Structural Materials: Properties and Selection [1st ed. 2019] 978-3-030-26160-3, 978-3-030-26161-0

Table of contents :
Front Matter ....Pages i-xii
Conversion of Units (José Antonio Pero-Sanz Elorz, Daniel Fernández González, Luis Felipe Verdeja)....Pages 1-6
Introduction to Structural Materials: Naturals, Metals, Ceramics, Polymers, and Composites (José Antonio Pero-Sanz Elorz, Daniel Fernández González, Luis Felipe Verdeja)....Pages 7-17
Structural Materials: Metals (José Antonio Pero-Sanz Elorz, Daniel Fernández González, Luis Felipe Verdeja)....Pages 19-76
Structural Materials: Ceramics (José Antonio Pero-Sanz Elorz, Daniel Fernández González, Luis Felipe Verdeja)....Pages 77-132
Organic Polymers (José Antonio Pero-Sanz Elorz, Daniel Fernández González, Luis Felipe Verdeja)....Pages 133-162
Composites (José Antonio Pero-Sanz Elorz, Daniel Fernández González, Luis Felipe Verdeja)....Pages 163-191
Selection of Structural Materials: Combined Mechanical Properties and Materials-Selection Charts (José Antonio Pero-Sanz Elorz, Daniel Fernández González, Luis Felipe Verdeja)....Pages 193-203
Materials for Beams (José Antonio Pero-Sanz Elorz, Daniel Fernández González, Luis Felipe Verdeja)....Pages 205-247
Materials for Columns and Struts (Compression–Tension) (José Antonio Pero-Sanz Elorz, Daniel Fernández González, Luis Felipe Verdeja)....Pages 249-273
Materials for Pressure Vessels (José Antonio Pero-Sanz Elorz, Daniel Fernández González, Luis Felipe Verdeja)....Pages 275-287
The Cost Factor for the Selection of Materials (José Antonio Pero-Sanz Elorz, Daniel Fernández González, Luis Felipe Verdeja)....Pages 289-300
Materials Resistant to Fatigue: Quenched and Tempered Steels (José Antonio Pero-Sanz Elorz, Daniel Fernández González, Luis Felipe Verdeja)....Pages 301-347
Considerations of Temperature in the Service and Behaviour of Materials (José Antonio Pero-Sanz Elorz, Daniel Fernández González, Luis Felipe Verdeja)....Pages 349-356
Additional Conclusions (José Antonio Pero-Sanz Elorz, Daniel Fernández González, Luis Felipe Verdeja)....Pages 357-362
Back Matter ....Pages 363-372

Citation preview

José Antonio Pero-Sanz Elorz  Daniel Fernández González  Luis Felipe Verdeja

Structural Materials Properties and Selection

Structural Materials “If we know the dimensions of a given structure and the properties of the materials from which it is made, we can at least try to predict how strong it ought to be and how much it will deflect. As a rule, both the shape and the materials of any structure which has evolved over a long period of time in a competitive world represent an optimization with regard to the loads which it has to carry and to the financial or the metabolic cost”. —J. E. Gordon, The philosophy of design–or the shape, the weight and the cost, in Structures or Why Things Don’t Fall Down (1978)

José Antonio Pero-Sanz Elorz
Daniel Fernández González
Luis Felipe Verdeja

Structural Materials Properties and Selection

123

José Antonio Pero-Sanz Elorz Escuela de Ingeniería de Minas, Energía y Materiales de Oviedo University of Oviedo Oviedo, Asturias, Spain

Daniel Fernández González Escuela de Ingeniería de Minas, Energía y Materiales de Oviedo University of Oviedo Oviedo, Asturias, Spain

Luis Felipe Verdeja Escuela de Ingeniería de Minas, Energía y Materiales de Oviedo University of Oviedo Oviedo, Asturias, Spain

ISBN 978-3-030-26160-3 ISBN 978-3-030-26161-0 https://doi.org/10.1007/978-3-030-26161-0

(eBook)

© Springer Nature Switzerland AG 2019 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

This book is dedicated to the memory of José Antonio Pero-Sanz Elorz (1934 to 2012).

Prologue and Acknowledgements

After the publication of Solidification and Solid State Transformations of Metals and Alloys (Elsevier, 2017) and Physical Metallurgy of Cast Irons (Springer, 2018), we now present the English version of the book, Structural Materials: Properties and Selection, which was originally written by Professor José Antonio Pero-Sanz Elorz (who died in 2012) and now has been revised and updated with examples, exercises, problems, and case studies by Daniel Fernández González (MSc) and Professor Luis Felipe Verdeja González (Head of the Department of Materials Science and Metallurgical Engineering, University of Oviedo, Asturias, Spain). In this book, we present a multidisciplinary approach to the understanding of metals, ceramics, polymers, and composites, which were previously considered independent fields of study; this broad approach will undoubtedly be welcomed. This book is aimed primarily at undergraduate students but also at anyone interested in the field of materials science and engineering, for whom this textbook will provide an invaluable reference. To achieve the broad approach above referred, it was necessary to rely on the cooperation of physicists, chemists, and engineers; however, the book was written with a primarily engineering focus. The science of the subject matter is tackled from an engineering perspective, which therefore influenced the types of materials considered, i.e., materials that might be classified as structural materials—those that are interesting either because of their bulk properties (stiffness, elasticity, mechanical resistance, density, toughness, behaviour under conditions of fatigue, creep, etc.) or their surface properties (behaviour when subjected to oxidation, corrosion, friction, abrasion, wear, etc.). In professional circles, the study of functional properties (thermal, magnetic, electric, optical, etc.) is still the domain of physicists, chemists, and physicist–chemists. From this viewpoint, it is understood that anyone beginning to investigate the field of materials science and engineering must start with the field that is traditionally known as “physical metallurgy” for two main reasons: • One reason is the type of education and training that engineering candidates receive before reaching university: Generally, the students have achieved a reasonable level in physics and chemistry—both organic and inorganic—which not only helps them understand polymeric and ceramic materials but also equips them to deal with the study of the properties (e.g., semiconductivity,

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Prologue and Acknowledgements

superconductivity, magnetism, etc.) connected with atomic structures. The same could be said with regard to the physical–structural characteristics (polycrystalline, vitreous, amorphous) of materials, i.e., the properties derived from their structure not only in metals but also in ceramics and composites. It is also assumed that students will have sufficient knowledge about solidification processes, phase diagrams, transformations in solid state, deformations, etc., to enable them to understand, for instance, ceramics engineering, which is far more complex than pottery. • The second reason is that the reality of industrial practice conditions this priority, which is not only temporary, on the study of metallic materials: This is due to the importance of metals in the energy industry (oil pipelines, gas pipelines, nuclear-power plants, thermal-power plants, electricity lines, etc.), the construction industry (civil construction, platforms, cryogenic vessels for gases), and the transport industry (naval, air, and ground). In this regard, it is worth noting that technological advances in the manufacture of, for instance, steel makes the product significantly more desirable, from both quantitative and qualitative perspectives, in the field of new materials […]. The foregoing considerations are not meant to contradict current thinking, either research and academic, regarding materials science in any way; the intention is merely to prevent the current demise of metallurgical science and engineering, which have been enriched from the study of other fields. In fact, the aim of this book precedes the publication of other books entitled Metallurgy of Iron, Steel, and Ferroalloys and Refractory and Ceramic Materials by Professor Luis Felipe Verdeja González and collaborators. We stress the importance of the information obtained from works listed in the Bibliography. We also want to express our gratitude for the permission granted to reproduce some tables and figures found in this book (the original source of each is indicated). We are sincerely grateful for the support provided during preparation of the text and the experimental work as well as the suggestions and encouragement received from several members of the Department of Materials Science and Metallurgical Engineering of the Escuela de Ingeniería de Minas, Energía y Materiales de Oviedo (University of Oviedo), particularly Senior Professor José Ignacio Verdeja González; the laboratory technicians José Ovidio García García and Lucio García-Pertierra Luiña; and Dr. Cristian Gómez Rodríguez and Professor Francisco Blanco Álvarez (Head of the Escuela de Ingeniería de Minas, Energía y Materiales de Oviedo). Oviedo, Asturias, Spain June 2019

José Antonio Pero-Sanz Elorz Daniel Fernández González Luis Felipe Verdeja

Contents

1

Conversion of Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Conversion of Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2

Introduction to Structural Materials: Naturals, Metals, Ceramics, Polymers, and Composites . . . . . . . . . . . . . . 2.1 Structural Materials: Naturals, Metals, Ceramics, Polymers and Composites . . . . . . . . . . . . . . . . . . . 2.2 Structural Materials: Natural Materials . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3

Structural Materials: Metals . . . . . . . . . . . . . . . . . . 3.1 Structural Metallic Materials . . . . . . . . . . . . . . 3.2 Ellingham’s Diagram DG0(T) for the Formation of Metallic Oxides . . . . . . . . . . . . . . . . . . . . . 3.3 The Five Main Metals . . . . . . . . . . . . . . . . . . . 3.3.1 Iron . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.2 Aluminium . . . . . . . . . . . . . . . . . . . . 3.3.3 Copper . . . . . . . . . . . . . . . . . . . . . . . 3.3.4 Zinc . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.5 Lead . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Metals for Which Annual Production Is Approximately 2 • 106 T . . . . . . . . . . . . . . . 3.4.1 Nickel . . . . . . . . . . . . . . . . . . . . . . . . 3.4.2 Magnesium . . . . . . . . . . . . . . . . . . . . 3.4.3 Tin . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Twelve Other Strategic Metals . . . . . . . . . . . . . 3.5.1 Silicon . . . . . . . . . . . . . . . . . . . . . . . . 3.5.2 Titanium . . . . . . . . . . . . . . . . . . . . . . 3.5.3 Refractory Metals . . . . . . . . . . . . . . . . 3.5.4 Metals for the Nuclear Industry . . . . . . 3.6 Precious Metals . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Structural Materials: Ceramics . . . . . . . . . . . . . . . . . . . . . . 4.1 Structural Materials: Ceramics . . . . . . . . . . . . . . . . . . . 4.1.1 Simple Crystalline Structures in Ceramics, Intermetallics, and Semiconductors . . . . . . . . . 4.2 Tough Ceramics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.1 Toughness in Refractories and Ceramics . . . . . 4.2.2 Mechanical Resistance in Ceramic Materials . . 4.3 Simple Components of Refractory Materials: SiO2, Al2O3, and CaO . . . . . . . . . . . . . . . . . . . . . . . . 4.3.1 Crystalline Silica . . . . . . . . . . . . . . . . . . . . . . 4.3.2 Silica Glass . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.3 Alumina: Alumina Refractories . . . . . . . . . . . . 4.3.4 The SiO2–Al2O3 Binary System . . . . . . . . . . . 4.3.5 Silicon-Based Refractories . . . . . . . . . . . . . . . 4.3.6 Silica–Alumina (%Al2O3 < 46%) and Alumina Refractories (46% < %Al2O3 < 72%) . . . . . . . 4.3.7 Mullite Refractories (75% Al2O3) . . . . . . . . . . 4.3.8 Lime (CaO) . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.9 SiO2–CaO Binary Diagram . . . . . . . . . . . . . . . 4.3.10 SiO2–Al2O3–CaO Ternary System: Slags and Refractories Used in Obtaining Steels . . . . . . . 4.3.11 Basic Slags . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.12 Basic Refractories . . . . . . . . . . . . . . . . . . . . . 4.3.13 Basicity Index . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Engineering Ceramics . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.1 Refractories: Thermal-Shock Resistance . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Organic Polymers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.1 Young’s Modulus in Polymers: Tensile Test and Application . . . . . . . . . . . . . . . . . . . 5.2 Thermoplastic Polymers . . . . . . . . . . . . . . . . . . . . . . . 5.3 Thermosetting Polymers . . . . . . . . . . . . . . . . . . . . . . . 5.4 Elastomers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Foams (Expanded Polymers) . . . . . . . . . . . . . . . . . . . . 5.6 Summary of Organic Polymer Properties . . . . . . . . . . . 5.6.1 Viscoelastic Behaviour and Creep in Polymers .

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Composites . . . . . . . . . . . . . . . . . . 6.1 Introduction . . . . . . . . . . . . . 6.2 Polymeric-Matrix Composites 6.3 Metallic-Matrix Composites . .

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Contents

6.3.1 MMCs . . . . . . . . . . . . . . . . . . . . . . . 6.3.2 Other Composites of Metallic Matrix 6.4 Ceramic-Matrix Composites . . . . . . . . . . . . . 6.4.1 Concrete . . . . . . . . . . . . . . . . . . . . . 6.4.2 Ceramics Reinforced with Fibre . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

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Selection of Structural Materials: Combined Mechanical Properties and Materials-Selection Charts . . . . . . . . . . . . . . 7.1 Selection of Structural Materials . . . . . . . . . . . . . . . . . . 7.2 Combined Mechanical Properties and Materials-Selection Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Materials for Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Materials for Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Geometry of the Section and Mechanical Behaviour . . . . 8.2.1 Geometry and Stiffness (Resistance to Elastic Deformation) . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.2 Geometry and Resistance to Plastic Deformation (Yielding) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Comparison Between Materials for Beams and Plates . . . 8.3.1 Minimizing Weight for Equal Stiffness . . . . . . . 8.3.2 Minimizing Weight with Equal Resistance to Plastic Deformation (Yielding) . . . . . . . . . . . 8.3.3 Selection Based on the Risk of Unstable Failure Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Materials for Columns and Struts (Compression–Tension) . 9.1 Materials for Columns . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Geometry of the Section and Buckling . . . . . . . . . . . . . 9.3 Comparison Between Materials for Columns . . . . . . . .

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10 Materials for Pressure Vessels . . . . . . . . . . . . . . . . . 10.1 Materials for Pressure Vessels . . . . . . . . . . . . . . 10.2 Geometry and Behaviour Under Pressure . . . . . . 10.3 Materials for Vessels . . . . . . . . . . . . . . . . . . . . . 10.3.1 Spherical Vessels: Minimum Weight for a Particular Stiffness . . . . . . . . . . . . 10.3.2 Spherical Vessels: Minimum Weight Without Plastic Deformation (Yielding) . 10.3.3 Spherical Vessels: Minimum Weight Without Risks of Unstable Failure . . . . . Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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11 The Cost Factor for the Selection of Materials . . . . . . . . . 11.1 The Cost Factor in the Selection of Materials . . . . . . . 11.2 Charts of Materials and Cost . . . . . . . . . . . . . . . . . . . 11.2.1 Other Assumptions . . . . . . . . . . . . . . . . . . . . 11.3 Selection by Multiple Properties: Attributive Analysis . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Materials Resistant to Fatigue: Quenched and Tempered Steels . . . . . . . . . . . . . . . . . . . . . . . . 12.1 Materials Resistant to Fatigue: Quenched and Tempered Steels . . . . . . . . . . . . . . . . . . . . 12.2 Heat Treatments of Steels . . . . . . . . . . . . . . . . 12.2.1 Cooling of Metals in Heat Treatments with Previous Austenization . . . . . . . . 12.2.2 Quenching . . . . . . . . . . . . . . . . . . . . . 12.2.3 Normalizing . . . . . . . . . . . . . . . . . . . . 12.2.4 Full Annealing . . . . . . . . . . . . . . . . . . 12.2.5 Subcritical Heat Treatments . . . . . . . . 12.2.6 Isothermal Heat Treatments . . . . . . . . 12.2.7 Thermal–Chemical Treatments . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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13 Considerations of Temperature in the Service and Behaviour of Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349 13.1 Considerations of Temperature in the Service and Behaviour of Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . 349 Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 356 14 Additional Conclusions . . . . . . . . . . . 14.1 Additional Conclusions . . . . . . . 14.2 Materials and Materials Science . References . . . . . . . . . . . . . . . . . . . . .

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Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369

1

Conversion of Units

Abstract

This introductory chapter shows the conversion between different units of measurement: temperature, length, power, energy, weight, etc. Moreover, several solved exercises are included at the end of the chapter.

1.1

Conversion of Units

This introductory section will be useful for readers because it will familiarize them with some units of measurement as well as the conversion between different units of measurement. Knowing this information while reading this book will prove useful to readers. – Temperature:

TC ¼ TK  273 TC ¼ ðTF  32Þ  TR ¼ TF þ 460

ð1:1Þ 5 9

ð1:2Þ ð1:3Þ

TC , temperature in degrees Celsius; TK , temperature in degrees Kelvin; TF , temperature in degrees Fahrenheit, and TR , temperature in degrees Rankine. The following tables show the conversion between different units of measurement: length (Table 1.1), mass (Table 1.2), force (Table 1.3), pressure (Table 1.4), toughness (Table 1.5), energy (Table 1.6), power (Table 1.7), electrical resistivity © Springer Nature Switzerland AG 2019 J. A. Pero-Sanz Elorz et al., Structural Materials, https://doi.org/10.1007/978-3-030-26161-0_1

1

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Conversion of Units

(inverse of conductivity, Table 1.8) and thermal conductivity (Table 1.9). Table 1.10 collects the usual constants, while Table 1.11 shows the multiple and submultiple of basic units. Table 1.1 Units of length

1 1 1 1 1 1 1 1

mile (ground mile) nautical mile yd (yard) ft (pie) in (inch) µm (micrometre, micron) nm (nanometre) Å (Angstrom)

1609.344 m 1853 m 0.9144 m 0.3048 m 0.0254 m 10−6 m 10−9 m 10−10 m

Table 1.2 Units of mass

1 1 1 1 1 1

oz (avoirdupois ounce) oz tr (troy ounce) lb (avoirdupois pound) t (metric ton) ton (long ton, imperial ton [UK]) sh ton (short ton [USA])

28.349527 g (gram) 31.103481 g 453.592 g 1000 kg 1016.047 kg 907.185 kg

Table 1.3 Units of force

1 kgf or kp (kilogram-force or kilopond) 1 lbf (pound force) 1 dyn (dyne)

Table 1.4 Units of pressure

1 kgf/mm2

Table 1.5 Units of toughness

1 kgf m/cm2 1 kgf m

9.8065 N (Newton) 4.448 N 10−5 N

9.80665 N/mm2 (or MN/m2 or MPa) 98.0665 bar (1 bar = 0.986923 atm) 1442.3343 lbf/in2 (pound square inches, psi) 1.442334 ksi (or Psi; 1 Psi = 1000 psi) 0.71116717 (USA) tonf/in2 0.63497067 (UK) tonf/in2 100 kgf/cm2 = 100 atm 1 bar 0.986923 atm 1  105 Pa = 0.10 MPa 1 Pa 1.4504  10−4 psi 1 atm 1.01325  105 Pa = 0.101 MPa 1 P (poise) 0.1 Pa s = 1 g cm−1 s−1 1 cP (centipoise) 1  10−3 Pa s

9.80665 J/cm2 9.80665 J

1.1 Conversion of Units Table 1.6 Units of energy

3 1 kWh 1 J (Joule)

1 1 1 1

F l atm calth hp

3.6  106 J 0.101972 kgf m 107 erg (erg) 0.2388 cal or calth (calory international, calIT) 6.24  1018 eV (electronvolt) 9.478  10−4 Btu (British thermal unit) 0.7375 ft lbf 1 V C = 1 N m = 107 erg 1 C/V 101.325 J 4.184 J 745.7 W

Table 1.7 Units of power

1W 1 kW (kilowatt)

Table 1.8 Units of electrical resistivity

1 (X mm2/m)

Table 1.9 Units of thermal conductivity

1 (calIT/m h °C)

1 J/s 1000 W 1010 erg/s 1.34 hp (horsepower) 737.5 ft lbf/s 0.9484 Btu/s

100 lX cm 39.37 lX in 10−4 X cm 10−6 X m

0.01163 (W/cm K) 8.063627 (Btu in/ft2 h °F)

Table 1.10 Physical constants Constant

Symbol

Value

Gas constant Boltzmann constant Avogadro’s number Planck constant Stefan-Boltzmann constant Elementary charge Mass of the electron Faraday constant Vacuum permittivity Vacuum permeability Speed of light (vacuum) Speed of sound (in dry air)

R k NA h r e−

8.31441 J K−1 mol−1 1.3807  10−23 J K−1 atom−1 6.02204  1023 atom mol−1 6.62618  10−34 J s atom−1 5.6703  10−8 W m−2 K−4 1.602  10−19 C 9.1094  10−31 kg 96,485 C mol−1 8.854  10−12 C2 J−1 m−1 4 p 10−7 H m−1 = N A−2 = C2 J−1 m−1 299,792,458 m s−1 331.45 m s−1 (at 273 K)

F e0 l0 c0

4

1

Table 1.11 Multiple and submultiple of basic units

Conversion of Units

Prefix

Symbol

Multiplication factor of basic unit

Tera– Giga– Mega– Kilo– Hecto– Deca– Deci– Centi– Milli– Micro– Nano– Pico– Femto– Atto–

T G M k h da d c m µ n p f a

1012 109 106 103 102 10 10−1 10−2 10−3 10−6 10−9 10−12 10−15 10−18

Exercise 1.1 Calculate the number of moles in 1000 g of Fe, Al, Cu, SiO2, Al2O3, and CH2=CH2 (ethylene) 1 mol Fe ¼ 17:91 mol Fe 55:85 g Fe

ð1:4Þ

1 mol Al ¼ 37:03 mol Al 27 g Al

ð1:5Þ

1 mol Cu ¼ 15:75 mol Cu 63:5 g Cu

ð1:6Þ

1 mol SiO2 ¼ 16:64 mol SiO2 60:1 g SiO2

ð1:7Þ

1000 g Al2 O3 

1 mol Al2 O3 ¼ 9:8 mol Al2 O3 102 g Al2 O3

ð1:8Þ

1000 g C2 H4 

1 mol C2 H4 ¼ 35:71 mol C2 H4 28 g C2 H4

ð1:9Þ

1000 g Fe 

1000 g Al  1000 g Cu  1000 g SiO2 

Exercise 1.2 Calculate the number of moles in 100 g of the alloy 75% Cu–25% Ni (in weight). Make the same calculations in the alloy Fe–0.08% C (in weight). 100 g  0:75 

1 mol Cu 1 mol Ni þ 100 g  0:25  ¼ 1:61 mol 63:5 g Cu 58:7 g Ni

ð1:10Þ

1.1 Conversion of Units

100 g  0:9992 

5

1 mol Fe 1 mol C þ 100 g  0:0008  ¼ 1:79 mol 55:85 g Fe 12 g C

ð1:11Þ

Exercise 1.3 Calculate the relation between the mole and the weight fraction in one alloy (substance formed by two or more metals or metalloids). If we call A and B the metals, PA and PB their atomic weights, wA and wB their weight percentages, and xA and xB their mole fractions, then we have: xA ¼

number of moles of A wA =PA ¼ wA wB number of moles of A + number of moles of B PA þ PB

ð1:12Þ

xB ¼

number of moles of B wB =PB ¼ number of moles of A + number of moles of B wPAA þ wPBB

ð1:13Þ

xA þ xB ¼ 1

ð1:14Þ

In fact:

And inversely: weight ðin 100 gÞ of moles of A weight ðin!100 gÞ of moles of A þ weight ðin 100 gÞ of moles of B xA =PA ¼ xA  100 xB PA þ PB

wA ¼

ð1:15Þ weight ðin 100 gÞ of moles of B weight ðin!100 gÞ of moles of A þ weight ðin 100 gÞ of moles of B xB =PB ¼ xA  100 xB PA þ PB

wB ¼

ð1:16Þ In fact: wA þ wB ¼ 1

ð1:17Þ

These relations between mole fractions and weight percentages are usually employed in the study of binary systems (diagrams). They also can be used in ternary systems (comprised of three elements or chemical compounds), quaternary systems (comprised of four elements or chemical compounds), etc. Examples: Mole fractions of the alloys Fe–4.3% C and Al–4.5% Cu: Fe–4.3% C

6

1

xFe ¼

Conversion of Units

0:957=55:85 ¼ 0:827 ð82:7%Þ 0:957 0:043 55:85 þ 12

0:043=12 xC ¼ 0:957 0:043 ¼ 0:173 ð17:3%Þ 55:85 þ 12

ð1:18Þ

ð1:19Þ

Al–4.5% Cu xAl ¼

0:955=26:98 ¼ 0:98 ð98%Þ 0:955 0:045 26:98 þ 63:55

ð1:20Þ

0:045=63:55 ¼ 0:02 ð2%Þ 0:955 0:045 26:98 þ 63:55

ð1:21Þ

xCu ¼

Exercise 1.4 Determine the charge (in coulombs) equivalent to the displacement of a mole of electrons (number of Faraday). 1:6  1019 coulombs 6:023  1023 e ¼ 96;368 coulombs ð’96,500 coulombsÞ  1e equiv g ð1:22Þ Exercise 1.5 Demonstrate that 1 eV/atom is equivalent to approximately 23,000 cal/mol. Data: 1 C 1 V = 1 J. 1:6  1019 coulombs  1V 6:023  1023 atoms J cal  ¼ 96,368  atom mol mol 4:18 J ¼ 23,055 cal=molð’23,000 cal=molÞ

ð1:23Þ

The bonding energy between atoms and molecules in the condensed matter is usually expressed in electronvolt (eV).

2

Introduction to Structural Materials: Naturals, Metals, Ceramics, Polymers, and Composites

Abstract

This introductory chapter provides a definition of structural and functional materials. The relation between different properties is established. The first part of this chapter discusses the differences between structural and functional properties for natural materials, metals, ceramics, polymers, and composites; the second part is devoted to discussing natural materials (properties and applications).

2.1

Structural Materials: Naturals, Metals, Ceramics, Polymers and Composites

From the numerous definitions of the word, here we use the terms “material” or “materials” to designate a substance or substances—natural or artificial—used in the manufacture or production of parts. Materials are a source of scientific and technical interest for the engineer because to the engineer falls the task of selecting the most suitable material for a certain application. Sometimes, say, in the case of non-natural materials, the engineer is the responsible for producing the actual materials for different applications. The word “structural” is used to designate materials with fundamentally bulk mechanic (stiffness, elasticity, mechanical resistance, toughness, etc.) and surface (behaviour under friction, wear, oxidation, corrosion, etc.) properties useful for construction in the widest scope of the word. The term “functional” is reserved for materials that are required as a consequence of their electric or electronic (conductivity and resistivity, superconductivity, semiconductivity), magnetic, thermal, ionic, radioactive, biocompatible, etc. properties. Materials oriented to perform a nonstructural function are abundant in number, although the production volume is much lower than that of the structural materials; the unitary price of functional materials is usually greater as well. © Springer Nature Switzerland AG 2019 J. A. Pero-Sanz Elorz et al., Structural Materials, https://doi.org/10.1007/978-3-030-26161-0_2

7

8

2 Introduction to Structural Materials: Naturals…

Structural materials are usually grouped in terms of naturals, metals, ceramics, polymers, and composites. Exercise 2.1 How do we usually measure the mechanical strength of structural materials? What units are used? We use the term “structural materials” to define all materials subjected to loads in service. These loads generate tension stresses, compression stresses, bending stresses, torsion stresses, fatigue, creep, etc. These mechanical properties, which could be designated as “bulk properties,” must sometimes be complemented with other properties that we might call “surface properties” like resistance to corrosion and oxidation, resistance to wear-abrasion, antifriction properties, chemisorption, etc. The most used test to measure the mechanical properties of a structural material (metals, ceramics, polymers and composites) is the tensile test. This test is based on subjecting a standardized specimen to a uniaxial test where load-displacements are continuously measured until the breaking point (Fig. 2.1). From this test, three fundamental parameters are obtained: the yield strength (ry), the ultimate tensile strength (ru), and the elongation (A). Another complementary parameter is the section located below the stress–strain curve, which represents the energy absorbed by the specimen before breaking (i.e., its toughness). Other mechanical tests are also used to test properties like hardness (resistance to penetration), toughness (impact bending test), fatigue (resistance to cyclic alternative loads), and creep (deformation under constant load and high temperature). The unit of resistance adopted by the International System of Units (SI) is the MPa: 1 MPa = 106 N/m2 = 1 N/mm2. Other units of engineering resistance are the kgf/mm2; the hectobar = 100 bar = 1 kgf/mm2; the long ton/inch2 = 1016 kg/(25.4 mm)2 = 1.57 kg/mm2 = 15.4 MPa; the ksi or 1000 lb/inch2 = 453 kg/(25.4 mm)2 = 0.7 kg/mm2 = 6.9 MPa (the last two units are still in use in the United States and the United Kingdom and the others in Europe).

