Materials and Process Selection for Engineering Design, Second Edition [2 ed.] 9781420063080, 9781420069174, 1420063081

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SOLUTIONS MANUAL FOR Materials & Process Selection Engineering Design, Second Edition

by Mahmoud M. Farag

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SOLUTIONS MANUAL FOR Materials & Process Selection for Engineering Design, Second Edition

by Mahmoud M. Farag

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1/11/08 9:53:03 AM

CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2008 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed in the United States of America on acid-free paper 10 9 8 7 6 5 4 3 2 1 International Standard Book Number-13: 978-1-4200-6917-4 (Softcover) This book contains information obtained from authentic and highly regarded sources Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The Authors and Publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC) 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

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Materials and process selection for engineering design 

Contents Chapter 1

Product design and development in the industrial enterprise

1

Chapter 2

Failure under mechanical loading

4

Chapter 3

Environmental degradation

6

Chapter 4

Selection of materials to resist failure

8

Chapter 5

The nature of engineering design

12

Chapter 6

Effect of material properties on design

15

Chapter 7

Effect of manufacturing processes on design

21

Chapter 8

Economic and environmental impact of materials and processes

27

Chapter 9

The materials selection process

30

Chapter 10

Materials substitution

37

   

 

Materials and process selection for engineering design 

Chapter 1 Product design and development in the industrial enterprise Problem 1.1 Although concept A, which is driven by an internal combustion engine, meets the current EPA emission test limits, it is expected to be under increasing pressure as the test limit are set tighter in the future. Governments may charge lower taxes for lower emission cars and the price of petrol is expected to continue its increase. Such financial pressure on concept A will gradually make it a less attractive option. On the other hand, concept D, which is battery driven and is currently less attractive, will become more viable as lighter and more powerful batteries are developed. The main technical problems that need to be overcome in this regard are: 1. Longer time between battery charges (more power/charge) 2. Longer total battery life 3. Lighter total battery weight 4. Better network of battery service and repair centers across the country 5. Lower initial cost of the battery Problem 1.2 Students may perform this task either individually or in groups. Possible topics for projects include garden and seaside furniture; aids for recycling, example can crushers; household gadgets like jar and can openers; children toys; etc. Problem 1.3 Students may perform this task either individually or in groups. Class discussions are also a possibility. Problem 1.4 Students may perform this task either individually or in groups. Class discussions are also a possibility. Problems 1.5 and 1.6 These problems are suited for class discussion, after which students may do their own search and submit a report. The following photograph gives three examples of milk packaging and can be useful in the discussion. The discussion may include durability of the packaging

   

 

Materials and process selection for engineering design 

material, weight of package, possibility of recycling, shelf life of the milk in the package, cost of the package, etc.

Problem 1.7 Students may perform this task either individually or in groups. Class discussions are also a possibility. Problem 1.8 Students may perform this task either individually or in groups. Class discussions are also a possibility. The discussion can cover: (a) location of the house, whether in the city or in the countryside, (b) the main materials for construction, whether wood, bricks, stone, concrete, or a combination, (c) independent and dependent activities involved and their sequence. Problem 1.9 Sustainable development does not harm the environment or deplete irreplaceable resources. It should depend on renewable resources. Recycling of materials helps reduce the depletion of irreplaceable resources. Students may be asked to identify examples of projects within this definition. Problem 1.10

2   

 

Materials and process selection for engineering design 

Of the materials used for tea cups, glass has the highest recyclability as it can be remelted and reused. Styrofoam has the lowest recyclability. China and melamine have intermediate recyclability as they can be crushed and used as fillers to virgin materials. Table 8.7 gives the energy content of some materials. Problem 1.11 The comparison between porcelain and plastics in making soup dishes may include the following items Aspect of comparison Cost Toughness/resistance to breakage Aesthetic value and appearance Recyclability

Porcelain higher lower

Plastics lower higher

better

lower

intermediate

Weight Thermal resistance Main uses

heavier higher household and more formal setting

Higher if thermoplastic and intermediate if thermosetting lighter Lower picnics and less formal setting

Problem 1.12 Composite laminates for fruit juice packaging consist of several layers including: (a) innermost plastic film on the inside to avoid contamination and resist acidity of the contents, (b) aluminum foil layer to stop oxygen reaching the contents, (c) carton layer for strength and stiffness, (d) paper layer for printing of information, (e) outermost plastic film for water resistance. With such combination of materials, the package can preserve the juice for a longer period but it is difficult to recycle. Problem 1.13 Students are asked to perform this task either individually or in groups. Weights may be set by the instructor on the different requirements.

3   

 

Materials and process selection for engineering design 

Chapter 2 Failure under mechanical loading Problem 2.1 According to Eq. (2.3), fracture stress σf can be related to the fracture toughness, KIC, and the flaw size, 2a. Y is a correction factor which depends on the geometry of the part: σ

= K IC Y (π a )

1/ 2

f

= 45 / 1 × ( 3 . 14 × 1 . 25 × 10 − 3 ) 1 / 2 = 718 MPa

Problem 2.2 From Eq. (2.3), the fracture stress σf can be related to the fracture toughness, KIC, and the flaw size, 2a. Y is a correction factor which depends on the geometry of the part: σ

= K IC Y (π a )

1/ 2

f

= 47 / 2 × ( 3 .14 × 0 . 5 × 10 − 3 ) 1 / 2 = 593 MPa

σf is larger than YS, the material will yield first at 462 MPa, which is about 18% higher than the applied stress of 390 MPa. This is probably not high enough if a factor of safety of 1.5 or higher is needed. Problem 2.3 For the definition of alternating and fluctuating stresses, please refer to Fig 2.8. The rear axel of the motorcar is subjected to fluctuating stresses but the connecting rod is subjected to alternating stresses. Problem 2.4 Welded joints in steel structures are possible sites for fatigue crack initiation because: a) They could be associated with internal stresses b) They could have internal cavities or small cracks c) They could be associated with stress concentrations d) They are usually associated with metallurgical inhomogeneities. Placing welded joints away from highly stressed areas, stress relieving using thermal or mechanical treatments, careful quality control by NDT, and adhering to codes of practice are among the ways of avoiding fatigue failures of welded structures. Problem 2.5 Fatigue strength of FRP is normally acceptable (endurance ratio above 0.3 as shown in Table 4.7). The elements of the frame have to be assembled using adhesives, which could present a source of weakness under fatigue loading. To ensure that the joint design and strength are 4   

 

Materials and process selection for engineering design 

acceptable, it is suggested that a testing program consisting of two groups of tests be devised. The first group of tests is to select the most suited adhesive. The second group of tests consists of different joint designs to select the optimum. Problem 2.6

Problem 2.7 According to Eq. (2.3), fracture stress σf can be related to the fracture toughness, KIC, and the flaw size, 2a, with Y as a correction factor that depends on the geometry of the part. For Maraging steel: σ

= K IC Y (π a )

1/ 2

f

= 90 / 1 × ( 3 .14 × 1 .5 × 10 − 3 ) 1 / 2 = 1311 MPa

Yield strength is 1730 MPa. Both yield strength and σf are higher than the applied stress of 600 MPa. For AISI 4340(T 260oC): σ

= K IC Y (π a )

1/ 2

f

= 50 / 1 × ( 3 . 14 × 1 . 5 × 10 − 3 ) 1 / 2 = 728 . 9 MPa

Yield strength is 1640 MPa

5   

 

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The design should be based on the fracture stress, which is only about 20% higher than the applied stress. This may not be sufficient if a factor of safety of 1.5 or higher is needed.

