management design manual

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Module1

Introduction

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Lecture1

Need Identification and ProblemDefinition

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Instructional objectivesThe primary objective of this lecture module is to outline how to identify the need and define the

problem so as to begin with the activities and steps involved in design for manufacturing process

Steps involved in Engineering Design processFigure 1.1.1 schematically outlines the typical steps involved in an engineering design process.

Define Problem Problem Statement

Benchmarking QFD PDS

Project Planning

Evaluation of

concepts Pugh concept selection

Decision matrix

Concept Generation Brainstorming

Functional decomposition

Mor phological chart

Conceptual

Desi n

Gather Information Internet Patents

Trade Literature

Parametric design Robust design

Tolerances Final dimensions

DFM

Configuration Design Preliminary selection of

material and manufacturing

Modeling/sizing of parts

Embodimdesign

Product architecture

Arrangement of

physical

elements to carry out

functions

Detail design Detail drawing and

specifications

Figure 1.1.1 Discrete steps involved in engineering design process. It also mentions the

important techniques used in each steps.

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Conceptual Design

It is a process in which we initiate the design and come up with a number of design concepts and

then narrow down to the single best concept. This involved the following steps.

(1) Identification of customer needs : The mail objective of this is to completely understand

the customers needs and to communicate them to the design team

(2) Problem definition : The mail goal of this activity is to create a statement that describes

what all needs to be accomplished to meet the needs of the customers requirements.

(3) Gathering Information : In this step, we collect all the information that can be helpful for

developing and translating the customers needs into engineering design.

(4) Conceptualization : In this step, broad sets of concepts are generated that can potentially

satisfy the problem statement

(5) Concept selection : The main objective of this step is to evaluate the various designconcepts, modifying and evolving into a single preferred concept.

Embodiment Design

It is a process where the structured development of the design concepts takes place. It is in this

phase that decisions are made on strength, material selection, size shape and spatial

compatibility. Embodiment design is concerned with three major tasks product architecture,

configuration design, and parametric design.

(1) Product architecture : It is concerned with dividing the overall design system into small

subsystems and modules. It is in this step we decide how the physical components of the

design are to be arranged in order to combine them to carry out the functional duties of

the design.

(2) Configuration design : In this process we determine what all features are required in the

various parts / components and how these features are to be arranged in space relative to

each other.

(3) Parametric design : It starts with information from the configuration design process and

aims to establish the exact dimensions and tolerances of the product. Also, final decisions

on the material and manufacturing processes are done if it has not been fixed in the

previous process. One of the important aspects of parametric designs is to examine if the

design is robust or not.

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Detail Design

It is in this phase the design is brought to a state where it has the complete engineering

description of a tested and a producible product. Any missing information about the

arrangement, form, material, manufacturing process, dimensions, tolerances etc of each part is

added and detailed engineering drawing suitable for manufacturing are prepared.

Need Identification and Problem DefinitionOut of all the steps in the engineering design process, the definition of the problem is by far the

most important step. A complete and thorough understanding the problem is prerequisite in

achieving the targeted solution. For example, the ultimate test of a product is how well it sells.

However, it is first essential to understand and provide what a customer wants in the product

which can only be achieved by defining the problem precisely at the first place.

A-priori ActivitiesIn majority of the situations, a significant amount of development work precedes the tight

definition of a design problem. These a-priori development works can generally be referred to

planning. The primary purpose of the planning stages is to collect all the necessary information

and to decide, for example, whether manufacturing a new product is feasible or what would be

the best time to market a new or modified product, or whether a specific company has the

adequate resource to manufacture a new product. Usually the initial design projects can be

categorized as follows.

Variation of an existing product

This includes minor changes in few parameters of an existing the product e.g. change in the

power of a motor or change in the design of a typical clamping bracket, and so on.

Improvement in an existing product

This involves major redesign of an existing product primarily to improve performance and

quality, update features ( may be due to competitions ), reduce cost in manufacturing and so on.

Development of a new product for a low-volume production run

This is primarily referred to new parts or products that would possibly be manufactured in

smaller number of units (e.g. < 10000). In many cases, a large manufacturing unit may wish to

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buy standard available components available from smaller manufacturing units rather than

actually making the same to avoid additional costs.

Development of a new product for mass production

These include products or parts which need to be produced in large volumes e.g. in the category

of automobiles, home appliance etc. Such design projects provide the design engineer the

flexibility in selecting appropriate material and manufacturing process through careful planning.

One-of-a-kind design

Such projects can vary from a simple, quick design requiring minimum of analysis like designing

of a welding fixture to hold parts to large exclusive projects such as building of a 200-MW steam

turbine.

Product Life cycleEvery product goes through a cycle from birth, followed by an initial growth stage, a relatively

stable matured period, and finally into a declining stage that eventually ends in the death of the

product as shown schematically in Figure 1.1.2 .

(1) Introduction stage : In this stage the product is new and the customer acceptance is low

and hence the sales are low.

(2) Growth stage : Knowledge of the product and its capabilities reaches to a growing number

of customers.(3) Maturity stage : The product is widely acceptable and sales are now stable, and it grows

with the same rate as the economy as a whole grows.

(4) Decline stage : At some point of time the product enters the decline stage. Its sales start

decreasing because of a new and a better product has entered the market to fulfill the

same customer requirements.

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Figure 1.1.2 Schematic outline of a product life cycle

Technology development cycleThe development of a new technology follows a typical S-shaped curve [ Figure 1.1.3 (a )]. In its

early stage, the progress is limited by the lack of ideas. A single good idea can make several

other god ideas possible, and the rate of progress is exponential. Gradually the growth becomes

linear when the fundamental ideas are in place and the progress is concerned with filling the gaps

between, the key ideas. It is during this time when the commercial exploitation flourishes. But

with time the technology begins to run dry and increased improvements come with greater difficulty. This matured technology grows slowly and approaches a limit asymptotically. The

success of a technology based company lies in its capabilities of recognizing when the core

technology on which the companys products are based begin to mature and through an active

R&D program, transfer to another technology growth curve [ Figure 1.1.3 (b)] which offers

greater possibilities.

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(a) (b)

Figure 1.1.3 Schematic outline of (a) technology development curve, (b) improved program to

develop new technology before the complete extinct of existing technology.

Identifying Customer NeedsIt is usually the desire of the customers that drive the development of a new product or

modification of an existing product. It is thus critical to collate the need or views of the

customers when starting a design project. The needs of the customers can be gathered through

multiple routes.

Interviewing with customers

An active team should constantly meet current and potential customers to identify the strength

and weakness of a product so as to examine if there is any need to upgrade.

Focus group

A focus group refers to a small sub-set of existing customers or potential customers. A

discussion is usually facilitated in many such groups separately to identify more closely the

merits and demerits of the product.

Customer survey

A written questionnaire is possibly the best way to know the pubic opinions for redesigning an

existing product or developing a new product.

Customer complaints

Complaints from customers provide a significant premise to identify the requisite improvement

for an existing product.

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Constructing a Survey InstrumentFollowing are some essential steps to prepare a survey document based on the views and feed-

backs from the customers.

1. Determine the purpose of the survey, its result and the how the result will be used.2. Determine the type of possible data collection method such as face to face interview or by

questionnaire or some other way.

3. Determine what specific information is needed. Each question should have a clear goal.

Also the number of question should be optimized and kept at as minimum as possible.

4. Design the questions in such a way that they are unambiguous, unbiased, clear, brief and

simple to understand and to answer too. There are usually three basic type of questions.

Attitude questions: how the customer thinks or feels about something, Knowledge questions: Questions asked to determine whether the customer know

the specifics about the product,

Behavior questions they usually contain phrases like how often, how much, or

when.

Following are some tips for developing the questions.

Use simple language and vocabulary. Each question should have a specific goal

and focus directly on one specific topic.

Questions may include yes no do not know or strongly disagree mildly

disagree neutral mildly agree strongly agree , etc.