Fig. 2.1 Stress–strain curve

2.1 Structural Materials: Naturals, Metals…

9

A ranking based on the resistance (ry), which is merely indicative, is as follows: – Low-resistance materials: 1 < ry < 250 MPa – Medium-resistance materials: 250 < ry < 750 MPa – High-resistance materials: 750 < ry < 1500 MPa – Extremely high-resistance materials: ry > 1500 MPa In this book, we will try to stablish a relationship between structure (number of phases, composition, size, distribution) and properties. More specifically, the relation between properties [physical, chemical, or mechanical (Table 2.1)] on one hand and chemical composition, microstructure, and defects of manufacture on the other hand will be stablished. The parameter microstructure (grain size in a polycrystalline aggregate) usually has relevance in structural metallic materials. In a general way, we can say that the physical properties (elastic constants, density, melting point, etc.) are independent of the microstructure (grain size and atomic defects) except for the electric resistivity. On the contrary, the chemical and mechanical properties, especially the mechanical properties, are very sensitive to the nature and state (microstructure) of the material, e.g., cold strength and high-temperature toughness, thermal shock resistance, fatigue resistance, etc.

Table 2.1 Relation between properties in the case of structural materials Nature ’ Composition Structure: structure (grain size and microstructural defects) Condition: manufacture defects Properties = F (Nature, State, Condition) Properties of the materials Physical Mechanical Elastic constants (E, G) Yield strength Ultimate tensile strength Melting point (TM ) Density Specific heat Expansion Electrical conductivity Thermal conductivity Magnetic and optic Radioactive Biocompatible Acoustic Properties not dependent on the structure

Chemical Adsorption, Wettability (Surface tension) Corrosion resistance

Elongation, ductility (reduction of area) Toughness, fracture toughness Oxidation resistance Hardness (wear and abrasion resistances) Creep resistance Fatigue resistance Properties dependent on the structure

10

2 Introduction to Structural Materials: Naturals…

We might say that the main objective of the first part of this book is to clarify the reason for the relation that exists between composition, structure, and properties in structural metallic materials and how the relation could be modified. Moreover, we strive to clarify what we can or cannot expect from metallic materials, including clarification of the advantages and disadvantages in relation to competing materials (structural ceramic, polymeric, and composite materials). The second part of this book is dedicated to the selection of materials based on compounded properties; weight and cost will decide in a rational way the domains of applications possessed by each material and possible available alternatives.

2.2

Structural Materials: Natural Materials

In nature exist such materials as natural stone, the structural properties of which are still currently interesting for stone blocks, floor tiles, tiles, etc. Due to their genesis, natural stones, i.e., rocks, can be classified as igneous—either plutonic or volcanic (with the latter being the most durable); sedimentary (which is less stable than igneous), and metamorphic. Igneous rocks come from the solidification of a molten magma. Some of them are called “plutonic” when the consolidation takes place in the deep zones of the Earth’s crust, generally in large masses without stratification as happens with granite (complex aggregate of feldspar, quartz, and mica), syenite, diorite, and olivine rock. Volcanic rock, such as basalt, porphyry, trachyte, and diabase, solidify in the exterior of the Earth’s crust or in zones close to it. Both types of igneous rocks are classified into three large groups depending on the quantity of quartz and feldspathoids and on the lack of one or the other. The average composition of igneous rocks expressed in oxides is, according to Clark and Washington, 59.1% SiO2, 15.3% Al2O3, 3.8% FeO, 3.1% Fe2O3, 3.5% MgO, 5.1% CaO, 3.8% Na2O, 3.1% K2O, and 3.2% others. These constituents do not vary arbitrarily; they are dependent. Exercise 2.2 Binary equilibrium diagrams and nonequilibrium solidification, as obtained in the laboratory, are tools used by geologists to understand the solidification or consolidation of magmas and plutons. Although the phenomena of the solidification of magmas is very complex (involving temperatures, pressures, and times, sometimes expressed in thousands of years), it was possible to reach some conclusions about segregation problems of different types (gravitational segregation, segregation by zones, microsegregation) that have influenced—and in many cases favoured—the formation of mineral substances, which can be useful for anthropological purposes. For instance, enclosed it is possible to find the diagram of the feldspars plagioclases, which are of great importance in petrography because, except for quartz, the silicates that form igneous rocks constitute all of them

2.2 Structural Materials: Natural Materials

11

“mixed” crystals: To a certain extent they are alloyed, thus replacing some elements because they have equivalent atomic radius and crystalline structures (isomorphism). The isomorphic series most typical and frequent in hypogenic rocks is the feldspars plagioclases, which are formed by solid solutions with variable quantities of two minerals—albite (NaAlSi3O8 [Ab]) and anorthite (CaAl2Si2O8 [An])— which sometimes contain small quantities of orthose (KAlSi3O8). The ends of the series, Ab and An, are completely soluble in both liquid and solid states (vid. Figure 2.2), and their mixes (petrous alloys) are obtained in a similar way as in the case of more simple chemical elements or compounds. In the lower part of the diagram, the corresponding ranges of composition are collected (in wt%) for the feldspars, which can appear in the case of a differentiated solidification (magmatic differentiation). For instance, we can demonstrate that an andesine (’48% Ab) with partition coefficient of k = 0.35 can give crystals or internal zones of bytownite surrounded by zones of more acidic other constituents like labradorite, andesine, oligoclase, and—even from the most external zones—Ab (Fig. 2.3). The required conditions for this differentiation would be stirred magma, long periods of time, and suitable temperatures. In the same way, if the magma would be in state of rest (lack of volatile phases) and in extended periods of time, a gravitational segregation of the most basic materials to the bottom of the magma (bytownite) could take place. Or, if the residual liquid could escape through cracks along the walls of the cavity, it would comprise, preferentially, the most acid minerals (Ab type). There are many other mineral systems whose equilibrium diagrams (more or less simplified) are analogous to that of the plagioclases, for instance, that of the olivine (ultrabasic minerals), which are total mixtures of forsterite [Mg2SiO4 (melting point 1890 °C)] and fayalite [Fe2SiO4 (melting point 1205 °C)]. An example of the mentioned diagram is shown in Exercise 2.3.

Fig. 2.2 Ab–An diagram (adapted from Roso de Luna 1947)

12

2 Introduction to Structural Materials: Naturals…

Fig. 2.3 Image of a crystal of Ab–An with microsegregation (directional solidification) during its freezing (layers). The inner zone is rich in An and the external zone in Ab

For the mineral 52% An–48% Ab, calculate: – The interval of solidification under equilibrium conditions. From the diagram of Fig. 2.2: TL ¼ 1462  C

ð2:1Þ

TS ¼ 1285  C

ð2:2Þ

DT ¼ TL  TS ¼ ð1462  1285Þ  C ¼ 177  C

ð2:3Þ

The interval of solidification under nonequilibrium conditions From the diagram of Fig. 2.2: TL ¼ 1462  C

ð2:4Þ

TS ¼ 1140  C

ð2:5Þ

DT ¼ TL  TS ¼ ð1462  1140Þ  C ¼ 322  C

ð2:6Þ

We calculate the partition coefficient (k): At 1462 °C: k¼

cS 16 ¼ 0:33 ¼ cL 48

ð2:7Þ

2.2 Structural Materials: Natural Materials

13

At 1420 °C: k¼

cS 25 ¼ 0:41 ¼ cL 61

ð2:8Þ

k¼

cS 36 ¼ 0:47 ¼ cL 76

ð2:9Þ

At 1367 °C:

The average partition coefficient is:
k ¼ 0:33 þ 0:41 þ 0:47 ¼ 0:40 3

ð2:10Þ

For the first temperature T1 ¼ 1420  C, cL ¼ 61% Ab, then (the partition coefficient is k ¼ 0:40): 1 1  10:40  1k c0 48 fL ¼ ¼ ¼ 0:6707ð67:07%Þ ! fS ¼ 0:3293ð32:93%Þ 61 cL

ð2:11Þ

Average composition of the solid:
cS ¼

c0  fL  cL 48  0:6707  61 ¼ 21:52% Ab ¼ 0:3293 fS

ð2:12Þ

For the second temperature, T2 ¼ 1367  C, cL ¼ 76% Ab, then: fL ¼

1 1  10:40  1k c0 48 ¼ ¼ 0:4649 ð46:49%Þ ! fS ¼ 0:5351 ð53:51%Þ ð2:13Þ 76 cL

Average composition of the solid:
cS ¼

c0  fL  cL 48  0:4649  76 ¼ 23:67% Ab ¼ 0:5351 fS

ð2:14Þ

For the third temperature, T3 ¼ 1285  C, cL ¼ 87% Ab, then: fL ¼

1 1  10:40  1k c0 48 ¼ ¼ 0:3711 ð37:11%Þ ! fS ¼ 0:6289 ð62:89%Þ ð2:15Þ 87 cL

Average composition of the solid: cS ¼

c0  fL  cL 48  0:3711  87 ¼ 24:99% Ab ¼ 0:6289 fS

ð2:16Þ

14

2 Introduction to Structural Materials: Naturals…

For the fourth temperature, T4 ¼ 1219  C, cL ¼ 94% Ab, then: fL ¼

1 1  10:40  1k c0 48 ¼ ¼ 0:3262ð32:62%Þ ! fS ¼ 0:6737ð67:37%Þ 94 cL

ð2:17Þ

Average composition of the solid:
cS ¼

c0  fL  cL 48  0:3262  94 ¼ 25:73% Ab ¼ 0:6737 fS

ð2:18Þ

And finally, at the Ab melting temperature, TM ¼ 1140  C, cL ¼ 100% Ab, then: 1   1  1k c0 48 10:40 fL ¼ ¼ ¼ 0:2942ð29:42%Þ ! fS ¼ 0:7057 ð70:57%Þ ð2:19Þ 100 cL

Average composition of the solid: cS ¼

c0  fL  cL 48  0:2942  100 ¼ 26:33% Ab ¼ 0:7057 fS

ð2:20Þ

Can the periphery of the crystals be formed by nearly pure Ab? The external zones would be comprised of crystals of pure Ab as under nonequilibrium conditions Ab is the last liquid to solidify (’25% Ab). cS 17 ¼ 0:35 ’ cL 48 

fL ð at 1140 CÞ ¼



48 100

1 10:35

ð2:21Þ

’ 0:32

ð2:22Þ

Note: For the study of the nonequilibrium solidification, we recommend reading Chap. 5 (nonequilibrium solidification and chemical heterogeneities) in the book Solidification and Solid-State Transformations of Metals and Alloys (Pero-Sanz et al. 2017). Exercise 2.3 In the process of the solidification of magma, the first materials that will solidify (from the most refractory) are the series of olivine, (Mg,Fe)2SiO4, whose diagram of total solubility is shown (Fig. 2.4). In fact, both forsterite Mg2SiO4, and fayalite Fe2SiO4 crystalize in the orthorhombic system with similar lattice parameters: The atomic radius of the cations Fe2+ and Mg2+ (rFe2 þ ¼ 0:87; rMg2 þ ¼ 0:78) differ by < 15% and they have the same valence. Calculate the

2.2 Structural Materials: Natural Materials

15

Fig. 2.4 Forsterite–fayalite binary diagram

number of molecules that will have forsterite and fayalite per unit cell and the trend that the nonequilibrium solidification system will have. Data: ˚ b ¼ 10:195 A; ˚ c ¼ 5:990 A; ˚ q ¼ 3:222 g/cm3 Forsterite : a ¼ 4:756 A; ˚ b ¼ 10:483 A; ˚ c ¼ 6:095 A; ˚ q ¼ 4:392 g/cm3 Fayalite : a ¼ 4:822 A; Forsterite:   x molecules  140:69ðg/molÞ lattice 3:222g/cm3 ¼ !x¼4 ˚ 3 Þ  ð108cmÞ3 6:023  1023  ð4:756  10:195  5:990Þ  ðA A ð2:23Þ Fayalite:   x molecules  203:6ðg/molÞ lattice 4:392g/cm3 ¼ !x¼4 ˚ 3 Þ  ð108cmÞ3 6:023  1023  ð4:822  10:483  6:095Þ  ðA A

ð2:24Þ That is to say, half that in the cubic lattice of the cristobalite, because each cation of Fe2+ or Mg2+, are linked to two anions of O2−. Regarding nonequilibrium solidification (k, partition coefficient ’0.2/0.5 = 0.4), it will be prone to slow directional solidification (micro-segregation) with internal zones being more refractories (rich in Mg) and external zones having greater Fe content. Moreover, there will be more tendency towards segregation through gravity because the most refractory crystals of forsterite will tend to float in the liquid magma if the stirring rate and the volatile matter content allow it (see Chap. 5 in Pero-Sanz et al. 2017).

16

2 Introduction to Structural Materials: Naturals…

Sedimentary rocks, also known as exogenous, have their origin in the interaction of erosion external agents with the Earth’s crust. These agents erode rocks due to mechanical disaggregation and chemical decomposition, transport the products or detritus and sediment them. They comprise stratums whose constituents have been grouped as a consequence of the mechanical sedimentation. For instance, this happens with clays, sand and gravel, sandstones, limes, etc. Later they can petrify due to the chemical action (sulphides as gypsum, limestone, dolostone, chlorides, marls, etc.) or due to the bacterial action either above the organic animal matter that they would have acquired during the transportation (v. gr. Crete, Tripoli) or above vegetal matter (lignite, anthracite, bituminous coal). Metamorphic rocks are the result of physicochemical changes due to the action of factors such as temperature, pressure, and chemically active fluids above both igneous (gneiss, serpentine, etc.) and sedimentary rocks (slate, marble). Ceramics, cements, and petrous conglomerates are manufactured using natural rocks. An example of these conglomerates or composites, where cement is used as adhesive, of noticeable interest in engineering is the concrete. Woods are another structural natural composite. Wood is one of the oldest and still the most widely used structural material. They are comprised—approximately 40–50%—of aligned fibres of cellulose that are agglomerated with lignin (20–30%) and hemicellulose (15–20%). They also contain resins and oils, and their composition varies depending on the species of wood. The mechanical properties of the woods depend on the arrangement of the fibre with respect to the applied load. Woods are used as construction materials despite several obvious drawbacks, such as their combustibility, volumetric instability, biological degradation, low mechanical resistance, anisotropy of properties, etc.; however, different current treatments have improved the behaviour of this material facing to these problems. The world production, which is significant, is difficult to quantify because world wood-production statistics include its utilization also as combustible or as raw material for the paper industry (1750 Mt in 2012, Earth Policy Institute from U.N. Food and Agriculture Organization, ForesSTAT). Industrially, woods can be classified into the following four categories: (1) soft timber (poplar, chestnut, willow, etc.), which is used in packaging, paper pulp, planking, simple woodwork, etc.; (2) hard timber (oak, beech, holm or evergreen oak, guayacan, walnut, etc.), formerly used in railway sleepers and some of them are used in mine shoring, tiles, doors, furniture industry, etc.; (3) fine timber (mahogany, ebony, olive, apple tree, etc.), used to make furniture, musical instruments, chess sets, etc.; and (4) resinous timber (firs, cypress, pine, etc.), used when resistance to humidity is required. In the Una Norma Española standards (from the Asociación Española de Certificación y Normalización), it is possible to check the naming of the main Spanish woods or those acclimatized to Spain, the scientific name, the official commercial name, other commercial names, and some information about the quality and applications of each timber (similar standards can be found in other countries).

2.2 Structural Materials: Natural Materials

17

Apart from stone and wood, many other natural materials—such as canes, fibres (wool, flax, etc.), corks, and straws—have been used since the origin of the humanity in their natural form. Progressive technological advancement has led humans to use, from ancient times, many materials processed from natural sources.

References Pero-Sanz, J. P., Quintana, M. J., & Verdeja, L. F. (2017). Solidification and solid-state transformations of metals and alloys. Boston: Elsevier. Roso de Luna, I. (1947). El Equilibrio Físico-químico en Metalogenia, Litogenia y Metalurgia. Ed. Servicio de Publicaciones de la ETSIMM, Madrid, Spain.

3

Structural Materials: Metals

Abstract

This chapter is dedicated to metallic structural materials. We review the main properties and applications of metallic materials and provide different exercises that enable the reader’s understanding of the utilization of metals in each application. For that reason, the Ellingham’s diagrams are described at the beginning of the chapter.

3.1

Structural Metallic Materials

Regarding all the elements of the periodic table, the main difference between metals and nonmetals is that metallic atoms have few electrons in the external orbitals, i.e., electrons that metals easily cede to form a complete and stable structure of external orbit. “Electropositivity” refers to the ease with which a metal loses electrons and then the atom is positively charged. Table 3.1 lists the metals in decreasing order of electropositivity (in increasing order of electronegativity), which is calculated as the Pauling’s criterion considering the energy required to gain an electron. The value of the most electronegative element, fluoride 3.98, can be used as reference. Note the small electropositive character of the metals subsequent to tin in this series. The values corresponding to the nonmetals are as follows: boron (2.04), phosphorus (2.19), hydrogen (2.20), carbon (2.55), sulphur (2.58), iodine (2.66), bromine (2.96), nitrogen (3.04), chlorine (3.16), oxygen (3.44), and fluorine (3.98). When a great difference of electronegativity exists between two metals, the chemical bond between them to form a molecule is of ionic type. Other types of chemical bonding include the covalent bond, the coordinate covalent bond, the polar covalent bond, etc.

© Springer Nature Switzerland AG 2019 J. A. Pero-Sanz Elorz et al., Structural Materials, https://doi.org/10.1007/978-3-030-26161-0_3

19

20 Table 3.1 Increasing electronegativity of metals as calculated according to Pauling’s criterion (eV)

3 Cs Rb K Ba Na Sr Li Ca La Ce Pr Nd Sn Gd Dy Y Ho Er Lu

0.79 0.82 0.82 0.89 093 0.95 0.98 1 1.10 1.12 1.13 1.14 1.17 1.20 1.22 1.22 1.23 1.24 1.27

Pu Mg Zr Sc Np U Ti Be Mn Al V Zn Cr Cd In Ga Fe Co Cu

Structural Materials: Metals

1.28 1.31 1.33 1.36 1.36 1.38 1.54 1.57 1.55 1.61 1.63 1.65 1.66 1.69 1.78 1.81 1.83 1.88 1.90

Si Ni Ag Sn Hg Ge Bi Sb Mo As Pd Ir Rh Pt Pb W Au Se

1.90 1.91 1.93 1.96 2 2.01 2.02 2.05 2.16 2.18 2.20 2.20 2.28 2.28 2.33 2.36 2.54 2.55

In the case of a metallic solid, each atom cedes peripheral electrons to an electron cloud. This electron cloud, because it is electronegative, sticks the positively charged atoms: It is not reasonable to talk about molecules. This bond between atoms of metallic crystal, or grain, is peculiar of metals and it is known by the name of metallic bond. The anonymous character of the atom bond between them is specific to this kind of bond: Each atom is not firmly bonded to the other in contrast to what occurs in other types of chemical bonds. The mobility of the electrons in the electronic cloud formed by the valence electrons is also characteristic of the metallic bond: This easy displacement of the electrons explains the thermal and electric conductivity of the metals as well. Another property of metals that also pertains to its atomic bond is the ability of deforming before breaking, which is in contrast to that observed with nonmetallic materials such as ceramics, glasses, ionic solids, etc. In all of them, the bond is broken, and the breakage occurs at the point where the energy needed for elastic separation between atoms of a molecule is reached. In metals, on the contrary, when the point of elastic deformation is passed, breakage of the bond does not take place: Some atomic planes can slide above others. This is macroscopically defined as “plastic deformation.” The metallic bond also explains the aptitude of metals to form types of alloys known as “substitutional solid solutions.” As a consequence of the anonymous character of this bond, some atoms can be replaced by atoms of another metal (solute) in the crystalline lattice of a metal (solvent) (see Figs. 4.9 and 4.10).

3.1 Structural Metallic Materials

21

Some atoms can be substituted with other atoms almost completely if: the atomic diameter of the solute, their electropositivity, their valence (the number of electrons that can be ceded to the electronic cloud), and their crystalline lattice are analogous to that of the solvent metal. In this way, for instance, copper and nickel can form alloys (it does not matter the percentage of one and another element): Examples include Monel alloy (75% Ni and 25% Cu), Constantan alloy (45% Ni and 55% Cu), and cupronickel alloy (75% Cu and 25% Ni), etc.

3.2

Ellingham’s Diagram ΔG0(T) for the Formation of Metallic Oxides

Metallic elements are usually found in nature as oxides and in general as oxides and sulphides. Ellingham’s diagram for oxides (see Fig. 3.1) represents the standard Gibbs free energy for the formation of metallic oxides as a function of temperature, referred to in mols of oxygen. In the general case: x 2 2
M þ O2 Mx Oy þ DG0 ðT Þ y y

ð3:1Þ

In the case of Ellingham’s diagram for sulphides (Fig. 3.2): x 2 2
M þ S2 Mx Sy þ DG0 ðT Þ y y

ð3:2Þ

In general, we emphasize the low resistance of metals to oxidation and corrosion (Ellingham) and their “relative” easiness to be extracted from their minerals (oxides and sulphides) through pyrometallurgy, hydrometallurgy, and electrometallurgy operations. They are abundant and relatively cheap. They occupy considerable positions regarding the production and consumption of structural materials (especially if they should resist loads in service). Exercise 3.1 Why can the equations of Ellingham’s formation energies be assimilated to straight lines? DG0 ðT Þ ¼ DH 0 ð298Þ  T
DS0 ð298Þ

ð3:3Þ

ΔH0(298) and ΔS0(298) are the enthalpies and entropies related to the reaction, i.e., the difference of the corresponding values of the products and the reagents: At 298 K: DG0 ðT Þ ¼ DH 0 ð298Þ  298
DS0 ð298Þ

ð3:4Þ

22

3

Structural Materials: Metals

Fig. 3.1 Variation of free energy for the formation of oxides as a function of temperature. F = melting point; E = boiling point

At temperature T: DG0 ðT Þ ¼ DH 0 ðT Þ  T
DS0 ðT Þ

ð3:5Þ

3.2 Ellingham’s Diagram ΔG0(T) for the Formation of Metallic Oxides

23

Fig. 3.2 Ellingham’s diagram for sulphides

However:

ZT

DH 0 ðT Þ ¼ DH 0 ð298Þ þ

Dcp
dT

ð3:6Þ

Dcp
dT T

ð3:7Þ

298

0

0

ZT

DS ðT Þ ¼ DS ð298Þ þ 298

Dcp is the difference of the specific heat of the products and the reagents. From where: DG0 ðT Þ ¼ DH 0 ð298Þ  298
DS0 ð298Þ þ

ZT

ZT Dcp
dT  T

298

298

Dcp
dT ð3:8Þ T

However, the difference between the last two terms is irrelevant with respect to the difference between the first two. In fact, considering that: Dcp ’ 2 cal=mol
K, for DT ¼ 2000 K: ZT Dcp
dT ’ 2 cal=mol
K
2000 K ’ 4 kcal=mol

ð3:9Þ

298

ZT T 298

    Dcp
dT T cal 2000 K
ln ’ T
Dcp
ln ’ 2000 K
2 298 mol
K 298 K T ’ 8 kcal=mol

ð3:10Þ

24

3

Structural Materials: Metals

The difference between them, 4 kcal=mol, is of approximately the experimental error of the thermodynamic data considered (1 kcal=mol); consequently, the following approximation can be made: DG0 ðT Þ ¼ DH 0 ð298Þ  298
DS0 ð298Þ

ð3:11Þ

which is the equation of a straight line with slope DS0 ð298Þ, of approximately value, in the case of the oxides, of 0:04 kJ=mol (i.e., very small). A second approximation can be made at room temperature (298 K) as a consequence of the low value of DS0 ð298Þ with respect to the H 0 ð298Þ or reaction heat: DG0 ð298Þ ¼ DH 0 ð298Þ  298
DS0 ð298Þ ’ DH 0 ð298Þ ’ DH 0 ð273Þ

ð3:12Þ

The free enthalpy and the heat of reaction are almost equal at room temperature. Now, we explain the origin of the Ellingham’s straight line for the formation of Al2O3 (alumina corundum): 4=3 Al(s) þ O2 ðg) 2=3Al2 O3 ðs)

ð3:13Þ

From Ellingham’s diagram for the formation of alumina (see Fig. 3.1): DG0 ð273 KÞ ’ 253 kcal=mol O2

ð3:14Þ

DG0 ð1273 KÞ ’ 207 kcal=mol O2

ð3:15Þ

207 ¼ DH 0 ð273Þ  1273
DS0 ð273Þ

ð3:16Þ

253 ¼ DH 0 ð273Þ  273
DS0 ð273Þ

ð3:17Þ

We subtract Eq. (3.17) from Eq. (3.16) to obtain: 46 ¼ 1000
DS0 ð273Þ

ð3:18Þ

DS0 ð273Þ ¼ 46=1000 ¼ 4:6
102 kcal=mol
K

ð3:19Þ

DH 0 ð273Þ ¼ 253  273
4:6
102 ’ 266 kcal=mol O2

ð3:20Þ

and:

Then, Ellingham’s straight line for alumina would be: DG0 ðT Þ ¼ 266 kcal=mol þ 4:6
102
T ðKÞ
kcal=mol
K

ð3:21Þ

3.2 Ellingham’s Diagram ΔG0(T) for the Formation of Metallic Oxides

25

We suggest repeating the calculations of Ellingham’s straight line for the formation of other oxides, for example, TiO2 , Cr2 O3 , NiO, Fe2 O3 . The following equations would be obtained: TiO2 ! DG0 ðT Þ ¼ 219 þ 0:044
T ðkcal=mol O2 Þ

ð3:22Þ

Cr2 O3 ! DG0 ðTÞ ¼ 178 þ 0:047
T ðkcal/mol O2 Þ

ð3:23Þ

NiO ! DG0 ðT Þ ¼ 115 þ 0:047
T ðkcal=mol O2 Þ

ð3:24Þ

Fe2 O3 ! DG0 ðT Þ ¼ 131 þ 0:044
T ðkcal=mol O2 Þ

ð3:25Þ

That is to say, Ellingham’s lines, which are standard Gibbs free energy for the formation of metallic oxides as a function of temperature, are straight lines, almost parallel between them (x-axis), of small slope ’ 0:04  0:05: The range of temperatures is more “closed” than that of the DG0 ðT Þ (negatives ordinates to subtract them). Exercise 3.2 Calculate the partial pressure of O2 required, in equilibrium, for the formation of an oxide. DGðT Þ ¼ DG0 ðT Þ þ R
T
ln k ¼ 0, where k is the equilibrium constant of the reaction, the quotient of the activities of the products to their stoichiometric coefficients power over the reagents to their stoichiometric coefficients power. If metal and oxide are found in the pure state: k ¼ 1=pO2 . That is to say: In Joules:  DG0 ðT Þ ¼  1:987

   cal J
4:184
T
ln k mol
K cal

¼ ð1:987Þ
ð4:184Þ
T
ln 10
log k

ð3:26Þ

¼ 19:14
T
log pO2 In calories:  DG0 ðT Þ ¼  1:987

 cal
T
ln k ¼ ð1:987Þ
ln 10
T
log k mol
K ¼ 4:575
T
log pO2

ð3:27Þ

Continuing with alumina: DG0 ðT Þ ¼ 266 þ 0:046
T ðkcal=mol O2 Þ

ð3:28Þ

26

3

Structural Materials: Metals

At T = 298 K (room temperature): DG0 ð298Þ ¼ 252:3 kcal=mol O2 log pO2 ¼ 

252,300 cal=mol ¼ 185 ! pO2 ¼ 10185 atm ! 0 cal
298 K 4:575 mol
K

ð3:29Þ ð3:30Þ

At T = 1000 K (727 °C): Aluminium would be in liquid state at this temperature, and we have disregarded the change of slope in Ellingham’s straight line due to the latent heat of fusion of aluminium when passing from the solid state to the liquid one: log pO2 ¼

ð266,000 þ 0:046
1000Þcal=mol ¼ 58 cal
1000 K 4:575 mol
K pO2 ’ 1058 atm ! 0

ð3:31Þ ð3:32Þ

The conclusions of the calculation have industrial relevance: Aluminium “naturally” tends (under a pressure of oxygen that might be disregarded) to be coated with a skin of alumina, which implies: – In as solid state, aluminium has excellent anticorrosive qualities (it is impermeable to oxygen) due to its isolating and adherent nature, except for saline aqueous mediums that dissolve the alumina skin (nanometric). – In a liquid state, skins of alumina (solid) can render difficult—or even impoverish —the castability of aluminium: The liquid phase should be skimmed off or filter before casting. The same adherent and protective behaviour is observed in the case of TiO2 with respect to titanium. From the positioning of the standard free enthalpies for the formation of metallic oxides in Ellingham’s diagram, we can deduce the following: – Some oxides are easily dissociable by simply increasing the temperature, DG0 ð273Þ [ 20 kcal=mol O2 (see Fig. 3.1). – Some oxides are easily reducible using solids and gases, such as C, CO, and H2, 20\DG0 ð273Þ\100 kcal/mol O2 (see Fig. 3.1). – Some oxides are easily reducible using gases with possible situations of equilibrium, 100\DG0 ð273Þ\200 kcal=mol O2 (see Fig. 3.1). – Some oxides are not reducible using CO and H2, DG0 ð273Þ\200 kcal=mol O2 , These are also known as “refractory oxides.” They are only reducible using electrometallurgical and metallothermic processes (see Fig. 3.1). The first two categories correspond to oxides that are easily dissociable; the last two correspond to oxides that are dissociable with difficultly.