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Chapter 3 Environmental degradation Problem 3.1 Galvanizing is covering the steel with a layer of Zn, while chrome plating covers the steel with a layer of Cr. According to Table 3.1, Zn is lower than iron in the galvanic series and the protection is cathodic, i.e. the steel is the cathode and Zn is the sacrificial anode. Protection continues even if the Zn layer is slightly damaged. Cr is higher than iron in the galvanic series and protection in this case is by isolation of the steel from the environment. The plated surface does not corrode as Cr has excellent corrosion resistance. If the Cr layer is damaged, the exposed steel surface corrodes quickly, as in the case of small anode with large cathode. Problem 3.2 Plastics are electrically insulating and, therefore, do not corrode by electrochemical action as in the case of metals. Plastics are attacked UV radiation, which has enough energy to break the molecular bonds and may cause cross linking, thus reducing the molecular weight and degrading the strength and ductility. Organic solvents also reduce the molecular weight. Adsorption of water molecules within the plastic molecular structure causes internal stresses and a reduction of the strength. Problem 3.3 Galvanizing is covering the steel with a layer of Zn, while tinning is covering the steel with a layer of tin. According to Table 3.1, Zn is lower than iron in the galvanic series and the protection is cathodic, i.e. the steel is the cathode and Zn is the sacrificial anode. Protection continues even if the Zn layer is slightly damaged. As tin is higher than iron in the galvanic series (Table 3.1), protection is normally by isolation. However, in a sealed food can, the potential reverses and tin acts as a sacrificial anode, thus protecting steel. Coating by polymeric paints isolates the surface of the steel from the environment and, thus, prevents corrosion. Problem 3.4 Tin is higher than steel in the galvanic series but in a sealed food can, the potential reverses and tin acts as a sacrificial anode, thus protecting steel. Tin is also nontoxic and is widely used for protecting steel in food cans. Zinc is lower than steel in the galvanic series and the protection is cathodic, i.e. the steel is the cathode and Zn is the sacrificial anode. Protection continues even if the Zn layer is slightly damaged. Zn is cheaper than tin and is, therefore, preferred for outdoor fencing where the requirements are not as stringent as in food cans. 7   

 

Materials and process selection for engineering design 

Problem 3.5 Section 3.7 lists 5 methods of corrosion control. These include: 1. Using galvanic protection, as in case of galvanizing of steel for use outdoors. 2. Using corrosion inhibitors, as in the case of motorcar radiators, reactors in chemical industries and water boilers. 3. Selecting the right material, as in the case of using stainless steels for food processing and household applications. 4. Using protective coatings, such as polymer-base paints for motorcar bodies. 5. Observing design rules, such as avoiding electrical contact between dissimilar metals. Points 1 and 2 are discussed in more detail in Section 3.7. Point 3 is discussed in Section 4.8, point 4 in Section 4.9, and point 5 in Section 6.7. Problem 3.6 Taking the density of steel as 7.8 g/cc, The weight loss = 0.050 g = 10 x 10 x 7.8 x t, t = the thickness loss per day = 6.41 x 10-5 cm = 6.41 x 10-4 mm Useful life = (10 - 6)/t = 6.24 x 103 days = 17 years. Problem 3.7 Weight loss during the lifetime of the tank = (50/1000) x12 x 10 = 6 g/dm2 Thickness loss = (6/7.8x100) =0.0077 cm = 0.077 mm Working stress σ = 350 x 0.5 x 0.7 =122.5 MPa = (pr)/t = (3.5 x 0.5)/t t =14.29 mm = minimum safe thickness Starting thickness = 14.29 + 0.077 = 14.367 mm. Adding chromate salts as inhibitor can reduce the corrosion rate. Other solutions include galvanizing and galvanic protection, as discussed in Section 3.7

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Materials and process selection for engineering design 

Chapter 4 Selection of materials to resist failure Problem 4.1 a) The main performance requirements for the wing structure of a two passenger training aircraft are light weight, high strength and high stiffness. The corresponding material properties are low density, high tensile strength and high elastic modulus. These properties can be combined in high specific strength (σ/ρ) and high specific modulus (E/ρ) b) From the given material properties:

Aluminum alloy Magnesium alloy Epoxy+56% glass fibers

E/ρ

σ/ρ

25.9 26.5 21.7

214.8 164.7 521.8

Scaled E/ρ (1) 98 100 82

Scaled σ/ρ (2) 41 32 100

(1)+(2) Rank 139 132 182

2 3 1

Scaled properties are obtained by giving the highest value a score of 100 and giving lower values proportionately lower scores. Problem 4.2 AISI 1050 steel has bcc structure and is unsuitable for serving at -50oC because this temperature is below its ductile –brittle transition temperature. Possible substitutes include austenitic stainless steels and aluminum alloys, which have fcc structures and do not suffer ductile-brittle transition.

Problem 4.3 From Equation (2.3) σ f = K IC Y (πa )1 / 2 = 87.4/[1x (3.14 x0.5x10-3)]=2205 MPa

σ f is higher than the yield strength and the latter value will be used in design. Taking a factor of safety of 2, the area of the bar A = (150,000 x 2)/1480 = 202.7 mm2 Diameter of the bar = 16 mm

9   

 

Materials and process selection for engineering design 

Problem 4.4 Ti-6Al-4V 55 1389

Aluminum 7075T6 24 606

σy (MPa)

830 use for design as it lower than σ f

495 use for design as it lower than σf

Compared with applied stress of 400 MPa

Safe to use as σy is higher by a good margin.

Not very safe to use as σy is only slightly higher than external stress.

1/2

KIC (MPa m ) σ f (MPa)

Problem 4.5 Hardening of steel increases it hardness and tensile strength but lowers it elongation percent and toughness. If the crankshaft is hardened throughout the whole section it would not be tough enough to function safely. Surface hardening of the crankshaft makes the surface wear resistant but preserves the tough interior, thus combining the advantages of high hardness and high toughness. Problem 4.6 The material requirements for the blades of household scissors include high hardness, good corrosion resistance, and reasonable toughness. Possible materials include martensitic stainless steels and high carbon chromium-containing steels. Some tough ceramics, such as silicon nitride, can be considered if the design is changed to avoid subjecting them to tensile stresses. Problem 4.7 The material requirements for the radiator of a motorcar include good thermal conductivity, good corrosion resistance, reasonable strength at operating temperature, and high ductility during manufacture. Possible materials include copper alloys and aluminum alloys. Problem 4.8 Endurance limit of steel = 316 MPa. Endurance limit of aluminum alloy = 97.7 MPa. Working stress of steel = 316 x 0.68 x 0.8 x 0.75 = 128.9 MPa Working stress of aluminum = 97.7 x 0.64 x 0.77 x 0.7 = 33.7 MPa Area of steel member = (6000 x 10)/128.9 = 465.5 mm2, diameter of steel member = 24.4 mm, weight of steel member = 3.63 kg Area of aluminum member = (6000 x 10)/33.7 = 1785.7 mm2, diameter of aluminum member = 47.6 mm, weight of aluminum member = 4.82 kg, which is heavier than steel.

10   

 

Materials and process selection for engineering design 

(Cost of steel member)/(cost of aluminum member) = (3.63 x 1)/(4.82x3.2) = 0.235 Aluminum alloy gives a heavier and more expensive tensile member, which makes it unsuitable. Problem 4.9 Stainless steels are corrosion resistant because the oxide layer that forms on their surfaces is strong and free from defects. When such layer is formed the material is passive. Conditions that lead to destruction of this layer, such as local depletion of chromium or service in conditions that do not allow the oxide layer to form, cause activation and sensitization of the material. Problem 4.10 The main material requirements for the electric heating wire include: creep and oxidation resistance at the operating temperature. A reasonable electrical resistance is needed for the wire to conduct electricity but also to get hot. Ductility at room temperature will allow the wire to be drawn to the required diameter and then shaped. Problem 4.11 Galvanizing is covering the steel with a layer of zinc and tinning is covering the steel with a layer of tin. Tin is higher than steel in the galvanic series but in a sealed food can, the potential reverses and tin acts as a sacrificial anode, thus protecting steel. Tin is also nontoxic and is widely used for protecting steel in food cans. Zinc is lower than steel in the galvanic series and the protection is cathodic, i.e. the steel is the cathode and Zn is the sacrificial anode. Protection continues even if the Zn layer is slightly damaged. Zn is cheaper than tin and is, therefore, preferred for outdoor fencing where the requirements are not as stringent as in food cans. Problem 4.12 Organic coatings are usually more ductile and, therefore, less likely to chip and crack than vitreous enamels. However, the latter are harder and more temperature resistant. More detailed discussion is given in Section 4.9. Problem 4.13 Aluminum has high affinity to oxygen and very quickly forms a layer of Al2O3 on its freshly exposed surface. Al2O3 is strong and impervious and even a very thin layer of it isolates the aluminum surface from the environment and prevents further corrosion. Iron also forms an oxide layer on its freshly exposed surface, but the iron oxide that forms is mechanically weaker and more porous than Al2O3 and, therefore, does not provide the same protection. Problem 4.14

