Instructors Manual for The Science & Engineering of Materials 7th Edition by Donald Askeland & Wende

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COMPLETE Elaborated Test Bank for Instructor’s Solutions Manual to accompany The Science and Engineering of Materials, 7Ed. by Donald R. Askeland Wendelin J. Wright.

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The Science and Engineering of Materials, 7Ed.by Donald R. Askeland & Wendelin J. Wright.

ISBN-10 0357447867 ISBN-13 978-0357447864 Instructor’s Solutions Manual to accompany for The Science and Engineering of Materials, 7Ed.by Donald R. Askeland & Wendelin J. Wright.

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INSTRUCTOR'S SOLUTIONS MANUAL TO ACCOMPANY

The Science and Engineering of Materials Seventh Edition

DONALD R. ASKELAND University of Missouri—Rolla, Emeritus

WENDELIN J. WRIGHT Bucknell University

Contents Chapter 1 .......................................................................................................................................... 1 Chapter 2 ........................................................................................................................................13 Chapter 3 ........................................................................................................................................25 Chapter 4 ........................................................................................................................................77 Chapter 5 ........................................................................................................................................97 Chapter 6 ......................................................................................................................................121 Chapter 7 ......................................................................................................................................153 Chapter 8 ......................................................................................................................................173 Chapter 9 ......................................................................................................................................197 Chapter 10 ....................................................................................................................................225 Chapter 11 ....................................................................................................................................251 Chapter 12 ....................................................................................................................................271 Chapter 13 ....................................................................................................................................293 Chapter 14 ....................................................................................................................................313 Chapter 15 ....................................................................................................................................325 Chapter 16 ....................................................................................................................................335 Chapter 17 ....................................................................................................................................349 Chapter 18 ....................................................................................................................................375 Chapter 19 ....................................................................................................................................385 Chapter 20 ....................................................................................................................................403 Chapter 21 ....................................................................................................................................415 Chapter 22 ....................................................................................................................................431 Chapter 23 ....................................................................................................................................443

Chapter 1: Introduction to Materials Science and Engineering

1-1

Define materials science and engineering (MSE).

Solution: Materials science and engineering (MSE) is an interdisciplinary field that studies and manipulates the composition and structure of materials across length scales to control materials properties through synthesis and processing.

1-2

What is the importance of the engineering tetrahedron for materials engineers?

Solution: Structure, properties and performance all depend on the route in which a material is processed. We cannot predict the end properties for a material until we have specified a process to produce the component. Using the same material, but changing the way it is processed will result in different structure, properties and performance of that material.

This is applicable to all material systems.

1-3

Define the following terms: (a) composition; (b) structure; (c) synthesis; (d) processing; and (e) microstructure.

Solution: (a) The chemical make-up of a material. (b) The arrangement of atoms, seen at different levels of detail. (c) How materials are made from naturally occurring or man-made chemicals. (d) How materials are shaped into useful components. (e) The structure of an object at the microscopic scale.

1-4

Explain the difference between the terms materials science and materials engineering.

Solution: Materials scientists work on understanding underlying relationships between the synthesis and processing, structure, and properties of materials. Materials engineers focus on how to translate or transform materials into useful devices or structures.

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1-5

The myriad materials in the world primarily fall into four basic categories; what are they? What are materials called that have one or more different types of material fabricated into one component? Give one example.

Solution: Metals, polymers and ceramics. The addition of one or more of these to a single system is called a composite. An example of a composite material is fiberglass.

1-6

What are some of the materials and mechanical properties of metals and alloys?

Solution: Metals and alloys have good electrical and thermal conductivity, high strength, ductility and formability, and high stiffness.

1-7

What is a ceramic, and what are some of the properties that you expect from a ceramic?

Solution: Ceramics tend to have very high compressive strengths, but behave in a brittle (glasslike) manner. They have very high melting temperatures. Poor thermal conductivity and electrical conductivity make ceramics behave as an insulator instead of a conductor.

1-8

Make comparisons between thermoplastics and thermosetting polymers (a) on the basis of mechanical characteristics upon heating, and (b) according to possible molecular structures.

Solution: Thermoplastics tend to soften with elevated temperature exposure with gradually decreasing viscosity. Thermosetting polymers do not soften with elevated temperature exposure; instead they will remain hard and will degrade, possibly charring with prolonged exposure.

Thermoplastics consist of long chain molecular arrangements of covalently bonded carbon atoms with various side groups. Thermosetting polymers tend to be a complex 3D arrangement usually deviating from the clearly defined long-chain molecular arrangement.

1-9

Give three examples of composites that can be fabricated.

Solution: Metal matrix composites (MMC) – A metal matrix reinforced with a ceramic material in the form of particles, whiskers or fibers. Example: Cobalt alloy reinforced with tungstencarbide particulates.

Polymer matrix composites (PMC) – A polymer matrix reinforced with a ceramic material in the form of whiskers or fibers. Example: Kevlar or fiberglass.

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Ceramic matrix composites (CMC) – A ceramic matrix reinforced with ceramic or metallic material in the form of whiskers or fibers. Example: Carbon fibers in an alumina (Al2O3) matrix.

1-10

For each of the following classes of materials, give two specific examples that are a regular part of your life: (a) metals; (b) ceramics; (c) polymers; and (d) semiconductors.

Specify the object that each material is found in and explain why the material is used in each specific application. Hint: One example answer for part (a) would be that aluminum is a metal used in the base of some pots and pans for even heat distribution.

It is also a lightweight metal that makes it useful in kitchen cookware. Note that in this partial answer to part (a), a specific metal is described for a specific application.

Solution: (a) Aluminum was described in the problem statement. Stainless steel is used for flatware. It is easily formed and has good corrosion resistance, strength, and hardness. (b) Two specific examples of polymers are polystyrene and polytetrafluoroethylene also known as Teflon. Styrofoam is polystyrene rigid foam insulation that is used for cups that keep hot drinks warm. Teflon is used as a coating on the inside of kitchen cookware such as frying pans because it prevents food from sticking to the pan while cooking. (c) Two examples of semiconductors are silicon doped with phosphorus (n–type) and silicon doped with boron (p–type). Both types of impurities convert silicon from a poor into a useful conductor. Both n and p–type semiconductors are contained in the semiconductor device called a diode, so that at the junction between both types, current is able to flow. A diode blocks current in one direction while allowing current flow in the other direction. A device that uses batteries, e.g. a remote control or a calculator, often contains a diode that protects the device if the batteries are inserted backward. The diode blocks the current from leaving the battery if it is reversed, protecting the sensitive electronics in the device. Another semiconductor is the compound semiconductor AlxGa1-xAs, which is used in lasers. (d) Ceramics are compounds that contain metallic and nonmetallic elements. Two specific examples are tungsten carbide and magnesia. Tungsten carbide is often bonded with cobalt and/or nickel. Tungsten carbide is used mainly in tips for metal cutting tools (knives can be made with this) because of its good wear resistant characteristics.

Magnesia is a heat resistant ceramic that is used in liners for ovens. Magnesia can resist high temperatures but is susceptible to thermal stress cracking.

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1-11

Describe the enabling materials property of each of the following and why it is so: (a) steel for I-beams in skyscrapers; (b) a cobalt chrome molybdenum alloy for hip implants; (c) polycarbonate for eyeglass lenses; and (d) bronze for artistic castings.

Solution: (a) Steel for I-beams in skyscrapers must be strong in order to bear large mechanical loads. (b) A cobalt chrome molybdenum alloy for hip implants must be biocompatible, meaning that it must not degrade when inserted into the body nor be toxic or otherwise dangerous. (c) Polycarbonate for eyeglass lenses must be transparent and impact-resistant. (d) Bronze can be melted and poured into molds to be shaped. It is also fairly corrosion resistant (which is important for outdoor sculptures). Over long periods of time when subjected to an outdoor environment, bronze will develop an oxide known as a patina.

The patina protects the bronze from further corrosion.

1-12

Describe the enabling materials property of each of the following and why it is so: (a) aluminum for airplane bodies; (b) polyurethane for teeth aligners (invisible braces); (c) steel for the ball bearings in a bicycle’s wheel hub; (d) polyethylene terephthalate for water bottles; and (e) glass for wine bottles.

Solution: (a) Aluminum has a high strength to weight ratio. Thus it has the strength needed to withstand the forces imposed on an airframe, but keeps the weight of the airplane low compared to other metals. The lighter the airplane body, the less force it takes to lift the plane into the air. This results in less fuel being used and a reduction in operating costs.

Aluminum also has good corrosion resistance. (b) Polyurethane for teeth aligners must be highly formable and transparent. Since teeth aligners are worn during the day, a transparent material is desirable to make them less conspicuous. A unique set of teeth aligners must be produced for each individual, and so the polyurethane must be easily molded. Computer software is used to produce computer models of a person’s teeth during various stages of the correction process. A rapid prototyping tool is used to create physical models of the person’s teeth at each stage. A sheet of polyurethane is then heated and formed onto the models to produce the teeth aligners.

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(c) Ball bearings in a bicycle’s wheel hub reduce friction between metal surfaces.

Therefore, steel is used because it has high strength and hardness. (d) Polyethylene terephthalate is easily formed by a blow molding process and is recyclable, critical properties for mass-produced water bottles. (e) Glass has high chemical resistance; thus, glass bottles are used to preserve the taste of the wine contained in them.

1-13

What properties should an engineer consider for a total knee replacement of a deteriorated knee joint with an artificial prosthesis when selecting the materials for this application?

Solution: Properties that should be considered are those relating to strength (fatigue, tensile and compressive) since the knee sustains static, dynamic and cyclic loads. Hardness will promote wear resistance. Modulus of elasticity similar to that of the human bone otherwise other problems will occur. Chemical stability in regards to corrosion resistance and cellular toxicity to prevent negative reactions to the material selected.

The material needs to have the ability to bond with the residual bone material and have longevity in order to avoid frequent replacement.

1-14

Write one paragraph about why single-crystal silicon is currently the material of choice for microelectronics applications. Write a second paragraph about potential alternatives to single-crystal silicon for solar cell applications. Provide a list of the references or websites that you used. You must use at least three references.

Solution: Answers will vary.

1-15

Coiled springs should be very strong and stiff. Silicon nitride (Si3N4) is a strong, stiff material. Would you select this material for a spring? Explain.

Solution: Springs are intended to resist high elastic forces, where only the atomic bonds are stretched when the force is applied. The silicon nitride would satisfy this requirement; however, we would like to also have good resistance to impact and at least some ductility (in case the spring is overloaded) to ensure that the spring will not fail catastrophically. We also would like to be sure that all springs will perform satisfactorily.

Ceramic materials such as silicon nitride have virtually no ductility, poor impact properties, and often are difficult to manufacture without introducing at least some small flaws that cause failure even for relatively low forces. The silicon nitride is NOT recommended.

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1-16

Temperature indicators are sometimes produced from a coiled metal strip that uncoils a specific amount when the temperature increases. How does this work; from what kind of material would the indicator be made; and what are the important properties that the material in the indicator must possess?

Solution: Bimetallic materials are produced by bonding two materials having different coefficients of thermal expansion to one another, forming a laminar composite. When the temperature changes, one of the materials will expand or contract more than the other material. This difference in expansion or contraction causes the bimetallic material to change shape; if the original shape is that of a coil, then the device will coil or uncoil, depending on the direction of the temperature change. In order for the material to perform well, the two materials must have very different coefficients of thermal expansion and should have high enough moduli of elasticity so that no permanent deformation of the material occurs.

1-17

What is the purpose of the classification for functional materials?

Solution: It specifically categorizes the types of materials used with specific applications. For instance, aerospace materials use lightweight materials such as aluminum alloys or carbon-composites for flight applications instead of using heavy materials such as stainless steel.

1-18

Explain the difference between crystalline and amorphous materials. Give an example of each that you use in your daily life.

Solution: Crystalline materials have long-range order arrangement of its atoms while amorphous materials have short-range order. One example of a crystalline material is metal and an example of an amorphous material is glass.

1-19

If you were given a material and were asked to determine whether it is crystalline or amorphous, how would you determine it?

Solution: Crystalline materials can be delineated from amorphous materials using diffraction techniques. Crystalline materials will produce diffraction patterns while amorphous materials will not.

1-20

List six materials performance problems that may lead to failure of components.

Solution: Excessive deformation (overload), fracture, wear, corrosion, fatigue and creep.

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1-21

Steel is often coated with a thin layer of zinc if it is to be used outside. What characteristics do you think the zinc provides to this coated, or galvanized, steel? What precautions should be considered in producing this product? How will the recyclability of the product be affected?

Solution: The zinc provides corrosion resistance to the iron in two ways. If the iron is completely coated with zinc, the zinc provides a barrier between the iron and the surrounding environment, therefore protecting the underlying iron. If the zinc coating is scratched to expose the iron, the zinc continues to protect the iron because the zinc corrodes preferentially to the iron (see Chapter 23). To be effective, the zinc should bond well to the iron so that it does not permit reactions to occur at the interface with the iron and so that the zinc remains intact during any forming of the galvanized material. When the material is recycled, the zinc will be lost by oxidation and vaporization, often producing a “zinc dust” that may pose an environmental hazard. Special equipment may be required to collect and either recycle or dispose of the zinc dust.

1-22

The relationship between structure and materials properties can be influenced by the service conditions (environmental conditions). Name two engineering disasters that have had tragic results and why they happened.

Solution: Answers will vary. Two sample answers: The Titanic sank when it hit an iceberg. Hull plate and rivets of the ship were made of the strongest material to use at that time, but at room temperature. The material failed due to its anisotropy and temperature sensitivity for when it was exposed to the frigid waters during its voyage, behaved like glass (brittle).

The Space Shuttle Challenger tragic flight when the rocket boosters exploded due to the solid rocket booster (SRB) O-ring failure. The freezing temperatures of that morning had the O-rings behave like glass and the vibrations during lift-off cracked the O-rings and the boosters exploded.

1-23

What is the difference between physical and mechanical properties? List three examples for each one.

Solution: Physical properties are the properties of a material as it is found in nature such as density, thermal conductivity, electrical conductivity, luster, color, corrosion resistance, oxidation resistance, coefficient of thermal expansion (CTE) and magnetic permeability.

Mechanical properties are properties that can be tested for such as strength (fatigue, impact, tensile, yield), ductility, stiffness, creep rupture, toughness and hardness.

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1-24

The type of jet engine used on most large commercial aircraft is called a turbofan jet engine because it has a large wheel at the front of the engine that propels air rearward.

Most of this air bypasses the engine, but bypass air significantly increases thrust and efficiency of these engines. Some engine manufacturers are now using carbon fiberepoxy composites rather than traditional aluminum blades. (a) What materials properties do you think an engineer must consider when selecting a material for this application? Be as specific as possible. (b) What benefits do you think carbon fiber epoxy composites have compared to aluminum alloys? What limitations or possible downsides could there be to using a carbon fiber epoxy composite?

Solution: (a) Properties that should be considered include the following: Tensile strength, impact toughness, fracture toughness, modulus of elasticity and resistance to ultraviolet radiation, which is a deteriorative property for polymers. (b) Potential benefits include significant weight reduction and greater stiffness over aluminum. The downsides include increase cost and most likely lower fracture toughness.

1-25

You are an engineer working for a manufacturer of land-based gas turbines. These turbines are similar to jet engines, but they are used on land to provide power for electricity generation and gas compression pipeline applications. Suppose that you would like to apply a ceramic-based thermal barrier coating to the turbine blades in the first-stage turbine to increase the operating temperature and efficiency of the engine. (a) What difficulties might engineers experience in trying to design a ceramic coating that will be applied to a super alloy metal blade? (b) What properties should be taken into consideration when choosing a suitable ceramic material for a coating? Be as thorough as possible.

Solution: (a) Difficulties include a suitable method to apply and bond the ceramic to the metal blades. Coefficients of thermal expansion (CTE) for metals and ceramics differ greatly which would result in spallation of the ceramic coating. (b) Properties to be considered should be coefficient of thermal expansion (CTE), thermal conductivity, specific heat, melting temperature, impact toughness and deteriorative properties.

1-26

We would like to produce a transparent canopy for an aircraft. If we were to use a traditional window glass canopy, rocks or birds might cause it to shatter. Design a material that would minimize damage or at least keep the canopy from breaking into pieces.

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Solution: We might sandwich a thin sheet of a transparent polymer between two layers of the glass. This approach, used for windshields of automobiles, will prevent the “safety” glass from completely disintegrating when it fails, with the polymer holding the broken pieces of glass together until the canopy can be replaced.

Another approach might be to use a transparent, “glassy” polymer material such as polycarbonate. Some polymers have reasonably good impact properties and may resist failure. The polymers can also be toughened to resist impact by introducing tiny globules of a rubber, or elastomer, into the polymer; these globules improve the energyabsorbing ability of the composite polymer, while being too small to interfere with the optical properties of the material.

1-27

You would like to design an aircraft that can be flown by human power nonstop for a distance of 30 km. What types of material properties would you recommend? What materials might be appropriate?

Solution: Such an aircraft must possess enough strength and stiffness to resist its own weight, the weight of the human “power source,” and any aerodynamic forces imposed on it. On the other hand, it must be as light as possible to ensure that the human can generate enough work to operate the aircraft. Composite materials, particularly those based on a polymer matrix, might comprise the bulk of the aircraft. The polymers have a light weight (with densities of less than half that of aluminum) and can be strengthened by introducing strong, stiff fibers made of glass, carbon, or other polymers. Composites, having the strength and stiffness of steel, but with only a fraction of the weight, can be produced in this manner.

1-28

You would like to place a three-foot diameter microsatellite into orbit. The satellite will contain delicate electronic equipment that will send and receive radio signals from earth. Design the outer shell within which the electronic equipment is contained. What properties will be required, and what kind of materials might be considered?

Solution: The shell of the microsatellite must satisfy several criteria. The material should have a low density, minimizing the satellite weight so that it can be lifted economically into its orbit; the material must be strong, hard, and impact resistant in order to ensure that any “space dust” that might strike the satellite does not penetrate and damage the electronic equipment; the material must be transparent to the radio signals that provide communication between the satellite and earth; and the material must provide some thermal insulation to ensure that solar heating does not damage the electronics.

One approach might be to use a composite shell of several materials. The outside surface might be a very thin reflective metal coating that would help reflect solar heat.

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The main body of the shell might be a light weight fiber-reinforced composite that would provide impact resistance (preventing penetration by dust particles) but would be transparent to radio signals.

1-29

What properties should the head of a carpenter’s hammer possess? How would you manufacture a hammer head?

Solution: The head for a carpenter’s hammer is produced by forging, a metal working process; a simple steel shape is heated and formed in several steps while hot into the required shape. The head is then heat treated to produce the required mechanical and physical properties.

The striking face and claws of the hammer should be hard—the metal should not dent or deform when driving or removing nails. These portions must also possess some impact resistance, particularly so that chips do not flake off the striking face and cause injuries.

1-30

You would like to select a material for the electrical contacts in an electrical switching device that opens and closes frequently and forcefully. What properties should the contact material possess? What type of material might you recommend? Would Al2O3 be a good choice? Explain.

Solution: The material must have a high electrical conductivity to ensure that no electrical heating or arcing occurs when the switch is closed. High purity (and therefore very soft) metals such as copper, aluminum, silver, or gold provide the high conductivity. The device must also have good wear resistance, requiring that the material be hard. Most hard, wear resistant materials have poor electrical conductivity.

One solution to this problem is to produce a particulate composite material composed of hard ceramic particles embedded in a continuous matrix of the electrical conductor.

For example, silicon carbide particles could be introduced into pure aluminum; the silicon carbide particles provide wear resistance while aluminum provides conductivity.

Other examples of these materials are described in Chapter 17.

Al2O3 by itself would not be a good choice—alumina is a ceramic material and is an electrical insulator; however, alumina particles dispersed into a copper matrix might provide wear resistance to the composite.

1-31

Aluminum has a density of 2.7 g/cm3. Suppose you would like to produce a composite material based on aluminum having a density of 1.5 g/cm3. Design a material that would have this density. Would introducing beads of polyethylene, with a density of 0.95 g/cm3, into the aluminum be a likely possibility? Explain.

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Solution: In order to produce an aluminum-matrix composite material with a density of 1.5 g/cm3, we would need to select a material having a density considerably less than 1.5 g/cm3.

While polyethylene’s density would make it a possibility, the polyethylene has a very low melting point compared to aluminum; this would make it very difficult to introduce the polyethylene into a solid aluminum matrix—processes such as casting or powder metallurgy would destroy the polyethylene. Therefore polyethylene would NOT be a likely possibility.

One approach, however, might be to introduce hollow glass beads. Although ceramic glasses have densities comparable to that of aluminum, a hollow bead will have a very low density. The glass also has a high melting temperature and could be introduced into liquid aluminum for processing as a casting.

1-32

You would like to be able to identify different materials without resorting to chemical analysis or lengthy testing procedures. Describe some possible testing and sorting techniques you might be able to use based on the physical properties of materials.

Solution: Some typical methods might include: measuring the density of the material (may help in separating metal groups such as aluminum, copper, steel, magnesium, etc.), determining the electrical conductivity of the material (may help in separating ceramics and polymers from metallic alloys), measuring the hardness of the material (perhaps even just using a file), and determining whether the material is magnetic or nonmagnetic (may help separate iron from other metallic alloys).

1-33

You would like to be able to physically separate different materials in a scrap recycling plant. Describe some possible methods that might be used to separate materials such as polymers, aluminum alloys, and steels from one another.

Solution: Steels can be magnetically separated from the other materials; steel (or carboncontaining iron alloys) are ferromagnetic and will be attracted by magnets. Density differences could be used—polymers have a density near that of water; the specific gravity of aluminum alloys is around 2.7; that of steels is between 7.5 and 8. Electrical conductivity measurements could be used—polymers are insulators while aluminum has a particularly high electrical conductivity.

1-34

Some pistons for automobile engines might be produced from a composite material containing small, hard silicon carbide particles in an aluminum alloy matrix. Explain what benefits each material in the composite may provide to the overall part. What problems might the different properties of the two materials cause in producing the part?

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Solution: Aluminum provides good heat transfer due to its high thermal conductivity. It has good ductility and toughness, reasonably good strength, and is easy to cast and process. The silicon carbide, a ceramic, is hard and strong, providing good wear resistance, and also has a high melting temperature. It provides good strength to the aluminum, even at elevated temperatures. There may be problems, however, producing the material—for example, the silicon carbide may not be uniformly distributed in the aluminum matrix if the pistons are produced by casting. We need to ensure good bonding between the particles and the aluminum—the surface chemistry must therefore be understood.

Differences in expansion and contraction with temperature changes may cause debonding and even cracking in the composite.

1-35

Investigate the origins and applications for a material that has been invented or discovered since you were born or investigate the development of a product or technology that has been invented since you were born that was made possible by the use of a novel material. Write one paragraph about this material or product. Provide a list of the references or websites that you used. You must use at least three references.

Solution: Answers will vary.

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Chapter 2: Atomic Structure

2-1

What is meant by the term composition of a material?

Solution: The chemical make-up of the material.

2-2

What is meant by the term structure of a material?

Solution: The spatial arrangement of atoms or ions in the material.

2-3

What are the different levels of structure of a material?

Solution: Atomic structure, short- and long-range atomic arrangements, nanostructure, microstructure, and macrostructure.

2-4

Why is it important to consider the structure of a material when designing and fabricating engineering components?

Solution: The structure of the material at all levels will affect the physical and mechanical properties of the final product.

2-5

What is the difference between the microstructure and macrostructure of a material?

Solution: A length scale of about 100,000 nm (100 μm) separates microstructure (less than 100,000 nm) from macrostructure (greater than 100,000 nm).

2-6

(a) Aluminum foil used for storing food weighs about 0.3 grams per square inch. How many atoms of aluminum contained in one square inch of the foil? (b) Using the densities and atomic weights given in Appendix A, calculate and compare the number of atoms per cubic centimeter in (i) lead and (ii) lithium.

Solution: (a) = 1 in.

0.3 g Al 1 mol Al 6.022 × 10 atoms

# 1 in. 26.982 g Al 1 mol

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= 6.7 × 10 % Al atoms (b) 11.36 g Pb 1 mol Pb 6.022 × 10 atoms =

# 1 cm 207.19 g Pb 1 mol = 33.0 × 10 %

=

0.534 g Li 1 mol Li 6.022 × 10 atoms

# 1 1 cm 6.94 g Li mol = 46.3 × 10 %

2-7

Pb atoms cm

Li atoms cm

(a) Using data in Appendix A, calculate the number of iron atoms in one ton (2000 pounds). (b) Using data in Appendix A, calculate the volume in cubic centimeters occupied by one mole of boron.

Solution: =

2000 lb 454 g 1 mol Fe 6.022 × 10 atoms

# 1 ton 1 lb 55.847 g Fe mol 1 = 9.8 × 10 .

Fe atoms ton

10.81 g B 1 cm B

/ = 1 mol B # 1 mol B 2.36 g B / = 4.58 cm B

2-8

In order to plate a steel part having a surface area of 200 in.2 with a 0.002 in.-thick layer of nickel: (a) how many atoms of nickel are required? (b) How many moles of nickel are required?

Solution: (a) We start with the volume required: / = 200 in.

0.002

2.54 cm in. = 6.55 cm 1 in.

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= 6.55 cm

8.902 g Ni 1 mol Ni 6.022 × 10 atoms

# 1 cm 58.71 g Ni 1 mol

= 598 × 10 % atoms Ni (b)

8.902 g Ni 1 mol Ni 23 = 6.55 cm

1 cm 58.71 g Ni 23 = 0.99 mol Ni

2-9

Define electronegativity.

Solution: Electronegativity is the tendency of an atom to accept an electron (which has a negative charge) and become an anion.

2-10

Write the electronic configuration of the following elements (a) tungsten, (b) cobalt, (c) zirconium, (d) uranium, and (e) aluminum.

Solution: (a) W: [Xe] 4f145d46s2 (b) Co: [Ar] 3d74s2 (c) Zr: [Kr] 4d24s2 (d) U: [Rn] 5f36d17s2 (e) Al: [Ne] 3s23p1 2-11

Write the electron configuration for the element Tc.

Solution: Since Technetium is element 43: [Tc] = 1s22s22p63s23p64s23d104p65s24d5 2-12

Assuming that the Aufbau Principle is followed, what is the expected electronic configuration of the element with atomic number Z = 116?

Solution: Using the Aufbau diagram produces: [116] = 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p67s25f146d107p4 Or in shorthand:

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[116] = [Rn] 5f146d107s27p4 2-13

Using the Aufbau Principle, what is the expected electronic configuration of the hypothetical element with atomic number Z = 123?

Solution: Using the Aufbau diagram produces: [123] = 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p67s25f146d107p68s25g3 Or in shorthand: [123] = [Rn] 5f145g36d107s27p68s2 Or assuming that [118] is another inert gas: [123] = [118] 5g38s2 2-14

Suppose an element has a valence of 2 and an atomic number of 27. Based only on the quantum numbers, how many electrons must be present in the 3d energy level?

Solution: We can let x be the number of electrons in the 3d energy level. Then: 1s22s22p63s23p63dx4s2 (must be 2 electrons in 4s for valence = 2) Since 27 - (2 + 2 + 6 + 2 + 6 + 2) = 7 = x there must be 7 electrons in the 3d level.

2-16

Bonding in the intermetallic compound Ni3Al is predominantly metallic. Explain why there will be little, if any, ionic bonding component. The electronegativity of nickel is about 1.8.

Solution: The electronegativity of Al is 1.5, while that of Ni is 1.9. These values are relatively close, so we wouldn’t expect much ionic bonding. Also, both are metals and prefer to give up their electrons rather than share or donate them.

2-17

Plot the melting temperatures of elements in the 4A to 8–10 columns of the periodic table versus atomic number (i.e., plot melting temperatures of Ti through Ni, Zr through Pd, and Hf through Pt). Discuss these relationships, based on atomic bonding and binding energies: (a) as the atomic number increases in each row of the periodic table and (b) as the atomic number increases in each column of the periodic table.

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Solution: The melting temperatures are taken from Figure 2-9 and are plotted below:

Melting temperature v. atomic number Melting temperature (°C)

4000 3500 3000 2500 2000 1500 1000 500 0 Ti - Ni

Zr - Pd

Hf - Pt

For each row, the melting temperature is highest when the outer “d” energy level is partly full. In Cr, there are 5 electrons in the 3d shell; in Mo, there are 5 electrons in the 4d shell; in W there are 4 electrons in the 5d shell. In each column, the melting temperature increases as the atomic number increases—the atom cores contain a larger number of tightly held electrons, making the metals more stable.

2-18

Plot the melting temperature of the elements in the 1A column of the periodic table versus atomic number (i.e., plot melting temperatures of Li through Cs). Discuss this relationship, based on atomic bonding and binding energy.

Solution: Using data from Figure 2-9:

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Melting temperature (°C)

Melting point as a function of atomic number 200 180 160 140 120 100 80 60 40 20 0 0

10

20

30

40

50

60

Atomic number (Z)

As the atomic number increases, the melting temperature decreases, in contrast to the trend found in Problem 2-17.

2-19

Compare and contrast metallic and covalent primary bonds in terms of (a) the nature of the bond, (b) the valence of the atoms involved, and (c) the ductility of the materials bonded in these ways.

Solution: (a) Metallic bonds are formed between the one or two free electrons of each atom. The free electrons form a gaseous cloud of electrons that move between atoms. Covalent bonds involve the sharing of electrons between atoms. (b) In metallic bonding, the metal atoms typically have one or two valence electrons that are given up to the electron “sea.” Covalent bonds form between atoms of the same element or atoms with similar electronegativities. (c) Metallic bonds are non–directional. The non–directionality of the bonds and the shielding of the ions by the electron cloud lead to high ductilities. Covalent bonds are highly directional – this limits the ductility of covalently bonded materials by making it more difficult for the atoms to slip past one another.

2-20

Differentiate the three principle bonding mechanisms in solids. What is Van der Waal’s bonding? What are the relative binding energies of the different mechanisms?

Solution: The principle bonding mechanisms are covalent, ionic and metallic. Van der Waals bonding is a weak secondary bonding mechanism that allows some gaseous or small molecules to condense. The relative binding energies of the primary bonds are reflected

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in the strength of the material. Covalent and ionic bonds produce the strongest bonds while metallic bonds have lower bonding energies than covalent or ionic materials.

2-21

What type of bonding does KCl have? Fully explain your reasoning by referring to the electronic structure and electronic properties of each element.

Solution: KCl has ionic bonding. The electronic structure of [K] = 1s22s22p63s23p64s1 = [Ar] 4s1. The electronic structure of [Cl] = 1s22s22p63s23p5 = [Ne] 3s23p5. Therefore, K wants to give up its 4s1 electron in order to achieve a stable s2p6 configuration, and Cl wants to gain an electron in order to gain the stable s2p6 configuration. Thus an electron is transferred from K to Cl, and the bonding is ionic.

2-22

The compound aluminum phosphide (AlP) is a compound semiconductor having mixed ionic and covalent bonding. Calculate the fraction of bonding that is ionic.

Solution: Starting with Equation 2-1, we determine the fraction of ionic bonding with this equation: 45 657 = 1 − 9 :;<. =∆? A @

The exponential term is the covalent fraction. Since there can be only ionic or covalent bonds, the two fractions must add to 1. In an equation form: 45 657 + 47 C 236 = 1 Or: 45 657 + 9 :;<. =∆? A = 1 @

Anyway, the electronegativities for Al and P are 1.5 and 2.1 respectively. The difference is 0.6: 45 657 = 1 − 9 :;<. =D<.EF A @

Doing the math: 45 657 = 0.086 So less than one tenth of the bonding is ionic. This should not be surprising when their positions in the periodic table are compared.

2-23

Calculate the fraction of bonding in MgO that is ionic.

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Solution: Starting with Equation 2-1, we determine the fraction of ionic bonding with this equation: 45 657 = 1 − 9 :;<. =∆? A @

The electronegativities of magnesium and oxygen are 1.2 and 3.5, so their difference is 2.3. Inserting: 45 657 = 1 − 9 :;<. =D . F A @

Math time!

45 657 = 0.734 The compound is held together mostly by ionic bonding.

2-24

Calculate the fraction of bonding that is covalent for silica (SiO2).

Solution: Using Equation 2-1, we determine the fraction of covalent bonding: 47 C 236 = 9 :;<. =∆? A @

The electronegativities of silicon and oxygen are 1.8 and 3.5, so their difference is 1.7.

Inserting: 47 C 236 = 9 :;<. =D%..F A @

47 C 236 = 0.486

2-25

Calculate the fraction of bonding that is ionic in nature for zirconia (ZrO2)?

Solution: 45 657 = 1 − 9 :;<. =∆? A @

The electronegativities of zircon and oxygen are 1.4 and 3.5, so their difference is 2.1.

Inserting: 45 657 = 1 − 9 :;<. =D .%F A @

45 657 = 0.668

2-26

What is the type of bonding in diamond? Are the properties of diamond commensurate with the nature of the bonding?

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Solution: In diamond, the carbon atoms are covalently bonded. Diamond is electrically insulating, which makes sense: each carbon is bonded to four other carbon atoms thus leaving no free valence electrons available to conduct electricity.

2-27

What are the bonding mechanisms in thermoplastics?

Solution: The primary bond in thermoplastics are the covalent bonds that hold the individual carbon atoms together along the polymer chains while van der Waals secondary bonds hold the polymer chains close together.

2-28

Why are covalently bonded materials generally less dense than those that are bonded ionically or metallically?

Solution: Covalently bonded materials are typically less dense than metallically or ionically bonded materials due to the nature of their bonding. The bonding in covalent materials is directional in nature which doesn’t allow the atoms to pack together in a dense manner like the ionic or metallic bonded materials. This results in a lower mass and lower density for covalently bonded materials.

2-31

Calculate the fractions of ionic bonds in silicon carbide (SiC) and silicon nitride (Si3N4).

Solution: We use Equation 2.1 and take the electronegativities from Figure 2-9: 45 657 = 1 − 9 :;<. =∆? A @

The electronegativities of silicon and carbon are 1.8 and 2.5, so their difference is 0.7.

Inserting: 45 657 = 1 − 9 :;<. =D<..F A @

Math time!

45 657 = 0.115 Repeating this for silicon nitride: 45 657 = 1 − 9 :;<. =∆? A @

The electronegativities of silicon and nitrogen are 1.8 and 3.0, so their difference is 1.2.

Inserting: 45 657 = 1 − 9 :;<. =D%. F A @

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45 657 = 0.302 2-32

One particular form of boron nitride (BN) known as cubic boron nitride is a very hard material and is used in grinding applications. Calculate the fraction of bonding that is covalent in this material.

Solution: Using Equation 2-1, we determine the fraction of covalent bonding: 47 C 236 = 9 :;<. =∆? A @

The electronegativities of boron and nitrogen are 2.0 and 3.0, so their difference is 1.0.

Inserting: 47 C 236 = 9 :;<. =D%.<F A @

47 C 236 = 0.779

2-34

Is there a trend in the number of electrons in the outermost energy shell of atoms or ions that have formed bonds?

Solution: Yes. They try to fill their valence shells.

2-35

In order to increase the operating temperature of an engine, it is suggested that some of the aluminum components be coated with a ceramic. What kinds of problems could this pose? How could you overcome these problems?

Solution: Creating a mechanical bond between the ceramic and the metallic component could pose a problem since the ceramic is ionic in nature and the component is metallically bonded. This can be overcome by creating a slightly roughened surface and choosing a ceramic that has a limited degree of chemical reactivity with the metal which would enhance bonding.

Another problem that we face is likely to spallation or debonding of the coating due to the differences in the coefficients of thermal expansion. To overcome this problem, it would be possible to select a ceramic that could have a compatible coefficients of thermal expansion with the aluminum for the given operating temperature range.

2-36

Aluminum and silicon are side by side on the periodic table. Compare the melting temperatures of the two elements and explain the difference in terms of atomic bonding.

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Solution: The melting temperature of silicon is 1410°C while aluminum is 660°C. The differences in melting temperatures can be explained by the types of bonds that bind the elements together. It is expected that since aluminum is a metal that it would have metallic bonding. Silicon on the other hand is a metalloid (between metal and non-metal) and has covalent bonding. Since covalent bonding has higher binding energy than metallic bonding, we can conclude that the silicon has the higher melting temperature due to the higher strength of the silicon bonds when compared to aluminum.

2-37

Titanium is stiffer than aluminum, has a lower thermal expansion coefficient than aluminum, and has a higher melting temperature than aluminum. On the same graph, carefully and schematically draw the potential well curves for both metals. Be explicit in showing how the physical properties are manifest in these curves.

Solution: The well of titanium, represented by A, is deeper (higher melting point), has a larger radius of curvature (stiffer), and is more symmetric (smaller thermal expansion coefficient) than the well of aluminum, represented by B. 2-38

Would you expect iron or silicon nitride (SiN) to have a higher modulus of elasticity?

Why?

Solution: It is expected that SiN would have the higher modulus of elasticity due to its bonding nature (covalent) compared to iron (metallic). Covalent bonds result in higher binding energies thus having a direct result for a higher modulus of elasticity.

2-39

Beryllium and magnesium, both in the 2A column of the periodic table, are lightweight metals. Which would you expect to have the higher modulus of elasticity? Explain, considering binding energy and atomic radii and using appropriate sketches of force versus interatomic spacing.

Solution: The smaller Be electrons are held closer to the core, therefore ∴ held more tightly, giving a higher binding energy: 4 Be 1s22s2 E = 42 × 106 psi rBe = 1.143 Å 12 Mg 1s22s22p63s2 E = 6 × 106 psi rMg = 1.604 Å 2-41

Would you expect MgO or magnesium to have the higher modulus of elasticity? Explain.

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Solution: MgO has ionic bonds. A higher force will be required to cause the same separation between the ions in MgO compared to the atoms in Mg. Therefore, MgO should have the higher modulus of elasticity. In Mg, E ≈ 6 × 106 psi; in MgO, E ≈ 30 × 106 psi.

2-43

Aluminum and silicon are side-by-side in the periodic table. Which would you expect to have the higher modulus of elasticity? Explain.

Solution: Silicon has covalent bonds; aluminum has metallic bonds. Therefore, Si should have a higher modulus of elasticity.

2-45

Steel is coated with a thin layer of ceramic to help protect against corrosion. What do you expect to happen to the coating when the temperature of the steel is increased significantly? Explain.

Solution: Ceramics are expected to have a low coefficient of thermal expansion due to strong ionic/covalent bonds; steel has a high thermal expansion coefficient. When the structure heats, steel expands more than the coating, which may crack and expose the underlying steel to corrosion.

2-48

Name at least four allotropes of carbon. Why is graphite electrically conductive while diamond is not if both are pure forms of carbon?

Solution: The four allotropes of carbon are diamond, graphite, nanotubes and buckminsterfullerene. In diamond, the carbon atoms are covalently bonded to four other carbon atoms thus leaving no free valence electrons available to conduct electricity. In graphite, the carbon is arranged in layers where the carbon atoms form 3 strong bonds with other carbon atoms, but have a fourth bond between layers which is a weak van der Waals bond. This results in the fourth electron for each of the carbon atom to be available to conduct electricity.

2-49

Bond hybridization in carbon leads to numerous crystalline forms. With only six electrons, how is this possible? Explain.

Solution: Carbon in graphite form has the electron configuration of 1s2 2s2 2p2 which only allows for 3 bonds. Additions of pressure and heat, hybridization occurs to the point that the electron moves from the 2s to the 2p orbital (1s2 2s1 2p3) allowing access for 4 bonds to occur.

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Chapter 3: Atomic and Ionic Arrangements

3-1

What is a crystalline material?

Solution: A uniform substance made up of a three dimensional, grid-like repeating pattern.

3-2

What is a single crystal?

Solution: A crystalline material that is composed of only one crystal.

3-3

State two applications in which single crystals are used.

Solution: Silicon integrated circuit (computer) chips and lithium niobate optoelectric devices.

3-4

What is a polycrystalline material?

Solution: A crystalline material which is composed of many smaller crystals or grains, randomly oriented.

3-5

What is a liquid crystal material?

Solution: A material that behaves as a liquid in one state, but forms small crystalline regions when an external stimulus is applied.

3-6

What is an amorphous material?

Solution: A material that displays only short range order. For solids, amorphous is antonymic to crystalline.

3-7

Why do some materials assume an amorphous structure?

Solution: The kinetics of the manufacturing process do not allow a regular or crystalline pattern to form.

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3-8

Explain how it is possible for a substance to exhibit short-range, but not long-range order.

Solution: A substance with only short-range order is similar to a scattered ream of paper or deck of cards. Each piece of paper represents a short-range organization of wood fiber, but the pieces have no order with respect to each other. If they were collected into a stack, they would have both short- and long-range order.

3-9

Can an alloy exist in both crystalline and amorphous forms?

Solution: Not at the same time. Forming amorphous alloys is frequently very difficult and requires extraordinary cooling rates.

3-10

Approximately how many grains are shown in Figure 3-2(b)?

Solution: By rough count, about 50. Count the areas separated by the darker lines, which are the grain boundaries.

3-11

Define the terms lattice, unit cell, basis, and crystal structure.

Solution: A lattice is a collection of points arranged in a spatial pattern so each point has a relation its surrounding points that is valid for all other points as well. A unit cell is the smallest part of a lattice that shows the repeating pattern of the lattice. The basis is composed of atoms with a certain relationship spatial relationship. The crystal structure is the combination of the lattice and basis and shows how the atoms of a crystal are arranged with respect to each other throughout the entire thing.

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3-12

List the seven different crystal systems and the types of Bravais lattices that are associated with their groups.

Solution: They are cubic (simple, face-centered, and body-centered), tetragonal (simple and bodycentered), orthorhombic (simple, body-centered, base-centered, and face-centered), rhombohedral/trigonal, hexagonal, monoclinic (simple and base-centered), and finally triclinic. The order of this list is random.

3-13

Explain why there is no face-centered tetragonal Bravais lattice.

Solution: The facial lattice points would not have identical surroundings, which is necessary for a unit cell.

3-14

Calculate the atomic radius in cm for the following: (a) BCC metal with a0 = 0.3294 nm; and (b) FCC metal with a0 = 4.0862 Å.

Solution: For BCC metals:

=

√3

4

√3 0.3294 nm = 0.1426 nm 4

For FCC metals:

=

3-15

=

= 1.426 × 10 cm =

√2

4

√2 4.0862 Å = 1.4447 Å 4

= 1.4447 × 10 cm

Calculate the lattice parameter of each of the following elements using their atomic radii: (a) iron, (b) aluminum, (c) copper, and (d) magnesium.

Solution: Iron is BCC and the a0 to r relationship is

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=

4

√3

=

4

√3

1.24

= 2.86 Å

Aluminum is FCC and the a0 to r relationship is =

4

√2

=

4

√2

1.43

= 4.04 Å

Copper is FCC and the a0 to r relationship is =

4

√2

=

4

√2

1.27

= 3.59 Å

Magnesium is HCP and the a0 to r relationship is = 2

= 1.633

= 2 1.6 = 3.2 Å

= 1.633 3.2 = 5.2 Å

3-16

Determine the crystal structure for the following: (a) a metal with a0 = 4.9489 Å, r = 1.75 Å and one atom per lattice point; and (b) a metal with a0 = 0.42906 nm, r = 0.1858 nm and one atom per lattice point.

Solution: We want to determine if “x” in the calculations below equals 2 (for FCC) or 3 (for BCC).

We know it must be one of these because we are only given a0. = For (a): 4.9489 Å = Solving:

4

√ 4

√

1.75 Å

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=2

So the crystal is face-centered cubic (FCC) For (b): 0.42906 nm = Solving:

4

√

0.1858 nm

=3

So the crystal is body-centered cubic (BCC).

3-17

The density of potassium, which has the BCC structure, is 0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate (a) the lattice parameter; and (b) the atomic radius of potassium.

Solution: Using Equation 3-5: Density % =

number of atoms/cell atomic mass volume of unit cell /0

g atom 32 4 339.09 4 g cell mol 0.855 = 2 atom cm 2 36.022 × 1052 4 mol 2 = 1.5189 × 10 55 cm2 = 5.3349 × 10 cm

For the second part, (b), we use the relationship between the atomic radius and the lattice parameter: =

√3 4

= 5.3349 × 10 cm

√3 4

= 2.3101 × 10 cm

3-18

The density of thorium, which has the FCC structure, is 11.72 g/cm3. The atomic weight of thorium is 232 g/mol. Calculate (a) the lattice parameter; and (b) the atomic radius of thorium.

Solution: Using Equation 3-5:

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Density % =

number of atoms/cell atomic mass volume of unit cell /0

g atom 32 4 3232 4 g cell mol 11.72 = cm2 2 36.022 × 1052 atom4

mol 2 = 1.315297 × 10 55 cm2 = 5.0856 × 10 cm

For the second part, (b), we use the relationship between the atomic radius and the lattice parameter: =

√2 4

= 5.0856 × 10 cm

√2 4

= 1.7980 × 10 cm

3-19

A metal having a cubic structure has a density of 2.6 g/cm3, an atomic weight of 87.62 g/mol, and a lattice parameter of 6.0849 Å. One atom is associated with each lattice point. Determine the crystal structure of the metal.

Solution: Using Equation 3-5: Density % =

number of atoms/cell atomic mass volume of unit cell /0

g 387.26 4 g mol 2.6 = 2 atom cm 6.0849 × 10 cm 2 36.022 × 1052 4 mol =4

atom cell

The crystal must be face-centered cubic (FCC).

3-20

A metal having a cubic structure has a density of 1.892 g/cm3, an atomic weight of 132.91 g/mol, and a lattice parameter of 6.13 Å. One atom is associated with each lattice point. Determine the crystal structure of the metal.

Solution: Using Equation 3-5:

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Density % =

number of atoms/cell atomic mass volume of unit cell /0

g 3132.91 4 g mol 1.892 = cm2 6.13 × 10 cm 2 36.022 × 1052 atom4 mol =2

atom cell

The crystal must be body-centered cubic (BCC).

3-21

Indium has a tetragonal structure with a0 = 0.32517 nm and c0 = 0.49459 nm. The density is 7.286 g/cm3, and the atomic weight is 114.82 g/mol. Does indium have the simple tetragonal or body-centered tetragonal structure?

Solution: Using Equation 3-5: Density % =

number of atoms/cell atomic mass volume of unit cell /0

g 3114.82 4 g mol 7.286 = 2 atom cm 3.2517 × 10 cm 5 4.9459 × 10 cm 36.022 × 1052 4 mol =2

atom cell

Therefore, the crystal is body-centered tetragonal (BCT).

3-22

Bismuth has a hexagonal structure, with a0 = 0.4546 nm and c0 = 1.186 nm. The density is 9.808 g/cm3, and the atomic weight is 208.98 g/mol. Determine (a) the volume of the unit cell; and (b) the number of atoms in each unit cell.

Solution: The volume, (a), of the unit cell is:

6 = 5 cos 30°

6 = 0.4546 nm 5 1.186 nm cos 30° 6 = 2.1226 × 10 55 cm2

Reprising Equation 3-5 for part (b): Density % =

number of atoms⁄cell atomic mass volume of unit cell /0

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g 3208.98 4 g mol 9.808 = cm2 2.1226 × 10 55 cm2 36.022 × 1052 atom4 mol =6

3-23

atom cell

Gallium has an orthorhombic structure, with a0 = 0.45258 nm, b0 = 0.45186 nm, and c0 = 0.76570 nm. The atomic radius is 0.1218 nm. The density is 5.904 g/cm3, and the atomic weight is 69.72 g/mol. Determine (a) the number of atoms in each unit cell; and (b) the packing factor in the unit cell.

Solution: The volume of the unit cell is:

6 = 9

6 = 0.45258 nm 0.45186 nm 0.76570 nm = 0.1556 nm2 6 = 1.566 × 10 55 cm2

Using Equation 3-5 for part (a): Density % =

number of atoms⁄cell atomic mass volume of unit cell /0

g 369.72 4 g mol 5.904 = cm2 1.566 × 10 55 cm2 36.022 × 1052 atom4 mol =8

atom cell

The packing factor (PF) is calculated using Equation 3-4: Packing factor =

number of atoms⁄cell volume of each atom volume of unit cell

Packing factor =

38

atom 4 4=/3 0.1218 nm 2 cell 0.1566 nm2

Packing factor = 0.387

3-24

Beryllium has a hexagonal crystal structure, with a0 = 0.22858 nm and c0 = 0.35842 nm.

The atomic radius is 0.1143 nm, the density is 1.848 g/cm3, and the atomic weight is 9.01 g/mol. Determine (a) the number of atoms in each unit cell; and (b) the packing factor in the unit cell.

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Solution: The volume of the unit cell is:

6 = 5 cos 30°

6 = 0.22858 nm 5 0.35842 nm cos 30° 6 = 16.22 × 10 5> cm2

Using Equation 3-5 for part (a): Density % =

number of atoms⁄cell atomic mass volume of unit cell /0

g 39.01 4 g mol = 1.848 2 atom cm 16.22 × 10 5> cm2 36.022 × 1052 4 mol =2

atom cell

The packing factor (PF) is calculated using Equation 3-4: Packing factor =

number of atoms⁄cell volume of each atom volume of unit cell

Packing factor =

32

atom 4 4=/3 0.1143 nm 2 cell 0.01622 nm2

Packing factor = 0.77

3-25

A typical paper clip weighs 0.59 g and consists of BCC iron. Calculate (a) the number of unit cells; and (b) the number of iron atoms in the paper clip. (See Appendix A for required data)

Solution: The lattice parameter for BCC iron is 2.866 × 10-8 cm. Therefore 6?@AB CDEE = 2.866 × 10 cm 2 = 2.354 × 10 52 cm2

The density is 7.87 g/cm3. The number of unit cells (a) is: /CDEE =

37.87

0.59 g

g cm2 4 F2.354 × 10 52 G 2 cell cm

/CDEE = 3.185 × 105H cell

There are 2 atoms/cell in BCC iron. The number of atoms (b) is

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/IBJK = 3.185 × 105H cell F2

atom G cell

/IBJK = 6.37 × 105H atom

3-26

Aluminum foil used to package food is approximately 0.001 inch thick. Assume that all of the unit cells of the aluminum are arranged so that a0 is perpendicular to the foil surface. For a 4 in. × 4 in. square of the foil, determine (a) the total number of unit cells in the foil; and (b) the thickness of the foil in number of unit cells. (See Appendix A.)

Solution: The lattice parameter for aluminum is 4.04958 × 10-8 cm. Therefore 6?@AB CDEE = 4.04958 × 10 cm 2 = 6.6409 × 10 52 cm2

The volume of the foil is:

6LJAE = 4 in. 4 in. 0.001 in. = 0.016 in.2 = 0.262 cm2

The number of unit cells (a) is: /CDEE =

0.262 cm2

6.6409 × 10 52

cm2 cell

/CDEE = 3.945 × 105H cell

The thickness of the foil, again in unit cells (b):

cm in. 4 /CDEE = 4.04958 × 10 cm 0.001 in. 32.54

/CDEE = 6.27 × 10> cell

3-27

Sodium has the body-centered cubic crystal structure and a lattice parameter (axial length) of 4.2906 × 10-8 cm. (a) Calculate the mass density (g/cm3) for sodium. (b) Calculate the atomic radius of a sodium atom according to the hard sphere model of the BCC crystal structure.

Solution: First we convert the lattice parameter to atomic radius using Equation 3-2:

=

4

√3

=

4

√3

4.2906 × 10 cm

= 1.7321 × 10 cm

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Using Equation 3-5 for part (a): Density % = Density % =

number of atoms⁄cell atomic mass volume of unit cell /0 32

g atom 4 322.99 4 cell mol

1.7321 × 10 cm 2 36.022 × 1052 % = 14.69

This was already done:

g cm2

atom 4 mol

= 1.7321 × 10 cm

3-28

Calculate the density for zinc (HCP) if the c/a ratio is 1.85 and r = 1.332 Å.

Solution:

= 2 1.332 Å = 2.664 Å

= 1.85 2.664 Å = 4.928 Å 6 = 5 cos 30° %=

M 6

g 4 mol N atom O 6.022 × 1052 mol %= 2.664 × 10 cm 5 4.928 × 10 cm cos 30° 2 atom 365.39 % = 7.169

3-29

g cm2

Thoria or thorium dioxide can be described as an FCC lattice with a basis of Th (0, 0, 0) and O (1/4, 1/4, 1/4) and O (1/4, 1/4, 3/4). Thorium and thorium dioxide are radioactive materials. Thorium dioxide is commonly used in tungsten electrodes for arc welding.

Thoria improves the high temperature properties of the electrodes and promotes emission of electrons from the electrode tip. (a) How many atoms of each type are there in the conventional unit cell of thorium dioxide? (b) Is this consistent with the chemical formula of thorium dioxide? Explain. (c) What is the coordination number (number of nearest neighbors) for each Th atom? (d) What is the coordination number (number of nearest neighbors) for each O atom? (e) What type of interstitial site do the oxygen atoms occupy?

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Solution: The Th atom at the origin contributes one-eighth of an atom, but it recurs at each corner. There are eight corners per unit cell, so corner Th atom 1 Th atom ×8 =1 F G unit cell unit cell 8 corner

The two oxygen atoms occur wholly within the unit cell and contribute two atoms each. The total then is three atoms per unit cell.

Yes. For each unit cell there is one thorium atom and two oxygen atoms, for a molecular formula of ThO2.

Each thorium atom has four oxygen atoms equally close to it, so its coordination number is 4.

Each oxygen atom has one atom closer to the others: the thorium atom. It has a coordination number of 1.

3-30

Zinc has the hexagonal close-packed crystal structure. The lattice parameters for zinc are a = 0.26648 nm and c = 0.49470 nm, and the atomic radius is 0.1332 nm. Note that zinc does not have the ideal atomic packing factor. (a) What is the number of atoms per unit cell in the hexagonal close-packed structure? (b) Determine the atomic packing factor of the Zn unit cell. (c) Is the c/a ratio for zinc greater or less than the ideal HCP c/a ratio? Will slip be harder or easier in zinc compared to the ideal HCP structure? Explain your answer fully.

Solution: Examining Table 3-2, there are two atoms per unit cell of HCP crystal.

For this, we repair to Equation 3-4: Packing factor =

number of atoms⁄cell volume of each atom volume of unit cell

First we find the volume of the unit cell:

6 = 5 cos 30°

6 = 0.26648 nm 5 0.49470 nm cos 30°

Packing factor =

32

atom 4 4=/3 0.1332 nm 2 cell 0.0304 nm2

Packing factor = 0.65

The ideal c/a ratio for HCP is 1.663, the c/a ratio for zinc is:

0.49470 nm 3 4 = = 1.856 Q@ 0.26648 nm

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So then:

3-31

3 4 >3 4 Q@ STU AVDIE

Rutile is the name given to a crystal structure commonly adopted by compounds of the form AB2, where A represents a metal atom and B represents oxygen atoms. One form of rutile has atoms of element A at the unit cell coordinates (0, 0, 0) and (1/2, 1/2, 1/2) and atoms of element B at (1/4, 1/4, 0), (3/4, 3/4, 0), (3/4, 1/4, 1/2), and (1/4, 3/4, 1/2).

The unit cell parameters are a = b ≠ c and a = b = γ = 90°. Note that the lattice parameter c is typically smaller than the lattice parameters a and b for the rutile structure. (a) How many atoms of element A are there per unit cell? (b) How many atoms of element B are there per unit cell? (c) Is your answer to part (b) consistent with the stoichiometry of an AB2 compound? Explain. (d) Draw the unit cell for rutile. Use a different symbol for each type of atom. Provide a legend indicating which symbol represents which type of atom. (e) For the simple tetragonal lattice a = b ≠ c and α = β = γ = 90°. There is one lattice point per unit cell located at the corners of the simple tetragonal lattice. Describe the rutile structure as a simple tetragonal lattice and a basis.

Solution: There are two atoms of element A per unit cell: one at the corners of the unit cell and one at the body–centered position.

There are four atoms of element B per unit cell: two atoms on faces and two located in the interior of the unit cell.

Yes, the answer to (b) is consistent with the stoichiometry of an AB2 compound. The ratio of A:B atoms is 2:4 or 1:2.

The unit cell is shown below.

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The rutile structure can be described as a simple tetragonal lattice with a basis of A at (0, 0, 0) and (1/2, 1/2, 1/2) and B at (1/4, 1/4, 0), (3/4, 3/4, 0), (3/4, 1/4, 1/2), and (1/4, 3/4, 1/2).

3-32

Consider the CuAu crystal structure. It can be described as a simple cubic lattice with a basis of Cu (0, 0, 0), Cu (1/2, 1/2, 0), Au (1/2, 0, 1/2), and Au (0, 1/2, 1/2). (a) How many atoms of each type are there per unit cell? (b) Draw the unit cell for CuAu. Use a different symbol for each type of atom. Provide a legend indicating which symbol represents which type of atom. (c) Give an alternative lattice and basis representation for CuAu for which one atom of the basis is Au (0, 0, 0). (d) A related crystal structure is that of Cu3Au. This unit cell is similar to the face-centered cubic unit cell with Au at the corners of the unit cell and Cu at all of the face-centered positions. Describe this structure as a lattice and a basis. (e) The Cu3Au crystal structure is similar to the FCC crystal structure, but it does not have the face-centered cubic lattice. Explain briefly why this is the case.

Solution: There is one lattice point per unit cell for the simple cubic lattice. There are two copper atoms in the basis per lattice point and two gold atoms per lattice point such that there are two copper atoms and two gold atoms per unit cell.

The unit cell is shown below.

An alternative representation is the simple cubic lattice with a basis of Au (0, 0, 0), Au (1/2, 1/2, 0), Cu (0, 1/2, 1/2), and Cu (1/2, 0, 1/2).

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By definition, each lattice point must have identical surroundings. Although the Cu and Au atoms occupy the sites of an FCC lattice, each site does not have identical surroundings and therefore cannot be a lattice point because these are atoms of different types.

Cu3Au has a simple cubic lattice with a basis of Au (0, 0, 0), Cu (0, 1/2, 1/2), Cu (1/2, 0, 1/2), and Cu (1/2, 1/2, 0).

3-33

Nanowires are high aspect-ratio metal or semiconducting wires with diameters on the order of 1 to 100 nanometers and typical lengths of 1 to 100 microns. Nanowires likely will be used in the future to create high-density electronic circuits.

Nanowires can be fabricated from ZnO. ZnO has the wurtzite structure. The wurtzite structure is a hexagonal lattice with four atoms per lattice point at Zn (0, 0, 0), Zn (2/3, 1/3, 1/2), O (0, 0, 3/8), and O (2/3, 1/3, 7/8). (a) How many atoms are there in the conventional unit cell? (b) If the atoms were located instead at Zn (0, 0, 0), Zn (1/3, 2/3, 1/2), O (0, 0, 3/8), and O (1/3, 2/3, 7/8), would the structure be different? Please explain. (c) For ZnO, the unit cell parameters are a = 3.24 Å and c = 5.19 Å. (Note: This is not the ideal HCP c/a ratio.) A typical ZnO nanowire is 20 nm in diameter and 5 mm long. Assume that the nanowires are cylindrical. Approximately how many atoms are there in a single ZnO nanowire?

Solution: On average, these are the contributions from each type of site: Zn (0, 0, 0): 8 corner sites per cell × 1/8 atom per site = 1 atom Zn (2/3, 1/3, 1/2): 1 site per cell × 1 atom per site = 1 atom O (0, 0, 3/8): 4 edge sites per cell × 1/4 atom per site = 1 atom O (2/3, 1/3, 7/8): 1 site per cell × 1 atom per site = 1 atom Thus, there are 4 atoms per unit cell. Alternatively, there is one lattice point per unit cell and four atoms per lattice point, again giving a total of 4 atoms per unit cell.

No, the structure would not be different. The (1/3, 2/3, 1/2) and (2/3, 1/3, 1/2) sites and (1/3, 2/3, 7/8) and (2/3, 1/3, 7/8) sites are equivalent.

The total volume of the nanowire, Vtotal, is given by 6BJBIE = =W 5 X

Where R is the radius and L the length of the nanowire.

6BJBIE = = Y

5

20 × 10 Z m[ 5 × 10 \ m 2

6BJBIE = 1.571 × 10 5H m2

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The volume of a unit cell is

6CDEE = ]^I_D

Where Abase is the area of the base of the unit cell and c the height of the unit cell.

The geometry of the unit cell is shown here:

6CDEE = ]^I_D

5 √3 1 √3 ]^I_D = 2 Y × × [= 2 2 2

6CDEE = Y

6CDEE = Y

5 √3 [ 2

`3.24 × 10 H ma5 √3 [ 5.19 × 10 H m 2 6CDEE = 4.718 × 10 5Z m2

6BJBIE atom /IBJK_ = F G F4 G 6CDEE cell /IBJK_ = 1.33 × 10 atom

3-34

Calculate the atomic packing fraction for the hexagonal close-packed crystal structure

for which = b2. Remember that the base of the unit cell is a parallelogram.

Solution: This is an exercise in studying the geometry of the HCP unit cell. The base of the HCP unit cell can be treated as two equilateral triangles each of side a as shown below. The B atom in the HCP structure is located at the (2/3, 1/3, 1/2) location. The distance from the coordinate (1, 0, 0) to (2/3, 1/3, 0), which is the centroid of the triangle in the base, c √2

is 2 .

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The c/a ratio is found by applying the Pythagorean Theorem to the triangle shown below. It is necessary to recognize that each B atom touches the three A atoms in the layer below it at a distance a = 2R, where R is the radius of the atom.

5

Solving for c/a:

√3 5 Y [ + 3 4 = 5 3 2 5 5 + = 5 3 4 5 =

8 5 3

8 =e ∎ 3

3-35

Magnesium has the hexagonal close-packed crystal structure. The lattice parameters for magnesium are a = 0.32087 nm and c = 0.5209 nm, and the atomic radius is 0.1604 nm.

Note that magnesium does not have the ideal atomic packing factor. (a) What is the number of atoms per unit cell in the hexagonal close-packed structure? (b) Determine the atomic packing factor of the Mg unit cell. (c) Is the c/a ratio for Mg greater or less than the ideal HCP c/a ratio? Will slip be harder or easier in Mg compared to the ideal HCP structure? Explain your answer fully.

Solution: Examining Table 3-2, there are two atoms per unit cell of HCP crystal.

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For this, we repair to Equation 3-4: Packing factor =

number of atoms⁄cell volume of each atom volume of unit cell

First we find the volume of the unit cell:

6 = 5 cos 30°

6 = 0.32087 nm 5 0.5209 nm cos 30° = 0.04645 nm2 Packing factor =

32

atom 4 4=/3 0.1604 nm 2 cell 0.04645 nm2

Packing factor = 0.744

The ideal c/a ratio for HCP is 1.663, the c/a ratio for magnesium is:

So then:

0.5209 nm

= 1.623 3 4 = gh 0.32087 nm

3 4 >3 4 gh STU AVDIE

Slip in Mg will be slightly harder to produce than in ideal HCP structures because the basal close packed planes where slip generally occurs in HCP metals will be marginally more closely packed.

3-36

What is the difference between an allotrope and a polymorph?

Solution: Allotropy describes the property of having multiple crystals forms when pure elements are described. Polymorphism is the same property, but occurring in compounds of elements.

3-37

What are the different polymorphs of zirconia?

Solution: Monoclinic, tetragonal, cubic and orthorhombic. These occur at different temperatures and pressures.

3-38

A number of metals undergo an allotropic transformation from one crystal structure to another at a specific temperature. There is generally a volume change that accompanies the transformation. What is the practical significance of such data?

Solution:

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The practical significance is that the metal will undergo a much larger volume change than normally expected according to simple thermal expansion. If this is not accounted for, large stresses can build up within the metal possibly causing the material to crack.

3-39

Above 882°C, titanium has a BCC crystal structure, with a = 0.332 nm. Below this temperature, titanium has a HCP structure, with a = 0.29503 nm and c = 0.46831 nm.

Determine the percent volume change when BCC titanium transforms to HCP titanium.

Is this a contraction or expansion?

Solution: We can find the volume of each unit cell. Two atoms are present in both BCC and HCP titanium unit cells, so the volumes of the unit cells can be directly compared.

6iTT = 0.332 nm 2 = 0.0366 nm2

6STU = 3 0.29503 nm 5 0.46831 nm cos 30° = 0.1059 nm2

6STU 6iTT 0.1059nm2 0.0366 nm2 − − 2 nlMo # lm nlMo # lm nlMo ∆6 = × 100% = 6 nlMo × 100% 6iTT 0.0366 nm2 # lm nlMo 2 nlMo ∆6 = −3.55%

The titanium contracts when it allotropically transforms from BCC structure to HCP structure as temperatures drop below 882°C. 3-40

α-Mn has a cubic structure with a0 = 0.8931 nm and a density of 7.47 g/cm3. β-Mn has a different cubic structure, with a0 = 0.6326 nm and a density of 7.26 g/cm3.The atomic weight of manganese is 54.938 g/mol and the atomic radius is 0.112 nm. Determine the percent volume change that would occur if α-Mn transforms to β-Mn.

Solution: First we need to find the number of atoms in each unit cell so we can determine the volume change based on equal numbers of atoms. From the density equation, we find for the α-Mn: g 354.938 4 g mol 7.47 = cm2 8.931 × 10 cm 2 36.022 × 1052 atom4 mol

= 58

For the β-Mn:

atom ; 6s g@ = 8.931 × 10 cm 2 = 7.12 × 10 55 cm2 cell

g 354.938 4 g mol 7.26 = cm2 6.326 × 10 cm 2 36.022 × 1052 atom4 mol

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= 20

atom ; 6t g@ = 6.326 × 10 cm 2 = 2.53 × 10 55 cm2 cell

The volume of the β-Mn can be adjusted by a factor of 58/20, to account for the different number of atoms per cell. The volume change is then: ∆6 =

58 3204 6t g@ − 6s g@ 6s g@

2.9 2.53 nm2 − 7.12 nm2 × 100% = × 100% 7.12 nm2 ∆6 = +3.05%

During the time of crystal change, it grows greater.

3-41

Calculate the theoretical density of the three polymorphs of zirconia. The lattice constants for the monoclinic form are a = 5.156, b = 5.191 and c = 5.304 Å, respectively.

The angle β for the monoclinic unit cell is 98.9°. The lattice constants for the tetragonal unit cell are a = 5.094 and c = 5.304 Å, respectively. Cubic zirconia has a lattice constant of 5.124 Å. Each unit cell contains for formula units of ZrO2.

Solution: For all three, the theoretical density is given by Equation 3-5: Density % =

number of atoms⁄cell atomic mass volume of unit cell /0

The atomic mass of zirconia (ZrO2) is 123.22 g/mol.

For monoclinic zirconia, there is 1 atom per unit cell. The volume of the monoclinic cell is given by: 6KJ@JCEA@AC ?@AB CDEE = 9 sin u

6KJ@JCEA@AC ?@AB CDEE = 0.5156 nm 0.5191 nm 0.5304 nm sin 98.9° 6KJ@JCEA@AC ?@AB CDEE = 0.1403 nm2

g atom 4 3123.22 4 mol cell %KJJ@CEA@AC Qvwx = atom 0.1403 × 10 5H cm2 36.022 × 1052 4 mol 31

%KJJ@CEA@AC Qvwx = 1.458

g cm2

For tetragonal, there is 1 atom per unit cell. The volume is: 6BDBvIhJ@IE ?@AB CDEE = 5 5

6BDBvIhJ@IE ?@AB CDEE = 5.094 Å 5.304 Å 6BDBvIhJ@IE ?@AB CDEE = 0.1376 nm2

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g atom 4 3123.22 4 mol cell %BDBvIhJ@IE Qvwx = atom 0.1376 × 10 5H cm2 36.022 × 1052 4 mol 31

%BDBvIhJ@IE Qvwx = 1.487

g cm2

For cubic, there is 1 atom per unit cell. The volume is: 6C?^AC ?@AB CDEE = 2

2

6C?^AC ?@AB CDEE = 5.124 Å

6C?^AC ?@AB CDEE = 0.1345 nm2

g atom 4 3123.22 4 cell mol %C?^AC Qvwx = atom 0.1345 × 10 5H cm2 36.022 × 1052 4 mol 31

%C?^AC Qvwx = 1.521

3-42

g cm2

From the information in this chapter, calculate the volume change that will occur when the cubic form of zirconia transforms into a tetragonal form.

Solution: From problem 3-41, we have the densities of the two forms: g cm2 g %BDBvIhJ@IE Qvwx = 1.487 cm2 %C?^AC Qvwx = 1.521

By assuming a constant mass of 1 g, we can take the inverse of the densities and take their differences: H %C?^AC Qvwx = 0.657

cm2 g

H %BDBvIhJ@IE Qvwx = 0.672

cm2 g

∆6 = 1 g ∆% H = 1 g Y−0.015 ∆6 = −0.015 cm2

3-43

cm2 [ g

Monoclinic zirconia cannot be used effectively for manufacturing oxygen sensors or other devices. Explain.

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Solution: Monoclinic zirconia is not volume stable and may fail in use.

3-44

What is meant by the term stabilized zirconia?

Solution: Dopants (deliberate impurities) are added to the crystal structure to make the cubic crystal form preferred energetically, so it does not change.

3-45

State any two applications of stabilized zirconia ceramics.

Solution: Thermal barrier coats for turbine blades and electrolytes for oxygen sensors.

3-46

Explain the significance of crystallographic directions using an example of an application.

Solution: Metal defamation is easier in directions of crystal orientation.

3-47

Why are Fe-Si alloys used in magnetic applications “grain oriented”?

Solution: The direction of the grains will facilitate or hinder the magnetization of the alloy.

3-49

Determine the Miller indicies of the directions for the following points: (a) from (1, 0, 2) to (2, 4, 1); (b) from (2, 1, 3) to (5, 4, 2); and (c) from (3, 1, 3) to (9, 1, 5).

Solution: [ 141 ], [ 331 ], [602] 3-50

Indicate the directions (a) [111], (b) [025], and (c) [414] within a unit cell.

Solution:

3-51

Determine the Miller indices for the directions in the cubic unit cell shown in Figure 337.

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Solution: A: 0, 1, 0 - 0, 1, 1 = 0, 0, -1 = [ 001 ] B: ½, 0, 0 - 0, 1, 0 = ½, -1, 0 = [ 120 ] C: 0, 1, 1 - 1, 0, 0 = -1, 1, 1 = [ 111 ] D: 1, 0, ½ = 0, ½, 1 = 1, -½, -½ = [ 211 ] 3-52

Determine the indices for the directions in the cubic unit cell shown in Figure 3-38.

Solution: A: 0, 0, 1 - 1, 0, 0 = -1, 0, 1 = [ 101 ] B: 1, 0, 1 - ½, 1, 0 = ½, -1, 1 = [ 122 ] C: 1, 0, 0 - 0, ¾, 1 = 1, -¾, -1 = [ 434 ] D: 0, 1, ½ = 0, 0, 0 = 0, 1, ½ = [021] 3-53

Indicate the planes (a) (100), (b) (134), and (c) (101) within a unit cell.

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Solution:

3-54

Determine the indices for the planes in the cubic unit cell shown in Figure 3-39.

Solution: A: x = 1; 1/x = 1 y = -1; 1/y = -1 z = 1; 1/z = 1 Therefore, ( 111 ) B: x = ∞; 1/x = 0 y = ⅓; 1/y = 3 z = ∞; 1/z = 0 Therefore, (030) C: x = 1 1/x = 1 y = ∞; 1/y = 0

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z = -½; 1/z = -2 Therefore, ( 102 ) with origin at (0, 0, 1) 3-55

Determine the indices for the planes in the cubic unit cell shown in Figure 3-40.

Solution: A: x = -1; 1/x = -1 × 3 = -3 y = ½; 1/y = 2 × 3 = 6 z = ¾; 1/z =

4 ×3=4 3

Therefore, ( 364 ) with origin at (1, 0, 0) B: x = 1; 1/x = 1 × 3 = 3 y = -¾; 1/y = -

4 × 3 = -4 3

z = ∞; 1/z = 0 × 3 = 0 Therefore, ( 340 ) with origin at (0, 1, 0) C: x = 2; 1/x = ½ × 6 = 3 y=

3 ; 1/y = ⅔ = 6 = 4 2

z = 1; 1/z = 1 × 6 = 6 Therefore, (346)

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3-56

Determine the indices for the directions in the hexagonal lattice shown in Figure 3-41, using both the three-digit and four-digit systems.

Solution: A: (1, -1, 0) – (0, 0, 0) = (1, -1, 0) = [ 110 ] h = ⅓(2 + 1) = 1 k = ⅓ (-2 - 1) = -1 i = -⅓ (1 - 1) = 0 l=0 Therefore, [ 1100 ] B: (1, 1, 0) – (0, 0, 1) = (1, 1, -1) = [ 111 ] h = ⅓(2 - 1) = ⅓ k = ⅓(2 - 1) = ⅓ i = -⅓(1 + 1) = -⅔ l = -1 Therefore, [ 1123 ] C: (0, 1, 1) – (0, 0, 0) = (0, 1, 1) = [011]

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h = ⅓(0 - 1) = -⅓ k = ⅓(2 - 0) = ⅔ i = -⅓(0 + 1) = -⅓ l=1 Therefore, [ 1213 ] 3-57

Determine the indices for the directions in the hexagonal lattice shown in Figure 3-42, using both the three-digit and four-digit systems.

Solution: A: (0, 1, 1) – (½, 1, 0) = (-½, 0, 1) = [ 102 ] h = ⅓(-2 - 0) = -⅔ k = ⅓(0 + 1) = ⅓ i = -⅓(-1 + 0) = ⅓ l=2 Therefore, [ 2116 ] B: (1, 0, 0) – (1, 1, 1) = (0,-1, -1) = [ 011 ] h = ⅓(0 + 1) = ⅓ k = ⅓(-2 + 0) = -⅔ i = -⅓(0 - 1) = ⅓

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l = -1 [ 1213 ] C: (0, 0, 0) – (1, 0, 1) = (-1, 0, -1) = [ 101 ] h = ⅓(-2 + 0) = -⅔ k = ⅓(0 + 1) = ⅓ i = -⅓(-1 + 0) = ⅓ l = -1 [ 2113 ] 3-58

Determine the indices for the planes in the hexagonal lattice shown in Figure 3-43.

Solution: A: a1 = 1; 1/a1 = 1 a2 = -1; 1/a2 = -1 a3 = ∞; 1/a3 = 0 c = 1; 1/c = 1 Therefore, ( 1101 ) with origin at a2 = 1.

B: a1 = ∞; 1/a1 = 0

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a2 = ∞; 1/a2 = 0 a3 = ∞ 1/a3 = 0 c = ⅔; 1/c =

3 2

Therefore, (0003) C: a1 = 1; 1/a1 = 1 a2 = -1; 1/a2 = -1 a3 = ∞; 1/a3 = 0 c = ∞; 1/c = 0 Therefore, ( 1100 ) 3-59

Determine the indices for the planes in the hexagonal lattice shown in Figure 3-44.

Solution: A: a1 = 1; 1/a1 = 1 a2 = -1; 1/a2 = -1 a3 = ∞; 1/a3 = 0 c = ½; 1/c = 2 Therefore, ( 1102 )

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B: a1 = ∞; 1/a1 = 0 a2 = 1; 1/a2 = 1 a3 = -1; 1/a3 = -1 c = 1; 1/c = 1 Therefore, ( 0111 ) C: a1 =-1; 1/a1 =-1 a2 = ½; 1/a2 = 2 a3 = -1; 1/a3 = -1 c = ∞; 1/c = 0 Therefore, ( 1210 ) 3-60

Sketch the following planes and directions within a cubic unit cell. (a) [101] (b) [ 010 ] (c) [ 122 ] (d) [301] (e) [ 201 ] (f) [ 213 ] (g) ( 011 ) (h) (102) (i) (002) (j) ( 130 ) (k) ( 212 ) (l) ( 312 )

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Solution:

3-61

Sketch the following planes and directions within a cubic unit cell. (a) [ 110 ] (b) [ 221 ] (c) [410] (d) [ 012 ] (e) [ 321 ] (f) [ 111 ] (g) ( 111 ) (h) ( 011 ) (i) (030) (j) ( 121 ) (k) ( 113 ) (l) ( 041 )

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Solution:

3-62

Sketch the following planes and directions within a hexagonal unit cell. (a) [ 0110 ] (b) [ 1120 ] (c) [ 1011 ] (d) (0003) (e) ( 1010 ) (f) ( 0111 )

Solution:

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3-63

Sketch the following planes and directions within a hexagonal unit cell. (a) [ 2110 ] (b) [ 1121 ] (c) [ 1010 ] (d) ( 1210 ) (e) ( 1122 ) (f) ( 1230 )

Solution:

3-64

What are the indices of the six directions of the form <110> that lie in the ( 111 ) plane of a cubic cell?

Solution: [ 110 ] [101] [011] [ 110 ] [ 101 ] [ 011 ]

3-65

What are the indices of the four directions of the form <111> that lie in the ( 101 ) plane of a cubic cell?

Solution: [111] [ 111 ] [ 111 ] [ 111 ]

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3-66

Determine the number of directions of the form <110> in a tetragonal unit cell and compare to the number of directions of the form <110> in an orthorhombic unit cell.

Solution: Tetragonal: [110], [ 110 ], [ 110 ], [ 110 ] = 4 Orthorhombic: [110], [ 110 ] = 2 Note that in cubic systems, there are 12 directions of the form <110>.

3-67

Determine the angle between the [110] direction and the (110) plane in a tetragonal unit cell; then determine the angle between the [011] direction and the (011) plane in a tetragonal cell. The lattice parameters are a0 = 4.0 Å and c0 = 5.0 Å. What is responsible for the difference?

Solution:

[110] ⊥ (110)

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tan(θ/2) = 2.5 / 2 = 1.25 θ/2 = 51.34° θ = 102.68° The lattice parameters in the x and y directions are the same; this allows the angle between [110] and (110) to be 90°. But the lattice parameters in the y and z directions are different!

3-68

Determine the Miller indices of the plane that passes through three points having the following coordinates. (a) (0, 0, 1); (1, 0, 0); and (½, ½, 0) (b) (½, 0, 1); (½, 0 ,0); and (0, 1, 0) (c) (1 ,0, 0); (0, 1, ½) ; and (1, ½, ¼) (d) (1, 0, 0); (0, 0, ¼) ; and (½, 1, 0)

Solution: The plane indices are (a) (111) (b) (210) (c) ( 012 ) (d) (218)

3-69

Calculate and compare the linear densities for the <100>, <110> and <111> directions in a BCC unit cell. Which direction is the most close-packed (dense)?

Solution: %〈H 〉 = %〈HH 〉 =

/IBJK_ 2 √3 = = = 0.866 Distance F 4 G 2 √3

/IBJK_ = Distance

%〈HHH〉 =

2 √3 = = 0.612 4 √2 F G 2√2 √3

/IBJK_ Distance

4 = 1.000 4 √3 F G √3

So <111> is the most closely packed in a BCC unit cell.

3-70

Calculate and compare the linear densities for the <100>, <110> and <111> directions in a FCC unit cell. Which direction is the most close-packed (dense)?

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Solution:

/IBJK_ 2 √2 = = = 0.707 Distance F 4 G 2 √2 /IBJK_ 4 = = 1.000 %〈HH 〉 = 4 Distance F G √2 √2 2 /IBJK_ %〈HHH〉 = = 0.408 4 Distance F G √3 √2 So <110> is the most closely packed in a FCC unit cell. %〈H 〉 =

3-71

Calculate and compare the planar densities for the {100}, {110} and {111} planes in a BCC unit cell. Which plane is the most close-packed (dense)?

Solution:

1 5 /IBJK_ A@ ~EI@D ]IBJK 4 344 = %|H } = = = 0.59 Area 4 5 F G √3 %|HH } =

/IBJK_ A@ ~EI@D ]IBJK = Area

2= 5

4 5 √2 F G √3

= 0.83

1 324 = 5 /IBJK_ A@ ~EI@D ]IBJK %|HHH} = = = 0.34 5 Area 1 4 √2 32 4 Y [ sin 60° √3 So {110} is the most dense in a BCC unit cell.

3-72

Calculate and compare the linear densities for the {100}, {110} and {111} planes in an FCC unit cell. Which plane is the most densely packed?

Solution:

/IBJK_ A@ ~EI@D ]IBJK 2 = 5 = = 0.785 Area 4 5 F G √2 /IBJK_ A@ ~EI@D ]IBJK 2= 5 %|HH } = = = 0.555 Area 4 5 √2 F G √3 /IBJK_ A@ ~EI@D ]IBJK 2= 5 %|HHH} = = = 0.907 5 Area 1 4 √2 32 4 Y [ sin 60° √2 %|H } =

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So {111} is the most dense in a FCC unit cell.

3-73

Determine the repeat distance, linear density, and packing fraction for FCC nickel, which has a lattice parameter of 0.35167 nm, in the [100], [110], and [111] directions. Which of these directions is close packed?

Solution: = For [100]:

√2 0.35167 nm = 0.1243 nm 4

Repeat distance = a0 = 0.35167 nm Linear density = 1/a0 = 2.84 nm-1 Linear packing fraction = (2)(0.1243 nm)(2.84 nm-1) = 0.707

For [110]: Repeat distance = 20.5a0/2 = 0.2487 nm Linear density = (2/20.5)a0 = 4.02 nm-1 Linear packing fraction = (2)(0.1243 nm)(4.02 nm-1) = 1.0

For [111]: Repeat distance = 30.5a0 = 0.6091 nm Linear density = 1/(30.5a0) = 1.642 nm-1 Linear packing fraction = (2)(0.1243 nm)(1.642 nm-1) = 0.408 Only the [110] is close packed; it has a linear packing fraction of 1.

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3-74

Determine the repeat distance, linear density, and packing fraction for BCC lithium, which has a lattice parameter of 0.35089 nm, in the [100], [110], and [111] directions.

Which of these directions is close packed?

Solution: = For [100]:

√3 0.35089 nm = 0.1519 nm 4

Repeat distance = a0 = 0.35089 nm Linear density = 1/a0 = 2.85 nm-1 Linear packing fraction = (2)(0.1519)(2.85) = 0.866

For [110]: Repeat distance = 20.5a0 = 0.496 nm Linear density = 1/(20.5a0) = 2.015 nm-1 Linear packing fraction = (2)(0.1519 nm)(2.015 nm-1) = 0.612

For [111]: Repeat distance 30.5a0/2 = 0.3039 nm Linear density = 2/(30.5a0) ? 3.291 nm-1 Linear packing fraction = (2)(0.1519 nm)(3.291 nm-1) = 1

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The [111] direction is close packed; the linear packing factor is 1.

3-75

Determine the repeat distance, linear density, and packing fraction for HCP magnesium in the [ 2110 ] direction and the [ 1120 ] direction. The lattice parameters for HCP magnesium are given in Appendix A.

Solution: a0 = 3.2087 Å; r = 1.604 Å For [ 2110 ]: Repeat distance = a0 = 3.2087 Å Linear density = 1/a0 = 0.3116 nm-1 Linear packing fraction = (2)(1.604)(0.3116) = 1 (Same for [ 1120 ])

3-76

Determine the planar density and packing fraction for FCC nickel in the (100), (110), and (111) planes. Which, if any, of these planes is close packed?

Solution: a0 = 3.5167 Å For (100): Planar density =

2 = 0.1617 × 10H\ cm 5 3.5167 × 10 cm 5

Packing fraction =

63

2= 5

4 5 F G √2

= 0.785

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For (110): Planar density =

2

√2 3.5167 × 10 cm 5

Packing fraction =

2= 5

4 5 √2 F G √2

= 0.1144 × 10H\ cm 5

= 0.555

For (111): From the sketch, we can determine that the area of the (111) plane is: Y

√2 √3 [Y [ = 0.866 5 2 2

There are (3)(½) + (3)(⅙) = 2 atoms in this area.

Planar density =

2 = 0.1867 × 10H\ cm 5 0.866 3.5167 × 10 cm 5

Packing fraction =

√2 2= Y 4 [ 0.866 5

5

= 0.907

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The (111) is close packed.

3-77

Determine the planar density and packing fraction for BCC lithium in the (100), (110), and (111) planes. Which, if any, of these planes is close packed?

Solution: a0 = 3.5089 Å For (100): Planar density =

1 = 0.0812 × 10H\ cm 5 3.5089 × 10 cm 5

Packing fraction =

For (110): Planar density =

√3 = Y 4 [

5

2

√2 3.5089 × 10 cm 5

Packing fraction =

= 0.589

5

√3 2= Y 4 [ √2 5

= 0.1149 × 10H\ cm 5

5

= 0.833

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For (111): There are only (3)(⅙) = ½ points in the plane, which has an area of 0.866a02 .

1 2 Planar density = = 0.0469 × 10H\ cm 5 0.866 3.5089 × 10 cm 5 Packing fraction =

= √3 2 Y 4 [ 0.866 5

5

= 0.34

There is no close-packed plane in BCC structures.

3-78

Suppose that FCC rhodium is produced as a 1-mm thick sheet, with the (111) plane parallel to the surface of the sheet. How many (111) interplanar spacings d111 thick is the sheet? See Appendix A for necessary data.

Solution: HHH =

thickness = 3-79

√15 + 15 + 15

=

3.796 Å √3

= 2.1916 Å

0.1 mm = 4.563 × 10\ × HHH 2.1916 × 10 cm

In an FCC unit cell, how many d111 are present between the (0,0,0) point and the (1,1,1) point?

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Solution:

The distance between the 0,0,0 and 1,1,1 points is √3 . The interplanar spacing is HHH =

√15 + 15 + 15

Therefore the number of interplanar spacings is / =

3-80

=

√3

√3 = 3

F G √3

What are the stacking sequences in the FCC and HCP structures?

Solution: For FCC, ABCABCABC… For HCP, ABABAB… 3-81

Determine the minimum radius of an atom that will just fit into (a) the tetrahedral interstitial site in FCC nickel; and (b) the octahedral interstitial site in BCC lithium.

Solution: For the tetrahedral site in FCC nickel (a0 = 3.5167 Å): A = F

√2 3.5167 Å = 1.2433 Å 4

G = 0.225 A BDBvI DVvIE _ABD = 1.2433 Å 0.225 = 0.2798 Å

For the octahedral site in BCC lithium (a0 = 3.5089 Å):

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A = F

√3 3.5089 Å = 1.519 Å 4

G = 0.414 A JCBI DVvIE _ABD = 3.5089 Å 0.414 = 0.6290 Å

3-82

What are the coordination numbers for octahedral and tetrahedral sites?