Open ended questions allow customers to express more explicitly, Arrange the question in such an order that it makes sense and provides content to

what you are trying to learn from the customer,

Pretest the survey on a small sample before distributing the survey. It helps to

identify questions that were poorly built, misunderstood, whether the rating scale

was adequate and whether the questionnaire is too long Administer the survey: Proper care should be taken that the sample of the survey

should constitute a representative from all the key areas.

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Evaluating Customer NeedsThe responses of the customer should be evaluated on a relative scale, say using a scale from 1

(low importance ) to 5 ( high importance ). Those responses with high average score should be

given a greater priority when redesigning an existing product or designing a new product. It isvery essential to divide the customer needs into two groups: hard constrains that should be

satisfied (must) and softer needs that can be traded off against other customer needs (wants).

Customer needs can best be identified from face to face interview, from a focus group survey or

from the higher-ranking items in the written survey.

Customer requirements

Customer requirements must be characterized on the basis of performance , time , cost and quality . The performance would refer to the specific or intended function of a product. The time

would include all the time aspects that would be involved in the design. A proper design should

be able to reduce the cycle time to market a new product. The cost includes all the monetary

aspects of the design and hence, quite crucial. The cost aspect also determines the buying

decisions of any product by the customers. The quality is a complex characteristic with many

aspects and definitions and can best be defined as the totality of features and characteristic of a

product that bears on its ability to satisfy its stated needs. Another important aspect of the

customers requirements is the value of a product that can be envisaged as the ratio of the

function (or the quality) provided and the cost. For example, the quality of a manufactured

product can be envisaged from the following eight basic dimensions.

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Dimensions Description

PerformanceDoes the product perform to its standards? Does the product perform for the

intended service?

Features What additional benefits will be added to the product? Will they be they

tangible or non-tangible benefits?

Reliability Is the product consistent? Will it perform well over its lifetime and perform

consistently?

Durability How durable is your product. Will it last with daily use?

Conformance Does your product meet with any agreed internal and national

specifications?

Serviceability Is the product easy to service.

Aesthetics Is the product appealing to the eye?

Perceived

Quality

What sort of quality perception does the marketing team want to convey in

their marketing message? Will price charged reflect the quality of the

product?

The dimensions of performance, features and conformance are often interrelated. We therefore

need to recognize that there are four levels of customer requirements as

(1) Expectations that refer to the basic attributes, which one would expect to be present in the

product as standard features,

(2) Spokens that refer to the specific features, which the customer would say and want as a

feature in the product.

(3) Unspokens that refer to the attributes of a product that the customers would not generally

ask for but are still important and hence, cannot be ignored.

(4) Exciters which are also known as delighters and are features that make the product unique

and distinguish the same from their competitors.These requirements must be satisfied at each level before we move and address those at the next

level. Not all customer requirements are equal and hence it becomes very essential to identify

these requirements which are important and ensure that they are delivered in the product. To do

this one must adopt a strategy for actively seeking the the voice of the customer .

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Quality Function deploymentQuality function deployment (QFD) is a planning and problem-solving tool that is used from

transforming customer requirement into the engineering characteristics of the product. QFDhelps to transform the customer needs (also referred to as voice of customer ) into engineering

characteristics (and appropriate test methods) for a product. It is a graphical technique, which

systematically looks at all the elements that are deemed important based on customers survey go

into the production definitions. A sample layout of the QFD diagram is shown below. Further

analysis of QFD with real-life examples is presented later in Lecture 3 of Module 5 ( Design for

Quality ).

Figure 1.1.2 Schematic presentation of quality function deployment (QFD ) house / table

Following are a brief outline of each section of the quality function deployment table.

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Customer requirements (whats)

These are typically the customer requirements.

Competitive assessment

It shows how the top two or three competitive products rank with respect to the customer

requirements. This starts with ranking each customer requirements on a scale of 1 to 5 and then

by considering planned improvement and any requirements that are planned for special

attentions.

Engineering characteristics (hows)

The engineering characteristics that enable satisfying the customer requirements are listed in this

column.

Correlation matrix

It shows the degree of interdependence of the engineering characteristics with each other in theroof of the house.

Relationship matrix

It represents the correlation between the engineering characteristics and the customer

requirements.

Absolute importance

To determine the absolute importance we need to multiply the numerical value in each of the

cells of the relationship matrix(6) by the importance rating (3) and then sum the numbers in the

cells of each column.

Relative importance

This represents the absolute importance but normalized on a scale of 1 to 100.

Technical competitive assessment

This refers to the benchmarking of the company performance against the top two or three

competitors for each of the engineering characteristics.

Technical difficulties

These depict the ease (or the extent of difficulty) to achieve each engineering characteristic.

Target values

This would depict the final target set based on the key engineering characteristics that are

deemed important and the assessment of the technical difficulty.

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Product Design SpecificationsThe product design specification is the basic control and reference document that would include

the outcomes of the product development exercise, and is the must to begin with and execute the

design and manufacturing of any specific part or product. The quality function deployment tool provides the most crucial inputs in writing the product design specifications . Following are some

of the important elements of a typical product design specification document. It is, however, not

necessary that the product design specification document of any product will contain all these

elements.

[A] In-use purposes and market requirements

(a) Title and Purpose or function of the product,

(b) Predictable unintended use of the product,

(c) Special features of the product,

(d) What would be the competitive products?

(e) What is the indented market and why there is a need for this product?

(f) Relationship of the product to the other company products,

(g) Anticipated market demand (units per year) and target price.

[B] Functional Requirements

(a) Functional performances such as flow of energy, information, materials, operational

steps, efficiency, accuracy, etc.,

(b) Physical requirements such as shape, size, weight, surface finish, etc.,

(c) Service environment such as storage and transportation requirement,

(d) Life-cycle issues including useful life, reliability (mean time to failure), robustness,

ease of installation, maintenance and repair, recyclability, etc.

(e) Human factors including importance of aesthetics, ergonomics and user-training.

[C] Corporate Constrains

(a)

Is there adequate time to design a quality product and its manufacturing process ( time tomarket )

(b) What are the requirements for manufacturing this product?

(c) Do existing relationships with the suppliers pose any constraint on manufacturing?

(d) Are there any constraints in using the trademark, logo, brand name ?

(e) What are the profitability and return on investment ( ROI ) that must be met?

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(f) The production team should follow professional ethics at every level of the design

process when they are dealing with suppliers, dealers, corporate officials, society etc.

[D] Social, Political and Legal Requirements

(a) The product design specification should meet / contain all the requisite safety and

environmental regulations,

(b) The product design specification should contain all the required standards,

(c) The product design specification must be completed with respect to all safety and

liability norms,

(d) The product design specification should consider all the information related to the

patents and intellectual property that are applicable.

Product Design Specifications (PDS) is explained with the following example in which the PDS

is done for an adjustable wheel chair.

Criterion Performance

Requirement Seat width

Primary Customer Patient

Metrics & Targets Metric Target

Seat width adjustable by

userInches

16 through 20 2 inch

increments

Target Basis Market research

Verification Method Prototyping

Similarly, other dimensions of the wheel chair can be set. However, the criteria are not only

based on the performance. There can be other criteria as well like aesthetics , life in service , legal

(Patents, Product Liability) and so on. One of them is shown below.

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Criterion Life in Service

Requirement Appropriate service life for capital asset

Primary Customer Hospitals

Metrics & Targets Metric Target

Life Years 15 for frame and 5 for seat

Target Basis Market Analysis

Exercise1. Write a survey to find what customers want in a refrigerator.

References1. G Dieter, Engineering Design - a materials and processing approach, McGraw Hill, NY,

2000.

2. http://web.cecs.pdx.edu/~far/me491/Sample%20Reports/Keen%20PDS%20Report.pdf.


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Module2

Selection of Materials and Shapes


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Lecture

1Physical and Mechanical Properties of Engineering Materials


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Instructional objectivesAt the of this lecture, the student should be able to appreciate

(a) general classification of engineering materials, and

(b) physical and mechanical properties of engineering materials

Engineering MaterialsMaterials play an important role in the construction and manufacturing of various parts and

components. An appropriate selection of a material for a given application adds to economy,

working and life of the final part and component.