3.2 Ellingham’s Diagram ΔG0(T) for the Formation of Metallic Oxides

27

Exercise 3.3 Calculate the ideal energy cost (in GJ/t) for the direct reduction of hematite (Fe2O3) using C. From Ellingham’s diagram, DG0298 ðFe2 O3 Þ ¼ 120 kcal=mol O2 ; DG0298 ðCOÞ ¼ 65 kcal=mol O2 . Atomic weight of Fe, 56 g=mol Fe. Reduction reaction: Fe2 O3 þ 3C 2Fe þ 3CO þ DG0298

ð3:33Þ

Formation of Fe2 O3 : 4=3 Fe þ O2 2=3 Fe2 O3  120 kcal=mol O2

ð3:34Þ

Formation of CO: 2C þ O2 2CO  65 kcal=mol O2

ð3:35Þ

Equation 3:33 ¼ 3=2½Eq: 3:35  Eq: 3:34 ¼ 3=2½65 þ 120 ¼ 82:5 kcal=mol ð3:36Þ Then the ideal energy consumption will be: 82,500 cal=mol 106 g 4:18 J 1 GJ GJ


’ 3:1 2
56 g Fe=mol t cal 109 J t Fe

ð3:37Þ

The real energy cost to obtain 1 t of steel (Fe–C alloy) from the blast furnace and the subsequent conversion with oxygen in the converter (integrated ironmaking and steelmaking) is approximately 17 GJ=t. Exercise 3.4 Calculate the ideal energy cost (in GJ/t) for obtaining metallic aluminium. From Ellingham’s diagram, DG0298 ðAl2 O3 Þ ’ 250 kcal=mol O2 ; the atomic weight of Al is 27 g=mol Fe. 4 2 Al þ O2 Al2 O3  250 kcal=mol O2 3 3

ð3:38Þ

That is to say, for the dissociation of alumina using heat, a heat supply DH 0298 ’ DG0298 of 250 kcal=mol O2 would be necessary. Then the ideal energy supply per ton of Al is equal to: 250

kcal 103 cal 1 mol O2 1 mol Al 4:18 J 106 g 1 GJ GJ



9 ’ 29 ð3:39Þ

mol O2 1kcal 4=3 mol Al 27 g cal t 10 J t Al

This value is nearly 10 times greater than the ideal energy consumption to obtain 1 t of steel.

28

3

Structural Materials: Metals

The real energy consumption for obtaining of 1 t of aluminium is 270 GJ=t, almost 10 times greater than the theoretical value: Obtaining aluminium requires greater quantities of electric power (electrometallurgy). For that reason, countries that produce aluminium have inexpensive electric power. Exercise 3.5 European and American criteria for the definition of G. In chemical reactions, it is only interesting to know the heat and energy exchanges (DQ; DW; DH; DU) between a system and the environment with independence of the frame of reference chosen for the elements that form it. We can use one of the two following criterions: – European: The enthalpy H at 273 K (0 °C) of any chemical element is 0. Example: C þ 3Fe ¼ Fe3 C þ DH ð273 KÞ DH ¼ H 0Fe3 C ð273 KÞ  3H 0Fe ð273 KÞ  H 0C ð273 KÞ ¼ H 0Fe3 C ð273 KÞ

ð3:40Þ ð3:41Þ

The heat of reaction coincides with the enthalpy of formation of cementite. – American: At any temperature, the free energy of formation for any chemical element is 0. Example: DGðT Þ ¼ G0Fe3 C ðT Þ  3G0Fe ðT Þ  G0C ðT Þ ¼ G0Fe3 C ðT Þ

ð3:42Þ

The only thermodynamic function whose absolute value is known is the entropy: ZT S¼ 0

dQrev ¼ T

ZT 0

ce
dT ¼ Sð T Þ T

ð3:43Þ

as Sð0 K) ¼ 0: For example, calculate the standard free energy of the Boudouard equilibrium at 1500 K under the following conditions: 1. Knowing the standard Gibbs free energy of the chemical compounds involved in the reaction. 2. Knowing the enthalpy and entropy of the compound at room temperature (298 K).

3.2 Ellingham’s Diagram ΔG0(T) for the Formation of Metallic Oxides

29

1. C þ CO2 ¼ 2CO þ DG0 ðT Þ European rule DG0 ð1500 KÞ ¼ 2
G0CO  G0C  G0CO2 ¼ 2
445,066 J þ 27,389 J þ 771,871 J ¼ 90,872 J ð3:44Þ 2. Applying Ellingham: DG0 ð1500 KÞ ¼ DH 0 ð298 KÞ  1500
DS0 ð298 KÞ

ð3:45Þ

DH 0 ð298 KÞ ¼ 2H 0CO  H 0C  H 0CO2 ¼ 2
110,528 J þ 0 þ 393,521 J ¼ 172,475 J

ð3:46Þ DS0 ð298 KÞ ¼ 2
S0CO  S0C  S0CO2 ¼ 2
197.648 J=K  213.794 J=K  5740 J=K ¼ 175.762 J=K ð3:47Þ T
DS0 ð1500 KÞ ¼ 175.762 J=K
1500 K ¼ 263,643 J

ð3:48Þ

DG0 ð1500 KÞ ¼ 172,475 J  263,643 J ¼ 91,178 J

ð3:49Þ

The difference between both results is 306 J (0.07 kcal): Ellingham’s rule is compatible with the European rule, i.e., it is in the margin of error of the experimental results. Exercise 3.6 Are the metals easily oxidized at high or at low temperature? Justify the answer. Considering a purely thermodynamic criterion (see Ellingham’s diagram [Fig. 3.1]), the stability of oxides decreases with increasing temperature (i.e., the straight lines have positive slope). They could even decompose as a consequence of heat, at high temperatures, as happens with copper and noble metals (Ag, Au), but the kinetics (growing rate) of the oxide layer exponentially depends on the temperature, i.e., on the rate at which O2 can pass through the layer of oxide until being in contact with the metal (or vice versa). The mass transfer (in this case O2 ) is thermally activated and depends on the temperature according to a law of Arrhenius type. It is expressed by the diffusion coefficient D ¼ D0
expðQ=RTÞ: D0 is the frequency factor of approximately 1cm2 =s ¼ 104 m2 =s; Q is the activation energy for diffusion; R is the universal constant of gases; and T is the absolute temperature. Moreover, it is possible to demonstrate that the thickness of the oxide layer formed is approximately equal to the square root of the diffusion coefficient multiplied by pffiffiffiffiffiffiffiffiffi the time x ’ D
t; then the increase in weight for the formation of the oxide layer

30

3

Structural Materials: Metals

pffiffiffiffi will be proportional to Dm / kp
t1=2 , being kp / D, which strongly depends (exponentially) on the temperature. The law for the parabolic growth of oxidation ðDmÞ2 ¼ kp
t.

3.3

The Five Main Metals

Only five metals are produced in quantities of several millions of tons every year: iron, aluminium, copper, zinc, and lead. From this total production, that of steel (iron with carbon content 1500 million tons. This is equivalent to >90% of the annual production of metallic materials and exceeds almost 9 times the annual production, in weight, of organic polymeric materials. After concrete, steel is the most used structural material. Regarding cast irons—the second-most used metallic material worldwide, after steel, the relation between the production of cast-iron parts and the production of steel is approximately 4–5% (Pero-Sanz et al. 2018a, b). In developing countries with low steel consumption, the above-mentioned relation is even lower as a consequence of the weak infrastructure for manufacturing metallic products and machine equipment. Ranking of the production and consumption of metals for those after steel are: aluminium (60 million t in 2017), copper (19.7 million t in 2017), zinc (13.2 million t in 2017), and lead (4.7 million t in 2017). These five metals and other three—eight in total: iron, aluminium, copper, zinc, lead, nickel, magnesium, and tin—represent 99% of the annual production (in weight) of metallic materials worldwide. With respect to the economic importance of this production (tons multiplied by the price/t) (Fig. 3.3) the ranking of importance would be iron, aluminium, copper, gold, nickel, tin, zinc, lead, and others. On another note, it is possible to indicate that the tendency in the production and consumption of the main structural metals has changed significantly in the last 20 years given the important economic growth of China. It is possible to indicate

Fig. 3.3 Economic importance (%) of the worldwide production of metals

3.3 The Five Main Metals

31

that the production and consumption of certain metals has grown faster than in the case of other metals.

3.3.1 Iron Iron is the fourth most abundant element in the Earth’s crust: only oxygen, silicon, and aluminium are more abundant. It is estimated that the proportion in weight of iron is 50,000 ppm in the Earth’s crust (Table 3.7) and that Earth’s nucleus, with 3000-km radius, is formed mainly of iron and nickel. This could mean that iron content in the planet could be as great as 40%. Iron ores, i.e., minerals from which it is extracted, include hematite (Fe2O3), magnetite (Fe3O4), limonite (FeO(OH)), and siderite (FeCO3). It is also available in pyrites as double sulphides that are more or less complexes. The main iron ore reserves (crude ore, iron content) with high iron content are located in (2017) the following countries: Australia (crude ore: 50,000 Mt [29.4%]; iron content: 24,000 Mt [28.9%]), Russia (crude ore: 25,000 Mt [14.7%]; iron content: 14,000 Mt [16.9%]), Brazil (crude ore: 23,000 Mt [13.5%]; iron content: 12,000 Mt [14.5%]), China (crude ore: 21,000 Mt [12.4%]; iron content: 7200 Mt [8.7%)], and India (crude ore: 8100 Mt [4.8%]; iron content: 5200 Mt [6.3%]) for total global reserves of 170,000 Mt (crude ore) and 83,000 Mt (iron content). The name “iron” for this mineral—except for the ultrapure element—does not seem adequate from a structural point of view. Although the quantity of carbon available was small—even at approximately 50 ppm—we might call it “steel”; and for carbon contents greater up to the saturation limit for austenite, the product is actually then a cast iron (Pero-Sanz et al. 2018a, b). It seems convenient, before ending this section, to make a short reference to the growing interest of the use of mini-mills. Integrated steel mills, which starting from iron ore obtain pig iron in blast furnaces and later steel in converters, are currently suffering serious difficulties in developed countries arising from the periodic requirement of large capital investments to renew the expensive but obsolete equipment. An important part of the difficulties also comes from the low cost of the workforce in third-world countries. However, the growing interest in using mini-mills mainly derives from energetic considerations as a consequence of the large amount of scrap existing worldwide. Iron scrap is usually a low-cost raw material compared with iron ore. In this way, obtaining a ton of steel from scrap in a mini-mill, in an electric furnace, means a consumption of 5:6
106 kJ. Obtaining the same quantity of steel, in a blast furnace with the subsequent conversion with oxygen, requires 16:96
106 kJ. Nevertheless, the growing interest in the mini-mills will not render integrated steel mills obsolete. Without blast furnaces, which start the sequence (iron ore ! pig iron ! steel ! products ! scrap), mini-mills— according to American forecasts—would close in 10 years.

32

3

Structural Materials: Metals

Exercise 3.7 The constant of the parabolic law for the oxidation of iron is 0:37
expð138,000=R
TÞ g2 =cm4 s. Calculate, at a temperature of 0:7
T M ðT M ¼ 1811 KÞ, the thickness of the oxidized layer after 1 day. Data: Atomic weight of oxygen ¼ 16 g=mol; atomic weight of iron ¼ 56 g=mol; density of iron ¼ 7:9 g=cm3 . The constant of the parabolic law for the oxidation of iron is: 0 kp ¼ 0:37
exp@

1 138,000 J=mol A 2 g =cm4 s ¼ 7:6
107 g2 =cm4 s J 8:31 mol
K
1268 K ð3:50Þ

The mass involved in oxidation is: g2 24 h 3600 s
¼ 0:066 g2 =cm4 ! Dm
4 cm
s 1 day 1 h ¼ 0:26 gO=cm2

ðDmÞ2 ¼ 7:6
107

NO ¼

0:26 0:26 56 NA ¼ NFe ; ðDmÞFe ¼ NA
¼ 0:91 g Fe=cm2 16 16 NA

ð3:51Þ ð3:52Þ

Crust thickness formed daily: 0:91

g g ¼ 7:9 3
Dxðcm) ! Dx ’ 0:12 cm ð1:2 mm) 2 cm cm

ð3:53Þ

3.3.2 Aluminium “Aluminium,” a centenarian metal, was obtained for first time in the laboratory in 1859. The beginning of its production was in France in 1889 using the Hérault method, the fundamentals of which are still currently valid (electrolysis of alumina [Al2O3] solved in molten cryolite [F6AlNa3]). Aluminium is the second most consumed metal worldwide before copper and lead (and other metals, such as gold, which falls below hydrogen in the electromotive series), whose first utilization goes back to prehistory. After oxygen and silicon, aluminium is the most abundant element in the Earth’s crust (98% of it is composed of the previously mentioned elements as well as iron, calcium, sodium, potassium, magnesium, and titanium [in decreasing order of importance]). Approximately two thirds of high-quality bauxite (Al2O3
2H2O) reserves are located in five countries (2017): Australia (83 Mt (27.7%), China (68 Mt [22.7%]), Guinea (45 Mt [15%]), Brazil (36 Mt [12%]), and India (27 Mt [9%]) for a total worldwide production of bauxite of 300 Mt.

3.3 The Five Main Metals

33

Table 3.2 Standard oxidation potentials (adapted from Revie and Uhlig 2008) Reaction

Standard oxidation potentials (E) in V at 25 °C

Reaction

Standard oxidation potentials (E) in V at 25 °C

Li = Li+ + e− K = K+ + e− Ca = Ca2+ + 2e− Na = Na+ + e− Mg = Mg2+ + 2e− Be = Be2+ + 2e− U = U3+ + 3e− Hf = Hf4+ + 4e− Al = Al3+ + 3e− Ti = Ti2+ + 2e− Zr = Zr4+ + 4e− Mn = Mn2+ + 2e− Nb = Nb3+ + 3e− Zn = Zn2+ + 2e− Cr = Cr3+ + 3e− Ga = Ga3+ + 3e− Fe = Fe2+ + 2e−

3.05 2.93 2.87 2.71 2.37 1.85 1.80 1.70 1.66 1.63 1.53 1.18 (approx. 1.1) 0.763 0.74 0.53 0.440

Cd = Cd2+ + 2e− In = In3+ + 3e− Ti = Ti+ + e− Co = Co2+ + 2e− Ni = Ni2+ + 2e− Mo = Mo3+ + 3e− Sn = Sn2+ + 2e− Pb = Pb2+ + 2e− H2 = 2H+ + 2e− Cu = Cu2+ + 2e− 2Hg = Hg22+ + e− Ag = Ag+ + e− Pd = Pd2+ + 2e− Hg = Hg2+ + 2e− Pt = Pt2+ + 2e− Au = Au3+ + 3e−

0.403 0.342 0.336 0.277 0.250 (approx. 0.2) 0.136 0.126 0.000 −0.337 –0.789 −0.800 −0.987 −0.854 (approx. −1.2) −1.50

In 2017, 130 Mt of alumina were produced worldwide. They were allocated in China (72.3 Mt [55.6%]), Australia (20.6 Mt [15.8%]), Brazil (11 Mt, 8[.5%]), and India (6.17 Mt [4.7%]), which were the main producers. In 2017, 60 Mt of aluminium were produced worldwide. The main producers were China (32.6 Mt [54.3%]), Russia (3.6 Mt [6%]), Canada (3.21 Mt [5.4%]), India (3.2 Mt [5.3%]), and United Arab Emirates (2.6 Mt [4.3%]). Aluminium has a low density, 2.69 g/cm3, almost a third that of iron. Alloys of this metal are called “light alloys.” Aluminium occupies a high position in the series of standard oxidation potentials (Table 3.2) and consequently has great tendency to pass to the ionic form and it is anodic for metals located below aluminium such as iron, copper, etc. The equilibrium potentials are shown in Table 3.2 for a concentration equal to activity 1. Considering the low solubility of some metals’ salts, the classification of standard oxidation potentials has limited utility when we want to know if a metal is anodic with respect to another metal. Aluminium is, in any case, a metal with good properties of corrosion resistance in nonalkaline aqueous mediums. The poor image that aluminium had with respect to this corrosion in the first nautical applications was a consequence of usually being riveted with copper.

34

3

Structural Materials: Metals

The resistance of aluminium in the face of intergranular corrosion, or the face of pit-corrosion, decreases when microstructurally it has precipitates that act as anodic or cathodic zones as, for instance, in the alloys Al–Cu and Al–Zn. Aluminium is easily oxidized in the presence of oxygen with heat release. For that reason, aluminium is used as reductant reagent in some processes and as heat source in others (for instance, welding of rails for railways by way of aluminothermic process). In Fig. 3.1, it is possible to see the energy for the formation of aluminium oxide, Al2O3, and the practical impossibility of obtaining aluminium by reduction of alumina with another element. Exercise 3.8 Aluminothermic reduction of chromium. Ellingham’s diagram describe the possibility of obtaining metallic chromium from chromite Cr2O3 using aluminium powder as reductant reagent. Data: Latent heat of fusion (L) of chromium: 4200 cal/mol; specific heat (ce) of chromium 0.15 cal/°C g; in the case of alumina (Al2O3): L = 26,000 cal/mol; ce = 0.27 cal/°C g. Reactions: 2 4 2 4 Cr2 O3 ðsÞ þ Al(s) ¼ Al2 O3 ðl) þ Cr(l) 3 3 3 3

ð3:54Þ

4 Cr(s) þ O2 ðgÞ ¼ Cr2 O3 ðlÞ ! DG0 ð273 KÞ ’ DH 0 ð273 KÞ ¼ 167 kcal/mol 3 ð3:55Þ 4 2 Al(s) + O2 ðg) = Al2 O3 ðl) ! DG0 ð273 K) =  253 kcal/mol 3 3

ð3:56Þ

DH 0 ðEq. 2:54Þ ¼ DH 0 ðEq. 2:56Þ  DH 0 ðEq. 2:55Þ ¼ 86 kcal

ð3:57Þ

This heat is used to heat the products of the reaction and also to melt them (Kirchoff cycle). Latent heat and melting of Cr (TM ¼ 1875  C): cal 52 g Cr 4 cal 4 
moles Cr þ 4200
moles Cr 
1875 C
mol Cr 3 mol Cr 3 g Cr
C ¼ ð19500 þ 5600Þ cal ¼ 25:1 kcal

0:15

ð3:58Þ Latent heat and “possible” melting of alumina (in fact it melts at 2020 °C): cal 102 g Al2 O3 2 cal 2 
molesAl2 O3 þ 26,000
moles Al2 O3 
1875 C
mol Al2 O3 3 mol Al2 O3 3 g Al2 O3
C ¼ ð34400 þ 17300Þcal ¼ 51:7 kcal 0:27

ð3:59Þ

3.3 The Five Main Metals

35

Heat used in the reaction: 76.8 kcal Available heat: 86.0 kcal The difference is small. If external energy is not supplied, the chromium aluminothermy will not be self-maintained. Exercise 3.9 Study the possibility of reducing chromium oxide (Cr2 O3 ) with carbon at a temperature of 1000 °C. What would be the CO partial pressure required under equilibrium conditions? Explain the consequences (use the principle of Le Châtelier-Braun). According to Ellingham’s diagram: DG0 ðT ÞðCr2 O3 Þ ¼ 178 þ 0:042
T ðKÞ kcal=mol O2

ð3:60Þ

DG0 ðT ÞðCOÞ ¼ 55  0:041
T ðKÞ kcal=mol O2

ð3:61Þ

Reduction of chromium oxide with carbon: Cr2 O3 ðsÞ þ 3CðsÞ ! 3COðgÞ þ 2CrðsÞ

ð3:62Þ

2 4=3 Cr þ O2 ! Cr2 O3 3

ð3:63Þ

2C þ O2 ! 2CO

ð3:64Þ

Calculation of free energy: 3 3 DG0 ðEq: 3:62Þ ¼ ½Eq: 3:64  Eq: 3:63 ¼ ½123  0:083T  kcal=mol Cr2 O3 2 2 ð3:65Þ DG0 ðAÞð1273 KÞ ¼ 26 kcal=mol Cr2 O3

ð3:66Þ

  DG ¼ DG0 þ R
T
ln P3CO ¼ 0

ð3:67Þ

In equilibrium:

26; 000 cal=mol ¼ 1:987

cal
1273K
lnP3CO ! P(CO) ¼ 0:032 atm mol
K ð3:68Þ

36

3

Structural Materials: Metals

Conclusions – Displacing the reaction of Eq. (3.62) to the right would require working at CO pressures reinforced concrete ’ aluminium Cu 2014 T6 > mild steel). However, if we consider cost, reinforced concrete is the best option (reinforced concrete > mild steel > aluminium Cu 2014 T6 > CFRP). Exercise 8.6 Four-point bending test: Consider a beam of length L ¼ 3  a and width b fixed and thickness t variable (Fig. 8.12). We want to select a material with sufficient stiffness and minimum cost from all those listed in Table 8.4. The equation of the elastic curve in the zone where the maximum deflection takes place (a\x  2  a with L ¼ 3  a) is: y¼

 Fa   3  L  x  3  x 2  a2 6EI

ð8:57Þ

The maximum deflection y ¼ dmax , is produced when x ¼ 3  a=2: dmax ¼

 Fa   3  3  a  3  a=2  3  ð3  a=2Þ2  a2 6EI

Table 8.3 Structural and economic efficiency (beam of thickness t variable)   Material E (GPa) q (g/cm3) cp (£/t) E1=3 =q E1=3 = cp  q Reinforced concrete 50 Mild steel 210 Aluminium Cu 2014 T6 72.4 CFRP 220

2.4 7.85 2.8 1.55

200 250 2050 20,000

1.5 0.8 1.5 3.9

(2.6) 7.5  10−3 (1) (4.9) 3.2  10−3 (2.3) (2.6) 7.3  10−4 (10.3) (1) 2  10−4 (37.5)

ð8:58Þ

224

8

Materials for Beams

Fig. 8.12 Beam subjected to four-point bending

Table 8.4 Materials to select for Exercise 8.6

E (GPa) q (g/cm3) cp (₤/t)

Material

Aluminium 1100 annealed (1) 69 Polypropylene (50) 1.4 Glass (66) 70 Grey cast iron (34) 145

dmax ¼

2.71 0.9 2.5 7.3

23  F  a3 24  E  I

1550 550 1000 125

ð8:59Þ

The second moment of the area of the cross-section about the central axis of the beam of thickness t variable is: I¼

b  t3 23  F  a3 23  F  a3 ! dmax ¼ ¼ 12 24  E  ðb  t3 =12Þ 2  E  b  t3

ð8:60Þ

 1=3 23 F  a3 23 F  a3   ! t  2 E  b  t3 2 E  b  dmax

ð8:61Þ

d  dmax ¼

The weight of the beam is: P¼q3abt

ð8:62Þ

Thus, the minimum weight is obtained as follows: P¼q3ab Then P # if E1=3 =q ". To minimize cost:

 1=3 23 F  a3 1  ¼ Constant  1=3 2 E  b  dmax E =q

ð8:63Þ

8.3 Comparison Between Materials for Beams and Plates

C ¼ P  cp ¼ constant 

1 E

1=3

=q

225

 cp ¼ constant 

E

1=3

1 =q  cp

ð8:64Þ

Similarly, C # if E1=3 =q  cp " : Considering the materials listed in Table 8.4, we obtain the results listed in Table 8.5. In the case of a beam of thickness t variable, the order of selection of the most suitable materials from all those listed in Table 8.4 to have the best structural efficiency (stiffness) is glass > aluminium 1100 annealed > polypropylene > grey cast iron. If we consider cost, the order is grey cast iron > polypropylene > glass > aluminium 1100 annealed. Exercise 8.7 We have a plate with length L and width b fixed and thickness t variable supported at its ends according to the width (Fig. 8.13). A force q is uniformly distributed in the opposite face to the supports. We want to select from the materials listed in Table 8.6 to (scenario 1) best maximize stiffness but simultaneously minimize both the weight and cost as well as (scenario 2) best maximize resistance to plastic deformation (yielding) but simultaneously minimize both the weight and cost. To maximize stiffness: First, we calculate the equation of the elastic curve: E  I  y00 ¼ M

ð8:65Þ

q  x q  x2 þ 2 2L

ð8:66Þ

q  x2 q  x3 þ þ C1 4 6L

ð8:67Þ

E  I  y00 ¼  We integrate for the first time: E  I  y0 ¼ 

Table 8.5 Structural and economic efficiency of the materials to be selected Materials

E (GPa) q (g/cm3) Cp (£/t) E1=3 =q

Aluminium 1100 annealed 69 Polypropylene 1.4 Glass 70 Grey cast iron 145

2.71 0.9 2.5 7.3

1550 550 1000 125

1.5 1.2 1.7 0.7

(1.1) (1.4) (1) (2.4)

E1=3 =q  cp 9.7 2.2 1.7 5.6

   

10−4 10−3 10−3 10−3

(5.8) (2.5) (3.3) (1)

226

8

Materials for Beams

Fig. 8.13 Plate with load uniformly distributed load ðq ¼ F=L  bÞ

Table 8.6 Materials to select for a plate with uniformly distributed load Material

E (GPa)

ry ðMPaÞ

q (g/cm3)

cp (£/t)

Reinforced concrete (76) Mild steel (37) Aluminium Cu 2014 T6 (6) CFRP (64)

50 210 72.4 220

40 280 414 670

2.4 7.85 2.8 1.55

200 250 2050 20,000

We calculate the constant of integration C 1 : x¼ EI0¼

L ! y0 ¼ 0 2

ð8:68Þ

q  ðL=2Þ2 q  ðL=2Þ3 q  L2 þ þ C1 ! C1 ¼ 4 6L 24

ð8:69Þ

q  x2 q  x3 q  L2 þ þ 4 6L 24

ð8:70Þ

q  x3 q  x4 q  L2 þ þ  x þ C2 12 24  L 24

ð8:71Þ

This way: E  I  y0 ¼  We integrate for a second time: EIy¼

We calculate the constant of integration C 2 : x¼0!y¼0 EIy¼

q  x3 q  x4 q  L2 þ þ  x þ C2 ! C2 ¼ 0 12 24  L 24

And the equation of the elastic curve is:

ð8:72Þ ð8:73Þ

8.3 Comparison Between Materials for Beams and Plates

y¼

227

 qx  3 L  2  L  x2 þ x3 24  E  I

ð8:74Þ

The maximum deflection ðy ¼ dmax Þ occurs when x ¼ L=2: dmax ¼

i q  ðL=2Þ h 3 5  q  L4  L  2  L  ðL=2Þ2 þ ðL=2Þ3 ¼ 24  E  I 384  E  I

ð8:75Þ

The second moment of the area of the cross-section about the central axis in the beam of thickness t variable is: I¼

b  t3 5  q  L4 ! dmax ¼ 12 384  E  ðb  t3 =12Þ

d  dmax ¼

ð8:76Þ

 1=3 5  q  L4 5  q  L4 ! t  32  E  b  t3 32  E  b  dmax

ð8:77Þ

The weight of the beam is: P¼qLbt

ð8:78Þ

Thus, the minimum weight is obtained as follows: 

5  q  L4 P¼qLb 32  E  b  dmax

1=3 ¼ Constant 

1 E

1=3

=q

ð8:79Þ

Then P # if E1=3 =q ". To minimize cost: C ¼ P  cp ¼ constant 

1 E1=3 =q

 cp ¼ constant 

1 E1=3 =q

 cp

ð8:80Þ

Similarly, C # if E1=3 =q  cp " : Considering the materials listed in Table 8.6, we obtain the results listed in Table 8.7.