Solution: Six and four, respectively.

3-83

The atomic packing fraction for the FCC crystal structure is 0.74. This is the densest arrangement of atoms achievable in three dimensions when all atoms have the same radius R. If atoms in the hard sphere model occupy 74% of the volume of the unit cell, then the other 26% is empty space. Imagine that the empty spaces located at the center of each edge and at the body–centered position of the FCC unit cell are filled with smaller spheres of radius r such that r 5 0.414R. The smaller atoms fit perfectly in between the atoms of radius R. (a) By counting in the usual way, how many atoms of radius r are there per FCC unit cell? (b) What are the coordinates of these atoms of radius r? Do not double count atoms. Provide the same number of coordinates as the number of atoms of radius r per unit cell. (c) What is the atomic packing fraction for this structure?

Solution:

1 atom *12 edges +1 body–center = 4 atoms of radius r 4 edge These atoms have the following coordinates: (½ , 0, 0)

(0, ½ , 0)

(0, 0, ½ )

(½, ½, ½ )

By definition, the APF is

APF =

number of atoms per unit cell ∗ atomic volume unit cell volume

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There are four atoms per unit cell with radius R and four atoms per unit cell with radius r. Since the smaller atoms fit perfectly in between the atoms of radius R, a 2 = 4R , where a is the lattice parameter for the FCC unit cell.

4 4 4 4 16 16 4 ∗ π R3 + 4 ∗ π r 3 4 ∗ π R3 + 4 ∗ π (0.414R)3 π + π (0.414)3 3 3 3 3 3 APF = = = 3 = 0.79 (4R / 2 )3 (4R / 2 )3 (4 / 2 )3 .

3-84

What is the radius of an atom that will just fit into the octahedral site in FCC copper without disturbing the crystal structure?

T? = 1.278 Å

F G = 0.414 T? JCBI DVvIE _ABD = 1.278 Å 0.414 = 0.529 Å

3-85

Using the ionic radii given in Appendix B, determine the coordination number expected for the following compounds: (a) Y2O3 (b) UO2 (c) BaO (d) Si3N4 (e) GeO2 (f) MnO (g) MgS (h) KBr

Solution:

0.89 Å = = 0.67; CN = 6 w x 1.32 Å 0.97 Å = = 0.67; CN = 8 w x 1.32 Å

w x 1.32 Å = = 0.99; CN = 8 iI x 1.34 Å 0.15 Å = = 0.36; CN = 4 A 0.42 Å

D 0.53 Å = = 0.40; CN = 4 w x 1.32 Å

g@ x 0.80 Å = = 0.61; CN = 6 w x 1.32 Å gh x x

=

0.66 Å 1.84 Å

= 0.36; CN = 6

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1.33 Å = = 0.68; CN = 6 iv 1.96 Å 3-86

A particular unit cell is cubic with ions of type A located at the corners and face-centers of the unit cell and ions of type B located at the midpoint of each edge of the cube and at the body-centered position. The ions contribute to the unit cell in the usual way (1/8 ion contribution for each ion at the corners, etc.). (a) How many ions of each type are there per unit cell? (b) Describe this structure as a lattice and a basis. Check to be sure that the number of ions per unit cell given by your description of the structure as a lattice and a basis is consistent with your answer to part (a). (c) What is the coordination number of each ion? (d) What is the name commonly given to this crystal structure?

Solution: For type A:

For type B:

ion 1 ion 1 × 8 corner + × 6 face center = 4 ion 2 face center 8 corner 1 ion × 12 edge center + 1 body center = 4 ion 4 edge center

The FCC lattice and basis A (0, 0, 0) and one of the following four B bases: (½, 0, 0), (0, ½, 0), (0, 0, ½), (½, ½, ½).

There are four lattice points per unit cell and two ions in the basis per lattice point for a total of eight ions per unit cell – which is consistent with part (a).

CN = 6 for both A and B. Common name is sodium chloride.

3-87

Would you expect NiO to have the cesium chloride, sodium chloride, or zincblende structure? Based on your answer, determine (a) the lattice parameter; (b) the density; and (c) the packing factor.

Solution:

A x 0.69 Å = = 0.52; CN = 6 w x 1.32 Å

A coordination number of 8 is expected for the CsCl structure, and a coordination number of 4 is expected for ZnS. But a coordination number of 6 is consistent with the NaCl structure. = 2 0.69 Å + 2 1.32 Å = 4.02 Å

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g g ion 4 358.17 + 16 4 mol mol cell %= atom 4.02 × 10 cm 2 36.022 × 1052 4 mol 34

% = 7.64

g cm2

2 2 4= ion 4 34 4 3 0.69 Å + 1.32 Å 4 3 cell PF = 2 4.02 Å

3

PF = 0.678

3-88

Would you expect UO2 to have the sodium chloride, zincblende, or fluorite structure?

Based on your answer, determine (a) the lattice parameter; (b) the density; and (c) the packing factor.

Solution:

0.97 Å = = 0.735; CN = 8 w x 1.32 Å

The radius ratio predicts a coordination number of 8; however, there must be twice as many oxygen ions as uranium ions in order to balance the charge. The fluorite structure will satisfy these requirements, with U = FCC position (4) and O = tetrahedral position (8). √3 = 4 + 4 w x = 4 0.97 Å + 1.32 Å = 9.16 Å → = 5.2885 Å g g 4 + 8 316 4 g mol mol %= = 12.13 atom cm2 5.2885 × 10 cm 2 36.022 × 1052 4 mol 4 3238.03

PF =

3-89

2 2 4= 3 3 4 34 0.97 Å + 8 1.32 Å 4 2

5.2885 Å

= 0.624

Would you expect BeO to have the sodium chloride, zincblende, or fluorite structure?

Based on your answer, determine (a) the lattice parameter; (b) the density; and (c) the packing factor.

Solution:

iD x 0.35 Å = = 0.265; CN = 4 w x 1.32 Å

This coordination number signifies a zincblende structure. √3 = 4 iD x + 4 w x = 4 0.35 Å + 1.32 Å = 6.68 Å → = 3.8567 Å

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%=

g g + 16 4 mol mol

4 39.01

g = 2.897 atom cm2 3.8567 × 10 cm 2 36.022 × 1052 4 mol 2 2 4= 4 4 3 0.35 Å + 8 1.32 Å 4 PF = 3 = 0.684 2 3.8567 Å

3

3-90

Would you expect CsBr to have the sodium chloride, zincblende, fluorite, or cesium chloride structure? Based on your answer, determine (a) the lattice parameter; (b) the density; and (c) the packing factor.

Solution:

T_ 1.67 Å = = 0.852; CN = 8 iv 1.96 Å

This coordination number signifies a cesium chloride structure. √3 = 2 T_ + 2 iv = 2 1.67 Å + 1.96 Å = 7.26 Å → = 4.1916 Å %=

379.904

g = 4.8 atom cm2 4.1916 × 10 cm 2 36.022 × 1052 4 mol PF =

3-91

g g + 132.91 4 mol mol

2 2 4= 3 3 4 3 1.96 Å + 1.67 Å 4 2

4.1916 Å

= 0.693

Sketch the ion arrangement on the (110) plane of ZnS (with the zincblende structure) and compare this arrangement to that on the (110) plane of CaF2 (with the flourite structure). Compare the planar packing fraction on the (110) planes for these two materials.

Solution: ZnS:

√3 = 4 Q@ x + 4 x = 4 0.074 nm + 0.184 nm → = 0.596 nm

PPF =

5 2 = Q@ + 2 = 5

√2

=

2= 0.074 nm 5 + 2= 0.184 nm 5 √2 0.596 nm 5

= 0.492

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CaF2:

√3 = 4 TI x + 4 = 4 0.099 nm + 0.133 nm → = 0.536 nm

PPF =

3-92

5 2 = TI + 4 = 5

√2

=

2= 0.099 nm 5 + 2= 0.133 nm 5 √2 0.536 nm 5

= 0699

MgO, which has the sodium chloride structure, has a lattice parameter of 0.396 nm.

Determine the planar density and the planar packing fraction for the (111) and (222) planes of MgO. What ions are present on each plane?

Solution: As described in the answer to Problem 3-72, the area of the (111) plane is 0.866a02. = 2 gh x + 2 w x = 2 0.66 Å + 1.32 Å = 3.96 Å

HHH =

2 Mg = 0.1473 × 10H\ cm 5 0.866 3.96 × 10 cm 5 PPF HHH =

5

2= 0.66 Å

5 = 0.202

0.866 3.96 Å

555 = 0.1473 × 10H\ cm 5

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PPF HHH =

3-96

5

2= 1.32 Å

5 = 0.806

0.866 3.96 Å

A diffracted x-ray beam is observed from the (220) planes of iron at a 2θ angle of 99.1° when x-rays of 0.15418 nm wavelength are used. Calculate the lattice parameter of the iron.

Solution:

sin

sin =

2 55

99.1° 0.15418 nm √25 + 25 + 05 = 2 2 =

0.15418 nm √8 2 sin 49.55°

= 0.2865 nm

3-97

A diffracted x-ray beam is observed from the (311) planes of aluminum at a 2θ angle of 78.3° when x-rays of 0.15418 nm wavelength are used. Calculate the lattice parameter of the aluminum.

Solution:

=

sin =

2 2HH

0.15418 nm √35 + 15 + 15 78.3° 2 sin 2 = 0.40497 nm

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3-98

Figure 3-45 shows the results of an x-ray diffraction experiment in the form of the intensity of the diffracted peak versus the 2θ diffraction angle. If x-rays with a wavelength of 0.15418 nm are used, determine (a) the crystal structure of the metal; (b) the indices of the planes that produce each of the peaks; and (c) the lattice parameter of the metal.

Solution: The 2θ values can be estimated from the figure: 1 2 3 4 5 6 7 8

2θ 17.5 20.5 28.5 33.5 35.5 41.5 45.5 46.5

sin2 θ 0.023 0.032 0.061 0.083 0.093 0.123 0.146 0.156

sin2 θ/0.0077 3 4 8 11 12 16 19 20

Planar indicies (111) (200) (220) (311) (222) (400) (331) (420)

d = λ/(2 sin θ) 0.5068 0.4332 0.3132 0.2675 0.2529 0.2201 0.2014 0.1953

a0 = (h2 + k2 +l2)0.5 0.8777 0.8665 0.8858 0.8872 0.8760 0.8805 0.8781 0.8734

The sin2 θ values must be divided by 0.077 (one third the first sin2 θ value) in order to produce a possible sequence of numbers).

The 3, 4, 8, 11, … sequence means that the material is FCC The average a0 = 0.8781 nm.

3-99

Figure 3–46 shows the results of an x-ray diffraction experiment in the form of the intensity of the diffracted peak versus the 2θ diffraction angle. If x-rays with a wavelength of 0.07107 nm are used, determine (a) the crystal structure of the metal; (b) the indices of the planes that produce each of the peaks; and (c) the lattice parameter of the metal.

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Solution: The 2θ values can be estimated from the figure: 1 2 3 4 5 6 7 8

2θ 25.5 36.5 44.5 51.5 58.5 64.5 70.5 75.5

sin2 θ 0.049 0.095 0.143 0.189 0.235 0.285 0.329 0.375

sin2 θ/0.0077 1 2 3 4 5 6 7 8

Planar indicies (110) (200) (211) (220) (310) (222) (321) (400)

d = λ/(2 sin θ) 0.1624 0.116 0.0947 0.0825 0.0739 0.0672 0.0625 0.0586

a0 = (h2 + k2 +l2)0.5 0.2297 0.2320 0.2319 0.2334 0.2338 0.2327 0.2339 0.2342

The sequence 1, 2, 3, 4, 5, 6, 7, 8 (which includes the “7”) means that the material is BCC. The average a0 = 0.2327 nm.

3-100 A sample of zirconia contains cubic and monoclinic polymorphs. What will be a good analytical technique to detect the presence of these two different polymorphs?

Solution: Transmission electron microscopy would work for this, if the budget and other considerations allow for it.

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Chapter 4: Imperfections in the Atomic and Ionic Arrangements 4-1

Gold has 5.82 × 108 vacancies/cm3 at equilibrium at 300 K. What fraction of the atomic sites is vacant at 600 K?

Solution: The number of vacancies per cm3 nv is given by

 −Q  nv = n exp  v  ,  RT  where n is the number of atoms per cm3, Qv is the energy required to produce one mole of vacancies, R is the gas constant, and T is the absolute temperature. Gold has a density of 19.302 g/cm3 and an atomic mass of 196.97 g/mol. Thus the number of atoms per cm3 is

19.302 g 1 mol 6.022 ×1023 atoms × × = 5.90 ×1022 atoms/cm3 3 cm 196.97 g 1 mol Solving for Qv, the energy is given by

 5.82 ×108  n  Qv = − RT ln  v  = −8.314 J/(mol − K) × (300 K) ln  22  n  5.90 ×10  = 80,438 J/mol.

The fraction of the vacant atomic sites is given by nv/n. At 600 K,

  nv 80,438 J/mol  Q  −8 = exp  − v  = exp   = 9.93 × 10 . n  RT   8.314 J/(mol − K) × 600 K  4-2

Calculate the number of vacancies per m3 for gold at 900 °C. The energy for vacancy formation is 0.86 eV/atom.

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Solution:

4-3

Calculate the number of vacancies per cm3 expected in copper at 1080°C (just below the melting temperature). The energy for vacancy formation is 20,000 cal/mol.

Solution:

n=

(4 atoms/unit cell) = 8.47 ×1022 atoms/cm3 (3.6151×10−8cm)3

nv = 8.47 ×1022exp[ − 20,000/(1.987)/(1353)] = 8.47 ×1022exp( − 7.4393) = 4.98 ×1019 vacancies/cm3 4-4

The fraction of lattice points occupied by vacancies in solid aluminum at 660°C is 10–3.

What is the activation energy required to create vacancies in aluminum?

Solution: nv/n = 10–3 = exp[–Q/(1.987)/(933)] ln(10–3) = –6.9078 = –Q/(1.987)/(933) Q = 12,800 cal/mol 4-5

The density of a sample of FCC palladium is 11.98 g/cm3, and its lattice parameter is 3.8902 Å. Calculate (a) the fraction of the lattice points that contain vacancies; and (b) the total number of vacancies in a cubic centimeter of Pd.

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Solution:

4-6

(x) (106.4 g/mol) (3.8902 × 10 cm)3 (6.022 × 1023atoms/mol) x = 3.9918 (a) 4.0 − 3.9918 fraction = = 0.00204 4 0.0082 vacancies/unit cell = 1.39 ×1020 vacancies/cm3 (b) number = −8 3 (3.8902 ×10 cm) 11.98 g/cm3 =

−8

The density of a sample of HCP beryllium is 1.844 g/cm3, and the lattice parameters are a0 = 0.22858 nm and c0 = 0.35842 nm. Calculate (a) the fraction of the lattice points that contain vacancies; and (b) the total number of vacancies in a cubic centimeter of Be.

Solution:

Vunitcell = (0.22858 nm)2 (0.35842 nm)cos30° = 0.01622 nm3 = 1.622 ×10−23cm3 (a) From the density equation:

(x) (9.01 g/mol) (1.622 ×10 cm 3 ) (6.02 ×1023atoms/mol) x = 1.9984 2 −1.9984 fraction = = 0.0008 2 0.0016 vacancies/unit cell = 0.986 ×1020 vacancies/cm3 (b) number = −23 3 1.622 ×10 cm 1.844 g/cm3 =

4-7

−23

BCC lithium has a lattice parameter of 3.5089 × 10–8 cm and contains one vacancy per 200 unit cells. Calculate (a) the number of vacancies per cubic centimeter; and (b) the density of Li.

Solution:

(a)

1 vacancy = 1.157 ×1020 vacancies/cm3 (200) (3.5089 ×10−8cm)3

(b) In 200 unit cells, there are 399 Li atoms. The atoms/cell are 399/200:

ρ= 4-8

(399/200) (6.94 g/mol) = 0.532 g/cm3 −8 3 23 (3.5089 ×10 cm) (6.02 ×10 atoms/mol)

FCC lead has a lattice parameter of 0.4949 nm and contains one vacancy per 500 Pb atoms. Calculate (a) the density; and (b) the number of vacancies per gram of Pb.

Solution: The number of atoms/cell = (499/500)(4 sites/cell)

ρ=

(499/500) (4) (207.19 g/mol) = 11.335 g/cm3 (4.949 ×10−8cm)3 (6.02 ×1023atoms/mol)

The 500 Pb atoms occupy 500 / 4 = 125 unit cells:

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1 vacancy × [(1/11.335 g/cm3 )] = 5.82 ×1018 vacancies/g   125 cells  (4.949 ×10−8cm)3    4-9

Cu and Ni form a substitutional solid solution. This means that the crystal structure of a Cu-Ni alloy consists of Ni atoms substituting for Cu atoms in the regular atomic positions of the FCC structure. For a Cu–30% wt% Ni alloy, what fraction of the atomic sites does Ni occupy?

Solution: In 100 g of a Cu–30% wt.% Ni alloy, Cu comprises 70 g, and Ni comprises 30 g. In 70 g of Cu, there are

1 mol 6.022 ×1023atoms 70 g Cu × × = 6.63 ×1023 Cu atoms.

63.54 g 1 mol In 30 g of Ni, there are

30 g Cu ×

1 mol 6.022 ×1023 atoms × = 3.08 ×1023 Ni atoms.

58.71 g 1 mol

Therefore, the Ni atoms occupy

3.08 ×1023 Ni atoms = 0.32 3.08 ×1023 Ni atoms + 6.63 ×1023 Cu atoms or 32% of the atomic sites.

4-10

Au and Ag form a substitutional solid solution. This means that the crystal structure of a Au-Ag alloy consists of Ag atoms substituting for Au atoms in the regular atomic positions of the FCC structure. For a Au-50 at% Ag alloy, what is the wt% Ag in the

alloy?

Solution: In 1 mol of a Au–50 at.% Ag alloy, Au comprises 0.5 mol, and Ag comprises 0.5 mol.

196.97 g Au = 98.485 g Au. 1 mol 107.868 g Au 0.5 mol Ag × = 53.934 g Ag. 1 mol 0.5 mol Au ×

Thus the wt% Ag is

53.934 g = 35.4 wt% Ag 53.934 g +98.485 g 4-11

A niobium alloy is produced by introducing tungsten substitutional atoms into the BCC structure; eventually an alloy is produced that has a lattice parameter of 0.32554 nm and a density of 11.95 g/cm3. Calculate the fraction of the atoms in the alloy that are tungsten.

Solution:

11.95 g/cm3 =

(xw ) (183.85 g/mol) + (2 − xw ) (92.91 g/mol) (3.2554 ×10−8cm)3 (6.022 ×1023 atoms/mol)

248.186 = 183.85xW + 185.82 – 92.91xW

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90.94xW = 62.366 or

xW = 0.687 W atoms/cell

There are 2 atoms per cell in BCC metals. Thus: fw = 0.687/2 = 0.344 4-12

Tin atoms are introduced into an FCC copper crystal, producing an alloy with a lattice parameter of 3.7589 × 10–8 cm and a density of 8.772 g/cm3. Calculate the atomic percentage of tin present in the alloy.

Solution:

(xSn ) (118.69 g/mol) + (4 − xSn ) (63.54 g/mol) (3.7589 ×10−8cm)3 (6.02 ×1023 atoms/mol) 280.5 = 55.15xSn + 254.16 or xSn = 0.478 Sn atoms/cell

8.772 g/cm3 =

There are 4 atoms per cell in FCC metals; therefore the at% Sn is (0.478/4) = 11.95% 4-13

We replace 7.5 atomic percent of the chromium atoms in its BCC crystal with tantalum.

X-ray diffraction shows that the lattice parameter is 0.29158 nm. Calculate the density of the alloy.

Solution:

4-14

ρ=

(2) (0.925) (51.996 g/mol) + 2(0.075) (180.95 g/mol) = 8.262 g/cm3 −8 3 23 (2.9158 ×10 cm) (6.022 ×10 atoms/mol)

Suppose we introduce one carbon atom for every 100 iron atoms in an interstitial position in BCC iron, giving a lattice parameter of 0.2867 nm. For this steel, find the density and the packing factor.

Solution:

There is one carbon atom per 100 iron atoms, or 1 C/50 unit cells, or 1/50 C per unit cell:

(2) (55.847 g/mol) + (1/50)(12 g/mol) = 7.89 g/cm3 −8 3 23 (2.867 ×10 cm) (6.02 ×10 atoms/mol) 2(4π /3) (1.241)3 + (1/50)(4π /3) (0.77)3 Packing Factor = = 0.681 (2.867)3

ρ=

4-15

The density of BCC iron is 7.882 g/cm3, and the lattice parameter is 0.2866 nm when hydrogen atoms are introduced at interstitial positions. Calculate (a) the atomic fraction of hydrogen atoms; and (b) the number of unit cells on average that contain hydrogen atoms.

Solution:

(a) 7.882 g/cm = 3

2(55.847 g/mol) + x(1.00797 g/mol) (2.866 ×10−8cm)3 (6.022 ×1023atoms/mol)

x = 0.0449 H atoms/cell The total atoms per cell include 2 Fe atoms and 0.0449 H atoms. Thus,

fH =

0.0449 = 0.02195 2.0449

(b) Since there is 0.0449 H/cell, then the number of cells containing H atoms is

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cells = 1 / 0.0449 = 22.3 or 1 H in 22.3 cells 4-16

Suppose one Schottky defect is present in every tenth unit cell of MgO. MgO has the sodium chloride crystal structure and a lattice parameter of 0.396 nm. Calculate (a) the number of anion vacancies per cm3; and (b) the density of the ceramic.

Solution: In 10 unit cells, we expect 40 Mg + 40 O ions, but due to the defect: 40 Mg – 1 = 39 40 O – 1 = 39 1 vacancy/(10 cells)(3.96 × 10–8 cm)3 = 1.61 × 1021 vacancies/cm3

ρ= 4-17

(39/40) (4) (24.312 g/mol) + (39/40) (4) (16 g/mol) = 4.205 g/cm3 −8 3 23 (3.96 ×10 cm) (6.02 ×10 atoms/mol)

ZnS has the zinc blende structure. If the density is 3.02 g/cm3 and the lattice parameter is 0.59583 nm, determine the number of Schottky defects (a) per unit cell; and (b) per cubic centimeter.

Solution: Let x be the number of each type of ion in the unit cell. There normally are 4 of each type. (a)

3.02 g/cm3 =

x(65.38 g/mol) + x(32.064 g/mol) (5.9583 ×10−8cm)3 (6.022 ×1023ions/mol)

x = 3.9478

4 – 3.9478 = 0.0522 defects/unit cell (b) # of unit cells/cm3 = 1/(5.9583 × 10–8 cm)3 = 4.728 × 1021 Schottky defects per cm3 = (4.728 × 1021)(0.0522) = 2.466 × 1020 4-18

Suppose we introduce the following point defects. (a) Mg2+ ions substitute for yttrium atoms in Y2O3; (b) Fe3+ ions substitute for magnesium ions in MgO; (c) Li1+ ions substitute for magnesium ions in MgO; and (d) Fe2+ ions replace sodium ions in NaCl. What other changes in each structure might be necessary to maintain a charge balance?

Explain.

Solution: (a) Remove 2 Y3+ and add 3 Mg2+ – create cation interstitial. (b) Remove 3 Mg2+ and add 2 Fe3+ – create cation vacancy. (c) Remove 1 Mg2+ and add 2 Li+ – create cation interstitial. (d) Remove 2 Na+ and add 1 Fe2+ – create cation vacancy.

4-19

Write down the defect chemistry equation for introduction of SrTiO3 in BaTiO3 using the Kröger-Vink notation.

Solution:

4-20

Do amorphous and crystalline materials plastically deform by the same mechanisms?

Explain.

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Solution: Without a crystalline structure, the concepts of dislocations and slip systems don’t apply to amorphous systems. Amorphous systems tend to be susceptible to flaws in the material.

4-21

What is the Burger’s vector orientation relationship with the dislocation axis for both edge and screw dislocations?

Solution: The edge dislocation axis is perpendicular to the Burgers vector while the screw dislocation axis is parallel to the Burgers vector.

4-22

What is a slip system and what role does it play in plastic deformation?

Solution: A slip system comprises a slip plane and slip direction, usually closed-packed, in which line defects can move. The movement of these defects or dislocations is referred to as slip. Slip is plastic deformation in crystalline solids.

4-23

Draw a Burgers circuit around the dislocation shown in Figure 4-18. Clearly indicate the Burgers vector that you find. What type of dislocation is this? In what direction will the dislocation move due to the applied shear stress t? Reference your answers to the coordinate axes shown.

Solution: A negative edge dislocation with its Burgers circuit and Burgers vector is shown in the diagram. The dislocation will move in the direction due to the applied shear stress τ.

4-24

What are the Miller indices of the slip directions: (a) on the (111) plane in an FCC unit cell? (b) on the (011) plane in a BCC unit cell?

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Solution:

4-25

(11 1), (1 11)

(111), (1 1 1)

What are the Miller indices of the {110} slip planes in BCC unit cells that include the [111] slip direction?

Solution: 4-27

[1 11], [ 11 1] [ 1 11], [11 1]

What are the Miller indices of the slip planes in FCC unit cells that include the [101] slip direction?

Solution: 4-26

[0 11], [01 1] [ 110], [1 10] [ 101], [10 1]

(1 10), (110)

(0 11), (01 1)

(10 1), (101)

Calculate the length of the Burgers vector in the following materials: (a) BCC niobium; (b) FCC silver; and (c) diamond cubic silicon.

Solution: (a) The repeat distance, or Burgers vector, is half the body diagonal, or:

b = repeat distance = (½)( 3)(3.294 Å) = 2.853 Å (b) The repeat distance, or Burgers vector, is half of the face diagonal, or:

b = (½)( 2a0 ) = (½)( 2)(4.0862 Å) = 2.889 Å (c) The slip direction is [110], where the repeat distance is half of the face diagonal:

b = (½) ( 2) (5.4307 Å) = 3.840 Å 4-28

Determine the interplanar spacing and the length of the Burgers vector for slip on the expected slip systems in FCC aluminum. Repeat, assuming that the slip system is a (110) plane and a [1 11] direction. What is the ratio between the shear stresses required for slip for the two systems? Assume that k = 2 in Equation 4-2.

Solution:

(a) For (111)/[1 10] ,

b = (½) ( 2) (4.04958 Å) = 2.863 Å d111 =

4.04958 Å 12 + 12 + 12

= 2.338 Å

(b) If (110)/[1 11] , then

b = 3 (4.04958 Å) = 7.014 Å d110 =

4.04958 Å 12 + 12 + O 2

= 2.863 Å

(c) If we assume that k = 2 in Equation 4-2, then

2.338 2.863 = 0.8166 (d /b)b = = 0.408 2.863 7.014 τ exp(−2(0.8166)) ∴ a = = 0.44 τ b exp(−2(0.408)) ( d / b) a =

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4-29

Determine the interplanar spacing and the length of the Burgers vector for slip on the (110)[111] slip system in BCC tantalum. Repeat, assuming that the slip system is a

(111)/[1 10] system. What is the ratio between the shear stresses required for slip for the two systems? Assume that k = 2 in Equation 4-2.

Solution:

(a) For (110)/[1 11] :

b = (½) ( 3) (3.3026 Å) = 2.860 Å d110 =

3.3026 Å 12 + 12 + 02

= 2.335 Å

(b) If (111)/[1 10] , then:

b = 2(3.3026 Å) = 4.671 Å d111 =

3.3026 Å 12 + 12 + 12

= 1.907 Å

(c) If we assume that k = 2 in Equation 4-2, then:

2.335 1.907 = 0.8166 (d /b)b = = 0.408 2.86 4.671 τ a exp(−2(0.8166)) = = 0.44 τ b exp(−2(0.408))

(d /b) a =

4-30

The crystal shown in Figure 4-19 contains two dislocations A and B. If a shear stress is applied to the crystal as shown, what will happen to dislocations A and B?

Solution: Under the action of the applied shear stress shown, dislocation A will move to the right, and dislocation B will move to the left. When the dislocations meet, they will annihilate because the combination of a negative edge dislocation and a positive edge dislocation will form perfect crystal.

4-31

Can ceramic and polymeric materials contain dislocations?

Solution: Yes. All crystalline materials can contain dislocations.

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4-32

Why is it that ceramic materials are brittle?

Solution: Dislocations do not move easily enough because of bonding strengths, so the materials fail due to flaws such as cracks and pores before any slip can occur.

4-33

What is meant by the terms plastic and elastic deformation?

Solution: Plastic deformation describes an irreversible change to the shape of an object when a force is applied. Elastic deformation is a temporary change in shape that is recovered when the force is removed.

4-34

Why is the theoretical strength of metals much higher than that observed experimentally?

Solution: Slip allows metallic bonds to be broken individually rather than requiring all bonds in a sample to be broken at once, as is predicted by simply counting metallic bond strengths.

4-35

How many grams of aluminum, with a dislocation density of 1010 cm/cm3, are required to give a total dislocation length that would stretch from New York City to Los Angeles (3000 miles)?

Solution: (3000 mi)(5280 ft/mi)(12 in./ft)(2.54 cm/in.) = 4.828 × 108 cm

(4.828 ×108cm) (2.699 g/cm3 ) = 0.13 g (1010 cm/cm3 ) 4-36

The distance from Earth to the Moon is 240,000 miles. If this were the total length of dislocation in a cubic centimeter of material, what would be the dislocation density?

Compare your answer to typical dislocation densities for metals.

Solution: (240,000 mi)(5280 ft/mi)(12 in./ft)(2.54 cm/in.) = 3.86 × 1010 cm/cm3 This is reasonable as dislocation densities range from 106 cm/cm3 to 1012 cm/cm3.

4-37

Why would metals behave as brittle materials without dislocations?

Solution: The dislocations allow slip, which in turn allows Without dislocations, the metal would behave as a ceramic and fail completely when its ultimate strength was reached.

4-38

Why is it that dislocations play an important role in controlling the mechanical properties of metallic materials, however, they do not play a role in determining the mechanical properties of glasses?

Solution: Glasses do not contain dislocations. Glasses are amorphous, and therefore, they do not have defects such as dislocations.

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4-39

Suppose you would like to introduce an interstitial or large substitutional atom into the crystal near a dislocation. Would the atom fit more easily above or below the dislocation line shown in Figure 4-7(c)? Explain.

Solution: The atom would fit more easily into the area just below the dislocation due to the atoms being pulled apart; this allows more space into which the atom can fit.

4-40

Compare the c/a ratios for the following HCP metals, determine the likely slip processes in each, and estimate the approximate critical resolved shear stress. Explain. (See data in Appendix A.) (a) Zinc; (b) Magnesium; (c) Titanium; (d) Zirconium; (e) Rhenium; and (f) Beryllium.

Solution: We expect metals with c/a > 1.633 to have a low τcrss:

4.9470 = 1.856 − low τ crss 2.6648 5.209 = 1.62 − medium τ crss (b) Mg: 3.2087 4.6831 = 1.587 − high τ crss (c) Ti: 2.9503 5.1477 = 1.593 − high τ crss (d) Zr: 3.2312 4.458 = 1.615 − medium τ crss (e) Rh: 2.760 3.5842 = 1.568 − high τ crss (f) Be: 2.2858 (a) Zn:

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4-41

Using Schmid’s Law equation, the resolved shear stress operating on a slip direction/plane is given by τ r = σ cos λ cos φ . If either of these angles equal 90°, what happens to the dislocations involved?

Solution: The dislocations will not slip.

4-42

A single crystal of an FCC metal is oriented so that the [001] direction is parallel to an applied stress of 5000 psi. Calculate the resolved shear stress acting on the (111) slip plane in the [110], [0 11] , and [10 1] slip directions. Which slip system(s) will become active first?

Solution: φ = 54.76° λ110 = 90° λ011 = 45° λ101 = 45°

4-43

τ = 5000 cos 54.76° cos λ τ=0 τ = 2040 psi active τ = 2040 psi active

A single crystal of a BCC metal is oriented so that the [001] direction is parallel to the applied stress. If the critical resolved shear stress required for slip is 12,000 psi, calculate the magnitude of the applied stress required to cause slip to begin in the [1 11] direction on the (110), (011), and (10 1) slip planes.

Solution: CRSS = 12,000 psi = σ cosφ cosλ λ = 54.76° 12, 000 psi

cos φ cos λ φ110 = 90o φ011 = 45° φ101 = 45°

=σ

σ=∞ σ = 29,412 psi σ = 29,412 psi

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4-44

A single crystal of silver is oriented so that the (111) slip plane is perpendicular to an applied stress of 50 MPa. List the slip systems composed of close-packed planes and directions that may be activated due to this applied stress.

Solution: There are four possible slip planes in an FCC crystal. They are the (111), (11 1), (1 11) , and (11 1) . Each slip plane contains three slip directions. Thus there are a total of twelve potential slip systems in an FCC single crystal. Some of these slip systems will not be activated, however, because either the slip plane or the slip direction is perpendicular to the applied stress. The twelve potential slip systems are listed below, and those that are not activated are crossed out. If the (111) slip plane is perpendicular to the applied stress of 50 MPa, then the plane normal – the [111] direction – is the direction of the applied stress. Thus there can be no shear stress acting on the (111) plane. All slip systems that include the (111) plane are eliminated. In the (111) system, the [01 1] is perpendicular to the [111] direction, and therefore, is eliminated. In the (1 11) system, the [101] is perpendicular to the [111] direction, and therefore, is eliminated. In the (11 1) , the [110] is perpendicular to the [111] direction, and therefore, is eliminated.

4-45

(111)[1 10]

(111)[01 1]

(1 11)[101]

(11 1)[011]

(111)[101]

(111)[110]

(1 11)[110]

(11 1)[101]

(111)[01 1]

(111)[101]

(1 11)[011]

(11 1)[110]

Why is it that single crystal and polycrystalline copper are both ductile; however, only single crystal, but not polycrystalline, zinc can exhibit considerable ductility?

Solution: Single crystal metals with the FCC structure are inherently ductile because FCC has a large number of slip systems (12). Thus regardless of the orientation of the applied stress, a number of slip systems will be active. The HCP structure has a relatively small number of slip systems (3), and these slip systems are parallel to each other. When the applied stress is oriented favorably for shear, the slip systems will be active and the single crystal will deform in a ductile fashion. When the HCP material is polycrystalline, it will behave in a brittle fashion because cross slip does not occur.

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4-46

Explain why hexagonal close-packed metals tend to have a limited ability to be strain hardened.

Solution: HCP metals with c/a ratios > 1.633 have limited ability to strain harden because of the limited number of parallel slip systems these metals possess. Since there are only three parallel slip systems, the opportunities for dislocations to interact and become tangled are low. In addition, since these metallic systems do not have the ability to cross-slip, ductility is also limited.

4-47

Why is it that cross slip in BCC and FCC metals is easier than in HCP metals? How does this influence the ductility of BCC, FCC, and HCP metals?

Solution: Cross slip is easier in BCC and FCC metals than in HCP metals because there are multiple non-parallel and intersecting slip systems in the BCC and FCC structures. This makes it possible for dislocations to move from one slip plane to another. All of the favored slip systems in HCP metals are parallel to each other making cross slip difficult.

4-48

Arrange the following metals in the expected order of increasing ductility: Cu, Ti, and Fe. Explain.

Solution: From least ductile to most ductile, the order is Ti, Fe, and Cu. Ti is HCP and thus will be the most brittle due to the limited number of slip systems. BCC metals (such as Fe) tend to have higher strengths and lower ductilities than FCC metals (such as Cu) due to their higher critical resolved shear strengths.

4-49

What are the imperfections in the atomic arrangements that have a significant effect on the material behavior? Give an example of each.

Solution: Imperfections include point (vacancies), line (dislocations) and surface defects (grain boundaries).

4-50

The strength of titanium is found to be 65,000 psi when the grain size is 6.7 ×10−4 in. −5

and 82,000 psi when the grain size is 3.15×10 in. Determine (a) the constants in the Hall-Petch equation; and (b) the strength of the titanium when the grain size is reduced to 8.00 ×10 –6 in. .

Solution:

65,000 = σ 0 + K 82,000 = σ 0 + K

1 6.7 × 10 −4 1 3.15 × 10−5

= σ 0 + 38.633 K = σ 0 + 178.17 K

(a) By solving the two simultaneous equations:

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K = 121.8 psi in. σ o = 60,293 psi (b) σ = 60,293+121.8/ 8.00 ×10−6 =103,400 psi =103.4 ksi 4-51

A copper-zinc alloy has the properties shown in the table below: Grain diameter (mm) Strength (MPa) d–½ 0.015 170 MPa 8.165 0.025 158 MPa 6.325 0.035 151 MPa 5.345 0.050 145 MPa 4.472 Determine (a) the constants in the Hall-Petch equation; and (b) the grain size required to obtain a strength of 200 MPa.

Solution: The values of d–½ are included in the table; the graph shows the relationship. We can determine K and σo either from the graph or by using two of the data points.

170 = σ o + K(8.165) (a) 145 = σ o + K(4.472)

25 = 3.693K

K = 6.77 MPa/ mm σ o = 114.7 MPa (b) To obtain a strength of 200 MPa:

200 = 114.7 + 6.77/ d 85.3 = 6.77/ d d = 0.0063 mm

4-52

If there were 50 grains per in2 in a photograph of a metal taken at 100× magnification, calculate the ASTM grain size number (n).

Solution: N = 2 (n-1) 50 = 2(n-1)

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ln 50 = (n–1)*ln 2 3.912 = (n–1) (0.6931) 4.605 = 0.6931n n = 6.64 4-53

If the area of a photograph measured 7.812 in2 and 23 grains were documented, what would the ASTM grain size number (n) be?

Solution:

4-54

You have the choice to either purchase a copper alloy that has an ASTM grain size of 5 or ASTM grain size of 8. You can’t decide if there is much of a difference between these two. Determine how many grains/in2 would appear on a photograph taken at 100× for a metal given these two ASTM grain size choices. Does this seem like a significant difference?

Solution:

This is a very significant difference because there is are 8 times more grains per unit area with an ASTM grain size of 8 compared to an ASTM grain size of 5.

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4-55

The mill test report (MTR) on a structural steel you are considering using on an elevated pedestrian bridge over a busy roadway shows an ASTM grain size of 1.5. Would you accept this steel for this particular project? Explain.

Solution: An ASTM grain size of 1.5 means very large grains. For the sake of this application, very large grains will provide low strength and toughness which are not desirable in structural applications.

4-56

For an ASTM grain size number of 8, calculate the number of grains per square inch (a) at a magnification of 100 and (b) with no magnification.

Solution: N = 2n–1 N = 28–1 = 27 = 128 grains/in.2 No magnification means that the magnification is “1”: (27)(100/1)2 = 1.28 × 106 grains/in.2 4-57

Determine the ASTM grain size number if 20 grains/square inch are observed at a magnification of 400.

Solution: (20)(400/100)2 = 2n–1 log(320) = (n–1)log(2) 2.505 (n–1)(0.301) or n = 9.3 4-58

Determine the ASTM grain size number if 25 grains/square inch are observed at a magnification of 50.

Solution: 25(50/100)2 = 2n–1 log(6.25) = (n–1)log(2) 0.796 (n–1)(0.301) or n = 3.6 4-59

Determine the ASTM grain size number for the materials in Figure 4-14 and Figure 4-20.

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Solution: (a) There are about 26 grains in the photomicrograph, which has the dimensions 2.375 in. × 2 in. The magnification is 100, thus

26 = 2n −1 log(5.47) = 0.738 = (n − 1)log(2) n = 3.5 (2.375)(2) (b) There are about 59 grains in the photomicrograph, which has the dimensions 2.25 in. × 2 in. The magnification is 500, thus

59(500/100) 2 = 2n −1 log(328) = 2.516 = (n − 1) log(2) n = 9.4 (2.25)(2) There are about 28 grains in the photomicrograph, which has the dimensions 2 in. × 2.25 in. The magnification is 200, thus

28(200/100) 2 = 2n −1 log(24.889) = 1.396 = (n − 1)log(2) n = 5.6 (2.25)(2) 4-61

The angle θ of a tilt boundary is given by sin(θ/2) = b/2D [see Figure 4-15(a)]. Verify the correctness of this equation.

Solution: From the figure, we note that the grains are offset one Burgers vector, b, only for two spacings D. Then it is apparent that sin(θ/2) must be b divided by two D. 4-62

Calculate the angle θ of a small-angle grain boundary in FCC aluminum when the dislocations are 5000 Å apart. (See Figure 4-15 and the equation in Problem 4-61.)

Solution:

b = (½)( 2)(4.04958) = 2.8635 Å and D = 5000 Å 2.8635 sin(θ /2) = = 0.000286 (2)(5000) θ/2 = 0.0164 θ = 0.0328°

4-63

For BCC iron, calculate the average distance between dislocations in a small-angle grain boundary tilted 0.50°. [See Figure 4-15(a).]

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Solution:

sin(0.5/2) =

½( 3)(2.866) 2D

0.004364 = 1.241/D D = 284 Å

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Chapter 5: Atom and Ion Movements in Materials 5-10

Atoms are found to move from one lattice position to another at the rate of 5 × 105 jumps/s at 400°C when the activation energy for their movement is 30,000 cal/mol.

Calculate the jump rate at 750°C.

Solution:

Rate =

c exp[ − 30, 000/(1.987)/(673)] 5 × 105 = o = exp( − 22.434 + 14.759) x co exp[ − 30, 000/(1.987)/(1023)]

5 ×105 = exp( − 7.675) = 4.64 ×10−4 x 5 ×105 x= = 1.08 ×109 jumps/s −4 4.64 ×10 5-11

The number of vacancies in a material is related to temperature by an Arrhenius equation. If the fraction of lattice points containing vacancies is 8 × 10–5 at 600°C, determine the fraction of lattice points containing vacancies at 1000°C.

Solution: 8 × 10–5 = exp[–Q/(1.987)/(873)] Q = 16,364 cal/mol f = nv/n = exp[–16,364/(1.987)/(1273)] = 0.00155 5-12

The Arrhenius equation was originally developed for comparing rates of chemical reactions. Compare the rates of a chemical reaction at 20 and 100°C by calculating the ratio of the chemical reaction rates. Assume that the activation energy for liquids in which the chemical reaction is conducted is 10 kJ/mol and that the reaction is limited by diffusion.

Solution: An Arrhenius relationship gives the rate of a process or reaction as a function of temperature. In general, for an Arrhenius relationship,

 Q  Rate = c0 exp  −  RT  , where c 0 is a constant, Q is the activation energy, R is the universal gas constant, and T is the absolute temperature. Taking the ratio of the rates of two reactions:

    10 ×10 3 J/mol Q  c0 exp −  exp −  R ( 293 K )   8.314 J/ ( mol ⋅ K )( 293 K )  Rate 20˚C = = = 0.41     Rate100˚C Q 10 ×10 3 J/mol c0 exp −  xp −   R ( 373 K )   8.314 J/ ( mol ⋅ K )( 373 K )  5-14

(a) Compare interstitial and vacancy atomic mechanisms for diffusion and (b) cite two reasons why interstitial diffusion is normally more rapid than vacancy diffusion.

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Solution: (a) Interstitial diffusion is generally limited to the atom movements of C, H, O and N from interstitial site to interstitial site. This diffusion mechanism, with lower activation energies, is comparatively faster than vacancy diffusion. Vacancy diffusion is limited to the availability of lattice vacancies. Solute atoms can diffuse if a vacancy occupies a neighboring lattice position. Thus, vacancy diffusion requires much higher activation energies and occurs much more slowly. (b) More interstitial sites than vacancy sites exist in lattice structures and lower activation energies.

5-17

Compare the diffusion coefficients of carbon in BCC and FCC iron at the allotropic transformation temperature of 912°C and explain the reason for the difference in their values.

Solution: DBCC = 0.011 exp[–20,900/(1.987)/(1185)] = 1.51 × 10–6 cm2/s –7

2

DFCC = 0.23 exp [–32,900/(1.987)/(1185)] = 1.92 × 10 cm /s

The packing factor of the BCC lattice (0.68) is less than that of the FCC lattice (0,74); consequently, atoms are expected to be able to diffuse more rapidly in the BCC iron at the same temperature.

5-18

Compare the diffusion coefficients for hydrogen and nitrogen in FCC iron at 1000°C and explain the reason for the difference in their values.

Solution:

DH in FCC = 0.0063 exp [–10,300/(1.987)/(1273)] = 1.074 × 10–4 cm2/s DN in FCC = 0.0034 exp [–34,600/(1.987)/(1273)] = 3.898 × 10–9 cm2/s

Nitrogen atoms have a larger atomic radius (0.71 Å) compared with that of hydrogen atoms (0.46 Å); the smaller hydrogen ions are expected to diffuse more rapidly.

5-19

Activation energy is sometimes expressed as (eV/atom), (e.g., as shown in Figure 5-15, which illustrates the diffusion coefficients of ions in different oxides). Convert eV/atom to J/mole.

Solution: Using the appropriate conversion factors:

1 eV 1.6021766 ×10 −19 J 6.022141×10 23 atoms * * = 96, 485 J/mole atom 1 eV 1 mole 5-20

In order to produce a certain semiconductor, a process called doping is performed in which phosphorus is diffused into germanium. If D0 = 2.0 cm2/s and the activation energy is 57,500 cal/mol, determine the diffusivity of P in Ge at 800°C.

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Solution:

5-21

The activation energy for the diffusion of copper into silver is 193,000 J/mol. Calculate the diffusion coefficient at 1200 K given that D at 1000 K is 1.0 × 10-14 m2/s.

Solution:

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5-22

Determine the diffusion coefficient D for the diffusion of hydrogen into FCC iron at 800°C.

Solution:

5-23

The diffusion coefficient for Cr3+ in Cr2O3 is 6 × 10–15 cm2/s at 727°C and 1 × 10–9 cm2/s at 1400°C. Calculate (a) the activation energy; and (b) the constant D0.

Solution:

6 × 10−15 D0 exp[−Q/(1.987)/(1000)] = 1× 10−9 D0 exp[−Q/(1.987)/(1673)] 6 × 10–6 = exp[–Q(0.00050327 – 0.00030082)] = exp[–0.00020245 Q]

–12.0238 = −0.00020245 Q or Q = 59,390 cal/mol 1 × 10–9 = D0 exp[–59,390/(1.987)/(1673)]

D0 = 0.057 cm 2 /s 5-24

The diffusion coefficient for O2- in Cr2O3 is 4 × 10–15 cm2/s at 1150°C and 6 × 10–11 cm2/s at 1715°C. Calculate (a) the activation energy; and (b) the constant D0.

Solution:

4 × 10−15 D0 exp[−Q /(1.987)/(1423)] = 6 × 10−11 D0 exp[−Q /(1.987)/(1988)] 6.67 × 10–5 = exp[–0.0001005 Q]

–9.615 = –0.0001005 Q or Q = 95, 700 cal/mol 4 × 10–15 = D0 exp[–95,700/(1.987)/(1423)] = D0(2.02 × 10–15) D0 = 1.98 cm2/s 5-28

Determine the maximum allowable temperature that will produce a flux of less than 2000 H atoms/(cm2·s) through a BCC iron foil when the concentration gradient is –5 × 1016 atoms/(cm3·cm). (Note the negative sign for the flux.)

Solution: 2000 H atoms/(cm2·s) = –0.0012 exp[–3600/(1.987T)][–5 × 1016 atoms/(cm3·cm)] ln(3.33 × 10–11) = –3600/(1.987T) T = –3600/[(–24.12)(1.987)] = 75 K = –198°C

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5-29

Calculate the diffusion coefficients for the diffusion of hydrogen through BCC iron and FCC aluminum at room temperature (25°C). For the diffusion of H into Al, D0 = 0.11 cm 2 / s and Q = 9780 cal/mol . Based on your calculations, which material would be better suited as the material for a high-pressure hydrogen storage tank?

Solution:

5-30

How would the internal pressure of hydrogen in a storage tank affect the rate of diffusion? Explain.

Solution: The concentration of the number of H atoms at the inside surface of the metal increases as the pressure increases. Assuming that the external concentration remains the same, the concentration gradient must increase leading to an increased net flux or increased rate of diffusion.

5-31

A 1-mm-thick BCC iron foil is used to separate a region of high nitrogen gas concentration of 0.1 at% from a region of low nitrogen gas concentration at 650°C. If the flux of nitrogen through the foil is 1012 atoms/(cm2 · s), what is the nitrogen concentration in the low concentration region?

Solution: The flux J is given by

J = −D

∆c , ∆x

where D is the diffusion coefficient, c is concentration, and x is position. The term ∆c/∆x is the concentration gradient. The diffusion coefficient D is given by

 Q  D = D0 exp  − ,  RT  where D0 is the pre–exponential term, Q is the activation energy, R is the gas constant, and T is the absolute temperature. From Table 5–1, the activation energy is 18,300 cal/mol, and is 0.0047 cm2/s. Thus,

 Q  ∆c J = − D0 exp  −  .  RT  ∆x 101 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

The concentration is given in at% and must be converted to a per unit volume basis.

For BCC iron, there are two atoms per unit cell with an axial length of 2.866 × 10–8

18,300   clow − 0.1  2  ,   −8 3  1.987 × 923  0.1  (2.866 ×10 )

12 cm. Substituting, 10 = −0.0047 exp  −

and solving for the low concentration of atoms clow: clow = 0.099995 at%.

5-32

A 0.2-mm-thick wafer of silicon is treated so that a uniform concentration gradient of antimony is produced. One surface contains 1 Sb atom per 108 Si atoms and the other surface contains 500 Sb atoms per 108 Si atoms. The lattice parameter for Si is 5.4307 Å (Appendix A). Calculate the concentration gradient in (a) at % per cm; and (b) Sb atoms/(cm3 · cm).

Solution:

(1/108 − 500/108 ) ∆c /∆x = ×100% = −0.02495 at %/cm 0.02 cm a0 = 5.4307Å Vunit cell = 160.16 ×10−24 cm3 (8 Si atoms/unit cell)(1 Sb/108 Si) = 0.04995×1016 Sb atoms/cm 3 −24 3 160.16 ×10 cm /unit cell (8 Si atoms/unit cell)(500 Sb/108 Si) c2 = = 24.975×1016 Sb atoms/cm 3 −24 3 160.16 ×10 cm /unit cell c1 =

∆c /∆x = 5-33

(0.04995 − 24.975) ×1016 = −1.246 ×1019 Sb atoms/(cm3 ⋅ cm) 0.02 cm

When a Cu-Zn alloy solidifies, one portion of the structure contains 25 at% zinc and another portion 0.025 mm away contains 20 at% zinc. The lattice parameter for the FCC alloy is 3.63 × 10–8 cm. Determine the concentration gradient in (a) at% Zn per cm; (b) wt% Zn per cm; and (c) Zn atoms/(cm3 · cm).

Solution: Note that either positive or negative values for the answers are acceptable here. (a)

dc 20 at.% − 25 at.% = = −2000 at% Zn / cm dx 0.0025 cm

(b)  65.38 g Zn   65.38 g Zn  0.20 mol Zn  0.25 mol Zn     mol Zn   mol Zn  −  65.38 g Zn   63.54 g Cu   65.38 g Zn   63.54 g Cu  0.20 mol Zn   + 0.80 mol Cu   0.25 mol Zn   + 0.75 mol Cu   dc  mol Zn   mol Cu   mol Zn   mol Cu  = ×100 dx 0.0025 cm

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dc = −2031 wt% Zn/cm dx

     4 atoms  unit cell unit cell  − 0.25 4 atoms   0.20    unit cell  ( 3.63×10 −8 cm )3   unit cell  ( 3.63×10 −8 cm )3  dc 1.67 ×10 24 Zn atoms     = =− dx 0.0025 cm cm 3 ⋅ cm

(c) 5-34

A 0.001 in. BCC iron foil is used to separate a high hydrogen content gas from a low hydrogen content gas at 650°C. 5 × 108 H atoms/cm3 are in equilibrium on one side of the foil, and 2 × 103 H atoms/cm3 are in equilibrium on the other side. Determine (a) the concentration gradient of hydrogen; and (b) the flux of hydrogen through the foil.

Solution:

∆c/∆x =

2 ×103 − 5×108 = −1.969 ×1011 H atoms/(cm3 ⋅ cm) (0.001 in.) (2.54 cm/in.)

J = −D(∆c/∆x) = −0.0012 exp[ − 3600/(1.987)/(923)]( −1969 ×108 ) J = 3.3×107 H atoms/(cm 2 ⋅ s) 5-35

Compute the number of kilograms of hydrogen that pass per hour through a 5-mm-thick sheet of palladium having an area of 0.20m2 at 500°C. Assume a diffusion coefficient of 1.0 × 10-8 m2/s and that the concentrations at the high and low pressure sides of the plate are 2.4 and 0.6 kg of hydrogen per m3 of palladium. Assume steady state conditions.

Solution: The flux J is the number of atoms being transported per unit area per unit time.

Therefore, the flux J must be calculated and multiplied by the area of the sheet. The flux J is given by

J = −D

dc , dx

where D is the diffusion coefficient and dc /dx is the concentration gradient. The diffusion coefficient is given in the problem statement. The concentration gradient is calculated as

 (0.6 − 2.4 kg H )    m3 Pd dc   . = dx 0.005 m Therefore,

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 (0.6 − 2.4 kg H )    m3 Pd dc   3600 s  −8 2 2  JA = −DA = −(10 m /s)(0.2 m )   = 2.59 × 10 −3 kg/hr  hr  dx 0.005 m 5-36

A 1-mm-thick sheet of FCC iron is used to contain nitrogen in a heat exchanger at 1200°C. The concentration of N at one surface is 0.04 at% and the concentration at the second surface is 0.005 at%. Determine the flux of nitrogen through the foil in N atoms/(cm2·s).

Solution:

∆c /∆x =

(0.00005 − 0.0004) (4 atoms per cell)/(3.589 ×10−8cm)3 (1 mm)(0.1cm/mm)

= −3.03 ×1020 N atoms/(cm3 ⋅ cm) J = − D ( ∆c /∆x) = −0.0034 exp[ − 334,600/(1.987)/(1473)]( − 3.03 × 1020 ) = 7.57 × 1012 N atom/(cm 2 ⋅ s) 5-37

A 4-cm-diameter, 0.5-mm-thick spherical container made of BCC iron holds nitrogen at 700°C. The concentration at the inner surface is 0.05 at% and at the outer surface is 0.002 at%. Calculate the number of grams of nitrogen that are lost from the container per hour.

Solution:

[0.00002 − 0.0005](4 atoms/cell)/(2.866 ×10−8cm)3 ∆c /∆x = (0.5 mm)(0.1cm/mm) = −8.16 ×1020 N/(cm3 ⋅ cm) J = −0.0047 exp[ − 18,300/(1.987)/(973)][ − 8.16 × 1020 ] = 2.97 × 1014 N atoms(cm 2 ⋅ s) Asphere = 4π r 2 = 4π (2 cm) 2 = (50.27 cm 2 ) t = 3600 s/h 14

19

N atoms/h = (2.97 × 10 ) (50.27) (3600) = 5.37 × 10 N atoms/h

N loss = 5-38

(5.37 ×1019 atoms) (14.007 g/mol) = 1.25 ×10−3g/h 23 (6.022 ×10 atoms/mol)

A BCC iron structure is to be manufactured that will allow no more than 50 g of hydrogen to be lost per year through each square centimeter of the iron at 400°C. If the concentration of hydrogen at one surface is 0.05 H atom per unit cell and 0.001 H atom per unit cell at the second surface, determine the minimum thickness of the iron.

Solution: c1 = 0.05 H/(2.866 × 10–8 cm)3 = 212.4 × 1019 H atoms/cm3 c2 = 0.001 H/(2.866 × 10–8 cm)3 = 4.25 × 1019 H atoms/cm3

4.25 ×1019 − 212.4 ×1019 −2.08 ×1021 = ∆x ∆x 2 23 (50 g/cm /y) (6.02 ×10 atoms/mol) J= = 9.47 ×1017 H atoms/(cm 2 ⋅ s) 6 (1.00797 g/mol)(31.536 ×10 s/y)

∆c /∆x =

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J = 9.47 ×1017 H atoms/(cm 2 ·s) = (−2.08 ×1021 /∆x)(0.0012) exp[−3600/1.987/673] ∆x = 0.179 cm 5-39

Determine the maximum allowable temperature that will produce a flux of less than 2000 H atoms/(cm2·s) through a BCC iron foil when the concentration gradient is –5 × 1016 atoms/(cm3·cm). (Note the negative sign for the flux.)

Solution: 2000 H atoms/(cm2·s) = –0.0012 exp[–3600/(1.987T)][–5 × 1016 atoms/(cm3·cm)] ln(3.33 × 10–11) = –3600/(1.987T) T = –3600/[(–24.12)(1.987)] = 75 K = –198°C 5-44

Use the diffusion data in the table below for atoms in iron to answer the questions that follow. Assume metastable equilibrium conditions and trace amounts of C in Fe. The gas constant in SI units is 8.314 J/(mol · K).

Diffusion Couple Diffusion Mechanism Q (J/mol) D0 (m2/s) C in FCC Iron Interstitial 1.38 × 105 2.3 × 10–5 C in BCC Iron Interstitial 8.74 × 104 1.1 × 10–6 Fe in FCC Iron Vacancy 2.79 × 105 6.5 × 10–5 Fe in BCC Iron Vacancy 2.46 × 105 4.1 × 10–4 (a) Plot the diffusion coefficient as a function of inverse temperature (1/T) showing all four diffusion couples in the table. (b) Recall the temperatures for phase transitions in iron, and for each case, indicate on the graph the temperature range over which the diffusion data is valid. (c) Why is the activation energy for Fe diffusion higher than that for C diffusion in iron? (d) Why is the activation energy for diffusion higher in FCC iron when compared to BCC iron? (e) Does C diffuse faster in FCC Fe than in BCC Fe? Support your answer with a numerical calculation and state any assumptions made.

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Solution: (a)

The diffusion coefficient D is plotted as a function of 1/T according to

 Q  D = D0 exp  −  , where D0 is the pre–exponential term, Q is the activation  RT  energy, R is the gas constant, and T is the absolute temperature.

(b) The temperature at which BCC iron transforms to FCC iron is 912°C (1185 K), and the temperature at which FCC iron transforms to δ -iron is 1394°C (1667 K). The graph in (a) shows the lines for each diffusion couple over the temperature ranges for which the diffusion data is valid. (c) C diffuses by interstitial diffusion and Fe by vacancy diffusion. The C atom is smaller than the Fe atom, and there are more interstitial sites available than vacancies, making interstitial diffusion easier than vacancy diffusion. This is reflected in the lower activation energy for C diffusion in Fe when compared to Fe diffusion in Fe. (d) FCC Fe has a higher packing fraction than BCC Fe; hence, the activation energy is expected to be higher for diffusion in FCC Fe. (e) We can calculate the value of D at 912°C for interstitial diffusion of C in both FCC and BCC Fe according to

 Q  D = D0 exp  − .  RT  We know the activation energies Q and diffusion coefficients D0 from the table.

DC in FCC Fe = 2.3 × 10–5 exp[–1.38 × 105/(8.314 × 1183)] = 1.85 × 10–11 m2/s DC in BCC Fe = 1.1 × 10–6 exp[–8.74 × 104/(8.314 × 1183)] = 1.52 × 10–10 m2/s At the transition temperature from BCC to FCC iron (912°C), C diffuses faster in BCC Fe than in FCC Fe as evidenced by the higher diffusivity for C in BCC Fe compared to FCC Fe. 5-45

The plot below has three lines representing grain boundary, surface, and volume selfdiffusion in a metal. Match the lines labeled A, B, and C with the type of diffusion. Justify your answer by calculating the activation energy for diffusion for each case.

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Solution: The expression for the diffusion coefficient D is given by

 Q  D = D0 exp  − ,  RT 

Line A A B B C C

5-49

where D0 is the pre–exponential term, Q is the activation energy, R is the gas constant, and T is the absolute temperature. Taking the natural logarithm of both sides, ln(D) = ln(D0) – (Q/R)(1/T).

The activation energy can be determined from the slope of a ln(D) versus (1/T) graph.

Slope = –(Q/R) = [ln(D2) – ln(D1)] / [(1/T2) – (1/T1)].

From the graph, the values for the diffusion coefficient and (1000/T) can be determined. The table below shows these values for the end points of each line in the graph and calculates the slopes and activation energies for each line.

1000/T (K– ln(D) 1 Diffusivity (m2/s) ) (m2/s) 1/T (K) Q/R (K) Q (J/mol) 10–12 0.8 –27.631 0.0008 10–15 1.0 –34.539 0.0010 34,539 2.87 × 105 10–10 1.1 –23.026 0.0011 5 × 10–12 1.4 –26.022 0.0014 9,986 8.30 × 104 10–8 1.5 –18.421 0.0015 10–9 1.8 –20.723 0.0018 7,675 6.38 × 104 We know that surface diffusion has the least constraints and hence is expected to have the lowest activation energy. In this case, line C has the lowest activation energy and represents surface diffusion. Line A has the highest activation energy and hence represents volume diffusion, and line B with the intermediate value of activation energy represents grain boundary diffusion.

Pure zinc is to be diffused into the copper by dipping copper into molten zinc at 450°C. Calculate how long it would take to obtain 10 wt% zinc depth of 0.5 mm beneath the copper surface. Is this commercially feasible? What practical problems might arise if we raise the temperature to 1000°C? Use Table 5-4 for error function values as needed. At 450°C, the diffusion coefficient D = 6.3 × 10-13 cm2/s.

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Solution:

  100 −10 0.05 cm   = erf   −13 2 100 − 0  2 ( 6.3×10 cm /s) ( t )  7.334 ×108 s = 23.3 years This process would not be feasible at this temperature because it takes too long for commercial use. At 1000°C, diffusion would be faster, but copper melts at 1080°C resulting in the possibility of copper diffusing into the molten zinc.

5-50

Nitriding is a process in which nitrogen is allowed to diffuse into the surface of steel for the purpose of increasing the surface hardness of a component. It has been determined that a satisfactory nitrogen case depth is produced in BCC iron after 1 hour at 700°C. How much time is required to produce a satisfactory case depth if nitriding is carried out at 600°C?

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Solution:

5-51

Determine the carburizing time necessary to achieve a carbon concentration of 0.30 wt% at a position 4-mm into a steel alloy that initially contains 0.10 wt% carbon. The surface concentration is to be maintained at 0.90 wt% carbon and the treatment is to be conducted at 1100°C. Use the diffusion data for FCC iron and the error function values in Table 5-4 as needed.

Solution: For a constant surface concentration condition, we can use Equation 5-7:

cs − c x  x  = erf  , cs − c0  2 Dt  where cs is a constant concentration of the diffusing atoms at the surface of the material, c0 is the initial uniform concentration of the diffusing atoms in the material, and cx is the concentration of the diffusing atom at location x below the surface after time t. Equation 5-7 becomes

 0.4 cm  0.9 − 0.3 = erf   0.9 − 0.1  2 Dt  DC in FCC = 0.23 exp [–32,900/(1.987)/(1373)] = 1.332 × 10–6 cm2/s and solving for t,

t = 45, 400 s =12.6 hours

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5-52

One method of fabricating complex aerospace components is diffusion bonding. We wish to join a nickel sheet, intended to rapidly transfer lightning strike data away from critical locations, to an FCC iron (stainless steel) plate. We find that we can accomplish a good bond in 8 hours when the two metals are joined at 800°C. How long will it take to obtain an equally good bond if we decide to perform the bonding at 600°C? (Assume that the quality of the bond is determined by the rate of diffusion into the FCC iron).

Solution:

5-53

Determine the carburizing time necessary to achieve a carbon concentration of 0.45 wt% at a position 2-mm into a steel alloy that initially contains 0.20 wt% carbon. The surface concentration is to be maintained at 1.30 wt% at 1000°C. Use Table 5-4 for error function values as needed.

Solution: For a constant surface concentration condition, we can use Equation 5-7:

cs − c x  x  = erf   cs − c0  2 Dt  , where cs is a constant concentration of the diffusing atoms at the surface of the material, c0 is the initial uniform concentration of the diffusing atoms in the material, and cx is the concentration of the diffusing atom at location x below the surface after time t. Equation 5-7 becomes

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 0.2 cm  1.3− 0.45 = erf   1.3− 0.2  2 Dt  DC in FCC = 0.23 exp [–32,900/(1.987)/(1273)] = 5.164 × 10–7 cm2/s and solving for t,

t = 26, 568 s = 7.4 hours 5-54

Nitrogen from a gaseous phase is to be diffused into pure iron (BCC) at 700°C. If the surface concentration is maintained at 0.10 wt% N, what will the concentration 1 mm from the surface be after 10 hrs? Use Table 5-4 for error function values as needed.

Solution: For a constant surface concentration condition, we can use Equation 5-7:

cs − c x  x  = erf   cs − c0  2 Dt  , where cs is a constant concentration of the diffusing atoms at the surface of the material, c0 is the initial uniform concentration of the diffusing atoms in the material, and cx is the concentration of the diffusing atom at location x below the surface after time t. Equation 5-7 becomes

 0.1 cm  0.1− cx = erf   0.1− 0  2 Dt  DN in BCC = 0.0047 exp [–18,300/(1.987)/(973)] = 3.642 × 10–7 cm2/s and solving for

cx ,

cx = 0.054 5-55

The surface of 0.10 wt% carbon steel is to be strengthened by carburizing. The steel is placed in an atmosphere that provides a maximum of 1.2 wt% carbon at the surface of the steel at a high temperature. Carbon then diffuses 0.2 cm into the surface of the steel giving a carbon concentration of 0.45 wt% carbon at that depth. How long will the process of carburizing take if the diffusion coefficient is 2 × 10-7 cm2/s? Use Table 5-4 for error function values as needed.

Solution: For a constant surface concentration condition, we can use Equation 5-7:

cs − c x  x  = erf   cs − c0  2 Dt  , where cs is a constant concentration of the diffusing atoms at the surface of the material, c0 is the initial uniform concentration of the diffusing atoms in the material, and cx is the concentration of the diffusing atom at location x below the surface after time t. Equation 5-7 becomes

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  1.2 − 0.45 = erf  1.2 − 0.1 2 

   2 ×10−7 cm 2 / s t   0.2 cm

(

)

DC in FCC = 0.23 exp [–32,900/(1.987)/(1373)] = 1.332 × 10–6 cm2/s and solving for t,

t = 100,361 s = 27.9 hours 5-56

Transistors are made by doping single crystal silicon with different types of impurities to generate n- and p-type regions. Phosphorus and boron are typical n- and p-type dopant species, respectively. Assuming that a thermal treatment at 1100°C for 1 h is used to cause diffusion of the dopants, calculate the constant surface concentration of P and B needed to achieve a concentration of 1018 atoms/cm3 at a depth of 0.1 mm from the surface for both n- and p-type regions. The diffusion coefficients of P and B in single crystal silicon at 1100°C are 6.5 × 10–13 cm2/s and 6.1 × 10–13 cm2/s, respectively.

Solution:

For a constant surface concentration condition, we can use Equation 5-7:

cs − c x  x  = erf  , cs − c0  2 Dt  where cs is a constant concentration of the diffusing atoms at the surface of the material, c0 is the initial uniform concentration of the diffusing atoms in the material, and cx is the concentration of the diffusing atom at location x below the surface after time t.

For P,

  cs − 1018 cm −3 0.1× 10−4 cm = erf  . −13 2 cs − 0  2 6.5 × 10 cm /s × 3600 s  Solving for cs:

1−

1018 = 0.1162 cs

cs = 1.1 × 1018 cm–3 For B,

  cs − 1018 cm −3 0.1× 10−4 cm = erf  . −13 2 cs − 0 2 6.1 × 10 cm /s × 3600 s   Solving for cs:

1−

1018 = 0.1199 cs

cs = 1.1 × 1018 cm–3

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5-57

Consider a 2-mm-thick silicon wafer to be doped using antimony. Assume that the dopant source (gas mixture of antimony chloride and other gases) provides a constant concentration of 1022 atoms/m3. We need a dopant profile such that the concentration of Sb at a depth of 1 micrometer is 5 × 1021 atoms/m3. What is the required time for the diffusion heat treatment? Assume that the silicon wafer initially contains no impurities or dopants. Assume that the activation energy for diffusion of Sb into silicon is 380 kJ/mole and D0 for Sb diffusion into Si is 1.3 × 10-3 m2/s. Assume T = 1250°C.

Solution: For a constant surface concentration condition, we can use Equation 5-7:

cs − c x  x  = erf  , cs − c0  2 Dt  where cs is a constant concentration of the diffusing atoms at the surface of the material, c0 is the initial uniform concentration of the diffusing atoms in the material, and cx is the concentration of the diffusing atom at location x below the surface after time t. The diffusion is calculated as

  380 ×103 J/mol  = 1.20 ×10 −16 m 2 /s D = 1.3×10 −3 m 2 /s * exp−    8.314 J/ ( mol ⋅ K )(1523 K )  Thus,

  10 22 − 5×10 21 1×10 −6 m   = erf 22   −16 2 10 − 0  2 (1.20 ×10 m /s) ( t )  and solving for t,

t = 9130 s = 2.5 hours 5-58

Consider doping of silicon with gallium. Assume that the diffusion coefficient of gallium in Si at 1100°C is 7 × 10-13 cm2/s. Calculate the concentration of Ga at a depth of 2.0 micrometer if the surface concentration of Ga is 1023 atoms/cm3. The diffusion times are 1, 2, and 3 hours.

Solution: For t = 1 hour,

  10 23 − cx 2 ×10 −4 cm   = erf 23   −13 2 10 − 0  2 ( 7 ×10 cm /s) (1* 3600 s)     2 ×10 −4 cm 23    = 4.84 ×10 20 cx = 10 * 1− erf    2 7 ×10 −13 cm 2 /s (1* 3600 s)    

(

)

For t = 2 hours,

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    2 ×10−4 cm   = 4.64 ×10 21 cx = 10 * 1− erf   −13 2   2 ( 7 ×10 cm /s) ( 2 * 3600 s)   23

For t = 3 hours,

   −4  2 ×10 cm  =1.04 ×10 22 cx = 10 23 * 1− erf    −13 2   2 ( 7 ×10 cm /s) ( 3* 3600 s)   5-59

Compare the rate at which oxygen ions diffuse in alumina (Al2O3) with the rate at which aluminum ions diffuse in Al2O3 at 1500°C. Explain the difference.

Solution:

5-60

DO–2 = 1900 exp[–152,000/(1.987)/(1773)] = 3.47 × 10–16 cm2/s DAl+3 = 28 exp[–114,000/(1.987)/(1773)] = 2.48 × 10–13 cm2/s The ionic radius of the oxygen ion is 1.32 Å, compared with the aluminum ionic radius of 0.51 Å; consequently it is much easier for the smaller aluminum ion to diffuse in the ceramic.

A carburizing process is carried out on a 0.10% C steel by introducing 1.0% C at the surface at 980°C, where the iron is FCC. Calculate the carbon content at 0.01 cm, 0.05 cm, and 0.10 cm beneath the surface after 1 h.

Solution: D = 0.23 exp[–32,900/(1.987)/(1253)] = 42 × 10–8 cm2/s

1 − cx = erf [ x /2 (42 × 10 −8 )(3600)] = erf [ x /0.0778] 1 − 0.1 (1 – cx ) x = 0.10 : erf [0.10/0.0778] = erf (0.285) = = 0.144 cx = 0.87% C 0.9 x = 0.05 : erf [0.05/0.0778] = erf (0.643) =

(1 – cx ) = 0.636 cx = 0.43% C 0.9

x = 0.01: erf [0.01/0.0778] = erf (1.1285) =

(1 – cx ) = 0.914 cx = 0.16% C 0.9

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5-61

Iron containing 0.05% C is heated to 912°C in an atmosphere that produces 1.20% C at the surface and is held for 24 h. Calculate the carbon content at 0.05 cm beneath the surface if (a) the iron is BCC; and (b) the iron is FCC. Explain the difference.

Solution: t = (24 h)(3600 s/h) = 86,400 s DBCC = 0.011 exp[–20,900/(1.987)/(1185)] = 1.54 × 10–6 cm2/s DFCC = 0.23 exp[–32,900/(1.987)/(1185)] = 1.97 × 10–7 cm2/s

1.2 − cx = erf [0.05/(2 (1.54 ×10−6 )(86, 400))] = erf [0.0685] = 0.077 1.2 − 0.05 cx = 1.11% C 1.2 − cx FCC : = erf [0.05/(2 (1.97 ×10−7 )(86, 400))] = erf [0.192] = 0.2139 1.2 − 0.05 cx = 0.95% C

BCC :

Faster diffusion occurs in the more loosely packed BCC structure, leading to the higher carbon content at point “x”.

5-62

What temperature is required to obtain 0.50% C at a distance of 0.5 mm beneath the surface of a 0.20% C steel in 2 h, when 1.10% C is present at the surface? Assume that the iron is FCC.

Solution:

1.1 − 0.5 = 0.667 = erf [0.05/(2 Dt )] 1.1 − 0.2 0.05/(2 Dt ) = 0.685 or Dt = 0.0365 or t = (2 h)(3600 s/h) = 7200 s

Dt = 0.00133

D = 0.00133/7200 = 1.85 × 10 −7 = 0.23 exp[ − 32,900/(1.987T )] exp( − 16,558/T) = 8.043 × 10 −7 T = 1180K = 907°C 5-63

A 0.15% C steel is to be carburized at 1100o C, giving 0.35% C at a distance of 1 mm beneath the surface. If the surface composition is maintained at 0.90% C, what time is required?

Solution:

0.9 − 0.35 = 0.733 = erf[0.1/(2 Dt )] 0.9 − 0.15 0.1/(2 Dt ) = 0.786 or Dt = 0.0636 or

Dt = 0.00405 –6

D = 0.23 exp[–32,900/(1.987)/(1373)] = 1.332 × 10 cm2/s t = 0.00405/1.332 × 10–6 = 3040 s = 51 min 5-64

A 0.02% C steel is to be carburized at 1200°C in 4 h, with the carbon content 0.6 mm beneath the surface reaching 0.45% C. Calculate the carbon content required at the surface of the steel.

Solution:

cs − 0.45 = erf [0.06/(2 Dt )] cs − 0.02 115

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D = 0.23 exp[–32,900/(1.987)/(1473)] = 3.019 × 10–6 cm2/s t = (4 h)(3600) = 14,400 s

Dt = (3.019 × 10−6 )(14, 400) = 0.2085 erf[0.06/((2)(0.2085))] = erf(0.144) = 0.161

cs − 0.45 = 0.161 or cs = 0.53% C cs − 0.02 5-65

A 1.2% C tool steel held at 1150°C is exposed to oxygen for 48 h. The carbon content at the steel surface is zero. To what depth will the steel be decarburized to less than 0.20% C?

Solution:

0 − 0.2 = 0.1667 ∴ x /(2 Dt ) = 0.149 0 − 1.2 D = 0.23 exp[–32,900/(1.987)/(1423)] = 2.034 × 10–6 cm2/s t = (48 h)(3600 s/h) = 17.28 × 104 s

Dt = 0.5929 Then from above, x = (0.149)(2)(0.5929) = 0.177 cm 5-66

A 0.80% C steel must operate at 950°C in an oxidizing environment for which the carbon content at the steel surface is zero. Only the outermost 0.02 cm of the steel part can fall below 0.75% C. What is the maximum time that the steel part can operate?

Solution:

0 − 0.75 = 0.9375 = erf [x /(2 Dt )] ∴ x /(2 Dt ) = 1.317 0 − 0.8 0.02/(2 Dt ) = 1.317 or Dt = 0.00759 or Dt = 5.76 ×10−5 D = 0.23 exp[–32,900/(1.987)/(1223)] = 3.03 × 10–7 cm2/s t = 5.76 × 10–5 / (3.03 × 10–7) = 190 s

5-67

A steel with the BCC crystal structure containing 0.001% N is nitrided at 550°C for 5 h. If the nitrogen content at the steel surface is 0.08%, determine the nitrogen content at 0.25 mm from the surface.

Solution:

0.08 − cx = erf [0.025/(2 Dt )] t = (5 h)(3600 s/h) = 1.8 × 10 4 s 0.08 − 0.001 D = 0.0047 exp[–18,300/(1.987)/(823)]

= 6.488 ×10 –8 cm 2 /s Dt = 0.0342 erf[0.025/((2)(0.0342))] = erf(0.3655) = 0.394

0.08 − cx = 0.394 or cx = 0.049% N 0.079 5-68

What time is required to nitride a 0.002% N steel to obtain 0.12% N at a distance of 0.002 in. beneath the surface at 625°C? The nitrogen content at the surface is 0.15%.

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Solution:

0.15 − 0.12 = 0.2027 = erf [x /(2 Dt )] ∴ x /(2 Dt ) = 0.1816 0.15 − 0.002 D = 0.0047 exp[–18,300/(1.987)/(898)] = 1.652 × 10–7 cm2/s x = 0.002 in. = 0.00508 cm

0.00508 2 (1.652 ×10−7 )t

= 0.1816

Dt = 1.956 × 10–4 or t = 1.956 × 10–4/(1.652 × 10–7) = 1184 s 5-69

We can successfully perform a carburizing heat treatment at 1200°C in 1 h. In an effort to reduce the cost of the brick lining in our furnace, we propose to reduce the carburizing temperature to 950°C. What time will be required to give us a similar carburizing treatment?

Solution: D1200 = 0.23 exp[–32,900/(1.987)/(1473)] = 3.019 × 10–6 cm2/s D950 = 0.23 exp[–32,900/(1.987)/(1223)] = 3.034 × 10–7 cm2/s t1200 = 1 h

t950 = D1200t1200 /D950 = 5-70

(3.019 ×10−6 )(1) = 9.95 h 3.034 ×10−7

During freezing of a Cu-Zn alloy, we find that the composition is nonuniform. By heating the alloy to 600°C for 3 hours, diffusion of zinc helps to make the composition more uniform. What temperature would be required if we wished to perform this homogenization treatment in 30 minutes?

Solution: D600 = 0.78 exp[–43,900/(1.987)/(873)] = 7.9636 × 10–12 t600 = 3 h tx = 0.5 h Dx = D600 t600/tx = (7.9636 × 10–12)(3)/0.5 Dx = 4.778 × 10–11 = 0.78 exp[–43,900/(1.987T)] ln (6.1258 × 10–11) = –23.516 = –43,900/(1.987 T) T = 940 K = 667°C

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5-71

To control junction depth in transistors, precise quantities of impurities are introduced at relatively shallow depths by ion implantation and diffused into the silicon substrate in a subsequent thermal treatment. This can be approximated as a finite source diffusion problem. Applying the appropriate boundary conditions, the solution to Fick’s second law under these conditions is

c ( x, t ) =

 − x2  Q exp   π Dt  4 Dt 

where Q is the initial surface concentration with units of atoms/cm2. Assume that we implant 1014 atoms/cm2 of phosphorus at the surface of a silicon wafer with a background boron concentration of 1016 atoms/cm3 and this wafer is subsequently annealed at 1100°C. The diffusion coefficient (D) of phosphorus in silicon at 1100°C is 6.5 × 10–13 cm2/s. (a) Plot a graph of the concentration c (atoms/cm3) versus x (cm) for anneal times of 5 minutes, 10 minutes, and 15 minutes. (b) What is the anneal time required for the phosphorus concentration to equal the boron concentration at a depth of 1μm?

Solution: (a) Using Q = 1014 atoms/cm2 and D = 6.5 × 10–13 cm2/s in the equation above, the concentration c for t = 5 min, 10 min, and 15 min can be calculated using a series of values for “x”. A graph of c versus x can be made with most standard spreadsheet software.

(b) From the graph, the time for the P concentration to match the background B concentration is slightly more than 10 minutes. Software packages are available that can solve this equation, but we can also find the solution by substituting t values close to 10 minutes. Rearranging the equation above and taking the natural logarithm of each side, ln(c) = ln[Q/√(πDt)] – x2/(4Dt) c = 1016 atoms/cm3, and the values of Q and D are known.

For t = 650 s, x = 0.974 μm, which is too small.

For t = 700 s, x = 1.008 μm, which is too large.

For t = 689 s, x = 1 μm.

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5-72

Arrange the following materials in increasing order of self-diffusion coefficient: Ar gas, water, single crystal aluminum, and liquid aluminum at 700°C.

Solution: In increasing order from the lowest to high self-diffusion coefficient, the answer is single crystal aluminum, water, liquid aluminum at 700°C, and argon gas. We expect higher diffusion coefficients in gases compared to solids. Since the diffusion coefficient depends exponentially on temperature, we expect the self-diffusion coefficient in liquid aluminum at 700°C to be higher than the self-diffusion coefficient of water at (presumably) room temperature.

5-77

A ceramic part made of MgO is sintered successfully at 1700°C in 90 minutes. To minimize thermal stresses during the process, we plan to reduce the temperature to 1500°C. Which of the following will limit the rate at which sintering can be performed: diffusion of magnesium ions or diffusion of oxygen ions? What time will be required at the lower temperature?

Solution:

Diffusion of oxygen is the slower of the two, due to the larger ionic radius of the oxygen.

D1700 = 0.000043 exp[–82,100/(1.987)/(1973)] = 3.455 × 10–14 cm2/s D1500 = 0.000043 exp[–82,100/(1.987)/(1773)] = 3.255 × 10–15 cm2/s

t1500 = D1700t1700 /D1500 = 5–78

(3.455 ×1014 )(90) = 955 min = 15.9 h 3.255 ×10−15

A Cu-Zn alloy has an initial grain diameter of 0.01 mm. The alloy is then heated to various temperatures, permitting grain growth to occur. The times required for the grains to grow to a diameter of 0.30 mm are shown below.

Temperature (°C) Time (min) 500 80,000 600 3,000 700 120 800 10 850 3 Determine the activation energy for grain growth. Does this correlate with the diffusion of zinc in copper? (Hint: Note that rate is the reciprocal of time.)

Temperature 1/T Time Rate (°C) (K) (K–1) (min) (min–1) 500 773 0.00129 80,000 1.25 × 10–5 600 873 0.00115 3,000 3.33 × 10–4 700 973 0.001028 120 8.33 × 10–3 800 1073 0.000932 10 0.100 850 1123 0.000890 3 0.333 A plot of the natural logarithm of the rate versus the inverse of the absolute temperature is shown below with a linear fit. The slope is equal to –Q/R. For a slope of 25,409 K, the activation energy Q is Q = 25,409 K × 1.987 cal/(mol–K) = 50,488 cal/mol.

Solution:

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5-80

A sheet of gold is diffusion-bonded to a sheet of silver in 1 h at 700°C. At 500°C, 440 h are required to obtain the same degree of bonding, and at 300°C, bonding requires 1530 years. What is the activation energy for the diffusion bonding process? Does it appear that diffusion of gold or diffusion of silver controls the bonding rate? (Hint: Note that rate is the reciprocal of time.)

Solution:

Temperature (°C) (K) 700 973 500 773 300 573

0.278 × 10−3 0.207 × 10

−10

=

1/T (K–1) 0.001007 0.001294 0.001745

Time (s) 3600 1.584 × 106 4.825 × 1010

Rate (s–1) 0.278 × 10–3 0.631 × 10–6 0.207 × 10–10

exp[−Q /(1.987)/(973)] exp[−0.0005172Q ] = exp[−Q /(1.987) / (573)] exp[−0.0008783Q ]

ln(1.343 × 107) = 16.413 = 0.0003611 Q Q = 45,450 cal/mol.

The activation energy for the diffusion of gold in silver is 45,500 cal/mole; thus, the diffusion of gold appears to control the bonding rate.

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Chapter 6: Mechanical Properties: Part One 6-14

Calculate the maximum force that a 0.2-in. diameter rod of Al2O3, having a yield strength of 35,000 psi, can withstand with no plastic deformation. Express your answer in lbs and N.

Solution: F = SA0 = (35,000 psi)(π/4)(0.2 in.)2 = 1100 lb F = (1100 lb)(4.448 N/lb) = 4891 N 6-15

A cylindrical specimen of a titanium alloy having an elastic modulus of 107 GPa and an original diameter of 3.8 mm should experience only elastic deformation when a tensile load of 2000 N is applied. Compute the maximum length of the specimen for a maximum allowable elongation of 0.42 mm.

Solution:

S=

 ∆l  F = Ee = E  A0  l0 

π  2 107 ×10 3 MPa   (3.8 mm ) ( 0.42 mm ) E A ∆l 4 l0 = = = 255 mm F 2000 N 6-16

A material with a diameter of 8 mm is pulled with a force of 2500 N. Calculate the stress.

If the original length is 50 mm, what is the strain under load if it is elongated to 50.15 mm?

Solution:

F 2500 N = = 49.7 MPa A0  π  8 mm 2  ( ) 4 ∆l 50.15 mm − 50.0 mm e= = = 0.003 l0 50.0 mm

S=

6-17

An aluminum plate 0.5 cm thick is to withstand a force of 50,000 N with no permanent deformation. If the aluminum has a yield strength of 125 MPa, what is the minimum width of the plate?

Solution:

The area is A0 = F/S = 50,000 N / (125 N/mm2) = 400 mm2

The minimum width is w = A0 /t = (400 mm 2 )(0.1 cm / mm) 2 /(0.5 cm) = 8 cm 6-18

A 850-lb force is applied to a 0.15-in. diameter nickel wire having a yield strength of 45,000 psi and a tensile strength of 55,000 psi. Determine (a) whether the wire will plastically deform; and (b) whether the wire will experience necking.

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(a) First determine the stress acting on the wire: S = F/A0 = 850 lb / [(π/4)(0.15 in.)2] = 48,100 psi Because S is greater than the yield strength of 45,000 psi, the wire will plastically deform. (b) Because S is less than the tensile strength of 55,000 psi, no necking will occur.

Solution:

6-19

A force of 100,000 N is applied to an iron bar with a cross-sectional area of 10 mm × 20 mm and having a yield strength of 400 MPa and a tensile strength of 480 MPa. Determine whether the bar will plastically deform and whether the bar will experience necking.

Solution:

6-21

First determine the stress acting on the wire: S = F/A0 = 100,000 N / [(10 mm)(20 mm)] = 500 N/mm2 = 500 MPa Because S is greater than the yield strength of 400 MPa, the wire will plastically deform.

Because S is greater than the tensile strength of 480 MPa, the wire will also neck.

At what point does necking typically initiate during a tensile test?