Classification of Engineering MaterialsEngineering materials can be broadly classified as metals such as iron, copper, aluminum and

their alloys, and non-metals such as ceramics (e.g. alumina and silica carbide), polymers (e.g.

polyvinyle chloride or PVC), natural materials (e.g. wood, cotton, flax, etc.), composites (e.g.

carbon fibre reinforced polymer or CFRP, glass fibre reinforced polymer or GFRP, metal matrix

composites or MMC, Concrete, Ceramic matrix composites, Engineering wood such as plywood,

oriented strand board, wood plastic composite etc.) and foams.

Properties of Engineering MaterialsMaterial property is the identity of material, which describes its state (physical, chemical) and

behavior under different conditions. The material properties can be broadly categorized as

physical, chemical, mechanical and thermal.

The physical properties define the physical state of material and are independent of its chemical

nature. The physical properties of engineering materials include appearance, texture, mass,

density, Melting point, boiling point, viscosity, etc. The chemical properties describe the

reactivity of a material and are always mentioned in terms of the rate at which the material

changes its chemical identity, e.g. corrosion rate, oxidation rate, etc. The mechanical properties

describe the resistance against deformation, in particular, under static and dynamic mechanical


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loading condition. The mechanical properties include elastic modulus, Poissons ratio, yield

strength and ultimate tensile strength, hardness and toughness, etc. The thermal properties

describe material behavior under thermal loading and include thermal conductivity, specific heat,

thermal diffusivity, coefficient of thermal expansion, etc. For a given application or service, an

engineering material is selected based on a set of appropriate material properties, often referred

to as attributes, that would be requisite to sustain various expected loads. Figure 2.1.1 depicts a

schematic representation of material family, which is utilized in selection of materials for a target

application.

Figure 2.1.1 Organized classification of materials and properties [1]

Physical PropertiesPhysical properties describe the state of material, which is observable or measurable. Color,

texture, density, melting point, boiling point, etc. are some of the commonly known physical

properties.

Color: Represents reflective properties of substance

Density: Amount of mass contained by unit volume of material. The higher the density

the heavier is the substance. (SI unit: kg/m 3)


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Melting point: Melting point is the temperature at which material changes its state from

solid to liquid. (SI units: K)

Boiling point: Boiling point is the temperature at which material changes its state from

liquid to gaseous. (SI units: K)

Chemical PropertiesChemical properties are the measure of reactivity of a material in the presence of another

substance or environment which imposes change in the material composition. These properties

are always mentioned in term of the rate of change in its composition. Corrosion rate, oxidation

rate, etc. are some of the chemical properties of material.

Corrosion rate: Corrosion rate is measured in terms of corrosion penetration for given

period of time at specific surrounding condition. Corrosion rate is given by length of

penetration per unit time. (Units: mm/year)

Oxidation rate: Oxidation rate is measured in terms of amount of material consumed

forming oxide or amount of oxide scale formed for given period of time at specific

surrounding temperature. Oxidation rate is given by amount of mass of material lost or

thickness of scale formed during oxidation per unit time. (Units: gms/min o r m/min ).

Mechanical PropertiesMechanical properties describe the behavior of material in terms of deformation and resistance to

deformation under specific mechanical loading condition. These properties are significant as they

describe the load bearing capacity of structure. Elastic modulus, strength, hardness, toughness,

ductility, malleability are some of the common mechanical properties of engineering materials.

Every material shows a unique behavior when it is subjected to loading. Figure 2.1.2 shows a

typical stress-strain curve of C-steel under uniaxial tensile loading. Point A indicates the

proportional limit. Stress strain behavior is linear only up to this point. Point B represents the

point at which the material starts yielding. Between point A and B, the stress strain behavior is

not linear, though it is in elastic region. Point C is referred to the upper yield point. The

material behavior after point D is highly nonlinear in nature. Point E is the maximum stress

that the material can withstand and the point F schematically indicates the point of rupture.


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Figure 2.1.2 Stress-strain curve for carbon-steel [3]

Stresses computed on the basis of the original area of the specimen are often referred to as the

conventional or nominal stresses . Alternately, the stresses computed on the basis of the actual

area of the specimen provide the so called true stress . Within the elastic limit , the material

returns to its original dimension on removal of the load. The elastic modulus is referred to the

slope of the stress-strain behavior in the elastic region and its SI unit is conceived as N.m -2

Figure 2.1.3(a) to (c) schematically shows the uniaxial tensile , shear and hydrostatic

compression on a typical block of material. When a sample of material is stretched in one

direction it tends to get thinner in the other two directions. The Poisson's ratio becomes

important to highlight this characteristic of engineering material and is defined as the ratio

between the transverse strain (normal to the applied load) and the relative extension strain , or

the axial strain (in the direction of the applied load). For an engineering material, the elastic

modulus (E), bulk modulus (K), and the shear modulus (G) are related as: G = E/2(1+ ) and K =

E/3(1-2 ), where refers to the Poissons ratio.

. The

elastic modulus is also referred to as the constant of proportionality between stress and strain

according to Hookes Law . Beyond the elastic limit , the materials retains a permanent,

irreversible strain (or deformation) even after the load is removed. The modulus of rigidity of a

material is defined as the ratio of shear stress to shear strain within the elastic limit . The bulk

modulus is referred to the ratio of pressure and volumetric strain within the elastic limit .


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(a) (b) (c)

Figure 2.1.3 Schematic presentation of (a) tensile, (b) shear and (c) hydrostatic compression [4]

The strength (SI units: Pa or N/m 2

The hardness is another important mechanical property of engineering material and refers to the

resistance of a material against abrasion / scratching / indentation. The hardness of a material is

always specified in terms of the particular test that is used to measure the same. For a measure of

resistance against indentation , Vickers, Brinell, Rockwell, Knoop hardness tests are common.

Alternately, for a measure of resistance against scratch , Mohrs hardness test is followed. The

basic principle used in these testing involves the pressing of a hard material against the candidate

material, whose hardness is to be measured. The Brinell hardness (figure 2.1.5) test method

consists of indenting the test material with a 10 mm diameter hardened steel or carbide ball

) is the property that enables an engineering material to resist

deformation under load. It is also defined as the ability of material to withstand an applied load

without failure. Based on the typical stress-strain behavior of an engineering material, a few

reference points are considered as important characteristics of the material. For example, the proportional limit is referred to the stress just beyond the point where the stress / strain behavior

of a material first becomes non-linear. The yield strength refers to the stress required to cause

permanent plastic deformation. The ultimate tensile strength refers to the maximum stress value

on the engineering stress-strain curve and is often considered as the maximum load-bearing

strength of a material. The rupture strength refers to the stress at which a material ruptures

typically under bending. Different material behaves differently when subjected to load. Figure

2.1.4 indicates the different in stress strain behavior of typical cast iron, low carbon steel, and

aluminum alloy. Cast iron, being a brittle material generates steeper curve than low carbon steel

or aluminum alloy. There is no sign of yielding prior to failure, so the yield point has to be found

out graphically. The yield point strength in the case of low carbon steel and aluminum alloys can

be identified easily.


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subjected to a load of 3000 kg. For softer materials the load can be reduced to 1500 kg or 500 kg

to avoid excessive indentation. The full load is normally applied for 10 to 15 seconds in the case

of iron and steel and for at least 30 seconds in the case of other metals. The diameter of the

indentation left in the test material is measured with a low powered microscope. The Brinell

harness number is calculated by dividing the load applied by the surface area of the indentation.

The typical Brinell hardness values of some of the commonly used engineering materials are as

follows: aluminum 15, copper 35, mild steel 120, austenitic stainless steel 250, hardened

tool steel 650, and so on.

Figure 2.1.4 Comparison of behavior of different material

Another important mechanical property of engineering materials is the toughness that provides a

measure of a material to withstand shock and the extent of plastic deformation in the event of

rupture. Toughness may be considered as a combination of strength and plasticity. One way to

measure toughness is by calculating the area under the stress strain curve from a tensile test. The

toughness is expressed in Joule to indicate the amount of energy absorbed in the event of failure

or rupture. Figure 2.1.6 shows the schematic set-ups of Izod impact test and Charpy impact test.