Table 8.7 Structural and economic efficiency of materials listed in Table 8.6 Material

E (GPa) q (g/cm3) cp (£/t) E1=3 =q

Reinforced concrete 50 Mild steel 210 Aluminium Cu 2014 T6 72.4 CFRP 220

2.4 7.85 2.8 1.55

200 250 2050 20,000

1.6 0.8 1.5 3.9

(2.4) (4.9) (2.6) (1)

E1=3 =q  cp 8  10−3 (1) 3.2  10−3 (2.5) 7.3  10−4 (11.0) 2.0  10−4 (40.0)

228

8

Materials for Beams

In the plate uniformly loaded with thickness t variable, the ranking of the most suitable materials, from those proposed in Table 8.6, regarding structural efficiency (stiffness) is CFRP > reinforced concrete > aluminium Cu 2014 T6 > mild steel. If we consider cost, the ranking is reinforced concrete > mild steel > aluminium Cu 2014 T6 > CFRP. To maximize resistance to plastic deformation: We use Navier’s law: r¼

My Mmax  y ! rmax ¼ I I

ð8:81Þ

The maximum moment takes place in x ¼ L=2: M max ¼ 

q  ðL=2Þ q  ðL=2Þ2 q  L þ ¼ 2 8 2L

ð8:82Þ

The value of y is: y¼

t 2

ð8:83Þ

Moreover, the following relation must be verified: rmax \ry

ð8:84Þ

To avoid plastic deformation of the beam:   ðq  L=8Þ  ðt=2Þ 3  q  L 1=2  r ! t  y 4  b  ry ðb  t3 =12Þ

ð8:85Þ

The weight of the beam is: P¼qLtb

ð8:86Þ

Thus, the minimum weight is obtained as follows:  P¼qLb

3qL 4  b  ry

1=2

¼ Constant 

1 ry

1=2 =q

ð8:87Þ

Then, P # if ry 1=2 =q " : To minimize cost: C ¼ P  cp ¼ constant 

1 ry

1=2 =q

 cp ¼ constant 

1 ry

1=2 =q

 cp

ð8:88Þ

8.3 Comparison Between Materials for Beams and Plates

229

Similarly, C # if ry 1=2 =q  cp " : Considering the materials listed in Table 8.6, we obtain the results listed in Table 8.8. In the case of a plate uniformly loaded with thickness t variable, the ranking of materials from those listed in Table 8.6 regarding resistance to plastic deformation (yielding) is CFRP > aluminium Cu 2014 T6 > reinforced concrete > mild steel. If we consider cost, the ranking is reinforced concrete > mild steel > aluminium Cu 2014 T6 > CFRP. Exercise 8.8 We consider a cantilever beam, with length L and width b fixed and thickness t subjected to a force F at its end (Fig. 8.14). We want to select the material (Table 8.9) with the greatest stiffness and minimum weight and cost. At the same time, we want to maximize the resistance to plastic deformation and minimize weight and cost. To maximize stiffness: To determine the material that could maximize stiffness, first we calculate the equation of the elastic curve (Fig. 8.15): E  I  y00 ¼ M

ð8:89Þ

M ¼ F  ðL  x Þ

ð8:90Þ

E  I  y00 ¼ F  ðL  xÞ

ð8:91Þ

where the moment, M, is:

This way:

We integrate for the first time: E  I  y0 ¼ 

F  ðL  x Þ2 þ C1 2

ð8:92Þ

We calculate the constant of integration C 1 at the fixed end:

Table 8.8 Structural and economic efficiency of materials listed in Table 8.6 (considering resistance to plastic deformation) Material

ry ðMPaÞ q (g/cm3) cp (£/t) ry 1=2 =q

Reinforced concrete 40 Mild steel 280 Aluminium Cu 2014 T6 414 CFRP 670

2.4 7.85 2.8 1.55

200 250 2050 20,000

2.6 (6.4) 2.1 (8.0) 7.3 (2.3) 16.7 (1)

ry 1=2 =q  cp 0.013 (1) 8.4  10−3 (1.6) 3.6  10−3 (3.6) 8.4  10−4 (15.5)

230

8

Materials for Beams

Fig. 8.14 Cantilever beam with force at its end

Table 8.9 Properties of materials to be selected for Exercise 8.8 Material

E (GPa)

ry ðMPaÞ

q (g/cm3)

cp (£/t)

Zinc CP rolled (29) Glass (66) PVC (53) 18/8 stainless steel (45)

97 70 2.6 208

120 50 48 230

7.18 2.5 1.4 7.94

1200 1000 500 1150

Fig. 8.15 Diagram of bending moments

8.3 Comparison Between Materials for Beams and Plates

x ¼ 0 ! y0 ¼ 0 ! E  I  0 ¼ 

231

F  ð L  0Þ 2 F  L2 þ C1 ! C1 ¼ 2 2

ð8:93Þ

F  ð L  x Þ 2 F  L2 þ 2 2

ð8:94Þ

Thus: E  I  y0 ¼  We integrate for a second time: EIy¼

F  ð L  x Þ 3 F  L2 þ  x þ C2 6 2

ð8:95Þ

We calculate the constant of integration C 2 at the fixed end: x¼0!y¼0!EI0¼

F  ð L  0Þ 3 F  L 2 F  L3 þ  0 þ C2 ! C2 ¼  6 2 6 ð8:96Þ

This way: EIy¼

F  ðL  xÞ3 F  L2 F  L3 þ x 6 2 6

ð8:97Þ

Finally, the equation of the elastic curve is: y¼

F  x2  ð3  L  x Þ 6EI

ð8:98Þ

The maximum deflection ðy ¼ dmax Þ is produced when x ¼ L: dmax ¼

F  L2 F  L3  ð3  L  L Þ ¼ 6EI 3EI

ð8:99Þ

The second moment of the area of the cross-section about the central axis in the beam of thickness t variable is: I¼

b  t3 F  L3 4  F  L3 ! dmax ¼ ¼ 3 12 3  E  ðb  t =12Þ E  b  t3

d  dmax

 1=3 4  F  L3 4  F  L3 ¼ ! t E  b  t3 E  b  dmax

The weight of the beam is:

ð8:100Þ

ð8:101Þ

232

8

Materials for Beams

P¼qLbt

ð8:102Þ

Thus, the minimum weight is obtained as follows:  P¼qLb

4  F  L3 E  b  dmax

1=3 ¼ Constant 

1 E1=3 =q

ð8:103Þ

1 =q  cp

ð8:104Þ

Then, P # if E1=3 =q " : To minimize cost: C ¼ P  cp ¼ constant 

1 E

1=3

=q

 cp ¼ constant 

E

1=3

Similarly, C # if E1=3 =q  cp " : Considering the materials listed in Table 8.9, we obtain the results listed in Table 8.10. In the case of the cantilever beam with thickness t variable, the ranking of materials, from those listed in Table 8.9, as a result of structural efficiency (stiffness), is glass > PVC > 18/8 stainless steel > zinc CP rolled. If we consider cost, the ranking is PVC > glass > 18/8 stainless steel >zinc CP rolled. To maximize resistance to plastic deformation (in the beam of side t variable): We use Navier’s law: r¼

My Mmax  y ! rmax ¼ I I

ð8:105Þ

The maximum moment is produced in the free end of the cantilever beam: Mmax ¼ F  L

ð8:106Þ

The value of y is: y¼

t 2

ð8:107Þ

Table 8.10 Structural and economic efficiency of materials listed in Table 8.9 Material

E (GPa)

q (g/cm3)

cp (£/t)

E1=3 =q

E1=3 =q  cp

Zinc CP rolled Glass PVC 18/8 stainless steel

97 70 2.6 208

7.18 2.5 1.4 7.94

1200 1000 500 1150

0.6 1.7 1.0 0.8

5.0 1.7 2.0 7.0

(2.8) (1) (1.7) (2.1)

   

10−4 10−3 10−3 10−4

(4.0) (1.2) (1) (2.9)

8.3 Comparison Between Materials for Beams and Plates

233

The following equation must be verified to avoid plastic deformation: rmax \ry

ð8:108Þ

  F  L  ðt=2Þ 6  F  L 1=2  r ! t  y b  ry ðb  t3 =12Þ

ð8:109Þ

The weight of the beam is: P¼qLbt

ð8:110Þ

Thus, the minimum weight is obtained as follows: P¼qLb

  6  F  L 1=2 1 ¼ Constant  1=2 b  ry ry =q

ð8:111Þ

Then P # if ry 1=2 =q " : To minimize cost: C ¼ P  cp ¼ constant 

1 ry

1=2 =q

 cp ¼ constant 

1 ry

1=2 =q

 cp

ð8:112Þ

Similarly, C # if ry 1=2 =q  cp " : Considering the materials listed in Table 8.9, we obtain the results listed in Table 8.11. In the case of the cantilever beam with thickness t variable, the ranking of materials, from those proposed in Table 8.9, regarding resistance to plastic deformation is PVC > glass > 18/8 stainless steel > zinc CP rolled. If we consider cost, the ranking is PVC > glass > 18/8 stainless steel > zinc CP rolled. Exercise 8.9 We have a cantilever beam of length L and width b fixed and thickness t variable (Fig. 8.16). In the upper face, a load, F, is uniformly distributed. Select the most suitable material from all those listed in Table 8.12 to (1) maximize stiffness and minimize weight, (2) maximize resistance to plastic deformation and minimize weight, and (3) minimize the risks of unstable cracking and minimize weight. The minimum cost must be considered in all situations.

Table 8.11 Structural and economic efficiency of materials listed in Table 8.9 Material

ry ðMPaÞ

q (g/cm3)

cp (£/t)

ry 1=2 =q

ry 1=2 =q  cp

Zinc CP rolled Glass PVC 18/8 stainless steel

120 50 48 230

7.18 2.5 1.4 7.94

1200 1000 500 1150

1.5 2.8 5.0 1.9

1.3  10−3 (7.7) 2.8  10−3 (3.6) 0.01 (1) 1.7  10−3 (5.9)

(3.3) (1.8) (1) (2.6)

234

8

Materials for Beams

Fig. 8.16 Cantilever beam with uniformly distributed load

Table 8.12 Materials to select for the manufacture of a cantilever beam with uniformly distributed load   Material E (GPa) ry ðMPaÞ KIC MPa  m1=2 q (g/cm3) cp (£/t) Reinforced concrete (76) Mild steel (37) Aluminium Cu 2014 T6 (6) CFRP (64)

50 210 72.4 220

40 280 414 670

0.3 140 31 45

2.4 7.85 2.8 1.55

200 250 2050 20,000

To minimize stiffness: The material with the greatest stiffness would be that where the maximum deflection ðdmax Þ does not exceed the value provided by the equation of the elastic curve ðdelas Þ. The q is the force F divided by section A (where A ¼ L  b). The equation of the elastic curve in this case is: y¼

 q  x2   6  L2  4  L  x þ x2 24  E  I

ð8:113Þ

8.3 Comparison Between Materials for Beams and Plates

235

The maximum deflection ðy ¼ dmax Þ takes place when x ¼ L: dmax ¼

 q  L2  q  L4  6  L2  4  L  L þ L2 ¼ 24  E  I 8EI

ð8:114Þ

The second moment of the area of the cross-section about the central axis in a beam of thickness t variable is: I¼

b  t3 q  L4 3  q  L4 ! dmax ¼ ¼ 12 8  E  ðb  t3 =12Þ 2  E  b  t3

d  dmax ¼

 1=3 3  q  L4 3  q  L4 ! t  2  E  b  t3 2  E  b  dmax

ð8:115Þ

ð8:116Þ

The weight of the beam is: P¼qLbt

ð8:117Þ

Thus, the minimum weight is obtained as follows: 

3  q  L4 P¼qLb 2  E  b  dmax

1=3 ¼ constant 

1 E1=3 =q

ð8:118Þ

Then P # if E1=3 =q " : To minimize cost: C ¼ P  cp ¼ constant 

1 E

1=3

=q

 cp ¼ constant 

E

1=3

1 =q  cp

ð8:119Þ

Similarly, C # if E1=3 =q  cp " : Considering the materials listed in Table 8.12, we obtain the results listed in Table 8.13. In the case of the cantilever beam, the ranking of materials, from those listed in Table 8.12, regarding structural efficiency (stiffness) is CFRP > reinforced concrete ’ aluminium Cu 2014 T6 > mild steel. If we consider cost, the ranking is reinforced concrete > mild steel > aluminium Cu 2014 T6 > CFRP. Table 8.13 Structural and economic efficiency of materials listed in Table 8.12 (stiffness) Material

E (GPa) q (g/cm3) cp (£/t) E1=3 =q

Reinforced concrete 50 Mild steel 210 Aluminium Cu 2014 T6 72.4 CFRP 220

2.4 7.85 2.8 1.55

200 250 2050 20,000

1.5 0.8 1.5 3.9

(2.6) (4.9) (2.6) (1)

E1=3 =q  cp 7.5 3.2 7.3 2.0

   

10−3 10−3 10−4 10−4

(1) (2.3) (10.3) (37.5)

236

8

Materials for Beams

To maximize resistance to plastic deformation (yielding): With respect to resistance to plastic deformation, we should consider the equation of Navier: r¼

My I

ð8:120Þ

When x ¼ 0 ! M ¼ Mmax . The total moment, MT , is: ðL  x Þ2 2L

ð8:121Þ

ð L  0Þ 2 F  L ¼ 2 2L

ð8:122Þ

MT ¼ F  This way: Mmax ¼ F 

At half the thickness of the neutral fibre: y¼

t Mmax  y ðF  L=2Þ  ðt=2Þ 12  F  L  t 3  F  L ! rmax ¼ ¼ ¼ ¼ ð8:123Þ 2 I 4  b  t3 b  t2 ðb  t3 =12Þ

The weight of the beam is: P¼qLbt

ð8:124Þ

The following equation must be verified to avoid plastic deformation of the beam: ry [ rmax

  3FL 3  F  L 1=2 ¼ ! t[ b  t2 ry  b

ð8:125Þ

Thus, the minimum weight is obtained as follows: P¼qLb

  3  F  L 1=2 1 ¼ constant  1=2 ry  b ry =q

ð8:126Þ

1 1  cp ¼ constant  1=2 ry 1=2 =q ry =q  cp

ð8:127Þ

Thus, P # if ry 1=2 =q " : To minimize cost: C ¼ P  cp ¼ constant 

Similarly, C # if ry 1=2 =q  cp ".

8.3 Comparison Between Materials for Beams and Plates

237

Considering the materials listed in Table 8.12, we obtain the results listed in Table 8.14. In the case of the cantilever beam, the ranking of materials, from all those listed in Table 8.12, regarding structural efficiency (resistance to plastic deformation) is CFRP > aluminium Cu 2014 T6 > reinforced concrete > mild steel. If we consider cost, the ranking is reinforced concrete > mild steel > aluminium Cu 2014 T6 > CFRP. To minimize the risks of unstable cracking: The stress-intensity factor is calculated as follows: K IC  cte  rmax 

pffiffiffiffiffiffiffiffiffi pa

ð8:128Þ

The relation shown below must be verified to avoid unstable failure: rmax ¼

3FL  ry b  t2

ð8:129Þ

The weight of the beam is: P¼qLbt

ð8:130Þ

Considering Eqs. (8.128) and (8.129):   3  F  L pffiffiffiffiffiffiffiffiffi 3  F  L pffiffiffiffiffiffiffiffiffi 1=2  pa KIC  cte   p  a ! t  cte  b  t2 KIC  b

ð8:131Þ

Thus, the minimum weight is obtained as follows:   3  F  L pffiffiffiffiffiffiffiffiffi 1=2 1 P ¼ q  L  b  cte   pa ¼ constant  1=2 K IC  b K IC =q

ð8:132Þ

Then P # if K IC 1=2 =q " : To minimize cost: C ¼ P  cp ¼ constant 

1 K IC

1=2 =q

 cp ¼ constant 

1 K IC

1=2 =q

 cp

ð8:133Þ

Table 8.14 Structural and economic efficiency of materials listed in Table 8.12 (resistance to plastic deformation) Material

ry ðMPaÞ q (g/cm3) cp (£/t) ry 1=2 =q

Reinforced concrete 40 Mild steel 280 Aluminium Cu 2014 T6 414 CFRP 670

2.4 7.85 2.8 1.55

200 250 2050 20,000

2.6 (6.4) 2.1 (8.0) 7.3 (2.3) 16.7 (1)

ry 1=2 =q  cp 0.013 (1) 8.4  10−3 (1.5) 3.6  10−3 (3.6) 8.4  10−4 (15.5)

238

8

Materials for Beams

Similarly, C # is K IC 1=2 =q  cp " : Considering the materials listed in Table 8.12, we obtain the results listed in Table 8.15. In the case of the cantilever beam, the ranking of materials, from all those listed in Table 8.12, regarding structural efficiency (minimize risks of unstable cracking) is CFRP > aluminium Cu 2014 T6 > mild steel > reinforced concrete. Considering cost, the ranking is mild steel > reinforced concrete > aluminium Cu 2014 T6 > CFRP. As a consequence of significant differences existing among the materials with respect to their structural and economic efficiencies, we should consider attributive analysis to weight their effects. All this is said without considering other questions such as availability, energy costs, time in service, aesthetic aspects, and environmental issues. Exercise 8.10 Some remarks about the utilization of charts for the selection of materials in the following situations: (scenario a) maximize stiffness, (scenario b) minimize the risk of unstable cracking, and (scenario c) maximize resistance to plastic deformation. Scenario (a) Minimize weight for the same stiffness. Consider a beam of length L and width b fixed and thickness t (variable). The deflection, regardless if how the load is applied (four-point bending, uniformly distributed, three-point bending, etc.) depends on 1=E  I. A structural stiffness C ¼ E  I is imposed, whilst the second moment of the area of the cross-section about the central axis and the weight of the beam are given by the following equations, respectively: I¼

b  t3 12

ð8:134Þ

P¼qLbt

ð8:135Þ

Replacing in the equation of structural stiffness: C¼E

b  t3 E  P3 ¼ 12 12  L3  b2  q3

ð8:136Þ

Table 8.15 Structural and economic efficiency for materials listed in Table 8.12 (risk of unstable cracking) Material

KIC (MPa m1/2) q (g/cm3) cp (£/t) KIC 1=2 =q

Reinforced concrete Mild steel Aluminium Cu 2014 T6 CFRP

0.3 140 31

2.4 7.85 2.8

45

1.55

KIC 1=2 =q  cp

200 0.2 (21.5) 10−3 (6.0) 250 1.5 (2.9) 6.0  10−3 (1) 2050 2.0 (2.2) 9.8  10−4 (6.1) 20,000 4.3 (1)

2.2  10−4 (27.3)

8.3 Comparison Between Materials for Beams and Plates

239

This way: E 12  L3  b2  C ¼ q3 P3

ð8:137Þ

Taking logarithms:   log E ¼ 3  log q þ 3  log 12  L3  b2  C  3  log P

ð8:138Þ

The graphical representation of Eq. (8.138) is shown in Fig. 8.17. It is possible to deduce from Fig. 8.17 that: – Two materials located over the same straight line of slope 3 have the same weight. – When P #, the ordinate in the origin increases. – If C (required structural stiffness), L; and, b increase, the criterion E=q3 becomes stricter ðE "; q #Þ. Scenario (b): Minimizing weight (cost) with the same resistance to plastic deformation (yielding) regardless the profile of the rectangular beam (plate): L, length; b, width; and, t, thickness (variable). The restriction in this case is that the material must not enter into plasticity: rmax ¼

Mmax 6  Mmax ¼ \ry W b  t2

ð8:139Þ

Fig. 8.17 Representation log E  log q for a plate of length L and width b (constant) and thickness t (variable)

240

8

Materials for Beams

The weight of the beam is: P¼qLbt

ð8:140Þ

This way: t¼

P qLb

ð8:141Þ

Replacing in Eq. (8.139): rmax ¼

6  Mmax 6  Mmax  L2  b 2  q \ry  2 ¼ P2 P b  qLb

ð8:142Þ

This way: pffiffiffiffiffi ry 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼  6  Mmax  L2  b P q

ð8:143Þ

  log ry ¼ 2  log q þ log 6  Mmax  L2  b  log P2

ð8:144Þ

Taking logarithms:

The graphical representation of Eq. (8.144) is shown in Fig. 8.18.

Fig. 8.18 Representation log ry  log q for a plate of length L and width b (constant) and thickness t (variable)

8.3 Comparison Between Materials for Beams and Plates

241

We can conclude from Fig. 8.18 that: – Two materials over the same straight line of slope 2 have the same weight. – The weight decreases if the ordinate in the origin increases. – If M; L; b ", it would be necessary to use materials with greater ry 1=2 =q. Scenario (c): Selection considering weight without risks of catastrophic failure (M; a are known). Beam of length L and width b constant and thickness t (variable). We use the equations that provide the stress and the weight: r¼

M 6M ¼ W b  t2

P¼qLbt !t ¼

ð8:145Þ P qLb

ð8:146Þ

Replacing: r¼

6  M  L2  b  q2 P2

ð8:147Þ

The criterion of unstable failure is given by the following equation: KIC  r 

pffiffiffiffiffiffiffiffiffi pa

ð8:148Þ

Replacing: KIC 

6  M  L2  b  q2 pffiffiffiffiffiffiffiffiffi  pa P2

ð8:149Þ

This way: pffiffiffiffiffiffiffiffiffi 1 KIC 1=2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 6  M  L2  b  4 p  a  P q

ð8:150Þ

Taking logarithms:   1 log KIC ¼ 2  log q þ log 6  M  L2  b þ  logðp  aÞ  2  log P 2

ð8:151Þ

The graphical representation of Eq. (8.151) is shown in Fig. 8.19. We can conclude from Fig. 8.19 that: – Two materials over the same straight line of slope 2 have the same weight. – The weight decreases if the ordinate in the origin increases.

242

8

Materials for Beams

Fig. 8.19 Representation log KIC  log q for a plate of length L, width b (constant), and thickness t (variable) Fig. 8.20 Cantilever beams: Front view (upper panel), top view (lower panel)

– If M; L; b; a increase, it would be necessary to choose materials with greater pffiffiffiffiffiffiffi KIC =q. Exercise 8.11 Consider solids (beams) of the same resistance as those manufactured in such a way that the maximum stress is the same in all cross-sections. This way, the section modulus is proportional in each section to the bending moment. If the thickness (height) h of the section is constant, the elastic curve will be a circle (circumference). That is to say: q¼

EI E I h h ðI=h=2Þ E  h=2 ¼   ¼E  ¼ M M ðh=2Þ 2 2 M rf

ð8:152Þ

8.3 Comparison Between Materials for Beams and Plates

243

where E, h, and rf (working stress) are constants; 1=q (curvature) and q (radius of curvature) will also be constants; and the elastic curve will be a circle (circumference). We calculate the conditions where a rectangular section beam (variable b width and thickness constant, h) loaded in cantilever with force F would be a solid of the same resistance (applied to the manufacture of springs). From the Navier equation, we can deduce that: b (variable along the beam, and we can call it y) yðbÞ ¼

6  F  ðL  xÞ rt  h 0 2

ð8:153Þ

At the fixed end: b0 ¼

6FL rt  h 0 2

ð8:154Þ

In Fig. 8.20, the ordinates of the line AC measure the values of b in the respective cross-sections. This way, eliminating F from the previous equations: Y¼

b0  ð L  x Þ L

ð8:155Þ

Y0 ¼

b0 ð L  x Þ  L 2

ð8:156Þ

Or:

taking the semi-ordinates from one and the other side (Fig. 8.20b). To achieve that the right end section could resist the shearing stress, we give a width to the beam equal to: bmin ’

F r t  h0

ð8:157Þ

We demonstrate that the equation of the elastic curve is a circle:   d2 y 1 EI 2 ¼EI ¼ F  ðL  x Þ dx q

ð8:158Þ

However, by hypothesis: I ¼ I0 

b b0

ð8:159Þ

244

8

Materials for Beams

And: b L ðprofile of the line ACÞ ¼ b0 L  x

ð8:160Þ

The differential equation of the elastic curve is: E  I0 

1 ¼ F  L ¼ cte q

ð8:161Þ

The elastic curve is a circumference with centre (O, E  I0 =F  L), tangent to the x-axis in O. It is possible to demonstrate that the deflection at its end is 3/2 that in the same part if it had constant width and half volume. Now we consider the case of a rectangular section with constant width and variable height (thickness). By definition of beam of equal resistance, the following relation is verified: M 6Fx 6FL ¼ ¼ ¼ cte W b0  h20 b0  h20

ð8:162Þ

where h0 is the height at the fixed end. For that reason: h2 ¼ h20 

x L

ð8:163Þ

The thickness of the beam varies according to a parabolic-law equation. In the loaded end, the area of the straight section would be zero. We obtain this result because we disregard the influence of shearing stress in beams of the same bending resistance. In practical applications, the influence of shearing by modifying the area in the loaded end is considered with the purpose of having enough section to resist the shearing stress. As a result of having constant curvature, the deflection of the beam will be: Zl d¼ 0

¼

1  x  M  dx ¼ EI

Zl 0

12  F  x2 12  F  L3=2  dx ¼  E  b  h3 E  b0  h30

Zl 0

pffiffiffi x  dx ð8:164Þ

3

2 FL  3 E  I0

where I0 ¼ b0  h0 3 =12, the second moment of the area of the cross-section at the fixed end. Comparing this equation with that of a beam of constant section, loaded in cantilever, we can deduce that the deflection is twice that of a beam of constant structural stiffness subjected to equivalent load. In other words, the beam has the same resistance, but not the same stiffness than the prismatic bar. To ensure the shearing resistance, we assume that sc ¼ 54  sf , thus we will have:

8.3 Comparison Between Materials for Beams and Plates

245

Situation 1: Beam of the same resistance, of variable width and constant thickness: Minimum cross-section: P 5 F 5F ¼  ¼ h0  yðbÞ ! yðbÞ ¼ s 4 rf 4  r f  h0

ð8:165Þ

Taking the value yðbÞ ¼ CC 00 to the meeting with AC (Fig. 8.20). Situation 2: Beam of the same resistance with constant width and variable thickness: Minimum cross-section: P 5 F 5F ¼  ¼ b0  yðhÞ ! yðhÞ ¼ s 4 rf 4  r f  b0

ð8:166Þ

Taking the value yðhÞ ¼ BB0 , and drawing through B0 a parallel B0 D to the axis, the profile FDB0 B results (see Fig. 8.21). When the height (thickness) is small and b0 (width) is large, springs of overlapped flat sheets (Fig. 8.22) are used, the effect of which is that of a triangular spring with base A  B ¼ n  b0 for n sheets. As in previous cases, the study is reduced to that of a solid in cantilever with a load at the free end. The Navier equation would be: PL¼

rf  n  b0  h20 6

ð8:167Þ

Or: P¼

rf  n  b0  h20 6L

ð8:168Þ

rf  L2 E  h0

ð8:169Þ

The maximum deflection is: f ¼

Fig. 8.21 Profile of a beam of equal resistance, constant width, and variable thickness

246

8

Materials for Beams

Fig. 8.22 Spring of overlapped sheets (b, large; h, small)

It can be seen from this equation that for this type of cantilever of uniform strength, the deflection at the end varies as the square of the length and inversely as the depth and E. The flexibility would be: f rf  L ¼ L E  h0

ð8:170Þ

1 1 rf 2  ðn  b0 Þ  h0  L Pf ¼  2 12 E

ð8:171Þ

The work of deformation is:

These results may be used to compute the approximate stresses and deflections in a spring of leaf type. The triangular shape mentioned above is thought of as being divided into strips arranged as shown in Fig. 8.22. The initial curvature and the friction between strips are neglected for a first approximation and Eq. (8.169) can then be considered sufficiently accurate. A spring of length L ¼ 400 mm has six sheets of 8-mm thickness and 60-mm width. Calculate the load that can support and the deflection.