Solution: Necking typically occurs when the maximum load or ultimate tensile strength is reached. After that point, the engineering stress required to continue deformation on the sample decreases due to the reduction in cross-sectional area present to support the load in the necked region. Calculation of the true stress in this region will indicate that the strength of the metal will actually continue to rise until fracture occurs.

6-23

Derive the expression ε = ln(1 + e), where ε is the true strain and e is the engineering strain. Note that this expression is not valid after the onset of necking.

Solution: True strain is given by

l  l0 

ε = ln   , where is l the instantaneous length and l0 is the initial length. The instantaneous length can be written l = l0 + Δl, where Δl is the change in length. Substituting,

l  l0 

 l + ∆l   ∆l   = ln 1 +  . l0   l0  

ε = ln   = ln  0

The engineering strain e is given by

e=

∆l , l0

such that

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

ε = ln 1 + 

6-25

∆l   = ln(1 + e) . l0 

Two rods in the form of single crystals for pure tin show wildly different yield strengths at 0.2% offset yield strength. What would be the cause of this phenomenon?

Solution: Since the rods are in the form of single crystals, the yield strength will be highly dependent on the crystallographic orientation. The critical resolved shear stress (τCRSS) should be used to determine the strength of single crystals rather than yield strength.

6-26

Develop an expression for the engineering strain along the loading axis of a bar subjected to tension prior to any necking that may occur. Your expression should be a function only of the bar’s initial and instantaneous diameters. Assume that the volume is constant.

Solution: The engineering strain e is defined as

e=

l − l0 l = − 1, l0 l0

where l0 is the initial length of the bar and l is the instantaneous length of the bar, respectively. Assuming a constant volume during deformation, then

A0 l0 = Al , where A is the instantaneous length of the bar. Rearranging,

π d02

l A0 d02 4 = = = , l0 A πd2 d2 4 such that

e= 6-27

l d2 − 1 = 02 − 1. l0 d

The cable of a hoist has a cross-section of 80 mm2. The hoist is used to lift a crate weighing 500 kg. The free length of the cable is 30 m. Assume all deformation is elastic. (a) What is the stress on the cable? (b) How much will the cable extend if it is made from steel (E = 200 GPa)? (c) How much will it extend if the cable is made from polypropylene (E = 1.2 GPa)?

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Solution: (a) S =

S=

F ∆l = Ee= E l0 A0

500 kg ( 9.81 m/s2 )

= 61.3 MPa 80 mm 2 S l 61.3125 MPa ( 30 m ) = 9.20 mm (b) ∆l = 0 = E 200 ×10 3 MPa S l 61.3125 MPa ( 30 m ) = 1.53 m (c) ∆l = 0 = E 1.2 ×10 3 MPa 6-28

A force of 20,000 N will cause a 1 cm × 1 cm bar of magnesium to stretch from 10 cm to 10.045 cm. Calculate the modulus of elasticity, both in GPa and psi.

Solution: The strain e is e = (10.045 cm – 10 cm)/10 cm = 0.0045 cm/cm

The stress S is S = 20, 000 N /[(10 mm)(10 mm)] = 200 N / mm 2 = 200 MPa E = S/e = 200 MPa / (0.0045 cm/cm) = 44,444 MPa = 44.4 GPa E = (44,444 MPa)(145 psi/MPa) = 6.44 × 106 psi 6-29

A polymer bar’s dimensions are 1 in. × 2 in. × 15 in. The polymer has a modulus of elasticity of 600,000 psi. What force is required to stretch the bar elastically from 15 in. to 15.25 in.?

Solution: The strain e is e = (15.25 in. – 15 in.) / (15 in.) = 0.01667 in./in.

The stress S is S = Ee = (600,000 psi)(0.01667 in./in.) = 10,000 psi The force is then F = SA0 = (10,000 psi)(1 in.)(2 in.) = 20,000 lb 6-30

A steel cable 1.25 in. in diameter and 50 ft long is to lift a 20 ton load without permanently deforming. What is the length of the cable during lifting? The modulus of elasticity of the steel is 30 × 106 psi.

Solution:

The stress is S = F /A0 =

(20 ton)(2000 lb/ton) = 32,595 psi (π /4)(1.25 in.)2

The strain is e = S/E = 32,595 psi / (30 × 106 psi) = 0.0010865 in./in. e = (ℓf – 50 ft) / 50 ft = 0.0010865 ft/ft ℓf = 50.0543 ft 6-31 (a) Carbon nanotubes are one of the stiffest and strongest materials known to scientists and engineers. Carbon nanotubes have an elastic modulus of 1.1 TPa (1 TPa = 1012 Pa). If a carbon nanotube has a diameter of 15 nm, determine the engineering stress sustained by the nanotube when subjected to a tensile load of 4 μN (1 μN = 10–6 N) along the length of the tube. Assume that the entire cross-sectional area of the nanotube is load bearing.

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(b) Assume that the carbon nanotube is only deformed elastically (not plastically) under the load of 4 μN. The carbon nanotube has a length of 10 mm (1 μm = 10–6 m). What is the tensile elongation (displacement) of the carbon nanotube in nanometers (1 nm = 10–9 m)?

Solution:

(a) The engineering stress is given by

F , A0

S=

where S is the engineering stress, F is the applied load, and A0 is the initial cross–sectional area of the nanotube. A0 is equal to

A0 =

π

D2 ,

4

where D is the nanotube diameter. Substituting into the expression for S,

S=

F

π 4

D2

=

4 × 10 −6 N

π 4

= 22.64 × 109 Pa = 22.64 GPa

(15 × 10−9 m) 2

(b) According to Hooke’s Law, S = Ee where E is the elastic modulus and the engineering strain e is given by

e=

∆l , l0

where Δl is the elongation and l0 is the initial nanotube length.

Rearranging,

Sl0 (22.64 × 109 Pa) (10 × 10−6 m) ∆l = = = 2.06 × 10−7 m = 206 nm .

12 E (1.1 × 10 Pa) 6-32

A force (load) of 4000 lbs is applied to a cylindrical metal bar that is 0.505 in. in diameter. This load produces a reduction in diameter of 0.0003 inches. The deformation is purely elastic. What is the Poisson’s ratio of this material if the elastic modulus is 15 × 106 psi?

Solution:

F = Eeaxial A e ∆d / d0 eaxial = − transverse = −

ν

ν

 ∆d / d0  F = −E    ν  A

ν =−

EA  ∆d    F  d0  125

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π  2 15×10 6 psi   ( 0.505 in.)  −0.0003 in.  4 ν =−   = 0.45  0.505 in.  4000 lbs Note that this value is unrealistically high for a metal.

6-33

A cylindrical bar of steel, 10 mm in diameter, is to be deformed elastically by application of a force along the bar axis (axial loading). Determine the force that will produce an elastic reduction of 0.003 mm in the diameter. The Poisson’s ratio is 0.30 and the modulus of elasticity is 207 GPa for carbon steel.

Solution:

F = Eeaxial A e ∆d / d0 eaxial = − transverse = −

ν

ν

 ∆d / d0  π  2  −0.003 mm /10 mm  3 F = −EA   = −207 ×10 MPa   (10 mm )    ν  4   0.3

F =16, 258 N 6-34

(a) A 0.4-in.-diameter, 12-in.-long titanium bar has a yield strength of 50,000 psi, a modulus of elasticity of 16 × 106 psi, and Poisson’s ratio of 0.30. Determine the length and diameter of the bar when a 500-lb load is applied. (b) When a tensile load is applied to a 1.5-cm-diameter copper bar, the diameter is reduced to 1.498-cm diameter.

Determine the applied load, using the data in Table 6–3.

Solution:

(a) The stress is S = F/A0 = 500 lb [(π /4) (0.4 in.)2] = 3,979 psi The applied stress is much less than the yield strength; therefore Hooke’s law can be used.

The strain is e = S/E = 3,979 psi/ (16 × 106 psi) = 0.00024868 in./in.

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ℓ f − ℓ0 ℓ0

=

ℓ f − 12 in.

12 in.

= 0.00024868 in./in.

ℓf = 12.00298 in.

From Poisson’s ratio, v = –elat/elong = 0.3 elat = – (0.3) (0.0002468) = – 0.0000746 in./in.

d f − d0 d0

=

d f − 0.4 in.

0.4

= −0.0000746 in./in.

df = 0.39997 in. (b) From Table 6–3, v = elat/elong = 0.36

elat =

1.498 − 1.5 = − 0.001333 1.5

elong = –elat/v = –(–0.001333)/0.36 = 0.0037 in./in.

S = Ee = (124.8 GPa) (1000 MPa/GPa) (0.0037 in./in.) = 462 MPa F = SA0 = (462 MPa) (π/4) (15 mm)2 = 81,640 N 6-35

A standard 0.505-in.-diameter tensile bar was machined to a 2.00-in.-gage length from a copper-nickel alloy and the following data were collected: Load (lbs) 0 1000 2000 3000 4000 6000 8000 10000 11000 9000

Gage Length (in.)

Engineering Stress (ksi) Engineering Strain 2.00000 0 0 2.00045 5.0 0.000225 2.00091 10.0 0.000455 2.00136 15.0 0.000678 2.0020 20.0 0.001 2.020 30.0 0.010 2.052 39.9 0.026 2.112 49.9 0.056 2.280 (max load) 54.9 0.140 2.750 (fracture) 44.9 0.375

After fracture, the gage length was 2.75 in. and the diameter was 0.365 in. Plot the engineering stress strain curve and calculate (a) the 0.2% offset yield strength; (b) the tensile strength; (c) the modulus of elasticity; (d) the % elongation; (e) the % reduction in area; (f) the engineering stress at fracture; (g) the true stress at necking; and (h) the modulus of resilience.

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Solution: 60 S

UTS

= 54.9 ksi S

Engineering Stress (ksi)

50

fracture

= 44.9 ksi

40

30

20 e 10

0 0.00

e

0.05

0.10

UTS

fracture

= 0.375

= 0.140

0.15

0.20

0.25

0.30

0.35

0.40

Engineering Strain 40

Engineering Stress (ksi)

35 30

(0.00068, 15 ksi)

25

S = 22 ksi y

20 15 10

E = 22 ksi

5 0 0.000

0.002

0.004

0.006

0.008

0.010

Engineering Strain

By inspection of the plots: (a) the 0.2% offset yield strength is 22 ksi; (b) the tensile strength is 54.9 ksi; and (c) the modulus of elasticity is 22 Mpsi from a linear fit to the data points from the elastic loading.

Also (d) % elongation =

2.75 − 2.00 ×100 = 37.5% ; 2.00

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π (e) percent reduction in area = 4

π

( 0.505 in.) − ( 0.365 in.)

2

π 4

4

( 0.505 in.)

2

×100 = 47.8%

2

By inspection of the plot, (f) the engineering stress at fracture is 44.9 ksi.

The true stress σ at necking can be written as

σ = SUTS (1+ e ) .

UTS

This is the last data point at which this expression for true stress, i.e., σ = S(1+ e) , is valid. Substituting,

σ = SUTS (1+ e ) = 54.9 ksi [1+ 0.140 ] = 62.6 ksi UTS

Thus (g) the true stress at necking is 62.6 ksi.

The modulus of resilience is the area under the elastic portion of a stress-strain curve.

The area under the elastic portion of the curve can be calculated as the area of a triangle with one data point being the stress and strain at the proportional limit.

modulus of resilience =

1 (15 ksi) * ( 0.00068) = 0.005 ksi = 5 psi 2

(h) the modulus of resilience is 5 psi.

6-36

A cylindrical specimen of an AISI-SAE type 416 stainless steel with a 0.505-in. diameter tensile bar is machined to a 2.00-in gage length. After fracture, the gage length was 2.20-in and the diameter was 0.325-in. Plot the engineering stress strain curve and calculate (a) the 0.2% offset yield strength; (b) the tensile strength; (c) the modulus of elasticity; (d) the % elongation; (e) the % reduction in area; (f) the engineering stress at fracture; (g) the true stress at necking; (h) the modulus of resilience; and (i) the elastic and plastic strain to fracture. (j) When the sample was loaded to 11,400 lbs, the diameter was measured to be 0.504 in. Calculate the tranverse and axial strains at this load. Compute the Poisson’s ratio. (k) Obtain the tensile properties for type 416 stainless steel that has been quenched and tempered and compare them to your answers. Discuss the similarities and differences.

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Gage Length (in.) Engineering Stress (ksi) Engineering Strain 2.000 0.0 0.0000 2.001 14.2 0.0005 2.002 28.5 0.0010 2.003 42.7 0.0015 2.004 56.9 0.0020 2.006 85.4 0.0030 2.008 99.9 0.0040 2.01 103.9 0.0050 2.015 114.8 0.0075 2.02 120.8 0.0100 2.03 133.8 0.0150 2.04 143.8 0.0200 2.08 168.0 0.0400 2.12 178.5 0.0600 2.14 179.7 0.0700 2.16 179.0 0.0800 2.2 170.0 0.1000 2.23 139.8 0.1150

Load (lbs) 0 2,850 5,710 8,560 11,400 17,100 20,000 20,800 23,000 24,200 26,800 28,800 33,650 35,750 36,000 35,850 34,050 28,000 Solution: 200

S

Engineering Stress (ksi)

UTS

= 180 ksi

150 S

fracture

= 140 ksi

100 E = 28.5 Mpsi 50 e 0 0.00

0.02

0.04

0.06

UTS

= 0.070 0.08

e

fracture

0.10

= 0.115 0.12

Engineering Strain

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Engineering Stress (ksi)

200

150 0.2% offset yield strength = 107 ksi

100 (0.003, 85 ksi) 50

0 0.000

0.005

0.010

0.015

0.020

Engineering Strain

By inspection of the plots: (a) the 0.2% offset yield strength is 107 ksi; (b) the tensile strength is 180 ksi; and (c) the modulus of elasticity is 28.5 Mpsi from a linear fit to the data points from the elastic loading.

Also

2.20 − 2.00 ×100 = 10%; 2.00 2 2 π π ( 0.505 in.) − ( 0.325 in.)

4 (e) percent reduction in area = 4 ×100 = 58.6% 2 π 0.505 in. ( ) 4 (d) % elongation =

By inspection of the plot, (f) the engineering stress at fracture is 140 ksi.

The true stress σ at necking can be written as

σ = SUTS (1+ e ) .

UTS

This is the last data point at which this expression for true stress, i.e., σ = S(1+ e) , is valid. Substituting,

σ = SUTS (1+ e ) =180 ksi [1+ 0.070 ] =192 ksi UTS

Thus (g) the true stress at necking is 192 ksi.

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The modulus of resilience is the area under the elastic portion of a stress-strain curve.

The area under the elastic portion of the curve can be calculated as the area of a triangle with one data point being the stress and strain at the proportional limit.

modulus of resilience =

1 (85 ksi) * ( 0.003) = 0.0128 ksi = 12.8 psi 2

(h) the modulus of resilience is 12.8 psi.

The total strain to failure is 0.115. The plastic strain to failure is the % elongation. Thus the elastic strain at failure is

elastic strain = total strain - plastic strain = 0.115 − 0.100 = 0.015 (Note that this value is not consistent with 140 ksi / 28.5 Mpsi = 0.005 as it should be.

This is an error in the problem statement for the specification of the length of the sample after fracture. Assuming that the total strain and elastic strain remain the same, the gage length after fracture should be 2.22 in. not 2.20 in.) (i) the elastic strain to fracture is 0.015 and the plastic strain to fracture is 0.100 (or the elastic strain is 0.005 and the plastic strain is 0.110.) (j) When the sample was loaded to 11,400 lbs, the diameter was measured to be 0.504 in. Calculate the tranverse and axial strains at this load. Compute the Poisson’s ratio.

2.004 − 2.000 = 0.002 2.000 0.504 − 0.505 etransverse = = −0.00198 0.505 eaxial =

e

ν = − transverse = − eaxial

0.00198 = 0.99 0.002

(Note that this value for a Poisson’s ratio for steel is unreasonably large. The diameter of the specimen was not 0.504 in. at a load of 11, 400 lbs. A reasonable value for the Poisson’s ratio is 0.3. Thus the diameter of the bar at 11,400 lbs was approximately 0.5047. Small errors in measurement produce large errors in elastic properties.) (k) Obtain the tensile properties for type 416 stainless steel that has been quenched and tempered and compare them to your answers. Discuss the similarities and differences.

The data sheet for 416 stainless steel can be found on MatWeb for quenched and tempered bar. The calculated modulus of elasticity and the modulus listed on the website are very similar. The calculations for the 0.2% offset yield and tensile strengths are roughly 10 ksi and 70 ksi greater than the tabulated data found on the website.

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6-37

The following data were collected from a test specimen of cold-rolled and annealed brass. The specimen had an initial gage length l0 of 35 mm and an initial cross-sectional area A0 of 10.5 mm2.

Load (N) Δl (mm) 0 0.0000 66 0.0112 177 0.0157 327 0.0199 462 0.0240 797 1.72 1350 5.55 1720 8.15 2220 13.07 2690 22.77 2410 25.25 (a) Plot the engineering stress strain curve and the true stress strain curve. Since the instantaneous cross-sectional area of the specimen is unknown past the point of necking, truncate the true stress true strain data at the point that corresponds to the ultimate tensile strength. Use of a software graphing package is recommended. (b) Comment on the relative values of true stress strain and engineering stress strain during the elastic loading and prior to necking. (c) If the true stress strain data were known past the point of necking, what might the curve look like? (d) Calculate the 0.2% offset yield strength. (e) Calculate the tensile strength. (f) Calculate the elastic modulus using a linear fit to the appropriate data.

Solution: Load (N) 0 66 177 327 462 797 1350 1720 2220 2690 2410

The table below contains the values for engineering stress and strain and true stress and strain.

Displacement Engineering Engineering True True Stress (MPa) (mm) Strain Stress (MPa) Strain 0.0000 0 0 0 0 0.0112 0.000320 6.29 0.000320 6.29 0.0157 0.000449 16.86 0.000448 16.87 0.0199 0.000569 31.14 0.000568 31.17 0.0240 0.000686 44.00 0.000685 44.03 1.72 0.049143 75.91 0.047974 79.64 5.55 0.15857 129 0.14719 149 8.15 0.23286 164 0.20933 202 13.07 0.37343 211 0.31731 290 22.77 0.65057 256 0.50112 423 25.25 0.72143 230 0.54315 ***** (a) A plot of the engineering stress and strain and true stress and strain is shown below. The true stress strain data is truncated at the point that corresponds to the ultimate tensile strength.

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(b) During the elastic loading, the values of true stress–strain and engineering stress–strain are nearly identical. Upon close inspection, however, the true stress values are greater than the engineering stress values, and the true strain values are less than the engineering strain values. These differences become much more pronounced past the point of yielding. In tension, the cross–sectional area becomes smaller as the test proceeds. Thus the true stress, which is computed using the instantaneous cross–sectional area rather than the initial cross–sectional area, is larger than the engineering stress. In tension, the length of the sample becomes longer during the test. Thus by accounting for instantaneous changes in length, the true strain is smaller than the engineering strain prior to necking. These trends hold for every tension test of a metallic material. (c) If the true stress strain data were known past the point of necking, the curve would not have a maximum. The true stress would continually increase. (d) The 0.2% offset yield strength is approximately 46 MPa. (e) The ultimate tensile strength is the maximum engineering stress sustained during a tension test. This value can be read from the table or the plot as 256 MPa. (f) The best estimate is obtained using a linear fit to the data during the elastic loading. The elastic modulus determined in this fashion is 105 GPa. 6-38

The following data were collected from a standard 0.505-in.-diameter test specimen of a copper alloy (ℓo = 2.0 in.): Load (lb) 0 3,000 6,000 7,500 9,000 10,500

∆ℓ (in.) Stress (psi) 0.0000 0 0.00167 15,000 0.00333 30,000 0.00417 37,500 0.0090 45,000 0.040 52,500

Strain (in./in.)

0.0 0.000835 0.001665 0.002085 0.0045 0.02

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12,000 0.26 12,400 0.50 (maximum load) 11,400 1.02 (fracture)

60,000 0.13 62,000 0.25 57,000 0.51

After fracture, the total length was 3.014 in. and the diameter was 0.374 in. Plot the engineering stress strain curve and calculate (a) the 0.2% offset yield strength; (b) the tensile strength; (c) the modulus of elasticity; (d) the % elongation; (e) the % reduction in area; (f) the engineering stress at fracture; and (g) the true stress at necking; and (h) the modulus of resilience.

Solution: S = F /[(π/4)(0.505)2] = F/0.2 e = (ℓ – 2) / 2

0.2% offset yield strength = 45,000 psi tensile strength = 62,000 psi E = (30,000 – 0) / (0.001665 – 0) = 18 × 106 psi

(3.014 − 2) ×100 = 50.7% 2 (π /4) (0.505) 2 − (π /4) (0.374) 2 % reduction in area = × 100 (π /4) (0.505) 2 % elongation =

= 45.2% engineering stress at fracture = 57,000 psi true stress at necking = SUTS (1 + eUTS) = 62,000 (1+0.25) = 77,500 psi From the graph, yielding begins at about 37,500 psi. Thus: ½(yield strength)(strain at yield) = ½ (37,500)(0.002085) = 39.1 psi 6-39

The following data were collected from a 0.4-in.-diameter test specimen of polyvinyl chloride (ℓo = 2.0 in.): Load (lb) 0 300 600 900

∆ℓ (in.)

0.00000 0.00746 0.01496 0.02374

Stress (psi) 0 2,387 4,775 7,162

Strain (in./in.)

0.0 0.00373 0.00748 0.01187

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1200 1500 1660 1600

0.032 0.046 0.070 (maximum load) 0.094

9,549 11,937 13,210 12,732

0.016 0.023 0.035 0.047

1420

0.12 (fracture)

11,300

0.06

After fracture, the total length was 2.09 in. and the diameter was 0.393 in. Plot the engineering stress strain curve and calculate (a) the 0.2% offset yield strength; (b) the tensile strength; (c) the modulus of elasticity; (d) the % elongation; (e) the % reduction in area; (f) the engineering stress at fracture; and (g) the modulus of resilience.

Solution:

S = F / [(π/4)(0.4)2] e = (ℓ – 2) / 2

0.2% offset yield strength = 11,600 psi tensile strength = 13,210 psi E = (7162 – 0) / (0.01187 – 0) = 603,000 psi

(2.09 − 2) × 100 = 4.5% 2 (π /4) (0.4) 2 − (π /4) (0.393) 2 ×100 = 3.5% % reduction in area = (π /4) (0.4)2 % elongation =

engineering stress at fracture = 11,300 psi true stress at necking = SUTS (1 + eUTS) = 13,210 (1+0.035) = 13,672 psi From the figure, yielding begins near 9550 psi. Thus: ½(yield strength)(strain at yield) = ½ (9550)(0.016) = 76.4 psi 6-40

The following data were collected from a 12-mm-diameter test specimen of magnesium (ℓo = 30.00 mm): Load (N) 0 5,000 10,000 15,000

∆ℓ (mm) 0.0000 0.0296 0.0592 0.0888

Stress (MPa) 0 44.2 88.4 132.6

Strain (mm/mm) 0.0 0.000987 0.001973 0.00296

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20,000 25,000 26,500 27,000 26,500 25,000

0.15 0.51 0.90 1.50 (maximum load) 2.10 2.79 (fracture)

176.8 221.0 234.3 238.7 234.3 221.0

0.005 0.017 0.030 0.050 0.070 0.093

After fracture, the gage length was 32.61 mm and the diameter was 11.74 mm. Plot the engineering stress strain curve and calculate (a) the 0.2% offset yield strength; (b) the tensile strength; (c) the modulus of elasticity; (d) the % elongation; (e) the % reduction in area; (f) the engineering stress at fracture; (g) the true stress at necking; and (h) the modulus of resilience.

Solution:

S = F / [(π/4)(12 mm)2] = F/113.1 e = (ℓ – 30)/30

0.2% offset yield strength = 186 MPa tensile strength = 238.7 MPa E = (132.6 – 0) / (0.00296 – 0) = 44,800 MPa = 44.8 GPa

(32.61 − 30) × 100 = 8.7% 30 (π /4) (12) 2 − (π /4) (11.74) 2 % reduction in area = ×100 = 4.3% (π /4) (12) 2 % elongation =

engineering stress at fracture = 221 MPa true stress at necking = SUTS (1 + eUTS) = 238.7 (1+0.05) = 251 MPa From the figure, yielding begins near 133 MPa. Thus: ½(yield strength)(strain at yield) = ½ (133)(0.00296) = 0.2 MPa

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6-41

The following data were collected from a 20-mm-diameter test specimen of a ductile cast iron (ℓo = 40.00 mm): True Stress (MPa) 0 79.6 159.3 239.1 288 339 397 459

Displacement Engineering Engineering (mm) Strain Stress (MPa) True Strain 0.0000 0 0 0 0.0185 0.000463 79.6 0.000462 0.0370 0.000925 159.2 0.000925 0.0555 0.001388 238.7 0.000139 0.20 0.005 287 0.005 0.60 0.015 334 0.015 1.56 0.039 382 0.038 4.00 (max.

0.100 417 0.095 load) 125,000 7.52 (fracture) 0.188 398 0.172 ***** After fracture, the total length was 47.42 mm and the diameter was 18.35 mm. Plot the engineering stress strain curve and the true stress strain curve. Since the instantaneous cross-sectional area of the specimen is unknown past the point of necking, truncate the true stress–true strain data at the point that corresponds to the ultimate tensile strength. Use of a software graphing package is recommended. Calculate (a) the 0.2% offset yield strength; (b) the tensile strength; (c) the modulus of elasticity, using a linear fit to the appropriate data; (d) the % elongation; (e) the % reduction in area; (f) the engineering stress at fracture; (g) the true stress at necking; and (h) the modulus of resilience.

Load (N) 0 25,000 50,000 75,000 90,000 105,000 120,000 131,000

Solution:

A plot of the engineering stress and strain and true stress and strain is shown below. The true stress strain data is truncated at the point that corresponds to the ultimate tensile strength.

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0.2% offset yield strength = 274 MPa tensile strength = 417 MPa E = (238.7 – 0) / (0.001388 – 0) = 172,000 MPa = 172 GPa

(47.42 − 40) × 100 = 18.55% 40 (π /4) (20)2 − (π /4) (18.35) 2 % reduction in area = ×100 = 15.8% (π /4) (20) 2 % elongation =

engineering stress at fracture = 397.9 MPa true stress at necking = SUTS (1 + eUTS) = 417 (1+0.10) = 459 MPa From the figure, yielding begins near 238.7 MPa. Thus: ½(yield strength)(strain at yield) = ½ (238.7)(0.001388) = 0.17 MPa

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6-42

Consider the tensile stress strain diagram in Figure 6–30 labeled 1 and 2. These diagrams are typical of metals. Answer the following questions, and consider each part as a separate question that has no relationship to previous parts of the question.

(a) Samples 1 and 2 are identical except for the grain size. Which sample has the smaller grains? How do you know? (b) Samples 1 and 2 are identical except that they were tested at different temperatures. Which was tested at the lower temperature? How do you know? (c) Samples 1 and 2 are different materials. Which sample is tougher? Explain. (d) Samples 1 and 2 are identical except that one of them is a pure metal and the other has a small percentage alloying addition. Which sample has been alloyed? How do you know? (e) Given the stress strain curves for materials 1 and 2, which material has the lower hardness value on the Brinell hardness scale? How do you know? (f) Are the stress strain curves shown true stress strain or engineering stress strain curves? How do you know? (g) Which of the two materials represented by samples 1 and 2 would exhibit a higher shear yield strength? How do you know?

Solution: (a) Sample 1 has the smaller grains. Smaller grains result in more grain boundaries. Grain boundaries hinder the motion of dislocations. A higher stress must then be applied to cause dislocations to continue to move and change the shape of the material. This makes the material stronger. Sample 1 is stronger than Sample 2; therefore, all other factors being identical, Sample 1 has the smaller grains. (b) Sample 1 was tested at the lower temperature because it has the higher strength and lower ductility. As temperature increases, it becomes easier for dislocations to move leading to lower strengths and higher ductilities. Since Sample 1 is stronger and less ductile than Sample 2, it was tested at the lower temperature. (c) Sample 2 is tougher. Toughness is the ability to absorb energy prior to fracture. Toughness is a combination of strength and ductility and is represented graphically by the area under the stress–strain curve. The area under the stress–strain curve of Sample 2 is larger; therefore, Sample 2 is tougher. (d) Sample 1 has been alloyed. Introducing a solute to the metal increases the strength. Solute atoms hinder the motion of dislocations. A higher stress must then be applied to cause dislocations to continue to move and change the shape of the

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material. This makes the material stronger. Sample 1 is stronger than Sample 2; therefore, Sample 1 has been alloyed. (e) Sample 2 has the lower hardness value on the Brinell hardness scale. Hardness is directly proportional to strength. Sample 2 is weaker than Sample 1; therefore, it has the lower hardness. (f) The stress strain curves shown represent true stress and true strain because they do not exhibit maxima as do tensile engineering stress and engineering strain curves. (g) Sample 1 will have the higher shear yield strength because it has the higher tensile yield strength 6-45

A square specimen of MgO is loaded using a three-point bend test. Compute the minimum possible thickness of the specimen without fracture, given that the applied load is 95 lbs, the flexural strength is 15 ksi, and the separation between load points is 2 in.

Solution:

flexural strength = 3FL/(2wh 2 ) = 3FL/(2t 3 ) where t is the specimen thickness. Solving for t,

(

)(

)

1/3  3 95 lbs 2 in.    3FL  = 0.27 in. t =  = 2 2*flexural strength   2*15,000 lbs/in.   

6-46

1/3

A bar of Al2O3 that is 0.25 in. thick, 0.5 in. wide, and 9 in. long is tested in a three-point bending apparatus, with the supports located 6 in. apart. The deflection of the center of the bar is measured as a function of the applied load. The data are shown below.

Determine the flexural strength and the flexural modulus.

Force (lb) 14.5 28.9 43.4 57.9 86.0

Solution:

Deflection (in.)

0.0025 0.0050 0.0075 0.0100 0.0149 (fracture)

Stress (psi) 4,176 8,323 12,499 16,675 24,768

stress = 3LF /(2 wh 2 ) = (3)(6 in.)F/[(2)(0.5 in.)(0.25 in.) 2 ]

(6–15)

= 288F

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The flexural strength is the stress at fracture, or 24,768 psi.

The flexural modulus can be calculated from the linear curve; picking the first point as an example:

FM = 6-47

FL3 (14.5 lb)(6 in.)3 = = 40 × 106 psi 3 3 4wh δ (4)(0.5 in.) (0.25 in.) (0.0025 in.)

A three-point bend test is performed on a block of ZrO2 that is 8 in. long, 0.50 in. wide, and 0.25 in. thick and is resting on two supports 4 in. apart. When a force of 400 lb is applied, the specimen deflects 0.037 in. and breaks. Calculate (a) the flexural strength; and (b) the flexural modulus, assuming that no plastic deformation occurs.

Solution:

flexural strength = 3FL /(2wh 2 ) =

(3)(400 lb)(4 in.) = 76,800 psi (2)(0.5 in.)(0.25 in.) 2

flexural modulus = FL3 / (4wh3δ ) =

(400 lb) (4 in.)3 (4)(0.5 in.) (0.25 in.)3 (0.037 in.)

= 22.14 ×106 psi 6-48

A three-point bend test is performed on a block of silicon carbide that is 10 cm long, 1.5 cm wide, and 0.6 cm thick and is resting on two supports 7.5 cm apart. The sample breaks when a deflection of 0.09 mm is recorded. The flexural modulus for silicon carbide is 480 GPa. Assume that no plastic deformation occurs. Calculate (a) the force that caused the fracture; and (b) the flexural strength.

Solution:

(a)

The force F required to produce a deflection of 0.09 mm is F = (flexural modulus)(4wh3δ)/L3 F = (480,000 MPa)(4)(15 mm)(6 mm)3(0.09 mm) / (75 mm)3 F = 1327 N flexural strength = 3FL /(2wh 2 ) = (3)(1327 N)(75 mm)/[(2)(15 mm)(6 mm) 2 ] (b) = 276 MPa

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6-49

(a) A thermosetting polymer containing glass beads is required to deflect 0.5 mm when a force of 500 N is applied. The polymer part is 2 cm wide, 0.5 cm thick, and 10 cm long.

If the flexural modulus is 6.9 GPa, determine the minimum distance between the supports. Will the polymer fracture if its flexural strength is 85 MPa? Assume that no plastic deformation occurs. (b) The flexural modulus of alumina is 45 × 106 psi, and its flexural strength is 46,000 psi. A bar of alumina 0.3 in. thick, 1.0 in. wide, and 10 in. long is placed on supports 7 in. apart. Determine the amount of deflection at the moment the bar breaks, assuming that no plastic deformation occurs.

Solution: The minimum distance L between the supports can be calculated from the flexural modulus.

L3 = 4wh3d(flexural modulus)/F L3 = (4)(20 mm)(5 mm)3(0.5 mm)(6.9 GPa)(1000 MPa/GPa) / 500 N

L3 = 69, 000 mm3

L = 41 mm

or

The stress acting on the bar when a deflection of 0.5 mm is obtained is

σ = 3FL /(2wh 2 ) = (3)(500 N)(41 mm)/[(2)(20 mm)(5 mm) 2 ] = 61.5 MPa The applied stress is less than the flexural strength of 85 MPa; the polymer is not expected to fracture.

6-57

A Brinell hardness measurement, using a 10-mm diameter-indenter and a 500-kg load, produces an indentation of 2.5 mm on a steel plate. (a) Determine the Brinell hardness number (HB) of the steel plate; and (b) approximate the tensile strength of the steel.

Solution:

HB =

HB =

(

2F

π D D − D 2 − Di2

)

2 ( 500 kg)

 π (10 mm ) 10 mm − 

 (10 mm ) − ( 2.5 mm)  2

2

= 100.2

TS = 500 *100.2 HB = 50.1 ksi 6-58

A Brinell hardness measurement, using a 10-mm-diameter indenter and a 500 kg load, produces an indentation of 4.5 mm on an aluminum plate. Determine the Brinell hardness number (HB) of the metal.

Solution:

HB =

(

2 500 kg

)

π (10 mm) [10 − 10 − 4.5 ] 2

2

= 29.8

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6-59

When a 3000 kg load is applied to a 10-mm-diameter ball in a Brinell test of a steel, an indentation of 3.1 mm diameter is produced. Estimate the tensile strength of the steel.

(

Solution:

HB =

2 3000 kg

)

π (10 mm) [10 − 102 − 3.12 ]

= 388

Tensile strength = 500 HB = (500)(388) = 194,000 psi 6-61

The elastic modulus of a metallic glass is determined to be 95 GPa using nanoindentation testing with a diamond Berkovich tip. The Poisson’s ratio of the metallic glass is 0.36. The unloading stiffness as determined from the load-displacement data is 5.4 × 105 N/m. The maximum load is 120 mN. What is the hardness of the metallic glass at this indentation depth?

Solution: The hardness of a material as determined by nanoindentation is

H=

Pmax , Ac

where Pmax is the maximum load imposed over the contact area. The maximum load is given in the problem statement, but the contact area must be determined. The reduced elastic modulus Er is related to the contact area by

Er =

π S , 2β A c

where S is the unloading stiffness and β is a constant for the shape of the indenter being used (β = 1.034 for a Berkovich indenter). The reduced elastic modulus is also given by

1 1 − v2 1 − v2 = + , Er E Ei where E and v are the elastic modulus and Poisson’s ratio of the material being indented, respectively, and Ei and vi are the elastic modulus and Poisson’s ratio of the indenter, respectively (for diamond, Ei = 1.141 TPa and vi = 0.07). Substituting to find the reduced elastic modulus,

1 1 − 0.362 1 − 0.07 2 = + Er 95 × 109 Pa 1.141 × 1012 Pa Er = 99.66 × 109 Pa. Solving for the contact area,

Ac =

πS2 π (5.4 × 105 N/m) 2 = = 2.157 × 10−11 m 2 .

2 2 2 9 2 4 β Er 4(1.034 )(99.66 × 10 Pa)

Thus, the hardness is

H=

Pmax 120 × 10−3 N = = 5.6 × 109 Pa = 5.6 GPa . −11 2 Ac 2.157 × 10 m

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6-62

The elastic modulus and hardness of MgO are determined to be 306 GPa and 9.2 GPa, respectively, using nanoindentation testing with a diamond tip. The Poisson’s ratio of the MgO is 0.17. The maximum load is 150 mN at this particular indentation depth.

Determine the contact stiffness S. Take β = 1.

Solution: The hardness of a material as determined by nanoindentation is

H=

Pmax , Ac

where Pmax is the maximum load imposed over the contact area. The maximum load and hardness are given in the problem statement, and we can solve for the contact area.

The reduced elastic modulus Er is related to the contact area by

Er =

π S , 2β A c

where S is the unloading stiffness and β is a constant for the shape of the indenter being used (β = 1 for this problem). The reduced elastic modulus is also given by

1 1 − v2 1 − v2 = + , Er E Ei where E and v are the elastic modulus and Poisson’s ratio of the material being indented, respectively, and Ei and vi are the elastic modulus and Poisson’s ratio of the indenter, respectively (for diamond, Ei = 1.141 TPa and vi = 0.07). Substituting to find the reduced elastic modulus,

1 1− 0.172 1− 0.072 = + Er 306 × 109 Pa 1.141 × 1012 Pa Er = 247 × 109 Pa Substituting,

Ac =

Pmax H

and solving for S,

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S=

6-63

2β Er Pmax

πH

( )( =

) 150 ×10 N = 1.12 × 10 N/m π ( 9.2 × 10 Pa ) −3

2 1 247 × 109 Pa

6

9

Plot the Charpy V-notch data below for commercial steel. Label the upper and lower energy shelves on your plot. Would you recommend this steel for service conditions below 10°C? Explain.

Temperature (°C) Impact energy (J) 50

75

40

75

30

70

20

58

10

37

0

24

-10

15

-20

10

-30

4

-40

1

Solution:

The steel should not be recommended for service below 10°C because the ductile-brittle transition temperature (DBTT) occurs roughly at 10°C. The steel will behave in a brittle manner below that temperature range and will not likely have enough toughness for service requirements.

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6-64

The following data were obtained from a series of Charpy impact tests performed on four steels, each having a different manganese content. Plot the data and determine (a) the transition temperature of each (defined by the mean of the absorbed energies in the ductile and brittle regions); and (b) the transition temperature of each (defined as the temperature that provides 50 J of absorbed energy).

Impact Energy (J)

150

1.55% Mn 1.01% Mn 0.39% Mn 0.30% Mn

100

50 J

50

0

-100

-80

-60

-40

-20

0

20

40

60

80

100

Temperature (˚C)

Solution: (a) The transition temperatures are defined by the mean of the absorbed energies in the brittle and ductile regions: 0.30% Mn: (2 + 130) / 2 = 66 J

T = 28˚C

0.39% Mn: (5 + 135) / 2 = 70 J

T = 8˚C

1.01% Mn: (5 +135) / 2 = 70 J

T = –3˚C

1.55% Mn: (15 + 140) / 2 = 78 J T = –20˚C

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(b) The transition temperatures defined by 50 J are 0.30% Mn: T = 17˚C 0.39% Mn: T = –5˚C 1.01% Mn: T = –18˚C 1.55% Mn: T = –44˚C 6-65

Plot the transition temperature versus manganese content using the data in Problem 664 and discuss the effect of manganese on the toughness of steel. What is the minimum manganese allowed in the steel if a part is to be used at 0oC?

Solution: 30

Transition Tem perature (˚C)

20 10

Average Absorbed Energy 0 -10

50 J Absorbed Energy

-20 -30 -40 -50

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

% Manganese

Increasing the manganese increases the toughness and reduces the transition temperature; manganese is therefore a desirable alloying element for improving the impact properties of the steel. If the part is to be used at 0˚C, we would want at least 1.0% Mn in the steel based on the mean absorbed energy criterion or 0.39% Mn based on the 50 J criterion.

6-66

The following data were obtained from a series of Charpy impact tests performed on four ductile cast irons, each having a different silicon content. Plot the data and determine (a) the transition temperature of each (defined by the mean of the absorbed energies in the ductile and brittle regions); and (b) the transition temperature of each (defined as the temperature that provides 10 J absorbed energy). (c) Plot the transition temperature versus silicon content and discuss the effect of silicon on the toughness of the cast iron. What is the maximum silicon allowed in the cast iron if a part is to be used at 25oC?

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Test temperature (oC) –50 –5 0 25 50 75 100 125

Impact energy (J) 2.55% Si 2.85% Si 3.25% Si 2.5 2.5 2 3 2.5 2 6 5 3 13 10 7 17 14 12 19 16 16 19 16 16 19 16 16

3.63% Si 2 2 2.5 4 8 13 16 16

Solution: 20

2.55% Si 2.85% Si 3.25% Si 3.63% Si

Impact Energy (J)

15

10 J

10

5

0

-60

-40

-20

0

20

40

60

80

100

120

Temperature (˚C)

Transition Temperature (˚C)

70

60

50

40

10 J Absorbed Energy 30

20

10

Average Absorbed Energy 2.4

2.6

2.8

3.0

3.2

3.4

3.6

3.8

Wt% Silicon (a) Transition temperatures defined by the mean of the absorbed energies are

2.55% Si: mean energy = (2.5 + 19)/2 = 9.8 J;

T = 13oC

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2.85% Si: mean energy = (2.5 + 16)/2 = 8.3 J; 3.25% Si: mean energy = (2 + 16)/2 = 9 J; 3.63% Si: mean energy = (2 + 16)/2 = 9 J;

T = 16oC T = 35oC T = 55oC

(b) Transition temperatures defined by 10 J are

2.55% Si: 2.85% Si: 3.25% Si: 3.63% Si:

T = 14oC T = 25oC T = 40oC T = 60oC

(c) Increasing the silicon decreases the toughness and increases the transition temperature; silicon, therefore, reduces the impact properties of the cast iron. If the part is to be used at 25oC, we would want a maximum of about 2.9% Si in the cast iron.

6-67

FCC metals are often recommended for use at low temperatures, particularly when any sudden loading of the part is expected. Explain.

Solution: FCC metals do not normally display a transition temperature; instead the impact energies decrease slowly with decreasing temperature and, in at least some cases (such as some aluminum alloys), the energies even increase at low temperatures.

The FCC metals can obtain large ductilities, giving large areas beneath the true stress strain curve.

6-68

A steel part can be made by powder metallurgy (compacting iron powder particles and sintering to produce a solid) or by machining from a solid steel block. Which part is expected to have the higher toughness? Explain.

Solution: Parts produced by powder metallurgy often contain considerable amounts of porosity due to incomplete sintering; the porosity provides sites at which cracks might easily nucleate. Parts machined from solid steel are less likely to contain flaws that would nucleate cracks, therefore improving toughness.

6-71

A number of aluminum-silicon alloys have a structure that includes sharp-edged plates of brittle silicon in the softer, more ductile aluminum matrix. Would you expect these alloys to be notch-sensitive in an impact test? Would you expect these alloys to have good toughness? Explain your answers.

Solution: The sharp-edged plates of the brittle silicon may act as stress-raisers, or notches, thus giving poor toughness to the alloy. The presence of additional notches, such as machining marks, will not have a significant effect, since there are already very large numbers of “notches” due to the microstructure. Consequently this type of alloy is expected to have poor toughness but is not expected to be notch sensitive.

6-76

A load versus displacement diagram is shown in Figure 6–31 for a metallic glass. A metallic glass is a non-crystalline (amorphous) metal. The sample was tested in compression. Therefore, even though the load and displacement values are plotted as positive, the sample length was shortened during the test. The sample had a length in the direction of loading of 6 mm and a cross-sectional area of 4 mm2. Numerical values for the load and displacement are given at the points marked with a circle and an X. The

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first data point is (0, 0). Sample failure is indicated with an X. Answer the following questions.

a. b. c. d.

Calculate the elastic modulus.

How does the elastic modulus compare to the modulus of steel?

Calculate the engineering stress at the proportional limit.

Consider your answer to part (c) to be the yield strength of the material. Is this a high yield stress or a low yield stress? Support your answer with an order of magnitude comparison for a typical polycrystalline metal. e. Calculate the true strain at the proportional limit. Remember that the length of the sample is decreasing in compression. f. Calculate the total true strain at failure. g. Calculate the work of fracture for this metallic glass based on engineering stress and strain.

Solution: a. The elastic modulus may be determined by taking the slope of an engineering stress – engineering strain curve. The slope can be based on the data points of (0, 0) and (0.100 mm, 7020 N) for the load and displacement. The engineering stress S is given by

S=

F , A0

where F is the applied force and A0 is the initial cross–sectional area. The engineering strain e is given by

e=

∆l , l0

where ∆l is the displacement and l0 is the initial length. The elastic modulus E is given by

E=

S , e

according to Hooke’s Law. Therefore, at the load and displacement values of (0.100 mm, 7020 N), the last point in the linear region,

E=

F l0 7020 N 6 mm = × = 105,300 MPa = 105.3 GPa .

2 A0 ∆l 4 mm 0.100 mm

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b. The elastic modulus of this metallic glass is lower than that of steel, which has an elastic modulus of 200 GPa. c. At the proportional limit, the applied load is 7020 N. The engineering stress S is

S=

F 7020 N = = 1755 MPa .

A0 4 mm 2

d. This is a high stress. Pure crystalline metals have yield strengths on the order of tens of MPa or less. Alloys have yield strengths on the order of hundreds of MPa. Some tool steels have yield strengths as high as 2 GPa, but strengths as high as 1755 MPa are rare for a metal. e. The true strain is given by



ε = ln(1 + e) = ln 1 + 

∆l   l0 

Substituting,

 

ε = ln 1 − f.

0.100 mm   = −0.01681 = −1.681% .

6 mm 

The true strain e at failure is calculated as

 

ε = ln 1 −

0.105 mm   = −0.01765 = −1.765% .

6 mm 

g. The work of fracture can be approximated as the area under an engineering stress strain curve. The area under the curve can be calculated as the area of a triangle plus the area of a rectangle.

The area of the rectangle will be approximated by assuming that the stress in the plastic region has a constant value of 1755 MPa. The engineering strain at the proportional limit is

e=

∆l −0.1 mm = = −0.0167 , l0 6 mm

and the engineering strain at failure is

e=

∆l −0.105 mm = = −0.0175 . l0 6 mm

Positive values of strain will be used for the purpose of finding the area under the stress–strain curve.

Work 1 = ∫ S de = × 1755 MPa × 0.0167 + 1755 MPa × (0.0175 − 0.0167) Volume 2 Work = 16.1× 106 J/m3 .

Volume If we calculate the total work, we simply multiply by the sample volume.

Work = 16 ×10 6 J/m 3 *Volume=16 ×10 6 J/m 3 ( 6 ×10 −3 m ) ( 4 ×10−6 m 2 ) = 0.4 J

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Chapter 7: Mechanical Properties: Part Two

7-1

Alumina (Al2O3) is a brittle ceramic with low toughness. Suppose that fibers of silicon carbide SiC, another brittle ceramic with low toughness, could be embedded within the alumina. Would doing this affect the toughness of the ceramic matrix composite?

Explain.

Solution: The SiC fibers may improve the toughness of the alumina matrix. The fibers may do so by several mechanisms. By introducing an interface (between the fibers and the matrix), a crack may be blocked; to continue growing, the crack may have to pass around the fiber, thus increasing the total energy of the crack and thus the energy that can be absorbed by the material. Or extra energy may be required to force the crack through the interface in an effort to continue propagating. In addition, the fibers may begin to pull out of the matrix, particularly if bonding is poor; the fiber pull-out requires energy, thus improving toughness. Finally, the fibers may bridge across the crack, helping to hold the material together and requiring more energy to propagate the crack.

7-2

What are the differences between the stress intensity factor (KI) and the plane strain fracture toughness (KIc)? Explain.

Solution: The stress intensity factor is a measure of the crack driving force in a material that has a pre-existing flaw. The plane-strain fracture toughness is the resistance of a material to crack propagation and is a materials property.

7-3

A wing component on an aircraft is fabricated from an aluminum alloy that has a plane strain fracture toughness of 26 MPa√ . It has been determined that the fracture results at a stress of 112 MPa when the maximum internal crack length is 8.6 mm. For this same component and alloy, compute the stress level at which fracture will occur for a critical internal crack length of 6.0 mm.

Solution: KIc = 26 MPa√ KIc = a= 7-4

σ = 112 MPa

f√

f=

= 3 mm

a= √

. = 4.3 mm

= 1.997

σ = = 134 MPa √

Calculate the maximum internal crack length allowable for a titanium alloy component that is loaded to a stress that is half of its yield strength. Assume that the geometry factor is 1.5. The yield strength of this alloy is 132 ksi, and the fracture toughness is 50 ksi in. .

Solution: σy = 132 ksi

σapplied = = 66 ksi

f = 1.50

KIc = 50 ksi√

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a = ( )2 = 0.081 in , so 2a = 0.162 in 7-5

A steel alloy with a plane strain fracture toughness of 55 MPa m is exposed to a stress of 1020 MPa. Will this component fracture if it is known that the largest surface crack is 0.5 mm long? Assume the geometry factor to be 1.0.

Solution: KIc = 55 MPa√ σ = 1020 MPa f = 1.0 a = 0.5 mm = 40.426 MPa√ KI = f√ KI < KIc therefore the component is not expected to fail.

7-6

A ceramic matrix composite contains internal flaws as large as 0.001 cm in length. The plane strain fracture toughness of the composite is 45 MPa m and the tensile strength is 550 MPa. Will the stress cause the composite to fail before the tensile strength is reached? Assume that f = 1.

Solution: Since the crack is internal, 2a = 0.001 cm = 0.00001 m. Therefore a = 0.000005 m

K Ic = f σ π a

or σ = K Ic /( f π a )

σ = (45 MPa m)/[(1) π (0.000005 m)] = 11,354 MPa The applied stress required for the crack to cause failure is much larger than the tensile strength of 550 MPa. Any failure of the ceramic should be expected due to an overload, not because of the presence of the flaws.

7-7

An aluminum alloy that has a plane strain fracture toughness of 25,000 psi in fails when a stress of 42,000 psi is applied. Observation of the fracture surface indicates that fracture began at the surface of the part. Estimate the size of the flaw that initiated fracture. Assume that f = 1.1.

Solution:

K Ic = f σ π a

or a = (1/π )[ K Ic /( f σ )]2

a = (1/π )[25, 000 psi in./((1.1)(42, 00 psi))]2 = 0.093 in.

7-8

A polymer that contains internal flaws 1 mm in length fails at a stress of 25 MPa. Determine the plane strain fracture toughness of the polymer. Assume that f = 1.

Solution: Since the flaws are internal, 2a = 1 mm = 0.001 m; thus a = 0.0005 m

K Ic = f σ π a = (1)(25 MPa) π (0.0005 m) = 0.99 MPa m 7-9

A ceramic part for a jet engine has a yield strength of 75,000 psi and a plane strain fracture toughness of 5,000 psi in . To be sure that the part does not fail, we plan to ensure that the maximum applied stress is only one third of the yield strength. We use a nondestructive test that will detect any internal flaws greater than 0.05 in. long.

Assuming that f = 1.4, does our nondestructive test have the required sensitivity?

Explain.

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Solution: The applied stress is s = (⅓)(75,000 psi) = 25,000 psi

a = (1/π )[( K Ic /( f σ ))]2 = (1/π )[5, 000psi in./((1.4)(25, 000 psi))]2 a = 0.0065 in.

The length of internal flaws is 2a = 0.013 in.

Our nondestructive test can detect flaws as small as 0.05 in. long, which is not smaller than the critical flaw size required for failure.

Thus our NDT test is not satisfactory.

7-10

A manufacturing process that unintentionally introduces cracks to the surface of a part was used to produce load-bearing components. The design requires that the component be able to withstand a stress of 450 MPa. The component failed catastrophically in service.

You are a failure analysis engineer who must determine whether the component failed due to an overload in service or flaws from the manufacturing process. The manufacturer claims that the components were polished to remove the cracks and inspected to ensure that no surface cracks were larger than 0.5 mm. The manufacturer believes the component failed due to operator error.

It has been independently verified that the 5-cm diameter part was subjected to a tensile load of 1 MN (106 N).

The material from which the component is made has a fracture toughness of

75 MPa m and an ultimate strength of 600 MPa. Assume external cracks for which f = 1.12. (a) Who is at fault for the component failure, the manufacturer or the operator? Show your work for both cases. (b) In addition to the analysis that you presented in (a), what features might you look for on the fracture surfaces to support your conclusion?

Solution: (a) The engineering stress imposed on the part is calculated as

S=

F 106 N = = 509 MPa , A0 π (0.05 m) 2 4

where S is the engineering stress, F is the applied load, and A0 is the initial cross– sectional area of the part. Thus, while the part may have yielded under the applied load, it should not have failed because the imposed stress was less than the 600 MPa ultimate tensile strength.

Given that the part failed at a stress of 509 MPa, which is above the yield stress but below the ultimate tensile strength of the material when no flaws are present, it seems that the part failed due to the propagation of a surface flaw.

The critical flaw size that would lead to failure can be determined from the following equation:

K IC = f σ π a , where KIC is the plane strain fracture toughness, f is the geometry factor, σ is the applied stress, and a is the crack length. Rearranging, 2

2

1K  1  75 MPa m  a =  IC  =   = 0.0055 m = 5.5 mm . π  f σ  π 1.12(509 MPa)  155 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

The critical flaw size that would lead to part failure at a stress of 509 MPa is 5.5 mm. Thus, the surface cracks must have been larger than the specified 0.5 mm, and the manufacturer is at fault. Whether the parts were actually polished should be investigated. (b) If examination of the part reveals that it experienced considerable deformation and necking prior to failure, then failure likely occurred due to tensile overload. The fracture surface may be dimpled in this case.

If the part failed due to the propagation of surface cracks, then a Chevron pattern may be visible.

7-27

List four measures that may be taken to increase the resistance to fatigue of a metal alloy.

Solution:

1.

2.

3.

4.

5.

7-28

A fatigue test was performed in which the mean stress was 70 MPa and the stress amplitude was 210 MPa. Compute the maximum and minimum stress.

Solution:

7-29

Improving surface finish – making surfaces smoother helps eliminate surface stress concentrations.

Shot peening – helps establish residual compressive stress layer.

Lower design stress – mean stress or stress amplitudes NDE – inspecting components of fatigue surface periodically to locate cracks in Stages I and II. Alternative material – seek material of higher fatigue strength.

σmean = 70 MPa =

140 MPa = σmax + σmin σmax = 140 MPa – σmin

σa = 210 MPa =

420 MPa = σmax - σmin 420 MPa = (140 – σmin) – σmin 420 MPa = 140 – 2σmin σmin = -140 MPa σmax = 140 MPa – (-140MPa) = 280 MPa

A cylindrical tool steel specimen that is 6 in. long and 0.25 in. in diameter rotates as a cantilever beam and is to be designed so that failure never occurs. Assuming that the maximum tensile and compressive stresses are equal, determine the maximum load that can be applied to the end of the beam. (See Figure 7–18.)

156 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Solution: The stress must be less than the endurance limit, 60,000 psi.

σ = 10.18 LF /d 3 or F = (endurance limit)d 3 /(10.18L) F = (60,000 psi)(0.25 in.)3 / [(10.18)(6 in.)] = 15.35 lb 7-30

A 2-cm-diameter, 20-cm-long bar of an acetal polymer (Figure 7–28) is loaded on one end and is expected to survive one million cycles of loading, with equal maximum tensile and compressive stresses, during its lifetime. What is the maximum permissible load that can be applied?

157 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Solution: From the figure, we find that the fatigue strength must be 22 MPa in order for the polymer to survive one million cycles. Thus, the maximum load is F = (fatigue strength)d3/(10.18L) F = (22 MPa)(20 mm)3 / [(10.18)(200 mm)] = 86.4 N 7-31

A nickel-based alloy (INCONEL® alloy 801) beam (Figure 7-29) is annealed and subjected to a cyclical load of 100 N with equal maximum and minimum stresses. The beam diameter is 5 mm, and the beam length is 50 mm. What is the expected lifetime of the beam in number of cycles to failure? If the cycle frequency is 10 Hz, what is the lifetime in hours?

Solution: The stress amplitude is given by

( ) 32 *100 N *50 mm = 407 MPa

σ = 32FL/ π d 3 =

π (5 mm)

3

According to Figure 7-29, the number of cycles to failure for this stress amplitude is approximately 9 ×10 5 .

7-32

A cyclical load of 1500 lb is to be exerted at the end of a 10-in.-long aluminum beam (Figure 7–18). The bar must survive for at least 106 cycles. What is the minimum diameter of the bar?

Solution: From the figure, we find that the fatigue strength must be 35,000 psi in order for the aluminum to survive 106 cycles. Thus, the minimum diameter of the bar is

d = 3 10.18LF /fatigue strength d = 3 (10.18)(10 in.)(1500 lb)/35, 000 psi = 1.634 in.

158 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7-33

A cylindrical acetal polymer bar 20 cm long and 1.5 cm in diameter is subjected to a vibrational load at one end of the bar at a frequency of 500 vibrations per minute with a load of 50 N. How many hours will the part survive before breaking? (See Figure 7–28.)

Solution: The stress acting on the polymer is σ = 10.18LF/d3 = (10.18)(200 mm)(50 N) / (15 mm)3 = 30.16 MPa From the figure, the fatigue life at 30.16 MPa is about 2 × 105 cycles.

Based on 500 cycles per minute, the life of the part is life = 2 × 105 cycles / (500 cycles/min) / (60 min/h) = 6.7 h 7-34

Suppose that we would like a part produced from the acetal polymer shown in Figure 7– 28 to survive for one million cycles under conditions that provide for equal compressive and tensile stresses. What is the fatigue strength, or maximum stress amplitude, required? What are the maximum stress, the minimum stress, and the mean stress on the part during its use? What effect would the frequency of the stress application have on your answers? Explain.

Solution: From the figure, the fatigue strength at one million cycles is 22 MPa. The maximum stress is +22 MPa, the minimum stress is –22 MPa, and the mean stress is 0 MPa. A high frequency will cause heating of the polymer. As the temperature of the polymer increases, the fatigue strength will decrease. If the applied stress is not reduced, then the polymer will fail in a shorter time.

7-35

The high-strength steel in Figure 7–20 is subjected to a stress alternating at 200 revolutions per minute between 600 MPa and 200 MPa (both tension). Calculate the growth rate of a surface crack when it reaches a length of 0.2 mm in both m/cycle and m/s. Assume that f = 1.0.

159 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Solution:

For the steel, C = 1.62 × 10–12 and n = 3.2. The change in the stress intensity factor ΔK is

∆K = f ∆σ π a = (1.0)(600 MPa – 200 MPa) π (0.0002 m) = 10.027 MPa m The crack growth rate is da/dN = 1.62 × 10–12(ΔK)3.2 da/dN = 1.62 × 10–12(10.027)3.2 = 2.590 × 10–9 m/cycle da/dt = (2.590 × 10–9 m/cycle)(200 cycles/min) / (60 s/min) da/dt = 8.63 × 10–9 m/s 7-36

The high strength steel in Figure 7–20, which has a critical fracture toughness of

80 MPa m , is subjected to an alternating stress varying from –900 MPa (compression) to +900 MPa (tension). It is to survive for 105 cycles before failure occurs.

Assume that f = 1. Calculate (a) the size of a surface crack required for failure to occur; and (b) the largest initial surface crack size that will permit this to happen.

Solution: (a) Only the tensile portion of the applied stress is considered in Δσ. Based on the applied stress of 900 MPa and the fracture toughness of 80 MPa m , the size of a surface crack required for failure to occur is

K = f σ π ac

or ac = (1/π )[K /( f σ )]2

ac = (1/π )[80 MPa m /((1)(900 MPa))]2 = 0.0025 m = 2.5 mm 160 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(b) The largest initial surface crack tolerable to prevent failure within 105 cycles is

N = 105 cycles =

2[(0.0025 m)(2–3.2)/ 2 – ai(2–3.2)/ 2 ] (2 – 3.2)(1.62 ×10 –12 )(1)3.2 (900)3.2 (π )3.2/ 2

105 =

2[36.41 – (ai ) –0.60 ] (–1.2)(1.62 × 10 –12 )(1)(2.84 × 109 )(6.244)

(ai)–0.60 = 1760 ai = 3.9 × 10–6 m = 0.0039 mm 7-37

The manufacturer of a product that is subjected to repetitive cycles has specified that the product should be removed from service when any crack reaches 15% of the critical crack length required to cause fracture.

Consider a crack that is initially 0.02 mm long in a material with a fracture toughness of

55 MPa m . The product is continuously cycled between compressive and tensile stresses of 300 MPa at a constant frequency. Assume external cracks for which f = 1.12.

The materials constants for these units are n = 3.4 and C = 2 × 10–11. (a) What is the critical crack length required to cause fracture? (b) How many cycles will cause product failure? (c) If the product is removed from service as specified by the manufacturer, how much of the useful life of the product remains?

Solution: (a) The critical flaw size that would lead to failure can be determined from the following equation:

K IC = f σ π a , where KIC is the plane strain fracture toughness, f is the geometry factor, σ is the applied stress, and a is the crack length.

Rearranging, 2

2

1K  1  55 MPa m  a =  IC  =   = 0.00853 m = 8.53 mm π  f σ  π 1.12(300 MPa)  The critical flaw size that would lead to part failure at a stress of 300 MPa is 8.53 mm. (b) The number of cycles that would cause product failure is given by

N=

2 (ac )(2− n )/ 2 − (ai )(2− n )/ 2  (2 − n)Cf n ∆σ nπ n / 2

,

where ai is the initial flaw size and ac is the flaw size required for fracture. The term ∆σ is the difference between the maximum and minimum cyclical stress; however, a crack does not propagate during compression. Therefore, ∆σ = 300 MPa. Substituting,

N=

2 (0.00853)(2–3.4)/2 − (0.00002)(2–3.4)/2  (2 – 3.4)(2 ×10 –11 )(1.12)3.4 (300)3.4π 3.4/2

= 50,366.

(c) If the product is removed from service when the critical crack length reaches 0.15(8.53 mm) = 1.28 mm, then the number of cycles remaining is calculated as in (b) using an initial crack length of 1.28 mm:

161 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

N=

2 (0.00853)(2–3.4)/2 − (0.00128)(2–3.4)/ 2  (2 – 3.4)(2 ×10 –11 )(1.12)3.4 (300)3.4π 3.4/ 2

= 2, 043,

and 2,043 / 50,366 or 4% of the useful product life remains.

7-38

A material containing cracks of initial length 0.010 mm is subjected to alternating tensile stresses of 25 and 125 MPa for 350,000 cycles. The material is then subjected to alternating tensile and compressive stresses of 250 MPa. How many of the larger stress amplitude cycles can be sustained before failure? The material has a fracture toughness of 25 MPa m and materials constants of n = 3.1 and C = 1.8 × 10–10 for these units.

Assume f = 1.0 for all cracks.

Solution: The critical flaw size that will lead to failure under a stress of 250 MPa can be determined from the following equation:

K IC = f σ π a , where KIC is the plane strain fracture toughness, f is the geometry factor, σ is the applied stress, and a is the crack length. Rearranging, 2

2

1K  1  25 MPa m  a =  IC  =   = 0.00318 m = 3.18 mm π  f σ  π 1.0(250 MPa)  The stress value of 250 MPa is of interest since this is the applied stress when the material fails. When subjected to the first round of cyclic testing, the crack grows to a length a1 given by

N=

2 (a1 )(2– n )/2 – (ai )(2– n )/2  (2 – n)Cf n ∆σ nπ n /2

,

where ai is the initial flaw size and N is the number of cycles. The term ∆σ is the difference between the maximum and minimum cyclical stress, which in this case is 125 – 25 = 100 MPa. Substituting,

350,000 =

2 (a1 )(2–3.1)/2 – (0.00001)(2–3.1)/ 2  (2 – 3.1)(1.8 ×10 –10 )(1.0)3.1 (100 MPa)3.1π 3.1/ 2

,

and solving for a1, a1 = 0.0000476 m = 0.0476 mm.

Then the part is subjected to tensile and compressive cyclic stresses of 250 MPa. The term ∆σ is the difference between the maximum and minimum cyclical stress; however, a crack does not propagate under compression. Therefore, ∆σ = 250 MPa. Solving for N with ai = 0.0000476 m and ac = 0.00318 m,

N=

2 (ac )(

2– n ) / 2

– (ai )(2– n )/ 2 

(2 – n)Cf n ∆σ nπ n /2

=

2  (0.00318)(2–3.1)/ 2 – (0.0000476)(2–3.1)/ 2  (2 – 3.1)(1.8 × 10 –10 )(1.0)3.1 (250 MPa)3.1π 3.1/ 2 = 13,560 cycles.

162 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7-39

The acrylic polymer from which Figure 7–30 was obtained has a critical fracture toughness of 2 MPa m . It is subjected to a stress alternating between –10 and +10 MPa. Calculate the growth rate of a surface crack when it reaches a length of 5 × 10–6 m if f = 1.0.

Solution:

Δσ = 10 MPa – 0 = 10 MPa,

since the crack doesn’t propagate for compressive loads.

∆K = f ∆σ π a = (1.0)(10 MPa) π (5 × 10 –6 m) = 0.04 MPa m From the graph, da/dN = 1.5 × 10–7 m/cycle 7-40

Calculate the constants C and n in Equation 7–18 for the crack growth rate of an acrylic polymer. (See Figure 7–30.)

Solution: Let’s pick two points on the graph:

da /dN = 2 ×10 –6 m/cycle when ∆K = 0.1 MPa m da /dN = 1× 10 –7 m/cycle when ∆K = 0.037 MPa m

2 ×10 –6 C (0.1) n = 1×10 –7 C (0.037) n 20 = (0.1 / 0.037)n = (2.703)n

ln(20) = n ln(2.703) 2.9957 = 0.994n n = 3.01

2 ×10 –6 = C (0.1)3.01 = 0.000970C C = 2.061×10 –3 7-41

The acrylic polymer from which Figure 7–30 was obtained is subjected to an alternating stress between 15 MPa and 0 MPa. The largest surface cracks initially detected by nondestructive testing are 0.001 mm in length. If the critical fracture toughness of the polymer is 2 MPa m , calculate the number of cycles required before failure occurs.

Let f = 1.0. (Hint: Use the results of Problem 7–32.)

Solution: From Problem 7–32, C = 2.061 × 10–3 and n = 3.01 The critical flaw size ac is

ac = (1/π )[ K Ic /( f σ )]2 = (1/π )[2 MPa m /((1.0)(15 MPa))]2 ac = 0.00566 m = 5.66 mm Then

N=

2[(0.00566 m)(2–3.01)/ 2 − (0.000001 m)(2–3.01)/ 2 ] (2 – 3.01)(2.061×10 –3 )(1.0)3.01 (15 MPa)3.01 (π )3.01/ 2

N = 52 cycles 7-43

Verify that integration of da/dN = C(ΔK)n will give Equation 7–20.

Solution: Given that

da = C ∆K n , dN and

∆K = f ∆σ π a = f ∆σ (π a )1/2 ,

163 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

then

dN =

da Cf ∆σ nπ n /2 a n /2 n

Integrating, N

da

ac

∫ dN = ∫ Cf ∆σ π a o

n

ai

n

n /2

n /2

,

where ac is the critical crack length at which failure occurs and ai is the initial crack length.

Recall that

a1+ p .

1+ p n If p = − , then 2 p ∫ a da =

ac

ac

i

i

da a1– n /2 2a (2– n )/ 2 2 = = = [ac(2– n )/2 – ai(2– n )/ 2 ] , ∫ai a n /2 (1 – n /2) (2 – n) a (2 – n) a ac

such that

N= 7-50

2[ac(2– n )/2 − ai(2– n ) /2] .

(2 – n)Cf n ∆σ nπ n /2

Approximate the temperature at which creep deformation becomes an important consideration for each of the following metals: tin, molybdenum, iron, gold, zinc and chromium.

Convert the melting temperature from °C to K, then take 0.4 of that value and convert back to °C. The critical values are listed in the table below.

Metal Melting Temperature, °C Critical Temperature, °C Tin (Sn) 232 -71 Molybdenum (Mo) 2623 885 Iron (Fe) 1536 451 Gold (Au) 1064 262 Zinc (Zn) 420 4 Chromium 1907 599

Solution:

Notice that tin and zinc can experience creep at room temperature since their critical temperatures are below 25°C. 7-51

The activation energy for self-diffusion in copper is 49,300 cal/mol. A copper specimen creeps at 0.002

in./in. when a stress of 15,000 psi is applied at 600°C. If the creep rate h

of copper is dependent on self-diffusion, determine the creep rate if the temperature is 800°C. Solution:

The creep rate is governed by an Arrhenius relationship of the form: rate = A exp(–Q/RT). From the information given,

164 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

x A exp[–49,300/(1.987)/(800 + 273)] 9.07 ×10 –11 = = 0.002 A exp[–49,300/(1.987)/(600 + 273)] 4.54 ×10 –13 in./in. x = (0.002)[9.07 × 10 –11 /(4.54 ×10 –13 )] = 0.4 . h 7-52

When a stress of 20,000 psi is applied to a material heated to 900°C, rupture occurs in 25,000 h. If the activation energy for rupture is 35,000 cal/mol, determine the rupture time if the temperature is reduced to 800°C.

Solution: The rupture time is related to temperature by an Arrhenius relationship of the form tr = A exp(+Q/RT); the argument of the exponential is positive because the rupture time is inversely related to the rate. From the information given

tr A exp [35, 000/(1.987)/(800 + 273)] 1.35 ×107 = = 25, 000 h A exp [35, 000/(1.987)/(900 + 273)] 3.32 ×106 tr = (25,000)[1.35 × 107 / (3.32 × 106)] = 101,329 h 7-53

The following data were obtained from a creep test for a specimen having a gage length of 2.0 in. and an initial diameter of 0.6 in. The initial stress applied to the material is 10,000 psi. The diameter of the specimen after fracture is 0.52 in.

Length Between Gage Marks (in.) Time (h) Strain (in./in.)

2.004 0 0.002 2.010 100 0.005 2.020 200 0.010 2.030 400 0.015 2.045 1000 0.0225 2.075 2000 0.0375 2.135 4000 0.0675 2.193 6000 0.0965 2.230 7000 0.115 2.300 8000 (fracture) 0.15 Determine (a) the load applied to the specimen during the test; (b) the approximate length of time during which linear creep occurs; (c) the creep rate

in./in. and in %/h; h

and (d) the true stress acting on the specimen at the time of rupture.

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Solution:

(a) The load is F = σA = (10,000 psi)(π/4)(0.6 in.)2 = 2827 lb. (b) The plot of strain versus time is linear between approximately 500 and 6000 hours, or a total of 5500 hours. (c) From the graph, the strain rate is the slope of the linear portion of the curve.

∆ε /∆t =

0.095 – 0.03 in./in. = 1.44 ×10 –5 = 1.44 × 10 –3 %/h 6000 – 1500 h

(d) At the time of rupture, the force is still 2827 lb, but the diameter is reduced to 0.52 in. The true stress is therefore σt = F/A = 2827 lb / [(π/4)(0.52 in.)2] = 13,312 psi 7-54

A stainless steel is held at 705°C under different loads. The following data are obtained: Applied Stress (MPa) Rupture Time (h) Creep Rate (%/h) 106.9 1200 0.022 128.2 710 0.068 147.5 300 0.201 160.0 110 0.332 Determine the exponents n and m in Equations 7–22 and 7–23 that describe the dependence of creep rate and rupture time on applied stress.

Solution: Equations 7–22 and 7–23 have the form Creep Rate = Aσn and Rupture Time = Bσm, where A and B depend on temperature but not stress, σ is the applied stress, and n and m are constants for the material. Taking the natural logarithm of both sides of both equations produces ln (Creep Rate) = ln A + nln σ and ln (Rupture Time) = ln B + m ln σ A plot of the natural logarithm of the creep rate versus the natural logarithm of the applied stress and a plot of the natural logarithm of the rupture time versus the

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natural logarithm of the applied stress are shown below. The slopes are the values of n and m, respectively. The value of n is 6.82, and the value of m is –5.7.

7-55

Using the data in Figure 7–26 for an iron-chromium-nickel alloy, determine the activation energy Qr and the constant m for rupture in the temperature range 980 to 1090°C.

167 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Solution:

The appropriate equation is tr = Kσmexp(Qr/RT).

From Figure 7–26(a), we can determine the rupture time versus temperature for a fixed stress, say σ = 1000 psi: tr = 2,400 h at 1090°C = 1363 K tr = 14,000 h at 1040°C = 1313 K tr = 100,000 h at 980°C = 1253 K From this data, the equation becomes tr = K′exp(Qr/RT) and we can find Qr by simultaneous equations or graphically.

Qr = 117,000 cal/mol

168 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

We can also determine the rupture time versus applied stress for a constant temperature, say 1090°C: tr = 105 h for σ = 450 psi tr = 104 h for σ = 800 psi 3 tr = 10 h for σ = 1200 psi tr = 102 h for σ = 2100 psi With this approach, the equation becomes tr = K″σm, where m is obtained graphically or by simultaneous equations: m = 3.9

7-56

A 1-in.-diameter bar of an iron-chromium-nickel alloy is subjected to a load of 2500 lb.

How many days will the bar survive without rupturing at 980°C? [See Figure 7–26(a).]

Solution: The stress is σ = F/A = 2500 lb / [(π/4)(1 in.)2] = 3183 psi From the graph, the rupture time is 700 h / (24 h/day) = 29 days 7-57

A 5 mm × 20 mm bar of an iron-chromium-nickel alloy is to operate at 1040°C for 10 years without rupturing. What is the maximum load that can be applied? [See Figure 7– 26(a).]

Solution: The operating time is (10 years)(365 days/year)(24 h/day) = 87,600 h From the graph, the stress must be less than 500 psi. The load is then F = σA = (500 psi)(5 mm/25.4 mm/in.)[(20 mm/(25.4 mm/in.)] = 77.5 lb 7-58

An iron-chromium-nickel alloy is to withstand a load of 1500 lb at 760°C for 6 years.

Calculate the minimum diameter of the bar. [See Figure 7–26(a).]

Solution: The operating time is (6 years)(365 days/year)(24 h/day) = 52,560 h From the graph, the stress must be less than 7000 psi. The minimum diameter of the bar is then d = (4/π )( F /σ ) = (4/π )(1500 lb/7000 psi) = 0.52 in.

7-59

A 1.2-in.-diameter bar of an iron-chromium-nickel alloy is to operate for 5 years under a load of 4000 lb. What is the maximum operating temperature? [See Figure 7–26(a).]

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Solution: The operating time is (5 years)(365 days/year)(24 h/day) = 43,800 h The stress is σ = F/A = 4000 lb / [(π/4)(1.2 in.)2] = 3537 psi From the figure, the temperature must be below 850°C in order for the bar to survive five years at 3537 psi.

7-60

A nickel-based alloy component (INCONEL alloy 601) is designed to withstand a service stress of 4000 psi at a temperature of 925°C. Determine the temperature at which the component should be operated in order to double the expected life of the component if the applied stress is to remain the same. Refer to Figure 7-31.

Solution: According to Figure 7-31, the Larsen-Miller parameter for a stress of 4 ksi (4000 psi) is approximately 37. Using the equation given in the figure and solving for the expected life t at 925˚C,

   9  9 37 =  492 + T  (14.883+ Log[t]) ×10−3 =  492 + ( 925)(14.883+ Log[t]) ×10−3    5  5

t = 9.7 hours To double the expected life at the same applied stress, the temperature must be lowered to

  9  9  37 =  492 + T  (14.883+ Log[t]) ×10−3 =  492 + T (14.883+ Log[19.4]) ×10−3   5  5 

T = 878˚C Please note that the term “Log” in the problem statement was meant to signify the natural logarithm.

170 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7-61

For a temperature of 300°C and a lifetime of three years, what is the maximum stress that the nickel-based component of Figure 7-31 can withstand?

Solution: Using the equation for the Larsen-Miller parameter in Figure 7-31 at a temperature of 300˚C for a lifetime of three years,

  9 365 days 24 hours  L.M. =  492 + ( 300 )14.883+ Log[3 years * * ] ×10 −3 = 25.9   5 year day  The maximum stress that a component can withstand with a Larsen-Miller parameter of 25.9 is 50 ksi or 345 MPa. Please note that the term “Log” in the problem statement was meant to signify the natural logarithm.

7-62

A 1 in. × 2 in. ductile cast-iron bar must operate for 9 years at 650°C. What is the maximum load that can be applied? [See Figure 7–27(b).]

Solution: The operating time is (9 years)(365 days/year)(24 h/day) = 78,840 h.

The temperature is 650 + 273 = 923 K LM = (923/1000)[36 + 0.78 ln(78,840)] = 41.35 From the graph, the stress must be no more than about 1000 psi. The load is then F = σA = (1000 psi)(2 in.2) = 2000 lb 7-63

A ductile cast-iron bar is to operate at a stress of 6000 psi for 1 year. What is the maximum allowable temperature? [See Figure 7–27(b).]

Solution: The operating time is (1 year)(365 days/year)(24 h/day) = 8760 h From the graph, the Larson-Miller parameter must be 34.4 at a stress of 6000 psi.

Thus 34.4 = (T / 1000)[36 + 0.78 ln(8760)] = 0.043T T = 800 K = 527°C

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Chapter 8: Strain Hardening and Annealing

8-5

What is the difference between extrusion and drawing?

Solution: Extrusion – the force is exerted on the larger original cross-section of the work-piece that is pushed through a die. Drawing – the force is applied on the smaller cross-section that has already passed through the die.

8-6

Consider the tensile stress strain curves in Figure 8–20 labeled 1 and 2 and answer the following questions. These curves are typical of metals. Consider each part as a separate question that has no relationship to previous parts of the question.

(a) Which material has the larger work hardening exponent? How do you know? (b) Samples 1 and 2 are identical except that they were tested at different strain rates.

Which sample was tested at the higher strain rate? How do you know? (c) Assume that the two stress strain curves represent successive tests of the same sample. The sample was loaded, then unloaded before necking began, and then the sample was reloaded. Which sample represents the first test: 1 or 2? How do you know?

Solution: (a) Sample 1 has the larger work hardening exponent. The slope of the stress versus strain curve is higher for 1 than 2, indicating a larger work hardening exponent. (b) Sample 1 was tested at the higher strain rate. At higher strain rates, metals typically exhibit higher strengths. Sample 1 is stronger than Sample 2; therefore, all other factors being identical, Sample 1 was tested at a higher strain rate. (c) Sample 2 represents the first test because it has lower strength and greater ductility. As the sample was loaded the first time, it was work–hardened, imparting higher strength and lower ductility on reloading. Since Sample 2 is weaker and more ductile, it must have been the first test.

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8-7

Figure 8–21 is a plot of true stress versus true strain for a metal. For total imposed strains of ε = 0.1, 0.2, 0.3 and 0.4, determine the elastic and plastic components of the strain. The modulus of elasticity of the metal is 100 GPa.

Solution: The total true strain εtotal is the sum of elastic and plastic components according to εtotal = εelastic + εplastic, where εelastic is the elastic component of the strain and εplastic is the plastic component. The elastic component is given by Hooke’s Law:

ε plastic =

σ E

,

where σ is the true stress and E is the elastic modulus. The table below shows the components of elastic and plastic true strain for the total true strain specified. The stress values are read from the stress strain diagram. Answers will vary slightly based on the selected stress values. εtotal σ(MPa) εplastic = εtotal – εelastic σ

ε elastic =

0.1 0.2 0.3 0.4 8-9

185 280 370 455

0.00185 0.00280 0.00370 0.00455

E 0.0982 0.1972 0.2963 0.3955

An annealed copper sample experiences a true strain of 0.07 caused by a true stress of 164 MPa. If this is in the plastic deformation region and the strain-hardening exponent is 0.44, what is the strength coefficient?

Solution: Starting with Equation 8-1: =

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=

=

164 MPa 0.07 .

= 528 MPa 8-10

A 0.505 in.-diameter metal bar with a 2-in. gage length l0 is subjected to a tensile test.

The following measurements are made in the plastic region: Force (lb) Change in Gage length (in.) (∆l) Diameter (in.)

27,500 0.2103 0.4800 27,000 0.4428 0.4566 25,700 0.6997 0.4343 Determine the strain-hardening exponent for the metal. Is the metal most likely to be FCC, BCC, or HCP? Explain.

Solution: Force (lb)

27,500 27,000 25,700

σ t = Kε tn

Gage length (in.)

2.2103 2.4428 2.6997

Diameter (in.)

True stress (psi)

0.4800 0.4566 0.4343

151,970 164,893 173,486

True strain (in./in.)

0.100 0.200 0.300

or ln σ = ln K + n ln ε

ln(151,970) = ln K + n ln(0.1) ln(173,486) = ln K + n ln(0.3) n = 0.12

11.9314 = ln K – n (2.3026) 12.0639 = ln K – n (1.2040) –0.1325 = –1.0986 n which is in the range of BCC metals

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8-11

A 1.33-cm-diameter metal bar with a 3-cm gage length (l0) is subjected to a tensile test.

The following measurements are made in the plastic region: Force (N) Change in Gage length (cm) (∆l) Diameter (cm) 16,240 0.6642 1.2028 19,066 1.4754 1.0884 19,273 2.4663 0.9848 Determine the strain-hardening coefficient for the metal. Is the metal most likely to be FCC, BCC, or HCP? Explain.

Solution: Force (N) 16,240 19,066 19,273

Gage length (cm) 3.6642 4.4754 5.4663

Diameter (mm) 12.028 10.884 9.848

True stress (MPa) 143 205 249

True strain (cm/cm) 0.200 0.400 0.600

σt = Kεtn

ln 143 = ln K + n ln 0.2 ln 249 = ln K + n ln 0.6 (4.962 – 5.517) = n(–1.609 + 0.511) n = 0.51 A strain hardening coefficient of 0.51 is typical of FCC metals.

8-12

A true stress-true strain curve is shown in Figure 8–22. Determine the strain hardening exponent for the metal.

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Solution: σt = Kεtn εt σt 0.05 in./in.

60,000 psi 0.10 in./in.

66,000 psi 0.20 in./in.

74,000 psi 0.30 in./in.

78,000 psi 0.40 in./in.

81,000 psi From graph: K = 93,000 psi n = 0.15

8-13

Figure 8–23 shows a plot of the natural logarithm of the true stress versus the natural logarithm of the true strain for a Cu-30% Zn sample tested in tension. Only the plastic portion of the stress strain curve is shown. Determine the strength coefficient K and the work-hardening exponent n.

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Solution: Power law strain–hardening is described by the equation σ = Kεn, where σ is the true stress, K is the strength coefficient, ε is the true strain, and n is the strain–hardening exponent. Taking the natural logarithm of both sides: ln (σ) = ln(K) + n ln (ε).

A linear fit of this equation to the data produces ln (σ) = 20.569 + 0.55645 ln(ε).

Therefore, n = 0.56, and ln(K) = 20.569, such that K = 857 MPa. Alternatively, two points on the line can be used to calculate the slope. When ln (ε) = –1, ln (σ) = 20.0, and when ln (ε) = –2, ln (σ) = 19.45. Thus the strain–hardening exponent is given by

n=

20.0 − 19.45 = 0.55 , −1 − (−2)

which is similar to the result from the curve fit. Using the data point ln (ε) = –1, ln (σ) = 20.0, 20.0 = ln (K) + 0.55(–1), or ln (K) = 20.55, such that K = 841 MPa, which is again similar to the result from the curve fit.

8-14

A Cu-30% Zn alloy tensile bar has a strain-hardening coefficient of 0.50. The bar, which has an initial diameter of 1 cm and an initial gage length of 3 cm, fails at an engineering stress of 120 MPa. At the moment of fracture, the gage length is 3.5 cm and the diameter is 0.926 cm. No necking occurred. Calculate the true stress when the true strain is 0.05 cm/cm.

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Solution: ε = ln(ℓf/ℓo) = ln(3.5/3.0) = 0.154

S = 120 MPa =

F (π /4)(10 mm)2

F = 9425 N

9425 N = 139.95 MPa (π /4)(9.26 mm) 2 σ = K (0.154)0.5 = 139.95 MPa or K = 356.6

σ=

The true stress at ε = 0.05 cm/cm is

σ = 356.6(0.05)0.5 or σ = 79.7 MPa 8-15

What would a strain hardening exponent of zero mean mathematically and practically?

Solution: Mathematically, raising a real number (like the strain exhibited in a material) to zero sets it equal to 1. This means the true stress of the material would equal K. Practically it means the stress would not be a function of strain. It does not mean the material would never strain, only that the strain would not be a function of the stress or vice versa.

This is not physically impossible, the same thing happens when the true strain is equal to 1.

8-23

A 0.25-in.-thick copper plate is to be cold worked 63%. Find the final thickness.

Solution: (See Figure 8–7.)

63 = 8-24

0.25

× 100% or t f = 0.0925 in.

A 0.25-in.-diameter copper bar is to be cold worked 63% in tension. Find the final diameter.

Solution:

63 = 8-25

0.25 − t f

(0.25)2 − d 2f (0.25) 2

× 100% or d 2f = 0.025 or d f = 0.152 in.

A 2-in.-diameter copper rod is reduced to a 1.5 in. diameter, then reduced again to a final diameter of 1 in. In a second case, the 2-in.-diameter rod is reduced in one step from 2 in. to a 1 in. diameter. Calculate the % CW for both cases.

Solution:

% CW =

(2) 2 − (1) 2 ×100 = 75% in both cases (2) 2

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8-26

If a 25-mm diameter steel bar is cold extruded first to 22 mm, then from 22 mm to a final diameter of 20 mm, determine (a) the amount of cold work for each step, and (b) compare the sum of the amounts of cold work for each step to the cold work process from 25 mm directly to 20 mm. (See the next problem to fully understand your solution.)

Solution: a.) For the two step process (25-mm to 22-mm, then 22-mm to 20-mm) the percent cold work for each step will be:

[ ]

=

"# "" = 0.2256 "#

=

"" " = 0.1736 ""

(25-mm to 22-mm): CW1 =

(22-mm to 20-mm): CW2 =

[ ]

By adding both cold work amounts together you will have the total amount of cold working performed in the two-step process.

CW1 + CW2 = 0.2256 + 0.1736 = 0.3992 X 100% = 39.92% b.) For the single step process (25-mm to 20-mm) the amount of cold work will be: (25-mm to 20-mm): CW1 =

[ ]

=

"# " = 0.36 X 100% = 36% "#

It is clear to see that the amount of cold work is not the same even though the initial and final diameters are the same.