In both the cases, impact loading is applied in notched specimen of predefined dimension.

Energy absorbed during the breakage of the specimen is the measure of the toughness. In a


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similar line, resilience of a material refers to the energy absorbed during elastic deformation and

is measured by the area under the elastic portion of the stress strain curve. Izod and charpy

tests are two important methods for evaluating toughness of a material.

Figure 2.1.5 Brinell Hardness Test [5]

Figure 2.1.6 Schematic set-up of (a) Izod Test and (b) Charpy Test [6]

Thermal PropertiesThe thermal properties of an engineering material primarily refer to the characteristic behaviors

of the material under thermal load. For example, thermal conductivity is a measure of the ability


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of material to conduct heat and is expressed as W.K -1.m -1 in SI unit. The specific heat refers to

the measure of energy that is required to change the temperature for a unit mass and is expressed

as J.kg -1.K -1. The product of density and specific heat is often referred to the heat capacity of a

unit mass of material. The thermal diffusivity refers to the ratio of thermal conductivity and heat

capacity of a material and provides a measure the rate of heat conduction. The thermal diffusivity

is expressed in terms of m 2.s-1

When a material is subjected to both thermal and mechanical loading, two more characteristics

of materials - coefficient of thermal expansion and thermal shock resistance - become significant.

.

The coefficient of thermal expansion provides a measure of unit change in strain of a material for

unit change in temperature and is expressed in terms of K -1

E)-(1K T

in SI unit. The thermal strain in

material is considered to be isotropic in nature. The thermal shock resistance provides a measure

to which a material can withstand an impact load which is either thermal or thermo-mechanical

in nature. The thermal shock resistance is expressed as , where K is the thermal

conductivity, T

maximal tension the material can resist, the thermal expansion coefficient, E

the Youngs modulus and the Poissons ratio.

Getting Familiar with Different Materials

Metals

Metals have free valance electrons which are responsible for their good thermal and electrical

conductivity. Metals readily loose their electrons to form positive ions. The metallic bond is held

by electrostatic force between delocalized electrons and positive ions. Metals are primarily used

in the form of alloys which depict a combination of two or more materials, in which at least one

is metal. The iron based alloys are characterized as ferrous alloys. For example, steel is an alloy

of iron, carbon and other alloying elements, brass is an alloy of copper and zinc, bronze is an

alloy of copper and tin, and so on. Metals and alloys are typically characterized by an excellent

blend of mechanical and thermal properties. Table 2.1.1 indicates the typical material properties

and common applications of some of the widely used metallic materials.


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Table 2.1.1 Common material properties of metallic materials [7]

Material

Properties

Iron Copper Aluminum C-Steel AA6061 Ti-6Al-4V

Type Pure Pure Pure Fe- Alloy Al-alloy Ti-Alloy

Density (kg.m -3 7870) 8930 2698 8000 2700 4420

Melting

Temperature (K)1808 1357 933 1750

Solidus = 855

Liquidus = 924

Solidus = 1877

Liquidus = 1933

Boiling

Temperature (K)3134 2835 2792 3300 3533

Youngs

Modulus(GPa)200 110 68 210 70-80 113.8

Shear Modulus(GPa)

77.5 46 25 79.3 26 44

Bulk

Modulus(GPa)166 140 76 160 40.7

Poissons Ratio 0.291 0.343 0.36 0.27-0.3 0.33 0.342

Yield Strength

(MPa)50 33.3 250 275 880

Ultimate Tensile

Strength (MPa)210 90-180 410 310 950

Coefficient of

Thermal

Expansion X 10 -6

(K -1

12.2

)

16.4 24 10.8 23.6 8.6

Thermal

Conductivity

(W.mm -1.K -176.2

)

400 210 35-55 180 6.7

Specific Heat

(J.kg -1.K -1440

)385 900 490 896 526.3

Application Heat

Exchanger

Aerospace,

Construction,

Electrical

conductors

Utensils,

Naval

Construction,

Chemical

transport,

Aircraft fittings,

Pistons, Bike

frames

Aerospace,

Marine, Power

generation,

Offshore

Industries


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Ceramics and Glasses

Ceramics are non-metallic in nature and refer to the carbide, boride, nitride and oxides of

Aluminum, silica, zirconium, etc. However, the ceramics possess excellent resistance to thermal

and chemical corrosion and wear resistant. Ceramics are also good thermal and electrical

insulators. Table 2.1.2 indicates the typical material properties and common applications of some

of the widely used ceramics.

Table 2.1.2 Material properties and applications of commonly used Ceramics [7]

Material

Properties

Alumina Silicon Carbide Silicon NitrideGlass

(Soda-lime glass)

Density (kg.m -3 3960) 3000 3290 2520

Melting

Temperature (K)2300 3000 2173 1313

Youngs

Modulus(GPa)370 410 310 72-74

Shear

Modulus(GPa)150 179 29.8

Bulk Modulus(GPa) 165 203

Poissons Ratio 0.22-0.27 0.14 0.27

Ultimate TensileStrength (MPa)

300 250

Coefficient of

Thermal Expansion

X 10 -6 (K -15.4

)

2.77 3.3 8.5

Thermal

Conductivity

(W.mm -1.K -130

)

33-155 30

Specific Heat (J.kg -

1.K -1 850) 715 840

Application

Cutting

wheels,

polishing

clothes

High temperature

furnace, Heat

shield

Balls and roller of

bearing, Cutting

tools, Engine valves,

Turbine blades

Windows, food

Preparation


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Polymer

Polymer is a chain of molecules connected by covalent (sharing of electrons) chemical bond.

Three types of polymers are most common: (a) thermoplastics which can be reworked on

heating, (b) thermosets which cannot be worked with after curing is over, and (c) elastomers,

which typically provide very high elastic deformation. The polymers cannot withstand high

temperature due to their low transition temperature Table 2.1.3 indicates the typical material

properties and common applications of some of the widely used polymers.

Table 2.1.3 Material properties and applications of commonly used Polymers

Material

Properties

Polyvineyl

chloride (PVC)

Bakelite Silicone

Type thermoplastic elastomer Elastomer

Density (kg.m -3 1350) 1300968-1290

High density silicone-2800

Melting Temperature (K) 373-530 588

Youngs Modulus(MPa) 1-5

Yield Strength (MPa)10-60

(Flexible-rigid)

Ultimate Tensile Strength

(MPa)2.6 21-47 11

Coefficient of Thermal

Expansion X 10 -6 (K -152

)8.1

Thermal Conductivity

(W.mm -1.K -10.14-0.28

)0.23 0.22

Specific Heat (J.kg -1.K -1 900) 1465

Application Plumbing ElectricalInsulators

Electrical appliances,

Structural application

(below 200C)


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Natural Materials

The most common examples of natural materials are wood, cotton, flax, wools, bamboo, jute

which primarily come from the plants or animals. Most of the natural materials are recyclable

and require considerable processing operations before use. Table 2.1.4 indicates the typical

material properties and common applications of some of the widely used natural materials.

Table 2.1.4 Material properties and applications of commonly used natural materials

Material

Properties

Oak Wood Wool Flax

Type

European

Oak

Density (kg.m -3 650) 22

Ignition Temperature (K) 523 873

Heat of Combution

(Kcal/g)4.9

Youngs Modulus (GPa) 9-13Longitudinal: 3.5

Transverse: 0.93

Shear Modulus(GPa)

Bulk Modulus(GPa)

Poissons Ratio

Yield Strength (MPa)


(MPa)50-180 163

Coefficient of Thermal

Expansion (K -134-54

)

Thermal Conductivity

(W.mm -1.K -1 0.3-0.35) 0.028

Specific Heat (J.kg -1.K -1 0.17)

Application Furniture,

Packaging

Fabric, Thermal

insulator

Fabrication

of twine


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Composite

Composite material is formed by combining one or more different materials. Unlike alloy

system each constituent is distinguishable and retain their properties. Composite materials

consist of matrix material with reinforcement to enhance its strength. Few common examples of

composites are FRP (Glass/carbon fiber reinforced polymers), Metal matrix composites. Using

composites one can combine attractive qualities of other materials and engineer properties to

demand. On the other side they are expensive and difficult to fabricate and join. Table 2.1.5

indicates common properties and applications of composites.

