipes or wires to reduce heat loss. More the thickness of insulation layer, lesser is the heat loss. From outer layer of the insulation, heat is dissipated to the surroundings by convection. As the thickness of insulation layer goes on increasing, the area of the outer surface, which is responsible for dissipating heat to the surrounding by convection, also increases. It is possible that some optimum thickness of insulation exist due to these opposing effects.

9.5.1

Critical Insulation Thickness for Pipe

Let the radius of pipe be R and outer radius of the insulating surface be r. Thickness of insulation surface = (r – R) Let the thermal conductivity of pipe material be k and convective film coefficient between the insulation surface and surroundings be h. The thermal conductivity of pipe material is much higher than that of insulation material; as such heat drop across the pipe wall has been neglected. Heat flow per unit length of the pipe through the insulation

2π (T − Ta ) q = 1 r R L + ln hr k =

2π (T − Ta ) q = 1 L r R + ln   hr  k 

where Ta = temperature of the ambient atmosphere.

r ln 1 R + is maximum. The quantity q/L will be minimum only where the denominator hr k

...(1)

Calculus Method for Optimization

185

1 D= + hr

Let

∴

r R k

ln

dD should be zero. dr −1 dD 1 = 2 + =0 dr hr kr

or;

⇒

rc =

k h

Thus optimum insulation thickness can be determined corresponding to this value, heat transfer is q = L

⇒

2π (T − Ta ) 1 + ln(r /R ) . h k /h

q 2π k (T − Ta ) = 1 + ln(r /R) L h

Insulation

R Pipe

r

r (variable)

T (constant)

Fig. 9.8 Critical thickness of pipe where:

rc = critical thickness of pipe

h = film coefficient (W/m2 k) 2 h0 = film coefficient (W/m k) R = radius of pipe (mm) r = outer radius of the insulating surface (mm) k = thermal conductivity of pipe material (W/mk)

186

Mechanical System Design q = heat flow (w/m) T = surface temperature of the pipe Ta = ambient temperature of atmosphere

9.5.2

Critical Thickness of Insulation for Sphere

If we consider that the inner radius of the sphere is fixed, the equation of heat flow through sphere with insulation is given by

4π (ti − to ) 1 r2 − r1 + 2 r1r2 k r2 ho

q=

The heat transfer will be maximum at the particular value of r2

r − r 1  d × 2 1 + 2 =0 dr2  r1r2 k r2 ho  1 1 1  d × − + 2 =0 dr2  r1k r2 k r2 ho 

or,

=

or,

1 r22 k

r2 = rc =

or,

9.5.3

−

2 r23 ho

=0

2k ho

Explanation of Insulation System

The addition of insulation on plane wall does not affect the face area of heat transfer while it increases the outer area in case of cylindrical or spherical walls. Therefore, the rate of heat transfer due to increase in thickness (by application of insulation) of a plane wall will always decrease, but it may not so in case of cylinders or spheres. According to radial heat flow in a tube formula

q=

2π kl (t1 − t2 ) r log e 2 r1

The value of q decreases with increasing value of

r1 or with increasing wall. r2

Thickness of the inside radius r, is kept constant. However, connective heat transfer at the inside usually accompanies the process of conduction through the cylinder wall and outside surface as the wall thickness or as the value of r2 increases the convective heat transfer is also increased. On the other

Calculus Method for Optimization

187

hand the rate of heat transfer by conduction through the wall decreases. If the overall heat transfer from inside surface to outside surface is considered, the rate of heat flow may first increase with increasing r 2 and after passing through the maximum may start decreasing. This argument, when applied to insulated pipes or cables show that: 1. The heat transfer rate from the cylindrical surface increases by addition of insulation (increase in outer radius) up to a certain value (rc). 2. At this value of the outer radius (rc) called critical radius the rate of heat transfer by conduction (equal to rate of heat transfer by convection at the surface) is maximum. 3. The heat transfer rate decrease if further insulation is applied to the surface beyond critical rc shown in Fig. 9.9. q

rc

Fig. 9.9

r

Critical radius of insulation pipe

Therefore in case of pipes the outside radius of pipe after applying insulation is kept greater than γ c while in case of current carrying wires and cables is kept lower than or equal to γ c (as this will dissipate more heat than bare pipe). The insulation of electrical wire is therefore advantageous in as much as it increases the dissipation of heat generated within the conductor. The value of r 2 obtained from this equation is the critical radius, rc given by k rc = ho The critical radius in case of spherical shell can be evaluated on the same lines as 2k rc = ho In practice, however, the selection of insulation will be based on the cost analysis and the compromise between desirability of heat dissipation and the cost involved. Cases will arise when rc < r1 . The addition insulation in such a case will always reduce the heat transfer. For electrical conductors therefore r1 (inner radius of insulation or outer radius of wire) should always be of the necessity of application of insulation again electrical current on one hand and decrease in the heat dissipation rate on the other hand, if r1 > rc .

188

Mechanical System Design

9.5.4

Heat Transfer

One Dimensional Heat Conduction – Resistance Concept: In heat transfer by conduction, heat flow dQ through an area dA in a plane normal to the direction of heat transfer in time dt is given by dQ = − kdA

dt dt dx

(Fourier’s law of conduction)

dT = Temperature gradient in plane of heat transfer, k = Thermal conductivity, which depends dx on the state of the substance (solid, liquid or gas), and for a given state it varies with temperature (unit of k are K cal/mhr °C). where

Rate of heat transfer

KA ∆T ∆x

θ =

where ∆x = length of heat flow path or thickness of material.

where

∆T R T = T1 − T2

and

R=

q=

∆x kA

∆T can be called as temperature potential and R can be termed as thermal resistance. Question 1: A 10 mm cable is to be laid in atmosphere of 20 °C, ( h0 = 8.5 W/m2 K). The surface temperature of the cable is likely to be 65 °C due to heat generated within. Discuss the effect of insulating the cable with rubber having k = 0.155 W/mk. Solution: The equation (1) giving the rate of heat dissipation per unit length, can be simplified for this case as:

q=

⇒

q=

2π (T − Ta ) r ln 1 + R hr k 2π (65 − 20) 1000 ln r /5 + 8.5r 0.155

=q=

(where r is in mm).

43.8 r 18.4 ln + 5 r

Calculus Method for Optimization

189

This will have a maximum value at critical radius:

rc =

k ho

= rc =

qmax =

0.155 × 1000 8.5

43.8 18.4 ln +1 5

rc = 18.4 mm

= 19 W/m Ans. If the pipe is bare the heat dissipation would have been:

q(bare) =

=

=

(T − Ta )2π 1 hr (65 − 20)2π 1000 8.5 × 5

45 × 2π × 8.5 × 5 1000

= 12 W/m It may be noted that bare pipe could dissipate heat @ 12 W/m. On application of insulation upto a thickness of (18.4–10/2) = 13.4 mm the heat dissipation is increased to 19 W/m i.e., an increase of 36%. Questin 2: A 60 watt lamp is buried in soil (k = 0.0084 J/S cm°c) at °C and burned until steady state is reached. Find the temperature 30 cm away if 1 watt lamp produces 1 J/s heat. Solution: We know that

q=

4π k (t1 − t ) r1r2 r2 − r1

t = t1 at r = r1 t = t2 at r = r2 In the present case, the boundary conditions are t = t1 at r = r1 t = 0 at r = ∞

So,

q = 4π k (t1 ) r1

190

Mechanical System Design

q 4π kr1

t1 =

or

q = 60 × J/s

t=

or

60 × 1 4π × 0.0084 × 30

⇒ t = 19 °C Question 3: A closed cubical box is made of perfectly insulating material and the only way for heat to 2 enter or leave the box is through two solid cylindrical metal plugs, each of cross-sectional area 12 cm and length 8 cm fixed in the opposite walls of the box. The outer surface of one plug is kept at a temperature of 100 °C while the outer surface of the plug is maintained at a temperature of 4 °C. The thermal conductivity of the material of the plug is 2.0 W/m °C. A source of energy generating 13 W is enclosed inside the box. Find the equilibrium temperature of the inner surface of the box assuming that it is the same at all points on the inner surface. Solution:

θ1 = 100 °C

S

θ2 = 4 °C

Fig. 9.10 The situation is shown in Fig. 9.10. Let the temperature inside the box be θ . The rate at which heat enters the box through the left plug is

∆Q KA(θ1 − θ ) = x ∆t The rate of heat generation in the box = 13 W. The rate at which heat flows out of the box through the rigid plug is ∆Q kA(θ − θ 2 ) = x ∆t In the steady state:

∆Q1 ∆θ 2 + 13 = ∆t ∆t or

KA KA (θ1 − θ ) + 13 = (θ − θ 2 ) x x

Calculus Method for Optimization

191 2 KA kA (θ1 + θ 2 ) + 13 θ = x x

or

θ =

or

=

θ1 + θ2 13x + 2 2kA 100 + 4 13 × 0.08 + 2 2 × 2 × 2 × 10 −4

= 52 °C + 216.67 °C

θ = 269 °C Given that x = 0.08 m

k = 2.0 W/m °C A = 2 * 10−4 m 2 θ1 = 100 °C θ 2 = 4 °C The rate of heat generation in the box = 13 W Question 4: In Fig. 9.11 is shown a large tank of water at a constant temperature θ 0 and a small vessel containing a mass m of water at an initial temperature θ1 (< θ 0 ). A metal rod of length L, area of cross section A and thermal conductivity k connects the two vessels. Find the time taken for the temperature of the water in the smaller vessel to become θ 2 (θ1 < θ 2 < θ 0 ). Specific heat capacity of water is s and all other heat capacities are negligible.

L

θ0

θ1

Fig. 9.11 Solution: Suppose, the temperature of the water in the smaller vessel is θ at time t. In the next time interval dt, a heat ∆θ is transferred to it where

∆Q =

kA (θ0 − θ )dt L

...(1)

This heat increases the temperature of the water of mass m to θ + dθ where

Q = msd θ From equation (1) and (2) we get:

...(2)

192

Mechanical System Design kA (θ 0 − θ ) dt = msd θ L

dt =

⇒

Lms dθ × (θ0 − θ ) kA

T

∫

Lms dt = kA o

θ2

∫θ

o1

dθ 0−θ

where T is the required temperature of the water to be where θ 2

T =

Lms  θo − θ1  ln   Ans. kA  θo − θ2 

Question 5: An electric water heater tank is insulated to reduce heat loss to surroundings. The electric water heater is controlled by a thermostat so as to maintain the temperature of water in the tank at θ . θi is the temperature of water supply to the heater, θ a is the ambient temperature of the surroundings. Develop a model to compute the rate of heat flow (q) of heating element if water flow from the heated tank is equal to n. Assume that c = thermal capacitance of insulation, n = water flow from the heater tank and s = specific heat of water. UPTU 2004 Solution: We know that

dQ = − k (θ − θ a ) dt where θ a is the temperature of the surrounding and θ is the temperature of water in the tank at time t. Suppose θ = θ1 at t = 0 then, θn

∫

θ1

or or

ln

t

dθ = − k dt θ − θa o

∫

θ − θn = − kt (Q log e N = x, N = e x ) θ1 − θ a θ − θa = (θ1 − θa )e − kt

...(1)

The electric water heater tank continues to lose heat till its temperature becomes equal to that of the surrounding. The loss of heat in the entire period is Qm = ms (θ1 − θ 0 ) Qm = θ1 − θ 0 ms

Calculus Method for Optimization

193 =

θ m θ1 − θ 0 = 2ms 2

If the water heat tank loses this heat in time t, the temperature at t1 will be

θ1 −

(θ1 − θ 0 ) θ1 + θ 0 = 2 2

Putting the values of time and temperature in Eq. (1), we get:

=

θ1 + θ 2 − θo = (θ1 − θo ) e − kt1 2

⇒ e − kt = ⇒ t1 =

1 2

ln 2 k

We also know that R = Thermal resistance =R=

x kA

By Fourier’s law of conduction dq = − kdA

dT dt1 dx

(Heat flow)

q = − kA

dt t1 x

q = − kA

(θ − θ i ) t1 x

q=

q=

Heat flow

q=

(θ − θ i ) t1  R    kA  (θ − θi ) × ln 2 R

(θ − θ i ) × ln 2 R

194

Mechanical System Design

Exercise

9 . . .

1. 2. 3. 4. 5.

Explain the model with one decision variable. Explain the model with two decision variables with no constraint. Explain the model with two variables with equality constraint by calculus methods for optimization. Explain the model with two variables with inequality constraints. Write a short note on (a) Model with one decision variable (b) Model with two decision variable (c) Model with two decision variable with equality constraint (d) Model with two decision variable with inequality constraint. 6. Briefly explain a model with two decision variable with no constraint. Analyse whether necessary and sufficient conditions apply. 7. Explain necessary and sufficient conditions for model with single variable. 8. What is the condition of Maxima and Minima?

A Case Study 1. 2. 3. 4.

Derive the critical insulation thickness for a pipe with suitable step. Derive critical thickness of insulation of a sphere. Explain an insulation system. What is heat transfer? Explain with a suitable steps in the language of MSD.

Decision Analysis

195

10 Decision Analysis 10.1

INTRODUCTION

This chapter presents methods for choosing the optimum plan or policy from a specific set of possible alternatives. The expected monetary value is not always a good criterion because of the personal values of the decision-maker which may be quite different from person to person, may vary with time, and is in general, a non-linear function of the monetary values. These personal values are indicated by the utility function, which is a powerful concept, but one that is difficult to assess precisely to indicate these personal values. The approach to decision-making in probabilistic situation should include: 1. Determining the possible results of each of the alternative; 2. Determining the probability of each outcome; 3. Determining the utility curve for the decision-maker and assigning utility values to the result; and 4. Calculating the expected utility values of the alternatives. The selection criterion is the maximum expected utility value. A decision tree, a special type of linear graph, can be used to graphically represent the decision problem. Baye’s theorem provides an easy means for computing conditional probabilities. The decisionmaker should continue to acquire new information as long as the increase in expected utility attributable to this new information is greater than the cost. If it is not possible to estimate the probabilities of the outcomes, then expected values cannot be used as a criterion in decision-making. However, if the problem is important enough to receive formal analysis, the decision-maker must either make a subjective estimate of the probability or obtain data that will permit an estimate to be made.

10.2

ELEMENTS OF A DECISION PROBLEM

The primary objective in a decision problem is to choose the optimum plan or policy from a specific set of possible alternatives. However, the decision must be made in a logical and subjective manner so that all possible alternatives and their consequences are considered and the final decision can be explained and justified to the interested parties. In addition, contingency plans must also be prepared for all potential consequences of decision.

196

Mechanical System Design

For example, consider the decision problem of the contractor discussed in example. 1. Decision node 2. Chance node 3. Outcome node The contractor must choose among three alternatives: 1. To move the equipment away from the river bank to avoid loss due to potential flooding. 2. To leave the equipment at the location and build a protective platform, or 3. To leave the equipment at the location and not build a protective platform. Each alternative results in certain cost to the contractor, and the cost of the last two alternatives will depend on the extent of flooding during the spring. Therefore, in making his decision, the contractor must consider the chances of flooding and the potential loss that may result. Furthermore, should he decide to leave the equipment on location, he must have available contingency plans for replacing the equipment in case of serious loss from flood. Example 1: Consider the problem encountered by a contractor who is performing some work in a river flats that has been subjected to high water conditions and occasional destructive flooding in the past. There is a four-month period which he has no use for the equipment either on this job or of others. He can keep the equipment on this job in the river flats, or else move it out, store it, and then move it back at a total cost of Rs 1800. If he keeps it in the river flats, he has the option of building a platform for the equipment at a cost of Rs 500, which will protect it against high water, but not against a destructive flood. The damage that would be caused by high water amounts to Rs 10,000. If there is no platform, a destructive flood would entail a loss of Rs 60,000; regardless of whether or not he builds a platform. The probability of high water in the four-month period is 0.25; the probability of destructive flood is 0.02. The contractor has to choose from three possible options. Solution: A decision tree is a graphical representation of elements of a decision problem. This figure considers the decision problem of the contractor. The contractor must choose among three alternatives. – Rs 500 – Rs 1800 Move equipment

Normal (.73) – Rs 500 Build platform

High water (.25) – Rs 60,500

Flood (.02)

0

Normal (.73) Decision node

No platfrom

Chance node Outcome node

Fig. 10.1

High water (.25) – Rs 10,000

Flood (.02)

– Rs (60,000)

Elements of decision problem.

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197

Two common elements that characterise all engineering decision problem: 1. Probabilistic events, and 2. Insufficient data. Probabilistic events are occurrences that are beyond the control of the decision-maker; although the probability or percentage chance of these occurrences can usually be predicted from either historical data or engineering foresight. For example, in element, weather is beyond the control of the engineer who is planning a construction schedule, yet it is a factor that he must take into consideration. Natural events such as flooding, earth quakes, and hurricanes are vital factors that must be considered in many engineering decision problems, including the structural design for high-rise buildings, location of dams and reservoirs, and location of transportation systems. The amount of technical data that is available to a decision-maker is usually limited by many factors. Among the more common factors are: technical skill of the decision-maker, state of the art in science and technology, limited financial resources, limited research capability, and limited time available for research and experimentation. Although technical data can usually be obtained at a certain cost to the decision-maker, the value of the additional information must be weighted against the cost of obtaining such information. For example, traffic serveys may be conducted to provide data on the volume of traffic flow, degree of congestion, computer traffic behavior, percentage of truck and thorough traffic and so on. Such surveys require time and money but provide in return information that traffic engineers may use to make sound decisions on the design of city traffic systems. A decision problem involves a set of alternative actions that are connected in some way to a set of possible outcomes. If the decision-maker knew which outcomes would occur with each act, he could choose the act that resulted in the outcome he most valued. Actions may lead to outcomes that depend on chance, and, therefore the outcome is known only in terms of probability of occurrence if certain actions are taken. The necessary first step in the decision-making process, therefore, is to identify the alternative actions and outcomes and their inter-relationships. The value of each outcome and the probabilities of its occurrence for the alternative action should be quantified. These parameters, along with their relationships constitute the problem model, which may then be analysed to determine the set of actions a decision-maker should take to achieve his most valued outcome.

10.3

DECISION MODEL

Problem Statement A contractor working on an outdoor construction project in a coastal area reviews his progress on August 1. He finds that if he keeps the normal speed and loses no time because of hurricanes, he will be able to complete the job on August 31. However, due to the poor weather conditions in the area after August 16, he will have only a 40 percent chance of finishing on time. He estimates that there is a 50 per cent chance of minor hurricane, which will cause a delay of 5 days, and a 10% chance of a major hurricane, which will cause of delay of 10 days. He has to decide now whether he should start a crash program on August 2 at an additional cost of Rs 75 per day and finish the project on August 16. As an alternative, he can maintain the normal schedule and review his progress on August 31. At that time if a hurricane has occurred and the project is delayed, he will have the choice of accepting the delay at a certain penalty cost or he can try to cash the program then. The penalty cost for delay of completion will be Rs 400 per day for the first 5-days and Rs 600 per day for the second 5-days. The additional cost of

198

Mechanical System Design

a crash program after the hurricane will be Rs 200 per day. The total additional cost is computed as sum of delay penalty cost and crash cost. He further estimates that the possible results (outcomes) of a crash program after a minor hurricane causes a 5-day delay will be as follows:

Table 10.1 Crash Program Result

Probability

Total Additional Cost

Save 1 day Save 2 days Save 3 days

0.5 0.3 0.2

Rs (4 × 400) + Rs (4 × 200) = Rs 2400 Rs (3 × 400) + Rs (3 × 200) = Rs 1800 Rs (2 × 400) + Rs (2 × 200) = Rs 1200

The possible results of a crash program after a major hurricane which causes a 10-days delay is estimated as follows:

Table 10.2 Crash Program Result

Probability

Save 2 days Save 3 days Save 4 days

0.7 0.2 0.1

Total Additional Cost Rs [(2000 + 3 × 600) + 8 × 200] = Rs 5400 Rs [(2000 + 2 × 600) + (7 × 200)] = Rs 4600 Rs [(2000 + 1 × 600) + (6 × 200)] = Rs 3800

The above problem description adequately outlines all the alternatives, the consequences, as well as all the relevant data needed to make the decision. Therefore, this description constitutes a descriptive model of the decision problem. It is obvious however, that even for problem as simple as this, it is difficult to obtain an overview of the problem from such a descriptive model.

10.4

THE SCIENTIFIC APPROACH TO THE DECISION PROCESS 1. The problem for analysis is defined and the condition for observation is determined. 2. Observation is made under different conditions to determine the behavior of the system containing the problem. 3. Based on the observations, a hypothesis that describes how the factors involved are thought to interest or what is the best solution to the problem, is conceived. 4. To test the hypothesis, an experiment is designed. 5. The experiment is executed and measurements are obtained and recorded. 6. The results of the experiment are analysed and the hypothesis is either accepted or rejected.

These six steps of the scientific method can be applied to decision-making. For example, the evaluation of alternatives is done scientifically through experimentation. The overall relationship of the scientific approach to the decision-making process is shown in Fig. 10.2.

Decision Analysis

199

Decision-making Process

Scientific Method

Problem defined

Problem defined

Step 1

Observation

Step 2

Search for alternatives

Hypothesis postulated

Step 3

Evaluation

Experimentation design

Step 4

Choice

Hypothesis accepted or rejected

Step 5

Fig. 10.2

Relationship of the scientific approach to the decision process. Guidelines for effective decision-making

The following guidelines may be followed for effective decision-making: 1. Define the goals. 2. Ensure that the decision will contribute to the goal. 3. Adopt a diagnostic approach to the decision-making. 4. Involve subordinates in decision-making process. 5. Ensure successful implementation of the decision. 6. Evaluate the results, and 7. Be flexible and revise the decision, which does not yield the desired results.

10.5

QUANTITATIVE METHODS IN DECISION-MAKING

Regardless of how much difficult the study of mathematics may be to some, the fact remains that the business manager today must deal with quantitative concepts in order to run a complex operation effectively. The increasing use of quantitative techniques has enabled managers to make more intelligent use of the vast amounts of data now available to the business organisation. As a result of the use of the many available quantitative techniques, this data can be processed to provide information valuable to the manager in carrying out planning, decision-making, and/or controlling functions. During World War II, certain mathematical techniques were developed to aid in the resolution of complex military problems and in recent years they have been applied to aid managers in their decision-making. The application of quantitative techniques to management decision-making can aid

200

Mechanical System Design

greatly in reducing the risk and uncertainty the manager must confront daily. Although some managers have used quantitative tools for decades, the emphasis on quantitative tools in recent years have been of Operations Research (O.R.), a concept which emerged during World War II. Used in connection with the computer, O.T. is the mathematical application of the scientific method to the solution of business problems. At the center of the O.R. is the mathematical model—a set of equations representing the actual problem situation. The specific technique used to construct and solve the model depends upon the nature of problem. Some of the O.R. techniques are Linear Programming, and Queuing theory. Break-even Analysis Linear Programming 1. Deterministic Models

Capital Budgeting Inventory Management Expected Value Model Decision Tree

2. Probabilistic Models

Simulation Waiting Line Theory Game Models Information Theory

3. Other Techniques

Utility Theory Heuristic Problem Solving

Fig. 10.3 A deterministic model is typically appropriate when a manager can obtain reliable data. If a manager knows precisely what revenues are for a particular year, this enables him to solve problems involving revenues. Probabilistic model does not require the exact value of the factors in a problem as a deterministic model needs. Probabilistic models on the other hand, can be used for solving problems involving chance or random factors. Managers rarely, if ever, know with complete certainty how demand will shift as the price of a product is raised. Typically, managers must make decisions under conditions of at least some uncertainty. The function of probabilistic models is to reduce uncertainty.

10.6

DECISION TREES

A decision tree is a graphical representation of the decision process indicating decision alternatives, states of nature, probabilities attached to the states of nature and conditional benefits and losses. It consists of a network of nodes and branches. Two types of nodes are used: decision node represented by a square and state of nature (chance or event) node represented by a circle. Alternative courses of action (strategies) originate from the decision node as main branches (decision branches). At the end of each decision branch, there is a state of nature node from which emanate chance events in the form of sub-branches (chance branches). The respective payoffs and the probabilities associated with alternative courses and the chance events are shown alongside these branches. At the terminal of the chance branches are shown with the expected values of the outcome.

Decision Analysis

201

The general approach used in decision tree analysis is to work backward through the tree from right to left, computing the expected value of each chance node. We then choose the particular branch leaving a decision node, which leads to the chance node with the highest expected value. This is known as roll back or fold back process. Illustration: Suppose we have the decision-making problem represented by the following Table 10.3.

Table 10.3

Conditional profits

States of nature

Probability

S1 (High demand) S2 (Low demand)

Alternative actions (A1) Produce 25 units Rs 4,000 Rs 2,000

0.6 0.4

(A2) Produce 75 units Rs 10,000 Rs 5,000

The decision tree for the above problem is shown in Fig. 10.4. For a decision alternative (strategy) the EMV is calculated by summing the products of pay off of each state and its probability. For example, EMV of decision alternative A1 (or node 2) is = Rs (4,000 × 0.6 + 2,000 × 0.4) = Rs. 3,200.

2

A1 State of Nature

S1 p = 0.6

Rs 4,000

S2

p = 0.4

Rs 2,000

1 Decision Node

A2

(Chance) Node 3

S1 p = 0.6

Rs 10,000

S2

p = 0.4

Rs 5,000

Fig. 10.4 Decision trees are useful for representing the inter-related, sequential and multi-dimensional aspects of decision-making problem. By drawing a decision tree, one is in a position to visualise the entire complexity of the decision problem in all its dimensions as also the actual processes and stages for arriving at the final decision.

Steps in Decision Tree Analysis 1. Identify the decision points and the alternative courses of action at each decision point systematically.

202

Mechanical System Design 2. At each decision point determine the probability and the pay off associated with each course of action. 3. Commencing from the extreme right end, compute the expected pay offs (EMV) for each course of action. 4. Choose the course of action that yields the best pay off for each of the decisions. 5. Proceed backwards to the next stage of decision points. 6. Repeat above steps till the first decision point is reached. 7. Finally, identify these courses of action to be adopted from the beginning to the end under different possible outcomes for the situation as a whole.

10.7

ADVANTAGES AND LIMITATIONS OF DECISION TREE APPROACH

Advantages 1. It structures the decision process and helps decision-making in an orderly, systematic and sequential manner. 2. It requires the decision-maker to examine all possible outcomes, whether desirable or undesirable. 3. It communicates the decision-making process to others in an easy and clear manner, illustrating each assumption about the future. 4. It displays the logical relationship between the parts of a complex decision and identifies the time sequence in which various actions and subsequent events would occur. 5. It is especially useful in situations wherein the initial decision and its outcome affect the subsequent decisions. It can be applied in various fields such as introduction of a new product, marketing, make or buy decisions, investment decisions, etc.

10.7.1

Limitations of Decision Tree Approach

1. Decision tree diagrams become more complicated as the number of decision alternatives increases and more variables are introduced. 2. It becomes highly complicated when interdependent alternatives and dependent variables are present in the problem. 3. It assumes that utility of money is linear with money. 4. It analyses the problem in terms of expected values and thus yields an “average” valued solution. 5. There is often inconsistency in assigning probabilities for different events. Example 2: A client asks an estate agent to sell three properties A, B, and C for him and agrees to pay him 5% commission on each sale. He specifies certain conditions. The estate agent must sell property. A first, and this he must do within 60 days. If and when A is sold the agent receives his 5% commission on that sale. He can then either back out at this stage or nominate and try to sell one of the remaining two properties within 60 days. If he does not succeed in selling the nominated property in that period, he is not given the opportunity to sell the other. If he does sell it in the period, he is given the opportunity to sell the third property on the same conditions. The following table summarises the prices, selling costs (incurred by the estate agent whenever a sale is made) and the state agent’s estimated probability of making a sale.

Decision Analysis

203

Table 10.4 Property

Price of property

Selling costs

Probability of sale

A B C

Rs 12,000 Rs 25,000 Rs 50,000

Rs 400 Rs 225 Rs 450

0.7 0.6 0.5

(i) Draw up an appropriate decision tree for the estate agent. (ii) What is the estate agent’s best strategy under EMV approach? Solution: The estate agent gets 5% commission if he sells the properties and satisfies the specified conditions. The amount he receives as commission on sale of properties A, B, and C will be Rs 600, Rs 1,250 and Rs 2,500 respectively. Since the selling costs incurred by him are Rs 400, Rs 225 and Rs 450, his conditional profits from sale of properties A, B, and C are Rs 200, Rs 1,025 and Rs 2,050 respectively. The decision tree for the problem is shown in Fig. 10.5.

Takes B

Sells B, Rs 1,025 B 0.6

Sells C, Rs 2,050

3

0.5

Does not sell

Does not sell Rs 0 0.5

Sells A, Rs 200

0.4

2 Takes C

0.7 1

Accepts A

A

Does not sell A

Sells B, Rs 1025 0.6

Sells C, Rs 2,050 4 C 0.5

Does not sell 0.4

0.3

0.5 Rs 0

Rs 0

Does not sell C Rs 0

Fig. 10.5 EMV of node 3 = Rs (0.5 × 2,050 + 0.5 × 0) = Rs 1,025. EMV of node 4 = Rs (0.6 × 1,025 + 0.4 × 0) = Rs 615. EMV of node B = Rs [0.6 (1,025 + 1,025) + 0.4 × 0] = Rs. 1,230. EMV of node C = Rs [0.5 (2,050 + 615) +0.5 × 0] = Rs 1,332.50. EMV of node 2 = Rs 1,332.50. (Higher of the EMV at B and C). EMV of node A = Rs [0.7 (200 + 1,332.50) + 0.3 × 0] = Rs 1,072.75. EMV of node 1 = Rs 1,072.75. The optimal strategy path is drawn in bold lines. Thus the optimum strategy for the estate agent is to sell A; if he sells A then try to sell C and if he sells C, then try to sell B to get an optimum expected amount of Rs 1,072.75.

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Mechanical System Design

Example 3: Mr Sinha had to decide whether or not to drill a well on his farm. In his village, only 40% of the wells drilled were successful at 200 feet of depth. Some of the farmers who did not get water at 200 feet, drilled further upto 250 feet but only 20% struck water at 250 feet. Cost of drilling is Rs 50 per foot. Mr Sinha estimated that he would pay Rs 18,000 during a 5-year period in the present value terms, if he continues to buy water from the neighbours rather than go for the well, which would have a life of 5 years. Mr Sinha has three decisions to make; (a) should he drill up to 200 feet and (b) if no water is found at 200 feet, should he drill up to 250 feet? (c) should he continue to buy water from his neighbour? Solution: Decision tree diagram for the above problem is shown in Fig. 10.6. The cost associated with each outcome is written on the decision tree. EMV of node B = Rs [0.2 × 12,500 + 0.8 × 30,500] = Rs [2,500 + 24,400) = Rs 26,900. EMV of node 2 = Rs 26,900 (Lesser of the two values of Rs 26,900 and Rs 28,000). EMV of node A = Rs [0.4 × 10,000 + 0.6 × 26,900] = Rs [4,000 + 16,140] = Rs 20,140 EMV of node 1 = Rs 18,000. [Lesser of the two values of Rs 20,140 and Rs 18,000.] Water Struck

Rs 10,000 Rs 12,500 Water struck 0.4 Drill up to 250 Drill up to 200 feet A Water not B struck 0.8 No water struck 1 Do not drill 0.6 2 Rs (12,500 + 18,000) Do not drill = Rs 30,500 Rs 18,000 Rs (10,000 + 18,000) = Rs 28,000

Fig. 10.6 Thus the optimal (least cost) course of action for Mr Sinha is not to drill the well and pay Rs 18,000 for water to his neighbour for five years. Example 4: A complex airborne navigating system incorporates a sub-assembly, which unrolls map of the flight plan synchronously with the movement of the aeroplane. This sub-assembly is bought on very good terms from a sub-contractor, but is not always in perfect adjustment on delivery. The sub-assemblies can be readjusted on delivery to guarantee accuracy at a cost of Rs 50 per sub-assembly. It is not, however, possible to distinguish visually those sub-assemblies that need adjustment. Alternatively, the sub-assemblies can each be tested electronically at a cost of Rs 10 per subassembly tested. Past experience shows that about 30% of those supplied are defective; the probability of the test indicating a bad adjustment when the sub-assembly is faulty is 0.8, while the probability that the test indicates a good adjustment when the sub-assembly is properly adjusted is 0.7. If the adjustment is not made, the sub-assembly is found to be faulty when the system has its final check and the cost of subsequent rectification will be Rs 140. Draw up an appropriate decision tree to show the alternatives open to the purchaser and use it to determine his appropriate course of action. Solution: The decision tree diagram for the problem is shown in Fig. 10.7.

Decision Analysis

205 O.K. As per test Assembly actually properly adjusted

Rs 0

0.7

B 0.3

Good, 0.7

Bad as per test

Rs 50 A O.K. As per test Rs 140 0.3 Rs 10 No Test 0.2 Assembly actually faulty C Bad as per test Rs 50 0.8 Rs 50

Test 1

Fig. 10.7 EMV of node A = Rs [0.7 × 15 + 0.3 × 68] = Rs 30.90. EMV of node B = Rs [0.7 × 0 + 0.3 × 50] = Rs 15. EMV of node C = Rs [0.2 × 140 + 0.8 × 50] = Rs 68 EMV of node 1 via the branch “Test” = Rs [10 + 30.90] = Rs 40.90. EMV of node 1 via the branch “No Test” = Rs 50. Optimal EMV of node 1 is Rs 40.90 corresponding to the optimal decision “Test the sub-assemblies”.

10.8

MATHEMATICAL PROGRAMMING

Besides the calculus, there are other management science techniques, which can be employed to resolve a variety of decision problems. One such technique is Mathematical Programming, which is useful whenever factors constrain the choice of status. Consider the inventory problem. If the objective is simply to minimize total cost, there are no constraints which limit out choice of strategies. If there are constraints, they might limit either the space in which inventory can be placed by the purchasing department. This being the case, it would have become a problem in constrained minimization and mathematical programming techniques could be used to find a solution. The constraints create the environment within which decision-makers strive to maximize or minimize the objective to be achieved. This is the essence of mathematical programming in constrained maximization or minimization. It becomes an intuitively appalling framework for the analysis of many types of business problems. The difficult task, however, is shouldered by the model builder, who must abstract from the environment those important elements that are to be incorporated in the mathematical model. Linear programming techniques such as simplex method, graphical method, etc. make the mathematical model to solve them.

10.9

PROBABILITY OF A DENSITY FUNCTION

The exact tensile strength of a given steel beam have value between 1.00 g/cm3 and 1.15 g/cm3 . In such cases, the pattern of occurrences be modeled by a mathematical function, f (x) which is called the probability density function of that variable. Then the probability that the variable x may take on a value between, say x1 and x2 can be computed by integration as follows. x1

P( x1 ≤ x ≤ x2 ) =

∫ f ( x) dx

x2

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Mechanical System Design

Figure 10.8 illustrates such a probability model for the tensile strength of a steel beam. x1

P( x1 ≤ x ≤ x2 ) =

∫ f ( x) dx

x2

f (x)

0.8

0.9 x1

1.0 Tensile strength (x)

Fig. 10.8

1.1 x2

1.2

Probability density function

Another basic property of probability may be learned from the above problem concerning the contractor. It has been assumed that there is no chance that more than 10 days can be lost due to in inclement weather. That is, P(delay longer than 10 days) = 0 In fact, it has been definitely assumed that there is to be either no delay, a delay of 5 days, or a delay of 10 days. That is, P(Either no delay, a delay of 5 days, or a delay of 10 days) = 1 Therefore, the summation of probability values of the above three mutually exclusive (no two events can occur simultaneously) events must be equal to 1. In general, let x1, x2 ... xn be n mutually exclusive events that could possibly occur at a chance node in the decision tree and let P(xi) denote the probability that event xi occurs; then n

∑ P( xi ) = 1

x =1

10.10

EXPECTED MONETARY VALUE (EMV)

Future cash inflow may be treated as a random variable that assumes various values with time along with some probability attached to the event that it will assume a particular value. Once the probabilities are assigned to the future events, the next step is to compute the expected monetary value. This is found out by multiplying the monetary values of the possible events by the probabilities. n

Thus

(EMV)t = At = ∑ Ait Pit i =1

...(1)

Decision Analysis

207

where Ait is the cash inflow for the ith event in the ith time period. For comparison of investment proposals, the expected NPV (Net Present Value) of each of them can then be computed as follows: Expected

NPV = ∑ ( PV of At ) − c

(2.10)

it

Consider the problem of the contractor when this evaluating the course of action he would take on the assumption that he is at decision node c. In using the EMV as a criterion, a decision-maker must always keep that approach which carries the following two implications. (a) The in decision-maker is betting on the law of averages; since the EMV of an alternative means that if he chooses this alternative many times under similar condition he would receive this much in return on the average. (b) The EMV is a completely objective measure of the value of money and implies that every rupee within a sum of money provides the same amount of satisfaction.