8.3 Comparison Between Materials for Beams and Plates

247

Data: rf ¼ 36 kg/mm2 ; E ¼ 20,500 kg/mm2 The load is: P¼

rf  n 36 kg/mm2  6  b0  h20 ¼  60 mm  ð8 mmÞ2 ¼ 345:6 kg 6L 6  400 mm

ð8:172Þ

The deflection is: f ¼

rf  L2 36 kg/mm2  ð400 mmÞ2 ¼ ¼ 35:1 mm E  h0 20,500 kg/mm2  8 mm

ð8:173Þ

Calculate the width and the deflection of a leaf spring for a car where P ¼ 2500 kg; h0 ¼ 10 mm; L ¼ 600 mm; and rf ¼ 50 kg/mm2 ; n ¼ 10: The width is: 2500 kg ¼

rf  n 50 kg=mm2  ð10  b0 Þ  ð10 mmÞ2  b0  h20 ¼ ! b0 6L 6  600 mm

2500 kg  6  600 mm ¼ ¼ 180 mm ð18 cmÞ 50 kg=mm2  100 mm2

ð8:174Þ

The deflection is: f ¼

50 kg=mm2  ð600 mmÞ2 ¼ 87:8 mm ð8:8 cmÞ 20,500 kg=mm2 10mm

ð8:175Þ

Reference Pero-Sanz, J. A. (2006). Ciencia e Ingeniería de Materiales. Estructura, Transformaciones, Propiedades y Selección, 5th edn. Madrid, Spain: Ed. CIE Dossat.

9

Materials for Columns and Struts (Compression–Tension)

Abstract

This chapter is dedicated to materials used for struts and columns. We describe in this chapter the selection of materials for columns when considering a design based on resistance to buckling. In the same way, we study the use of struts subjected to tension efforts. Additionally, we include two cases of application: tower cranes and beverage cans.

9.1

Materials for Columns

When a column works under axial load, its behaviour is like that of a bar subjected to compression. However, in the case of a column, the answer to the applied effort, when the column is tall and slender, is of special interest. We assume a column (Fig. 9.1), of uniform cross-section, made from a homogeneous material that behaves elastically. When a load F is applied, and if this load does not exceed a certain value, the column is laterally stable. That is to say, if the loaded column is pushed at the upper extreme, a deflection d is produced, but the initial shape is recovered when the lateral force is removed. In contrast, from a certain value of applied axial load—or critical load Fcr —if the column is laterally pushed, the column will remain laterally deformed when the force is removed. This instability of the column is called “lateral buckling.”

9.2

Geometry of the Section and Buckling

If we apply the equation of elastic curve to columns of length L1 manufactured from the same material, critical loads, Fcr , that would produce buckling can be © Springer Nature Switzerland AG 2019 J. A. Pero-Sanz Elorz et al., Structural Materials, https://doi.org/10.1007/978-3-030-26161-0_9

249

250

9

Materials for Columns and Struts (Compression–Tension)

Fig. 9.1 Slender columns

calculated. It is possible to deduce—for slender columns, for columns embedded at their bases, and for columns free at the upper extreme—that the critical load is: Fcr ¼ k1  E  I

ð9:1Þ

where E is Young’s modulus; I is the second moment of the area of the cross-section of the column or strut; and k1 is a constant equal to p2 =L21 (L1 is the length of the column). This is the Euler equation: Fcr ¼

p2  E  I L21

ð9:2Þ

where Fcr is the Euler buckling load. If I, the second moment of the area, is written as A  k2 , where A is the area of the cross-section and k is the radius of gyration, Euler’s equation can be put into terms of stresses: rE ¼

p2  E ðl=kÞ2

ð9:3Þ

where rE is Euler’s buckling stress. The k in Euler’s equation is the radius of gyration or the distance from the neutral axis at which the area of the cross-section may be considered to be acting. It takes the following values depending on the cross-section: – Squared section: k ¼ 0:29  t, where t is the side. – Circular section (bar): k ¼ 0:5  r, where r is the radius. – Tubular section (thin): k ¼ 0:707  r, where r is the radius. The ratio l=k is known as the slenderness ratio of the strut and Euler’s equation only coincides with the measured failure load (or stress) of a strut when the

9.2 Geometry of the Section and Buckling

251

Fig. 9.2 Ideal behaviour against buckling for steel and aluminium. Line 1, steel; line 2, aluminium

slenderness ratio is rather high. When it is very low, i.e., when the strut is short and stubby, Euler’s buckling stress becomes greater than the yield strength in compression of the material from which the strut is made, and it is obvious that failure will then occur by simple compression rather than by buckling. The relationship between the Euler buckling stress, rE ; and the slenderness ratio is shown in Fig. 9.2 where ry is yield strength. For steel, the Young’s modulus is 200 GPa and the yield strength is 250 MPa, l=k ’ 89. For aluminium, Young’s modulus is taken as 68 GPa and yield strength as 230 MPa, so that l=k ’ 54, but this idealized relationship overestimates the buckling stress actually measured in practice. The discrepancy is due to manufacturing imperfections; the fact that no strut is ever perfectly straight; and the difficulty of obtaining precise alignment between the direction of the compressive load and the axis of the strut. The critical load Fcr that a column can support without buckling does not depend on the yield strength under compression of the material—rather, it depends on the stiffness E  I of the column and on the length L—and, for that reason, for the same material and the same length, it be can increased if the moment of inertia of the section increases. This way, for equal area of the cross-section of the column (equal weight of the column), this critical load Fcr could be greater the more distant the material is from the axis of the column (greater I). For this reason (equal Fcr ), and for the former reason, for the same stiffness, tubular sections of the same material will weigh less—they are less expensive—than solid circular sections. For that reason, also, for equal weight of empty column, decreasing the thickness of the walls and increasing, on the contrary, the cross-dimensions, the stability of the column in the face buckling can be increased. As example, we might consider how an empty can, like those used to held carbonated beverages, can almost resist the weight of a person without suffering buckling problems. In addition, the critical load, Fcr , also depends on the stiffness E  I for other types of columns. Other typical columns can be observed in Fig. 9.3:

252

9

Materials for Columns and Struts (Compression–Tension)

Fig. 9.3 Columns with extremes supported or jointed. a column with one extreme fixed and the other simply supported, b column with both extremes supported or jointed and obliged to move in the direction of the axis, c column with one extreme embedded/fixed and the other guided, and d column with both extremes embedded/fixed

– Column with both extremes supported or jointed and obliged to move in the direction of the axis (Fig. 9.3b), whose F cr ¼ k2  E  I, being k2 ¼ 4  k1 . – Column with one extreme embedded/fixed and the other guided as in the previous case (Fig. 9.3c), whose F cr ¼ k3  E  I, being k3 ¼ 8  k1 . – Column with both extremes embedded/fixed (Fig. 9.3d), with F cr ¼ k4  E  I, being k4 ¼ 16  k1 . Exercise 9.1 Cans for beverages, a technologically advanced product. Few people think that simple cans for storing carbonated beverages are a product of high technology in terms of materials engineering. However, the design requirements are very demanding: Cans must be manufactured in two parts, body and cover, to avoid lateral welding; they must be light, liquid-tight, nontoxic, resistant to corrosion, and recyclable. The manufacture of cans in aluminium alloy is important from an economical point of view. 100,000 million cans are manufactured every year in the United States—300 million cans are used daily—comprising 10% of the annual production of primary aluminium. The most typically used alloy is 3004-Al–1% Mn–1% Mg–0.3% Si, which allows forming through drawing and wall ironing (DWI) in a single operation. Hardening mechanisms involved in this process are solid-solution, precipitation, and temper. However, perhaps the “secret” of can manufacture is closer to the process annealing or partial recrystallization, which produces—for the same resistance of the plate—more ductility, isotrophy, and better texture—it decreases the drawing ears and losses as a consequence of chips

9.2 Geometry of the Section and Buckling

253

Table 9.1 Characteristics of cans Material

Length, diameter, thickness (all in mm)

Weight (g [empty can])

Gross weight (g)

HVN

ru (compression in kg)

Aluminium Steel

115.5  66  0.18 115.5  66  0.12

16 32

356 372

’100 ’ 200

70

and crops—than the cold-rolled plate. It should be added that the surface quality observed in subsequent forming is better in plates partially recrystallizated from that state (cold deformation) than when a plate of the same hardness (resistance) to that recovered or partially recrystallized, but obtained directly through cold deformation, is used. We could think that aluminium has replaced steel (tinplate) in the manufacture of beverage cans if we pay attention to the mass media. Nothing could be further from reality: In both the United States and the rest of the world fierce competition between aluminum and tinplate exists. Aluminium is light, but it has a worse formability than steel and energetically is much more expensive: 270 GJ/Tm for aluminium compared with 17 GJ/Tm for steel [see Table 3.5 (Chap. 3)]. Table 9.1 lists some geometric and mechanic characteristics for cans manufactured from both materials. From all that, the following conclusions can be deduced: – The thickness of a steel can is 1.5 times lower than that of aluminium (iron has a density, approximately, 3 times greater than that of aluminium). – The decrease in weight achieved, for a beverage net weight of 340 g, is 5% (favourable for aluminium). The question is this: Can the greater price of aluminium compensate for the decrease of 16 g/can? – The hardness of steel (200 HVN) is approximately double that of aluminium (100 HVN). The loads that a can might support, for the considered thicknesses, would be—both theoretically and experimentally—greater in steel. The compression tests performed in empty cans confirm such an appraisal. With the exception of the tests performed with slower compression speed [5 mm/min (Fig. 9.4)], the load supported by a steel can is 2–3 times greater than that supported by an aluminium one, which does not exceed 70 kg with the exception of the mentioned case: Aluminium cans cannot support the weight of a person. – The compression stress supported by both materials in the plastic
collapse is too small. Assuming a load of F ¼ 70 kg—for steel r ¼ 70kg= p  ðð33mmÞ2 
ð33mm  0:12mmÞ2 ÞÞ ’ 2:8 kg/mm2 ; for aluminium r ¼ 70kg= p  ðð33mmÞ2

ð33mm  0:18mmÞ2 ÞÞ ’ 1:9 kg/mm2 . In contrast, the slenderness of the cans is small: l=k ’ 3:5. Then the failure did not take place as a consequence of buckling; it occured due to compression: However, it is interesting that both values are very small compared with the ultimate tensile strength (’60 kg/mm2 for steel and

254

9

Materials for Columns and Struts (Compression–Tension)

Fig. 9.4 Results of compression tests in cans

30 kg/mm2 for aluminium). If there were any reason to replace Young’s modulus with the tangent modulus (at the beginning of the plastic deformation, EL ¼ dr=de ’ E=1000Þ, there would be reason to think that the mechanism of deformation and collapse of cans would happen partially due to buckling and partially due to pure compression. – Cans, regardless of the material they are manufactured from—steel or aluminium— are, structurally speaking, small pressure vessels. They receive part of their strength from the internal pressure produced by carbon dioxide (beer or carbonated beverages) or nitrogen (fruit juices). If the pressure of the gas is approximately 6 atm, the circumferential tension supported would be rc ¼ p  D=2  t; that is, rsteel ’ 165 MPa and rAl ’ 110 MPa, which do not justify the small tension of compression supported by the materials (the empty cans). The use of tubes is extremely popular within engineers and tubular struts are very widely used for all sorts of purposes. However, a tube under compression has a choice of two modes of buckling. It may buckle, in a long-wave mode, over its whole length in Euler-fashion. Alternatively, it may buckle in a short-wave mode locally by getting a sort of crease or crumple into the walls of the tube. If the radius of the tube is large and the wall is thin, the strut may well be safe against Euler, or long-wave, buckling; however, it will fail by load crumple of the skin. This is easily shown with a thin-walled paper tube. One form of this local buckling or crumpling is called “Brazier buckling.” This effect sets a limit to the use of simple tubes and thin-walled cylinders in compression as would be the case for beverage cans. In a thin-walled circular tube, local buckling will generally occur when stress in the skin reaches a value equivalent to 1/4E(t/r), which, in our case, well lies on the order of magnitude of applied loads (P = 475 kg for iron; P = 350 kg for aluminium).

9.3 Comparison Between Materials for Columns

9.3

255

Comparison Between Materials for Columns

The reasoning to select materials destined to be stable columns, resistant to buckling, is similar to that discussed in previous sections. This way, for cylindrical columns or bars—we should consider F1 and the length L1 of the column fixed; and the radius variable—the weight will be: P1 ¼ vol  q ¼ p  r 2  L1  q ¼ constant  r 2  q

ð9:4Þ

To avoid the phenomenon of buckling in the cases described in Figs. 9.1 and 9.3 for a force F1 , the following relation should be verified: F1 ¼ constant  E  I ¼ constant  E  r 4 ¼ constant  ðP1 =qÞ2 That is to say: E

1=2

=q ¼ cte 

1=2 F1 =P1

rffiffiffiffiffiffiffiffiffiffiffi 2  F1 1 ¼L   P1 p 2

ð9:5Þ

ð9:6Þ

which can be graphically relisted. The demonstration is shown in the following lines. The weight of the circular column is: P ¼ q  p  r2  L

ð9:7Þ

The buckling critical load is: p2  E  I p2  E  p  r 4 2  Fcr  L2 4 ! r2 ¼ Fcr ¼ ¼ ! r ¼ L2 2  L2 p3  E

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2  Fcr  L2 p3  E

ð9:8Þ

Thus: rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffi 2  Fcr  L2 2  Fcr 2 P¼qpL ¼ q  L  p3  E pE

ð9:9Þ

rffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffi E 2  Fcr 1 2 ¼L   P p q

ð9:10Þ

This way:

Taking logarithms we would have the straight lines obtained in Fig. 7.1 (Chap. 7): rffiffiffiffiffiffiffiffiffiffiffiffi 2  Fcr log E ¼ 2  log q þ 2  log þ 4  log L  log P p

ð9:11Þ

256

9

Materials for Columns and Struts (Compression–Tension)

Fig. 9.5 Representation log E  log q (buckling)

Now we plot log E versus log q (Fig. 9.5). Thus: pffiffiffiffi – P # if E=q " : – If two materials are located over the same straight line of slope 2, P ¼ cte. pffiffiffiffi – If Fcr "; L "! E=q " for the same P. Consequently, if we choose a material A, any point in Fig. 7.1 (Chap. 7), the relation E1=2 =q will allow locating in the straight line of slope 2—which passes through this point—all materials that could be used in the manufacture of columns of the same weight, PA , and with equal resistance to buckling. Materials situated on another parallel line, but with greater ordinate in the origin, would give columns of equal weight PA but more resistant than A or columns of lower weight for the same load F1 . Exercise 9.2 A column of length L known and radius r variable (Fig. 9.6) is subjected to compression loads of value F. We want to select the material that could maximize buckling resistance with minimum weight and minimum cost. The moment of inertia of the square section is: I¼

p  r4 p  r4 r ! A  k2 ¼ p  r 2  k2 ¼ ! k ¼ ¼ 0:5  r 2 4 4

ð9:12Þ

The force F must not exceed the maximum force ðFmax Þ that produces buckling:  1=4 4 EI 4  Fmax  L2 2 E  ðp  r =4Þ Fmax  p  2 ! Fmax  p  ! r L L2 p3  E 2

ð9:13Þ

9.3 Comparison Between Materials for Columns

257

Fig. 9.6 Circular column subjected to buckling

Table 9.2 Selection of materials with resistance to buckling

Material

E (GPa)

q (g/cm3)

E1/2/q

Aluminium Steel Beryllium Titanium

70 210 250 103

2.7 7.9 1.8 4.5

3.0 1.8 9.0 2.3

(3) (5) (1) (4)

The method of minimizing weight is: P¼qLp

 1=4 !2 4  Fmax  L2 1 ¼ Constant  1=2 p3  E E =q

ð9:14Þ

In the text, we explain that the selection would be performed using the compounded property E1=2 =q of the selected materials. Table 9.2 lists the value of the E1=2 =q for the four selected materials: Structural efficiency in the face of buckling of aluminium, steel (iron), and titanium is very similar. The efficiency of beryllium is noticeable higher: It is a hexagonal compact metallic material, very anisotrope, and it melts at 1277 °C. It is the metal with the highest specific heat. For that reason, beryllium is used in aeronautics as “heat sink” or as “heat shield” for the atmospheric entry of spacecrafts. It is also used in other aerospace applications due to its stiffness (its Young’s modulus is high, approximately 1.5 times that of steel) and lightness (density 1.84 g/cm3). Its yield strength is approximately 700 MPa, the ultimate tensile strength 900 MPa, and the elongation between 1 and 10%. However, its toughness is low [see Sect. 3.5.4.2 (Chap. 3)].

258

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Materials for Columns and Struts (Compression–Tension)

Fig. 9.7 Scheme of the latticework

Exercise 9.3 What would be the maximum length that latticework should achieve so that bars subjected to compression did not suffer buckling? Use a safety factor of 1 and a safety factor of 2. What would happen if we enlarge the lattice? (Fig. 9.7). The degree of static indeterminacy (DSI) is: DSI ¼ b  2  n0 ¼ 8  8 ¼ 0

ð9:15Þ

This is a statically determinate girder (lattice, truss). Equilibrium condition: In knot A (Fig. 9.8): pffiffiffi pffiffiffi pffiffiffi 2 2 þ 1 ¼ 0 ! T 7 ¼  pffiffiffi ¼  2 ! T 7 ¼  2 T7  2 2

ð9:16Þ

pffiffiffi pffiffiffi pffiffiffi 2 ¼ 0 ! T 8 ¼ T 7 ¼ 2 ! T 8 ¼ 2 2

ð9:17Þ

pffiffiffi T8 þ T7  2 

Fig. 9.8 Knot A

9.3 Comparison Between Materials for Columns

259

In knot B (Fig. 9.9): T 7 ¼ T 5 ! T 5 ¼

pffiffiffi 2

pffiffiffi pffiffiffi 2 2 ¼ T5  þ T6 T7  2 2 pffiffiffi pffiffiffi 2 þ T 6 ! T 6 ¼ 1 1 ¼ 2  2

ð9:18Þ ð9:19Þ ð9:20Þ

In knot C (Fig. 9.10): pffiffiffi T 3 ¼ T 5 ! T 3 ¼  2 T8 þ T5 

pffiffiffi pffiffiffi 2 2 ¼ T3  þ T4 2 2

pffiffiffi pffiffiffi 2 þ 1 ¼ 1 þ T 4 ! T 4 ¼ 2 þ 2

ð9:21Þ ð9:22Þ ð9:23Þ

In knot D (Fig. 9.11): pffiffiffi pffiffiffi 2 ! T1 ¼ 2

ð9:24Þ

pffiffiffi pffiffiffi pffiffiffi 2 2 þ T2 ¼  2  1 T1  2 2

ð9:25Þ

T 1 ¼ T 3 ¼

Fig. 9.9 Knot B

Fig. 9.10 Knot C

260

9

Materials for Columns and Struts (Compression–Tension)

Fig. 9.11 Knot D

1 þ T 2 ¼ 1  1 ! T 2 ¼ 3

ð9:26Þ

Bar 4 in tension (steel of 40 kg/mm2, S = 1):
pffiffiffi 1; 000 kg d2 ¼ 40 kg=mm2  p  ! d ’ 10:42 mm F ¼ 2þ 2  1t  1t 4

ð9:27Þ

Bar 2 (−3000 kg, possible buckling). Applying Euler’s formula (Young’s modulus of the steel: 21000 kg/mm2): 3;000 kg ¼ p2 

  21;000 kg=mm2 d2 d 2  0:5   p  2 4 L2

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p3  21;000 kg=mm2  ð10:42 mmÞ4  0:52 ’ 283 mm L¼ 3;000 kg  8

ð9:28Þ

ð9:29Þ

Using a factor of safety S ¼ 2:
pffiffiffi 1;000 kg d2 2  2þ 2  1t  ¼ 40 kg=mm2  p  ! d ’ 26:1 mm 1t 4

ð9:30Þ

gives a length of: sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p3  21;000 kg=mm2  ð26:1mmÞ4  0:52 L¼ ’ 1774 mm 3;000 kg  8

ð9:31Þ

The length of the structure is multiplied by 6. If we enlarge one element in the structure (Fig. 9.12): DSI ¼ 12  2  6 ¼ 0

ð9:32Þ

pffiffiffi Tb ¼  2

ð9:33Þ

The problem is static. In knot E (Fig. 9.13):

9.3 Comparison Between Materials for Columns

261

Fig. 9.12 Latticework enlarged with one element

Fig. 9.13 Knot E

Fig. 9.14 Knot F

pffiffiffi Ta  2 

pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi 2 2 ¼ 2þ 2þ 2  ¼ 3 þ 2 ! Ta ¼ 4 þ 2 2 2

ð9:34Þ

In knot F (Fig. 9.14): T c ¼ T b ¼ pffiffiffi Td þ 2 

pffiffiffi 2

pffiffiffi pffiffiffi pffiffiffi 2 2 ¼ 2  3 ! T d ¼ 5 2 2

ð9:35Þ ð9:36Þ

Exercise 9.4 Decreased-scale models do not always predict adequately the stresses supported by the parts (or equipment at real scale). The square-cube law was

262

9

Materials for Columns and Struts (Compression–Tension)

described by Galileo and it indicates that the supported stresses grow linearly with the dimensions of the parts (see demonstration at the end of the exercise). The square-cube law is a good reason, among others, to believe that vehicles, ships, aircrafts, and machinery must be designed by suitable modern analytical methods. For that reason, these last ones have experienced a late development, at least in their modern form. Another application of the square-cube law of Galileo is the following: – If an element of section A supports a load P:

r¼

P ¼ cte ! P / A A

ð9:37Þ

And its weight is: W / A3=2 ! W ¼ P3=2 ðsquare-cube lawÞ

ð9:38Þ

– If the same load P is supported by two elements:  3=2 P ’ 0:7  P3=2 W /2 2

ð9:39Þ

– If it is supported by three elements:  3=2 P W /3 ’ 0:57  P3=2 3

ð9:40Þ

– If it is supported by four elements:

W /4

 3=2 P ’ 0:5  P3=2 4

ð9:41Þ

The weight of the structure decreases with the number of elements supporting the load P. Note: Square-cube law (Galileo). In the case of a cube of side a, the weight is P / a3 and the stress is r ¼ P=a2 / a: In the case of a cube of side 2a, the weight is P / ð2aÞ3 ¼ 8a3 and the stress is r ¼ P=ð2aÞ2 ¼ 8a3 =4a2 / 2a:

9.3 Comparison Between Materials for Columns

263

The stress grows linearly with the dimension. This way, the failure of any structure susceptible to failure by breaking cannot be predicted using scale models based on previous experience. Application under tension. It can be shown that the total weight of the end fittings of two tension bars, operating in parallel, is less than that of the end fittings of a single bar of equivalent cross-section. This because the cross-section of a tension bar is proportional to the load (Eq. 9.37), whereas the volume of the end fittings increases as the power of 3/2 of the load (Eq. 9.38). It follows that, in general, weight is saved by subdividing a tensile load between two or more tension members instead of carrying it in a single one. We can put the problem of carrying a load P, over a length L in n parallel tension bars, in the following form: P Z ¼q  s

k pffiffiffi  1þ W L n

rffiffiffi! P ðCox-GordonÞ s

ð9:42Þ

where: Z is the total weight of all tension members per unit length (between end fittings); P is the total load carried; s is the safe working stress; k is a coefficient connected to the skill of the designer; W is the work of fracture of the material; n is the number of tension members employed; and q is the density of materials. A steel rod or cable must support, over a distance of 10 m, a load of 1000 kg (ton). The working stress is 33 kg/mm2. Considering the end fittings, the total weight (per unit length) is approximately 3.5 kg/m. What weight is saved for n (number of tension members) equal to 4? Data: W = 100 kgm/m2, q = 8000 kg/m3, Z ðwith end fittingsÞ ¼ 0:35 kg/m. For n ¼ 1, weight of cable (without end fittings): Z0 ¼ q 

P kg 1000 kg ¼ 8000 3  ’ 0:24 kg/m s m 33  106 kg/m2

ð9:43Þ

Solving for k: kg k ð0:35  0:24Þ ¼ m 100 kgm  10 m m2 / 83;260 kg=m

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1000 kg kg  0:24 ! k 2 6 m 33  10 kg/m ð9:44Þ

264

9

Materials for Columns and Struts (Compression–Tension)

Fig. 9.15 Representation of Z  L versus L

For n ¼ 4: 2

3 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u kg 6 83;260 kg=m kg u 1;000 kg 7 pffiffiffi  t Z ¼ 0:24  41 þ 5 ’ 0:3 33 kg 106 mm2 m m 100 kg  m=m2  10 m  4 mm2  1m2 ð9:45Þ Weight saved = 0:35  0:30 ’ 0:05 kg/m ð0:5 kg=10 mÞ. Z # if k # W " L " n " We represent Z  L as a function of L when n ¼ 1 and n ¼ 4 (see Fig. 9.15). Application under compression: To try to carry such a load in compression (1000 kg) over such a distance (10 m) by means of a solid steel rod would be silly because if a solid rod was thick enough to avoid buckling, it would need to be very heavy indeed*. In practice, we might well use a steel tube, which would be approximately 16 cm in diameter with a wall thickness of, say, 5 mm. Such a tube withstands a compression load of: F¼

p2  E  I L2

I ¼ A  ð0:707  r Þ2

ð9:46Þ ð9:47Þ

For a thin-wall annulus tube, the buckling critical load would be:   2 p2  2:05  1010  kg/m2  ð0:707  8  102 mÞ p  16  102 m  5  103 m Fcr ¼ 100 m2 ’ 16;300 kg ð1000 kg supported by rod, enough safetyÞ ð9:48Þ

9.3 Comparison Between Materials for Columns

265

Such a tube would weigh approximately 200 kg. In other words, it would weigh between 50 and 60 times as much as the tension rod. The cost might well grow in the same proportion. Furthermore, if we want to subdivide a compression structure, the situation gets no better but much worse. If we wanted to support a load of 1 ton, not by a simple strut but by a table-like arrangement of four struts, each of them 10 m long, then the total weight of the struts would be twice as great, e.g., approximately 400 kg. The weight continues to increase the more the structure is pffiffiffi subdivided: In fact, as n, where n is the number of columns. *If we apply Euler’s formula to the mentioned bar: Fcr ¼

p2  E  I p2  E  A  k2 ¼ L2 L2

ð9:49Þ

E ’ 20;500 kg/mm2 , r ¼ 3:1 mm (section in tension ’30 mm2), we would have: 2

Fcr ¼

p2  20;500  106 kg/m2  30  106 m2  ð0:5  3:1  103 mÞ ’ 0:15 kg 100 m2  1000 kg ðsevere bucklingÞ ð9:50Þ

The minimal section r required ðk ¼ 0:5  rÞ would be (bar): p2  20;500  106  kg/m2  p  r 2  ð0:5  r Þ2 ! 100 m2 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 100;000 m4 !r¼4 3 ’ 0:0282 m ’ 2:82 cm p  20;500  106  0:52

1;000 kg ¼

ð9:51Þ

ð9:52Þ

And the weight: p  ð2:82 cmÞ2  8

g 1 kg 100 cm  ’ 20 kg/m ð0:35 kg/m in tensionÞ  3 cm 1000 g 1 m ð9:53Þ

This means that would be at 50–60 times heavier, as previously stated. Exercise 9.5 Efficiency of a panel, subjected to a load F, under compression stresses (buckling). The thickness of the panel can vary only in one dimension. We assume a second moment of area of the cross-section per unit of panel width (C; C 0 and C 00 are constants): I ¼ C  t3 ¼

F  L2 F  L2 ! t3 ¼ C 0  2 2 p E p E

ð9:54Þ

266

9

Materials for Columns and Struts (Compression–Tension)

The weight of the panel per unit of width is: sffiffiffiffiffiffiffiffiffiffiffiffi 2 pffiffiffiffi 3 F  L q ffiffiffiffi  L5=3  3 F P ¼ t  L  q ¼ C 00  q  L  ¼ C 00  p 3 2 p E E

ð9:55Þ

p ffiffiffiffi 3 E 1 1 ¼ cte   P q L5=3  F 1=3

ð9:56Þ

p ffiffiffiffi 3 E 1 F 1=3 ¼ cte   P q L5=3

ð9:57Þ

This way, the efficiency (multiplying and dividing by F 2=3 in both sides of the equality) is: p ffiffiffiffi 2=3 3 E F F ¼ cte   P q L5=3

ð9:58Þ

pffiffiffiffi Thus, F 2=3 =L5=3 is a “load coefficient of the structure” and 3 E=q is a “criterion that measures the efficiency of the material.” pffiffiffiffi If F=P ", 3 E=q ", F ", L #. The weight-cost " if the efficiency of the material " and the load coefficient of the structure ". Exercise 9.6 Aeronautics case: The criterion of control of the deformations is less important in the selection of materials than the buckling of the part. As we have seen, the relation weight–cost (and usually the cost) in a part working under compression is governed not by E/q (strut in tension) but by E1/2/q (strut in compression). The relation weight–cost in a panel is controlled by E1/3/q [panel in compression (see Exercise 9.5)]. These requirements are listed in Table 9.3 (where we include wood as material of selection). We will see that density is a key parameter and thus steel is bad if it is compared with bricks and concrete.