11   

 

Materials and process selection for engineering design 

The main material requirement for a kitchen-knife blade include: a) high hardness, b) good corrosion resistance, and c) reasonable toughness. Possible materials include martensitic stainless steels and high carbon chromium-containing steels. Ceramics can be suitable candidates in view of their high hardness and excellent corrosion resistance. Some silicon nitride ceramics can now be manufactured with sufficient toughness to make them suitable candidates for knife blades. Problem 4.15 The main material requirements for gas turbine blade includes: a) high specific tensile strength at operating temperature, b) high specific creep rupture strength at operating temperature, c) good oxidation resistance at operating temperature, d) good resistance to thermal and mechanical fatigue, and e) low cost. Superalloys (Fe-Ni-base, Ni-base, and Cobase) are most suited for service at 900oC. More details are given in Section 4.7. Problem 4.16 A stronger material in a given class of materials is usually achieved by adding alloying elements and/or heat treatment. Such increases in strength are usually accompanied by a decrease in ductility and toughness. Figure 4.4 illustrates this point for steels, aluminum alloys and titanium alloys. Materials exhibiting lower ductility and lower toughness usually fail catastrophically in a brittle manner. On the other hand, more ductile and tougher materials usually plastically deform before fracture, thus giving some opportunity of taking preventive action. In addition, a stronger material in a given class of materials is usually more expensive. These points are discussed in more detail in Section 4.3. Problem 4.17 Low carbon steels are a very versatile and inexpensive group of materials that are widely used in many applications ranging from steel structures and concrete reinforcing bars to motorcar body sheets and chassis elements. Their strength can be controlled by controlling the carbon content, heat treatment and cold work. They are also usually ductile and tough. Their main limitations include heavy weight and low corrosion resistance. They are also prone to the ductile-brittle transition at low temperatures. Problem 4.18 The main material requirements for motorcar exhaust manifold include: corrosion resistance, moderate temperature resistance, good formability and weldability. Ferritic and austenitic stainless steels give good service life. The main material requirements for coil for electrical resistance heater include: oxidation and creep resistance at operating temperature, moderate electrical resistivity, and good formability at room temperature. Fe-Ni-base alloys, such as Inconel and Incoloy are suitable, see appendix Table A.16.

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Materials and process selection for engineering design 

The main material requirements for railway line include: high wear and impact resistance, moderate corrosion resistance, good weldability and formability. Suitable materials include Mn steels.

Chapter 5 The nature of engineering design

Problem 5.1 For a thin-wall tube under internal pressure, the stress in the wall (σ) when the pressure is P, diameter is d and thickness is t : σ = Pd/2t = 8.4 [(75+73)/2]/2x1 = 310.8 MPa According to Appendix Table A.11, the yield strength of Al 2014T6 is 420 MPa. Factor of safety = 420/310.8 = 1.35 Problem 5.2 Factor of safety is a number, which is greater than unity. It is used to reduce (by division) a value of a material strength, as measured in the laboratory or taken from suppliers’ information, to a working value that can safely be used in design calculations. The main factors that affect the value of the factor of safety include: 1. Uncertainties associated with material properties due to variations in composition, heat treatment and processing conditions as well as environmental variables such as temperature, time, humidity, and ambient chemicals. 2. Parameters related to manufacturing processes also contribute to the uncertainties of component performance. These include variations in surface roughness, internal stresses, sharp corners, identifying marks, and other stress raisers. 3. Uncertainties in loading and service conditions. A derating factor is a number less than unity and is used to reduce a material strength value (by multiplication) to take into account manufacturing imperfections and the expected severity of service conditions. Problem 5.3 a) From Fig. 5.4, S − L = 200 - 150 = 50 MPa,

[

Standard deviation of the curve (S − L ) = (σ S ) + (σ L ) 2

From Eq. 5.8, z = - 50/30.4 = -1.64 13   

]

2 1/ 2

= [(30)2 + (5)2]1/2 = 30.4.

 

Materials and process selection for engineering design 

From Table 5.2 and by interpolation, probability of failure is 0.069 (6.9%)

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Materials and process selection for engineering design 

b) For probability of failure 1% and from Table 5.2, z = -2.33 = -( S − L )/30.4 S − L = 2.33 x 30.4 = 70.8 MPA 

L = 200 – 70.8 = 129.2 Factor of safety = 200/129.3 = 1.55 Problem 5.4 The first step is to decide on the main selling points, which could be environmental, cost of operation, and styling is this case. The next step is to make a preliminary and conceptual design, define the system, and identify the different parts of the solar heater. Developing detail designs and selecting materials of the different parts then follows. Attempting to use as many of the materials and processes used for the gas heater, would reduce to cost of conversion. Problem 5.5 From Fig. 5.4, S − L = 2400 - 1800 = 600 MPa,

[

Standard deviation of the curve (S − L ) = (σ S ) + (σ L ) 2

]

2 1/ 2

= [(380)2 + (330)2]1/2 = 503.3

From Eq. 5.8, z = - 600/503.3 = -1.19 From Table 5.2 and by interpolation, probability of failure is 0.119 (11.9%) Problem 5.6 Students may be asked to select a simple product that they are familiar with, to analyze how it works, identify possible means in which it would fail and the effect of failure on performance and safety, and to devise ways of avoiding such failure. Assigning a factor of safety can be based on uncertainties in loading conditions and material and manufacturing defects. Problem 5.7 The students may continue the assignment in problem 5.6 by emphasizing the changes they introduced to avoid failure of the product. Problem 5.8 This question is best answered in an open discussion in class. The main points of discussion could include: types of joints, joint design (clearance in the joint, which affects the thickness of the adhesive, amount of overlap, etc.), types and properties of adhesives (refer to Appendix Table A.22), and position of the joint which affects the stress system in the joint.

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Problem 5.9 Shop 1: average of hardness values = 49.5 Shop 2: average of hardness values = 49.2 The difference between the average values is less than the difference between individual readings in each shop, i.e. the difference is not significant. Problem 5.10 The many designs for carjack can be divided into mechanically operated (screw and nut) and hydraulically operated (piston in a cylinder). Both types of design lend themselves to power operation from the cigarette lighter in the car. Students may be asked to select a design and modify it for power operation from the cigarette lighter in the car.

16   

 

Materials and process selection for engineering design 

Chapter 6 Effect of material properties on design Problem 6.1 Deflection (y) of a cantilever beam under a load (L) acting on its tip is given by the relationship: y = (L l3) / (3 E I), where: l = the length of the cantilever = 2000 mm E = elastic modulus of steel = 210 GPa, I = the second moment of area of the cross section. From Fig. 4.3, I =

b (3b ) 3 L l3 1000 × ( 2000 ) 3 = = 12 3 y E 3 × 25 . 4 × 210 × 10 3

The above equation gives the width of the beam. b = 21.71 mm. Taking a factor of safety n = 2 and using Eq. (6.3),

Z=

21.71(3 × 21.71) 3 n M 2 × L × 2000 = 15348.67 mm3 = = 12 × 32.565 1420 YS

The safe value of L = 5448.78 N.

Problem 6.2 From Eq. (6.8), KIC = 87.4 = Y σf (π a)1/2 Taking Y = 1 and a = 1.5 mm, then σf = 1273.2 MPa. From Eq. (6.3), L = (σf x Z)/(n x l) = (1273.2 x 15348.67)/(2 x 2000) = 4885.5 N Conclusion This load is less than the “safe load” calculated in problem 6.1, which means that had the presence of cracks been ignored, failure would have taken place.