10.11

UTILITY VALUE

In order to provide a personal measure of the relative worth of the decision results, the monetary values may be transformed to their equipment utility value by a utility function. Each decision-maker has his own utility functions, and it must reflect his risk behavior and his outlook. Utility (utilizes) 100

50

–100

–50

50

100 Rupees

–50

–100

Fig. 10.9

Utility function

Example 5: A loss of Rs 20,000 has an equivalent utility value of –30 utilities, and a gain of Rs 1,00,000, Rs 5,000 and Rs 6,000 have an equivalent utility value of 75,5 and 6 utilities respectively. Utilities are defined as units that express personal values. By substituting these utility values for their corresponding monetary values in the decision tree, the expected utility values for alternatives A and B can now be computed as follows: Expected utility value of alternatives A = 0.7 × (–30) + (0.3) × (75) = 1.5 utilities

208

Mechanical System Design

Expected utility value of alternative B = (0.5) × (5) + 0.5 × (6) = 5.5 utilities The results are illustrated in Fig. 10.10. Thus based on the expected utility values alternative B is the better choice for this conservative investor. 0.7 1.5 Utilities

30 Utilities

Alternative 0.3

0.5

75 Utilities

5 Utilities

Alternative

0.5

5.5 Utilities

Fig. 10.10

10.12

6 Utilities

Decision analysis using expected utility value

FUNDAMENTAL OF A PROBABILITY

It is beyond the scope of this book to present a detailed discussion of the theories of probability. However, since probability is such an integral part of the method of decision analysis, it is useful to review here some of the basic fundamentals of the subject. P (No delay) = 0.4

P(Minor hurricane) = 0.5 P(Major hurricane) = 0.1 P(No delay) =

number of years with no hurricane 20 = = 0.4 Total number of years counted 50

P(Minor hurricane) =

number of years with a minor hurricane 25 = = 0.5 Total number of years counted 50

P(Major hurricane) =

number of years with a major hurricane 5 = = 0.1 Total number of years counted 50

P( x rainy days) =

number of years with x rainy days Total number of years counted

The greater the number of years of weather records used in the analysis, the more reliable will be the computed probability value. However, it is important to realize that these probability values indicate the chance occurrence of these events during the past years.

Decision Analysis

10.12.1

209

Conditional Probability

A word of explanation is necessary to understand conditional probability. Suppose a pair of dice is thrown. Here the sample space S contains 36 points (1,1), (1,2), ..., (1,6), (2,1), (2,2), ..., (2,6), ..., (6,1), ..., (6,6). Suppose that we now ask the question, if a pair of dice shows an event sum, what is the probability that this sum is less than 6? Here we are restricting the sample space to subject of points corresponding to even sum only (18 such points) and asking “Which of these possible points (outcomes) represents a sum, less than 6?”. There are four such points, viz. (1,1), (1,3), (3,1), (2,2). Since all these outcomes are equally likely, the required probability

P=

4 2 = 18 9

Observe that here we have imposed the condition that the sum x + y is even (event A) and asked the probability for x + y to be less than 6 (event B). We denote this conditional probability by the symbol P(B/A), which is read “the probability of the event B under the condition that the event A has already happened”. A little consideration will show that the probability 2/9 was obtained when we divided the number n( A ∩ B ) of points in the subset.

A ∩ B = {(x, y): x + y is even and < 6} By the number, n(A), of points in the subset A = {(x, y)}: x + y is even}, so that

P( B / A) =

n( A ∩ B) n( A ∩ B )/n( S ) = n( A) n( A)/n( S )

P( B / A) =

P( A ∩ B) P( A)

...(1)

where n(S) is the number of points in the entire sample space. Note that here it is assumed that P ( A) ≠ 0

Although the equation (1) has been obtained for the special case where all points of S have the same probability 1/36, it will be found to hold in those cases as well as where the equality of probability does not hold. Thus equation (1) constitutes our formula definition of conditional probability of B on the Hypothesis A (or for given A). 1. e.g. Multiplication theorem P( A ∩ B) = P( A) P( B / A) 2. Mutually independent P ( A ∩ B ) = P ( A) P ( B )

Theorem 1: Total Probability for Compound Events Let { A1 , A2 , ..., An } be ‘n’ partition of a sample space and let A be any event. Then

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Mechanical System Design n

P( A) = ∑ P( Ai ) P( A/ Ai ) i =1

...(1)

provided (PAi) ≠ 0; i = 1, 2, 3, ..., n. Proof: If S is the sample space, then

S=

n

∑A

i

and therefore

i =1

A = A ∩ S •A ∪

n

∑A

i

i =1

=

n

∑ ( A ∩ A ) (By distributive law) i

i =1

Since Ai ∩ Aj = φ for i ≠ j , it follows that ( A ∩ Ai ) ∩ ( A ∩ Ai ) φ for i ≠ j . n

Hence

P( A) = P ∑ ( A ∩ Ai ) i =1

n

= ∑ P( A ∩ Ai ) i =1 n

= ∑ P( Ai ) P( A/ Ai ) i =1

Theorem 2: Baye’s theorem Let the set up events { A1 , A2 , ..., An } form a partition of the sample spaces S where P ( Ai ) ≠ 0; i =1, 2, ..., n. Then for any event A for which P ( A) ≠ 0 and for 1 ≤ k ≤ n .

P ( AK / A) =

P ( PAK / A) P ( AK ) n

∑ P( A/ Ai ) P( Ai )

i =1

(First method) Can be put in the following useful form. Let S = A1, ∪ A2 ∪, ..., ∪ An where Ai are simple events. Then clearly Ai form a partition of S. If A is any non-empty subset of S, then for echo integer:

K (1 ≤ k ≤ n);

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211

P ( AK / A) =

P( PAK / A) P( AK ) n

∑ P( A/ Ai ) P( Ai )

i =1

Proof: By theorem (1) Theory of total probability for compounds events n

P( A) = ∑ P( Ai ) P( A/ Ai )

...(1)

P( A ∩ AK ) ) = P( AK ) P( A/ AK )

...(2)

i =1

Also

P( AK / A) = P( AK ∩ A) =

P( A ∩ AK ) P( A)

P( AK ) ⋅ P( A/ AK )

=

n

∑ P( Ai ) P( A/ Ai )

i =1

By Equation (1) and (2) we get,

P( AK ) ⋅ P( A/ AK )

P( AK / A) =

n

∑ P( Ai ) P( A/ Ai )

Proved.

i =1

(Second Method) Proof of Bayes’ theorem Let A and B be two events such that the occurrence of one event influences directly the occurrence of the other event. Then the conditional probability that event A would occur after knowing that event B has already occurred is denoted as P(A/B), which reads, the probability A gives B. Bayer’s theorem states that

P ( A /B ) =

P( B / A) P( A) P( B)

c a

d

b

Fig. 10.11 P ( A) =

a c

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Mechanical System Design

P( B) =

b c

P( A/ B) =

d b

P ( B / A) =

d a

P( B / A) ⋅ P( A) d /a × a /c = P( B) b /c = ∴ P ( A /B ) =

d c d × = c b b P( B / A) ⋅ P( A) d = P( B) b

Proved

(Third Method) Proof of Baye’s theorem If event A can be realized only when one of the events B and C which form a complete group of exclusive events occurs, the probability of event A is computed from the formula

P( A) = P( B) ⋅ P( A/B) + P(C ) ⋅ P( A/C ) Proof: Event A can occur if one of the compatible events (B and A), (C and A) is realized. By the theorem of additional probability, we get,

P( A) = P( B and A) + P(C and A) = P( B) ⋅ P( A/B) + P(C ) ⋅ P( A/C ) Suppose that event A has taken place. Then we can determine P(B/A) and P(C/A) P ( A) and P ( B ) = P ( B ) ⋅ P ( A /B ) = P ( A) ⋅ P ( B / A)

P( B / A) =

P ( B ) ⋅ P ( A /B ) P( A)

P( B / A) =

P ( B ) ⋅ P ( A /B ) P( B) ⋅ P( A/B) + P(C ) ⋅ P( A/C )

Substituting for P(A), P(C/A) is determined in similar fashion. This is known as Baye’s Theorem. Question 1: Bag A contains 2 white and 3 red balls and bag B contains 4 white and 5 red balls. One ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from bag B. Solution: Let event L be drawing a red ball Event M choosing bag A and Event N choosing bag B.

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213

We have

Now

P(M ) =

1 2

P( N ) =

1 2

P(L/M) = Probability of drawing a red ball from bag A

= And

3 5

P(L/N) = Probability of drawing a red ball from bag B =

5 9

The required probability = P(N/L)

=

P( N ) ⋅ P( L / N ) (By Baye’s theorem) P ( N ) ⋅ P ( L / N ) + P ( M ) ⋅ P ( L /M )

1 × 2 = 1 5 × + 2 9 =

5 9 1 3 × 2 5

25 Ans. 52

Question 2: Given the probability that A can solve a problem is 2/3 and the probability that B can solve the problem is 3/5, find the probability that (1) at least one of A and B will be able to solve the problem. (2) None of the two will be able to solve the problem. Solution: Let E1 be the event that A can solve a problem and E2 the event that B can solve the same problem.

2 3

Then

P( E1 ) =

And

P ( E1 ) = 1 −

P( E2 ) =

2 1 = 3 3

3 3 2 and P( E2 ) = 1 − = 5 5 5

1. P ( E1 ∪ E2 ) = P ( E1 ) + P ( E2 ) − P ( E1 ∩ E2 )

=

2 3 2 3 + − × 3 5 3 5

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Mechanical System Design =

13 Ans. 15

2. P( E1 ∩ E2 ) = P( E1 ) ∩ P( E2 )

=

1 2 × 3 5

=

2 Ans. 15

Question 3: There are 4 machines. The probability that a machine is in operation at an arbitrary time “t” is equal to 0.9. Find the probability that at a time “t” at least one machine is working. Solution: There are 4 machines. The probability that a machine is in operation at an arbitrary time “t” = 0.9. The probability that a machine is not in operation at time “t” = 1–0.9 = 0.1 The probability that all 4 machines are not working at “t” = (0.1) (0.1) (0.1) (0.1) = 0.0001 Hence, the probability that at time “t” at least one machine is working = 1 – 0.0001 = 0.9999 Ans. Question 4: A speaks truth in 75% cases and B in 80% cases. In what percent of cases are they likely to contradict each other in narrating the same incident? Solution: The probability that A speaks truth is

75 3 = 100 4

3 1 = 4 4

The probability that A utters lie is

1−

The probability that B speaks truth is

80 4 = 100 5

The probability that B speaks lie is

1−

1 1 = 4 5

There are two cases in which they are likely to contradict each other: 1. A speaks truth and other lie 2. A utters lie and B speaks truth The probability of contradiction in the first case is

3 1 3 × = 4 5 20

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215

And the probability of contradiction in the second case is

1 4 1 × = 4 5 5 Hence the total probability of contradiction =

3 1 + 20 5

=

7 20

= 35 % Ans. Question 5: Explain what is meant by conditional probability. Give an example of a situation where you would use a knowledge of conditional probability. UPTU 2004 Suppose we have a machine. The past experience shows that the machine is incorrectly set 25% of the time and when it is wrong it produces 30% defectives. When it is right, it produces 5% defectives. Suppose the machine has been used to produce 5 parts out of which 1 part was defective. (i) What is the probability of producing 1 defective if the machine is set correctly? (ii) What is the probability that the machine was set incorrectly and produces one defective? Solution: Conditional Probability: Multiplication theorem is not applicable in the case of dependent events. Two events A and B are said to be dependent when B can occur only when it is known to have occured (or vice versa). The probability attached to such an event is called the conditional probability. It is denoted by P(A/B), in other words, probability of A given B has occured.

P( A/B) =

P ( AB ) P( B)

We can write multiplication theorem in terms of conditional probability P(A and B) = P(B) × P(A/B) P(A and B) = P(A) × P(B/A) For three events A, B and C, P(ABC) = P(A) × P(B/A) × P(C/AB) Now let us consider the probability that the machine is incorrectly set is P(A1) The past experience shows that the machine is set incorrectly at 25% of the time P ( A2 ) =

25 1 = 100 4

and P ( A2 ) when correctly set up P ( A2 ) = 1 −

25 3 = 100 4

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Mechanical System Design

Let D be the event that a part is defective, then machine when it is set (wrong) incorrectly it produces 30 % defectives

P( D/ A1 ) =

30 3 = 100 10

When it is set (right) correctly, it produce to 5% defective

P( D/ A2 ) =

5 1 = 100 20

By using Baye’s theorem the probability of producing 1 defective if the machine is set correctly (right) (i)

P( A2 /D) =

P( A2 ) ⋅ P( D/ A2 ) P( A2 ) ⋅ P( D/ A2 ) + P( A1 ) ⋅ P( D/ A1 )

3 1 × 4 20 =  3  1   1  3     +     4  20   4  10  =

1 3

= 0.33 Ans. (ii) The probability of producing 1 defective if the machine is set incorrectly (wrong).

P( A1/D) =

P( A1 ) ⋅ P( D/ A1 ) P( A1 ) ⋅ P( D/ A1 ) + P( A2 ) ⋅ P( D/ A2 )

1 3 × 4 10 =  1  3   3  1     +     4  10   4  20  =

2 3

= 0.66 Ans.

10.13

DECISION ANALYSIS MAKING UNDER CONDITION

Decision theory provides rational approach in dealing with such problems confronted with partial, imperfect or uncertain future conditions.

Steps in Decision Theory Approach 1. List all the variable alternatives. 2. Identify the expected future events.

Decision Analysis

217

3. Construct a pay off table. 4. Select optimum decision criterion.

Decision Making Environment 1. Decision-making under condition of certainty. 2. Decision-making under condition of uncertainty. 3. Decision-making under conditions of risk.

Maximax Criterion This criterion provides the decision-maker with optimistic criterion. He finds the maximum possible pay off for each possible alternative and then chooses the alternative with maximum pay off within this group.

Table 10.5 Alternatives

Expand Construct Sub construct

States of nature (Product demand) High

Moderate

Rs 50,000 Rs 70,000 Rs 30,000

Rs 25,000 Rs 30,000 Rs 15,000

Maximum of row

Low Rs –25,000 Rs –40,000 Rs –1,000

Nil Rs –45,000 Rs –80,000 Rs –10,000

Rs 50,000 Rs 70,000 Rs 30,000

Maximax = 70,000 First choose or select maximum of row, then maximum of each.

Maximin Criterion This criterion provides the decision with pessimistic criterion. To use this criterion, the decision-maker maximizes his minimum possible pay off. He finds first the minimum possible pay off for each alternative and then chooses the alternative with maximum payoff within this group.

Table 10.6 Alternatives

Expand Construct Sub construct

States of nature (Product demand) High

Moderate

Rs 50,000 Rs 70,000 Rs 30,000

Rs 25,000 Rs 30,000 Rs 15,000

Low Rs –25,000 Rs –40,000 Rs –1,000

Maximin = Rs –10,000 First choose or select minimum of row then maximum of each

Minimum of row Nil Rs –45,000 Rs –80,000 Rs –10,000

Rs –45,000 Rs –80,000 Rs –10,000

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Mechanical System Design

Minimax Regret Criterion This decision criterion was developed by L.J. Savage. He pointed out that the decision-maker might experience regret after the decision has been made and the states of nature, i.e. events have occurred. Thus the decision-maker should attempt to minimize regret before actually selecting particular alternative. This is the strategy.

Table 10.7 Alternatives Expand Construct Sub construct

High

Moderate

Low

High

Rs 50,000 Rs 70,000 Rs 30,000

Rs 25,000 Rs 30,000 Rs 15,000

Rs –25,000 Rs –10,000 Rs –1,000

Rs –45,000 Rs –80,000 Rs –10,000

Table 10.8 Alternatives

Expand Construct Sub construct

States of nature (Product demand) High

Moderate

Rs 20,000 Rs 0 Rs 40,000

Rs 5,000 Rs 0 Rs 15,000

Minimax Regret = Rs 35,000 The required steps are provided High 70,000 – 50,000 = 20,000 70,000 – 70,000 = 0 70,000 – 30,000 = 40,000 Low –10,00 – (–25,000) = Rs 24,000 –1,000 – (–40,000) = Rs 39,000 –1,000 – (–1,000) = Rs 0 Thus the minimax regret = Rs 35,000 Ans.

Low Rs 24,000 Rs 39,000 Rs 0

Maximum of row Nil Rs 35,000 Rs 70,000 Rs 0

Rs 35,000 Rs 70,000 Rs 40,000

Moderate 30,000 – 25,000 = 5,000 30,000 – 30,000 = 0 30,000 – 15,000 = 15,000 Nil –10,000 – (–45,000) = Rs 35,000 –10,000 – (–80,000) = Rs 70,000 –10,000 – (–10,000) = Rs 0

Hurwitz’s Criterion Also called the weighted average criterion, it is a compromise between the maximax (optimistic) and maximin (pessimistic) decision criteria. This criteria is based on Hurwitz’s concept of coefficient of optimum (or steps in a Hurwitz’s concept criterion pessimism). 1. Choose an appropriate degree of optimism. 2. So that (1 − α ) represents the degree of pessimism.

Decision Analysis

219

3. Determine the maximum as well as minimum of each alternative and obtain. P = α (Maximum) + (1 − α ) × Minimum For each alternative 4. Choose the alternative that yields the maximum value of P. Where α = 0.8

Table 10.9 Alternative

States of Nature High

Expand Construct Sub construct

Rs 50,000 Rs 70,000 Rs 30,000

Moderate

Low

Rs 25,000 Rs –25,000 Rs –45,000 Rs 30,000 Rs –40,000 Rs –80,000 Rs 15,000 Rs –1,000 Rs –10,000

Expand P = α (maximum) + (1 − α ) minimum

= = = =

0.8 × 50,000 + (1 – 0.8) × (–45,000) 40,000 + 0.2 × (–45,000) 40,000 – 9000 Rs 31,000

Construct P = α (maximum) + (1 − α ) minimum

= = = =

Nil

0.8 × 70,000 + (1 – 0.8) × (–80,000) 56,000 – 0.2 × 80,000 56,000 – 16,000 Rs 40,000

Sub construct P = α (maximum) + (1 − α ) minimum

= 0.8 × 30,000 + (1 – 0.8) × (–10,000) = 24,000 – 0.2 × (–10,000) = 24,000 – 2,000 = Rs 22,000 Maximum = Rs 40,000 Ans.

Maximum of Row

Minimum of Row

P= α (max) + (1 – α ) min.

Rs 50,000 Rs 70,000 Rs 30,000

Rs –45,000 Rs –80,000 Rs –10,000

Rs 31,000 Rs 40,000 Rs 22,000

220

Mechanical System Design

Laplace Criterion This criterion is based upon what is known as the principles of insufficient reason since the probability associated with the occurrence of various events are unknown. P’s denote the payoffs, then expected value for strategy, say, s, is =

1 [ P1 + P2 + P3 + ... + Pn ] n

Table 10.10 Alternative

High Rs

Moderate Rs

Low Rs

Nil

Expected payoff

Expand

50,000

25,000

–25,000

–45,000

1000 (50 + 25 – 25 – 45) = 1250 4

Construct

70,000

30,000

–40,000

–80,000

1000 (70 + 30 – 40 – 80) = –5000 4

Sub construct

30,000

15,000

–10,00

–10,000

1000 (30 + 15 – 1 – 10) = 8500 4

Thus the alternative subcontract results in maximum average payoff Rs 8,500. Expected value criterion Example: A newspaper boy has the following probabilities of selling a magazine:

Table 10.11 No. of Copies Sold

Probability

10 11 12 13 14

0.10 0.15 0.20 0.25 0.30 1.00

Total

Cost of copy is 30 paise and sale price is 50 paise. He cannot return unsold copies. How many copies should be ordered? Solution: Payoff = 20 × copies sold – 30 copies unsold

Decision Analysis

221

Condition: Profitable Paise Possible demand and of no. copies

Probability

10 11 12 13 14

0.10 0.15 0.20 0.25 0.30

10

11

200 200 200 200 200

170 220 220 220 220

Possible stock action 12 13 140 190 240 240 240

14

110 160 210 260 260

80 130 180 230 280

Expected Profit Table Possible demand 10 11 12 13 14

Probability

Expected profit table from stocking paise 10 20 30 40 50 60 200

0.10 0.15 0.20 0.25 0.30 Total

11 17 33 44 55 66 215

12 14 28.5 48 60 72 222.5

13 11 24 42 65 78 220

14 8 19.5 36 57.5 84 205

Ans. Total expected profit = 222.5 paise. Question: Explain, in what ways decision-making models under conflict differ from models for decisionmaking under risk and uncertainty. The following matrix gives the payoffs in utilities (a measure of utility) for 3 alternatives and 3 possible states of nature. UPTU 2004

Table 10.12 Alternative

State of Nature

A1 A2 A3

S1

S2

S3

50 60 90

80 70 30

80 20 60

Which of the following alternatives are to be chosen? 1. Laplace principle 2. The maximax rule and 3. The minimax regret rule 1. Laplace principle A1 =

50 + 80 + 80 210 = = 70 3 3

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Mechanical System Design

A2 =

60 + 70 + 20 150 = = 50 3 3

A3 =

90 + 30 + 60 180 = = 60 3 3

Ans. A1 alternative (ii) Maximax Rule A1 = 80 A2 = 70 A3 = 90

Ans. A3 alternative (iii) Minimax regret rule Alternative

S1

S2

S3

A1

40

0

0

40

A2

30

10

60

60

A3

0

50

20

50

Minimax = 40 Ans. A1 Alternative

10.14

A CASE STUDY: INSTALLATION OF A MACHINERY

10.14.1

Introduction

Machine A machine is defined as combination of rigid and resistant bodies having definite motion and capable of performing some useful work. The definition of machine contains following important features: 1. A machine must be capable of doing some useful work, otherwise it cannot be called a machine. For example, a car transports passengers, pump raises water from well, a washing machine cleans clothes while a drilling machine makes holes. 2. A machine consists of a number of fixed and moving parts called “Links”. The moving parts of the machine must have controlled and constrained motion. For example, when the handle of the screw jack is rotated through one revolution, the load is raised along the axis of the screw, through a distance equal to the pitch of the screw with single start threads. 3. The various parts of the machine are interposed between the source of power and the work to be done for the purpose of adapting one to another. This concept of machine is illustrated in Fig. 10.12. The source of power can be electric motor, engine or even manual power as in case of hand operated machines. The output work can be turning in case of lathe or blanking in case of press.

Decision Analysis

223

4. A machine transforms and transfers energy. A car converts chemical energy of fuel into heat energy and finally into mechanical energy. An electric motor transforms electrical energy into mechanical energy. A generator in hydro power station transforms potential energy of water into electrical energy. 5. The terms such as motion, power, force, torque and work are predominant concepts in study of machines. Input (source of power)

Kinematic arrangement of links

Output (useful work)

Fig. 10.12 Concept of machine There are two terms that are frequently used in relation to machines, viz. mechanism and structure. A mechanism is a simplified model, frequently in the form of a line diagram, which will reproduce exactly the same motion that takes place in actual machine. The fundamental objective in case of mechanism is to achieve a desired motion. The examples of mechanism are as follows: (i) Slider crank mechanism (ii) Pantograph linkage (iii) Cam and follower mechanism, and (iv) Epicyclic gear train. A structure is also a combination of rigid and resistant bodies, but there is no relative motion between its various parts. The examples of structure are as follows: (i) Roof truss, (ii) Machine tool bed and column, and (iii) Automobile chassis. The purpose of structure is not to do some useful work, but to support external load. A structure can be moved from one place to another and is “movable” in that sense, however, there is no internal movement within the structure. On the other hand, there is always relative motion between various parts of the machine or mechanism.

Machine Design Machine design is a creation of plans for machine to perform the desired functions. The machine may be entirely new in concept performing new type of work or it may perform more economically the work that can be done by existing machine. It may be an improvement or enlargement of an existing machine for better economy and capability. Machine design is defined as the use of scientific principles, technical information and imagination in the description of a machine or a mechanical system to perform specific functions with maximum economy and efficiency. This definition of machine design contains following important features: 1. A designer uses principles of basic and engineering sciences such as physics, mathematics, stastics and dynamics, thermodynamics and heat transfer, vibrations and liquid mechanics. Some of the examples for these principles are: (i) Newton’s laws of motion

224

Mechanical System Design

2.

3.

4. 5.

(ii) D’ Alembert’s principle (iii) Boyle’s and Charles’ laws of gases (iv) Carnot cycle, and (v) Bernoulli’s principle. The designer has technical information of the basic elements of machine. These elements include fastening devices, chain, belt and gear drives, bearings, oil seals and gaskets, springs, shafts, keys, couplings and so on. A machine is a combination of these elements. The designer knows the relative advantage and disadvantages of these basic elements and their suitability application. The designer uses his skill and imagination to produce a configuration, which is a combination of these basic elements. However, this combination is unique and different in different situations, this intellectual part of selection of proper configuration is creative in nature. The final outcome of design process consists of description of the machine. The descriptions are in the form of drawings of assembly and individual components. A design is created to satisfy a recognized need for the customer. The need may be to perform specific functions with maximum economy and efficiency.

Machine design establishes and defines solutions and patent structures for problems not solved before and provide new solutions to problems that have previously been solved in a different way. Design should not be confused with the word “Discovery”. Discovery means getting the first sight of, or the first knowledge of something. For example, Columbus discovered America. We can discover what has already existed but has not been known before. On the other hand, machine design creates a machine that has not existed; instead, it is created expressly to satisfy the need of the customer.

10.14.2

Analysis of the Case Study: Installation of Machinery

The first thing to be done on receipt of any machine tool is the initial inspection, i.e. to open the pickings immediately and to check for any transport damages and shortages. Generally a packing list is enclosed, according to which all the items are checked. While unpacking, care must be taken that nothing gets lost especially with the boxes containing accessories. In case of any complaint, it should be immediately conveyed to the nearest manufacturer’s representative quoting the type of machine and serial number and copies sent to the head office, servicing center, etc. At the time of packing at the works, all bright parts of the machine are coated with rust preventives, which should be carefully removed before moving any of the sliding parts of the machine. Kerosene can be used for this purpose but the solvents that dissolve paints should be avoided. Soft cotton waste is generally used for washing away the preventive coating and dry cloth to wipe the machine dry. All bright parts should then the immediately oiled. All nipples should also be cleaned carefully to prevent any dirt from being forced in when lubricating with an oil gun. There are certain moving parts which are generally clamped to the fixed parts and these should be first unclamped; e.g. in the case of lathe the saddle is clamped to the bed side ways and, therefore, before traversing the saddle, it must be unclasped. When moving the machine for erection purpose, such parts should be kept clamped. The next step is the preparation of foundation plan. In preparing it, the location of the machine should be selected in such a way that sufficient working space is provided for the operator and floor space for workpiece to be machined and the machined workpieces awaiting removal. The foundation is best made up with damped uncrate which is fully unreeled before the machine is placed on it. The foundation plan of a lathe machine is shown in Fig. 10.13.

226

Mechanical System Design

seat in the blind holes given in each plate. The machine is leveled provisionally by means of the leveling screws making use of a sprit-level with bubble gauge sensitivity of 0.03 to 0.06 mm/m. Final leveling is undertaken only after foundation bolts are set hard in the cement grounding. The four leveling screws may be adjusted to obtain suitable transverse levels at three positions near the headstock, near the tailstock and in the middle positions. The middle screws are adjusted to achieve the desired convexity. The transverse level is to be rechecked at the three measuring positions. The longitudinal leveling is also to be carried out carefully especially for machine with intermediate supports. When the level is adjusted correctly as above, the machine is run for about 2 hours at various speeds and the levels rechecked after this period of running. If the same previous accuracy of level is not indicated, then the leveling screws are adjusted again. After ensuring that correct level is thus obtained by manipulating the leveling screws, the foundation bolts are evenly tightened taking care that previously obtained accuracy of level is being maintained. The machines again run for an hour on medium speed. The level is again checked and readjusted if necessary. Now the machine is ready for grouting.

Initial Lubrication Initial lubrication is then carried out as per the schedule supplied by the manufacturer of machine tool after thorough cleaning of the machine. All the moving parts like saddle cross and top slides, etc. are traversed along their length for a few times. The tailstock spindle is drawn in and out to spread the oil. All the levers are manipulated to ensure easy movement. All the machines are generally supplied with complete electrical equipment (main motors, switch-gear and control circuitry) and simply have to be connected to the power mains. The incoming voltage supply (voltage level), permitted voltage variation and frequency variation should be as per the recommendation of the supplier so that the motors can give guaranteed performance. For starting the machine first, the main isolating switch is to be switched on and then the spindle starting lever be turned downward and rotation of main spindle checked. If it is clockwise, it is all right. If direction of rotation is anti clockwise, simply two-phase connections be interchanged. Pressing the lever upwards makes the spindle rotate in anti clockwise direction. The machine can be stopped by pressing down the brake pedal.

Description of Operating Mechanism The following description is based on the center lathe supplied by H.M.T. Fig. 10.15 shows the various operating controls provided on the machine tool. 1. Head stock casting: The head stock casting is compact and rigid to carry the spindle assembly. The two ribs apart from contributing to rigidity, carry the pulley on a sleeve, which runs in ball bearings, housed in the ribs. Since the sleeve carrying the pulley is independently mounted concentric with the main spindle but clearing it, the belt tension is not directly transmitted to the main spindle. A counter shaft below the main spindle carries a cam, which operates a plunger pump for lubrication. The main spindle runs on a high precision double row taper roller bearing in the front and a single row taper roller bearing in the rear end. The outer race of the rear bearing spring loaded to allow for thermal expansion of the main spindle. 2. Speed box unit: Both speed box and main motor are mounted on a heavy cast iron plate which can be easily assembled into the cabinet leg of the lathe as a single unit and do not offer

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227

any difficulty in extracting it out for repair and maintenance work. The first shaft of the speed box is directly coupled to the motor. 3. Pedal brake: A brake pedal (3) shown in Fig. 10.14 running the length between the cabinet and end legs ensures sensitive foot control of the main spindle. The pedal when pressed actuates a limit switch to cut off power supply to the main motor and in addition to this applies its brake band over the brake drum on the motor shaft thereby bringing the main spindle to a quick stop.

12

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Fig. 10.14 4. Inching: The push button control “inching” is provided for easy shifting of gears. The control push button for this is housed on the front side of the headstock. 5. Feed gear box: The gear box is of totally enclosed design. A piston pump located inside the feed box and driven by an eccentric on one of the shafts raises oil to a reservoir at the top. From there, oil flows by gravity on the rotating gears. 6. Feed selection: Feed selection is done by the manipulation of feed selector levers 6 to 10 shown in figure. The lever 9 has two positions neutral and engaged, power feed is obtained only when the lever is in the engaged position. 7. Screw cutting: The selection of pitch is done by manipulation of the levers 6 to 10 shown in figure. The thread clad-figure shows the corresponding positions of the lever for any particular pitch (metric, inch, module or diametric pitches).

228

Mechanical System Design 8. Thread chasing dial: The thread chasing dial is provided for re-aligning the position on the lead screw and thereby the position of the tool in subsequent cuts when the split not is disengaged at the end of the thread and the saddle is brought back for re-engagement of the split unit. 9. Micrometer and screw cutting shop: The cross slide movement incorporates a micrometer dead stop that can be set in either direction and it can be adjusted with a dial calibrated to a least count of 0.005 mm. A worm on which this dial is fixed is engaged to mesh with a worm wheel on the crossfeed screw, by means of a lever provided for the purpose. A few runs of the hand wheel in either direction brings the cross feed screw in engagement with the worm wheel and stop dead. 10. Quick change tool post: The quick change tool post is mounted on the top slide and holders can be clamped within seconds by turning the level A which work on eccentric locking device giving high rigidity to the tool holder. This kind of tool post and present tool holder arrangement eliminates loss of the time. 11. Tailstock: The tailstock is of sturdy design and can be firmly clamped to the bed by means of two easily accessible hexagonal nuts. The taper bore in the spindle is Morse 5. The spindle moves forward when the hand wheel is rotated in the anticlock-wise direction. The linear movement of the spindle is read on the graduated sleeve on the hand wheel. 12. Taper turning attachment: The taper turning attachment to the machine is supplied as a special accessory. For taper turning the two nuts are loosened, the guide is adjusted to the required angle and clamped by tightening the nuts. 13. Lubrication: Regular lubrication with carefully selected lubricants is an essential factor in satisfactory functioning of the machine tool for a long time. Before putting the machine into operation, the various oil reservoirs should be filled with oil up to the oil sight glass. The various parts should be lubricated as per the lubrication plan supplied by the manufacturer of the machine tool. The lubricant represented in the case of lathe, for head stock feed box, speed gear box gears and general lubrication, is mobil oil heavy medium of Indian Oil Company, and for grooved pulley bearing and, motor bearing is Mobile greases BRB of Indian Oil Company. 14. Lubrication of head stock: Initially the oil is filled in the head stock after removing the screw cap provided on the top of the head stock. The oil lever is visible on the right glass generally provided on the rear side of the lathe. Oil can be drained through a drain plug provided at the bottom. Usually the oil is pumped by a piston pump operated by a cam and the oil is first passed through a wire mesh filter designed to trap all impurities and then to the front bearing and also to the various rotating head stock parts. 15. Lubrication of speed box: Speed box is lubricated by a continuous splash of oil. The speed box is generally housed in the cabinet leg, directly below the head stock and to gain access to it the rear cover of the cabinet leg is removed. The screw plug on the speed box is removed and oil is filled in, up to the set level in oil sight glass. The speed box pulley bearings are generally filled with grease while assembling at the work. To replenish the pulley with grease, and plates are dismantled cleaned well and refilled with bearing greases having a melting point greater than 140 °C. 16. Feed box: The lubrication in the feed box is effected by a piston pump driven by an eccentric fitted to one of the rotating shafts. Oil is pumped to a reservoir on the top of the feed box and

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from there oil is fed through pipes and channels, which ensure oiling, by capillary action through wicks. 17. Saddle and apron: These are lubricated from a hand pump fixed at the right hand side of the saddle. Oil is drawn from the reservoir and is pumped to various points (bed slide ways, cross feed unit, apron gears). The screw plug on the top of the saddle is removed for filling oil into the reservoir. The hand pump should be operated 4 to 6 times every day before operating the saddle movement. 18. Worm housing and lead screw end bearing: Worm housing is lubricated through the oil nipples under the levers of the apron with an oil gun. The end bearing of the lead screw is wick lubricated from an oil pump in this bearing housing. Oil is filled in every three days into this pump by removing the screw plug on the top of the end bearing.

Maintenance 1. General: Careful maintenance will increase the life and efficiency of the machine. A thorough weekly cleaning is very necessary for any machine in continuous use. It plays to check and adhere to the lubricating instructions from time to time and a defect however small it is, should be immediately rectified. Otherwise this will lead to a major breakdown of the machine, resulting in loss of production and high cost of repairs. It is necessary to check the electrical outfit once in every six months and clean particularly the contact points. Melting or burnt strips should be removed and the links lubricated carefully. The ball bearings of the motor require cleaning and refilling with grease once in a year. To perform this: (a) The covers are removed (b) The bearing shields are removed to leave them both accessible on the shaft. When replacing the damaged bearings, the new bearings should be oil heated and mounted on the shaft and greased. It is important to prevent any damage to the parts. 2. Adjustment of main axis in horizontal plane: The head stock is secured on the lathe bed by five fixing studs and nuts. The studs on the right side of the head stock acts as pivot. The other four ends have 1 mm clearance around them in the respective head stock holes. The headstock can be moved around the pivot for aligning to spindle axis in the horizontal plane. 3. Adjusting feed shaft clutch: To feed shaft drives is thorough control friction clutch which is spring loaded and is set at the works. If the longitudinal or transverse feed drive is overloaded, the clutch will slip. In case of clutch adjustment, the nut c is set that a weight of 25 kg attached to the hand wheel of the apron, just stops the feed movement by slipping of the clutch. Replacement of V belt: For replacing the V belts the main spindle has to be removed and then the pulley sleeve is to be removed. For removing the main spindle (a) The top cover of the headstock is removed. (b) The 8 socket head cap screws B fixing the front cover m is removed. (c) From the front gear the locking wire b and grub screw C are removed. (d) The rear nut D is unscrewed and removed and the collar is taken out.