Table 9.3 Efficiency of several materials for different applications Mild steel (37) Titanium CP Degree 2 (27) Aluminium 1100 annealed (1) Magnesium AZ91B moulded (11) CFRP (64) Timber

E (GPa)

q

E/q

E1/2/q

E1/3/q

210 103 69 45 220 14

7.85 4.5 2.71 1.81 1.55 0.5

26.8 22.9 25.5 24.9 141.9 28

1.8 2.3 3.1 3.7 9.6 7.5

0.76 1.0 1.5 2.0 3.9 4.8

9.3 Comparison Between Materials for Columns

267

Table 9.4 Materials to select for strut manufacture Material

E (GPa) KIC (MPa m1=2) ry ðMPaÞ q (g/cm3) cp (£/t)

Reinforced concrete (76) Mild steel (37) Aluminium Cu 2014 T6 (6) CFRP (64)

50 210 72.4 220

0.3 140 31 45

40 280 414 670

2.4 7.85 2.8 1.55

200 250 2050 20,000

Furthermore, for many light-weight applications—such as airships or artificial limbs—wood is even better than carbon-fibre materials in addition to being cheaper. The following conclusions can be deduced from Table 9.3: – E/q (specific Young modulus), criterion of selection in bars working under tension stresses. For steel, titanium, aluminium, magnesium the value of E/q is nearly constant (’25). – E1/2/q, criterion of selection in bars subjected to compressive loads. – E1/3/q, criterion of selection for panels subjected to compressive loads. The best material is that with the lowest density: Timber is even better than carbon-fibre materials in addition to being cheaper. Exercise 9.7 We consider a strut with length L and square section of side t variable, which is subjected to tension stress F. We want to select the most suitable material to maximize stiffness, maximize resistance to plastic deformation, and minimize the risk of unstable cracking. We should also minimize the weight and cost in all previously indicated situations. The proposed materials are listed in Table 9.4 (Fig. 9.16). To maximize stiffness and minimize weight and cost: To maximize stiffness, we use Hooke’s law in the elastic domain: r¼Ee

ð9:59Þ

Because the initial elongations are small, then we can assume that the true and engineering deformations are similar and, in this way, e ’ e. Thus: e¼

L0  L dL ¼ L L

ð9:60Þ

268

9

Materials for Columns and Struts (Compression–Tension)

Fig. 9.16 Square-section strut

where the elastic elongation of the bar is dL. Because stress is the force divided by the area, then: F dL E  2 t L

ð9:61Þ

Because the weight of the strut is: P ¼ q  L  t2

ð9:62Þ

Because the side t is variable:  t

FL E  dL

1=2

ð9:63Þ

The method to minimize weight is:  !2 F  L 1=2 1 ¼ Constant  E  dL E=q

 P¼qL Then, P # if E=q " :

ð9:64Þ

9.3 Comparison Between Materials for Columns

269

Table 9.5 Square strut with side t: structural and economic efficiency E (GPa) q (g/cm3) cp (£/t) E=q

Material

Reinforced concrete 50 Mild steel 210 Aluminium Cu 2014 T6 72.4 CFRP 220

2.4 7.85 2.8 1.55

200 250 2050 20,000

E=q  cp

20.8 (6.8) 26.8 (5.3) 25.9 (5.5) 141.9 (1)

0.104 (1) 0.107 (1) 0.013 (8.2) 7.1  10−3 (15.1)

The minimum cost is obtained when: C ¼ P  cp ! C ¼ Constant 

1 E=q  cp

ð9:65Þ

In the same way, C # if E=q  cp " : Considering the materials listed in Table 9.4, we obtain the results listed in Table 9.5. If we only take into account structural efficiency for the square strut with side t variable, CFRP would be the best option from all those listed in Table 9.4 (CFRP > mild steel > aluminium Cu 2014 T6 > reinforced concrete). However, if we consider cost, reinforced concrete and mild steel are the most suitable materials (reinforced concrete ’ mild steel > aluminium Cu 2014 T6 > CFRP). Maximize resistance to plastic deformation and minimize weight and cost: The stress must not exceed the yield strength: F  ry t2

ð9:66Þ

As in the previous case: t

 1=2 F ry

ð9:67Þ

Thus, the minimum weight is obtained when: P¼qL

 1=2 !2 F 1 ¼ Constant  ry ry =q

ð9:68Þ

r

Then, P # if qy : The cost is minimum when: C ¼ P  cp ! Cost ¼ Constant 

1 ry =q  cp

ð9:69Þ

270

9

Materials for Columns and Struts (Compression–Tension)

Table 9.6 Square strut with side t: Structural and economic efficiency Material

ry ðMPaÞ q (g/cm3) cp (£/t) ry =q

Reinforced concrete 40 Mild steel 280 Aluminium Cu 2014 T6 414 CFRP 670

2.4 7.85 2.8 1.55

200 250 2050 20,000

16.7 (25.9) 35.7 (12.1) 147.9 (2.9) 432.3 (1)

ry =q  cp 0.0835 0.1428 0.0721 0.0216

(1.7) (1) (2.0) (6.6)

Similarly, C # if ry =q  cp " : Considering the materials listed in Table 9.4, we obtain the results listed in Table 9.6. If we only consider the structural efficiency (resistance to the plastic deformation), in the case of the square strut with side t variable, the CFRP would be the best option from all those listed in Table 9.4 (CFRP > aluminium Cu 2014 T6 > mild steel > reinforced concrete). However, if we consider cost, the mild steel would be the most suitable material (mild steel > reinforced concrete > aluminium Cu 2014 T6 > CFRP). To minimize the risks of unstable cracking and minimize weight and cost: In the case of a column subjected to a load F, we admit into the design the existence of micro-cracks in the material, which do not exceed a half-length a(we do not admit the catastrophic propagation of cracks): K IC  ðconstantÞ  rt 

pffiffiffiffiffiffiffiffiffi F pffiffiffiffiffiffiffiffiffi p  a ¼ ðconstantÞ  2  p  a t  pffiffiffiffiffiffiffiffiffi1=2 ðconstantÞ  F  p  a ! t K IC

ð9:70Þ

Thus, the minimum weight is achieved when: P¼qL

 pffiffiffiffiffiffiffiffiffi1 !2 pffiffiffiffiffiffiffiffiffi ðconstantÞ  F  p  a 2 1 ¼ ðconstantÞ  F  L  p  a  K IC K IC =q ð9:71Þ

Then, P # if K IC =q ". The minimum cost is: C ¼ P  cp ! C ¼ Constant 

1 K IC =q  cp

ð9:72Þ

Similarly, C # if KIC =q  cp " : Considering the materials listed in Table 9.4, we obtain the results listed in Table 9.7.

9.3 Comparison Between Materials for Columns

271

Table 9.7 Square strut with side t: structural and economic efficiency Material

KIC (MPa m1=2) q (g/cm3) cp (£/t) KIC =q

KIC =q  cp

Reinforced concrete Structural steel Al 2014-T6 CFRP

0.3 140 31 45

6.3  10−4 (113) 0.0712 (1) 5.4  10−3 (13.2) 1.5  10−3 (47.5)

2.4 7.85 2.8 1.55

200 250 2050 20,000

0.125 (232) 17.8 (1.6) 11.1 (2.6) 29.0 (1)

Table 9.8 Materials to select for the square strut Material

E=q

E=q  cp

ry =q

ry =q  cp

KIC =q

KIC =q  cp

Reinforced concrete Mild steel

20.8 (6.8) 26.8 (5.3) 25.9 (5.5) 141.9 (1)

0.104 (1)

16.7 (25.9) 35.7 (12.1) 147.9 (2.9) 432.3 (1)

0.0835 (1.7) 0.1428 (1) 0.0721 (2.0) 0.0216 (6.6)

0.125 (232) 17.8 (1.6) 11.1 (2.6) 29.0 (1)

6.3  10−4 (113) 0.0712 (1)

Aluminium Cu 2014 T6 CFRP

0.107 (1) 0.013 (8.2) 7.1  10−3 (15.1)

5.4  10−3 (13.2) 1.5  10−3 (47.5)

If we only consider structural efficiency (minimize the risks of unstable cracking) for the square strut with side t variable, CFRP would be the most suitable material from all those listed in Table 9.4 (CFRP > mild steel > aluminium Cu 2014 T6 > reinforced concrete). However, if we consider cost, mild steel is the best option (mild steel > aluminium Cu 2014 T6 > CFRP > reinforced concrete). We are going to perform an attributive analysis where stiffness would have an importance of 45%, resistance to plastic deformation would have a relevance of 20%, and the risk of unstable cracking would have an importance of the 35%. Table 9.8 lists the functions to be maximized for the four materials. When we require maximum stiffness, CFRP is the most suitable material followed by mild steel, aluminium Cu 2014 T6, and reinforced concrete. In the case of economic efficiency, reinforced concrete is the best material together with mild steel, followed by aluminium Cu 2014 T6 and CFRP. When we require maximum resistance to plastic deformation, CFRP is the most adequate material followed by aluminium Cu 2014 T6, mild steel, and reinforced concrete. In the case of economic efficiency, mild steel is the best material followed by reinforced concrete, aluminium Cu 2014 T6, and CFRP. With respect to minimizing the risk to propagate the cracking, CFRP would be also the most suitable material followed by mild steel, aluminium Cu 2014 T6, and finally reinforced concrete. If we consider cost, mild steel would be the most adequate material followed by aluminium Cu 2014 T6, CFRP, and reinforced concrete.

272

9

Materials for Columns and Struts (Compression–Tension)

Table 9.9 Results of attributive analysis Material

E=q

Reinforced concrete 20.8 Mild steel 26.8 Aluminium Cu 2014 T6 25.9 CFRP 141.9

ðE=qÞi ðE=qÞmax

ry =q

ðry =qÞi KIC =q ðry =qÞmax

ðKIC =qÞi ðKIC =qÞmax

Q

0.147 0.189 0.183 1

16.7 35.7 147.9 432.3

0.039 0.083 0.342 1

0.004 0.614 0.383 1

0.075 0.317 0.285 1

0.125 17.8 11.1 29.0

In the attributive analysis:   ry =q i ðE=qÞi ðKIC =qÞi   0:2 þ  0:45 þ   0:35 Q¼ ðE=qÞmax ðKIC =qÞmax ry =q max

ð9:73Þ

The results are listed in Table 9.9. We can check that the most adequate material considering the results of the attributive analysis would be CFRP. If we had considered cost in the attributive analysis, we would have seen that the results would have been different because of the cost of CFRP. The pattern of materials use is constantly changing, and the rate of change is increasing. Whilst the succession of the Stone, Bronze, and Iron ages can be measured in millennia, the flow of current materials development causes changes in decades [see Fig. 14.1 (Chap. 14)]: There may also be changes in the criteria that determine whether or not a particular material can be put into large-scale use. In the past, these criteria have been simply the availability of the basic raw materials and the technological skills of the chemist, physician, metallurgist, and engineer in converting them into useful parts at acceptable cost, leading to the present situation in which the most important materials are still steel, concrete, and timber [Table 14.1 (Chap. 14)], but these supplemented by a constantly increasing usage rate of other material including metals (aluminium, copper, zinc, and titanium), plastics (thermoplastic and thermosets), ceramics, and composites (mainly based at present on resins but with possible extension into ceramics and metals as well). However, two additional criteria may assume increasing importance in the future arising from the concept of “Spaceship Earth” (meaning the limited resources of the planet on which we live): These are the total energy costs of a given material and the ease with which it may be recycled. Concrete is a low-energy material but cannot be recycled. In contrast, titanium is a high-energy material that is difficult and expensive to recycle. Only some plastics are recycled to a limited extent at present, but steels and many other metals can be recycled with relative ease. Some authors have calculated the energy content of various materials in relation to the delivered level of a given mechanical property. If, for example, the energy content regards tensile strength, the appropriate parameter is W  q=rTS where W is the energy in kWh required to produce 1 kg of the material; q is the density in kg/m3;

9.3 Comparison Between Materials for Columns

273

and rTS is the tensile stress in MPa. Timber has been found to be the most energy-conserving material with a value of 24, whilst rebar concrete is also low 145. Steels lie within the range 100–500, plastics in the range of 475–1002, and aluminium, magnesium and, titanium alloys in the range 710–1029. Concrete and timber are considered to retain their predominant position into the twenty-first century, but an intense rivalry between metallic and nonmetallic materials is foreseen. The exhaustion of oil would require polymers to be extracted from coal or biomass, and the scarcity of some metals [Table 3.6 (Chap. 3)] will limit their use to certain especially suitable applications. Composite materials will continue to be  widely developed. In any case, materials with minimum W= ry =q , W=ðE=qÞ, W=ðKIC =qÞ are desirable.

Materials for Pressure Vessels

10

Abstract

According to the modern view of fracture mechanics, when we break a structure by tension load, we ought not regard fracture as being caused directly by the action of the applied load pulling on the chemical bonds between the atoms in the material. The direct result of increasing the load on the structure is only to cause more strain energy to be stored within the material. Modern fracture mechanics is therefore less concerned with forces and stresses than with how, why, where, and when strain energy can be turned into fracture energy. In large or complicated structures—such as bridges, ships, or pressure vessels— when they are subjected to a sudden blow, it is possible for this strain energy to be converted into fracture energy so as to create or propagate cracks. The best materials for pressure vessels are studied in this chapter. Three criterions are chosen: rigidity, plasticity, and propagation of cracking as discussed in previous chapters. The theoretical section is complemented with exercises about the selection of materials for a pipe and for a cylindrical vessel.

10.1

Materials for Pressure Vessels

Pressure vessels are those recipients with a surface of revolution shape and a very small wall thickness, t—approximately 10 times smaller than the radius of curvature of meridians and parallels—subjected to inner pressure.

10.2

Geometry and Behaviour Under Pressure

We assume that the recipient is subjected to a uniform internal pressure, p. This pressure produces perpendicular forces, acting over any differential of surface dS, equal to the product of p by dS. © Springer Nature Switzerland AG 2019 J. A. Pero-Sanz Elorz et al., Structural Materials, https://doi.org/10.1007/978-3-030-26161-0_10

275

276

10

Materials for Pressure Vessels

Fig. 10.1 Pressure vessels (Pero-Sanz 2006)

Because the walls are of small thickness with respect to the principal curvatures, this differential of wall behaves as a membrane subjected to tangential efforts to the average surface of the wall and they distribute uniformly in all thickness. Figure 10.1 corresponds to an element limited by two meridians and two parallel circles. Over this element, two efforts act: r2 (tension stress in the direction of the parallel) and r1 (tension stress in the direction of the meridian). The forces resulting from these stresses will take the values of r2
dS1
t (force F2 , tangent to the circle of the parallel) and r1
dS2
t (force F1 , tangent in the direction of the meridian), respectively. Summing up the two forces F2 , the resulting force will be perpendicular to the element of surface and will take the following value: 2
F2
cosð90  dh2 =2Þ ¼ 2
F2
senðdh2 =2Þ ¼ 2
F2
dh2 =2 ¼ F2
dh2 ¼ F2
dS2 =r2 ¼ ðr2
dS1
tÞ
dS2 =r2 ð10:1Þ Analogously, the resulting force of the two forces F1 will be also perpendicular to the surface element and will take ðr1
dS2
tÞ
dS1 =r1 . The addition of both resulting forces—F2 and F1 —should be equal to the perpendicular force due to the acting pressure; that is to say, p
dS1
dS2 . Consequently, the following relation will be verified:

10.2

Geometry and Behaviour Under Pressure

r1 r2 p þ ¼ t r1 r2

277

ð10:2Þ

For that reason, in spherical deposits ðr1 ¼ r2 Þ, of radius r1 and small thickness of wall t with respect to r1 , when subjected to internal pressure p, the stress r1 (equal to r2 ) is equal to p
r1 =2
t: Analogously, in cylindrical deposits (r1 equal to infinite) of radius r2 , thin-thickness walls t, length L1 , and subjected to uniform pressure p, the stress on the wall of the cylinder is r2 ¼ p
r2 =t: The lengthwise stress r1 is, on the contrary, half the r2 ; that is to say, p
r=2
t. In fact, in each base of the cylinder, the resulting force is equalized with the lengthwise stress r1 throughout the circumference; and the following relation is verified r1
2
p
r
t ¼ p
p
r 2 . For the selection of materials to be used in spherical or cylindrical deposits, the procedure is similar as that described in other chapters; the charts shown in Figs. 7.1 through 7.4 (Chap. 7) can be used as we will see in the following paragraphs.

10.3

Materials for Vessels

The following sections will be dedicated to decrease the weight in three different situations: Preserving the same stiffness of the deposit, maintaining the same resistance to plastic deformation, and avoiding the risks of unstable failure.

10.3.1 Spherical Vessels: Minimum Weight for a Particular Stiffness Elastic deformation d1 produced by the internal pressure p1 in the spherical vessel is obtained using the following equation: E¼

r d1 =r1

ð10:3Þ

The stiffness of a deposit is the relation p1 =d1 . For that reason—considering that r ¼ p1
r1 =2
t—the stiffness of a spherical vessel of constant radius r1 is equal to the product of E
t by a constant c1 . On the other hand, for a density of the material q, the weight P of this spherical deposit will be equal to 4
p
r12
t
q. Consequently, for r1 constant, t ¼ cte
P=q is verified; for that reason, expressing stiffness as a function of weight P, we will have: Stiffness ¼ c1
E
P=q

ð10:4Þ

278

10

Materials for Pressure Vessels

Or (which is the same): E c2
Stiffness ¼ q P

ð10:5Þ

where c1 and c2 are constants that depend exclusively on the value of r1 . For that reason, when choosing materials for spherical vessels of radius r1 and fixed and calculated the stiffness, if we select a material that would make maximum the value E=q, this is equivalent to minimize the weight of the deposit for this stiffness. Or, as is equally obvious, it is equivalent to obtain the maximum stiffness for the same weight of the deposit. Choose, as example, a certain material A in the chart shown in Fig. 7.1. All of those materials corresponding to the points situated on the straight line of slope 1, which passes through A, constitutes the same family; all have the same value of E=q and, for that reason, they provide deposits with the same stiffness p1 =d1 and the same weight PA . As is logical, the greater the Young’s modulus E of the material chosen from all those of this family, the greater should be also the density q and, consequently, the smaller the wall thickness t of the vessel manufactured with this material. On the contrary, it would be possible to manufacture spherical vessels of equal stiffness to those materials of family A, but with less weight, using materials situated on a straight line of slope 1 but whose origin ordinate is greater than the line that passes through A. It would be also possible to say that, for equal weight of the vessel PA , this vessel would be stiffer if it was manufactured with materials of family B. From the relation E=q, it is also possible to determine the propagation speed for longitudinal waves in the mass, vL ¼ ðE=qÞ1=2 , simply by using in a chart log E  log q the parametric straight lines of slope 1 (that is to say, log E ¼ log q þ cte).

10.3.2 Spherical Vessels: Minimum Weight Without Plastic Deformation (Yielding) For r1 constant, r ¼ p1
r1 =2
t ¼ cte
q
p1 =P is verified. Consequently, the relation to select materials is: r p1 ¼ cte
q P

ð10:6Þ

For that reason, on any straight line of slope 1, in the chart shown in Fig. 7.2, we can place a group of materials that would provide equal weight P of the spherical vessel of radius r1 for the same internal pressure p. The greater the origin ordinate of the chosen line, the materials situated on it will provide deposits of lower total weight for the same internal pressure p. Equally, we could discuss that, for the same weight, the deposit would resist bigger pressures without plastic deformation (yielding).

10.3

Materials for Vessels

279

10.3.3 Spherical Vessels: Minimum Weight Without Risks of Unstable Failure If we want to apply a stress r without catastrophic failure of the vessel, the following equation should be verified on the assumption that the pre-existing internal cracks in the material would not be >2
c1 : KIC ¼ r
ðp
c1 Þ1=2

ð10:7Þ

On another note, the relation between r and the internal pressure of the vessel, p1 , is r ¼ p1
r1 =2
t; t can be expressed as a function of the weight P1 of the vessel. Consequently, and definitely, the equation of the last paragraph can be expressed as follows: 1=2

KIC c ¼ cte
1 q P1

ð10:8Þ

For that reason, KIC =q will be the relation to optimize when we want to minimize weight when choosing the material for a spherical vessel of radius r1 , subjected to pressure p1 , without risks of catastrophic failure when the pre-existing cracks have a length  2
c1 . For that—drawing straight lines of slope 1 over the chart of Fig. 7.3—we proceed as in previous cases. Any possible risk of unstable, explosive or without signal, failures must be avoided when designing vessels of medium or big dimensions and highly pressurized. And, for that reason, for security reasons, cracks of critical length equal to the thickness t of the vessel must be avoided; the same as, for instance, holes that could produce corrosion or any other problems in the future. From these holes, sporadic leaks in the vessel could be admitted always if they can be repaired with guaranties; but it is never admitted that—because of being of critical size for the internal pressure of the deposit—a crack could bring the failure of the deposit without previous warning, in an explosive way. That is to say, the material for this recipient should be that where, facing r resulting from internal pressure p1 , cracks of length equal to twice the thickness of the wall t would not be critical for the vessel. As a consequence, to compare materials with the support of charts, like the one shown in Fig. 7.3, with the purpose of minimizing weight—or to improve the confidence facing the stable failure by means of tougher materials—we should take into account that c1 should be  t; for that reason, the equation used to calculate the value of KIC should verify the following: KIC [


q1=2
p
r  p1 1 1
ðp
tÞ1=2 ¼ cte
1=2 ¼ cte
p1
P 2
t t

ð10:9Þ

280

10

Materials for Pressure Vessels

Consequently, the following relation should be verified: 2 KIC p2 [ cte
1 q P

ð10:10Þ

Consequently, to minimize the weight of a spherical vessel of radius r1 —or with 2 =q will be suitable the same weight increasing the inner pressure—the relation KIC to select materials that would not produce a catastrophic failure if cracks would reach a size equal to the thickness of the vessel. As in previous cases, the selection could be performed using charts, as that shown in Fig. 7.3 but placing the materials over straight lines in parallels of slope 1:2. In cylindrical vessels of thin walls, of known radius r1 and length L1 , it is possible to understand that in order to minimize the weight for a certain stiffness, the relation to optimize is E=q. In fact, the weight P is equal to 2
p
r1
L1
t
q; and r ¼ p
r1 =t ¼ E
d=r1 . For that reason, the stiffness p=d is equal to the product of a constant by E
P=q; consequently: E Stiffness ¼ cte
q P

ð10:11Þ

To minimize the weight without plastic deformation, the relation is r=q in cylindrical vessels; as r ¼ p
r1 =t ¼ p
r1
2
p
r1
q=P. For that reason: r cte ¼ q P

ð10:12Þ

In the same way, in cylindrical deposits—to minimize weight without risks of unstable failure in the case of notches of length 2
c1 —the relation that must be maximized is KIC =q as follows: 1=2

KIC c ¼ cte
1 q P1

ð10:13Þ

To finish this section, we should finally remember that Fig. 7.4 includes, in solid lines, the critical size of cracks that produces unstable failure for different values of r. This facilitates, in the first approximation, the selection of materials paying attention to toughness (the greater the toughness should be, the larger the critical crack size is because the resistance to propagation of a crack is measured by the fracture toughness, KIC ). Thus, thermoplastics and aluminium might be advantageous for vessels of little responsibility (e.g., aerosols). When the requirement is stability to resist fracture, stiffness (Ref. Fig. 7.1), and yielding (Ref. Fig. 7.2), steels might be the best option. In any case, for the selection of materials it is necessary to consider, apart from the aforementioned, other factors, among which cost could be included as the most important.

10.3

Materials for Vessels

281

p

Fig. 10.2 Diagram of stresses in the cross-section of a pipe

Table 10.1 Materials for selection Ref. Table 7.1 Metal 6 35 37 41 64 76 53 Note As stated

ry (MPa) q (g/cm3) cp (£/t)

Al Cu 2014 T6 414 Spheroidal ductile cast iron 289 Mild steel 280 823 M30 quenched and tempered steel 850 CFRP 670 Reinforced concrete 40 PVC 48 at the end of Table 7.1, cost values are approximate

2.8 7.1 7.85 7.83 1.55 2.4 1.4

2050 190 250 670 20,000 200 500

Exercise 10.1 Design based on yield strength (ry). We suggest selecting one material to manufacture a tube of radius r and thickness t (variable) along which a fluid at pressure p circulates (Fig. 10.2). The material must not yield. From the materials listed in Table 10.1, select the most adequate for the application mentioned and stablish a ranking based on the weight and the cost. From the cross-section force equilibrium (imagine the pipe sliced in two according to the diameter, the result of all pressure forces acting on the inside of each half of the shell must equal the addition of all of the stresses that would have acted on the cut surface, the area of which is 2
p
r
tÞ: 2
r
t ¼p
r !t ¼

p
r r

Pðper unit lengthÞ ¼ 2
p
r
t
q ¼ 2
p
r
q


ð10:14Þ

p
r 2
p
r
2
p ¼ ð10:15Þ r ðr=qÞ

The structural efficiency grows ðP #Þ when ðr=qÞ" (maximum value of r is equal to the ry of the material). Analogously, the economic efficiency would increase ðC #Þ when ðr=cp
qÞ increases.