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Problem 6.3 From problem 6.1, Z = 15348 .67 mm3, Mean stress = Sm = M/Z = 1000 x 2000/ 15348 .67 = 130.3 MPa Endurance limit = Se = 1800 x 0.56 = 1008 MPa. Taking a factor of safety of 2 for the static load and 3 for the dynamic load, and k f as 0.7, from Eq. (6.12),

(2 ×130.3 /1800) + (3 × Sa / 0.7 ×1008) = 1 Sa = 201.14 MPa, Alternating load = (201.4 x 15348 .67 )/2000 = 1545.6 N Problem 6.4 From eq. (6.12):

(nm K t S m / UTS ) + (na K f S a / S e ) = 1

S m = 150 Take nm and Kt as 2 and 1.5 respectively.

Se = 600 x 0.4 = 240 MPa Take na and Kf as 3 and 2 respectively. Sa = 10 MPa Problem 6.5 Stiffness of a component is related to its deflection under load. Deflection under load is a function of both the geometry of the component and the elastic modulus of the material from which the component is made. For example the deflection (y) of a cantilever beam under a load (L) acting on its tip is given by: y = (L l3) / (3 E I), where: l = the length of the cantilever (geometry), I = the second moment of area of the cross section (geometry) E = elastic modulus of the cantilever material. Problem 6.6 As discussed in problem 6.5, the stiffness of a component made out of low elastic modulus materials, such as plastics, can be increased by increasing the second moment of area of the 18   

 

Materials and process selection for engineering design 

critical cross section. This is illustrated in Fig. 6.2. Students may be asked to identify similar items. Problem 6.7 The main functional requirements of the supersonic aircraft wing are high stiffness and light weight. Considering the supersonic wing as a flat plate in bending, Table 4.5, shows that the performance index is: E1/3/ρ. From table 4.4: For aluminum alloys, E = 71 GPa, ρ = 2.7 Mg/m3, E1/3/ρ = 71.2 For titanium alloys, E = 120 GPa, ρ = 4.5 Mg/m3, E1/3/ρ = 50.9 The above figures show that aluminum alloys have higher performance index. However, because the temperature of the supersonic wing-tip gets too hot for aluminum alloys, titanium alloys are used as they retain their strengths at the high temperatures encountered in supersonic flights. Problem 6.8 •

•

•

•

The main material requirements for a motorcar exhausted manifold are to retain strength at the high temperature of the exhaust gases, resist the corrosive action of the hot exhaust gases, and to resist corrosion of the condensed exhaust gases when the engine is turned off. Stainless steels are most suitable. See appendix Table A.7. Coated carbon steels give limited service life. The main material requirements for coil for electrical resistance heater are reasonable electrical resistance (not too low as not to get hot on passing the electrical current and not too high as to obstruct the current completely), good creep resistance at the operating temperature, and excellent oxidation resistance at the operating temperature. Fe-Ni-base alloys, such as Inconel and Incoloy are suitable, see appendix Table A.16. The main material requirements for kitchen knife blade are high hardness in order to retain its sharp edge and corrosion resistance in order not to contaminate the food. Reasonable toughness is also required. Martensitic stainless steels are most suited. See appendix Table A.7. The main material requirements for railway line are wear resistance and toughness. Reasonable corrosion resistance is also an asset. Manganese steels have traditionally been used for this function.

Problem 6.9 The main requirements for gas turbine blades are high creep, wear and oxidation resistance as well as toughness at operating temperatures. Fe-Ni-base, Ni-base, and Co-base alloys are suitable. See appendix Tables A.16, A.17. Problem 6.10 The main requirements for materials used for the sleeves of lubricated journal bearings are high compressive and fatigue strengths, hardness that is lower than the shaft material, lower elastic modulus, high wear resistance, high corrosion resistance, good thermal conductivity, and reasonable cost. Suitable materials include white metals (Babbitt alloys), Cu-base alloys, Albase alloys, and some plastics. The material requirements and suitable materials are discussed in more detail in Section 11.3. 19   

 

Materials and process selection for engineering design 

Problem 6.11 The main requirements for making motorcar bodies include buckling resistance, stiffness and light weight. Considering the body to be made of panels, the main performance index for materials is E1/3/ρ. High elastic modulus (E) and low density (ρ) are desirable. Suitable materials include carbon steels, HSLA steels, aluminum alloys and sheet molding compounds. The material requirements and suitable materials are discussed in more detail in Section 11.5. Problem 6.12 The main requirements for motorcar bumpers are stiffness, impact resistance and light weight. Steels are stiffer and tougher than fiber reinforced plastics, but the latter are lighter. Introducing outward curvature and steel or oriented fiber ribs in FRP bumpers overcomes their weak points but maintains their light weight. Problem 6.13

a) The main material requirements for a paper clip include elasticity, in order to grip the paper, and corrosion resistance. b) Materials normally used include Cr plated carbon steel wire( on the right of the above figure), plastic coated carbon steel wire (second from the right), and injection molded plastic ( the two examples on the left). c) When plastics are used, a cantilever design is used together with roughened surface is used to improve paper gripping.

20   

 

Materials and process selection for engineering design 

Problem 6.14 The main material requirements for a small aircraft wing structure, is high stiffness and strength as well as light weight. Considering the structure as cylinders subjected to tension and compression, the material requirements can be combined into specific stiffness (E/ρ) and specific strength (S/ ρ), as given in Tables 4.2 and 4.5. For the 3 candidate materials: Aluminum alloy Magnesium alloy Polyester+65% glass fibers

E/ρ 25.9 26.4 10.9

S/ρ 214.8 164.7 188.9

Chapter 9 discusses quantitative methods for comparing materials, but for the present problem qualitative comparison will be sufficient. From the above table Al alloy has the highest S/ρ and the second highest E/ρ. Mg alloy has the highest E/ρ and the least S/ρ. FRP has the lowest E/ρ and the second highest S/ρ. This qualitative comparison favors Al ally with Mg alloy as second best. Problem 6.15 According to Eq. (6.10):

Se = ka kb kc kd ke kf kg kh Se'

where Se = endurance limit of the material in the component Se'= endurance limit of the material as determined by laboratory fatigue test = 0.5 x 950 ka = surface finish factor = 0.68 kb = size factor = 0.8 kc = reliability factor = 0.75 kf = stress concentration factor = 0.7 The rest of the derating factors are defined under Eq. (6.10) but will be ignored here. Se = 0.5 x 950 x 0.68 x 0.8 x 0.75 x 0.7 = 135.66 MPa Area of shaft = 50000 x 2/135.66 = 737.14 mm2 ,

Diameter of shaft = 30.6 mm

Problem 6.16

( )

Deflection under load = y = Ll 3 /(48EI ) a) Maximizing stiffness, i.e. minimizing y, entails selecting a material with maximum elastic modulus (E) and maximizing the second moment of area (I). b) Steel has higher E and would result in stiffer beam if space limitations require same I. If the designer is free to increase I for CFRP it would result in a lighter but equally stiff beam. It is expected that the steel beam would be less expensive in both cases.

21   

 

Materials and process selection for engineering design 

Problem 6.17 a) The material properties that are usually measured in a tensile test include elastic modulus, elastic limit (limit of proportionality), yield and ultimate tensile strengths, elongation and reduction in area percent. b) The area under the stress-strain diagram can be taken as an indication of toughness. c) The main design requirement of members of the tensile testing machine is stiffness. Deflection under load in the members of the machine should be kept to a minimum. d) High elastic modulus is the main required property of the material used in making the tensile testing machine. Normally steel and cast iron are used in making such members. Problem 6.18 The main material requirements for airplanes are high strength and elastic modulus, light weight, high fatigue strength, and high reliability. Aluminum alloys are traditionally used for making most of the load-bearing parts of the fuselage and wing structures because of its good specific strength and specific stiffness combined with reliability. Titanium alloys have lower specific stiffness (see problem 6.7) and is also much more expensive compared aluminum. Magnesium and FRP are less desirable than aluminum as was shown in problem 6.14. Reliability of aluminum alloys is also higher and most of the design codes for civilian subsonic aircraft have been developed with aluminum in mind.