230

Mechanical System Design (e) The two gear blocks are blocked with wooden pieces at E and F against the nearest fixed support. (f) A mild steel rod of φ 25 mm threaded at both ends along with a stirrup with a hole as shown in the diagram is used for extracting the main spindle. Removing the pulley sleeve (a) The locking wire and the screw G are removed. (b) The ring nut and the locking washer I are then removed. (c) From the top the plug J (bearing positioning plug) is removed. (d) The retaining rings R and the deflector L are dismantled. (e) With a mallet or plastic hammer, the sleeve is tapped out towards the right till the front bearing is clear. (f) Now the sleeve is tapped out towards the right until it is free. Caution: Rough handling must be avoided while extracting and replacing. The spindle and pulley sleeve should release without any difficulty. The spindle can be removed and replaced without altering the bearing adjustments. Re-assembly of the main spindle: When the main spindle is being re-assembled the following points are to be noted: (a) The bearings should be thoroughly cleaned. (b) The reduction gear is greased lightly to ease refilling on the spindle. (c) Care is taken to see that the direct drive dog N does not join against the sleeve. This is done by moving the lever operating the front gear continuously before. This is very important. (d) The alignment of the outer face of the front bearing should be checked before fitting it in the housing. (e) The posting screw O should enter the notch in the collar freely. 5. Adjustment of main spindle bearing: These high precision bearing are carefully adjusted at the works. The machine is test-run and performance tests are carried out in the test floor. The final testing before a machine is dispatched normally, this adjustment should remain constant during several years of uninterrupted running, even at the highest permitted speeds. Temperature Check: The abnormal increase in temperature of the main bearings is a reliable guide to check the bearing adjustment as well as the lubrication. To observe the temperature rise, the cavity provided in the top of the front plate is to be filled with oil and industrial mercury thermometer is placed in it. A non-stop run of ninety minutes at a spindle speed of 1600 rpm should cause a rise in temperature of 40 °C above the ambient temperature. A higher temperature increase indicates that the spindle bearings are constrained due to over-tightening. 6. Replacing of broken spear pin lead screw: The lead screw is connected to the feed box through a safety shear pin coupling. When the pin is sheared due to any overload, the lead screw will not transit motion from the feed box. This pin can be seen by sliding out the knurled collar E span found on the left end of the lead screw. 7. Adjustment of V-belt tension: Belt tension is adjusted by slackening the nut A, completely allowing the muter and speed box weight to tension the belts. Now the nut A is hand tightened

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V

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M

Stirrup

I E D

M.S. Bar

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Wooden piece

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Wooden piece

and nut B is fully tightened by spanner. The speed box base plate is now held rigidly between the nut A and B.

Fig. 10.16 Main spindle dismantling 8. Adjustment of brake: The bearing force could be set as follows: Slacken the nut C. Keeping the brake pedal pressed down firmly, lighten the nut D, so that the brake band grips the drum. Now the pedal is released. To suit the degree of braking desired the nut C is slightly tightened.

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C B

A

Fig. 10.17

Exercise

Adjustment of V-belt tension and brake

10 . . .

1. Explain elements of a decision problem on the basis of decision analysis. 2. Explain, in what ways decision-making models under conflict differ from models for decision-making under risk and uncertainty. UPTU 2004 3. What is difference between decision problem and decision model? Briefly explain in the language of decision analysis. 4. Why do you use probability of a density function. Where is it applicable? 5. What is the meaning of EMV? Briefly explain it. 6. What is utility value function? What is the importance of utility value in the form of decision analysis? 7. What is the fundamental probability? Write down Bayes’ theorem. 8. Explain what is meant by conditional probability. UPTU 2004 Give an example of a situation where you would use a knowledge of conditional probability. 9. Write a short note on: (a) Decision model (b) Decision analysis (c) Bayes’ theorem (d) Conditional probability (e) Fundamental of probability (f) Probability of a density function (g) Utility value (h) EMV 10. Briefly explain expected monetary value with suitable example. UPTU 2004 11. Explain clearly the various ingredients of decision problem. What are the basic steps of a decision making process?

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A Case Study 1. What is the main factors of a case study in the installation of a machinery? 2. Provide the description of an operating mechanism. 3. Write a short note on: (a) Headstock (b) Speedbox unit (c) Inching (d) Feed gear box (e) Thread chasing dial (f) Tail stock 4. What are the steps of maintenance of a machinery? 5. How can re-assemble of the main spindle done? 6. Write down of the V-belt tension and brake of adjustment. 7. Explain lubrication, stock, lubrication of head stock and feed box.

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Mechanical System Design

11 System Simulation 11.1

INTRODUCTION

Simulation involves the modeling of a complex situation into a simpler and more convenient form that can be studied in isolation without the troublesome complex side effects that usually accompany a real engineering situation. The purpose of the simulation is to explore the various outputs that might be obtained from the real system by subjecting the model to environments that are in some way representative of the situations it is desired to understand. Simulation invariably involves the use of the computer to perform often laborious computations and to follow the dynamics of the situation. Simulation generally refers to using a digital computer to perform experiments on a model of real system. These are experimental, to aid in its design, to see how the system might lead to change in its structure. Simulation is particularly appropriate to situations in which the size or complexity of the problem makes the use of optimizing techniques difficult or impossible. The job shops, which are characterized by complex queuing problems, have been studied extensively via simulation, as have certain types of inventory layout and maintenance problem. The designers and analysts in physical sciences have long used the technique of simulation and it promises to become an important tool for tackling the complicated problems of managerial decisionmaking. Scale models of machines have been used to simulate the plant layouts and models of aircrafts have been tested in wind tunnels to determine their aerodynamic characteristics. Simulation, which can appropriately be called management laboratory, determines the effect of a number of alternate policies without disturbing the real system. It helps in selecting the best policy with the prior assurances that its implementation will be beneficial. Probably John Von Neumann and Stan Slaw Ulam after studying the tedious behavior of neutrons in a nuclear shielding problem, which was too complex for mathematical analysis, made the first important application of simulation. With the remarkable success of the technique on neutron problem, it became popular and found many applications in business and industry. Development of digital computer in early 1950s is further responsible for the rapid progress made by the simulation techniques. The range of simulation application varies from simple queuing models to models of large integrated systems of production.

System Simulation

11.1.1

235

When to Use Simulation?

In the foregoing chapters we have discussed a number of operations research tools and techniques for solving various types of managerial decision-making problems. Techniques like linear programming, dynamic programming, queuing theory, network models, etc., are not sufficient to tackle all the important managerial problems requiring data analysis. Each technique has its own limitations. Linear programming models assume that the data do not alter over the planning horizon. It is a one time decision process and assumes average values for the decision variables. If the planning horizon is long, say 10 years, the multiperiod linear programming model may deal with the yearly averaged data, but will not take into account the variations over the months and weeks, with the result the month to month and week to week operations are left implicit. Other important limitation of linear programming is that it assumes the data to be known with certainty. In many real situations, the uncertainties about the data are such that they cannot be ignored. In case the uncertainty relates to only a few variables, the sensitivity analysis can be applied to determine its effect on the decision. But, in situations, where uncertainty pervades the entire model, the sensitivity analysis may become too cumbersome and computationally difficult to determine the impact of uncertainty on the recommended plan. Dynamic programming models, however, can be used to determine optimal strategies, by taking into account the uncertainties and can analyse multiperiod limitations of linear programming. But it has its own shortcomings. Dynamic programming models can be used to tackle very simple situations involving only a few variables. If the number of state variables is a bit larger, the computation task becomes quite complex and involved. Similar limitations hold good for other mathematical techniques like dynamic stochastic models such as inventory and waiting line situations. Only small-scale systems are amenable to these models; moreover, by making a number of assumptions the systems are simplified to such an extent that in many cases the results obtained are only rough approximations. From the above discussion, we conclude that when the characteristics such as uncertainty, complexity, dynamic interaction between the decision and time intervals, combine together in one situation, it becomes too complex to be solved by any of the techniques of mathematical programming and probabilistic models. It must be analyzed by some other kind of quantitative technique, which may give quite accurate and reliable results. Many new techniques are coming up, but, so far, the best available is simulation. In general, the simulation technique is a dependable tool in situations where mathematical analysis is either too complex or too costly.

11.1.2

What is Simulation?

Simulation is an imitation of reality. A children cycling park, with various crossings and signals, is a simulated model of the city traffic system. In the laboratories, a number of experiments are performed on simulated models to determine the behavior of the real system in true environments. A simple illustration is the testing of an aircraft model in wind tunnel from which we determine the performance of the actual aircraft under real operating conditions. Planetarium shows represent a beautiful simulation of the planet system. Environments in a geological garden and in a museum of natural history are other examples of simulation. In all these examples, it has been tried to imitate the reality to see what might happen under real operation conditions. This imitation of reality, which may be in the physical form or on the form of mathematical equations, may be called simulation.

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The simple examples cited above are of simulating the reality in physical form, and are referred to as (environmental) analogue simulation. For the complex and intricate problems of managerial decision making, the analogue simulation may not be practicable, and actual experimentation with the system may be uneconomical. Under such circumstances, the complex system is formulated into a mathematical model for which a computer program is developed, and using high-speed electronic computer solves the problem, and hence it is named as computer simulation or system simulation. With this background, it will now be in order to define simulation. According to one definition “simulation is a representation of reality through the use of a model or other device which will react in the same manner as reality under a given set of conditions.” Simulation has also been defined as “the use of a system model that has the designed characteristics of reality in order to produce the essence of actual operation.” According to Donald G Malcolm, a simulated model may be defined as one which depicts the working of a large scale system of men, machines, materials and information operating over a period of time in a simulated environment of the actual real world conditions. The example of the simulation of an engineering system or process by mathematical modeling and computer simulation would be the simulation of a traffic control problem or the solidification of large steel casting.

11.2

SIMULATION CONCEPT

The role of models in system analysis is illustrated in Fig. 11.1. Because engineering systems are so complex, a system model is first constructed to represent the real system and its environment. The model should include all the relevant system components and clearly define the component interrelationships, as well as indicate the constraints within those imposed on the system. The designers have complete control over the components, structure, and constraints within their model. It is with this model that they test their designs and study the behavior of the system under various conditions.

ata

o acti bstr

A System environment System

Ap

pli

Solve , man ip tests ulate

nd

Real world

Model: Analytical, iconic, or analog

ca

tio

n Results: Optimum design; System understanding

Fig. 11.1

Role of models in systems analysis

Simulation is the process of conducting experiments with a model of the system that is being studied or designed. It is a powerful technique for both analyzing and synthesizing engineering systems. In an analysis problem, the system model is generally fixed and the objective is to determine the system

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237

response when a set of input variables is allowed to take on different values. The simulation process is then basically an iterative procedure and may be described as an input-output study with feedbacks provided to guide the changes in the input parameters, as illustrated in Fig. 11.2. The inputs define the set of events and conditions to which the system can be subjected in the real world and the outputs predict the system response. By studying the outputs at the end of each iteration, the designer learns more and more about the system and may then use his newly acquired knowledge to define new sets of inputs to be processed through the model. Bridge model

Input variables

Geometry of structure Interaction of forces Member characteristics size, strength, shape

Loading, wind, current flow, earthquake, etc.

Output system response

Feedback To define the next set of values for the input variables

Fig. 11.2

Analysis by simulation

As an example, consider the problem of determining the failure conditions of a steel bridge. The simulation model may be in the form of a set of mathematical equations relating the interaction of the forces among the members, the geometric structure of the bridge, the size and tensile strength of the members, and so on. The input variables may include loading conditions, velocity and direction to wind, the flow velocity of water and debris in the channel and the occurrence of earthquakes. For each specific set of combination of the input parameters, the model may be used to determine the stresses in the members and the deflection of the structure, which provide a direct measure of the performance of the system under the given set of conditions. In system synthesis, the designer is interested in determining how the system components can best be put together so that the system can meet the performance standard. In this case, the system model itself is a variable, but a set of input-output characteristics has been specified as a design standard. The simulation process is again an iterative procedure, but the output of a simulation study is now used to decide which system parameter can best be changed in order to improve the performance of the system. As shown in Fig. 11.3. the feedback is now directed to changes within the model itself. For example, the problem may be to choose an optimum combination of sizes and strength of the total structure, its weight, as well as its cost if one design fails to meet the performance standard according to one simulation result. These system parameters may be changed and the simulation repeated until a set of feasible alternatives have been established and an optimum solution is identified. The major advantage of the simulation approach lies in the fact that once a simulation model has been constructed, it may be used for both system analysis and synthesis and to test the design under a wide spectrum of environmental conditions. Moreover, after the design has been completed and the system implemented in the real world, the simulation model may be used to locate the sources of unpredicted system problems and to plan for system improvement.

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Mechanical System Design Bridge model Input variables

Geometry of structure Interaction of forces member characteristics, size, strength, shape

Output system response Member stresses, deflection, etc.

Feedback To change the attributes of system components

Fig. 11.3

Synthesis by simulation

11.3 SIMULATION MODELS—ICONIC, ANALOG, AND ANALYTICAL Simulation models can take many forms and can be of many different levels of complexity. A good model should represent the characteristics of the system so that the problem under consideration can be solved. Simulation models can be broadly grouped into three types: 1. Iconic 2. Analog 3. Analytical. 1. Iconic: Iconic models are physical replicas of the real systems on a reduced scale. This type of model is common in engineering. In aircraft design, wind tunnels are used to simulate the environment around an aircraft, in flight. By subjecting a model of the aircraft to a well chosen range of aerodynamic conditions, the designer can gain insight into the performance characteristics of the design. In the design of large engineering structures, such as skyscrapers, dams, bridges and airports, three dimensional architectural models are often prepared to provide a realistic view of the design. Such models are extensively useful both as a design tool and as a visual aid in presenting the project to the interested public. At the University of Illinois for example, an indoor watershed experimentation system is present. Iconic model looks like the original; iconic model is a physical model such as used in globes. 2. Analog: Simulation is used to determine the response of reinforced concrete structures to earthquake shocks. Such a simulation model, in which the real system is modeled through a completely different physical media, is called an analog model. Analog models behave original, e.g.—man, machine, chart or germination structure. 3. Analytical model: In problems in which the characteristics of the system components and system structure can be mathematically defined, an analytical model constitutes a powerful simulation tool. However, to construct an analytical model requires that fundamental properties of the system components and their interactions be understood. With the availability of high-speed, large memory electronic computers, analytical models are becoming versatile design aids in all disciplines of engineering. Example: Analytical simulation models are playing an increasingly important role in air pollution control.

System Simulation

11.4

239

WAITING LINE SIMULATION OR QUEUING THEORY

Waiting line simulation is an (operation research) technique, which aids the manager in decision makinginvolving the establishment of service facilities to meet irregular demands. Cost problem arise when there are more service facilities available than are needed, or when too few facilities are available and consequently long waiting form. Quening theory is applied to any simulation producing a felt need to balance the cost of increasing available service against the cost of letting unit wait to arrive at the best number of service facilities the manager and operation research must first determine.

11.4.1

Benefits/Advantages of Waiting Line (Queuing) Theory

The benefits of queuing theory are as follows: 1. Queuing theory provides models that are capable of influencing arrival pattern of customers or determines the most appropriate amount of service or number of service stations. 2. Queuing models help in striking balance between the two opportunity costs to achieve near optimisation or correct ratio of waiting cost and service costs. 3. Queuing theory attempts to formulate, interpret and predict purposes of better understanding of the queues and for the scope to introduce remedies such as adequate service with tolerate waiting.

11.5

SIMULATION PROCESS

The example of a concrete plant operation demonstrated the general procedure of system simulation. The major procedure of system simulation study may be summarized as follows: To illustrate the method of simulation in general and that of analytical simulation in particular, consider a problem involving the operations of a company producing ready-mixed concrete. The company now owns a batching plant to mix the appropriate quantities of cement, sand, gravel, water, and special additives to produce concrete mixes that are then loaded into delivery trucks for delivery to the customer. The company now owns five delivery trucks. Their problem is to determine the optimum number of delivery trucks that can be added to their fleet so that its batching plant can be fully utilized.

11.6

PROBLEM DEFINITION

A simulation study is usually conducted to serve the following primary objectives. 1. To measure the system response under a wide range of system inputs. 2. To measure the system response when the system components and their inter-relationship themselves are altered. Vital to clearly defined input parameters. 3. System model construction. 4. Solution process. 5. Analyzing simulation results. 6. System input model construction.

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11.7

Mechanical System Design

INPUT MODEL CONSTRUCTION

Models must be constructed to simulate the behavior and pattern of the input parameters in the real system. The purpose of these models is to generate fictitious input data in the simulation process. The models may be composed of mathematical functions, logical steps, or numerical data. They are usually constructed from a set of sample data collection from the real system. Solution Process The solution process comprises two major cyclic steps, as shown in Fig. 11.4. In the inner cycle, the system response is measured with respect to a specifics of values for the design parameters within the system model. In order to test the design rigorously under the complete feasible range of the input parameters, a large number of independent solutions must be connected within this cycle. Specify design parameters

Generate input data

System solution

System response

Fig. 11.4 The solution process in simulation

11.8

LIMITATIONS OF SIMULATION APPROACH

The simulation approach provides a powerful tool for analyzing complex systems that cannot be easily studied by any other means. However the accuracy and reliability of the simulation results depend largely on how well the simulation model can truly represent the response characteristics of the system in the real world. On the other hand, the validity of the models and the reliability of the results are extremely difficult to evaluate because of the very complexity of the system. It is a advisable, therefore, that the simulation results be always analyzed with a certain degree of reservation, and that the response of the system in the real world be continuously monitored so that the feedback from the real world may be used to update and validate the simulation models. Because of the above limitations, the simulation approach should, wherever possible be supported concurrently by a theoretical study of the system. The latter serves to provide a hypothesis or simplified mathematics models of the system response characteristics. The results from the two approaches then serve as mutual checks. Moreover, even a simple theoretical analysis can help to narrow the values of the input and design parameters to test in the simulation process and thus effectively reduce the simulation calculations.

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Simulation Methodology

Start

Define Problem

Construct Simulation Model

Specify Values of Variables and Parameter

Procedure Simulation

Evaluate the Result

Evaluation

Propose New Experiment

Stop

Fig. 11.5 In spite of all the advantages claimed by the simulation technique, many operations research analysts consider it a method of last resort and use it only when all other techniques fail. If a particular type of problem can be well represented by a mathematical model, the analytical approach is considered to be more economical, accurate and reliable. On the other hand, in very large and complex problems, simulation may suffer from the same deficiencies as other mathematical models. In brief, the simulation technique suffers from the following limitations: 1. Simulation does not produce optimum result, when the model deals with uncertainties; the results of simulation are only reliable approximations subject to statistical errors. 2. Quantification of the variables is another difficulty. In a number of situations, it is not possible to quantify all the variables that affect the behavior of the system. 3. In very large and complex problems, the large number of variables and the inter-relationships between them make the problem very unwieldy and hard to program. The number of variables may be too large and may exceed the capacity of the available computer.

242

Mechanical System Design 4. Simulation is by no means a cheap method of analysis. In a number of situations, simulation is comparatively costlier and time consuming. 5. Other important limitations stem from too much tendency to rely on the simulation models. This results in application of the technique to some simple problems, which can more appropriately be handled by other techniques of mathematical programming.

11.9

SIMULATION PROGRAMS AND LANGUAGES

The efficiency of programming and execution of a simulation project depends upon the programming language used. In addition to the general-purpose languages such as FORTRAN and PL1, a large number of specialized computer languages are used to write the simulation programs. FORTRAN, being highly general in nature, can be used for any simulation project. Being well known and commonly available on computer systems, FORTRAN is quite often used to write the simulation programs. It is generally considered to be more efficient in computer time and storage requirements. However, programming in FORTRAN is more difficult and time consuming, as compared to the special simulation languages. When the complexities of the simulation project increase, the book keeping of the intricate details of the simulation becomes difficult and this makes the programming in FORTRAN harder. Thus for realistic situations, simulation programs should be written in specialized simulation languages, which are designed to meet the following objectives: l

l

l

l l

To conveniently describe the elements, which commonly appear in simulation, such as the generation of random variable for most of the statistical distributions. Flexibility of changing the design configuration of the system so as to consider alternate configurations. Internal timing and control mechanism for book keeping of the vital information during the simulation run. To obtain conveniently the data and statistics about the behavior of the system. To provide simple operational procedures, such as altering the initial state of the system, and kind of output data to be generated, etc.

GASP and SIMSCRIPT are two widely used general simulation languages, which can easily do the job of FORTRAN or PL1. These are FORTRAN based languages and hence the knowledge of FORTRAN is a pre-requisite for learning GASP and SIMSCRIPT. The most commonly used simulating language is GPSS (General purpose simulation system), which was developed by IBM. It is easy to learn and incorporates all the features, which are unique to simulation. GPSS is a problem-oriented language, but has a wide range of applications. It employs the next event incrementing time flow mechanism and uses integral time units. The calculations in integer arithmetic help to keep the round of errors to minimum. In GPSS, the system to be simulated is flow charted in the form of block diagrams, and the blocks are then written in GPSS statements. The simulation programming languages are very economical in respect of the users programming time, though they take a slightly larger CPU time in execution of the program. Some of the many simulation languages are, DYNAMO, SIMPAC, SIMULATE, SIMULA, CSMP, GSP, ESP and CSL. Simulation programs can be categorized as general purpose and special purpose. General-purpose software is really a language that allows programmers to build their own models.

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Examples: SLAM II, SIMCRIPTII.S, SIMAN, GPSS\H, GPSS\PC, PC-MODEL, RESQ, etc. Special purposes software like Lotus programs are specially beneficial to simulate specific application, such as MPA, simfactory, etc.

11.9.1 Desirable Features of Simulation Software 1. 2. 3. 4. 5. 6. 7. 8. 9.

11.10

Be capable of being used indirectly as well as along complete runs. Be user friendly and easy to understand. Allow modules to be benefit and then connected. Allow users to write and incorporate their own routines. Have building blocks that contain benefit in command such as speciality analysis or decision rules of where to go next. Output specifics such as cycle times, utilization and wait times. Allow a variety of data analysis alternative for both input and output data. Have animation capabilities to display graphically the product flow through the system. Permit interactive debugging of the model so the user can trace flows through the model easily and find errors.

ADVANTAGES AND DISADVANTAGES OF SIMULATION

Advantages: The simulation technique, when compared with the mathematical programming and standard probability analysis, offers a number of advantages over these techniques; a few important among them can be summarized as follows: 1. Many important managerial decision problems are too intricate to be solved by mathematical programming and experimentation with the actual system; even if possible, is to too costly and risky. Simulation offers the solution by allowing experimentation with a model of the system without interfering with the real system. Simulation is, thus, often a bypass for complex mathematical analysis. 2. Through simulation, management can foresee the difficulties and bottlenecks, which may come up due to the introduction of new machines, equipment or process. It, thus, eliminates the need of costly trial and error methods of trying out the new concept on real methods and equipments. 3. Simulation has the advantage of being relatively free from mathematics and, thus, can be easily understood by the operating personnel and non-technical managers. This helps in getting the proposed plans accepted and implemented. 4. Simulation models are comparatively flexible and can be modified to accommodate the changing environments of the real situation. 5. Computer simulation can compress the performance of a system over several years and involving large calculations into a few minutes of computer running time. 6. The simulation technique is easier to use than mathematical models and is considered quite superior to the mathematical analysis. 7. Simulation has advantageously been used for training the operating and managerial staff in the operation of compiled plans. It is always advantageous to train people on simulated models before putting into their hands the real system. Simulated exercises have been developed to impart the trainee sufficient exercise and experience. A simulated exercise familiarizes the

244

Mechanical System Design

8. 9. 10. 11. 12. 13. 14. 15. 16.

trainee with the data required and helps in judging what information is really important. Due to his personal involvement into the exercise, the trainee gains sufficient confidence, and moreover becomes familiar with data processing on electronic computer. Simulation can be used to analyze transient conditions whereas mathematical techniques usually cannot. Many standard packages, covering a wide range of topics, are available commercially. Simulation answers a lot of questions. Developing a model of a system leads to a better understanding of the real system. Time can be compressed in simulation. Simulation does not describe ongoing activities of the real system. Simulation is far more general than mathematical models and cannot be used where conditions are not suitable for mathematical models. Simulation can be used as a game for training experience. Simulation provides a more realistic specification of a system than mathematical analysis.

Disadvantages: 1. While a great deal of time and effort may be spent to develop a model for simulation, there is no guarantee that model will provide good answers. 2. There is no way to prove that a simulation model’s performance is completely reliable. 3. Depending on the system to be simulated, building a simulation model can take anywhere from an hour to 100 work years. Complicated system can be very costly and take a long time. 4. Simulation may be less accurate than mathematical analysis because it is randomly based. If a system can be represented by a mathematical model, it may be better to use than simulation. 5. A significant amount of computer time may be needed to run complex models. 6. The technique of simulation, while making progress still lacks a standardized approach. Therefore, model of the same system built by different individuals may differ widely.

11.11 MONTE CARLO METHOD The Monte Carlo method of simulation owes its development to the two mathematicians, John Von Neumann and Stan Slaw Ulam, during World War II when the physicists were faced with the puzzling problem of behavior of neutrons. How far would neutrons travel through different materials? The hit and trial experimental solution would have been very costly and time consuming, and the problem was too complicated for theoretical analysis. The two mathematicians suggested a solution by submitting the problem to a roulette wheel or wheel of chance. The basic data regarding the occurrence of various events were known, into which the probabilities of separate events were merged in a step-by-step analysis to predict the outcome of the whole sequence of events. The technique provided an approximate but quite workable solution to the problem. The mathematical techniques they applied had been known for many years, but it was at this stage that it was given the name Monte Carlo. With the remarkable success of the techniques on neutron problem, it soon became popular and found many applications in business and industry. At present it forms a very important tool of operation researcher’s tool kit. Monte Carlo technique has been used to tackle a variety of problems involving stochastic situations and mathematical problems, which cannot be solved with mathematical techniques and where physical experimentation with the actual system is impracticable. The stochastic situations are usually a long

System Simulation

245

sequence of probabilistic events or steps. We may be able to write mathematical formulae for probability of a particular event, but to write a mathematical relationship for the probabilities of all events in the sequence is a difficult task. In contrast to mathematical modeling where the results of the analysis yield a direct and overall solution to the problem, in simulation, the behavior of the system is observed over a sufficiently long period of time, and in the process, the relevant information is collected. The system is first described by listing the various events in the order of their occurrence. An event representing a point in time signifies the end of one or more activities and the beginning of the next activity. As each event occurs, certain actions are taken, resulting in the generation of new events, which are further, considered in sequence. For example, the arrival of a customer at a service facility is the occurrence of an event, and action taken depends upon the availability of facility, which may be free or occupied. If free, service begins, if occupied, wait in line, which further generates new activities. If service begins, compute service time; and if to wait in queue, compute waiting time and length of queue, etc. The process is repeated for the next arrival. The experimentation is preformed on a simulated model of the real system. It is a sort of sampling technique in which, instead of drawing samples from a real population, the samples are drawn from a theoretical equivalent of the real population. By making use of roulette wheel or random numbers, Monte Carlo approach determines the probability distribution of the occurrence of the event under consideration, and then samples the data from this distribution. The procedure and the concept of Monte Carlo technique can well be illustrated with the help of examples. A few very simplified examples have been solved here for this purpose. Example 1: Find the value of π experimentally by simulation. Solution: Draw the coordinate axes OX and OY. With center O, draw an arc PR of unit radius as shown in Fig. 11.6 and complete the square OPQR. Equation of the circle is x 2 + y 2 = 1. From random number Table 11.1, select any two random numbers, say 0.2068 and 0.7295 (first two four-digit numbers from the second column) and let x = 0.2068 and y = 0.7295. Plot the point P1 (0.2068, 0.7295). Obviously, if x 2 + y 2 = 1, P1 will lie inside or on the circle but if x 2 + y 2 > 1, the point P1 will lie outside the circle but within the square. In this manner, hundreds or thousands of pairs of random numbers are selected and it is ascertained whether the points representing them lie in/on the arc or beyond the arc but inside the square. Suppose N is the total number of points considered, out of which π lie in/on the arc. Then

n area enclosed by the arc = N area of the square π 2 (1) π = 4 = 1 4 or

n π 4n = , i.e. π = N 4 N

246

Mechanical System Design Y Q

P y

P1

r=1

O

x

R

X

Fig. 11.6 The above equation gives the experimental value of π . Obviously, the larger the sample size N, closer will be the obtained value to the true value of π . Example: Three points are chosen at random on the circumference of a circle. Find by Monte Carlo methods the probability that they lie on the same semi-circle.

Table 11.1

Random Numbers Table

2181922396 1128105582 7112077556 6557477468 4199520858 3545174840 174920382 9103161011 0764238934 3493969525 1292054466 4397426117 3807950579 4984768758 1340145867 0590453442 9566554338 7615116284 8508771938

2068577984 7295088579 3440672486 5435810788 3090908872 2275698645 4832630032 7413686599 7666127259 0272759769 0700014629 6488888550 9564268448 2389278610 9120831830 4800088084 5585265145 9172824179 4035554324

8262130892 9586111652 1882412963 9670852913 2039593181 8416549348 5670984959 1198757695 5263097712 0385998136 5169439659 40316525526 3457416988 3859431781 7228567652 1165628554 5089052204 5544814334 0840126299

8374856049 7055508767 0684012006 1291265730 5973470495 4676463101 5432114610 0414294470 5133648980 9999089966 8408705169 8123543276 1531027886 3643768456 1267173884 5407921254 9780623691 0016943666 4942059208

4837657422 5172382962 0633147925 2290031331 8076135523 5629367907 0666095693 9240121544 5111966912 1344056826 6574373193 6027534501 5116633717 5041314549 1320651658 9468932498 5795448061 2628538741 7875623913 (Contd)

System Simulation 6970024586 5799997185 6364375912 4165492027 0354683246 9130555058 5826984369 6285048453 7527791048 8976470693 2327991661 1182095589 3659397698 5924779358 3941092295 1955104281 2117595534 7471564123 8674262892 9061122871 6438538678 1834182305 1884227732

11.12

247 9324732696 0135968939 8383232768 6349104233 4765104877 5255147182 4725370390 3646121751 5765558107 0441608934 7936797054 5535798279 1056981450 1333750468 5726289716 0903099163 1634107293 7344613447 2809456764 0746980892 3049068967 6203476893 29237275501

1186263397 7678931194 1892850701 3382569662 8149224168 3519287786 9641916289 8436077768 8762562043 8749472723 9527542791 4764439855 8416606706 9434074212 3420847871 6827824899 8521057472 1128117244 580655509 9285305274 6955157269 5937802079 8044389132

4425143189 1351031403 2323673751 4579426926 5468631609 2481675649 5049082870 2928794356 6185670830 2202271078 4711871173 6279247618 8234013222 5273692238 1820481234 6383872737 1471300754 3208461091 8224980942 6331989649 5482964330 3445280195 4611203081

3316653259 6002561840 3188881718 1513082455 6474393896 8907598697 7463807244 9956043516 6363845920 5897002653 8300978148 4446895088 6426813469 5902177065 0318831723 5901682626 3044151557 1699403490 5738031833 8764467686 2161984904 3694915658 6072112445

QUEUING MODEL WITH RANDOM INPUT OR POISSON’S ARRIVAL

The following assumptions about the pertinent features of queue are made of the analysis of the model. 1. System Input: The system input is assumed to be random, i.e. the units requiring service arrive randomly and become a part of the waiting line. Random input means that the chance of next arrival’s occurrence is independent of time that has passed since last arrival. In this case, it is the mean arrival rate, the probability of an arrival in time interval “t” to t + ∆t is λ ∆ t and is independent of time t. Hence, we get probability, P (on arrival) occurs in between time t and (t + ∆t ) − λ ∆t. ...(1) Equation (1) is called Poisson’s arrival or random systems input. 2. The number of service unit is one. 3. The queue discipline is FIFO, i.e. first in first out. 4. Service output: It is assumed that the service time is exponentially distributed. The following notations are used in the analysis of the model.

λ µ

= Arrival rate units/unit time = Service rate units/unit time

248

Mechanical System Design Pn(t) = Probability of n units in line at time t = Mean number (Average number) of units in the system including the one being served. Now let us evaluate the probability that there would be n units in the system of time (t + ∆t ) . It is assumed that the probability of more than one unit arriving or being serviced during the interval ∆t is negligible, even that there are n units ( n > 0) in the system at time (t + ∆t ) can occur in four different ways. (a) There are n units in the system at time t and no units arrive in the time interval ∆t and no units are serviced during time ∆t. The probability of occurrence of this compound event

n

= Pn (t ) (1 − λ ∆t ) ⋅ (1 − µ ∆ t ).

(b) There are n units in system at time t and one unit arrives during time interval ∆t and one unit is serviced during time ∆t. The probability of occurrence of this compound event = Pn (t ) λ ∆ t ⋅ ( µ ∆ t ).

(c) There are (n + 1) units in the system at time t and no unit arrives in time interval ∆t and one unit is serviced during time ∆t. The probability of occurrence of this compound event: = Pn + 1 (t ) ⋅ (1 − λ ∆t ) ⋅ µ ∆ t (d) There are (n – 1) units in the system at time t and one unit arrives in time interval ∆t and no unit is serviced during time ∆t. The probability of occurrence of this compound event: Pn − 1 (t ) λ ∆t (1 − µ ∆t ) The probability of n units, hence, can be obtained by adding the four probabilities, and thus we have Pn Pn (t + ∆ t ) Pn (t ) = Pn (t ) (1 − λ ∆ t ) (1 − µ ∆ t ) + Pn (t ) (λ ∆ t ) ( µ ∆ t ) +

Pn + 1 (t ) (λ ⋅ ∆ t ) µ ∆ t + Pn − 1 (t ) λ ∆ t (1 − µ ∆ t ) As ∆t is small quantity, neglecting square and higher powers of t, we get Pn (t + ∆ t ) − Pn (t ) = Pn (t ) (1 − µ ∆t − λ ∆t ) + Pn + 1 (t ) µ ∆t + Pn − 1λ ∆ t ) t

→ (e )

Pn (t + ∆ t ) − Pn (t ) = λ Pn − 1 (t ) + µ Pn +1 (t ) − (λ + µ ) Pn (t ) , at t → 0 we get d n (t ) = λ Pn − 1 (t ) + µ Pn +1 (t ) − ( λ + µ ) Pn (t ) ... ( n > 0) dt

→ (f)

Similarly, the event that there are zero units in the system at time (t + ∆t ) can occur in two different ways. 1. There are zero units in the system at time t and no units arrive during time interval ∆t.

System Simulation

249 = Po (t ) (1 − λ ∆ t )

2. There are one unit in the system at time t. = P1 (t ) (1 − µ ∆ t ) ⋅ µ ∆ t .

Adding the probability 1 and 2, we get, Po (t + ∆ t ) = Po (t )(1 − λ ∆ t ) + P1 (t ) (1 − µ ∆ t ) ⋅ µ ∆ t Po

(t + ∆ t ) − Po (t ) = λ Po (t ) + µ P1 (t ) t

dPo (t ) = − λ Po (t ) + µ P1 (t ) dt If Pn(t) is independent of time t and equals Pn, then equation (h) and (f) lead us to t →0

−λ Po + µ P1 = 0

λ Pn − 1 + µ Pn + 1 − (λ + µ ) Pn = 0 From equation (i), we get,

∂Po (t ) = − λ Po (t ) + µ P1 (t ) ∂t ⇒

∂Po = − λ Po + µ P1 ∂t

⇒

∂Po =0 ∂t = − λ Po + µ P1 = 0

⇒ P1 =

λ Po µ

Putting n = 1 in equation (j) and value P1 in terms of Po, we get, 2

λ P2 =   Po µ 3

Similarly

(g)

λ  P3 =   Po µ 4

λ P4 =   Po µ

(h)

(i)

250

Mechanical System Design 5

λ P5 =   Po µ M n

λ  Pn =   Po µ n

∑ Pi = Po + P1 + P2 + ... + Pn = 1

i=0

Substituting the value of P1 , P2 in terms of Po we get, ∝

2

λ λ Pi = Po +   Po +   Po + ... + λ n Po µ µ i=o

∑

λ Pn −1 + µ Pn +1 − (λ + µ ) Pn = 0

(j)

Put n = 1, we get

λ P1−1 + µ1+1 − (λ + µ ) P1 = 0 ⇒ λ Po + µ P2 − (λ + µ ) P1 = 0

⇒ λ Po + µ P2 − (λ + µ ) ⇒ µ P2 = (λ + µ )

λ P0 = 0 µ

λ Po − λ Po µ

⇒ µ 2 P2 + λ 2 Po 2

λ P2 =   Po µ 2   λ λ P2 = Po 1 +   +   + ...    µ µ  

Now: Let us assume that

λ < 1, i.e. µ > λ , µ

Mean arrival rate is less than mean service rate. ∝

∑ i=o

  1 Pi =  Po + λ  1−  µ  = Po = 1 −

λ µ

   =1   

(k)

System Simulation

251

 λλ Pn = 1 −    µµ 

n

(l)

Mean Number (Average Number) in the system. n = ∑ nPn = P1 + 2 P2 + 3 P3 + 4 P4 + ... n = 0 n =0

2

3

λ λ λ = Po   + 2   + 3   + ... µ µ     µ 2  (λ )  λ λ 1 + 2   + 3   + ... = Po µ   µ µ  = Po

(λ ) s µ

(m) 2

where

λ λ λ s = 1 + 2   + 3   + ... +   µ µ µ

n −1

2

s=

(λ ) λ λ = λ + 2   + ... + (n − 1)   µ µ µ 2

So,

3

n

λ + (n + 1)   + ... µ n −1

 λ  λ λ λ s 1 −  = 1 +  +   +   + ... + µ  µ µ  µ s=

1  λ 1 −  µ 

2

Subtracting the value of s in equation (m). we get

λ  n = Po   s µ λ 1 n = Po   2  µ  λ 1 −  µ    λ λ 1 − µ  µ  n=  λ 1 −  λ µ 

2

λ + n   ... µ n

λ   + ... µ

252

Mechanical System Design

n=

λ µ−λ

tn =

1 1 − µ −λ µ

Expected waiting time

=

λ µ (µ − λ )

Q. 1. Auto vehicles arrive at a petrol pump, having one petrol unit, in Poisson’s fashion with an average of 10 units per hour. The service time is distributed exponentially with a mean of 3 minutes. Find out: given that µ = 20 units/hours (a) Average number of units in the system. (b) Average waiting time for customer. (c) Average length of queue. (d) The probability that a customer arriving at the pump will have to wait. (e) The utilization of factor pump unit. (f) Probability that the number of units in the system is 2. Solution: Mean arrival rate λ = 10 units/hour Mean service rate µ = 20 units/hour

λ µ−λ

(a) Average number of units in the system n =

=

10 20 − 10

= n = 1 unit (b) Average waiting time for customer tw.

tw =

λ µ (µ − λ )

=

10 20(20 − 10)

=

1 hours 20

(c) Average length of queue. = λ ⋅ tw = 10 ×

3 60

System Simulation

253

= 0.5 (d) The probability that a customer arriving at the pump will have to wait. = 1 – Po

 λ  λ 10 = 1 − 1 −  = = = 0.5 µ  µ 20  (e) Utilization factor for pump unit = Proportion of time for which the pump is busy = (1 – Po) = 0.5 (f) Probability that number of unit in system is 2.