282

10

Materials for Pressure Vessels

Table 10.2 Economic–structural efficiency Metal

ry (MPa) q (g/cm3) cp (£/t) ry/q

Al Cu 2014 T6

414

2.8

2.050

ry/(cp
q)

147.9 (2.9) 0.07 (3)

Spheroidal ductile cast iron

289

7.1

190

40.7 (10.6) 0.21 (1)

Mild steel

280

7.85

250

35.7 (12.1) 0.14 (1.5) 108.6 (4.0) 0.16 (1.3)

823 M3 quenched and tempered steel 850

7.83

670

CFRP

670

1.55

20,000 432.3 (1)

Reinforced concrete

40

2.4

200

16.7 (25.9) 0.08 (2.6)

PVC

48

1.4

500

34.3 (12.6) 0.07 (3)

0.02 (10.5)

Materials selected according to Table 10.1 include Al Cu 2014 T6 (6), spheroidal ductile cast iron (35), mild steel (37), 823 M3 quenched and tempered steel (41), CFRP (64), reinforced concrete (76), and PVC (53) (properties are shown in Table 10.1). Structural efficiency (from best to worst) (Table 10.2) is CFRP, Al Cu 2014 T6, 823 M3 quenched and tempered steel, spheroidal ductile cast iron, mild steel, PVC and reinforced concrete. Economic efficiency (from best to worst) (Table 10.2) is spheroidal ductile cast iron, 823 M3 quenched and tempered steel, mild steel, reinforced concrete, Al Cu 2014 T6, PVC, and CFRP. As a consequence of the large differences existing within the materials with respect to their structural and economic efficiencies, we could use attributive analysis to weight their effects. If cost was the most important parameter, we would choose either spheroidal ductile cast iron or 823 M3 quenched and tempered steel. On the contrary, if the weight was the main parameter, we would choose either composites or light alloys despite the cost. This selection would be made without considering other parameters such as availability, energy cost, time in service, aesthetics, or environment. Exercise 10.2 Choose the most suitable material from all those listed in Table 10.3 to manufacture a cylindrical pipe of radius r and small thickness t variable (r  tÞ. This pipe carries gas, which is at a pressure p. The following requirements must be satisfied: – The pipe must be rigid/stiff and have the minimum weight. – The pipe must have maximum resistance to deformation and minimum weight. – The pipe must have minimum risk to unstable cracking and minimum weight. The pipe is used in an application of responsibility and thus the cost is not a problem. Applying Hooke’s law: relastic ¼ E
e ’ E


delastic r

ð10:16Þ

10.3

Materials for Vessels

283

Table 10.3 Materials to select for manufacture of the pipe Ref. Table 7.1

Material

E (GPa)

ry (MPa)

KIC (MPa m1/2)

q (g/cm3)

3

Aluminium Mg 5083 annealed Annealed brass Grey cast iron 18/8 stainless steel PVC

70.3

145

35

2.66

105 145 208

78 260 230

35 6 110

8.52 7.3 7.94

2.6

48

2.4

1.4

15 34 45 53

The next equation must be verified: d  delastic

ð10:17Þ

p
r delastic p
r2 E
! t t r E
delastic

ð10:18Þ

In this way:

The weight of the pipe is: P¼q
L
2
p
r
t

ð10:19Þ

This way, to minimize the weight for maximum stiffness: P¼q
L
2
p
r
t


p
r2 1 ¼ cte
E=q E
delastica

ð10:20Þ

Now the maximum resistance to the plastic deformation is: r\ry

ð10:21Þ

This way: ry [

p
r p
r ! t[ t ry

ð10:22Þ

The weight of the pipe is: P¼q
L
2
p
r
t

ð10:23Þ

284

10

Materials for Pressure Vessels

This way, the minimum weight for maximum resistance to plastic deformation is: P¼q
L
2
p
r


p
r 1 ¼ cte
ry ry =q

ð10:24Þ

Finally, the minimum risk of unstable cracking is obtained as follows: KIC  cte
rwork


pffiffiffiffiffiffiffiffiffiffiffi cte
p
r
p
ac pffiffiffiffiffiffiffiffiffiffiffi p
r pffiffiffiffiffiffiffiffiffiffiffi
p
ac ! t  p
ac ¼ cte
ð10:25Þ t KIC

The weight of the pipe is: P¼q
L
2
p
r
t

ð10:26Þ

This way, the minimum risk of unstable cracking for minimum weight is: pffiffiffiffiffiffiffiffiffiffiffi cte
p
r
p
ac 1 ¼ cte
P¼q
L
2
p
r
KIC KIC =q

ð10:27Þ

In this way, we list in Table 10.4 the merit indices for each material. We see that the Al–Mg 5083 annealed and the 18/8 stainless steel are the best options. Exercise 10.3 We want to design a spherical vessel that could be used in the chemical industry. The diameter of the vessel is 3000 mm and it supports an inner pressure of 40 bar. The vessel must leak before breaking in an unstable way. We want to select the most suitable material for the vessel from those listed in the Table 10.5. Is this material still that with the lowest weight? And does this material have the lowest cost? The equation that must be used is: KIC  1:12
r


pffiffiffiffiffiffiffiffiffiffiffi p
ac

ð10:28Þ

where ac is the critical size of a crack for the considered stress; KIC is the fracture toughness; and r is the applied stress. Table 10.4 Merit indices for each material Material

E/q

ry/q (MPa)

KIC/q (MPa m1/2)

Al–Mg 5083 annealed Annealed brass Grey cast iron 18/8 Stainless steel PVC

26.4 12.3 19.9 26.2 1.9

54.5 9.2 35.6 29.0 34.3

13.2 4.1 0.8 13.9 1.7

10.3

Materials for Vessels

285

Table 10.5 Materials for manufacture of a pressure vessel Ref. Table 7.1 Material

E (GPa) ry (MPa) KIC (MPa m1/2) cp (£/t) q (g/cm3)

1

69

34

45

1550 2.71

105 110

78 825

35 75

1340 8.52 29,000 4.43

208 220 145

230 670 260

110 45 6

1150 7.94 20,000 1.55 125 7.3

Aluminium 1100 annealed Annealed brass Titanium 6Al 4 V annealed 18/8 Stainless steel CFRP Grey cast iron

15 28 45 64 34

The vessel must tolerate a crack length of ac equal to the wall thickness: ac  t

ð10:29Þ

where t is the wall thickness. In the case of a spherical vessel, the stress is calculated as follows: r¼

p
r 2
t

ð10:30Þ

where p is the inner pressure; and r is the radius of the vessel. This way: 2 KIC  1:122



p
r 2 2
t

2
p
t ! KIC  1:122


2 KIC  1:122


p2
r 2
p
t 4
t2

p2
r 2
p 4
t

ð10:31Þ ð10:32Þ

The yield strength ðry Þ must not be exceed: p
r p
r  ry ! t  2
t 2
ry

ð10:33Þ

Replacing: 2 KIC  1:122


p2
r 2

p p
r 4
2
r y

ð10:34Þ

286

10

Materials for Pressure Vessels

This way: 2 KIC 1:122
p
r
p  2 ry

ð10:35Þ

Then the materials that satisfy the relation in Eq. (10.39) can be used in the application indicated in the exercise: p ¼ 40 bar


105 Pa ¼ 4 MPa 1 bar

ð10:36Þ

r ¼ 1500 mm ¼ 1:5 m

ð10:37Þ

2 KIC 1:122
p
r
p 1:122
4
1:5
p ¼  2 2 ry

ð10:38Þ

2 KIC  11:82 ry

ð10:39Þ

2 The greatest pressure is carried by the vessel with the largest values of KIC =ry . 2 A guideline of KIC =ry is shown in the chart of Fig. 7.4 (slope ½). Only aluminium 1100 annealed, annealed brass, and 18/8 stainless steel could be used in the manufacture of the pressure vessel (Table 10.6) from those materials indicated in Table 10.5. To minimize weight (Table 10.7):

2  1:122
KIC

P ¼ q
4
p
r2
t

ð10:40Þ

p2
r 2 p2
r 2
p ! t  1:122

p 2 4
t 4
KIC

ð10:41Þ

P ¼ q
4
p
r 2
1:122


2 Table 10.6 Value of KIC =ry for materials indicated in Table 10.5

p2
r 2 1
p ¼ cte
2 2 KIC =q 4
KIC

Material

2 KIC =ry

Aluminium 1100 annealed Annealed brass Titanium 6Al 4 V annealed 18/8 Stainless steel CFRP Grey cast iron

59.6 15.7 6.8 52.6 3.0 0.1

ð10:42Þ

10.3

Materials for Vessels

2 Table 10.7 Value of KIC =ry for materials (to minimize weight)

2 Table 10.8 Value of KIC =ry for materials (to minimize cost)

287 Material

2 KIC =q

Aluminium 1100 annealed Annealed brass 18/8 Stainless steel

747.2 (2) 143.8 (10.6) 1523.9 (1)

Material

2 KIC =q
cp

Aluminium 1100 annealed Annealed brass 18/8 Stainless steel

0.482 (2.6) 0.107 (12.4) 1.325 (1)

With respect to the cost: Cost ¼ P
cp ! Cost ¼ Constant


1 2 =q KIC


cp

ð10:43Þ

From the materials listed in Table 10.5, 18/8 stainless steel is the most adequate because it satisfies the condition of leaking before breaking and also it minimizes both weight and cost (Table 10.8). In metals, the energy absorbed by a fracture is vastly greater that in ceramics and polymers, almost always because of the plasticity associated with crack propagation. It is this characteristic that gives steel its pre-eminence as the material for highly stressed structures when weight is not important.

Reference Pero-Sanz, J. A. (2006). Ciencia e Ingeniería de Materiales. Estructura, Transformaciones, Propiedades y Selección (5th ed.). Madrid, Spain: CIE Dossat.

The Cost Factor for the Selection of Materials

11

Abstract

In this chapter, we consider cost, although cost was used as a factor of selection in several exercises included in previous chapters. Charts for the comparison of materials based on one criterion of selection, apart from cost, are included in this section. They are very useful because we can define regions that comprise materials with the desired requirements. We conclude this chapter with a new concept, that of attributive analysis, which consists of giving weight to each criterion of selection and, in this way, selecting suitable material considering more than one property.

11.1

The Cost Factor in the Selection of Materials

Cost is a fundamental factor in the selection of materials, but it is a complex factor that is difficult to include in a chart of materials. It comprises both the price of the raw material (cost of extraction from the mineral) and the cost associated with the manufacturing process until having the finished part. We should not forget that the prices in the market depends mainly on the economic law of supply and demand. Regarding raw materials, these will be logically more expensive when (1) the mineral from which the raw material is extracted is more stable (i.e., greater energetic cost is spent to obtain it); (2) the mineral ore is less rich (commonly 60– 65% for iron, 1–1.5% for copper, and 0.001–0.0001% for gold); (3) the world’s resources are less abundant and recycling is less probable; (4) the material is purer (for instance, 99.99% aluminium costs 1.8 times more than 99.95% aluminium and that with 1 part per million [ppm] of impurities is as expensive as gold); (5) the greater the content of an expensive element in an alloy (e.g., considering that the price of the nickel is 10 times greater than that of the extra-mild steel; austenitic stainless steel with 8% Ni would cost two times more than the other steel), etc. © Springer Nature Switzerland AG 2019 J. A. Pero-Sanz Elorz et al., Structural Materials, https://doi.org/10.1007/978-3-030-26161-0_11

289

290

11

The Cost Factor for the Selection of Materials

On another note, value added to the raw material by the process of manufacture of the part comprises a group technical sequences derived from the particular material: forming (solidification or sintering, hot and/or cold deformation), heat treatments, surface finishing (coatings, mechanizing, holes, etc.), joints (riveting, welding, sealing), etc. In this way, it seems reasonable indicating that, currently, with the purpose of decreasing cost, the process engineering attempts to directly achieve parts almost finished (near net shape) without requiring mechanizing processes. Along this line, the solidification at pressure, semi-solid forming or liquid forging, and sintering of powders offer large possibilities in the obtaining of almost finished parts. Assume that—despite the complexity involved—costs per unit of weight, cp , for different materials, which allows comparing them, are known. In fact, for the exercises proposed in this book, we will use the values listed in Table 7.1 (Chap. 7), although they are purely indicative. Admitting that the cp of a material is known, the total cost of a part made from this material would be equal to cp multiplied by the total weight of the part, P1 . For that reason, the relation for minimizing weight deduced in previous sections would be also adequate to minimize cost only by changing q by the product q
cp . This way, for instance, we consider the relation E1=3 =q. We should remember that this relation, applied to plates subjected to bending, means that all materials for which that relation would be the same will have the same stiffness for the same weight P1 of the plate, as: E1=3 cte ¼ P1 q

ð11:1Þ

If in this equality the denominators are multiplied by cp , we would have: E1=3 cte cte ¼ ¼ q
cp P1
cp total cost

ð11:2Þ

For that reason, for equal stiffness and weight of the plate, the material with the greater value of E1=3 =q
cp would be more economical. Analogously, the criterion of lightness for equal bending moment M1 translates into materials that have high value of r1=2 =q. For that reason, for equal bending moment M1 , the most economic material would be that with the greatest value of r1=2 =q
cp . The same procedure could be used to deduce other relations. This way, for instance, considering the risks of catastrophic breaking, a suitable light material for 1=2 plates would be that with the greatest KIC =q and the most economic would that 1=2

with the greatest KIC =q
cp .

11.1

The Cost Factor in the Selection of Materials

291

As application exercise, we choose the most suitable material for plates of car bodywork from one of the five materials listed in Table 7.1: mild steel (37), aluminium 1100 annealed (1), ferritic–pearlitic HSLA steel (43), glass-fibre reinforced plastic (GFRP) (64), and titanium 6Al 4V annealed (28). Moreover, we assume that for this selection the following requirements are necessary: (1) The material must be stiff (e.g., someone could sit on the bonnet of the car without denting it); (2) the material must be resistant to plastic deformation (i.e., if the bonnet of the car dents, it should recover immediately the original shape when removing the force that caused the deformation); (3) the material must be light (that would mean savings of combustible); and (4) the material must be economic. With all the considerations that the simplifications involve, it is possible to estimate that the material will work as a beam with the shape of a plate (Fig. 8.1 [Chap. 8]) with length L1 and width w1 fixed (both values are fixed during the design) and variable thickness t. From the data listed in Table 7.1, we calculated for the five materials previously mentioned some of the relations that classify and determine the selection: E1=3 =q, E1=3 =q
cp , r1=2 =q, and r1=2 =q
cp . From them, we can define (see Table 11.1) an order of interest for their selection depending on one or another criterion. This way, if the principal criterion is stiffness with minimal weight, the order for the selection of those five materials would be (from greatest to smallest E1=3 =qÞ: CFRP > aluminium 1100 annealed > titanium 6Al 4V annealed > steels (the type of steel does not matter because the Young’s modulus, E, is independent of structure). However, if equal stiffness with minimal cost of the plate is required, the order of selection would be (from greatest to smallest E1=3 =q
cp ): mild steel > HSLA steel > aluminium 1100 annealed > CFRP > titanium 6Al 4V annealed. At the same time, if the imperative criterion for selection is minimal weight without risks of deformation, we would have—by determining the order of values of r1=2 =q— CFRP > titanium 6Al 4V annealed > HSLA steel > aluminium 1100 annealed > mild steel. However, if the preferential criterion for selection is the most economical material without risks of plastic deformation, the resulting order would be (from greatest to smallest r1=2 =q
cp Þ: mild steel > HSLA steel > aluminium 1100 annealed > CFRP > titanium 6Al 4V annealed.

Table 11.1 Comparison between materials Material Mild steel (37) HSLA steel (43)

E1=3 =q

E1=3 =q
cp

r1=2 =q −3

2.1316 (x7.8) (x1) 8.526
10−3

0.7571 (x5.1) (x1) 3.0287
10

−3

3.1203 (x5.4) (x1.4) 5.9436
10−3

−3

2.1516 (x7.8) (x6.1) 1.3882
10−3

0.7571 (x5.1) (x2.1) 1.4423
10

Aluminium 1100 annealed (1)

1.5135 (x2.6) (x3.1) 0.9764
10

CFRP (64)

3.8947 (x1)

Titanium 6Al 4V annealed (28)

r1=2 =q
cp

(x15.6) 0.1947
10−3 16.6995 (x1) −3

1.0815 (x3.6) (x81.2) 0.0373
10

(x10.2) 0.8349
10−3

6.4837 (x2.6) (x38.1) 0.2236
10−3

292

11.2

11

The Cost Factor for the Selection of Materials

Charts of Materials and Cost

If we draw a chart of materials (using logarithmic coordinates) in such way that for each material the ordinate in the origin belongs to the Young’s modulus, E, and the x-axis corresponds to the product of the density, q, and the cost, cp of the material, the conclusions defined from this chart ½log E; logðq
cp Þ with respect to cost are similar to that deduced with respect to weight with the support of charts ðlog E; log qÞ like that shown in Fig. 7.1 (Chap. 7). This way, once elaborated on the chart ½log E; logðq
cp Þ we could see that, e.g., in plates subjected to bending, materials situated on the same straight line of slope 3 would cost the same for the same stiffness. To achieve the same behaviour with lower cost, e.g., half of the cost, it would be necessary to choose one of the materials situated on a straight line parallel to that previously mentioned but with greater ordinate in the origin. Concretely, for half the cost, its ordinate in the origin should exceed, in decimal logarithm of 2, that of the previous one. As practical exercise we propose to graphically determine the order of selection for plates working under the force of bending, among the five materials previously proposed, the same could be done with all materials listed in Table 7.1: The first two solved cases can be used—(1) stiffness of a plate and lightness or (2) stiffness and minimum cost of the plate. In the first case, stiffness and lightness of the plate, identify in the diagram ðlog E; log qÞ—Fig. 7.1—those five materials as is done in Fig. 11.1a. In this diagram, from all the parallel straight lines of slope 3, that of larger ordinate in the origin would be the one passing through CFRP, then that passing through aluminium 1100 annealed, then that passing through titanium 6Al 4V annealed, and— finally—that passing through steels. In the second case—stiffness of the plate and minimal cost—the
 diagram we should use is the one that relates (see Fig. 11.1b) log E and log q
cp . Determining in it the straight lines of slope 3 that pass through each material, the order of preference would be also the same as that previously defined: mild steel > HSLA steel > aluminium 1100 annealed > CFRP > titanium 6Al 4V annealed. Analogously, the two exercises regarding the plate’s resistance to plastic deformation and lightness, or the plate’s resistance to plastic deformation and economy, can be graphically solved in Fig. 11.2a, b. For resistance of the plate to plastic deformation and lightness, the order is CFRP > titanium 6Al 4V annealed > HSLA steel > aluminium 1100 annealed > mild steel. For resistance to plastic deformation and economy, the order is mild steel > HSLA steel > aluminium 1100 annealed > CFRP > titanium 6Al 4V annealed. Complementarily, the order of selection based on the resistance of a plate without catastrophic breaking when prioritizing lightness or economy can be observed in Fig. 11.3a, b. When prioritizing lightness, the order is: CFRP > aluminium 1100 annealed > titanium 6Al 4V annealed > mild steel > HSLA steel. However, when prioritizing economy, the order is: mild steel > HSLA steel > aluminium 1100 annealed > CFRP > titanium 6Al 4V annealed.

11.2

Charts of Materials and Cost

293

Fig. 11.1 Stiffness, lightness, and cost (adapted from Pero-Sanz 2006). a log E–log q, b log E– log q
cp

In Sect. 11.3, with regard to the attributive analysis, a new order of preference for the selection from these five materials will be defined. However, in this new order, it is not considered that the selection of a material can be conditioned by the possibility of repairing a component if it deteriorates when operating. The owner of the vehicle, for instance, usually prefers repairing small dents produced in the metallic bodywork of the car than replacing the whole plate; as would be the case of the bodywork of the car manufactured with carbon fibre–reinforced polymer (whose failure occurs without previous plastic deformation).

11.2.1 Other Assumptions In other assumptions, different from those mentioned until this moment, and depending on if the weight or cost of the material to be chosen should be prioritized, it is possible to graphically operate in the same way as previously was done, i.e., determining the corresponding bundle of parallel straight lines in the suitable chart. This way, for instance, in the selection of materials to be used in plates that should resist buckling efforts, the equation that defines the critical load is: Fcr ¼ p2
E
I=L21 . Value that can be expressed as a function of the weight of the

294

11

The Cost Factor for the Selection of Materials

Fig. 11.2 Lightness, cost, and resistance to plastic deformation (adapted from Pero-Sanz 2006). a log r–log q, b log r–log q
cp

Fig. 11.3 Stable breaking, lightness, and cost (adapted from Pero-Sanz 2006). a log KIC–log q, b log KIC–log q
cp

11.2

Charts of Materials and Cost

295

plate by means of P1 ¼ L1
w1
t
q ¼ cte
t
q, from which we can deduce that for selecting materials that should be resistant to buckling and also be light, the relation would be E1=3 =q, whilst for those to be resistant to buckling and economy, the relation would be E1=3 =q
cp . In the case of cylindrical bars of equal length L1 and variable radius r working under bending force, the equation C1 ¼ E
I ¼ E
p
r 4 =4 ¼ cte
E
ðP=qÞ2 is verified and the relation is E1=2 =q ¼ cte=P. Consequently, less weight for a certain stiffness is obtained by maximizing the relation E1=2 =q. For that reason, lowest cost for a certain stiffness is determined by the relation E1=2 =q
cp . With regard to the resistance of cylindrical bars facing to plastic deformation, with bending moment M1 , we have that: r ¼ M1 =W1 ¼ 4
M1 =p
r 3 ¼ cte
ðq=PÞ3=2 or (which is the same) r2=3 =q ¼ cte=P. Less weight without risks of plastic deformation, for the same bending moment M1 , is obtained by maximizing the relation r2=3 =q. Analogously, to obtain lowest cost, the relation r2=3 =q
cp should be maximized. As is possible to deduce, in cylindrical bars subjected to bending without risks of catastrophic breaking for cracks of length c1 , the relation to minimize weight is: 2=3 KIC =q. In the case of avoiding unstable breaking and economy, the relation is: 2=3

KIC =q
cp . In columns or cylindrical bars working under buckling efforts, the relations are: E1=2 =q in the case of searching stability and lightness and E1=2 =q
cp in the case of attaining stability and economy. In the case of cylindrical bars subjected to tension, the relations are E=q for stiffness and lightness (E=q
cp for stiffness and economy); r=q for lightness and avoiding risks of plastic deformation (r=q
cp for resistance to plastic deformation and economy); and KIC =q for resistance to unstable breaking and lightness (KIC =q
cp for resistance to unstable breaking—with cracks of critical size c1 and economy). For the selection of materials to be used in pressure vessels, the relations are E=q for stiffness and lightness (E=q
cp for stiffness and economy); r=q for resistance to plastic deformation and lightness (r=q
cp for resistance to plastic deformation and economy); and KIC =q for lightness and resistance to catastrophic breaking (KIC =q
cp for resistance to unstable breaking and economy). For the selection of other elements, such as thin walls tubes, profiles, or others— in the face of tension, bending, buckling, stiffness, to avoid plastic deformation, and to avoid unstable breaking—it is possible to calculate in the same way the relations of properties that will make possible drawing up the corresponding bundle of parallel straight lines in the chart depending on whether lightness or economy is most important. In the case of elements subjected to torsion, e.g., a shaft of length L and polar moment of its cross-section equal to I0 ; the stiffness of a material, whose shearing modulus would be G, is specified by means of the total angle of torsion if the

296

11

The Cost Factor for the Selection of Materials

external torque couple (moment) is Mt , and thus we have: Mt
L=G
I0 . It is verified for a circular section of diameter, d, that I0 ¼ p
d 4 =32. Analogously, the maximum shearing stress under torsion, s, of a circular shaft and solid can be calculated through Mt
r=I0 . For that reason, we have s ¼ cte
Mt =d3 . One and another value (those mentioned in the previous paragraph)—in bars subjected to torsion—allow defining relations that minimize weight for the same stiffness ðG1=2 =qÞ, or for the same resistance to plastic deformation ðs2=3 =qÞ, or for the same 2=3 stability in the face of breaking ðKIC =qÞ. It is necessary to replace the value of q in these relations with the value of q
cp when, instead of considering weight, minimum cost should be prioritized. It is convenient considering that perpendicular and shearing stresses approximately verify the relation s ¼ r=2 and that G is equal to 3
E=8. Exercise 11.1 Demonstrate that, for a hollow (tube) shaft with de = 2
di , the maximum torsion angle and the maximum stress—if compared with those of a solid shaft of the same external diameter d—increase only approximately 7%, whilst the reduction in weight is almost 25%. – For the solid shaft: We take as references its torque couple (moment), Mt , its rotation angle h per unity of length, and the maximum shearing stress:

Mt ¼ G
h
Ip Ip ¼

p
d4 32

ð11:3Þ ð11:4Þ

16
Mt ¼G
h
r p
d3

ð11:5Þ

d ! t ¼ ðd  di Þ=2 2

ð11:6Þ

Mt ¼ G
h0
Ip

ð11:7Þ

smax ¼

– For the hollow (tube) shaft:

di ¼

"  4 #  p
4 p d 15
p
d4 4 4
de  di ¼
d  IP ¼ ¼ 32 32 2 32
16

ð11:8Þ

11.2

Charts of Materials and Cost

297

p
15
d4 p
15
d4 p
d4 16
h ! G
h0
¼G
h
! h0 ¼ 15 32
16 32
16 32 ¼ 1:07
h ð11:9Þ

Mt ¼ G
h0


s0max ¼ G
h0
r ! s0max ¼ G
1:07
h
r ! s0max ¼ 1:07
smax

ð11:10Þ

Decreased weight (1-Phollow/Psolid) per unity of length is calculated as the relation of the areas of the cross-sections (the following assumptions were made: qhollow = qsolid; Lhollow = Lsolid): 

Phollow Psolid

 h 2   2 i h 2   i   p
di p
de p
ðd=2Þ2 p
d d2 2 d 
  4 4 4 4 4  

p
d2  ¼ ¼ ¼ 2 Lsolid p
d2 d
4 4  qsolid 1 ¼ 1 ¼ 0:75 ! Phollow ¼ 0:75
Psolid 4 Lhollow qhollow

ð11:11Þ It would be enough to increase the diameter of the hollow shaft by approximately 2% to obtain the same performance. Note: It is important to use hollow shafts in applications where decreased weight is of remarkable importance, e.g., in aeronautic applications. Crankshafts, as explained in Pero-Sanz et al. (2018), have complex shape, although this question does not impede obtaining them through as-cast processes. Crankshafts have been traditionally manufactured from forged steel to maintain their functional integrity. Crankshafts began to be manufactured by casting a few decades ago for engines (Otto and diesel) of small and medium sizes. The work performed by crankshafts subjected to a torque couple (moment) of variable intensity makes them vulnerable to failure due to fatigue. Their manufacture and treatment depends on the requirements. Ductile iron crankshafts, for engines with capacities >1000 cm3, require careful control of their microstructure and hardness. Their microstructure should be pearlitic. Forged-steel crankshafts can be supplied, after forging, in either normalizing or as quenched and tempered (Chap. 12). The success of spheroidal cast iron has limited the utilization of forged-steel crankshafts to engines with special or severe requirements (competition vehicles): The crankshaft is obtained by machining and, after quenching and tempering, is subjected to nitriding surface treatment to increase resistance to fatigue. This kind of steel usually contains 0.3–0.4% C, alloyed (low alloy) with Mn–Cr–Mo. The maximum value of engine power (horsepower [hp]) is calculated as follows: PðhpÞ ¼

Cðkg
mÞ
N ðrpmÞ 716

ð11:12Þ

298

11

The Cost Factor for the Selection of Materials

The engine torque is calculated as: C ¼ 2
p
smax


r03 3

ð11:13Þ

where smax is the maximum shearing stress; and r0 is the radius of the part. This way, in the case of a 100-hp engine that worked between 3000 and 6000 rpm, with r0 ¼ 1:5 cm, the maximum shearing stress varies within:
3 1:5 cm
1001 m 716 cm ’ 24 kg
m ! 24kg
m ¼ 2
p
smax
3000rpm 3 ¼ 33 MPaðrmax Þ

ð11:14Þ


3 1:5 cm
1001 m 716 cm ’ 12 kg
m ! 12kg
m ¼ 2
p
smax
6000rpm 3 ¼ 17 MPa ðrmin Þ

ð11:15Þ

C1 ¼ 100hp
! smax

C2 ¼ 100hp
! smax

That is to say, a 33–17 MPa of fluctuating stress can be origin of fatigue and thus the origin of catastrophic failure. Forged crankshafts can be manufactured in a wide variety of steels (the steel shown in this example is the most simple and economic because it does not require quenching and tempering; however, it does require a nitriding treatment) depending on the application. The success of spheroidal graphite cast irons (cheap) in the manufacture of crankshafts limits the utilization of forged steels to high-performance engines. The crankshafts of diesel engines are usually obtained by forging. For instance, steel 722 M24 HT (B5970), 0.24% C–3% Cr–0.5% Mo is a nitrided steel and it would be adequate for competition engines: The crankshaft would be mechanized, subjected to heat treatment (quenching and tempering), and then nitrided to provide maximum resistance to fatigue.