22   

 

Materials and process selection for engineering design 

Chapter 7 Effect of manufacturing processes on design

Problem 7.1 a) Bottle for mineral water: PET, blow molding b) Fiberglass bath tub: Glass fiber reinforced epoxy or polyester resin, spray molding c) Inner lining of refrigerator door: ABS, polyethylene, polypropylene, or PVC, thermoforming, vacuum forming d) Steel bars for reinforcing concrete: low or medium carbon steel, hot rolling Problem 7.2 material

YS

UTS

elastic modulus

ductility

hardness

toughness

AA 2014 O

L

L

Same

H

L

H

AA 2014 T6

H

H

Same

L

H

L

AISI 1015

L

L

Same

H

L

H

AISI 1040

H

H

Same

L

H

L

AISI 1060 as quenched

H

H

Same

L

H

L

AISI 1060 quenched and tempered

L

L

Same 

H

L

H

Low density polyethylene

L

L

L

H

L

H

High density polyethylene

H

H

H

L

H

L

Problem 7.3 The crankshaft can either be made of steel or nodular cast iron. If made from steel, the primary processes would be die-forging. If made from nodular cast iron, the primary process would be sand or shell molding. In either case, the secondary manufacturing processes involve machining on the lathe and surface hardening of the sections of the crankshaft that fit 23   

 

Materials and process selection for engineering design 

in the bearings and the connecting rods. The finishing process involves grinding of the sections that were machined and surface hardened. To ensure prolonged fatigue life, ensure generous fillets between the machined sections and webs of the crankshaft. Problem 7.4 a) The main material requirements for railway line are wear resistance and toughness. Manganese steels have traditionally been used for this function. The main manufacturing process is hot rolling. b) The main material requirements for coil for electrical resistance heater are reasonable electrical resistance (not too low as not to get hot on passing the electrical current and not too high as to obstruct the current completely), good creep resistance at the operating temperature, and excellent oxidation resistance at the operating temperature. Fe-Ni-base alloys, such as Inconel and Incoloy are suitable, see appendix Table A.16. The primary manufacturing process is hot rolling to produce a rod. Secondary processes involve wire drawing to produce the required diameter. c) The main material requirements for a small aircraft wing structure, is high stiffness and strength as well as light weight. In the solution of problem 6.14 it was shown that aluminum alloys are most suited for this function. Plates and sheets are manufactured by hot rolling followed by cold rolling. These are then riveted to the structural members of the wing

Problem 7.5 a) Pistons for internal combustion engine are normally made of aluminum alloys, which can be manufactured by die casting followed by turning on a lathe. b) Connecting rods for internal combustion engines can either be made of steel or cast iron. Steel con rods can be manufactured by closed-die forging followed by machining, while cast iron con rods can be shell molded followed by machining. c) Cylinder heads are normally made of aluminum alloys, which can be manufactured by die casting followed by machining. d) Cam shafts are normally made of steel, which can be manufactured by closed die forging, followed by machining and surface hardened, and finally finished by grinding. Problem 7.6 Components made of plastics can usually be manufactured in one step with no need for machining, finishing, or coating. This saves time and cost. Metallic components, on the other hand, usually require more than one step of primary, secondary, and sometimes tertiary manufacturing processes. Coating may also be needed. Students may be asked to select a component which can be made from plastics or metals and compare the required manufacturing processes. 24   

 

Materials and process selection for engineering design 

Problem 7.7 a) Telephone set: possible material is ABS and possible manufacturing process is injection molding. b) Two-liter lubricating oil container: possible material is polyethylene and possible process is blow molding. c) Safety-shield for a mechanical press needs to be transparent and can be made from polycarbonate processed by injection molding or extrusion depending on the shape. d) Hardhat for construction workers: possible material is ABS processed by injection molding. Problem 7.8 The frame of a milling machine needs to be rigid, dimensionally stable, and dampens vibrations. Both steel and cast iron can provide the required rigidity and stability, but cast iron provides better vibration damping. In the case of steel, welding of plates and structural shapes can be used for manufacturing the frame, followed by heat treatment and machining. In the case of cast iron, sand casting can be used for making the frame, followed by heat treatment and machining. Problem 7.9 In solving the problem of the tie rod, the cross sectional area is calculated based on yield and then based on deflection. The larger area is used to calculate the weight and relative cost. Some of the calculations are given in Table 6.1. Material ASTM A675 grade 60 High strength steel Aluminum 5052 H38 Polyester-65% glass fibers

Area based on yield strength (mm2) 244 103 193 147

Area based on deflection (mm2) 131 131 392 1417

Mass based on the larger area (kg) 19 10.2 10.6 25.5

Relative cost 19 15.3 53 255

The high strength steel gives the lightest and least expensive tie rod. This tie rod can be manufactured by hot rolling followed by some cold rolling or cold drawing to hold the diameter to closer tolerances and to also control the strength. Problem 7.10 Heat treatment of steel normally involves fast cooling, to achieve martensitic structure, followed by tempering to improve toughness and to eliminate the internal stresses that arise from the fast cooling. With non-uniform thicknesses, thinner sections are expected to have faster cooling rates, which would lead to non-uniform hardness, thermal stresses, and cracking. Steels with high hardenability require less severe cooling rate to achieve martensite 25   

 

Materials and process selection for engineering design 

and would be more suitable for components of non-uniform sections. Providing gradual change in sections and having larger fillets would also help in eliminating the problem that may arise when heat treating such components.

Problem 7.11 Apart from some limitations on the minimum cross section, casting gives the designer the freedom to introduce hollow sections, curved surfaces, and a variety of thicknesses and fillet diameters. On the other hand, welded structures are made of standard sections and plates and would require an extensive amount of welding in order to obtain complex shapes. In addition, welded joints have to be accessible to the welder and changes in section thickness and fillets are usually limited. The cylinder block of a water-cooled internal combustion engine is an example of a cast component that would be very difficult to produce by welding. Problem 7.12 Powder metallurgy is best suited for making large numbers of small components that need little finishing after sintering. Small gears, if needed in large numbers can economically be produced from a wide variety of metallic powders. Investment or die casting such gears would be possible, but the process would be slower and the surface finish would not be as good. There would also be limitations on the melting point of the gear material in the case of die casting. Problem 7.13 From Eq. 7.2, the load carrying capacity of the two fillet welds is: 270000 = 2 x 0.3 x 414 x 0.707 x 12 x L L = 128 mm on each side parallel to the axis of the angle. Problem 7.14 Both welding and casting are possible manufacturing processes for the given gear, provided that the weldability and castability of the materials allow it. Care should be taken to ensure that the quality and position of the welded joints do not cause fatigue failure in service. With the required number of 10 units, however, it is possible that sand casting would be more economical. Problem 7.15 Spot welding has been traditionally used for joining different sheet metal parts of the motorcar body. With the increasing use of FRP for this application, adhesive bonding has gained more acceptance as a joining method. The design of the adhesive-bonded joints is discussed in Section 7.10. 26   

 

Materials and process selection for engineering design 

Problem 7.16 As shown in Fig. 7.11, butt joints should be avoided in the case of brazed and adhesivebonded joints. Lap, scarf and double-scarf, and double strap joints can be used to increase the bond area.