 λ  λ  = P2 =  1 −   µ  µ  = (0.5) 2 (1 − 0.5) = (0.5) 2 (1 − 0.5) = 0.125

11.13

A CASE STUDY: INVENTORY CONTROL IN A PRODUCTION PLANT

11.13.1

Introduction

Inventory is a list of movable items required for manufacturing the product and to maintain plant facilities in working condition. It is investment in the form of raw materials, tools, gauges, supplies, etc.

Classification of Inventory 1. Direct Inventory: l Raw materials l In-process inventories l Purchased parts l Finished goods. 2. Indirect Inventories: l Tools l Supplies

Inventory Control Inventory control means making the desired items of required quality and in scientifically determined quantity to meet the production demands available to various departments when needed.

Inventory High Inventory: Problem of storage/maintenance/damage/pilferage/money blocked.

254

Mechanical System Design

Low Inventory: Chances of stoppage of production/idle machine hours.

Inventory Functions: Need for Inventories 1. 2. 3. 4. 5. 6.

11.13.2

To ensure in-time deliveries/better customer relations. To maintain smooth and efficient production flow. To take advantage of quantity discounts. To utilize to advantage price fluctuations. To have effective utilization of men and machinery. To guard against scarcity of materials in the market.

Inventory Control and its Feasibility

An inventory is a detailed list of goods and materials, which can be moved from one place to another such as raw material, material in process, equipments and finished products. Such a list along with the items listed are the available quantity and also the price of each. Each such item is usually needed for the manufacture of a product being produced by the running factory and keeping that in working order. Inventory can be classified for the sake of convenience on the one hand and to have effective control of the other.

Raw Inventory It includes all such items, which are supplied by some firm in the form of raw material, for the concern to which these are resupplied. These raw material is used in the finished product in some form or the other.

Process Inventory These are used in process of manufacture and as such they are neither raw material nor finished goods.

Finished Inventories These are finished goods and ready to be taken in store and stocked for sale purpose.

11.13.3

Purpose of Inventory Control

Inventory controls are needed to have proper track of the inventory, i.e. that the available movable material, to find out that the material of required size and quality in required quantity is available so that manufacturing work does not suffer at all. It is to be seen that each department has and is likely to get the desired material without loss of time and with least labour. The control is observed and exercised to ensure obsolete items are separated from used items and also space shortage problem is solved. It is also to be seen that inventory has been divided in different categories so that there is no problem about laying hand on the required material. Such control aims at having periodic inventory check up and those who are supposed to exercise control are to insure that the needed material is purchased and stocked when it is available at economical prices, so that the manufacturer is not required to unnecessarily waste money. Inventory means keeping up-to-date and accurate record of the available material and the objective of this control is to insure that this record is properly and carefully kept.

System Simulation

11.13.4

255

Advantages of Inventory Control

(i) It is ensured that there is no shortage in the supply of raw materials and as such to a large extent delays in production schedules can be avoided. (ii) Since efforts are made to purchase the inventory at the appropriate time, the manufacturer can take advantage of price fluctuations. (iii) When time comes the firm can purchase inventory in large quantities at low prices and maintain the supply of raw materials. There is no disruption in supplies and no disturbance in time schedule. The goods are then timely supplied and because of timely delivery reputation of the company goes up. (iv) Since the goods are pushed in economic lots, financial gain can be achieved and good quality material can be supplied in the market. (v) To find out which of the two alternatives is better, i.e. each department working separately from the other or clubbing together of different departments. (vi) To see that there is proper storage and scientific utilization of the storage space. (vii) Since material in sufficient quantity is available and in case other factors are favorable, production and output can be increased to some extent. (viii) Inventory control has a certain specific function. It helps in estimating material cost of the product manufactured by a firm because the inventory controller knows prices of all the materials supplied to the departments and shops.

11.13.5

Process in Inventory Control

There is need to fix minimum and maximum quantities of inventories in stock. Terms commonly used in inventory control: (i) Standard Order: It means quantity to be purchased at any time. It is the difference between maximum and minimum quantity. (ii) Dead Time: It is a time, which is calculated by a firm on the basis of past experience. It is the time that lapse between preparing of invoices for the placement of orders and the time taken for the supply of materials. (iii) Maximum Store: It is the upper limit of the inventory and is the largest quantity, which should be kept in stores taking interests of the company into consideration. As against minimum quantity or minimum stores, which indicates lower limit of the inventory, the safety margin can be used in case of emergency. This quantity also includes abnormal requirements of the firm. (iv) Stock Holding: It means holding of stock so that the inventory does not go out of stock. It is also means a buffer stock that can be utilised when material falls even below the minimum level. It takes care of shortage of inventory in the market and helps in reducing the variety of items to be handled. It also helps in deciding quick materials at the time of need and necessity. (v) Ordering Point: It means a point, which indicates time to initiate a purchase order. It is quantity of material required between the exhaustion of available stocks and the supply during the interval between the placements of an order and delivery of materials.

256

Mechanical System Design (vi) Codification: It helps in objective description of various articles and describes an article from general to particular. A good codification has enough flexibility and can cover all the materials received in the concern. Not only this but it has enough provision for unforeseen contingencies. It becomes easy to specify materials at every stage. These days codifications as applied to all components and products as well as raw materials, semi-finished goods.

Inventory Costs: Deterministic Model In determining the “economic order quantity” it is assumed that the cost of managing the inventory consists of two parts: (i) Ordering cost: Ordering cost per piece reduces as quantity/order increases. (ii) Inventory carrying cost (Holding cost): It is the cost of storage and are based on average inventory. Carrying costs increase directly with the number of units. Interest on a average inventory on stock, insurance etc. all increase with the order quantity. (ii) Cost/piece of material purchased is consistent for a considerable range of quantities. Under these circumstances there will be some order size Q that will be most economical. It can be determined as follows:

11.13.6

Inventory Model

1. Static inventory model. 2. Dynamic model—deterministic (fixed demand) (i) Purchased inventory items. (ii) Self produced inventory items. Y = Yearly cost. Q = Economic order quantity. S = Set up or ordering cost. C = Commodity cost/piece. I = Inventory carrying cost. M = Annual usage (Usage Rates) rate/year. Max (EOQ) No. of items

Average

Time

Fig. 11.7

Probabilistic variation in demand

System Simulation

257 Total cost

Fixed cost Ordering cost

No. of cost

c

q

Fig. 11.8

No. of pieces

Economic order quantity

Total cost Y will be the sum of: Y = carrying cost + ordering cost + commodity cost

Y =

No. of order =

µ Q × C × I + × S + µ /C 2 Q µ Q

⇒

dY CI µ S = − 2 =0 2 dQ Q

⇒

CI µ S = 2 2 Q

⇒ Q=

2µ S CI

Q = Economic order quantity Maximum inventory level = (Total production during the production run) – (Total used during the producing run) = Pt – dt But Q = total produced = Pt, thus t = dt = consumption rate

Q P

258

Mechanical System Design

Part of inventory cycle during which production is taking place Inventory level

Demand part of cycle with no production

d  Q 1 −  P  Maximum

t

Time

Fig. 11.9 Set up cost =

Production order quantity model

US Q

P = Production rate U = Consumption rate (usage rate) Y = Holding cost + set up cost + material cost Y =

Q d U × S + UC  1 −  CI + 2 P Q

dY 1  d US = 1 −  CI − 2 2 dQ p Q  ⇒ Q=

2US  d CI 1 −   P

Total cost 1. Holding cost =

Q × CI 2

2. Ordering cost =

U ×s Q

d  1 −  P 

3. Material cost = UC Q. 1. The annual demand for an item is 3200 parts. The unit cost is Rs 6 and inventory carrying charges are estimated as 25 per cent per annum. If the cost of one procurement is Rs 150, Find 1. Economic order quantity.

System Simulation

259

2. Time between two-consecutive orders. 3. Number of orders per year. 4. The optimum cost. Solution: given that µ = 3200 C =6 I = 0.25 S = 150 1. Economic order quantity Q is given by 2µ S CI

Q=

3200 × 150 6 × 0.25

=

3200 × 50

=

0.25

Q = 800 pieces/order 2. Orders per year

=

µ Q

=

3200 800

=4

3. Time between order 3 months 4. Total cost of purchasing Y = Carrying cost + ordering cost + commodity cost

=

µ Q CI + × S + µ c 2 Q

=

800 3200 × 6 × 0.25 + × 150 + 3200 × 6 2 800

= 600 + 600 +19,200 = Rs 20, 400 Ans.

(Production order quantity model) Question 2: Nathan manufacturing Inc. makes and sells specialty hub cups for the retail automobile after market. Nathan’s forecast for its wire-wheel hubcup is 1,000 unit next year with an average daily demand of 6 units. However the production process is most efficient at 8 units/day. But the company

260

Mechanical System Design

produces 10 (Ten) hub cups/day but uses only 6/day. Given the following values solve for the optimum number of units/order. (Note the plant operates to produce hub cups only 167 days/year) Solution: Given: Annual demand = U = 1000 units/year Set-up cost (S) = Rs 10 Holding cost (CI) = 50 per unit per year Daily production rate (p) = 8 units/day Daily consumption demand = (d) = 6 units per day

QP =

=

2US  d CI 1 −   P 2 × 1000 × 10 6  50 1 −  8 

= 400 hubcups Also note that

d= =

D Number of days plant operates/year 1000 167

d = 6 units Question 3: Using the simulation in the study of inventory problem, find the following. A automobile company manufactures around 150 scooters. The daily production varies from 146 to 154 depending upon the availability of raw materials and other working conditions. Production per day: Probability:

146 0.04

147 0.09

148 0.12

149 0.14

150 0.11

151 0.10

152 0.20

153 0.12

154 0.08

The finished scooters are transported in a specially arranged lorry accommodating 150 scooters. Using the following random numbers. 80, 81, 76, 75, 64, 43, 18, 26, 10, 12, 65, 68, 69, 61, 57. Simulate the process to find out (a) What will be the average number of scooters waiting in the factory? (b) What will be the average number of empty space on the lorry? Solution: The random number are established as in the given table below:

System Simulation

261

S.No.

Production per day

Probability

Commutative probability

Random numbers assigned

1 2 3 4 5 6 7 8 9

146 147 148 149 150 151 152 153 154

0.04 0.09 0.12 0.14 0.11 0.10 0.20 0.12 0.08

0.04 0.13 0.25 0.39 0.50 0.60 0.80 0.92 1.00

00–03 04–12 13–24 25–38 39–49 50–59 60–79 80–91 92–99

Based on the 15 random numbers given, we simulate the probability per day: Simulation Sheet S.No.

Random numbers

Probability per day

No. of scooter waiting

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

80 81 76 75 64 43 18 26 10 12 65 68 69 61 57

153 153 152 152 152 150 148 149 147 147 152 152 152 152 151

3 3 2 2 2 0

0 2 1 3 3

2 2 2 2 1

Total

21

(a) Average number of scooter waiting =

No. of empty space into lorry

21 15

= 1.4 per day Ans. (b) Average number of empty spaces on the lorry =

9 15

= 0.6 per day Ans.

9

262

Mechanical System Design

Question 4: For an infinite population, single channel waiting line system, the probabilities associated with time between arrivals and service time are given by Tables (x) and (y) respectively. Using simulated sampling for 10 periods, determine the average idle time of the server. For simulating, use the following digit random numbers. UPTU 2004 The first 2 digits of the random numbers be used for simulation of the arrivals whereas the last 2 digits of the random numbers be used for service times. 48867, 32267, 27345, 55753, 93124, 98658, 68216, 17901, 88124, 83464, Table x

Table y

Time between arrivals (min)

Probability

Service time (min)

Probability

4 5 6 7 8 9

0.075 0.200 0.300 0.225 0.150 0.050

2 3 4 5 6 7 8

0.0062 0.0606 0.2417 0.3830 0.2417 0.0606 0.0062

Time between arrivals (min)

Probability

R. N. Interval

Commutative time

4 5 6 7 8 9

0.075 0.200 0.300 0.225 0.150 0.050

00–06 07–26 27–56 57–79 80–94 95–99

0.075 0.275 0.575 0.800 0.950 1.000

Solution:

Unit 5 Service Time

Service Probability

2 3 4 5 6 7 8

0.0062 0.0606 0.2417 0.3830 0.2417 0.0606 0.0062

Commutative prop 0.0062 0.0668 0.3085 0.6915 0.9332 0.9938 1.000

R. N. Interval 00 01–06 07–30 31–68 69–92 93–98 99

System Simulation

263

S.No

Arrival Time

Time

R.No.

Service time

Start

End

48 32 27 55 93 98 68 17 88 83

6 6 6 6 8 9 7 5 8 8

6 12 18 24 32 41 48 53 61 69

67 67 45 53 24 58 16 01 24 64

5 5 5 5 4 5 4 3 4 4

6 12 18 24 32 41 48 53 61 69

11 17 23 29 36 46 52 56 65 73

Idle time 6 12 – 11 = 1 18 – 17 = 1 24 – 23 = 1 32 – 29 = 3 41 – 36 = 5 48 – 46 = 2 53 – 52 = 1 61 – 56 = 5 69 – 65 = 4

Total

29

Average ideal time =

29 10

= 2.9 Ans. Question 5: Customers arrive at a milk booth for the required service. Assume that inter-arrival and service times are constant and given by 1.8 and 4 times units, respectively. Simulate the system by hand computations for 14 time units. What is the average waiting time per customer? What is the percentage idle time of the facility? [Assume that the system starts at t = 0.] Solution: In the beginning, when the facility is free, first customer starts service. Its departure time is t = 0 + 4 = 4. Next event (arrival) occurs at t = 0 + 1.8 = 1.8, which is served before Ed at t = 4. Now since the facility is still busy, customer 2 is put in the queue and is first to be considered in this queue. A new arrival event (customer 3) occurs at t = 1.8 + 1.8 = 3.6 which precedes Ed at t = 4. Again customer 3 is put in the queue and a new arrival event Ea (customer 4) occurs at t = 3.6 + 1.8 = 5.4. This event succeeds E4 at t = 4. At this point, first customer departs which leaves the facility free. Customer 2, who was the first to join the queue, not gets service. The waiting time is computed as the time period from the instant he joined the queue until he commences service. The procedure is repeated until the simulated period is completed. The results of simulation are given in the following Table. Time

Event

Customer

Waiting time

0.0 1.8 3.6 4.0 5.4 7.2 8.0

Ea Ea Ea Ed Ea Ea Ed

1 2 3 1 4 5 2

4 – 1.8 = 2.2 (customer 2)

8 – 3.6 = 4.4 (customer 3) (Contd)

264

Mechanical System Design Time 9.0 10.8 12.0 13.6 14

Event

Customer

Ea Ea Ed Ea Ed

6 7 3 8 –

Waiting time

12 – 5.4 = 6.6 (customer 4) 14 – 7.2 = 6.8 (customer 5) 14 – 9.0 = 5.0 (customer 6) 14 – 10.8 = 3.2 (customer 7) 14 – 13.6 = 0.4 (customer 8)

It is evident from this simulation that the average waiting per customer is

=

0 + 2.2 + 4.4 + 6.6 + 6.8 + 5.0 + 3.2 + 0.4 = 3.57 8

Average waiting time per customer for those who must wait is 28.6/7 = (4.08) and % Idle time of the facility = 0% Ans. Question 6: At a telephone booth, customers arrive with an average time of 1.2 time units between one arrival and the next. Service time are assumed to be 2.8 time units. Simulate the system for 12 times units by assuming the system starts at t = 0. What is the average waiting time per customer? (Assume that the system starts at t = 0) UPTU 2005 Solution: Time 0.0 1.2 2.4 2.8 3.6 4.8 5.6 6.0 7.2 8.4 9.6 10.8 11.2 12.0

Event

Customer

Ea Ea Ea Ed Ea Ea Ed Ea Ea Ed Ea Ea Ed End

1 2 3 1 4 5 2 6 7 3 8 9 4 –

Waiting time

2.8 – 1.2 = 1.6 (customer 2)

5.6 – 2.4 = 3.2 (customer 3)

8.4 – 3.6 = 4.8 (customer 4)

11.2 – 4.8 = 6.4 (customer 5) 12 – 6 = 6.0 (customer 6) 12 – 7.2 = 4.8 (customer 7) 12 – 9.6 = 2.4 (customer 8) 12 – 10.8 = 1.2 (customer 9)

System Simulation

265

Average waiting time per customer

=

0 + 1.6 + 3.2 + 4.8 + 6.4 + 6.0 + 4.8 + 2.4 + 1.2 9

= 3.37 time unit per customer Ans.

Exercise

11 . . .

1. What is simulation concept on the basis of system simulation? 2. What is simulation? Explain its purpose. List different advantages and limitations of simulation. UPTU 2005 3. What is the role of simulation on the basis of mechanical system design? 4. How many types of simulation models are there? Briefly explain. 5. Briefly explain waiting line simulation. 6. Briefly explain simulation process. 7. What is simulation process? Also explain problem definition. 8. Write down step to step input model construction. 9. What is simulation process of system simulation? Also write down the limitation of simulation approach. 10. Write down simulation language and explain simulation approach.

A Case Study 1. 2. 3. 4. 5. 6.

What is inventory control and inventory function in a production plant? Briefly explain inventory costs; deterministic model. Explain inventory model. What is probabilistic variation in demand? Derive a expression of Eoθ (Economic order quantity.) Explain production order quantity model.

266

Mechanical Systems and Design

12 Application of Mechanical System Design to Control System 12.1

INPUT-OUTPUT RELATIONSHIP

Introduction The control system is very important for all engineers. The first significant control device was James Watt’s flyball governor. This was invented in 1767 to keep the speed of the engine constant by regulating the supply of steam to the engine. In control system, the behaviour of the system is described by the differential equations. Minor Sky in 1922 showed how to determine the stability from the differential equations describing the systems. The differential equations may be ordinary differential equations or the difference equations. The control system can be classified as open loop control system and closed loop control systems.

12.1.1

Open Loop Control System

The open loop control system is also known as control system without feedback or non feedback control systems. In open loop systems, the control actions are independent of the desired output. In this system, the output is not compared with the reference input. The components of the open loop systems are controller and controlled process. The controller may be amplifier, filter, etc. depends upon the system. An input is applied to the controller and the output of the controller gives to the controlled process and we get the output (desired). Input

Controller

Controlled process

Output

Fig. 12.1 Examples 1. Automatic washing machine is the example of the open loop systems. In the machine the operating time is set manually. After the completion of set time the machine will stop with the

Application of Mechanical System Design to Control System

267

result we may or may not get the desired (output) amount of cleanliness of washed clothes because there is no feedback provided to the machine for desired output. 2. Immersion rod is another example of open loop system. The rod heats the water but how much heating is required is not sensed by the rod because of no feedback to the rod. 3. A field control d.c. motor is the example of open loop system. Field control d.c. motor

Input

Speed or

Torque Load

Output Load position

Fig. 12.2 4. For automatic control of traffic, the lamps of three different colors (red, yellow and green) are used. The time for each lamp is fixed. The operation of each lamp does not depend upon the density of the traffic but depends upon the fixed time. Thus we can say that the control system, which operates on the time basis, is an open loop system.

Advantages l l l l

Open loop control systems are simple. Open loop control systems are economical. Less maintenance is required and not difficult. Proper calibration is not a problem.

Disadvantages l l l l

Open loop systems are inaccurate. These are not reliable. These are slow. Optimization is not possible.

12.1.2

Closed Loop Control System

Closed loop control systems are also known as feedback control systems. In closed loop control systems the control action is dependent on the desired output. If any system having one or more feedback paths in forms a closed loop system. In closed loop systems, the output is compared with the reference input and error signal is produced. The error signal is fed to the controller to reduce the error and desired output is obtained. Error signal Reference Input

Amplifier

Controller

– +

Feedback

Fig. 12.3

Process

Controlled output

268

Mechanical Systems and Design

Example In a room, we need to regulate the temperature and humidity for comfortable living. Air conditioners are provided with thermostat. By measuring the actual room temperature and comparing it with desired temperature an error signal is produced, the thermostat turns on or turns off the compressor. The block diagram is shown in Fig. 12.4. Desired temp.

Room Temp.

Comparison

Controller

Compressor

Error signal

Fig. 12.4

Advantages (i) (ii) (iii) (iv)

These systems are more reliable. Closed loop systems are faster. A number of variables can be handled simultaneously. Optimization is possible.

Disadvantages (i) Closed loop systems are expensive. (ii) Maintenance difficult. (iii) Complicated installation. Comparison between open loop and closed loop

Table 12.1 S. No. 1. 2. 3. 4. 5.

Open loop system These are not reliable. It is easier to build. If calibration is good they perform accurately. Open loop systems are generally more stable. Optimization is not possible.

Closed loop system These are reliable. It is difficult to build. They are accurate because of feedback. These are less stable. Optimization is possible.

Elements or components of closed loop system The various components of closed loop system are shown in Fig. 12.5.

Application of Mechanical System Design to Control System Command Input

Reference input element

269

Control element

– +

Controlled system

Controlled output

Feedback element

Fig. 12.5

Command The command is the externally produced input and independent of the feedback control system. Reference input element: This produces the standard signals proportional to the command. Error detector: The error detector receives the measured signal and compares it with reference input. The difference of two signals produces the error signal. Control element: This regulates the output according to the signal obtained from error detector. Controlled system: This represents what we are controlling by the feedback loop. Feedback element: This element feedback the output to the error detector for comparison with the reference input.

12.2

TRANSFER FUNCTION

Transfer function for single input single output system The transfer function is defined as the ratio of Laplace transform of the output to the Laplace transform of input with all initial conditions are zero. Consider a linear system having input r(t) and c(t) is the output system, the input-output relation can be described by the following nth order differential equation:

Ei (t ) = Ri1 (t ) +

1 di (t ) di (t ) i1 dt + L1 1 − M 2 C1 dt dt

∫

(12.1)

E2 (t ) = R2 i2 (t )

0 = R2 i2 (t ) + ( L2 + L3 )

(12.2)

1 d i2 (t ) + dt c1

∫i

2

dt − M

d ii (t ) dt

(12.3)

Take Laplace transform of Eqs. (12.1), (12.2) and (12.3)

  1 E1 ( s ) = I1 ( s )  R + + sL1  C1 s  

(12.4)

E2 ( s ) = R2 I 2 ( s )

(12.5)

0 = I 2 ( s) [ R2 + s( L2 + L3 ) + 1/sC2 ] − s M I1 ( s)

Solving the equation (12.4), (12.5) and (12.6) the required transfer function

(12.6)

270

Mechanical Systems and Design

G (s ) =

12.3 12.3.1

s 3 R2 C1 C2 M [ sR2 C2 + s 2 C2 ( L2 + L3 ) + 1] [ s 2 L1 C1 + sC1 R1 + 1] − M 2 s 4 C1 C2 Ans.

MECHANICAL SYSTEM Translational Systems

The motion that takes place along a straight line is known as translational motion. There are three types of forces that resist motion. 1. Inertia Force: Consider a body of mass M and acceleration “a”. Then according to Newton’s second law of motion, the inertia force will be equal to the product of mass M and acceleration “a”. FM (t ) = Ma (t )

(12.7)

In the terms of velocity the Eq. (12.7) becomes FM (t ) = M

dv (t ) dt

(12.8)

In terms of displacement the Eq. (12.7) can be expressed as FM (t ) = M

d 2 x (t ) dt 2

(12.9)

x (t) M

f (t )

Fig. 12.6 2. Damping Force: For viscous friction, we assume that the damping force is proportional to the velocity. FD (t ) = Bv(t ) = B

dx (t ) dt

where B = Damping coefficient Unit of B = N/m/sec. We can represent “B” by a dashpot which consists of piston and cylinder. x

fD(t)

0

B

Fig. 12.7

(12.10)

Application of Mechanical System Design to Control System

271

3. Spring force: A spring stores the potential energy. The restoring force of a spring is proportional to the displacement. Fk (t ) = K x (t )

(12.11)

x(t) fk(t) K

Fig. 12.8 where K = spring constant or stiffness Unit of K = N/m The stiffness of a spring can be defined as restoring force per unit displacement.

12.4

ROTATIONAL SYSTEM

The rotational motion of a body can be defined as the motion of a body about a fixed axis. There are three types of torques that resist the rotational motion. 1. Inertia torque: Inertia is the property of an element that stores the kinetic energy of rotational motion. The inertia torque T1 is the product of moment of inertia and angular acceleration α (t ). T1 (t ) = J α (t )

T1 (t ) = J =J

d ω (t ) dt d2 θ (t ) dt 2

T1(t )

(12.12)

θ (t )

J

Fig. 12.9 where

ω (t ) = angular velocity θ (t ) = angular displacement

Unit of Torque = N–m 2. Damping Torque: The damping torque TD (t ) is the product of damping coefficient B and angular velocity ω.

272

Mechanical Systems and Design TD (t ) = B ω (t ) TD (t ) = B

d θ (t ) dt

(12.13)

3. Spring Torque: Spring torque Tθ (t ) is the product of torsional stiffness and angular displacement. Tθ (t ) K

θ (t )

Fig. 12.10 Tθ (t ) = K θ (t )

Unit of K = N–m/rad. (12.14) If we compare the equations, we get an analogous system. The analogous quantities are tabulated as follows.

Table 12.2 S.No. 1. 2. 3. 4. 5. 6. 7.

12.5

Translational

Rotational

Force, F Acceleration, a Velocity, v Displacement, x Mass, M Damping coefficient B Stiffness

Torque, T Angular acceleration, α Angular velocity, ω' Angular displacement, θ Moment of inertia, J Rotational damping coeff. B Torsional stiffness

D’ ALEMBERT’S PRINCIPLE

The principle states that “for any body the algebraic sum of externally applied forces resisting motion in any given direction is zero”. D’ Alembert principle is useful in writing the equation of motion of mechanical system. Consider a system shown in Fig. 12.11 consisting of a mass M, spring and dashpot. First choose a reference direction. All the forces in the direction of reference direction are considered as positive and the forces opposite to the reference direction are taken as negative.

Application of Mechanical System Design to Control System

273

Fig. 12.11 External Force: F(t) Resisting Forces: (a) Inertia force Fm (t ) = − M

d2 x(t ) dt 2

(b) Damping force FD (t ) = − B

d x(t ) dt

(c) Spring force FK (t ) = − K x (t ) According to D’alembert principle F (t ) + Fm (t ) + FD (t ) + FK (t ) = 0 F (t ) − M

d2 d x (t ) − B x (t ) − K x (t ) = 0 2 dt dt

F (t ) = M

d2 d x(t ) + B x(t ) + K x (t ) 2 dt dt

Consider a rotational system shown in Fig. 12.12. T K

K B

Fig. 12.12

(12.15)

274

Mechanical Systems and Design

External Torque: T(t) Resisting Torque: (a) Inertia Torque T1 (t ) = −

Jd ω (t ) dt

(b) Damping Torque TD (t ) = − B

d θ (t ) dt

(c) Spring Torque Tθ (t ) = − K θ According to D’ Alembert principle T (t ) + T1 + TD + TK = 0

or

T (t ) − J

d d ω (t ) − B θ (t ) − Kθ (t ) dt dt

T (t ) = J

d d ω (t ) + B θ (t ) + K θ (t ) dt dt

(12.16)

So D’ Alembert principle for rotational motion is: “For any body the algebraic sum of externally applied torques and the torques resisting rotation about any axis is zero”. Procedure of writing the models of Mechanical System 1. Assume system is in equilibrium. 2. Assume that the system is given some arbitrary displacement if number of distributing forces are present. 3. Draw the free body diagram of forces exerted on each mass in the system. 4. Apply Newton’s law of motion to each diagram; using the convention that any force acting in the direction of assumed displacement is positive. 5. Rearrange the equations in suitable form to be solved by any means. Example 1: Draw the free body diagram and write the differential equation of the given system shown in Fig.12.13. Solution: Differential equation for M1 External force: F(t) Resisting forces: 1. Inertia force FM = − M 1

d2 x1 dt 2

2. Damping force FD = − B1

d ( x1 − x2 ) dt

3. Spring force FK = K1 ( x1 − x2 )

Application of Mechanical System Design to Control System

K2

275

B2

M2 x2 K1

B1

M1 x1

F (t)

Fig. 12.13 M

d2 dt

x 2 1

B1

K1(x1 − x 2 )

M2

d (x − x 2 ) dt 1

dt 2

x2

B2

K 2x 2

M1

d x dt 2

M2

K1( x1 − x 2 )

F (t )

Fig. 12.14 F (t ) = M 1

d2

B1

d (x1 − x 2 ) dt

Free body diagram

d2 d x1 + B1 ( x1 − x2 ) + K1 ( x1 − x2 ) dt dt 2

(i)

Similarly for mass M 2 K ( x1 − x2 ) + B1

d d 2 x2 dx ( x1 − x2 ) = M 2 + B2 2 + K 2 x2 dt dt dt 2

(ii)

Equation (i) and (ii) are the required equations. Example 2: Write the differential equations describing the dynamics of the system shown in Fig. 12.15 and find the ratio

X 2 ( s) . F ( s)

276

Mechanical Systems and Design x1

x2 K2

K1

F(t )

M1

M2

B1 = 0

B2 = 0

Fig. 12.15 Solution: Free body diagram. For mass M1

F(t )

K1(x1 − x 2 ) M1 M1

d 2 x1 dt 2

Fig. 12.16(a) Free body diagram. For mass M 2 K1(x1 − x 2 )

K 2x 2 M2 M2

d2 dt 2

x2

Fig. 12.16(b) Differential equation for mass M1 F (t ) = M 1

d2 x1 + K1 ( x1 − x2 ) dt 2

K1 ( x1 − x2 ) = K 2 x2 + M 2

d 2 x2 dt 2

(1)

(2)

Take Laplace transform of Eq. (1), assume initial conditions zero. F (t ) = M 1

d2 x1 + K1 x1 − K 2 x2 dt 2

F ( s) = M 1 s 2 X1 ( s) + K1 X1 ( s ) − K1 X 2 ( s) Laplace transform of equation in Equation (2) K1 X 1 − K1 X 2 = K 2 X 2 + M 2

d2 X2 dt 2

(3)

Application of Mechanical System Design to Control System

277

K1 X1 ( s ) − K1 X 2 ( s ) = K 2 X 2 ( s ) + M 2 s 2 X 2 ( s) X1 (s ) =

X2 ( s) [ s 2 M 2 + K1 + K 2 ] K1

(4)

Put the value of X 1 ( s) in Eq. (3)

F ( s) =

X 2 ( s) 2 [ s M 2 + K1 + K 2 ] [ s 2 M1 + K1 ] − K1 X 2 ( s) K1

X 2 (s ) K1 = 2 Ans. F (s ) ( s M 2 + K1 + K 2 ) ( K1 + s 2 M1 ) − K12 Example 3: A mass spring system under equilibrium condition is shown in Fig.12.17. Derive the system equation where M = 10 kg x K F

M

Fig. 12.17 B = 30 N/M/sec. and K = 20 N/M. Solution: Free body diagram. Kx F

M

dx B dt

M

d 2x dt 2

Fig. 12.18 Differential equation, F =M

d 2x dx +B + Kx 2 dt dt

Put the given values F = 10

d 2x dx + 30 + 20 x 2 dt dt

Ans.

Example 4: Derive the system equations and find the value of X 2 ( s )/F ( s) for the system shown in Fig.12.19.

278

Mechanical Systems and Design x2

x1

F(t )

B12

K1

M1

M2

fc1

K2

fc2

Fig. 12.19 Solution: Free body diagram for mass M 1 d (x − x ) dt 1 2 d2 M1 x1 dt 2 dx1 fc1 dt

B12

K1x1 M1 F (t )

Free body diagram for mass M 2 B12

d (x − x ) dt 1 2 dx fc 2 dt 2

K 2x 2 M2

M2

d 2x dt 2

System equation for mass M 1 F (t ) = M 1

d 2 x1 dx d ( x1 − x2 ) + K1 x1 + fc1 1 + B12 2 dt dt dt

(1)

System equation for mass M 2 B12

d d 2 x2 dx ( x1 − x2 ) = M 2 + fc2 2 + K 2 x2 2 dt dt dt

(2)

Laplace transform of equation (1)

F ( s) = X1 ( s) [ s 2 M 1 + sB12 + sfc1 + K1 ] − B12 sX 2 ( s)

(3)

Laplace transform of equation (2)

B12 sX1 ( s ) = ( s 2 M 2 + B12 s + sfc2 + K 2 ) X 2 ( s) X1 (s ) =

X 2 ( s) ( s 2 M 2 + B12 s + sfc2 + K 2 ) B12 s

Put the value of X 1 ( x) for Eq. (1) in equation (3)

(4)

Application of Mechanical System Design to Control System

F ( s) = X 2 ( s ) =

279

[ s 2 M 2 + B12 s + sfc2 + K 2 ] [ s 2 M1 + sB12 + sfc1 + K1 ] − B12 sX 2 ( s) B12 S

X 2 (s ) sB12 = 2 2 2 Ans. F (s ) ( s M 1 + sB12 + sfc1 + K1 ) ( s M 2 + sB12 + sfc2 + K 2 ) − s 2 B12

12.6

ANALOGOUS SYSTEM

Consider a series RLC circuit applies Kirchoff’s voltage law

di 1 + idt dt C

∫

E = Ri + L

L

R

C

i

E

(12.17)

Fig. 12.20 In terms of charge Eq. (12.17) becomes E=R

dq d 2q 1 +L 2 + q dt C dt

R

i

L

C

(12.18)

E

Fig. 12.21 Now consider a parallel RLC circuit. Now apply Kirchoff’s current law

1=

E 1 dE + Edt + C R L dt

∫

In terms of magnetic flux linkage the Eq. (12.19) becomes

∫

Since ϕ = Edt

(12.19)

280

Mechanical Systems and Design

I =

1  dϕ  1 d 2ϕ  + ϕ +C 2 R  dt  L dt

(12.20)

Now compare the equation (12.15) with the equation (12.18), both equations are differential equations of same order, i.e. identical. Such type of systems whose differential equations are in the same form are called analogous systems. On comparison of Eq. (12.15) with Eq. (12.18) we can see that in mechanical system the force F is analogous to voltage E in electrical system. Such type of analogy is known as force voltage analogy. From equation (12.15) and (12.18) mass M is analogous to inductance L, coeff. of viscous friction B is analogous to resistance R, spring stiffness K is analogous to 1/C and soon.

Table 12.3 S.No. 1. 2. 3. 4. 5.

Mechanical translation system

Electrical system

Force (F) Mass (M) Stiffness (K) (Elastance,1/K) Damping coeff. (B) Displacement (x)

Voltage (E) Inductance (L) Reciprocal of capacitance (1/C) Capacitance (C) Resistance (R) Charge (q)

Now compare the Eq. (12.15) with Eq. (12.18) since the force F is analogous to the current source. Such type of analogy is known as force current analogy. The analogous quantities are tabulated as:

Table 12.4 S.No.