Table 2.1.5 Material properties and applications of commonly used composites

Material

Properties

Carbon fiber

reinforced polymer

matrix

Aluminummatrix

Titaniummatrix

Aluminamatrix

Cermet

Density (kg.m -3 1800) 2650 3860

Youngs Modulus(GPa) 210 300 85 100

Poissons Ratio 0.295

Yield Strength (MPa) 350 500


(N.mm -27000

)1500 1750 385 500

Application

Mechanical

components,

Protection

screen,

Sporting

equipments

Aerospace,

Sporting

equipments,

Electronic

packaging

Aerospace

Turbines

High

temperature

Mechanical

Application

Cutting

tools,

Polishing

materials

Foams

Foam is a substance formed by trapping many gaseous bubbles in liquid or solid. Solid foams are

very important class of structure due to its light weight. The foams can be metallic (eg. Titanium

foam), ceramic (alumina foam) or based on polymer (polyurethane foam). The metallic foams

are commonly used for medical implants. The ceramic foams are used typically as insulators

while the polymer based foams are primarily used for packaging and acoustic insulators.


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Exercise

Choose the correct answer.

1. Hydrostatic stress results in

(a) linear strain (b) shear strain (c) both linear and shear strain (d) None2. Toughness of a material is equal to the area under ____ part of the stress-strain curve.

(a) Elastic (b) Plastic (c) Both elastic and plastic (d) None

3. During a tensile loading, the length of a steel rod is changed by 2 mm. If the original length

of the rod has been 20 mm, what is the amount of strain induced

(a) 0.1 (b) 2 (c) 0.9 (d) 0.22

4. ____ is an example of a chemical property.

(a) Density (b) Mass (c) Acidity (d) Diffusivity

Answers:

1. (d) 2. (c) 3. (a) 4. (c)

References1. M F Ashby, Material Selection in Mechanical Design, Butterworth-Heinemann, 1999.

2. G E Dieter, Mechanical Metallurgy, McGraw-Hill, 1961.

3. http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-

ROORKEE/strength%20of%20materials/homepage.html, (28.05.2012).

4. http://www.grantadesign.com/education/datasheets/sciencenote.html, (28.05.2012).

5. http://www.azom.com/article.aspx?ArticleID=2765, (28.05.2012).

6. http://www.azom.com/article.aspx?ArticleID=2763, (28.05.2012).

7. http://www.matweb.com, (28.05.2012).
http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-ROORKEE/strength%20of%20materials/homepage.htmlhttp://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-ROORKEE/strength%20of%20materials/homepage.htmlhttp://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-ROORKEE/strength%20of%20materials/homepage.htmlhttp://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-ROORKEE/strength%20of%20materials/homepage.htmlhttp://www.grantadesign.com/education/datasheets/sciencenote.htmlhttp://www.azom.com/article.aspx?ArticleID=2765http://www.azom.com/article.aspx?ArticleID=2763http://www.matweb.com/http://www.matweb.com/http://www.azom.com/article.aspx?ArticleID=2763http://www.azom.com/article.aspx?ArticleID=2765http://www.grantadesign.com/education/datasheets/sciencenote.htmlhttp://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-ROORKEE/strength%20of%20materials/homepage.htmlhttp://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-ROORKEE/strength%20of%20materials/homepage.html


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Module2

Selection of Materials andShapes


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Lecture

2Selection of Materials - I


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Instructional objectivesBy the end of this lecture, the student will learn

(a) what is a material index and how does it help in selection of material for a given

application, and(b) how to develop material indices considering the appropriate material properties for an

intended service.

Selection of MaterialsAppropriate selection of material is significant for the safe and reliable functioning of a part or

component. Engineering materials can be broadly classified as metals such as iron, copper,

aluminum, and their alloys etc., and non-metals such as ceramics (e.g. alumina and silica

carbide), polymers (e.g. polyvinyle chloride or PVC), natural materials (e.g. wood, cotton, flax,

etc.), composites (e.g. carbon fibre reinforced polymer or CFRP, glass fibre reinforced polymer

or GFRP, etc.) and foams. Each of these materials is characterized by a unique set of physical,

mechanical and chemical properties, which can be treated as attributes of a specific material. The

selection of material is primarily dictated by the specific set of attributes that are required for an

intended service. In particular, the selection of a specific engineering material for a part or

component is guided by the function it should perform and the constraints imposed by the properties the material.

The problem of selection of an engineering material for a component usually begins with setting

up the target Function, Objective, Constraints, and Free Variables. The Function refers to the

task that the component is primarily expected to perform in service for example, support load,

sustain pressure, transmit heat, etc. The Objective refers to the target such as making the

component functionally superior but cheap and light. In other words, the Objective refers to

what needs to be minimized or maximized. The Constraints in the process of material selection

are primarily geometrical or functional in nature. For example, the length or cross-sectional area

of a component may be fixed. Similarly, the service conditions may demand a specific

component to operate at or beyond a critical temperature that will prohibit use of materials with

low melting temperature. The Free Variables refer to the available candidate materials.


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Material Index (M)The Material Index (M) refers to an attribute (or a combination of attributes) that

characterizes the performance of a material for a given application. The material index allows

ranking of a set of engineering materials in order of performance for a given application.Development of a M aterial Index (M) for an intended service includes the following steps.

Initial Screening of Engineering Materials. Identification of Functions, Constrains, Objectives and Free Variables. Development of a Performance Equation. Use constraints to eliminate the free variable(s) from the performance equation and

develop the material index .

Rank a suitable set of materials based on the material index .

Example 1: Selection of Material for a Light and Strong Tie-Rod [Fig.2.2.1]

Figure 2.2.1 Schematic presentation of a Tie-Rod with an axial tensile load, F

Function: Tie-rod to withstand an axial tensile load of F

Objective: Minimise mass (m) where = ALm , where is the material density.

Constraints: (i) Length L is specified, (ii) Must not yield under axial tensile load, F

Free variable: (i) Cross-sectional area, A, (ii) Material

Performance Equation: yAF

, where y is the yield strength of any material,

The Performance Equation can be rewritten by substituting the cross-sectional area, A, as

yy

)L)(F(m FL

m (1)

So to minimize mass, we have to minimize the term, y . Or other way, we can maximize the

term y for the sake of our convenience (as the available material property charts are y vs.


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format) . So the material index, M 1 y, in this case becomes and a material with higher

value of M 1 is expected to perform better in comparison to a material with lower value of M 1 . It

should be noted that the Material Index in this case provides a ratio between the ultimate tensile

strength and the density of the material. Thus, the Material Index (M 1

) would provide a premiseto examine if a material with higher weight (density) has to be selected to ensure that the same

has sufficient strength to avoid failure.

Example 2: Selection of Material for a Light and Stiff Beam [Fig. 2.2.2]

Figure 2.2.2 Schematic presentation of a beam with a bending load, F

Function: Beam to withstand a bending load of F

Objective: Minimise mass (m) where = L bm 2 , where is the material density.

Constraints: (i) Length L is specified, (ii) Must not bend under bending load, F

Free variable: (i) Edge length, b, (ii) Material

Performance Equation:

The Performance Equation can be developed considering the fact that the beam must be stiff

enough to allow a maximum critical deflection, , under the bending load, F. Thus, thePerformance Equation can be given as

31 LEI

)C(F

(2)

where is the maximum permissible deflection, E is the youngs modulus, I is the second

moment of area. The stiffness, S, of the beam, can be written as, = FS and the second moment

of area, I, can be written as, 12 bI 4= .