8-27

If the amount of extrusion in the previous problem were expressed in true strains, (a) determine the true strain for each of the extrusion steps, and (b) compare the sum of the true strains for each extrusion to the true strain from 25 mm to 20 mm.

Solution: a.) Using the true strain equation: ε = ln

$ %

= ln = ln = 2 ln ( $ %

"#

""

(25-mm to 22-mm): ε1 = 2 ln ( = 2 ln ("" = 0.2557

(22-mm to 20-mm): ε2 = 2 ln ( = 2 ln (" = 0.1906 By adding both true strain amounts together you will have the total true strain amount to be ε1+ ε2 = 0.2557 + 0.1906 = 0.4463 b.) For the single step process (25-mm to 20-mm) the amount of true strain will be:

"#

(25-mm to 20-mm): ε = 2 ln ( = 2 ln (" = 0.4463

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It is shown that the total strain from the two-step process is the same as the single step process.

8-28

A 2-mm-diameter, pure copper wire is reduced to 1.75-mm diameter by cold working.

What is its tensile strength? (See Figure 8-6.)

Solution: Referring to Equation 8-5, we need to calculate the cross sectional areas of the wire before and after the operation. & = '( "

& = ' 2 mm " & = ' 1.75 mm " Inserting into Equation 8-5: %CW = -

& − & / × 100 &

' 2 mm " − ' 1.75 mm " %CW = 1 2 × 100 ' 2 mm " Factoring and cancelling the π’s. %CW = 1

2 mm " − 1.75 mm " 2 × 100 2 mm "

4 mm" − 3.0625 mm" %CW = 1 2 × 100 4 mm" %CW = 23.4 %

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Repairing to Figure 8-6, we see that SUTS is about 59 ksi.

8-29

A 3105 aluminum plate is reduced from 1.75 in. to 1.15 in. Determine the final properties of the plate. Note 3105 designates a special composition of aluminum alloy.

(See Figure 8–24.)

Solution:

8-30

1.75 − 1.15 × 100 = 34.3% 1.75 TS = 25 ksi YS = 21 ksi % elongation = 6% % CW =

A Cu-30% Zn brass bar is reduced from a 1 in. diameter to a 0.45 in. diameter.

Determine the final properties of the bar. (See Figure 8–25.)

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Solution:

(1)2 − (0.42) 2 ×100 = 79.75% (1) 2 TS = 105 ksi YS = 68 ksi % elongation = 1% % CW =

8-31

A 3105 aluminum bar is reduced from a 1 in. diameter, to a 0.8 in. diameter, to a 0.6 in. diameter, to a final 0.4 in. diameter. Determine the % CW and the properties after each step of the process. Calculate the total percent cold work. Note 3105 designates a special composition of aluminum alloy. (See Figure 8–24.)

Solution: If we calculated the percent deformation in each step separately, we would find that 36% deformation is required to go from 1 in. to 0.8 in.

The deformation from 0.8 in. to 0.6 in. (using 0.8 in. as the initial diameter) is 43.75%, and the deformation from 0.6 in. to 0.4 in. (using 0.6 in. as the initial diameter) is 55.6%. If we added these three deformations, the total would be 135.35%. This would not be correct.

Instead, we must always use the original 1 in. diameter as our starting point. The following table summarizes the actual deformation and properties after each step.

2

2

(1) − (0.8) = 100 = 36% (1) 2 (1) 2 − (0.6)2 = 100 = 64% (1) 2 (1) 2 − (0.4)2 = 100 = 84% (1) 2

TS ksi YS ksi % elongation 25 21 6 28

25

4

30

28

3

The total percent cold work is actually 84%, not 135.35%.

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8-32

We want a copper bar to have a tensile strength of at least 70,000 psi and a final diameter of 0.375 in. What is the minimum diameter of the original bar? (See Figure 8– 6.)

Solution: % CW ≥ 50% to achieve the minimum tensile strength

50 =

d o2 − (0.375) 2 × 100 d o2

0.5 d o2 = 0.140625 or 8-33

d o = 0.53 in.

We want a Cu-30% Zn brass plate originally 1.2 in. thick to have a yield strength greater than 50,000 psi and a % elongation of at least 10%. What range of final thicknesses must be obtained? (See Figure 8–25.)

Solution: YS > 50,000 psi requires CW > 20% % E > 10% requires CW < 37%

1.2 − t f 1.2

= 0.20

1.2 − t f 1.2

= 0.37

tf = 0.96 in. tf = 0.76 in. tf = 0.76 to 0.96 in.

8-34

A rolled Cu-30 wt% Zn plate 0.500-in.-thick has a 2% elongation as-received by the supplier. The desired specifications for the final sheet are a thickness of 0.125 in., minimum tensile strength of 70,000 psi, and minimum elongation of 7%. Assume that the rolling is conducted so that the width of the sheet is unchanged. Specify all steps in the procedure, including heat treatments. (See Figure 8-25.)

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Solution: Since the material already has a 2% elongation, the first step would be to anneal to rid the material of any previous cold-work.

The desired properties call for a minimum tensile strength of 70,000 psi (70 ksi) which looking at the figure provided gives a %CW of 30% and 45% for a minimum of 7% elongation. The best solution is to choose 37%CW which is between both values and satisfies the desired properties.

The intermediate thickness for which 37% cold-work would give the desired properties 4 .5"# would be 0.37 = 4 = 0.198 inches The procedure would be as follows:

Anneal 8-35

CW to t=0.19"

Anneal

CW to tf = 0.125"

A piece of cold-worked Cu-30% Zn brass is identified, but the extent to which it has been cold-worked is unknown. Tensile testing indicates that the tensile strength is 60 ksi with an elongation of 30%. Determine the amount of cold-work and tensile strength of the brass in the annealed condition. (See Figure 8-25).

Solution: A 60 ksi tensile strength and 30% elongation gives an 18% CW condition. The tensile strength in the annealed condition is 42 ksi.

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8-36

We want a copper sheet to have at least 50,000 psi yield strength and at least 10% elongation, with a final thickness of 0.12 in. What range of original thickness must be used? (See Figure 8–6.)

Solution: YS > 50 ksi requires CW ≥ 25% % E > 10% requires CW ≤ 30%

to − 0.12 = 0.25 to

to − 0.12 = 0.30 to

to = 0.16 in. to = 0.17 in. to = 0.16 to 0.17 in.

8-37

A 3105 aluminum plate previously cold worked 20% is 2 in. thick. It is then cold worked further to 1.3 in. Calculate the total percent cold work and determine the final properties of the plate. (Note: 3105 designates a special composition of aluminum alloy.) (See Figure 8–24.)

Solution: The original thickness (before the 20% cold work) must have been to = 2.5 in. to − 2

to

= 0.20

The total cold work is then based on the prior 2.5 in. thickness: TS = 26 ksi 2.5 − 1.3 CW = × 100% = 48% YS = 23 ksi 2.5 %e = 4% 8-38

An aluminum-lithium (Al-Li) strap 0.2 in. thick and 2 in. wide is to be cut from a rolled sheet as described in Figure 8–9. The strap must be able to support a 35,000 lb load without plastic deformation. Determine the range of orientations from which the strap can be cut from the rolled sheet.

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Solution:

σ=

35, 000 (0.2)(2)

≥ 70, 000 psi

The properties can be obtained at angles of 0 to 20° from the rolling direction of the sheet.

8-49

What is cold working? What features does a material have after it has been coldworked?

Solution: Cold-working is the plastic deformation of a metal below its recrystallization temperature. A cold-worked metal exhibits higher yield strength than an annealed metal and has very good surface finish and very close dimensional tolerances as compared to a hot-worked part. The latter arises from having passed through working dies with controlled tolerances.

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8-53

We want to draw a 0.3-in.-diameter copper wire having a yield strength of 20,000 psi into 0.25-in. diameter wire. (a) Find the draw force, assuming no friction; (b) Will the drawn wire break during the drawing process? Show why. (See Figure 8–6.)

Solution: (a) Before drawing (0% CW), the yield strength is 20 ksi = 20,000 psi. which gives YS = 53,000 psi in the (0.3)2 − (0.25) 2 = CW = 30.6% drawn wire 2

(0.3)

The force needed to draw the original wire is

20, 000 psi = F / [(π /4)(0.3)2 ] or

F = 1414 lb

(b) The stress acting on the drawn wire is: S = 1414/[(π/4)(0.25)2] = 28,800 psi < 53,000 psi Since the actual stress (28,800 psi) acting on the drawn wire is less than the yield strength (53,000 psi) of the drawn wire, the wire will not break during manufacturing.

8-54

A 3105 aluminum wire is to be drawn to give a 1 mm diameter wire having a yield strength of 20,000 psi. Note 3105 designates a special composition of aluminum alloy. (a) Find the original diameter of the wire; (b) calculate the draw force required; and (c) determine whether the as-drawn wire will break during the process. (See Figure 8–24.)

Solution: (a) We need to cold work 25% to obtain the required yield strength:

d o2 − 12 = 0.25 d o = 1/0.75 = 1.1547 mm = 0.04546 in. d o2 (b) The initial yield strength of the wire (with 0% cold work) is 8000 psi, so the force required to deform the initial wire is

F = 8000[(π /4)(0.04546) 2 ] = 12.98 lb (c) The stress acting on the drawn wire (which has a smaller diameter but is subjected to the same drawing force) is

S=

12.98 lb = 10, 662 psi < 20, 000 psi (π /4)(1 mm/25.4 mm/in) 2

Since the actual stress is less than the 20,000 psi yield strength of the drawn wire, the process will be successful and the wire will not break.

8-55

The electrical conductance of a Cu-40 wt% Zn wire is measured. How much less conductive is it than a 60% cold-worked pure Cu wire? (See Figure 8-11.)

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Solution: Referring to Figure 8-11 (b) we read the conductivity of the alloy as being about 18 μS/m.

Referring to Figure 8-11 (a) the conductivity of the worked wire is about 57 μS/m.

A simple division gives us a factor of 3.2× less conductive.

8-64

A titanium alloy contains a very fine dispersion of Er2O3 particles. What will be the effect of these particles on the grain growth temperature and the size of the grains at any particular annealing temperature? Explain.

Solution: These particles, by helping pin the grain boundaries, will increase the grain growth temperature and decrease the grain size.

8-68

Samples of cartridge brass (Cu-30% Zn) were cold rolled and then annealed for one hour. The data shown in the table below were obtained.

Annealing Temperature (°C) Grain Size (μm) Yield Strength (MPa) 400 15 159 500 23 138 600 53 124 700 140 62 800 505 48 (a) Plot the yield strength and grain size as a function of annealing temperature on the same graph. Use two vertical axes, one for hardness and one for grain size. (b) For each temperature, state which stages of the annealing process occurred, Justify your answers by referring to features of the plot.

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Solution: (a) The plot below shows the grain size and yield strength as a function of annealing temperature.

(b) The yield strength of samples annealed at 500°C is less than the yield strength of samples annealed at 400°C. Thus the recrystallization temperature must be less than 500°C. At temperatures above 500°C, grain growth occurs.

8-69

The following data were obtained when a cold-worked metal was annealed. (a) Estimate the recovery, recrystallization, and grain growth temperatures; (b) recommend a suitable temperature for a stress-relief heat treatment; (c) recommend a suitable temperature for a hot-working process; and (d) estimate the melting temperature of the alloy.

Annealing Temperature (oC) 400 500 600 700 800 900 1000 1100

Electrical Conductivity (ohm–1 · cm–1) 3.04 × 105 3.05 × 105 3.36 × 105 3.45 × 105 3.46 × 105 3.46 × 105 3.47 × 105 3.47 × 105

Yield Strength (MPa) 86 85 84 83 52 47 44 42

Grain Size (mm) 0.10 0.10 0.10 0.098 0.030 0.031 0.070 0.120

Solution: (a) recovery temperature ≈ 550°C recrystallization temperature ≅ 750°C grain growth temperature ≅ 950°C (b) Stress relief temperature = 700°C (c) Hot working temperature = 900°C (d) 0.4 Tmp ≅ 750°C = 1023 K Tmp ≅ 1023 / 0.4 = 2558 K = 2285°C

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8-70

The following data were obtained when a cold-worked metallic material was annealed. (a) Estimate the recovery, recrystallization, and grain growth temperatures; (b) recommend a suitable temperature for obtaining a high-strength, high-electrical conductivity wire; (c) recommend a suitable temperature for a hot-working process; and (d) estimate the melting temperature of the alloy.

Annealing Temperature (oC) 250 275 300 325 350 375 400 425

Residual Stresses (psi) 21,000 21,000 5,000 0 0 0 0 0

Tensile Strength (psi) 52,000 52,000 52,000 52,000 34,000 30,000 27,000 25,000

Grain Size (in.)

0.0030 0.0030 0.0030 0.0030 0.0010 0.0010 0.0035 0.0072

Solution: (a) recovery temperature ≈ 280°C recrystallization temperature ≅ 330°C grain growth temperature ≅ 380°C (b) For a high strength, high conductivity wire, we want to heat into the recovery range. A suitable temperature might be 320°C. (c) Hot working temperature = 375°C (d) 0.4 Tmp ≅ 330°C = 603 K

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Tmp ≅ 603 / 0.4 = 1508 K = 1235°C

8-78

Plot the recrystallization temperatures in Table 8-4 as a function of the melting temperatures. Are they proportional and, if they are, what is the proportionality constant? Make sure to convert the temperatures to kelvin first.

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Solution: The graph:

Recrystallization temperature (K)

Recrystallization temperature as a function of melting temperature for some pure elements 1600 1400 1200 1000 800 600

y = 0.3903x + 47.613 R² = 0.9716

400 200 0 0

500

1000

1500

2000

2500

3000

3500

4000

Melting temperature (K)

As mentioned in the text, there is an apparent proportionality. The proportionality constant relating the two temperatures is about 0.4.

8-79

Add another column to Table 8-4 containing the range (°C) of warm working temperatures. Remember to convert to and from degrees kelvin in performing your calculations.

Solution: This is best done in a spreadsheet. The melting temperature is converted to kelvins, multiplied by 0.3 and 0.6, and the resulting temperatures converted back into degrees centigrade. In a table: Elements Warm working range (°C) Sn

-122 - 30

Pb

-93 - 87

Zn

-65 - 143

Al

7 - 287

Mg

4 - 281

Ag

98 - 468

Cu

134 - 542

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8-80

Fe

270 - 814

Ni

245 - 763

Mo

592 - 1457

W

832 - 1937

Consider the tensile stress strain curves in Figure 8–20 labeled 1 and 2 and answer the following questions. These diagrams are typical of metals. Consider each part as a separate question that has no relationship to previous parts of the question.

(a) Which of the two materials represented by samples 1 and 2 can be cold rolled to a greater extent? How do you know? (b) Samples 1 and 2 have the same composition and were processed identically, except that one of them was cold worked more than the other. The stress strain curves were obtained after the samples were cold worked. Which sample has the lower recrystallization temperature: 1 or 2? How do you know? (c) Samples 1 and 2 are identical except that they were annealed at different temperatures for the same period of time. Which sample was annealed at the higher temperature: 1 or 2? How do you know? (d) Samples 1 and 2 are identical except that they were annealed at the same temperature for different periods of time. Which sample was annealed for the shorter period of time: 1 or 2? How do you know?

Solution: (a) Sample 2 can be cold–rolled to a greater extent because it is more ductile as evidenced by the larger strain to failure. (b) Sample 1 was cold–worked more than Sample 2 as evidenced by its higher strength and lower ductility. Because it was cold–worked more than Sample 2, Sample 1 has the lower recrystallization temperature. (c) Sample 2 was annealed at a higher temperature than Sample 1. The higher annealing temperature led to either a larger reduction in the dislocation density or larger grain sizes (or both), thereby decreasing strength and increasing ductility. (d) Sample 1 was annealed for a shorter period of time. It is stronger and less ductile because its dislocation density was not reduced as much as for Sample 2 and its grains are smaller.

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8-81

We wish to produce a 0.3-in. thick plate of 3105 aluminum having a tensile strength of at least 25,000 psi and a % elongation of at least 5%. The original thickness of the plate is 3 in. The maximum cold work in each step is 80%. Describe the cold working and annealing steps required to make this product. Compare this process with what you would recommend if you could do the initial deformation by hot working. (See Figure 8– 24.)

Solution:

For TS ≥ 25000

CW ≥ 30%;

ti − 0.3 = 0.30 ti

or

Cold work/anneal treatment CW 75% from 3.0 to 0.75 in. anneal CW 42.8% from 0.75 to 0.429 in. anneal CW 30% from 0.429 to 0.3 in.

8-82

For % elongation ≥ 5%

CW ≤ 30% ∴ required CW = 30%

ti = 0.429 in.

Hot work treatment HW 85.7% from 3.0 to 0.429 in.

CW 30% from 0.429 to 0.3 in.

We wish to produce a 0.2-in.-diameter wire of copper having a minimum yield strength of 60,000 psi and a minimum % elongation of 5%. The original diameter of the rod is 2 in. and the maximum cold work in each step is 80%. Describe the cold working and annealing steps required to make this product. Compare this process with that you would recommend if you could do the initial deformation by hot working. (See Figure 8– 6.)

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Solution:

For YS > 60 ksi,

d i2 − (0.2) 2 = 0.42 di2

CW ≥ 40%; or

Cold work/anneal treatment CW 75% from 2 to 1 indiameter anneal CW 75% from 1 to 0.5 in. anneal CW 72.3% from 0.5 to 0.263 in. anneal CW 42% from 0.263 to 0.2 in.

For % elongation > 5

CW ≤ 45% For cold work = 42%

di = 0.04/0.58 = 0.263 in.

Hot work treatment HW 98.3% from 2 to 0.263 in.

CW 42% from 0.263 to 0.2 in.

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Chapter 9: Principles of Solidification 9-5

What is the difference between homogenous nucleation and heterogeneous nucleation?

Solution: Homogenous nucleation occurs when nuclei of the resulting phase have the same probability of forming at every location. Heterogeneous nucleation of the resulting phase occurs at certain preferred locations in the original phase. In practice, homogenous nucleation does not occur. Heterogeneous nucleation commonly occurs at the surfaces for molds of castings, surfaces of inclusions and/or induced by inoculation.

9-6

From the thermodynamic point of view, what two things must occur for solidification from liquid to solid to proceed?

Solution: Thermodynamically speaking, the first thing that must occur is the formation of a critical-sized nucleus. At this critical size, there is an activation energy that must be overcome. At the melting or freezing point, this activation energy will approach infinity.

Thus the second thing that must occur is an undercooling in the liquid that is to solidify.

A greater undercooling will lower the activation energy resulting in a greater tendency to solidify.

9-9

Derive the formula for the change in total energy change (i.e. the derivative of ΔG with respect to r) as a function of the radius r of the solid nucleus. Begin with Equation 9-1.

Numerically, how is the critical point at which solidification proceeds defined?

Solution: This is a calculus problem. Starting with the equation: 4 ∆ = ∆ + 4 3 Differentiating: ∆ 4

= ∆ + 4

3

∆ 4 = ∆ 3 + 4 2

3 ∆ = 4 ∆ + 8

The critical point for solidification is when the radius reaches the critical radius. This is also when ΔG reaches an inflection point, so: ∆ = 0; > 0

9-10

Using Equation 9-2, prove that interfacial energy (σsl) has units of J/m2 in SI.

Solution: Equation 9-2 is

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∗ = Solving for the interfacial energy: =

In SI, the units are as follows: Name Critical radius Freezing temperature Heat of fusion Temperature difference

2 ∆ ∆

∗ ∆ ∆ 2 SI units m K J/m3 K

Inserting these into the previous equation: m J K 1 = " $ $ 1 m 1 K J = m 9-11

Calculate the total interfacial surface energy for 1016 spheres of copper, each with a radius of r*.

Solution: Example 9-1 found the critical radius of copper to be 12.51 × 10-8 centimeters.

Calculating the total surface area of a single sphere: & = 4 = 4 12.51 × 10*+ cm = 1.9110*/ cm The total surface area for all spheres: 0 & = 10/1 1.9210*/ cm = 2310 cm

Using the interfacial energy from Table 9-1: 2410 cm 177710*8 J 5 = 35 mJ 1 1 cm 9-12

Of the ferrous elements, which has the lowest undercooling required for homogeneous nucleation? Does this have any practical significance?

Solution: Between iron and nickel, the lowest ΔT in Table 9-1 belongs to iron at 420 °C. This does not have a great practical significance in itself because homogeneous nucleation is not seen in industry or commerce. Heterogeneous nucleation is seen in production. *With apologies, as written, this problem statement does not make sense. Perhaps it should have been of the “ferromagnetic elements.”

9-13

If the total change in free energy of a molten metal being cooled is 5.34 × 10-17 J upon formation of the first stable solid and the free energy per unit volume is -17.7 J/cm3, approximate the radius of the spherical solid. The surface free energy of the solid-liquid interface is 100 × 10-7 J/cm2.

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Solution: This is a straightforward use of Equation 9-1, however making it explicit in r is difficult algebraically. Instead a numerical solution using a spreadsheet’s goal seek function or calculator’s SOLVE application is advised. The result will be 1.16 × 10-6 cm.

9-14

Using the densities in Appendix A, convert the heats of fusion in Table 9-1 from units J/cm3 to kJ/kg.

Solution: While doing the conversions, note that [kJ/kg] are the same as [J/g] Element Heat of fusion (kJ/kg) Ga 83 Bi 55 Pb 21 Ag 92 Cu 183 Ni 310 Fe 195 9-15

Assume that instead of a spherical nucleus, we have a nucleus in the form of a cube of length x. Calculate the critical dimension x* of the cube necessary for nucleation. Write an equation similar to Equation 9-1 for a cubical nucleus, and derive an expression for x* similar to Equation 9-2.

Solution:

∆G = x 3∆Gv + 6x 2σ sl ∂G = 3x 2 ∆Gv +12xσ sl = 0 ∂x 2 ∂G = 3 ( x * ) ∆Gv +12x * σ sl = 0; x > 0 ∂x x=x*

x* = −

12 σ 4σ = − sl 3∆Gv ∆Gv

Note that ∆Gv is negative such that x * is positive.

A similar treatment for Equation 9-1 for a spherical nucleus produces 2 ∂G = 4π ( r * ) ∆Gv + 8π r* σ sl = 0; r > 0 ∂r r=r*

r* = −

2 σ sl ∆Gv 199

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Comparison to Equation 9-2 shows that

∆Gv = −

∆H f ∆T Tm

Thus,

x* = −

9-16

4 σ sl 4 σ sl Tm = ∆Gv ∆H f ∆T

Why is undercooling required for solidification? Derive an equation showing the total free energy change as a function of undercooling when the nucleating solid has the critical nucleus radius r*.

Solution: Undercooling is required for solidification because the energy to create a new solid-liquid interface needs to be overcome for a solid phase to begin forming. The undercooling below the melting temperature is the driving force that overcomes this barrier.

From Equation 9–1, ∆G = (4/3) πr3 ∆Gv + 4πr2σsl, where ∆G is the total change in free energy upon solidification, r is the radius of spherical solid, ∆Gv is the free energy change per unit volume for solidification, and σsl is the surface energy per unit area of the solid–liquid interface. From Equation 9–2, the critical radius for solidification r* is given by r* = 2σslTm/(∆Hf∆T), where Tm is the melting temperature, ∆Hf is the latent heat of fusion, and ∆T is the undercooling. At r = r*,

∆G = (4/3)π [2σ slTm /(∆H f ∆T )]3 ∆Gv + 4π [2σ slTm /(∆H f ∆T )]2 σ sl ∆G = (32π /3)(σ sl3 Tm3 ∆Gv /∆H 3f )(1/∆T 3 ) + 16π (σ sl3 Tm2 /∆H 2f )(1/∆T 2 ) 9-17

Why is it that nuclei seen experimentally are often sphere-like but faceted? Why are they sphere-like and not like cubes or other shapes?

Solution: Nuclei are sphere-like because spheres have the largest volume to surface area ratio.

Nuclei are faceted to favor crystallographic planes with low surface energies.

9-18

Explain the meaning of each term in Equation 9-2.

Solution:

∆H f is the latent heat of fusion per unit volume, Tm is the equilibrium solidification temperature in kelvin, and ∆T is the undercooling when the liquid temperature is T. The latent heat of fusion represents the heat given up during the liquid-to-solid transformation. σ sl Is the surface energy per area of the solid–liquid interface.

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9-19

Suppose that liquid nickel is undercooled until homogeneous nucleation occurs.

Calculate (a) the critical radius of the nucleus required and (b) the number of nickel atoms in the nucleus. Assume that the lattice parameter of the solid FCC nickel is 0.356 nm.

Solution:

From Table 9–1, ∆Tmax = 480°C

r* =

(2)(255 ×10−7 J/cm 2 )(1453 + 273) = 6.65 ×10−8 cm (2756 J/cm3 )(480)

V = 45.118 × 10–24 cm3 Vnucleus = (4π/3)(6.65 × 10–8 cm)3 = 1232 × 10–24 cm3 number of unit cells = 1232/45.118 = 27.3 atoms per nucleus = (4 atoms/cell)(27.3 cells) = 109 atoms

ao = 3.56 Å

9-20

Suppose that liquid iron is undercooled until homogeneous nucleation occurs. Calculate (a) the critical radius of the nucleus required and (b) the number of iron atoms in the nucleus. Assume that the lattice parameter of the solid BCC iron is 2.92 Å.

Solution:

r* =

(2)(204 ×10−7 J/cm 2 )(1538 + 273) = 10.128 ×10−8 cm 3 (1737 J/cm )(420)

V = (4π/3)(10.128)3 = 4352 Å3 = 4352 × 10–24 cm3 Vuc = (2.92 Å)3 = 24.897 Å3 = 24.897 × 10–24 cm3 number of unit cells = 4352/24.897 = 175 atoms per nucleus = (175 cells)(2 atoms/cell) = 350 atoms 9-21

Suppose that solid nickel was able to nucleate homogeneously with an undercooling of only 22°C. How many atoms would have to group together spontaneously for this to occur? Assume that the lattice parameter of the solid FCC nickel is 0.356 nm.

Solution:

r* =

(2)(255 ×10−7 J/cm 2 )(1453 + 273) = 145.18 ×10−8 cm 3 (2756 J/cm )(22)

Vuc = 45.118 × 10–24 cm3 (see Problem 9–11) Vnucleus = (4π/3)(145.18 × 10–8 cm)3 = 1.282 × 10–17 cm3 number of unit cells = 1.282 × 10–17 / (45.118 × 10–24) = 2.84 × 105 atoms per nucleus = (4 atoms/cells)(2.84 × 105 cells) = 1.136 × 106 9-22

Suppose that solid iron was able to nucleate homogeneously with an undercooling of only 15°C. How many atoms would have to group together spontaneously for this to occur? Assume that the lattice parameter of the solid BCC iron is 2.92 Å.

Solution:

r* =

(2)(204 ×10−7 J/cm 2 )(1538 + 273) = 283.6 ×10−8 cm 3 (1737 J/cm )(15)

Vuc = 24.897 × 10–24 cm3 (see Problem 9–12) Vnucleus = (4π/3)(283.6 × 10–8 cm)3 = 95,544,850 × 10–24 cm3

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number of unit cells = 95,544,850/24.897 = 3.838 × 106 atoms per nucleus = (2 atoms/cells)(3.838 × 106 cells) = 7.676 × 106 9-23

Explain the term inoculation.

Solution: Inoculation is the addition of particles to the liquid to induce the nucleation of equiaxed grains in a casting. The common inoculants are refractory elements with melting points much higher than the liquid melt. The inoculants provide surfaces on which heterogeneous nucleation of solid crystals may occur.

9-35

What is a dendrite and why do dendrites form during solidification?

Solution: A dendrite is a “tree-like” formation of solid crystals from liquid. The primary dendrite (main trunk) is due to a preferred growth of crystals in specific crystallographic directions which occurs opposite the direction of heat extraction. From the main trunk, branches (secondary dendrites) or dendritic arms form and grow also along preferred directions into the liquid. For cubic crystals, the easiest growth path are along the <100> directions.

9-36

Use the data in Table 9–1 and the specific heat data given below to calculate the undercooling required to keep the dendritic fraction at 0.5 for each metal.

Metal Specific heat [J/(cm3–K)] Bi 1.27 Pb 1.47 Cu 3.48 Ni 4.75

Solution: The dendritic fraction f is given by Equation 9–3: f = c∆T/∆Hf, where c is the specific heat, T is the temperature, and ∆Hf is the latent heat of fusion.

For f = 0.5, the undercooling required is ∆T = 0.5 ∆Hf/c.

The results for the various metals are shown in the table.

Metal Heat of Fusion ∆Hf (J/cm3) Bi 543 Pb 237 Cu 1628 Ni 2756

Specific heat c [J/(cm3–K)] 1.27 1.47 3.48 4.75

∆T for f = 0.5 (K) 213.8 80.6 233.9 290.1

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9-37

Calculate the fraction of solidification that occurs dendritically when silver nucleates (a) at 10°C undercooling; (b) at 100°C undercooling; and (c) homogeneously. The specific heat of silver is 3.25 J/(cm3·°C).

Solution:

c∆T [3.25 J/cm3 ⋅°C](10°C) = = 0.0337 ∆Hf 965 J/cm c∆T [3.25 J/(cm 3 ⋅°C)](100°C) (b) = = 0.337 ∆H f 965 J/cm3 (a) f =

(c)

9-38

c∆T [3.25 J/(cm3 ⋅°C)](250°C) = = 0.842 ∆H f 965 J/cm3

Calculate the fraction of solidification that occurs dendritically when iron nucleates (a) at 10°C undercooling; (b) at 100°C undercooling; and (c) homogeneously. The specific heat of iron is 5.78 J/(cm3 · °C).

Solution:

(a) f =

c∆T [5.78 J/(cm3 ⋅ °C)](10°C) = = 0.0333 ∆H f 1737 J/cm3

c∆T [5.78 J/(cm 3 ⋅°C)](100°C) (b) = = 0.333 ∆H f 1737 J/cm3 (c)

c∆T [5.78 J/(cm 3 ⋅°C)](420°C) = = 1.40 1737 J/cm3 ∆H f

Therefore, essentially all of the solidification occurs dendritically.

9-39

Analysis of a nickel casting suggests that 28% of the solidification process occurred in a dendritic manner. Calculate the temperature at which nucleation occurred. The specific heat of nickel is 4.1 J/(cm3 · °C).

Solution:

f =

c∆T [4.1 J/(cm3 ⋅°C)]( ∆T ) = = 0.28 ∆H f 2756 J/cm3 ∆T = 188°C or Tn = 1453 − 188 = 1265°C

9-40

It is desired to increase the dendritic fraction from 0.012 to 0.025. Assuming that the specific heat of the metal being solidified is constant, determine the increase in the amount of undercooling required.

Solution: Equation 9-3 is linear, so we can simply divide the two fractions to find the ratios of ΔT. :∆ 9= ∆ :∆ ∆ 9 = 9/ :∆ / ∆

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9 ∆ = 9/ ∆ / 0.025 ∆ = 2.1 = 0.012 ∆ / So, the undercooling is to be increased by a factor of 2.1. Note that this means the temperature difference, not the actual temperature the metal is being undercooled to.

9-42

Find the mold constant B and exponent n in Chvorinov’s rule using the following data and a log–log plot.

Solution: Chvorinov’s rule gives the solidification time ts as n

V  ts = B   ,  A where B is the mold constant, V is the volume of the casting, A is the surface area of the casting in contact with the mold, and n is a constant.

Taking the natural logarithm of both sides, ln (ts) = ln (B) + n ln (V/A) The values of n and B can be found by plotting ln (ts) versus ln (V / A).

The volume and surface area of each shape is given in the table.

Shape Dimensions (cm) Cylinder Radius = 10, Length = 30 Sphere Radius = 9 Cube Length = 6 Plate Length = 30, Width = 20, Height = 1

Volume (cm3) Area (cm2) Volume / Area (cm) 9425 2513 3.75 3054 216 600

1018 216 1300

3 1 0.46

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From the graph, the slope n = 2.21 and ln (B) = 5.3407 such that B = 209 s/cm2.

9-43

A 2-in. cube solidifies in 4.6 min. Assume that n = 2. Calculate (a) the mold constant in Chvorinov’s rule; and (b) the solidification time for a 0.5 in. × 0.5 in. × 6 in. bar cast under the same conditions.

Solution: (a) We can find the volume and surface area of the cube: V = (2)3 = 8 in.3 A = 6(2)2 = 24 in.2 t = 4.6 = B(8/24)2 B = 4.6/(0.333)2 = 41.48 min/in.2 (b) For the bar, assuming that B = 41.48 min/in.2: V = (0.5)(0.5)(6) = 1.5 in.2 A = 2(0.5)(0.5) + 4(0.5)(6) = 12.5 in.2 t = (41.48)(1.5/12.5)2 = 0.60 min 9-44

A 5-cm diameter sphere solidifies in 1050 s. Calculate the solidification time for a 0.3 cm × 10 cm × 20 cm plate cast under the same conditions. Assume that n = 2.

Solution:

2

 (4π /3)(2.5)3  t = 1050 s = B  = B[2.5/3]2 or B = 1512 s/cm 2  2  4π (2.5)  (1512)(0.3 ×10 × 20) 2 t= = 1512[60/418]2 = 31.15 s 2 [2(0.3)(10) + 2(0.3)(20) + 2(10)(20)]

9-45

Find the constants B and n in Chvorinov’s rule by plotting the following data on a log-log plot:

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Solution: V (in.3) A (in.2) V/A (in.)

48 212 0.226 60 112 0.536 15.6 37.5 0.416 36 98 0.367 From the graph, we find that B = 48 min/in.2 and n = 1.72

9-46

Find the constants B and n in Chvorinov’s rule by plotting the following data on a log-log plot:

Solution: V (cm3) A (cm2) V/A (cm) 6 26 0.23 32 64 0.5 64 96 0.67 240 236 1.02 From the graph, we find that B = 305 s/cm2 and n = 1.58

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9-47

A 3-in.-diameter casting was produced. The times required for the solid-liquid interface to reach different distances beneath the casting surface were measured and are shown in the following table: Distance from surface d (in.) Time t (s)

t

0.1 32.6 5.71 0.3 73.5 8.57 0.5 130.6 11.43 0.75 225.0 15.00 1.0 334.9 18.22 Determine (a) the time at which solidification begins at the surface and (b) the time at which the entire casting is expected to be solid. (c) Suppose the center of the casting actually solidified in 720 s. Explain why this time might differ from the time calculated in part (b).

Solution: We could plot d versus

t , as shown, finding tsurface from where the

plot intersects the x-axis and

tcenter where the plot intersects d = 1.5

in. Or we could take two of the data points and solve for c and k.

d = k t −c

0.1 = k 32.6 − c 0.5 = k 130.6 − c −0.4 = k[ 32.6 − 130.6] = −5.718 k k = 0.070 c = 0.070 32.6 − 0.1 = 0.30

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d = 0 = 0.070 t − 0.30 tsurface = (0.3/0.07)2 = 18.4 s

1.5 = 0.070 t − 0.3 tcenter = (1.8/0.07)2 = 661 s The mold gets hot during the solidification process, and consequently heat is extracted from the casting more slowly. This in turn changes the constants in the equation and increases the time required for complete solidification.

9-48

What solidification time is required for a casting of 9.5 cm3 and a mold contact area of 10.0 cm2? The mold constant is 838 s/cm2.01.

Solution: This is an easy application of Chvorinov’s rule. The only curveball is that the exponential constant must be read from the units of the mold constant. > @ ; = <= ? & .B/ s 9.5 cm ; = 838

cm .B/ 10.0 cm ; = 756 s 9-49

An aluminum alloy plate with dimensions 20 cm × 10 cm × 2 cm needs to be cast with a secondary dendrite arm spacing of 10–2 cm (refer to Figure 9–6). What mold constant B is required (assume n = 2)?

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Solution: From Figure 9–6, for a secondary dendrite arm spacing of 0.01 cm, the solidification time is 500 s. Chvorinov’s rule for solidification time (Equation 9–4) is ts = B(V/A)n, where B is the mold constant, V is the volume of the casting, A is the surface area of the casting in contact with the mold, and n is a constant. In this case, the volume of the casting is V = 20 × 10 × 2 = 400 cm3, and the surface area of the casting is A = 2 × 20 × 10 + 2 × 2 × 10 + 2 × 2 × 20 = 520 cm2.

Substituting into the equation above and taking n = 2, ts = B(V/A)n 500 = B(400/520)2 B = 845 s/cm2.

9-50

Figure 9–5(b) shows a micrograph of an aluminum alloy. Estimate (a) the secondary dendrite arm spacing and (b) the local solidification time for that area of the casting.

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Solution: The distance between adjacent dendrite arms can be measured.

Although most people doing these measurements will arrive at slightly different numbers, the author’s calculations obtained from four different primary arms are 16 mm / 6 arms = 2.67 mm 9 mm / 5 arms = 1.80 mm 13 mm / 7 arms = 1.85 mm 18 mm / 9 rms = 2.00 mm average = 2.08 mm = 0.208 cm Dividing by the magnification of ×50: SDAS = 0.208 cm / 50 = 4.16 × 10–3 cm From Figure 9–6, we find that the local solidification time (LST) = 90 s.

9-51

Find the constants k and m relating the secondary dendrite arm spacing to the local solidification time by plotting the following data on a log-log plot:

Solution: The secondary dendrite arm spacing (SDAS) is given by SDAS = ktsm , where k and m are constants and ts is the solidification time. Taking the natural logarithm of both sides, ln (SDAS) = ln (k) + m ln (ts).

The values of k and m can be found by plotting ln (SDAS) versus ln (ts).

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From the graph, the slope m = 0.35 and ln (k) = –5.8013 such that k = 0.003 cm / s.

9-52

Figure 9–25 shows dendrites in a titanium powder particle that has been rapidly solidified. Assuming that the size of the titanium dendrites is related to solidification time by the same relationship as in aluminum, estimate the solidification time of the powder particle.

Solution: The secondary dendrite arm spacing can be estimated from the photomicrograph at several locations. The author’s calculations, derived from measurements at three locations, are 11 mm / 8 arms = 1.375 mm 13 mm / 8 arms = 1.625 mm 13 mm / 8 arms = 1.625 mm average = 1.540 mm Dividing by the magnification of 2200: SDAS = (1.540 mm)(0.1 cm/mm) / 2200 = 7 × 10–5 cm The relationship between SDAS and solidification time for aluminum is SDAS = 8 × 10–4 t0.42 = 7 × 10–5 t = (0.0875)1/0.42 = 0.003 s 9-53

The secondary dendrite arm spacing in an electron-beam weld of copper is 9.5 × 10–4 cm. Estimate the solidification time of the weld.

Solution: From Figure 9–6, we can determine the equation relating SDAS and solidification time for copper: m = 20/50 = 0.4 k = 4 × 10–3 cm Then for the copper weld: 9.5 × 10–4 = 4 × 10–3(LST)0.4 (Note: LST is local solidification time) 0.2375 = (LST)0.4 or –1.4376 = 0.4 ln LST ln LST = –3.5940 or LST = 0.03 s

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9-54

A zinc alloy has a SDAS of 0.01 cm and m = 0.40. What is the solidification time?

Solution: The m value is a distraction here. By referring to Figure 9-6, we directly read the solidification time as being about 120 seconds.

9-55

In Figure 9-7, what is the equation (slope intercept form) of the tensile strength line, and what SDAS would appear to give a tensile strength of zero if the line is extrapolated? Does extrapolating the trend like this make sense?

Solution: Since the graph is purely linear, we can easily take two points and find the equation of the line. Points that land on a gridline can be more accurately read, such as (0.005 cm, 44 ksi) and (0.010 cm, 39.5 ksi) that both fall on vertical gridlines.

Taking the difference to find the slope: 39.5 ksi − 44 ksi ksi ∆D = = −900 3= cm ∆E 0.010 cm − 0.005 cm Applying the slope to find the x-intercept: ∆D 39.5 ksi − E ksi 3= = = −900 ∆E 0.010 cm − 0 cm 39.5 ksi − E ksi = −900 0.010 cm cm

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3950

Together:

E ksi ksi − = −900 cm cm 0.010 cm ksi E 4850 = cm 0.010 cm E = 48.5 ksi

D = =−900

ksi ? E + 48.5 ksi cm

Where x is in cm and y is in ksi.

To find out what SDAS (x) results in a strength of zero, we set y to zero and solve: ksi 0 ksi = =−900 ? E + 48.5 ksi cm ksi 48.5 ksi = =900 ? E cm E = SDAS = 0.0539 cm The tensile strength for the aluminum alloy will never be zero regardless of how large the SDAS spacing is. Thus we see that extrapolation can be dangerous since trends change. The curve will plateau at a value for which the SDAS becomes sufficiently large that the presence of dendrites no longer causes strength to be increased relative to a casting that solidified without dendrites.

9-60

In Figure 9-8, what do the slopes of the lines A-B and E-onward represent? Do these quantities represent physical properties?

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Solution: No. The slopes of these lines represent how fast the temperature is falling in the solid and liquid phases respectively. They can be changed at will by increasing or decreasing the rate of heat removal from the samples.

9-61

A cooling curve is shown in Figure 9–26. Determine (a) the pouring temperature; (b) the solidification temperature; (c) the superheat; (d) the cooling rate, just before solidification begins; (e) the total solidification time; (f) the local solidification time; and (g) the probable identity of the metal. (h) If the cooling curve was obtained at the center of the casting sketched in the figure, determine the mold constant, assuming that n = 2.

Solution: (a) Tpour = 475°C (b) Tsol = 320°C (c) ∆Ts = 475 – 320 = 155°C (d) ∆T /∆t = 9-62

475 − 320 = 1.0°C/s 160 − 0

(e) ts = 470 s (f) LST = 470 – 160 = 310 s (g) Cadmium (Cd) (h) ts = 470 = B[38.4/121.6]2 B = 4713 s/cm2

A cooling curve is shown in Figure 9–27. Determine (a) the pouring temperature; (b) the solidification temperature; (c) the superheat; (d) the cooling rate, just before solidification begins; (e) the total solidification time; (f) the local solidification time; (g) the undercooling; and (h) the probable identity of the metal. (i) If the cooling curve was obtained at the center of the casting sketched in the figure, determine the mold constant, assuming n = 2.

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Solution:

(a) Tpour = 900°C (b) Tsol = 420°C (c) ∆Ts = 900 – 420 = 480°C (d) ∆T /∆t =

900 − 400 = 250°C/min 2−0

(i) ts = 9.7 = B[8/24]2 9-63

(e) ts = 9.7 min (f) LST = 9.7 – 2.5 = 7.2 min (g) 420 – 360 = 60°C (h) Zn

or B = 87.3 min/in.2

Figure 9–28 shows the cooling curves obtained from several locations within a cylindrical aluminum casting. Determine the local solidification times and the SDAS at each location, then plot the tensile strength versus distance from the casting surface.

Would you recommend that the casting be designed so that a large or small amount of material must be machined from the surface during finishing? Explain.

Solution: The local solidification times can be found from the cooling curves and can be used to find the expected SDAS values from Figure 9–6. The SDAS values can then be used to find the tensile strength, using Figure

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9–7.

Surface: LST = 10 s ⇒ SDAS = 1.5 × 10–3 cm ⇒ TS = 47 ksi Mid-radius: LST = 100 s ⇒ SDAS = 5 × 10–3 cm ⇒ TS = 44 ksi Center: LST = 500 s ⇒ SDAS = 10 × 10–3 cm ⇒ TS = 39.5 ksi If high strength is desirable, you prefer to machine as little material off the surface of the casting as possible; the surface material has the finest structure and highest strength; any excessive machining simply removes the “best” material.

9-70

Calculate the volume, diameter, and height of the cylindrical riser required to prevent shrinkage in a 1 in. × 6 in. × 6 in. casting if the H/D of the riser is 1.0.

Solution: The riser should take longer to solidify than the casting: tr > tc, or

V  V  B  > B  ,  A r  A c where tr is the solidification time for the riser, tc is the solidification time for the casting, and ( VA )r and ( VA )c are the volume (V) to surface area (A) ratios of the riser (r) and casting (c), respectively. The mold constant B is the same for the riser and casting. ( VA )c is given by

1× 6 × 6 36 V  = ,   =  A c 2 ×1× 6 + 2 ×1× 6 + 2 × 6 × 6 − π D 2 96 − π D 2 4 4 V  and   is given by  A r

π

D2 H V   4 ,   =π 2  A r D + π DH 4 where D is the diameter of the riser and H is its height. Note that the riser area in contact with the casting is not included in either the riser or casting surface area; no heat is lost across this interface. Since H / D is 1.0,

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V  D   = .  A r 5 Thus,

   D  36 > , 5  96 − π D 2  4   and solving for D, D > 1.93 in.

For this minimum diameter, the minimum height is 1.93 in., and the minimum volume is

Vr = 9-71

π 4

(1.93 in.) 2 × (1.93 in.) = 5.7 in.3

Calculate the volume, diameter, and height of the cylindrical riser required to prevent shrinkage in a 4 in. × 10 in. × 20 in. casting if the H/D of the riser is 1.5.

Solution: The riser should take longer to solidify than the casting: tr > tc, or

V  V  B  > B  ,  A r  A c where tr is the solidification time for the riser, tc is the solidification time for the casting, and ( VA )r and ( VA )c are the volume (V) to surface area (A) ratios of the riser (r) and casting (c), respectively. The mold constant B is the same for the riser and casting. ( VA )c is given by

4 ×10 × 20 800 V  =   =  A c 2 × 4 ×10 + 2 × 4 × 20 + 2 ×10 × 20 − π D 2 640 − π D 2 4 4 V  and   is given by  A r

π

D2 H V  4 ,  = π A 2   D + π DH 4 where D is the diameter of the riser and H is its height. Note that the riser area in contact with the casting is not included in either the riser or casting surface area; no heat is lost across this interface. Since H / D is 1.5,

 V  3D .   =  A r 14 Thus,

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3D 800 . > 14 640 − π D 2 4 Therefore, solving for D, D > 6.11 in.

For this minimum diameter, the minimum height is 9.17 in., and the minimum volume is

Vr = 9-72

π 4

(6.11 in.) 2 × (9.17 in.) = 269 in.3

Figure 9–29 shows a cylindrical riser attached to a casting. Compare the solidification times for each casting section and the riser and determine whether the riser will be effective.

Solution:

(8)(6)(3) = 0.889 (3)(6) + 2(3)(8) + 2(6)(8) (6)(6)(6) (V /A) thick = = 1.13 (6)(3) + 5(6)(6) − (π /4)(3) 2 (V /A) thin =

(V /A) riser =

(π /4)(3)2 (7) = 0.68 π (3)(7) + (π /4)(3)2

Note that the riser area in contact with the casting is not included in either the riser or casting surface area; no heat is lost across this interface. In a like manner, the area of contact between the thick and thin portions of the casting are not included in the calculation of the casting area.

The riser will not be effective; the thick section of the casting has the largest V/A ratio and therefore requires the longest solidification time. Consequently the riser will be completely solid before the thick section is solidified; no liquid metal will be available to compensate for the solidification shrinkage.

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9-73

Figure 9–30 shows a cylindrical riser attached to a casting. Compare the solidification times for each casting section and the riser and determine whether the riser will be effective.

Solution:

(4)(4)(4) = 0.73 5(4)(4) + 1(2)(4) (2)(2)(4) (V /A) thin = = 0.50 3(2)(4) + 2(2)(2)

(V /A) thick =

(V /A) riser =

(π /4)(42 )(8) = 0.8 π (4)(8) + 2(π /4)4 2

The area between the thick and thin sections of the casting are not included in calculating casting area; no heat is lost across this interface.

The riser will not be effective; the thin section has the smallest V/A ratio and therefore freezes first. Even though the riser has the longest solidification time, the thin section isolates the thick section from the riser, preventing liquid metal from feeding from the riser to the thick section. Shrinkage will occur in the thick section.

9-74

A hollow cylindrical mold for casting aluminum ingots has a 300 mm inside diameter and is 2 m high. If the mold is filled with liquid aluminum at 935 K, what is the largest spherical cavity that may form in the ingot?

Solution: Since aluminum has 7% volume shrinkage, if the top of the ingot freezes first before the entire ingot solidifies, there will be a cavity within the ingot. The largest cavity forms when the entire shrinkage volume is converted to one cavity, which we shall assume to be spherical.

L L Volume of initial liquid = M D2H = M (0.3)2(2) = 0.1414 m3 Shrinkage volume = 0.1414 (0.07) = 0.009896 m3 Assuming a spherical cavity: M 0.009896 m3 = πr3 ; we find that r = 0.1332 m or 13.32 cm, so the diameter of the cavity would be 26.64 cm.

If we assume that the solidification process proceeds from the bottom up with no cavities formed within the ingot, the top of the ingot will be lower than the inside height of the ingot mold. Assuming the outside diameter of the ingot to be the same as the

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inside diameter of the ingot mold, the height of the ingot produced would be 93% (100 – 7% from shrinkage) of the mold height.

Ingot height = 0.93 (2 m mold height) = 1.86 m Diameter of the ingot = 0.3 m 9-75

A 4-in.-diameter sphere of liquid copper is allowed to solidify, producing a spherical shrinkage cavity at the center of the casting. Compare the volume and diameter of the shrinkage cavity in the copper casting to that obtained when a 4-in. sphere of liquid iron is allowed to solidify.

Solution: Cu: 5.1% Fe: 3.4% rsphere = 4/2 = 2 in.

Cu: Vshrinkage = (4π/3)(2)3 (0.051) = 1.709 in.3 (4π/3)r3 = 1.709 in.3 or r = 0.742 in. dcavity = 1.48 in.

Fe: Vshrinkage = (4π/3)(2)3 (0.034) = 1.139 in.3 (4π/3)r3 = 1.139 in.3 or r = 0.648 in. dcavity = 1.30 in.

9-76

A 4-in. cube of a liquid metal is allowed to solidify. A spherical shrinkage cavity with a diameter of 1.49 in. is observed in the solid casting. Determine the percent volume change that must have occurred during solidification.

Solution: Vliquid = (4 in.)3 = 64 in.3 Vshrinkage = (4π/3)(1.49/2)3 = 1.732 in.3 Vsolid = 64 – 1.732 = 62.268 in.3

% volume change = 9-77

64 − 62.268 × 100 = 2.7% 64

A 2 cm × 4 cm × 6 cm magnesium casting is produced. After cooling to room temperature, the casting is found to weigh 80 g. Determine (a) the volume of the shrinkage cavity at the center of the casting and (b) the percent shrinkage that must have occurred during solidification.

Solution: The density of magnesium is 1.738 g/cm3. (a) Vinitial = (2)(4)(6) = 48 cm3 Vfinal = 80 g/(1.738 g/cm3) = 46.03 cm3 (b) % shrinkage = 9-78

48 − 46.03 × 100% = 4.1% 48

A 2 in. × 8 in. × 10 in. iron casting is produced and, after cooling to room temperature, is found to weigh 43.9 lb. Determine (a) the percent shrinkage that must have occurred during solidification and (b) the number of shrinkage pores in the casting if all of the shrinkage occurs as pores with a diameter of 0.05 in.

Solution: The density of the iron is 7.87 g/cm3

Vactual =

(43.9 lb)(454 g/lb) = 2532.5 cm3 3 7.87 g/cm

Vintended = (2)(8)(10) = 160 in.3 × (2.54 cm/in)3 = 2621.9 cm3

shrinkage =

2621.9 − 2532.5 × 100% = 3.4% 2621.9

Vpores = 2621.9 – 2532.5 = 89.4 cm3

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rpores = (0.05 in./2)(2.54 cm/in.) = 0.0635 cm

# pores = 9-79

89.4 cm3 = 83,354 pores (4π /3)(0.0635 cm)3

If you cool an open vessel of liquid gallium until a thin solid “cake” forms, will the cake form on the top or the bottom? Assume the air above the liquid gallium is stagnant and does not cool the surface at all.

Solution: The cake will float on top. Table 9-2 shows that gallium expands when it solidifies. This means that a given volume of solid gallium is lighter than the same volume of liquid gallium. In other words, it is less dense.

9-80

Give examples of materials that expand upon solidification.

Solution: From Table 9-2, gallium, gray cast iron and water.

9-81

What is Sievert’s Law? How can gas porosity in molten alloys be removed or minimized?

Solution: Sievert’s Law expresses the solubility of gases in liquid metals as a function of the partial pressure of the gas over the liquid. In vacuum degassing of the melt, the partial pressure of the dissolved gas over the melt is reduced and therefore the concentration of the gas in the liquid is reduced.

9-83

Liquid magnesium is poured into a 2 cm × 2 cm × 24 cm mold and, as a result of directional solidification, all of the solidification shrinkage occurs along the 24-cm length of the casting. Determine the length of the casting immediately after solidification is completed.

Solution: Vinitial = (2)(2)(24) = 96 cm3

% contraction = 4 or 0.04 × 96 = 3.84 cm3 Vfinal = 96 – 3.84 = 92.16 cm3 = (2)(2)(L) Length (L) = 23.04 cm 9-84

A liquid cast iron has a density of 7.65 g/cm3. Immediately after solidification, the density of the solid cast iron is found to be 7.71 g/cm3. Determine the percent volume change that occurs during solidification. Does the cast iron expand or contract during solidification?

Solution:

1/7.71−1/7.65 0.1297 cm 3 − 0.1307 cm 3 ×100% = ×100% = −0.78% 1/7.65 0.1307 cm3 The casting contracts.

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9-85

Molten copper at atmospheric pressure contains 0.01 wt% oxygen. The molten copper is placed in a chamber that is pumped down to 1 Pa to remove gas from the melt prior to pouring into the mold. Calculate the oxygen content of the copper melt after it is subjected to this degassing treatment.

Solution: According to Sievert’s law, the amount of gas that can be dissolved in a molten metal is given by

Percent of gas = K pgas , where K is a constant for a particular metal-gas system at constant temperature and pgas is the partial pressure of the gas in contact with the metal. Taking the ratio of the oxygen concentrations,

%O atm K patm = = %O1 Pa K p1 Pa

patm

.

p1 Pa

Solving for the oxygen content at 1 Pa,

%O1 Pa = %Oatm

p1 Pa . patm

Recalling that 1 atm = 1.013 × 105 Pa,

%O1 Pa = 0.01 9-86

1 = 3.14 ×10−5 wt% oxygen .

5 1.013 ×10

From Figure 9–14, find the solubility of hydrogen in liquid aluminum just before solidification begins when the partial pressure of hydrogen is 1 atm. Determine the solubility of hydrogen (in cm3/100 g Al) at the same temperature if the partial pressure were reduced to 0.01 atm.

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Solution: 0.46 cm3 H2/100 g aluminum

1 0.01 x = 0.46 0.01 = 0.046 cm 3 /100 g Al

0.46/x =

9-87

The solubility of hydrogen in liquid aluminum at 715°C is found to be 1 cm3/(100 g Al). If all of this hydrogen precipitated as gas bubbles during solidification and remained trapped in the casting, calculate the volume percent gas in the solid aluminum.

Solution: (1 cm3 H2/100 g Al)(2.699 g/cm3) = 0.02699 cm3 H2/cm3 Al = 2.699% 9-88

Review Example 9-7. The usual method of measuring a vacuum in the United States vacuum pump industry is in inches of mercury. For example, atmospheric pressure is 29.92 inches of mercury. Covert the pressure found in the example to in. Hg.

Solution: The pressure required is 10-6 atm. An atmosphere conversion factor is given, so: 10*1 atm 29.92 in. Hg N= 5 = 0.00002992 in. Hg 1 1 atm Note that the method usually used in industry is more confusing: A perfect vacuum is 29.92 in. Hg and atmospheric pressure is zero. This is the absolute value of the reading of a vacuum gage.

9-98

Why has continuous casting of steels and other alloys assumed increased importance?

Solution: The advantages of the continuous casting process are 1) more uniform composition and less segregation across the section of the ingot, 2) produces higher yield (less cropping) in primary shapes, 3) the shapes may be immediately deformed after casting with minimal reheating of the slabs, blooms or billets to form other desired shapes and 4), saves energy cost and increases productivity.

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Chapter 10: Solid Solutions and Phase Equilibrium 10-10 How much of a temperature change is required to bring liquid magnesium from solidliquid equilibrium to vapor-liquid equilibrium at 1 atmosphere? (See Figure 10-2.)

Solution: Reading the graph in Figure 10-2, we read about 433 °C difference between the two lines at 1 atm. If the Y axis were scaled, we could find out the temperature difference at any pressure.

10-11 The unary phase diagram for SiO2 is shown in Figure 10–18. Locate the triple point where solid, liquid, and vapor coexist and give the temperature and the type of solid present. What do the other “triple” points indicate?

Solution: (a) The solid-liquid-vapor triple point occurs at 1713°C; the solid phase present at this point is β-cristobalite. (b) The other triple points describe the equilibrium between two solids and a vapor phase.

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10-13 Natural diamond is formed approximately 120 to 200 km below the earth’s surface under high pressure and high temperature conditions. Assuming that the average density of the earth is 5500 kg/m3, use this information and the unary phase diagram for C (Figure 10–19) to calculate the range of the earth’s geothermal gradient (rate of increase of temperature with depth). Estimate the pressure below the earth’s surface as ρgh where ρ is density, g is gravity, and h is depth. Note that 10 kbar = 109 Pa.

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Solution:

The pressure P in the given depth range needs to be determined.

This can be estimated by P = ρgh, where ρ is the density, g is the acceleration due to gravity, and h is the depth. At 120 km, P120 = 5500 kg/m3 × 9.81 m/s2×120 × 103m = 6.47 × 109 Pa = 65 kbar.

At 200 km, P200 = 5500 kg/m3 × 9.81 m/s2 × 200 × 103m = 1.08 × 1010 Pa = 108 kbar.

The temperature range for which diamond is stable at 65 kbar and 108 kbar of pressure can be determined according to the unary phase diagram for carbon (Figure 10–20). At 108 kbar (200 km), the temperature range is approximately 4100°C to 4200°C and at 65 kbar (120 km), the temperature range is 4150°C to 4400°C. The range of the geothermal gradient is

At 200 km : (4100/200) to (4200/200), which is 20.5 to 21°C/km.

At 120 km : (4150/120) to (4400/120), which is 34.6 to 36.7°C/km.

Thus the range is from 20.5°C / km to 36.7°C / km.

It should be noted that tectonic and volcanic activity can significantly affect local temperature and pressure and also cause movement of material to different depths. Most natural diamond that is mined is brought closer to the surface by deep volcanic eruptions.

10-22 A 15 kg block of copper is being alloyed with zinc and brought to the solubility limit at 200°C. How much zinc is required? (See Figure 10-4)

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Solution: Reading the graph, we see that the solubility limit at 200 °C is about 35 wt% Zn. Thus,

= 0.35 + 15 kg Cu = 8.1 kg Zn 10-23 A 115 kg total of copper and zinc (4 parts Zn to 6 parts Cu) is mixed at 400°C. How much of the zinc is in excess of solubility? (See Figure 10-4.)

Solution: First we separate the 115 kg mathematically into 46 kg Zn and 69 kg Cu. In this case we know that there is more zinc than the copper can absorb at this temperature. Figure 104 shows that the solubility limit at 400°C is about 38%. We set up a relation of the amount of zinc and copper in solution:

= 0.38 69 kg Cu + = 42.3 kg Cu Taking the difference between this amount in solution and the amount supplied, we find the excess is 46 kg Cu − 42.3 kg Cu = 3.7 kg Cu 10-26 What is the difference between limited and unlimited solid solubility?

Solution: Unlimited solid solubility is when two phases are mixed together producing one single phase in both liquid and solid form. The liquid would have the same composition, properties and structure everywhere and if cooled would produce a single uniform structure with no existent interface. Limited solid solubility is when two phases are mixed together producing a two-phase solid upon solidification (similar to having too much salt in water) due to the saturation of one alloy into the other. Solubility increases with temperature.

10-27 Copper and nickel have unlimited solid solubility in each other whereas lead and iron are completely insoluble. Using the Hume-Rothery rules, justify these conditions. Show your work.

Solution: Based on Hume-Rothery’s Rules, we would predict that Cu-Ni are 100% soluble while FePb are not soluble:

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Element

Structure

%Δr

Valence

Unlimited solubility

Cu

FCC

2.82% +2

Yes

Ni

FCC

2.82% +1

Yes

Pb

FCC

41.1% +4

No

Fe

BCC/FCC

41.1% +2, +3

No

10-28 Which of the following elements would be expected to have unlimited solid solubility in gold? (a) Ag; (b) Al; and (c) Fe. Solution: Of the elements, silver is the only element to potentially have unlimited solubility with gold since all values are predominately the same. Aluminum is structurally close, but the electronic state is very different from gold. Iron electronic state and crystal structures differ from gold making it insoluble in gold.

Element

Structure

%Δr

Electronegativity Valence

Unlimited solubility

Au

FCC

0

2.4

+1 Yes

Ag

FCC

0

1.9

+1 Yes

Al

FCC

0.69%

1.5

+3 No

Fe

BCC/FCC

13.80%

1.8 +2, +3

No

10-29 Based on Hume-Rothery’s conditions, which of the following systems would be expected to display unlimited solid solubility? Explain. (a) Au–Ag (b) Al–Cu (c) Al–Au (d) U–W (e) Mo–Ta (f) Nb–W (g) Mg–Zn (h) Mg–Cd Solution: (a) rAu = 1.442 rAg = 1.445 Δr = 0.2 % (b) rA1 = 1.432 rCu = 1.278 Δr = 10.7 % (c) rAl = 1.432 rAu = 1.442 Δr = 0.7 % (d) rU = 1.38 rW = 1.371 Δr = 0.7 %

v = +1 FCC v = + 1 FCC Yes v = +3 FCC v = +1 FCC No v = +3 FCC v = +1 FCC No v = +4 Ortho v = +4 FCC No

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(e) rMo = 1.363 v = +4 BCC rTa = 1.43 v = +5 BCC Δr = 4.7 % No (f) rNb = 1.426 v = +4 BCC rW = 1.371 v = +4 BCC Δr = 3.6 % Yes (g) rMg = 1.604 v = +2 HCP rZn = 1.332 v = +2 HCP Δr = 17 % No (h) rMg = 1.604 v = +2 HCP rCd = 1.490 v = +2 HCP Δr = 7.1 % Yes The Au–Ag, Mo–Ta, and Mg–Cd systems have the required radius ratio, the same crystal structures, and the same valences. Each of these might be expected to display complete solid solubility. (The Au– Ag and Mo–Ta do have isomorphous phase diagrams. In addition, the Mg–Cd alloys all solidify like isomorphous alloys; however, a number of solid-state phase transformations complicate the diagram.)

10-30 Identify which of the following oxides when added to BaTiO3 are likely to exhibit 100% solid solubility: (a) SrTiO3; (b) CaTiO3; (c) ZnTiO3; and (d) BaZrO3. All of these oxides have a perovskite crystal structure.

Solution:

The Hume-Rothery rules must be checked for the various combinations.

Crystal Structure: All of these oxides have the perovskite crystal structure.

Values for the valence and atomic radii of the elements are given in Appendix B and shown in the table below.

Atom Valence Atomic Radius (Å) Ba +2 2.176 Sr +2 2.151 Ca +2 1.976 Zn +2 1.332 Ti +4 1.475 Zr +4 1.616

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Note that all of the oxides have a formula ABO3 where A = atom with valence +2 and B = atom with valence +4. Thus we need to compare the atoms of the same valence for their size factor.

Valence: All oxides pass the criterion for valence of ions.

Size: The combinations of Ba:Sr and Ba:Ca meet the 15% rule while the combination of Ba:Zn does not. The combination of Ti:Zr meets the 15% rule.

Electronegativity: All systems meet the electronegativity criterion.

Based on the information above, SrTiO3, CaTiO3 and BaZrO3 are likely to exhibit 100% solid solubility while ZnTiO3 may not.

10-31 Suppose 1 at% of the following elements is added to copper (forming a separate alloy with each element) without exceeding the solubility limit. Which one would be expected to give the higher strength alloy? Are any of the alloying elements expected to have unlimited solid solubility in copper? (a) Au (b) Mn (c) Sr (d) Si (e) Co Solution:

For copper: rCu = 1.278 Å (a) Au: r = 1.442

May be unlimited r −r ∆r = Au Cu = +12.8% solubility rCu

(b) Mn: r = 1.12 ∆r = –12.4 % Different structure (c) Sr: r = 2.151 ∆r = +68.3 % Highest Strength (d) Si r = 1.176 ∆r = –8.0 % Different structure (e) Co: r = 1.253 ∆r = –2.0 % Different structure The Cu–Sr alloy would be expected to be strongest (largest size difference). The Cu–Au alloy satisfies Hume-Rothery’s conditions and might be expected to display complete solid solubility — in fact, it freezes like an isomorphous series of alloys, but a number of solidstate transformations occur at lower temperatures.

10-32 Suppose 1 at% of the following elements is added to aluminum (forming a separate alloy with each element) without exceeding the solubility limit. Which one would be expected to give the higher strength alloy? Why? (a) Mn; (b) Mg; (c) Cu; and (d) Zn.

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Solution: For aluminum: r = 1.432 Å (FCC structure with valence of 3) (a) Mn: r = 1.12

∆r = –21.8%

(b) Mg: r = 1.604

∆r = 12.0%

(c) Cu: r = 1.278

∆r = –10.8%

(d) Zn: r = 1.332

∆r = 7.0%

Mn is expected to give the higher strength alloy since the Mn atoms are most different in size from the Al atoms.

10-33 Suppose 1 at% of the following elements is added to aluminum (forming a separate alloy with each element) without exceeding the solubility limit. Which one would be expected to give the smallest reduction in electrical conductivity? Are any of the alloy elements expected to have unlimited solid solubility in aluminum? (a) Li (b) Ba (c) Be (d) Cd (e) Ga Solution: For aluminum: r = 1.432 Å (FCC structure with valence of 3) (e) Li: r = 1.519 ∆r = 6.1% BCC valence = 1 (f) Ba: r = 2.176 ∆r = –52.0% BCC valence = 2 (g) Be: r = 1.143 ∆r = –20.2% HCP valence = 2 (h) Cd: r = 1.49 ∆r = 4.1% HCP valence = 2 (i) Ga: r = 1.218 ∆r = 14.9% Orthohombic valence = 3 The cadmium would be expected to give the smallest reduction in electrical conductivity, since the Cd atoms are most similar in size to the aluminum atoms. None are expected to have unlimited solid solubility, due either to difference in valence, atomic radius, or crystal structure.

10-34 Which of the following oxides is expected to have the largest solid solubility in Al2O3? (a) Y2O3 (b) Cr2O3 (c) Fe2O3 Solution: The ionic radius of Al3+ = 0.51 Å 0.63 − 0.51 (a) rY3+ = 0.89

∆r =

0.51

(b) rCr3+ = 0.63

Δr = 23.5%

(c) rFe3+ = 0.64

Δr = 25.5%

× 100 = 74.5%

We would expect Cr2O3 to have a high solubility in Al2O3; in fact, they are completely soluble in one another.

10-38 What is a binary phase diagram and what information can be learned from it? In terms of thermodynamics, what is meant by the term “equilibrium phase diagram”?

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Solution: A binary phase diagram shows the different equilibrium phase that may exist between two elements when mixed or alloyed together. The diagram may be used as a guide in the processing of an alloy consisting predominately of the two elements. These are called equilibrium diagrams because they pertain to very slow processes that allow the ultimate phases to be established. In terms of thermodynamics, we consider these phases to be stable with the lowest energy.

10-39 What is the significance of the liquidus curve? Solidus curve?

Solution: The liquidus curve indicates the highest temperature the solid phase may exist in an alloy system. The solidus curve indicates the lowest temperature the liquid phase may exist in an alloy system.

10-40 What is an alloy? In general, what is the difference in the melting behavior of an alloy from that of a pure component?

Solution: An alloy is a material composed of two or more elements. While a pure component has a definite melting point, an alloy generally has a temperature range where the solid and liquid phases co-exist before complete melting is attained.

10-41 Determine the degrees of freedom under the following conditions: (a) Tl–20 wt% Pb at 325°C and 400°C; (b) Tl–40 wt% Pb at 325°C and 400°C; (c) Tl–90 wt% Pb at 325°C and 400°C. Refer to the phase diagram in Figure 10–8(d).

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Solution: The Gibbs Phase Rule is C + 1 = F + P, where C is the number of components, F is the number of degrees of freedom, and P is the number of phases present for an equilibrium phase diagram at constant pressure.

For Tl–20 wt% Pb at 325°C, P = 2 (solid solution and liquid) and C = 2 (Pb and Tl), hence F = C–P + 1 = 1 There is one degree of freedom.

For Tl–20 wt% Pb at 400°C, P = 1 (liquid) and C = 2 (Pb and Tl), hence F = C–P + 1 = 2–1 + 1 = 2 There are two degrees of freedom.

For Tl–40 wt% Pb at 325°C, P = 1 (solid solution) and C = 2 (Pb and Tl), hence F = C–P + 1 = 2–1 + 1 = 2 There are two degrees of freedom.

For Tl–40 wt% Pb at 400°C, P = 1 (liquid) and C = 2 (Pb and Tl), hence F = C–P + 1 = 2–1 + 1 = 2 There are two degrees of freedom.

For Tl–90 wt% Pb at 325°C, P = 2 (solid solution and liquid) and C = 2 (Pb and Tl), hence F = C–P + 1 = 1 There is one degree of freedom.

For Tl–90 wt% Pb at 400°C, P = 1 (liquid) and C = 2 (Pb and Tl), hence F = C–P + 1 = 2–1 + 1 = 2 There are two degrees of freedom.

10-42 Locate the following points in the Bi-Sb phase diagram and indicate the phases present and their relative amounts (a) 60 at% Bi at 250°C; (b) 30 at% Bi at 500°C; and (c) 50 at% Bi at 600°C. (See Figure 10-20.)

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Solution: (a) 60 at% Bi = 40 at% Sb at 250˚C S: 40 at% Sb

100% S

(b) 30 at% Bi = 70 at% Sb There are two phases present: S and L. S: 86 at% Sb L: 55 at% Sb

at% S =

70 − 55 ×100% = 48% 86 − 55

at% L =

86 − 70 ×100% = 52% 86 − 55

(c) 50 at% Bi = 50 at% Sb L: 50 at% Sb

100% L

10-43 Determine the composition range in which the Tl–Pb alloy at 350°C is (a) fully liquid; (b) fully solid; and (c) partly liquid and partly solid. Refer to Figure 10–8(d) for the Tl–Pb phase diagram. Further, determine the amount of liquid and solid solution for Tl–25 wt% Pb and Tl–75 wt% Pb at 350°C and also the wt% Pb in the liquid and solid solution for both of the alloy compositions.

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Solution: At 350°C, the Tl–Pb system exists as various phases as follows: Fully liquid: From 0 wt% to 22 wt% Pb and from 82 wt% to 100 wt% Pb Fully solid or solid solution: From 30 wt% to 68 wt% Pb Partly liquid and partly solid: From 22 wt% to 30 wt% Pb and 68 wt% to 82 wt% Pb. Using the lever rule, the amount of liquid and solid solution can be determined.

For a Tl-25 wt% Pb alloy at 350°C, % Liquid = [(30–25)/(30–22)]] × 100 = 62.5% % Solid solution = [(25–22)/(30–22)] × 100 = 37.5% For a Tl-75 wt% Pb alloy at 350°C, % Liquid = [(75–68)/(82–68)]] × 100 = 50% % Solid solution = [(82–75)/(82–68)] × 100 = 50% The wt% Pb in the solid solution and liquid are determined by the point where the tie line intersects the solidus and liquidus, respectively.

For a Tl-25 wt% Pb alloy at 350°C, there is 22 wt% Pb in the liquid and 30 wt% Pb in the solid solution.

For a Tl-75 wt% Pb alloy at 350°C, there is 82 wt% Pb in the liquid and 68 wt% Pb in the solid solution.

10-44 Determine the liquidus temperature, solidus temperature, and freezing range for the following NiO–MgO ceramic compositions. [See Figure 10–8(b).] (a) NiO–30 mol% MgO (b) NiO–45 mol% MgO (c) NiO–60 mol% MgO (d) NiO–85 mol% MgO

Solution: (a) TL = 2330°C (b) TL = 2460°C (c) TL = 2570°C (d) TL = 2720°C

Ts = 2150°C Ts = 2250°C Ts = 2380°C Ts = 2610°C

FR = 180°C FR = 210°C FR = 190°C FR = 110°C

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10-45 Determine the liquidus temperature, solidus temperature and freezing range for the following Al2O3 – Cr2O3 ceramic compositions: (a) Al2O3 – 30wt% Cr2O3, (b) Al2O3 – 50wt% Cr2O3 and (c) Al2O3 – 75wt% Cr2O3. (See Figure 10-9.)

Solution: a) TL = 2125°C TS = 2075°C

Freezing range = 50°C

b) TL = 2175°C TS = 2100°C

Freezing range = 75°C

c) TL = 2225°C

Freezing range = 65°C

TS = 2260°C

10-46 Determine the phases present, the compositions of each phase, and the amount of each phase in wt% for the following Al2O3 – Cr2O3 ceramic at 2150°C: (a) Al2O3 – 30wt% Cr2O3, (b) Al2O3 – 50wt% Cr2O3 and (c) Al2O3 – 75wt% Cr2O3. (See Figure 10-9.)

Solution:

10-47 For an Ag-60 wt% Pd alloy determine the (a) liquidus temperature; (b) solidus temperature; and (c) freezing range. (See Figure 10-21.)

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Solution: (a) The liquidus temperature is 1270°C (b) The solidus temperature is 1230°C (c) The freezing range is 40°C 10-48 Locate the following points in the Ag-Pd phase diagram and indicate the phases present and their relative amounts (a) 50 wt% Pd at 1300°C; (b) 80 wt% Pd at 1425°C; and (c) 90 wt% Ag at 1100°C. (See Figure 10-21.)

Solution:

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10-49 Determine the liquidus temperature, solidus temperature, and freezing range for the following MgO–FeO ceramic compositions. (See Figure 10–22.) (a) MgO–25 wt% FeO (b) MgO–45 wt% FeO (c) MgO–65 wt% FeO (d) MgO–80 wt% FeO

Solution: (a) TL = 2600°C (b) TL = 2340°C (c) TL = 2000°C (d) TL = 1750°C

Ts = 2330°C Ts = 1900°C Ts = 1610°C Ts = 1480°C

FR = 370°C FR = 440°C FR = 390°C FR = 270°C

10-50 Determine the phases present, the compositions of each phase, and the amount of each phase in mol% for the following NiO–MgO ceramics at 2400°C. [See Figure 10–8(b).] (a) NiO–30 mol% MgO (b) NiO–45 mol% MgO (c) NiO–60 mol% MgO (d) NiO–85 mol% MgO Solution: (a) L: NiO–30 mol % MgO 100% L (b) L: 38% MgO 62 − 45

%L=

S: 62% MgO (c) L: 38% MgO S: 62 % MgO

×100% = 70.8% 62 − 38 45 − 38 %S= × 100% = 29.2% 62 − 38 62 − 60 %L= × 100% = 8.3% 62 − 38 60 − 38 %S= × 100% = 91.7% 62 − 38

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(d) S: 85% MgO

100% S

10-51 Determine the phases present, the compositions of each phase, and the amount of each phase in wt% for the following MgO–FeO ceramics at 2000°C. (a) MgO–25 wt% FeO (b) MgO–45 wt% FeO (c) MgO–60 wt% FeO (d) MgO–80 wt% FeO (See Figure 10–21.)

Solution:

(a) S: 25 % FeO (b) S: 39% FeO L: 65% FeO (c) S: 39% FeO L: 65% FeO (d) S: 80% MgO

100% S

65 − 45 ×100% = 76.9% 65 − 39 45 − 39 %L = × 100% = 23.1% 65 − 39 65 − 60 %S= ×100% = 19.2% 65 − 39 60 − 39 %L= × 100% = 80.8% 65 − 39 %S=

100% L

10-52 Consider a ceramic composed of 30 mol% MgO and 70 mol% FeO. Calculate the composition of the ceramic in wt%.

Solution: MWMgO = 24.312 + 16 = 40.312 g/mol MWFeO = 55.847 + 16 = 71.847 g/mol

(30)(40.312) × 100% = 19.4% (30)(40.312) + (70)(71.847) (70)(71.847) wt% FeO = × 100% = 80.6% (30)(40.312) + (70)(71.847) wt% MgO =

10-53 A NiO–20 mol% MgO ceramic is heated to 2200°C. Determine (a) the composition of the solid and liquid phases in both mol% and wt%; (b) the amount of each phase in both mol% and wt%; and (c) assuming that the density of the solid is 6.32 g/cm3 and that of the liquid is 7.14 g/cm3, determine the amount of each phase in vol% [See Figure 10– 8(b).]

Solution:

MWMgO = 24.312 + 16 = 40.312 g/mol MWNiO = 58.71 + 16 =74.71 g/mol (a) L: 15 mol% MgO

wt % MgO =

(15)(40.312) × 100% = 8.69% (15)(40.312) + (85)(74.71)

S: 38 mol % MgO

wt% MgO =

(38)(40.312) × 100% = 24.85% (38)(40.312) + (62)(74.71)

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(b) mol% L =

38 − 20 × 100% = 78.26% mol% S = 21.74% 38 − 15

The original composition, in wt% MgO, is

(20)(40.312) × 100% = 11.9% (20)(40.312) + (80)(74.71) 24.85 − 11.9 wt% L = × 100% = 80.1% wt% S = 19.9% 24.85 − 8.69 80.1/ 7.14 (c) vol% L = × 100% = 78.1% (80.1 / 7.14) + (19.9 / 6.32) vol% S = 21.9% 10-54 A Nb–60 wt% W alloy is heated to 2800°C. Determine (a) the composition of the solid and liquid phases in both wt% and at%; (b) the amount of each phase in both wt% and at%; and (c) assuming that the density of the solid is 16.05 g/cm3 and that of the liquid is 13.91 g/cm3, determine the amount of each phase in vol%. (See Figure 10–23.)

Solution: (d) L: 49 wt % W

at% W =

49/183.85 × 100% = 32.7% (49/183.85) + (51/92.91)

α: 70 wt % W

(70/183.85) × 100% = 54.1 (70/183.85) + (30/92.91) 70 − 60 (e) wt% L = × 100% = 47.6% wt% α = 52.4% 70 − 49 at% W =

The original composition, in wt% W is

60/183.85 × 100% = 43.1% (60/183.85) + (40/92.91) 54.1 − 43.1 at% L = ×100% = 51.4% wt% α = 48.6% 54.1 − 32.7 47.6/13.91 (f) vol % L = × 100% = 51.2% (47.6/13.91) + (52.4/16.05) vol% α = 48.8% 10-55 How many grams of nickel must be added to 500 grams of copper to produce an alloy that has a liquidus temperature of 1350°C? What is the ratio of the number of nickel atoms to copper atoms in this alloy? [See Figure 10–8(a).]

Solution: We need 60 wt% Ni to obtain the correct liquidus temperature.

x × 100% or x = 750g Ni x + 500g Ni atoms (750g)(N A )/(58.71g/mol) = = 1.62 Cu atoms (500g)(N A )/(63.54g/mol)

% Ni = 60 =

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10-56 How many grams of nickel must be added to 500 grams of copper to produce an alloy that contains 50 wt% α at 1300°C? [See Figure 10–8(a).]

Solution: At 1300°C, the composition of the two phases in equilibrium are L: 46 wt % Ni and α: 58 wt% Ni The alloy required to give 50% α is then

x − 46 × 100 = 50% α 58 − 46

or

x = 52 wt % Ni

The number of grams of Ni must be

x × 100% = 52 or x + 500

x = 541.7 g Ni

10-57 How many grams of MgO must be added to 1 kg of NiO to produce a ceramic that has a solidus temperature of 2200°C? [See Figure 10–8(b).]

Solution: MWMgO = 40.312 g/mol MWNiO = 74.71g/mol 38 mol% MgO is needed to obtain the correct solidus temperature.

wt% MgO =

(38)(40.312) × 100% = 24.85% (38)(40.312) + (62)(74.71)

The number of grams required is

x × 100% = 24.85% or x + 100

x = 331 g of MgO

10-58 How many grams of MgO must be added to 1 kg of NiO to produce a ceramic that contains 25 mol% solid at 2400°C? [See Figure 10–8(b).]

Solution: L:38 mol% MgO S: 62 mol% MgO

MWMgO = 40.312g/mol MWNiO = 74.71g/mol

x − 38 × 100% = 25%S or x = 44 mol% MgO 62 − 38 (44)(40.312) wt% MgO = × 100% = 29.77% (44)(40.312) + (56)(74.71) The number of grams of MgO is then

x × 100% = 29.77% or x + 1000

x = 424g MgO

10-59 We would like to produce a solid MgO–FeO ceramic that contains equal mol percentages of MgO and FeO at 1200°C. Determine the wt% FeO in the ceramic. (See Figure 10–22.)

Solution:

Only solid is present at 1200°C.

50 mol% FeO:

MWMgO = 40.312g/mol MWFeO = 71.847g/mol

(50)(71.847) = 64.1 wt % FeO (50)(40.312) + (50)(71.847)

10-60 We would like to produce a MgO–FeO ceramic that is 30 wt% solid at 2000°C. Determine the composition of the ceramic in wt%. (See Figure 10–22.)

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Solution:

L: 65 wt% FeO

30 wt % =

S: 38 wt% FeO

65 − x ×100% or 65 − 38

x = 56.9 wt % FeO

10-61 A Nb–W alloy held at 2800°C is partly liquid and partly solid. (a) If possible, determine the composition of each phase in the alloy, and (b) if possible, determine the amount of each phase in the alloy. (See Figure 10–23.)

Solution:

(a) L:49 wt% W α:70 wt% W (b) Not possible unless we know the original composition of the alloy.

10-62 A Nb–W alloy contains 55% α at 2600°C. Determine (a) the composition of each phase, and (b) the composition of the alloy. (See Figure 10–23.)

Solution: L:22 wt% W

x − 22 0.55 = 42 − 22

α: 42 wt% W

or

x = 33wt% W

10-63 Suppose a 1200 lb bath of a Nb–40 wt% W alloy is held at 2800°C. How many pounds of tungsten can be added to the bath before any solid forms? How many pounds of tungsten must be added to cause the entire bath to be solid? (See Figure 10–23.)

Solution: Solid starts to form at 2800°C when 49 wt% W is in the alloy. In 1200 lb of the original Nb–40% W alloy, there are (0.4)(1200) = 480 lb W and 720 lb Nb. The total amount of tungsten that must be in the final alloy is

0.49 =

x x + 720

or

x = 692 lb W total

or 692–480 = 212 additional pounds of W must be added To be completely solid at 2800°C, the alloy must contain 70 wt% W. The total amount of tungsten required in the final alloy is

0.70 =

x x + 720

or

x = 1680 lb W total

or 1680–480 = 1200 additional pounds of W must be added 10-64 A fiber-reinforced composite material is produced, in which tungsten fibers are embedded in a Nb matrix. The composite is composed of 70 vol% tungsten. (a) Calculate the wt% of tungsten fibers in the composite, and (b) suppose the composite is heated to 2600°C and held for several years. What happens to the fibers? Explain. (See Figure 10– 23.)

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Solution: (a) wt =

(70cm3 )(19.254 g / cm3 ) = 83.98 wt % W (70)(19.254) + (30)(8.57)

(b) The fibers will dissolve. Since the W and Nb are completely soluble in one another and the temperature is high enough for rapid diffusion, a single solid solution will eventually be produced.

10-65 Suppose a crucible made from pure nickel is used to contain 500 g of liquid copper at 1150°C. Describe what happens to the system as it is held at this temperature for several hours. Explain. [See Figure 10–8(a).]

Solution: Cu dissolves Ni until the Cu contains enough Ni that it solidifies completely. When 10% Ni is dissolved, freezing begins.

0.10 =

x or x = 55.5 g Ni x + 500

When 18% Ni dissolved, the bath is completely solid.

0.18 =

x or x = 109.8 g Ni x + 500

10-68 A Bi-60 at% Sb alloy is cooled under equilibrium conditions from the liquid state.

Determine (a) the temperature at which the first solid nucleus forms and its composition; (b) the compositions and relative amounts of the phases at 450°C; and (c) the temperature at which the last liquid is present and its composition. (See Figure 1020.)

Solution: (a) 510˚C, 88% at% Sb (b) There are two phases present: S and L. S: 80 at% Sb L: 42 at% Sb

at% S =

60 − 42 ×100% = 47% 80 − 42

at% L =

80 − 60 ×100% = 53% 80 − 42

(c) 360˚C, 20 at% Sb 10-69 Equal moles of MgO and FeO are combined and melted. Determine (a) the liquidus temperature, the solidus temperature, and the freezing range of the ceramic, and (b) determine the phase(s) present, their composition(s), and their amount(s) at 1800°C. (See Figure 10–22.)

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Solution: MWMgO = 40.312 g/mol

wt % FeO =

(1mol FeO)(71.847 g/mol) = 64.1% (1 mol FeO)(71.847) + (1 mol MgO)(40.312)

(a) TLiq = 2000°C (b) L: 75% FeO

%L=

MWFeO = 71.847 g/mol

Ts = 1620°C S: 50% FeO

FR = 380°C

64.1 − 50 × 100% = 56.4% % S = 43.6% 75 − 50

10-70 Suppose 75 cm3 of Nb and 45 cm3 of W are combined and melted. Determine (a) the liquidus temperature, the solidus temperature, and the freezing range of the alloy, and (b) determine the phase(s) present, their composition(s), and their amount(s) at 2800°C. (See Figure 10–23.)

Solution:

(45cm3 )(19.254 g/cm3 ) wt % W = ×100 = 57.4wt % (45)(19.254) + (75)(8.57) (a) TLiq = 2900°C (b) L: 49% W

Tss = 2690°C

α: 70% W

% α = 40%

FR = 210°C

70 − 57.4 %L = = 60% 70 − 49

10-71 A NiO–60 mol% MgO ceramic is allowed to solidify. Determine (a) the composition of the first solid to form, and (b) the composition of the last liquid to solidify under equilibrium conditions. [See Figure 10-8(b).]

Solution: 1st α: 80% MgO Last L:35% MgO 10-72 A Nb–35% W alloy is allowed to solidify. Determine (a) the composition of the first solid to form, and (b) the composition of the last liquid to solidify under equilibrium conditions. (See Figure 10–23.)