Mechanical rotational

Electrical system

1. 2. 3.

Torque (T) Moment of Inertia (I) Damping coeff. (B)

4.

Stiffness (K) (Elastance,1/K) Angular displacement (θ )

Current (I) Capacitance (C) Reciprocal of resistance (1/R) i.e. conductance (G) Reciprocal of inductance (1/L)

5.

Flux linkage (ϕ )

Application of Mechanical System Design to Control System

281

Table 12.5 S.No.

Mechanical translation system

Electrical system

1. 2. 3.

Force (F) Mass (M) Damping coeff. (B)

4.

Stiffness (K) (Elastance 1/K)

5.

Displacement (x)

Current (I) Capacitance (C) Reciprocal of resistance (1/R) i.e. Conductance (G) Reciprocal of inductance (1/L) Inductance (L)

6.

Velocity (x)

Flux linkage (ϕ ) Voltage (E)

Now consider the rotational system and compare the Eq. (12.16) with (12.18). On comparison the analogous quantities we get Torque current analogy and can be tabulated as:

Table 12.6

Torque voltage analogy

S.No.

Mechanical rotational system

Electrical system

1. 2. 3. 4.

Torque (T) Moment of Inertia (I) Damping coeff. (B) Stiffness (K) (Elastance, 1/K)

Voltage (E) Inductance (L) Resistance (R) Reciprocal of capacitance

5.

Angular displacement (θ )

Capacitance (1/C)

6.

Angular velocity (ω )

Charge (q) Current (I)

Following change the rules for drawing the analogous circuit can be stated as: (a) For force voltage analogy: “Each junction in the mechanical system corresponds to a closed loop which consists of electrical excitation sources and passive elements analogous to the mechanical driving sources and passive elements connected to the junction. All points on a rigid mass are considered as the junction.” (b) For force current analogy: “Each junction in the mechanical system corresponds to a node which joins electrical excitation sources and passive elements analogous to the mechanical driving sources and passive elements connected to the junction. All points on a rigid mass are considered as the same junction and one terminal of the capacitance analogous to a mass is always connected to the ground.” The reason for connecting one terminal of capacitance analogous to a mass is always connected to the ground is that the velocity (or displacement) of a mass is always referred to the earth.

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Consider the system shown in Fig. 12.11 corresponding to x in mechanical system. We will have one loop in F–V analogous electrical circuit. The circuit consists of a voltage source inductance L(M), resistance R(B), the reciprocal of capacitance (C) and charge q(x). L q

V

1/C

R

Fig. 12.22 The equation for the mechanical system shown in Fig.12.11 is d 2x dx +B + Kx 2 dt dt For the electrical circuit the equation will be F =M

d 2q dq 1 +R + q 2 dt C dt Since q = idt the equation (1.65) will become V =L

∫

di 1 di 1 + Ri + idt ⇒ or V = Ri + L + idt dt C dt C The above equation is the voltage equation for series RLC circuit. Series RLC circuit is the analogous electrical circuit of the mechanical system shown in Fig.12.11. V =L

∫

∫

Example 5: Draw the analogous electrical network of the given system. Use F–V analogy. Solution: Now corresponding to x1 , x2 , x3 we have three loops in electrical circuit.

K3

B3

M3

x3

K2

B2

M2 x2 B1 M1

x1

F

Fig. 12.23

Application of Mechanical System Design to Control System

283

First loop consists of a voltage source V(F), Inductance L1 ( M1 ) , resistance R1 ( B1 ). The resistance VR1 is common in between first and second loop.

Second loop consists of inductance L2 ( M 2 ) , resistance R2 ( B2 ) and 1/C2 ( K 2 ) . Resistance R2 and 1/C2 are common in second loop and third loop. Similarly third loop consists of inductance L3 ( M 3 ) , 1/C3 ( K 3 ) and R3 ( B3 ) , q1 , q2 and q3 are analogous quantities of x1 , x2 , x3 respectively. L1

L3

L2

1 C3

1 C2

V

R1 q1

Fig. 12.24

R2

q2

R3

q3

Analogous electrical circuit of Fig. 12.23

Example 6: Draw the electrical analogous circuit of the system shown in Fig. 12.25. x2

x1

B F

K2

M2

M1

Frictionless

Fig. 12.25 Solution: By inspection and with the help of table we can draw the electrical circuit shown in Fig. 12.26. L1

L2

R

V

q1

q2

1 C2

Fig. 12.26 Example 7: Draw the analogous electrical circuit of the system shown in Fig. 12.27.

284

Mechanical Systems and Design x1

x2

B1

K1

F

M2

M1 K2

B2

B3

Fig. 12.27 Solution: Corresponding to x1 and x2 we have two loops in electrical circuit. The analogous electrical circuit is shown in Fig. 12.28. L1

V

L2

R

q1 1 C2

q2

1 C1

Fig. 12.28 Example 8: Draw the analogous electrical circuit of the given system shown in Fig. 12.27, use F–I analogy. Assume frictionless wheels. Solution: Since wheels are frictionless, B2 = 0 and B3 = 0 Corresponding to x1 and x2 we have two independent nodes. The analogous electrical circuit is shown in Fig. 12.29. 1/ R1(B1)

V1

1 L1 (K1)

(K 2 )

V2

1/ L2

C1

C2

M1

M2

i (t ) (f )

Fig. 12.29 Example 9: Draw the analogous electrical circuit of the system shown in Fig. 12.25. Use F–I analogy. Solution: Corresponding to x1 and x2 we have two independent nodes. The analogous circuit is shown in Fig. 12.30.

Application of Mechanical System Design to Control System V1

1/ R

285

V2

B i (t)

K2

C2

C1 (M1)

1/ L2

(M2 )

Fig. 12.30 Example 10: Draw the analogous electrical circuit of a rotational mechanical system shown in Fig. 12.31, use F–I analogy. J2

J1

K3

K2

K1 T

Frictionless

Fig. 12.31 Solution: The analogous circuit shown in Fig. 12.32. 1/ L1

i (t)

C1 (J1)

1/ L2

C2 (J 2 )

K3

1/ L3

Fig. 12.32

12.7

MECHANICAL EQUIVALENT NETWORK

The equivalent mechanical network is useful to determine the transfer function of the system. The following procedure is applied to draw the mechanical equivalent network: Step 1: Draw a reference line, consider the fixed points as reference. Step 2: Corresponding to the displacement x1 , x2 , ..., select the nodes. Step 3: Connect one end of masses to the reference line. Step 4: Connect other elements of the system to the modes. Step 5: Apply modal analysis, write the system equations. Consider a simple system shown in Fig. 12.11. The mechanical equivalent network is shown in Fig. 12.33.

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Mechanical Systems and Design

M

d2 dt

f (t)

x

2

B

Kx

dx dt

B

M

K

Fig. 12.33 Apply the nodal analysis, the system equation will be

F (t ) = M

d 2x dx +B + Kx 2 dt dt

(12.21)

With the help of mechanical equivalent network we can draw the analogous electrical circuit (shown in Fig. 12.34).

C (M)

i (t)

1/R (B)

1/L (K)

Fig. 12.34 Example 11: Draw the mechanical network of the system shown in Fig. 12.27. Solution: Since there are two displacements x1 , x2 , we have two nodes. B1

x1

K1

M1

K2

B2

B3

x2

M2

f (t)

Fig. 12.35

Example 12: Draw the mechanical equivalent network. Write the system equations and find the system shown in Fig. 12.36.

f (s ) of x2 ( s )

Application of Mechanical System Design to Control System F (t)

x1

K1

287

x2 K2

M1

M2

K3

Frictionless wheels

Fig. 12.36 Solution: Mechanical network is shown in Fig. 12.37. x1 K 2

f (t)

K1

M1

x2

M2

K3

Fig. 12.37 Apply nodal analysis at x1 and x2 F (t ) = M 1

d 2 x1 + K1 x1 + K 2 ( x1 − x2 ) dt 2

K 2 ( x2 − x1 ) + M 2

d 2 x2 + K 3 x2 = 0 dt 2

(1)

(2)

Laplace transform of Eqs. (1) and (2)

F ( s) = s 2 M 1 X1 ( s) + ( K1 + K 2 ) X1 ( s) − K 2 X 2 ( s )

(3)

s 2 M 2 X 2 ( s ) + ( K 2 + K 3 ) X 2 ( s) − K 2 X 1 ( s) = 0

(4)

From Eq. 4 we get,

X1 (s ) =

( s 2 M 2 + K 2 + K3 ) X 2 ( s ) K2

(5)

Put the value of X 1 ( s ) in Eq. (3) and solve for F2 ( s )/ X 2 ( s)

( s 2 M1 + K1 + K 2 ) ( s 2 M 2 + K 2 + K3 ) − K 22 F (s ) = X 2 (s ) K2 Eq. (6) is the required answer.

(6)

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Mechanical Systems and Design

Question 13: A disk rotating in a viscous medium and supported by a shaft is shown in Fig. 12.38. Applying torque tending to rotate the disk is T. Develop an input-output model of the system to compute UPTU-2004 “ θ (t)” for a given “T”. Td = Torque required to overcome viscous friction Ts = Torque tending to twist the shaft = K s θ Ks = Tensional stiffness of the shaft,

θ = Angular displacement of the shaft. θ

T

ω

Fixed end

Ts Td

Fig. 12.38 (e)

T =−J Td = −

dw (t ) dt

B dθ (t ) dt

Ts = –Ks θ (t )

T (t ) + T + Td + K( s ) = 0 T (t ) − J

dw dθ (t ) − B (t ) − K sθ (t ) dt dt

T (t ) = J

dw dθ (t ) + B (t ) + K sθ (t ) dt dt

⇒ K s θ (t ) = T (t ) − J ⇒ θ (t ) =

1 Ks

dw dθ (t ) − B (t ) dt dt

dw dθ   T (t ) − J dt (t ) − B dt (t )   

Application of Mechanical System Design to Control System

θ (t ) =

289

dw dθ   T (t ) − J dt (t ) − B dt (t )   

1 Ks

Question 14: For the mechanical system shown in Fig. 12.39 determine the differential equation relating f and x and equation relating f and y and equation relating x and y. UPTU-2004 f

x K1

M

y

B

y

K2

Fig. 12.39 Ans. K1 (x − y )

f

Md 2 dt 2

(x − y )

M

K 2y

Fig. 12.40

dy dt

Free body diagram of mass M

By D’ Alembert’s principle of mass M Given that

M = mass

B

290

Mechanical Systems and Design dy = velocity dt

x and y are displacement. K1 and K 2 are spring constraints

B = damping coefficient f =

Md 2 ( x − y ) dy + K1 ( x − y ) + K 2 y + B dt dt 2

This is in the form of differential equation.

Exercise 12 . . . 1. 2. 3. 4.

5. 6. 7. 8.

Briefly explain input-output relationship. What is difference between open loop control system and close loop control system? What is transfer function? Write a short not on. (a) Mechanical system (b) Translation system (c) Inertia force (d) Damping force (e) Spring force (f) Inertia torque (g) Damping torque (h) Spring torque What is D’ Alemberts principle? What is difference between mechanical translational system and electrical system? What is difference between translational system and rotational system? Write down analog of mechanical rotational system to electrical system.

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13 The Product Design Process 13.1

INTRODUCTION

What is design? If you search the literature for an answer to that question, you will find about as many definitions as there are designs. Perhaps the reason is that the process of design is such a common human experience. Webster’s dictionary says that to design is “to fashion after a plan”, but that leaves out the essential fact that to design is to create something that has never been. Certainly an engineering designer practices design by that definition, but so does an artist, a sculptor, a composer, a playwright or many other creative member of our society. Thus, although engineers are not the only people who design things it is true that the professional practice of engineering is largely concerned with design; it is frequently said that design is the essence of engineering. To design is to pull together something new or arrange existing things in a new way to satisfy a recognized need of society. An elegant word for “pulling together” is synthesis. We shall adopt the following formal definition of design: “Design establishes and defines solutions to and pertinent structures for problems not solved before or new solutions to problems which have previously been solved in a different way.” The ability to design is both a science and an art. The science can be learned through techniques and methods to be covered in this course, but doing design best learns the art. It is for this reason that your design experience must involve some realistic project experience. The emphasis that we have given to the creation of new things in our discussion of design should not unduly alarm you. To become proficient in design is a perfectly attainable goal for an engineering student, but its attainment requires the guided experience that we intend this course to provide. Design should not be confused with discovery. Discovery is getting the first sight of or the first knowledge of something, as when Columbus discovered America. We can discover what has already existed but has not been known before. But a design is the product of planning and work. We will present a structured design process to assist you in doing design. We should note that a design may or may not involve invention. To obtain a legal patent on an invention requires that the design be a step beyond the limits of the existing knowledge (beyond the state of the art). Some designs are truly inventive, but most are not. Good design requires both analysis and synthesis. Typically, we approach complex problems like design by decomposing the problem into manageable parts. Because we need to understand how the part will perform in service, we must be able to calculate as much about the part’s behavior as possible

292

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by using the appropriate disciplines of science and engineering science and the necessary computational tools. This is called analysis. It usually involves the simplification of the real world through models. Synthesis involves the identification of the design elements that will comprise the product, its decomposition into parts and the combination of the part solution into a total workable system. At your current stage in your engineering education, you are much more familiar and comfortable with analysis. You have dealt with courses that were essentially disciplinary. For example, you were not expected to use thermodynamics and fluid mechanics in a course in mechanics of materials. The problems you worked in the course were selected to illustrate and reinforce the principles. If you could construct the appropriate model you usually could solve the problem. Most of the input data and properties were given and there usually was a correct answer to the problem. However, real world problems rarely are that neat and circumscribed. The real problem that your design is expected to solve may not be readily apparent. You may need to draw on many technical disciplines (solid mechanics, fluid mechanics, electromagnetic theory etc.) for the solution and usually on no engineering disciplines as (economics, finance, law, etc.). The input data may be fragmentary at best and the scope of the project may be so huge that no individual can follow it all. If that is not difficult enough usually the design must proceed under severe constraints of time and/or money. There may be major societal constraints imposed by environmental or energy regulations. Finally in the typical design, you rarely have a way of knowing the correct answer. Hopefully your design works but is it the best most efficient design that could have been achieved under the conditions? Only time will tell. We hope that this has given you some idea of the design environment and the design process, the design environment and the design process. One way to summarize the challenges presented by the design environment is to think of the Four ‘C’s of design. One thing that should be clear by now is how engineering design extends well beyond the boundaries of science. The expanded boundaries and responsibilities of engineering create almost unlimited opportunities for you. In your professional career, you may have the opportunity to create dozens of designs and have the satisfaction of seeing them become working realities. A scientist will be lucky if he makes one creative addition to human knowledge in his whole life and many never do so. A scientist can discover a new star but he cannot make one. He would have to ask an engineer to do it for him.

The Four ‘C’s of Design Creativity l Requires creation of something that has not existed before or not existed in the designer’s mind before Complexity l Requires decisions on many variables and parameters Choice l Requires making choices between many possible solutions at all levels from basic concepts to smallest detail of shape Compromise l Requires balancing multiple and sometimes conflicting requirements

The Product Design Process

13.2

293

PRODUCT DESIGN PROCESS

Importance of Product Design The engineering design process can be applied to several different ends. One is the design of products whether they be consumer goods and appliances or highly complex products such as missile systems or jet planes. Another is a complex engineered system such as an electric power generating station or a petrochemical plant while yet another area is the design of a building or bridge. The principles and methodology of design can be usefully applied in each of these situations. However the emphasis in this book is on product design because it is the area in which many engineers will apply their design skills. Moreover examples taken from this area of design are easier to grasp without extensive specialized knowledge. U.S. manufacturers began to lose their world domination of markets gained after world war II in the late 1960 and 1970 as overseas producers entered the large attractive U.S. marketplace. Cameras, typewriters, electronic products (radios, television sets, hi-fi sets), and copiers were the first products to be displaced but as new products were developed (tape recorder and fax machine) a strong U.S. producer never emerged. In the 1980’s Japanese automotive producers gained an appreciable share of the U.S. market and German companies did the same for the high end of the auto market. Initially the competitive advantage for foreign manufacturers was that you should adopt the practice of making all your design calculations in a bound notebook. In that way you won’t be missing a vital calculation when you are forced by an error to go back and check things out. Just draw a line through the part in error and continue. It is of special importance to ensure that every equation is dimensionally consistent. Engineering sense checks have to do with whether the answer “feels right.” Even though the reliability of your feeling of rightness increases with experience you can now develop the habit of staring at your answer for a full minute rather than rushing on to do the next calculation. If the calculated stress is 106 psi you know something went wrong. Limit checks are a good form of engineering sense check. Let a critical parameter in your design approach some limit (zero, infinity, etc.), and observe whether the equation behaves properly. We have stressed the iterative nature of design. An optimization technique aimed at producing a robust design that is resistant to environmental influences (water vapor, temperature, vibration, etc.), most likely will be employed to select the best values of key design parameters. The management decision as to when to “freeze the design” will be dictated chiefly by considerations of time and money.

Communication of the Results It must always be kept in mind that the purpose of the design is to satisfy the needs of a customer or client. Therefore, the finalized design must be properly communicated, or it may lose much of its impact or significance. The communication is usually by oral presentation to the sponsor as well as by a written design report. A recent survey showed that design engineers spend 60 per cent of their time in discussing designs and preparing written documentation of designs, while only 40 per cent of the time is spent in analyzing designs and doing the designing. Detailed engineering drawings, computer programs and working models are frequently part of the “deliverables” to the customer. It hardly needs to be emphasized that communication is not a one-time occurrence to be carried out at the end of the project. In a well-run design project there is continual oral and written dialog between the project manager and the customer.

294

13.3

Mechanical Systems and Design

CONSIDERATIONS OF A GOOD DESIGN

Design is a multifaceted process. To gain a broader understanding of engineering design, we group various considerations in good design into three categories:(1) design requirements (2) life cycle issues and (3) regulatory and social issues.

13.3.1

Design Requirements

It is obvious that to be satisfactory the design must demonstrate the required performance. Acceptable performance is the first but far from being only design requirement. Performance measures both function and behavior of the design i.e. how well the device does what it is designed to do. Performance requirements can be divided into functional performance requirements and complementary performance requirements. Functional requirements address such capacity measures as force, strength, energy or material flows power and defection. They also are concerned with the efficiency of the design, its accuracy and sensitivity. Complementary performance requirements are concerned with the useful life of the design, its robustness to factors in the service environment, its reliability and ease economy and safety of maintenance. Issues such as built-in safety features and the noise level in operation must be considered. Finally the design must conform to all legal requirements and design codes. A variety of analysis techniques must be employed in arriving at the features of a component in the design. By feature we mean specific physical attributes such as shape dimensions or material properties. The digital computer has had a major impact in this area by providing powerful analytical tools based on finite element analysis and finite difference. Calculations of stress temperature and other field-dependent variables can be made rather handily for complex geometry and loading conditions. When these analytical methods are coupled with interactive computer graphics we have the exciting capability known as Computer Aided Engineering (CAE). Next to performance requirements, we have physical requirements. These pertain to such issues as size, weight, shape and surface finish. Environmental requirements deal with two separate aspects. The first concerns the service conditions under which the product must operate. The extremes of temperature, humidity, corrosive conditions, dirt vibration, noise etc. must be predicted and allowed for in the design. The second aspect of environmental requirements pertains to how the product will behave with regard to maintaining a safe and clean environment i.e., green design. Among these issues is the disposal of the product when it reaches its useful life. Aesthetic requirements refer to “the sense of the beautiful.” They are concerned with how the product is perceived by a customer because of its shape, color, surface, texture and also such factors as balance, utility and interest. This aspect of design usually is the responsibility of the industrial designer as contrasted with the engineering designer. The industrial designer is an applied artist. Decisions concerning the appearance of the product should be an integral part of the initial design concept. Manufacturing technology must be intimately connected with product design. There may be restrictions on the manufacturing processes that can be used because of either selection material or availability of equipment within the company. The final major design requirement is cost. Every design has requirements of an economic nature. These include such issues as product development cost, initial product cost, life cycle product cost, tooling cost and return on investment. In many cases, cost is the most important design requirement, for if preliminary estimates of product cost look unfavorable the design project may never be initiated.

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Cost enters into every aspect of the design process. Therefore, we have considered the subject of economic decision-making (engineering economics).

13.3.2

Total Life Cycle

The total life cycle of a part starts with the conception of a need and ends with the retirement and disposal of the product. Material selection is a key element in the total life cycle in selecting materials for a given application. The first step is evaluation of the service conditions. Next the properties of materials that relate most directly to the service requirements must be determined. Except in almost trivial conditions, there is never a simple relation between service performance and material properties. The design may start with the consideration of static yield strength but properties that are more difficult to evaluate such as fatigue, creep, toughness, ductility and corrosion resistance may have to be considered. We need to know whether the material is stable under the environmental conditions. Does the microstructure change with temperature? Does the material corrode slowly or wear at an unacceptable rate? Material selection cannot be separated from productivity. There is an intimate connection between design and material selection and the production processes. The objective in this area is a trade off between the opposing factors of minimum cost and maximum durability. Durability is concerned with the number of cycles of possible operation i.e. the useful life of the product. Current societal issues of energy conservation, material conservation and protection of the environment result in new pressures in selection of materials and manufacturing processes. Energy costs once nearly totally ignored in design are now among the most prominent design considerations. Design for materials recycling also is becoming an important consideration.

13.4

REGULATORY AND SOCIAL ISSUES

Specifications and standards have an important influence on design practice. The standards produced by such societies as ASTM and ASME represent voluntary agreement among many elements (users and producers) of industry. As such they often represent minimum or least common denominator standards. When good design requires more than that, it may be necessary to develop your own company or agency standards. On the other hand because of the general nature of most standards, a standard sometimes requires a producer to meet a requirement that is not essential to the particular function of the design. The code of ethics of all professional engineering societies requires the engineer to protect public health and safety. Increasingly, legislation has been passed to require federal agencies to regulate many aspects of safety and health. The requirements of the Occupational Safety and Health Administration (OSHA), the Consumer Products Safety Commission (CPSC), and the Environmental Protection Agency (EPA) place direct constraints on the designer. Several aspects of the CPSC and the regulations have far reaching influence on product design. Although the intended purpose of a particular product normally is quite clear the unintended uses of that product are not always obvious under the CPSC regulations. The designer has the obligation to foresee as many unintended uses as possible, then develop the design in such a way as to prevent hazardous use of the product in an unintended but foreseeable manner. When unintended use can not be prevented by functional design, clear and complete unambiguous warning must be permanently attached to the products. In addition the designer must be cognizant of all advertising material, owner manuals and operating instructions that relate to the product to ensure that the contents of the material are consistent with safe operating procedures and do not promise performance characteristics that are beyond the capability of the design.

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Mechanical Systems and Design

An important design consideration is adequate attention to human factors in engineering, which uses the sciences of biomechanics, ergonomics, and humans can use engineering psychology to assure that these are designed efficiently by humans. It applies physiological and anthropometrics data to such design features as visual and auditory display of instruments and control systems. It is also concerned with human muscle power and response times. Morris Asimov1 was among the first to give a detailed description of the complete design process in what he called the morphology of design.

13.5

CONCEPTUAL DESIGN

Conceptual design is the process by which the design is initiated. Carried to the point of creating a number of possible solutions and narrowed down to a single best concept, it is sometimes called the feasibility study. Conceptual design is the phase, which requires the greatest creativity, involves the most uncertainty and requires coordination among many functions in the business organization; the following are the discrete activities that we consider under conceptual design: l

l

l

Identification of customer need: The goal of this activity is to completely understand the customer’s needs and to communicate them to the design team. Problem definition: The goal of this activity is to create a statement that describes what has to be accomplished to satisfy the needs of the customer. This involves analysis of competitive products, the establishment of target specifications and the listing of constraints and trade offs. Quality Function Deployment (QFD) is a valuable tool for linking customer needs with design requirements. A detailed listing of the product requirements is called a Product Design Specification (PDS). Gathering information: Engineering design presents special requirements over engineering research in the need to acquire a broad spectrum of information. This subject is covered in Chapter 4. 1. Decide what performance characteristics and features you want the product to have. 2. Operate the product for several hours and observe its performance. Does, it meet the criteria you have established? Is it a quality product? 3. Disassemble each product into its components, for each component describe its intended function, the material from which it is made and the most likely process used for its manufacture. 4. What physical principles were used to achieve the function of each component? Refer to the TROZ tables in Chapter 5 as an aid. 5. Suggest how the design of the product could be improved with respect to performance of human factors design for the environment or cost. 6. Take photographs of common consumer products or tear pictures out of old magazines to build a display of industrial designs that appeal to you and designs that you consider need improvement. Be able to defend your decisions on the basis of aesthetic values. 7. Look at the website http;/www.baddesigns.com/examples.html for examples of poor userfriendly design. Then from your everyday environment identify five other examples. How would you change these designs to be more users friendly?

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8. Read the boxed example on the comeback of the diesel engine. Dig deeper into the story by learning more about diesel engines. What do you think is the likelihood of having a diesel engine in your sport utility vehicle?

13.6

HUMAN FACTORS IN DESIGN/ERGONOMICS

Ergon + Nomon = Work + Natural Laws (Greek) (English) The area of study covered under it comprises two groups–one deals with anatomy, physiology, psychology and the other group of engineering sciences such as physics, mathematics, material science and design. So, human factors engineering or ergonomics brings together two groups of specialists: l l

Those of machines and process (Engineering Sciences) Those of human capabilities (Social Sciences)

The knowledge of human aspects is essential for a designer while designing any system wherein human being also constitutes one of the components. Man applies himself both physically as well as mentally when he is part of such man-machine systems. For example, when he winds a wristwatch, dials a telephone, uses an electric iron, operates micro-oven, drives a car, and tunes a radio and so on. The user (man) of the system is expected to be compatible with the rest of the components in the system which is largely outside the designer’s control. Hence the compatibility between the components of the system can be ensured only when the system is designed around the user. Such a design is possible only when we understand the manner in which man and machine interact in a system.

Anthropometrics Data One of the primary responsibilities of ergonomics is to provide data about body size and dimension of people operating with given space. Such a study forms a part of an anatomist and is called Anthropometrics (man as occupant of space). When variation between individuals is unusually large then statement of average cannot be suitably employed, e.g., most adult are between 1.5 and 1.85m tall but some are 2.5m tall. If all doorways are made to cater to a select few (2.5m), it will not be a wise decision. So, anthropometrics data are expressed in the form of 5th, 50th and 95th percentile. Things to be considered include body size and posture (forward, upward and side reach), posture of sitting, etc. People vary in size, weight, strength and skill. Considerable anthropometrics data on the dimensions of various parts of the human body have been collected and serve as a basis for the design of tools and systems. Investigators have tabulated data on the reach, range of motion, speed of response, strength, sitting height, working height, and other variables in the machine-worker relationship. The reach required to operate a piece of equipment should be no greater than the shortest reach of all persons who are expected to operate it. All controls and information displays should be located for clear visibility. Switches should be located so that all the off position are in the same direction. Gauges should be arranged so that the indicators point in approximately the same direction when the devices are in operation normally. Thus the operator can quickly spot deviation from normal. Levers and hand wheels should be of the proper size and located so that sufficient operating force may be applied in the appropriate direction.

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The Work Environment The environment in which people work can effect their comfort, health, and productivity. Some environmental variables to be considered are temperature, noise, and lighting.

Temperature Humans can perform under various combinations of temperature, humidity and air movements. The effects of these variables depend on the strenuousness of the work task and individual adaptation to the conditions.

13.7

A CASE STUDY: HUMAN FACTOR CONSIDERATIONS IN MAN-MACHINE ENVIRONMENT SYSTEM DESIGN

Abstract This paper deals with the ergonomic considerations in man-machine environment system design. It is very well established that a man’s working environment has a profound influence on his working efficiency and safety. This system environment has to be adapted to the human characteristics, capabilities and limitations. The basic issues relating environment with manufacturing have been addressed. A brief discussion on the physiological requirements of artificial lighting, lighting for fine work and for offices using VDTs has been included. Physiological and psychological effects of noise, protection against noise and how noise and vibrations affect performance in manufacturing have also been discussed. Recommendations have been made regarding indoor climate; comfort indoors, day light, colours and music for pleasant work environment. Data available regarding environmental factors, which have been found to give beneficial results, have also been included.

Key Words Work environment, man-machine system, human factor, and workplace

13.7.1 Introduction Efficiency of a man-machine environment system depends upon the system design, which in turn is based upon the principles of industrial engineering with special emphasis on human factors. These fields are so wide and the volume of data available is so large and scattered that it is very difficult to consider all the relevant information regarding the environmental considerations around man-machine interface. In the present work, an endeavor has been made to offer such data at one place and present it in a concise form, for the use of ergonomists, work design engineers and all those who are interested in environmental issues in manufacturing.

13.7.2 Lighting in the Environment Illumination levels in the open vary between 2000 to 100,000 Lx in the day while at night artificial light of 50–500 Lx is normal. Among the light sources, direct filament lamps cause hard shadows, discomfort and headaches while the fluorescent tubes provide high output of light, long life, and low luminance when adequately shielded. Fluorescent tubes are, therefore, preferred.

13.7.2.1 Physiological Requirements of Artificial Lighting Illumination levels: Current recommendations regarding illumination levels by European Standards for different types of activities are given in Table 13.1 (Mc Cormik, E.J. et al. 1987)

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Table 13.1

299

European standards for intensity of light in Lux

Type of activity Assembly Work: Rough/Precise/Delicate Work on tool making machine: Rough/Fine/Very Precise Book keeping and Office work Technical Drawing VDT Workstations: Well printed/data Reduced readability source documents and data entry tasks

Recommended intensity D/N 250/1000/1500 250/500/1000 500 1000 300–500 500–700

If a strong light is necessary at some stations, a good general illumination of the order of 30% of working light is recommended. Designers of VDT workstations or other workplaces are advised to select colours of similar brightness for the different surfaces. Renounce eye-catching effects, avoid reflecting materials and give preference to matt colours. Arrangement and overall distribution of lights The following recommendations should be considered: 1. Source of lights should be shaded or covered with glare shields to keep the luminance within 200 Cd/sq.m. 2. The line of eye should be at 30 deg. to the source of light while fluorescent tubes should be at right angles to the line of sight. Use more lamps of lower power than a few of higher power. 3. Avoid glare due to reflection, if high-powered lamps are used. 4. Avoid use of reflective colours and materials on machines and apparatus.

13.7.2.2 Lighting for Fine Work For precise assembly work or delicate mechanical tasks, use of frontal lighting with diffusive or frosted glass screen is recommended. The lamps should not be directly visible. The source should emit light from a large area. The diffusing screens should be broad and deep so as to make the illumination of the workbench as uniform as possible. Phase shifted fluorescent tubes are preferable to filament lamps since the latter give off more heat.

13.7.2.3 Lighting for VDT Offices The contrast ratio between the screen and the source document should not exceed 1:10. Recommended illumination levels at VDT workstations are given in Table 13.1. All other surfaces in the visual environment should have luminance (reflectances) lying between those of the screen and the source documents. The VDT workstations must be placed at right angles to the window. The light fixtures should be installed parallel and on either side of the operator-screen axis.

13.7.3 Noise Noises can significantly and adversely affect operator performance and health (Kalpokiean and Schmid, 2001). Excessive noise, cause messages to be misunderstood, hearing loss or permanent deafness. In the following paragraphs, this aspect will be discussed in detail.

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13.7.3.1 Measurement of Noise Noise is any disturbing sound, which is discordant and is not in harmony with the intentions at that moment. Noise load is the extent of noise and is measured in two units. Equivalent level of sustained noise that expresses the average level of sound energy during a given period of time. The summated frequency level is measured with a sound level indicator and a frequency counter operating over a given time. Human ear is capable of sensing sound pressures in the range of 20 Pa, to 20 × 106 Pa. Such a wide range cannot be accommodated on a normal scale. Decibel unit, a logarithmic scale, is used to express the sound level, according to the formula: LdB = 20 log (Px/P0) Where LdB is sound level in decibel, Px = sound pressure in Pa and P0 = basic sound pressure internationally fixed at 20 Pa. The apparent loudness of a sound depends a good deal on its pitch or frequency. Human ear is sensitive to sounds in the frequency range between 16 Hz–20,000 Hz. The greatest sensitivity lies in the range 2000–5000 Hz. At 4000 Hz noise a hearing loss of 50–60 dB occurs. Pitch of the human voice is between 300 and 700 Hz.

13.7.3.2 Source of Noise There are basically two types of noise that affect the industrial workers (External and Internal). External noise is disturbing in design offices, schools, manufacturing centers, along side of a road. Internal noise is the noise produced within the industry. Nowadays the so called weighted sound levels have come into use as measures of loudness (Grandjean E. 1988). External (Street) noise (dB) (a) Traffic: Heavy, in day By night Moderate, in day By night Light, in day By night

(b) Industry-soft drink plant Pneumatic hammer/chisel (c) Neighbouring industries e.g. spinning and weaving

Internal (Industrial) noise (dB) 65–75 55–65 60–65 50–55 50–55 40–45

90–100 115–120 105

(a) Machines, motors saw mills running. (b) Stamping presses milling, weaving (c) Rocking sieve, Pneumatic riveter (d) Pneumatic chiselPneumatic hammer

90–95 100–105 105–115 120

Noise in offices and from office machines (a) Very small offices, drawing offices 40–45 (b) Large 46–52 (c) Large noisy offices 53–60 (d) Copying machine 55–70 (e) VDT-cooling fan 30–60 (f) Matrix printer with hood 60–62

13.7.3.3 Physiological and Psychological Effects of Noise Stimulation from noise may initially lead to temporary deafness (threshold shift) but repeated stimulation causes slow, progressive degeneration of sound sensitive cells of the inner ear. Threshold of hearing

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shifts by 10dB up to noise levels of 90dB. At 100 dB however this shift amounts to 50–60 dB. LdB. for an eight hour day should not exceed 85dB (A) to safeguard against damage to hearing. Noise may cause impaired concentration, and alertness, disturbance in sleep and annoyance. If background noise level is at least 10dB below the level of the speaking voice, speech comprehension is considered to be unimpaired. During industrial activity if the voice is used to convey information it should not exceed 65–70 dB at 1m distance and the background noise level should not exceed 55–60 dB. If the verbal communication contains unfamiliar names and strange words the background noise should not exceed 45–50 dB.

13.7.4 Effect of noise on performance Noise impairs performance and output. A decline in performance may be attributed to certain kinds of jobs requiring skill and dexterity. Many studies (Jerison, H.J. 1959) have shown that a level of noise above 90 dB, discontinuous or unexpected, can impair mental performance. Noise can be, in the right circumstances, positively stimulating. Performance improves in boring jobs and situations where there are many other distractions. Sometimes, if these distractions can be concentrated on the dominant noise even mental performance may be improved. Conversations of other people are distracting not only because of loudness but also because of their information content. In manufacturing industries the following experiences have been reported (Grandjean E., 1988): 1. In machine shop a reduction of about 25 dB in the noise led to 50% fewer rejected pieces. 2. In an assembly shop a reduction in noise level by 20 dB raised production by 30%. 3. In a typing pool a reduction in noise level by 25 dB was accompanied by a 30% reduction in typing errors. It has also been reported (Grandjean E., 1988) that exposure to noise produces an increase in blood pressure, acceleration of heart rate, increase in metabolism and muscular tension, slowing down of digestive organs, and contraction of blood vessels of the skin.

13.8

PROTECTION AGAINST NOISE

In manufacturing environment noise is a common source of disturbance. Some common methods used to counter noise are discussed in the following paragraph.

Protective Planning This is done at architect’s drawing board by using proper building materials and in planning sub-divisions of the buildings. For example, planning department, works requiring concentration and skill and offices requiring mental work may be located away from the sources of noise. Various building items dampen the noise. Normal single doors, double doors and heavy special doors dampen the noise by 25, 35 and 42 dB (A) respectively. Windows: Single glazing, double-glazing and double-glazing with felt packing dampen the noise level by 22, 25 and 32 dB(A) respectively. Dividing brick walls: 90 mm, 300 mm and double wall 120 mm, reduce noise evel by 40, 52 and 62 dB(A) respectively.

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Tackling Noise at Source Noise can be tackled at source, for example, metal wheels may be replaced by rubber wheels, fabric belts may be used for belt drives etc. Enclosing the source of noise by using an airtight housing of suitable material may reduce radiated noise by 20–30 dB (A). Sound absorbing panels absorb part of the sound and so reduce reflection back into the room. Sound reductions of 5–10 dB can be expected at such places. This will be of no value to the worker working close to a source of noise.