The Performance Equation can now be rewritten by substituting one of the free

variables (edge length, b) as


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5.03

5.0

1 E)L(

LCS12

m (3)

The material index, M 2 ( )5.0E, in this case becomes and a material with higher value of M 2 is

expected to perform better in comparison to a material with lower value of M 2 . In other words,the Material Index (M 2

) will depict if a material with higher weight (density) has to be selected

to ensure that the same has sufficient stiffness (i.e. E) to avoid bending during service.

Example 3: Selection of Material for a Light and Strong Beam [Fig. 2.2.3]

Figure 2.2.3 Schematic presentation of a beam with a bending load, F

Function: Beam to withstand a bending load of F

Objective: Minimise mass (m) where = L bm 2 , where is the material density.

Constraints: (i) Length L is specified, (ii) Must not fail under bending load, F

Free variable: (i) Edge length, b, (ii) MaterialPerformance Equation:

The Performance Equation can be developed considering the fact that the beam must be strong

enough so that it does not fail due to an applied bending moment, M, due to the load, F. Thus,

the Performance Equation can be given as

L2/ bI

)C(LM y

2 (4)

where y12 bI 4=

is the yield strength of the material and I is the second moment of area. The secondmoment of area, I, can be written as, .

The Performance Equation can now be rewritten by substituting one of the free

variables (edge length, b) as


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)L(LC

F6mor )L(

LCM6

m3/2

y

332

22

3/2y

332

2

(5)

The material index, M 3 ( ) 3/2y, in this case becomes and a material with higher value of M 3 isexpected to perform better in comparison to a material with lower value of M 3 . In other words,

the Material Index (M 3 ) allows the examination if a material with higher weight (density) has to

be selected to ensure that the same has sufficient strength (i.e. f

) to avoid failure during service.

Example 4: Selection of Material for a Light and Stiff Panel [Fig. 2.2.4]

Figure 2.2.4 Schematic presentation of a panel with a bending load, F

Function: Panel to withstand a bending load of F

Objective: Minimise mass (m) where = Ltwm , where is the material density.

Constraints: (i) Length L is specified, (ii) Must not bend under bending load, F

Free variable: (i) Panel Thickness, t, (ii) Material


The Performance Equation can be developed considering the fact that the stiffness of the panel

is sufficient to allow a maximum critical deflection, , under the bending load, F. Thus, thePerformance Equation can be given as

33

L

EI)C(

F(6)

where is the maximum permissible deflection, E is the youngs modulus, I is the second

moment of area. The stiffness, S, of the beam, can be written as, = FS and the second moment

of area, I, can be written as, 12wtI 3= . The Performance Equation can now be rewritten by

substituting one of the free variables (panel thickness, t) as


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3/12

3/1

3

2

E)L(

LCSw12

m (7)

The material index, M 4 ( )3/1E, in this case becomes and a material with higher value of M 4 is

expected to perform better in comparison to a material with lower value of M 4The above four examples depict the simple procedure to develop Material Indices for the

selection of suitable material for various structural requirements. These Material Indices can be

used subsequently to shortlist a range of suitable materials from appropriate Material Property

Charts in a graphical manner. The Material Property Charts display the combination of material

properties like Youngs modulus and density, strength and density, Youngs modulus and

strength, thermal conductivity and electrical resistivity, strength and cost, and so on. Figure 2.2.5

shows a typical Material Property Chart that displays Youngs modulus (in GPa) vis--vis

density (in Mg/m

.

3

) for a range of engineering materials in a log-log scale.

Figure 2.2.5 Material Property Chart of Youngs Modulus vis--vis Density [2]


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Exercise

Choose the correct answer.

1. The Material Index that can be used to select a suitable material for a light, stiff panel is

(a) ( )3/1E (b) ( )E3/1 (c) ( )E (d) ( )3E2. The Material Index that can be used to select a suitable material for a light, stiff tie-rod is

(a) ( )E (b) ( )E2 (c) ( )E (d) ( )f 3. The Material Index that can be used to select a suitable material for a light, stiff beam is

(a) ( )2/1E (b) ( )E2/1 (c) ( )E (d) ( )2E4. The Material Index that can be used to select a suitable material for a light, strong beam is

(a) ( ) 3/2f (b) ( )f 3/2 (c) ( )f (d) ( )3E5. The Material Index that can be used to select a suitable material for a light, cheap and strong

beam is

(a) ( ) m3/2f C (b) ( )f m3/2 C (c) ( ) mf C (d) ( )f 3/23/2mC

Answers:

1. (a) 2. (a) 3. (a) 4. (a) 5. (a)

References

1. G Dieter, Engineering Design - a materials and processing approach, McGraw Hill, NY,

2000.

2. M F Ashby, Material Selection in Mechanical Design, Butterworth-Heinemann, 1999.


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Module2



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Lecture

3Selection of Materials - II


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Instructional objectivesThis is a continuation of the previous lecture. By the end of this lecture, the student will further

learn

(a) how to develop material index for structural application.(b) how to use the typical material indices for the selection of material for common

engineering parts.

Example 5: Selection of Material for a Cheap and Stiff Column

Figure 2.3.1 Schematic presentation of a cylindrical column with a compressive load, F

Figure 1 shows the schematic picture of a typical cylindrical column with uniform cross-section

subjected to a compressive load, F. The task is to develop a suitable material index that can help

in the selection of the material for the column that is cheap and sufficiently stiff to avoid a

buckling failure. The above problem can be translated into functional requirement , objective ,

constraints to be considered , and the free variables that the designers are allowed to change.

Function: Column to withstand a compressive load of F

Objective: Minimise cost (C) where mC)AL(C = , where is the material density.

L is the length of the column, A is the uniform circular cross-section of the

column, and mC is the cost per unit mass of the material.

Constraints: (i) Length L is specified, (ii) Must not buckle under compressive load, F

Free variable: (i) Cross-sectional area, A, (ii) Material


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The Performance Equation can be developed considering the fact that the compressive load, F,

must not exceed the critical buckling load (F CR

) for the given column considering its important

dimension and material property. Usually, the length of typical columns is always a constraint.

Thus, the Performance Equation can be given as

2

2

CR CR L

EInF where FF

= (1)

where r is the radius of the cylindrical column, E is the youngs modulus, I is the second moment

of area and can be given as

=

=4A

4r

I22

, and n depends on the typical end constraints of the

column.

The Performance Equation can now be rewritten by substituting the expression of I in

equation (1) and eliminating the term A thereafter from equation (1) as

2/1

m3/12/1

2

2/1

E

C)L(

L

Fn4

C (2)

The material index, M, in this case becomesm

2/1

C

Eand a material with higher value of M would

be a better candidate both in terms stiffness and cost in comparison to a material with lower

value of M. Table 2.3.1 illustrates the evaluated values of the above material index for a range of

common engineering materials.

Table 2.3.1 Estimated values of ( m2/1 CE ) for common engineering materials

Materials E (MPa) C m (kg/m(Rs./kg) (app.) 3 m2/1 CE)

Mild Steel 205000 60.00 7870 9.59 x 10 -

Aluminium 68000 250.00 2698 3.87 x 10 -

Concrete 20000 10 2010 70.35 x 10 -

Stainless Steel 193000 350.00 7860 1.59 x 10 -

The above example in addition to the four examples explained in the previous lecture show the

general approach to develop Material Indices for the selection of suitable material for various


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structural requirements. Once a typical material index is developed for an intended application,

the same can be used next to choose a range of suitable materials either from a Material Property

Chart in a graphical manner or analytically by computing and ranking the values of the material

index for a range of engineering materials. The actual worked out examples on how to evaluate

the material indices for common engineering materials are presented in subsequent chapter.

General Approach to develop a Material Index

The general approach to develop a Material Index for an intended application lies in the

appropriate realization of the functional requirements , constraints , objective , and the free

variables (unconstrained). It is easy to understand these aspects through a set of queries. For

example, the function requirement can be realized by asking a question: what does the

component intend to do ? The constraints can be realized by asking the query: what specific

constraints must be met e.g. stiffness and / or strength and / or dimensions ? The constraints can

often be specified as hard constrains that are non-negotiable. For example, a component must

carry a certain load without elastic deflection or plastic deformation or failure. The constraints

can also soft that are primarily relation to the aesthetic aspects or cost and hence, negotiable The

objective primarily stands for what is to be minimized or maximized ? The free variables refer to

the features that the designers are free to change (e.g. dimensions, materials, etc. ).