Problem 7.17 Such chairs usually have a hard frame and a soft seat and back. The frame is usually made of tubular sections for light weight. Materials used can be aluminum or coated steel tubes. Plastics may be used for the less stressed areas, such as the arms of an armchair. The soft material can either be made of natural fibers, such as cotton or linen, or synthetic fibers, such as polyester or nylon. There are several designs of folding chairs that use different mechanisms for folding the chair. The method of assembly depends on the folding mechanism. Students may be asked to prepare neat sketches of one of the designs. Problem 7.18 a) Material requirements for the water tank material include: non-toxicity, corrosion resistance, tensile strength, light weight, resistance to ultraviolet light, and toughness at low temperatures in cold climates. Processability and cost should also be considered. b) Among the materials that can be used are galvanized carbon steel sheets, polyethylene, and glass fiber reinforced epoxy or polyester. c) There are several possible design including cylindrical (horizontal or vertical axis), conical (with base of smaller diameter than the top), or cubical (in the case of steel sheets). Students may be asked to select a design and prepare neat sketches of their selected design. d) The manufacturing process depends on the material used. In the case of steel sheets and cubical design, the sheets can be joined by brazing, soldering or riveting. Inlet and outlet tubes can be joined by threaded joints. Blow or rotational molding are suitable for plastics. Chopped glass in liquid epoxy or polyester can be sprayed on a core, which can be taken out after the resin hardens. Inlet and outlet pipes can be joined by adhesive bonding. Problem 7.19 The most important properties for high pressure steam pipes are creep resistance at the operating temperature and corrosion resistance to condensed steam. Ductility and weldability are also important for manufacturing. Welded joints must be inspected by a reliable NDT method, such as X-ray to ensure soundness. 27   

 

Materials and process selection for engineering design 

Problem 7.20 a) Nodular cast iron is increasingly used for making crankshafts because of its strength, which is close to steal and its ease of manufacture by casting. Sand or shell molding may be used in this case. b) There are several types of superalloys that can be used for making gas turbine blades. Tables A.16 and A.17 in the Appendix give the composition and properties of a number of such alloys. Some of the superalloys are wrought and can be manufactured by closed die forging or machined by EDM. Cast superalloys are manufactured by shell molding or investment casting (lost wax method). In some cases directional solidification is used to obtain single crystal or elongated grains in the axial direction for better creep resistance. c) Two-liter polyethylene water bottles can be made by blow molding. d) Cemented carbide cutting tools are manufactured by powder metallurgy in the form of triangular or square tablets and then mechanically attached or brazed to a steel shank.

28   

 

Materials and process selection for engineering design 

Chapter 8 Economic and environmental impact of materials and processes

Problem 8.1 Stainless steels are more expensive than plain carbon steels for three main reasons: a) Stainless steels contain expensive alloying elements such as Cr and Ni b) Standards of quality control on stainless steels are usually more stringent as they are special steels c) Stainless steels are produced in smaller quantities are require special treatment in melting, forming, and finishing Problem 8.2 The price of titanium has decreased in the last 20 years because better and more efficient technology was introduced and because increasing demand has made it possible to produce larger quantities, which is known to reduce the unit cost. Problem 8.3 As shown in Figs. 4.1, 4.2 and 9.2 (a, b) CFRP have higher specific stiffness and specific strength than most engineering materials. This means that structures made of CFRP are lighter for the same level of performance. In applications such as aerospace and aviation this weight saving can be translated into economic advantage as more passengers and cargo can be transported. In many cases, the economic advantage is more than the extra cost paid for the more expensive CFRP. Sports equipment is another area where better performance outweighs the extra cost. In addition, increasing use of carbon fibers has caused their price to decrease as it was possible to employ large-scale production methods. Problem 8.4 The cost of a finished product is made up of the cost of materials, the cost of manufacturing, and the cost of finishing. Although the cost per unit weight of plastics is higher than many metals, their low density means that less weight of plastics may be used. In addition, the costs of manufacturing and finishing of plastics are less than those of many metals as plastics are easy to process and need little or no finishing. Problem 8.5 The cost of materials in a product can be viewed as the cost of material in the finished product and the cost material lost in trimming of fins in forgings or recycling of gates and 29   

 

Materials and process selection for engineering design 

risers in castings. The material utilization factor, m, see section 8.6, is known to be much higher in the case of powdered materials, which offsets part of the cost difference between bulk and powder materials. In addition, powder metallurgy products can be produced to better tolerances and surface-finish and, therefore, need less finishing costs. In many cases, the faster rate of production of powder metallurgy products gives them an added economic advantage. Problem 8.6 As figure 8.1 shows, the cost of aluminum ore is less than the cost of iron ore as it is more abundant in nature. However, the figure also shows that the cost of ore beneficiation is higher as Al2O3 is seldom found without gangue materials. The sharp increase in cost when the beneficiated ore is converted into ingot is a result of the large amount of energy needed to break the bond between Al and O2, which is about 15 KWh/kg of Al, and the high cost of auxiliary materials needed for extraction. Problem 8.7 For steel shelf: Volume = 420 cc, weight = 3.276 kg, material cost = $2.2932, Total cost per shelf = 2.2932 + 20 (1/60) +30,000/200,000 = $2.77 For GFRP shelf: Volume = 1080, weight = 2.052 kg, material cost = $3.07 Total cost per shelf=3.07 + 20(3/60) +15,000/200,000 +500,000/5 x 200,000 – 1.2 = $3.45 The steel shelf is more economical. The substitution is not economically justified. Problem 8.8 The savings per piece, Sp, can be calculated from Eq. (8.3): S p = (R t + Ro t ) − (R ' t '+ Ro t ')

Where R = labor rate per hour without tooling; R' = labor rate per hour using tooling; t = production time per piece without tooling, hours; t' = production time per piece with tooling, hours; Ro = machine cost per hour, including overheads. 25 25 ⎞ ⎛ 15 15 ⎞ ⎛ S p = ⎜15 × + 8 × ⎟ − ⎜11× + 8 × ⎟ = $4.83 60 60 ⎠ ⎝ 60 60 ⎠ ⎝

Tooling cost per piece = 1500/500 = $3.00 The cost saving per piece ($4.83) is more than the tooling cost per piece ($3.00). Therefore, the use of jig is economically justifiable.

30   

 

Materials and process selection for engineering design 

Problem 8.9 The jig should perform two functions: a) fix the work piece and b) guide the drill to the correct location of the hole. Different designs that perform the two functions can be developed by the students independently or in a class discussion. Problem 8.10 The savings per piece, Sp, can be calculated from Eq. (8.3): S p = (R t + Ro t ) − (R ' t '+ Ro t ')

where R = labor rate per hour without tooling; R' = labor rate per hour using tooling; t = production time per piece without tooling, hours; t' = production time per piece with tooling, hours; Ro = machine cost per hour, including overheads. 15 ⎞ ⎛ 10 10 ⎞ 15 ⎛ S p = ⎜12 × + 20 × ⎟ − ⎜10 × + 20 × ⎟ = $3.00 60 60 ⎠ ⎝ 60 60 ⎠ ⎝

The total tooling cost, CT, is the sum of the initial tooling cost, Ct, plus the interest on the tooling cost. Taking the number of years over which the tooling will be used as (n), the rate of interest as (i), and assuming straight line depreciation, Eq. (8.4) gives: CT = Ct + (Ct n i ) / 2 = 500 +

500 × (7 / 12) × 0.12 = 517.5 = S p × N 2

The number of parts that will justify introducing the jig N =173. Problem 8.11 The total time required to get a batch of 100 components through the workshop is:

Problem 8.12 a) Total time on the engine lathe Tte

b) Total time for turret lathe Ttt

c) Average production time for engine lathe = (493.3 x60)/500 = 59.2 min Average production time for turret lathe = (478.87 x60)/500 = 57.5 min

31   

 

Materials and process selection for engineering design 

32   

 

Materials and process selection for engineering design 

Chapter 9 The materials Selection Process Problem 9.1 Product Milk containers

Performance requirements Inertness with respect to milk Resistance to transport and handling conditions Shelf life Sales appeal Ease of manufacture Cost

Gas turbine blades

Creep behavior Structural stability Mechanical shock resistance Thermal shock and fatigue resistance Light weight Resistance to working environment Ease of manufacture

Sleeve for sliding journal bearing (see case study 11.3 for more details)

Piston for internal combustion engine

Cost Support of load on bearing Conformability Embeddability Long life Ability to conduct heat Resistance to working environment Cost Strength under operating conditions Thermal shock and fatigue resistance Resistance to working environment Light weight Cost