Personal Sound Protection If a worker is forced to work in noisy environment and technological means fail to reduce noise to a safe level of below 85 dB (A), personal ear protection is used as a last resort with the following possibilities: 1. Earplugs made of cotton wool, wax or synthetic material (select one) are used for noise levels 85–100 dB (A). They reduce noise level by about 30 dB. If used regularly, they may cause irritation (such as aural eczema). 2. Protective caps (ear-muffs) cover the whole of the external ear and are used for noise levels above 100 dB (A). If correctly fitted, earmuffs can reduce noise levels by about 40–50 dB in the frequency range of 1000–8000 Hz. Periodic audiometric tests should be carried out for workers working in noisy places, to detect noise damage at its onset to devise protective measures against noise and for the introduction of personal protection in the factories.

13.8 VIBRATIONS Vibrations are mechanical oscillations produced by periodic movements of a body about its resting position. The effects of vibrations on people have been discussed in detail by Grether W.F. (1971), Grandjean E. (1988) and Griffin M.J., (1973). Human body does not vibrate as a simple mass; different body parts have their own natural frequencies. As the forcing frequency approaches the natural frequency, the amplitude of forced vibrations increases tremendously. The body of a sitting person responds to vertical vibrations as shown in Table 13.2.

Table 13.2

Response of human body parts to vertical vibrations

Forcing frequency range (Hz) 3–4 4 5 20–30 60–90 100–200

Human body parts affected (sitting person) Strong resonance in the cervical vertebrae Peak of resonance in the lumbar vertebrae Very strong resonance in the shoulder trunk and neck Resonance between head and shoulders Resonance in eyeballs Resonance in the lower jaw

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13.8.1.1 Physiological Effects Vibrations affect visual perception, psychomotor performance, generate muscle reflexes and result in increased energy consumption, heat rate and respiratory rate. Vibrations impair visibility and visual perception, mental processing of information and mechanical oscillations may lead to risk of error and accidents. Vibrations cause annoyance and nuisance depending upon the inducing frequency, rate of acceleration of the oscillations and the length of time they continue. The most intense subjective frequency lies between 4–8 Hz. The threshold of “very severe” intensity comes at an acceleration of about 10 m/s2. At an acceleration of about 15 m/s2, the vibrations become dangerous and intolerable.

13.8.1.2 Damage to Health Exposure to vibrations at the workplace can, if repeated daily, lead to morbid changes in the organs affected. Vertical vibrations when standing or sitting down on a seat can cause degenerative changes in the spine whereas the vibrations of power tools affect mainly the hands and arms. Tractor drivers suffer from disc troubles and arthritic complaints in the spine and intestinal ailments, prostrate troubles and hemorrhoids (Guignard J.C. et al 1972). Heavy and prolonged vibrations cause wear and tear of intervertebral discs and joints. Workers using hand power tools for years together may suffer ailments of hands and arms. Tools with a frequency of vibration below 40Hz (like heavy pneumatic hammer can cause degenerative symptoms in bones, joints and tendons of hands and arms leading to arthritis in the wrist, elbow and in the shoulders. Power tools with frequencies between 40 and 300 Hz usually have very small amplitude of oscillation (0.2–5 mm). These vibrations are dampened in the tissues. Such vibrations have ill effects on the blood vessels and nerves of the hands resulting in one or more of the fingers going “dead”—becoming white or bluish, cold and numb. After a while the finger becomes pink again and is painful. Vibrations become intolerable in the following circumstances: 1–2 Hz at an acceleration of 3 to 4 g, 4–14 Hz at acceleration of 1.2 to 3.2 g and above 14 Hz at an acceleration of 5 to 9 g.

13.8.2 Indoor Climate Indoor climate means the physical conditions under which the work is carried out. Components of indoor climate include: air temperature, surrounding surfaces temperature, air humidity and air movements.

13.8.2.1 Thermal Regulation in Man A constant body core temperature fluctuating a little around 37 °C ( ± 1 °C) is necessary in brain, heart and abdominal organs for vital functions. Skin temperature may go down to 35 °C up to 20 mm below the skin in cool surroundings. In warmer surroundings, however, it may be 35–36 °C a few mm below the skin. The muscles get a few degrees warmer during heavy work compared to the resting temperature.

13.8.2.2 The Control Process Heat control center in brain acts as a thermal regulator. Information about body temperature is received by nerve cells of the control center or from heat sensitive nerves in the skin. Heat control center sends impulses to regulatory mechanism to keep core temperature constant. Heat is diffused by circulatory system, and is lost by sweat. When human body is exposed to a sudden cooling, shivering takes place, increasing metabolic heat in the muscle instantaneously.

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13.8.2.3 Heat Exchange The human body converts chemical energy into mechanical energy and heat. This heat is used to maintain a constant core temperature. Any excess heat dissipates to the surroundings. There is, thus a constant heat exchange between body and surroundings regulated by the physiological control and the ordinary laws of conduction, convection and radiation. Of the total heat exchange 25% goes by convection, 25% (600 Kcal. per day) goes by evaporation and the remaining heat is lost by radiation. Radiation loss in summer is 1000–1500 Kcal. representing about 40% of the total heat lost from body. A relative humidity of about 40–50% is desirable in heated rooms and a relative humidity of 30% or less is unhygienic and affects mucus membrane of the nose.

13.8.2.4 Recommendations for Comfort Indoors (a) Air temperatures of 21 °C in winter and 20–24 °C in summer are comfortable. (b) Surface temperature of adjacent objects should be at air temperature or should not be ± 2– 3 °C different. (c) Relative humidity of room air should be 30% or more in winter and 40–60% in summer. (d) Draughts should be at the head or knee level and should not exceed 0.2 m/s. At a relative humidity of 50% the comfortable room temperature is 21 °C for sedentary manual work, 19 °C for sedentary light manual work, 18 °C for standing light manual work, 17 °C for standing heavy manual work and 15–16 °C for extremely heavy manual work.

13.8.2.5 Air Pollution and Ventilation Recommendations regarding volume of air available per person around workplace and fresh air flow in meter cube per hour are shown in Table 13.3 (Grandjean E., 1988).

Table 13.3

Air pollution and ventilation

Volume/person in meter cube

5 10 15

Fresh air flow/person in meter cube/h Minimal

Desirable

35 20 10

50 40 30

13.8.2.6 Guidelines for Workers Working in Very Hot Conditions (a) A worker should be acclimatized to heat by stages, beginning with 50% of working time in heat and increase it by 10% each day. (b) The more the heat stress and the greater the physical effort under heat stress the longer and more frequent should be the cooling pauses. If the limit of heat tolerance exceeds, working day be shortened. (c) Workers should drink small amounts of fluid at frequent intervals but never more than 1/4 liter every 10–15 minutes.

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(d) Recommended drinks are: lightly sweetened tea or coffee or soup. If large quantities of fluid are needed, plain water is best. (e) Iced drinks, fruit juices and alcoholic drinks are not recommended. (f) Drinks should be available close to the worker so that he can drink whenever he feels the need. (g) Where radiant heat is excessive, the worker must be protected by special goggles, screens, protective clothings against the risk of burning his eyes and hands. In addition, attempt should be made to reduce the impact of heat on the worker by improving ventilation, artificial drying of air and screens to protect against radiant heat. Physiological means are adopted to express heat load and degree of physical work (in Kcal, it can not be easily expressed). Upper limits of these physiological parameters for working in heat for entire day are: (a) Heart rate (b) Rectal temperature (c) Evaporation of sweat

100–110 beats/min. 38 °C 0.5 lit/h

CONCLUSION 1. Use of extra artificial lighting is necessary for jobs requiring precision. A general illumination of 30% of working light is recommended. Renounce the use of reflecting materials and eyecatching effects. 2. Noise impairs performance and causes damage to hearing. At 4000 Hz noise a hearing loss of 50–60 dB (A) occurs. Noise should also not exceed 85 dB (A) to safeguard against damage to hearing. Music improves performance in boring jobs or situations having distractions. 3. Methods used to counter noise include: protective planning at architect’s board, tackling it at source or providing personal sound protectors. 4. Vibrations affect visual perception and mental processing of information. Mechanical oscillations may lead to risk of error and accidents. Tools with a vibration frequency below 40 Hz (e.g. pneumatic hammer) can affect bones, joints, tendons of hands and arms leading to arthritis in wrist, elbow and shoulder. Between 40–300 Hz (0.2–5 mm amplitude) the vibrations may cause one or more fingers going dead, cold and numb. Vibrations are intolerable in the range: 1–2 Hz, 4–14Hz and above 14Hz accelerations of 3–4 g, 1.2–3.2 g and 5–9 g respectively. 5. Air temperatures in winter 21 °C (R.H. 30%) and in summer 20–24 °C (R.H. 40–60%) are comfortable and air draughts above head or below knee level should not exceed 0.2 m/s. 6. Upper limits of physiological parameters for working in heat for entire day are: Heart rate100– 110 beats min; Rectal temperature 38 °C; Evaporation of sweat 0.5 lit/h. Example 1: A travelling salesman of a company is to carry his samples around. He may either travel by a taxi in which case the cost will be reimbursed by the company or he may buy and use his own car in which case the company will give a travelling allowance at the rate of Rs 1.50 per km travelled. The expenditure involved in owning a car. Depreciation Rs 5,000 per year. Tax and insurance Rs 1200 per year. Maintenance Rs 1000 per year.

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Plus Rs 0.20 per km. Fuel and oil Rs 0.90 per km. Plot the per km costs and compute the distance travelled per month which will make owning a car an difficult proposition. Solution: First, we determine the fixed and variable costs in the given list of cost elements. Fixed costs: (i) Depreciation Rs (ii) Tax and insurance (iii) Maintenance Total fixed costs

: : : :

5000 per year 1200 per year 1000 per year 7200 per year

Variable costs: (i) Maintenance Rs (ii) Fuel and oil Total variable costs

0.20 per km 0.90 per km Rs 1.10 per km

The total cost (TC) incurred per km of travel, if he used his car, is calculated as below, for various arbitrary distance of travel. If he travels, (i) (ii) (iii) (iv) (v)

5,000 kms TC/km = (7200/5000) + 1.1 = Rs 2.54 10,000 kms TC/km = (7200/10000) + 1.1 = Rs 1.32 15,000 kms TC/km = (7200/15000) + 1.1 = Rs 1.58 17,000 kms TC/km = (7200/17500) + 1.1 = Rs 1.61 20,000 kms TC/km = (7200/20000) + 1.1 = Rs 1.46

(Note: More the number of points, smoother the curve will be.)

Per KM cost or cost revenue

We know that the travelling salesman gets Rs 1.50 per km of travel. We plot the total cost per km and the income per km for various distances of travel. (see graph below)

BEP Income VC

1,84,000 5000

10000 15000 20000 Distance travelled (Kms)

The distance to be travelled (Q), which makes owning a car on attractive proposition, is calculated as follows: At the break-even point, the total cost/km is equal to the income per km.

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Hence, (7200/Q) + 1.1 = 1.5 He gets, Q = 18,000 kms per year or = (18,000/12 months) = 1500 kms per month on an average

Designing for Profit We know there can be more than one way of producing a given product. If there exists a production method (process) wherein both fixed as well as variable costs are lower relative to that of others while the rest of the things are equal, it is exactly the process what the designer is looking for. But seldom the designer encounters such situations. In most of the situations, one of the methods may call for larger fixed costs than that of others, but may result in lower variable costs per piece produced. However, a designer is interested in reducing the total costs to the maximum possible extent, retaining the utility of the product. He chooses that method which satisfies this requirement. Example 2: Consider two production processes A and B, available to a designer, with cost elements as follows: A B Fixed costs Rs 10,000 Rs 4,000 Variable costs per piece Rs 5.00 Rs 7.00 The designer is interested to know the range of production volume in which each of the processes is superior to the other, the production volume at which he can be indifferent between the processes, and the process he should choose for a given production volume. It is a simple task. We know total cost = fixed cost + variable cost Total cost (A) = 10000 + 5.00 × Q Total cost (B) = 4000 + 7.00 × Q where Q is production volume Plot the total cost lines corresponding to the production process A and B, against production volume as shown in the figure below: B

Total cost in Rs

A

10000

4000

1000 2000 3000 4000 Production volume in units (Q)

Solution: At the point of intersection, total cost of A and total cost B are equal, i.e. 10000 + 5.00 × Q = 4000 + 7.00 × Q Hence Q = (10000 – 4000)/(7.00 – 5.00) = 3,000 units As we can see from the graph, If Q is less than 3000 units, process B is better.

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If Q is greater than 3000 units, process A is better, and At Q = 3000, he can be indifferent between the processes. Example 3: A company makes curtain rods of size 2 mts in length. Three materials A, B and C are available. Each material calls for a different process and machine for manufacturing and their cost data is given as below. UPTU 2006 based Question Materials Items A B C Raw material cost Rs/mtr 2.25 2.75 3.00 Equipment cost Rs/year 6000 5000 3000 Labour cost Rs/rod 0.55 0.62 0.25 Plot the total cost vs. yearly production volume. If a sales volume of 10,000 rods/year is expected, which material should be used?

Total cost in Rs (X 1000)

Solution: We identify the fixed costs are variable costs in the list of items given as shown in the table: The total cost lines for materials A, B, C associated with quantity Q are given by, Material A: 6000 + (2 × 2.25 + 0.55) × Q Material B: 5000 + (2 × 2.75 + 0.62) × Q Material C: 3000 + (2 × 3 + 0.25) × Q When these straight lines are plotted for various arbitrary values of Q on a graph, we get

24

B C

18

A

12 6 5 3 934

2500 2 3 Production volume in units

4

The point of intersection between A and B is given by, 6000 + 5.05 × Q = 5000 + 6.12 × Q, we get Q = 934.5 rods The point of intersection between A and C is given by, 6000 + 5.05 × Q = 3000 + 6.25 × Q, we get Q = 2500 rods The point of intersection between B and C is given by, 5000 + 6.12 × Q = 3000 + 6.25 × Q, we get Q = 15,384.5 rods For a sales volume below 2500 rods, material C is better but for a sales volume greater than 2500 rods, material A is better. As a sales volume of 10,000 rods/year is expected, the material A should be used. The total cost associated with it is 6000 + 5.05 × 10,000 = Rs 56,500.

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Example 4: A small company makes certain rods of standard sizes 2 m in length. These are three materials A, B and C which can be used. Each material calls for a different process and machines for manufacturing and their cost data, may be summarized as below. Item

Material

Raw material (Rs/m) Equipment cost (Rs/year) Labour cost (Rs/Rod)

A

B

C

2.20 5,000 0.50

2.50 3,000 0.60

2.60 4,000 0.20

Draw the total cost versus the yearly production volume curve for each of the three materials. If a sales volume of 10,000 rods per year is anticipated, which material should be used? UPTU 2006 Solution: We identify the fixed costs and variable costs in the list of items given as shown in the table. The total costs lines for material A, B and C associated with Quantity Q are given by, Material A: 5000 + (2 × 2.20 + 0.50) × Q = 5000 + (4.40 + 0.50) × Q = 5000 + (4.90) × Q Material B: 3000 + (2 × 2.50 + 0.60) × Q = 3000 + (5.0 + 0.60) × Q = 3000 + (5.60) × Q Material C: 4000 + (2 × 2.60 + 0.20) × Q = 4000 + (5.20 + 0.20) × Q = 4000 + (5.40) × Q When these straight lines are plotted for various arbitrary values of Q, on a graph, we get;

Total cost in Rs (X 1000)

10

C

9

B

8

A

7 6 5 4 3

2000 2857.1 Production volume in units

The point of intersection between A and B is given by 5000 + (4.9)Q = 3000 + (5.6)Q ⇒ 5000 – 3000 = (5.6 – 4.9)Q ⇒ 2000 = 0.7Q

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⇒

Q=

2000 = 2857.1 rods 0.7

The point of intersection between A and C is given by 5000 + (4.9)Q = 4000 + (5.4)Q ⇒ 5000 – 4000 = (5.4 – 4.9)Q ⇒ 1000 = 0.5Q

⇒

Q=

1000 = 2000 rods 0.5

The point of intersection between B and C is given by 3000 + (5.6)Q = 4000 + (5.4)Q ⇒ (5.6 – 5.4)Q = 4000 – 3000 ⇒ 0.2Q = 1000

⇒

Q=

1000 0.2

⇒ Q = 5000 rods For a sale volume below 2857.1 rods, material B is better but for a sales volume greater than 2857 rods, material A is better. As a sales volume of 10,000 rod/year is anticipated the material A should be used. The total cost associated with it is 5000 + 4.9 × (10,000) = Rs 54,000

Exercise 13 . . . 1. 2. 3. 4. 5. 6.

Write down the human factor in mechanical system design/ergonomics. Briefly explain product design process. What is the importance of product design? In mechanical system design what is necessary for requirements of design? What is total life cycle? What is conceptual design?

Computer System Concept

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14 Computer System Concept 14.1

INTRODUCTION

A computer is more than a high-powered collection of electronic devices performing a variety of information processing chores. A computer is a system, an interrelated combination of components that performs the basic system function of input, processing, output, storage, and control, thus providing end users with a powerful information-processing tool. Understanding the computer as a computer system is vital to the effective use and management of computers. There are a wide variety of computer systems available in terms of size, complexity, and power. Regardless of size, every computer system includes hardware and software to perform processing functions. A computer is a system of hardware devices organized according to the following system functions. 1. Input: The input devices of a computer system include keyboards, touch screens, pens, electronic mouse, optical scanners, and so on. They convert data into electronic machinereadable form for direct entry or through telecommunication links into a computer system. 2. Processing: The Central Processing Unit (CPU) is the main processing component of a computer system (In microcomputers, it is the main microprocessor). In particular, the electronic circuits of the arithmetic-logic unit, one of the CPU’s major components, perform the arithmetic and logic functions required in computer processing. 3. Output: The output devices of a computer system include video display units, printers, audio response units, and so on. They convert electronic information produced by the computer system into human-intelligible form for presentation to end-users. 4. Storage: The storage function of a computer system takes place in the storage circuits of the computer’s primary storage unit, or memory, and in secondary storage devices such as magnetic disk and tape units. These devices store data and program instructions needed for processing. 5. Control: The control unit of the CPU is the control component of a computer system. Its circuits interpret computer program instructions and transmit directions to the other components of the computer system.

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Central Processing Unit

Input devices enter data and instructions into the CPU

Control unit interprets instructions and directs processing

Special purpose processors

Keyboard Mouse Touch Screen Optical Screen Voice recognition etc.

Cache memory

Output devices communicate and record information

Primary storage (memory) stores data and program instruction during processing

Secondary storage devices strore data and programmes for processing

Fig. 14.1

14.2

Arithmetic-logic unit performs arithmetic operations and makes comparisons

Visual display unit Printer Audio-response Physical control devices etc.

Magnetic disk and tape units, optical disks, etc.

The computer system concept

DATA FLOW DIAGRAM

The structured systems analysis develops a conceptual, logical, and graphical model of the system. It is developed with reference to the objective and taking into consideration the constraints under which the system operates. The model is developed with four symbols as given in figure below.

Flow of data

Process of transforming the data

Storage of data

Fig. 14.2

Entity

Computer System Concept

313

For example, the logical model of the customer order processing and issuing the order acceptance can be shown in the e model, using these symbols. Order Acceptance Customer Process Customer

Order

Customer

Product data

Customer data

Acceptance Order

Price data

Fig. 14.3 This model illustrates the following: 1. 2. 3. 4. 5.

Documents in the system. Sources of documents. Process center converting the customer order into the order acceptance. Use of stored data in the process center. Output or documents provided by the process center and its destination.

Such Data Flow Diagrams (DFDs) provide a logical clarity in terms of input, output, and use of stored data or master data already available in the organization or in the system. Detailing the system in a hierarchical manner can make the use of DFDs. In this process, we are detailing the activities, which are performed in the system in its logical order shown in Fig. 14.4.

Customer

Customer Order

Order acceptance document

Validate customer order

Commercial processing of customer order

Process for customer order acceptance

Print order acceptance

Customer data

Product data

Price data

Order acceptance data

DFD of customer order processing

Fig. 14.4

System flowchart

314

Mechanical System Design

The system flowchart defines the same operation in a pictorial form. It looks like a program flowchart, which can describe any organizational procedure. The symbols used are different from the programming flowchart. Each symbol specifies an operational task. (a) The system flowchart starts with a document symbol where the job applications are received daily. (b) The proceeding symbol is a manual operations’ symbol indicating the application done by the clerk. (c) The triangular symbol (point downward) indicates the temporary offline storage. (d) The next application is batch application to be input and verified. The result of these operations are the cards held for processing and for returning the original application to personnel department where the valid request are keyed into the system. The result is displayed on the monitor as a soft copy. These results can be printed out with the help of a printer. It is called proof listing used to verify the result of the process data. Proof listing and visual display occur at the place and information is transferred back to the user. The systems flow chart is a vital documentation tool that system analyst very commonly use and relies upon it. The different symbols used in the system flowchart are described in Table 14.1.

Table 14.1 Symbol

Flowchart symbols Description This is a punch card symbol and it denotes the use of punch card. It is a magnetic type model. It defines the use of magnetic tape. Online keyboard symbol defines the use of online keyboard to input data directly into the computer. Manual operations symbol and describes the offline manual operations in which data is handled without mechanical assistance. These are known as flow symbols. They indicate the direction of the processing of the data. This is a document symbol and it identifies any printed document used in processing of output. It is an offline storage symbol and it identifies the non computerized type of storage media including the tapes and printed materials etc. It is the display symbol; it defines the use of monitor, where the data is presented in soft copy format. Magnetic disk symbol defines the use of magnetic disk device. (Contd)

Computer System Concept

315

Symbol

Description Processing symbol denotes a program used in processing. Online storage symbol used for online storage device whose data is directly accessible to the computer. Keying operation symbol denotes the use of offline keyboard device to prepare data. Communication link symbol denotes the online transmission of data by telephonic line. Auxiliary operation symbol defines an offline mechanical operation that supplements the handling of data often by use of office machine.

14.3

CONVERSION OF MANUAL TO COMPUTER-BASED SYSTEMS

To increase our understanding of computer-based management information systems, we continue our transition from manual to computer system by describing the steps involved in making a conversion or changeover from the inventory accounting system of Fig. 14.5, assuming as we do so that a feasibility study has been made and that the system conversion is economical and feasible. The steps involved in the conversion are: Processor (Inventory Clerk)

Input

Orders and issues

Multiply unit cost by units issued

Outputs

Deduct units issued from balance on hand Deduct gross value of issue from dollar value of inventory

Updated records

Inventory status report Internal

Procedures manual working storage

External

Inventory records and files Storage

Fig. 14.5 Manual inventory accounting system

316

Mechanical System Design

(i) (ii) (iii) (iv) (v) (vi) (vii) (viii)

14.4

System description Input documents Output documents File design The program flowchart Computer program System verification Documentation

FUNDAMENTALS OF CAD/CAM

CAD/CAM involves the use of the digital computer to accomplish certain functions in design and production. CAD is concerned with the use of the computer to support the design engineering functions, and CAM is concerned with the use of the computer to support manufacturing engineering activities. The combination of CAD and CAM in the term CAD/CAM is symbolic of efforts to integrate the design and manufacturing functions in a firm into a continuum of activates, rather than to treat them as two separate and disparate activities, as they have been considered in the past. In this chapter, we discuss computer-aided design, computer-aided manufacturing, and how CAD/ CAM fits within the scope of CIM. A more detailed discussion of CAD/CAM is given below.

14.4.1

Computer-Aided Design

Computer-Aided Design (CAD) can be defined as any design activity that involves the effective use of the computer to create, modify, or document an engineering design. CAD is most commonly associated with the use of an interactive computer graphics system, referred to as CAD system. The term CAD/ CAM system is also used if it supports manufacturing as well as design application. There are several important reasons for using a computer-aided design system to support the engineering design function [1]. 1. To increase the productivity of the designer: This is accomplished by helping the designer to conceptualize the product and its components. In turn this helps to reduce the time required by the designer to synthesize, and document in design. 2. To improve the quality of the design: The use of CAD system with appropriate hardware and software capabilities permits the designer to do a more complete engineering analysis and to consider a larger number and variety of design alternatives. The quality of the resulting design is thereby improved. 3. To improve design documentation: The graphical output of a CAD system results in better documentation of the design than what is practical with manual drafting. The engineering drawings are superior, and there is more standardization among the drawings, fewer drafting errors and greater legibility. 4. To create a manufacturing database: In the process of creating the documentation for the product design (geometric specification of the product, dimensions of the components, materials specification, bill of materials, etc.) much of the required database to manufacture the product is also created.

Computer System Concept

317

The Design Process The general process of design is characterized by Shingley as an interactive process which consists of the following six phases: 1. 2. 3. 4. 5. 6.

Recognition of need Definition of the problem Synthesis Analysis and optimization Evaluation Presentation

These six design steps, and the interactive nature of the sequence in which they are performed, are depicted in Fig. 14.6. Recognition of need involves the realization by someone that a problem exists for which some corrective action can be taken in the form of a design solution. This recognition might mean identifying some defect in current machine design by an engineer or perceiving some new product opportunity by a salesperson. Definition of the problem involves a through specification of the item to be designed. This specification includes the physical characteristics function, cost quality, and operating performance. Synthesis and analysis are closely related and highly interactive in the design process. Consider the development of a certain product design. Each of the subsystems of the product would be conceptualized by the designer, analyzed, and improved through this computer-aided design. Recognition of need

Definition of problem

Synthesis

Analysis and optimization

Evaluation

Presentation

Fig. 14.6

Design process as defined by Shingley ([7]

Reprinted by permission from Mikell P. Grover and Emory W. Zimmers, CAD/CAM; Computer-Aided Designs and Manufacturing (Englewood Cliffs, NJ: Prentice-Hall Inc. 1984), p. 57 [1].

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Mechanical System Design

Analysis procedure, redesigned, analyzed again, and so on, the process would be repeated until the design has been optimized within the constraints imposed on the designer. The individual components would be synthesized and analyzed into the final product in a similar manner. Evaluation is concerned with measuring the design against the specification established in the problem definition phase. This evaluation often involves the fabrication and testing of prototype model to assess operating performance, quality, reliability, and other criteria. The final phase in the design procedure is the presentation of the design. Presentation is concerned with the documentation of the design by means of drawing, material specifications, assembly lists, and so on. In essence, documentation means that the design database is created.

14.5

APPLICATION OF COMPUTERS IN DESIGN

A computer-aided design system can beneficially be used in four phases of the design process. The four, together with the CAD activity that corresponds to the phase given in parentheses, are: l l l l

Synthesis (geometric modeling) Analysis and optimization (engineering analysis) Evaluation (design review and evaluation) Presentation (automated drafting) The design process

Computer-aided design

Recognition of need

Definition of problem

Synthesis

Geometric modeling

Analysis and optimization

Engineering analysis

Evaluation

Design review and evaluation

Presentation

Automated drafting

Fig. 14.7

Design process using CAD. [7]

Reprinted by permission from Mikell P. Grover and Emory W. Zimmers, CAD/CAM; Computer-Aided Designs and Manufacturing (Englewood Cliffs, NJ: Prentice-Hall Inc. 1984), p. 57 [1].

Computer System Concept

319

These CAD activities are illustrated in Fig. 14.7 as an overlay on the design process of Shingley. Geometric modeling is concerned with the use of a CAD system to develop a mathematical description of the geometry of an object. The mathematical description, called a model, is contained in computer memory. This permits the user of the CAD system to display an image of the model on graphics terminal and to perform certain operation on the model. These operations include creating new geometric models from basics building blocks available in the system, moving the images around on the screen, zooming in on certain features of the images, and so on. These capabilities permit the designer to design effectively. Shop floor control refers to production management techniques for collecting data from factory operations and using the data to help control production and inventory in the factory. Computerized factory data collection techniques and computer process monitoring represent the preferred means of implementing a shop floor control system today.

14.6

COMPUTER INTEGRATED MANUFACTURING

The term Computer Integrated Manufacturing (CIM) is often used interchangeably with CAD/CAM. The terms are closely related; however, CIM is interpreted to possess a slightly broader meaning than CAD/CAM. In this section, we attempt to define the meaning and scope of CAD/CAM and CIM.

CAD/CAM As described in our introduction, computer-aided design and computer-aided manufacturing are concerned principally with the engineering functions in design and manufacturing respectively. Product design, engineering analysis, and documentation of the design (e.g., drafting) represent engineering activities in the design function. Process planning, NC part programming, and many other activitites associated with CAM represent engineering activities in manufacturing. The CAD/CAM systems developed during the 1970s and early 1980s were designed primarily to address these types of engineering problems. In addition, CAM has evolved including many other functions in manufacturing, such as material requirements planning, productions scheduling, computer production monitoring, and computer process control. It should also be noted that CAD/CAM denoted an integration of design and manufacturing activities by means of computer systems. The method of manufacturing a product is a direct function of its design. With conventional procedures practiced for so many years in industry, engineering drawing were prepared by design draftsmen and then used by manufacturing engineers to develop the process plan (i.e., the route sheets). The activities involved in designing the product were separated from the activities associated with process planning. Essentially, a two-step procedure was employed. This was time consuming and involved duplication of effort by design and manufacturing personnel. With CAD/ CAM, a direct link is established between product design and manufacturing engineering. It is the goal of CAD/CAM not only to automate certain phases of design and certain phases of manufacturing, but also to automate the transition from design to manufacturing. In the ideal CAD/CAM system it is possible to take the design specification of the product as it resides in the CAD database and convert it into a process plan for making the product, this conversion being done automatically by the CAD/CAM system. A large portion of the processing might be accomplished on numerically controlled machine tool. As part of the process, the NC part program would be generated automatically by CAD/CAM. The CAD/CAM system would then download the NC program directly to the machine tool by means of a telecommunications network. Hence, under this arrangement (which is certainly within the capabilities

320

Mechanical System Design

of today’s CAD/CAM technology), product design, NC programming, and physical production can all be implemented by computer. Computer integrated manufacturing includes all of the engineering function of a CAD/CAM, but it also includes the business functions of the firm as well. The ideal CIM system applies computer technology to all of the operational function and information processing functions in manufacturing from order receipt, through design and production, to product shipment stage. The scope of computer-integrated manufacturing, compared with the more limited scope of CAD/CAM, is depicted in Fig. 14.8. Scope of CIM Scope of CAD/CAM

Design

Business Functions

Factory Operations

Mfg. Planning

Mfg. Control

Fig. 14.8

Scope of CAD/CAM and CIM.

The CIM concept is that all of firm’s operation related to the production function is incorporated in an integrated computer system to assist, augment, and/or automate the operation. The computer system is persuasive throughout the firm, touching all activates that support manufacturing. In this integrated computer system, the output of one activity serves as the input to the next activity, through the chain of events that starts with the sales order and culminates with shipment of the product. The components of the integrated computer system, and their relationship to our model of manufacturing, are illustrated in Fig. 14.9 Customer orders are initially entered by the company’s sales force into computerized orderentry system. The orders contain the specifications describing the product. The specifications serve as the input to the product design department. New products are designed on a CAD system. The components that comprise the product are designed, the bill of materials is completed, and assembly drawing is prepared. The output of the design department serves as the input to manufacturing engineering, where process planning, tool design, and similar activities are accomplished to prepare for production. Many of these manufacturing engineering activities are supported by the CIM computer system: Process planning is performed using computer-aided process planning, and tool design is done by a CAD system, making use of the product model generated during product design. The output from manufacturing

Computer System Concept

321

engineering provides an input to production planning and control, where material requirements planning and scheduling is performed during the computer system. CAD Geometric modelling Engineering analysis Design review and evaluation Automated drafting

Design

Computerized Business systems Order entry accounting Payroll Customer billing

Business Functions

Factory Operations

Mfg. Planning

CAM Cost estimating CAPP NC part programming computerized work stds. MRP, capacity planning

Mfg. Control

CAM Process control Process monitoring Shop floor control Computer aided inspection

Fig. 14.9

Computerized elements of a CIM system

A subroutine in a robot program is a group of statements that are to execute separately when called from the main program. In a preceding example, the subroutine SAFESTOP was named in the REACT statement for use in safety monitoring. Other uses of subroutines include making calculations or performing repetitive motion sequences at a number of different places in the program. Rather than write the same steps several times in the program, the use of subroutine is more efficient.

14.7

SIMULATION AND OFF-LINE PROGAMMING

Off line programming permits the robot program to be prepared at a remote computer terminal and downloaded to the robot controller for execution. In true offline programming there would be no need to physically locate the positions in the workspace for the robot as required with present textual programming languages. The programming procedure would be similar to the offline producers used in NC part programme. The significant advance of true offline programming is that the downtime for reprogramming would be minimized, thus permitting the robot to continue in production uninterrupted. The offline programming procedures being doped and commercially offered use a graphical simulation of a CAD/CAM system. Examples of currently available CAD/CAM simulation packages that can be

322

Mechanical System Design

used for offline robot programming include PLACE [4] (McDonnell Manufacturing Industry Systems Company), Robographix (Computer Vision Corporation), Robot-SIM (General Electric CALMA Company), and ROBO-CAM [2] (Silma, Inc.) The PLACE system is one of the more widely used of the simulation packages. The acronym PLACE stands for positioned layout and cell evaluation. It is a computer graphics model builder that is used to construct a three dimensional model of a robot cell for evaluation and offline programming. The cell might consist of the robot, machine tools conveyors, and other hardware. PLACE permits their components of the cell to be displayed on the graphics monitor and for the robot to perform its work cycle in animated computer graphics. The associated software module used to write the robot program is called CAM-MAND. The motion sequences for the robot are developed using the PLACE and CAM-MAND modules. After the program has been developed using the simulation procedure, it is then converted into the textual language corresponding to the particular robot employed in the cell. This is a step in the offline programming procedure that is equivalent to post processing in NC part programming. In the current commercial offline programming packages, some adjustment must be performed to account for geometric differences between the three-dimensional model in the computer system and the actual physical cell. For example, the position of a machine tool in the physical layout might be slightly different than in the model used to do the offline programming. For the robot to reliably load and unload the machine, it must have an accurate location of the load/unload point recorded in its control memory. The PLACE system allows for these adjustment in position to be made by means of a module called ADJUST. This module is used to calibrate in three-dimensional computer model by substituting location data from the actual cell for the approximate values involved in the original PLACE model. The disadvantage with calibrating the cell is that time is lost in performing this procedure. In future programming systems, the offline procedure described above will probably be argumented by means of a machine vision system (and other sensors) located in the cell. The vision and sensor systems would be used to update the three-dimensional model of the workplace and thus avoid the necessity for the calibration step in current offline programming methods. The term sometimes used to describe future programming systems in which the robot possess accurate knowledge of its threedimensional workplace in world modeling. Associated with the concept of world modelling is the use of very high-level language statements, in which the programmer would specify a task to be done without giving details of the procedure used to perform the task.

Exercise 14 . . . 1. 2. 3. 4. 5. 6. 7. 8.

What is computer system concept? What is data flow diagram? Write down symbols used in preparing system flowchart. Write down conversion of manual to computer based systems. What are the fundamentals of CAD/CAM? Write down application of computers in design. What is computer integrated manufacturing? Briefly explain simulating and off line programming.

Bond Graph

323

15 Bond Graph 15.1

INTRODUCTION

Bond graph is an explicit graphical tool for capturing the common energy structure of systems. It increases one’s insight into systems behavior. In the vector form, they give concise description of complex systems. Moreover, the notations of causality provides a tool not only for formulation of system equations, but also for intuition based discussion of system behavior, viz. controllability, observability, fault diagnosis, etc. In 1959, Prof. H.M. Paynter gave the revolutionary idea of portraying systems in terms of power bonds, connecting the elements of the physical system to the so called junction structures which were manifestations of the constraints. This power exchange portray of a system is called Bond Graph, which can be both power and information oriented. Later on, bond graph theory has been further developed by many researchers like Karnopp, Rosenberg, Thoma, Breedveld, etc. who have worked on extending this modelling technique to power hydraulics, mechatronics, general thermodynamic systems and recently to electronics and non-energetic systems like economics and queuing theory. By this approach, a physical system can be represented by symbols and lines, identifying the power flow paths. The lumped parameter elements of resistance, capacitance and inertance are interconnected in an energy conserving way by bonds and junctions resulting in a network structure. From the pictorial representation of the bond graph, the derivation of system equations is so systematic that it can be algorithmized. The whole procedure of modeling and simulation of the system may be performed by some of the existing software e.g., ENPORT, Camp-G, SYMBOLS, COSMO, LorSim, etc. Bond graph bibliography by F.E. Cellier and Bondgraphs.com Bibliography give a thorough account of the work carried out in the particular field. The following sections will briefly explain the reader to understand the mnemonics of the bond graph. However, the explanation given here is not a complete one; it is only limited to the extent of initiating new bond graphers.

15.2

POWER VARIABLES OF BOND GRAPHS

The language of bond graphs aspires to express general class physical systems through power interactions. The factors of power, i.e. Effort and Flow, have different interpretations in different physical domains.