Once the functional requirements , constraints , objective , and the free variables are identified for

a typical application, all the constraints related to the task should be listed and if possible, the

constraints can be presented in the form of a set of a single or multiple expressions. Next, the

objective of the design must be expressed in terms of functional requirements, geometry and

materials properties. This expression is referred to as the performance equation as shown in the

examples 1 to 5. If the performance equation contains a free variable , we have to identify the

constraint that limits the free variable . Next, we use this constrain to eliminate the free variable

in the performance equation . Lastly, we should be able to select the combination of the material

properties , referred to as the material index , which would maximize the performance. A set of

examples are given below to show how a suitable material index can be developed for the

selection of material for different applications.


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Example 6: Selection of Material for legs of a typical Reading Table .

Figure 2.3.2 Schematic picture of a Reading Table with four supporting legs

Figure 2 shows the schematic view of a typical table with four supporting legs. For simplicity,

we can assume that the supporting lags confirm to a uniform circular cross-section and each leg

must support a compressive load, F, without buckling distortion. Thus, the problem can be

envisaged as to develop a suitable material index to aid to the selection of material for a slender

and light supporting leg that will be able to support an applied design load without buckling

distortion and will not break if struck accidentally. Thus, the nature of the problem is similar to

the Example-5 , as outlined above. The above problem can be translated into functional

requirement , objective , constraints to be considered , and the free variables that the designers are

allowed to change as follows.Function: Column to withstand a compressive load of F

Objective: Minimise the mass (m) and Maximize the slenderness

Constraints: (i) Length L is specified,

(ii) Must not buckle under a compressive load, F, which is envisaged as

the design load.

(iii) Must not fracture if struck accidentally.

Free variable: (i) Cross-sectional area, A, or Diameter of legs, (2r) (ii) Material


Considering that the supporting leg to be a slender column of any material with density andlength L, the mass, to be minimized, can be given as

= Lr m 2 (3)


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The maximum buckling load that each leg can support without a buckling distortion can be given

as

4r

I whereL4

Er

L

EIF

2

2

43

2

2

CR === (4)

Substituting the free variable, r, from equation (4) to equation (3), we can write

1

221

2/122

1

M1

)L(F4

m E

)L(F4

m (5)

where M 1 represents the material index . It is easy to understand from equation (5), that the mass,

m, of a supporting table leg can be minimized as the material index, M 1

A comparison of equations (5) and (3) further shows that thinnest possible leg that will not buckle under a designed compressive load, F, can be given as

, will be maximized for a

given set of candidate materials.

4/1

2

221

4/12/14

1

3 M1

)L(F4

r E1

)L(F4

r

(6)

It is clear from equation (6), that the diameter (2r) of the supporting table legs with uniform

cross-section would be minimized, which will in turn enhance the slenderness, with the increase

in the value of the material index, M 2 . To meet the constraint that the table legs should not

fracture if struck accidentally, a third material index, M 3 , may be considered corresponding tothe fracture toughness (K 1C

) of the material.

Thus, the problem of selecting a suitable material for the table leg can be envisaged as an

optimization problem where all the three material indices E

M2/11

= , ( ) EM 2 = and

( ) K M C13 = are required to be maximized within a set of candidate materials. Based on initial

screening, an active set of candidate materials may be considered as wood, steel, aluminum,titanium, composite (e.g. CFRP), etc. The values of all the three material indices for the initially

screened materials can be evaluated either from material handbook or material property charts

and accordingly, a ranking of these materials based on the corresponding values of the material

indices can be prepared. At this stage, the manufacturability and the cost evaluation of each


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material should be undertaken. The final selection of material at the end of such an exercise

would be a trade-off among greater values of the corresponding material indices,

manufacturability aspects and cost evaluation.

Exercise

1. Develop a suitable material index for the selection of material for Oars used for rowing.

2. Develop a suitable material index for the selection of material for Spatula used for cooking.

References


2000.2. M F Ashby, Material Selection in Mechanical Design, Butterworth-Heinemann, 1999.


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Module2



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Lecture

4Case Studies - I


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Instructional objectivesThis is a continuation of the previous lecture. By the end of this lecture, the student will further

learn how to develop and use typical material indices for the selection of material for common

engineering parts. Two examples are illustrated here with regard to the development of materialindices for corresponding applications.

Example 7: Selection of Material for an efficient flywheel

Figure 2.4.1 Schematic presentation of a typical flywheel

Figure 2.4.1 shows the schematic picture of a typical flywheel that is used to store rotationalenergy in applications such as automotive transmissions. An efficient flywheel should be able to

store the maximum energy per unit volume or per unit mass at a specified angular velocity. The

task is to search for a suitable material index for the selection of material for an efficient

flywheel that should have adequate toughness and can store the maximum kinetic energy per unit

volume or mass. The above problem can be translated into functional requirement , objective ,

constraints to be considered , and the free variables that the designers are allowed to change as

follows.

Function: Flywheel for energy storage

Objective: Maximize kinetic energy per unit mass.

Constraints: (i) Outer radius, R, may be fixed, (ii) Must not burst, and (iii) Should have

adequate toughness to avoid catastrophic failure.

Free variable: Choice of material


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Performance Equation

The Performance Equation can be developed as follows. The energy (U) stored in a flywheel

can be estimated as

2tR

J where J21

U4

2

== (1)

where is the density of the material, is the angular speed of the flywheel, R is the radius and tis the thickness of the flywheel disc. The mass (m) of the flywheel disc can be given as

t Rm 2 = (2)Hence, the energy per unit mass can be given as,

22R 4

1

m

U= (3)

The maximum principal stress ( max

22max R

2

1=

) on the flywheel disc as a function of the rotational velocity

can be expressed as,

(4)

Since the maximum principal stress should not exceed the failure strength ( f ) of the material,we can develop the material index, M 1

1f M

21

21

mU

=

, to maximize the energy stored per unit mass by

rearranging equations (3) and (4) as,

(5)

It is clear from equation (5) that greater values of M 1 will tend to maximize the energy stored per

unit mass for a given angular speed, radius and thickness of the flywheel disc. A second material

index, M 2 , can be considered in terms of fracture toughness (K 1C

Figure 2.4.2 depicts a typical chart of material properties (strength, ) of the material.

f , vis--vis density, ) in alog-log scale. The advantage of log-log scale over decimal scale is that the constant material

index lines will appear as a straight line which makes the selection and representation easier. The

black line represents the constant material index line, for M 1,.

+= logMloglog f

Since the chart is plotted in log-log

scale, the black line confirms to a straight line [denoted as: ]. Thus, any

engineering material falling around the line will confirm to a similar value of material index and

by sliding the line, it is possible to select a set of suitable candidate materials considering M 1 .

Figure 2.4.2 depicts that aluminum alloys, titanium alloys, engineering composites and


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engineering ceramic would provide excellent values of M 1 . However, the engineering ceramics

would provide very low fracture toughness [M 2

] and hence, may be eliminated. Further selection

must be made based on the cost and the energy storage capacity of specific materials.

Figure 2.4.2 Schematic material chart of strength vis--vis density of engineering materisls

Example 8: Selection of Material for pressure vessel

Figure 2.4.3 shows the schematic picture of the cross-section of a spherical pressure vessel and

typical circumferential stresses experienced by the vessel wall with a presumed crack. The safe

design of a small sized pressure vessel would require that the material yields before a final

fracture. Similarly, the safe design of large pressure vessels typically calls for a criterion that any

small crack opens as a leak prior to a catastrophic failure. The above problem can be translated

into functional requirement , objective , constraints to be considered , and the free variables that

the designers are allowed to change as follows.

Function: Pressure vessel to contain an internal pressure of p safely.

Objective: Maximize kinetic energy per unit mass.