11.3

Selection by Multiple Properties: Attributive Analysis

The selection could be also carried out—increasing the number of factors—as a function of cost and several properties. However, with the purpose of optimizing this selection of materials, the attributive parameter of importance that should be given to each property should be included when considering several properties. This way, for instance, assume three specific properties—K, L, M—ordered in decreasing importance (for instance, whose weight factors were 1/2, 1/3, 1/6, respectively), the valuation obtained by each material, from those available, would be given by the equation: 

1 Ki 1 Li 1 Mi
Q ¼ 100
þ
þ
2 Kmax 3 Lmax 6 Mmax

 ð11:16Þ

11.3

Selection by Multiple Properties: Attributive Analysis

299

where the subscript “max” refers to the material with maximum value for the considered property; and the subscript i indicates the value of the properties K, L, M in the considered material. Also, if instead of maximizing a property the objective is minimizing some property (e.g., cost, resistivity, damping, corrosion resistance, coefficient of expansion, thermal capacity, etc.), the relation Rmin =Ri should be considered in the previous equation (Rmin belongs to the available material with minimum cost, resistivity, thermal expansion, etc.). In the hypothetic situation of a material that would have maximum value in the three considered properties, its score would reach the upper limit value of 100 (or 1 if expressed as a decimal). The contribution of each property would be quantified as follows: that of the property K in 50 points; that of the property L in 33.3 points; and that of the property M in 16.7 points. Returning to the case previously considered—the selection from five materials for plates subjected to bending efforts—we assume that we give a contribution of one half to the cost, one third to materials that provide required stiffness with minimum weight, and one sixth to the material resistant to plastic deformation with minimum weight. This would allow conferring to the five materials taken as example the contributions listed in Table 11.2. That is to say, 58 points for mild steel, 38 points for HSLA steel, 23 points for aluminium 1100 annealed, 55 points for CFRP, and 22 points for titanium 6Al 4V annealed. The previous examples only allow initiating the reader into a systematic process for the selection of materials considering the properties and cost; however, some of the previous examples could seem set out for the selection of the best material to be used in the bodywork of a vehicle, nothing could be further from the truth. Because many factors must be considered for the rational selection (including cost) of materials, it is not easy to obtain a final solution. In the case of manufacturing, in series, a new complex factor for determining cp is simply the total number, N, of units that will be manufactured. Generally, the value of cp noticeable decreases with increasing N. This number is precisely and frequently the one that defines the material as well as process that must be employed (searching not only for uniformity of the product—and, for that reason, reliability as happens sometimes when preferring metallic materials—but also decreasing cost). We should think, for instance, in terms of the small economic importance that— in the final product—means a significant investment in tools and dies for deep-drawing operations when manufacturing bodyworks for vehicles using steel Table 11.2 Attributive analysis Mild steel

¼ 12


HSLA steel

¼


Aluminium 1100 annealed

¼


CFRP

¼ 12


Titanium 6Al 4V annealed

1 2 1 2

¼ 12



250

þ 13



0:7571

þ 16



2:1316 

16:6965 ¼ 0:58
3:1203  1 1 575 þ 3
3:8947 þ 6
16:6965 ¼ 0:38
250  1
1:5134 1
2:1516  1550 þ 3
3:8947 þ 6
16:6965 ¼ 0:23 250


250  



3:8947


0:7571

250 20;000

þ 13


250 29;000

þ 13





3:8947 3:8947

þ 16


3:8947

þ 16



1:0815


16:6965 16:6965

¼ 0:55

16:6965

¼ 0:22


6:4837 

300

11

The Cost Factor for the Selection of Materials

sheet. Regarding investments that can be only justified for large series of cars, these investments would not be reasonable when considering only thousands of cars. In the case of equipment for bodywork of polymer (which is least expensive but supports fewer unities), this carries more importance in the price of the product. From other economic considerations, the value of N probably justifies that most car manufacturers continue using steel sheets for car bodywork. The selection of materials is not always about just choosing a material but also a certain manufacture process. It seems convenient to advise that simplifications of processes sometimes involve microstructural changes that, at the same time, usually require a redesign of the nature of the material. Continuous annealing of cold-rolled steel sheet destined to drawing provides an example. The utilization of continuous-annealing furnaces means considerable savings compared with box annealing: decreases of 19% in energy consumption, 73% in personal costs, 90% in process times, and 48% in losses due to scraps. However, and as opposed to box annealing, obtaining textures , which favourable for drawing, cannot be well controlled (see Pero-Sanz 2006, pp. 502–504). To solve this question, other steel compositions must be used. The procedure for the selection of materials is interesting when we have a wide collection of data referring to the combined properties of different materials, with a weight given to each one of them, and the illustrative costs estimated. This selection methodology, which is linked to other considerations (technical or not)—like sometimes the prioritizing factor is basically aesthetic and fashion requirements—, can be complemented with the answer to this final difficult question: Simply, what is the best material? In any case, as we will conclude in Chap. 14, concerning the properties mentioned in this chapter, the concepts and methodology described in this book should be the first criterion for the selection of structural materials (conceptual design).

References Pero-Sanz, J. A. (2006). Ciencia e Ingeniería de Materiales. Estructura, Transformaciones, Propiedades y Selección, (5th ed.). Madrid (Spain): CIE Dossat. Pero-Sanz, J. P., Fernández-González, D., & Verdeja, L. F. (2018). Materiales para Ingeniería. Fundiciones Férreas. Madrid (Spain): Pedeca Press Publicaciones S. L. U.

Materials Resistant to Fatigue: Quenched and Tempered Steels

12

Abstract

This chapter is entirely dedicated to fatigue resistance as a very important factor for the selection of materials. From the materials resistant to fatigue, steels are one of the best options. The chapter, complemented with exercises, focuses on heat treatments, particularly the quenching and tempering treatments, applied to steels with the purpose of increasing their resistance to fatigue.

12.1

Materials Resistant to Fatigue: Quenched and Tempered Steels

In previous sections, we considered several examples of material selection as a function of the costs and the mechanical properties such as stiffness, resistance to plastic deformation (yielding), and resistance to catastrophic brittle fracture. All previously indicated referred to the behaviour of structural materials—either naturals, ceramics, metals, polymers, or composites—against static stresses. However, the most common failure of materials in service is that, when subjected to alternative loads, they break due to fatigue. This failure can take place for stresses well below yield strength. Other factors, for instance, the geometrical part design or other extrinsic factors, as the chosen material, are also important in the behaviour in the face of fatigue. The selection of materials for service under alternative loads probably constitutes one of the most difficult tasks for the materials engineer. Behaviour against fatigue of ceramic, polymeric and composite materials is not extremely satisfactory. All of them have low values of KIC . The best reliability facing to fatigue—and particularly in the case of a large number of cycles—corresponds to metallic materials. However, many metals and alloys do not resist indefinitely alternative loads no matter the value of the load (they do not have endurance limit). This justifies, for example, the requirement of periodically reviewing aircrafts to replace those © Springer Nature Switzerland AG 2019 J. A. Pero-Sanz Elorz et al., Structural Materials, https://doi.org/10.1007/978-3-030-26161-0_12

301

302

12

Materials Resistant to Fatigue: Quenched and Tempered Steels

elements, such as those made from aluminium, which presumably have incubated cracks due to a high number of cycles in service. Steels offer the most interesting possibilities from all materials for resistance to fatigue. High resistance and toughness are required for a good behaviour against fatigue. In steels with ferritic–pearlitic structure, increasing carbon content increases the ultimate tensile strength, ru ; on the contrary, toughness decreases. In Fig. 12.1, it is possible to compare the resistance ranges and carbon percentages of structural steels, medium carbon steels, railway steel, and wires for pretensioned concrete (all of them with ferritic–pearlitic structure). In Fig. 12.1, it is also possible to see several of these steels, with the same carbon content, but with the structure of tempered martensite. They are more resistant to fracture and simultaneously more tough than with ferritic–pearlitic structure. We can compare, for instance, structural steels (ferritic–pearlitic) with steel for pressure boilers (quenched and tempered) or medium-carbon steels (ferritic–pearlitic) with high-strength and ultra-high–strength quenched and tempered steels. This simultaneous increase of resistance and toughness in steels, due to the structure of precipitated cementite, is the result of a high tempering of martensite. Complementarily, when steel must support alternative loads during a large number of cycles, this finely dispersed cementite also provides a structure more favourable than lamellar pearlitic cementite because it enlarges the steps of clustering and propagation of cracks during fatigue (see Pero-Sanz et al. 2017).

Fig. 12.1 Some types of steels (elaborated with data from Pero-Sanz 2006)

12.1

Materials Resistant to Fatigue: Quenched and Tempered Steels

303

In fact, only steels with a medium carbon content (of approximately 0.40% C), quenched and tempered (quenched and tempered steels)—and some austempered ductile cast irons—are materials suitable for parts with responsibility exposed to fatigue of a large number of cycles (as happens in the crankshafts of vehicles) (see Pero-Sanz et al. 2018a, b). Fundamentals for the selection of steels destined to quenched and tempered parts, as a function of the equivalent round bar diameter and of the cooling medium used for quenching, defines oil as the best refrigerant to minimize the risks of cracks. Complementarily, it is possible to mention as habitual—consider, e.g., a large series of parts for the automotive industry that are quenched and tempered for resistance to fatigue—that the equivalent round bar should not exceed 120 mm in diameter. The range of resistances is usually 900–1300 MPa. Defined requirements, as a function of the dimensions of the parts and the loads they should support, should rationally guide the selection of suitable steels if we have available a group of steels as those shown in Fig. 12.2 and Table 12.1. Steels that, after tough quenching and tempering, allow obtaining, at the point of slowest cooling of the part, a structure of tempered martensite and with the required resistance. It is obvious that suitable steel for a part with 120-mm–equivalent diameter and resistance of 1300 MPa would be also adequate to manufacture a part of equivalent diameter equal to 20 mm and resistance ru equal to 900 MPa; only with increasing the temperature and tempering time. Proceeding this way, we would be wasting the characteristics of this steel (as it would be more expensive due to being more alloyed).

1300

1100

Resistance (MPa)

Fig. 12.2 Selection of steels for alternative loads

900

30

60

90

Equivalent diameter (mm)

120

304

12

Materials Resistant to Fatigue: Quenched and Tempered Steels

Table 12.1 Characteristics of steels shown in Fig. 12.2 Steel C

Mn

Si

P (max) S (max) Ni

A B C D E F G H I J K L

0.60/0.90 0.60/0.90 0.60/0.90 0.60/0.80 0.60/0.80 0.60/0.80 0.60/0.80 0.60/0.80 0.60/0.80 0.60/0.80 0.60/0.80 0.60/0.80

0.10/0.40 0.10/0.40 0.10/0.40 0.10/0.40 0.10/0.40 0.10/0.40 0.10/0.40 0.10/0.40 0.10/0.40 0.10/0.40 0.10/0.40 0.10/0.40

0.035 0.035 0.035 0.035 0.035 0.035 0.035 0.035 0.035 0.035 0.035 0.035

0.27/0.33 0.32/0.38 0.37/0.43 0.27/0.33 0.32/0.38 0.37/0.43 0.27/0.33 0.32/0.38 0.37/0.43 0.24/0.30 0.30/0.36 0.37/0.43

0.035 0.035 0.035 0.035 0.035 0.035 0.035 0.035 0.035 0.035 0.035 0.035

– – – 0.80/1.10 0.80/1.10 0.80/1.10 1.80/2.25 1.80/2.25 1.80/2.25 2.75/3.25 2.75/3.25 2.75/3.25

Cr

Mo

0.85/1.15 0.85/1.15 0.85/1.15 0.70/0.90 0.70/0.90 0.70/0.90 0.70/0.90 0.70/0.90 0.70/0.90 0.70/0.90 0.70/0.90 0.70/0.90

0.15/0.25 0.15/0.25 0.15/0.25 0.20/0.30 0.20/0.30 0.20/0.30 0.25/0.35 0.25/0.35 0.25/0.35 0.30/0.40 0.30/0.40 0.30/0.40

Obviously, the selected steel should have more content in alloying elements— i.e., it should be more alloyed—when the equivalent round bar diameter (Fig. 12.2) is greater with the purpose of increasing the hardenability due to solid solution of the alloying elements in the austenite. With that mentioned, resistance after tough tempering will be greater. All of this can be also used to make critical assessments about steels with resistance to fatigue fixed in each country by national standardization organizations; that is to say, to evaluate if there are among these standards several steels that can be used for the same requirements of equivalent diameters and resistances and—on the contrary—if there are requirements not fulfilled by the typified steels.

12.2

Heat Treatments of Steels

The set of controlled heating and cooling cycles to which a solid alloy can be subjected with the purpose of modifying the microstructure (and consequently the properties) without changing the chemical composition of the alloy is called “heat treatment.” The phrase “thermomechanical treatment” indicates treatments that produce structural modification in the alloy due to heating and, simultaneously, mechanical forming of the solid alloy. The phrase “thermochemical treatment” indicates treatments that involve a chemical modification in the periphery of the alloy. As previous treatment, it is possible to consider the “homogenization annealing.” Homogenization annealing has as objective providing a uniform chemical composition in the “as-cast” structure for any solid solution (see Chap. 5 in Pero-Sanz et al. 2017).

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Heat Treatments of Steels

305

Other treatments that could be mentioned as a consequence of presenting only transformations in solid state through solubility changes include hyperquenching, aging, and hyperquenching + aging (i.e., aluminium alloys) (see Chap. 7 in Pero-Sanz et al. 2017). Other important heat treatments, whose purpose is eliminating the “work-hardening” in parts deformed at low temperature, are called “recovery annealing” and “process annealing” (see Chap. 1 in Pero-Sanz et al. 2017). Exercise 12.1 A cold-drawn copper wire recrystallizes by 50% at a temperature of 400 °C in 5 min and at 300 °C in 40 h. How long would take partial recrystallization (50%) at room temperature? Define the activation energy for the diffusion Q and its relation to melting temperature. If copper is used in electrical applications, what would be the requirements? The time required for recrystallization follows an equation of Arrhenius type as corresponds to a process of transformation by clustering and growth: cold worked matrix ! recrystallized matrix:
 1 Q / A  exp  tx T

ð12:1Þ

Qdif ¼ R  Q ’ 18  R  TM , where TM is the melting point of copper in K; tx is the time required to have a fraction x recrystallized; and R is the Boltzmann constant. Using the time in minutes:
 1 Q ¼ A  exp  5 min 400  C þ 273

ð12:2Þ


 1 Q ¼ A  exp  2400 min 300  C þ 273

ð12:3Þ


 2400 min Q  100 K ¼ exp 5 min 673 K  573 K

ð12:4Þ

If we take Napierian logarithms: Qdif 6:17  573  673 K ¼ 23;808 K ’ 100 R

ð12:5Þ

Thus, Qdif (activation energy required for recrystallization [elimination of defects by self-diffusion of vacancies in copper]) is: Qdif ’ 18  R  TM ! 2 

cal cal  23; 808 K ’ 2   18  TM ! TM ¼ 1323 K mol  K mol  K ðTM ÞCu ¼ 1050  Cð’1083  CÞ

ð12:6Þ ð12:7Þ

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Materials Resistant to Fatigue: Quenched and Tempered Steels

At room temperature (300 K), copper does not recrystallize: 1 t300


 23;808 K /A  300 K

ð12:8Þ

Comparing with t50% at 300 °C or 400 °C, we see: ðt50% Þ300 K ’ 6:3  1019 min ðinviableÞ

ð12:9Þ

Copper for electrical applications must have a high purity (99.99%) because impurities reduce electrical conductivity. Ag (’0.1%) increases the conductivity of pure copper. Exercise 12.2 The time required to achieve the recrystallization of 50% of a cold-worked metallic material (98%) is determined by an equation like that mentioned as follows: 1 t50%


 Q ¼ A  exp  T

ð12:10Þ

For another cold reduction of 80%, at the same temperature T, the value of Q would be smaller, the same, or greater? The activation energy for the recrystallization process is equivalent to the self-diffusion energy of the corresponding material: The transformation is of Arrhenius type. The clustering of recrystallized grains is very fast and the growth depends on T and Q. It is possible to demonstrate that: Q’

Qdif ’ 18  TM R

ð12:11Þ

TM is the melting point of the material in K. Q does not depends on T. Increasing temperature implies that the recrystallization time decreases exponentially. Increase (decrease) of the cold deformation, at constant T of annealing, would affect the value of the false constant A. If the degree of deformation increases, A also increases (tx , recrystallization time decreases).

12.2.1 Cooling of Metals in Heat Treatments with Previous Austenization Several heat treatments of steels require first achieving by heating an austenitic structure; the objective is, from these high temperatures, transforming the structure by cooling in the desired constituents. This is the case of quenching, normalizing, supercritical annealings, patenting, austempering, etc.

12.2

Heat Treatments of Steels

307

Cooling can be continuous and performed in different ways: cooling inside the furnace, in air, in water, in oil, etc. In other cases, cooling is isothermal; achieved, for instance, by means of submerging the part in a molten salt bath. The microstructure steel will have at the end of heat treatment will be the result of the interaction between the cooling curves—of the periphery and nucleus of the part—and the time–temperature–transformation (TTT) curve. Depending on the cooling rate from the austenitic state, the same part of steel can have in an inner point martensitic, bainitic, fine pearlite, or coarse pearlite structure. Factors that conditionate the microstructure of a part at the end of heat treatment with previous austenization can be summarized in three factors: – The TTT curve of the considered steel. – The heat-transfer factor of the cooling medium (i.e., the aptitude to draw heat from the surface of the part). – The size of the part to be treated.

12.2.1.1 The TTT Curve The TTT curve of steel (Fig. 12.3), and concretely its position with respect to the origin of times, depends on austenite grain size and composition of the steel. The curve will be far from the origin of times the greater the grain size and the greater the quantity of elements in solid solution the austenite possesses. For that reason, we say that steel has more “hardenability,” i.e., the easier the quenching, the greater the distance of the TTT curve from the origin of times. This way, the TTT curve for a highly alloyed steel will be very displaced from the origin of times and martensite can be obtained by simply air-cooling from the austenization temperature. Steels that can be quenched by air-cooling are called “self-hardening steels.” The efficiency of alloying elements in quenching requires their previous solubilization in austenite. 12.2.1.2 Heat-Transfer Factor of the Cooling Agent The heat-transfer factor of a cooling agent can be evaluated not only qualitatively (water medium has greater heat-transfer factor than oil and oil more than the air) but also quantitatively. This question can be observed when cooling a steel surface. Heat flow through the interface metal-cooling agent (medium) follows the Newton law: q ¼ A  ð TM  TS Þ

ð12:12Þ

where: TS is the temperature in the surface of the part; TM , is the temperature of the cooling medium; and A is the coefficient of thermal conductivity in the interface. Coefficient A changes during cooling of the surface of the part. This happens, for instance, if the cooling medium is water at rest. In fact—as indicated in Fig. 12.4— in a first moment, when the contact between water and the steel surface (which is at

308

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Materials Resistant to Fatigue: Quenched and Tempered Steels

Fig. 12.3 The TTT curve of steel 0.57% C, 0.70% Mn, 0.20% Si, 0.70% Cr, 1.70% Ni, 0.30% Mo, 0.10% V; austenization temperature, 880 °C (Pero-Sanz et al. 2017)

the austenization temperature) is established, the water is vaporized and the vapour bubbles render difficult the heat transfer (the value of A is small). In a second step, when the bubbles detach from the surface and the heat-convection mechanism in the liquid begins, then heat transfer is faster (A increases). Finally, in a third step of direct contact between water at rest and the surface, the decrease of temperature in the steel surface is slow (A decreases). In the considered situation—three different values of the heat-transfer coefficient A—we can consider an average value M. We call the heat-transfer factor (H) of the cooling agent to the quotient (M/2k) where k is the thermal conductivity of the steel (that can be considered constant for all steels with the exception of those that have high chromium contents). The value of H is greater when the capacity of heat absorption of the cooling medium is greater. This way, the value of H is greater for water than for air. And, as is logical, the value of H increases for the same coolant if the cooling medium is stirred. The values of H for different coolants and different stirring degrees can be calculated by means of experimental methods. Considering an “ideal” cooling medium, we could say that it has infinite heat-transfer factor if the part periphery instantaneously reaches the temperature of the coolant when the part is submerged in it. It should be noted that this would not imply that the temperature of the part’s nucleus would cool at the same rate: The part’s nucleus would decrease the temperature progressively at the same time that heat was flowing from the part’s inside to the surface in contact with the ideal cooling medium.

Heat Treatments of Steels

309

Temperature

12.2

Time Fig. 12.4 Cooling of the steel surface

Exercise 12.3 The heat-transfer factor H is expressed by the quotient H ¼ M=2k, where M is the coefficient of heat transfer in the interface steel/coolant; and k is the thermal conductivity of steel. Deduce the value of H for quenching in water assuming that the diameter of the steel round bar is expressed in inches. Data : k ’ 0:07 cal=cm   C  s; M ¼ 101 cal=cm2   C  s H¼

M 101 cal=cm2  s   C 2:54 cm ¼ ¼ 1:8 in.1  2  k 2  0:07 cal=cm   C  s 1 in.

ð12:13Þ

It is possible to demonstrate that: – If quenching is performed in brine water with high-velocity stirring ; H ’ 5. – If quenching is performed in oil, H ’ 0:5 (see Table 12.2).

12.2.1.3 Round Size (Diameter) The size of the steel part is another factor that has influence on the microstructure achieved at the end of heat treatment by cooling from the austenitic state. We

310

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Materials Resistant to Fatigue: Quenched and Tempered Steels

Table 12.2 Heat-transfer factor of the quenching medium (Pero-Sanz 2004, p. 102) Stirring

Air

Salts

Oil

Water

Saltwater

Without Weak Moderate Average Strong Very strong

0.02 – – – – 0.08

0.25–0.30 0.30–0.35 0.35–0.40 0.40–0.50 0.50–0.80 –

0.25–0.30 0.30–0.35 0.35–0.40 0.40–0.50 0.50–0.80 0.80–1.10

0.90–1.00 1.00–1.10 1.20–1.30 1.40–1.50 1.60–2.00 4.00

2.00 2.00–2.20 – – – 5.0

assume a cylinder of diameter D1 and so long that the heat release would be only radial (not through the extremes of the cylinder) when cooling in a coolant of heat-transfer factor H1 (for instance, no-motion in still water [see Fig. 12.5a]). A gradient of temperatures is established between the periphery and the axis of the cylinder when cooling. This gradient changes with time because the calories (the heat stored inside of the cylinder) that must be released are proportional to the volume. On the contrary, heat that is released to the coolant is proportional to the lateral surface. The ratio of volume to surface is equal to D1 =4: For that reason, the relation between the stored heat and the heat that can be released increases with increasing diameter. This way, for a same coolant of heat-transfer factor H1 (for instance, no-motion in still water, calm water), when comparing the cooling curves of periphery and nucleus of two steel cylinders whose diameters were D1 and D2 ðD1 \D2 Þ, we can see (Fig. 12.5a, b) that for the round bar of diameter, D2 , both the nucleus and the periphery cool slower than they do in the round bar of diameter D1 . After a time t1 , not only is the temperature of the periphery greater in the round bar of diameter D2 , its gradient of temperature between the periphery and the nucleus is greater than that in the round bar of diameter D1 . On the other hand, if a coolant with less heat-transfer factor than H1 (e.g., air) is used, these gradients would be smaller in both cases. Compare Fig. 12.5a, c , which correspond to round bar D1 . Or compare Fig. 12.5b, d , which correspond to round bar D2 . Exercise 12.4 Continuous-cooling transformation curve of a steel SAE–1038 (comprising 0.38% C, 0.70% Mn, and 0.25% Si) is showed in the Fig. 12.6. (1) Explain why the percentages of ferrite and pearlite depend on the cooling rate (centre of the bar). (2) With a fast cooling, calculate the critical rate for quenching in oil. (1) The temperature A3 (start of the transformation c ! a) moves towards lower temperatures when increasing the cooling rate (Fig. 12.7). This way, in round bars of different diameter (inches) cooled in air (Table 12.3).

12.2

Heat Treatments of Steels

311

Fig. 12.5 Cooling of the surface (V p ) and the nucleus (V n ) when the rod size is changed (D1 \D2 ) the same as the heat-transfer factor of the coolant: a round bar of diameter D1 cooled in water; b round bar of diameter D2 cooled in water; c round bar of diameter D1 air-cooled; d round bar of diameter D2 air-cooled

That is to say, in round bars of greater diameter, the ferrite content (pearlite) approximates to that deduced in an equilibrium cooling of the metastable Fe–C diagram: %Pearlite ’ 0:38=0:77 ’ 50%

ð12:14Þ

In smaller round bars (faster cooling), the proportion of pearlite is 3 mm, for oil quenching, in the round bar of diameter D1 ? Yes, increasing the austenization temperature will do this because the hardenability of the steel moves the TTT curves (CCT) from the origin of times. 6. Would you use this steel to build up welded metallic structures? No, because in the heat-affected zones (HAZ) close to weld-bead stresses, deformations and cracks, derived from the increase of volume caused by martensitic transformation, could appear. 7. Would you use the steel for case-hardening in the manufacture of gears? Probably yes. As we will see in the last exercise of this chapter, the thickness required for a cemented layer is approximately 2 mm and the steel, in the referred terms, will satisfy the requirements of the quenching penetration.

12.2.5 Subcritical Heat Treatments The term “subcritical heat treatments” indicates heat treatments in which the allotropic transformation of alpha iron into gamma iron does not take place. Subcritical annealing and tempering are two examples of this group of treatments. In this section, we also include “stress-relieving” annealing and tempering treatments that, as their name indicates, have as purpose relieving internal stresses in steels by heating them to 150 °C without changing the micrographic constituents. The recrystallization annealing of cold-worked ferritic steels—due to continuous annealing or in batch annealing—is another subcritical treatment.

12.2.5.1 Subcritical Annealing Subcritical annealing is other type of heat treatment used to soften steels. This treatment is performed by heating the steel part to a temperature close to but lower than the Ae and holding it for a time at that temperature (Fig. 12.18c). The cooling rate after subcritical annealing is not important because there will not be any allotropic transformations. The structure obtained at the end of holding at this constant temperature is of globular cementite distributed in a ferrite matrix. Obtaining his structure softens the steel no matter what the initial structure, whether it was martensite, bainite, or pearlite. The softening, after being at the subcritical temperature of the treatment, is enough for the purpose required in parts to be machined (Verdeja et al. 2009). The softening rate, as a function of the holding time at that temperature, is fast at the beginning and decreases asymptotically. A certain time later, the duration of the treatment can be increased, but the softening is not increased. This type of annealing is also usually known as “spheroidizing annealing,” due to the shape that the cementite

12.2

Heat Treatments of Steels

329

adopts. Thus, this name is reserved for intercritical annealing. For some alloyed self-hardening and tool steels, the softening achieved using a subcritical annealing is not enough and full or intercritical annealing heat treatments are required. With the limitations indicated in the previous paragraph, subcritical annealing offers several advantages with respect to the annealing treatments described in Sect. 12.2.4. First, energy is saved because temperatures greater than Ae are not required. Moreover, deformation and/or cracking risks due to allotropic transformations are avoided. The dimensional changes, due to the allotropic transformation alpha ! gamma during heating until reaching the austenization temperature, are inherent to the process and they make necessary that the heating until reaching the austenization temperature is slow. The larger the size of the part, the slower the heating must be. Risks of deformation, stress, and cracking during heating to austenization temperature are common in full-annealing treatments, in quenching, in normalization, in forging in gamma state, etc.

12.2.6 Isothermal Heat Treatments Despite the fact that these treatments are usually called “isothermal treatments,” it would be more reasonable calling them “heat treatments with isothermal cooling.” All of them, including isothermal annealing (see Fig. 12.19), have in common that the steel parts are heated to the austenization temperature and, from that temperature, they are isothermally cooled by submerging them in molten salts, molten lead, or other coolant liquid that allows constancy of temperature during the transformation of austenite.

12.2.6.1 Patenting Patenting is an isothermal treatment that is usually given, as final operation, to steel wires of 0.7–0.9% C, which require high mechanical resistance under tension stress because they are used in pre- and post-stressed concrete. To achieve these characteristics, the wire—after being austenitized—is submerged in a bath of molten lead (or in salts) at a temperature corresponding to the lower part of the TTT curve of the steel (Fig. 12.22). This has as objective transforming the austenite into a very fine pearlite, with distances between lamellas of cementite, S0 , from 0.1 to 0.2 µm. With this treatment, ultimate tensile strength ru of approximately 1600 MPa and elongations At from 5 to 10% are reached. Sometimes patenting is used as intermediate treatment against work-hardening during wire drawing, to (apart from finally achieving the mentioned high values of ru or greater) facilitate the operation of wire drawing. In fact, to wire drawing is convenient for a structure of fine cementite, also previously achieved by patenting. However, while we are drawing the wire (cold-forming), and the section of the wire is decreased, ferrite is acquiring a work-hardening that progressively makes more difficult the process and would end up making the process unfeasible. A new patenting, to eliminate the cold-worked ferrite, allows continuing with the cold reduction of the section.

330

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Materials Resistant to Fatigue: Quenched and Tempered Steels

Fig. 12.22 Patenting

The ultimate tensile strength that can be obtained in eutectoid binary steels, by means of a suitable combination of fine pearlite and cold-forming due to drawing in wires with final diameter of 0.08 mm, exceeds 3000 MPa. Exercise 12.12 The drawing of wires of circular section—axiometric reduction of the section passing the wire through a drawing die, generally conic, and pulling the wire—is one of the simplest operations of forming by plastic deformation. The diameter of the initial hot-rolled round bar is 5–20 mm. The carbon content comprises all the range from the extra-mild steel to the eutectoid, or even slightly hypereutectoid composition. Limits to the content of Mn (1.65%), Si (0.60%), and Cu (0.40%) are usually established to keep the denomination of the carbon steel round bar. Wires of mild steel are drawn with the structure resulting from the rolling or after subcritical annealing treatment of intermediate recrystallization after a series of drawing passes. Wires of medium and high carbon content, especially for the manufacture of high-resistance steel (tires wires, piano wires), are drawn from a structure of lamellar pearlite—as fine, homogenous, and perfect as possible—obtained by means of isothermal transformation at approximately 500 °C habitually in lead bath (“patented”) or by means of controlled cooling from the temperature of the rolling end (“Stelmor” technology or similar). The spacing between lamellas of pearlite achieved is