324

Mechanical System Design

Yet, power can always be used as a generalized co-ordinate to model coupled systems residing in several energy domains. One such system may be an electrical motor driving a hydraulic pump or an thermal engine connected with a muffler; where the form of energy varies within the system. Power variables of bond graph may not be always realizable (viz. in bond graphs for economic systems); such factual power is encountered mostly in non-physical domains and pseudo bond graphs. In the following table, effort and flow variables in some physical domains are listed.

Table 15.1 Systems

Effort (e)

Flow ( f )

Mechanical

Force (F) Torque (τ ) Voltage (V) Pressure (P) Temperature (T) Pressure (P) Chemical potential ( µ ) Enthalpy (h) Magneto-motive force ( em )

Velocity (v) Angular velocity (ω ) Current (i) Volume flow rate (dQ/dt) Entropy change rate (ds/dt) Volume change rate (dV/dt) Mole flow rate (dN/dt) Mass flow rate (dm/dt) Magnetic flux (φ )

Electrical Hydraulic Thermal Chemical Magnetic

15.3

BOND GRAPH STANDARD ELEMENTS

In bond graphs, one needs to recognize only four groups of basic symbols, i.e., three basic one port passive elements, two basic active elements, two basic two port elements and two basic junctions. The basic variables are effort (e), flow ( f ), time integral of effort (P) and the time integral of flow (Q).

Basic 1-P-elements A 1-port element is addressed through a single power port, and at the port a single pair of effort and flow variables exists. Ports are classified as passive ports and active ports. The passive ports are idealized elements because they contain no sources of power. The inertia or inductor, compliance or capacitor, and resistor or dashpot are classified as passive elements.

R-Elements The 1-port resistor is an element in which the effort and flow variables at the single port are related by a static function. Usually, resistors dissipate energy. This must be true for simple electrical resistors, mechanical dampers or dashpots, porous plugs in fluid lines, and other analogous passive elements. The bond graph symbol for the resistive element is shown in Fig. 15.1. e f

Fig. 15.1

R

Bond Graph

325

The half arrow pointing towards R means that the power, i.e., product of F and V (or e * f ) is positive and flowing into R, where e represents effort or force, and f represents flow or velocity. The constitutive relationship between e, f and R is given by: e = R * f, Power = e * f = R * f 2.

C-Elements Consider a 1-port device in which a static constitutive relation exists between an effort and a displacement. Such a device stores and gives up energy without loss. In bond graph terminology, an element that relates effort to the generalized displacement (or time integral of flow) is called a one port capacitor. In the physical terms, a capacitor is an idealization of devices like springs, torsion bars, electrical capacitors, gravity tanks, and accumulators, etc. The bond-graphic symbol, defining constitutive relation for a Celement is shown in Fig. 15.2. e f

C

Fig. 15.2 In a spring, the deformation (Q) and the effort (e) at any moment is given by,

Q=

∫

t

f dt ,

−∞

e=K

∫

t

−∞

f dt.

Here, flow is the cause and deformation (and hence effort) is the consequence. In a capacitor, the charge accumulated on the plates (Q) or voltage (e) is given by,

Q=

∫

t

−∞

i dt , e = C −1

∫

t

−∞

i dt.

Here, the current is the cause and the total charge (and hence voltage) is the consequence.

I-Elements A second energy storing 1-port arises if the momentum, P, is related by a static constitutive law to the flow, f. Such an element is called an inertial element in bond graph terminology. The inertial element is used to model inductance effects in electrical systems and mass or inertia effects in mechanical or fluid systems. The bond graph symbol for an inertial element is depicted as shown in Fig. 15.3. e f

I

Fig. 15.3 If the mechanics of mass point is examined by considering the impulse-momentum equation, then we have

P=

∫

t

−∞

e dt , f = m−1

∫

t

−∞

e dt.

Here, effort is the cause and velocity (and hence momentum) is the consequence. Similarly the current in an inductor is given by

326

Mechanical System Design

i = L−1

∫

t

−∞

e dt .

Effort and Flow Sources The active ports are those, which give reaction to the source. For example, if we step on a rigid body, our feet reacts with a force or source. For this reason, sources are called active ports. Force is considered as an effort source and the surface of a rigid body gives a velocity source. They are represented as an half arrow pointing away from the source symbol. The effort source is represented by Loading system

SE

Fig. 15.4 and the flow source is represented as shown in Fig. 15.5. Loading system

SF

Fig. 15.5 In electrical domain, an ideal shell would be represented as an effort source. Similarities can be drawn for source representations in other domains.

Basic 2-Port Elements There are only two kinds of two port elements, namely “Transformer” and “Gyrator”’. The bond graph symbols for these elements are TF and GY, respectively. As the name suggests, two bonds are attached to these elements.

The Transformer The bond graphic transformer can represent an ideal electrical transformer, a massless lever, etc. The transformer does not create, store or destroy energy. It conserves power and transmits the factors of power with proper scaling as defined by the transformer modulus (discussed later). The meaning of a transformer may be better understood if we consider an example given here. In this example, a massless ideal lever is considered. Standard and bondgraphic nomenclature of a lever are shown in Fig. 15.6. It is also assumed that the lever is rigid, which means a linear relationship can be established between power variables at both the ends of the lever. F2 V2 e2 f2 a

b

ei fi

f1 e1 V1 f1

Fig. 15.6

.r . TF

ej fj

Bond Graph

327

From the geometry, we have, V2 = (b /a ) V1

The power transmission implies F2 = ( a /b ) F1 , so that V2 F2 = V1 F1.

In bondgraphs, such a situation may be represented as shown in the above figure. The ‘r’ above the transformer denotes the modulus of the transformer, which may be a constant or any expression (like ‘b/a’). The small arrow represents the sense in which this modulus is to be used.

f j = r fi , and e j = (1/r ) e j . Thus the following expression establishes the conservation of power,

e j f j = ei fi .

The Gyrator A transformer relates flow-to-flow and effort-to-effort. Conversely, a gyrator establishes relationship between flow-to-effort and effort-to-flow, again keeping the power on the ports same. The simplest gyrator is a mechanical gyroscope, as shown in Fig. 15.7.

Y

ω zz

Z X

ei fi

µ GY

ej fj

Fig. 15.7 A vertical force creates additional motion in horizontal direction and to maintain a vertical motion, a horizontal force is needed. So the force is transformed into flow and flow is transformed into force with some constant of proportionality. In this example, I zz stands for moment of inertia about z axis. ω x , ω y and ω z stand for angular velocities about respective axes; Tx , Ty and Tz represent torque acting about the corresponding axis.

Tx = I zz ωz ω y . The power transmission implies

Ty = I zz ωz ω x , so that Txω x = Ty ω y . Such relationship can be established by use of a Gyrator as shown in the figure above. The µ above the gyrator denotes the gyrator modulus, where µ = I zz ω z . This modulus does not have a direction sense associated with it. This modulus is always defined from flow to effort.

e j = µ f i , ei = µ f j .

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Mechanical System Design

Thus the following expression establishes conservation of power, ei fi = e j f j . In the electrical domain, an ideal DC motor is represented as an gyrator, where the output torque is proportional to the input current and the back emf is proportional to the motor angular speed. In general, gyrators are used in most of the cases where power from one energy domain is transferred to another, viz. electrical to rotational, electrical to magnetic, and hydraulic to rotational.

The 3-Port Junction Elements The name 3-port used for junctions is a misnomer. In fact, junctions can connect two or more bonds. There are only two kinds of junctions, the 1 and the 0 junction. They conserve power and are reversible. They simply represent system topology and hence the underlying layer of junctions and two-port elements in a complete model (also termed the Junction Structure) is power conserving. 1-junctions have equality of flows and the efforts sum up to zero with the same power orientation. They are also designated by the letter S in some older literature. Such a junction represents a common mass point in a mechanical system, a series connection (with same current flowing in all elements) in a electrical network and a hydraulic pipeline representing flow continuity, etc. Two such junctions with four bonds are shown in Fig. 15.8. Element

e1 f1

Element

1

e4 f4

4

1

2

3

e2 f2

Element

e2 f2

Element

e3 f3 Element Element

e1 f1

Element

1

e4 f4

4

1

2

3

e3 f3 Element

Fig. 15.8 Using the inward power sign convention, the constitutive relation (for power conservation at the junctions) for the figure in the left may be written as follows; e1 f1 + e2 f 2 + e3 f3 + e4 f 4 = 0.

As 1-junction is a flow equalizing junction, f1 = f 2 = f3 = f 4 .

Bond Graph

329

This leads to, e1 + e2 + e3 + e4 = 0. Now consider the above bond graph shown on the right. In this case, the constitutive relation becomes e1 f1 − e2 f 2 + e3 f3 − e4 f 4 = 0 , and f1 = f 2 = f3 = f 4 .

Thus,

e1 − e2 + e3 − e4 = 0.

So, a 1-junction is governed by the following rules: The flows on the bonds attached to a 1-junction are equal and the algebraic sum of the efforts is zero. The signs in the algebraic sum are determined by the half-arrow directions in a bond graph. 0-junctions have equality of efforts while the flows sum up to zero, if power orientations are taken positive toward the junction. The junction can also be designated by the letter P. This junction represents a mechanical series, electrical node point and hydraulic pressure distribution point or pascalian point. Element

e1 f1

Element

e4 f4

1 4

0

2

3

e2 f2

Element

e2 f2

Element

e3 f3 Element Element

e1 f1

Element

e4 f4

1 4

0

2

3

e3 f3 Element

Fig. 15.9 In case of the model in the left, the constitutive relation becomes e1 f1 + e2 f 2 + e3 f3 + e4 f 4 = 0.

whereas, the model in the right is governed by the following relation, e1 f1 − e2 f 2 + e3 f3 − e4 f 4 = 0 ,

As 0-junction is an effort equalizing junction, e1 = e2 = e3 = e4 .

This leads to, f1 + f 2 + f 3 + f 4 = 0 and f1 − f 2 + f3 − f 4 = 0 , for the left and the right models, respectively.

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Mechanical System Design

So, a 0-junction is governed by the following rules: The efforts on the bonds attached to a 0-junction are equal and the algebraic sum of the flows is zero. The signs in the algebraic sum are determined by the half-arrow directions in a bond graph.

15.4

POWER DIRECTIONS ON THE BONDS

When one analyses a simple problem of mechanics, say, the problem of a single mass and spring system as shown in the figure below, one initially fixes a co-ordinate system. One may take positive displacement, x, towards right and all its time derivatives are then taken positive towards right. The force acting on the mass may also be taken positive towards right. The system, however, in the course of motion may attain such a state that when it is displaced towards the right, the force on the mass happens to be towards the left. .. x . x x K m

Force

Fig. 15.10 This phenomenon may be interpreted from the results obtained by solving the system of equation(s) when a positive value of displacement and a negative value of force are seen. So, the initial fixing of a positivity is arbitrary. However, further analysis is related to this fixation. In practice, bond graphs are drawn for general systems. Thus left and right, up and down, clockwise or counter-clockwise, etc. may not be of general relevance. One has to then create a view point which is general and any particular system interpretation should be easily derivable. This is done by assigning the bonds with Power directions. This may be as arbitrary as fixing co-ordinate systems in classical analysis. Say in a bond graph, the power is directed as shown in Fig. 15.11 below, where J E

: :

Junction, Element, J

E

Half arrow: direction of power.

Fig. 15.11 This assignment means, such variables are chosen for effort and flow, so that whenever both these variables acquire positive values, then the power goes from J to E. However, for mixed signs of the variables, the power direction is reversed. As has been discussed earlier, the interpretation of the relative orientation of positive effort and flow may be subjective depending on whether the analysis is carried out from the stand point of J or E. For instance, we can say the downward motion is positive for a particular mass and the upward motion is positive for another mass according to our co-ordinate system; but for a spring, upward and downward

Bond Graph

331

motion do not convey any meaning. In such cases, either the compression or the tension of the spring must be identified as a co-ordinate. Consider, the following example which belongs to the field of mechanics. The mechanical system with its bond graph model are displayed in the Fig. 15.12. Here the variables are so chosen that for positive force and velocity, the power goes to the spring. This may be as per the instrumentation arranged as shown in the two illustrations given below: C 1

1

ω

2 SF

1 Direction of positive displacement

2 K

3

Direction of positive force

4 5 6

6 Direction of positive displacement

5 4 3

K

Direction of positive force W

2 1

Fig. 15.12 If one assumes the stand point of the junction-1, then characteristic of the spring should be as determined by the arrangement as shown above, and may appear as plotted in the right.

332

Mechanical System Design K

Force

Displacement

Fig. 15.13

15.5

ASSIGNING NUMBERS TO BONDS

The bonds in a bond graph may be numbered sequentially using integers starting with 1. However, one need not follow any fixed rule. Assignment of bond number also fixes the name of the elements or junctions. This is the best book-keeping technique adopted by most of the existing software products. Some software though follow numbering of elements according to their instance. However, in models using fields, where many bonds are connected to an element, such a nomenclature cause difficulty in book-keeping. For example, the two 1-junctions in the bond graph shown in the right can be uniquely identified as (S 1 2 3 4) and (S 5 8 9); similarly symbols like C 3, C 9 can be used to identify a particular element. R

I

2 SE

1

1

C

7 4

0

3

6

C

R

9 5

1

8

R

Fig. 15.14 The use of characters S and P instead of numerals 1 and 0 allows to represent models in tele-type code in the list form, which can be exchanged across platforms. The power direction and causal information may also be appended to the list form to create a globally adoptable model code, which may read something like (S + –1 –2 –3 – +4) (P + –4 ...).

15.6

CAUSALITY

Causality establishes the cause and effect relationships between the factors of power. In bond graphs, the inputs and the outputs are characterized by the causal stroke. The causal stroke indicates the direction in which the effort signal is directed (by implication, the end of the bond that does not have a causal stroke is the end towards which the flow signal is directed).

Bond Graph

333

To illustrate the basic concept of the causality, let us assume that there is a prime mover with a smart speed governor as shown in Fig. 15.15.

Prime Mover

Load

Fig. 15.15 The prime mover is driving the load, i.e., the power is going from the prime mover to the load. Apart from sending the power, the prime mover also decides that the load should run at a particular speed depending on the setting of the governor. So, it may be said that from the prime mover, the information of flow is generated which goes to the load. The load generates the information of torque (effort) which the prime mover receives and adjusts the inner mechanisms to compensate for it. The wavy lines in the following figure indicate the direction of flow of the particular information. f

Prime Mover

Load

e

Fig. 15.16 The selected causality is generally indicated by a cross bar or causal bar at the bond end to which the effort receiver is connected. In expressing causal relations between the effort and the flow, the choice of causality has an important effect. In the following section, the causality of the different elements are discussed. l For inertance I type storage elements, the flow ( f ) is proportional to the time integral of the effort.

f = m −1

∫

t

−∞

e dt

An example of this would be a mass subjected to a force, causing it to accelerate.

v = m −1

l

∫

t

−∞

F dt

In the above relationship effort history is integrated to generate contemporary flow. Therefore, an I element receives effort (cause) and generates flow (effect). For capacitive C type storage elements, the effort (e) is proportional to the time integral of the flow ( f ).

e=k

∫

t

−∞

f dt

334

Mechanical System Design Examples of this would be a spring being deformed by a force or a capcitor being charged.

F = K −1 V = C −1

l

∫

t

−∞

∫

f dt

t

−∞

i dt

The above integral relationships show that a C element receives flow and generates effort. The resistive or dissipative elements do not have time integral form of constitutive laws. e = R f or f = e/R The flow and the effort at this port are algebraically related and can thus have any type of causal structure, either with an open-ended bond (causal stroke is away from the element, i.e. at the junction end) indicating a resistive causality or a stroke ended bond indicating a conductive causality. As per the above discussions, the causal strokes for I, C and R elements are shown in Fig. 15.17. J

I

J

C

J

R

J

R

Fig. 15.17 l

Sources impose either an effort or a flow on a system, but not both. SE

SF

Fig. 15.18 The bond graphic sources are assumed to be robust, i.e., as providers of active and infinite energy. The Effort Source (SE), imposes an effort on the system which is independent of flow. An example of this would be an electrical cell that decides the terminal voltage and the attached load decides the current that the cell has to take and adjust its chemical reactions to maintain the rated terminal voltage. The Flow Source (SF), imposes a flow on the system independent of the effort. Examples of this would be cams, constant displacement hydraulic pumps, road excitation, etc. Thus the causality of the source elements are mandatory as shown in the figure below. The elements I, C, R, SF, and SE are classified as single port, since they interact with the system through one bond only. However, the I, C and R elements can be connected to many bonds to represent tensorial nature, such as the spatial motion of a free body or the stress-strain relationships in a compressible material; in which case they are termed as field elements. The transformer, by its elemental relation, receives either flow or effort information in one bond and generates the same in its other bond. Thus, one of its port is open-ended with the other end stroked as shown in Fig. 15.19. ei fi

.r . TF

ej

ei fi

fj

Fig. 15.19

.r . TF

ej fj

Bond Graph

335

For the first case, the constitutive equations would be,

f j = r fi and ei = r e j , whereas for the second, the relations are

f j = 1/r fi and ei = 1/r e j . As mentioned earlier, the gyrator relates flow-to-effort and effort-to-flow, and therefore, both of its ports have either open-ended or stroke-ended causality as shown in Fig. 15.20. ei fi

.r . GY

ej

ei fi

fj

.r . GY

ej fj

Fig. 15.20 For the first case, the constitutive equations would be

e j = r fi and ei = r f j , whereas for the second, the relations are

f j = r ei and fi = r e j . At a 1-junction, only one bond should bring the information of flow; i.e., only one bond should be open ended and all others should be stroked as displayed in the Fig. 15.21. This uniquely causalled bond at a junction is termed as Strong bond. In any case two bonds cannot be causalled away from the 1junction, since this would lead to violation of rules of information exchange (the two bonds may not impart equal flows). Strong bond

1

Fig. 15.21 Similarly at a 0-junction, only one bond should be stroked nearer to the junction. This strong bond determines the effort at the junction, which the weak bonds (other bonds besides the strong bond) carry around. The proper causality, for a storage element (I or C), is called Integral Causality, where the cause is integrated to generate the effect. For example, in C element, a study of the constitutive equations reveal that flow is integrated and multiplied with the stiffness to generate effort. Sometimes the causal strokes will have to be inverted, which means the constitutive relationship for the corresponding element is written as a differential equation. For example, flow in a spring is the time derivative of the ratio of effort and stiffness. Such causality pattern is called Differential Causality.

336

Mechanical System Design Strong bond

0

Fig. 15.22 Implications of an integral causality means the past data of cause or history function is integrated to arrive at the effect felt at present, whereas, differential causality needs differentiation of the cause at present (that cannot be found properly, since the future is not known) to arrive at the effect. Hence differential causality makes the system dependent on future, as if the system is being dragged towards a predestined configuration or in other terms adds specific constrains on the dynamics of the system. Genuine differential causality is not commonly encountered during system modeling except in certain cases of modeling mechanisms, robotics, etc. where link flexibilities or other aspects are neglected in the model. Such causality is a common occurrence owing to certain direct manipulations in the model. However, spurious occurrences are not ruled out in certain cases such as modelling of a superfluous element (try bond graph of a single degree freedom system, where another mass is placed directly in contact with the main mass), or leaving out certain important element from system (neglecting contact point stiffness etc.). The occurrence of differential causalities in a system may indicate serious violations of principles of conservation of energy, as illustrated in the following example. Let us consider the charging of a capacitor through a constant voltage battery source. The system and its bond graph model are shown in Fig. 15.23.

SE

1

C

Fig. 15.23 From the causalled bond graph model, we observe the differential causality in the C-element. If we consider the energy exchange, the energy (E) stored in the capacitor after complete charging to a voltage V is

E = Q2 /2C , where C is the capacitance and Q is the charge stored in the capacitor. Also

Q=

∫

t

−∞

i dt and V = Q /C.

The energy spent by the shell during charging is

Bond Graph

337

E=

∫

t

−∞

V i dt = V

∫

t

−∞

i dt = Q 2 /C.

The later result is anomalous with the energy stored in the capacitor. The loss of half the energy is unaccounted for. This can be attributed as one of the implications of the differential causality. The half energy lost is always through dissipation in the system which wrongly has been neglected. If we introduce the resistance in the model, the causality problem is automatically corrected and proper energy conservation is obtained as shown in the Fig. 15.24. R

SE

1

C

Fig. 15.24

Causality Assignment Procedure 1. Assign fixed causalities to sources. 2. Propagate the causality through junctions, if possible, i.e., if any bond has got a causality such that it has become the strong bond for a junction, the causality for all other bonds (weak bonds) is determined by laws for causality of junctions and if all other bonds of junction are causalled, the last bond should be the strong bond. Similarly, if any port of a two port is causalled (of TF and GY), the causality of the other can be assigned. 3. Assign integral causality to one of the storage elements and propagate the causality through junctions. Continue the procedure with other storage elements. This should normally result in complete causalling of the graph. 4. If the graph is not completely causalled yet, start assigning a resistive causality to a R-element and propagate it. Continue till the entire graph is causalled. In cases, where the model is determined through causalities of R-elements, there may be several possible causal models. It is always advisable to maximize resistive causalities and minimize the conductive causalities in R-elements. 5. If the system develops differential causalities in some storage elements, try minimizing its number of occurrence through assigning initial integral causalities to other storage elements than those selected before. 6. Try to avoid differential causalities by suitable changes to the model, such as introducing some compliance or resistance or both. 7. Discard all models, which result in a causal structure, that violates junction causality rules. Two animated examples of causalling a model in different ways are illustrated below:

338

Mechanical System Design R

C

SE

I

1

1

0

R

C

SF R

C

SE

C

I

1

1

0

R

R

R

SF

Fig. 15.25

15.7

GENERATION OF SYSTEM EQUATIONS

In this section, the method of generation of system equations is discussed. From an augmented bond graph, using a step-by-step procedure, system equations may be generated. We take the model of a simple single degree of freedom mass-spring-damper system as the starting point. F(t )

I 3

M

K

SE

1

1

2

C

4

R

R

Fig. 15.26 The differential equations describing the dynamics of the system are written in terms of the states of the system. All storage elements (I and C) correspond to stored state variables (P for momentum and Q for displacement, respectively) and equations are written for their time derivatives (i.e., effort and flow). These equations are derived in four steps as described below:

Bond Graph

339

1. Observe what the elements (sources, I’s, C’s, and R’s) are giving to the system and write down their equations looking at the causalities and using variables for strong bonds. 2. Write down equations for the junctions and the two-port elements for the variables for the strong bonds. 3. Replace the variables, which are expressed in terms of states in other equations. Continue sorting and replacement till the right side of the entire set of equations are expressed in terms of states and system parameters only. 4. If some equations are still not completely reduced, there is the existence of some kind of a loop (algebraic loop, causal loop or differential causality, as shall be discussed later). Try solving those as a set of linear equations either through substitution or matrix inversion. And finally, erase all trivial equations other than those for derivatives of state variables and write them in terms of state variables. Thus using the following steps, the equations for the above system would be: Step 1: e1 = SE1 f 3 = P3/M3 e2 = K2 * Q2 e4 = R4 * f 4 = R4 * f 3 (by junction rule, f 4 = f 3 and strong bond number is 3) Step 2: e1 – e2 – e3 – e4 = 0 or e3 = e1 – e2 – e4 Step 3: e1 = SE1 f 3 = P3/M3 e2 = K2 * Q2 e4 = R4 * P3/M3 e3 = SE1 – K2 * Q2 – R4 * P3/M3 Step 4: DQ2 = f 2 = f 3 DP3 = e3 where prefix ‘D’ stand for time derivative d/dt which leads to DQ2 = P3/M3 DP3 = SE1 – K2 * Q2 – R4 * P3/M3 The difference between equations derived from bond graphs and otherwise are that there will be ‘N’ sets of first order differential equations, where ‘N’ is the number of states. The term, number of states means the number of lumped parameter storage elements I and C with integral causality present in a system. The equation of motion for the system discussed above by traditional method is:

m d 2 x /dt 2 + r * dx /dt + k * x = F (t ).

340

Mechanical System Design

The two state equations derived from a bond graph model (dropping suffixes) are dP/dt = –r/m * P – k * Q + SE, dQ/dt = 1/m * P, where, P is momentum of m * dx/dt, Q is displacement of x and SE is F(t). From the second equation, P = m * dQ/dt, which when replaced in the first leads to

m * d 2 Q /dt 2 = − r * dQ /dt − k * Q + F (t ). This equation corresponds to that derived through traditional method after rearrangement. However, manual derivation of equations for larger systems is not all that simple. For instance, derivation of system differential equations for the animated bond graph model discussed in the causality section after proper causalling would lead to formation of the so called algebraic loops. Similarly, complexities and errors of various types, like Causal loops, Power loops, and Differential Causalities may exist in the model of a system. The method for derivation of their equations is described in corresponding sections. Example 2: Here we take another example of a system with two-ports, whose model is shown in Fig. 15.27. Ideal Bearings

~

Mass-less Rigid shaft

Flexible rope

g

I 2 SE

1

C

I 6

µ

4

1

..

GY

5

11

7

..

TF

3

12

R

9

r

1

1

8

0

10

1 14 SE

R

Fig. 15.27 Then we follow the normal equation derivation steps: Step 1: e1 = SE1 f 2 = P2/M2

13

I

Bond Graph

341 e3 = R3 * f 3 = R3 * f 2 = R3 * P2/M2 f 6 = P6/M6 e11 = K11 * Q11 e12 = R12 * f 12 = R12 * f 9. f13 = P13/M13

Step 2: f 1 = f 2 = f 4 = f 3 = P3/M3 e2 = e1 – e3 – e4 = SE1 – R3 * P2/M2 – e4 e4 = MU * f 5 = MU * f 6 = MU * P6/M6 e5 = MU * f 4 = MU * f 2 = MU * P2/M2 f 5 = f 7 = f 6 = P6/M6 e6 = e5 – e7 f 8 = r * f 7 = r * P6/M6 e7 = r * e8 = r * e9 e8 = e10 = e9 f 9 = f 8 – f 10 = r * P6/M6 – f 13 = r * P6/M6 – P13/M13 f 10 = f 14 = f 13 = P13/M13 e13 = e10 + e14 = e9 + SE14 f 11 = f 12 = f 9 e9 = e11 + e12 = K11 * Q11 + R12 * f 9 Step 3: e1 = SE1 f 2 = P2/M2 e3 = R3 * f3 = R3 * f 2 = R3 * P2/M2 f 6 = P6/M6 e11 = K11 * Q11 e12 = R12 * f 9 = R12 * (r * P6/M6 – P13/M13) f 13 = P13/M13 f 1 = f 2 = f 4 = f 3 = P3/M3 e2 = SE1 – R3 * P2/M2 – e4 = SE1 – R3 * P2/M2 – MU * P6/M6 e4 = MU * f 5 = MU * f 6 = MU * P6/M6 e5 = MU * f 4 = MU * f 2 = MU * P2/M2 f 5 = f 7 = f 6 = P6/M6 e6 = e5 – e7 = MU * P2/M2 – r * (e11 + e12) = MU * P2/M2 – r * (K11 * Q11 + R12 * (r * P6/M6 – P13/M13) f 8 = r * f 7 = r * P6/M6 e7 = r * e8 = r * e9 = r * (e11 + e12) = r * (K11 * Q11 + R12 * (r * P6/M6 – P13/M13) e8 = e10 = e9 = e11 + e12 = K11 * Q11 + R12 * (r * P6/M6 – P13/M13) f 9 = f 8 – f 10 = r * P6/M6 – f 13 = r * P6/M6 – P13/M13

342

Mechanical System Design f 10 = f 14 = f 13 = P13/M13 e13 = e10 + e14 = e9 + SE14 = K11 * Q11 + R12 * (r * P6/M6 – P13/M13) + SE14 f 11 = f 12 = f 9 = r * P6/M6 – P13/M13 e9 = e11 + e12 = K11 * Q11 + R12 * f 9 = K11 * Q11 + R12 * (r * P6/M6 – P13/M13)

Step 4: DP2 = SE1 – R3 * P2/M2 – MU * P6/M6 DP6 = MU * P2/M2 – r * (K11 * Q11 + R12 * (r * P6/M6 – P13/M13) DP13 = K11 * Q11 + R12 * (r * P6/M6 – P13/M13) + SE14 DQ11 = r * P6/M6 – P13/M13 In matrix form, these equations may be written as d{Y}/dt = [A]{Y} + [B]{u}, where, {Y} is a vector of states (P2, P6, P13 and Q11), [A] is square matrix, {u} is the array of sources (SE1 and SE14} and [B] is a matrix of dimension N × M; N being the number of states and M being the number of sources. The matrices [A] and [B] are:

0.0 0.0  −MU /M 6  − R3/M 2   2 2 MU /M 2 − R12 * r /M 6 R12 * r /M 13 r * K11  , A=  0.0 R12 * r /M 6 K11  − R12/M 13   0  r /M 6 −1/M 13  0.0 1 0 B= 0   0

15.8

0 0  1  0 

ACTIVATION

Some bonds in a bond graph may be only information carriers. These bonds are not power bonds. Such bonds, where one of the factors of the power is masked are called activated bonds. As an example, let the system as shown in Fig. 15.28 to be considered. µ

Amplifier ( µ )

4

5

GY

e=0

m Velocity pick-up

f=0 1

Electromagnetic exciter K

..

2

R

1 3 I:m

R:R C:K

Fig. 15.28

Bond Graph

343

Here the velocity pick-up only carries the information of velocity to the amplifier through which an electro-magnetic exciter applies force proportional to velocity on the mass and the exciter does not carry the information of velocity and impose it back on the mass as a reactive force. So on the bond representing the velocity pick-up the information of force must be masked and on the bond representing the exciter the information of the flow must be masked. A full arrow somewhere on the bonds shows that some information is masked and what information is masked may be written near that full arrow. According to this convention the bond graph of the system is shown to the right of the system drawn above. The bond number 4 in the figure is the pick-up bond where information of effort is masked and in bond number 5, which is the exciter bond, the flow is activated. The concept of activation is very significant to depict feedback control systems. The term activation initially seems a misnomer. However, Paynter’s idea was based on the fact that though the information of a factor of power is masked on one end, an activated bond on the other end can impart infinite power which is derived from a tank circuit used for both the measurement or actuation device (for instance, the pick-up, the amplifier and the exciter, all have external power sources). R : Armature

Kt 2

Kt 1

I

J2

J1

+

+ Amplifier

–

–

Fig. 15.29 In the system shown above, a D.C. motor with externally energized field is driving an elastic shaft with two disks and with a fluctuating resistive load. The speed, picked up by a tachometer is fedback to control the speed of the motor by adjusting the armature current (increasing voltage alternatively). The bond graph for this system is drawn as shown in Fig. 15.30. I

C : Kt1

I : J1

C : Kt 2

I : J2

0

1

0

1

V : SE

µ

R

..

1

GY

f

α && GY

R

e

Fig. 15.30

Observers Additional states can be added for measurement of any factor of power on a bond graph model using the Observer storage elements. An effort activated C-element would observe the time integral of flow (and

344

Mechanical System Design

consequently flow), whereas a flow activated I-element would observe the generalized momentum (and consequently effort). Activated elements are perceived conceptual instrumentations on a model. They don’t interfere in the dynamics of the system (i.e., their corresponding states never appear on the righthand side of any state equation). A system with N states, M sources and L observers would have the state equations of the form d/dt{Y} = [A]{Y} + [B]{u}, d/dy{Z} = [C]{Y} + [D] {u}, where {Y} is a vector of true states, {Z} is the vector of observed states, {u} is a vector of sources, [A] is M × M matrix, [B] is M × N matrix, [C] is Z × M matrix and [D] is Z × N matrix. Here is an example of a single-degree-of-freedom system with instrumentations, whose model with observer elements as shown in Fig. 15.31. mu

..

GY Exciter Amplifier m Conceptual spring (velocity sensor)

f 11 Velocity pick-up mg

Conceptual inductor (force sensor)

10 e

I 12

C

e 8

1

9

SE

R

7 0

5 2

1 3

1 SF

C

4

0 f 6 I

Fig. 15.31 The conceptual instrumentations shown here, assumes a spring of zero stiffness (thus measuring displacement and not generating any reactive effort) and an inductive coil within which the segments of the damper move to induce emf (a measurement of force) without any reaction on the damper segment. The equations for this model can be derived as shown below: Step 1: f 1 = SF1 e3 = K3 * Q3 e5 = R5 * f 5 f6=0 e9 = SE9 e8 = 0 e10 = 0

Bond Graph

345 f 12 = P12/M12 e11 = MU * f 10 = MU * f 12 = MU * P12/M12

Step 2: e12 = e7 + e9 – e8 + e11 – e10 = e2 + SE9 + MU * P12/M12 f 2 = f 1 – f 7 = SF1 – P12/M12 f 5 = f 4 – f 6 = f 4 = f 2 = SF1 – P12/M12 e2 = e3 + e4 = e3 + e5 = K3 * Q3 + R5 * f 5 = K3 * Q3 + SF1 – P12/M12 Step 3: e12 = K3 * Q3 + R5 * (SF1 – P12/M12) + SE9 + MU * P12/M12 e5 = R5 * (SF1 – P12/M12) Step 4: The state equations are DP12 = e12 = K3 * Q3 + R5 * (SF1 – P12/M12) + SE9 + MU * P12/M12 DQ3 = f 3 = f 2 = SF1 – P12/M12 and the observer equations are DP6 = e6 = e5 = R5 * (SF1 – P12/M12) DQ8 = f 8 = f 12 = P12/M12

15.9

BOND GRAPH MODELLING

A simple mechanical system as shown in Fig. 15.32 is considered here for explaining the modelling procedure. V, i

τ,ω Electric motor

Pump

P , Vs

Fig. 15.32 The output from the pump would be flow and pressure, while the input would be torque and angular velocity applied to the pump shaft. The input power is provided by an electric motor. The power flow diagram are shown in the Fig. 15.33. Power Source

V i

Electric Motor

τ ω

Drive Shaft

τ ω

Pump

P Vs

Fig. 15.33 Two transformations are identified. First from electric power to mechanical power through the electric motor; followed by the mechanical power to the fluid power through the pump. In the next step to construct the model, the components are to be looked into more details. In this simple analysis of the pump it has its own moving parts (inertia); oil flow has got its own compressibility (capacitance) because

346

Mechanical System Design

of which it can store energy; pump has leakage due to its geometric tolerance and friction between the moving parts (resistance) through which it can dissipate energy. The bond graph of the system with power directions and causality is shown in Fig. 15.34. C

I

SE

µ1 ..

V i

1

GY

τ1 ω1

1

µ2

..

τ2 ω2

TF

P1 V1

I

P2 V2

0

C

1

R

R

Fig. 15.34 A two degrees of freedom mechanical system and its corresponding Bond graph model as shown in Fig. 15.35 can be explained in the following manner. F(t )

SE

X2 m2

10 I

9

C

1

F F2

F1 K2

8

R2 0

5

1

X1

F

7

R

3

m1 I

K1

6

1

1

2

C

4

R1

R

Fig. 15.35 The bottom mass velocity is same as the rate of stretching of the spring and damper connected to the ground. So, these are modeled as being connected to the 1-junction. The middle spring and damper experience components of force, say F1 and F2, respectively such that F = F1 + F2, i.e., the force felt by the bottom mass, the spring and damper in combination and the top mass is equal. Alternatively, the distortion of the spring and the damper is dependent on relative motion of the masses. Hence, one can easily smell the presence of a force pass or flow summation junction, in other words the 0-junction here. Further, this 0-junction has three branches corresponding to three components of distribution of effort.

Bond Graph

347

Another 1-junction appears in the right, that signifies the equal relative velocity (i.e., rate of compression here considering the power directions) felt by the middle spring and the damper. The 1-junction at the top depicts the coherent motion of the forcing device with the top mass. For the above system, alternative methods can be applied to arrive at a bond graph model, that may apparently look different. However, the final equations derived from different models lead to the same set. It may be noted that bond graphs for mechanical systems may be drawn using the method of flow balance (kinematics) or force transmission, or a combination of the both. Since the basic external elements used are the same and only the energy conserving junction structure may vary, considering one factor of power automatically takes care of the effect of the other factor. This is perhaps the greatest contribution of the bond graph theory.

Exercise 15 . . . 1. 2. 3. 4. 5. 6. 7.

Write down an introduction about bond graphs. What is power variables? Briefly explain example of modes. What is difference between causality and activation? What is difference between multi and vector bond graphs? What is simulation and control? Briefly discuss about the father of bond graph.