33   

Corresponding material requirements Corrosion resistance and non-toxicity Tensile strength, tear resistance, and toughness Permeability to O2 and ease of sterilization, Color and printability and light weight Processability Cost of materials, processing, and transportation. Creep rate at operating temperature, rupture strength, and stress rupture ductility Stability of microstructure at operating temperature Toughness Thermal conductivity, thermal expansion coefficient, Young’s modulus High specific strength (σ/ρ) Oxidation resistance at operating temperature Castability for cast alloys, ductility for forged blades Materials and processing costs Compressive strength, fatigue strength Low Young’s modulus Low hardness Wear resistance Thermal conductivity Corrosion resistance Cost of materials and processing Hot tensile strength and fatigue strength Thermal conductivity, thermal expansion coefficient, Young’s modulus Corrosion resistance at operating and room temperatures High specific strength (σ/ρ) Cost of materials and processing

 

Materials and process selection for engineering design 

 

Airplane wing structure

Mechanical strength and stiffness Light weight

Yield strength, fracture toughness, Young’s modulus High specific strength (σ/ρ), high specific stiffness (E/ρ) Weldability if welding is used Cost of materials and processing

Ease of manufacture Cost Problem 9.2

As shown in Fig. 4.2, CFRP has very high specific strength (σ/ρ). Similarly, it also exhibits very high specific stiffness (E/ρ). These two properties are important for many of the sports equipment, as discussed in Case Study 11.4 for tennis rackets. Problem 9.3 The main design requirements for an elevator hoisting cable are: high yield strength, fatigue resistance (can be related to tensile strength) and ductility at low cost. In the following table It is assumed that the required properties are equally important. Material ASTM-A675, 45 ASTM-A675, 70 ASTM-A242 Type I ASTM-A717, Grade 70

Scaled tensile strength 64 98 82 100

Scaled yield strength 32 50 66 100

Scaled Scaled elongation% cost

Total score

Rank

100 55 64 58

296 270 260 278

1 3 4 2

100 67 48 20

Problem 9.4 Material Weldability Tensile strength Fatigue strength Corrosion resistance Cost Performance index Rank

A 5×0.15 3×0.15 5×0.25 3×0.20 2×0.25 3.55 2

B 1×0.15 5×0.15 3×0.25 5×0.20 5×0.25 3.90 1

34   

C 3×0.15 2×0.15 3×0.25 3×0.20 3×0.25 2.85 3

 

Materials and process selection for engineering design 

Problem 9.5 Material Weldability Tensile strength Fatigue strength Corrosion resistance Cost Performance index Rank

A 5×0.25 3×0.10 5×0.20 3×0.25 2×0.20 3.7 1

B 1×0.25 5×0.10 3×0.20 5×0.25 5×0.20 3.6 2

C 3×0.25 2×0.10 3×0.20 3×0.25 3×0.20 2.9 3

Problem 9.6 Scaled property×(weight)

Al-2014T6

Scaled yield strength×(0.2) Scaled Young’s modulus×(0.3) Scaled weldability index×(0.2) Scaled specific gravity×(0.3) Performance index Rank

37×0.2 34×0.3 60×0.2 75×0.3 52.1 3

Steel AISI 1015 48×0.2 100×0.3 100×0.2 27×0.3 67.7 2

Epoxy-70% glass fabric 100×0.2 11×0.3 80×0.2 100×0.3 69.3 1

Cost can be used as one of the properties and given an appropriate weight, which can be relatively low in the case of racing cars, since performance is much more important than cost. Problem 9.7 The main functional requirements and corresponding material properties of the material used for boiler shell include: a) Functional requirements b) Material properties c) Weights d) Mode of failure

Strength at operating temperature creep resistance at operating temperature 0.3 Creep rupture

Toughness

Resistance to service environment Corrosion Impact and resistance and fracture oxidation resistance toughness 0.15 0.2 Brittle Pitting fracture corrosion or unsafe reduction of thickness

Processability

Cost

Weldability

Cost of materials and processing 0.15 Uneconomical performance

0.2 Defects and internal stresses in welded joints leading to fracture

e) Low alloy steel, for example 0.2C-1Cr-1Mo-0.25V. See Section 4.7 for more details. 35   

 

Materials and process selection for engineering design 

Problem 9.8 Material

Al-770

Scaled yield strength× 0.2 Scaled wear resistance ×0.11 Scaled fatigue strength×0.14 Scaled corrosion resistance×0.11 Scaled thermal conductivity ×0.2 Performance index γ Rank

20 4.4 14 6.6 20 65 1

Bronze ASTM B22 19.4 11 11.2 4.4 5 51 2

Tin-alloy ASTM B23 4.6 4.4 2.9 11 6 28.9 3

Problem 9.9 Students may be asked to find a suitable pair of kitchen scissors and to make the necessary free-hand sketches complete with dimensions. The main functional requirements of the blades are to remain sharp, resist corrosion, and to resist fracture on impact. Corresponding material requirements include high hardness, corrosion resistance to water and weak acids, and high toughness. Possible materials for the blades include martensitic stainless steels formed by forging. The handles need to be soft with nice texture, preferably colorful, and corrosion resistant. The corresponding material properties are low elastic modulus, good processability, and corrosion resistance to water and weak acids. Possible material for the handles include PVC, ABS or nylon formed by injection molding. Problem 9.10 Students may be asked to find a suitable screw driver set and to make the necessary free-hand sketches complete with dimensions. The main functional requirements for the screw driver blade is to maintain its shape and sharpness for a long time. The corresponding material properties include high strength, hardness and elastic modulus. Possible materials include hardenable or tool steels formed by forging followed by machining. The main functional requirements for the handles include electrical insulation, resistance to fracture on impact and nice feel and color. The corresponding material properties include electrical resistance, toughness, low elastic modulus, and good processability. Possible materials include ABS or hard rubber. Possible method of assembly include injection molding of the handle over the blade. Problem 9.11 Students may be asked to find a suitable suitcase and to make the necessary free-hand sketches complete with dimensions. The main structural elements include outer skin, an

36   

 

Materials and process selection for engineering design 

internal frame, handle, and wheels (optional). The main functional requirements and corresponding material properties for the different elements are given in the following table: Functional requirements Mechanical resistance

Sales appeal Light weight Cost Mechanical resistance Light weight Cost Mechanical resistance Light weight Sales appeal Cost

Material properties (weight) Outer skin Yield strength (0.1) Tear resistance (0.15) Scratch resistance (0.15) Toughness (0.15) Color and texture (0.2) Specific strength (0.15) Cost of material and processing (0.1) Frame Yield strength (0.15) Toughness (0.2) Elastic modulus (0.15) Specific strength (0.2) Specific stiffness (0.2) Cost of material and processing (0.1) Handle Yield strength (0.25) Toughness (0.25) Specific strength (0.2) Color and texture (0.2) Cost of material and processing (0.1)

Possible materials for the skin include: polystyrene, PVC, ABS. Possible materials for the frame include: Aluminum or magnesium alloys and FRP Possible materials for the handle include: ABS, nylon and acrylics Problem 9.12 For the present case, it is assumed that the overhead pedestrian crossing will be used to cross a road 16 m wide with an island in the middle. The main structural elements include two sets of stairs and a horizontal overhead passage with span of 20 m, which is supported at the ends and in the middle. Possible materials include structural steel, concrete, bricks, timber, or FRP. Combinations of these materials are also possible. The following table gives a summary of the functional requirements, corresponding material properties, and weights. The selection of the final material is also influenced by the surrounding architectural features and materials, required delivery time, and constraints of the building site.