Solution: (a) 57 wt% W at 2675˚C (b) 15 wt% W at 2560˚C 10-73 For equilibrium conditions and a MgO–65 wt% FeO ceramic, determine (a) the liquidus temperature; (b) the solidus temperature; (c) the freezing range; (d) the composition of the first solid to form during solidification; (e) the composition of the last liquid to solidify; (f) the phase(s) present, the composition of the phase(s), and the amount of the phase(s) at 1800°C; and (g) the phase(s) present, the composition of the phase(s), and the amount of the phase(s) at 1600°C. (See Figure 10–22.)

Solution: (a) Liquidus = 2000°C (b) Solidus = 1605°C (c) Freezing range = 2000–1605 = 395°C (d) First Solid: 40% FeO (e) Last liquid: 88% FeO

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(f) L: 75% FeO α: 51% FeO (g) α: 65% FeO

%L=

65 − 51 × 100% = 58% 75 − 51

% α = 42% 100% α

10-74 Figure 10–24 shows the cooling curve for a NiO–MgO ceramic. Determine (a) the liquidus temperature; (b) the solidus temperature; (c) the freezing range; (d) the pouring temperature; (e) the superheat; (f) the local solidification time; (g) the total solidification time; and (h) the composition of the ceramic.

Solution: (a) Liquidus = 2690°C (b) Solidus = 2570°C (c) Freezing range = 2690–2570 = 120°C (d) Pouring temperature = 2775°C (e) Superheat = 275–2690 = 85°C (f) Local Solidification time = 27–5 = 22 min (g) Total solidification time = 27 min (h) 80% MgO 10-75 For equilibrium conditions and a Nb–80 wt% W alloy, determine (a) the liquidus temperature; (b) the solidus temperature; (c) the freezing range; (d) the composition of the first solid to form during solidification; (e) the composition of the last liquid to solidify; (f) the phase(s) present, the composition of the phase(s), and the amount of the phase(s) at 3000°C; and (g) the phase(s) present, the composition of the phase(s), and the amount of the phase(s) at 2800°C. (see Figure 10–23.)

Solution: (a) Liquidus = 3100°C (b) Solidus = 2920°C (c) Freezing range = 3100–2920 = 180°C (d) First solid: 90% W (e) Last liquid: 64% W

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(f) L: 70% W

%L =

85 − 80 × 100% = 33.3% 85 − 70

α: 85% W % α = 66.7% (g) α: 80% W 100% α 10-76 Figure 10–25 shows the cooling curve for a Nb–W alloy. Determine (a) the liquidus temperature; (b) the solidus temperature; (c) the freezing range; (d) the pouring temperature; (e) the superheat; (f) the local solidification time; (g) the total solidification time; and (h) the composition of the alloy.

Solution:

Liquidus = 2900°C Solidus = 2710°C Freezing range = 2900–2710 = 190°C Pouring temperature = 2990°C Superheat = 2990–2900 = 90°C Local Solidification time = 340–40 = 300 s Total solidification time = 340 s Nb–60 wt% W

10-77 Cooling curves are shown in Figure 10–25 for several Mo–V alloys. Based on these curves, construct the Mo–V phase diagram.

Solution: 0% V 20% V 40% V 60% V 80% V 100% V

TLiquidus 2630°C 2500°C 2360°C 2220°C 2100°C 1930°C

TSolidus 2320°C 2160°C 2070°C 1970°C

10-79 For the nonequilibrium conditions shown for the MgO–65 wt% FeO ceramic, determine (a) the liquidus temperature; (b) the nonequilibrium solidus temperature; (c) the freezing range; (d) the composition of the first solid to form during solidification; (e) the composition of the last liquid to solidify; (f) the phase(s) present, the composition of the phase(s), and the amount of the phase(s) at 1800°C; and (g) the phase(s) present, the

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composition of the phase(s), and the amount of the phase(s) at 1600°C. (See Figure 10– 22.)

Solution: (a) Liquidus = 2000°C (b) Solidus = 1450°C (c) Freezing range = 2000–1450 = 550°C (d) First solid: 40% FeO (e) Last liquid: 92% FeO (f) L: 75% FeO 65 − 46

%L=

75 − 46

S:46% FeO % S = 34.5% (g) L: 88% FeO 65 − 55

%L=

S: 55%FeO

88 − 55

×100% = 65.5%

×100% = 30.3%

% S = 69.7%

10-80 For the nonequilibrium conditions shown for the Nb–80 wt% W alloy, determine (a) the liquidus temperature, (b) the nonequilibrium solidus temperature, (c) the freezing range, (d) the composition of the first solid to form during solidification, (e) the composition of the last liquid to solidify, (f) the phase(s) present, the composition of the phase(s), and the amount of the phase(s) at 3000°C, and (g) the phase(s) present, the composition of the phase(s), and the amount of the phase(s) at 2800°C. (See Figure 10– 23.)

Solution: (a) Liquidus = 3100°C (b) Solidus = 2720°C (c) Freezing range = 3100–2720 = 380°C (d) First solid: 90% W (e) Last liquid: 40% W (f) L:70% W 88 − 80

%L=

α:88% W (g) L:50% W

88 − 70

× 100% = 44.4%

% α = 55.6%

%L=

83 − 80 × 100% = 9.1% 83 − 50

α: 83% W % α = 90.9% 10-84 A copper–nickel alloy that solidifies with a secondary dendrite arm spacing (SDAS) of 0.001 cm requires 15 hours of homogenization heat treatment at 1100°C. What is the homogenization time required for the same alloy with a SDAS of 0.01 cm and 0.0001 cm? If the diffusion coefficient of Ni in Cu at 1100°C is 3 × 10–10 cm2/s, calculate the constant c in the homogenization time equation. What assumption is made in this calculation?

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Solution:

The relationship between homogenization time and the SDAS is given

by Equation 10-6: t = c (SDAS)2/Ds. For SDAS = 0.001 cm, t = 15 hr = 15 × 3600 = 54,000 s, c = (54,000 × 3 × 10–10)/(0.001)2 c = 16.2 s.

For the same alloy, for an SDAS of 0.01 cm, the homogenization time is given by t = 16.2(0.01)2/(3 × 10–10) t = 5,400,00 s t = 1500 hrs (or 62.5 days).

For an SDAS of 0.0001 cm, the homogenization time is given by t = 16.2 (0.0001)2/(3 × 10–10) t = 540 s (or 9 minutes).

One of the assumptions made in this calculation is that the diffusion coefficient does not change as the composition of the copper-nickel matrix changes during homogenization.

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Chapter 11: Dispersion Strengthening and Eutectic Phase Diagrams 11-9

Under what circumstances is the total composition the same as the phase composition or compositions? Under what circumstances is it different?

Solution: In a one phase region the alloy or total composition is the same as the phase composition. When we have more than one phase and are not exactly on the solubility (saturation) line, the phase compositions are given by the termination (end points) of the tie lines. The alloy or total composition lies somewhere between the phase compositions and is only useful to determine phase quantities.

11-10 Why is it common practice to only label one-phase regions in a binary phase diagram?

Solution: We know by simple observation that each two phase region is bounded by corresponding single phase regions. There is no reason to “clutter” the phase diagrams, especially the more complex diagrams because they would look rather busy. Therefore, only the single phase regions in the Cu-Zn diagram need to be identified and labeled for consistency among users. Also note that each single phase identified has unique characteristics such as crystal structure.

11-11 A hypothetical phase diagram is shown in Figure 11–26. (a) Are any intermetallic compounds present? If so, identify them and determine whether they are stoichiometric or nonstoichiometric. (b) Identify the solid solutions present in the system. Is either material A or B allotropic? Explain. (c) Identify the three-phase reactions by writing down the temperature, the reaction in equation form, the composition of each phase in the reaction, and the name of the reaction.

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Solution: (a) θ = non-stoichiometric intermetallic compound. (b) α, η, γ, and β; material B is allotropic, existing in three different forms at different temperatures (c) 1100°C: γ + L → β; peritectic; L:82% B γ: 97% B β: 90% B L1: 28% B 900°C: L1 → L2 + α monotectic; L2: 50% B α: 5% B L: 60% B 690°C: L → α + β; eutectic; 600°C: α + β → θ;

peritectoid;

300°C: β → θ + η;

eutectoid;

α: 5% B β: 90% B α: 5% B β: 80% B θ: 37% B β: 90% B θ: 40% B η: 95% B

11-12 Consider Figure 11-5. For an equal component alloy of A and B starting at 1400°C, write each reaction as the crucible is cooled to room temperature (25°C).

Solution: Note that the miscibility gap does not have a reaction associated with it, only a separation is occurring.

At the monotectic: L1 + L2 → γ + L2 At the eutectic: L2 + γ → γ + β At the eutectoid: γ + β → α + β At the peritectoid: α + β → μ

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These reaction equations do not eliminate the “spectator” components. The phase diagram does not indicate the stoichiometric ratios of each phase in the reactions, so some of the other phases may be consumed or generated. For example, at the eutectic, the coefficients could be L2 + 2γ → γ + β.

11-13 The Cu–Zn phase diagram is shown in Figure 11–27. (a) Are any intermetallic compounds present? If so, identify them and determine whether they are stoichiometric or nonstoichiometric. (b) Identify the solid solutions present in the system. (c) Identify the three-phase reactions by writing down the temperature, the reaction in equation form, and the name of the reaction.

Solution: (a) β, β′, γ, δ, ∈: all nonstoichiometric. (b) α, θ (c) 900°C: α + L → β; peritectic 830°C: β + L → γ; peritectic 700°C: γ + L → δ; peritectic 600°C: δ + L → ε; peritectic 550°C: δ → γ + ε; eutectoid 420°C: ε + L → θ; peritectic 250°C: β′ → α + γ; eutectoid 11-14 The Al–Li phase diagram is shown in Figure 11–28. (a) Are any intermetallic compounds present? If so, identify them and determine whether they are stoichiometric or nonstoichiometric. Determine the formula for each compound. (b) Identify the threephase reactions by writing down the temperature, the reaction in equation form, the composition of each phase in the reaction, and the name of the reaction.

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Solution: (a) β is non-stoichiometric @ 21 wt% Li:

at% Li =

21g /(6.94 g/mol) × 100% = 50 at% Li ∴ AlLi 21/6.94 + 79/26.981

γ, is stoichiometric @ 34 wt% Li:

at% Li =

34 g/(6.94 g/mol) × 100% = 66.7% Li ∴ AlLi 2 34/6.94 + 66/26.981

(b) 600°C: L → α + β

eutectic

L: 9.9% Li

α : 4% Li β : 20.4% Li 510°C: β + L → γ

peritectic β: 25% Li

L : 47% Li γ : 34% Li 170°C: L → γ + α (Li)

eutectic

L: 98% Li

γ : 34% Li α (Li) : 99% Li

11-15 An intermetallic compound is found for 38 wt% Sn in the Cu–Sn phase diagram.

Determine the formula for the compound.

Solution:

at% Sn =

38 g/(118.69 g/mol) = 0.25 or Cu 3Sn 38/118.69 + 62/63.55

11-16 An intermetallic compound is found for 10 wt% Si in the Cu–Si phase diagram.

Determine the formula for the compound.

Solution:

at% Si =

10 g/(28.08 g/mol) = 0.20 or SiCu 4 10/28.08 + 90/63.55

11-17 Use the Gibbs phase rule to determine the number of degrees of freedom in each region of the phase diagram in Figure 11-6.

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Solution: For the α region, 1 + C = F + P. There are two components (lead and tin), and there is one phase, therefore: 1 + 2 = F + 1. F must be 2.

For the α + L region, 1 + C = F + P. The difference between this and the previous is the number of phases. 1 + 2 = F + 2. Therefore, F is 1.

These two results are general for the α, β, L regions and the α + L, L + β, and α + β regions respectively.

11-18 Consider a Pb–15% Sn alloy. During solidification, determine (a) the composition of the first solid to form; (b) the liquidus temperature, solidus temperature, solvus temperature, and freezing range of the alloy; (c) the amounts and compositions of each phase at 260°C; (d) the amounts and compositions of each phase at 183°C; and (e) the amounts and compositions of each phase at 25°C. Solution: (a) 8% Sn (b) liquidus = 290°C, solidus = 240°C, solvus = 170°C, freezing range = 50°C (c) L: 30% Sn α: 12% Sn;

15 − 12 × 100% = 17% % α = 83% 30 − 12 (d) α :15% Sn 100% α (e) α : 2% Pb β :100% Sn 100 − 15 %α= × 100 = 87% % β = 13% 100 − 2 %L =

11-19 Consider an Al–12% Mg alloy (Figure 11–29). During solidification, determine (a) the composition of the first solid to form; (b) the liquidus temperature, solidus temperature, solvus temperature, and freezing range of the alloy; (c) the amounts and compositions of each phase at 525°C; (d) the amounts and compositions of each phase at 450°C; and (e) the amounts and compositions of each phase at 25°C.

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Solution: (a) 2.5% Mg (b) liquidus = 600°C, solidus = 470°C, solvus = 400°C, freezing range = 130°C (c) L : 26% Mg α : 7% Mg;

26 − 12 × 100% = 74% % L = 26% 26 − 7 (d) α :12% Mg 100% α (e) α :1% Mg β : 34% Mg 34 − 12 %α= × 100% = 67% % β = 33% 34 − 1 %α=

11-20 Consider a Pb–35% Sn alloy. Determine (a) if the alloy is hypoeutectic or hypereutectic; (b) the composition of the first solid to form during solidification; (c) the amounts and compositions of each phase at 184°C; (d) the amounts and compositions of each phase at 182°C; (e) the amounts and compositions of each microconstituent at 182°C; and (f) the amounts and compositions of each phase at 25°C. Solution: (a) hypoeutectic (b) 14% Sn (c) α :19% Sn L : 61.9% Sn

61.9 − 35 × 100% = 63% % L = 37% 61.9 − 19 (d) α :19% Sn β :97.5% Sn 97.5 − 35 %α= × 100% = 80% % β = 20% 97.5 − 19 primary α :19% Sn % primary α = 63% %α=

(e)

eutectic : 61.9% Sn % eutectic = 37% β :100% Sn 100 − 35 %α= × 100% = 66% % β = 34% 100 − 2

(f) α : 2% Sn

11-21 Consider a Pb–70% Sn alloy. Determine (a) if the alloy is hypoeutectic or hypereutectic; (b) the composition of the first solid to form during solidification; (c) the amounts and compositions of each phase at 184°C; (d) the amounts and compositions of each phase at 182°C; (e) the amounts and compositions of each microconstituent at 182°C; and (f) the amounts and compositions of each phase at 25°C. Solution: (a) hypereutectic (b) 98% Sn (c) β : 97.5% Sn

L : 61.9% Sn 70 − 61.9 %β = × 100% = 22.8% % L = 77.2% 97.5 − 61.9 (d) α :19% Sn β :97.5% Sn 97.5 − 70 %α= × 100% = 35% % β = 65% 97.5 − 19

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% primaryβ = 22.8% primaryβ : 97.5% Sn % eutectic = 77.2% eutectic : 61.9% Sn (f) α : 2% Sn β :100% Sn 100 − 70 %α= × 100% = 30% % β = 70% 100 − 2 (e)

11-22 (a) Sketch a typical eutectic phase diagram with components A and B having similar melting points. B is much more soluble in A (maximum = 15%) than A is in B (maximum = 5%), and the eutectic composition occurs near 40% B. The eutectic temperature is 2/3 of the melting point. Label the axes of the diagram. Label all the phases. Use α and β to denote the solid phases. (b) For an overall composition of 60% B, list the sequence of phases found as the liquid is slowly cooled to room temperature.

Solution:

(b) Liquid ⇒ β + Liquid ⇒ α + β 11-23 The copper–silver phase diagram is shown in Figure 11–30. Copper has a higher melting point than silver. Refer to the silver-rich solid phase as gamma (γ) and the copper-rich solid phase as delta (δ). Denote the liquid as L.

(a) For an overall composition of 60% B (40% A) at a temperature of 800°C, what are the compositions and amounts of the phases present?

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(b) For an overall composition of 30% B (70% A) at a temperature of 1000°C, what are the compositions and amounts of the phases present? (c) Draw a schematic diagram illustrating the final microstructure of a material with a composition of 50% B (50% A) cooled to 200°C from the liquid state. Label each phase present.

Solution: (a) At a temperature of 800°C for an overall composition of 60% B, the two phases of and liquid are in equilibrium. As shown in the diagram, their compositions may be read directly from the phase diagram as the δ being 93% B (7% A) and the liquid being 34% B (66% A). (These are approximations that may vary slightly.) The amounts of and liquid are determined using the lever rule as follows and as shown in the phase diagram:

93 − 60 × 100 = 56% 93 − 34 60 − 34 × 100 = 44% %δ = 93 − 34 %L =

(b) For a composition of 30% B and 70% A and a temperature of 1000°C, there is a single liquid phase. Since it is a single phase, it must have the overall composition 30% B and 70% A. (c) The diagram below shows δ particles surrounded by a lamellar microstructure of δ (black) and γ (white).

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11-24 Determine the phases that are present and the compositions for each phase in Cu-55 wt% Ag at 600°C. (See Problem 11-23 and Figure 11-30.)

Solution: Ag is element A and Cu is element B. Thus, the composition of interest is 45 wt% Cu. At 600˚C, the solid solution of Cu in Ag is approximately 3 wt% Cu, and the solid solution of Ag in Cu is approximately 97 wt% Cu. 11-25 Determine the phases that are present and the compositions for each phase in Cu-85 wt% Ag at 800°C. (See Problem 11-23 and Figure 11-30.)

Solution: Both β and L phases are present and their compositions are 800°C: β = 93 wt% Ag – 7 wt% Cu ଼ହି଻଺

800°C: % β = ଽଷି଻଺ = 52.9%

L = 76 wt% Ag – 24 wt% Cu

%L = 100 – 52.9% = 47.1%

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11-26 Calculate the total % β and the % eutectic microconstituent at room temperature for the following lead-tin alloys: 10% Sn, 20% Sn, 50% Sn, 60% Sn, 80% Sn, and 95% Sn. Using Figure 11–18, plot the strength of the alloys versus the % β and the % eutectic and explain your graphs.

%β

Solution: 10% Sn 20% Sn 50% Sn 60% Sn 80% Sn 95% Sn

10 − 2 = 8.2% 99 − 1 20 − 2 = 18.6% 99 − 2 50 − 2 = 49.5% 99 − 2 60 − 2 = 59.8% 99 − 2 80 − 2 = 80.4% 99 − 2 95 − 2 = 95.9% 99 − 2

% eutectic 0%

20 − 19 = 2.3% 61.9 − 19 50 − 19 = 72.3% 61.9 − 19 60 − 19 = 95.6% 61.9 − 19 97.5 − 80 = 49.2% 97.5 − 61.9 97.5 − 95 = 7.0% 97.5 − 61.9

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11-27 Consider the Al-Si phase diagram. What are the percentages of α-phase and liquid for an Al-5 wt% Si alloy at 620, 600, and 578°C? What are the percentages of α and β phases in this Al-5 wt% Si alloy at 576 and 550°C? (See Figure 11-19.)

Solution:

620°C: % α =

଺.ଶିହ.଴ = 22.2% ଺.ଶି଴.଼

%L = 100 – 22.2% = 77.8%

600°C: % α =

ଽ.଴ିହ.଴ = 51.3% ଽ.଴ିଵ.ଶ

%L = 100 – 51.3% = 48.7%

ଵଶ.଺ିହ.଴

%L = 100 – 69.4% = 30.6%

ଽଽିହ.଴

%β = 100 – 96.6% = 3.4%

ଽଽିହ.଴

%L = 100 – 96.2% = 3.8%

578°C: % α = ଵଶ.଺ିଵ.଺ହ = 69.4% 576°C: % α = ଽଽିଵ.଺ହ = 96.6% 550°C: % α = ଽଽିଵ.ଷ = 96.2%

11-28 At the eutectic in the Al-Si phase diagram, what phase(s) is (are) present? Give a chemical analysis of the phase(s). (See Figure 11-19.)

Solution:

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The phases present are α-phase with a composition of 1.65%Si, 98.35%Al, β-phase with a composition of 99%Si, 1%Al and liquid with a composition of 12.6%Si, 87.4%Al which are all found at 577°C. 11-29 We discussed the primary phase or primary constituent. Why would we be interested in the percentage of the primary phase in the Al-Si alloy system?

Solution: When the alloy is hypoeutectic, the primary phase is α which an aluminum lattice with small amounts of silicon. This particular alloy is soft and ductile When the alloy is hypereutectic, the primary phase is β, which is a silicon lattice containing small amounts of aluminum. This alloy would be hard and brittle.

The primary phase and its quantity in conjunction with the amount of silicon will determine the overall mechanical and physical properties of the microstructure. The size, distribution, shape and orientation of the phases will also influence these properties.

11-30 Consider an Al–4% Si alloy (Figure 11–19). Determine (a) if the alloy is hypoeutectic or hypereutectic; (b) the composition of the first solid to form during solidification; (c) the amounts and compositions of each phase at 578°C; (d) the amounts and compositions of each phase at 576°C, the amounts and compositions of each microconstituent at 576°C; and (e) the amounts and compositions of each phase at 25°C. Solution: (a) hypoeutectic (b) 1% Si (c) α :1.65% Si

L:12.6% Si 12.6 − 4 %α= = 78.5% % L = 21.5% 12.6 − 1.65 (d) α :1.65% Si β : 99.83% Si 99.83 − 4 %α= = 97.6% % β = 2.4% 99.83 − 1.65 primary α :1.65% Si % primary α = 78.5% (e)

eutectic :12.6% Si

% eutectic = 21.5% 100 − 4 (f) α : 0% Si β :100% Si % α = = 96% % β = 4% 100 − 0 11-31 Consider an Al–25% Si alloy. (See Figure 11–19.) Determine (a) if the alloy is hypoeutectic or hypereutectic; (b) the composition of the first solid to form during solidification; (c) the amounts and compositions of each phase at 578°C; (d) the amounts and compositions of each phase at 576°C; (e) the amounts and compositions of each microconstituent at 576°C; and (f) the amounts and compositions of each phase at 25°C. Solution:

hypereutectic 100% Si

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β : 99.83% Si L :12.6% Si 99.83 − 25 %L= = 85.8% % β = 14.2% 99.83 − 12.6 α :1.65% Si β : 99.83% Si 99.83 − 25 %α= = 76.2% % β = 23.8% 99.83 − 1.65 primary β : 99.83% Si % primary β = 14.2% eutectic :12.6% Si % eutectic = 85.8% 100 − 25 α : 0% Si β :100% Si % α = = 75% % β = 25% 100 − 0 11-32 A Pb–Sn alloy contains 45% α and 55% β at 100°C. Determine the composition of the alloy. Is the alloy hypoeutectic or hypereutectic?

Solution:

% α = 45 =

98.0 − x ×100 or 98.0 − 5

x = 56.15% Sn Hypoeutectic

11-33 An Al–Si alloy contains 85% α and 15% β at 500°C. Determine the composition of the alloy. Is the alloy hypoeutectic or hypereutectic? (See Figure 11-19.)

Solution:

% α = 85 =

100 − x × 100 or 100 − 1

x = 15.85% Si Hypereutectic

11-34 A Pb–Sn alloy contains 23% primary α and 77% eutectic microconstituent immediately after the eutectic reaction has been completed. Determine the composition of the alloy.

Solution:

% primary α = 23 =

61.9 − x × 100 or 61.9 − 19

x = 52% Sn

11-35 An Al–Si alloy contains 15% primary β and 85% eutectic microconstituent immediately after the eutectic reaction has been completed. Determine the composition of the alloy.

Solution:

% eutectic = 85 =

100 − x × 100 or 100 − 12.6

x = 25.71% Si

11-36 Observation of a microstructure shows that there is 28% eutectic and 72% primary β in an Al–Li alloy (Figure 11–28). Determine the composition of the alloy and whether it is hypoeutectic or hypereutectic.

Solution:

28 =

20.4 − x ×100 or 20.4 − 9.9

x = 17.46% Li Hypereutectic

11-37 Write the eutectic reaction that occurs, including the compositions of the three phases in equilibrium, and calculate the amount of α and β in the eutectic microconstituent in the Mg–Al system (Figure 11–29).

Solution:

L32.3 → α12.7 + γ 40.2

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∴ % α Eut =

40.2 − 32.3 × 100% = 28.7% and % γ Eut = 71.3% 40.2 − 12.7

11-38 Calculate the total amount of α and β and the amount of each microconstituent in a Pb– 50% Sn alloy at 182°C. What fraction of the total α in the alloy is contained in the eutectic microconstituent?

Solution:

97.5 − 50 × 100% = 60.5% 97.5 − 19 61.9 − 50 α Primary = × 100% = 27.7% 61.9 − 19 α in eutectic = 60.5 − 27.7 = 32.8%

α total =

βTotal = 39.5% Eutectic = 72.3%

f = 32.8 / 60.5 = 0.54 11-39 Figure 11–31 shows a cooling curve for a Pb–Sn alloy. Determine (a) the pouring temperature; (b) the superheat; (c) the liquidus temperature; (d) the eutectic temperature; (e) the freezing range; (f) the local solidification time; (g) the total solidification time; and (h) the composition of the alloy.

Solution: (a) pouring temperature = 360°C (b) superheat = 360–250 = 110°C (c) liquidus temperature = 250°C (d) eutectic temperature = 183°C (e) freezing range = 250–183 = 67°C (f) local solidification time = 600–110 = 490 s (g) total solidification time = 600 s (h) approximately 32% Sn

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11-40 Figure 11–32 shows a cooling curve for an Al–Si alloy and Figure 11–19 shows the binary phase diagram for this system. Determine (a) the pouring temperature; (b) the superheat; (c) the liquidus temperature; (d) the eutectic temperature; (e) the freezing range; (f) the local solidification time; (g) the total solidification time; and (h) the composition of the alloy.

Solution: (a) pouring temperature = 1150°C (b) superheat = 1150–1000 = 150°C (c) liquidus temperature = 1000°C (d) eutectic temperature = 577°C (e) freezing range = 1000–577 = 423°C (f) local solidification time = 11.5–1 = 10.5 min (g) total solidification time = 11.5 min (h) approximately 45% Si 11-41 Draw the cooling curves, including appropriate temperatures, expected for the following Al–Si alloys: (a) Al–4% Si; (b) Al–12.6% Si; (c) Al–25% Si; and (d) Al–65% Si.

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Solution:

11-42 Cooling curves are obtained for a series of Cu–Ag alloys (Figure 11–33). Use this data to produce the Cu–Ag phase diagram. The maximum solubility of Ag in Cu is 7.9%, and the maximum solubility of Cu in Ag is 8.8%. The solubilities at room temperature are near zero.

Solution: 0% Ag 8% Ag 20% Ag 50% Ag 71.9% Ag 90% Ag 100% Ag

→ → → → → → →

Tliq 1085°C 1030°C 975°C 860°C 780°C 870°C 961°C

Tsol 950°C 780°C 780°C 780°C 780°C

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11-46 The binary phase diagram for the silver (Ag) and germanium (Ge) system is shown in Figure 11–34.

(a) Schematically draw the phase diagram and label the phases present in each region of the diagram. Denote α as the Ag-rich solid phase and β as the Ge-rich solid phase.

Use L to denote the liquid phase. (b) For an overall composition of 80% Ge (20% Ag) at a temperature of 700 K, what are the compositions and amounts of the phases present? (c) What is the transformation in phases that occurs on solidification from the melt at the point marked with a circle? What is the special name given to this transformation? (d) Draw a schematic diagram illustrating the final microstructure of 15% Ge (85% Ag) cooled to 300 K from the liquid state. (e) Consider two tensile samples at room temperature. One is pure Ag and one is Ag with 2% Ge. Which sample would you expect to be stronger?

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Solution:

(b) At 700 K, the composition of the α and β may be read directly from the phase diagram as the α being 5% Ge (95% Ag) and the β being 100% Ge (0% Ag). The amount of α and β is determined using the lever rule as follows:

100 − 80 = 21% 100 − 5 80 − 5 %β= = 79% 100 − 5 %α=

(c) Eutectic: L → α + β (d) The diagram shows α particles surrounded by a lamellar microstructure of α and β.

(e) The alloyed Ag will be stronger (Ag–2% Ge.) The Ge atoms in the a phase will hinder dislocation motion, and the two–phase microstructure of α and β contains phase boundaries that also hinder dislocation motion. Obstructions to dislocation motion strengthen a material.

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11-47 The copper–silver phase diagram is shown in Figure 11–30. Copper has a higher melting point than silver.

(a) Is copper element A or element B as labeled in the phase diagram? (b) Schematically draw the phase diagram and label all phases present in each region (single phase and two phase) of the phase diagram by writing directly on your sketch. Denote the silver-rich solid phase as gamma (γ) and the copper-rich solid phase as delta (δ). Denote the liquid as L. (c) At 600°C, the solid solution of element A in element B is stronger than the solid solution of element B in element A. Assume similar processing conditions. Is a material cooled from the liquid to 600°C with a composition of 90% A and 10% B likely to be stronger or weaker than a material with the eutectic composition?

Explain your answer fully. (d) Upon performing mechanical testing, your results indicate that your assumption of similar processing conditions in part (c) was wrong and that the material that you had assumed to be stronger is in fact weaker. Give an example of a processing condition and a description of the associated microstructure that could have led to this discrepancy.

Solution: (a) Copper is element B because element B has the higher melting point.

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(c) According to the lever rule, the material with the eutectic composition contains more of the stronger phase and therefore is likely to be stronger than the material with a composition of 90% A and 10% B. (d) The composition of 90% A and 10% B may have a smaller grain size or a finer lamellar microstructure than the eutectic composition. Both microstructures would result in a strength increase relative to the assumed processing conditions because grain boundaries and phase boundaries hinder dislocation motion. Both microstructures could be achieved with a higher cooling rate relative to the assumed processing conditions.

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Chapter 12: Dispersion Strengthening by Phase Transformations and Heat Treatment 12-1

(a) Determine the critical nucleus size r* for homogeneous nucleation for precipitation of phase β in a matrix of phase α. Hint: The critical nucleus size occurs at the maximum in the expression for ΔG(r) in Equation 12–1. (b) Plot the total free energy change ΔG as a function of the radius of the precipitate. (c) Comment on the value of r* for homogeneous nucleation for solid-state precipitation when compared to the liquid to solid transformation.

Solution: (a) From Equation 12–1, the total change in free energy for nucleation of a spherical solid precipitate ΔG is

4 4 ∆G = π r 2 ∆Gv (α → β ) + 4π r 2σ αβ + π r 3ε , 3 3 where r is the radius of spherical solid, ΔGv(α→β) is the free energy change per unit volume for precipitation of phase β from phase α, σαβ is the surface energy per unit area of the α–β interface, and is the strain energy per unit volume required for a precipitate to fit into the surrounding matrix. Rearranging,

4 ∆G = π r 3 [ ∆Gv (α → β ) + ε ] + 4π r 2σ αβ .

3 The critical radius for nucleation r* is the value at which ΔG is a maximum. At this point, the slope is zero. Thus,

∂∆G = 4π r 2 [ ∆Gv (α → β ) + ε ] + 8π rσ αβ , ∂r and

∂∆G = 4π (r*) 2 [∆Gv (α →β ) + ε ] + 8π r *σ αβ = 0 . ∂r r = r* Solving for r*,

r* = −

2σ αβ ∆Gv (α → β ) + ε

.

(b) A graph of total free energy versus the radius of the β phase will have the form shown. The volume free energy change is the driving force, while the strain energy and interface energy terms need to be overcome for precipitation.

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(c) The difference between the critical radius for homogeneous nucleation in a solid phase system versus a liquid to solid transformation is the strain energy term, which is the energy required to fit a solid precipitate in a solid matrix; additional energy is required to accommodate the precipitate.

12-3

Determine the constants c and n in Equation 12–2 that describe the rate of crystallization of polypropylene at 140°C. (See Figure 12–30.)

Solution: Equation 12–2 gives the fraction f transformed in a solid-state phase transformation as a function of time t as f = 1–exp(–ctn), where c and n are constants. Rearranging and taking the natural logarithm of both sides, ln(1–f) = –ctn, and again rearranging and taking the natural logarithm of both sides, ln [–ln(1–f)] = ln(c) + n ln(t).

The values of c and n can be found by plotting ln [–ln(1–f)] versus ln(t).

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The values of f and t determined from Figure 12–30 are given in the table.

From the graph, the slope n = 2.81, and ln(c) = –11.63 such that c = 8.9 × 10–6. f t (min) 0.1 28 0.2 37 0.3 44 0.4 50 0.5 55 0.6 60 0.7 67 0.8 73 0.9 86

12-6

If a FCC sample has an observed incubation time of 10 seconds and is fully transformed at 3.5 × 104 s, what are its c and n values in Equation 12-2?

Solution: To solve this problem, the equations must be manipulated with algebra to be explicit in one of the variables: ݂ = 1 − expሺ−ܿ‫ ݐ‬௡ ሻ 0 = 1 − expሺ−ܿሾ10 sሿ௡ ሻ 1 = expሺ−ܿሾ10 sሿ௡ ሻ 0 = −ܿሾ10 sሿ௡ This implies that c is zero, which would mean the sample is always fully transformed. To solve this, it must be assumed that at the incubation time a small fraction has transformed: 0.0001 = 1 − expሺ−ܿሾ10 sሿ௡ ሻ 0.9999 = expሺ−ܿሾ10 sሿ௡ ሻ −0.0001 = −ܿሾ10 sሿ௡ 0.0001 = ܿሾ10 sሿ௡ ܿ = 0.0001ሾ10 sሿି௡

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Similarly for the fully transformed condition, an assumption of a slight remaining untransformed amount is required because of the exponential function.

0.9999 = 1 − expሺ−ܿሾ3.5 × 10ସ sሿ௡ ሻ 0.0001 = expሺ−ܿሾ3.5 × 10ସ sሿ௡ ሻ −9.21 = −ܿሾ3.5 × 10ସ sሿ௡ ܿ = 9.21ሾ3.5 × 10ସ sሿି௡ Combining and solving for n: 9.21ሾ3.5 × 10ସ sሿି௡ = 0.0001ሾ10 sሿି௡ ௡ ሾ3.5 × 10ସ sሿ௡ 9.21 3.5 × 10ସ s = =ቆ ቇ ሾ10 sሿ௡ 0.0001 10 s ௡ 9.21 3.5 × 10ସ s =ቆ ቇ 0.0001 10 s 92,100 = ሺ3500ሻ௡ ݊ = 1.4 Returning and solving for c: ܿ = 0.0001ሾ10 sሿିଵ.ସ ܿ = 4 × 10ି଺ s ିଵ.ସ 12-7

For cold-worked copper, construct a plot for the time to 50% recrystallization as a function of the annealing temperature. (See Figure 12-2 for data.)

Solution: This requires the use of the graph in Figure 12-2. We read the t-T data from the four sigmoid curves: Time [min] 9 22 78 230 Temperature [°C] 135 119 102 88 And then plot these on a curve:

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Time to τ as a function of temperature.

250

Time [min]

200 y = 2E+17x-7.648 R² = 0.9993

150 100 50 0 0

20

40

60

80

100

120

140

160

Temperature [°C]

Viewing the graph, we see that there is not a linear relationship, so one cannot say that for every centigrade degree added to the temperature the time decreases by x minutes.

Instead, the relationship must be used to find out the new time to half recrystallization.

12-14 Electromigration (diffusion of atoms/ions due to momentum transfer from high energy electrons) leads to voids in aluminum interconnects used in many semiconductor metallization processes and thus is a leading cause of device reliability issues in the industry. Propose an additive to Al that can help mitigate this issue. Refer to the appropriate phase diagram to justify your answers.

Solution: Several additives can help make aluminum more robust against electromigration; however, for interconnects for semiconductor metallization, electrical conductivity is an important parameter. Thus copper with its excellent electrical conductivity is a good choice. According to Figure 12–9, the copper content must be kept lower than 5.65% (typically 4% or lower), so that the temperatures experienced in backend semiconductor processing result in a structure that leads to fine precipitates of Cu in an a matrix, thereby strengthening the interconnects.

12-15 For what dihedral angle will the interfacial energy of the matrix-precipitate boundary and the grain boundary energy of the matrix be equal?

Solution: The formula is given by: ߛ୫,୥ୠ = 2ߛ୫୮ cos

Setting the two energies equal to each other: ߠ 1 = 2 cos 2 ߠ 0.5 = cos 2 ߠ 1.05 = 2

ߠ 2

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ߠ = 2.10 radians = 120° 12-16 What is the relationship between the interfacial surface energies of the matrixprecipitate boundary and grain boundary when the precipitate forms a tetrahedron?

Solution: Equilateral triangles have three equal angles of 60°. Using the unnumbered equation on page 457: ߠ ߛ୫,୥ୠ = 2ߛ୫୮ cos 2 60° ߛ୫,୥ୠ = 2ߛ୫୮ cos 2 ߛ୫,୥ୠ = 1.73 ߛ୫୮ 12-17 45.7 kg of aluminum comprise 35 wt% of a Cu-Al alloy. The temperature is halfway between the upper and lower bounds of the L + η phase region at this composition. How much more aluminum must be added and how much must the resulting liquid mixture be cooled to reach the eutectic point? (See Figure 12-5.)

Solution: Examining Figure 12-5, 35 wt% Al is 65 wt% Cu and the alloy are halfway between the two essentially horizontal phase boundaries. Averaging 591 °C and 626 °C gives 608.5 °C. To reach the eutectic point at 33.2 wt% Cu: 45.7 kg Al ݉୲୭୲ୟ୪ = = 130.6 kg 0.35 kg Al kg ݉େ୳ = 130.6 kg ሺAl + Cuሻ − 45.7 kg Al = 84.9 kg Cu Now this mass of copper has to become 33.2 wt% of the new total:

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84.9 kg Cu = 255.6 kg kg Cu 0.332 kg ݉′୅୪ = 255.6 kg − 84.9 kg Al = 171 kg Al The eutectic line is at 548 °C, so the temperature difference is: ∆ܶ = 548 ℃ − 608 ℃ = −60℃ ݉′େ୳ =

12-29 (a) Recommend an artificial age-hardening heat treatment for a Cu–1.2% Be alloy (see Figure 12–31). Include appropriate temperatures. (b) Compare the amount of the γ 2 precipitate that forms by artificial aging at 400°C with the amount of the precipitate that forms by natural aging.

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Solution: (a) For the Cu–1.2% Be alloy, the peritectic temperature is 870°C; above this temperature, liquid may form. The solvus temperature is about 530°C. Therefore, 1) Solution treat between 530°C and 870°C (780°C is typical for beryllium copper alloys) 2) Quench 3) Age below 530°C (330°C is typical for these alloys) (b) We can perform lever law calculations at 400°C and at room temperature. The solubility of Be in Cu at 400°C is about 0.6% Be and that at room temperature is about 0.2% Be:

1.2 − 0.6 × 100 = 5.4% 11.7 − 0.6 1.2 − 0.2 γ 2 (room T ) = × 100 = 8.5% 12 − 0.2

γ 2 (at 400°C) =

12-30 Suppose that age hardening is possible in the Al–Mg system (see Figure 12–10). (a) Recommend an artificial age-hardening heat treatment for each of the following alloys; and (b) compare the amount of the β precipitate that forms from your treatment of each alloy: (i) Al–4% Mg; (ii) Al–6% Mg; and (iii) Al–12% Mg. (c) Testing of the alloys after the heat treatment reveals that little strengthening occurs as a result of the heat treatment. Which of the requirements for age hardening is likely not satisfied?

Solution:

(a) The heat treatments for each alloy might be Al–4% Mg Al–6% Mg TEutectic = 451°C 451°C TSolvus = 210°C 280°C Solution Treat at: 210–451°C 280–451°C Quench Quench Age at: <210°C <280°C

Al–12% Mg 451°C 390°C 390–451°C Quench <390°C

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(b) Answers will vary depending on the aging temperature selected. If all three are aged at 200°C, as an example, the tie line goes from about 3.8 to 35% Mg: Al–4% Mg: % β = (4–3.8)/(35–3.8) × 100 = 0.6% Al–6% Mg: % β = (6–3.8)/(35–3.8) × 100 = 7.1% Al–12% Mg: % β = (12–3.8)/(35–3.8) × 100 = 26.8% (c) Most likely, a coherent precipitate is not formed; simple dispersion strengthening, rather than age hardening, occurs.

12-31 An Al–2.5% Cu alloy is solution-treated, quenched, and overaged at 230°C to produce a stable microstructure. If the θ precipitates as spheres with a diameter of 9000 Å and a density of 4.26 g/cm3, determine the number of precipitate particles per cm3. (See Figure 12–5.)

Solution:

53 − 2.5 = 97.12% wt% θ = 2.88% 53 − 1 2.88 g/(4.26 g/cm 3 ) vol fraction θ = = 0.0182 cm3θ /cm3alloy 2.88/4.26 + 97.12/2.669 –10 dθ = 9000 × 10 m = 9 × 10 –5 cm rθ = 4.5 × 10 –5 cm wt% α =

Vθ = (4π/3) (4.5 × 10–5 cm)3 = 382 × 10–15 cm3

# of particles =

0.0182 cm3 = 4.76 × 1010 particles 382 ×10 −15cm3

12-32 Examine Figure 12-10 and state the phases that will be present in each stage if this process path is followed: (500 °C, 20 wt% Mg) to (500 °C, 80 wt% Mg) to (200 °C, 80 wt% Mg) Solution: At the first point on the phase diagram, α phase and the single liquid phase L are present. At the second point, δ phase and the liquid phase L. Finally we cool down the process so the two solid phases γ and δ exist.

12-41 Based on the principles of age hardening of Al–Cu alloys, rank the following Al–Cu alloys from highest to lowest for maximum yield strength achievable by age hardening and longest to shortest time required at 190°C to achieve the maximum yield strength: Al–2 wt% Cu, Al–3 wt% Cu, and Al–4 wt% Cu. Refer to the Al–Cu phase diagram (See Figure 12–9).

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Solution: The ranking from highest to lowest maximum yield strength is Al–4wt% Cu > Al–3wt% Cu > Al–2wt% Cu As the weight percentage of Cu is increased, a larger volume fraction of precipitates forms, thereby increasing the maximum yield strength. The time to achieve the maximum yield strength at 190°C from longest to shortest is Al–2wt% Cu > Al–3wt% Cu > Al–4wt% Cu As the weight percentage of Cu increases, Cu atoms have to diffuse shorter distances to form GP zones and θ’ precipitates, and hence the maximum yield strength can be achieved faster for the higher wt% Cu alloys.

12-46 Figure 12–33 shows a hypothetical phase diagram. Determine whether each of the following alloys might be good candidates for age hardening, and explain your answer.

For those alloys that might be good candidates, describe the heat treatment required, including recommended temperatures. (a) A–10% B; (b) A–20% B; (c) A–55% B; (d) A–87% B; and (e) A–95% B. Solution: (a) A–10% B is a good candidate:

Solution Treatment @ T = 290 to 400°C Quench Age @ T < 290°C (b) A–20% B: Some age hardening may occur when the alloy is solution treated below 400°C and quenched; however, eutectic is

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also present and the strengthening effect will not be as dramatic as in (a). (c) A–55% B: almost all θ is formed. The alloy is expected to be very brittle. (d) A–87% B: the alloy cools from a two-phase (β + θ) region to a onephase (β) region, opposite of what we need for age hardening. (e) A–95% B: the alloy is single phase (β) at all temperatures and thus cannot be age hardened.

12-54 Draw the eutectoid portion of the Fe- Fe3C phase diagram. Be sure to indicate all of the compositions and temperatures.

Solution:

12-55 Draw the peritectic portion of the Fe-C phase diagram. Be sure to indicate all of the compositions and temperatures and write the relevant reaction.

Solution:

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12-61 Austenite and Fe3C are in equilibrium at 1000°C and 3 wt% C. What is the weight percent of iron carbide in the two-phase alloy?

Solution: Finding the point (3 wt% C, 1000 °C) on Figure 12-14, the lever rule may be used to find the amount of iron carbide: 3 − 1.6 × 100 % = 27.6 % Feଷ C 6.67 − 1.6 12-62 For an Fe–0.35%C alloy, determine (a) the temperature at which austenite first begins to transform on cooling; (b) the primary microconstituent that forms; (c) the composition and amount of each phase present at 728°C; (d) the composition and amount of each phase present at 726°C; and (e) the composition and amount of each microconstituent present at 726°C. Solution:

(a) 795°C (b) primary α-ferrite (c)

α : 0.0218% C γ : 0.77% C

(d) α: 0.0218% C Fe3C: 6.67% C

(e)

primary α : 0.0218% C pearlite : 0.77% C

0.77 − 0.35 ×100 = 56.1% 0.77 − 0.0218 % γ = 43.9% 6.67 − 0.35 %α= ×100 = 95.1% 6.67 − 0.0218 % Fe 2C = 4.9% % primary α = 56.1% % Pearlite = 43.9% %α=

12-63 For an Fe–1.15% C alloy, determine (a) the temperature at which austenite first begins to transform on cooling; (b) the primary microconstituent that forms; (c) the composition and amount of each phase present at 728°C; (d) the composition and amount of each phase present at 726°C; and (e) the composition and amount of each microconstituent present at 726°C. Solution: 880°C primary Fe3C Fe3C: 6.67% C γ: 0.77% C

% Fe3C =

1.15 − 0.77 ×100 = 6.4% 6.67 − 0.77

% γ = 93.6% α: 0.0218% C Fe3C: 6.67% C

%α=

6.67 − 1.15 × 100 = 83% 6.67 − 0.0218

% Fe3C = 17%

primary Fe3C : 6.67% C

% primary Fe3C = 6.4%

pearlite : 0.77% C

% Pearlite = 93.6%

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12-64 A steel contains 8% cementite and 92% ferrite at room temperature. Estimate the carbon content of the steel. Is the steel hypoeutectoid or hypereutectoid?

Solution:

α = 0.92 =

6.67 − x 6.67 − 0

x = 0.53% C, ∴ Hypoeutectoid

12-65 A steel contains 18% cementite and 82% ferrite at room temperature. Estimate the carbon content of the steel. Is the steel hypoeutectoid or hypereutectoid?

Solution:

α = 0.82 =

6.67 − x 6.67 − 0

x = 1.20% C, ∴ Hypereutectoid

12-66 A steel contains 18% pearlite and 82% primary ferrite at room temperature. Estimate the carbon content of the steel. Is the steel hypoeutectoid or hypereutectoid?

Solution:

0.77 − x 0.77 − 0.0218 x = 0.156% C, ∴ Hypoeutectoid

primary α = 0.82 =

12-67 A steel contains 94% pearlite and 6% primary cementite at room temperature. Estimate the carbon content of the steel. Is the steel hypoeutectoid or hypereutectoid?

Solution:

Pearlite = 0.94 =

6.67 − x 6.67 − 0.77

x = 1.124% C, ∴ Hypereutectoid

12-68 A steel contains 55% α and 45% γ at 750°C. Estimate the carbon content of the steel.

Solution:

α = 0.02% C and γ = 0.6% C (from the tie line at 750°C) 0.6 − x %α = 55 = × 100 x = 0.281% C 0.6 − 0.02

12-69 A steel contains 96% γ and 4% Fe3C at 800°C. Estimate the carbon content of the steel.

Solution:

γ = 0.92% C and Fe3C = 6.67% C (from the tie line at 800°C) γ = 0.96 =

6.67 − x 6.67 − 0.92

x = 1.15% C

12-70 A steel is heated until 40% austenite, with a carbon content of 0.5%, forms. Estimate the temperature and the overall carbon content of the steel.

Solution: In order for γ to contain 0.5% C, the austenitizing temperature must be about 760°C (from the tie line). At this temperature

0.4 =

x − 0.02 0.5 − 0.02

x = 0.212% C

12-71 A steel is heated until 85% austenite, with a carbon content of 1.05%, forms. Estimate the temperature and the overall carbon content of the steel.

Solution: In order for γ to contain 1.05% C, the austenitizing temperature must be

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about 845°C (from the tie line). At this temperature

0.84 =

6.67 − x 6.67 − 1.05

x = 1.893% C

12-72 The carbon steels listed in the table below were soaked at 1000°C for 1 hour to form austenite and were cooled slowly, under equilibrium conditions to room temperature.

Refer to the Fe–Fe3C phase diagram to answer the following questions for each of the carbon steel compositions listed in the table below. (a) Determine the amounts of the phases present; (b) determine the C content of each phase; and (c) plot the C content of pearlite, α, and Fe3C versus yield strength. Based on the graph, discuss the factors influencing yield strength in steels with <1% C. Carbon % Yield Strength (MPa) 0.2 295 0.4 353 0.6 372 0.8 376 0.95 379 (a) The amounts of phases and microconstituents present can be determined using the lever rule: % α/100 = (6.67–%C)/(6.67–0) % Fe3C/100 = (%C–0)/(6.67–0) For C < 0.77%, % Pearlite/100 = (%C–0.0218)/ (0.77–0.0218) % Primary α/100 = (0.77–%C)/ (0.77–0.0218) For C > 0.77%, % Pearlite/100 = (6.67–%C)/ (6.67–0.77) % Primary Fe3C/100 = (%C–0.77)/ (6.67–0.77) The table shows the summary of the phases and microconstituents present. % Carbon α Fe3C Pearlite Primary α Primary Fe3C 0.2 97.0 3.0 23.8 76.2 0.4 94.0 6.0 50.5 49.5 0.6 91.0 9.0 77.3 22.7 0.8 88.0 12.0 99.5 0.5 0.95 85.8 14.2 96.9 3.1 (b) At room temperature, α consists of pure iron and Fe3C contains 6.67 wt% C. (c) The graph shows that as the %C increases, the amount of hard Fe3C phase increases and the amount of relatively soft a decreases, leading to increased yield strength. The yield strength also increases as the amount of pearlite increases.

Solution:

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12-73 A faulty thermocouple in a carburizing heat-treatment furnace leads to unreliable temperature measurement during the process. Microstructure analysis from the carburized steel that initially contained 0.2% carbon revealed that the surface had 93% pearlite and 7% primary Fe3C, while at a depth of 0.5 cm, the microconstituents were 99% pearlite and 1% primary Fe3C. Estimate the temperature at which the heat treatment was carried out if the carburizing heat treatment was carried out for four hours. For interstitial carbon diffusion in FCC iron, the diffusion coefficient D0 = 2.3 × 10– 5 m2/s and the activation energy Q = 1.38 × 10-5 J/mol. It may be helpful to refer to Sections 5–6 and 5–8.

Solution: The %C at the surface and at a depth of 0.5 cm needs to be determined using the lever rule. The presence of primary Fe3C indicates a hypereutectoid steel at the surface and at a depth of 0.5 cm.

At the surface where there is 93% pearlite, the %C is determined according (% Pearlite/100) = (6.67–%C)/ (6.67–0.77) = 0.93 %C = 6.67–0.93 × (6.67–0.77) = 1.18% At a depth of 0.5 cm, where there is 99% pearlite, the %C is determined according (% Pearlite/100) = (6.67–%C)/ (6.67–0.77) = 0.99 %C = 6.67–0.99 × (6.67–0.77) = 0.829% For a carburizing treatment, the carbon profile is given by Equation 57,

cs − c x  x  = erf  , cs − c0  2 Dt  where cs is a constant concentration of the diffusing atoms at the surface of the material, c0 is the initial uniform concentration of the diffusing atoms in the material, and cx is the concentration of the diffusing atom at location x below the surface after time t. In this case, x = 0.5 cm, cs = 1.18%, cx = 0.829%, and c0 = 0.2%; therefore,

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1.18 − 0.829  0.5  = 0.36 = erf  , 1.18 − 0.2  2 Dt  such that

0.5 = 0.33 , 2 Dt and Dt = 0.571.

For t = 4 hours = 14,400 s, D = 3.966 × 10–5 cm2/s = 3.966 × 10–9 m2/s.

The diffusion coefficient D is given by

 Q  D = D0 exp  − ,  RT  where D0 is the pre–exponential term, Q is the activation energy, R is the gas constant, and T is the absolute temperature. Thus,

 1.38 ×105  −9 D = 2.3 ×10 exp  −  = 3.99 ×10 .  8.314T  −5

Solving for T, T = 1915 K or 1642 °C. 12-74 Determine the eutectoid temperature, the composition of each phase in the eutectoid reaction, and the amount of each phase present in the eutectoid microconstituent for the following systems. For the metallic systems, comment on whether you expect the eutectoid microconstituent to be ductile or brittle. (a) ZrO2–CaO (See Figure 12–34) (b) Cu–Al at 11.8%Al [See Figure 12–31(c)] (c) Cu–Zn at 47%Zn [See Figure 12–31(a)] (d) Cu–Be [See Figure 12–31(d)]

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Solution: (a) @900°C: Tetragonal12% CaO → Monoclinic3% CaO + Cubic16% CaO

% Monoclinic =

16 − 12 × 100 = 31% % Cubic = 69% 16 − 3

The eutectoid microconstituent (and the entire material, for that matter) will be brittle because the materials are ceramics (b) @565°C: β11.8% Al → α9.4% Al + γ2 15.6% Al

%α =

15.6 − 11.8 × 100 = 61.3% % β = 38.7% 15.6 − 9.4

Most of the eutectoid microconstituent is α (solid-solution strengthened copper) and is expected to be ductile. (c) @250°C: β′47% Zn → α36% Zn + γ59% Zn

%α=

59 − 47 ×100 = 52.2% % γ = 47.8% 59 − 36

Slightly more than half of the eutectoid is the copper solid solution; there is a good chance that the eutectoid would be ductile. (d) @605°C: γ1 6% Be → α1.5% Be + γ2 11% Be

%α =

11 − 6 × 100 = 52.6% %β = 47.4% 11 − 1.5

Slightly more than half of the eutectoid is the copper solid solution; we might then expect the eutectoid to be ductile.

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12-76 Compare the interlamellar spacing and the yield strength when a eutectoid steel is isothermally transformed to pearlite at (a) 700°C; and (b) 600°C. Solution: We can find the interlamellar spacing from Figure 12–19 and then use this spacing to find the strength from Figure 12–18. (a) λ = 7.5 × 10 –5 cm 1/λ = 13,333 YS = 200 MPa (29, 400 psi) (b) λ = 1.5 × 10−5 cm 1/λ = 66, 667

YS = 460 MPa (67, 600 psi)

12-78 Plot the yield and tensile strengths of the ratio of coarse to fine pearlite as a function of carbon composition according to Table 12-1.

Solution: The figure below shows the differences for both physical properties are similar and only slightly affected by the carbon content in this range. A statistical analysis would be required to determine if the dip at 0.8 % was significant.

Ratio of strengths of coarse over fine pearlite 1

0.75

0.5

0.25

0 0

0.2

0.4

0.6

0.8

1

Carbon %

12-79 In hypoeutectoid steel, both eutectoid and proeutectoid ferrite exist. Explain the difference between them. What will be the carbon concentration in each?

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Solution: Eutectoid ferrite forms after the eutectoid temperature is reached during cooling.

Proeutectoid ferrite forms between the A3 and A1 temperatures and is present before the eutectoid temperature is reached. The carbon content will be very limited (on the order of 0.022 wt% or less for both forms of ferrite.

12-86 An isothermally transformed eutectoid steel is found to have a yield strength of 410 MPa. Estimate (a) the transformation temperature and (b) the interlamellar spacing in the pearlite.

Solution: We can first find the interlamellar spacing from Figure 12–18; then using this interlamellar spacing, we can find the transformation temperature from Figure 12–19. (a) transformation temperature = 615°C (b) 1/λ = 60, 000 or λ = 1.67 × 10−5 cm 12-87 Determine the required transformation temperature and microconstituent if a eutectoid steel is to have the following hardness values: (a) HRC 38; (b) HRC 42; (c) HRC 48; and (d) HRC 52.

Solution: (a) 600°C pearlite (b) 400°C bainite (c) 340°C bainite (d) 300°C bainite 12-88 Describe the hardness and microstructure in a eutectoid steel that has been heated to 800°C for 1 h, quenched to 350°C and held for 750 s, and finally quenched to room temperature.

Solution: HRC = 47 and the microstructure is all bainite.

12-89 Describe the hardness and microstructure in a eutectoid steel that has been heated to 800°C, quenched to 650°C and held for 500 s, and finally quenched to room temperature.

Solution: HRC = 25 and the microstructure is all pearlite.

12-90 Describe the hardness and microstructure in a eutectoid steel that has been heated to 800°C, quenched to 300°C and held for 10 s, and finally quenched to room temperature.

Solution: HRC = 66 and the microstructure is all martensite.

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12-91 Describe the hardness and microstructure in a eutectoid steel that has been heated to 800°C, quenched to 300°C and held for 10 s, quenched to room temperature, and then reheated to 400°C before finally cooling to room temperature again.

Solution: HRC = 42 and the microstructure is all tempered martensite.

12-92 A 1080 steel was austenitized at 775°C for one hour and quenched to 550°C until it isothermally transformed at 550°C. How long must the austenite be held at 550°C to complete the transformation? What is the expected microstructure and hardness?

Solution: When isothermally transformed at 550°C, the TTT diagram indicates that the transformation was completed in 10 seconds. The microstructure is a very fine pearlite with a hardness of 40 HRC. A hardness of 40 HRC is approximately 370 BHN giving the tensile strength for this Brinell number from 1150-1325 MPa. 12-93 What is the primary source of hardening (strengthening) during heat-treating for steels?

Solution: The primary source of strengthening of steels is the martensitic transformation and the presence of carbon in the composition that distorts the martensite structure. The strength of the as-quenched martensite depends solely on the carbon content and not on alloy content.

12-102 A steel containing 0.3% C is heated to various temperatures above the eutectoid temperature, held for 1 h, and then quenched to room temperature. Using Figure 12– 35, determine the amount, composition, and hardness of any martensite that forms when the heating temperature is (a) 728°C; (b) 750°C; (c) 790°C; and (d) 850°C.

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Solution:

0.3 − 0.0218 ×100% = 37.2% HRC 65 0.77 − 0.0218 0.3 − 0.02 (b) γ : 0.60% C % M = ×100% = 48.3% HRC 65 0.6 − 0.02 0.3 − 0.02 (c) γ : 0.35% C % M = × 100% = 84.8% HRC 58 0.35 − 0.02 (d) γ : 0.3% C % M = 100% HRC 55 (a) γ : 0.77% C

%M =

12-103 A steel containing 0.95% C is heated to various temperatures above the eutectoid temperature, held for 1 h, and then quenched to room temperature. Using Figure 12– 35, determine the amount and composition of any martensite that forms when the heating temperature is (a) 728°C; (b) 750°C; (c) 780°C; and (d) 850°C. Solution:

6.67 − 0.95 × 100% = 96.9% HRC 65 6.67 − 0.77 6.67 − 0.95 (b) γ = 0.82% C % M = ×100% = 97.8% HRC 65 6.67 − 0.82 6.67 − 0.95 (c) γ = 0.88% C % M = × 100% = 98.8% HRC 65 6.67 − 0.88 (d) γ = 0.95% C % M = 100% HRC 65 (a) γ = 0.77% C

%M =

12-104 A steel microstructure contains 75% martensite and 25% ferrite; the composition of the martensite is 0.6% C. Using Figure 12–35, determine (a) the temperature from which the steel was quenched and (b) the carbon content of the steel.

Solution: In order for γ (and therefore martensite) to contain 0.6% C, the austenitizing T = 750°C. Then

M = γ = 0.25 =

0.6 − x x = 0.455% C 0.6 − 0.02

12-105 A steel microstructure contains 92% martensite and 8% Fe3C; the composition of the martensite is 1.10% C. Using Figure 12–35, determine (a) the temperature from which the steel was quenched and (b) the carbon content of the steel.

Solution: In order for γ (and therefore martensite) to contain 1.10% C, the austenitizing T = 865°C. Then

M = γ = 0.92 =

6.67 − x x = 1.55% C 6.67 − 1.10

12-106 A steel containing 0.8% C is quenched to produce all martensite. Estimate the volume change that occurs, assuming that the lattice parameter of the austenite is 3.6 Å. Does the steel expand or contract during quenching?

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Solution: Vγ = (3.6 Å)3 = 46.656 × 10–24 cm3 VM = a2c = (2.85 × 10–8 cm)2 (2.96 × 10–8) = 24.0426 × 10–24 cm3 To ensure that we have the same number of atoms, we need to consider two unit cells of martensite (2 atoms/cell) for each cell of FCC austenite (4 atoms/cell)

 (2)(24.0426) − 46.656  % ∆V =   × 100% = 3.06%, ∴ expansion 46.656  12-107 Describe the complete heat treatment required to produce a quenched and tempered eutectoid steel having a tensile strength of at least 125,000 psi. Include appropriate temperatures.

Solution: Austenitize at approximately 750°C, Quench to below 130°C (the Mf temperature) Temper at 620°C or less.

12-108 Describe the complete heat treatment required to produce a quenched and tempered eutectoid steel having an HRC hardness of less than 50. Include appropriate temperatures.

Solution: Austenitize at approximately 750°C, Quench to below the Mf (less than 130°C) Temper at a temperature higher than 330°C, but less than 727°C. 12-109 In eutectic alloys, the eutectic microconstituent is generally the continuous one, but in the eutectoid structures, the primary microconstituent is normally continuous. By describing the changes that occur with decreasing temperature in each reaction, explain why this difference is expected.

Solution: In a eutectoid reaction, the original grain boundaries serve as nucleation sites; consequently the primary microconstituent outlines the original grain boundaries and isolates the eutectoid product as a discontinuous constituent.

In a eutectic reaction, the primary phase nucleates from the liquid and grows. When the liquid composition approaches the eutectic composition, the eutectic constituent forms around the primary constituent, making the eutectic product the continuous constituent.

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Chapter 13: Heat Treatment of Steels and Cast Irons 13-3

A manufacturing plant requires 15 lb of AISI 8620 per final product. The plant makes 700 units per day (three shifts) six days a week, 50 weeks a year. How many tons of yttrium is this plant consuming per year if nothing is wasted?

Solution: Referring to Table 13-1, we see that AISI 8620 contains from 0.15 to 0.25 % yttrium, so the answer will cover a range. The average value (0.20 %) is also acceptable since the question did not say if we should use the minimum, maximum or range.

15 lb 700 unit 6 day 50 week 0.0015 lb Y ton ton Y = = 2.36

unit day week year lb 2000 lb year 15 lb 700 unit 6 day 50 week 0.0025 lb Y ton ton Y = = 3.94

unit day week year lb 2000 lb year The plant consumes between 2.36 and 3.94 tons of yttrium per year.

13-4

Calculate the amounts of ferrite, cementite, primary microconstituent, and pearlite in the following steels: (a) 1015; (b) 1035; (c) 1095; and (d) 10130.

Solution: (a) 1015:

6.67 − 0.15 × 100 = 97.8% Fe3C = 2.2% 6.67 − 0 0.77 − 0.15 primary α = × 100 = 82.9% pearlite = 17.1% 0.77 − 0.0218

α=

(b) 1035:

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6.67 − 0.35 × 100 = 94.8% Fe3C = 5.2% 6.67 − 0 0.77 − 0.35 primary α = ×100 = 56.1% pearlite = 43.9% 0.77 − 0.0218

α=

(c) 1095:

6.67 − 0.94 × 100 = 85.8% Fe3C = 14.2% 6.67 − 0 0.95 − 0.77 primary Fe3C = × 100 = 3.1% pearlite = 96.9% 6.67 − 0.77

α=

(d) 10130:

6.67 − 1.30 × 100 = 80.5% Fe3C = 19.5% 6.67 − 0 1.30 − 0.77 primary Fe3C = × 100 = 9.0% pearlite = 91.0% 6.67 − 0.77

α=

13-5

Estimate the AISI-SAE number for steels having the following microstructures. (a) 38% pearlite–62% primary ferrite; (b) 93% pearlite–7% primary cementite; (c) 97% ferrite–3% cementite; and (d) 86% ferrite–14% cementite.

Solution: (a) 38% pearlite–62% primary ferrite

62% =

0.77 − x × 100 x = 0.306% C 1030 steel 0.77 − 0.0218

(b) 93% pearlite–7% primary cementite

93% =

6.67 − x × 100 x = 1.183% C 10120 steel 6.67 − 0.77

(c) 97% ferrite–3% cementite

97% =

6.67 − x × 100 6.67 − 0

x = 0.200% C 1020 steel

(d) 86% ferrite–14% cementite

86% = 13-7

6.67 − x × 100 x = 0.934% C 1095 steel 6.67 − 0

Two samples of steel contain 93% pearlite. Estimate the carbon content of each sample if one is known to be hypoeutectoid and the other hypereutectoid.

Solution: The hypoeutectoid steel comprises 93% pearlite and 7% ferrite while the hypereuctectoid steel comprises 93% pearlite and 7% cementite.

For the hypoeutectoid steel (using the lever rule):

(% Pearlite/100) = (% C − 0.0218)/(0.77 − 0.0218) 0.93(0.7482) = % C − 0.0218 % C = 0.72 For the hypereutectoid steel (using the lever rule):

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(% Pearlite/100) = (6.67 − % C)/(6.67 − 0.77) 0.93(5.9) = 6.67 − % C % C = 1.18% 13-9

Define austempering and explain its purpose.

Solution: Austempering is process of quenching the metal to a temperature above the MS line and hold it until the formation of bainite. The purpose of austempering is to achieve a bainite structure in one step instead of the two step quench and temper process.

13-10 Complete the following table: Solution:

1035 steel 10115 steel A1 temperature 727°C 727°C A3 or Acm temperature 790°C 880°C Full annealing temperature 820°C 757°C Normalizing temperature 845°C 935°C Process annealing temperature 557–647°C — Spheroidizing temperature — 697°C

13-12 Determine the constants c and n in the Avrami relationship (Equation 12–2) for the transformation of austenite to pearlite for a 1050 steel. Assume that the material has been subjected to an isothermal heat treatment at 600°C and make a log–log plot of f versus t given the following information: f = 0.2 at t = 2s; f = 0.5 at t = 4s; and f = 0.8 at t = 7s.

Solution: The fraction transformed f is given by the equation, f = 1–exp (–ctn), where t is the elapsed time and c and n are constants for a particular temperature. Taking the natural logarithm of both sides, ln (1–f) = –ctn and rearranging and taking the natural logarithm again, ln [– ln (1–f)] = ln(c) n ln(t) Therefore, the slope of a plot of ln[–ln(1–f)] versus ln(t) is the value of n.

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For this case, n = 1.578 and ln(c) =–2.5828 such that c = 0.076.

13-13 In a pearlitic 1080 steel, the cementite platelets are 4 × 10–5 cm thick, and the ferrite platelets are 14 × 10–5 cm thick. In a spheroidized 1080 steel, the cementite spheres are 4 × 10–3 cm in diameter. Estimate the total interface area between the ferrite and cementite in a cubic centimeter of each steel. Determine the percent reduction in surface area when the pearlitic steel is spheroidized. The density of ferrite is 7.87 g/cm3 and that of cementite is 7.66 g/cm3.

Solution: First, we can determine the weight and volume percents of Fe3C in the steel:

0.80 − 0.0218 × 100 = 11.705 6.67 − 0.0218 11.705/7.66 vol% Fe3C = × 100 = 11.987 (11.705/7.66) + (88.295/7.87) wt% Fe3C =

Pearlite: Based on the thicknesses of the ferrite and cementite platelets in pearlite, there are two interfaces per (4 × 10–5 cm + 14 × 10–5 cm) = 18 × 10–5 cm, or 2 interfaces/(18 × 10–5cm) = 1.1 × 104 interfaces/cm

If all of the platelets are parallel to one another, then in 1 cm3 of pearlite, there is a total of A = (1.1 × 104/cm) (1 cm3) = 11,000 cm2 of interface/cm3

Spheroidite: The volume of an Fe3C sphere with r = 2 × 10–3 cm is V = (4π/3) (2 × 10–3 cm)3 = 3.35 × 10–8 cm3 The volume of Fe3C in 1 cm3 of spheroidite is given by the volume fraction of cementite, or 0.11987. The number of spheres in 1 cm3 of spheroidite is number = 0.11987 cm3/(3.35 × 10–8 cm3) = 3.58 × 106 spheres/cm3

The surface area of the spheres is therefore

A = 4π (2 ×10–3 cm) 2 (3.58 ×106 spheres/cm3 ) = 180 cm 2 of interface/cm3 The percent reduction in surface area during spheroidizing is then

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%=

(11, 000 − 180) cm 2 ×100 = 98.4% 11, 000 cm 2

13-14 Describe the microstructure present in a 1050 steel after each step in the following heat treatments: (a) heat at 820°C, quench to 650°C and hold for 90 s, and quench to 25°C; (b) heat at 820°C, quench to 450°C and hold for 90 s, and quench to 25°C; (c) heat at 820°C and quench to 25°C; (d) heat at 820°C, quench to 720°C and hold for 100 s, and quench to 25°C; (e) heat at 820°C, quench to 720°C and hold for 100 s, quench to 400°C and hold for 500 s, and quench to 25°C; (f) heat at 820°C, quench to 720°C and hold for 100 s, quench to 400°C and hold for 10 s, and quench to 25°C; and (g) heat at 820°C, quench to 25°C, heat to 500°C and hold for 103 s, and air cool to 25°C. Solution: (a) Austenite is present after heating to 820°C; both ferrite and pearlite form during holding at 650°C; ferrite and pearlite remain after cooling to 25°C. (b) Austenite is present after heating to 820°C; bainite forms after holding at 450°C; and bainite remains after cooling. (c) Austenite is present after heating to 820°C; martensite forms due to the quench. (d) Austenite is present after heating to 820°C; ferrite forms at 720°C, but some austenite still remains. During quenching, the remaining austenite forms martensite; the final structure is ferrite and martensite. (e) Austenite is present after heating to 820°C; ferrite begins to form at 720°C, but austenite still remains. At 400°C, the remaining austenite transforms to bainite; the final structure contains ferrite and bainite. (f) Austenite is present after heating to 820°C; ferrite begins to form at 720°C; some of the remaining austenite transforms to bainite at 400°C, but some austenite still remains after 10 s; the remaining austenite transforms to martensite during quenching. The final structure is ferrite, bainite, and martensite. (g) Austenite is present after heating to 820°C. The austenite transforms to martensite during quenching. During reheating to 500°C, the martensite tempers. The final structure is tempered martensite.

13-15 Describe the microstructure present in a 10110 steel after each step in the following heat treatments: (a) heat to 900°C, quench to 400°C and hold for 103 s, and quench to 25°C; (b) heat to 900°C, quench to 600°C and hold for 50 s, and quench to 25°C; (c) heat to 900°C and quench to 25°C; (d) heat to 900°C, quench to 300°C and hold for 200 s, and quench to 25°C; (e) heat to 900°C, quench to 675°C and hold for 1 s, and quench to 25°C;

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(f) heat to 900°C, quench to 675°C and hold for 1 s, quench to 400°C and hold for 900 s, and slowly cool to 25°C; (g) heat to 900°C, quench to 675°C and hold for 1 s, quench to 300°C and hold for 103 s, and air cool to 25°C; and (h) heat to 900°C, quench to 300°C and hold for 100 s, quench to 25°C, heat to 450°C for 3600 s, and cool to 25°C. Solution: (a) Austenite forms at 900°C. At 400°C, all of the austenite transforms to bainite. The final structure is all bainite. (b) Austenite forms at 900°C. At 600°C, all of the austenite transforms to cementite and pearlite, which gives the final structure. (c) Austenite forms at 900°C. All of the austenite transforms to martensite during quenching. (d) Austenite forms at 900°C. None of the austenite transforms within 200 s at 300°C; consequently all of the austenite forms martensite during quenching. This is a martempering heat treatment. (e) Austenite forms at 900°C. Cementite begins to form at 675°C; the remainder of the austenite transforms to martensite during quenching to 25°C. The final structure is cementite and martensite. (f) Austenite forms at 900°C. Cementite begins to form at 675°C. The remaining austenite transforms to bainite at 400°C. The final structure is cementite and bainite. (g) Austenite forms at 900°C. Cementite begins to form at 675°C. At 300°C, some of the remaining austenite transforms to bainite, but the Bf line is not crossed. The remaining austenite forms martensite during air cooling. The final structure is cementite, bainite, and martensite. (h) Austenite forms at 900°C. No transformation occurs at 300°C, since the time is too short. Consequently all of the austenite transforms to martensite during quenching. Reheating to 450°C for 3600 s (1 hour) tempers the martensite. The final structure is tempered martensite.

13-16 Recommend appropriate isothermal heat treatments to obtain the following, including appropriate temperatures and times: (a) an isothermally annealed 1050 steel with HRC 23; (b) an isothermally annealed 10110 steel with HRC 40; (c) an isothermally annealed 1080 steel with HRC 38; (d) an austempered 1050 steel with HRC 40; (e) an austempered 10110 steel with HRC 55; (f) an austempered 1080 steel with HRC 50.

Solution: (a) Austenitize at 820°C Quench to 600°C and hold for more than 10 s Cool to room temperature (b) Austenitize at 900°C Quench to 640°C and hold for more than 10 s Cool to room temperature

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(c) Austenitize at 780°C Quench to 600°C for more than 10 s Cool to room temperature (d) Austenitize at 820°C Quench to 390°C and hold for 100 s Cool to room temperature (e) Austenitize at 900°C Quench to 320°C and hold for 5000 s Cool to room temperature (f) Austenitize at 780°C Quench to 330°C and hold for 1000 s Cool to room temperature 13-17 Compare the minimum times required to isothermally anneal the following steels at 600°C. Discuss the effect of the carbon content of the steel on the kinetics of nucleation and growth during the heat treatment. (a) 1050; (b) 1080; and (c) 10110.

Solution: (a) 1050: The Pf time is about 5 s, the minimum time (b) 1080: The Pf time is about 10 s, the minimum time (c) 10110: The Pf time is about 3 s, the minimum time The carbon content has relatively little effect on the minimum annealing time (or the Pf time). The longest time is obtained for the 1080, or eutectoid, steel.

13-19 Typical media used for quenching include air, brine (10% salt in water), water, and various oils. (a) Rank the four media in order of the cooling rate from fastest to slowest. (b) Describe a situation when quenching in air would be undesirable. (c) During quenching in liquid media, typically either the part being cooled or the bath is agitated. Explain why.

Solution: (a) The cooling rate of the media from fastest to slowest is Brine > Water > Oils > Air The presence of gases on the surface of the quenched part can reduce the cooling rate significantly. Brine has lower dissolved gas content than water, and hence, reduced bubble formation leads to better wetting of the surface and subsequently to faster cooling in brine than in water. Air, on the other hand, is the least effective cooling medium as the gas absorption on the surface is a maximum. (b) An air quench is a mild quench. When fast cooling rates are desired to form certain phases (to obtain certain desirable mechanical properties), air quenching is not suitable. Typically, non-ferrous metals and alloys are not air quenched. (c) Good heat transfer is critical to obtain a good quench. When a hot part is dipped into a liquid medium, heat needs to be carried away

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from the surface effectively. Agitation of the bath or the part being quenched helps break up the insulating gas blanket that typically forms between the part being quenched and the liquid and allows fresh liquid to contact the surface.

13-20 We wish to produce a 1050 steel that has a Brinell hardness of at least 330 and an elongation of at least 15% (a) Recommend a heat treatment, including appropriate temperatures, that permits this to be achieved. Determine the yield strength and tensile strength that are obtained by this heat treatment. (See Figure 13-9.) (b) What yield and tensile strengths would be obtained in a 1080 steel using the same heat treatment? (See Figure 12–27.) (c) What yield strength, tensile strength, and % elongation would be obtained in the 1050 steel if it were normalized? (See Figure 13–4.)

Solution: (a) It is possible to obtain the required properties; the Brinell hardness is obtained if the steel is quenched and then tempered at a temperature below 480°C, and the % elongation can be obtained if the tempering temperature is greater than 420°C. Therefore a possible heat treatment would be Austenitize at 820°C Quench to room temperature Temper between 420°C and 480°C Cool to room temperature The quench and temper heat treatment will also give a yield strength between 140,000 and 160,000 psi, while the tensile strength will be between 150,000 and 180,000 psi. The higher strengths are obtained for the lower tempering temperatures. (b) If a 1080 steel is tempered in the same way (Figure 12–23), the yield strength would lie between 130,000 and 135,000 psi, and the tensile strength would be 175,000 to 180,000 psi. The higher strengths are obtained for the lower tempering temperatures. (c) If the 1050 steel were normalized rather than quenched and tempered, the properties would be about (from Figure 13–4) 100,000 psi tensile strength 65,000 psi yield strength 20% elongation 13-21 We wish to produce a 1050 steel that has a tensile strength of at least 175,000 psi and a reduction in area of at least 50%. (a) Recommend a heat treatment, including appropriate temperatures, that permits this to be achieved. Determine the Brinell hardness number, % elongation, and yield strength that are obtained by this heat treatment. (See Figure 13-9.) (b) What yield strength and tensile strength would be obtained in a 1080 steel using the same heat treatment? (See Figure 12-27.) (c) What yield strength, tensile strength, and % elongation would be obtained in the 1050 steel if it were annealed? (See Figure 13-4.)

Solution: (a) Using a quench and temper heat treatment, we can obtain the minimum tensile strength by tempering below 430°C, and the minimum reduction in area by tempering above 400°C. Our heat treatment is then

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Austenitize at 820°C Quench to room temperature Temper between 400°C and 430°C Cool to room temperature This heat treatment will also give 390 to 405 BH 14 to 15% elongation 160,000 to 165,000 psi yield strength (b) If the same treatment is used for a 1080 steel, the properties would be 140,000 psi yield strength 180,000 psi tensile strength (c) If the 1050 steel is annealed, the properties are (From Figure 13– 4) 52,000 psi yield strength 85,000 psi tensile strength 25% elongation 13-22 A 1030 steel is given an improper quench and temper heat treatment, producing a final structure composed of 60% martensite and 40% ferrite. Estimate the carbon content of the martensite and the austenitizing temperature that was used. What austenitizing temperature would you recommend?

Solution: We can work a lever law at several temperatures in the α + γ region of the iron-carbon phase diagram, finding the amount of austenite (and its composition) at each temperature. The composition of the ferrite at each of these temperatures is about 0.02% C. The amount and composition of the martensite that forms will be the same as that of the austenite:

at 800°C : γ : 0.33% C % γ = (0.30 – 0.02) / (0.33 – 0.02) = 90% at 780°C : γ : 0.41% C % γ = (0.30 – 0.02) / (0.41 – 0.02) = 72% at 760°C : γ : 0.54% C % γ = (0.30 – 0.02) / (0.54 – 0.02) = 54% at 740°C : γ : 0.68% C % γ = (0.30 – 0.02) / (0.68 – 0.02) = 42% at 727°C : γ : 0.77% C % γ = (0.30 – 0.02) / (0.77 – 0.02) = 37% The amount of austenite (equal to that of the martensite) is plotted versus temperature in the graph. Based on this graph, 60% martensite forms when the austenitizing temperature is about 770°C. The carbon content of the martensite that forms is about 0.48% C. The A3 temperature of the steel is about 805°C. A proper heat treatment might use an austenitizing temperature of about 805°C + 55°C = 860°C.

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13-23 A 1050 steel should be austenitized at 820°C, quenched in oil to 25°C, and tempered at 400°C for an appropriate time. (a) What yield strength, hardness, and % elongation would you expect to obtain from this heat treatment? (See Figure 13-9.) (b) Suppose the actual yield strength of the steel is found to be 125,000 psi. What might have gone wrong in the heat treatment to cause this low strength? (c) Suppose the Brinell hardness is found to be HB 525. What might have gone wrong in the heat treatment to cause this high hardness?

Solution: The properties expected for a proper heat treatment are 170,000 psi yield strength 190,000 psi tensile strength 405 HB 14% elongation If the yield strength is 125,000 psi (much lower than expected), then the tempering process might have been done at a tempering temperature greater than 400°C (perhaps as high as 500°C). Another possible problem could be an austenitizing temperature that was lower than 820°C (even lower than about 770°C, the A3), preventing complete austenitizing and thus not all martensite during the quench.

If the hardness is HB 525 (higher than expected), the tempering temperature may have been too low or, in fact, the steel probably was not tempered at all.

13-24 A part produced from a low-alloy, 0.2% C steel (Figure 13–15) has a microstructure containing ferrite, pearlite, bainite, and martensite after quenching. What microstructure would be obtained if we had used a 1080 steel? What microstructure would be obtained if we used a 4340 steel? [See Figures 13-14, 13-15, and 13-16(b).]

Solution: To produce ferrite, pearlite, bainite, and martensite in the same microstructure during continuous cooling, the cooling rate must have been between 10 and 20°C/s. If the same cooling rates are used for the other steels, the microstructures are 1080 steel: fine pearlite 4340 steel: martensite

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13-25 Fine pearlite and a small amount of martensite are found in a quenched 1080 steel.

What microstructure would be expected if we had used a low alloy, 0.2% C steel? What microstructure would be expected if we had used a 4340 steel? [See Figures 13–14, 13– 15, and 13–16.]

Solution: A cooling rate of about 50°C/s will produce fine pearlite and a small amount of martensite in the 1080 steel. For the same cooling rate, the microstructure in the other steels will be low-alloy, 0.2% C steel: ferrite, bainite, and martensite 4340 steel: martensite 13-26 Predict the phases formed when a bar of 1080 steel is quenched from slightly above the eutectoid temperature under the following conditions: (a) oil (without agitation); (b) oil (with agitation); (c) water (with agitation); and (d) brine (no agitation).

Suggest a quenching medium if we wish to obtain coarse pearlite.

[See Figure 13–14 and Table 13–2 for this problem.]

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Solution: The cooling rates for the various conditions in the problem, assuming a 1 in. diameter bar, are obtained from Table 13-2.

Oil (without agitation) 18°C/s Oil (with agitation) 45°C/s Water (with agitation) 190°C/s Brine (no agitation) 90°C/s Using Figure 13–14, the phases formed for the 1080 steel are identified as follows: Oil (without agitation) 18°C/s Fine pearlite Oil (with agitation) 45°C/s Pearlite + Martensite Water (with agitation) 190°C/s Martensite Brine (no agitation) 90°C/s Pearlite + Martensite None of these cooling rates will result in coarse pearlite as the cooling rates are too high. To obtain coarse pearlite, the cooling rate needs to be lower than 5°C/s. This cannot be achieved by any quenching treatment but can be achieved by slowly annealing the 1080 steel as the temperature in the furnace is ramped down at less than 5°C/s.

13-27 Define ausforming and explain its purpose.

Solution: Ausforming is similar to austempering for the exception that the metal undergoes deformation prior to quenching. The purpose of ausforming is to make metals into desired shapes while they are soft (γ-matrix) instead of trying to shape the metal when they are hard (martensite).

13–29 We have found that a 1070 steel, when austenitized at 750°C, forms a structure containing pearlite and a small amount of grain-boundary ferrite that gives acceptable strength and ductility. What changes in the microstructure, if any, would be expected if the 1070 steel contained an alloying element, such as Mo or Cr? Explain.

Solution: The alloying element may shift the eutectoid carbon content to below 0.7% C, making the steel hypereutectoid rather than hypoeutectoid.

This in turn means that grain boundary Fe3C will form rather than grain boundary ferrite. The grain boundary Fe3C will embrittle the steel.

13-30 Using the TTT diagrams, compare the hardenabilities of 4340 and 1050 steels by determining the times required for the isothermal transformation of ferrite and pearlite (Fs, Ps, and Pf) to occur at 650°C. [See Figures 13-8 and 13-16(a).]

Solution: From the diagrams, we can find the appropriate times: 4340 steel: Fs = 200 s Ps = 3,000 s Pf = 15,000 s 1050 steel: Fs = 3 s Ps = 10 s Pf = 50 s Because the transformation times are much longer for the 4340 steel, the 4340 steel has the higher hardenability.

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13-31 Water hardenable 1095 steel is supposed to be tempered at 500 °C, but a negligent operator does not check the automatic process control and so it is tempered at 450 °C. What is the hardness of the as-produced steel compared to the specified steel?

Solution: Referring to Figure 13-19, water hardenable 1095 has a Rockwell C hardness of about 42 at 500 °C. At 450 °C it has a Rockwell C hardness of about 43. The steel is actually harder than it should be by an increment of 1 HR. 13-32 Determine the cooling rate for AISI 9310, 4320 and 1080 steels to have the same hardness. (See Figure 13-21.)

Solution: Referring to Figure 13-21, we see the hardness curves for these steels intersecting at a Jominy distance of about 4/16 of an inch. This corresponds (Table 13-3) to a cooling rate of 36 °C/s.

13-33 We would like to obtain a hardness of HRC 38 to 40 in a quenched steel. What range of cooling rates would we have to obtain for the following steels? Are some steels inappropriate for achieving these levels of hardness? (See Figure 13-21 and Table 13-3.) (a) 4340; (b) 8640; (c) 9310; (d) 4320; (e) 1050; and (f) 1080.

Solution: (a) 4340: not applicable; the hardnesses are always much higher than the desired range. (b) 8640: a Jominy distance of about 18/16 to 20/16 is required to

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give the desired hardness; this corresponds to a cooling rate of about 3 to 4°C/s (c) 9310: a Jominy distance of 10/16 to 12/16 is required to give the desired hardness; this corresponds to a cooling rate of 8 to 10°C/s (d) 4320: a Jominy distance of about 6/16 is required to give the desired hardness; this corresponds to a cooling rate of 22°C/s (e) 1050: a Jominy distance of 4/16 to 4.5/16 is required to give the desired hardness; this corresponds to a cooling rate of 32 to 36°C/s (f) 1080: a Jominy distance of 5/16 to 6/16 is required to give the desired hardness; this corresponds to a cooling rate of 16 to 28°C/s 13-34 A steel part must have an as-quenched hardness of HRC 35 in order to avoid excessivewear rates during use. When the part is made from 4320 steel, the hardness is only HRC 32. Determine the hardness if the part were made under identical conditions, but with the following steels. Which, if any, of these steels would be better choices than 4320? (a) 4340; (b) 8640; (c) 9310; (d) 1050; and (e) 1080.

Solution:

The Jominy distance that gives a hardness of HRC 32 in the 4320 steel is 9/16 in. The cooling rates, and hence Jominy distances, will be the same for the other steels. From the hardenability curves, the hardnesses of the other steels are (a) 4340: HRC 60 (b) 8640: HRC 54 (c) 9310: HRC 40 (d) 1050: HRC 28 (e) 1080: HRC 36 All of the steels except the 1050 steel would develop an as-quenched hardness of at least HRC 35 and would be better choices than the 4320 steel. The 1080 steel might be the best choice, since it will likely be the least expensive (no alloying elements present).