348

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Question Papers

B.TECH. EIGHTH SEMESTER EXAMINATION, 2003–2004 MECHANICAL SYSTEM DESIGN Time: 3 Hours

Total Marks: 100

Note: Attempt ALL questions. Assume suitably any missing data/information, if any. 1. Attempt any FOUR of the following: (5 × 4 = 20) (a) Why is the systems approach becoming popular in the study of engineering problems? (b) Explain the basic concept of Concurrent Engineering. What are major advantages of implementing the concurrent engineering for product design and production? (c) Describe a basic framework to be used for analyzing an engineering system. Give a list of optimization techniques used for the analysis of the system. (d) Prepare a checklist for carrying out the need analysis of the design of a product. Give suitable explanations wherever necessary. (e) A company uses flat leather belt in its high-speed machine. The belt has been found to fail repeatedly. The belt was designed on the same basis used successfully on earlier machine models running at comparatively lower speeds. The specifications of the belt under investigation are: Flat leather belt density = 1.08 × 10–3 kg/mm3. σ Working = 198 × 10–3 (kgf/mm2) Pulley Diameter = 150 mm Angle of contact = 180° Speed = 3000 rpm Power to be delivered = 0.2 KW Coeff. of friction = 0.2 Preliminary computations indicate that a belt of size 6.25 × 1.6 mm should be satisfactory. However, this is not OK, investigate the problem of belt failure and suggest suitable steps to be taken.

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349

(f) Discuss the basic problems concerning systems. Illustrate how these problems can be made use of in case of an input-output system. 2. Attempt any FOUR parts of the following: (5 × 4 = 20) (a) Discuss different types of models, used commonly, for mechanical systems design. Illustrate your answer suitably. (b) Describe the steps involved in modeling a mechanical system. (c) An electric water heater tank is insulated to reduce heat loss to surroundings. The electric water heater is controlled by a thermostat so as to maintain the temperature of water in the tank at θ , θ1 is the temperature of water supply to the heater, θ a is the ambient temperature of the surroundings. Develop a model to compute the rate of heat flow (q) of heating element of water. Flow from the heated tank is equal to n. Assume that C = thermal capacitance of water in tank, R = Thermal resistance of insulation, n = water flow from the heater tank and s = specific heat of water. (d) Explain what is meant by Systems Analysis. List and explain the important types of models used in manufacturing systems analysis/design. (e) A disk rotating in a viscous medium and supported by a shaft is shown in Fig. 1. Applied torque tending to rotate the disk is T. Develop an input-output model of the system to compute “θ ” for a given “T”. Td = Torque required to overcome viscous friction Ts = Torque tending to twist the shaft = Ks. θ (Ks = tensional stiffness of the shaft,

θ = Angular displacement of the shaft). θ T

ω

Ts Fixed end

Td

Fig. 1 ( f ) For the mechanical system shown in Fig. 2, determine the differential equation relating f and x and equation relating f and y and equation relating x and y. f

x M

y

B

Fig. 2

y

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3. Attempt any two of the following: (10 × 2 = 20) (a) For the network shown in Fig. 3, find the shortest path from node 1 to node 8. What is the longest path? The figures (values) adjacent to the arcs denote their lengths. 3 5 Start 1

4

3 2

2

7 4

1

2 2

3

4

8 End 4

1 5

6

4

Fig. 3 (b) Consider a function in two variables x1 and x2, given as:

F ( x1 , x2 ) = x12 − 8x2 + 2 x22 − 6x1 + 30. Find the optimal values of x1 and x2 and verify the values obtained for maxima/minima. (c) Explain the difference between goal and objective. Give examples of each. Suppose 2 products are to be produced with a single unit of “A” requiring 2.4 mins of moulding time and 5.0 mins. of assembly time. The profit for product “A” is 0.6 per unit. A single unit of product B requires 3.0 mins. of moulding time and 2.5 mins of finishing time. The unit profit for B is 0.7. Suppose the production manager is to determine the number of units of each product to be produced per week in consideration of the following equally ranked goals: (i) A profit target of Rs 350 per week should be met. (ii) Overtime should not exceed 20% of the available production time in moulding department and should be avoided if possible. (iii) Available production time in assembly and finishing departments should be fully utilized but not exceeded. Assume: (i) Available production time in moulding department is equal to 1200 mins/week. (ii) Available time in assembly shop = 1500 mins/week (iii) Available time in finishing shop = 600 mins/week. Formulate the production problem in goal programming model form. 4. Attempt any two of the following. (10 × 2 = 20) (a) Discuss the elements of Feasibility Analysis. (b) Suppose that 2 robot alternatives are being considered for a certain industrial application. The cost and life data of the robots are given below:

Investment Life Salvage value Annual disbursements

Robot Type-1

Robot Type –2

Rs 1,50,000 3 years 0 Rs 80,000

Rs 1,25,000 5 years 0 Rs 50,000

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351

Which alternative should be selected? Use MARR of 20%. Assume a salvage value for the robot Type- 2 at the end of 3 years as Rs 50,000. (c) Find the dimensions of a cylindrical tin (with top and bottom) made up of sheet metal to maximize its volume such that the total surface area is equal to A: = 24π . Check that your solution corresponds to maximum volume. 5. Attempt any two of the following: (10 × 2 =20) (a) For an infinite population, single channel waiting line system, the probabilities associated with time between arrivals and service time are given by Tables (x) and (y) respectively. Using simulated sampling for 10 periods, determine the average idle time of the server. For simulation, use the following 5 digit random numbers. The first 2 digits of the random numbers be used for simulating the arrivals whereas the last 2 digits of the random numbers are used for service times: 48867, 32267, 27345, 55753, 93124, 98658, 68216, 17901, 88124, 83464. Table (x) Time between arrivals (min.)

Probability

4 5 6 7 8 9

0.075 0.200 0.300 0.225 0.150 0.050 Table (y)

Service time (min.)

Probability

2 3 4 5 6 7 8

0.0062 0.0606 0.2417 0.3830 0.2417 0.0606 0.0062

(b) Explain, in what ways decision-making models under conflict differ from models for decision making under risk and uncertainty. The following matrix gives the pay offs in utils (a measure of utility) for 3 alternatives and 3 possible states of nature: Alternative

A1 A2 A3

State of nature S1

S2

S3

50 60 90

80 70 30

80 20 60

352

Question Papers Which alternatives are to be chosen as given under? (i) Laplace principle (ii) The maximax rule and (iii) The minimax regret rule? (c) Explain what is meant by conditional probability? Give an example of a situation where you would use knowledge of conditional probability. Suppose we have a machine. The past experience shows that the machine is incorrectly set 25% of the time and when it is wrong it produces 30% defectives. When it is right, it produces 5% defectives. Suppose the machine has been used to produce 5 parts out of which 1 part was defective. (i) What is the probability of producing 1 defective if the machine is set correctly? (ii) What is the probability that the machine was set incorrectly and produced one defective?

Question Papers

353

B.TECH. EIGHTH SEMESTER EXAMINATION, 2004–2005 MECHANICAL SYSTEM DESIGN

Time: 3 Hours

Total Marks: 100

Note: (i) Attempt all questions. (ii) Assume suitable values for any missing data. (iii) In case of numerical problems assume data wherever not provided. 1. Attempt any four of the following: (5 × 4 = 20) (a) What are the characteristics of a system? Give their importance in system design. (b) Give the basic aspects of concurrent engineering. How does it help in product design and development? Illustrate with suitable example. (c) What are the attributes of a system? Explain with suitable example. (d) Explain engineering activities matrix. In what way it helps in system design? (e) For each of the following products, give at least two need statements of different generality, for which the listed product is a solution: (i) Electric iron (ii) Oil pipe line (iii) Solar cooker (iv) Telephone (v) Envelope (vi) Handkerchief. (f) Explain a system design where environment and safety is of prime consideration. 2. Attempt any four of the following: (5 × 4 = 20) (a) Explain the black box approach of system analysis and design. (b) Explain the decision process approach for system analysis. (c) Write in brief Iconic, Analogue and Mathematical models. (d) Model building is the essence of system design. Discuss. (e) Formulate maximization of a Cobb-Douglas type utility function under budget Limitation C, if unit capital and unit labour cost are Pk and Pl respectively. (f) What is mathematical model of real situation? Discuss the graphical model in system design. 3. Attempt any two of the following: (10 × 2 = 20) (a) A truck delivers concrete from the readymix plant to construction site. The network in Fig. 1 represents the available routes between the plant and the site. The distance from node to node are given along the route lines in km. What is the best route from plant to site?

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Question Papers 2

2

4 5

3

Plant

3

7

1

6

Site

6

1

4

4 3 1

5

Fig. 1 (b) Explain clearly the various ingredients of decision problem. What are the basic steps of a decision making process? (c) What is EMV? How it is computed to be used as a criterion of decision-making? 4. Attempt any two of the following: (10 × 2 = 20) (a) Explain the various factors, which are considered, while feasibility analysis of a system. (b) In a company there are two machines M1 and M2 the former being replaced after every three years while the latter is replaced in six years. Given below are the yearly costs of both the machines: Year

1

2

3

4

5

6

M1 M2

1000 1700

200 100

400 200

1000 300

200 400

400 500

Take the value of money as 10% and find out which machine should be purchased? (c) A beam of uniform rectangular cross-section is to be cut from a log of circular cross-section of dia 2a. The beam has to be used as cantilever beam of fixed length to carry a point load at free end. Find the dimensions of the beam that corresponds to maximum tensile (bending) stress carrying capacity. 5. Attempt any two of the following: (10 × 2 = 20) (a) What is simulation? Explain its purpose. List different advantages and limitations of simulation. (b) Simulate a waiting line with mean arrival rate of 6 min. and mean service time of 5 min. The probability distribution for arrival and service time is observed to follow the following pattern: Arrival time (min.) Probability Service time (min.) Probability

3

4

5

6

7

8

0.02

0.2

0.4

0.3

0.1

0.08

3

4

5

6

7

0.1

0.2

0.4

0.28

0.02

(c) At a telephone booth, customers arrive with an average time of 1.2 time units between one arrival and the next. Service times are assumed to be 2.8 time units. Simulate the system for 12 time units by assuming the system starts at t = 0. What is the average waiting time per customer?

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355

B. TECH. EIGHTH SEMESTER EXAMINATION, 2005–2006 MECHANICAL SYSTEM DESIGN

Time: 3 Hours Note: (i) Answer all questions. (ii) All questions carry equal marks. (iii) In case of numerical problems assume data wherever not provided. (iv) Be precise in your answer.

Total Marks: 100

1. Attempt any four parts of the following: (5× 4 = 20) (a) Make a systematic search of fresh design ideas using morphological analysis for the following (i) Wall clock (ii) A liquid ink writing instrument. (b) To prevent the excessive injuries and damages by automobile head-on collisions with other vehicles and stationary objects, shock absorbing bumpers can be installed in all vehicles. Carry out a detailed analysis of need for this problem (c) Write need statements for each of the following products. (i) Electric Iron (ii) Solar Cooker (d) What are the principles of concurrent engineering? Write its advantages. (e) Describe the environmental factors that affect the performance of a system under design. (f) Explain in brief Design Morphology. 2. Attempt any two parts of the following: (10 ×2 = 20) (a) How can the systems be classified? (b) What are the advantages of mathematical modelling? Explain its use through an example of your choice. (c) For the mechanical system shown in figure below. (i) Determine the differential equation relating f and x. (ii) Construct the direct (force-voltage) analog. (iii) Construct the inverse (force-current) analog. y

x k

M

f B

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Question Papers

3. Attempt any two parts of the following: (10 × 2 = 20) (a) List various optimization methods which may be used in the design of a mechanical system. (b) Explain, giving suitable examples, different methods of optimization. (c) For the network shown in figure below find the shortest and the longest path from node 1 to node 9. The values adjacent to the arcs denote their lengths. 9

4 2 1

7 11

7 6

23

12

2

9 9

5

8

13

5

7 3

8

9

6

4. Attempt any two parts of the following: (10 × 2 = 20) (a) What is the objective of feasibility analysis in system design? How it is carried out? (b) Write a note on time value for money. (c) A small company makes certain rods of standard sizes, 2 m in length. There are three materials, A, B and C which can be used. Each material calls for a different process and machines for manufacturing and their cost data may be summarized as below: Item

Raw material (Rs/m) Equipment Cost (Rs/year) Labour Cost (Rs/Rod)

Materials A

B

C

2.20 5,000 0.50

2.50 3,000 0.60

2.60 4,000 0.20

Draw the total cost versus the yearly production volume curve for each of the three materials. If a sales volume of 10,000 rods per year is anticipated, which material should be used? 5. Attempt any two parts of the following: (a) What is rational decision making? Describe the problems in making rational decisions. (b) Explain the methods of conflict resolution in organizational decision-making. (c) (i) Write a note on simulation programs and languages. (ii) What are the desirable features of a simulation software?

Tutorials

357

Tutorials Mechanical System Design (TME-802)

TUTORIAL 1 Question 1: There are different kinds of constraints present for the designing of the system. Explain. Question 2: It is said, that designing of “mechanical system” is a creative process. Justify the statement. Question 3: When any system is designed, engineering functions are used. Describe in detail. Question 4: Write down different functions used in system design and express them in matrix form. Question 5: Classify the different constraints in the mechanical system of design. Question 6: Define the 0–1 matrices and explain its importance in mechanical system design. Question 7: Explain the term “system” and explain its basic concept. Question 8: Explain all attributes of the “system”. Question 9: Write down the different aspects of engineering problems. Question 10: Explain the terms “input”, “output” and “process”.

TUTORIAL 2 Question 1: Write down the general procedure of optimum system design. Question 2: Differentiate between the inductive and deductive design. Question 3: Why hierarchical method is required for the system design? Question 4: Explain how concurrent engineering is useful for system design. Question 5: With a suitable example breakdown the different activities of a system design and mention the advantages.

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Tutorials

Question 6: Explain the limitation and application of the concurrent engineering. Where can it be used practically? Question 7: Explain the role of system analysis in mechanical system design. Question 8: How will you proceed with the design of high speed belt? Mention all assumption involved there in. Question 9: Explain the system frame for the design of any model. Question 10: Explain the role of hierarical method in the system approach.

TUTORIAL 3 Question 1: Explain the role of modelling in mechanical system design. Question 2: Explain the objective of the modelling in the system design. Question 3: Classify the models and explain the objective and purpose of each of them. Question 4: Explain the method for construction of any model. Question 5: A thermal system consists of an electric fire in a room. The fire emits heat at the rate of q1 and loses heat at the rate of q2. Assuming that the air in the room is a uniform temperature T and that there is no heat storing in the walls of the room, derive an equation describing how the room temperature will change with time. Question 6: Suppose a machine mounted on the ground with suitable spring, and dashpot system. Make a differential mathematical model and study the behavior. Question 7: Study the behavior of the system using the mathematical models in a wheel of a car moving along a road. Question 8: A motor is used to rotate a load. Device a model and obtain the differential equation for it. Question 9: What systems are called linear and nonlinear? Explain with example. Question 10: Explain the system analysis and its importance. Question 11: Explain the system according to state theory approach.

TUTORIAL 4 Question 1: How can this problem be formulated using state concept? What are the possible states and transformation? Question 2: Many problem faced by engineers are so complex that there is no readily apparent equation of the state or they explain so many state variables that all the possible states can not be defined. Question 3: Explain the method of formation of decision tree structure with suitable example. Question 4: Explain the tools, which are required for decision process approach. Question 5: Differentiate between the certainty, uncertainty, risk and conflict. Question 6: How does the professional solve the engineering problem Question 7: Explain the qualities in the professional for taking the decision.

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359

Question 8: How to solve the problem by expected monetary value. Question 9: Explain “expected monetary value” and “pay off matrix.” Question 10: Explain the present worth factor.

TUTORIAL 5 Question 1: Explain the principle of network flow process and its application. Question 2: Explain the different terms used in path problems. Question 3: Explain the process of selection of best machine used by time value of money. Question 4: How to take decisions by dynamic programming? Question 5: State the theory of black box approach. J=0

J=1

J=2 3

1

1

6

7

J=3 9

6

3 8

7

2

1

7

4 1

J=4

2

6

9

2

1

4 7

10

5 3

10

8

5 3

3

3

Question 6: In the above problem find out the shortest path from vertex A to vertex B along arcs joining various vertices lying between A and B length of each path is given. (Min. length = 21 1–1–2–1–1) Question 7: A truck can carry a total of 10 tons of product. Three type of the product are available for shipment. Their weights and values are tabulated. Assuming that at least one of each type must be shipped, determine the loading which will maximize the total value. Type

Value (Rs)

Weight (Tones)

A B C

20 50 60

1 2 2

(Rs 270, A–3, B–1, C–2) Question 8: Explain the principle of Optimality. And write down the application. Question 9: Explain the method of forward and backward calculation for optimum result.

TUTORIAL 6 Question 1: The cost of the machine is Rs 6100 and its scrap value is only Rs 100. Maininence cost is found from experience to be as follows:

360

Tutorials Year Maintenance cost

1

2

3

4

5

6

7

8

100

250

400

600

900

1250

1600

2000

When should the machine be replaced?

(At the 6th year-end)

Question 2: Machine A cost Rs 3600. Annual operating cost is Rs 40 for first year and then increase by Rs 360 every year. Assuming machine A has no resale value, determine the best replacement age. (A at the end 5th year) Question 3: Another machine B that is similar to machine A costs Rs 4000. Annual running cost is Rs 200 for first year and then increase by Rs 200 every year. It has resale value of Rs 1500, Rs 1000 and Rs 500, if replaced at the end of first, second and third year respectively. It has no resale value during fourth year onwards. Which machine would you prefer to purchase? Future costs are not discounted (replace A by B) Question 4: An engineering company is offered two types of material handling equipment A and B. A is priced at Rs 60,000 including cost of installation. The cost of operational cost and maintt. cost are estimated at Rs 10,000 for each of the first five years, increasing by Rs 3000 per year in the sixth and subsequent years. Equipment B with the rated capacity same as A required an initial investment of Rs 30,000 but operations and maintenance cost Rs 13,000 per year for first six years, increasing by Rs 4000 per year for each year from 7th year onwards. Company expects a return of Rs 10%, Neglect scrap value and determine company policy, which it to be bought and when it should be replaced. (Purchase machine B and replace it after 6 year) Question 5: Explain the role of planning horizon in system design. Question 6: Differentiate between the goal and objective; how these are related with each other? Explain with the suitable examples. Question 7: Why proper freedom is required for the obtaining of the best solution and how is it achieved. Question 8: Write down the different steps for setting up Maize Starch Plant. Question 9: How to decide volume of the production in manufacturing plant from economical point of view? Question 10: Explain the breakeven analysis for feasibility.

TUTORIAL 7 Question 1: A small manufacturing plant has two products A and B. A group of type 1 machine and in succession must work each product, by a group of type 2 machines. Product A requires 2 hrs on type 1 machine and 1 hr required on type 2 machine. Product B requires 1 hr on type 1 machine and 4 hrs require on type 2 machine. A total of 6000 hrs is available per week on type 1 machine while 10,000 hrs is available on type 2 machines. The net profit is Rs 5.00 per unit for product B and Rs 3.50 per unit for the product A. How to maximize the profit by the production? Question 2: Explain the method to find out the thickness of insulation in the pipe carrying the steam. Question 3: A total of 300 mtr of tube are to be installed in the heat exchanger in order to provide the necessary surface area. The total rupee cost for insulation include:

Tutorials

361

(a) Total cost of the tube Rs 700/(b) The cost of shell Rs 20D2.5L where as L length of the shell and D diameter of the shell. The spacing between tubes is such that 20 tubes will fit in a cross sectional area of 1 m2 inside the shell. Determine the diameter and length of the shell to minimize the overall cost. Question 4: Explain Lagrange multiplier method for the calculation of optimization. Question 5: A positive quantity b is to subdivided into n parts in such a way that the product of the n parts is to be maximum. Use Lagrange’s multiplier technique to obtain the optimal solution. (b/n) Question 6: Obtain the necessary and sufficient conditions for the optimum solution of the following non-linear program problem min Z = f (x1, x2) = 3e2x1 + 1 + 2ex2 + 5 subjection to the constrains x1 + x2 = 7 and x1 x2 > 0 x1 = 1/3[11-log3] x2= 7–1/3(11-log3) Question 7: Determine the maximum and minimum values of the function F(x) = 12x5 – 45x4 + 40x3 + 5 Question 8: A beam of uniform rectangular cross section is to be cut from a log having a circular cross section of diameter 2a. The beam hast to be used as a cantilever beam (the length is fixed) to carry a concentrated load at the free end. Find the dimensions of the beam that correspond to the maximum tensile stress carrying capacity.

( a/ 3 , y = 2/3a) Question 9: Define objective function, criterion function and constraints in the optimization techniques. Question 10: Describe the optimization techniques used by calculus method.

TUTORIAL 8 Question 1: How to find out best solution by calculus method in optimization? Question 2: Explain the replacement theory with suitable example. Question 3: Explain with example that feasible assessment is necessary for setting up of new projects. Question 4: Describe the principle of model of equal constraint and unequal constraint with the example. Question 5: There are two points A and B in a given vertical plane. Find the path from A to B along which a particle of mass m will slide under the force of gravity without friction in the shortest time by optimization technique. Question 6: Differentiate single and multiple objective function. Question 7: With a example describe the differential method of optimization.

TUTORIAL 9 Question 1: Explain the Baye’s theorem. How it is used in decision process?

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Tutorials

Question 2: Bag “A” contains 2 white and 3 red balls and bag “B” contains 4 white and 5 red balls. One ball is drawn at random from one of the bags and found to be red. Find probability whether it came from bag B. (25/32) Question 3: Consider a bag consisting of 3 red and 7 black balls. In a draw of two trails find the probability of getting (a) 2 red (b) 2 black (c) one red and one black ball. When the drawn balls are replaced each trail is independent to the previous one. (0.21) Question 4: How to formulate the problem of decision-making? Question 5: Differentiate between the iconic, analog and analytical models. Question 6: Explain the different method for the calculation of expected monetary value. Question 7: Simulation is a real concept for engineering application. Justify the statement.

TUTORIAL 10 Question 1: Explain all definitions of Simulation. Question 2: Explain all the phases of simulation with block diagram and different type of simulation. Question 3: Records of 100 truckload of finished jobs arriving in department, check-out area show the following: Checking out takes 5 min and checkers take care of only one truck at a time. The data is summarized in the following table Truck inter arrival Time (minutes)

1

2

3

4

5

6

7

8

9

10

Frequency

1

4

7

17

31

23

7

5

3

2

As soon as the trucks are checked out the truck driver takes them to the next department. Using Monte-Carlo simulation determine (a) What is the average waiting time before service? (b) What is likely to be the longer wait? (0.8 min, 3 min) Question 4: Differentiate between the analogue and analytical simulation. Question 5: Describe advantages and disadvantages of the simulation techniques. Question 6: Customer arrives at a milk booth for the required service. Assume that inter-arrival and service times are constant and given by 1.8 and 4 time unit respectively. Simulate the system by hand computation for 14 time unit. What is the average waiting time per customer? What is percentage idle time of the facility (assume that the system start at t = 0). (3.57 times unit, 4.08, 0%) Question 7: Explain the limitation of Simulation. Question 8: How to classify the simulation? Explain with example. Question 9: Write down a practical application of simulation in engineering. Question 10: Describe the various terms used in waiting line simulation.

References

363

References 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25.

L.H. Shames, Mechanics of Fluids, McGraw-Hill, Int Student Edition. Modi and Seth, Fluid Mechanics. Garde, R.S., Fluid Mechanics through Problems, Wiley Eastern Limited, New Delhi, 1989. Hunter Ronse, Elementary Mechanics of Fluids, John Wiley & Sons, 1946. W. Leonhard, System, Allied Publishers Pvt. Ltd, 1976. I.J. Nagrath and M. Gopal, Control System Engineering, New Age Int. Pvt. Ltd. D.P. Feleman, Automatic Process Control, Wiley Eastern Ltd. B.C. Kuo, Automatic Control System, Prentice-Hall of India, 1976. E.J. Mocormick, Human Factor Engg, McGraw-Hill. A.K. Chitale and R.C. Gupta, Product Design and Manufacturing, Prentice-Hall of India. M.K. Star, Product Design and Decision Theory, Prentice-Hall. Shigley, Machine Design, McGraw-Hill. Bayzitoughe and Ozisik, Element of Heat Transfer, McGraw-Hill Book Company. James R. Welty, Fundamentals of Momentum, Heat and Mass Transfer, 4th Edition, Low cost edition, John Wiley & Sons (Pvt.) Ltd. Richard M. Phelan, Fundamentals of Machine Design, Tata McGraw-Hill Publ. (1978). D.D. Reredith, K.V. Wong, R.W. Woodhead and R.R. Worthman, Design and Planning of Engineering Systems, Prentice-Hall, Englewood Clifts, New Jersey. J.R. Dixon, Design Engineering, Tata McGraw-Hill Publishing Company, New Delhi. V. Gupta and P.N. Murphy, An Introduction to Engineering Design Method, Tata McGraw-Hill. Robert Maiousck, Engineering Design, Blackish and Son Ltd, Glasgow. S.S. Rao, Optimizations Techniques. David Cleland and William Raking, System Analysis and Project Management, McGraw-Hill. Allen, J. Scale, Model in Hydraulic Engineering, London, Longman Green and Co., 1947. Barabasch, S.M. Contt. L. Gentilini, and A math’s, A Heat Exchange Simulator, Automatic, 11, no. 1 (1964). K. Bloch, and G. Gordon, Systems Simulation with Digital Computers, IBM Systm, J III no. 1 (1964). A. Bloch, Electrometrical Analogies and their use for the Analysis of Mechanical and Electromechanical Systems. I Inst. Elects, Engg. X cll, no S2 (1945).

364

References

26. Percy W. Bridgeman, Dimensional Analysis, New York: AMS Press, Inc, 1976. 27. D.N. Chorafas, Systems and Simulation, New York: Academic Press, Inc, 1965. 28. C. West Churchman, Russell L. Ackoff, and E. Leonard Arnoff, Introduction to Operations Research, New York: John Wiley and Sons, Inc, 1957. 29. James R. Emshoff, and Roger L. Session, Design and use of Computer Simulation Models, New York: The Macmillan Company, 1970. 30. Harold Josheph Highland, A Taxonomy of Models Simulator, IV no.2, (1973). 31. P.G. Hubbard, Applications of the Electrical Analogy in Fluid Mechanics Research. Rev sic. In strum, x x, no. 11, (1949). 32. K.K. Keropyyan, ed, Electrical Analogues of Pin-jointed Systems, New York: The Macmillan Company, 1965. 33. George Jan Klir, Approach to General Systems Theory, New York: Van Nostrum and Reinhold Co., 1969. 34. Henry Lewis Longhair, Dimensional Analysis and Theory of Models, New York: John Wiley and Sons, Inc, 1951. 35. Danielle Mihram, and G. Arthur Mihram, Human Knowledge: The Role of Models, Metaphors and Analogy, International J. General Systems L. no.1 (1974). 36. J.R. Ragaini, and L.A. Aden, The Analysis of Sample Data Systems, trans, Am. Inst. Elecr. Engg. Lx x1, Pc, 11 (1952). 37. Patrick Rivet, and Russell Lincoln Aackolf, Managers Guide to Operations Research, New York: John Wiley and Sons Inc, 1963. 38. Paul A. Samuelson, Economics, New York: McGraw-Hill Book Company, 1976. 39. James D. Watson, The Double Helix, New York: Athenam Publishers 1968. 40. Bernard P. Zeigler, Theory of Modeling and Simulation, New York: John Wiley and Sons, Inc, 1976. 41. R.K. Jain, Production Technology, Khanna Publishers. 42. Kanti Swaroop, Operation Research, Sultan Chand and Sons, New Delhi. 43. Gupta and Parkas, Engineering Heat Transfer, New Chand and Bros, Rorkie 2477667. 44. Konz, Stephan Grip Publishing Inc. Columbus Ohio. 45. J.L. Riggs, Production Systems, Planning Analysis and Control, 4th Edition, New York: John Wiley and Sons, 1987. 46. C.D.J. Waters, An Introduction to Operation Management, Addison Wesley Publishing Company, 1991. 47. Kostas N. Dervitsotis, Operations Management, New York: McGraw-Hill Book Company, 1981. 48. C.K. Mutafi, Operation Research, 3rd Edition, New Delhi: New Age Int., 2002. 49. M. Mangan, Industrial Engineering and Production Management, Delhi: Dhanpat Rai and Co. (P) Ltd, 2002. 50. J. Collins, Failure of Material-Mechanical Design, Wiley Eastern, 1992. 51. George B. Denting, Linear Programming and Extensions, Princeton, UP Princeton N.J., 1963. 52. Hasan Sayeed, Automatic Control Engineering. 53. M.L. Khanna, I.I.T. Mathematics, Jaiprakash Nath Publications, Meerut. 54. H.C. Verma, Concept of Physics, Bharti Bhawan Publishers, Patna. 55. H.K. Das, Advance Engg. Mathematics, S. Chand Publishers. 56. Classical Optimization Technical.

Index

365

Index A biomedical system 82 A case study e.g. A automobile instrumentation panel system 63 A Case study e.g. compound bar system model 90 A case study e.g. heating duct insulation system 42 A case study e.g. viscous lubrication system in wire drawing 29 A case study: heating duct insulation system 42 A case study: inventory control in a production plant 253 A case study: material handling system 113 A case study: optimization of an insulationsystem 184 A case study:- Installation of a machinery 222 A liquid level system 82 A system design example 2 Activation 342 Adjusting feed shaft clutch 229 Adjustment of brake 231 Adjustment of main spindle bearings 230 Adjustment of V-belt tension 230 Advantage and disadvantage of simulation 243 Advantages of inventory control 255 Advantages of the simulation technique 243 Alloys of aluminium 127 Aluminium extrusion system 126 Ammeter 67 Amount of an annuity 138

Analog 238 Analog model 72 Analogous system 279 Analytical simulation model 238 Analytical approach 124 Angle of contact in belt drive-centrifugal stress in belt or rope 53 Anthropometrics data 297 Application of computers in design 318 Application of system concepts in engineering 13 Assigning numbers to bonds 332 Automobile suspension system 81 Baye’s theorem (First to third method) 210 Benefit/ advantage of waiting line queuing theory 239 Black box 61 Bond graph 323 Bond graph modelling 345 Bond graph standard elements 324 Bonds 136 CAD/CAM 319 Calculus method for optimization 154 Case study-high speed belt drive system Casting aluminium alloys 128 Causality assignment procedure 337 Chain drives 54 Characteristics of a system in MSD 16

45

366 Characteristics/Features of material handling 113 Clad aluminium alloys 128 Classification of system in MSD 9 Closed loop control system 267 Combinatorial approach 125 Communication of the results 293 Component behavior 91 Component Integration approach 63 Compute-aided design 316 Computer integrated manufacturing 319 Computer system concept 311 Concept of a system 2 Conceptual design 296 Concurrent Engineering 25 Conditional probability 209 Considerations of a good design 294 Continuous time function 19 Conversion of manual and computer-based systems 315 Cost of product 148 Critical insulation thickness for pipe 184 Critical thickness of insulation for sphere 186 D’ Alembert’s principle 272 Data flow diagram 312 Dead zone 19 Decision analysis 197 Decision model 197 Decision process approach 61 Decision trees 200 Description of operating mechanism 226 Descriptive model 72 Design 26 Design of Transmission roller chains 54 Design requirement 294 Designation of aluminium and alloys 129 Designing for profit of manufacture of a maize starch system 148 Desirable features of simulation software 243 Deterministic model 72 Different types of model 71 Discrete time function 20 Distinction between engineering and design 33

Index Distributed parameter and lumped parameter system 20 Duraluminium 128 Annuity 138 Economic order quantity 257 Effect of alloying elements on aluminium 129 Effect of centrifugal tension on the power transmitted 53 Effort and flow source 326 Elements of a decision problem 196 Engineering activities matrix, 21 Engineering problem solving 24 Evaluation and generation 49 Evaluation, generalization and communication 51 Expected monetary value 206 Expected value criterion 220 Explanation insulation system 186 Factor considers 115 Feasibility assessment 132 Feed box 228 Feed gear box 227 Feed selection 227 Financial analysis 136 Flow chart symbol 314 Friction viscous 18 Fuel sensor 67 Full level gauge 67 Function of material handling 114 Fundamentals of CAD/CAM 316 Fundamentals of probability 208 General model 73 Generation of a system equations 338 Goal and objective criteria 122 Goal programming 108 Graph modeling and analysis process 97 Graphical method for linear programming problem solving 75 Hurwitz’s criterion 218 Head stock casting 226

Index

367

Heat transfer 188 Hierarchical nature of system, hierarchical nature of problem environment 37 Human factors in design/ergonomics 297 Hysteresis 18 Iconic simulation model 238 Iconic model 72 Identification and analysis of need 35 Identification of egg. function 15 Importance of product design 293 In car temp sensor 67 Inching 227 Indian standard specifications aluminium and its alloys 128 Initial lubrication 226 Input model construction 240 Input-output relationship 266 Instruments and their function 66 Inventor control 254 Inventory control and its feasibility 254 Inventory costs: deterministic model 256 Inventory function (need for inventories) 254 Inventory model 256 Laplace criterion 220 Level of a system 3 Limitation of decision tree approach 202 Limitation of simulation approach 240 Linear and non linear system 16 Linear graph analysis 95 Linear graph analysis, a path problem 100 Linear graph analysis network flow problem Linear graph modeling concepts 74 Linear programming model 75 Linear programming problem 76 Lubrication 228 Lubrication of head stock 228 Lubrication of speed box 228 Lubrication oil pressure guage 71 Machine (Installation) Machine design 223

222

102

Maintenance 229 Manufacture of a maize starch system 148 Material handling equipments 115 Material handling principles 114 Mathematical modelling concepts 79 Mathematical programming 205 Maximax criterion 217 Maximin criterion 217 Mechanical equivalent network 285 Mechanical system 270 Method of optimization 124 Micrometer and screw cutting shop 228 Minimax regret criterion 218 Model building is the essence of system design 74 Model by function 72 Model by nature of environment 72 Model by structure 74 Model by the extent of generality 73 Model formulation 43, 49 Model with equality constraint 167 Model with inequality constraint 181 Model with one decision variable 154 Model with two variables optimization with no constraints 158 Modeling types and purposes 71 Modelling and simulation 84 Monte Carlo method 244 Motivation and freedom of choice 122 Nature of engineering problem Need for modeling 70 Needs statement 34 Normalizing model 76

34

Objective of material handling 113 Observers 343 Open loop control system 266 Optimization concepts 124 Pedal brake 227 Place of Manufacture of a maize-Starch system 147

368

Index

Planning horizon 133 Power directions on the bonds 330 Power transmitted by Belt or rope 53 Power variables of bond graph 323 Predictive model 72 Preliminary need statement 36 Present value 138 Pressure sensor 67 Probabilistic model 73 Probabilistic variation in demand 256 Probability of a density function 205 Problem area and its environment 34 Problem definition 239 Problem formulation 33 Problem scope and constraints 41 Procedure of writing the models of mechanical system 274 Process in inventory control 255 Product analysis 132 Product design process 293 Production order quality model 258 Prop. and uses of aluminium 126 Proposed effort 23 Purpose of inventory control 254 Quantities methods in decision making 199 Querying up model with random input or poisson arrival 247 Quick change tool post 228 Random number table 246 Raw inventory 254 Realization of need 36 Re-assembley of the main spindle 230 Regulatory and social issues 295 Removing of the pulley sleeve 230 Repaying the loan 136 Replacement of V-belt 229 Replacing of broken shear pinlead screw Role of engineer 23 Rotational system 271

230

Saddle and apron 229 Sadle point 163 Scientific appraoch to decision process 198 Screw cutting 227 Selection between alternatives 140 Service output 247 Shortest path algorithm 96 Simulation and off-line programming 321 Simulation concept 236 Simulation languages 242 Simulation models 238 Simulation process 239 Simulation programs and languages 242 Solution by network technique 96 Specific model 73 Speed box unit 226 State theory approach 62 Statement of the problem 90 Static and dynamic system 17 Steps involved in modelling a mechanical system 73 Subjective approach 126 Sufficient condition 159 Sunlight sensor 67 Symbolic model 72 System analysis view point 59 System approach 13 System compatibility 92 System constraints 91 System equilibrium 92 System evaluation 135 System evaluation 131 System input 247 System modeling 69 System simulation 234 System theories 59 Tachometer 66 Tail stock 232 Taper turning attachment

228

Index Temp gauge 67 Temperature 298 Terminology 64 The 3-port junction elements 328 The computer system concept 311 The design process 317 The four’s of design 292 The Gyrator 327 The optimization process 121 The product design process 293 The role of models in engineering design The scientific approach to the decision process 198 The work environment 298 The world of automobiles 63 Thread chasing dial 228

369 Time invariant and time varying system Time value of money 135 Total life cycle 295 Transfer function 269 Transnational system 270 Trip computer 67

19

Utility value 207

84

Waiting line simulation or queuing theory 239 What is simulation? 235 When to use simulation 235 Wiring diagram and electrical symbols 65 Wiring system 65 Worm housing and lead screwend bearing 229 Wrought aluminium alloys 127