37   

 

Materials and process selection for engineering design 

Functional requirements

Material properties (weight) Overhead passage Yield strength (0.15) Compressive strength (0.10) Elastic modulus (0.10) Corrosion resistance (0.10) Wear resistance (0.2) Fire resistance (0.10) Specific strength (0.05) Cost of material and processing (0.2) Stairs Yield strength (0.15) Compressive strength (0.10) Elastic modulus (0.10) Corrosion resistance (0.10) Wear resistance (0.2) Fire resistance (0.1) Specific strength (0.05) Cost of material and processing (0.2) Supporting columns Yield strength (0.1) Compressive strength (0.20) Elastic modulus (0.20) Corrosion resistance (0.10) Wear resistance (0.05) Fire resistance (0.1) Specific strength (0.05) Cost of material and processing (0.2)

Mechanical properties Durability Light weight Cost Mechanical properties Durability Light weight Cost Mechanical Properties Durability Light weight Cost Problem 9.13 Property

Weight

YS 0.15 Wear 0.20 resistance Fatigue 0.20 strength Corrosion 0.25 resistance Processability 0.10 Cost 0.10 Performance index Rank

Scaled properties Al alloy Bronze S. Steel 62 60 100 40 100 100

Scaled properties x weight Al alloy Bronze S. Steel 9.3 9 15 8 20 20

47.6

38

100

9.5

7.6

20

50

100

100

12.5

25

25

60 100

100 50

100 33

6 10 55.3 3

10 5 76.6 2

6 3.3 89.3 1

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Problem 9.14 This problem is suitable for class discussion and student projects. Students may be asked to draw free-hand sketches of the bicycle, preferably from life, and then proceed from there. Materials that can be explored include steel, aluminum and plastics. Problem 9.15 For each of the candidate materials, the area of the member needs to be calculated once based yield strength and another based on the extension. The larger area is then selected for weight and cost calculations. The factor of safety will be ignored in the calculations as it will not affect the selection results. The density of the steel is taken as 7.8 g/cc. To resist yield, the area Ays= 50000/YS. To resist extension beyond the specified limit, the area Ae=(50000×2000)/(2.5×E) Material ASTM-A675, 45 ASTM-A675, 70 ASTM-A242, Type 1 ASTM-A717, Grade 70

Ays (mm2) 323 208 156 103

Ae (mm2) 188.7 188.7 188.7 188.7

Weight (kg) 2.5 1.6 1.47 1.47

Relative cost 2.5 2.4 3.1 7.35

Rank 2 1 3 4

Problem 9.16 Scaled property×(weight) Scaled yield strength×(0.2) Scaled Young’s modulus×(0.3) Scaled weldability index×(0.2) Scaled specific gravity×(0.3) Performance index Rank

Al-2014T6 37×0.2 34×0.3 60×0.2 75×0.3 52.1 3

Steel AISI 1015 48×0.2 100×0.3 100×0.2 27×0.3 67.7 2

Epoxy-70% glass fabric 100×0.2 11×0.3 80×0.2 100×0.3 69.3 1

Cost can be used as one of the properties and given an appropriate weight, which can be relatively low in the case of racing cars, since performance is much more important than cost. Problem 9.17 Students can be divided into groups with each group making the design, deciding on performance requirements, developing required material properties, and selecting materials and manufacturing processes for one of the given 5 cases. A class discussion can then be held to analyze the effect of package size on the selected materials and processes.

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Chapter 10 Materials substitution

Problem 10.1 Fruit juice package Attribute Disposable/returnable Shelf-life Weight Recyclability Cost Environmental impact Sales appeal

Glass Disposable/ Returnable Shortest Heaviest Good Intermediate Best Least

Plastic Disposable Intermediate Lightest Good Least expensive Intermediate Intermediate

Carton laminate Disposable Longest Intermediate Poor More expensive Worst Best

From the above table, it is clear that each of the candidate materials has its strong and weak points. Whether the replacement of glass is viable and which of the two candidates to choose depends on the size of the package, method of distribution, the type of user, etc. At present carton laminate is used extensively for small and intermediate size packages (up to one or two liters). Larger packages for restaurants and catering firms may be best made out of plastic. Glass may still be best for fresh juice for serving in restaurants. Problem 10.2 Milk package Attribute Disposable/returnable Shelf-life Weight Recyclability Cost Environmental impact Sales appeal

Glass Disposable/ Returnable Shortest Heaviest Good Intermediate Best Intermediate

Plastic Disposable Intermediate Lightest Good Least expensive Intermediate Intermediate

Carton laminate Disposable Longest Intermediate Poor More expensive Worst Best

At present glass is the standard material for door delivery of milk, in the UK. Plastic is most used for fresh milk packages of all sizes for supermarket distribution and carton laminate packages are used for long-life milk.

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Problem 10.3 Property

Weight A 100 60

Weldability 0.15 Tensile 0.15 strength Fatigue 0.25 strength Corrosion 0.20 resistance Cost 0.25 Performance index Rank

Scaled properties B C 20 60 100 40

Scaled properties x weight A B C 15 3 9 9 15 6

100

60

60

25

15

15

60

100

60

12

20

12

40

100

60

10 71 2

25 78 1

25 57 3

Problem 10.4 Property

YS Young’s modulus Weldability Specific gravity Cost Performance index Rank

Weight

0.20 0.30 0.10 0.30 0.10

Steel AISI 1015 48 100 100 26.9 100

Scaled properties EpoxyAl70% 2014 glass T6 36.5 100 33.8 10.6 60 80 75 100 25 20

Scaled properties x weight EpoxyAlSteel 2014 70% AISI glass T6 1015 9.6 7.3 20 30 10.1 3.2 10 6 8 8.1 22.5 30 10 2.5 2 67.7 48.4 63.2 1 3 2

Problem 10.5 The following table is based on Tables 10.2 and 10.3, with the addition of the last two columns. When the weight given to power is equal to that given to damping, γ (50-50), Epoxy+50%CF becomes better with the other three materials giving equal performance. This kind of weighting obviously favors the composites with lower carbon fiber content, since they are less expensive. On the other hand, when the weight given to power is much higher γ (90-10), The trend which was observed in the case of γ (70-30) becomes even stronger and Epoxy+65%CF becomes a clearer preference.

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Material

NP

ND

NC

γ (70-30)

Δγ%

ΔC%

(Δγ%/ ΔC%) γ (50-50)

Epoxy+ 50%CF 82 100 100 87.4 ---91 Epoxy+ 55%CF 88 93 92 89.5 2.40 8.6 0.28 90.5 Epoxy+ 60%CF 94 87 85 91.9 5.15 17.2 0.3 90.5 Epoxy+ 65%CF 100 81 80 94.3 7.9 25.8 0.31 90.5 NP = normalized power, ND = normalized damping, NC = normalized cost

γ (90-10) 83.8 88.5 93.3 98.1

Problem 10.6 Students may be asked to find a suitable bicycle and to make the necessary free-hand sketches complete with dimensions. The main functional requirements of the frame are rigidity, strength and light weight. Corresponding material requirements include high specific stiffness (E/ρ) and high specific strength (σ/ρ). Students may then compare various substitute materials (aluminum, magnesium, titanium, CFRP) after allocating appropriate weights to (E/ρ) and (σ/ρ). Analysis similar to that followed in Case Study 10.1 and scenarios similar to those in Problem 10.5 can be developed. Problem 10.7 As the solution to problem 10.5 shows, the cost of performance method, which separately compares the performance index (γ) and total cost (Ct), allows different scenarios to be developed based on the different weights that are allocated to the technical performance requirements. For the case of the tennis racket, allocating various weights to power and damping allows manufacturers to tailor-make their products to meet the technical preferences of different players. The compound performance function method (CPF), which combines technical performance and cost functions in one parameter, is simpler since only one figure is used for comparing different materials. Different scenarios showing the effect of allocating different weights to the technical performance against the cost can be easily developed. For example, lower weight can be allocated to the cost for the case of advanced and professional players than in the case of beginners. Case study 11.4, gives a detailed analysis of the materials substitution in the tennis racket. Problem 10.8 The term-project can be started early in the semester and the group can work on various aspects of the project as the course develops. The term project has proven to be a very useful part of a course on materials selection and can be considered as a capstone activity that sums up a lot of the knowledge gained in the course. It also emphasizes teamwork and

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communication skills (oral and written). The rest of the class may be asked to comment on the points of strength and weaknesses after a group presentation.

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69179 ISBN 1-420-06917-9

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