13-35 A part produced from a 4320 steel has a hardness of HRC 35 at a critical location after quenching. Determine (a) the cooling rate at that location, and (b) the microstructure and hardness that would be obtained if the part were made of a 1080 steel.

Solution: (a) To obtain the HRC 35 in a 4320 steel, the Jominy distance must be about 7.5/16 in., corresponding to a cooling rate of 16°C/s. (b) If the part is produced in a 1080 steel, the cooling rate will still be about 16°C/s. From the CCT diagram for the 1080 steel, the part will contain all pearlite, with a hardness of HRC 38.

13-36 A 1080 steel is cooled at the fastest possible rate that still permits all pearlite to form.

What is this cooling rate? What Jominy distance and hardness are expected for this cooling rate?

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Solution: The fastest possible cooling rate that still permits all pearlite is about 40°C/s. This cooling rate corresponds to a Jominy distance of about 3.5/16 in. From the hardenability curve, the hardness will be HRC 46.

13-37 Determine the hardness and the microstructure at the center of a 1.5-in.-diameter 1080 steel bar produced by quenching in (a) unagitated oil; (b) unagitated water; and (c) agitated brine.

Solution: (a) unagitated oil: the H-factor for the 1.5-in. bar is 0.25. The Jominy distance will be about 11/16 in., or a cooling rate of 9°C/s. From the CCT diagram, the hardness is HRC 36 and the steel is all pearlite. (b) unagitated water: the H-factor for the bar is 1.0. The Jominy distance will be about 5/16 in., or a cooling rate of 28°C/s. From the CCT diagram, the hardness is HRC 40 and the steel will contain pearlite. (c) agitated brine: the H-factor is now 5.0. The Jominy distance is about 3.5/16 in., or a cooling rate of 43°C/s. The steel has a hardness of HRC 46 and the microstructure contains both pearlite and martensite.

13-38 A 2-in.-diameter bar of 4320 steel is to have a hardness of at least HRC 35. What is the minimum severity of the quench (H coefficient)? What type of quenching medium would you recommend to produce the desired hardness with the least chance of quench cracking?

Solution: The hardness of HRC 35 is produced by a Jominy distance of 7.5/16 in.

In order to produce this Jominy distance in a 2-in. diameter bar, the Hcoefficient must be greater or equal to 0.9. All of the quenching media described in Table 13–2 will provide this Jominy distance except unagitated oil. To prevent quench cracking, we would like to use the least severe quenchant; agitated oil and unagitated water, with H = 1.0, might be the best choices.

13-39 A steel bar is to be quenched in agitated water. Determine the maximum diameter of the bar that will produce a minimum hardness of HRC 40 if the bar is (a) 1050; (b) 1080; (c) 4320; (d) 8640; and (e) 4340.

Solution: (a) 1050 steel: The H-coefficient for the agitated water is 4.0. For the 1050 steel, the Jominy distance must be at least 3/16 in. to produce the desired hardness. Therefore the maximum diameter that will permit this Jominy distance (or cooling rate) is 1.3 in. (b) 1080 steel: Now the Jominy distance must be at least 5/16 in., with the same H-coefficient. The maximum diameter allowed is 1.9 in.

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(c) 4320 steel: The minimum Jominy distance is 5/16 in., and the maximum diameter of the bar is 1.9 in. (d) 8640 steel: The minimum Jominy distance is 18/16 in.

Consequently bars with a maximum diameter of much greater than 2.5 in. will produce the desired cooling rate and hardness. (e) 4340 steel: Bars with a maximum diameter of much greater than 2.5 in. produce the required cooling rate.

13-40 The center of a 1-in.-diameter bar of 4320 steel has a hardness of HRC 40. Determine the hardness and microstructure at the center of a 2-in. bar of 1050 steel quenched in the same medium.

Solution: To obtain HRC 40 in the 4320 steel, we need a Jominy distance of 5/16 in. For a 1-in.-diameter bar, the quenching medium must have a minimum H-coefficient of 0.4. Therefore, if a 2-in. diameter bar is quenched in the same medium (i.e. H = 0.4), the Jominy distance will be about 11/16 in.; this Jominy distance produces a hardness of HRC 27 in a 1050 steel.

13-41 You encounter a specification that lists a Jominy distance of 1.27 cm and a bar diameter of 3.81 cm. What is the H coefficient for this?

Solution: To use the Grossman chart, which relates these three variables, we must convert the specifications into inches. There are exactly 2.54 cm per inch. This gives a Jominy distance of 0.5 in., or 8/16 in. The 3.81 cm bar is 1.5 inches in diameter. The chart is now easily read and we see the H coefficient is approximately 0.45.

13–44 A 1010 steel is to be carburized using a gas atmosphere that produces 1.0% C at the surface of the steel. The case depth is defined as the distance below the surface that contains at least 0.5% C. If carburizing is done at 1000°C, determine the time required to produce a case depth of 0.01 in. (See Chapter 5 for review.)

Solution: The diffusion coefficient for carbon in FCC iron at 1000°C is D = 0.23 exp [–32,900/1.987/1273] = 5.16 × 10–7 cm2/s The case depth “x” is to be 0.01 in. = 0.0254 cm.

From Fick’s law:

1.0 − 0.5 = 0.556 = erf[ x /(2 Dt )] 1.0 − 0.1 x /(2 Dt ) = 0.541 0.0254/[2 (5.16 × 10−7 )t ] = 0.541 t = 1068 s = 0.3 h 13–45 A 1015 steel is to be carburized at 1050°C for 2 h using a gas atmosphere that produces 1.2% C at the surface of the steel. Plot the percent carbon versus the distance from the surface of the steel. If the steel is slowly cooled after carburizing, determine the amount

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of each phase and microconstituent at 0.002-in. intervals from the surface. (See Chapter 5.)

Solution: The diffusion coefficient for carbon in FCC iron at 1050°C is D = 0.23 exp[–32,900/1.987/1323] = 8.44 × 10–7cm2/s t = 2 h = 7200 s

From Fick’s law:

1.2 − cx = erf[ x /2 (8.44 × 10−7 )(7200)] 1.2 − 0.15 1.2 − cx or = erf (6.41x) 1.05 If x = 0.002 in. = 0.00508 cm, then

1.2 – cx = erf (0.0326) = 0.037 1.05 cx = 1.161% C These calculations can be repeated for other values of x, with the results shown below:

x = 0.010 in. = 0.0254 cm cx = 1.009% C x = 0.020 in. = 0.0508 cm cx = 0.838% C x = 0.050 in. = 0.1270 cm cx = 0.413% C x = 0.100 in. = 0.2540 cm cx = 0.178% C The graph shows how the carbon content varies with distance.

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13-48 A 1050 steel is welded. After cooling, hardnesses in the heat-affected zone are obtained at various locations from the edge of the fusion zone. Determine the hardnesses expected at each point if a 1080 steel were welded under the same conditions. Predict the microstructure at each location in the as-welded 1080 steel.

Distance from Edge of Fusion Zone Hardness in 1050 Weld 0.05 mm HRC 50 0.10 mm HRC 40 0.15 mm HRC 32 0.20 mm HRC 28 Solution: We can take advantage of the fact that the cooling rate in the two steels will be virtually identical if the welding conditions are the same. Thus at a distance of 0.05 mm from the edge of the fusion zone, the HRC 50 hardness of the 1050 steel is obtained with a Jominy distance of 3y16 in., or a cooling rate of 50°C/s. At the same point in a 1080 steel weldment, the 3y16-in.

Jominy distance gives a hardness of HRC 53 (from the hardenability curve), and the 50°C/s cooling rate gives a microstructure of pearlite and martensite (from the CCT curve). The table below shows the results for all four points in the weldment.

Distance 0.05 mm 0.10 0.15 0.20

Jominy distance 3/16 in.

4/16 7/16 10/16

Cooling rate 50°C/s 36 17 10

Hardness HRC 53 HRC 46 HRC 38 HRC 36

Structure P+M pearlite pearlite pearlite

13-50 Surplus stock inventory contains 15 ton 304, 5.2 ton 304L, 1 ton 316 and 4.7 ton 440C stainless steel. A spike in molybdenum prices makes it worthwhile to sell off the steels that contain it. How much molybdenum is contained in all the steels and how much total mass is being sold?

Solution: Molybdenum content of selected stainless steels can be read from Table 13-4. Only 316 and 440C contain it. Calculating the total amount: ton Mo ton Mo ! = 1 ton 316 × 0.025 + 4.7 ton 440C × 0.007 ton 316 ton 440C ! = 0.0579 ton Mo = 115.8 lb Mo The total mass of the steels being sold off would be 1 ton + 4.7 ton = 5.7 ton.

13-51 A food processing plant that is designed to use 5.75 metric tons of 316 stainless steel will actually use 115 % of design specification. How many moles of chromium must be refined for this? Assume the refinery loses 5 % of what it refines.

Solution: First, the total mass required to be refined, including both percentages:

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'( =

5.75 tonne 316 1000 kg 316 0.19 kg Cr 1.15 kg Cr needed 1 kg Cr refined

1 1 1 tonne 316 1 kg 316 1 kg Cr specified 0.95 kg Cr sent = 1322.5 kg Cr 1322.5 kg Cr 1 kmol Cr .'( = = 25.4 kmol Cr

1 1 51.996 kg Cr

13-52 We wish to produce a martensitic stainless steel containing 17% Cr. Recommend a carbon content and austenitizing temperature that would permit us to obtain 100% martensite during the quench. What microstructure would be produced if the martensite were then tempered until the equilibrium phases formed?

Solution: We must select a combination of a carbon content and austenitizing temperature that puts us in the all-austenite region of the Fe–Cr–C phase diagram. One such combination is 1200°C and 0.5% C. If a 0.5% C steel is held at 1200°C to produce all austenite, and then is quenched, 100% martensite will form.

If the martensite is tempered until equilibrium is reached, the two phases will be ferrite and M23C6. The M23C6 is typically Cr23C6.

13-53 Occasionally, when an austenitic stainless steel is welded, the weld deposit may be slightly magnetic. Based on the Fe–Cr–Ni–C phase diagram [Figure 13–28(b)], what phase would you expect is causing the magnetic behavior? Why might this phase have formed? What could you do to restore the nonmagnetic behavior?

Solution: The magnetic behavior is caused by the formation of a BCC iron phase, in this case the high temperature δ-ferrite. The δ-ferrite forms during solidification, particularly when solidification does not follow equilibrium; subsequent cooling is too rapid for the δ-ferrite to transform to austenite, and the ferrite is trapped in the microstructure. If the steel is subsequently annealed at an elevated temperature, the δ-ferrite can transform to austenite and the steel is no longer magnetic.

13-55 Compare the eutectic temperatures of a Fe–4.3% C cast iron with a Fe–3.6% C–2.1% Si alloy. Which alloy is expected to be more machinable and why?

Solution: The carbon equivalent of the Fe-C-Si alloy is given by CE = %C + (1/3) % Si CE = 3.6 + (1/3) (2.1)% CE = 4.3% Both alloys have equivalent carbon contents; however, the presence of Si stabilizes the graphite phase and the phase diagram will follow the solid lines in Figure 13-31 more closely while the Fe–4.3%C alloy will tend to follow the dashed lines in Figure 13-31. Based on this, the eutectic temperature for the Fe-3.6%C–2.1% Si alloy is expected to be 1146°C and that of the Fe–4.3%C alloy will be 1140°C, a difference of 6°C. The Si containing alloy has the potential to be more machinable as the presence of silicon stabilizes the graphite phase, but the alloy may

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need some heat treatment to ensure proper graphitization.

13-57 A bar of a class 40 gray iron casting is found to have a tensile strength of 50,000 psi.

Why is the tensile strength greater than that given by the class number? What do you think is the diameter of the test bar?

Solution: The strength of gray iron depends on the cooling rate of the casting; faster cooling rates produce finer microstructures and more pearlite in the microstructure. Although the iron has a nominal strength of 40,000 psi, rapid cooling can produce the fine graphite and pearlite that give the higher 50,000 psi strength.

The nominal 40,000 psi strength is expected for a casting with a diameter of about 1.5 in.; if the bar is only 0.75 in. in diameter, a tensile strength of 50,000 psi might be expected.

13-58 You would like to produce a gray iron casting that freezes with no primary austenite or graphite. If the carbon content in the iron is 3.5%, what percentage of silicon must you add?

Solution: We get neither primary phase when the carbon equivalent (CE) is 4.3%. Thus

CE = %C + (1/3)% Si 4.3 = 3.5 + (1/3)% Si or % Si = 2.4

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Chapter 14: Nonferrous Alloys 14-1

In some cases, we may be more interested in cost of a material per unit volume than in the cost per unit weight. Rework Table 14–1 to show the cost in terms of $/cm3. Does this change/alter the relationship between the different materials?

Solution: We can find the density (in g/cm3) of each metal from Appendix A. We can convert the cost in $/lb to $/g (using 454 g/lb) and then multiply the cost in $/g by the density, giving $/cm3. The left hand side of the table shows the results of these conversions, with the metals ranked in order of cost per volume. The right hand side of the table shows the cost per pound.

Cost/volume Rank Cost/lb Rank 3 Al $0.0048/cm 1 Fe $0.30 1 Fe $0.0052/cm3 2 Al $0.80 2 Mg $0.0105/cm3 3 Zn $0.95 3 3 Zn $0.0149/cm 4 Pb $1.00 4 Pb $0.0250/cm3 5 Mg $2.75 5 3 Cu $0.0629/cm 6 Cu $3.20 6 Ti $0.0795/cm3 7 Ni $8.00 7 Ni $0.1568/cm3 8 Ti $8.00 7 3 W $1.0176/cm 9 W $24.00 9 Be $1.7318/cm3 10 Be $425.00 10 The relationship is changed; for example, aluminum is second based on weight, but first on the basis of volume.

14-2

Determine the specific strength of the following metals and alloys (use the densities of the major metal component as an approximation of the alloy density where required): Alloy/metal Tensile strength (MPa) 1100-H18 165 5182-O 290 2024-T4 469 2090-T6 552 201-T6 483 ZK40A-T5 276 Age hardened Cu 2% Be 1310 Alpha Ti alloy 862 W 455 Ta 186

Solution: To obtain specific strength, the tensile strength must first be converted from MPa to kg/m2 by dividing by the acceleration due to gravity: 9.81m/s2. The calculations are shown in the table below.

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Alloy/metal 1100-H18 5182-O 2024-T4 2090-T6 201-T6 ZK40A-T5 Age hardened Cu 2% Be Alpha Ti alloy W Ta 14-3

TS (MPa) TS (kg/m2) × 106 165 16.8 290 29.6 469 47.8 552 56.3 483 49.2 276 28.1 1310 133.5 862 455 186

87.9 46.4 19.0

Density (kg/m3) 2700 2700 2700 2700 2700 1740 8930

Specific Strength (m) 6.23 × 103 1.09 × 104 1.77 × 104 2.08 × 104 1.82 × 104 1.62 × 104 1.50 × 104

4510 19250 16,600

1.95 × 104 2.41 × 103 1.14 × 103

How much would 50 kg of the hypothetical alloy 5074-O cost in raw materials prices using the estimates given in Table 14-1?

Solution: We note that the alloy is 99.74 % aluminum (x074) and alloyed with magnesium (5xxx).

We set up two equations to solve for the individual prices: lb USD = 50 kg 0.9974 2.2 0.80 = 87.77 USD kg lb lb USD = 50 kg 1 − 0.9974 2.2 2.75 = 00.79 USD kg lb = 88.56 USD

14-4

Structural steels have traditionally been used in shipbuilding; however, with increasing fuel costs, it is desirable to find alternative lower weight materials. Some of the key properties required for shipbuilding materials are high yield strength, high corrosion resistance, and low cost. Discuss the benefits and disadvantages of aluminum alloys as a replacement for structural steels in ships.

Solution: The main advantage of Al in shipbuilding is its specific strength (tensile strength to density ratio), leading to lower weight structures that can result in significant fuel cost savings.

Al generally has good corrosion resistance due to the formation of a stable surface oxide; however, it could be susceptible to galvanic corrosion in salt water as it is anodic with respect to most other materials used in shipbuilding.

Potential disadvantages of Al include its low melting temperature, which could compromise the ship’s structural integrity due to accidents related to high temperature conditions / fires and relative difficulty to weld when compared to steels. Despite these disadvantages, with continuing developments in aluminum alloys, more of these light weight alloys are being used in shipbuilding.

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14-5

Assuming that the density remains unchanged, compare the specific strength of the 2090–T6 aluminum alloy to that of a die cast 443–F aluminum alloy. If you considered the actual density, do you think the difference between the specific strengths would increase or decrease? Explain.

Solution:

2090 − T6: Tensile strength = 80, 000 psi Spec. strength =

(80,000 psi) (454 g/lb) (2.7 g/cm 3 ) (2.54 cm/in.)3

= 8.2 × 105 in.

443 − F: Tensile strength = 33, 000 psi (33,000 psi) (454 g/lb) Spec. strength = (2.7 g/cm 3 ) (2.54 cm/in.)3 = 3.39 × 105in.

Both should increase since both Li and Si (the major alloying elements) are less dense than Al. 14-6

Explain why aluminum alloys containing more than about 15% Mg are not used.

Solution: When more than 15% Mg is added to Al, a eutectic microconstituent is produced during solidification. Most of the eutectic is the brittle intermetallic compound β, and it will likely embrittle the eutectic.

14-7

Would you expect a 2024–T9 aluminum alloy to be stronger or weaker than a 2024–T6 alloy? Explain.

Solution: The T9 treatment will give the higher strength; in this temper, cold working and age hardening are combined, while in T6, only age hardening is performed.

14-8

Estimate the tensile strength expected for the following aluminum alloys: (a) 1100–H14; (b) 5182–H12; and (c) 3004–H16.

Solution: (a) The tensile strength for 1100–H14 is the average of the O and H18 treatments.

1100 − O (0%CW): TS = 13 ksi 1100 − H18(75% CW) : TS = 24 ksi 13 + 24 TSH14 = = 18.5 ksi 2 (b) The tensile strength for 5182–H12 is the average of the O and H14 treatments, and H14 is the average of the O and H18 treatments.

We do not have data in Table 14–5 for 5182–H18; however, 5182– H19 has a tensile strength of 61,000 psi and H18 should be 2000 psi less, or 59,000 psi.

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5182 − O (0% CW): TS = 42 ksi 5182 − H18 (75% CW): TS = 61 − 2 = 59 ksi 42 + 59 TSH14 = = 50.5 ksi 2 42 + 50.5 TSH12 = = 46.25 ksi 2 (c) The tensile strength for 3004–H16 is the average of the H14 and H18 treatments, and H14 is the average of the O and H18 treatments.

3004 − O: TS = 26 ksi 3004 − H18: TS = 41 ksi 26 + 41 = 33.5 ksi TSH14 = 2 41 + 33.5 = 37.25 ksi TSH16 = 2 14-9

A 1-cm-diameter steel cable with a yield strength of 480 MPa needs to be replaced to reduce the overall weight of the cable. Which of the following aluminum alloys could be a potential replacement? (a) 3004–H18 (Sy = 248 MPa); (b) 1100–H18 (Sy = 151 MPa); (c) 4043–H18 (Sy = 269 MPa); and (d) 5182–O (Sy = 131 MPa).

The density of the steel used in the cable was 7.87 g/cm3, and assume that the density of all the aluminum alloys listed above is 2.7 g/cm3.

Solution: The force F sustained by the 1 cm diameter steel cable is

π  F = S y A0 = 480 × 106 N/m 2  (0.01 m) 2  = 37, 700 N , 4  where Sy is the yield strength and A0 is the initial cross–sectional area.

In order for the Al alloys listed to carry the same load, the diameter of the cable required can be calculated using

πd2 4

=

F Sy

or

d=

4F , π Sy

where d is the cable diameter. Per unit length, the weight of the cable is given by

Weight/Length =

πd2 4

(ρ )

where ρ is the density of the material. The results for each alloy are given in the table below.

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Alloy Sy (MPa) d(cm) Weight/Length (g/cm) Steel 480 1.00 6.18 3004–H18 248 1.39 4.10 1100–H18 151 1.78 6.74 4043–H18 269 1.34 3.78 5182–O 131 1.91 7.77 From the table, 3004–H18 and 4043–H18 are potential replacements for the steel cable.

14-10 Suppose, by rapid solidification from the liquid state, that a supersaturated Al–7% Li alloy can be produced and subsequently aged. Compare the amount of β that will form in this alloy with that formed in a 2090 alloy.

Solution: The 2090 alloy contains 2.4% Li; from the Al–Li phase diagram, the composition of the β is about 20.4% Li and that of the α is approximately 2% Li at a typical aging temperature or at room temperature:

7−2 × 100% = 27% 20.4 − 2 2.4 − 2 2090 : % β = ×100% = 2.2% 20.4 − 2

Al − 7% Li: % β =

14-11 Determine the amount of Mg2Al3 (β) expected to form in a 5182–O aluminum alloy. (See Figure 14–2.)

Solution: The 5182 alloy contains 4.5% Mg. Thus from the Mg–Al phase diagram, which shows the α contains about 0% Mg and β contains about 35% Mg:

%β =

4.5 − 0 × 100% = 12.9% 35 − 0

14-12 A 5182–O aluminum alloy part that had been exposed to salt water showed severe corrosion along the grain boundaries. Explain this observation based on the expected phases at room temperature in this alloy. (See Figure 14–2.)

Solution: Since 5182–O is an Al–4.5% Mg alloy, the Mg3Al2 intermetallic phase may have precipitated along the grain boundaries. This material is strongly anodic, and this probably led to galvanic corrosion in salt water.

14-13 From the data in Table 14–6, estimate the ratio by which the yield strength of magnesium can be increased by alloying and heat treatment and compare with that of aluminum alloys.

Solution: The exact values will differ depending on the alloys we select for comparison. The table below provides an example. Strengthening of

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Mg is only about 1/10 as effective as in Al. Magnesium YS YS/YSMg 13 ksi — Pure Al 17 ksi 1.3 CW 1100–O 28 ksi 2.2 5182–O Alloy

Pure Mg: Cold Worked Casting & T6 (ZK61A–T6) Wrought & T5 40 ksi 3.1 (AZ80A–T5)

2090–T6

Aluminum YS YS/YSAl 2.5 ksi — 22 ksi 8.8 19 ksi 7.6 75 ksi

30.0

14-14 Suppose a 24-in.-long round bar is to support a load of 400 lb without any permanent deformation. Calculate the minimum diameter of the bar if it is made from (a) AZ80A–T5 magnesium alloy and (b) 6061–T6 aluminum alloy.

Calculate the weight of the bar and the approximate cost (based on pure Mg and Al) in each case.

Solution: A = F/Yield Strength (a) AZ80A – T5 : YS = 40 ksi A = 400/40,000 = 0.01 in.2

d = 4 A/π = 0.113 in.

Weight = (24 in.) (0.01 in.2 ) (0.063 lb/in.3 ) = 0.0151 lb Cost = ($1.5/lb) (0.0151 lb) = $0.023 (b) 6061 – T6 : YS = 40 ksi therefore, A = 0.01 in.2 d = 0.113 in. as in part (a), but Weight = (24 in.) (0.01 in.2 ) (0.063 lb/in.3 ) = 0.0233 lb Cost = ($0.60/lb) (0.0233 lb) = $0.014 Al is less costly than Mg, even though Mg is lighter.

14-15 A 10-m rod, 0.5 cm in diameter, must elongate no more than 2 mm under load.

Determine the maximum force that can be applied if the rod is made from (a) aluminum; (b) magnesium; and (c) beryllium.

Solution:

F ∴ F = EA0 e diameter = 0.5 cm = 0.1969 in.

A0 e 10.002 m − 10.000 m e= = 0.0002 m/m = 0.0002 in./in.

10.000 m FAl = (10 × 106 psi) (π /4) (0.1969 in.) 2 (0.0002 in./in.)

E = S /e =

= 60.9 lb = 271 N FMg = (6.5 × 106 psi) (π /4) (0.1969 in.) 2 (0.0002 in./in.) = 39.6 lb = 176 N FBe = (42 × 106 psi) (π /4) (0.1969 in.) 2 (0.0002 in./in.) = 256 lb = 1138 N 318 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

14-16 A new process is proposed to harden an Mg-Al alloy. Part of the patent filing suggests that a 35 wt% Al liquid be solidified at 250°C. Is this realistic or should this application be thrown out as physically impossible?

Solution: Examining Figure 14-4 we find ourselves in the α+γ region, which is definitely solid. This is physically realistic.

14-18 We say that copper can contain up to 40% Zn or 9% Al and still be single phase. How do we explain this statement in view of the phase diagrams for the Cu-Zn system? [See Figure 14–5(a).]

Solution: This is possible due to slow kinetics of transformation at low temperatures.

14-19 Compare the percent increase in the yield strength of commercially pure annealed aluminum, magnesium, and copper by strain hardening. Explain the differences observed.

Solution:

Al:

H18 − 1100-O  22,000 − 5,000  =  × 100 = 340% 100-O 5,000  

Mg: Cu:

C.W. − Annealed 17,000 − 13,000  =  × 100 = 31% Annealed 13,000  

70% C.W.  53,000 − 4,800  =  × 100 = 1004% Annealed  4,800 

Both Al and Cu (with an FCC structure) have high strain-hardening coefficients and can be cold worked a large amount (due to their good ductility). Mg has the HCP structure, a low strain-hardening coefficient, and a limited ability to be cold worked.

14-20 What temperature ranges correspond to the phases of Cu-Sn alloys? (a) η @ 60 wt% Sn; (b) ζ @ 34 wt% Sn; (c) γ @ 30 wt% Sn; and (d) α @ 0 wt% Sn. Solution: All answers are obtained from Figure 14-5 (b).

150 °C down, potentially to absolute zero, but we cannot know unless we have a phase diagram down that low.

580 °C to 620 °C 600 °C to 720 °C 160 °C to 1075 °C

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14-21 We would like to produce a quenched and tempered aluminum bronze containing 13% Al. Recommend a heat treatment, including appropriate temperatures. Calculate the amount of each phase after each step of the treatment.

Solution: Heat to above about 710°C to get all β; 100% β, β: 13% Al Quench; still all β containing 13% Al. Reheat; temper at 400°C to allow g2 to form.

% γ2 =

13 − 9.4 = 54.5% γ 2 :16% Al, α : 9.4% Al 16 − 9.4

We want to be sure to temper above 400°C so we obtain γ2 in a matrix of a rather than a structure containing γ + γ2.

14-22 A number of casting alloys have very high lead contents; however, the Pb content in wrought alloys is comparatively low. Why isn’t more lead added to the wrought alloys?

What precautions must be taken when a leaded wrought alloy is hot worked or heat treated?

Solution: The lead-rich phase may melt during hot working or may form stringers during cold working.

We must be sure that the temperature is low enough to avoid melting of the lead phase.

14-23 Would you expect the fracture toughness of quenched and tempered aluminum bronze to be high or low? Would there be a difference in the resistance of the alloy to crack nucleation compared with crack growth? Explain.

Solution: The fracture toughness should be relatively good. The acicular, or Widmanstätten, microstructure forces a crack to follow a very tortuous path, which consumes a large amount of energy.

This microstructure is less resistant to crack nucleation. The acicular structure may concentrate stresses that lead to easier formation of a crack.

14-24 Based on the phase diagrams, estimate the solubilities of Ni, Zn, Al, Sn, and Be in copper at room temperature. Are these solubilities expected in view of Hume-Rothery’s conditions for solid solubility? Explain.

Solution: Cu–Ni

Solubility 100% Ni

Structure FCC

Valence 1

Cu–Zn

30% Zn

HCP

2

Cu–Al

8% Al

FCC

3

Cu–Be

0.2% Be

HCP

2

Cu–Sn

0% Sn

DC

4

Atom size difference

1.278 − 1.243 × 100 = 1.278 1.278 − 1.332 × 100 = 1.278 1.278 − 1.432 × 100 = 1.278 1.278 − 1.143 × 100 = 1.278 1.278 − 1.405 ×100 = 1.278

2.7% –4.2% –12.1% 10.6% –9.9%

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Hume-Rothery’s conditions do help to explain the differences in solubility. Solubilities tend to decrease as atom size difference increases.

14-25 Hastelloy is used in chemical reactors, pipes, and other vessels that contain high-value, highly corrosive materials. Hastelloy is mostly nickel. How much more load can Hastelloy B-2 sustain compared to pure, cold-worked nickel?

Solution: Referring to Table 14-8: !" #$%&' () *+, 130 ksi = = 1.37 × !" -( .+/(01'. 23 95 ksi 14-26 Based on the photomicrograph in Figure 14 –7(a), would you expect the γ′ precipitate or the carbides to provide a greater strengthening effect in superalloys at low temperatures? Explain.

Solution: The γ′ phase is more numerous and also more uniformly and closely spaced; consequently, the γ′ should be more effective than the smaller number of coarse carbides at blocking slip at low temperatures.

14-27 The density of Ni3Al is 7.5 g/cm3. Suppose a Ni–5 wt% Al alloy is heat treated so that all of the aluminum reacts with nickel to produce Ni3Al. Determine the volume percent of the Ni3Al precipitate in the nickel matrix.

Solution: In 100 g of the alloy, the total atoms present are

(95 g Ni)N A (5 g Al)N A + 58.71 g/mol 26.981 g/mol = 1.6181 N A + 0.1853 N A = 1.803 N A

atoms =

If all of the Al reacts to form Ni3Al, then the number of atoms in the compound is 0.1853 NA of Al and (3)(0.1853 NA) = 0.5559 NA of Ni. The mass of the Ni3Al is then

mass =

(0.1853 N A of Al) (26.981 g/mol) (0.5559 N A of Ni) (58.71 g/mol) + NA NA

= 37.64 g of Ni3 Al The mass of the Ni matrix is thus 62.36 g. The vol% Ni3Al is thus:

vol% Ni3Al =

37.64 g(7.5 g/cm3 ) 37.64 g/(7.5 g/cm3 ) + 62.36 g/(8.902 g/cm3 ) ×100% = 42%

Even a small amount (5 wt% aluminum) produces a very large volume percent of precipitate in the microstructure.

14-28 When steel is joined using arc welding, only the liquid-fusion zone must be protected by a gas or flux. When titanium is welded, both the front and back sides of the welded metal must be protected. Why must these extra precautions be taken when joining titanium?

Solution: The titanium may be contaminated or embrittled anytime the temperature is above about 535°C. Therefore, the titanium must be

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protected until the metal cools below this critical temperature. Since both sides of the titanium plate will be heated by the welding process, special provisions must be made to shield all sides of the titanium until the metal cools sufficiently.

14-29 Both a Ti–15% V alloy and a Ti–35% V alloy are heated to the lowest temperature at which all β just forms. They are then quenched and reheated to 300°C. Describe the changes in microstructures during the heat treatment for each alloy, including the amount of each phase. What is the matrix and what is the precipitate in each case?

Which is an age-hardening process? Which is a quench and temper process? [See Figure 14–12(a).]

Solution: Ti–15% V: 100% β → 100% α′ → β precipitates in α matrix.

% α 300C =

46 − 15 × 100% = 76% β = 24% 46 − 5

This is a quench and temper process.

Ti–35% V: 100% β → 100% βss → α precipitates in β matrix.

% α 300C =

46 − 35 × 100% = 27% β = 73% 46 − 5

This is an age hardening process.

14-30 A Ti-Mn alloy is composed of 20 wt% manganese and is kept at 550 °C. What phase(s) are present? See Figure 14-8(d).

Solution: Figure 14-8 (d) shows that this is right on a phase border and to the right of the eutectoid point. The phases present are α, β and θ.

14-31 Determine the specific strength of the strongest Al, Mg, Cu, Ti, and Ni alloys. Use the densities of the pure metals, in lb/in.3, in your calculations. Try to explain their order.

Solution: Strength Density Strength-to-weight (psi) ratio Ti 176,000 4.51 g/cm3 = 0.163 lb/in.3 10.8 × 105 in.

Al 75,000 2.7 g/cm3 = 0.098 lb/in.3 7.7 × 105 in.

Mg 40,000 1.74 g/cm3 = 0.063 lb/in.3 6.3 × 105 in.

3 3 Cu 175,000 8.93 g/cm = 0.322 lb/in.

5.4 × 105 in.

3 3 Monel 110,000 8.90 g/cm = 0.322 lb/in.

3.4 × 105 in.

Titanium is strong with a relatively low density. Cu and Ni are strong but dense. Al and Mg have modest strength but light weight.

The yield strengths were used in these calculations.

14-33 The temperature of a coated tungsten part is increased. What happens when the protective coating on a tungsten part expands more than the tungsten? What happens when the protective coating on a tungsten part expands less than the tungsten?

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Solution: If the protective coating expands more than tungsten, compressive stresses will build up in the coating and the coating will flake.

If the protective coating expands less than tungsten, tensile stresses will build up in the coating and the coating will crack and become porous.

14–35 List the refractory metals by their suitability for cryogenic service.

Solution: Going by the transition temperatures in Table 14-10: Ta, Nb, Mo, W. 14–36 Nanoparticles (3 nm in diameter) of platinum with a total weight of 1 milligram are used in automobile catalytic converters to facilitate oxidation reactions. As Pt is an expensive metal, a method to coat iron nanoparticles with Pt is being explored to reduce the cost of the catalyst metal. If 3 nm diameter Fe particles are being coated with 1 nm of Pt to achieve the same effective surface area with fewer particles and less Pt, calculate the difference in cost and weight of the catalyst. Assume that the cost of Pt is $40/g and that of Fe is $0.005/g. The density of Fe is 7.87 g/cm3 and that of Pt is 21.45 g/cm3.

Solution: The volume VPt of one 3 nm diameter Pt particle is VPt = (4πr3/3) = (4π/3)(1.5 × 10–7)3 cm3 = 14.137 × 10–21 cm3, where r is the nanoparticle radius. One milligram of Pt with a density of 21.45 g/cm3 at a cost of $40/g (such that 10–3 g costs $0.04) corresponds to a total volume of (10–3 g)/ (21.45 g/cm3) = 4.662 × 10–5 cm3.

The total number of Pt particles is then

Number of Pt particles = 4.662 ×10−5 /(14.137 ×10−21 ) = 3.2977 ×1015 .

The total surface area of the Pt particles is

Total Surface Area = (4π r 2 ) (Number of Pt particles) = 4π (1.5 × 10−7 )2 (3.2977 × 1015 ) = 932.4 cm 2 .

In order to achieve the same total surface area with 1 nm Pt coated on 3 nm Fe, the number of Fe-Pt particles required is 932.4 = 4π(2 × 10–7)2 (Number of Fe-Pt particles)

Number of Fe-Pt particles = 932.4/[4π (2 ×10−7 )2 ] = 1.8550 ×1015.

The volume VFe-Pt of one 4 nm diameter Fe-Pt particle is the sum of the volumes of the Fe (VFe) and Pt portions (VPt on Fe). The volume of the Fe portion VFe is given by VFe = Number of Fe-Pt particles (Volume of Fe portion for each particle) VFe = 1.8550 × 10–15(4π/3) (1.5 × 10–7)3 = 2.6224 × 10–5 cm3.

Multiplying by the density of iron, the total mass of Fe mFe is mFe = 2.6224 × 10–5(7.87 g/cm3) = 2.064 × 10–4 g, and the cost of the iron is given by

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Cost of Fe = ($0.005/g) (2.064 × 10–4 g) = $0.001 × 10–3, which is insignificant. The volume of the Pt portion VPt on Fe is given by VPt on Fe = Number of Fe-Pt particles Volume of Pt portion for each particle)

VPt on Fe = 1.8550 ×1015 (4π /3) [(2 ×10−7 )3 − (1.5 ×10−7 )3 ] = 3.5937 ×10−5cm3 .

Multiplying by the density of platinum, the total mass of Pt on Fe mPt on Fe is mPt on Fe = 3.5937 × 10–5(21.45 g/cm3) = 7.71 × 10–4 g Cost of Pt = ($40/g) (7.71 × 10–4g) = $0.031.

Combining the Fe and Pt values, for the Pt coated Fe particles, Total weight = 2.064 × 10–4 g + 7.71 × 10–4 g = 0.98 mg and the Total cost = $0.031 (as the cost of Fe is insignificant.)

Compared to the Pt particles, the Pt coated Fe particles weigh 0.02 mg less and cost $0.009 less. Also it should be noted that the costs here are the materials cost only. If the manufacturing costs for producing Pt coated Fe particles are calculated, then it is likely that they will cost more than the Pt particles.

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Chapter 15: Ceramics

15-4

Zirconia (cubic) and diamond are tested to determine their expansion when their temperatures increase by a set interval. Assuming both are pure, how much more should the ZrO2 expand than the diamond sample? Does this give you any insight about distinguishing between these materials?

Solution: Referring to Table 15-2, we can simply take the ratio of the two thermal expansion coefficients: cm 10.5 cm ∙ ℃ = 10.3 1.02 cm cm ∙ ℃

This is one way to tell a cubic zirconia from a diamond, but there are many others.

15-5

Convert the properties of Al2O3 from Table 15-3 into values with SI units.

Solution: Density, the three strengths and Young’s modulus are easily converted manually or using unit conversion functions of many calculators or software: g kg = 3980

cm m 30,000 psi = 207 MPa 80,000 psi = 552 MPa 400,000 psi = 2.76 GPa 56 × 10 psi = 386 GPa 3.98

5000 psi√in. 101.325 kPa √1 √m ! = 5.49 kPa√m " 1 14.696 psi √39.37 √in.

15-6

Table 15-4 describes the sodium chloride structure as a face-centered cubic lattice with a basis of Cl (0, 0, 0) and Na (1/2, 0, 0). Using Cl (0, 0, 0) as one ion of the basis, give three equivalent descriptions for the position of the Na ion.

Solution:

(

)

 1  2

 

(

)

 

1  2 

FCC lattice + basis Cl 0, 0, 0 and Na  0, , 0  OR

FCC lattice + basis Cl 0, 0, 0 and Na  0, 0, ,  OR

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(

 1 1 1 , ,  2 2 2 

)

FCC lattice + basis Cl 0, 0, 0 and Na  15-7

Table 15-4 describes the sodium chloride structure as a face-centered cubic lattice with a basis of Cl (0, 0, 0) and Na (1/2, 0, 0). Using Na (0, 0, 0) as one ion of the basis, give three equivalent descriptions for the position of the Cl ion.

Solution:

(

)

1 2

 

(

)

 1  2

 

(

)

 

1  2 

(

)

 1 1 1 , ,  2 2 2 

FCC lattice + basis Na 0, 0, 0 and Cl  , 0, 0  OR

FCC lattice + basis Na 0, 0, 0 and Cl  0, , 0  OR

FCC lattice + basis Na 0, 0, 0 and Cl  0, 0, ,  OR

FCC lattice + basis Na 0, 0, 0 and Cl  15-8

Table 15-4 describes the cubic zincblende structure as a face-centered cubic lattice with a basis of S (0, 0, 0) and Zn (1/4, 1/4, 1/4). Using S as one ion of the basis, give three other equivalent descriptions for the position of the Zn ion.

Solution: Any of the coordinates of the tetrahedral sites in the FCC structure are acceptable answers for the second atom of the basis (except for the ¼ ¼ ¼ position that was already given). Thus the correct answer is S (0, 0, 0) and Zn at any one of the following positions: (¾, ¾, ¼); (¾, ¼, ¾); (¼, ¾, ¾); (¾, ¼, ¼); (¼, ¾, ¼) (¼, ¼, ¾) (¾, ¾, ¾) 15-9

If a face-centered cubic crystal structure has all of its tetrahedral sites filled with atoms that fit perfectly into each tetrahedral site, what is its atomic packing factor?

Solution:

APF =

number of atoms per unit cell ∗ atomic volume unit cell volume

There are four atoms per unit cell in the FCC structure with radius R. There are eight tetrahedral sites in the FCC structure with radius r.

The unit cell has volume a 3 (a is the lattice parameter or axial length of the unit cell).

For the FCC structure,

a=

4R 2 326

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Therefore, 3 4 4 4 4 4 * π R3 + 8* π r 3 4 * π R 3 + 8* π ( 0.225R) 3 3 3 3 APF = = 3 3 a  4R     2

16 π (1+ 2 * 0.2253 ) π APF = 3 = 1+ 2 * 0.2253 ) = 0.757 ( 3 2 16 2

(

)

15-10 If a face-centered cubic crystal structure has all of its octahedral sites filled with atoms that fit perfectly into each octahedral site, what is the atomic packing factor?

Solution:

APF =

number of atoms per unit cell ∗ atomic volume unit cell volume

There are four atoms per unit cell in the FCC structure with radius R. There are four octahedral sites in the FCC structure with radius r.

The unit cell has volume a 3 (a is the lattice parameter or axial length of the unit cell).

For the FCC structure,

a=

4R 2

Therefore, 3 4 4 4 4 4 * π R 3 + 6 * π r 3 4 * π R 3 + 4 * π ( 0.414R) 3 3 3 3 APF = = 3 3 a  4R     2

16 π (1+ 0.4143 ) π APF = 3 = 1+ 0.4143 ) = 0.793 ( 3 2 16 2

(

)

15-11 If a hexagonal close-packed crystal structure has all of its tetrahedral sites filled with atoms that fit perfectly into each tetrahedral site, what is the atomic packing factor?

Note that each HCP unit cell has four tetrahedral sites with the same radius ratio as the tetrahedral sites in the FCC structure.

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Solution: There are two atoms per unit cell in the HCP structure with radius R, and there are four tetrahedral sites in the HCP structure with radius r. The radius ratio for tetrahedral sites in both the FCC and HCP structures is r/R = 0.225. The atomic packing factor (APF) is defined as

APF =

number of atoms per unit cell ∗ atomic volume unit cell volume

The geometry of the unit cell is shown here:

The unit cell has volume

 3  8  a 2c 3 = a 2   a = 2 a 3 Also a = 2R, such that the 2  2  3 

unit cell has volume 8 2R 3 (1 point).

Therefore, 3 4 4 4 4 2 * π R 3 + 4 * π r 3 2 * π R3 + 4 * π ( 0.225R ) 3 3 3 3 APF = = = 0.757 3 8 2R 8 2R 3

15-16 The density of Al2O3 is 3.96 g/cm3. A ceramic part is produced by sintering alumina powder. It weighs 80 g when dry, 92 g after it has soaked in water, and 58 g when suspended in water. Calculate the apparent porosity, the true porosity, and the closed porosity.

Solution: From the problem statement, ρ = 3.96, Wd = 80 g, Ww = 92, and Ws = 58. From the equations,

Apparent porosity =

Ww − Wd 92 − 80 × 100 = × 100 = 35.29% Ww − Ws 92 − 58

The bulk density is B = Wd/(Ww – Ws) = 80/(92 – 58) = 2.3529 g/cm3.

Therefore,

True porosity =

ρ−B 3.96 − 2.3529 ×100 = ×100 = 40.58% ρ 3.96

Closed porosity = 40.58 − 35.29 = 5.29%

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15-17 Silicon carbide (SiC) has a density of 3.1 g/cm3. A sintered SiC part is produced, occupying a volume of 500 cm3 and weighing 1200 g. After soaking in water, the part weighs 1250 g. Calculate the bulk density, the true porosity, and the volume fraction of the total porosity that consists of closed pores.

Solution: The appropriate constants required for the equations are

ρ = 3.1 g/cm3

B = 1200 g/500 cm3 = 2.4 g/cm3 Ww = 1250 g Wd = 1200 g Therefore,

B = 2.4 = Wd /(Ww − Ws ) = 1200/(1250 − Ws ) or Ws = 750 g W − Wd 1250 − 1200 × 100 = ×100 = 10% Apparent porosity = w Ww − Ws 1250 − 750 ( ρ − B)

(3.1 − 2.4) ×100 = 22.58% ρ 3.1 Closed porosity = 22.58 − 10 = 12.58% True porosity =

=

f closed = 12.58/22.58 = 0.557 15-18 A sintered zirconium oxide (ZrO2) ceramic has a true porosity of 28%, and the closed pore fraction is 0.5. If the weight after soaking in water is 760 g, what is the dry weight of the ceramic? The specific gravity of ZrO2 is 5.68 g/cm3.

Solution: Using Equation 15-2, True porosity = (ρ – B)/ρ × 100 = 28,

where ρ is the specific gravity and B is the bulk density of the ceramic. Rearranging, (ρ – B)/ρ = 0.28.

Substituting with the values given to find B, (5.68 – B)/5.68 = 0.28 3 B = (1 – 0.28) 5.68 = 4.09 g/cm .

The bulk density B is also given by Equation 15-3: B = Wd /(Ww – Ws),

where Wd is the dry weight, Ww is the weight after soaking in water, and Ws is the weight when suspended in water. Rearranging, Wd /B = (Ww – Ws).

The fraction of closed pores is given by Fractional Closed Pores = (True porosity – Apparent porosity ) / True porosity, and in this case is equal to 0.5. Thus, 0.5 = (0.28 – Apparent porosity)/0.28

Apparent porosity = 14%.

Equation 15-1 for apparent porosity is Apparent porosity = (Ww – Wd)/ (Ww – Ws) × 100.

From above, (Ww – Ws) = Wd/B, and substituting into the expression for apparent porosity to find Wd, 0.14 = B (Ww – Wd)/Wd Wd = Ww/[1 + (0.14/B)].

From the problem statement, Ww = 760 g and B = 4.09 g/cm3 such that Wd = Ww / [1 + (0.14 / B)] = 760 / [1 + (0.14 / 4.09)] = 735 g.

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15-19 The quality control team at a ceramics processing facility takes a rather large sample of a ceramic coming off the line and weighs the sample when dry, when immersed in water, and after removing it from the water. The weights are 1.589 lb, 1.575 lb, and 1.633 lb, respectively. Find the apparent porosity and the volume of the interconnected pores.

Solution: Using Equation 15-1 we calculate the apparent porosity: 1.633 lb − 1.589 lb #$ − #& × 100 = × 100 = 75.9 % #$ − #' 1.633 lb − 1.575 lb The volume of the interconnected pores is given by the numerator of the apparent porosity equation converted from mass to volume using the density of water: +#$ − #& ,- = +1.633 lb − 1.589 lb, .

1 ft 12 in.

1 . 1 = 1.22 in.

1 ft 62.3 lb

15-20 The following physical properties are known for a sample: Ww = 50.0 g, ρ = 2.170 g/cm3, and true porosity = 39.0%. Find the bulk density.

Solution: Using Equation 15-2 we can obtain B from the density and the true porosity: -−4 × 100 4 23 = 51 − 6 × 100 4 39.0 = 51 − 6 × 100 2.170 4 0.39 = 1 − 2.170 4 0.61 = 2.170 4 = 1.324 23 =

15-35 Part of a silica process is outlined in the following steps: 1.

2.

3.

4.

5.

Heat 15 mol% soda-85 mol% silica to 1800°C. Cool to 1200°C. At constant temperature, increase soda content to 25 mol%.

Cool to 1000 °C. At constant temperature, increase soda content to 35 mol%.

What, if any, solids are formed during this process? (See Figure 15-17.)

Solution: Following the process in Figure 15-8, we see that at step 2 in the above list, tridymite is formed in solution with the liquid glass. All other steps and their pathways are above the liquidus line.

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15-36 At 1200°C, list the viscosities of the glasses in Figure 15-18 in centipoise (cP).

Solution: Reading the graph and converting poise to centipoise (1 P = 100 cP): fused silica, 1013 cP; 96% silica, 1011 cP; borosilicate, 105 cP; soda-lime, 104 cP. The line for lead glass does not extend to 1200 °C, and extrapolation, especially of double log plots, is dangerous.

15-37 How many grams of BaO can be added to 1 kg of SiO2 before the O:Si ratio exceeds 2.5 and glass-forming tendencies are poor? Compare this to the case when Li2O is added to SiO2.

Solution: We can first calculate the mole fraction of BaO required to produce an O:Si ratio of 2.5:

(1O/BaO)f BaO + (2 O/SiO 2 )(1 − f BaO ) (1 Si/SiO 2 )(1 − f BaO ) f BaO = 1/3 and f silica = 2/3

OSi = 2.5 =

The molecular weight of BaO is 137.3 + 16 = 153.3 g/mol, and that of silica is 60.08 g/mol. The weight percent BaO is therefore,

wt% BaO =

(1/3 mol)(153.3 g/mol) × 100 = 56.06% (1/3)(153.3) + (2/3)(60.08)

For 1 kg of SiO2, the amount of BaO is

0.5606 =

xgBaO xgBaO + 100gSiO 2

or

x = 1276 g BaO

The mole fraction of Li2O required is

O/Si = 2.5 =

(1 O/Li 2 O)f Li2 O + (2 O/SiO 2 )(1 − f Li2O ) (1 Si/SiO 2 )(1 − f Li 2O )

f Li 2O = 1/3 and

f silica = 2/3

The molecular weight of Li2O is 2(6.94) + 16 = 29.88 g/mol, and that of silica is 60.08 g/mol. The weight percent Li2O is therefore,

wt% Li 2 O =

(1/3 mol)(29.88 g/mol) × 100 = 19.91% (1/3)(29.88) + (2/3)(60.08)

For 1 kg of SiO2, the amount of Li2O is

0.1991 =

x g Li 2 O x g Li 2 O + 1000 g SiO 2

or

x = 249 g Li 2 O

Much larger amounts of BaO can be added compared to Li2O and still retain the ability to form a glass.

15-38 Calculate the O:Si ratio when 30 wt% Y2O3 is added to SiO2. Will this material have good glass-forming tendencies?

Solution: MWyttria = 2(88.91) + 3(16) = 225.82 g/mol MWsilica = 60.08 g/mol

The mole fraction of yttrium is (assuming a base of 100 g of ceramic):

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f yttria =

30 g/(225.82 g/mol) = 0.102 30/225.82 + 70/60.08

The O/Si ratio is then

O/Si =

(3 O/Y2 O3 )(0.102) + (2 O/SiO 2 ) (0.898) = 2.34 (1 Si/SiO 2 )(0.898)

The material will produce a glass.

15-39 A silicate glass has 10 mol% Na2O and 5 mol% CaO. Calculate the O:Si ratio and the composition in wt% for this glass.

Solution: The number of O ions is given by Number of O ions per mole of glass = 1 O ion per Na2O × 0.1 mol fraction Na2O + 1 O ion per CaO × 0.05 mol fraction CaO + 2 O ion per SiO2 × 0.85 mol fraction SiO2 Number of O ions per mole of glass = 1 × 0.1 + 1 × 0.05 + 2 × 0.85 = 1.85.

The number of Si ions is given by Number of Si ions per mole of glass = 1 Si ion per SiO2 × 0.85 mol fraction SiO2 = 1 × 0.85 = 0.85.

Therefore, O/Si = 1.85/0.85 = 2.18.

The weight % for each component is as follows: wt% Na2O = (0.1 × 61.98 g/mol Na2O)/ (0.1 × 61.98 g/mol Na2O + 0.05 × 56.08 g/mol CaO + 0.85 × 60.08 g/mol SiO2) = 0.1032 or 10.32%.

wt% CaO = (0.05 × 56.08 g/mol CaO)/ (0.1 × 61.98 g/mol Na2O + 0.05 × 56.08 g/mol CaO + 0.85 × 60.08 g/mol SiO2) = 0.0467 or 4.67%.

The rest of the weight is SiO2 = 100 – 10.32 – 4.67 = 85.01%.

15-40 A borosilicate glass (82% SiO2, 2% Al2O3, 4% Na2O, 12% B2O3) has a density of 2.23 g/cm3, while a fused silica glass (assume 100% SiO2) has a density of 2.2 g/cm3. Explain why the density of the borosilicate glass is different from the weighted average of the densities of its components. The densities of Al2O3, Na2O, and B2O3 are 3.98 g/cm3, 2.27 g/cm3 and 2.5 g/cm3, respectively.

Solution: A weighted average of the densities gives 3

0.82 × 2.2 + 0.02 × 3.98 + 0.04 × 2.27 + 0.12 × 2.5 = 2.27 g /cm , 3

which is higher than the actual density of 2.23 g/cm . The weighted average calculation makes the implicit assumption that the oxides being added to SiO2 maintain their crystal structure and hence their density. In a glass, these additives are incorporated into the glass network and lose their original crystal structure such that the density is lower than the weighted average of the individual components.

15-41 Rank the following glasses by softening temperature (highest to lowest): (a) Pyrex™ (81% SiO2, 2% Al2O3, 4% Na2O, 12% B2O3); (b) Lime glass (72% SiO2, 1% Al2O3, 10% CaO, 14% Na2O); and (c) Vycor™ (96% SiO2, 4% B2O3).

Solution: (a) Pyrex™ is a borosilicate glass. From Figure 15-9, the softening temperature for this glass is expected to be around 900°C. (b) Lime glass is a soda lime glass. From Figure 15-9, the softening temperature is expected to be below 700°C.

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(c) Vycor™ has 96% silica, and from Figure 15-9, is expected to have a softening temperature in excess of 1500°C. Ranking the three glasses from highest to lowest softening temperature: Vycor™ > Pyrex™ > Lime Glass.

15-47 What is the maximum difference between the masses of MgO required to make 10 kg of chromite refractory and 10 kg of chromite-magnesite?

Solution: The minimum composition of chromite refractory is 10 wt%. The mass of MgO required to make 10 kg of this composition is 1 kg. The maximum MgO composition for chromite magnesite refractory is 39 wt%. The mass of MgO required to make 10 kg of this composition is 3.9 kg.

Thus the maximum difference between the masses required is 3.9 kg – 1 kg = 2.9 kg MgO. 15-48 What is the lowest temperature at which a mixture of SiO2-Al3O3 could still be completely liquid? (See Figure 15-24.)

Solution: The lowest point on the liquidus as read from Figure 15-17 is 1590˚C.

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Chapter 16: Polymers

16-6

Kevlar (C14H10N2O2) is used in various applications from tires to body armor due to its high strength-to-weight ratio. The polymer is produced from the monomers paraphenylene diamine (C6H8N2) and terephthaloyl chloride (C8H4Cl2O2) by the following reaction:

(a) What type of polymerization does the above reaction represent? (b) Assuming 100% efficiency, calculate the weight of terephthaloyl chloride required to completely combine with 1 kg of paraphenylene diamine. (c) How much Kevlar is produced?

Solution: (a) The reaction is an example of condensation polymerization. (b) The molecular weight of each reactant can be calculated.

For paraphenylene diamine (C6H8N2): (6 C atoms)(12 g/mol) + (8 H atoms)(1 g/mol) + (2 N atoms)(14 g/mol) = 108 g/mol.

For terephthaloyl chloride (C8H4Cl2O2): (8 C atoms)(12 g/mol) + (4 H atoms)(1 g/mol) + (2 Cl atoms)(35.5 g/mol) + (2 O atoms)(16 g/mol) = 203 g/mol.

In 1 kg of paraphenylene diamine, there are

1000 g(6.022 ×1023 monomers/mol) = 5.576 ×1024 monomers.

108 g/mol Assuming 100% efficiency, an equal number of terephthaloyl chloride monomers is needed with a weight of

(5.576 × 1024 monomers)(203 g/mol) = 1880 g.

6.022 × 1023 monomers/mol (c) The reaction produces 5.576 × 1024 molecules of Kevlar (C14H10N2O2). The molecular weight of Kevlar is (14 C atoms)(12 g/mol) + (10 H atoms)(1 g/mol) + (2 N atoms)(14 g/mol) + (2 O atoms)(16 g/mol) = 238 g/mol.

Thus, the mass of Kevlar produced is

(5.576 × 1024 monomers)(238 g/mol) = 2204 g = 2.2 kg.

6.022 × 1023 monomers/mol 16-7

Examine Figure 16-6. If 100 kg each of the two reactants were combined and the reaction proceeded to 100% completion, how much methanol would be generated?

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Solution: This question is a routine reaction analysis. The masses of each reactant must be converted to molecular amounts: =

100 kg DMTE 1 kmol DMTE = 0.5155 kmol DMTE

194 kg DMTE 1

=

100 kg EG 1 kmol EG = 1.61 kmol EG

1 62 kg EG

Examining the reaction equation, every molecule of DMTE has two reaction sites, one at each end. For 100% reaction, each of these will condense one molecule of methyl alcohol (methanol). DMTE is the limiting reactant. Thus, =

0.5155 kmol DMTE 2 kmol MeOH = 1.031 kmol MeOH

1 1 kmol DMTE

# = 16-8

1.031 kmol MeOH 32 kg MeOH = 33 kg MeOH

1 kmol MeOH 1

What would be the molecular weight of the polymer shown below where n = 106?

H | H –[ U –]nH | H Note that uranium can have a +4 oxidation state.

Solution: In a tabular form, the problem may be solved: Atom

Molecular mass Occurrences Subtotal

Hydrogen

1.01

Uranium

238.03

2000002

2020002

1000000 238030000

Summing the subtotals, one finds the molecular mass to be 2.40 × 108.

16-10 Revisit Example 16-5. What are the degrees of polymerization for the weight average and number average molecular weights?

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Solution: The hard part here, calculating the molecular masses, has been completed already.

Examining the repeat unit of polyethylene, its molecular mass is 28 g/mol. Therefore: g 9160 'mol g = 327 'mol $%& = g 28 'mol g 11,331 'mol g $%* = = 405 'mol g 28 'mol 16-11 Calculate the number of chains in a 5-m-long PVC pipe with an inner diameter 5 cm and a thickness of 0.5 cm if the degree of polymerization is 1000. Assume that the chains are equal in length. The density of PVC is 1.4 g/cm3. The repeat unit of PVC is shown in Table 16–3.

Solution: The molecular weight of PVC monomer is (2 C atoms)(12 g/mol) + (3 H atoms)(1 g/mol) + (1 Cl atom)(35.5 g/mol) = 62.5 g/mol.

For a degree of polymerization of 1000, the molecular weight of each chain MWchain is MWchain = 1000 (62.5) = 62,500 g/mol.

The volume V of the PVC in the pipe is 2 2  π douter  π dinner V = − l , 4   4

where l is the pipe length. Substituting,

 π (6 cm) 2 π (5 cm) 2  V = − (500 cm) = 4320 cm3 .  4 4   Using the density of PVC (1.4 g/cm3), the mass m of the PVC pipe is m = (1.4 g/cm3)(4320 cm3) = 6048 g.

The number of chains in the pipe is given by

Number of chains =

=

m(6.022 × 1023 chains/mol) MWchain

6048 g(6.022 ×1023chains/mol) = 5.8 ×1022 .

62,500 g/mol

16-12 The molecular weight of polymethyl methacrylate (see Table 16–3) is 250,000 g/mol. If all of the polymer chains are the same length, calculate (a) the degree of polymerization, and (b) the number of chains in 1 g of the polymer.

Solution:

The molecular weight of methyl methacrylate is MW = 5 C + 2 O + 8 H = 100 g/mol Degree of polymerization = 250,000/100 = 2,500

In 1 g of the polymer:

(1 g) (6.02 ×1023chains/mol) = 2.408 ×1018 chains 250,000 g/mol 337 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

16-13 The degree of polymerization of polytetrafluoroethylene (see Table 16–3) is 7500. If all of the polymer chains are the same length, calculate (a) the molecular weight of the chains, and (b) the total number of chains in 1000 g of the polymer.

Solution:

The molecular weight of tetrafluoroethylene is MW = 2 C + 4 F = 100 g/mol MWchains = (7500) (100) = 750,000 g/mol

In 1000 g of the polymer:

(1000 g) (6.02 ×1023 chains/mol) = 8.03 ×1020 chains 750,000 g/mol 16-14 A polyethylene rope weighs 0.25 lb per foot. If each chain contains 7000 repeat units, calculate (a) the number of polyethylene chains in a 10 ft length of rope, and (b) the total length of chains in the rope, assuming that carbon atoms in each chain are approximately 0.15 nm apart and the length of one repeat unit is 0.24495 nm.

Solution: The molecular weight of ethylene is 28 g/mol, so the molecular weight of the polyethylene is 7000 × 28 = 196,000 g/mol. The weight of the 10 ft length of rope is (0.25 lb/ft)(10 ft)(454 g/lb) = 1135 g.

The number of chains is

(1135 g)(6.02 ×1023 chains/mol) = 34.86 ×1020 chains 196,000 g/mol The length of one repeat unit is 0.24495 nm. Therefore, the length of each chain, which contain 7000 repeat units, is

one chain = (7000) (0.24495 nm) = 1715 nm = 1.715 × 10−4 cm all chains = (1.715 × 10 −4 cm/chain) (34.86 × 10 20 chains) = 5.978 × 1017 cm = 3.7 × 1012 miles 16-15 Analysis of a sample of polyacrylonitrile (see Table 16–3) shows that there are six lengths of chains, with the following number of chains of each length. Determine (a) the weight average molecular weight and degree of polymerization and (b) the number average molecular weight and degree of polymerization.

Solution: Number of chains 10,000 18,000 17,000 15,000 9,000 4,000 73,000

Mean Molecular weight of chains (g/mol) 3,000 6,000 9,000 12,000 15,000 18,000

xi 0.137 0.247 0.233 0.205 0.123 0.055 sum =

xiMi 411 1482 2097 2460 1845 990 9,285

weight 30 × 106 108 × 106 153 × 106 180 × 106 135 × 106 72 × 106 678 × 106

fi 0.044 0.159 0.226 0.265 0.199 0.106

fiMi 132 954 2034 3180 2985 1908 11,193

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The molecular weight of the acrylonitrile monomer is MWacrylonitrile = 3 C + 1 N + 3 H = 53 g/mol The weight average molecular weight and degree of polymerization are

MWw = 11,193 g/mol DPw = 11,193/53 = 211 The number average molecular weight and degree of polymerization are

MWn = 9, 285 g/mol DPn = 9285/(53 g/mol) = 175 16-16 Calculate the molecular weights of the repeat units in Table 16-3. Show your work for full credit.

Solution: NAME

PE

PVC

PP

PS

PAN

PMMA

PCTFE

PTFE

MW

28

62

42

104

53

100

116

100

It should be noted that PMMA and PTFE – two wildly different monomers – have the same molecular mass. One should not use molecular mass of the monomer as an indicator of anything about the resulting polymer.

16-17 Plot the tensile strength of the polymers in Table 16-2 as a function of the number of aromatic (benzene) rings in the monomer.

Solution: This yields the following graph.

Tensile strength v. aromatic rings in monomer Polymer tensile strength (psi)

30000 y = 3885.4x + 4918.4 R² = 0.726

25000 20000 15000 10000 5000 0 0

1

2

3

4

5

6

Number of aromatic rings in monomer

Tensile strength appears roughly correlated with the number of aromatic rings.

Aromatic rings are stiffer than most straight chains, but this graph has too small a sample size to draw conclusions from.

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16-19 Using Table 16–5, plot the relationship between the glass-transition temperatures and the melting temperatures of the addition thermoplastics. What is the approximate relationship between these two critical temperatures? Do the condensation thermoplastics and the elastomers also follow the same relationship?

Solution: Converting temperature to Kelvin: Tm LD polyethylene 388 HD polyethylene 410 PVC 448–485 Polypropylene 441–449 Polystyrene 513 PAN 593 Acetal 454 6,6 Nylon 538 Polycarbonate 503 Polyester 528 Polybutadiene 393 Polychloroprene 353 Polyisoprene 303

Tg 153 153 360 257 358–398 380 188 323 416 348 183 223 200

Most of the polymers fall between the lines constructed with the relationships Tg = 0.5Tm and Tg = 0.75Tm. The condensation thermoplastics and elastomers also follow this relationship.

16-20 List the addition polymers in Table 16–5 that might be good candidates for making a bracket that holds a side-view mirror onto the outside of an automobile, assuming that temperatures frequently fall below zero degrees Celsius. Explain your choices.

Solution: Because of the mounting of the sideview mirror, it is often subject to being bumped; we would like the mounting material to have reasonable ductility and impact resistance so that the mirror does not break off the automobile. Therefore we might want to select a material that has a glass-transition temperature below 0°C. Both polyethylene and polypropylene have low glass-transition temperatures and might be acceptable choices. In addition, acetal (polyoxymethylene) has a low glass-

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transition temperature and (from Table 16–5) is twice as strong as polyethylene and polypropylene. Finally all of the elastomers listed in Table 16–8 might be appropriate choices.

16-21 Based on Table 16–5, which of the elastomers might be suited for use as a gasket in a pump for liquid CO2 at –78°C? Explain.

Solution: We wish to select a material that will not be brittle at very low temperatures, that is, the elastomer should have a glass transition temperature below –78°C. Of the materials listed in Table 16–8, only polybutadiene and silicone have glass transition temperatures below –78°C. 16-22 How do the glass-transition temperatures of polyethylene, polypropylene, and polymethyl methacrylate compare? Explain their differences, based on the structure of the monomer of each.

Solution:

From Table 16–5:

polyethylene . . . . . . . . . . .

Tg = –120°C polypropylene . . . . . . . . . .

Tg = –16°C polymethyl methacrylate ..

Tg = +90 to 105°C The side groups in polyethylene are small hydrogen atoms; in polypropylene, more complex side groups are present; in PMMA, the side groups are even more extensive (see Table 16–5). As the complexity of the monomers increases, the glass transition temperature increases.

16-23 Which of the addition polymers in Table 16–5 are used in their leathery condition at room temperature? How is this condition expected to affect their mechanical properties compared with those of glassy polymers?

Solution: Both polyethylene and polypropylene have glass-transition temperatures below room temperature and therefore are presumably in the leathery condition. As a consequence, they are expected to have relatively low strengths compared to most other thermoplastic polymers.

16-25 Describe the relative tendencies of the following polymers to crystallize. Explain your answer. (a) Branched polyethylene versus linear polyethylene; (b) polyethylene versus polyethylene–polypropylene copolymer; (c) isotactic polypropylene versus atactic polypropylene; and (d) polymethyl methacrylate versus acetal (polyoxymethylene).

Solution: (a) Linear polyethylene is more likely to crystallize than branched polyethylene. The branching prevents close packing of the polymer chains into the crystalline structure. (b) Polyethylene is more likely to crystallize than the polyethylene– propylene copolymer. The propylene monomers have larger side groups than polyethylene and, of course, different repeat units are present in the polymer chains. These factors make close packing of the chains more difficult, reducing the ease with which

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crystallization occurs. (c) Isotactic polypropylene is more likely to crystallize than atactic polypropylene. In isotactic polypropylene, the side groups are aligned, making the polymer chain less random, and permitting the chains to pack more closely in a crystalline manner. (d) Acetal, or polyoxymethylene, is more symmetrical and has smaller side groups than polymethyl methacrylate; consequently, acetal polymers are more likely to crystallize.

16-26 The crystalline density of polypropylene is 0.946 g/cm3, and its amorphous density is 0.855 g/cm3. What is the weight percent of the structure that is crystalline in a polypropylene that has a density of 0.900 g/cm3?

Solution: From Equation 16-4, the weight percent crystallinity is given by

% Crystalline =

ρc ( ρ − ρa ) × 100 , ρ ( ρc − ρa )

where ρ is the measured density of the polymer, ρa is the density of amorphous polymer, and ρc is the density of completely crystalline polymer. Substituting as given,

% Crystalline =

0.946(0.9 − 0.855) × 100 = 52.0% 0.9(0.946 − 0.855)

16-27 If the strain rate in a polymer can be represented by dε/dt = σ/η + (1/E) dσ/dt where ε = strain, σ = stress, η = viscosity, E = modulus of elasticity, and t = time, derive Equation 16-5, assuming constant strain (i.e., dε = 0). What is relaxation time (λ) in Equation 16-5 a function of?

Solution: Under constant strain conditions (d∈ = 0), the above equation can be rearranged as σ/η = –(1/E) dσ/dt or –(E/η) dt = (1/σ) dσ.

Integrating this equation from t = 0 to t and σ = σo to σ, –(E/η) t = ln (σ/ σ0).

Rearranging, σ = σ0 exp [–t/(η/E)].

Let λ = η/E, then σ = σ0 exp (–t/λ).

By inspection, the relaxation time λ in Equation 16-5 is a function of the modulus of elasticity E and viscosity η of the material.

16-28 HDPE is loaded to 400 psi and left. What temperature would cause a rupture at about 100 hours of this constant load? See Figure 16-18.

Solution: Examining Figure 16-18, a temperature slightly above 90 °C, possibly 91 or 92 °C, would cause failure as requested.

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16-29 A polymer component that needs to maintain a stress level above 10 MPa for proper functioning of an assembly is scheduled to be replaced every two years as preventative maintenance. If the initial stress was 18 MPa and dropped to 15 MPa after one year of operation (assuming constant strain), will the part replacement under the specified preventative schedule successfully prevent a failure?

Solution: From Equation 16-5, the stress after time t is given by σ = σ0 exp (–t/λ), where σ0 is the original stress and λ is the relaxation time. In this case, σ0 = 18 MPa and σ = 15 MPa at t = 1 year. Substituting and solving for λ, 15 = 18 exp (–1/λ) λ –1 / [ln (15/18)] = 5.48 years.

With the relaxation constant known, the time for the stress level to decrease to 10 MPa can be determined. Substituting and solving for t, 10 = 18 exp (–t/5.48) t =–(5.48) [ln (10/18)] = 3.22 years.

This time is longer than the two-year preventative maintenance schedule; hence, a part failure is unlikely.

16-31 A stress of 2500 psi is applied to a polymer serving as a fastener in a complex assembly.

At a constant strain, the stress drops to 2400 psi after 100 h. If the stress on the part must remain above 2100 psi in order for the part to function properly, determine the life of the assembly.

Solution: First we can determine the relaxation constant λ: σ = σo exp (–t/λ)

2400 = 2500 exp (–100/λ ) or ln(2400/2500) = –100/λ −0.0408 = −100/λ

or λ = 2451 h

Then we can determine the time required before the stress relaxes to 2100 psi:

2100 = 2500 exp (– t /2451) or ln(2100/2500) = – t /2451 −0.1744 = −t /2451

or t = 427 h

16-32 A stress of 1000 psi is applied to a polymer that operates at a constant strain; after six months, the stress drops to 850 psi. For a particular application, a part made of the same polymer must maintain a stress of 900 psi after 12 months. What should be the original stress applied to the polymer for this application?

Solution: First we can determine the relaxation constant λ, using a time of 6 months = 4380 h; σ = σo exp (–t/λ)

850 = 1000 exp(–4380/λ ) or ln(850/1000) = −4380/λ −0.1625 = −4380/λ

or λ = 26,954 h

Then we can determine the initial required stress that will ensure a

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stress of 900 psi after 12 months = 8760 h:

900 = σ o exp(–8760/26,954) = σ o (0.722)

σ o = 1246 psi 16-33 Data for the rupture time of polyethylene are shown in Figure 16-18. At an applied stress of 600 psi, the figure indicates that the polymer ruptures in 0.2 h at 90°C but survives for 10,000 h at 65°C. Assuming that the rupture time is related to the viscosity, calculate the activation energy for the viscosity of polyethylene and estimate the rupture time at 23°C.

Solution:

We expect the rupture time to follow the expression: tr = a exp[Qη/(RT)] For T = 90°C = 363 K, tr = 0.2 h, while for T = 65°C = 338 K, tr = 10,000 h.

By solving simultaneous equations, we can find the constant “a” and the activation energy Q:

0.2 = a exp{Qη /[(1.987)(363)]} = a exp(0.0013864 Qη ) 10,000 = a exp{Qη /[(1.987)(338)]} = a exp[0.0014890 Qη ) 0.00002 = exp [(0.0013864–0.0014890)Qη] = exp(–0.0001026 Qη)

ln(0.00002) = –10.8198 = –0.0001026 Qη Qη = 1.055 × 105 cal/mol 0.2 = a exp{105, 456/[(1.987)(363)]} = a exp(146.21) a = 0.2/(3.149 × 1063 ) = 6.35 × 10−65 The rupture time at 23°C = 296 K is therefore

tr = 6.35 ×10−65 exp{105, 456/[(1.987)(296)]} tr = 4.70 ×1013 h The polyethylene will essentially not rupture at 23°C.

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16-34 The polymer ABS can be produced with varying amounts of styrene, butadiene, and acrylonitrile monomers, which are present in the form of two copolymers: BS rubber and SAN. (a) How would you adjust the composition of ABS if you wanted to obtain good impact properties? (b) How would you adjust the composition if you wanted to obtain good ductility at room temperature? (c) How would you adjust the composition if you wanted to obtain good strength at room temperature?

Solution: (a) Improved impact properties are obtained by increasing the amount of butadiene monomer; the elastomer provides large amounts of elastic strain, which helps to absorb an impact blow. (b) The styrene helps to provide good ductility; the butadiene provides good “elastic” strain but not “plastic” strain. Acrylonitrile has poor ductility when polymerized. (c) Higher acrylonitrile will help produce higher strengths.

16-35 Figure 16–23 shows the stress-strain curve for an elastomer. From the curve, calculate and plot the modulus of elasticity versus strain and explain the results.

Solution: We obtain the modulus of elasticity by finding the slope of the tangent drawn to the stress-strain curve at different values of strain. Examples of such calculations are shown below:

ε = 0 ∆σ /∆ε = (4000 − 0)/(2 − 0)

= 2000 psi

=1

= (5500 − 500)/(6 − 0) = 833 psi

=2 =3

= (5000 − 200)/(6 − 0) = 800 psi = (6500 − 0)/(6 − 0.6) = 1204 psi

=4

= (8700 − 0)/(6 − 1.7) = 2023 psi

=5

= (9300 − 0)/(6 − 2.2) = 2450 psi

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The modulus of elasticity is plotted versus strain in the sketch. Initially the modulus decreases as some of the chains become untwisted.

Eventually the modulus increases again as the chains become straight and higher stresses are required to stretch the bonds within the chains.

16-36 Explain the term thermosetting polymer. A thermosetting polymer cannot be produced using only adipic acid and ethylene glycol. Explain why.

Solution: Polymers that are heavily cross-linked to produce a strong three dimensional network structure are called thermosetting polymers. Unsaturated bonds are introduced into the linear polymer chain through the maleic acid. If the maleic acid were not present, cross-linking could not occur.

16-37 Explain why the degree of polymerization is not usually used to characterize thermosetting polymers.

Solution: Individual chains are no longer present after the polymer is completely cross-linked and polymerized; instead the entire polymer should be considered continuous.

16-38 Defend or contradict the choice to use the following materials as hot-melt adhesives for an application in which the assembled part is subjected to impact loading: (a) polyethylene; (b) polystyrene; (c) styrene-butadiene thermoplastic elastomer; (d) polyacrylonitrile; and (e) polybutadiene.

Solution: (a) Polyethylene is expected to have relatively good impact resistance due to the ease with which chains can move; the polyethylene is well above its glasstransition temperature. (b) Polystyrene is expected to have relatively poor impact resistance due to the resistance to chain sliding by the large benzene ring side groups. (c) Styrene-butadiene thermoplastic elastomers are expected to have good impact resistance; although the styrene portion may be rather brittle, the high energy absorbing capability of the butadiene component provides good impact properties. (d) Polyacrylonitrile will have relatively poor impact properties due to the presence of the side groups. (e) Polybutadiene, an elastomer, will provide good impact properties.

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16-44 Polymer Molecular Weight Distribution. The data below were obtained for polyethylene.

Determine the average molecular weight and degree of polymerization. Write a computer program or use a spreadsheet program to solve this problem.

Solution:

Molecular Weight Range (g/mol) 0–3,000 3,000–6,000 6,000–9,000 9,000–12,000 12,000–15,000 15,000–18,000 18,000–21,000 21,000–24,000

xi 0.03 0.10 0.22 0.36 0.19 0.07 0.02 0.01

fi 0.01 0.08 0.19 0.27 0.23 0.11 0.06 0.05

Mi fiMi xiMi 1,500 15 45 4,500 360 450 7,500 1425 1650 10,500 2835 3780 13,500 3105 2565 16,500 1815 1155 19,500 1170 390 22,500 1125 225 sum = 11,850 10,260 The molecular weight of the ethylene monomer is 28 g/mol.

Therefore, the weight average molecular weight and degree of polymerization are

MWw = 11,850 g/mol DPw = 11,850/(28 g/mol) = 423 The number average molecular weight and degree of polymerization are

MWn = 10,260 g/mol DPn = 10,260/(28 g/mol) = 366

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Chapter 17: Composites: Teamwork and Synergy in Materials 17-7

In Figure 17-2, find the largest and smallest difference in strength between Borsic and SAP and also the temperatures at which these differences occur.

Solution: The greatest difference occurs at approximately 25°C and is about 120,000 psi. The smallest difference occurs at the highest temperature on the chart, 550°C and is about 70,000 psi.

17-8

A tungsten matrix with 20% porosity is infiltrated with silver. Assuming that the pores are interconnected, what is the density of the composite before and after infiltration with silver? The density of pure tungsten is 19.25 g/cm3 and that of pure silver is 10.49 g/cm3.

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Solution: Before silver infiltration, by the rule of mixtures, the density of the composite ρ is given by ρ = fW ρW + fpore ρpore, where fW and fpore are the volume fractions of tungsten and the pores, respectively, and ρW and ρpore are the densities of tungsten and the pores, respectively. Since initially the pores are filled with air, ρpore can be approximated as zero such that ρ = 0.8(19.25) + 0.2(0) = 15.4g/cm3.

After silver infiltration, using the rule of mixtures and assuming that the pores are interconnected: ρ = fW ρW + fAg ρAg, where fAg and ρAg are the volume fraction and density of silver, respectively. Substituting, ρ = 0.8 (19.25) + 0.2 (10.49) = 17.5 g/cm3.

17-9

Nickel containing 2 wt% thorium is produced in powder form, consolidated into a part, and sintered in the presence of oxygen, causing all of the thorium to produce ThO2 spheres 80 nm in diameter. Calculate the number of spheres per cm3. The density of ThO2 is 9.69 g/cm3.

Solution: In 100 g of material, there are 98 g/(8.902 g/cm3) = 11.0088 cm3 of nickel. From the reaction Th + O2 = ThO2, 2 g Th/(232 g/mol) = x g ThO2/(264 g/mol) x = 2.2759 g ThO2 The total volume of the oxide is Voxide = 2.2759 g/(9.69 g/cm3) = 0.2349 cm3 The volume fraction of the oxide is

f oxide =

0.2349 = 0.0209 0.2349 + 11.0088

The volume of each oxide sphere is Vsphere = (4π/3)r3 = (4π/3)(40 × 10–7cm)3 = 2.68 × 10–16 cm3 The total number of oxide particles in 1 cm3 is

particles = 0.0209 cm3ThO 2 /(2.68 ×10 –16 cm3 /particle) = 7.8 ×1013 particles/cm3 17-10 Spherical aluminum powder (SAP) 0.002 mm in diameter is treated to create a thin oxide layer and is then used to produce an SAP dispersion-strengthened material containing 10 vol% Al2O3. Calculate the average thickness of the oxide film prior to compaction and sintering of the powders into the part.

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Solution: The volume of an aluminum powder particle is VAl = (4π/3)(0.002 mm/2)3 = 4.19 × 10–9 mm3 The volume fraction Al2O3 is

0.10 =

Voxide Voxide = Voxide + VAl Voxide + 4.19 ×10 −9

Voxide = 4.654 × 10−10 mm3 We can then calculate the radius of the particle after oxidation has occurred:

Voxide = (4π /3)r 3 − 4.19 ×10 –9 = 4.65 ×10 –10 r = 1.036 ×10−3 mm = 0.001036 mm The thickness of the oxide layer must therefore be

thickness = 0.001036 − 0.001 = 0.000036 mm = 3.6 × 10−5 mm 17-11 Yttria (Y2O3) particles 750 Å in diameter are introduced into tungsten by internal oxidation. Measurements using an electron microscope show that there are 5 × 1014 oxide particles per cm3. Calculate the wt% Y originally in the alloy. The density of Y2O3 is 5.01 g/cm3.

Solution: The volume of each particle is Voxide = (4π/3)[750 × 10–8/(2 cm)]3 =2.2089 × 10–16 cm3 The total volume of oxide particles per cm3 is given by Vyttria = (2.2089 × 10–16 cm3)(5 × 1014 particles) = 0.1104 cm3 The volume fraction of yttria is therefore foxide = 0.1104 The weight percentages of oxide and tungsten are

(0.1104)(5.01 g/cm3 ) ×100 = 3.13% (0.1104)(5.01) + (0.8896)(19.254) wt % W = 96.87

wt % Y2 O3 =

In 1 g of material, there is 0.0313 g of oxide. From the equation 2Y + (3/2)O2 = Y2O3 x g of Y/[2(88.91 g/mol)] = 0.0313 g of Y2O3/(225.82 g/mol) x = 0.0246 g of Y The weight percent Y in the original alloy was therefore

wt % Y =

0.0246 g Y × 100 = 2.48% 0.0246 g Y + 0.9687 g W

17-12 With no special treatment, aluminum is typically found to have an Al2O3 layer that is 3 nm thick. If spherical aluminum powder prepared with a total diameter of 0.01 mm is used to produce SAP dispersion-strengthened aluminum, calculate the volume percent Al2O3 in the material and the number of oxide particles per cm3. Assume that the oxide breaks into disk-shaped flakes 3 nm thick and 3 × 10–4 mm in diameter. Compare the number of oxide particles per cm3 with the number of solid solution atoms per cm3 when 3 at% of an alloying element is added to aluminum.

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Solution: The total volume of the powder particle is Vtotal = (4π/3) (0.01/2)3 = 5.235988 × 10–7 mm3 The volume of the oxide layer is

Voxide = 5.235988 ×10−7 − (4π /3)(0.005 − 3 ×10−6 )3 = 0.009419 ×10−7 mm3 The volume fraction of the oxide is foxide = 0.009419 × 10–7/(5.235988 × 10–7) = 0.001799 The volume of one disk-shaped oxide flake is

Vflake = (π /4)(3 ×10−4 mm)2 (3 ×10−6 mm) = 2.12 ×10−13 mm3 = 2.12 ×10−16 cm3 In one cm3 of SAP, there is 0.001799 cm3 of oxide. The number of oxide particles per cm3 is therefore

number = 0.001799 cm3 /(2.12 ×10−16 cm3 /particle) = 8.49 ×1012 flakes/cm3 The number of solid solution atoms per cm3 in an Al–3 at% alloying element alloy is calculated by first determining the volume of the unit cell: Vcell = (4.404958 × 10–8 cm)3 = 66.41 × 10–24 cm3 In 25 unit cells of FCC aluminum, there are 100 atom sites. In the alloy, three of these sites are filled with substitutional atoms, the other 97 sites by aluminum atoms. The number of solid solution atoms per cm3 is therefore

number = 3 atoms/[(25 cells)(66.41×10−24 cm3 /cell)] = 18.1×1020 substitutional atoms/cm3 The number of substitutional point defects is eight orders of magnitude larger than the number of oxide flakes.

17-15 If the volume fractions of parts in a composite are all identical (f1 = f2 = … = fn), but the density of each part varies, what does the rule of mixtures become algebraically?

Solution: The rule of mixtures is = If it is known that the fractions are all constant, they can be factored out of the summation: = The fractions may be further defined as a fraction of the total number of components in the mixture: =

1

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17-16 Calculate the density of a cemented carbide, or cermet, based on a titanium matrix if the composite contains 50 wt% WC, 22 wt% TaC, and 14 wt% TiC. (See Example 17–2 for densities of the carbides.)

Solution:

We must find the volume fractions from the weight percentages.

Using a basis of 100 g of the cemented carbide:

f WC =

50gWC/(15.77 g/cm3 ) = 0.298 (50/15.77) + (22/14.5) + (14/4.94) + (14/4.507)

f TaC =

22 g TaC/(14.5g/cm3 ) = 0.143 (50/15.77) + (22/14.5) + (14/4.94) + (14/4.507)

f TiC =

14 g TiC/(4.94 g/cm3 ) = 0.267 (50/15.77) + (22/14.5) + (14/4.94) + (14/4.507)

14 g TiC/(4.507 g/cm3 ) f Ti = = 0.292 (50/15.77) + (22/14.5) + (14/4.94) + (14/4.507) The density is then found from the rule of mixtures: ρc = (0.298)(15.77) + (0.143)(14.5) + (0.267)(4.94) + (0.292)(4.507) = 9.408 g/cm3 17-17 Spherical silica particles (100 nm in diameter) are added to vulcanized rubber in tires to improve stiffness. If the density of the vulcanized rubber matrix is 1.1 g/cm3, the density of silica is 2.5 g/cm3, and the tire has a porosity of 4.5%, calculate the number of silica particles lost when a tire wears down 0.4 cm in thickness. The density of the tire is 1.2 g/cm3; the overall tire diameter is 63 cm; and it is 10 cm wide.

Solution:

The volume change of the tire ΔV due to 0.4 cm wear along the circumference is given by

 π d 2 π d 2f  ∆V =  i −  w ,  4 4   where di and df are the initial and final tire diameters, respectively, and w is the tire width.

Substituting accordingly,

 π d 2 π d 2f   π (63) 2 π (62.2) 2  ∆V =  i − w = − (10) = 787cm3     4 4  4  4   From the rule of mixtures, the density of the tire ρ is given by ρ = fpore ρpore + fsilica ρsilica + frubber ρrubber, where fpore, fsilica, and frubber are the volume fractions of the pores, silica, and rubber in the tire, respectively, and ρpore, ρsilica, and ρrubber are the densities of the pores, silica, and rubber, respectively. The density of the pores is assumed to be zero. The volume fraction of silica is unknown. Substituting and solving for fsilica,

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1.2 = 0.045(0) + fsilica (2.5) + (1 − 0.045 − f silica )(1.1) 1.2 = (2.5 − 1.1) f silica + (1 − 0.045)(1.1) f silica = [1.2 − (1 − 0.045)(1.1)] / (2.5 − 1.1) f silica = 0.107.

The volume of silica lost when the tire wears 0.4 cm is 787(0.107) = 84 cm3.

The volume of one silica particle Vparticle is Vparticle = (4πr3)/3 = (4π(50 × 10–7)3]/3 = 5.24 × 10–16cm3, where r is the particle radius. Hence, the number of silica particles lost is

(Volume of silica lost/Volume of one silica particle) = 84/(5.24 ×10 −16 ) = 1.6 × 1017 particles.

17-18 A typical grinding wheel is 9 in. in diameter, 1 in. thick, and weighs 6 lb. The wheel contains SiC (density of 3.2 g/cm3) bonded by a silica glass (density of 2.5 g/cm3); 5 vol% of the wheel is porous. The SiC is in the form of 0.04 cm cubes. Calculate (a) the volume fraction of SiC particles in the wheel and (b) the number of SiC particles lost from the wheel after it is worn to a diameter of 8 in.