Constraints: (i) maximize safety using yield-before-break criterion (small vessel), or


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(ii) maximize safety using leak-before-break criterion (large vessel)

Free variable: (i) Choice of material

Figure 2.4.3 Schematic presentation of the cross-section of a spherical pressure vessel


The Performance Equation can be developed as follows. Stress ( ) in the wall of a thin-walledspherical pressure vessel of radius R, wall thickness t and with internal pressure, p, can be given

as

t2

pR = (6)

The minimum stress required for a presumed circumferential crack of diameter 2a c

c

C1

a

CK

=

to propagate

can be given as

(7)

where C is a constant and K 1C the plane-strain fracture toughness of the pressure vessel material.

Hence, the largest pressure (for a given vessel radius, R, wall thickness, t, and initial

circumferential crack diameter, 2a c) would be carried by the material with the greatest value of

K 1C

C11 K M =and hence, we can write

(8)

However, the material index M 1

f =alone cannot ensure a fail-safe design, which further requires

, where f is the failure strength of the material, which translates to an appropriate

material index, M 1 , as


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f

C11

f

C12c

K M

K Ca

=

(9)

For typical large pressure vessels, we can estimate the minimum stress required for a crack to

penetrate the wall thickness, t, thereby making a leak can be estimated as

2/t

CK C1f

= (10)

Substituting for t using equations (1) and (10), we can further express the safe internal pressure

as

f

2C1

2 K R

C4 p (11)

Thus, the maximum pressure would be contained safely by the material with the largest value of

=f

2C1

2K

M (12)

Lastly, we must ensure that the material with the minimum thickness would offer the maximum

strength and hence, we must consider another material index, M 3

f 3M =, as

(13)

Figure 2.4.4 depicts a typical chart of material properties (strength, f , vis--vis fracture

toughness, 1C) in a log-log scale with three black lines confirming to the material indices, M 1,M 2 and M 3 . The safe region would be the one which is above all the three lines as indicated in

the figure. It can be noticed that one of the most suitable material appears to be stainless steel

[M 1 ~ 0.35 m 1/2, M 3 ~ 300 MPa], which is actually used for all critical pressure vessels. For

example, some special grade of stainless steel is widely used for nuclear pressure vessels. A

second candidate is low-alloy steel [M 1 ~ 0.20 m 1/2, M 3 ~ 800 MPa], which is a standard material

used for manufacturing pressure vessels. A third candidate is copper [M 1 ~ 0.50 m 1/2 , M 3 ~ 200

MPa], and hard drawn copper is often used to manufacture small boilers and pressure vessels.The pressure tanks of rockets and aluminum are often made of aluminum alloys which confirm

to M 1 ~ 0.15 m 1/2 and M 3 ~ 200 MPa. Titanium alloys are another choice with M 1 ~ 0.13 m 1/2

and M 3

~ 800 MPa, which are often used for light pressure vessels while they are relatively

expensive.


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Figure 2.4.4 Schematic chart of strength vis--vis fracture toughness of engineering materials

Exercise

1. Develop a suitable material index for the selection of material for Oars used for rowing.2. Develop a suitable material index for the selection of material for Spatula used for cooking.

References


2000.

2. M F Ashby, Material Selection in Mechanical Design, Butterworth-Heinemann, 1999.


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Module2



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Lecture

5Selection of Shapes


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Instructional objectivesBy the end of this lecture, the student will learn

(a) what is shape factor and how it can be used to enhance the mechanical efficiency of a

material, and(b) how to develop shape factors considering appropriate load and different cross section.

Selection of ShapesSo far we have learned how the combination of material properties can be used to develop a

material index for the selection of a suitable material for a given application under different

loading conditions. Similarly, the cross-sectional shape of a part can be used to enhance the load

bearing capacity. An engineering material confirms to a modulus and strength, but it can be madestiffer and stronger when loaded under bending or twisting by shaping it into an I-beam or a

hollow tube, respectively. It can be made less stiff by flattening it into a leaf or winding it, in the

form of a wire, or into a helix. Shaped sections (i.e. cross-section formed to a tube, a box-

section, an I-section or the like) carry bending, torsional, and axial-compressive loads more

efficiently (i.e. for a given loading conditions, the section uses as little material as possible)

than solid sections. The efficiency can be enhanced by introducing sandwich panels of the same

or different materials. But when choosing shapes one has to be careful so the basic functional

requirement is not violated.

Shape Factor ( ) Shape Factor is a dimensionless number that characterizes the efficiency of the shape, regardless

of its scale, for a given mode of loading, e.g. bending, torsion, twisting, etc. The four primary

shape factors of our consideration are,

eB Macro shape factor for elastic bending eT Macro shape factor for elastic torsion f B Macro shape factor for onset of failure in bending f T Macro shape factor for onset of plasticity or failure in torsion


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All the shape factors are defined equal to 1 for a solid cylinder i.e. our reference cross sectional

shape is circular. Shape Factor of all other cross section will be evaluated w.r.t this one.

Example 9: Selection of shape factor in elastic bending of beam

The bending stiffness S of a beam is proportional to the product EI and can be given as

EIS (1)where E is Youngs modulus and I is the second moment of area of the beam about the axis of

bending (the x axis), which can be written as

= dAyI 2 (2)where y is measured normal to the bending axis and dA is the differential element of area at y.

The values of the moment I and of the area A for the common sections are listed in the first twocolumns of Table 2.5.1. The second moment of area, I 0

==

4A

4r

I24

0

, for a reference beam of circular section

with radius r is simply

(3)

The bending stiffness of any shaped section differs from that of a circular one with the same area

A by the factor eB where

200eB

AI4

EIEI

SS === (4)

Note that the factor eB is dimensionless and depends only on the shape. For example, the big

and small beams have the same value of eB if their section shapes are the same. Figure 2.5.1

shows three different shapes with their corresponding values of shape factor to be considered for

elastic bending of beams. It can be noted that the values of the shape factor do not change with

the size of the shape.

Figure 2.5.1 Schematic pictures of a set of (a) rectangular sections with eB = 2; (b) I-sections

with eB = 10; and (c) tubes section with eB = 12.


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Table 2.5.1 Area (A), Second moment of Area (I), Torsional moment of Area (K), Section

modulus in bending (Z), and in torsion (Q) for common engineering shapes

Section

Shape

A

(m 2)

I

(m 4)

K

(m 4)

Z

(m3)

Q

(m3)

2r 4

4r

4

2r

3

4r

3

2r

2b 12

4b 414.0 b

6

3b 321.0 b

ab ba 34

)( 2233

ba

ba

+

ba 24

)(2

2

ba

ba

6

2bh

)(

8.13

22

bh

bhhb

>+

2

43

a 332

4a

8034a

32

3a

20

3a

rt r r io

2)(

22

t r

r r io3

44

)(4

t r

r r io

3

44

2

)(2

t r

r r r ioo2

44

)(4

t r

r r r ioo2

44

2

)(2

bt 4 t b332

4

3 1

b

t t b t b 2

34

2

2 12

bt

t b

t ba )( +

+

a

bt a

31

43

22

2/5)(4ba

t ab

+

+

abt a 3

14

2

)(

)(2 2/13

ab

bat >

bt

hhbio

2

)(

2/

)(12

2

33

o

io

bth

hhb

------

( )o

io

o

bth

hhh

b

336

(2 bht +

)4(6

23 bt ht

+

+

b

hbt

41

32 3

+

h

bt h 31

3

2

+

bh

bt 4

132 2


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the beam. The shape factor in this case is considered through the section modulus, Z , and is

measured by the ratio, 0ZZ , where Z 0

2/330

f B

A

Z4

4r

ZZZ =

==

is the section modulus of a reference beam of circular

section with the same cross-sectional area, A. Hence, the shape factor to be considered against

failure in bending can be given as

(9)

Similarly, in case of a circular rod subjected to a torque T , the maximum shear stress max occursat the maximum radial distance r max

QT

JTr max

max ==

from the axis of twisting and can be given as

(10)

Figure 2.5.2 Schematic picture of (a) bending of beam, and (b) cross-section of beam

where M is the bending moment and Z is the section modulus. Failure in bending can occur when

b

0ZZ

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