Solution:

To find the volume fraction of SiC:

Vwheel = (π /4) D 2 h = (π /4)(9 in.) 2 (1 in.) = 63.617in.3 = 1042.5 cm3 Wwheel = 6 lb = 2721.5 g

ρ wheel = 2721.5 g/(1042.5 cm3 ) = 2.6106 g/cm3 From the rule of mixtures:

2.6106 = f pore ρ pore + fSiC ρSiC + f glass ρ glass 2.6016 = (0.05)(0) + fSiC (3.2) + (1 − 0.05 − fSiC )(2.5) fSiC = 0.337 First we can determine the volume of the wheel that is lost; then we can find the number of particles of SiC per cm3.

Vlost = (π /4)(9) 2 (1) − (π /4)(8) 2 (1) = 13.352 in.3 = 218.8 cm 3 Vparticles = (0.04 cm)3 = 6.4 × 10 −5 cm 3 In 1 cm3 of the wheel, there are (0.337)(1 cm3) = 0.337 cm3 of SiC. The number of SiC particles per cm3 of the wheel is 0.337 cm3/(6.4 × 10–5 cm3/particle) = 5265.6 particles/cm3 The number of particles lost during use of the wheel is particles lost = (5265.6/cm3)(218.8 cm3) = 1.15 × 106 particles

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17-19 An electrical contact material is produced by infiltrating copper into a porous tungsten carbide (WC) compact. The density of the final composite is 12.3 g/cm3. Assuming that all of the pores are filled with copper, calculate (a) the volume fraction of copper in the composite; (b) the volume fraction of pores in the WC compact prior to infiltration; and (c) the original density of the WC compact before infiltration.

Solution: (a)

ρc = 12.3 g/cm3 = f Cu ρCu + f WC ρ WC = f Cu (8.93) + (1 − f Cu )(15.77) f Cu = 0.507

(b) The copper fills the pores. Therefore the volume fraction of the pores prior to infiltration is equal to that of the copper, or fpores = 0.507. (c) Before infiltration, the composite contains tungsten carbide and pores (which have zero density):

ρcompact = f WC ρ WC + f pore ρpore = (0.493)(15.77) + (0.507)(0) = 7.77 g/cm3 17-20 An electrical contact material is produced by first making a porous tungsten compact that weighs 125 g. Liquid silver is introduced into the compact; careful measurement indicates that 105 g of silver is infiltrated. The final density of the composite is 13.8 g/cm3. Calculate the volume fraction of the original compact that is interconnected porosity and the volume fraction that is closed porosity (no silver infiltration).

Solution:

First we can find the volumes of the tungsten and silver: VW = 125 g/(19.254g/cm3) = 6.492 cm3 VAg = 105 g/(10.49 g/cm3) = 10.010 cm3 The volume fractions of each constituent are

fW =

6.492 6.492 = 6.492 + 10.010 + Vpore 16.502 + Vpore

f Ag =

10.010 10.010 = 6.492 + 10.010 + Vpore 16.502 + Vpore

f pore =

Vpore

6.492 + 10.010 + Vpore

=

Vpore

16.502 + Vpore

From the rule of mixtures:

13.8 = [6.492/(16.502) + Vpore )](19.254) +[10.010/(16.502 + Vpore )](10.49) + 0 Vpore = 0.165cm3 The total volume is 6.492 + 10.010 + 0.165 = 16.667 cm3. The fraction of the contact material that is interconnected porosity prior to silver infiltration is equal to the volume fraction of silver; the volume fraction of closed porosity is obtained from Vpore.

f interconnected = 10.010 /16.667 = 0.6005 f closed = 0.165 /16.667 = 0.0099

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17-21 How much clay must be added to 10 kg of polyethylene to produce a low-cost composite having a modulus of elasticity greater than 120,000 psi and a tensile strength greater than 2000 psi? The density of the clay is 2.4 g/cm3 and that of polyethylene is 0.92 g/cm3. (See Figure 17–6.)

Solution: From Figure 17–6, we find that fclay must be greater than 0.3 if the modulus is to exceed 120,000 psi; however, the fclay must be less than 0.46 to ensure that the tensile strength exceeds 2000 psi. Therefore, any clay fraction between 0.30 and 0.46 should be satisfactory.

f clay =

Wclay /(2.4g/cm3 )

(Wclay /2.4) + (10, 000 g/0.92)

If f clay = 0.3, then Wclay = 11,180g = 11.18 kg If f clay = 0.46, then Wclay = 22, 222g = 22.22 kg The overall cost of the composite will be reduced as the amount of clay added to the composite increases.

17-22 We would like to produce a lightweight epoxy part to provide thermal insulation. We have available hollow glass beads for which the outside diameter is 1/16 in. and the wall thickness is 0.001 in. Determine the weight and number of beads that must be added to the epoxy to produce a one-pound composite with a density of 0.65 g/cm3. The density of the glass is 2.5 g/cm3 and that of the epoxy is 1.25 g/cm3.

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Solution: First we can find the total volume of a glass bead, the volume of the glass portion of the bead, the weight of the glass in the bead, and finally the overall density (weight of glass divided by the total volume) of the bead. The air in the hollow bead is assumed to be weightless.

Vbead = (4π /3)(1/32)3 = 1.27832 × 10 −4 in.3 = 2.0948 ×10 −3 cm 3 Vglass = 1.27832 × 10−4 − (4π /3)[0.03125 − 0.001]3 = 0.1188 × 10−4 in.3 = 1.947 × 10−4 cm3 Wglass = (1.947 × 10−4 cm3 )(2.5 g/cm3 ) = 4.8675 × 10−4 g/bead

ρ bead = 4.8675 × 10 −4 g/(2.0948 × 10 −3 ) = 0.232 g/cm3 Now we can use the rule of mixtures to determine the volume fraction of beads that must be introduced into the epoxy.

ρ c = 0.65 = f bead ρ bead + (1 − f bead ) ρepoxy = f bead (0.232) + (1 − f bead )(1.25) f bead = 0.59 We want to produce 1 lb = 454 g = 454 g/(0.65 g/cm3) = 698.46 cm3 of composite material. The volume of beads required is (698.46 cm3) (0.59) = 412 cm3 of beads wt of beads = (412 cm3) (0.232 g/cm3) = 95.58 g of beads The number of beads needed is number = 95.58 g/(4.8675 × 10–4 g/bead) = 1.96 × 105 beads 17-28 Determine the elastic modulus and tensile strength parallel to the fibers of Borsicreinforced aluminum for 25 volume percent fibers. (See Figure 17-9.)

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Solution: This is a simple enough graph to read, provided one exercises caution in matching the lines to the axes. This is especially true when the units are of the same dimension, as here.

TS = 83,000 psi, E = 21,000 ksi 17-29 Five kilograms of continuous boron fibers are introduced in a unidirectional orientation into 8 kg of an aluminum matrix. Calculate (a) the density of the composite; (b) the modulus of elasticity parallel to the fibers; and (c) the modulus of elasticity perpendicular to the fibers.

Solution:

fB =

5kg/(2.36g/cm3 ) = 0.417 5kg / 2.36 + 8kg/2.699

f Al = 0.583

ρc = fBρB + fAlρAl = (0.417)(2.36) + (0.583(2.699)) = 2.558 g/cm3

Ec = f B EB + f Al EAl = (0.417)(55 ×106 ) + (0.583)(10 ×106 ) = 29 ×106 psi 1/Ec = f B /EB + f Al /EAl = 0.417/(55 ×106 ) + 0.583/(10 ×106 ) = 0.0659 ×10−6 Ec = 15.2 × 106 psi 17-30 We want to produce 10 pounds of a continuous unidirectional fiber-reinforced composite of HS carbon in a polyimide matrix that has a modulus of elasticity of at least 25 × 106 psi parallel to the fibers. How many pounds of fibers are required? See Chapter 16 for properties of polyimide.

Solution: The modulus for HS carbon is 40 × 106 psi and for polyimide is 300,000 psi. From the rule of mixtures, we can determine the required volume fraction of fibers:

25 ×106 psi = f carbon (40 ×106 psi) + (1 − f carbon )(0.3 ×106 psi) f carbon = 0.622 Then we can find the weight of fibers required to produce 10 lbs of composite :

0.622 =

Wcarbon / (1.75g/cm3 ) Wcarbon / 1.75 + (10 − Wcarbon ) / 1.39

or Wcarbon = 6.75 lbs

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17-31 Carbon nanotubes (CNTs) with low weight (density = 1.3 g/cm3), high tensile strength (50 GPa), and high modulus of elasticity (1 TPa) in the axial direction have been touted as the strongest material yet discovered. Calculate the specific strength for CNTs. How does this value compare to that of graphite whiskers given in Table 17–2? If a composite was made using an alumina (Al2O3) matrix with 1% volume CNT fibers, what fraction of the load would the CNT fibers carry? In practice, ceramic matrix CNT composites do not exhibit the expected improvement in mechanical properties. What could be some possible reasons for this?

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Solution:

The definition of specific strength is Specific strength = Tensile Strength/ρ, where ρ is the mass density. For carbon nanotubes, Tensile Strength = 50 × 109 Pa = 7252 ksi = 7252 × 103 psi, and ρ = 1.3 g/cm3 = 0.047 lb/in.3 Substituting,

Specific strength = Tensile Strength/ρ = (7252 × 103 psi)/(0.047 lb/in.3 ) = 154 × 106 in.

This value is about three times greater than that of graphite whiskers listed in Table 17–2.

From Table 17–2, the elastic modulus of Al2 O 3 ( EAl 2 O 3 ) is

EAl 2 O 3 = 56 ×106 psi = 386 GPa, and the elastic modulus of carbon nanotubes Ecnt is Ecnt = 1 TPa = 1000 GPa. If we assume good bonding between the fibers (f) and matrix (m), then the matrix and fibers experience equal strain; hence, assuming elastic strain and according to Hooke’s law, σm/Em = σf/Ef, where σm and σf are the stresses imposed on the matrix and fibers, respectively, and Em and Ef are the elastic moduli of the matrix and fibers, respectively. Rearranging, (σm/σm) = (Em/Ef) = 386/1000 = 0.386.

The fraction of the load carried by the fibers is given by Fraction of Force on Fiber = Ff/(Ff + Fm), where Ff and Fm are the loads borne by the fiber and matrix, respectively. Recalling that the force F is equal to the stress σ multiplied by the cross–sectional area A,

Fraction of Force on Fiber = Ff /( Ff + Fm ) = σ f Af /(σ f Af + σ m Am ) = Af /[ Af + Am (σ m /σ f )] = Af /[ Af + Am ( Em /Ef )].

If the fibers have a uniform cross–section, the area fraction of the fibers equals the volume fraction of the fibers and the area fraction of the matrix equals the volume fraction of the matrix. Substituting,

Fraction of Force on Fiber = 0.01/[0.01 + 0.99(0.386)] = 0.026.

One possible reason why ceramic matrix composites with carbon nanotubes do not exhibit the expected improvement in mechanical properties is that the bonding between the fibers and the matrix is weak leading to slippage.

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17-32 An epoxy matrix is reinforced with 40 vol% E-glass fibers to produce a 2-cm-diameter composite that is to withstand a load of 25,000 N that is applied parallel to the fiber length. Calculate the stress acting on each fiber. The elastic modulus of the epoxy is 0.4 × 106 psi.

Solution: We can assume that the strains in the composite, matrix, and fibers are equal. Thus εc = εm = εf = σm/Em = σf/Ef The modulus for the E-glass is 10.5 × 106 psi and that for the epoxy is 0.4 × 106 psi.

Therefore the ratio of the stresses is σm/σm = Ef/Em = 10.5 × 106/(0.4 × 106) = 26.25 The fraction of the force carried by the fibers (as described in Example 17–8) is (assuming that area and volume fractions are equal)

f = =

σ f Af Af = σ f Af + σ m Am Af + (σ m /σ f ) Am 0.4 = 0.9459 0.4 + (1/26.25)(0.6)

Since the total force is 25,000 N, the force carried by the fibers is Ff = (0.9459) (25,000 N) = 23,650 N The cross-sectional area of the fibers is Af = (ff)(π/4)d2 = (0.4) (π/4) (20 mm)2 = 125.66 mm2 Thus the stress is σf = 23,650 N/(125.66 mm2) = 188 MPa 17-33 A titanium alloy with a modulus of elasticity of 16 × 106 psi is used to make a 1000-lb part for a manned space vehicle. Determine the weight of a part having the same modulus of elasticity parallel to the fibers, if the part is made of (a) aluminum reinforced with boron fibers and (b) polyester (with a modulus of 650,000 psi) reinforced with highmodulus carbon fibers. (c) Compare the specific modulus for all three materials.

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Solution:

The titanium alloy has a density of about 4.507 g/cm3 = 0.163 lb/in.3. The volume of the 1000 lb part is therefore Vpart = 1000/0.163 = 6135 in 3

Ec = 16 ×106 = f B EB + (1 − f B ) EAl = f B (55 ×106 ) + (1 − f B )(10 ×106 ) f B = 0.133

ρc = (0.133)(2.36 g/cm3 ) + (0.867)(2.699 g/cm3 ) = 2.654 g/cm3 = 0.0958 lb/in.3 To produce a 6135 in.3 part of the composite, the part must weigh Weight = (6135 in.3) (0.0958 lb/in.3) = 587.7 lb

Ec = 16 × 106 = f C EC + (1 − f C ) EPET = fC (77 ×106 ) + (1 − f C )(0.65 ×106 ) f C = 0.201

ρC = (0.201)(1.9 g/cm 3 ) + (0.799)(1.28 g/cm3 ) = 1.405 g/cm3 = 0.0507 lb/in.3 To produce a 6135 in.3 part of the composite, the part must weigh Weight = (6135 in.3) (0.0507 lb/in.3) = 311 lb The specific modulii of the three materials are

Ti : E /ρ = 16 ×106 psi/(0.163 lb/in.3 ) = 9.82 × 107 in.

B − Al: E /ρ = 16 ×106 psi/(0.0958 lb/in.3 ) = 16.7 × 107 in.

C − PET: E /ρ = 16 ×106 psi/(0.0507 lb/in.3 ) = 31.6 × 107 in.

17-34 Short, but aligned, Al2O3 fibers with a diameter of 20 µm are introduced into a 6,6-nylon matrix. The strength of the bond between the fibers and the matrix is estimated to be 1000 psi. Calculate the critical fiber length and compare with the case when 1 µm alumina whiskers are used instead of the coarser fibers. What is the minimum aspect ratio in each case?

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Solution: The critical fiber length is given by ℓc = TSfd/(2τi) For the alumina fibers, τi = 1000 psi = 6.897 MPa; d = 20 × 10–6 m; and TSf = 300,000 psi = 2069 MPa. Thus, for alumina fibers:

ℓ c = (2069 MPa)(20 × 10 −6 m)/[(2)(6.897 MPa)] = 0.003 m = 0.3 cm

ℓ c /d = 0.3 cm/(20 × 10 −4 cm) = 150 For alumina whiskers, d = 1 × 10–6 m. The strength of the whiskers can be much higher than that of the fibers; 3,000,000 psi = 20,690 MPa can be achieved. Thus, for alumina whiskers:

ℓ c = (20, 690 MPa)(1× 10−6 m)/[(2)(6.897 MPa)] = 0.0015 m = 0.15 cm

ℓ c /d = 0.15cm/(1× 10−4 cm) = 1500 17-35 We prepare several epoxy-matrix composites using different lengths of 3-μm- diameter ZrO2 fibers and find that the strength of the composite increases with increasing fiber length up to 5 mm. For longer fibers, the strength is virtually unchanged. Estimate the strength of the bond between the fibers and the matrix.

Solution: We do not expect much change in the strength when ℓ > 15 ℓc.

Therefore

15ℓ c = 5mm or ℓ c = 0.333 mm In addition, ℓc = TSfd/(2τi) = 0.333 mm For ZrO2 fibers, the tensile strength = 300,000 psi = 2069 MPa. Therefore,

τ i = TS f d / (2ℓ c ) = (2069 MPa)(0.003 mm) / [(2)(0.333 mm) = 9.32 MPa 17-36 Glass fibers with a diameter 50 μm are introduced into a 6,6 nylon matrix. Determine the critical fiber length at which the fibers behave as if they are continuous. The strength of the bond between the fibers and matrix is 10 MPa. The relevant materials properties are as follows: Eglass = 72 GPa, Enylon = 2.75 GPa, TSglass = 3.4 GPa, and TSnylon = 827 MPa. Solution: The critical fiber length lc is given by lc = TSfd/(2τi), where TSf is the tensile strength of the fiber, d is the fiber diameter, and τi is related to the strength of the bond between the fiber and the matrix. For the glass fibers, TSf = 3.4 GPa, d = 50 × 10–6m, and τi = 10 MPa. Substituting,

lc = (3.4 × 109 )(50 × 10 −6 ) / [2(10 × 106 )] = 8.5 × 10−3 m = 0.85 cm.

For the fiber to behave as if it is continuous, it needs to be at least 15 times longer than lc, that is 15(0.85 cm) = 12.75 cm.

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17-37 A copper-silver bimetallic wire, 1 cm in diameter, is prepared by co-extrusion with copper as the core and silver as the outer layer. The desired properties along the axis parallel to the length of the bimetallic wire are as follows: (a) Thermal conductivity > 410 W/(m ⋅ K); (b) Electrical conductivity > 60 × 106 Ω–1m–1; and (c) Weight < 750 g/m.

Determine the allowed range of the diameter of the copper core.

Copper Silver Density (g/cm3) 8.96 10.49 Electrical conductivity (Ω–1m–1) 59 × 106 63 × 106 Thermal conductivity(W/(m ⋅ K)) 401 429

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Solution: The thermal conductivity of the composite wire kc parallel to the wire length is given by kc = fCu kCu + fAg kAg, where fCu and fAg are the volume fractions of the copper and silver, respectively, and kCu and kAg are the thermal conductivities of the copper and silver, respectively. The thermal conductivity must be a minimum of 410 W/(m–K). Since the thermal conductivity of copper is less than 410 W/(m–K), there is a maximum volume fraction of copper that will permit this thermal conductivity criterion to be met. The sum of the volume fractions of copper and silver must equal one, i.e., fCu + fAg = 1, such that fAg = 1–fCu. Substituting and solving for fCu, 410 < fCu(401) + (1–fCu)(429) (429–401) fCu < 429–410 fCu < 0.68.

The electrical conductivity of the composite wire σc parallel to the wire length is given by σc = fCu σCu + fAg σAg, The electrical conductivity must be a minimum of 60 × 106Ω–1m–1. Since the electrical conductivity of copper is less than 60 × 106Ω–1m–1, there is a maximum volume fraction of copper that will permit this electrical conductivity criterion to be met. Substituting and solving for fCu, 60 × 106 < fCu (59 × 106) + (1–fCu) (63 × 106) (63 × 106–59 × 106) fCu < (63 × 106–60 × 106) fCu < 0.75.

A 1 m length of the composite wire with a diameter of 1 cm has a volume V of V = [π (1 cm)2/4] (100 cm) = 78.5 cm3.

Thus the density of the composite wire must not exceed ρc = 750 g/(78.5 cm3) = 9.55 g/cm3.

The density of the composite wire is given by ρc = fCu ρCu + fAg ρAg. Since the density of silver is greater than 9.55 g/cm3, there is a maximum volume fraction of silver that will permit this density criterion to be met.

Substituting and solving for fCu, 9.55 > fCu (8.96) + (1–fCu) (10.49) (10.49–8.96)) fCu > (10.49–9.55) fCu > 0.61 Based on the three criteria listed in this problem, the allowed range for the volume fraction of copper is 0.61 < fCu < 0.68. Since the volume fraction of the copper is proportional to the square of the diameter of the copper core divided by the square of the wire diameter, 0.61 < [d2/(1 cm)2] < 0.68 or 0.78 cm < d < 0.82 cm.

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17-38 Find the materials with the highest numerical values for each of the six properties listed.

For example, tungsten (W) has the highest density at 19.25 g/cm3.

Solution: Property

Material

Value

ρ (g/cm3)

W

19.25

TS (ksi)

Al2O3, Graphite, SiC 3000

E (×106 psi)

Graphite

102.0

Tm (°C)

Graphite

3700

Specific Modulus (×107 in.) Graphite

170.0

Specific Strength (×106 in.)

50.2

Graphite

This should drive home some of the interesting qualities of graphite!

17-39 If a fiber-reinforced composite contains two separate regions of unidirectional fibers and is tested as shown, what would be the formula for the tensile strength?

Solution: In this case, the fibers do not provide any strengthening effect at all because there is a complete discontinuity in the middle of the sample. The strength is only that of the matrix: =

17-40 Plot seven equally spaced points from the curve in Figure 17-11 and find a nonpolynomial equation to fit the line that has an R2 value of at least 0.900.

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Solution: An exponential fit works well here:

Tensile strength of E-glass as a function of angle of stress to fiber direction.

Tensile strength (MPa)

400 350 y = 308.17e-0.024x R² = 0.9179

300 250 200 150 100 50 0 0

30

60

90

Angle (°)

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As before, caution must be exercised. Just because there is a certain well fit correlation here does not mean it is the governing equation. Correlation does not mean causation.

17-45 In one polymer-matrix composite, as produced, discontinuous glass fibers are introduced directly into the matrix; in a second case, the fibers are first “sized.” Discuss the effect this difference might have on the critical fiber length and the strength of the composite.

Solution: By sizing the glass fibers, the surface is conditioned so that improved bonding between the fibers and the matrix is obtained. From Equation 17–9, we expect that improved bonding (τi) will reduce the length of fibers required for achieving good strength. Improved bonding will also reduce pull-out of the fibers from the matrix.

Therefore, the sizing improves the strength and allows small fibers to still be effective.

17-46 Explain why bonding between carbon fibers and an epoxy matrix should be excellent, whereas bonding between silicon nitride fibers and a silicon carbide matrix should be poor.

Solution: In the carbon/epoxy composite, we are interested in developing high strength, with the stress carried predominantly by the strong carbon fibers. In order to transfer the applied loads from the weak epoxy to the strong carbon fibers, good bonding is required.

In the Si3N4/SiC composite, we are interested primarily in developing improved fracture toughness. The microstructure must absorb and dissipate energy. By ensuring that bonding is poor, the silicon nitride fibers can pull out of the silicon carbide matrix. This pull-out requires energy, thus improving the fracture toughness of the ceramic matrix composite.

17-47 A polyimide matrix with an elastic modulus of 0.3 × 106 psi is to be reinforced with 70 vol% carbon fibers to give a minimum modulus of elasticity of 40 × 106 psi. Recommend a process for producing the carbon fibers required. Estimate the tensile strength of the fibers that are produced.

Solution: The modulus of polyimide is 0.3 × 106 psi. The required modulus of the carbon fibers can be found from the rule of mixtures:

Ecomposite = f carbon Ecarbon + f PI EPI 40 ×106 = (0.7) Ecarbon + (0.3)(0.3 ×106 ) Ecarbon = 57.0 ×106 psi From Figure 17–19, we find that, to obtain this modulus in the carbon fibers, they must be pyrolized at 2500°C. This in turn means that the tensile strength will be about 250,000 psi.

17-48 Determine the ratio of the cross-sectional area of fiber to filament in both images in Figure 17-18.

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Solution: This requires use of a ruler to determine the diameters of the circles in the micrographs.

For the Borsic: =

= 1 mm = 1 mm 4 4 4

Next, the area of the fiber: =

− 4

=

14 mm − 1 mm = 195 mm 4 4 4

Taking the ratio is done by inspection: /

= 195

For the SiC on carbon:

= 4 mm = 16 mm 4 4 4

= 16 mm − 16 mm = 240 mm 4 4 4

= =

− 4

/

= 15

17-51 Calculate the steepest and the shallowest apparent (i.e., linear) slopes of the lines in Figure 17-26 and identify the materials they belong to.

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Solution: By inspection the steepest drop belongs to carbon-epoxy, which loses its strength the quickest with heating. The linearized slope is ∆' 1.2 − 1.6 × 10+ in. in. %= = = −3.3 × 10/ 160 − 40 ℃ ℃ ∆( The shallowest slope is that of titanium: %=

∆' 0.59 − 0.8 × 10+ in. in. = = −5.4 × 10 430 − 40 ℃ ∆( ℃

17-53 A microlaminate, Arall, is produced using five sheets of 0.4-mm-thick aluminum and four sheets of 0.2-mm-thick epoxy reinforced with unidirectionally aligned Kevlar™ fibers.

The volume fraction of Kevlar™ fibers in these intermediate sheets is 55%. The elastic modulus of the epoxy is 0.5 × 106 psi. Calculate the modulus of elasticity of the microlaminate parallel and perpendicular to the unidirectionally aligned Kevlar™ fibers.

What are the principle advantages of the Arall material compared with those of unreinforced aluminum?

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Solution:

First we can find the volume fractions of each material. The volumes (expressed in a linear direction) are

VAl = (5 sheets)(0.4 mm/sheet)

= 2mm

VKevlar = (0.55)(4 sheets)(0.2 mm/sheet) = 0.44mm Vepoxy = (0.45)(4 sheets)(0.2 mm/sheet) = 0.36mm total = 2.8 mm f Al = 2/2.8

= 0.714

f Kevlar = 0.44/2.8 = 0.157 f epoxy = 0.36/2.8 = 0.129 From the rule of mixtures, the modulus parallel to the laminate is Eparallel = (0.741) (10 × 106) +(0.157) (18 × 106) + (0.129) (0.5 × 106) = 10.04 × 106 psi Perpendicular to the laminate:

1/Eperpendicular = 0.714/(10 × 106 ) + 0.157/(18 × 106 ) + 0.129/(0.5 × 106 ) = 0.338 × 10−6 Eperpendicular = 2.96 × 106 psi 17-54 A laminate composed of 0.1-mm-thick aluminum sandwiched around a 2-cm-thick layer of polystyrene foam is produced as an insulation material. Calculate the thermal conductivity of the laminate parallel and perpendicular to the layers. The thermal conductivity of aluminum is 0.57 cal/(cm · s · K) and that of the foam is 0.000077 cal/(cm · s · K).

Solution: First we find the volume fractions:

f Al = 2(0.01 cm)/[(2)(0.001 cm) + 2 cm] = 0.0099 f foam = 0.9901 The thermal conductivity parallel to the laminate is

kparallel = (0.0099)(0.57) + (0.9901)(0.000077) = 0.00572 cal/(cm ⋅ s ⋅ K) Perpendicular to the laminate:

1/kperpendicular = 0.0099/0.57 + 0.9901/0.000077 = 12,858 kperpendicular = 0.000078 cal/(cm ⋅ s ⋅ K) 17-55 A 0.01-cm-thick sheet of a polymer with a modulus of elasticity of 0.7 × 106 psi is sandwiched between two 4-mm-thick sheets of glass with a modulus of elasticity of 12 × 106 psi. Calculate the modulus of elasticity of the composite parallel and perpendicular to the sheets.

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Solution: The volume fractions are

f polymer = 0.01cm/(0.01cm + 0.4cm + 0.4 cm) = 0.01234 f glass = 0.98765 The modulus parallel to the laminate is

Eparallel = (0.01234)(0.7 × 106 ) + (0.98765)(12 × 106 ) = 11.86 ×106 psi Perpendicular to the laminate:

1/Eperpendicular = 0.01234/(0.7 × 106 ) + 0.98765/(12 ×106 ) = 0.09994 × 10 −6 Eperpendicular = 10.0 ×106 psi This material is “safety” glass and is used in automobile windshields to keep the windshield from shattering.

17-56 A U.S. quarter is 15/16 in. in diameter and is about 1/16 in. thick. Assuming copper costs about $1.10 per pound and nickel costs about $4.10 per pound, compare the material cost in a composite quarter versus a quarter made entirely of nickel.

Solution:

In a quarter, the thickness (and hence the volume) ratio is 1/6 Ni: 2/3 Cu: 1/6 Ni. The volume fraction of each is

f Cu = 0.667

f Ni = 0.333

The volume of the quarter, copper, and nickel are

Vquarter = (π /4)(15/16 in.)2 (1/16 in.) = 0.04314 in.3 = 0.707 cm 3 VCu = (0.707 cm 3 )(0.667) = 0.4716 cm3 VNi = (0.707 cm 3 )(0.333) = 0.2354 cm3 The weights of copper and nickel in the coin are

WCu = (0.4716 cm3 )(8.93 g/cm3 ) = 4.211 g = 0.00928 lb WNi = (0.2354 cm3 )(8.902 g/cm3 ) = 2.0955 g = 0.004616 lb The cost of each material in the coin is $/Cu = (0.00928 lb)($1.10/lb) = $0.0102 $/Ni = (0.004616 lb)($4.10/lb) = $0.0189 The total cost of the composite (for materials only) = $0.0291 If the entire coin were made from nickel, then (0.707 cm3)(8.902 g/cm3) [1/(454 g/lb)] ($4.10/lb) = $0.0568 By using the composite coin, the cost of the materials is about half that of a pure nickel coin, yet the coin appears to have the nickel (or silvery) color.

17-57 Calculate the density of a honeycomb structure composed of the following elements: The two 2-mm-thick cover sheets are produced using an epoxy matrix prepreg containing 55 vol% E-glass fibers. The aluminum honeycomb is 2 cm thick; the cells are in the shape of 0.5 cm squares and the walls of the cells are 0.1 mm thick. The density of the epoxy is 1.25 g/cm3. Compare the weight of a 1 m × 2 m panel of the honeycomb with a solid aluminum panel with the same dimensions.

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Solution: Each cell of aluminum can be considered to be a hollow square shape where the dimensions of the cell are 0.5cm × 0.5 cm × 2 cm, with a wall thickness belonging uniquely to that cell of 0.1 mm/2 = 0.05 mm = 0.005 cm.

VAl = (4 sides) (0.005 cm) (0.5 cm) (2cm) = 0.02 cm3 The cover sheet dimensions that just cover the single cell described above are 0.5 cm× 0.5 cm × 2 mm. The volume is Vcover = (2 sheets) (0.5 cm) (0.5 cm) (0.2 cm) = 0.1 cm3 The total “height” of the cell, including the cover sheets, is 2 cm + 2(0.2 cm) = 2.4 cm. The total volume of the cell is Vtotal = (0.5 cm) (0.5 cm) (2.4 cm) = 0.6 cm3 The volume fractions of these constituents are

f Al in Cell = 0.02/0.6 = 0.0333 f cover = 0.1/0.6 = 0.1667 f void = 1 − ( f Al in Cell + f cover ) = 1 − (0.0333 + 0.1667) = 0.80 The densities of the three constituents can be determined. The density of the aluminum in the cells is 2.699 g/cm3, and the density of the void space within the cells is zero. The cover sheets are themselves composites ρcover = fglass ρglass + fepoxy ρepoxy = (0.55) (2.55 g/cm3) + (0.45) (1.25 g/cm3) = 1.965 g/cm3 Therefore the overall density of the honeycomb structure is

ρ honeycomb = f Al in Cell ρ Al in Cell + f cover ρcover + f void ρ void = (0.0333)(2.699) + (0.1667)(1.965) + (0.80)(0) = 0.417 g/cm3 The weight of a 1 m × 2 m panel of the honeycomb is

Whoneycomb = (2.4 cm)(100 cm)(200 cm)(0.417 g/cm3 ) = 20, 016 g = 20.016 kg = 44.1 lb If the panel were made from solid aluminum with the same dimensions, the panel would weigh

Wsolid = (2.4 cm)(100 cm)(200 cm)(2.699 g/cm3 ) = 129,552 g = 129.552 kg = 285 lb The weight savings using the honeycomb are enormous.

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Chapter 18: Construction Materials

18-1

Table 18–1 lists the densities for typical woods. Calculate the densities of the woods after they are completely dried and at 100% water content.

Solution:

The percentage of water in a given wood is given by Equation 18–1: % Water = (weight of water/dry weight) × 100.

Furthermore, the weight of the water is given by weight of water = green weight – dry weight, where the green weight is the weight of the wood when wet. Substituting accordingly,

(% Water/100) = [ (green weight – dry weight) / dry weight] = (green weight / dry weight) –1, or dry weight = green weight/ [1 + (% Water/100) ].

Assuming no volume change, the above equation can be applied to density. The densities in Table 18–1 are the green densities for 12% water content. Therefore, the density of the completely dry wood can be calculated using dry density = green density/(1 + 12/100) = green density/1.12.

The following table lists the calculated dry densities for the various woods.

Wood Density 12% Water (g/cm3) Dry Density (g/cm3) Cedar 0.32 0.29 Pine 0.35 0.31 Fir 0.48 0.43 Maple 0.48 0.43 Birch 0.62 0.55 Oak 0.68 0.61 At 100% water content, the density can be calculated using dry density = green density/ [1 + (100/100) ] or green density = 2(dry density).

Using this equation and values for dry densities in the above table, produces the table below.

Wood Dry Density (g/cm3) Density 100% water (g/cm3) Cedar 0.29 0.58 Pine 0.31 0.62 Fir 0.43 0.86 Maple 0.43 0.86 Birch 0.55 1.10 Oak 0.61 1.22 18-2

Use Table 18.1 to determine if there is a relationship between density and the elastic modulus of the woods listed.

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Solution: Generating the graph:

E as a function of ρ for woods Modulus of elasticity (psi)

2,500,000 2,000,000 1,500,000 1,000,000

y = 2E+06x + 505032 R² = 0.6554

500,000 0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

Density for 12% water (g/cm3)

There does appear to be a positive correlation between the modulus of elasticity and the density at constant water content.

18-3

A sample of wood with dimensions 3 in. × 4 in. × 12 in. has a dry density of 0.35 g/cm3. (a) Calculate the number of gallons of water that must be absorbed by the sample to contain 120% water. (b) Calculate the density after the wood absorbs this amount of water.

V = 3 × 4 × 12 = 144 in.3 = 2359.7 cm3 dry weight = 0.35 × 2359.7 = 825.9 g

Solution:

weight of water × 100 weight of dry wood water = (1.2)(825.9) = 991 g = 2.183 lb

120% water = (a)

= (2.183 lb)(7.48 gal / ft 3 )/(62.4 lb / ft 3 ) = 0.262 gal

(b) If the volume remains the same, then

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density = 18-4

825.9 g of dry wood + 991 g of water = 0.77 g/cm 3 3 2359.7 cm

The density of a sample of oak is 0.90 g/cm3. Calculate (a) the density of completely dry oak and (b) the percent water in the original sample.

Solution:

ρ12% water = 0.68 g/cm3 (Table 18–1) (a) Therefore, in 100 cm3 of wood at 12% H2O, there are 68 g.

 green weight – dry weight  12%water =   × 100 dry weight   dry weight = 68 / 1.12 = 60.71 g (b) When the density is 0.90 g/cm3, there are 90 g of green wood per 100 cm3. The water is therefore 90 – 60.71 g, or 29.29 g.

% H2O = 18-5

90 g – 60.71 g × 100 = 48.2% 60.71 g

A green wood with a density of 0.82 g/cm3 contains 150% water. The compressive strength of this wood is 27 MPa. After several days of drying, the compressive strength increases to 41 MPa. What is the water content and density of the dried wood? (See Figure 18-4.)

Solution:

The percentage of water in a given wood is given by Equation 18–1: % Water = (weight of water/weight of dry wood) × 100.

Furthermore, the weight of the water is given by weight of water = green weight – dry weight, where the green weight is the weight of the wood when wet. Substituting accordingly, 150 = [(green weight – dry weight)/dry weight] × 100.

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Rearranging, 1.5 = (green weight/dry weight) – 1 such that dry weight = green weight /2.5.

Assuming a constant volume for the wood whether it is wet or dry, the densities will follow a similar relationship: dry density = green density/2.5.

Hence, the density of completely dry wood is dry density = 0.82/2.5 = 0.328 g/cm3.

The water content can be calculated using the provided strength information and Figure 18–4. After some drying, the compressive strength of 41 MPa (5947 psi) corresponds to about 15% water. Using the approach outlined earlier to calculate the density of completely dry wood, the density of the wood with 15% water will be 1.15 times the density of the completely dry wood: density of wood with 15% water = 0.328 × 1.15 = 0.377 g/cm3.

18-6

Determine the expansion in the radial direction of cedar wood when it changes from 12% to 27% water content. The initial thickness is 1 ft.

Solution: Using Equation 18-2:

∆ = − From Table 18-4, the coefficient c for cedar in the radial direction is 0.00111 inches per inch-percent water content. in. ∆ = 1 ft 0.00111 27 % − 12 % in.∙ % H O in. in. ∆ = 1 ft 12 0.00111 15 % ft in.∙ % H O ∆ = 0.1998 in.

18-7

Boards of oak 0.5 cm thick, 1 m long, and 0.25 m wide are used as flooring for a 10 m × 10 m area. If the floor was laid at a moisture content of 25% and the expected moisture could be as high as 45%, determine the dimensional change in the floor parallel to and perpendicular to the length of the boards. The boards were cut from logs with a tangential–longitudinal cut.

Solution: In the tangential direction, which in this case is perpendicular to the length of the boards and to the thickness of the boards, the dimensional coefficient c of oak is 0.00462 in./(in. % H2O) (Table 18–4).

The dimensional change can be Δx calculated using Equation 18–2: ¢x = x0[c(Mf – Mi) ], where x0 is the initial dimension and Mf and Mi are the initial and final water contents, respectively. Substituting,

∆x = 0.25[0.00462(45 – 25) ∆x = 0.0231 m.

A 10 m wide area requires 40 boards that are each 0.25 m wide; hence, the total dimensional change in the tangential direction is

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40(0.0231) = 0.924 m.

In the longitudinal direction, the change in dimension is expected to be less than 0.2% or 0.002(10) = 0.02 m.

18-8

Boards of maple 1 in. thick, 6 in. wide, and 16 ft long are used as the flooring for a 60 ft × 60 ft hall. The boards were cut from logs with a tangential-longitudinal cut. The floor is laid when the boards have a moisture content of 12%. After some particularly humid days, the moisture content in the boards increases to 45%. Determine the dimensional change in the flooring parallel to the boards and perpendicular to the boards. What will happen to the floor? How can this problem be corrected?

Solution: Perpendicular: ctangential = 0.00353 in./(in. % H2O) for maple Δx = xo[c(Mf – Mi) ] = 6[0.00353(45 – 12) ] = 0.699 in. per 6 in.

Over a 60 ft span : ∆x =

(60 ft)(12 in./ft)(0.699 in.) = 83.9 in.

6 in.

The floor will therefore buckle due to the large amount of expansion of the boards perpendicular to the flooring.

Parallel: For most woods, only about a 0.2% change in dimensions occurs longitudinally. Thus the total change in the length of the boards will be about Δy = (0.002)(60 ft)(12 in./ft) = 1.44 in.

18-9

A wall 30 feet long is built using radial-longitudinal cuts of 5-in. wide pine with the boards arranged in a vertical fashion. The wood contains a moisture content of 55% when the wall is built; however, the humidity level in the room is maintained to give 45% moisture in the wood. Determine the dimensional changes in the wood boards and estimate the size of the gaps that will be produced as a consequence of these changes.

Solution: ctangential = 0.00141 in./(in.·% H2O) for pine

∆x = (30 ft)(12 in. / ft) [0.00141 in./(in. ⋅ % H 2 O)(45 – 55) ] = –5.076 in.

The total number of boards in the width of the wall is # of boards = (30 ft)(12 in./ft)/(5 in./board) = 72 boards Therefore there are 71 gaps between the boards. The average width of the gaps is gap = 5.076 in./71 gaps = 0.0715 in.

18-10 What is the maximum water to concrete ratio acceptable for a structure that must have a final compressive strength of at least 5 ksi? (See Figure 18-7.)

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Solution: Referring to Figure 18-7, the final compressive strength is achieved at about 28 days.

This line intersects the 5000 psi gridline at approximately 0.48. For safety, a lower value would be used in practice.

18-11 Determine the amounts of water, cement, and sand in 10 m3 of concrete if the cement– sand–aggregate ratio is 1:2.5:4.5 and the water–cement ratio is 0.4. Assume that no air is entrained into the concrete. The sand used for this mixture contains 4 wt% water, and the aggregate contains 2 wt% water.

Solution:

For a weight ratio of 1:2.5:4.5 of cement–sand–aggregate, per sack of cement (from Table 18–6): cement: (94 lb/sack)/(190 lb/ft3) = 0.495 ft3/sack sand: (2.5)(94 lb/sack)/(160 lb/ft3) = 1.469 ft3/sack aggregate: (4.5)(94 lb/sack)/(170 lb/ft3) = 2.488 ft3/sack water: (0.4)(94 lb/sack)/(62.4 lb/ft3) = 0.603 ft3/sack Volume per sack = 0.495 + 1.469 + 2.488 + 0.603 = 5.05 ft3/sack Total volume of concrete = 10 m3 = 353 ft3 Number of sacks of cement needed = 353 ft3/(5.05 ft3/sack) = 70 sacks sand = (70 sacks)(2.5)(94 lb/sack) = 16,450 lb aggregate = (70 sacks)(4.5)(94 lb/sack) = 29,610 lb water = (70 sacks)(0.4)(94 lb/sack) = 2632 lb We must also adjust for the water present in the wet sand (4%) and wet aggregate (2%). To find the amount of wet sand, divide by 0.96, and to find the amount of wet aggregate, divide by 0.98.

wet sand = 16, 450 lb/0.96 = 17,135 lb; water = 17,135 – 16, 450 = 685 lb

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wet aggregate = 29,610 lb/0.98 = 30, 214 lb; water = 30, 214 – 29,610 = 604 lb The actual amount of water needed should be reduced to reflect the water content of the sand and aggregate: water = 2632 – 685 – 604 = 1343 lb In summary, the ingredients for the concrete mix are cement = 70 sacks sand = 17,135 lb = 8.57 tons aggregate = 30,214 lb = 15.1 tons water = 1343 lb = 1343 lb(7.48 gal/ft3)/(62.4 lb/ft3) = 161 gal 18-12 Calculate the amount of cement, sand, aggregate, and water needed to create a concrete mix with a 28-day compressive strength of 34 MPa for a 10 m × 10 m × 0.25 m structure given the following conditions: allowed slump = 10 cm and only 3.8 cm (1.5 in.) aggregate with 2% moisture and coarse sand with 4% moisture are available for this project. Assume no air entrainment in your calculations.

Solution: The amount of water required to give the desired workability is determined from Figure 18–10. A slump of 10 cm (3.94 in.) is required and 1.5 inch aggregate is being used, so the required water content is about 305 lb/yd3 of concrete. To obtain a 28 day compressive strength of 34 MPa (4931 psi), assuming no air entrainment, a water/cement ratio of 0.49 is required according to Figure 18–9. The weight of cement required per cubic yard of concrete is 305 lb/0.49 = 622 lb cement.

For 1.5 inch aggregate with coarse sand, the volume ratio of aggregate to concrete is 0.75 using Figure 18-11. As noted in Figure 18–11, the bulk density of the aggregate is used to determine the volume ratio, but the true density is required for these calculations. The bulk density ρbulk is about 60% of the true density ρtrue. Thus it can be written, ρbulk = 0.6ρtrue, and from the definition of mass density,

m m = 0.6 , Vbulk Vtrue where the mass m is the same for the bulk and true densities, but the volumes Vbulk and Vtrue are different. Rearranging, Vtrue = 0.6 Vbulk.

Therefore, the amount of aggregate needed per cubic yard of concrete is 0.6(0.75) = 0.45 yd3.

For a structure 10 m × 10 m × 0.25 m volume = 25 m3 = 883 ft3 = 32.7 yd3.

The breakdown of the components in the concrete is as follows: water: 305 lb/yd3 (32.7 yd3) = 9974 lb, cement: 622 lb/yd3 (32.7 yd3) = 20,339 lb, aggregate: 0.45(32.7 yd3) = 14.72 yd3 = 397 ft3 by volume.

The density of the aggregate is 170 lb/ft3 from Table 18–6. Thus the weight of the aggregate is 397(170 lb/ft3) = 67,490 lb.

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volume of sand = (concrete volume – water volume – cement volume – aggregate volume) = 32.7 yd – [9974 lb/(62.4 lb/ft 3 )(0.037 yd 3 /ft 3 )] 3

– [20,339 lb/(190 lb / ft 3 )(0.037 yd 3 / ft 3 ) –14.72 yd 3 = 32.7 – 5.91 – 3.96 – 14.72 = 8.11 yd 3or 219 ft 3 weight of sand = 219 ft 3 (160 lb / ft 3 )(from Table 18 – 6) weight of sand = 35, 040 lb.

Since the sand and aggregate are wet, we need to account for the water in these components: aggregate (2% moisture) = 67,490 lb/0.98 = 68,867 lb (water = 1377 lb) sand (4% moisture) = 35,040 lb/0.96 = 36,500 lb (water = 1460 lb) Hence the actual amount of water needed is water = 9974 – 1377 – 1460 = 7137 lb.

In summary, the amount of each component needed is

water = 7137 lb = 7137 lb/(62.4 lb / ft 3 )(7.48 gal/ft 3 ) = 856 gal cement = 20,339 lb aggregate = 68,867 lb sand = 36,500 lb 18-13 We have been asked to prepare 100 yd3 of normal concrete using a volume ratio of cement–sand–coarse aggregate of 1:2:4. The water–cement ratio (by weight) is to be 0.5. The sand contains 6 wt% water and the coarse aggregate contains 3 wt% water. No entrained air is expected. (a) Determine the number of sacks of cement that must be ordered, the tons of sand and aggregate required, and the amount of water needed. (b) Calculate the total weight of the concrete per cubic yard. (c) What is the weight ratio of cement–sand–coarse aggregate?

Solution:

First determine the volume of each material on a “sack” basis, keeping in mind the 1:2:4 volume ratio of solids and the 0.5 water–cement weight ratio:

cement = (94 lb/sack)/(190 lb/ft 3 )

= 0.495 ft 3 /sack

sand = (2)(94 lb/sack)/(190 lb/ft 3 )

= 0.989 ft 3 /sack

aggregate = (4)(94 lb/sack)/(190 lb/ft 3 )

= 1.979 ft 3 /sack

water = (0.5)(94 lb/sack)/(62.4 lb/ft 3 ) = 0.753 ft 3 /sack total volume of concrete/sack = 4.216 ft 3 /sack In 100 yd3, or (100 yd3)(27 ft3/yd3):

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cement = 2700 ft 3 /(4.216 ft 3 /sack) = 640 sacks sand = (640 sacks)[(2)(94 lb/sack)/(190 lb/ft 3 )] (160 lb/ft 3 ) = 101, 322 lb = 50.7 tons aggregate = (640 sacks)[(4)(94 lb/sack) / (190 lb/ft 3 )](170 lb/ft 3 ) = 215,309 lb = 107.7 tons water = (640 sacks)(0.5)(94 Ib / sack) = 30,080 lb or

= (640 sacks)(0.5) (94 lb / sack) / (62.4 lb / ft 3 )(7.48 gal/ft 3 ) = 3606 We must make adjustments for the water that is already present in the sand and aggregate. There is 6% water in the sand and 3% water in the aggregate. Divide the dry sand by 0.94 to obtain the amount of wet sand that we need to order. Divide the dry aggregate by 0.97 to obtain the amount of wet aggregate we need to order.

wet sand = (101,322 lb)/0.94 = 107,789 lb = 53.9 tons water in sand = 107,789 – 101,322 = 6467 lb wet aggregate = (215,309 lb)/(0.97) = 221,969 lb = 111.0 tons water in aggregate = 221,969 – 215,309 = 6659 lb The actual amount of water that should be added to the concrete mix is

water = 30, 080 – 6467 – 6659 = 16,954 lb gal water = (16,954 lb)(7.48 gal/ft 3 )/(62.4 lb/ft 3 ) = 2032 gal Therefore, (a) The ingredients of the concrete mix are 640 sacks of cement 53.9 tons of sand 111.0 tons of aggregate 2032 gal of water (b) The total weight per yd3 is

wt /yd 3 =

(640 sacks)(94 lb/sack) +107,789 + 221,969 +16,954 100 yd 3

= 4069 lb / yd 3 (c) The cement-sand-aggregate ratio, on a weight basis, is

ratio = (640 sacks)(94 lb/sack) :107,789 lb : 221,969 lb = 60,160 :107,789 : 221,968 = 1:1.79 : 3.69

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18-14 We plan to prepare 10 yd3 of concrete using a 1:2.5:4.5 weight ratio of cement–sand– coarse aggregate. The water–cement ratio (by weight) is 0.45. The sand contains 3 wt% water; the coarse aggregate contains 2 wt% water; and 5% entrained air is expected.

Determine the number of sacks of cement, tons of sand and coarse aggregate, and gallons of water required.

Solution: First, determine the volume of each material required, using the 1:2.5:4.5 ratio to determine the weights per sack of cement and dividing by the density to determine the volume. Per sack of cement:

cement :

(94 lb/sack)/(190 lb/ft 3 ) = 0.495 ft 3 /sack

sand :

(2.5)(94 lb/sack)/(160 lb/ft 3 ) = 1.469 ft 3 /sack

aggregate:

(4.5)(94 lb/sack)/(170 lb/ft 3 ) = 2.488 ft 3 /sack

water : (0.45)(94 lb/sack)/(62.4 lb/ft 3 ) = 0.678 ft 3 /sack Volume per sack = 5.130 ft 3 / sack But 5% of the concrete is expected to be entrained air. The volume of air “x” per sack of cement is x/(5.130 + x) = 0.05 or x = 0.27 ft3 Therefore, the total volume of concrete per sack is Volume of concrete = 5.130 + 0.27 = 5.400 ft3/sack

In 10 yd3 = 270 ft 3 : cement = 270 ft 3 /(5.400 ft 3 /sack) = 50 sacks sand = (50 sacks)(2.5)(94 lb/sack) = 11, 750 lb aggregate = (50 sacks)(4.5)(94 lb/sack) = 21,150 lb water = (50 sacks)(0.45)(94 lb/sack) = 2,115 lb We must also adjust for the water present in the wet sand (3%) and wet aggregate (2%). To find the amount of wet sand, divide by 0.97, and to find the amount of wet aggregate, divide by 0.98: wet sand = 11,750 lb/0.97 = 12,113 lb; H2O = 363 lb wet aggregate = 21,150 lb/0.98 = 21,582 lb; H2O = 432 lb Therefore, the ingredients for the concrete mix include

cement = 50 sacks sand = 12,113 lb = 6.06 tons aggregate = 21,582 lb = 10.8 tons water = 2,115 lb – 363 lb – 432 lb = 1320 lb = (1320 lb) / (62.4 lb / ft 3 ) × (7.48 gal/ft 3 ) = 158 gal

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Chapter 19: Electronic Materials 19-1

Of the metals in Table 19-1, which are the most and least conductive?

Solution: The most conductive is silver (Au) and the least conductive is gallium (Ga).

19-2

Find the resistance of a fiber with a cross-sectional area of a 1.34 mm2 and a length of 10 cm that is subjected to a voltage of 225 V for which the current density is 1.25 A/cm2.

Assume the resistor is uniform in material and properties.

Solution: To begin, the formula called for is =

Solving for the conductivity:

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=

1.25 A 10 cm 1 =

1 cm

1 225 V = 0.0556

A V ∙ cm

The units are best not simplified at this time. The next equation is =

10 cm 1 V ∙ cm 1 10 mm =

1 0.0556 A 1.34 mm 1 cm = 13,433

19-3

V = 13,433 ohm A

A wire two microns in diameter is made from 10 cm silver, 0.1 cm gallium, and 10 cm silver all joined end to end (i.e., in series). Assume the joints are perfect and have no additional resistance. What is the total resistance of the wire?

Solution: Since the sections of the wire are in series, their resistances sum, i.e.

Rtotal = RAg + RGa + RAg = 2RAg + RGa In general,

R=

ρl A

,

where ρ is the resistivity. Substituting with values from Table 19–1,

Rtotal = 2RAg + RGa =

(2 ρ Ag ) (10 cm )  2 ×10 cm  π  2   −4

2

+

( ρGa ) ( 0.1 cm )  2 ×10 −4 cm  π  2  

2

      1 1 10 cm ) +  0.1 cm ) 2  5 −1 −1  ( 5 −1 −1  (  0.66 ×10 Ω cm    6.8 ×10 Ω cm   Rtotal = = 984Ω 2  2 ×10−4 cm  π  2  

19-4

A current of 10 A is passed through a 1-mm-diameter wire 1000 m long. Calculate the power loss if the wire is made from (a) aluminum and (b) silicon (see Table 19–1).

Solution: 2

2

2

2

Power = I R = I ℓ/(σA) = (10 A) (100,000 cm)/[(π/4) (0.1 cm) σ] 9 Power = 1.273 × 10 /σ

The electrical conductivity of each material is given in Table 19–1: (a) PAl = 1.273 × 109/(3.77 × 105) = 3377 watt

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(b) PSi = 1.273 × 109/(4 × 10–6) = 3.183 × 1014 watt (c)

19-6

PSiC = 1.273 ×109 /10 –10 = 1.273 ×1019 watt

The power lost in a 2-mm-diameter copper wire is to be less than 250 W when a 5 A current is flowing in the circuit. What is the maximum length of the wire?

Solution:

P = I 2 R = I 2 ℓ /(σ A) = 250 W ℓ = 250σ A/I 2 = (250)(5.98 × 105 )(π /4)(0.2) 2 /(5) 2 = 1.88 ×105 cm = 1.88 km

19-7

A current density of 100,000 A/cm2 is applied to a gold wire 50 m in length. The resistance of the wire is found to be 2 ohm. Calculate the diameter of the wire and the voltage applied to the wire.

Solution:

J = I /A = σ V /ℓ = 100, 000 A/cm 2 V = 100, 000ℓ /σ = (100, 000)(5000 cm)/(4.26 ×105 ) = 1174 V From Ohm’s law, I = V/R = 1174/2 = 587 A A = I/J = 587/100,000 = 0.00587 cm

2

(π/4)d2 = 0.00587 or d2 = 0.00747 or d = 0.0864 cm 19-8

We would like to produce a 5000-ohm resistor from boron-carbide fibers having a diameter of 0.1 mm. What is the required length of the fibers?

Solution: The electrical conductivity is 1 to 2 ohm–1 · cm–1.

R = ℓ/σA = 5000 ohm If the conductivity is 1 ohm–1 · cm–1: ℓ = RσA = (5000)(1 ohm · cm)(π/4)(0.01 cm)2 = 0.393 cm If the conductivity is 2 ohm–1 · cm–1: ℓ = RσA = (5000)(2 ohm · cm)(π/4)(0.01 cm)2 = 0.785 cm The fibers should be 0.393 to 0.785 cm in length.

19-9

Ag has an electrical conductivity of 6.80 × 105 Ω–1 · cm–1. Au has an electrical conductivity of 4.26 × 105 Ω–1 · cm–1. Calculate the number of charge carriers per unit volume and the electron mobility in each in order to account for this difference in electrical conductivity. Comment on your findings.

Solution: The electrical conductivity σ is given by σ = nqμ, where n is the density of free electrons, q is the charge on the electron, and μ is the electron mobility. The electronic structure of silver is [Kr] 4d105s1. Thus there is one valence electron contributed per silver atom, and the number of free electrons per cubic centimeter for silver is

10.49 g mol 6.022 ×1023 atoms 1 electron × × × = 5.86 ×1022 electrons/cm3 .

3 cm 107.868 g mol 1 atom Solving for the mobility,

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µ=

σ nq

=

6.80 ×105 Ω−1cm −1 = 73 cm 2 V −1s −1 22 3 −19 (5.86 ×10 electrons/cm )(1.60 ×10 C) The electronic structure of gold is [Xe] 4f145d106s1. Thus there is one valence electron contributed per gold atom, and the number of free electrons per cubic centimeter for gold is

19.302 g mol 6.022 ×1023atoms 1 electron × × × = 5.90 ×1022 electrons/cm3 .

3 cm 196.97 g mol 1 atom Solving for the mobility,

µ=

σ nq

=

4.26 ×105Ω −1cm −1 = 45 cm2 V −1s −1.

22 3 −19 (5.90 ×10 electrons/cm )(1.60 ×10 C)

Therefore, the electrical conductivity of silver is superior to that of gold primarily due to the higher mobility for electrons in silver compared to gold.

19-10 If the current density is 1.25 A/cm2 and the drift velocity is 107 cm/s, how many charge carriers are present?

Solution: This applies the following definition of current density: = ̅ = =

̅

1.25 A 1 carrier 1 s C

( 1 cm 1.6 × 10 !" C 10 cm A ∙ s = 7.8 × 10!!

carrier cm,

19-11 A current density of 5000 A/cm2 flows through a magnesium wire. If half of the valence electrons serve as charge carriers, calculate the average drift velocity of the electrons.

Solution: The total number of valence electrons is

nT =

(2 atoms/cell) (2 electrons/atom) = 8.612 × 1022 −8 2 −8 (3.2087 × 10 ) (5.209 × 10 ) cos 30°

The actual number of charge carriers is then 4.306 × 1022.

v = J /nq = (5000 A/cm 2 )/[(4.306 × 10 22 )(1.6 × 10−19 )] = 0.7257 cm/s 19-12 We apply 10 V to an aluminum wire 2 mm in diameter and 20 m long. If 10% of the valence electrons carry the electrical charge, calculate the average drift velocity of the electrons in km/h and miles/h.

Solution:

The total number of valence electrons is

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nT =

(4 atoms/cell) (3 electrons/atom) = 1.812 ×1023 /cm3 −8 3 (4.04958 ×10 cm)

The number of electrons carrying the electrical charge is one-tenth of the total number, or 1.812 × 1022. The electric field is

E = V /ℓ = 10 V/(2000 cm) = 0.005 V/cm

σ E = nqv or v = σ E /(nq ) v = (3.77 ×105 )(0.005)/[(1.812 ×1022 )(1.6 ×10−19 )] = 0.6503 cm/s v = (0.6503 cm/s)(3600 s/h)/(105cm/km) = 0.0234 km/h v = (0.6503 cm/s)(3600 s/h)(1 in./2.54 cm)(1 ft/12 in.)(1 mile/5280 ft) = 0.0145 miles/h 19-13 In a welding process, a current of 400 A flows through the arc when the voltage is 35 V. The length of the arc is about 0.1 in., and the average diameter of the arc is about 0.18 in. Calculate the current density in the arc, the electric field across the arc, and the electrical conductivity of the hot gases in the arc during welding.

Solution: R = V/I = 35 V/(400 A) = 0.0875 ohm The electrical conductivity of the gases in the arc is

σ = ℓ /( RA) =

(0.1 in.) (2.54 cm/in.)

(0.0875 ohm) (π /4) (0.18 in. × 2.54 cm/in.)2

= 17.68 ohm −1 ⋅ cm −1 The current density J is 2

J = I/A = 400 A/[(π/4) (0.18 in. × 2.54 cm/in.) ] = 2436 A/cm

2

The electric field is E = V/ℓ = 35 V/[(0.18 in.) (2.54 cm/in.)] = 76.6 V/cm 19-15 A typical thickness for a copper conductor (known as an interconnect) in an integrated circuit is 250 nm. The mean free path of electrons in pure, annealed copper is about 40 nm. As the thickness of copper interconnects approaches the mean free path, how do you expect conduction in the interconnect is affected? Explain.

Solution: The electrical conductivity of the interconnect decreases as its thickness approaches the mean free path. The surfaces of the interconnect act as additional scattering sites for electrons, thereby decreasing the electron mobility and the electrical conductivity. (Additionally as film thickness decreases, grain size typically decreases, which also leads to increased scattering and decreased electrical conductivity.)

19-16 Calculate the resistivities of cobalt and beryllium at 505 K. Solution: First, note that room temperature or 25 °C is 298 K, and that one kelvin represents the same temperature difference as one centigrade degree. Using Equation 19-7: - = -./ 01 + 2. ∆45

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From Table 19-3, the room temperature resistivity and temperature resistivity coefficient are 6.24 × 10-6 ohm·cm and 0.0053 ohm/ohm·°C, respectively.

-67 = 06.24 × 10 8 ohm ∙ cm5 91 + :0.0053

ohm ; <505 K − 298 K@A ohm ∙ ℃

-67 = 1.3 × 10 B ohm ∙ cm For beryllium: -CD = 04 × 10 8 ohm ∙ cm5 91 + :0.0250

ohm ; <505 K − 298 K@A ohm ∙ ℃

-CD = 2.5 × 10 B ohm ∙ cm

19-17 Calculate the electrical conductivity of platinum at –200°C. Solution:

Conductivity σ is the inverse of resistivity ρ. Resistivity of metals increases linearly with temperature according to ρ = ρRT(1 + αR∆T), where ρRT is the room temperature resistivity, αR is the temperature resistivity coefficient, and ∆T is the temperature difference between the temperature of interest and room temperature (25 °C). ρRT and αR are tabulated materials properties, and from Table 19–3 in the text, for platinum, ρRT = 9.85 × 10–6 Ω · cm and αR = 0.0039 ohm/(ohm · °C).

Therefore, the conductivity at is

ρ = ρ RT (1 + α R ∆T ) = 9.85 ×10−6 Ω ⋅ cm[1 + 0.0039°C−1 (−200°C)] = 1.21×10−6 Ω ⋅ cm σ=

1

ρ

=

1 = 8.3 × 105 Ω −1 ⋅ cm −1 1.21× 10−6 Ω ⋅ cm

19-18 Calculate the electrical conductivity of nickel at –50°C and at +500°C. Solution:

ρ room = 6.84 ×10−6 ohm ⋅ cm α = 0.0069 ohm ⋅ cm/°C ρ500 = (6.84 ×10−6 )[1 + (0.0069)(500 − 25)] = 29.26 ×10−6 ohm ⋅ cm

σ 500 = 1/ρ = 1/(29.26 ×10−6 ) = 0.34 ×105ohm −1 ⋅ cm −1 ρ−50 = (6.84 ×10−6 )[1 + (0.0069) ( − 50 − 25)] = 3.3003 × 10−6 ohm ⋅ cm

σ −50 = 1/(3.3003 ×10−6 ) = 3.03 ×105ohm −1 ⋅ cm −1 19-19 The electrical resistivity of pure chromium is found to be 18 × 10–6 ohm · cm. Estimate the temperature at which the resistivity measurement was made.

–6 Solution: ρ = 12.9 × 10 ohm · cm at 0°C α = 0.0030 ohm/(ohm · °C) –6

–6

18 × 10 = (12.9 × 10 ) [1 + (0.0030) (T – 0)] 1.3954 – 1 = 0.0030T T = 131.8°C

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19-20 After finding the electrical conductivity of cobalt at 0°C, we decide we would like to double that conductivity. To what temperature must we cool the metal?

Solution:

ρ room = 6.24 ×10 −6 ohm ⋅ cm α = 0.0053 ohm/(ohm ⋅ °C) –6

–6

ρzero = (6.24 × 10 ) [1 + (0.0053) (0 – 25)] = 5.413 × 10

We wish to double the conductivity, or halve the resistivity to 2.707 × 10–6. The required temperature is 2.707 × 10–6 = (6.24 × 10–6) [1 + (0.0053)(T – 25)]

−0.566 = 0.0053(T − 25) or T = −81.8°C 19-21 From Figure 19–9(b), estimate the defect resistivity coefficient for tin in copper.

Solution:

The conductivity and resistivity of pure copper are, from Table 19–1: σ = 5.98 × 105 ohm −1 ⋅ cm −1 ρ = 1/σ = 0.1674 × 10−5 ohm ⋅ cm For 0.2 wt% Sn in copper:

xSn =

(0.2/118.69) = 0.00107 (0.2/118.69) + (99.8/63.54)

xSn(1 – xSn) = (0.00107)(1 – 0.00107) = 0.00107 For 0.2% Sn, Figure 19–10(b) shows that the conductivity is 92% that of pure copper, or σ = (5.98 × 105)(0.92) = 5.50 × 105 ρ = 1/σ = 0.182 × 10–5 ∆ρ = 0.182 × 10–5 – 0.167 × 10–5 = 0.015 × 10–5 The following table includes the calculations for other compositions:

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wt% Sn 0 0.2 0.4 0.6 0.8 1.0

xSn 0 0.00107 0.00215 0.00322 0.00430 0.00538

xSn(1 – xSn) 0 0.00107 0.00215 0.00321 0.00428 0.00535

%σ 100 92 78 69 61 54

σ 5.98 × 105 5.50 × 105 4.66 × 105 4.13 × 105 3.65 × 105 3.23 × 105

ρ 0.167 × 10–5 0.182 × 10–5 0.215 × 10–5 0.242 × 10–5 0.274 × 10–5 0.310 × 10–5

∆ρ 0 0.015 × 10–5 0.048 × 10–5 0.075 × 10–5 0.107 × 10–5 0.143 × 10–5

These data are plotted. The slope of the graph is “b”:

0.274 × 10−5 − 0.215 × 10−5 0.00428 − 0.00215 = 3 × 10−4 ohm ⋅ cm

b=

19-23 (a) Copper and nickel form a complete solid solution. Draw a schematic diagram illustrating the resistivity of a copper and nickel alloy as a function of the atomic percent nickel. Comment on why the curve has the shape that it does. (b) Copper and gold do not form a complete solid solution. At the compositions of 25 and 50 atomic percent gold, the ordered phases Cu3Au and CuAu form, respectively. Do you expect that a plot of the resistivity of a copper and gold alloy as a function of the atomic percent gold will have a shape similar to the plot in part (a)? Explain.

Solution: (a) The increase in the resistivity due to solid solution atoms for dilute solutions is ρd = b(1 – x)x = bx – bx2, where ρd is the increase in resistivity due to the defects, b is the defect resistivity coefficient, and x is the atomic fraction of the impurity or solid solution atoms present. For small x, ρd ~ bx.

As the atomic percent nickel in copper increases, the resistivity increases approximately linearly. The resistivity must reach a maximum, and as the atomic percent nickel increases further, the resistivity must approach the value of pure nickel (also in a linear fashion). This schematic behavior is shown in the figure.

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(b) At the compositions corresponding to the ordered phases, the electrical resistivity will drop significantly. The increased order relative to a solid solution decreases the number of scattering sites for electrons, thereby increasing mobility and increasing conductivity.

19-24 The electrical resistivity of a beryllium alloy containing 5 at% of an alloying element is found to be 50 × 10–6 ohm · cm at 400°C. Determine the contributions to resistivity due to temperature and due to impurities by finding the expected resistivity of pure beryllium at 400°C, the resistivity due to impurities, and the defect resistivity coefficient.

What would be the electrical resistivity if the beryllium contained 10 at% of the alloying element at 200°C? Solution:

From the data in Table 19–3, the resistivity at 400°C should be ρT = (4.0 × 10–6) [1 + (0.025)(400 – 25)] = 41.5 × 10–6 ohm·cm Consequently the resistance due to impurities is

ρ = ρT + ρ d 50 × 10 = 41.5 ×10 −6 + ρ d −6

ρ d = 8.5 × 10−6 ohm ⋅ cm Since there are 5 at% impurities present, x = 0.05, and the defect resistivity coefficient is

ρ d = bx(1 – x) or b = ρ d /[ x(1 − x)] b = 8.5 × 10–6/[(0.05)(1 – 0.05)] = 178.9 × 10–6 ohm·cm The resistivity at 200°C in an alloy containing 10 at% impurities is ρ200 = ρT + ρd –6

–6

= (4.0 × 10 ) [1 + (0.025) (200 – 25) ] + 178.9 × 10 (0.1) (1 – 0.1) –6

–6

–6

= 21.5 × 10 + 16.1 × 10 = 37.6 × 10 ohm · cm

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19-28 For germanium and silicon, compare, at 25°C, the number of charge carriers per cubic centimeter, the fraction of the total electrons in the valence band that are excited into the conduction band, and the constant n0.

Solution: For germanium:

nGe =

(8 atoms/cell) (4 electrons/atom) = 1.767 × 1023 /cm3 −8 3 (5.6575 ×10 cm)

From Table 19–5, we can find the conductivity and mobilities for germanium. The number of excited electrons is then

nconduction = σ /[q(µ e + µ h )] = 0.0233/[(1.6 ×10−19 )(3900 +1900)] = 2.506 ×1013 fraction = 2.506 ×1013 /(1.767 ×1023 ) = 1.42 ×10−10 n0 = n/exp [ − Eg /(2kT )] = 2.506 ×1013 /exp{ − 0.67/[(2)(8.617 ×10−5 )(298)]} = 1.16 ×1019 For silicon:

nSi =

(8 atoms/cell) (4 electrons/atom) = 1.998 × 1023 /cm3 (5.4307 × 10−8cm)3

nconduction = σ /[q(µ e + µ h )] = 4 ×10−6 /[(1.6 ×10−19 )(1350 + 480)] = 1.366 ×1010 fraction = 1.366 ×1010 /(1.998 ×1023 ) = 6.84 ×10−14 n0 = n/ exp[ − Eg /(2kT )] = 1.366 ×1010 / exp{-1.11/[(2)(8.617 ×10−5 )(298)]} = 3.23×1019 19-29 For germanium and silicon, compare the temperature required to double the electrical conductivities from the room temperature values.

Solution: For germanium, we wish to increase the conductivity from 0.02 to 0.04 ohm–1 · cm–1.

From Problem 19–21, n0 = 9.76 × 1018: σ = nq(μe + μh) = n0q(μe + μh) exp (–Eg/2kT) 18 –19 –5 0.04 = (9.79 × 10 ) (1.6 × 10 ) (3900 + 1900) exp {–0.67/[(2)(8.63 × 10 )T]} T = 315K = 42°C

For silicon, we wish to increase the conductivity from 4 × 10–6 to 8 × 10–6 ohm–1 · cm– 1 . From Problem 19–21, n0 = 3.22 × 1019: σ = nq(μe + μh) = n0q(μe + μh) exp (–Eg/2kT) –6 19 –19 –5 8 × 10 = (3.22 × 10 ) (1.6 × 10 ) (1350 + 480) exp {–1.11/[(2)(8.63 × 10 )T]}

T = 308 K = 35°C 19-30 Determine the electrical conductivity of silicon when 0.0001 at% antimony is added as a dopant and compare it to the electrical conductivity when 0.0001 at% indium is added.

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Solution: 0.0001 at % = 1 impurity atom per 106 host atoms.

For antimony additions (an n-type semiconductor):

(8 atoms/cell) (1 Sb atom/106Si atoms) n= = 5 ×1016 −8 3 (5.4307 ×10 cm) 16

–19

–1

–1

σ = nqμe = (5 × 10 ) (1.6 × 10 ) (1350) = 10.8 ohm · cm

For indium additions (a p-type semiconductor):

(8 atoms/cell) (1 In atom/106Si atoms) n= = 5 ×1016 −8 3 (5.4307 ×10 cm) σ = nqμ = (5 × 1016) (1.6 × 10–19) (480) = 3.84 ohm–1 · cm–1 19-31 We would like to produce an extrinsic germanium semiconductor having an electrical conductivity of 2000 ohm–1 · cm–1. Determine the amount of phosphorus and the amount of gallium required to make n- and p-type semiconductors, respectively.

Solution: For phosphorus (an n-type semiconductor): –19

n = σ/(qμe) = 2000/[(1.6 × 10 ) (3900)] = 3.205 × 10

3.205 ×1018 =

18

(8 atoms/cell) (x P atoms/106 Ge atoms) (5.6575 ×10−8cm)3 6

x = 72.5 P atoms/10 Ge atoms = 0.00725 at% P

For gallium (a p-type semiconductor): n = σ/(qμh) = 2000/[(1.6 × 10–19) (1900)] = 6.579 × 1018

6.579 ×1018 =

(8 atoms/cell) (x Ga atoms/106 Ge atoms) (5.6575 ×10−8cm)3

x = 148.9 Ga atoms/106 Ge atoms = 0.01489 at% Ga 19-32 Estimate the electrical conductivity of silicon doped with 0.0002 at% arsenic at 600°C, which is above the plateau in the conductivity-temperature curve.

Solution:

(8 atoms/cell) (2 As atoms/106 Si atoms) Nd = = 9.99 ×1016 −8 3 (5.4307 ×10 cm) From Problem 19–21, n0 = 3.22 × 1019

σ 600 = N d q µe + q ( µe + µ h )n0 exp[− Eg /(2kT )] = (9.99 × 1016 )(1.6 ×10 −19 )(1350) + (1.6 × 10 −19 )(1350 + 480)(3.22 × 1019 ) exp{−1.11/[(2)(8.63 × 10−5 )(873)]} = 21.58 + 5.96 = 27.5 ohm −1 ⋅ cm −1 19-33 Determine the amount of arsenic that must be combined with 1 kg of gallium to produce a p-type semiconductor with an electrical conductivity of 500 ohm–1 · cm–1 at 25°C. The lattice parameter of GaAs is about 5.65 Å and GaAs has the zincblende structure.

–19 18 Solution: p = σ/(qμh) = 500/[(1.6 × 10 ) (400)] = 7.81 × 10

7.81× 1018 =

(4 Ga atoms/cell) (x As atoms/Ga atom) (5.65 ×10 −8cm)3

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x = 0.0003523 As atom/Ga atom

In 1 kg of Ga, there are

1 kg Ga ×

1000 g Ga 1 mol Ga 6.022 ×1023 atoms × × = 8.64 ×1024 Ga atoms.

1 kg Ga 69.72 g Ga 1 mol Ga Thus for 1 kg of Ga,

8.64 × 1024 Ga atoms ×

0.0003523 As atoms 1 mol As 74.9216 g As × × , 23 Ga atom 6.022 × 10 As atoms 1 mol As

there is 0.379 g of As. 19-34 Calculate the intrinsic carrier concentration for GaAs at room temperature. Given that the effective mass of electrons in GaAs is 0.067me, where me is the mass of the electron, calculate the effective mass of the holes.

Solution: Table 19–5 gives the resistivity ρ of GaAs at room temperature as 4 × 108 Ω · cm. For an intrinsic semiconductor, the conductivity σ is given by σ = qni(μn + μp), where ni is the intrinsic carrier concentration and μn and μp are the electron and hole mobilities, respectively. Recalling that resistivity is the inverse of conductivity,

1

ρ

= qni ( µn + µ p ) ,

such that

ni =

1 pq( µ n + µ p )

.

For GaAs, the electron mobility is 8500 cm2 V–1 s–1, and the hole mobility is 400 cm2 V–1 s–1. Therefore,

ni =

1

ρ q(µn + µ p )

=

1 (4 × 10 Ω ⋅ cm)(1.60 × 10 C)(8500 + 400 cm 2 V −1s −1 ) −19

8

= 1.756 ×106 cm −3 ni = 1.756 ×1012 m −3 .

The intrinsic carrier concentration is also given by

 − Eg  ni = n0 exp  ,  2k BT  where Eg is the band gap energy, kB is the Boltzmann constant, T is temperature, and

 2π k BT  n0 = 2   2  h 

3/ 2

( mn* m*p )3/ 4 . *

In this expression, h is Planck’s constant, and mn* and m p are the effective masses of electrons and holes in the semiconductor, respectively. Thus,

 − Eg   2π k BT  ni = n0 exp   = 2  2  h   2k BT 

3/ 2

 − Eg  ( mn* n*p )3/ 4 exp  ,  2k BT 

and

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m*p =

ni4/3   2π k BT 3/ 2  − Eg   m 2    exp  2    h  2k BT  

4/ 3

* n

(1.756 × 1012 ) 4/3

=

  2π (1.38 × 10−23 )(300)  3/ 2    −1.43 (0.067)(9.11× 10 ) 2  exp   (2)(8.63 × 10−5 )(300)   (6.63 × 10−34 ) 2       m*p = 3.86 × 10−31 kg = 0.42me .

4/3

−31

19-35 Calculate the electrical conductivity of silicon doped with 1018 cm–3 boron at room temperature. Compare the intrinsic carrier concentration to the dopant concentration.

Solution: Table 19–5 gives the resistivity ρ of silicon at room temperature as 2.5 × 105 Ω · cm.

For an intrinsic semiconductor, the conductivity σ is given by σ = qni(μn + μp), where ni is the intrinsic carrier concentration and μn and μp are the electron and hole mobilities, respectively. Recalling that resistivity is the inverse of conductivity,

1

ρ

= qni ( µn + µ p ) ,

such that

ni =

1

ρ q( µn + µ p )

.

For Si, the electron mobility is 1350 cm2 V–1 s–1, and the hole mobility is 480 cm2 V–1 s–1. Therefore,

ni =

1 1 = 5 −19 ρ q(µn + µ p ) (2.5×10 Ω⋅ cm)(1.60 ×10 C)(1350 + 480 cm 2 V −1s−1 )

ni = 1.37 ×1010 cm −3 Thus the number of intrinsic carriers is small compared to the dopant concentration of 1018 cm–3. Therefore, for the p–type semiconductor, the conductivity is σ = Na q μp = (1018 cm–3)(1.60 × 10–19 C)(480 cm2 V–1 s–1) = 77 Ω–1 · cm–1 where Na is the number of acceptor atoms.

19-36 At room temperature, will the conductivity of silicon doped with 1017 cm–3 of arsenic be greater than, about equal to, or less than the conductivity of silicon doped with 1017 cm– 3 of phosphorus?

Solution: For an extrinsic semiconductor at room temperature, the number of free carriers is approximately equal to the number of dopant atoms. The electron mobility is the same for both cases. Thus the conductivities will be approximately the same since the dopant concentrations are the same.

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19-37 When a voltage of 5 mV is applied to the emitter of a transistor, a current of 2 mA is produced. When the voltage is increased to 8 mV, the current through the collector rises to 6 mA. By what percentage will the collector current increase when the emitter voltage is doubled from 9 mV to 18 mV? Solution: First we can find the constants I0 and B in Equation 19–14.

2 mA I 0 exp (5 mV/B ) = 6 mA I 0 exp (8 mV/B ) 0.333 = exp (–3/B) –1.0986 = –3/B or B = 2.73 mV I0 = 2/ exp (5/2.73) = 0.32 mA At 9 mV, I = 0.32 exp (9/2.73) = 8.647 mA At 18 mV, I = 0.32 exp (18/2.73) = 233.685 mA

Therefore the percentage increase in the collector current is

∆= 19-42

233.685 − 8.647 × 100 = 2600% 8.647

If we want resistivity of a nylon to be 104 ohm · cm, what weight percent carbon fibers should be added?

Solution: Referring to Figure 19-20, the curve reaches the desired resistivity at about 22 wt% carbon fibers.

19-43 Calculate the electrical conductivity of a fiber-reinforced polyethylene part that is reinforced with 20 vol% of continuous, aligned nickel fibers.

Solution: From Table 19–1, σPE = 10–15 and σNi = 1.46 × 105

σ composite = f PEσ PE + f Niσ Ni = (0.8)(10−15 ) + (0.2)(1.46 ×105 ) = 0.292 × 105 ohm −1 ⋅ cm −1 19-49 Calculate the displacement of the electrons or ions for the following conditions: (a) electronic polarization in nickel of 2 × 10–7 C/m2; (b) electronic polarization in aluminum of 2 × 10–8 C/m2; (c) ionic polarization in NaCl of 4.3 × 10–8 C/m2; and (d) ionic polarization in ZnS of 5 × 10–8 C/m2.

Solution:

n is the number of charge centers per m3: (a) For FCC nickel, a0 = 3.5167 Å and the atomic number is 28:

(4 atoms/cell) (28 electrons/atom) = 2.575 × 1030 −10 3 (3.5167 × 10 m) d = P /(nq ) n=

= (2 × 10−7 C/m 2 )/[(2.575 × 1030 m3 )(1.6 × 10−19 C)] d = 4.85 × 10−19 m (b) For FCC aluminum, a0 = 4.04958 Å and the atomic number is 13:

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n=

(4 atoms/cell) (13 electrons/atom) = 0.783 × 1030 −10 3 (4.04958 × 10 m) –8

30

–19

d = P/(nq) = (2 × 10 )/[(0.783 × 10 ) (1.6 × 10 )] –19 d = 1.60 × 10 m

(c) For NaCl, a0 = 5.56 Å, there is one charge per ion. There are four of each type of ion per cell.

n=

(4 Na ions/cell) (1 charge/ion) = 2.33 × 1028 −10 3 (5.56 × 10 m) –8

28

–19

d = P/(nq) = (4.3 × 10 )/[(2.33 × 10 ) (1.6 × 10 )] –17 d = 1.15 × 10 m

(d) For ZnS, a0 = 5.96 Å and there are two charges per ion. There are four of each type of ion per cell. The lattice parameter is

a0 = (4rZn + 4rS )/ 3 = [(4)(0.74) + (4)(1.84)]/ 3 = 5.96 Å (4 ZnS ions/cell) (2 charge/ion) = 3.78 × 1028 n= −10 3 (5.96 × 10 m) –8

28

–19

d = P/(nq) = (5 × 10 )/[(3.78 × 10 ) (1.6 × 10 )]

d = 8.26 × 10

–18

m.

19-50 A 2-mm-thick alumina dielectric is used in a 60 Hz circuit. Calculate the voltage required to produce a polarization of 5 × 10–7 C/m2.

–7 –12 Solution: P = (κ – 1)ε0 E = (κ – 1)ε0V/ℓ where ℓ = 2 mm = 0.002 m 5 × 10 = (9 – 1) (8.85 × 10 )V/(2 × –3

10 )

V = 14.1 volts 19-51 Suppose we are able to produce a polarization of 5 × 10–8 C/m2 in a cube (5 mm side) of barium titanate. Assume a dielectric constant of 3000. What voltage is produced?

–8 – Solution: P = (κ – 1)ε0 E = (κ – 1)ε0V/ℓ where ℓ = 5 mm = 0.005 m 5 × 10 = (3000 – 1) (8.85 × 10 12

)V/0.005

V = 9.4 mV 19-52 For the 14 materials with both a dielectric strength and resistivity listed in Table 19-7, plot the dielectric strength as a function of resistivity. Assume log average values for the resistivity ranges.

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Solution:

Dielectric constant as a function of resistivity Dielectric strength (106 V/m)

45 40 35 30 25 20 15 10 5 0 1E+00

1E+02

1E+04

1E+06

1E+08

1E+10

1E+12

1E+14

1E+16

1E+18

Resistivity (ohm·cm)

There is no obvious relationship between the two.

19-55 Calculate the capacitance of a parallel-plate capacitor containing five layers of mica for which each mica sheet is 1 cm × 2 cm × 0.005 cm. The layers are connected in parallel.

Solution:

The capacitance C of a parallel plate capacitor is given by

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C=

kε 0 A , t

where k is the dielectric constant, ε0 is the permittivity of vacuum, and A is the surface area of the plates, which are separated by a distance t. Thus, for mica with a dielectric constant of seven, the capacitance per layer is

(7)(8.85 × 10−12 F/m)(0.01 m × 0.02 m) = 2.478 × 10−10 F. Clayer = 0.00005 m Capacitances in parallel add. All layers have the same geometric dimensions in this case.

Thus,

Ctotal = (number of layers) (capacitance per layer) = 5(2.478 ×10−10 F) Ctotal = 1.239 ×10−9 F = 0.001239 µ F. 19-57 A force of 20 lb is applied to the face of a 0.5 cm × 0.5 cm × 0.1 cm thickness of quartz crystal. Determine the voltage produced by the force. The modulus of elasticity of quartz is 10.4 × 106 psi.

Solution:

d = (2.3 × 10−12 m/V) (100 cm/m) (1 in./2.54 cm) = 9.055 × 10−11in./V A = [0.5 cm/(2.54 cm/in.)]2 = 0.03875 in.2 t = thickness = 0.1 cm = 0.03937 in.

V /t = σ /Ed = F /(AEd ) E is the elastic modulus.

V = Ft /(AEd ) =

(20 lb) (0.03937 in.)

(0.03875 in. ) (10.4 ×106 psi) (9.055 × 10−11 in./V) 2

V = 21,578 volts 19-58 For the example hysteresis loop in Figure 19-24, what is the polarization (C/m2) if the electric field starts at 0.2 MV/m and drops to zero?

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Solution: Because the field is a positive value initially and declines to zero, the polarization curve followed is the upper curve. At E = 0 V/m, P = 0.26 C/m2.

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Chapter 20: Magnetic Materials

20-6

Calculate and compare the maximum magnetization we would expect in iron, nickel, cobalt, and gadolinium. There are seven electrons in the 4f level of gadolinium.

Compare the calculated values with the experimentally observed values.

Solution: Iron: The number of atoms/m3 is

2 atoms/cell = 0.0850 × 030 atoms/m3 10 3 (2.866 ×10 m) M = (0.0850 ×10 30 )(4 magnetons/atom)(9.274 ×10−24 A ⋅ m 2 ) = 3.15 ×10 6 A/m = 39, 604 oersted Nickel: The number of atoms/m3 is

4 atoms/cell = 0.09197 ×1030 atoms/m3 −10 3 (3.5167 ×10 m) M = (0.09197 ×10 30 )(2 magnetons/atom)(9.274 ×10 −24 A ⋅ m 2 ) = 1.706 ×10 6 A/m = 21, 437 oersted Cobalt: The number of atoms/m3 is

2 atoms/cell = 0.0903×1030 atoms/m 3 2 −10 (2.5071×10 m) (4.0686 ×10 )cos 30° M = (0.0903×10 30 )(3 magnetons/atom)(9.274 ×10−24 A ⋅ m 2 ) −10

= 2.512 ×10 6 A/m = 31, 573 oersted Gadolinium: The number of atoms/m3 is

2 atoms/cell = 0.0303 × 1030 atoms/m3 (3.6336 × 10 m) 2 (5.781× 10−10 m) cos 30° M = (0.0303×10 30 )(7 magnetons/atom)(9.274 ×10 –24 A ⋅ m 2 ) −10

=1.964 ×10 6 A/m = 24, 683 oersted 20-7

Calculate the mass of an electron by using the mass of the Bohr magneton and Planck’s constant.

Solution: Starting with Equation 20-1: =

ℎ 4

Solving for the mass of the electron: =

ℎ 4

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=

1.6 × 10 C 6.63 × 10 J ∙ s 1 1 1 A∙s

! ! % % 1 1 electron 1 4 9.274 × 10 A∙m C

Simplifying: =

9.1 × 10 % J ∙ s% N ∙ m kg ∙ m

! % 1 electron ∙ m J N ∙ s% = 9.1 × 10

kg electron

20-12 Derive the equation μr = 1 + χm using Equations 20–4 through 20–7.

Solution: In Equations 20-4 – 20-7, B is the inductance, μ is the permeability of the material in the field, μ0 is the permeability of free space, H is the magnetic field strength, M is the magnetization, and χm is the magnetic susceptibility. Substituting B = μH (Equation 20-4) into B = μ0H + μ0M (Equation 20-6), μH = μ0H + μ0M. Rearranging, (μ – μ0)H = μ0M and (μ/μ0) = 1 + (M/H).

Substituting Equations 20-7 and 20-5 into the above equation, χm = (M/H) and μr = (μ/μ0) μr = 1 + χm.

20-13 A 4-79 permalloy solenoid coil needs to produce an inductance of 1.5 Wb/m2. If the maximum allowed current is 5 mA, how many turns are required in a wire 1 m long?

Solution: The magnetic field strength H is given by H = B/μ = B/(μr μ0), where B is the inductance, μ is the permeability of the material in the field, μr is the relative permeability, and μ0 is the permeability of free space. Substituting, H = 1.5/[200,000 (4π × 10–7) ] = 5.97 A/m.

The number n of required turns of coil is calculated according to n = H l/I, where l is the length of wire and I is the current. Substituting, n = 5.97 (1)/(5 × 10–3). n = 1194.

20-14 An alloy of nickel and cobalt is to be produced to give a magnetization of 2 × 106 A/m.

The crystal structure of the alloy is FCC with a lattice parameter of 0.3544 nm.

Determine the atomic percent cobalt required, assuming no interaction between the nickel and cobalt.

Solution: Let fNi be the atomic fraction of nickel; 1 – fNi is then the atomic

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fraction of cobalt. The number of Bohr magnetons per cubic meter due to nickel and to cobalt atoms are Ni: (4 atoms/cell)(2 magnetons/atom) fNi/(3.544 × 10–10 m)3 = 0.1797 × 1030fNi Co: (4 atoms/cell)(3 magnetons/atom)(1 – fNi)/(3.544 × 10–10 m)3 = 0.2696 × 1030 (1 – fNi) M = [(0.1797 × 1030) fNi + (0.2696 × 1030)(1 – fNi)] (9.27 × 10–24) M = –0.833 × 106fNi + 2.499 × 106 = 2 × 106

f Ni = 0.60

f Co = 0.40

20-15 Estimate the magnetization that might be produced in an alloy containing nickel and 70 at% copper, assuming that no interaction occurs.

Solution:

We can estimate the lattice parameter of the alloy from those of the pure nickel and copper and their atomic fractions: ao = (0.3)(3.5167) + (0.7)(3.6151) = 3.5856 Å If the copper does not provide magnetic moments that influence magnetization, then

M=

(4 atoms/cell)(0.3 fraction Ni)(2 magnetons/Ni atom)(9.27 × 10 –24 ) (3.5856 × 10 –10 m)3

M = 0.48 ×106 A/m = 6065 Oe 20-16 An Fe–80% Ni alloy has a maximum permeability of 300,000 when an inductance of 3500 gauss is obtained. The alloy is placed in a 20-turn coil that is 2 cm in length. What current must flow through the conductor coil to obtain this field?

Solution: Since B = μH, H = B/μ = 3500 G/(300,000 G/Oe) = 0.0117 Oe = 0.928 A/m Then, I = Hℓ/n = (0.928 A/m) (0.02 m) /20 turns = 0.00093 A 20-17 An Fe–49% Ni alloy has a maximum permeability of 64,000 when a magnetic field of 0.125 oersted is applied. What inductance is obtained and what current is needed to obtain this inductance in a 200-turn, 3-cm long coil?

Solution:

B = μH = (64,000 G/Oe)(0.125 Oe) = 8000 G If we convert units, H = 0.125 Oe/[4π × 10–3 Oe/(A/m)] = 9.947 A/m I = H/n = (9.947 A/m)(0.03 m) /200 turns = 0.00149 A = 1.49 mA

20-29 The following data describe the effect of the magnetic field on the inductance in a silicon steel. Calculate the initial permeability and the maximum permeability for the material.

Solution: H (A/m) 0 A/m = 20 A/m = 40 A/m = 60 A/m =

B (Tesla) 0 Oe 0.25 Oe 0.50 Oe 0.75 Oe

0T= 0.08 T = 0.30 T = 0.65 T =

0G 800 G 3,000 G 6,500 G

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80 A/m = 100 A/m = 150 A/m = 250 A/m =

1.01 Oe 1.26 Oe 1.88 Oe 3.14 Oe

0.85 T = 0.95 T = 1.10 T = 1.25 T =

8,500 G 9,500 G 11,500 G 12,500 G

The data is plotted; from the graph, the initial and maximum permeabilities are initial permeability = 800 G/0.25 Oe = 3200 G/Oe maximum permeability = 6500 G/0.75 Oe = 8667 G/Oe 20-30 A magnetic material has a coercive field of 167 A/m, a saturation magnetization of 0.616 tesla, and a residual inductance of 0.3 tesla. Sketch the hysteresis loop for the material.

Solution: Msat = Bsat = 0.616 T = 6160 G Br = 3000 G

H c = 167 A/m × 4π × 10 –3 Oe/(A/m) = 2.1Oe

20-31 A magnetic material has a coercive field of 10.74 A/m, a saturation magnetization of 2.158 tesla, and a remanance induction of 1.183 tesla. Sketch the hysteresis loop for the material.

Solution: Bsat = Msat = 2.158 T = 21,580 G Br = 1.183 T = 11,830 G Hc = 10.74 A/m = 0.135 Oe

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20-32 Using Figure 20–16, determine the following properties of the magnetic material: (a) remanance (b) saturation magnetization (c) coercive field (d) initial permeability (e) maximum permeability (f) power (maximum BH product)

Solution: (a) remanance = 13,000 G (b) saturation magnetization = 14,000 G (c) coercive field = 800 Oe (d) initial permeability = 7000 G/1200 Oe = 5.8 G/Oe (e) maximum permeability = 14,000 G/900 Oe = 15.6 G/Oe (f) We can try several BH products in the 4th quadrant:

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12, 000 G × 450 Oe = 5.4 × 106 G·Oe 10, 000 G × 680 Oe = 6.8 × 106 G·Oe 8, 000 G × 720 Oe = 5.76 × 106 G·Oe The maximum BH product, or power, is about 6.8 × 106 G · Oe. 20-33 Using Figure 20–17, determine the following properties of the magnetic material: (a) remanance (b) saturation magnetization (c) coercive field (d) initial permeability (e) maximum permeability (f) power (maximum BH product)

Solution:

(a) remanance = 5500 G (b) saturation magnetization = 5800 G (c) coercive field = 44,000 A/m (d) (e)

initial permeability = 2, 000 G / {(40, 000 A/m)[4π × 10 –3 Oe/(A/m)]} = 4.0 G/Oe maximum permeability = 5500 G/{(40, 000 A/m)[4π × 103 Oe/(A/m)]} = 10.9 G/Oe

(f) We can try several BH products in the 4th quadrant: 4500 G × 24,000 A/m × 4π × 10–3 Oe/(A/m) = 1.36 × 106 G · Oe 4000 G × 30,000 A/m × 4π × 10–3 Oe/(A/m) = 1.51 × 106 G · Oe 3500 G × 34,000 A/m × 4π × 10–3 Oe/(A/m) = 1.50 × 106 G · Oe 3000 G × 37,000 A/m × 4π × 10–3 Oe/(A/m) = 1.39 × 106 G · Oe The maximum BH product, or power, is about 1.51 × 106 G · Oe. 20-34 Give the electron configuration of the pure elements in Table 20-3.

Solution: The four elements are gadolinium, nickel, iron and cobalt. Their configurations are: Gd: .Xe0 4f 3 5d 6s% Ni: .Ar0 3d6 4s %

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Fe: .Ar0 3d8 4s %

Fe: .Ar0 3d3 4s % 20-35 In Figure 20-9(b), what materials have the highest and lowest saturation magnetizations and coercivities?

Solution: Highest saturation magnetization: CoFe, max 2.4 T. Lowest saturation magnetization: Hard ferrites, min 0.05 T. Highest coercivity: RCO, max 106 A/m.

Lowest coercivity: High Ni, min 10-1 A/m.

20-36 Sketch the M–H loop for Fe at 300 K, 500 K, and 1000 K. Solution: Thermal energy at temperatures below the Curie temperature causes domains to randomize more easily, thereby reducing saturation magnetization, remanence, and coercive field.

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20-42 Use the data in Table 20-4 to correlate maximum permeability with coercivity. Is there a potential trend?

Solution:

Permeability v. Coercivity for selected materials 120

Coercivity (A/m)

100 y = 424887x-1.004 R² = 0.8452

80 60 40 20 0 0

100000

200000

300000

400000

500000

600000

700000

Permeability (max)

A power law does appear to describe the relationship, but this is not a very large data set. Students should be wary of drawing conclusions from it. Supermendur is the outlier data point.

20-43 What is the maximum magnetic power for Co-γ-Fe2O3?

Solution: The maximum power of magnet is the BH product. Using SI units: 9: =

0.48 Wb 75,000 A V ∙ s W J ! ! ! ! % 1 m 1 m Wb A ∙ V W ∙ s 9: = 36,000

J m

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Note that the dimensions of magnetic power are different from those of mechanical or electrical power.

20-44 Estimate the power of the Co5Ce material shown in Figure 20–14.

Solution: H

B BH 0 Oe 7500 G 0 G · Oe 2000 Oe 7500 G 15 × 106 G · Oe 2500 Oe 6000 G 15 × 106 G · Oe 3000 Oe 0G 0 G · Oe

20-45 Use the data in Table 20-6 to see if there is a possible relationship between BHmax and the Curie temperature.

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Solution:

BHmax (kJ/m3)

Maximum magnetic power v. the Curie temp.

1000 900 800 700 600 500 400 300 200 100 0 0

50

100

150

200

250

300

350

400

Curie temperature (°C)

Prudent analysis is that there are far too few data points to justify any claims. Any number of possible correlations could be developed from this.

20-46 Take the maximum power from the problem in Example 20-3 and convert it to SI units.

Solution: The maximum power is already found in cgs emu: 9:@AB = 4.2 × 108 G ∙ Oe Repairing to Table 20-2, we find conversion factors for both gauss and oersted and use them: 9:@AB =

A 4.2 × 108 G ∙ Oe 1 1 T

!

1 4 × 10 Oe ∙ m 10 G 1 9:@AB = 33

kA ∙ T = 33 kJ/m m

Note that the unit gauss can be used for both volume magnetization and inductance.

Picking the wrong one will results in a confusing and incorrect answer.

20-47 Refer to Figure 20-13. What are the coordinates Msat? Use the units of A/m for both values.

Solution: The coordinates appear to be (4.2 × 104 A/m, 1.65 × 106 A/m). Allowance should be made for difficulty in reading the graph.

20-48 What advantage does the Fe–3% Si material have compared with permalloy for use in electric motors?

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Solution: Fe–3% Si has a larger saturation inductance than permalloy, allowing more work to be done. Fe–3% Si does require larger fields, however, since the coercive field for Fe–3% Si is large and the permeability of Fe–3% Si is small compared with that of permalloy.

20-49 The coercive field for pure iron is related to the grain size of the iron by the relationship

H c = 1.83 + 4.14/ A , where A is the area of the grain in two dimensions (mm2) and Hc has units of A/m. If only the grain size influences the 99.95% iron (coercivity 0.9 oersted), estimate the size of the grains in the material. What happens to the coercivity value when the iron is annealed to increase the grain size?

Solution: Hc = 0.9 Oe /[4π × 10–3 Oe/(A/m)] = 71.62 A/m Thus, from the equation,

71.62 = 1.83 + 4.14/ A A = 4.14/69.79 = 0.0593 or

A = 0.0035 mm 2

When the iron is annealed, the grain size increases, A increases, and the coercive field Hc decreases.

20-51 Suppose we replace 10% of the Fe2+ ions in magnetite with Cu2+ ions. Determine the total magnetic moment per cubic centimeter.

Solution: From Example 20–6, the lattice parameter is 8.37 Å.

Vunit cell = (8.37 ×10 –8 cm)3 = 5.86 ×10 –22 cm3 = 5.86 ×1028 m3 In the octahedral sites, the fraction of copper atoms is 0.1, while the fraction of Fe2+ ions is 0.9. The magnetic moment is then

(8 subcells)[0.1Cu(1 magneton) + 0.9 Fe(4 magnetons)](9.27 ×10−24 A ⋅ m 2 ) 5.86 ×10−28 m3 moment = 4.68 ×105 A ⋅ m 2 /m3 = 4.68 ×105 A/m = 0.468 A ⋅ m 2 /cm3 moment =

20-52 Suppose that the total magnetic moment per cubic meter in a spinel structure in which Ni2+ ions have replaced a portion of the Fe2+ ions is 4.6 × 105 A/m. Calculate the fraction of the Fe2+ ions that have been replaced and the wt% Ni present in the spinel.

Solution: From Example 20–6, the volume of the unit cell is 5.86 × 10–28 m3. If we let x be the fraction of the octahedral sites occupied by nickel, then (1 x) is the fraction of the sites occupied by iron. Then

(8)[( x)(2 magnetons) + (1 − x)(4 magnetons)](9.27 ×10−24 A ⋅ m 2 ) moment = 4.6 ×10 = 5.86 ×10−28 m3 x = 0.185 5

Thus the number of each type of atom or ion in the unit cell is oxygen: (4 atoms/subcell)(8 subcells) = 32 Fe3+: (2 ions/subcell)(8 subcells) = 16 Fe2+: (0.815)(1 ion/subcell)(8 subcells) = 6.52 Ni2+: (0.185)(1 ion/subcell)(8 subcells) = 1.48

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The total number of ions in the unit cell is 56; the atomic fraction of each ion is

f oxygen = 32/56 = 0.5714

f Fe3+ = 16/56 = 0.2857

f Fe2+ = 6.52/56 = 0.1164

f Ni2+ = 1.48/56 = 0.0264

The weight percent nickel is (using the molecular weights of oxygen, iron and nickel):

(0.0264)(58.71) (0.5714)(16) + (0.2857)(55.847) + (0.1164)(55.847) + (0.0264)(58.71) = 4.68 wt%

%Ni =

20-60 Design a permanent magnet to lift a 1000 kg maximum load under operating temperatures as high as 750°C. Which material(s) listed in Table 20–6 will meet the above requirement?

Solution: The force F needed to lift a 1000 kg load is F = 1000 kg (9.81 m/s2) = 9810 N. The magnetization M needed to lift this load is given by the equation F = μ0 M2A/2, where μ0 is the permeability of free space and A is the cross–sectional area of the magnet. Assuming A = 1 m2, M = [2F/(μ0 A)]0.5 M = [(2 × 9810)/(4π × 10–7 × 1)]0.5 M = 124,952 A/m.

The required inductance B is thus B = μ0 M = 4π × 10–7(124,952) = 0.157 T, where μ0 is the permeability of free space. According to Table 20–6, Co-steel and Alnico-5 are below their Curie temperatures at 750 °C and meet the inductance requirement.

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Chapter 21: Photonic Materials

21-10 Write the formula for wavelength as a function of frequency, and find the frequency of a 560-nm light wave.

Solution: Beginning with Equation 21-1: h

=

h =

=

3 × 10 m 1 10 nm

s 560 nm 1 m 1 = 5.4 × 10 Hz

where the unit hertz is the same as an inverse second of periodic events. Do not confuse the hertz with the becquerel, even though they are both s-1.

21-11 What material has a wavelength of about 0.5 μm? 0.88 μm? What is the band gap energy of lead sulfate?

Solution: All of these are answered on Figure 21-1. The wavelength of 0.5 μm corresponds to cadmium sulfate, 0.88 μm is gallium arsenide. PbS has a band gap energy of about 0.5 eV. 21-12 Find the speed of light through quartz. If the light has a wavelength of 5890 Å in a vacuum, what wavelength does it have in the quartz?

Solution: Equation 21-4 is used to solve both parts of this problem. = =

→ =

3 × 10 m 1 1 s 1.55

= 1.94 × 10

m s

The refractive index of quartz was found in Table 21-1.

Knowing the wavelength, we apply the remainder of Equation 21-4: =

!!" !!" → =

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=

5890 Å 1.55

= 3800 Å 21-13 A beam of photons strikes a material at an angle of 25° to the normal of the surface.

Which, if any, of the materials listed in Table 21–1 could cause the beam of photons to continue at an angle of 18 to 20° from the normal of the material’s surface?

Solution: Assuming that the beam originally is passing through air or a vacuum, n = sin θi/sin θt = sin 25°/sin θ To exit at an angle of 18°: n = sin 25°/sin 18° = 0.4226/0.3090 = 1.367 To exit at an angle of 20°: n = sin 25°/sin 20° = 0.4226/0.3420 = 1.236 In Table 21–1, only ice, water, and Teflon have an index of refraction between 1.236 and 1.367.

21-14 A laser beam passing through air strikes a 5-cm-thick polystyrene block at a 20° angle to the normal of the block. By what distance is the beam displaced from its original path when the beam reaches the opposite side of the block?

Solution:

The index of refraction for polystyrene is 1.60. Since the incident angle θi is 20°, the angle of the beam as it passes through the polystyrene block will be

n = sinθi /sinθ t = sin20°/sin θ t = 1.6 sin θ t = 0.3420/1.6 = 0.2138

θt = 12.35°

From the sketch, we can find the expected displacement of the beam if no refraction occurs:

tan 20° = x /5 or

x = 5 tan 20° = (5)(0.3640) = 1.820 cm

We can also find the displacement of the beam when refraction occurs:

tan 12.35° = y/5 or

y = 5 tan 12.35° = (5)(0.2189) = 1.095 cm

Because of refraction, the beam is displaced 1.820–1.095 = 0.725 cm from its path had no refraction occurred.

21-15 A length of 6000 km of fiber-optic cable is laid to connect New York to London. If the core of the cable has a refractive index of 1.48 and the cladding has a refractive index of

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1.45, what is the time needed for a beam of photons introduced at 0° in New York to reach London? Assume that dispersion effects can be neglected for this calculation.

What is the maximum angle of incidence at which there is no leakage of light from the core?

Solution: The speed of the photons in the fiber is related to the speed of the photons in air by cfiber = (nairCair)/nfiber = (1.00) (3 × 108)/1.48 = 2.03 × 108 m/s where cfiber and cair are the speeds of the photons in the fiber and air, respectively, and nair and nfiber are the indices of refraction of air and the fiber, respectively. For the light to travel 6000 km, the time t needed is t = 6000 × 103 m/(2.03 × 108 m/s) = 0.03 s.

θt= 90° represents the point of zero transmission out of the core. ncledding /nfiber = sin θi /sin 90° = sin θi / 1 = sin θi sin θi = ncladding/nfibe = 1.45/1.48 θi = 78.44 The maximum allowed angle at which photons can be introduced is θ = 90°–78.44° = 11.55° 21-16 A block of glass 10-cm-thick with n = 1.5 transmits 90% of light incident on it. Determine the linear absorption coefficient (α) for this material. If this block is placed in water, what fraction of the incident light will be transmitted through it?

Solution:

The intensity of light It transmitted through a material for an incident beam with intensity I0 is given by It = (1–R)2I0 exp (–αx), where R is the reflectivity of the material, α is its linear absorption coefficient, and x is the path through which the photons move in the material. Substituting as given, 0.9 = (1–R)2 exp (–αd), and rearranging, α =–In [0.9/(1–R)2]/d.

When the block is placed in air, R is calculated as R = (n–1)2/(n + 1)2 = [(1.5–1)/(1.5 + 1)]2 = 0.04,

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where n is the index of refraction of the material (n = 1.5 for glass.) For d = 10 cm, the linear absorption coefficient for the material is α =–In [0.9/(1–0.04)2]/10 = 0.0024 cm–1.

If this block is placed in water (n = 1.333), the reflectivity will be R = (1.5–1.333)2/(1.5 + 1.333)2 = 0.0035.

The fraction of transmitted light will be

(1 − R )2 I 0 exp(−α d )/I 0 = (1 − 0.0035) 2 exp[−0.0024(10)] = 0.970 or 97% 21-17 A beam of photons passes through air and strikes a soda-lime glass that is part of an aquarium containing water. What fraction of the beam is reflected by the front face of the glass? What fraction of the remaining beam is reflected by the back face of the glass?

Solution: The fraction of the beam reflected by the front face is 2

 nglass − nair   1.50 − 1.00  2 R= = = 0.04 = 4%  n + n   1.50 + 1.00  air   glass The fraction of the remaining beam reflected from the back face of the glass is 2

 nwater − nglass   1.333 − 1.50 2 R=  =   = 0.0035 = 0.35% n + n 1.333 + 1.50   water glass   21-18 We find that 20% of the original intensity of a beam of photons is transmitted from air through a 1-cm-thick material having a dielectric constant of 2.3 and back into air.

Determine the fraction of the beam that is (a) reflected at the front surface; (b) absorbed in the material; and (c) reflected at the back surface. Determine the linear absorption coefficient of the material.

Solution:

The dielectric material has an index of refraction of

n = k = 2.3 = 1.5166 (a) The fraction of the beam reflected at the front surface is 2

2

n − n   1.5166 − 1.00  R =  material air  =   = 0.04214  nmaterial + nair   1.5166 + 1.00  (b) The fraction transmitted through the material is 0.2; therefore, the linear absorption coefficient of the material is

I t /I 0 = (1 − R ) 2 exp(−α x) = (1 − 0.04214) 2 exp[−α (1 cm)] = 0.20 exp(−α ) = 0.21798 −α = In(0.21798) = −1.5233

α = 1.5233 cm −1 After reflection, the intensity of the remaining beam is Iafter reflection = (1–0.04214)I0 = 0.95786I0 Before reflection at the back surface, the intensity of the beam is

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Iafter absorption = 0.95786 exp [(–1.5233)(1)]I0 = 0.2088I0 The fraction of the beam that is absorbed is therefore Iabsorbed = (0.95786–0.2088)I0 = 0.74906I0 (c) The fraction of the beam reflected off the back surface is I0 = Ireflected, front + Iabsorbed + Ireflected, back + I transmitted I0 = 0.04214I0 + 0.74906I0 + Ireflected, back + 0.20I0 Ireflected, back = 0.0088I0 21-19 A beam of photons in air strikes a composite material consisting of a 1-cm-thick sheet of polyethylene and a 2-cm-thick sheet of soda-lime glass. The incident beam is 10° from the normal of the composite. Determine the angle of the beam with respect to the normal of the composite as it (a) passes through the polyethylene; (b) passes through the glass; and (c) passes through air on the opposite side of the composite. By what distance is the beam displaced from its original path when it emerges from the composite?

Solution:

The figure shows how the beam changes directions and the amount that the beam is displaced from the normal to the point of entry, as it passes through each interface.

As the beam passes from air into polyethylene (which has an index of refraction of 1.52),

sin θt = sin θ i /n = sin10°/1.52 = 0.1736/1.52 = 0.1142

θt = 6.56° When the beam enters the glass (which has an index of refraction of 1.50), the new angle is

ng /nPE = sin θ t / sin γ 1.50/1.52 = sin 6.56°/ sin γ sin γ = 0.11577 or γ = 6.65° When the beam emerges from the glass back into air, the final angle is

nair /ng = sin γ / sin x 1.00/1.50 = sin 6.65°/ sin x sin x = 0.1714 or

x = 10°

When the beam reaches the polyethylene-glass interface, it has been displaced

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tan 6.56° = x / 1cm or

x = 0.115 cm

When the beam then reaches the glass-air interface, it has been displaced an additional

tan 6.65° = y / 2cm or

y = 0.233 cm

The total displacement is therefore x + y = 0.348 cm. If the beam had not been refracted, the displacement would have been

tan10° = z /3 cm or

z = 0.529 cm

The beam has therefore been displaced 0.529–0.348 = 0.181 cm from its original path.

21-20 A glass fiber (n = 1.5) is coated with Teflon. Calculate the maximum angle that a beam of light can deviate from the axis of the fiber without escaping from the inner portion of the fiber.

Solution: To keep the beam from escaping from the fiber, the angle of transmission must be at least 90°. Therefore, the maximum angle that the incoming beam can deviate from the fiber axis is

nTeflon /nglass = sinθ i /sinθt 1.35/1.50 = sinθ i /sin 90° sinθi = 0.90 or θ i = 64.16° The maximum angle is therefore 90–64.16 = 25.84°.

21-21 A beam of light in water reflects off a polished germanium sphere. What is its reflectivity?

Solution: Applying the definition of reflectivity: − ( + * %=& + ( Inserting values from Table 21-1: 4.0 − 1.333 + %=& * 4.0 + 1.333 % = 0.25

21-22 If a 10-cm-thick prism reduces the intensity of a light by 25%, what is its linear absorption coefficient?

Solution: Refer to Equation 21-12: - = - exp1−234 If the final intensity is 25% lower than the initial, it is 75% of the strength of the initial: 0.75- = - exp1−234

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Solving: 0.75 = exp1−234 −0.288 = −23 0.288 =2 10 cm

2 = 0.0288 cm7 21-23 A material has a linear absorption coefficient of 591 cm–1 for a particular wavelength.

Determine the thickness of the material required to absorb 99.9% of the photons.

Solution:

I /I 0 = 0.001 = exp(–α x) = exp(–591x) ln(0.001) = –6.9078 = –591x x = 0.0117 cm

21-24 Plot Kα as a function of the absorption edge and see if there is any likely trend. (See Table 21-2.) Try and explain the trend(s), if you find any.

Solution: Plotting the points from Table 21-2:

Kalpha v. absorption edge for various elements Absorption edge (angstrom)

9 8 7 6 5 4 3

y = 0.9595x - 0.0949 R² = 0.9998

2 1 0 0

1

2

3

4

5

6

7

8

9

Kalpha (angstrom)

This is a very marked trend. The possible confusion to the students would be in recognizing that these both increase as a function of the structure of the elements, not because the other rises. In other words, even though there is a good correlation, one is not a function of the other.

21-33 A scanning electron microscope has three settings for the acceleration voltage (a) 5 keV; (b) 10 keV; and (c) 20 keV. Determine the minimum voltage setting needed to produce Kα peaks for the materials listed in Table 21–2.

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Solution:

Given the available acceleration voltage E, the associated wavelength λ is given by E = hc/ λ, where h is Planck’s constant and c is the speed of light.

Rearranging, λ = hc/E. For the 5 keV setting, the wavelength is

λ = (6.626 ×10−34 J ⋅ s)(3 ×108 m/s)/[(5000 eV)( 1.6 ×10−19 J/eV)] = 2.48 Å. (b) For the 10 keV setting, the wavelength is

λ = (6.626 ×10−34 J ⋅ s)(3 ×10 m/s)/[(10, 000 eV)(1.6 ×10 − J/eV)] 8

19

= 1.24 Å. (c) For the 20 keV setting, the wavelength is

λ = (6.626 ×10−34 J ⋅ s)(3 ×108 m/s)/[(20, 000 eV )(1.6 ×10−19 J/eV)] = 0.62 Å.

Based on the values in Table 21-2, the minimum settings required to produce Kα peaks are 5 keV for Al, Si, and S 10 keV for Cr, Mn, Fe, Co, Ni, Cu 20 keV for Mo, W 21-34 A phosphorescent material has a relaxation time of 75 s. What intensity will it have at exactly 75 seconds after the removal of the energy source?

Solution: Referring to Equation 21-16: 9 ln & * = − - : In this case, both t and tau are identical, so: ln & * = −1 - = 0.37 -

At the relaxation time, any phosphorescent material will emit light at 37% of the initial intensity.

21-35 The relaxation time of a phosphor used for a video display is 5 × 10–2 seconds. If the refresh frequency is 60 Hz, then what is the reduction in intensity of the luminescence before it is reset to 100% by the refresh?

Solution: The intensity I as a function of time t and the initial intensity I0 is given by I = I0 exp (– t/τ),

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where τ is the relaxation time for the material. Rearranging, I/I0 = exp (–t/τ).

At 60 Hz, the refresh time t = (1/60) = 0.0167 s.

Hence, the reduction in intensity is I/I0= exp {–1/[(60)(0.05)]} I/I0 = 0.717.

Thus the intensity reduces to 71.7% of its initial value, and the reduction in intensity is 100–71.7 = 28.3%.

21-36 Calcium tungstate (CaWO4) has a relaxation time of 4 × 10–6 s. Determine the time required for the intensity of this phosphorescent material to decrease to 1% of the original intensity after the stimulus is removed.

Solution:

ln(I /I 0 ) = – t /τ ln(0.01) = – t / (4 × 10 –6 ) –4.605 = – t / (4 × 10 –6 ) t = 18.4 ×10 −6 s

21-37 The intensity of radiation from a phosphorescent material is reduced to 90% of its original intensity after 1.95 × 10–7 s. Determine the time required for the intensity to decrease to 1% of its original intensity.

Solution: We can use the information in the problem to find the relaxation time for the material.

ln(I /I 0 ) = ln(0.9) = –(1.95 × 10 −7 )/τ –0.1054 = −(1.95 ×10−7 )/τ

τ = 1.85 × 10 −6 s Then we can find the time required to reduce the intensity to I/I0 = 0.01:

ln(0.01) = – t /(1.85 × 10−6 ) −4.605 = – t /(1.85 × 10−6 )

τ = 8.52 × 10−6 s 21-38 A phosphor material with a bandgap of 3.5 eV with appropriate doping will be used to produce blue (475 nm) and green (510 nm) colors. Determine the energy level of the donor traps with respect to the conduction band in each case.

Solution:

The bandgap corresponding to the wavelengths of blue and green can be determined using the equation, E = hc/λ, where E is the bandgap energy, h is Planck’s constant, c is the speed of light, and λ is the wavelength. The bandgap required to produce blue photons Eblue is

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Eblue = (6.626 ×10−34 J ⋅ s)(3×108 m/s)/[(475×10−9 m)][(1.6 ×10−19 J/eV)] = 2.62 eV For green photons,

Egreen = (6.626 ×10−34 J ⋅ s)(3 ×108 m/s)/[(510 ×10−9 m)(1.6 ×10−19 J/eV)] = 2.44 eV The bandgap for the phosphor is 3.5 eV; hence, in both cases, a donor doping that will provide a trap at lower energy is required. For blue, the donor trap should be 3.5–2.62 = 0.88 eV below the conduction band edge, and in the case of green, the donor trap should be 3.5– 2.44 = 1.06 eV below the conduction band edge.

21-40 Determine the wavelength of photons produced when electrons excited into the conduction band of indium-doped silicon (a) drop from the conduction band to the acceptor band; and (b) then drop from the acceptor band; to the valence band (see Chapter 19).

Solution: The acceptor energy in Si–In is 0.16 eV; the energy gap in pure Si is 1.11 eV. The difference between the energy gap and the acceptor energy level is 1.11–0.16 = 0.95 eV. (a) The wavelength of photons produced when an electron drops from the conduction band to the acceptor level, an energy difference of 0.95 eV, is

(6.63×10−34 J ⋅ s)(3×1010 cm/s) λ = hc/E = = 1.31×10−4 cm −19 (0.95 eV)(1.60 ×10 J/eV) (b) The wavelength of photons produced when the electron subsequently drops from the acceptor level to the valence band, an energy difference of 0.16 eV, is

λ = hc/E =

(6.626 ×10−34 J ⋅ s)(3×1010 cm/s) = 7.76 ×10−4 cm −19 (0.16 eV)(1.60 ×10 J/eV)

21-41 Which, if any, of the semiconducting compounds listed in Chapters 19 and 21 are capable of producing an infrared laser beam?

Solution: Infrared radiation has a wavelength between 10–2 and 10–4 cm. Thus the semiconducting compound must have an energy gap that lies between the energies corresponding to these wavelength limits:

Eg =

(6.63 ×10−34 J ⋅ s)(3 ×1010 cm/s) = 0.0124eV (10−2 cm)(1.60 ×10−19 J/eV)

Eg =

(6.63 ×10−34 J ⋅ s)(3 ×1010 cm/s) = 1.24eV (10−4 cm)(1.60 ×10−19 J/eV)

The following compounds (among others) have energy gaps between 0.0124 and 1.24 eV and can therefore act as infrared lasers: InSb HgCdTe PbS

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21-42 What type of electromagnetic radiation (ultraviolet, infrared, visible) is produced when an electron recombines with a hole in pure germanium and what is its wavelength?

Solution: For pure germanium, the energy gap is 0.67 eV; the wavelength is

(6.626 ×10−34 J ⋅ s)(3×1010 cm/s) = 1.854 ×10−4 cm −19 (0.67 eV)(1.60 ×10 J/eV)

λ=

This corresponds to the infrared region of the spectrum.

21-43 Which, if any, of the dielectric materials listed in Chapter 19 would reduce the speed of light in air from 3 × 1010 cm/s to less than 0.5 × 1010 cm/s?

Solution: To reduce the speed of light to less than 0.5 × 1010 cm/s, the index of refraction must be greater than n = c/v = 3 × 1010/(0.5 × 1010) = 6 Consequently the dielectric constant k of the material must be greater than k = n2 = 62 = 36 From Table 19–7, only H2O, BaTiO3, and TiO2 have dielectric constants greater than 36.

21-44 What filter material would you use to isolate the Kα peak of the following x-rays: iron, manganese, or nickel? Explain your answer.

Solution: Iron: use a manganese filter. The absorption edge for Mn is 1.896 Å, which lies between the iron Kα peak of 1.937 Å and the Kβ peak of 1.757 Å.

Manganese: use a chromium filter. The absorption edge for Cr is 2.070 Å, which lies between the manganese Kα peak of 2.104 Å and the Kβ peak of 1.910 Å.

Nickel: use a cobalt filter. The absorption edge for Co is 1.608 Å, which lies between the nickel Kα peak of 1.660 Å and the Kβ peak of 1.500 Å.

21-45 What voltage must be applied to a tungsten filament to produce a continuous spectrum of x-rays having a minimum wavelength of 0.09 nm?

Solution:

E=

hc

λ

=

(6.626 × 10−34 J ⋅s)(3 ×1010cm/s) = 2.21×10−15J −9 (0.09 × 10 m) (100 cm/m)

E = (2.21×10−15 J)/(1.60 ×10−19 J/eV) = 13,804 V 21-46 A tungsten filament is heated with a 12,400 V power supply. What is (a) the wavelength; and (b) frequency of the highest energy x-rays that are produced?

Solution:

E = (12,400 eV)( 1.60 × 10–19 J/eV) = 1.984 × 10–15 J

1.984 × 10 –15 = hc /λ =

(6.63 × 10−34 J ⋅ s)(3 × 1010 cm/s) λ

λ = 1.00 × 10–8 cm = 1.00 Å = 0.100 nm v = c/λ = (3 × 1010 cm/s)/(1.00 × 10–8 cm) = 3.0 × 1018 s–1

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21-47 What is the minimum accelerating voltage required to produce Kα x-rays in nickel?

Solution: The wavelength of Kα x-rays in nickel is 1.66 Å = 1.66 × 10–8 cm

E=

(6.626 ×10−34 J ⋅ s)(3 ×1010cm/s) = 7,484 V (1.66 × 10−8 cm) (1.60 × 10−19 J/eV)

21-48 Based on the characteristic x-rays that are emitted, determine the difference in energy between electrons in tungsten for (a) the K and L shells; (b) the K and M shells; and (c) the L and M shells.

Solution: The energy difference between the K and L shells produces Kα x-rays.

The wavelength of these x-rays is 0.211 Å:

(6.63 × 10−34 J ⋅ s)(3 × 1010 cm/s) E ( K − L) = = 58,916 eV (0.211 × 10−8 cm) (1.60 × 10−19 J/eV) The energy difference between the K and M shells produces Kβ x-rays.

The wavelength of these x-rays is 0.184 Å:

E(K − M ) =

(6.63 × 10−34 J ⋅ s)(3 ×1010 cm/s) = 67,561eV (0.184 × 10−8cm) (1.60 ×10−19 J/eV)

The energy difference between the L and M shells produces Lα x-rays. The wavelength of these x-rays is 1.476 Å:

E(L − M ) =

(6.63 × 10−34 J ⋅ s)(3 ×1010 cm/s) = 8,422 eV (1.476 × 10−8cm) (1.60 × 10−19 J/eV)

21-49 Figure 21–18 shows the results of an x-ray fluorescent analysis in which the intensity of x-rays emitted from a material are plotted relative to the wavelength of the x-rays.

Determine (a) the accelerating voltage used to produce the exciting x-rays; and (b) the identity of the elements in the sample.

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Solution: (a) The highest energy x-rays produced have a wavelength (λswl) of about 0.5 Å. The accelerating voltage is therefore

E=

(6.626 × 10−34 J ⋅ s)(3×1010 cm/s) = 24,848 V (0.5 ×10−8 cm)(1.60 ×10−19 J/eV)

(b) The wavelengths of the characteristic x-rays are listed below. By comparison with the wavelengths of characteristic x-rays from different elements, Table 21-2, we can match the observed x-rays with the x-rays of the elements to obtain the composition of the sample.

Observed Expected Element 1.4 Å 1.392 Å — Cu Kβ 1.55 1.542 — Cu Kα 1.9 1.910 — Mn Kβ 2.1 2.104 — Mn Kα 6.7 6.768 — Si Kβ 7.1 7.125 — Si Kα The alloy must contain copper, manganese, and silicon.

21-50 Figure 21–19 shows the intensity as a function of energy of x-rays produced from an energy-dispersive analysis of radiation emitted from a specimen in a scanning electron microscope. Determine the identity of the elements in the sample.

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Solution: The energy of the first observed peak is about 2200 eV; the wavelength corresponding to this energy is

(6.63 × 10−34 J ⋅ s)(3 × 1010 cm/s) λ = hc /E = (2200 eV) (1.60 × 10−19 J/eV) = 5.651 × 10−8 cm = 5.651 Å Similarly we can find the wavelength corresponding to the energies of the other characteristic peaks. The table below lists the energies and calculated wavelengths for each peak and compares the wavelength to the characteristic radiation for different elements, from Table 21– 2.

Energy Calculated λ Expected λ Element 2,200 eV 5.651 Å 5.724 Å — MoLα 5,250 2.368 2.291 — Cr Kα 6,000 2.072 2.084 — Cr Kβ 7,000 1.776 1.790 — Co Kα 7,800 1.594 1.621 — Co Kβ 17,300 0.719 0.711 — Mo Kα 19,700 0.631 0.632 — Mo Kβ The sample must contain molybdenum, chromium, and cobalt.

21-51 Refer to Figure 21-17. Assume the material is being slowly cooled from 5000°C to absolute zero. What wavelength of photons will be emitted last before the object stops emitting light altogether?

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Solution: This can be found by drawing a line through the four temperature peaks and noting where it intersects the x-axis. This is at about 2 × 105 Å.

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Chapter 22: Thermal Properties of Materials

22-3

If the gas constant was exactly 1.987 cal/(mol·K), what is the percent error in assuming 3R is 6 cal/(mol·K)?

Solution: Beginning with the exact value:

3 = 5.961

Calculating percent difference:

cal mol ∙ K

6 − 5.961 × 100 = 0.65 % 5.961

The approximation is very close and suitable for all but highly precise work.

22-4

In older references in the English language, you may find specific heat in units of BTU/(lbm·°R), where BTU is the British thermal unit, lbm is pounds, and °R is degrees Rankine. Look up the conversion factors for these units and determine what the specific heat of water is in these units.

Solution: From any source:

1 BTU = 1055 J

1 lbm = 0.454 kg ∆ ° = 1.8∆ !

The last indicates that a temperature difference of 1 Rankine degree is 1.8 times the temperature difference in kelvins. Converting: 4814 J 1 BTU 0.454 kg 1.8 K BTU = 3.73 " " " 1 kg ∙ K 1055 J 1 lbm 1 °R lbm ∙ °R

Modern day material science is metric, but old literature frequently needs to be consulted, so being able to do these conversions is a material science skill.

22-5

Of the metals listed in Table 22-1, which have the highest and lowest specific heats?

Also list their atomic masses.

Solution: The highest specific heat is that of boron (B), whose atomic mass is 10.81.

The lowest specific heat belongs to tungsten (W), whose atomic mass is 183.85.

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22-6

Calculate the increase in temperature for 1 kg samples of the following metals when 20,000 joules of heat is supplied to the metal at 25°C: (a) Al; (b) Cu; (c) Fe; and (d) Ni.

Solution: Heat supplied = Mass × Specific Heat × ΔT, where ΔT is the temperature change. Rearranging, ΔT = Heat supplied/(Mass × Specific Heat) Metal Specific Heat [J/(kg K) ] ΔT (K) Al 899.56 22.23 Cu 384.93 51.96 Fe 443.5 45.10 Ni 443.5 45.10 *Specific heat values from Table 22-1, converted using 1 cal/(g · K) = 4184 J/(kg · K) 22-7

Calculate the heat (in calories and joules) required to raise the temperature of 1 kg of the following materials by 50°C: (a) lead; (b) nickel; (c) Si3N4; and (d) 6,6–nylon.

The heat is the product of the specific heat, weight, and temperature change. Calories can be converted to joules by multiplying by 4.184. cPb = [0.038 cal/(g · K)] (1000 g)(50 K) = 1900 cal = 7950 J

Solution:

22-8

cNi

= [0.106 cal/(g ·K)](1000 g)(50 K) = 5300 cal = 22,175 J

csilicon nitride

= [0.170 cal/(g. K)] (1000 g)(50 K) = 8500 cal = 35,564 J

c6,6 nylon

= [0.400 cal/(g. K) ] (1000 g)(50 K) =20,000 cal = 83,680 J

Calculate the temperature of a 100 g sample of the following materials (originally at 25°C) when 3000 calories are introduced: (a) tungsten; (b) titanium; (c) Al2O3; and (d) low-density polyethylene.

Solution:

(a)

W: 3000 cal = [0.032 cal/(g · °C) ](100 g)(T –25) or TW = 962.5°C

(b)

Ti: 3000 cal = [0.125 cal/(g · °C) ] (100 g)(T – 25) or TTi = 265°C

(c) Al2O3: 3000 cal = [0.200 cal/(g · °C)](100 g)(T –25) or Talumina =175°C (d) LDPE: 3000 cal = [0.550 cal/(g · °C) ](100 g)(T –25) or TLDPE = 79.5°C

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22-9

An alumina insulator for an electrical device is also to serve as a heat sink. A 10°C temperature rise in an alumina insulator 1 cm × 1 cm × 0.02 cm is observed during use.

Determine the thickness of a high density polyethylene insulator that would be needed to provide the same performance as a heat sink. The density of alumina is 3.96 g/cm3.

Solution:

The heat absorbed by the alumina is

heat = [0.200 cal/(g ⋅°C)(10°C)(3.96 g/cm3 )(1 cm ×1 cm × 0.02 cm) = 0.1584 cal The same amount of heat must be absorbed by the polyethylene, which has a density of about : 0.96 g/cm3

heat = [0.440 cal/(g ⋅°C)](10°C)(0.96 g/cm 3 )(1 cm ×1 cm × t) 0.1584 = 4.224t t = 0.0375 cm 22-10 A 200 g sample of aluminum is heated to 400°C and is then quenched into 2000 cm3 of water at 20°C. Calculate the temperature of the water after the aluminum and water reach equilibrium. Assume no heat loss from the system.

Solution:

The amount of heat gained by the water equals the amount lost by the aluminum. If the equilibrium temperature is Te,

[0.215 cal/(g ⋅°C)](400 – Te )(200 g) = [1.0 cal/(g ⋅ °C)] (Te – 20)(2000 g) 17, 200 – 43Te = 2000Te – 40, 000 Te = 28°C 22-11 A lead pipe in the basement of an old house travels 15 feet from the water heater outlet elbow to the riser up to a sink. The basement is cool and dank at 20°C, and the water that is suddenly flowing through it is 53°C. How much linear expansion does the pipe undergo?

Solution: Apply Equation 22-4:

%=

∆& =

∆& &' ∆

∆& = &' %∆

15 ft 12 in. 29.0 × 10-. 33℃ / " " 1 1 ft ℃ 1 ∆& = 0.172 in.

22-12 A 2-m-long soda-lime glass sheet is produced at 1400°C. Determine its length after it cools to 25°C.

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Solution:

∆ℓ = ℓ 0α∆T = (2 m)[9 × 10 –6 m/ ( m ⋅°C )] (25 –1400) ∆ℓ = –0.02475 m ℓf = ℓ 0 + ∆ℓ = 2 + (–0.02475) = 1.97525 m

22-13 A copper casting requires the final dimensions to be 2.5 cm × 50 cm × 10 cm. Determine the size of the pattern that must be used to make the mold into which the liquid copper is poured during the manufacturing process.

Solution: To produce a casting having particular final dimensions, the mold cavity into which the liquid copper is to be poured must be oversized. After the liquid solidifies, which occurs at 1085°C for pure copper, the solid casting contracts as it cools to room temperature. If we calculate the amount of contraction expected, we can make the original pattern used to produce the mold cavity that much larger.

The linear coefficient of thermal expansion for copper is 16.6 × 10–6/°C. The temperature change from the freezing temperature to 25°C is 25 – 1085 =– 1060°C. The change in any dimension Δl is given by Δl = αl0ΔT, where α is the thermal expansion coefficient, l0 is the initial dimension, and ΔT is the temperature change. Rearranging with Δl = lf – l0, where lf is the final dimension, lf = (1 + αΔT)l0.

Substituting, lf = (1 + αΔT)l0 = [1 + 16.6 × 10–6/°C(–1060°C) ]l0 = 0.9824 l0 or l0 = lf/0.9824.

For final dimensions of 2.5 cm × 50 cm × 10 cm, the mold needs to be 2.5/0.9824 = 2.545 cm 50/0.9824 = 50.896 cm 10/0.9824 = 10.179 cm.

22-14 A copper casting requires the final dimensions to be 1 in. × 12 in. × 24 in. Determine the size of the pattern that must be used to make the mold into which the liquid copper is poured during the manufacturing process.

Solution: From the solution of 22–9, for copper, l0 = lf/0.9824.

For final dimensions of 1 in. × 12 in. × 24 in, the mold needs to be 1/0.9824 = 1.0179 in.

12/0.9824 = 12.215 in.

24/0.9824 = 24.430 in.

22-15 An aluminum casting is made using the permanent mold process. In this process, the liquid aluminum is poured into a gray cast iron mold that is heated to 350°C. We wish to produce an aluminum casting that is 15 in. long at 25°C. Calculate the length of the cavity that must be machined into the gray cast iron mold.

Solution: The aluminum casting shrinks between the solidification temperature (660.4°C) and

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room temperature (25°C); however, the gray cast iron mold expands when it is heated from 25°C to 350°C during the casting process. The linear coefficient of thermal expansion for aluminum is 25 × 10–6/°C. The temperature change from the freezing temperature to 25°C is 25 – 660.4 =–635.4°C. The change in dimension Δl is given by Δl = αl0ΔT, where α is the thermal expansion coefficient, l0 is the initial dimension, and ΔT is the temperature change. Rearranging with Δl = lf – l0, where lf is the final dimension, lf Al = (1 + α ΔT)l0 Al. Substituting,

lf Al = (1+ α∆T )l0 Al = [1+ 25×10 –6 /°C (–635.4°C)]l0 Al

l

lf Al = 0.9841 l0 Al or l0 Al = lf Al/0.9841.

For the gray cast iron, the linear coefficient of thermal expansion is 12 × 10–6/°C. The temperature change from 25°C to 350°C is 350 – 25 = 325°C such that lf gray iron = [1 + 12 × 10–6/°C(325°C) ]l0 gray iron = 1.0039 l0 gray iron or l0 gray iron = lf gray iron/1.0039.

The initial dimension of the gray iron mold is the quantity of interest, and lf gray iron = l0 Al when the aluminum is poured. Therefore, l0 gray iron = lf gray iron/1.0039 = l0 Al/1.0039 = lf Al/0.9841/1.0039.

The final length of the aluminum casting must be 15 in. Therefore, l0 gray iron = lf Al/0.9841/1.0039 = 15/0.9841/1.0039 = 15.183 in.

22-16 A Ti–alloy strip and stainless steel strip are roll bonded at 760°C. If the total length of the bimetallic strip is 1 m long when hot rolled, calculate the length of each metal at 25°C. What is the nature of the stress (compressive or tensile) just below the surface on the Ti–alloy side and on the stainless steel side?

Solution: When the bimetallic strip is cooled to 25°C, the Ti and stainless steel will contract based on each material’s coefficient of thermal expansion. If the two metals are not bonded to one another, the change in length Δl is given by Δl = αl0ΔT, where α is the thermal expansion coefficient, l0 is the initial length, and ΔT is the temperature change. aTi = 8.6 × 10–6/°C and astainless steel = 17.3 × 10–6/°C. Substituting with ΔT = 25°C – 760°C =–735°C ΔlTi = (8.6 × 10–6/°C)(1 m)(–735°C) =–0.00632 m, and Δlstainless steel = (17.3 × 10–6/°C) (1 m)(–735°C) =–0.0127 m.

The surface of the Ti strip will be under compressive stress while that of the stainless steel will be under tensile stress.

22-17 We coat a 100-cm-long, 2-mm-diameter copper wire with a 0.5-mm-thick epoxy insulation coating. Determine the length of the copper and the coating when their temperature increases from 25°C to 250°C. What is likely to happen to the epoxy coating as a result of this heating?

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Solution: Both the copper and the epoxy expand when heated. The final length of each material, assuming that they are not bonded to one another, is

ℓ Cu = (100 cm)(16.6 × 10 –6 )(250 – 25) + 100 = 100.3735 cm ℓ epoxy = (100 cm)(55 ×10 –6 )(250 – 25) + 100 = 101.2375 cm The epoxy expands nearly 1 cm more than does the underlying copper. If the copper and epoxy are well-bonded, the epoxy coating will buckle, debond, and perhaps even flake off.

22-18 We produce a 10-in.-long bimetallic composite material composed of a strip of yellow brass bonded to a strip of Invar. Determine the length to which each material would expand when the temperature increases from 20°C to 150°C. Draw a sketch showing what will happen to the shape of the bimetallic strip.

Solution: If the two metals are not bonded to one another, the amount each would like to expand is ℓbrass = 10 + (10)(18.9 × 10–6)(150 – 20) = 10.0246 in. ℓInvar = 10 + (10)(1.54 × 10–6)(150 – 20) = 10.0020 in.

The brass expands more than the Invar; if the two are bonded together, the bimetallic strip will bend, with the Invar on the inside radius of curvature of the strip

22-21 A nickel engine part is coated with SiC to provide corrosion resistance at high temperatures. If no residual stresses are present in the part at 20°C, determine the thermal stresses that develop when the part is heated to 1000°C during use. The elastic modulus of SiC is 60 × 106 psi.

Solution: The net difference in the amount of expansion is given by Δα = αnickel – αSiC = (13.0 – 4.3) × 10–6 = 8.7 × 10–6 The thermal stresses σthermal in the SiC coating are σthermal = EΔaΔT = (60 ×106 psi)[8.7 ×10–6 in./(in.°C)] (1000 – 20) = 511,560 psi The nickel expands more than the SiC; therefore, the stresses acting on the SiC are tensile stresses. The tensile strength of SiC is only on the order of 25,000 psi (Table 15–3), so the coating will likely crack.

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22-22 Alumina fibers, 2-cm-long, are incorporated into an aluminum matrix. Assuming good bonding between the ceramic fibers and the aluminum, estimate the thermal stresses acting on the fiber when the temperature of the composite increases 250°C. Are the stresses on the fiber tensile or compressive? The elastic modulus of alumina is 56 × 106 psi.

Solution: The net difference in the expansion coefficients of the two materials is Δα = αAl – aalumina = (25 – 6.7) × 10–6 = 18.3 × 10–6 The thermal stresses on the alumina are σthermal = EΔaΔT = (56 × 106 psi) [18.3 × 10–6 in./(in.°C) ] (250°C) = 256,200 psi The aluminum expands more than the alumina; thus, the alumina fibers are subjected to tensile stresses. The alumina has a tensile strength of only about 30,000 psi (Table 15–3); consequently, the fibers are expected to crack.

22-23 A 24-in.-long copper bar with a yield strength of 30,000 psi is heated to 120°C and immediately fastened securely to a rigid framework. Will the copper deform plastically during cooling to 25°C? How much will the bar deform if it is released from the framework after cooling?

Solution: If room temperature is 25°C, then the thermal stress that develops in the restrained copper as it cools is

σ thermal = Ea∆T = (18.1×106 psi)(16.6 × 10 –6 )(120 – 25) σ thermal = 28,544 psi The thermal stress is less than the yield strength; consequently, no plastic deformation occurs in the copper as it cools. When the copper is released from its restraint, the residual stresses will be relieved by elastic deformation. The strain stored in the material will be ε = σ/E = 28,544/(18.1 × 106) = 0.001577 in./in.

The decrease in length of the copper bar is Δℓ = (24 in.)(0.001577 in./in.) = 0.0378 in.

22-28 Determine the minimum heat flux (heat flow per unit area) when a 5-mm-thick sodalime glass is presented with boiling water on one side and freezing water on the other.

Solution: Using Equation 22-6:

0 ∆ = 2″ = 3 1 ∆4

Some sources use the “q-double-prime” notation for heat flow per area.

0 0.96 W 100 ℃ 1 = / " 1 1 m∙K 1 0.005 mm 0 W = 19,200 7 1 m

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22-29 If a 1-m-long silver bar is heated at one end to 300°C and the temperature measured at the other end is 100°C, calculate the heat transferred per unit area.

Solution: The heat Q transferred across a given plane of area A per second when a temperature gradient ΔT/Δx exists is given by Q/A = k(ΔT/Δx), where k is the thermal conductivity [430 W/(m · K) for silver], T is temperature, and x is position. Substituting as given, Q/A = 430[ (300 – 100)/1] Q/A = 430(200) = 86,000 W/m2.

22-30 A 3-cm-plate of silicon carbide separates liquid aluminum (held at 700°C) from a watercooled steel shell maintained at 20°C. Calculate the heat Q transferred to the steel per cm2 of silicon carbide each second.

Solution: The temperature change through the Δx = 3 cm SiC is ΔT = 700 – 20 = 680°C. The temperature gradient is thus ΔT/Δx = 680/3 = 226.7°C/cm The thermal conductivity is 0.21 cal/(cm · s · K) ; thus, Q/A = (0.21)(226.7) = 47.6 cal/(cm2 · s) 22-31 A sheet of 0.01 in. polyethylene is sandwiched between two 3 ft × 3 ft × 0.125 in. sheets of soda-lime glass to produce a window. Calculate (a) the heat lost through the window each day when the room temperature is 25°C and the outside air is 0°C and (b) the heat entering through the window each day when the room temperature is 25°C and the outside air is 40°C. Solution:

The volume fractions of each constituent are determined from the thicknesses: fPE = 0.01 in./(0.01 + 0.125 + 0.125) = 0.01/0.26 = 0.03846 fg = (2)(0.125)/0.26 = 0.96154 The thermal conductivity of this laminar composite perpendicular to the layers is 1/k = 0.03846/0.33 + 0.96154/0.96 k = 0.894 W/(m · K) The surface area of the glass is 3 ft × 3 ft, or A = (3 ft)2(12 in./ft)2(2.54 cm/in.)2 = 8361 cm2 The thickness of the composite is Δx = (0.26 in.)(2.54 cm/in.) = 0.6604 cm (a) The heat loss to the outside is

Q = kcomposite A∆T /∆x = [0.894 W/(m ⋅ K)(1 m/100 cm)](8361 cm 2 )(25 K/0.6604 cm) = 2830 W or Q = (2830 W)(3600 s/h)(24 h/day) = 2.45 × 108 J/day (b) The heat entering the room from outside is

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Q = [(0.894 W/ ( m ⋅ K ) (1 m/100 cm)](8361 cm 2 )(40 – 25)/(0.6604 cm) = 1698 W = 1.47 ×108 J /day 22-32 We would like to build a heat-deflection plate that permits heat to be transferred rapidly parallel to the sheet but very slowly perpendicular to the sheet. Consequently, we incorporate 1 kg of unidirectional copper wires, each 0.1 cm in diameter, into 5 kg of a polyimide polymer matrix. Estimate the thermal conductivity parallel and perpendicular to the wires.

Solution: We can first calculate the volume fractions of the two constituents in the composite. The volume of each material is

VCu = 1000 g/(8.93 g/cm3 ) = 111.98 cm3 VPI = 5000 g/(1.39 g/cm3 ) = 3597.12 cm3 f Cu = 111.98/(111.98 + 3597.12) = 0.030 f PI = 0.970 Parallel to the wires: k = 0.030/ [400 W/(m · K) ] + 0.970/ [0.21 W/(m · K) ] k = 12.2 W/(m · K) Perpendicular to the wires:

1/k = 0.030[400 W/(m ⋅ K)] + 0.970[0.21 W/(m ⋅ K)] k = 0.22 W/(m ⋅ K) The thermal conductivity is much higher parallel to the conductive copper wires than perpendicular to the wires.

22-33 An exothermic reaction at a battery electrode releases 80 MW of power per square meter. The electrode consists of a 5-μm-thick graphite layer on a 20-μm-thick copper foil; the reaction occurs at the graphite surface; and heat needs to be transferred out through the graphite to the copper foil. The temperature on the outer side of the copper foil is maintained at 30°C. What is the temperature on the inner surface of the electrode?

Solution: Based on the thickness of the graphite (gr) and copper (Cu) layers, the volume fraction of each is fgr = 5/(20 + 5) = 0.2 and fCu = 20/25 = 0.8.

The thermal conductivities for graphite and copper from Table 22-3 are kgr = 335 W/(m · K) and kCu = 400 W/(m · K). The thermal conductivity of the composite graphite-copper electrode (Chapter 17) is given by (1/k) = (fgr/kgr) + (fCu/kCu).

Substituting, (1/k) = (0.2/335) + (0.8/400) = 0.002597, such that k = 385 W/(m · K).

The heat Q transferred per second across a given plane of area A when a temperature gradient ΔT/Δx exists is given by Q/A = k(ΔT/Δx),

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where T is temperature, ΔT = Tinner – Touter, and x is position.

Substituting as given, 80 × 106 = 385 (Tinner – 30)/(25 × 10–6) Tinner = 35.2°C. 22-34 Suppose we just dip a 1-cm-diameter, 10-cm-long rod of aluminum into one liter of water at 20°C. The other end of the rod is in contact with a heat source operating at 400°C. Determine the length of time required to heat the water to 25°C if 75% of the heat is lost by radiation from the bar.

Solution: The heat required to raise the temperature of the water by 5°C is Heat = [1 cal/(g · K) ] (1000 g)(25 – 20) = 5000 cal Since 75% of the heat is lost by radiation, we must supply a total of Heat = 4 × 5000 = 20,000 cal The heat flux Q is energy per area per time; thus, Heat/t = k A ΔT/Δx

20, 000 [0.57 cal/(cm ⋅ s ⋅ k)](π /4)(1 cm)2 (400 – 20) = t 10 cm t = 1176 s = 19.6 min 22-35 Determine the Lorenz number using the data for silver found in this book. What is your percent difference from the given value and your explanation for any difference?

Solution: Using Equation 22-7:

8=

3 9

From Table 22-3, silver’s thermal conductivity is 430 W/m·K. From Table 19-1, the conductivity is 6.80 × 105 Ω–1·cm–1 at 300 K. Inserting values: 8=

430 W Ω ∙ cm 1 1 m / " " : 1 m ∙ K 6.80 × 10 300 K 100 cm

8 = 2.1 × 10-;

W∙Ω cal ∙ Ω -< = 5.04 × 10 K7 s ∙ K7

Calculating the percent difference, or percent error in this case since one value is known to be more accurate than the other: >

cal ∙ Ω cal ∙ Ω − 5.5 × 10-< s ∙ K7 s ∙ K 7 > × 100 % = 8.4 % cal ∙ Ω 5.5 × 10-< s ∙ K7

5.04 × 10-<

The Lorenz number is an empirically-, not theoretically-, derived calculation and should be interpreted as providing an approximation. It also slightly varies as a function of temperature.

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22-37 Gray cast iron has a higher thermal conductivity than ductile or malleable cast iron.

Review Chapter 13 and explain why this difference in conductivity might be expected.

Solution:

The thermal conductivities of the constituents in the cast irons are kgraphite = 335 W/(m.K) kferrite = 75 W/(m · K) kcementite = 50 W/(m · K) The gray cast iron contains interconnected graphite flakes, while the graphite nodules in ductile and malleable iron are not interconnected.

Graphite, or carbon, has a higher thermal conductivity than does the “steel” matrix of the cast iron. Consequently heat can be transferred more rapidly through the iron-graphite “composite” structure of the gray iron than through the ductile and malleable irons.

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Chapter 23: Corrosion and Wear

23-1

Explain how it is possible to form a simple electrochemical cell using a lemon, a penny, and a galvanized steel nail.

Solution: The penny serves as the cathode (copper is the cathode). The steel nail with a zinc coating serves as the anode (zinc is the anode), and the lemon serves as the electrolyte. A voltmeter or wire connecting the penny and nail completes the circuit.

23-3

A gray cast iron pipe is used in the natural gas distribution system for a city. The pipe fails and leaks, even though no corrosion noticeable to the naked eye has occurred.

Offer an explanation for why the pipe failed.

Solution: Because no corrosion is noticeable, the corrosion byproduct apparently is still in place on the pipe, hiding the corroded area. The circumstances suggest graphitic corrosion, an example of a selective chemical attack. The graphite flakes in the gray iron are not attacked by the corrosive soil, while the iron matrix is removed or converted to an iron oxide or hydroxide trapped between the graphite flakes.

Although the pipe appears to be sound, the attacked area is weak, porous, and spongy. The natural gas can leak through the area of graphitic corrosion and accumulate, eventually leading to an explosion.

23-4

A brass plumbing fitting produced from a Cu-30% Zn alloy operates in the hot water system of a large office building. After some period of use, cracking and leaking occur.

On visual examination, no metal appears to have been corroded. Offer an explanation for why the fitting failed.

Solution: The high zinc brasses are susceptible to dezincification, particularly when the temperature is increased, as in the hot water supply of the building. One of the characteristics of dezincification is that copper is redeposited in the regions that are attacked, obscuring the damage. The redeposited copper is spongy, brittle, and weak, permitting the fitting to fail and leak. Therefore dezincification appears to be a logical explanation.

23-5

Plot the electronegativity of the elements listed in Table 23–1 as a function of electrode potential. (Refer to Chapter 2 for electronegativity values.) What is the trend that you observe in this data? Using an example, explain how the electronic structure of a metal is consistent with its anodic or cathodic tendency.

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Solution:

In general, as electronegativity increases, the tendency for the elements to corrode decreases. Anodic elements are more electropositive (they have a low electronegativity), meaning that they want to give up their electrons. Cathodic elements are electronegative, i.e., they want to gain electrons. For example, consider lithium. Li has the electronic structure [Li] = [He] 2s1. Lithium wants to give up the 2s1 electron to achieve a stable configuration and has anodic tendencies.

23-6

Write the anodic or cathodic reaction (as appropriate) for aluminum, germanium and chromium.

Solution: → + 3

→ + 4 → + 3

23-7

Suppose 10 g of Sn2+ are dissolved in 1000 mL of water to produce an electrolyte.

Calculate the electrode potential of the tin half-cell.

Solution: The concentration of the electrolyte is C = 10 g/(118.69 g/mol)/(1 L) = 0.0843 M The electrode potential from the Nernst equation is E = –0.14 + (0.0592/2) log (0.0843) = –0.172 V Note that the logarithm is to the base 10 in this equation.

23-8

Use the Nernst equation to determine the concentration of chromium in water if the difference from standard potential is +0.016 volts.

Solution: The Nernst equation:

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= +

0.0592 log

With data from the problem, and for chromium: +0.016 ! =

0.0592 log 3

Rearranging and solving: 0.811 = log

= 6.5 M

23-9

A half-cell produced by dissolving copper in water produces an electrode potential of +0.32 V. Calculate the amount of copper that must have been added to 1000 mL of water to produce this potential.

Solution: From the Nernst equation, with E0 = 0.34 V and the molecular weight of copper of 63.54 g/mol, we can find the number of grams “x” added to 1000 ml of the solution.

For Cu2+, n = 2.

0.32 = 0.34 + (0.0592/2) log (x/63.54) log (x/63.54) = (–0.02)(2)/0.0592 = –0.67568

x /63.54 = 0.211 or

x = 13.4 g Cu per 1000 mL H 2 O

23-10 An electrode potential in a platinum half-cell is 1.10 V. Determine the concentration of Pt4+ ions in the electrolyte.

Solution: From the Nernst equation, with E0 = 1.20 V and the molecular weight of platinum of 195.09 g/mol, we can find the amount “x” of Pt added per 1000 mL of solution. For Pt, n = 4.

1.10 = 1.20 + (0.0592/4) log (x/195.09) log (x/195.09) = –6.7568 x/195.09 = 0.000000175 x = 0.000034 g Pt per 1000 mL H2O 23-11 Use Faraday’s equation to determine the time for a 5 A current to completely corrode away a block of silver weighing 50 grams.

Solution: Using Faraday’s equation: $=

%&' F

Rearranged: $ F =& %' Solving:

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&=

50 g 96,500 C 1 1 mol A ∙ s ) ) ) ) 1 1 mol 5 A 107.868 g C & = 8946 s = 2.5 hr

23-12 A current density of 0.05 A/cm2 is applied to a 150 cm2 cathode. What period of time is required to plate out a 1-mm-thick coating of silver onto the cathode?

Solution: The current in the cell is I = iA = (0.05 A/cm2)(150 cm2) = 7.5 A The weight of silver that must be deposited to produce a 1 mm = 0.1 cm thick layer is w = (150 cm2)(0.1 cm)(10.49 g/cm3) = 157.35 g From the Faraday equation, with n = 1 for silver:

157.35 g = (7.5 A)(t )(107.868 g/mol)/[(1)(96,500 C)] t = 18, 769 s = 5.21 h 23-13 We wish to produce 100 g of platinum per hour on a 1000 cm2 cathode by electro plating. What plating current density is required? Determine the current required.

Solution: In the Faraday equation, n = 4 for platinum, which has an atomic weight of 195.09 g/mol. The weight of platinum that must be deposited per second is 100 g/(3600 s/h) = 0.02778 g/s.

0.02778 g/s = (i )(1000 cm 2 )(195.09 g/mol)/[(4)(96,500 C)] i = 0.055 A/cm 2 The current must then be I = iA = (0.055 A/cm2)(1000 cm2) = 55 A 23-14 A 1-m-square steel plate is coated on both sides with a 0.005-cm-thick layer of zinc. A current density of 0.02 A/cm2 is applied to the plate in an aqueous solution. Assuming that the zinc corrodes uniformly, determine the length of time required before the steel is exposed.

Solution: The surface area includes both sides of the plate: A = (2 sides) (100 cm) (100 cm) = 20,000 cm2 The weight of zinc that must be removed by corrosion is w = (20,000 cm2)(0.005 cm)(7.133 g/cm3) = 713.3 g From Faraday’s equation, where n = 2 for zinc:

(0.02 A/cm 2 )(20, 000 cm 2 )(t )(65.38 g/cm3 ) (2)(96,500 C) t = 5264 s = 1.462 h

713.3 g =

23-15 A 2-in.-inside-diameter, 12-ft-long copper distribution pipe in a plumbing system is accidentally connected to the power system of a manufacturing plant, causing a current of 0.05 A to flow through the pipe. If the wall thickness of the pipe is 0.125 in., estimate the time required before the pipe begins to leak, assuming a uniform rate of corrosion.

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Solution: If the pipe corroded uniformly, essentially all of the pipe would have to be consumed before leaking. The volume of material in the pipe, which has an inside diameter of 2 in. and an outside diameter of 2.25 in., is V = (π/4)[(2.25 in.)2 – (2 in.)2](12 ft)(12 in./ft) = 120.17 in.3 = 1969.2 cm3 The density of copper is 8.93 g/cm3. The weight of material to be corroded is w = Vρ = (1969.2 cm3) (8.93 g/cm3) = 17,585 g From Faraday’s law, with n = 2 for copper: 17,585 g = (0.05 A) (t) (63.54 g/mol)/[(2) (96,500 C)] t = 1.07 × 109 s = 34 years 23-16 A steel surface 10 cm × 100 cm is coated with a 0.002-cm-thick layer of chromium. After one year of exposure to an electrolytic cell, the chromium layer is completely removed.

Calculate the current density required to accomplish this removal.

Solution: The volume and weight of chromium that must be removed are V = (10 cm) (100 cm) (0.002 cm) = 2 cm3 w = (2 cm3) (7.19 g/cm3) = 14.38 g There are 31.536 × 106 s per year. The surface area of the chromium is (10 cm) (100 cm) = 1000 cm2. From Faraday’s law, with n = 3 for chromium:

14.38 g =

(i )(1000 cm 2 )(31.536 × 106 s)(51.996 g/mol) (3)(96, 500 C)

i = 2.54 ×10 –6 A/cm 2 The current is (2.54 × 10–6A/cm2)(1000 cm2) = 2.54 × 10–3A = 2.54 mA 23-17 A corrosion cell is composed of a 300 cm2 copper sheet and a 20 cm2 iron sheet, with a current density of 0.6 A/cm2 applied to the copper. Which material is the anode? What is the rate of loss of metal from the anode per hour?

Solution: In the Cu-Fe cell, the iron is the anode.

I = iCuACu = iFeAFe (0.6 A/cm2) (300 cm2) = iFe(20 cm2) iFe = 9 A/cm2 The weight loss of iron per hour (3600 s) is

w = (9 A/cm 2 )(20 cm 2 )(3600 s)(55.847 g/mol)/[(2)(96, 500 C)] = 187.5 g of iron lost per hour 23-18 A corrosion cell is composed of a 20 cm2 copper sheet and a 400 cm2 iron sheet, with a current density of 0.7 A/cm2 applied to the copper. Which material is the anode? What is the rate of loss of metal from the anode per hour?

Solution:

In the Cu-Fe cell, the iron is the anode.

I = iCu ACu = iFeAFe (0.7 A/cm2) (20 cm2) = iFe(400 cm2) iFe = 0.035 A/cm2 The weight loss of iron per hour (3600 s) is

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w = (0.035 A/cm 2 )(400 cm 2 )(3600 s)(55.847 g/mol)/[(2)(96,500 C)] = 14.58 g of iron lost per hour Note that the rate of iron lost per hour when the anode area is large is much smaller than the rate of iron loss when the anode area is small (Problem 23-14).

23-19 List the only the pure elements of the galvanic series in seawater from cathodic to anodic in Table 23-2.

Solution: This comes from Table 23-3. Working up the table: Pt, Au, C, Ti, Ag, Cu, Sn, Pb, Cd, Zn and Mg. 23-20 Provide at least two reasons as to why steel alone in water with no other metal present will corrode.

Solution: Steel alone in water will corrode because it consists of two phases: ferrite and cementite. The ferrite is anodic to the cementite, and therefore, corrodes.

Additionally, steel is polycrystalline and contains grain boundaries. The grain boundaries are anodic to the interior of the grains and corrode preferentially.

23-21 Write the anode and cathode half-cell reactions caused by the precipitation of chromium carbide from austenitic stainless steels.

Solution: → + 3

3 3 + 1.554 3 + 3 → 36357 2 4 23-22 Explain why titanium is a material that is often used in biomedical implants even though it is one of the most anodic materials listed in Table 23-1.

Solution: Titanium rapidly corrodes in air forming an oxide on the surface. This is why it appears at the anodic end of the galvanic series. The oxide, however, acts as a barrier to further corrosion, thereby making titanium a relatively corrosion-resistant metal and safe to use in biomedical implants over long periods of time.

23-23 Alclad is a laminar composite composed of two sheets of commercially pure aluminum (alloy 1100) sandwiched around a core of 2024 aluminum alloy. Discuss the corrosion resistance of the composite. Suppose that a portion of one of the 1100 layers was machined off, exposing a small patch of the 2024 alloy. How would this affect the corrosion resistance? Explain. Would there be a difference in behavior if the core material were 3003 aluminum? Explain.

Solution: The Alclad composed of 2024 and 1100 alloys should have good corrosion resistance under most circumstances. The 1100 alloy has good corrosion resistance, since it is nearly pure aluminum, when it completely covers the underlying 2024 alloy.

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Furthermore, if the 1100 layer is disturbed by machining, scratching, or other means, the 1100 alloy serves as the anode and protects the 2024 alloy. The surface area of the 1100 alloy is large, and even corrosion of the 1100 alloy will be slow.

When the 3003 alloy is coated with 1100 alloy, a disturbance of the surface is more critical. The 3003 alloy will serve as the anode; since the exposed surface area of the 3003 anode is likely to be small compared to the surface area of the 1100 alloy, corrosion may occur rapidly.

23-24 The leaf springs for an automobile are formed from a high-carbon steel. For best corrosion resistance, should the springs be formed by hot working or cold working?

Explain. Would corrosion still occur even if you use the most desirable forming process?

Explain.

Solution: If we form the springs cold, residual stresses are likely to be introduced into the product, leading to a stress cell and corrosion of the spring. Forming the springs hot will reduce the level of any residual stresses introduced into the spring and minimize the stress cell.

The steel will contain ferrite and pearlite (forming a composition cell), not to mention grain boundaries, inclusions, and other potential sites for corrosion cells.

Corrosion is still likely to occur even if the springs were produced by hot working.

23-25 Several types of metallic coatings are used to protect steel, including zinc, lead, tin, aluminum, and nickel. In which of these cases will the coating provide protection even when the coating is locally disrupted? Explain.

Solution: Aluminum and zinc are anodic compared to iron; consequently these elements should provide protection (serving as sacrificial anodes) to the iron even if the coating is disrupted.

Nickel, tin, and lead are cathodic compared to iron; when these coatings are disrupted, small anodic regions of iron are exposed and corrosion may occur rapidly.

23-26 An austenitic stainless steel corrodes in all of the heat-affected zone (HAZ) surrounding the fusion zone of a weld. Explain why corrosion occurs and discuss the type of welding process or procedure that might have been used. What might you do to prevent corrosion in this region?

Solution: Since the entire heat-affected zone has corroded, the entire heat affected region must have been sensitized during the welding process. Sensitization means that chromium carbides have precipitated at the austenite grain boundaries during joining, reducing the chromium content in the austenite near the carbides. The low chromium content austenite serves as the anode and corrosion occurs.

Based on our observation of the corrosion, we can speculate that the austenitic stainless steel is not a low-carbon steel (that is, the steel contains more than 0.03%C). The welding process was such that the heat-affected zone experienced long exposure times and slow rates of cooling. Nearest the fusion zone, the steel was all austenite at the peak temperatures developed during welding; however, the slow rate of cooling provided ample time for carbide precipitation as the region cooled between 870 and 425°C. Farther from the fusion zone, the steel was exposed to the 870 to 425°C temperature range for a long time, permitting carbides to precipitate

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and sensitize the steel.

The long times and slow cooling rates may have been a result of the welding process—a low rate of heat-input process, or inefficient process, will introduce the heat slowly, which in turn heats up the surrounding base metal, which then acts as a poor heat sink. Preheating the steel prior to welding would also result in the same problems.

The stainless steel should be welded as rapidly as possible, using a high rate of heatinput process, with no preheat of the steel prior to welding.

The steel should be a low-carbon steel to ensure that carbides do not precipitate even when the cooling rates are slow. If problems persist, a quench anneal heat treatment might be used to redissolve the carbides.

23-27 A steel nut is securely tightened onto a bolt in an industrial environment. After several months, the nut is found to contain numerous cracks. Explain why cracking might have occurred.

Solution: When the nut is tightly secured onto the bolt, residual stresses are likely to be introduced into the assembly. The presence of numerous cracks suggests that stress corrosion cracking may have occurred as a result of these stresses.

23-28 The shaft for a propeller on a ship is carefully designed so that the applied stresses are well below the endurance limit for the material. Yet after several months, the shaft cracks and fails. Offer an explanation for why failure might have occurred.

Solution: The propeller is under a cyclical load during operation, but it is also in a marine environment, which may provide a relatively aggressive electrolyte. Failure, it is noted, requires some time to occur. Corrosion- fatigue sounds like a strong possibility in this case. Even though the stress is nominally below the endurance limit for the shaft, corrosion encouraged by the stress will lead to loss of material or development of pits in the shaft. This will increase the stress acting on the shaft and further encourage corrosion. The result is the eventual formation of fatigue cracks, encouraged by corrosion, which cause the shaft to fail.

23-29 An aircraft wing composed of carbon fiber-reinforced epoxy is connected to a titanium forging on the fuselage. Will the anode for a corrosion cell be the carbon fiber, the titanium, or the epoxy? Which will most likely be the cathode? Explain.

Solution: Titanium is expected to serve as the anode and corrode, while carbon is expected to be the cathode. Titanium is more anodic than carbon, or graphite. Both are electrical conductors, they are in physical contact with one another at the junction between the wing and the fuselage, and both can be exposed to the environment.

The epoxy should not participate in the electrochemical cell; epoxy is an electrical insulator.

23-30 The inside surface of a cast iron pipe is covered with tar, which provides a protective coating. Acetone in a chemical laboratory is drained through the pipe on a regular basis.

Explain why, after several weeks, the pipe begins to corrode.

Solution: During use, the acetone serves as a solvent for the tar; the protective tar coating is eventually dissolved and the cast iron pipe is then exposed to any corrosive material

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that is drained through the pipe.

23-31 A cold-worked copper tube is soldered, using a lead-tin alloy, into a steel connector.

What types of electrochemical cells might develop due to this connection? Which of the materials would you expect to serve as the anode and suffer the most extensive damage due to corrosion? Explain.

Solution: Several cells may develop. Composition cells include those between the solder and the steel, with the steel serving as the anode and the solder as the cathode. The steel then corrodes.

A composition cell may also develop between the copper and the solder. In this case, the solder will act as the anode.

Microcomposition cells may also develop. The steel contains ferrite and cementite; the ferrite will act as the anode. In addition, the lead-tin solder is a two-phase alloy containing nearly pure tin (β) and a solid solution of tin in lead (α). Lead is most likely to serve as the anode with respect to tin.

A concentration cell is also possible, particularly in the crevice between the copper tube and the steel. The material adjacent to the crevice will act as the anode.

Finally, the copper tube is cold worked; the cold working may cause a stress cell to develop. This may be accentuated by the soldering process; during soldering, the copper tube at the soldered joint will heat, perhaps to a high enough temperature to allow stress relieving to occur. This again helps to provide the stress cell between the cold worked and stress relieved portions of the tube.

23-32 Pure tin is used to provide a solder connection for copper in many electrical uses. Which metal will most likely act as the anode?

Solution: From the galvanic series, we find that tin is anodic to copper; therefore, the tin anode is expected to corrode while the copper cathode is protected.

23-33 Sheets of annealed nickel, cold-worked nickel, and recrystallized nickel are placed into an electrolyte. Which is the most likely to corrode? Which is the least likely to corrode?

Explain.

Solution: The cold-worked nickel sheet is expected to have the poorest corrosion resistance due to the residual stresses introduced during the cold working process.

The annealed nickel sheet should be most resistant to corrosion; the grain size is expected to be particularly large and no residual stresses are expected; consequently, a stress cell is unlikely. In addition, the annealed sheet is expected to have the most uniform composition, that is, the least segregation, so a composition cell is also unlikely.

The recrystallized nickel sheet should have intermediate corrosion resistance; the residual stresses should have been eliminated as a result of the heat treatment but the grain size may be smaller than in the annealed sheet.

23-34 A pipeline carrying liquid fertilizer crosses a small creek. A large tree washes down the creek and is wedged against the steel pipe. After some time, a hole is produced in the pipe at the point where the tree touches the pipe, with the diameter of the hole larger

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on the outside of the pipe than on the inside of the pipe. The pipe then leaks fertilizer into the creek. Offer an explanation for why the pipe corroded.

Solution: One possibility for the corrosion is a concentration cell caused by microbial corrosion. The point of contact between the tree and the pipe produces a low oxygen environment and also a location at which various microbes may grow. As the microbes grow on the pipe, a low oxygen environment is further encouraged. A galvanic cell is produced, with the pipe beneath the fallen tree (and thus the microbes) serving as the anode and the remainder of the pipe acting as the cathode.

Localized corrosion will then continue until a hole is corroded through the wall of the pipe.

23-35 Two sheets of a 1040 steel are joined together with an aluminum rivet (Figure 23–20).

Discuss the possible corrosion cells that might be created as a result of this joining process. Recommend a joining process that might minimize corrosion of these cells.

Solution: Composition cells: Aluminum may act as the anode in comparison to the steel, thus causing corrosion of the aluminum. In addition, ferrite may serve as the anode to cementite within the steel.

Stress cells: The aluminum rivet is deformed when the joint is produced, causing the most highly cold-worked portion of the rivet to act as the anode. In addition, grain boundaries in both the steel and the aluminum may act as anodes for a stress cell.

Concentration cells: Crevice corrosion may occur between the two steel sheets and also between the aluminum rivet and the steel sheets.

A fusion welding process, in which a filler material having a composition similar to that of the 1040 steel, might be the best way to minimize most of these cells.

23-36 Figure 23–21 shows a cross-section through an epoxy-encapsulated integrated circuit, including a microgap between the copper lead frame and the epoxy polymer. Suppose chloride ions from the manufacturing process penetrate the package. What types of corrosion cells might develop? What portions of the integrated circuit are most likely to corrode?

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Solution: Composition cells can develop between gold and aluminum (with the aluminum serving as the anode and corroding); gold and copper (with the copper serving as the anode and corroding); and aluminum and silicon (with aluminum serving as the anode and corroding).

23-37 A current density of 0.1 A/cm2 is applied to the iron in an iron-zinc corrosion cell.

Calculate the weight loss of zinc per hour (a) if the zinc has a surface area of 10 cm2 and the iron has a surface area of 100 cm2 and (b) if the zinc has a surface area of 100 cm2 and the iron has a surface area of 10 cm2.

Solution: (a) I = iFeAFe = (0.1 A/cm2)(100 cm2) = 10 A

wZn = (10 A)(3600 s)(65.38 g/mol)/[(2)(96,500 C)] = 12.2 g of Zn lost per hour (b) I = iFeAFe = (0.1 A/cm2)(10 cm2) = 1 A

wZn = (1 A)(3600 s)(65.38 g/mol)/[(2)(96, 500 C)] = 1.22 g of Zn lost per hour The loss of zinc is accelerated when the zinc anode area is small.

23-38 Determine the Pilling-Bedworth ratio for the following metals and predict the behavior of the oxide that forms on the surface. Is the oxide protective, does it flake off the metal, or is it permeable? (See Appendix A for the metal density.)

Solution: The Pilling-Bedworth relation is P-B = Moxideρmetal/(n Mmentalρoxide) From metal density data listed in Appendix A, the calculated P-B ratios are shown in the following table.

Oxide density (g/cm3) Metal density (g/cm3) n P-B ratio Mg-MgO 3.60 1.738 1 0.80 Na-Na2O 2.27 0.967 2 0.57 Ti-TiO2 5.10 4.507 1 1.47 Fe-Fe2O3 5.30 7.870 2 2.12 Ce-Ce2O3 6.86 6.6893 2 1.14 Nb-Nb2O3 4.47 8.570 2 2.41 W-WO3 7.30 19.254 1 3.33

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The oxides that form on magnesium and sodium have smaller volumes than the metals from which they form. The coatings are therefore porous, and oxidation continues rapidly. A P-B ratio of less than one indicates this condition.

The oxides that form on titanium and cerium have similar volumes to the metals from which they form; therefore, adherent, non-porous, protective films form. A P-B ratio of 1-2 indicates this condition.

The oxides that form on iron, niobium, and tungsten have larger volumes that the metals from which they form. The oxides may flake from the surface, exposing fresh metal that continues to oxidize. A P-B ratio greater than two indicates this condition.

23-39 Oxidation of most ceramics is not considered to be a problem. Explain.

Solution: Most ceramics are already oxides—thus materials such as MgO and Al2O3 are expected to be inert in an oxidizing atmosphere. Non-oxide ceramics, however, may sometimes be subjected to oxidation problems.

23-40 A sheet of copper is exposed to oxygen at 1000°C. After 100 h, 0.246 g of copper are lost per cm2 of surface area; after 250 h, 0.388 g/cm2 are lost; and after 500 h, 0.550 g/cm2 are lost. Determine whether oxidation is parabolic, linear, or logarithmic, then determine the time required for a 0.75-cm sheet of copper to be completely oxidized.

The sheet of copper is oxidized from both sides.

Solution: Let’s assume that the rate is parabolic. We can determine the constant “k” in the oxidation equation y = kt for each time, first converting the rate in g/cm2 to thickness y in cm: y1 = V/A = [0.246 g/(8.93 g/cm3)]/(l cm2) = 0.0275 cm y2 = V/A = [0.388 g/(8.93 g/cm3)]/(l cm2) = 0.0434 cm y3 = V/A = [0.550 g/(8.93 g/cm3)]/(l cm2) = 0.0616 cm If oxidation is parabolic, the value for k should be the same for each time:

0.0275 cm = k(100 h)

or k = 7.6 ×10−6 cm 2 /h

0.0434 cm = k(250 h) or k = 7.6 ×10−6 cm 2 /h 0.0616 cm = k(500 h) or k = 7.6 ×10−6 cm 2 /h Therefore, the rate of oxidation must be parabolic.

To completely oxidize the copper, assuming that the rate of oxidation is the same from both sides of the sheet, we need to determine the time required to oxidize 0.75 cm/(2 sides) = 0.375 cm:

y = 0.375 cm = (7.6 × 10−6 cm 2 /h)(t )

or t = 18, 503 h

23-41 At 800°C, iron oxidizes at a rate of 0.014 g/cm2 per hour; at 1000°C, iron oxidizes at a rate of 0.0656 g/cm2 per hour. Assuming a parabolic oxidation rate, determine the maximum temperature at which iron can be held if the oxidation rate is to be less than 0.005 g/cm2 per hour.

Solution:

The rate is given by an Arrhenius equation, rate = A exp [– Q/(RT)]. We can find the constants A and Q from the data provided.

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0.014 g/(cm2·h) = A exp {–Q/[1.987 cal/(mol·K)]/(800 + 273 K)} 0.0656 g/(cm2·h) = A exp {– Q/[1.987 cal/(mol·K)]/(1000 + 273 K)} Taking logarithms of both sides: – 4.2687 = ln A – 0.0004690Q – 2.7242 = ln A – 0.0003953 Q

1.5445 = 0.0000737Q or Q = 20,960 cal/mol – 4.2687 = ln A – (0.000469) (20,960) ln A = 5.56 A = 260 To keep the oxidation rate below 0.005 g/(cm2 · h), the maximum temperature is rate = 260 exp [–20,960/(RT)] = 0.005 ln(0.005/260) = –20,960/1.987/T ln(0.00001923) = –10.859 = –10,549/T T = 971 K = 698°C

455 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

456 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.